trureturingGitHub MATHEMATICAL ATLAS

THE OMEGA INSTITUTE / CONTRIBUTIONS

Conjectures

What the release settles, why it matters, and what it opens next.

Research highlights ↗

THE OMEGA INSTITUTE / TRUTH RELEASE

What we have resolved

One release, multiple areas of mathematics, two kinds of answer.

Each entry below records the exact question, the contribution that settles its released scope, and the next boundary it leaves open.

1,701resolved results
1,079proved
622refuted
Resolved conjectures by mathematical area1701 resolved results flow through 6 mathematical areas into proved and refuted outcomes.Quantum information: 287 provedQuantum information: 287 refutedDiscrete structures: 506 provedDiscrete structures: 183 refutedNumber theory: 167 provedNumber theory: 52 refutedAnalysis & dynamics: 73 provedAnalysis & dynamics: 67 refutedGeometry & algebra: 33 provedGeometry & algebra: 19 refutedFurther frontiers: 13 provedFurther frontiers: 14 refutedResolved release1701 Quantum information574 Discrete structures689 Number theory219 Analysis & dynamics140 Geometry & algebra52 Further frontiers27Proved1079Refuted622

PROBLEMS & CONTRIBUTIONS

Resolved conjectures

The question, its history, and the idea that resolves it.

Rocky Mountain J. Math. 15 (1985)47Editorial selectionErdős suggested 23 might be the only counterexample. 47 is another.Only the proposed uniqueness of 23 is refuted; conjecture (22) and the classification of all counterexamples remain outside this result.Refuted

Source excerpt

“I would not be surprised if 23 is the only counterexample.”

P. Erdős

The questionPut (21) ∏{i=1}^{k} (x + i) = uk(x) vk(x), (uk(x), vk(x)) = 1 where vk(x) is squarefree and all prime factors of uk(x) occur with an exponent greater than 1.

Why it mattersThe paper identifies 23 as a counterexample and proposes that it could be the only one.

The ideaFor 1 ≤ k ≤ 78, split the interval into 1..30, 31..60, and 61..78.

Our contributionOnly the proposed uniqueness of 23 is refuted; conjecture (22) and the classification of all counterexamples remain outside this result.

Scope & next questionIt makes no claim about conjecture (22), an estimate of k0(x), the assertion that 23 and 47 are the only counterexamples, or the statement that 23 itself is a counterexample.

Verification details
Release source recordRefuted in Lean

Open problem

Starting value 47 refutes Erdős's proposed unique counterexample

Erdős, Rocky Mountain Journal of Mathematics 15(2) (1985), page 361 (verbatim from the rendered scan at https://users.renyi.hu/~p_erdos/1985-27.pdf):

Put (21) ∏_{i=1}^{k} (x + i) = u_k(x) v_k(x), (u_k(x), v_k(x)) = 1 where v_k(x) is squarefree and all prime factors of u_k(x) occur with an exponent greater than 1. The representation in (21) is clearly unique. Clearly for k > k_0(x), u_k(x) > v_k(x). Perhaps one can estimate the smallest k_0(x) so that u_k(x) > v_k(x) for all k > k_0(x) quite well. I have not done this. For small values of k usually v_k(x) > u_k(x). I thought that for every x there is a k for which v_k(x) > u_k(x). x = 7 seemed a likely counterexample but if k = 7, u_7(7) = 2^7 3^3 < v_7(7) = 5·7·11·13. On the other hand a simple computation shows that n = 23 is a counterexample, i.e., for every k, v_k(23) < u_k(23). The reason for this is the existence of 24, 25, 27 and 32. I would not be surprised if 23 is the only counterexample. Perhaps in fact there is a k_0 so that for every k > k_0 and all n > n_0(k) (22) v_k(n) > u_k(n). (22) is perhaps too optimistic.

The literal refuted statement is

∀ x ≥ 1, x ≠ 23 → ∃ k ≥ 1, u(x,k) < v(x,k).

For P(x,k) = ∏_{i=1}^k (x+i), v(x,k) is the product of the primes whose exponent in P(x,k) is exactly one. It is not the squarefree part: a prime occurring to exponent at least two is absent from v. The factor u(x,k) is the product of every prime power whose exponent in P(x,k) is at least two, with its full exponent. Thus u(x,k)·v(x,k)=P(x,k).

The result refutes only the sentence proposing 23 as the unique counterexample. It makes no claim about conjecture (22), an estimate of k_0(x), the assertion that 23 and 47 are the only counterexamples, or the statement that 23 itself is a counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/ErdosConsecutiveProductSquarefreeFactorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 1985.

    Read source ↗
  • Scholarly context

    Paul Erdős made foundational contributions to number theory and combinatorics and posed many influential problems. This background does not enlarge the particular claim resolved here.

    Read source ↗
Why this order
  • Editorial pin · position 1

    Erdős proposes, in print, that 23 may be the only counterexample; a second one is exhibited and kernel-checked.

  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 4/4 · 10% weight

    Year in the primary source attribution: 1985. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Scholarly context · 4/4 · 5% weight

    Paul Erdős made foundational contributions to number theory and combinatorics and posed many influential problems. This background does not enlarge the particular claim resolved here.

    Source 1 · Source 2
Annals of Probability 51 (2023)NoEditorial selectionThe noisy voter model: the repetition variance is never sublinearSynchronous noisy voter dynamics on connected locally finite simple graphs with a neighbor at every vertex, fair independent opinions, noise 0 < ε < 1 and positive lag.Proved

Source excerpt

“May Var(ζ₁ + … + ζ_t) be sublinear? We do not have an answer.”

I. Benjamini, H. Helman Tov, M. Zhukovskii

The questionZhukovskii, Global information from local observations of the noisy voter model on a graph, Annals of Probability 51(5), 1963–1992 (2023), Section 2.2, asks:

Why it mattersThe question distinguishes the linear and logarithmic branches of Lemma 2.3 for a local statistic observed through time.

The ideaPairing its future fair coin with its flip gives a contribution of half this mass.

Our contributionSynchronous noisy voter dynamics on connected locally finite simple graphs with a neighbor at every vertex, fair independent opinions, noise 0 < ε < 1 and positive lag.

Verification details
Release source recordProved in Lean

Open problem

The repetition-count variance of the noisy voter model

I. Benjamini, H. Helman Tov and M. Zhukovskii, Global information from local observations of the noisy voter model on a graph, Annals of Probability 51(5), 1963–1992 (2023), Section 2.2, asks:

May Var(ζ₁ + … + ζ_t) be sublinear? We do not have an answer. However, we prove that it is either linear in t or O(ln t).

The quotation is from arXiv:2207.01224v2, p. 6. The answer is no. For a connected locally finite simple graph with a neighbour at every vertex, noise $0<\varepsilon<1$, observation vertex $u$ of degree $D$ and lag $d\ge1$, put $\zeta_i=\mathbf{1}{X_i(u)=X_{i+d}(u)}$ and $W_t=\sum_{i=1}^{t}\zeta_i$. The formal settlement is D5.S3.StatisticalMechanics.NoisyVoter.RepetitionVarianceLinear.result : claim. Preregistration: #16646. The Library note supplies the publication locator and quotations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Connected graph, possibly infinite | G : SimpleGraph V, G.LocallyFinite, Model.connected | No finite-vertex restriction; finite neighbour sets make uniform choices meaningful | | A vertex has a neighbour to copy | Model.hasNeighbor; Value G (Coord.choice k v) is the neighbour subtype | Excludes an isolated one-vertex graph whose uniform copying law is undefined | | Independent fair initial opinions | Coord.initial, coordinateLaw, Model.independent, Model.laws | Bernoulli one-half | | Independent fair fresh opinions and Bernoulli noise controls | Coord.fresh, Coord.refresh, coordinateLaw, Model.ε, its strict bounds | True means refresh | | Independent uniform neighbour choices | Coord.choice, neighbour-subtype value and uniform finite law | The choice is an actual neighbour | | Synchronous recursion | nvm | Initial, refresh and copying cases mirror Section 1 | | Repetition indicator at lag $d$ | W, sum over Finset.Icc 1 t, positive lag in claim | No normalization; the observation reaches time $t+d$ | | Every graph, vertex and probability model | The closed quantifiers of claim | Arbitrary vertex type and measurable probability space | | Variance is never sublinear | result, negated little-o in claim | Private model_explicitBound supplies the explicit lower bound for every natural $t$ | | Encoding requirements | Decidable equality, measurable-space and local-finiteness instances | Decidable equality is representational; no additional graph hypothesis |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/NoisyVoter/RepetitionVarianceLinear.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 2

    An explicit unanswered question in the Annals of Probability, answered under the source model and the stated graph hypotheses.

  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Phys. Rev. A 111 (2025)Yes: 41/144Editorial selectionA fully symmetric local triangle model beats 1/4Four outcomes per party, independent local triangle sources, and symmetry under party permutations and joint outcome relabelings. The witness refutes the universal 1/4 bound.Refuted

Source excerpt

“Open problem. Does there exist a distribution p(A,B,C) that is local in the triangle network and fully symmetric such that p(A=B=C) > 1/4?”

E. Bäumer, V. Gitton, T. Kriváchy, N. Gisin, R. Renner

The questionA 111, 052453 (2025)) study classical (local) models in the triangle network with four outcomes per party.

Why it mattersFully symmetric distributions with four outcomes include the distribution of the Elegant Joint Measurement, the main candidate for noise-robust quantum nonlocality in the triangle network, and the local value of p(A = B = C) calibrates the Bell-type inequalities used to certify such nonlocality.

The ideaEach source sends one of the 12 ordered pairs x = (x₁, x₂) of distinct outcomes, uniformly; every party applies f(x, y) = x₂ if x₂ ∈ {y₁, y₂} and x₁ otherwise, with a = f(β, γ), b = f(γ, α), c = f(α, β).

Our contributionFour outcomes per party, independent local triangle sources, and symmetry under party permutations and joint outcome relabelings. The witness refutes the universal 1/4 bound.

Verification details
Release source recordRefuted in Lean

Open problem

A fully symmetric triangle-local distribution with p(A = B = C) above 1/4

E. Bäumer, V. Gitton, T. Kriváchy, N. Gisin and R. Renner (arXiv:2405.08939, quant-ph; Phys. Rev. A 111, 052453 (2025)) study classical (local) models in the triangle network with four outcomes per party. Their best local fully symmetric construction has p(A = B = C) = 1/4, and their conclusion states:

Open problem. Does there exist a distribution p(A,B,C) that is local in the triangle network and fully symmetric such that p(A=B=C) > 1/4?

Issue #11256 fixes the reading. Locality is eq. trilocal: sources uniform on [0, 1] and measurable conditional distributions p_A(a|β,γ), p_B(b|γ,α), p_C(c|α,β). Full symmetry is invariance under every permutation of the parties and every joint relabelling of the four outcomes. The formal claim is the negative answer, s₁₁₁(p) ≤ 1/4 for every local fully symmetric p; the result refutes it, so the answer is yes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TriangleSymmetricLocalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 3

    The open problem that closes the paper; answered affirmatively with an explicit local model.

  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
J. Math. Phys. 50 (2009)No: a corner at s = ½Editorial selectionHiai, Mosonyi and Hayashi’s main open question: ψ need not be differentiableA finite-group counterexample with faithful states and the stated asymptotic convergence premise; the question is differentiability of ψ on (0, 1).Refuted

Source excerpt

“The main open question of the present work is to show that this is indeed the case.”

F. Hiai, M. Mosonyi, M. Hayashi

The questionHayashi, Quantum hypothesis testing with group symmetry, J.

Why it mattersDifferentiability would provide a route to matching asymptotic hypothesis testing bounds.

The ideaBoth diagonal matrices are positive definite and have trace one.

Our contributionA finite-group counterexample with faithful states and the stated asymptotic convergence premise; the question is differentiability of ψ on (0, 1).

Verification details
Release source recordRefuted in Lean

Open problem

A corner in the group-invariant asymptotic overlap

F. Hiai, M. Mosonyi and M. Hayashi, Quantum hypothesis testing with group symmetry, J. Math. Phys. 50 (2009) 103304, Section 7, verbatim excerpt with the indicated omissions:

To get a complete solution of the problems of the Chernoff and the Hoeffding bounds and of Stein's lemma, … the key point is to show that (4.5) holds with equality … This can be done, for instance, … [by showing] that the function ψ exists and is differentiable on the interval (0, 1). As the examples of Section 6 suggest, one can expect the differentiability of ψ to hold in our setting of group-invariant state discrimination. The main open question of the present work is to show that this is indeed the case.

| Source object or quantifier | Formal reading | |---|---| | Finite-dimensional Hilbert space | Complex coordinate space of arbitrary dimension d | | Compact group and unitary representation | Finite group G and a homomorphism into unitary complex matrices; finite groups with the discrete topology are compact | | Faithful states | Positive definite matrices with trace one | | Equation (2.1), tensor conditional expectation | Defs.twirlPower, the uniform finite-group Haar average of tensor conjugates, indexed by words Fin n → Fin d | | Finite logarithmic overlap | Defs.psiN, the real logarithm of the real part of the trace of the product of CFC powers | | Equation (3.1), asymptotic overlap | Defs.psi, the selected limit of psiN / n, with convergence at every s in (0,1) an explicit premise | | Differentiability assertion | Defs.claim, universal differentiability on (0,1) under these premises |

Refuting this finite-group restriction with faithful states refutes the compact-group assertion on its faithful domain.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/GroupInvariantPsi/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 4

    The authors call it the main open question of their paper; refuted by an explicit four-level faithful pair.

  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    Year in the primary source attribution: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
PRX Quantum 3 (2022)Proved for every d ≥ 5Editorial selectionThe Torpedo Game has a perfect classical strategy in every dimension d ≥ 5Perfect classical winning probability for the dimension-d Torpedo Game, for every integer d ≥ 5; the source had checked 5 ≤ d ≤ 23.Proved

Source excerpt

“Conjecture: θ^C_{d≥5} = 1.”

P.-E. Emeriau, M. Howard, S. Mansfield

The questionMansfield (arXiv:2007.15643, quant-ph; PRX Quantum 3, 020307 (2022)) introduce the dimension-d Torpedo Game: Alice receives x, z ∈ Zd and sends one dit; Bob is asked q ∈ {∞, 0, …, d − 1} and must answer a ≠ x for q = ∞ and a ≠ qx − z otherwise.

Why it mattersThe quantum Torpedo Game is won perfectly for d ≥ 3 by a qudit strategy built from stabilizer states, while d = 2, 3 show a classical–quantum gap.

The ideaA deterministic strategy is a colouring E of the d × d grid with d colours and a decoding D; it is perfect iff D(E(x, z), q) differs from the forbidden label for every (x, z, q).

Our contributionPerfect classical winning probability for the dimension-d Torpedo Game, for every integer d ≥ 5; the source had checked 5 ≤ d ≤ 23.

Verification details
Release source recordProved in Lean

Open problem

A perfect classical strategy for the Torpedo Game

P.-E. Emeriau, M. Howard and S. Mansfield (arXiv:2007.15643, quant-ph; PRX Quantum 3, 020307 (2022)) introduce the dimension-d Torpedo Game: Alice receives x, z ∈ Z_d and sends one dit; Bob is asked q ∈ {∞, 0, …, d − 1} and must answer a ≠ x for q = ∞ and a ≠ qx − z otherwise. They find the classical values 3/4 and 11/12 for d = 2, 3, perfect classical strategies for 5 ≤ d ≤ 23, and state

Conjecture: θ^C_{d≥5} = 1.

Issue #11350 fixes the reading: classical strategies use finite shared randomness with stochastic encodings and decodings, and θ^C_d = 1 means that 1 is the greatest classical winning probability.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/TorpedoGamePerfectClassical.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 5

    A named conjecture in PRX Quantum, checked by the authors for 5 ≤ d ≤ 23; proved for every d ≥ 5.

  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
IEEE Trans. Inf. Theory 51 (2005)YesEditorial selectionTwo singlets are more than twice as strong as CHSHStatistical strength means KL divergence in bits, under the uniform, product and correlated setting distributions defined in the source.Proved

The question · paraphrased

Can a 2×4×4 experiment using joint measurements on pairs of Bell singlets be more than twice as strong as CHSH?

W. van Dam, R. D. Gill, P. D. Grünwald

The questionGrünwald, “The Statistical Strength of Nonlocality Proofs”, IEEE Transactions on Information Theory 51(8), 2812–2835 (2005), arXiv:quant-ph/0307125v2, Section VI.D, p. 11, Conjecture 5 states:

The ideaThe commuting rows and columns of the Mermin–Peres square provide the first three PVM settings.

Our contributionStatistical strength means KL divergence in bits, under the uniform, product and correlated setting distributions defined in the source.

Verification details
Release source recordProved in Lean

Open problem

Two singlets give more than twice the statistical strength of CHSH

W. van Dam, R. D. Gill and P. D. Grünwald, “The Statistical Strength of Nonlocality Proofs”, IEEE Transactions on Information Theory 51(8), 2812–2835 (2005), arXiv:quant-ph/0307125v2, Section VI.D, p. 11, Conjecture 5 states:

There is an experiment on pairs of Bell singlets, of the $2\times4\times4$ type, more than twice as strong as CHSH, and involving joint measurements on the pairs.

The counts are two parties, four settings per party and four outcomes per setting. The source measures strength by KL divergence in bits, minimized over every local response mixture, with uniform, product or arbitrary correlated setting laws (Section IV.B, Definitions 1–3).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/StatisticalStrengthMagicSquare.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2005.

    Read source ↗
Why this order
  • Editorial pin · position 6

    Conjecture 5 in IEEE Transactions on Information Theory; proved for all three statistical-strength conventions in the paper.

  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 3/4 · 10% weight

    Year in the primary source attribution: 2005. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
arXiv:2502.12240 (2025)No: a witness at p = 2Editorial selectionSpacetime density matrices violate Schatten subadditivityA three-qubit witness at Schatten exponent p = 2 refutes Conjecture 1 for spacetime density matrices.Refuted

Source excerpt

“Conjecture 1 For any T_AB on a finite-dimensional H_A ⊗ H_B and p > 1: 1 + ||T_AB||_p ≥ ||Tr_A T_AB||_p + ||Tr_B T_AB||_p.”

A. Milekhin, Z. Adamska, J. Preskill

The questionPreskill, Observable and computable entanglement in time, arXiv:2502.12240v2, define the spacetime density matrix TAB of a subsystem A at time 0 and a subsystem B at time t. - The definition. Tr(TAB (OA ⊗ 1B)(1A ⊗ OB)) = ⟨OA(0) OB(t)⟩ = Tr(ρ OA U† OB U) (Eqs. (4), (9)); A and B "do not have to be related to each other in any particular way".

Why it mattersThe spacetime density matrix extends reduced density matrices to subsystems at different times, and its Rényi-type quantities continue timelike entanglement in field theory.

The idea1. The witness. Three qubits Q, R and S, with A = B = Q, the same qubit at times 0 and t.

Our contributionA three-qubit witness at Schatten exponent p = 2 refutes Conjecture 1 for spacetime density matrices.

Scope & next questionThe numerical evidence of the paper is consistent with the whole-system case, which remains open.

Verification details
Release source recordRefuted in Lean

arXiv

The spacetime density matrix violates the Audenaert-type Schatten subadditivity

A. Milekhin, Z. Adamska and J. Preskill, Observable and computable entanglement in time, arXiv:2502.12240v2, define the spacetime density matrix T_AB of a subsystem A at time 0 and a subsystem B at time t.

  • The definition. Tr(T_AB (O_A ⊗ 1_B)(1_A ⊗ O_B)) = ⟨O_A(0) O_B(t)⟩ = Tr(ρ O_A U† O_B U) (Eqs. (4), (9)); A and B "do not have to be related to each other in any particular way". The Schatten norm is ||M||_p = (Tr (M†M)^{p/2})^{1/p} (footnote a).
  • The conjecture. After Audenaert's inequality for density matrices (Theorem 1, Eq. (24)), §IV item 9 states: "Conjecture 1 For any T_AB on a finite-dimensional H_A ⊗ H_B and p > 1: 1 + ||T_AB||_p ≥ ||Tr_A T_AB||_p + ||Tr_B T_AB||_p. (25)"

The verbatim statements are in the literature note.

Issue #17385 fixes the reading: for every finite dimension, every density matrix ρ, every unitary U between the factorizations H_A ⊗ H_Ā and H_B ⊗ H_B̄ of the whole system, every T satisfying Eq. (9) for all O_A and O_B, and every real p > 1, inequality (25) holds.

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SpacetimeDensitySchattenSubadditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 7

    Conjecture 1 of a 2025 paper, supported there by numerical evidence; refuted by a three-qubit witness.

  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
arXiv:2605.21457 (2026)No, already at d = 2Editorial selectionPurify-and-clone is not optimal beyond m = nThe counterexample has n = 1, m = 2 and d = r = 2, under the one-site worst-case squared-infidelity reading.Refuted

Source excerpt

“This channel is optimal for m ≤ n (see Ref. [S2]), and we conjecture that it is also optimal for m > n.”

Z. Li, E. Theil, A. W. Harrow, I. Chuang

The questionChuang, An Exponential Sample-Complexity Advantage for Coherent Quantum Inference, arXiv:2605.21457v2, Section S3 A, p. 9, states:

Why it mattersCloning the entire system-environment register and cloning the system before adjoining a symmetric environment have different one-site risks.

The ideaThe proof uses convexity.

Our contributionThe counterexample has n = 1, m = 2 and d = r = 2, under the one-site worst-case squared-infidelity reading.

Verification details
Release source recordRefuted in Lean

Open problem

Purify-and-clone fails one-site worst-case optimality

Z. Li, E. Theil, A. W. Harrow and I. Chuang, An Exponential Sample-Complexity Advantage for Coherent Quantum Inference, arXiv:2605.21457v2, Section S3 A, p. 9, states:

This channel is optimal for m ≤ n (see Ref. [S2]), and we conjecture that it is also optimal for m > n.

Preregistration #19410 fixes $n=1$, $m=2$ and the one-site worst-case squared-infidelity reading. The universal optimality assertion is refuted by $d=r=2$. A strict-improvement theorem for every $d\ge r\ge2$ is not part of this settlement. The Library note identifies the source and conventions.

Statement echo

| Source clause or binding convention | Lean counterpart | | --- | --- | | Squared fidelity and infidelity | sqFidelity is the squared real trace of the positive square root; the loss uses one minus that value. | | All purifications, equation (S34) | purifications contains every positive trace-one rank-one matrix with the specified environment partial trace. | | One joint system-environment site | oneSiteMarginal retains both registers of either output pair. | | One common target before the maximum, equation (S13) | oneSiteLoss takes the infimum over the convex hull of purifications after forming the maximum of the two infidelities. | | Worst-case rank-at-most-$r$ domain, equation (7) | worstCaseOneSiteRisk takes the supremum over every canonical density state with matrix rank at most $r$. Infima and suprema use EReal. | | Totally symmetric output restriction | TotallySymmetricSupport asserts $S\Xi S=\Xi$, with $S=(I+F)/2$, on both sides. | | Werner cloner and the one-input RP map | wernerOneToTwo, randomPurifyOne and purifyAndClone retain the source matrix expressions, with cloning dimension $dr$. | | Genuine channel competitors | AdmissibleChannel uses QuantumChannel and the support restriction on every admissible input. | | Positive dimensions $1\le r\le d$ | claim retains these quantifiers; the counterexample instantiates them at two. | | Claimed optimality | claim compares the two worst-case risks for every admissible channel; result proves its negation. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/LiTheilHarrowChuangPurifyCloneRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 8

    An explicit optimality conjecture; refuted in dimension two under the preregistered one-site reading.

  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
J. Math. Phys. · arXiv:1508.03551All three equalities holdEditorial selectionHiai and Ruskai's exact contraction coefficients holdAll three equalities (45c–45e) for the specified non-unital qubit CQ channel, on the full parameter domain stated in Conjecture 6.3.Proved

Source excerpt

“Conjecture 6.3. Equality holds in (45c) through (45e) above.”

F. Hiai, M. B. Ruskai

The questionA quantum channel makes states harder to distinguish. What is its exact contraction under different information-geometric metrics?

Why it mattersHiai and Ruskai proposed three optimal bounds in their 2015 preprint. The question asks whether each bound is attained when all admissible states and tangent directions are considered.

The ideaA common input estimate and positive kernel representations control all three metrics. The maximally mixed state and a Pauli direction attain the bounds.

Our contributionAll three equalities (45c–45e) for the specified non-unital qubit CQ channel, on the full parameter domain stated in Conjecture 6.3.

Scope & next questionThis resolves the three metrics in Conjecture 6.3. A formula for every kernel in the broader metric class remains outside this result.

Verification details
Release source recordProved in Lean

Open problem

Hiai–Ruskai Conjecture 6.3: exact qubit CQ contraction coefficients

Hiai and Ruskai, Contraction coefficients for noisy quantum channels, arXiv:1508.03551v1, Section 6, Theorem 6.2 and Conjecture 6.3 (PDF p. 19), state:

Although the bounds in the above theorem are sufficient to disprove two conjectures as remarked below, we believe that they are optimal, i.e.,

Conjecture 6.3. Equality holds in (45c) through (45e) above.

The map on complex qubit matrices is $\Phi_{\alpha,\tau}(w_0I+\mathbf w\cdot\boldsymbol\sigma) =w_0I+\alpha w_1\sigma_1+\tau w_0\sigma_3$. For every $\alpha\geq0$, $0<|\tau|<1$ and $\alpha^2+\tau^2\leq1$, the conjecture asks for the three equalities

$$ \begin{aligned} \eta_{\widehat{\mathrm{WY}}}^{\mathrm{Riem}}(\Phi_{\alpha,\tau}) &=\frac{\alpha^2(1+\sqrt{1-\tau^2})}{2(1-\tau^2)},\ \eta_{x^{-1/2}}^{\mathrm{Riem}}(\Phi_{\alpha,\tau}) &=\frac{\alpha^2}{\sqrt{1-\tau^2}},\ \eta_{\mathrm{BKM}}^{\mathrm{Riem}}(\Phi_{\alpha,\tau}) &=\frac{\alpha^2}{2\tau}\log\frac{1+\tau}{1-\tau}. \end{aligned} $$

The kernels are $(1+\sqrt{x})^2/(4x)$, $x^{-1/2}$ and $\log(x)/(x-1)$, with the BKM value $1$ at $x=1$. The contraction coefficient is the literal supremum of $\langle\Phi(A),\Omega_{\Phi(\rho)}^\kappa(\Phi(A))\rangle/ \langle A,\Omega_\rho^\kappa(A)\rangle$ over all positive definite trace-one $2\times2$ complex matrices $\rho$ and all nonzero traceless Hermitian matrices $A$. The spectral action is $(\Omega_\rho^\kappa(X)){ij}=\kappa(\lambda_i/\lambda_j)X{ij}/\lambda_j$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/HiaiRuskaiRiemannianContraction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The dual WY, geometric and BKM Riemannian contraction coefficients of the non-unital qubit CQ channel equal the three bounds in Hiai and Ruskai's Conjecture 6.3.

    Read source ↗
  • Historical context

    First preprint year: 2015.

    Read source ↗
  • Scholarly context

    Mary Beth Ruskai made foundational contributions to quantum entropy inequalities, quantum channels and mathematical quantum information.

    Read source ↗
Why this order
  • Editorial pin · position 9

    Conjecture 6.3 of Hiai and Ruskai; all three equalities proved on the full stated domain.

  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Scholarly context · 4/4 · 5% weight

    Mary Beth Ruskai made foundational contributions to quantum entropy inequalities, quantum channels and mathematical quantum information.

    Source 1 · Source 2
J. Symbolic Comput. · arXiv:2401.08481(10.10) holds for all n ≥ 3Editorial selectionKoutschan, Krattenthaler and Schlosser's equation (10.10) is provedThis result proves the complete chain of equalities in equation (10.10), for all n ≥ 3. The other equations of Conjecture 23 are separate results.Proved

Source excerpt

“Moreover, the following identities are conjectured to hold for all n ≥ 3:”

C. Koutschan, C. Krattenthaler, M. Schlosser

The questionKoutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”.

Why it mattersThe determinant families share a rational column transformation.

The ideaThe multiplier has constant coefficient one, so its lower triangular coefficient matrix preserves determinants.

Our contributionThis result proves the complete chain of equalities in equation (10.10), for all n ≥ 3. The other equations of Conjecture 23 are separate results.

Verification details
Release source recordProved in Lean

Open problem

Koutschan–Krattenthaler–Schlosser Conjecture 23, equation (10.10)

Koutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”. Equation (10.10) is

$G_{3,0,3,3}(n)=\frac23G_{0,1,-2,-4}(n+1)=-\frac1{672}G_{1,3,-2,-8}(n+1)=\frac1{63}G_{5,4,3,-5}(n)=\frac4{1002001}G_{6,6,3,-9}(n)=-\frac85G_{9,5,8,-2}(n-1)$.

The preregistration is #19236. The Library note records the source definition and all three equations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (1.2): falling factorial divided by $p!$ for $p\ge0$, zero for $p<0$ | ShiftedG.binom | Rational Ring.choose on the nonnegative branch, integer lower argument. | | Section 10 determinant | ShiftedG.G | Integer parameters, rational entries, integer powers of 2 and zero-based Fin n indices. | | “for all n ≥ 3” | TenTen.claim | Natural size and the original lower bound. | | Every consecutive equality of (10.10) | TenTen.claim and TenTen.result | Original coefficients, signs and size shifts; a chain is a conjunction of adjacent equalities. |

Natural subtraction is truncated at zero; within the source range it agrees with integer subtraction. The rational/integer coercions express the same source entries. There are no additional assumptions on the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KoutschanKrattenthalerSchlosserTenTen.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 10

    Conjecture 23, equation (10.10): the entire equality chain is proved for every n ≥ 3.

  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Erdős on Mahler (1989)521² = 111101₁₂Editorial selectionMahler's 0/1-digit square was not uniqueA counterexample to the suggested uniqueness of a nontrivial coprime 0/1-digit square, as recorded by Erdős; not Mahler’s fixed-base finiteness conjecture.Refuted

Source excerpt

“In fact, the only nontrivial solution he found was 7^3 + 7^2 + 7 + 1 = 20^2, and perhaps there are no others”

P. Erdős, recalling K. Mahler

The questionErdős's 1989 reminiscence, page 2 (verbatim):

Why it mattersThe source presents the base-7 identity as Mahler's only known nontrivial example and immediately suggests that no others may exist.

The ideaThe finite certificate first proves Nat.Coprime 12 521 and 12^5 + 12^4 + 12^3 + 12^2 + 1 = 271441 = 521^2.

Our contributionA counterexample to the suggested uniqueness of a nontrivial coprime 0/1-digit square, as recorded by Erdős; not Mahler’s fixed-base finiteness conjecture.

Scope & next questionThis refutation makes no claim about Mahler's finiteness conjecture for any fixed k > 4 and no claim about infinitude for k = 2, 3, 4.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Erdős's report of Mahler's binary-digit square uniqueness

Erdős's 1989 reminiscence, page 2 (verbatim):

He told me about his last paper. Let e_i = 0 or 1. Then Σ_{i=1}^{n} e_i k^i = x^2 has infinitely many solutions for k = 2, 3 and 4. He conjectured that it has only a finite number of solutions for k > 4. In fact, the only nontrivial solution he found was 7^3 + 7^2 + 7 + 1 = 20^2, and perhaps there are no others (1 + (k^2 − 1) = k^2 counts as trivial).

The literal refuted statement is the proposition claim:

∀ (k x : ℕ) (S : Finset ℕ),
  5 ≤ k → 1 < x → BinaryBaseSquare k x S →
    (k = 7 ∧ x = 20) ∨ k = x ^ 2 - 1

Here BinaryBaseSquare k x S means Nat.Coprime k x ∧ ∑ i ∈ S, k ^ i = x ^ 2. The exponent set includes zero because the printed example contains k^0 = 1. "Trivial" means the two-digit family x^2 = k + 1, equivalently k = x^2 − 1, matching Erdős's parenthetical after renaming the base variable. The coprimality condition gcd(k, x) = 1 comes from Mahler's setting and excludes digit-shift scalings x ↦ k·x. The counterexamples (12, 521) and (8, 11677) are in no trivial family under either reading.

This refutation makes no claim about Mahler's finiteness conjecture for any fixed k > 4 and no claim about infinitude for k = 2, 3, 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/ErdosMahlerBinaryDigitSquareRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 1989.

    Read source ↗
  • Scholarly context

    Paul Erdős made foundational contributions to number theory and combinatorics and posed many influential problems. This background does not enlarge the particular claim resolved here.

    Read source ↗
Why this order
  • Editorial pin · position 11

    Mahler's suggested uniqueness, as recorded by Erdős; refuted by an explicit coprime square. Note: 11677 also appears in OEIS A342546 under a different framing.

  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    Year recorded in the source bibliography: 1989. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Scholarly context · 4/4 · 5% weight

    Paul Erdős made foundational contributions to number theory and combinatorics and posed many influential problems. This background does not enlarge the particular claim resolved here.

    Source 1 · Source 2
arXiv:2010.14682 (2020)No such ηEditorial selectionNo uniform contraction for trace-norm conditional mutual informationA qubit channel with local trace-norm contraction ratio at most 1/√2 refutes the claimed uniform contraction of trace-norm conditional mutual information.Refuted

The question · paraphrased

Must a locally strictly contracting channel uniformly contract trace-norm conditional mutual information?

C.-F. Chen, K. Kato, F. G. S. L. Brandão

The questionBrandão, Matrix Product Density Operators: when do they have a local parent Hamiltonian?, arXiv:2010.14682v3, Section III.C (PDF p. 17), Conjecture III.2:

Why it mattersThe source studies whether matrix product density operators are Gibbs states of quasi-local parent Hamiltonians through the decay of conditional mutual information.

Our contributionA qubit channel with local trace-norm contraction ratio at most 1/√2 refutes the claimed uniform contraction of trace-norm conditional mutual information.

Verification details
Release source recordRefuted in Lean

arXiv

A completely contractive data-processing inequality for the trace-norm CMI

C.-F. Chen, K. Kato and F. G. S. L. Brandão, Matrix Product Density Operators: when do they have a local parent Hamiltonian?, arXiv:2010.14682v3, Section III.C (PDF p. 17), Conjecture III.2:

For any channel $\mathcal E: C \rightarrow C'$ with local contraction ratio $\eta_{1,C}:=\sup_{\rho_C,\rho'C}\frac{\lVert\mathcal E_C[\rho{C}]-\mathcal E_C[\rho'{C}]\rVert_1 }{\lVert\rho{C}-\rho'{C}\rVert_1} <1$ There exists a global constant $\eta < 1$ such that for any tripartite system $ABC$ and any state $\rho{ABC}$, it holds that $I_1(A:C'|B){\mathcal E(\rho)} \le \eta I_1(A:C|B)\rho$.

Here $I_1(A:C|B)=\lVert\rho_{ABC}-\rho_A\otimes\rho_{BC}\rVert_1-\lVert\rho_{AB}-\rho_A\otimes\rho_B\rVert_1$ (Definition 2, PDF p. 14). The verbatim texts are in the literature note. Issue #14728 preregisters the reading over finite-dimensional Kraus channels and density matrices, and the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ChenKatoBrandaoTraceNormCMIRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Editorial pin · position 12

    Conjecture III.2 of Chen, Kato and Brandão; refuted by a qubit channel with local contraction ratio at most 1/√2.

  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2021Research highlightA universal matrix bound reveals a hidden geometryEvery complex matrix satisfying the proposed bound for all positive block completions is essentially Hermitian: it is an affine complex transform of a Hermitian matrix.Essentially Hermitian off-diagonal blocksProved

The questionWhen does a bound on every positive block matrix force its off-diagonal block to have a special structure?

Why it mattersBourin and Lee posed this question in 2021. It connects a norm inequality to the geometry of a matrix, and includes an earlier normality question of Hayashi.

The ideaFinite rank-one perturbations expose the extreme eigenvalues. Their symmetry forces normality; a second identity reconstructs the affine Hermitian structure in every dimension.

Our contributionEvery complex matrix satisfying the proposed bound for all positive block completions is essentially Hermitian: it is an affine complex transform of a Hermitian matrix.

Scope & next questionThe result gives a structural implication for the operator norm. An algorithm producing violating completions and extensions to other norms remain open.

Verification details
Release source recordProved in Lean

Open problem

Essentially Hermitian off-diagonal blocks

J.-C. Bourin and E.-Y. Lee, Eigenvalue inequalities for positive block matrices with the inradius of the numerical range, arXiv:2111.15180v1, printed page 6, Conjecture 3.3, states:

Let $X\in\mathbb{M}_n$. If the inequality

$$\left|\begin{bmatrix}A&X\X^*&B\end{bmatrix}\right|\infty\le|A+B|\infty$$

holds for all positive block-matrix with $X$ as off-diagonal block, then $X$ is essentially Hermitian.

The source says: “If W(T) is line segment, then T is a so-called essentially Hermitian matrix.” The binding convention of #13752 is the literal affine expression $X=\alpha H+\beta I$, with $H=H^*$ and $\alpha,\beta\in\mathbb C$, including points. The norm is the Euclidean operator norm, and positive means positive semidefinite. The same question is Conjecture 7.3.3 of Bourin's A journey into Matrix Analysis, arXiv:2307.03064v1, printed page 114.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/BlockNorm/EssentiallyHermitian.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2021The sharp worst-case MATEN process fidelity is $4^{-N}$ for $N\ge2$The product of single-site process marginals has squared fidelity at least four to the minus N, with equality for cyclic wire permutations on at least two qubits.Proved

The questionWudarski, Dual-map framework for noise characterization of quantum computers, arXiv:2112.04414v1 (2021); Phys.

Why it mattersMATEN retains each single-qubit process marginal and discards correlations between different process coordinates.

The ideaZero weights contribute zero to the arithmetic sum and one to the geometric product.

Our contributionThe product of single-site process marginals has squared fidelity at least four to the minus N, with equality for cyclic wire permutations on at least two qubits.

Verification details
Release source recordProved in Lean

Open problem

The sharp worst-case MATEN process fidelity is $4^{-N}$ for $N\ge2$

J. Sud, J. Marshall, Z. Wang, E. Rieffel and F. A. Wudarski, Dual-map framework for noise characterization of quantum computers, arXiv:2112.04414v1 (2021); Phys. Rev. A 106, 012606 (2022), Section IV, “Local vs Non-Local Channels”, state:

We conjecture that the effective channel with the maximally entangled $\chi$ matrix is the worst case scenario for the proposed MATEN protocol.

However, this conjecture requires more rigorous treatment, which we leave as an open problem.

These sentences occur in main.tex, lines 289 and 290, respectively, in the downloaded v1 source (709 lines; MD5 b2275e7db7e3a63cc4c871c172f7943a). The source specifies $\chi=|\Psi\rangle\langle\Psi|$ with $|\Psi\rangle=\tfrac12\sum_{i=0}^3|ii\ldots i\rangle$ on $N$ four-dimensional process factors (lines 286–289), and evaluates $F(\chi,\chi_{\mathrm{MA}})=4^{-N}$ (lines 292–296). Its MATEN definition is at lines 226–232. See also the literature note.

Issue #20689 fixes the settlement as a universal fidelity lower bound and attainment by cyclic wire permutations. The source's particular GHZ witness and its trace-preservation assertion require the qualification in Triage.

Write $a\in{0,1,2,3}^N$ for a Pauli string and $P_a=\bigotimes_k\sigma_{a_k}$, with $\sigma_0=I$, $\sigma_1=X$, $\sigma_2=Y$ and $\sigma_3=Z$, without Hilbert–Schmidt normalization. The formal definitions are:

  • IsProcessCPTP means $\chi\succeq0$ and $\sum_{a,b}\chi_{ab}P_bP_a=I$. This is the process representation of $\mathcal E_\chi(\rho)=\sum_{a,b}\chi_{ab}P_a\rho P_b$, matching the source's CP and TP conditions at lines 93–100.
  • marginal is the single-site partial trace: $(\chi_k){ij}=\sum{y:y_k=i}\chi_{y,y[k\leftarrow j]}$. productMap forms $\chi_{\mathrm{MA}}=\bigotimes_k\chi_k$, entrywise $(\chi_{\mathrm{MA}}){ab}=\prod_k(\chi_k){a_kb_k}$.
  • The fidelity used is squared Uhlmann fidelity, $F(\rho,\sigma)=(\operatorname{Re}\operatorname{tr}\sqrt{\sqrt\rho,\sigma\sqrt\rho})^2$, exactly sqFidelity. This convention agrees with the source's projector calculation and distinguishes squared fidelity from root fidelity.
  • Put $U_N=(\mathtt{qubitPermutation}(\mathtt{finRotate},N))^{\mathsf T}$. Here the untransposed permutation matrix has entries $1$ when $y=x\circ\mathtt{finRotate},N$ and $0$ otherwise. shiftCoeff is $\psi_N(a)=2^{-N}\operatorname{tr}(P_aU_N)$, and rankOneDensity gives $\chi_N=\psi_N\psi_N^\dagger$, with entries $\psi_N(a)\overline{\psi_N(b)}$.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/MarginalApproximationWorstFidelity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2024Bonetto–Jardón Kojakhmetov Conjecture 1The spectral signature of a graph Laplacian plus a positive adjacency weight is nonincreasing.Proved

The questionBonetto and Jardón Kojakhmetov, On the Eigenvalues of Graphs with Mixed Algebraic Structure, arXiv:2408.00487v1, Section 3, Conjecture 1 states:

Why it mattersThe source uses signature to compare stable and unstable directions in linear systems with diffusive and saddle interactions.

The ideaThe partition is retained for the original graph interpretation, although the proof uses only symmetry and positive semidefiniteness of LG.

Our contributionThe spectral signature of a graph Laplacian plus a positive adjacency weight is nonincreasing.

Verification details
Release source recordProved in Lean

arXiv

Bonetto–Jardón Kojakhmetov Conjecture 1

Bonetto and Jardón Kojakhmetov, On the Eigenvalues of Graphs with Mixed Algebraic Structure, arXiv:2408.00487v1, Section 3, Conjecture 1 states:

The signature of the matrix M(ε) is a monotonic decreasing function of ε.

For every universe-polymorphic finite vertex type V with Fintype and DecidableEq instances, let F,G,H be simple undirected loopless graphs on V with disjoint G,H and G union H = F. Set M(u)=L_G+u A_H over the reals, using the ordinary combinatorial Laplacian and 0/1 adjacency matrix. For every real 0<s≤t, the entire assertion is

signature(M(t)) ≤ signature(M(s)).

Signature is the integer number of positive roots minus negative roots of the actual characteristic polynomial, counted with algebraic multiplicity; zero roots contribute zero. For these symmetric matrices, the full root multiset equals the indexed spectral eigenvalue multiset. Decreasing means nonincreasing, allowing equality and constant intervals. Singular matrices, persistent kernels, disconnected graphs, empty edge classes, the empty vertex type, s=t and the whole positive parameter axis remain included. Transition counts, limiting signatures, strict decrease and substitution of zero for a positive parameter are separate assertions and are not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/BonettoSignature.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2013Linking quantum uncertainty to the best possible guessWe prove a stronger linear bound, Q(X|B) ≤ 1 − pguess, which implies the proposed logarithmic bound and includes mixed and singular conditional states.Datta–Dorlas–Jozsa–Benatti: conditional subentropy and guessingProved

The questionHow is conditional subentropy limited by the optimal probability of guessing a classical label from a quantum state?

Why it mattersConjecture 2 in the source proposes a logarithmic upper bound for arbitrary finite ensembles.

The ideaA convex pinching gap and Gram representations connect conditional subentropy to every allowed measurement, before taking the optimal guessing probability.

Our contributionWe prove a stronger linear bound, Q(X|B) ≤ 1 − pguess, which implies the proposed logarithmic bound and includes mixed and singular conditional states.

Scope & next questionThis does not supply strong subadditivity or a general multipartite conditional chain rule.

Verification details
Release source recordProved in Lean

Open problem

Datta–Dorlas–Jozsa–Benatti: conditional subentropy and guessing

Conjecture 2. For a c-q state ρXB, the conditional subentropy Q(X|B) satisfies the following bound: Q(X|B) ≤ Hmin(X|B) = − ln pguess(X|B).

The supplied locator is arXiv:1310.1312v1, §2.5.1, Conjecture 2. Let p be any finite probability distribution and rho_x any finite-dimensional density states. Conditional subentropy is Q(recordState p rho) minus Q(mixtureState p rho), using the canonical states. The strengthened target is

$$ Q(X|B)\le1-p_{\rm guess}(X|B)\le-\log p_{\rm guess}(X|B). $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Subentropy/POVMGram.cq_conditional_subentropy_guessing

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2023Finding the hidden mirror structure in a random wordWe prove the limiting expected largest block-palindrome width for every alphabet with at least two letters, including its exact coefficient series.Expected largest block-palindrome widthProved

The questionHow many matching outer blocks can a word contain, on average, when the word becomes very long?

Why it mattersGabrić and Shallit conjectured a limiting series after identifying its coefficients experimentally. A proof must explain both the limit and the sequence behind it.

The ideaPeeling off the shortest matching borders gives a counting recursion. A geometric tail bound controls the limit, while a composition recurrence identifies the coefficients.

Our contributionWe prove the limiting expected largest block-palindrome width for every alphabet with at least two letters, including its exact coefficient series.

Scope & next questionThe theorem concerns the limiting average of the largest factorization width, with the source’s block-palindrome conventions.

Verification details
Release source recordProved in Lean

Open problem

Expected largest block-palindrome width

D. Gabrić and J. Shallit, Smallest and Largest Block Palindrome Factorizations, arXiv:2302.13147v2, §3, p. 5, Conjecture 5; WORDS 2023, DOI 10.1007/978-3-031-33180-0_14:

Let $k\geq2$. Then $E_k=1+\sum_{i=1}^{\infty}a_i k^{-i}$ where the sequence $(a_n/2)_{n\geq1}$ is A274199 in the On-Line Encyclopedia of Integer Sequences (OEIS).

A274199 has offset zero. The paper's displayed coefficients $2,4,6,10,16,\ldots$ fix $a_i=2,\mathrm{A274199}(i)$ for $i\geq1$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A word is over a $k$-letter alphabet | Lists over Fin k, enumerated by DepthBudgetIidGreedyOptimality.words | Uniform cardinality is $k^n$. | | $w=w_m\cdots w_1w_0w_1\cdots w_m$, with nonempty outer blocks and possibly empty centre | GreedyRenewal.IsBPFactorization | bs lists the outer blocks from the centre outwards; the two flattenings have the source order. | | Width $2m$ or $2m+1$ according to the centre | GreedyRenewal.bpWidth | The empty word has width zero, extending the source without affecting a limit at infinity. | | Largest width over all valid factorizations | GreedyRenewal.largestBPWidth | The supremum is over valid widths; its boundedness is proved. | | $E_{n,k}=k^{-n}\sum_{i=1}^n i,\mathrm{LBP}k(n,i)$ | LargestFactorizationExpectedWidth.expectedWidth | The definition is the uniform function-space average. The private expectedWidth_weighted_count identity gives the source's weighted width-count expression. | | A274199 is the limiting reverse row of A274190 | CompositionInversion.a274190_stable and a274190_diagonal_limit | For $N\geq2d$, the diagonal equals $g(2d,d)$. The triangle includes $g(n,0)=1$ and the recurrence for every positive column. | | Every integer $k\geq2$ and convergence to $1+2\sum{i\geq1}\mathrm{A274199}(i)k^{-i}$ | LargestFactorizationExpectedWidth.claim and result | claim states existence of every reverse-row limit, summability of the positively indexed series, and convergence of the uniform average to one plus that series. result applies coefficient_limit_results to all three clauses. |

The auxiliary parameters in border and peeling statements range over arbitrary list alphabets. They strengthen the supporting statements and add no hypothesis to the settling result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/BlockPalindromes/LargestFactorizationExpectedWidth.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresEquivalence provedFirst preprint 2026Why do two parking rules lead to the same count?We give an explicit bijection for every number of cars, every allowed increment choice and every prescribed vacancy.A canonical fixed-fiber bijection for circular two-choice parkingProved

The questionCan circular parking with two choices per car be matched explicitly to classical parking functions, while keeping the prescribed vacancy?

Why it mattersThe source already knew the counts agreed. Its open problem asks for a reversible construction explaining that agreement.

The ideaRotate the circular preferences to align their vacancy, then cut the circle there. The inverse restores each car’s ordered pair of choices.

Our contributionWe give an explicit bijection for every number of cars, every allowed increment choice and every prescribed vacancy.

Scope & next questionThe result uses the stated two-choice rule and its scan convention, including checking the second choice itself.

Verification details
Release source recordProved in Lean

Open problem

A canonical fixed-fiber bijection for circular two-choice parking

Open Problem 1 of the linked paper asks:

Give a bijective proof of the factorization of Corollary 1 for d=1, refining Theorem 2 to a canonical bijection between each anchor class {E=j} intersect C_kappa and the set of classical parking functions.

The relevant specialization has n>=1, m=n+1, s=2, and d=1. Every car has an ordered pair consisting of an anchor and a distinct second choice. The anchor is tried first. If it is occupied, the car begins a clockwise scan at the second choice itself, so offset zero must be inspected before later spots. The matrix kappa fixes the positive clockwise increment from anchor to second for every car, and j fixes the actual unique vacancy.

The exact formal target is result n hn increments j: for every natural n with 1<=n, every per-car increment matrix with entries in 1..n, and every circular spot j, it certifies Function.Bijective (fixedFiberEquiv n increments j). The referenced equivalence remains the explicit forward map and inverse between the corresponding literal actual fiber and classical length-n parking functions; result is its sole source-level settlement theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Parking/CircularTwoChoiceParkingBijection.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Cutting the literal circular process at its observed vacancy gives an explicit fixed-fiber equivalence with classical parking functions, while a separate global equivalence retains all original observables.

    Read source ↗
  • Historical context

    First preprint year: 2026.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Turning increasing trees into permutations—and backWe construct the forward and inverse maps for the entire stated tree family and prove both inverse identities for every positive parameter pair.Testart Question 1: the native tree side and a type-C permutation modelProved

The questionCan Testart’s increasing trees be matched explicitly to another family counted by type-C poly-Bernoulli numbers?

Why it mattersThe source asks for a bijective explanation. The target permutation family has a prescribed set of strict excedance positions.

The ideaRead each tree component into a directed cycle. In reverse, recover every parent from the rightmost preceding smaller label.

Our contributionWe construct the forward and inverse maps for the entire stated tree family and prove both inverse identities for every positive parameter pair.

Scope & next questionThe result answers the tree-side question. The published count of the permutation family remains an attributed literature result.

Verification details
Release source recordProved in Lean

arXiv

Testart Question 1: the native tree side and a type-C permutation model

Benjamin Testart, On minimal pattern-containing inversion sequences, arXiv:2602.12130v2, Section 7, Question 1, printed page 17, asks:

Find bijections between the families $\mathrm{IP}{n,k}$ or $\mathcal T{n,k}$ of Section 6 and other families of combinatorial objects known to be counted by the poly-Bernoulli numbers of type C.

The question offers an IP-side or a T-side answer. The target here is the entire T-side family, rather than the source's already published IP/T correspondence. Section 6 defines unordered rooted increasing trees. For every pair of integers $a,b\ge1$, write $N=a+b-1$. The source $T(a,b)$ consists of trees on distinct labels $0,\ldots,N$, rooted at zero, whose internal labels are exactly $0,\ldots,a-1$ and whose leaves are exactly $a,\ldots,N$. Every parent label is smaller than its child label. The target $P(a,b)$ consists of permutations of $[N]$ whose strict excedance-position set is exactly $[b-1]$, allowing fixed points. Here $[0]=\varnothing$.

The fixed forward map deletes zero, reads each component root-first by DFS with decreasing child-root labels, and makes each word a directed cycle, sending each letter to the next and the last to the first. If the resulting permutation is $\sigma$, the output is $F(t)=D_N(\sigma)$, where

$$ D_N(\sigma)(i)=N+1-\sigma(N-i)\quad(1\le i<N),\qquad D_N(\sigma)(N)=N+1-\sigma(N). $$

The fixed reverse map applies $D_N$, starts each directed cycle at its minimum without reversing direction, attaches every later letter to the rightmost preceding smaller letter, and attaches each cycle minimum to a restored zero root. Forgetting traversal order gives $H$. The target is both landings and both identities $H(F(t))=t$ and $F(H(\pi))=\pi$, for every source tree and target permutation and every positive $a,b$. There is no bound on $N$, no plane or binary restriction, and no omission of $a=1$, $b=1$, $N=1$ or singleton cycles. The unchanged fixed-map statement is specified in #22578.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PriorityLattice/TestartTraversal.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationStructural characterizationFirst preprint 2021Which spectra can detect symmetric two-qubit entanglement?Two sharp inequalities characterize the three ordered eigenvalues of symmetric two-qubit entanglement witnesses.The eigenvalues of symmetric two-qubit entanglement witnesses satisfy µ3 ≥ −√(µ1µ2)Proved

The questionWhat eigenvalues can a symmetric two-qubit entanglement witness have?

Why it mattersThe source knew the spectral region for decomposable witnesses. The full symmetric case left a gap.

The ideaSpectral transport and determinant bounds establish necessity; explicit witnesses establish sufficiency.

Our contributionTwo sharp inequalities characterize the three ordered eigenvalues of symmetric two-qubit entanglement witnesses.

Scope & next questionThe characterization concerns symmetric two-qubit entanglement witnesses.

Verification details
Release source recordProved in Lean

Open problem

The eigenvalues of symmetric two-qubit entanglement witnesses satisfy µ3 ≥ −√(µ1µ2)

G. Champagne, N. Johnston, M. MacDonald and L. Pipes, Spectral properties of symmetric quantum states and symmetric entanglement witnesses, Linear Algebra Appl. 649, 273–300 (2022), arXiv:2108.10405v1, study the eigenvalues of symmetric entanglement witnesses.

  • Witnesses. A symmetric entanglement witness is a Hermitian W on C^d ∨ C^d with (v ⊗ v)*W(v ⊗ v) ≥ 0 for all v (§2.3); positive semidefinite W are included (footnote 1).
  • Theorem 3. For d = 2 and real µ1 ≥ µ2 ≥ µ3, the inequalities (a) µ2 ≥ 0 and (b) µ3 ≥ −√(µ1µ2) are sufficient for the existence of a symmetric witness with these eigenvalues, and necessary for decomposable ones. Every symmetric witness satisfies (a) and a weaker inequality (c).
  • The conjecture. After Theorem 3 and in §6(c) (page 20): "We conjecture that inequalities (a) and (b) from that theorem provide the correct characterization."

The verbatim statements are in the literature note.

Issue #17536 fixes the reading:

  • C²∨C² is identified with C³ through the orthonormal Dicke basis e₀₀, (e₀₁ + e₁₀)/√2, e₁₁, in which v ⊗ v has coordinates (v₀², √2 v₀v₁, v₁²).
  • "Eigenvalues µ1 ≥ µ2 ≥ µ3" means that the characteristic polynomial is (X − µ1)(X − µ2)(X − µ3).
  • The conjecture: for all real µ1 ≥ µ2 ≥ µ3, a symmetric witness with these eigenvalues exists if and only if µ2 ≥ 0 and µ3 ≥ −√(µ1µ2).

The conjecture is proved in this reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsoluteSeparability/SymmetricQubitWitnessSpectrum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The spectra of symmetric two-qubit entanglement witnesses are characterized by two sharp inequalities on their three ordered eigenvalues.

    Read source ↗
  • Historical context

    First preprint year: 2021.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Connectivity of prefix normal generatorsValid flip-to-zero generators are connected by finite adjacent-position exchanges.Proved

The questionDuncan Adamson, Moritz Dudey, Pamela Fleischmann and Annika Huch, Word Chain Generators for Prefix Normal Words, arXiv:2508.19619v1, Conclusion, pp. 17–18, ask:

Why it mattersPublished Lemma 17 supplies ascent-exchange preservation; the global argument applies it through actual contextual swaps to a common descending normal form.

The ideaExchanging an adjacent ascent a<b changes only the state after the first of the two flips.

Our contributionValid flip-to-zero generators are connected by finite adjacent-position exchanges.

Verification details
Release source recordProved in Lean

Open problem

Connectivity of prefix normal generators

Duncan Adamson, Moritz Dudey, Pamela Fleischmann and Annika Huch, Word Chain Generators for Prefix Normal Words, arXiv:2508.19619v1, Conclusion, pp. 17–18, ask:

Is there a path from any prefix normal generator to all others, just by swapping neighbouring numbers?

The institutional 18-page accepted author manuscript (AAM) poses the same question. For every positive n, the vertices are permutations of 1,...,n whose complete flip-to-zero chains are prefix normal. Starting at the all-one word, each entry names the position to flip. All n+1 states, including both endpoints, must satisfy the comparison of every contiguous factor with the prefix of the same length. A move exchanges entries at adjacent positions; their distinct values can be arbitrary. Every intermediate permutation must remain a valid generator. The requested path is finite and may have zero moves. No shortest-path, direction or complexity requirement is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Permutations/PrefixNormalGeneratorConnectivity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Factorized ladders for generalized quantum Zernike HamiltoniansThe generalized polynomial quantum Zernike Hamiltonian has ladders with the factorized structure function for every N.Proved

The questionThe binding polynomial formulation is proved by D5.S3.Quantum.CampoamorStursbergZernikeLadder.result.

Why it mattersThe source computes low orders and conjectures the arbitrary-order quantum ladder factorization.

The ideaThis identity is an explanation of the mechanism, not a separately exported theorem.

Our contributionThe generalized polynomial quantum Zernike Hamiltonian has ladders with the factorized structure function for every N.

Verification details
Release source recordProved in Lean

Open problem

Factorized ladders for generalized quantum Zernike Hamiltonians

Campoamor-Stursberg, Herranz, Latini, Marquette and Blasco, Nonlinearity 39, 055008, arXiv:2502.02491v2, Conjecture 1(iii), asks for number and ladder operators constructed from the quantum symmetries (2.8), satisfying (2.9)–(2.10) with the factorization (5.1) for every $N\geq1$. Preregistration: #19957.

The binding polynomial formulation is proved by D5.S3.Quantum.CampoamorStursbergZernikeLadder.result. Operators are complex linear endomorphisms of $\mathbb C[q_1,q_2]$. For every $N\geq1$ and every complex coefficient sequence there are two symmetries with the kinetic terms and correction span of (2.6), and number and ladder operators in the unital algebra generated by $K_1,K_2,K_3$, satisfying all four identities of the claim.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Coordinate multiplication and $-i$ times partial differentiation, (2.1) | Core q, p | Fin 2 indices 0 and 1 encode source coordinates 1 and 2. | | Hamiltonian and angular momentum, (2.3), (2.5) | Core dot, hamiltonian, angularMomentum | Composition of polynomial endomorphisms; arbitrary complex coefficients. | | Normal-ordered correction monomials of positive degree at most $2N$ and grading zero, (2.6) | Core monomial, corrections | The defining span uses exactly these conditions. | | Two commuting symmetries with prescribed kinetic terms, (2.6)–(2.7) | Core IsSymmetry; both first conjuncts of claim | Momentum-order upper bound; no exact-order assertion at degenerate coefficients. | | $K_1=C$, $K_2=(I'-I)/2$, $K_3=[K_1,K_2]$, (2.8) | Core k1, k2, k3 | Literal binding definitions. | | Operators constructed from these three generators | Three Algebra.adjoin memberships in claim | The unital generated algebra; no extra hypotheses. | | Raising and lowering commutators, (2.9) | First two final conjuncts of claim | Equalities of endomorphisms. | | $K_+K_-=\Phi(H,K)$ and the shifted commutator, (2.9)–(2.10) | Last two conjuncts of claim | Core phi1, phi2, phi give exactly (5.1). | | Every order and every coefficient choice | Universal binders of claim | $N\geq1$; only indices 1 through $N$ are used. |

The divided-difference definitions are auxiliary polynomial expressions. The claim has no localization, invertibility, nondegeneracy or analytic-domain hypothesis. No clause of the binding claim is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/CampoamorStursbergZernikeLadder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationStructural characterizationFirst preprint 2026When can a matrix mean reveal commutation?We give necessary and sufficient scalar conditions for the stated characterization over complex matrices, uniformly in matrix size.Characterizing rigid alternative meansProved

The questionWhich alternative matrix means detect whether two positive matrices commute through their linearizability?

Why it mattersThe question links scalar functions to matrix commutation and spectral structure.

The ideaInjectivity and strict convexity or concavity turn the matrix question into conditions on scalar functions.

Our contributionWe give necessary and sufficient scalar conditions for the stated characterization over complex matrices, uniformly in matrix size.

Scope & next questionThe formal result is over complex matrices. Extensions to other scalar fields are discussed separately.

Verification details
Release source recordProved in Lean

Open problem

Characterizing rigid alternative means

R. Dumitru, J. Franco, and A. Merino, On The Linearization of Alternative Means, arXiv:2606.25150v1, §1, ask in the supplied verbatim text:

Question : Can we characterize all functions f : (0, ∞) → (0, ∞) such that Aσ̂_f B is linearizable if and only if AB = BA and |Spec(A⁻¹♯B)| ≤ 2 ?

Remark 4.4 restates this question. Definition 2.3 specifies $A\widehat\sigma_fB=f(Y)Af(Y)$ with $Y=A^{-1}\sharp B$, for continuous positive $f$ with $f(1)=1$. Remark 2.2 calls a mean linearizable when it equals $aA+bB$ for real $a,b$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/AlternativeMeanLinearization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    A scalar characterization of alternative means whose linearizability detects commutation and at most two spectral values.

    Read source ↗
  • Historical context

    First preprint year: 2026.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPosed in 2026A132393: Kurkov's nested Stirling identityKurkov's full nested rational sum equals the unsigned Stirling number.Proved

The questionConjecture: T(n+m+1,m+1) = (n+m)! (Product{i=1..m} (n+i)^i) Sum{j1=0..n} Sum{j2=0..j1} ...

Why it mattersThis is the first-tier 2026 external conjecture selected by programme 8654.

The ideaA decreasing source chain gives 1 <= l1 < ... < lm <= N.

Our contributionKurkov's full nested rational sum equals the unsigned Stirling number.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A132393: Kurkov's nested Stirling identity

Conjecture: T(n+m+1,m+1) = (n+m)! * (Product_{i=1..m} (n+i)^i) * Sum_{j_1=0..n} Sum_{j_2=0..j_1} ... Sum_{j_m=0..j_{m-1}} (-1)^(n*m + j_1 + j_2 + ... + j_m) * (Product_{1 <= q < p <= m} (j_q - j_p + p - q)^2) * Product_{t=1..m} (n+t-1){j_t} / ((n-j_t+t)^(m+1) * (j_t+m-t)!) where (x){n} is the falling factorial defined as (x){n} = Product{k=0..n-1} (x-k). - Mikhail Kurkov, May 24 2026

The target quantifies over every natural n and every m >= 1, with equality in the rationals. T is the unsigned Stirling number of the first kind. The chain is n >= j_1 >= ... >= j_m >= 0. Falling factorial means the displayed descending product. This is the exact source assertion registered before formal probes in https://github.com/the-omega-institute/trureturing/issues/9456.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KurkovNestedStirlingIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedWhy integer coefficients survive every positive powerWe prove closure under every positive pointwise power over every characteristic-zero field, at every degree including zero.Hanna's universal logarithmic integer-series closureProved

The questionIf a sequence gives an integer series after a logarithmic change of coordinates, does raising its terms to a positive power preserve that property?

Why it mattersThree sequence entries record the same general question. Its scope is every admissible sequence and every positive power, beyond the individual examples.

The ideaReconstruct the sequence in an integer basis, prove that products stay integral degree by degree, and use induction for all positive powers.

Our contributionWe prove closure under every positive pointwise power over every characteristic-zero field, at every degree including zero.

Scope & next questionThe result concerns formal series. It requires no analytic convergence or finite-support assumption.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's universal logarithmic integer-series closure

Conjecture: For all integers m > 0, Sum_{n>=0} L(n)^m * log(1+x)^n/n! is an integer series whenever Sum_{n>=0} L(n)*log(1+x)^n/n! is an integer series.

A167138, A167139 and A167141 repeat this one general Hanna question. An integer series has integer ordinary coefficients. Let K be any characteristic-zero field and L any function from natural numbers to K. Put E_L(t)=sum L(n)t^n/n! and ell(X)=log(1+X). The exact conclusion is integrality of E_(L^m)(ell(X)) for every positive integer m whenever E_L(ell(X)) is integral, at every degree including zero. The natural positive exponent in Lean is the equivalent representation of a positive integer exponent. L(0) is arbitrary. No extra integral-L assumption, analytic convergence, finite support or degree/power bound is introduced.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaLogarithmicPowerClosure.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Even-legged hyperlinks reconstruct the entanglement entropies with N-independent leg-factorsUniversal rational leg factors reconstruct every real complement-symmetric set function vanishing at the empty set from its nonempty even hyperlinks.Proved

The questionRodríguez-Laguna, The hyperlink representation of entanglement and the inclusion-exclusion principle, arXiv:2601.17926v2, Phys.

Why it mattersUniversal leg-factors also make the reconstruction usable at every system size without solving a new linear system.

The ideaThe theorem is proved for every real set function S on the subsets of N parties with S(∅) = 0 and S(A) = S(Ω ∖ A).

Our contributionUniversal rational leg factors reconstruct every real complement-symmetric set function vanishing at the empty set from its nonempty even hyperlinks.

Verification details
Release source recordProved in Lean

Open problem

Even-legged hyperlinks reconstruct the entanglement entropies with N-independent leg-factors

S. N. Santalla, S. Singha Roy, G. Sierra and J. Rodríguez-Laguna, The hyperlink representation of entanglement and the inclusion-exclusion principle, arXiv:2601.17926v2, Phys. Rev. A 114, 022431 (2026), define the entanglement hyperlinks J_I = Σ_{A⊆I} (−1)^{|I|−|A|} S_A of the entanglement entropies S_A of a pure state on N parties (Eq. (5)).

Appendix B proposes the even-legged reconstruction S_A = Σ_{|I| even} Λ_{|I|,|I∩A|} J_I (B2), with Λ_{2l,0} = Λ_{2l,2l} = 0 and Λ_{2l,p} = Λ_{2l,2l−p}. It lists the leg-factors for 2l ≤ 8 (B5) and states: "It is still to be proved that the leg-factors Λ_{2l,p} exist and do not depend on N" and "the validity of Eq. (B2), with leg-factors that do not depend on N, has only been checked up to N = 9". The verbatim statements are in the literature note.

Issue #16381 fixes the reading: there are leg-factors, independent of N, with the zero and symmetry laws, such that (B2) holds for every N. It is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/EvenLeggedHyperlinkReconstruction.result

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Discrete structuresQuestion resolvedBinomial perfect-power counts over GF(2)Proved

The questionKagey's OEIS A350532 COMMENT states verbatim:

Why it mattersSquaring in characteristic two doubles exponents and preserves support size.

The ideaThe proof uses the sparse forms of that power: divisibility of its derivative forces the derivative to vanish.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Binomial perfect-power counts over GF(2)

P. Kagey's OEIS A350532 COMMENT states verbatim:

Conjecture: T(2*n,2) = n.

The settled statement is $T(2n,2)=n$ for every $n\in\mathbb N$, including zero.

The source's NAME defines the count verbatim:

Triangle read by rows: T(n,k) is the number of degree-n polynomials over Z/2Z of the form f(x)^m for some m > 1 with exactly k nonzero terms; 1 <= k <= n + 1.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Degree-n polynomials over Z/2Z, exactly k nonzero terms, some exponent m > 1 | KageyA350532Binomials.T | The literal cardinality subtype; distinct nonzero polynomials, with witnesses for base and exponent. | | Conjecture: T(2*n,2) = n. | KageyA350532Binomials.claim and KageyA350532Binomials.result | Literal universal assertion under the preregistered conventions. | | No restriction to positive degree in the definition | p ≠ 0 together with p.natDegree | The preregistered nonzero convention prevents zero from entering the degree-zero count. | | Natural-number variables | The natural-number binders and arithmetic | Subtraction and division in the trinomial assertion are natural-number operations. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Combinatorics/KageyA350532Binomials.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture provedA398542 fixed-bottom polynomial conjectureRational polynomial witnesses for every actual fixed-bottom completion count.Proved

The questionCharles Cornell Norton, OEIS A398542, August 1, 2026; revision 18, August 30, 2026, 18:45:04, exact selected %F:

Why it mattersThe target is the whole fixed-bottom assertion, not a bounded fit or a recursively defined count.

The ideaInsert upper permutations into weakly increasing gaps of the same complete bottom b.

Our contributionRational polynomial witnesses for every actual fixed-bottom completion count.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A398542 fixed-bottom polynomial conjecture

Charles Cornell Norton, OEIS A398542, August 1, 2026; revision 18, August 30, 2026, 18:45:04, exact selected %F:

Conjecture: for every m >= 1 and every b in S(m), d(b,k) = p(k)*binomial(2*k,k) + q(k)*4^k for polynomials p and q depending on b, with deg p <= m-1 and deg q <= m-2. Verified for m <= 6 and k <= 12.

S(m)=Av_m(132). For every m>=1 and b in S(m), choose p_b,q_b in Q[X] independently of k such that the displayed identity holds for every k>=0. The degree bounds are m-1 and m-2 for m>=2, with q_b=0 when m=1. Here d(b,k) counts actual permutations of 1,...,m+k avoiding 1324 whose lower-value subsequence is exactly b and whose standardized upper-value subsequence avoids 213. The value cut is fixed, the empty upper cell is allowed, and d(b,0)=1. Rational coefficients are an explicit sufficient choice beyond the source's unspecified coefficient field.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/A398542Polynomial.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsConjecture provedFirst preprint 2012Adamaszek's Conjecture 7.4 on necklace cycle lengthsAdamaszek's necklace transformation combines simultaneous jumps, vector changes, and corrections at facing gaps of length three.Proved

The questionAdamaszek, Hard squares on cylinders revisited, arXiv:1202.1655v2, Section 7, Conjecture 7.4, states:

Why it mattersThis proves the source's equivalent return statement on the isometry classes used as the vertices of Neck(k, n).

The ideaSort the occupied integer sites and extend them to positions xᵢ on ℤ, with xᵢ₊₂ₖ=xᵢ+n and periodic vectors vᵢ.

Our contributionAdamaszek's necklace transformation combines simultaneous jumps, vector changes, and corrections at facing gaps of length three.

Verification details
Release source recordProved in Lean

arXiv

Adamaszek's Conjecture 7.4 on necklace cycle lengths

M. Adamaszek, Hard squares on cylinders revisited, arXiv:1202.1655v2, Section 7, Conjecture 7.4, states:

The length of every cycle in the graph Neck(k, n) divides n−3k. In other words, for every (k, n)-necklace N we have Tⁿ⁻³ᵏN = N.

The section assumes even positive circumference n and positive k. A necklace has 2k distinct stones, each with signed vector −2, −1, 1 or 2. Consecutive stones have opposite directions. Away-facing gaps are odd; toward-facing gaps plus both vector lengths are odd. A toward-facing gap is at least three, and at gap three both vectors have length one. Necklaces are identified under circle isometries. T performs simultaneous JUMP, then TURN (−2→1, −1→2, 1→−2, 2→−1), then FIX of facing gaps three by shortening length-two vectors to one. Issue #11582 fixes this reading before the probe.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/HardCore/HardSquareNecklacePeriod.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2019A regular subset of a four-dimensional hyperdisk has no finite hyperdisk coverA regular subset of a four-dimensional hyperdisk has no finite hyperdisk cover.Refuted

The questionHu, Masking quantum information on hyperdisks, Phys.

Why it mattersHyperdisks retain the coefficient moduli of pure states while allowing all independent phases, and regular subsets are the linear-span intersections used in the paper's description of target states of masking machines.

The ideaThe lemma ambienthyperdisk proves that ambient is a hyperdisk using the standard coordinate family and the positive radii.

Our contributionA regular subset of a four-dimensional hyperdisk has no finite hyperdisk cover.

Verification details
Release source recordRefuted in Lean

Open problem

A regular subset of a four-dimensional hyperdisk has no finite hyperdisk cover

F. Ding and X. Hu, Masking quantum information on hyperdisks, Phys. Rev. A 102, 042404 (2020), arXiv:1909.11256v3, state in the conclusion at line 651 of masking.tex:

Nevertheless, it is an open question for a nondegenerate masking machine whether the number of hyperdisks in the set of maskable states is always finite. A related open question is whether a regular subset of a hyperdisk can be covered with finite number of hyperdisks.

The settled question is the second sentence. In the module, a hyperdisk is a set of unit vectors in EuclideanSpace ℂ (Fin n) represented by an orthonormal family φ, positive real coefficient moduli r, equality of the span of φ with the span of the set, and equality with the full unit-vector level set having those coefficient moduli. A regular subset c of s satisfies IsHyperdisk s and (Submodule.span ℂ c : Set _) ∩ s = c. The claim asks, for every natural n and every such c and s, for a natural k and a map h : Fin k → Set (EuclideanSpace ℂ (Fin n)) such that every h i is a hyperdisk contained in c and c = ⋃ i, h i.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Masking/HyperdiskRegularSubsetCover.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture refutedFirst preprint 2022Kumar–Peng–Tran Conjecture 4.4 fails at modulus 17Three unit bases modulo seventeen have greater inefficiency than base two.Refuted

The questionTran, Integer Complexity Generalizations in Various Rings, arXiv:2211.04379v1, Section 4, Conjecture 4.4:

Why it mattersThe settlement supplies a counterexample to the individual-base reading of the source's third-place assertion.

The ideaThe theorem memViff proves the recurrence exactly characterizes expression values.

Our contributionThree unit bases modulo seventeen have greater inefficiency than base two.

Verification details
Release source recordRefuted in Lean

Open problem

Kumar–Peng–Tran Conjecture 4.4 fails at modulus 17

A. Kumar, S. Peng and V. Tran, Integer Complexity Generalizations in Various Rings, arXiv:2211.04379v1, Section 4, Conjecture 4.4:

E(m, 2) is the third most inefficient for odd m.

The source defines expressions from copies of $x$ using addition, multiplication and parentheses, evaluates them in $\mathbb{Z}/m\mathbb{Z}$ at a base $s$, and defines $E(m,s)$ as the maximum expression complexity over residues. The binding proposition reads “third” by individual unit bases: for every odd $m \ge 3$, at most two units $s$ satisfy $E(m,2)<E(m,s)$. Bases $3$ and $16$ tie in the witness, so dense ranking by distinct inefficiency values is outside this proposition.

Statement echo

  • The universal modulus binder is $m : \mathbb{N}$ with an anonymous $[\operatorname{NeZero}(m)]$ instance.
  • The hypotheses are $3 \le m$ and $\operatorname{Odd}(m)$.
  • The base is the residue $2 : \mathbb{Z}/m\mathbb{Z}$.
  • The comparison domain is all units $s : (\mathbb{Z}/m\mathbb{Z})^\times$, coerced to $\mathbb{Z}/m\mathbb{Z}$.
  • The conclusion is that the cardinality of the filtered units satisfying $E(m,2)<E(m,s)$ is at most $2$.
  • The expression, leaf-count, evaluation, complexity and inefficiency clauses are represented by Expr, Expr.leaves, Expr.eval, complexity and inefficiency in the settling module.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/KumarPengTranModularComplexity.result

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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresQuestion resolvedFirst preprint 2024The even-parameter descent dichotomy for Lucas sequencesA machine-checked proof of Conjecture 5.2 on Lucas descent at all integer indices.Proved

The questionConjecture 5.2 of arXiv:2408.14632v2, as printed:

Why it mattersThe doubling hypotheses are the standard entry into descent arguments for Lucas sequences, and the even-parameter case is the one the paper's own Convention 3.20 excludes from its main development.

The ideaWork in the unit group with the companion matrix M = !![p, -q; 1, 0], whose integer powers carry U and V as entries and trace.

Our contributionA machine-checked proof of Conjecture 5.2 on Lucas descent at all integer indices.

Verification details
Release source recordProved in Lean

arXiv

The even-parameter descent dichotomy for Lucas sequences

Conjecture 5.2 of arXiv:2408.14632v2, as printed:

Set U_n := U_n(p, q), and let m >= 1 and n in Z. Assume that both p and m are even. If U_{2n} = 0 (mod m) and U_{2n+1} = q^n (mod m), then exactly one of the following two conclusions occur: (i) U_n = 0 (mod m), or (ii) U_n = m/2 (mod m) where n = c * e_U(m)/2, where c is some odd integer and e_U(m) is the entry point of m in (U_n).

The paper adds a second sentence: "Can we give conditions for when Conclusion (i) versus Conclusion (ii) occurs?"

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LucasEvenDescent.conjecture_five_two

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2024The K-number and extreme-point K-sets of real finite-dimensional sup-norm spacesFor real sup-norm spaces of dimension n ≥ 2, an extreme-point set is a K-set exactly when it spans, and the K-number is n.Proved

The questionPaul, On local preservation of orthogonality and its application to isometries, arXiv:2407.08900v1, Linear Algebra and its Applications (2024), states:

Why it mattersLocal preservation of Birkhoff–James orthogonality can force a global scalar isometry.

The ideaA nonzero norm-scaling linear endomorphism is a scalar multiple of a linear isometry.

Our contributionFor real sup-norm spaces of dimension n ≥ 2, an extreme-point set is a K-set exactly when it spans, and the K-number is n.

Scope & next questionThe classification of all such minimal K-sets remains open.

Verification details
Release source recordProved in Lean

arXiv

The K-number and extreme-point K-sets of real finite-dimensional sup-norm spaces

D. Sain, J. Manna, K. Paul, On local preservation of orthogonality and its application to isometries, arXiv:2407.08900v1, Linear Algebra and its Applications (2024), states:

Definition 1. Let X be a Banach space. A set A ⊂ S_X is said to be a Koldobsky-set, for the sake of brevity, a K-set, if any operator T ∈ L(X) that preserves Birkhoff-James orthogonality at each point of A, is necessarily a scalar multiple of an isometry.

Definition 3. Let X be a Banach space having a finite K-set. Then the K-number of X is defined by min{|A| : A is a K-set of X}. We denote the K-number of X by κ(X).

The Open Question at the end of §2 is:

Open Question: Characterize the minimal K-sets in ℓ^n_∞ and hence find κ(ℓ^n_∞).

The spaces are real, and minimality is under inclusion. The result determines the K-number for $n\ge2$ and characterizes the minimal K-sets consisting of extreme points; it leaves the classification involving non-extreme points open.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BanachGeometry/LinfKNumber.result

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Quantum informationSharp bound provedSource 2000How much entanglement can four qubits share?We prove the exact maximum average two-qubit marginal entropy, 1 + ½ log₂ 3 bits, and show that the Higuchi–Sudbery state attains it.Higuchi Sudbery Entropy MaximumProved

The questionFour qubits cannot be maximally entangled across every two-versus-two split. What is the largest possible average?

Why it mattersHiguchi and Sudbery asked this extremal question in 2000. It concerns the limits of distributing entanglement across a four-qubit pure state.

The ideaAn entropy majorant reduces the optimization to a coupled polynomial inequality. An exact sum-of-squares identity proves the bound for all normalized four-qubit pure states.

Our contributionWe prove the exact maximum average two-qubit marginal entropy, 1 + ½ log₂ 3 bits, and show that the Higuchi–Sudbery state attains it.

Scope & next questionThe result concerns the average entropy of the three complementary two-qubit cuts of a pure four-qubit state.

Verification details
Release source recordProved in Lean

Open problem

higuchi sudbery 2000 four qubit average entropy maximum

For every normalized pure state $\psi\in(\mathbb C^2)^{\otimes4}$, the average two-qubit marginal von Neumann entropy in bits satisfies

$$\frac{S(\rho_{AB})+S(\rho_{AC})+S(\rho_{AD})}{3}\le 1+\frac12\log_2 3.$$

Equality is attained at the explicit Higuchi–Sudbery state

$$|M_4\rangle=\frac1{\sqrt6}\bigl[|0011\rangle+|1100\rangle+\omega(|1010\rangle+|0101\rangle)+\omega^2(|1001\rangle+|0110\rangle)\bigr],\qquad\omega=e^{2\pi i/3}.$$

Higuchi–Sudbery, Section 3: “Given that a four-qubit state cannot have maximal entropy of entanglement for every two-qubit subset, we now ask what is the greatest possible average for such entropies, i.e. we seek to maximise”. Their average over the six pairs equals the displayed average over the three complementary cuts.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/HiguchiSudbery/HiguchiSudberyEntropyMaximum.result

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About this highlight
  • Contribution

    The explicit Higuchi–Sudbery state attains the sharp upper bound for the average two-qubit marginal entropy.

    Read source ↗
  • Historical context

    Year recorded in the source bibliography: 2000.

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Quantum informationQuestion resolvedFirst preprint 2013Datta–Dorlas–Jozsa–Benatti: subadditivity of subentropySubentropy is subadditive for every finite bipartite density state.Proved

The questionSubentropy is subadditive: Q(ρAB) ≤ Q(ρA) + Q(ρB).

Why it mattersThe spectral formula and its unique continuous extension are represented by the stage-1 SubentropyLike family; the determinant-integral subentropy is its canonical member.

The ideaFor nonnegative finite vectors the entropy Jensen gap

Our contributionSubentropy is subadditive for every finite bipartite density state.

Verification details
Release source recordProved in Lean

Open problem

Datta–Dorlas–Jozsa–Benatti: subadditivity of subentropy

Conjecture 1. Subentropy is subadditive: Q(ρAB) ≤ Q(ρA) + Q(ρB).

The supplied locator is arXiv:1310.1312v1, §2.3.1, Conjecture 1. The related restatement in arXiv:1510.08859v2, §7.1 is:

At present, we do not know wether the subadditivity inequality of subentropy is true or not

The assertion concerns every finite-dimensional bipartite density state, with no commutation, rank or separability restriction. In the canonical partial-trace interface, marginalRight retains A and marginalLeft retains B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Subentropy/Subadditivity.subentropy_subadditive

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Discrete structuresQuestion resolvedFirst preprint 2017Davis's coprime width-descent difference formulaCoprime-width descent differences have a shifted Eulerian distribution.Proved

The questionRobert Davis, Width-k Generalizations of Classical Permutation Statistics, JIS 20 (2017), gives the following descent-set line, generating function, and MacMahon identification together at the top of printed page 2.

Why it mattersThe question asks for a uniform identity over every coprime pair, rather than a finite verification or a coefficientwise approximation.

The ideaThe source proof has two live local deductions inside result.

Our contributionCoprime-width descent differences have a shifted Eulerian distribution.

Verification details
Release source recordProved in Lean

Open problem

Davis's coprime width-descent difference formula

Robert Davis, Width-k Generalizations of Classical Permutation Statistics, JIS 20 (2017), gives the following descent-set line, generating function, and MacMahon identification together at the top of printed page 2. Printed page 1 ends with the preceding descent-count formula:

where Des σ = {i ∈ [n − 1] | a_i > a_{i+1}}. Given any statistic st, one may form the generating function F_n^{st}(q) = Σ_{σ∈S_n} q^{st σ}. A famous result due to MacMahon [6] states that F_n^{des}(q) = F_n^{exc}(q), and that both are equal to the Eulerian polynomial A_n(q). The Eulerian polynomials themselves may be defined via the identity Σ_{j≥0} (1 + j)^n q^j = A_n(q)/(1 − q)^{n+1}.

Printed page 2 defines width-k descents:

For each of the following definitions, we assume n ∈ Z_{>0}, k ∈ [n − 1], ∅ ≠ K ⊆ [n − 1], and σ = a_1 a_2 ··· a_n ∈ S_n. We define a width-k descent of σ to be an index i ∈ [n − k] for which a_i > a_{i+k}. Thus the width-1 descents are the usual descents of a permutation. Let Des_k(σ) = {i ∈ [n − k] | a_i > a_{i+k}} denote the set of all width-k descents of σ, and set des_k(σ) = |Des_k(σ)|.

Printed page 6 introduces the Laurent polynomial and the named conjecture:

We now show that interesting behavior occurs when considering the function G_{n,k}(q) = Σ_{σ∈S_n} q^{des_k(σ)−des_{n−k}(σ)}. According to computational data, the following conjecture holds for all n ≤ 9 and 1 ≤ k < n for which gcd(k, n) = 1.

Conjecture 9. If gcd(k, n) = 1, then G_{n,k}(q) = n q^{1−k} A_{n−1}(q).

The formal theorem makes the surrounding range explicit: for all natural n and k, if 1 <= k, k < n, and Nat.Coprime k n, then

G(n,k) = n * q^(1-k) * eulerian(n-1).

Here eulerian(m) is the descent generating function that the paper identifies with A_m by MacMahon's theorem. The paper's separate infinite-series definition of the Eulerian polynomials is not formalized. One-based positions are transported to Fin n, and signed exponents live in the Laurent polynomial ring over the integers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/DavisWidthDescentDifferenceCoprime.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2022Resolved-count coherent-state weak valuesEvery resolved coherent-state weak value differs from the no-click value by the count times the single-photon weak value.Proved

The questionWiseman, Steinberg and Hallaji, “Obtaining a single-photon weak value from experiments using a strong (many-photon) coherent state”, AVS Quantum Science 5, 024401 (2023), arXiv:2211.15761v2, Section V, page 6:

Why it mattersThe public conclusion concerns exact coherent states with unbounded occupation, rather than a finite photon-number cutoff.

The ideaPassiveFockModel.exponentialtotal uses induction on the number of modes and uniqueness of analytic coefficients.

Our contributionEvery resolved coherent-state weak value differs from the no-click value by the count times the single-photon weak value.

Scope & next questionIts formalization and independent verification remain open.

Verification details
Release source recordProved in Lean

Open problem

Resolved-count coherent-state weak values

Wiseman, Steinberg and Hallaji, “Obtaining a single-photon weak value from experiments using a strong (many-photon) coherent state”, AVS Quantum Science 5, 024401 (2023), arXiv:2211.15761v2, Section V, page 6:

This supports earlier approximate results which suggested that if a photon-number resolving detector were used, the weak value post-selected on detecting m photons would exceed the no-click weak value by precisely m times the single-photon number. We leave a rigorous proof of this conjecture for future work.

Issue #17772 fixes the finite-mode Fock conventions. For every $k+1$-mode passive network $A,B$, coherent amplitude $\alpha\ne0$, and natural count $m$ of nonzero resolved probability, the conclusion is $W_m-W_0=mw$ when $m>0$. For $m=0$ the difference is zero without requiring a single-photon denominator. The detector acts on mode zero and ignores the remaining modes; the probed mode is also indexed zero, with arbitrary passive evolution between probing and detection.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Arbitrary passive linear network | Every natural $k$ and $A,B:\operatorname{EuclideanSpace},\mathbb C,(\operatorname{Fin}(k+1))\simeq_{\mathbb C}^{\mathrm{iso}}\operatorname{EuclideanSpace},\mathbb C,(\operatorname{Fin}(k+1))$ | No fixed dimension, network or intensity | | Coherent input in one mode, vacuum in the others | coherentInput k α, EuclideanSpace.single (0 : Fin (k+1)) α | Literal binding conventions | | Weak measurement of number, with other modes ignored | number, numberProjector m, generalizedWeakValue | Unbounded occupation; exact number-domain proofs; projector retains ignored-mode carrier | | Detect exactly $m$ photons | Every natural $m$ with coherentProbability A B α m ≠ 0 | Conditional values retain nonzero-denominator certificates | | Compare with no click | Existential nonzero vacuum probability and coherentWeakValue ... 0 | Vacuum condition is proved, not an added hypothesis | | Precisely $m$ times the single-photon value | claim, proved by result | Same $A,B$ and probed mode; the positive-count branch proves nonzero single-photon probability | | Zero count | The explicit if m = 0 branch of claim | Difference zero; no single-photon value assigned when its probability vanishes |

All public model definitions and the claim retain the mathematical bodies fixed by #17772; its mode-amplitude carrier is written as $\operatorname{EuclideanSpace},\mathbb C,(\operatorname{Fin} k)$ without a forwarding declaration. BeamSplitter.exponentialCoeff and BeamSplitter.exponentialVector supply the one-mode coefficients and vectors. The arbitrary finite-mode completion and totalness are supplied by PassiveFockModel.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/ResolvedCountWeakValue.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument gives a structural characterization or normal form.

    Source 1
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2024Degree generating function of Associated Mersenne graphsDegree enumeration for labelled circular run-constrained words.Proved

The questionWei and Yang, Associated Mersenne graphs, arXiv:2407.08237v1, Question 6.2: “For given n and k, how many vertices of Associated Mersenne graph ℳₙ have degree k?” The vertices are labelled binary words for which every circular run of ones is followed by a strictly longer run of zeros.

Why it mattersThe Library entry is Library/Words/wei2024associatedmersenne.md.

The ideaThe settling theorem combines these relations.

Our contributionDegree enumeration for labelled circular run-constrained words.

Verification details
Release source recordProved in Lean

Open problem

Degree generating function of Associated Mersenne graphs

Wei and Yang, Associated Mersenne graphs, arXiv:2407.08237v1, Question 6.2: “For given n and k, how many vertices of Associated Mersenne graph ℳₙ have degree k?” The vertices are labelled binary words for which every circular run of ones is followed by a strictly longer run of zeros. Adjacent vertices differ at one position. The conventions are $\mathbf M_0={\lambda}$ and $\mathbf M_1={0}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/AssociatedMersenne/DegreeGeneratingFunction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2016Lin: the g-property of trace augmentationTrace augmentation gives the geometric-mean comparison in every dimension.Proved

The questionThe original locator is Conjecture 3.1, printed page 124, of the paper identified by DOI 10.13001/1081-3810.3108.

Why it mattersIt asks for each individual singular value, beyond a bound on only the largest singular value or on a sum.

The ideaGram factorization W=ZZ with row blocks U,V gives A=UU, B=VV, X=UV.

Our contributionTrace augmentation gives the geometric-mean comparison in every dimension.

Verification details
Release source recordProved in Lean

Open problem

Lin: the g-property of trace augmentation

The original locator is Conjecture 3.1, printed page 124, of the paper identified by DOI 10.13001/1081-3810.3108. The fixed restatement is arXiv:1609.08231v1, §3, Example 3.5. Its setup states:

we assume that [A X; X∗ B], where A, X, B are n × n, is positive semidefinite.

Definition 1.1(iv) states:

M is said to have g-property if sj (M12 ) ≤ λj (M11 ♯ M22 ), j = 1, . . . , n.

Example 3.5 states:

[Φ(A) Φ(X); Φ(X∗) Φ(B)] where Φ : X ↦ X+(trX)In , is PPT … Numerical experiments suggest that the matrix [Φ(A) Φ(X); Φ(X∗) Φ(B)] has g-property, which we haven't been able to prove yet.

The concluding §5 refers to:

Besides the challenging problems described in Example 3.5 and Example 3.6, …

For every natural dimension, all complex matrices A, X, B with positive semidefinite block matrix W=[A X; X* B], and every decreasing index j, the target is

$$ s_j(\Phi(X))\le\lambda_j(\Phi(A)#\Phi(B)), \qquad \Phi(T)=T+(\operatorname{Tr}T)I. $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/LinGProperty/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2015Relative entropy coherence is bounded by l1 coherenceThe l1 norm of coherence bounds relative entropy coherence in every finite dimension.Proved

The questionLewenstein, Trace-distance measure of coherence, Phys.

Why it mattersThe comparison bounds the entropy gained by complete dephasing using only the absolute values of off-diagonal entries.

The ideaAudenaert's source note locates Theorems 8, 9 and 14 of Quantum Skew Divergence.

Our contributionThe l1 norm of coherence bounds relative entropy coherence in every finite dimension.

Verification details
Release source recordProved in Lean

Open problem

Relative entropy coherence is bounded by l1 coherence

S. Rana, P. Parashar and M. Lewenstein, Trace-distance measure of coherence, Phys. Rev. A 93, 012110, arXiv:1511.01854v2, Section III.2, Conjecture 7:

Conjecture 7. For all states ρ, C_{l1}(ρ) ≥ C_r(ρ).

The resolution is Proved for every finite dimension and every complex density matrix, including singular states. Preregistration #16965 fixes the conventions. The source note locates the conjecture and the definitions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every finite-dimensional state | claim quantifies over every natural n and every complex matrix on Fin n; IsState is positive semidefiniteness and trace one | No invertibility or rank hypothesis. In dimension zero the state premise is impossible. | | Section II, Eq. (3b), off-diagonal absolute-value sum | D5.S3.Quantum.Information.CorrelatedCoherenceMonogamyRefutation.l1Coherence | The complex norm gives the absolute value; both indices range over the full basis. | | Section III after Eq. (13), base-two entropy | vonNeumann2 | Negative real trace of continuous functional calculus for $x\log_2x$, with its value zero at zero. | | $C_r(\rho)=S_2(\Delta\rho)-S_2(\rho)$ | coherenceRel | The diagonal map retains exactly the diagonal entries. | | $C_{l1}(\rho)\ge C_r(\rho)$ | claim, proved by result | Literal preregistered inequality. |

IsState and vonNeumann2 have the unchanged defining bodies shared with QuantumTriangularJensenShannonRefutation; the latter imports their general owner. The claim uses the frozen D5.S3.Quantum.Information.CorrelatedCoherenceMonogamyRefutation.l1Coherence for $\sum_{i\ne j}|\rho_{ij}|$; exchanging the branches of its diagonal test gives the same off-diagonal sum term by term. The state and entropy definitions retain the source conventions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/CoherenceL1RelativeEntropyBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2021The minimal separable numerical range ratio of one two-qubit observableSzymański and K. Życzkowski (arXiv:2107.04365): the minimal volume ratio mu{2,1} equals 1/2.Proved

The questionSzymański, K. Życzkowski, Confident entanglement detection via the separable numerical range, Phys.

Why it mattersThe conjecture fixes the best possible ratio for one two-qubit observable.

The ideaThe proof uses spectral estimates.

Our contributionSzymański and K. Życzkowski (arXiv:2107.04365): the minimal volume ratio mu{2,1} equals 1/2.

Verification details
Release source recordProved in Lean

Open problem

The minimal separable numerical range ratio of one two-qubit observable

T. Simnacher, J. Czartowski, K. Szymański, K. Życzkowski, Confident entanglement detection via the separable numerical range, Phys. Rev. A 104, 042420 (2021), arXiv:2107.04365v1, Conjecture 8:

For a single measurement on a two-qubit system, the minimal volume ratio is $\mu_{2,1} = \frac{1}{2}$.

Here $\mu_{2,1}$ is the minimum over Hermitian two-qubit observables $A$ of $\operatorname{vol}L_{\mathrm{Sep}}(A)/\operatorname{vol}L(A)$, where $L_X(A)={\operatorname{Tr}\rho A\mid\rho\in X}$ for the set of all states ($L$) or of separable states ($L_{\mathrm{Sep}}$), and vol is the Euclidean volume (Definitions 1 and 2). The paper proves $\mu_{2,1}\ge\sqrt2-1$ (Proposition 7) and $\mu_{2,1}\le\frac12$ with the projector onto $|\phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SeparableNumericalRangeRatio.result

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Quantum informationQuestion resolvedFirst preprint 2024Exact classification of elementary channel divisibilityExact Kraus-space criterion for elementary right division of finite quantum channels.Proved

The questionFrederik vom Ende, A Sufficient Criterion for Divisibility of Quantum Channels, arXiv:2407.17103v3, Section V, asks:

Why it mattersThe necessary test and the stronger sufficient test offer different Kraus-space descriptions of elementary right division.

The ideaUse the source's input-first Choi convention and the complex Hilbert–Schmidt complement.

Our contributionExact Kraus-space criterion for elementary right division of finite quantum channels.

Verification details
Release source recordProved in Lean

Open problem

Exact classification of elementary channel divisibility

Frederik vom Ende, A Sufficient Criterion for Divisibility of Quantum Channels, arXiv:2407.17103v3, Section V, asks:

An open question at this point is whether our sufficient condition is also necessary, and whether our necessary condition is also sufficient.

The source is arXiv v3, with publication identifier DOI 10.1063/5.0231812. Definition 1 and Lemmas 2–3 specify the elementary channel and inverse; Proposition 1 and Theorem 1 specify the necessary and sufficient conditions. This dossier anchors the combined classification of the two reversals, including their fixed-pair and existential-pair readings.

Let the complex dimension be any finite n ≥ 2, let Φ be an actual CPTP map, and let (Kⱼ) be any finite Kraus representation of Φ. For an ordered pair of orthogonal unit vectors (x,y), E(x,y) means that Φ composed on the right with the inverse of the elementary channel Ψₜ(x,y) is completely positive for some real 0 < t < 1. N(x,y) means ⟨y,KⱼGx⟩ = 0 for every index j and every complex matrix G in the Hilbert–Schmidt orthogonal complement of the whole complex Kraus span. S(x,y) means N(x,y) together with the additional diagonal tests ⟨x,KⱼGx⟩ = 0 over the same indices and complement.

The classification established by the designated result has four cells:

| Reading | Answer | | --- | --- | | Fixed ordered pair: N(x,y) iff E(x,y) | Proved for every map and representation in the stated scope. | | Existential ordered pair: some N pair iff some E pair | Proved for the same map and representation. | | Fixed ordered pair: E(x,y) implies S(x,y) | Refuted in every dimension n ≥ 2. | | Existential ordered pair: some E pair implies some S pair | Refuted in every dimension n ≥ 2; the counterexample has no S pair in any finite Kraus representation. |

The Scribe claim's Proved kind refers to this combined classification, whose statement includes the counterexample. It does not label either E ⇒ S assertion as proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ElementaryChannelDivisibility.result

Frozen module record

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2026An explicit membership graph outside G₂Cuts in arbitrary words reconstruct their two-letter projections.Proved

The questionThe source is the unnumbered request in the conclusion of the paper identified by adamson2026twoword in the Library.

Why it mattersThis Tier-1 external question asks for an explicit graph, whereas Remark 30 already gives nonconstructive existence.

The ideaThe recursive cut definition scans the actual word; reconstruction inserts unit markers at those cuts and then projects onto a and the unit marker.

Our contributionCuts in arbitrary words reconstruct their two-letter projections.

Verification details
Release source recordProved in Lean

arXiv

An explicit membership graph outside G₂

The source is the unnumbered request in the conclusion of the paper identified by adamson2026twoword in the Library. Its literal wording is:

We did not find an example of a graph G such that G ∉ G₂. By Remark 30 we know that such a graph exists. Finding such a graph remains an open problem.

Let V = Fin 24 ⊕ Finset (Fin 24). The candidate U₂₄ has an edge from inl a to inr S exactly when a ∈ S, the same adjacency in reverse, and no within-part edges. Its endpoint states that there do not exist lists w,v over V such that (∀ z : V, count z w = 2), (∀ z : V, count z v = 2), and ∀ x,y : V, x ≠ y → (U₂₄.Adj x y ↔ filter (z = x ∨ z = y) w = filter (z = x ∨ z = y) v). These are Definitions 1 and 14 at k = 2. Positive counts give the full vertex set as the alphabet of both words.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/GraphRepresentation/ExplicitNonTwoUniform.u24_not_in_g2

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedFirst preprint 2026Mináč–Nguyen–Tân Question 3.5The exact n-sum property with restricted augmentation transfers from a finite field to every finite group algebra.Proved

The questionSuppose that F has the n-sum property with respect to U.

Why it mattersGardam's group-ring work provides the repository's existing native group algebra and genuine-unit context.

The ideaThe standard augmentation kills the Jacobson radical.

Our contributionThe exact n-sum property with restricted augmentation transfers from a finite field to every finite group algebra.

Verification details
Release source recordProved in Lean

arXiv

Mináč–Nguyen–Tân Question 3.5

The primary source is Question 3.5:

Suppose that F has the n-sum property with respect to U. Is it true that F[G] also has the n-sum property with respect to U_{ε_G}?

The complete statement quantifies over a finite field F with more than two elements, every finite group G, a subgroup U of F's units containing minus one, and the original natural number n at least two. If every field element is a sum of exactly n elements of U, every element of the actual group algebra must be a sum of exactly n genuine units whose coefficient-sum augmentations lie in U. Repetition is allowed. Modular characteristic, noncommutative groups, characteristic two with more than two field elements, and the trivial group remain in scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/MinacNguyenTanQuestion35.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Discrete structuresQuestion resolvedPosed in 2003Stephan's A059894 complement-reversal conjecturesStephan's binary complement-reversal recurrences and digit-count identities hold for all positive inputs.Proved

The questionComplement and reverse the order of all but the most significant bit in binary expansion of n. n = 1ab..yz -> 1ZY..BA = a(n), where A = 1-a, B = 1-b, ... .

Why it mattersThe independent question is whether the two published Stephan formula lines hold for every positive input, starting from the literal bit transformation.

The ideaEvaluate the definition at one using the binary digit lemmas.

Our contributionStephan's binary complement-reversal recurrences and digit-count identities hold for all positive inputs.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Stephan's A059894 complement-reversal conjectures

OEIS A059894, NAME (%N, verbatim):

Complement and reverse the order of all but the most significant bit in binary expansion of n. n = 1ab..yz -> 1ZY..BA = a(n), where A = 1-a, B = 1-b, ... .

Ralf Stephan's recurrence conjecture (%F, verbatim):

a(1) = 1, a(2n) = a(n) + 2^(floor(log_2(n))+1), a(2n+1) = a(n) + 2^floor(log_2(n)) (conjectured). - Ralf Stephan, Aug 21 2003

Ralf Stephan's digit-count conjecture (%F, verbatim):

A000120(a(n)) = A000120(A054429(n)) = A023416(n) + 1 (conjectured). - Ralf Stephan, Oct 05 2003

AUTHOR (%A, verbatim):

Marc LeBrun, Feb 06 2001

The definition a applies the quoted transformation to the little-endian list Nat.digits 2 n: drop its last entry, reverse the remaining list, map natural subtraction 1 - d, append [1], and evaluate with Nat.ofDigits 2. The definition complementRest omits only the reversal and represents A054429 on positive inputs. A000120 counts one bits and A023416 counts zero bits in the positive canonical binary expansion.

The literal theorem statement is:

a 1 = 1 ∧
(∀ n : ℕ, 0 < n → a (2 * n) = a n + 2 ^ (Nat.log 2 n + 1)) ∧
(∀ n : ℕ, 0 < n → a (2 * n + 1) = a n + 2 ^ (Nat.log 2 n)) ∧
(∀ n : ℕ, 0 < n →
  (Nat.digits 2 (a n)).count 1 = (Nat.digits 2 (complementRest n)).count 1 ∧
  (Nat.digits 2 (a n)).count 1 = (Nat.digits 2 n).count 0 + 1)

Here Nat.log 2 n agrees with the floor of the base-two logarithm on the positive domain. The counting and recurrence clauses have the explicit guard 0 < n; no such clauses at zero are asserted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/StephanComplementReverseRecurrence.result

Frozen module record

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    The exact recorded statement is proved.

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    Dated conjecture statement: 2003. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2026Enumeration of the Arrow Pattern (32; 1 → 3)The (32; 1 to 3) avoidance series satisfies a cubic equation and is its distinguished formal branch.Proved

The questionZhou and Xinyang Yu, Arrow-Wilf equivalences and enumerative results for short arrow patterns, arXiv:2609.29392v1, abstract:

Why it matters1 + (3x − 2)F + (1 − x)(1 − 2x)F² + x³F³ = 0.

The ideaAn avoider is characterized by its inverse Foata edges: whenever a < c < π̂(a), the value c must precede π̂(a) in the one-line word.

Our contributionThe (32; 1 to 3) avoidance series satisfies a cubic equation and is its distinguished formal branch.

Verification details
Release source recordProved in Lean

Open problem

Enumeration of the Arrow Pattern (32; 1 → 3)

Robin D.P. Zhou and Xinyang Yu, Arrow-Wilf equivalences and enumerative results for short arrow patterns, arXiv:2609.29392v1, abstract:

Together with earlier work of Archer and Laudone, this leaves only (32;1→3) unresolved for |ν|≤2, which we pose as an open problem.

Here S_n(32; 1 → 3) consists of permutations of 1, …, n avoiding the arrow pattern in the paper's sense: the selected values have the required order in the one-line word, and the prescribed arrow holds in π̂ = θ⁻¹(π).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ArrowThirtyTwoOneThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026The exact maximum purity of absolutely PPT qutrit–qudit statesProved

The questionAhiable, Kothakonda and Winter, arXiv:2608.03390v2, Conjecture 6.7, asks:

The ideaThe literal unitary definition implies both Hildebrand boundary LMIs through explicit Bell unitaries and principal compression.

Verification details
Release source recordProved in Lean

Open problem

The exact maximum purity of absolutely PPT qutrit–qudit states

Ahiable, Kothakonda and Winter, arXiv:2608.03390v2, Conjecture 6.7, asks:

Let $\mathcal P_{m,n}\subseteq \mathrm{APPT}{m,n}$ be the inscribed absolute PPT polytope with $2\leq m\leq n,; n>2$. Then $$\max{\lambda\in \mathrm{APPT}{m,n}}\sum{i=1}^{mn}\lambda_i^2=\max_{\lambda\in\mathcal P_{m,n}}\sum_{i=1}^{mn}\lambda_i^2$$ and occurs at the spectra given by Eq. (44).

The resolved sector is $m=3$, every integer $n\geq3$. For a positive semidefinite, trace-one matrix $\rho$ on $\mathbb C^3\otimes\mathbb C^n$, APPT means $(U\rho U^\dagger)^\Gamma\succeq0$ for every unitary $U$, with $\Gamma$ the second-factor partial transpose. The purity is $\operatorname{Re}\operatorname{Tr}(\rho^2)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsolutePPT/QutritQuditMaximumPurity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedSource 2025Divisors of All Reversed MultiplesDivisors of B squared minus one are exactly the positive divisors of all their reversed base-B multiples.Proved

The questionMohamed Ayad and Rachid Bouchenna, Which Integer Divides the Reverse of Any of Its Multiples?, INTEGERS 25 (2025), A37, printed p. 9, Problem 1:

Why it mattersThe paper's Theorem 3 characterizes the base-ten case by the divisors of 99.

The ideaA positive multiple whose leading base-B digit is one forces gcd(n,B)=1 whenever n has the reversal property.

Our contributionDivisors of B squared minus one are exactly the positive divisors of all their reversed base-B multiples.

Verification details
Release source recordProved in Lean

Open problem

ayad bouchenna base b reverse multiple divisors

Mohamed Ayad and Rachid Bouchenna, Which Integer Divides the Reverse of Any of Its Multiples?, INTEGERS 25 (2025), #A37, printed p. 9, Problem 1:

Let B ≥ 2 be any number base. Are the divisors of B² − 1 the only positive integers satisfying the property P*_B ?

Section 4.1 on printed p. 8 defines the base-B reverse m*_B by reversing the positional digits of the positive integer m. The property P*_B means that n divides m*_B for every positive multiple m of n. The exact statement is: for all natural B ≥ 2 and all natural n > 0, (∀ m : ℕ, 0 < m → n ∣ m → n ∣ Nat.ofDigits B (Nat.digits B m).reverse) ↔ n ∣ B ^ 2 - 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/AyadBouchennaReverseMultipleDivisors.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2025.

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Discrete structuresStructural characterizationFirst preprint 2025The fixed CDK image of improper partition matricesThe fixed Claesson-Dukes-Kubitzke image of improper partition matrices is characterized by the gaps in the whole alphabet of an inversion sequence.Proved

The questionShane Chern and Shishuo Fu, Signed counting of partition matrices, arXiv:2508.21318v2, Section 5, Question 5.5, page 26:

Why it mattersThe source's natural CDK map does not send the improper matrices to the restricted inversion sequences in its Theorem 1.4.

The ideaInversion bounds alone make this an upper triangular native partition matrix, with every row and column nonempty.

Our contributionThe fixed Claesson-Dukes-Kubitzke image of improper partition matrices is characterized by the gaps in the whole alphabet of an inversion sequence.

Verification details
Release source recordProved in Lean

Open problem

The fixed CDK image of improper partition matrices

Shane Chern and Shishuo Fu, Signed counting of partition matrices, arXiv:2508.21318v2, Section 5, Question 5.5, page 26:

Find a characterization (possibly in terms of pattern avoidance) of the inversion sequences produced by improper partition matrices under the action of the Claesson–Dukes–Kubitzke bijection.

For each integer $n\geq1$, write $[n]={1,\ldots,n}$ and let $\mathbf I_n={e:[n]\to\mathbb N:0\leq e_j\leq j-1\text{ for all }j\in[n]}$. A native partition matrix on $[n]$ is an upper triangular square matrix $P$ of sets, of any permitted positive dimension $k$, whose nonempty cells partition $[n]$, every row and column is nonempty, and $\operatorname{col}_P(a)\leq\operatorname{col}_P(b)$ whenever $a<b$. The row and column of a label are those of its unique cell.

An adjacent pair $i,i+1$, with $1\leq i<n$, is an ascent or descent only when its labels share a column and have respectively increasing or decreasing row indices. Such a pair is proper precisely when $i$ has the parity of the least label of that column. An improper matrix has no proper ascent or descent; equal-row pairs are neither. Write $\operatorname{IPPM}_n$ for this class.

The fixed source map is

$$ \Pi_n(P)j=\sum{d=1}^{\operatorname{row}P(j)-1} n_d, \qquad n_d=\sum{r=1}^{k}|P_{r,d}|. $$

Thus the summands are column cardinalities, and the sum is zero in the first row. The requested image uses this map and all permitted dimensions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PartitionMatrix/ChernFuImage.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    The fixed Claesson-Dukes-Kubitzke image of improper partition matrices is characterized by the gaps in the whole alphabet of an inversion sequence.

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    First preprint year: 2025.

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Quantum informationExact optimum establishedFirst preprint 2025Finding the best steady-state measurement in a quantum oscillatorWe prove that the aligned pure measurement attains the global minimum among all physical general-dyne measurements and stationary solutions in the model.The aligned general-dyne measurement attains the global steady-state optimumProved

The questionWhich general-dyne measurement minimizes the stationary covariance determinant in the source’s oscillator model?

Why it mattersThe covariance optimum controls the proposed steady-state ergotropy optimum within the stated physical model.

The ideaAlgebraic bounds exclude the competing stationary branch; an explicit aligned solution attains the remaining lower bound.

Our contributionWe prove that the aligned pure measurement attains the global minimum among all physical general-dyne measurements and stationary solutions in the model.

Scope & next questionThe result uses the stable thermal optical parametric oscillator model and does not classify every equality case.

Verification details
Release source recordProved in Lean

Open problem

The aligned general-dyne measurement attains the global steady-state optimum

K. H. Kua, A. Serafini and M. G. Genoni, Daemonic ergotropy of Gaussian quantum states and the role of measurement-induced purification via general-dyne detection, arXiv:2506.22288v2; Quantum Sci. Technol. 11, 015014 (2026), DOI 10.1088/2058-9565/ae2200, Section VI after Eq. (66), state:

We have also numerically verified that non-zero phases ϑ_m lead to worse result, and we can thus conjecture that this fully characterizes the optimal steady-state general-dyne monitoring.

The Library note gives the source locators and Eq. (66). Write $r=2\chi/\kappa$, $\nu=2n_{\rm th}+1$ and $c=\sqrt{1-r^2}$, with $\kappa>0$, $0\le\chi<\kappa/2$ and $n_{\rm th}\ge0$. For every finite physical general-dyne measurement covariance $M$ and physical stationary conditional covariance $S$, the result proves

$$\sqrt{\det S}\ge v_* = \frac{(\nu-1)c+\sqrt{(\nu-1)^2c^2+4\nu}}{2}.$$

The pure aligned covariance

$$M_*=\operatorname{diag}\left(\frac{1-r}{1+r},\frac{1+r}{1-r}\right)$$

has a physical stationary solution attaining this value. No uniqueness of the minimizing measurement is asserted by the public theorem.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Stable thermal OPO, Eqs. (57)–(58) | claim parameters κ, χ, nth and its four hypotheses; driftA | All damping, pump and thermal parameters in the stated regime | | Thermal input and diffusion | Literal scalar multiples of the identity inside eMat, bMat, steadyEq | $\sigma_{\rm in}=\nu I$ and $D=\kappa\nu I$ | | Passive coupling in physical q,p coordinates | coupling | $C=\sqrt\kappa,\Omega$, with $\Omega=\begin{pmatrix}0&1\-1&0\end{pmatrix}$ | | Source E and B, Eqs. (21)–(25) | eMat, bMat | The inverse matrix square root is CFC.sqrt of the inverse positive matrix | | Stationary Riccati equation, Eq. (55) | steadyEq | Every matrix product, transpose and diffusion term is retained | | Every physical measurement and stationary covariance | claim first conjunct; frozen IsGaussianMeasurementCov (m := 1) | Symmetry and the complex uncertainty matrix being positive semidefinite; arbitrary phase, squeezing and impurity | | Global optimum | vStar; result first conjunct | Universal determinant-square-root lower bound | | Aligned pure measurement, Eq. (66) | optMeas; result second conjunct | Existence of a physical stationary attainer with equality |

The symplectic, thermal-input and diffusion expressions are inlined in the public formulas. Proofs use the frozen PositiveOneModeWilliamson.physicalJ2 through omega_sq, and the frozen physical covariance predicate and public GaussianThermometryJointMeasurementRefutation.physical_data. The finite $2\times2$ real matrix carrier is the source's single-mode covariance representation. Requiring $S$ to be physical restricts the conclusion to conditional covariance solutions; it adds no hypothesis to the source's physical statement. The damping scale cancels without restricting its positive value.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/GeneralDyneSteadyStateOptimality.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    The aligned pure general-dyne measurement attains the smallest stationary covariance determinant for a stable thermal optical parametric oscillator, among all physical general-dyne measurements and physical stationary solutions.

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  • Historical context

    First preprint year: 2025.

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Discrete structuresQuestion resolvedFirst preprint 2026Finding the self-similar pieces of a priority latticeWe characterize the qualifying forests and prove that their number is the sum of factorials from 0! to n!.Principal filters of the priority latticeProved

The questionWhich principal filters in a priority lattice reproduce a smaller priority lattice, and how many are there?

Why it mattersThe problem relates the structure of priority forests to the intervals of their lattice.

The ideaContracting the last increasing tree produces the order isomorphism; counting the trees gives the formula.

Our contributionWe characterize the qualifying forests and prove that their number is the sum of factorials from 0! to n!.

Scope & next questionThe stated count applies for n ≥ 1.

Verification details
Release source recordProved in Lean

Open problem

Principal filters of the priority lattice

Adrián Lillo and Mercedes Rosas, The Priority Lattice, arXiv:2603.28905v1, Section 6, item 3, pp. 24–25, ask for the number of principal filters $[P,\hat1]$ isomorphic to a priority lattice $\Pi(m)$ with $m\le n$. Their data $2,4,10,34,154,874$ suggest the left factorials, OEIS A003422. The displayed formula is $\theta_n=\sum_{i=0}^{n-1}k!$, whose summation variable and bound do not match the data.

The question and its source conventions are preregistered in #14857. The Library note quotes the source definitions: increasing rooted trees with consecutive label sets, ordered by their root labels; edge inclusion; and an adjoined greatest element.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PriorityLattice/PrincipalFilters.result

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Further frontiersCounterexample foundFirst preprint 2025Tang–Zhang: real-order reciprocal critical-point momentsThe reciprocal critical-point moment has the sharp Tang–Zhang coefficient for every real p at least one.Refuted

The questionZhang, Sharp Schoenberg type inequalities and the de Bruin–Sharma problem, arXiv:2508.10341v3, §1.3, Conjecture 1.13.

Why it mattersConjecture 1.13 asks for a sharp real-order extension of reciprocal critical-point inequalities.

The ideaThe polynomial and its multiplicity-preserving reciprocal-root spectral reduction are reused from TangZhangOrderMinusTwo.

Our contributionThe reciprocal critical-point moment has the sharp Tang–Zhang coefficient for every real p at least one.

Verification details
Release source recordRefuted in Lean

Open problem

Tang–Zhang: real-order reciprocal critical-point moments

Q. Tang and T. Zhang, Sharp Schoenberg type inequalities and the de Bruin–Sharma problem, arXiv:2508.10341v3, §1.3, Conjecture 1.13. The source is tang2025schoenberg.

Conjecture 1.13. Let P(z) = z ∏_{j=1}^n (z − z_j), z_j ∈ C \ {0}, and let w_1, …, w_n be the zeros of P′(z). Then for every real number p ≥ 1 we have Σ_{k=1}^n 1/|w_k|^p ≤ ((n+1)^p + n − 1)/n · Σ_{j=1}^n 1/|z_j|^p. Moreover, equality holds if and only if z_1 = ⋯ = z_n.

For every natural $n\ge1$ and every family $z_1,\ldots,z_n\in\mathbb C\setminus{0}$, put

$$ P(Z)=Z\prod_{j=1}^n(Z-z_j),\qquad c(n,p)=\frac{(n+1)^p+n-1}{n}. $$

Let $w_1,\ldots,w_n$ be all roots of $P'$, counted with algebraic multiplicity. No distinctness, simplicity, reality or common argument hypothesis is imposed on the nonzero roots. The result proves the following three clauses:

  1. For every real $p\ge1$, $$\sum_{k=1}^n|w_k|^{-p}\le c(n,p)\sum_{j=1}^n|z_j|^{-p}.$$
  2. For every real $p>1$, equality holds if and only if $z_i=z_j$ for every $i,j$.
  3. Conjecture 1.13 as stated is false: at $p=1$, $n=2$ and $(z_1,z_2)=(1,2)$ give equality although $z_1\ne z_2$.

The formal claim is the conjunction of (i), (ii), and ¬ conjecture, where the original conjecture has its full dimension, root and order quantifiers:

def conjecture : Prop :=
    ∀ (n : ℕ) (hn : 1 ≤ n) (z : Fin n → ℂ), (∀ j, z j ≠ 0) → ∀ p : ℝ, 1 ≤ p →
      (((Polynomial.derivative (P z)).roots.map (fun w => ‖w‖ ^ (-p))).sum ≤
        c n p * ∑ j, ‖z j‖ ^ (-p)) ∧
      ((((Polynomial.derivative (P z)).roots.map (fun w => ‖w‖ ^ (-p))).sum =
        c n p * ∑ j, ‖z j‖ ^ (-p)) ↔ ∀ i j, z i = z j)

Every norm power is a real Real.rpow, including nonintegral orders. The named conjecture is refuted by its endpoint equality clause. The sharp inequality for all $p≥1$ and the strict-order equality case are proved in the same result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/CriticalPoints/TangZhangRealOrder.result

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Quantum informationQuestion resolvedFirst preprint 2023Spin-1 tensor spin-K/2 precession separable boundThe spin-1 tensor spin-K/2 precession protocol has separable bound ½ [1 + 2^{−(K−1)} binom(K−1, (K−1)/2) (K−1)/(K+1)] for every odd K ≥ 7.Proved

The questionKhoi-Nguyen Huynh-Vu, Lin Htoo Zaw and Valerio Scarani, Certification of genuine multipartite entanglement in spin ensembles with measurements of total angular momentum, arXiv:2311.00806v2, §III, Conjecture 2, Eq. (32), PDF p. 8:

Why it mattersThe public spin matrices, precession observable and spectral projector follow the source definitions; the compression is a proved identity, not a definition of the observable.

The ideaThe proof constructs the binomial/Krawtchouk eigenbasis of Jx and its inverse by polynomial homogenization.

Our contributionThe spin-1 tensor spin-K/2 precession protocol has separable bound ½ [1 + 2^{−(K−1)} binom(K−1, (K−1)/2) (K−1)/(K+1)] for every odd K ≥ 7.

Verification details
Release source recordProved in Lean

arXiv

Spin-1 tensor spin-K/2 precession separable bound

Khoi-Nguyen Huynh-Vu, Lin Htoo Zaw and Valerio Scarani, Certification of genuine multipartite entanglement in spin ensembles with measurements of total angular momentum, arXiv:2311.00806v2, §III, Conjecture 2, Eq. (32), PDF p. 8:

The separable bound for {ȷ̃, ȷ̃′} = {1, K/2} with K ≥ 7 is Psep_K({1, K/2}) = ½ [1 + 2^{−(K−1)} binom(K−1, (K−1)/2) (K−1)/(K+1)].

The protocol takes odd K, J_k = cos(2πk/K)J_x + sin(2πk/K)J_y, and Q_K = (1/K) Σ_k pos(J_k), where 2 pos acts on an eigenvalue m as 1+sgn(m), including half weight at zero. Eqs. (7) and (10) express the separable bound as the maximum over unit product pure states. The two factors here have dimensions 3 and K+1. Preregistration: https://github.com/the-omega-institute/trureturing/issues/11873.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PrecessionSpinOneSeparableBound.result

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Quantum informationQuestion resolvedFirst preprint 2022The prime-dimensional stabilizer commutation-rank boundGenuine entanglement forces a lower bound on the dimension of the Pauli commutation space over a prime field.Proved

The questionAugusiak, Fully non-positive-partial-transpose genuinely entangled subspaces, arXiv:2203.16902v2; Quantum 7, 915 (2023), Section 3, Conjecture 1 (final5.tex, lines 458–463):

Why it mattersThe commutation space records how global commutativity is assembled from local noncommutativity.

The ideaNontriviality of the stabilizer subspace is derived, not assumed.

Our contributionGenuine entanglement forces a lower bound on the dimension of the Pauli commutation space over a prime field.

Verification details
Release source recordProved in Lean

Open problem

The prime-dimensional stabilizer commutation-rank bound

O. Makuta, B. Kuzaka and R. Augusiak, Fully non-positive-partial-transpose genuinely entangled subspaces, arXiv:2203.16902v2; Quantum 7, 915 (2023), Section 3, Conjecture 1 (final_5.tex, lines 458–463):

Given a stabilizer $\mathbb{S}=\langle G_{1},\dots,G_{k}\rangle$, if the local dimension $d$ is a prime number and the corresponding stabilizer subspace $V$ is genuinely entangled, then

$\dim K(\mathbb{S}) \geqslant \left\lceil \frac{N-1}{d-1} \right\rceil.$

The source defines $K(\mathbb{S})$ using the local commutation exponents (lines 375–391). In the Lean coordinates, write the prime local dimension as $p$. A Pauli word is a phase residue together with $x,z:\mathrm{Fin},N\to\mathbb F_p$, acting as $\omega^{\mathrm{phase}}\bigotimes_n X^{x(n)}Z^{z(n)}$, with $\omega=\exp(2\pi i/p)$. For words $a,b$, the commutation vector has coordinates $z_a(n)x_b(n)-x_a(n)z_b(n)$; commutationSpace G is the $\mathbb F_p$-span of all generator-pair vectors.

IsStabilizer G requires pairwise commuting generator matrices and excludes every scalar matrix other than the identity from their generated matrix submonoid. stabilizerSubspace G is the intersection of the generators' fixed subspaces. genuinelyEntangled G says that every nonzero vector in this intersection fails to factor across every bipartition with two nonempty sides. Factorization means an amplitude of the form $u(x|Q)v(x|{Q^c})$; normalization is not required.

Issue #20105 fixes the statement for every $N\ge1$, prime $p$, and finite generator family. claim uses natural-number division:

$(N-1+(p-2))/(p-1)\le\operatorname{finrank}_{\mathbb F_p}(\operatorname{commutationSpace}G)$.

For $N\ge1$ and prime $p$, the left side is $\lceil(N-1)/(p-1)\rceil$. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/StabilizerGenuineEntanglementCommutationRank.result

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Quantum informationQuestion resolvedFirst preprint 2024Weighted Böttcher–Wenzel inequality, case (ii)The squared weighted Bottcher-Wenzel conjecture holds for every positive definite weight.Proved

The questionChruściński, “Böttcher-Wenzel inequality for weighted Frobenius norms and its application to quantum physics”, arXiv:2403.04199v2, Conjecture 1, equation (15), ask whether

Why it mattersThe source applies this mixed weighted/unweighted bound to relaxation rates of quantum Markovian dynamics.

The ideaNumerical-range convexity for compressions gives a pure state maximizing Cartesian variance.

Our contributionThe squared weighted Bottcher-Wenzel conjecture holds for every positive definite weight.

Verification details
Release source recordProved in Lean

Open problem

Weighted Böttcher–Wenzel inequality, case (ii)

A. Mayumi, G. Kimura, H. Ohno and D. Chruściński, “Böttcher-Wenzel inequality for weighted Frobenius norms and its application to quantum physics”, arXiv:2403.04199v2, Conjecture 1, equation (15), ask whether

$$|[A,B]|\omega\leq\sqrt{\frac{\lambda_m+\lambda_M}{\lambda_m}},|A|\omega|B|$$

holds for all $A,B\in M_n(\mathbb C)$ and every positive definite weight $\omega$. Here $[A,B]=AB-BA$, $|A|_\omega^2=\operatorname{tr}(A^*A\omega)$, $|B|^2=\operatorname{tr}(B^*B)$, and $\lambda_m,\lambda_M$ are the attained minimum and maximum eigenvalues of $\omega$. Issue #13715 preregisters this case as a tier-3 research line.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/WeightedBotcherWenzel.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundSource 2025Zhang–Jing Question 2.14: block positivity and normalityAn admissible scalar pair gives a positive block for a nonnormal complex matrix.Refuted

The questionFeng Zhang and Hefang Jing, More inequalities for positive semidefinite matrices, Electronic Journal of Linear Algebra 41 (2025), 595–601, Question 2.14, p. 601:

Why it mattersIt refutes the universal equivalence using one fixed two-dimensional matrix.

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionAn admissible scalar pair gives a positive block for a nonnormal complex matrix.

Verification details
Release source recordRefuted in Lean

Open problem

Zhang–Jing Question 2.14: block positivity and normality

Feng Zhang and Hefang Jing, More inequalities for positive semidefinite matrices, Electronic Journal of Linear Algebra 41 (2025), 595–601, Question 2.14, p. 601:

Let A ∈ M_n. Then [[f(|A|), A*], [A, g(|A|)]] ≥ 0 if and only if A is normal under the conditions that f(x)g(x) = x² for x ≥ 0, f(x) ≥ g(x) (x ≥ 1) and g(x) ≥ f(x) (0 ≤ x ≤ 1).

The notation on p. 595 is “|A| = (AA)^{1/2} and A is the conjugate transpose of A”. The assertion quantifies over every admissible pair of real functions, every natural dimension, and every complex matrix in that dimension. Preregistration: #18699.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $|A|=(A^*A)^{1/2}$ | CFC.sqrt of the conjugate-transpose product, inlined in block | Literal expression; the existing CFC.abs construction is not redeclared. | | Four displayed blocks, in their given order | block, using Matrix.fromBlocks and cfc | Every block is retained. | | $f(x)g(x)=x^2$ for $x\ge0$ | First quantified premise of claim | Same range and equality. | | $f(x)\ge g(x)$ for $x\ge1$ | Second quantified premise | Same inequality and range. | | $g(x)\ge f(x)$ for $0\le x\le1$ | Third quantified premise | Both endpoints included. | | Every $A\in M_n$, block nonnegative iff $A$ normal | Final quantifiers and equivalence in claim | Fin n labels the indices; positivity is Matrix.PosSemidef and normality is $A^A=AA^$. | | The universal assertion is false | result | Its conclusion is the literal negation of claim. |

No continuity, invertibility or spectral injectivity premise is added. The counterexample functions are continuous and satisfy the three required scalar conditions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/ZhangJingBlockNormalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2018A rank-one qutrit MIC with nine orthogonal pairsNine positive semidefinite rank-one effects on C^3 sum to the identity and span every Hermitian matrix over the reals.Refuted

The questionStacey, “The Varieties of Minimal Tomographically Complete Measurements”, arXiv:1812.08762v5 (quant-ph), Conjecture 1 on printed page 6 states:

Why it mattersThe zero entries of a MIC Gram matrix constrain the geometry of minimal quantum measurements.

The ideaPositive scaling of an outer product gives positive semidefiniteness and rank at most one; a nonzero diagonal entry proves rank at least one.

Our contributionNine positive semidefinite rank-one effects on C^3 sum to the identity and span every Hermitian matrix over the reals.

Scope & next question- Whether every rank-one qutrit MIC with the added equal-trace hypothesis tr(Ea)=1/3 has at most seven orthogonal pairs remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

A rank-one qutrit MIC with nine orthogonal pairs

J. B. DeBrota, C. A. Fuchs and B. C. Stacey, “The Varieties of Minimal Tomographically Complete Measurements”, arXiv:1812.08762v5 (quant-ph), Conjecture 1 on printed page 6 states:

A rank-1 MIC in dimension 3 can have no more than 7 pairs of orthogonal elements.

A MIC is a POVM of exactly d² positive semidefinite effects summing to the identity whose real span is the Hermitian operators. Orthogonality is tr(E_a E_b) = 0; distinct unordered pairs are counted once. Issue #11474 preregisters the literal reading for d = 3 before Lean implementation. The conjecture imposes no unbiasedness condition, although its preceding seven-pair example is explicitly called unbiased.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/QutritRankOneMicOrthogonalPairs.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2021Optimal calibrated visibility for joint quantum work measurementsThe fluctuation condition forces the optimal calibrated visibility bound.Proved

The questionStrunz, Joint measurability in nonequilibrium quantum thermodynamics, arXiv:2111.02854v2, Physical Review E 106, L022101 (2022), p. 4, state:

Why it mattersThe theorem removes the restriction to the inverse-Lüders ansatz in the necessity direction.

Our contributionThe fluctuation condition forces the optimal calibrated visibility bound.

Verification details
Release source recordProved in Lean

Open problem

Optimal calibrated visibility for joint quantum work measurements

K. Beyer, R. Uola, K. Luoma and W. T. Strunz, Joint measurability in nonequilibrium quantum thermodynamics, arXiv:2111.02854v2, Physical Review E 106, L022101 (2022), p. 4, state:

The ansatz for the effects W^U_ab in Eq. (6) is a priori not the most general one [54]. However, numerical simulations suggest that our visibility bounds are tight and we conjecture that our construction is optimal to satisfy condition (ii) for the given model Eq. (5)

The Supplemental Material repeats:

Therefore, we conjecture that the proposed form of the joint POVM is optimal for the given noise model if one wants to satisfy the fluctuation condition (ii).

For $d\ge2$ and $0<\lambda,\gamma<1$, set $A_a=\lambda\Pi_a+(1-\lambda)I/d$ and $B_b=\gamma\Pi'_b+(1-\gamma)I/d$, with $B_b^U=U^\dagger B_bU$. If every unitary admits a parent POVM satisfying the average and fluctuation conditions, then

$$\gamma\le\frac{2\kappa}{d+2\kappa-d\kappa},\qquad \kappa=2\sqrt{\lambda+(1-\lambda)/d}\sqrt{(1-\lambda)/d} +(d-2)(1-\lambda)/d.$$

The Library note records conditions (i)–(ii), Eq. (5), Eqs. (8)–(9), and both conjecture passages. Issue #16507 is the first-tier preregistration of the literal conventions and universal necessity statement.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Eq. (5), initial white-noise effects | effA | Literal scalar-weighted matrix unit plus identity | | Eq. (5), final white-noise effects | effB | Literal scalar-weighted matrix unit plus identity | | Arbitrary relative energy bases | effBU and unitary quantifier in claim | Standard matrix units with arbitrary unitary rotation cover every relative basis | | Lüders instrument | luders | The positive square root is CFC.sqrt | | Condition (i), all energy assignments | AverageCondition | All real functions on both outcome sets | | Condition (ii), every diagonal state and outcome | FluctuationCondition | PSD, trace one, zero off-diagonals, and every outcome pair | | A parent POVM for every unitary | claim and frozen ConcealmentKernelNecessityRefutation.IsPOVM | Every effect is PSD and the effects sum to identity | | $d\ge2$ and strict visibility bounds | Binders and antecedents in claim | Every natural dimension and every pair of real visibilities in the source's range | | Eqs. (8)–(9) | kappa, visibilityBound, result | Necessity of the literal bound |

The finite index type represents $d$ outcomes and a $d$-dimensional Hilbert space. The universal unitary quantifier absorbs the choice of the final energy basis. PSD and trace-one hypotheses define a density matrix; the zero-entry hypothesis defines its diagonality. These are source conditions, not additional restrictions on admissible parents. The public effects are literal source expressions; the proof uses the frozen Tomiyama noise map phi through the private consumed correspondence source_effect_eq_phi.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/CalibratedJointMeasurabilityOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument gives a structural characterization or normal form.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

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    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2012Every three-site improper marginalization is birational for open-boundary binary bond-two MPSEvery three-site improper marginal of an open-boundary binary bond-two matrix product state has a rational inverse on a nonempty Zariski-open subset.Proved

The questionMorton, Algebraic geometry of matrix product states, SIGMA 10 (2014), 095, state Conjecture 8 in the source TeX sigma14-095.tex, lines 430–434:

Why it mattersThe source's Theorem 7 identifies the three-site open-boundary binary bond-two model with a quartic hypersurface (lines 378–406).

The idea3. chartdenominator (n) (t) (hg : MvPolynomial.eval t chartPoly ≠ 0) proves the denominator used by rationalInverse is nonzero. inverseformula (n a) (hn : 4 ≤ n) (h : a + 3 ≤ n) (t) (hg : MvPolynomial.eval t chartPoly ≠ 0) identifies that rational map with an MPS state whose boundaries contain the inverse powers of the reconstructed sum matrix.

Our contributionEvery three-site improper marginal of an open-boundary binary bond-two matrix product state has a rational inverse on a nonempty Zariski-open subset.

Scope & next questionThe exceptional locus where chartPoly vanishes is not classified, and whether a larger chart or a simpler exceptional-set description exists remains open.

Verification details
Release source recordProved in Lean

Open problem

Every three-site improper marginalization is birational for open-boundary binary bond-two MPS

A. Critch and J. Morton, Algebraic geometry of matrix product states, SIGMA 10 (2014), 095, state Conjecture 8 in the source TeX sigma14-095.tex, lines 430–434:

For $N\geq 4$, a~generic~$N$-qubit state can be recovered from its improper margina-li-zation to any three consecutive states. That is, each improper marginalization map $\overline\OB(2,2,N)\to\overline\OB(2,2,3)$ is a~birational equivalence of varieties.

The paper defines an open-boundary state by $\Psi_{i_1\dots i_N}=b_0^TA_{i_1}\cdots A_{i_N}b_1$ and defines $\overline\OB(2,2,N)$ as the Zariski closure of these states (source lines 349–363). Its improper marginalization sums amplitudes over all indices outside a chosen consecutive triple (lines 419–426). The Lean reading uses closure n for the affine zero locus of the vanishing ideal and improperMarginal n a h for this coherent amplitude sum, where $h:a+3\le n$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/OpenBoundaryMPSMarginalBirationality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedSource 2012Two centered abelian period lengthsTwo centered abelian period lengths force ordinary periodicity.Proved

The questionAvgustinovich, Karhumäki and Puzynina, On abelian versions of critical factorization theorem, RAIRO–ITA 46(1), Open problem 6.2, ask:

Why it mattersThe source states that one centered period length forces periodicity and constructs a non-periodic word using three lengths in Theorem 3.2.

The ideaThe centered-square alternative makes the product of the two second differences zero.

Our contributionTwo centered abelian period lengths force ordinary periodicity.

Verification details
Release source recordProved in Lean

Open problem

Two centered abelian period lengths

Avgustinovich, Karhumäki and Puzynina, On abelian versions of critical factorization theorem, RAIRO–ITA 46(1), Open problem 6.2, ask:

Let k be an integer. Does there exist a non-periodic word w and integers l₁, l₂, such that w contains a central abelian 2k-power of length l₁ or l₂ at every position?

The answer is negative for every $k\geq1$, every finite alphabet and every pair of block lengths. The public theorem is D5.S1.Words.TwoCenteredAbelianPeriods.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Biinfinite word over a finite alphabet | $w:\mathbb Z\to\mathrm{Fin}(r)$, every natural $r$ | Alphabet labels are immaterial. | | Position between $w_i$ and $w_{i+1}$ | Integer cut $i$ | The first block starts at $i-lk+1$. | | Central abelian $2k$-power with period length $l$ | CentralAbelianPower w k l i | All $2k$ consecutive length-$l$ blocks have the same letter counts; $l>0$. | | Two lengths, one available at every position | The disjunction inside the universal integer-position hypothesis of claim | Both lengths are fixed globally. | | No non-periodic example | result : claim concludes Periodic w | A positive natural ordinary period is produced. | | Positive exponent | $1\leq k$ | Nonempty centered powers; no assertion about $k=0$. | | Natural lengths may equal zero | Positivity is part of each central-power predicate | A zero length never satisfies that predicate, so only the other length remains. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/TwoCenteredAbelianPeriods.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2012.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year recorded in the source bibliography: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2016The A001108 triangular-square indices have two odd divisor sumsA positive triangular number is square exactly when its index and value have odd divisor sums.Proved

The questiona(n)-th triangular number is a square: a(n+1) = 6a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Why it mattersKrizek's comment proposes an intrinsic divisor-sum test for membership among the triangular-square indices.

The ideaProve the classical characterization that a positive natural number has odd divisor sum exactly when it is a square or twice a square.

Our contributionA positive triangular number is square exactly when its index and value have odd divisor sums.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A001108 triangular-square indices have two odd divisor sums

OEIS A001108, NAME (%N, verbatim):

a(n)-th triangular number is a square: a(n+1) = 6*a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Jaroslav Krizek's COMMENT conjecture (%C, verbatim, with its preceding context sentence):

The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016

The literal proved statement is forall n : Nat, 0 < n -> (IsSquare (n * (n + 1) / 2) iff (Odd (sigma 1 n) and Odd (sigma 1 (n * (n + 1) / 2)))). The recurrence for A001108, Pell-type completeness, the NSW/A084703 interleaving structure, and all other A001108 comments are NOT claimed. The guard excludes n=0: although T_0=0 is a square, Mathlib defines sigma(0)=0, which is even, so the unguarded parity equivalence is false.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/KrizekTriangularSquareSigmaParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument gives a structural characterization or normal form.

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Quantum informationQuestion resolvedFirst preprint 2021A closed formula for the bubble-eigenstate configuration countA marked-gap count gives the closed formula for labelled bubble configurations on a periodic chain.Proved

The questionSen, Minimal model for Hilbert space fragmentation with local constraints (Phys.

Why it mattersAn energy-resolved count quantifies the explicitly constructed bubble sector and distinguishes its contribution from the full integer-energy spectrum.

The ideaThus the natural division in the formula is exact on its summation range.

Our contributionA marked-gap count gives the closed formula for labelled bubble configurations on a periodic chain.

Verification details
Release source recordProved in Lean

Open problem

A closed formula for the bubble-eigenstate configuration count

B. Mukherjee, D. Banerjee, K. Sengupta and A. Sen, Minimal model for Hilbert space fragmentation with local constraints (Phys. Rev. B 104, 155117, 2021; arXiv:2106.14897v3), section “Construction of bubble eigenstates”, TeX fragmentation_prb_v2.tex, lines 729–731, immediately following Table I:

While we have not been able to find a closed form expression for the degeneracy of the different bubble eigenstates for an arbitrary $L$, we can still place a lower bound on the degeneracy of such states when $L \gg 1$.

The counting prescription is explicit in the same source, lines 670–674:

The number of bubble eigenstates at an integer energy $E=n_1-n_2$ can be calculated for a given system size from the number of configurations that satisfy the hard core constraint of at least two consecutive $\downarrow$ spins separating any of the $n_1$ $X$ and $n_2$ $Y$ units on the periodic one-dimensional lattice.

Here $X$ and $Y$ are three-site units with energies $+1$ and $-1$, respectively (the source's construction, lines 634–647). Issue #26367 fixes the combinatorial reading: sites are $\mathrm{Fin},L$, a centre set $B$ is admissible when every ordered pair of distinct centres has clockwise separation at least five, and a subset $x\subseteq B$ specifies the $X$ labels, with the other centres labelled $Y$. Its energy is $2|x|-|B|$. Translations are counted separately, and the empty placement contributes one configuration at energy zero. Let $D_B(L,E)$ denote this count, defined by bubbleCount.

For $1\le n\le\lfloor L/5\rfloor$, write

$$C(L,n)=\left(L\binom{L-4n-1}{n-1}\right)\mathbin{\mathrm{div}}n,$$

where $\mathrm{div}$ is natural-number division, and put

$$A(n,E)=\begin{cases}\displaystyle\binom{n}{(n+E)/2},&n+E\text{ is even and }|E|\le n,\0,&\text{otherwise}.\end{cases}$$

The module proves claim: for every natural $L\ge5$ and every integer $E$,

$$D_B(L,E)=[E=0]+\sum_{n=1}^{\lfloor L/5\rfloor}C(L,n)A(n,E).$$

Here $[E=0]$ is one when $E=0$ and zero otherwise; the equality is evaluated in the integers by casting the natural count and placement coefficients. The guarded index $(n+E)/2$ is a nonnegative integer. This settles the stated closed-form gap for labelled configurations; identifying these configurations with eigenstates of $H_3$ uses the source's construction and is not part of the Lean theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/BubbleEigenstateDegeneracy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The motivation identifies an earlier mathematical result or partial case.

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Number theoryQuestion resolvedFirst preprint 2026A uniform family of reduced-coordinate cube recursionsOne uniform autonomous degree-three rational family has cube sums for the actual positive reduced pairs of every positive rational orbit.Proved

The questionMatijasevick, Rodriguez and Salazar, Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields, arXiv:2610.00765v1, Section 7, question 2, asks:

Why it mattersThe question asks whether its square phenomenon has a cubic analogue beyond a coordinate copy of the power map.

The ideaFor a,b positive coprime integers let u=a+tb and w=u+b.

Our contributionOne uniform autonomous degree-three rational family has cube sums for the actual positive reduced pairs of every positive rational orbit.

Verification details
Release source recordProved in Lean

Open problem

A uniform family of reduced-coordinate cube recursions

Matijasevick, Rodriguez and Salazar, Perfect squares in reciprocal-sum sequences and primes that are inert in quadratic fields, arXiv:2610.00765v1, Section 7, question 2, asks:

For the recursions of this paper the exponent m=2 is forced. Is there a natural family of rational recursions for which c a_n+b_n is always a cube, other than those conjugate to z -> z^3?

The reduced-fraction convention is inherited from Problem 1.1: a_n,b_n are coprime positive integers, starts are positive integers, c is positive rational, and indices are n>=2. “Natural” is undefined. The explicit minimal reading here is a uniform autonomous rational family over Q of fixed degree, parameterized by a nonnegative integer, with one formula and a single c independent of the parameter, initial value and index. An unstated stricter author intent is not resolved by this reading.

For every t in N, including zero, put

$$F_t(x)=\frac{x+t}{(x+t+1)^3-(x+t)},\qquad x_n=F_t^{\circ n}(x_0).$$

The proved endpoint is: for every positive rational x_0, the intrinsic reduced numerator a_n and denominator b_n are positive and coprime for every n>=0; for every n>=1,

$$a_n+b_n=\bigl(a_{n-1}+(t+1)b_{n-1}\bigr)^3.$$

Thus c=1 works uniformly and the inherited integer starts and n>=2 follow. Each map has degree exactly 3 and no M in PGL_2(C) has M F_t M^{-1}(z)=z^3 on the entire projective line. No assertion at n=0 of a cube sum, or classification of all cube recursions, is made.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CubeRecursionFamily.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Generating Function of Perfect Matchings Avoiding {123, 132, 213}The Catalan-Fibonacci series gives the generating function of all perfect matchings avoiding 123, 132 and 213.Proved

The questionSucharita Biswas, Umesh Shankar and Sivaramakrishnan Sivasubramanian, Matchings and shape-Wilf-Equivalence of sets of patterns of length three I: Triples, arXiv:2609.08562v1, Section 6, Question 1: "Can we enumerate the matchings that avoid the set of patterns P1 = {123, 132, 213} and P13 = {132, 213, 321}?" Matching patterns follow Section 4 and Figure 1: three arcs form an occurrence only when all three left endpoints precede all three right endpoints, and the label records the complement of the order of the right endpoints.

Why it mattersThis is a structural construction consumed by the all-size enumeration.

The ideaScanning the 2n positions from left to right, a P1-avoiding matching can only close the oldest or the second-oldest open arc; closing a third-oldest or newer arc creates the forbidden closing order 312 or 321.

Our contributionThe Catalan-Fibonacci series gives the generating function of all perfect matchings avoiding 123, 132 and 213.

Verification details
Release source recordProved in Lean

Open problem

Generating Function of Perfect Matchings Avoiding {123, 132, 213}

Sucharita Biswas, Umesh Shankar and Sivaramakrishnan Sivasubramanian, Matchings and shape-Wilf-Equivalence of sets of patterns of length three I: Triples, arXiv:2609.08562v1, Section 6, Question 1: "Can we enumerate the matchings that avoid the set of patterns P1 = {123, 132, 213} and P13 = {132, 213, 321}?" Matching patterns follow Section 4 and Figure 1: three arcs form an occurrence only when all three left endpoints precede all three right endpoints, and the label records the complement of the order of the right endpoints. This dossier settles the P1 clause and records an all-size structural scan bridge, finite continuation recurrence and exact all-size enumeration for the P13 clause. The P13 result concerns the original ordinary matching carrier, including empty and disconnected matchings.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PatternMatchings/TripleAvoidingMatchings.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026A Product Formula for Narayana Strip Series of Heights 4m and 4m + 1The product of two signed Narayana strip series is the Narayana strip series with squared parameters at heights 4m and 4m + 1.Proved

The questionJohann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 4, Conjecture 2, equation (79) (Conjecture 1, equation (62), in v1): for m ≥ 1,

The ideaThe first-return decomposition expresses each bounded series as a quotient p{h+1}/q{h+1} of continuants, the denominator having constant term 1 in z.

Our contributionThe product of two signed Narayana strip series is the Narayana strip series with squared parameters at heights 4m and 4m + 1.

Verification details
Release source recordProved in Lean

Open problem

A Product Formula for Narayana Strip Series of Heights 4m and 4m + 1

Johann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 4, Conjecture 2, equation (79) (Conjecture 1, equation (62), in v1): for m ≥ 1,

C^{(4m)}(t², z²) = c^{(4m)}(t, z) c^{(4m)}(−t, −z), C^{(4m+1)}(t², z²) = c^{(4m+1)}(t, z) c^{(4m+1)}(−t, −z).

Here C^{(h)}(t, z) = Σ_n C_n^{(h)}(t) zⁿ and c^{(h)}(t, z) = Σ_n c_n^{(h)}(t) zⁿ, where C_n^{(h)}(t) and c_n^{(h)}(t) are the weighted counts of Dyck paths of semilength n in the strip 0 ≤ y ≤ h, a down-step to height k having weight τ_k with τ = (1, t, 1, t, …) and τ = (1, t, −1, −t, …) respectively. The identity is the bounded analogue of the unrestricted identity c(t, z) c(−t, −z) = C(t², z²), equation (6) of the source.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/NarayanaStrip/CiglerStripProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedTrinomial perfect-power counts over GF(2)Conjecture: T(2n,3) = (n^2 - n)/2 for n >= 1.Proved

The questionKagey's OEIS A350532 COMMENT states verbatim:

Why it mattersSquaring in characteristic two doubles exponents and preserves support size.

The ideaThe proof uses the sparse forms of that power: divisibility of its derivative forces the derivative to vanish.

Our contributionConjecture: T(2n,3) = (n^2 - n)/2 for n >= 1.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Trinomial perfect-power counts over GF(2)

P. Kagey's OEIS A350532 COMMENT states verbatim:

Conjecture: T(2*n,3) = (n^2 - n)/2 for n >= 1.

The settled statement is $T(2n,3)=(n^2-n)/2$ for every $n\in\mathbb N$ with $1\leq n$.

The source's NAME defines the count verbatim:

Triangle read by rows: T(n,k) is the number of degree-n polynomials over Z/2Z of the form f(x)^m for some m > 1 with exactly k nonzero terms; 1 <= k <= n + 1.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Degree-n polynomials over Z/2Z, exactly k nonzero terms, some exponent m > 1 | KageyA350532Binomials.T | The literal cardinality subtype; distinct nonzero polynomials, with witnesses for base and exponent. | | Conjecture: T(2*n,3) = (n^2 - n)/2 for n >= 1. | KageyA350532Trinomials.claim and KageyA350532Trinomials.result | Literal universal assertion under the preregistered conventions. | | No restriction to positive degree in the definition | p ≠ 0 together with p.natDegree | The preregistered nonzero convention prevents zero from entering the degree-zero count. | | Natural-number variables | The natural-number binders and arithmetic | Subtraction and division in the trinomial assertion are natural-number operations. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Combinatorics/KageyA350532Trinomials.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Koprowski–Lewis Conjecture 3.1: the original masked hypermatrix countOriginal masked nondegenerate hypermatrix countProved

The questionFor every finite field F of cardinality q, every positive integer k, and nonincreasing nonnegative masks lambda, mu : Fin k -> Nat satisfying mu j <= lambda j, lambda j <= k - j, and mu j < k - j, count the pairs of actual (k+1) x k matrices (M1, M2) that respect the two zero regions and are nondegenerate.

Why it mattersKoprowski and Lewis state this as Conjecture 3.1 in Enumeration of Nondegenerate 2 x (k+1) x k Hypermatrices, arXiv:2602.22129v1.

The ideaThe coefficient polynomial is the exact ordinary-monomial determinant specification used by the source.

Our contributionOriginal masked nondegenerate hypermatrix count

Verification details
Release source recordProved in Lean

Open problem

Koprowski–Lewis Conjecture 3.1: the original masked hypermatrix count

For every finite field F of cardinality q, every positive integer k, and nonincreasing nonnegative masks lambda, mu : Fin k -> Nat satisfying mu j <= lambda j, lambda j <= k - j, and mu j < k - j, count the pairs of actual (k+1) x k matrices (M1, M2) that respect the two zero regions and are nondegenerate. The source's nondegeneracy condition is the nonvanishing of Cayley's second hyperdeterminant; equivalently, every nonzero linear combination of the two faces has rank k over the algebraic closure.

The resolved formula is

q^(k*k) * (q-1)^(2*k)
  * product (q-bracket (k+1-j-lambda j))
  * product (q-bracket (k-j-mu j)),

where the products range over j : Fin k and [r]_q = sum_{e=0}^{r-1} q^e. The count is over the literal matrix-pair subtype, with no count, bijection, inverse, normalization, or rational-combination premise added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hypermatrix/MaskedHypermatrixCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Finite Gilbreath extension component maximumThe exact finite Gilbreath extension component question over every integer extension.Proved

The questionLeila Muney, Holes in Valid-Extension Sets of Finite Gilbreath Sequences, arXiv:2606.23721v2, Section 16(4), asks:

The ideaFinite absolute-value preimages preserve finiteness and a single parity.

Our contributionThe exact finite Gilbreath extension component question over every integer extension.

Verification details
Release source recordProved in Lean

arXiv

Finite Gilbreath extension component maximum

Leila Muney, Holes in Valid-Extension Sets of Finite Gilbreath Sequences, arXiv:2606.23721v2, Section 16(4), asks:

Is max_{S∈G_n} #comp(K_S)=2^(n-4) for all n≥5? Verified for n≤10.

The integer triangle is s_a^0=s_a and s_a^b=|s_(a+1)^(b-1)-s_a^(b-1)| at every valid index. A prefix is Gilbreath when every positive-row left entry is one. The family G_n consists of every strictly increasing Gilbreath integer prefix of length n beginning with two and three. K_S contains every integer k for which appending k is Gilbreath; no increasing condition is imposed on this appended integer. Components are literal maximal runs of common difference two in the finite one-parity extension set.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/FiniteDifferences/GilbreathExtension.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe value-one terms of OEIS A334595 are exactly the powers of twoFor every positive natural number, the right edge of its binary XOR triangle has value one exactly when the input is a power of two.Proved

The questionPeter Kagey's OEIS A334595, revision 18, gives the sequence name "Binary interpretation of the right diagonal of the XOR-triangle with first row generated from the binary expansion of n." Its fourth %C comment states:

Why it mattersThe result is a classification of all positive inputs, rather than a bounded table or a statement about only record terms.

The ideaThe proof reconstructs a fixed-width source row from its edge, so the edge map is injective at that width.

Our contributionFor every positive natural number, the right edge of its binary XOR triangle has value one exactly when the input is a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The value-one terms of OEIS A334595 are exactly the powers of two

Peter Kagey's OEIS A334595, revision 18, gives the sequence name "Binary interpretation of the right diagonal of the XOR-triangle with first row generated from the binary expansion of n." Its fourth %C comment states:

Conjecture: a(n) = 1 if and only if n is a power of two.

The entry's definition uses the unpadded binary expansion of n. The literal formal claim is forall n : Nat, 1 <= n -> (a n = 1 <-> exists k : Nat, n = 2 ^ k). The exponent is allowed to be zero, so n = 1 is included; n = 0 is outside the positive-index domain. The right edge is read from the upper-right corner to the apex with every position retained, including zero bits.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/XorTriangle/ValueOne.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationExact counting formulaCounting the zero-energy states of a quantum spin chainFor a chain of twice an odd prime number of sites, we prove an exact formula for the dimension of the zero-energy subspace.The nullspace of the periodic XX chain on twice a prime number of sitesProved

The questionHow many distinct states of a periodic XX spin chain have exactly zero energy?

Why it mattersThe source turns this physics question into a counting problem about vanishing sums of roots of unity.

The ideaAn explicit bijection classifies the zero-sum choices and counts them uniformly for every odd prime.

Our contributionFor a chain of twice an odd prime number of sites, we prove an exact formula for the dimension of the zero-energy subspace.

Scope & next questionThe theorem covers chain lengths of the form 2p with p an odd prime.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The nullspace of the periodic XX chain on twice a prime number of sites

OEIS A392387 (Thore Posske, 2026) is the nullspace dimension of the periodic spin-1/2 XX Heisenberg chain on n > 1 sites. Its COMMENTS give the count

Also, the number of subsets K of {1,...,n} such that the sum of cosines of the angles in {(2j + (1 + (-1)^|K|)/2 )*Pi/n | j in K} is zero.

and its FORMULA section records

Conjecture: a(2*p) = 2*(6^((p-1)/2)+1) for odd prime p.

Issue #10063 fixes the readings: the formal object is this subset count, the combinatorial description common to the entry and to Hu, Gerken and Posske (arXiv:2602.15098, appendix on the XX model); its identification with the dimension of the zero-energy subspace is the source's Jordan–Wigner reading and is not formalized; the empty set is counted; p ranges over odd primes and n = 2p.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/XXChainNullspaceCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    For every odd prime p, the number of subsets K of {1, ..., 2p} whose cosines at the angles (2j + (1 + (-1)^|K|)/2) pi/(2p), j in K, sum to zero is 2 (6^((p-1)/2) + 1).

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The n = 3 collapse of the genuine four-party multi-entropy for stabilizer statesFor a normalized pure qubit stabilizer state, ternary replica contractions are positive real numbers.Proved

Why it mattersIts original partitioncount now uses a structural counting argument: ternary mode orbits classify anchored partitions, and a quotient-colouring bijection transports the merger sums before the finite partition coefficients cancel.

The ideaTernaryPartitionOrbit proves that every nonzero anchored ternary mode has exactly one representative with the same partition in any chosen opposite-pair representative family.

Our contributionFor a normalized pure qubit stabilizer state, ternary replica contractions are positive real numbers.

Verification details
Release source recordProved in Lean

Open problem

The n = 3 collapse of the genuine four-party multi-entropy for stabilizer states

S. Akella, N. Iizuka and A. Miyata, Genuine Multi-Entropy in the Toric Code, arXiv:2607.06050v2, state for stabilizer states the reduction $\mathrm{GM}^{(4)}_{3}(A:B:C:D)=-\left(a-\tfrac19\right)I_3$ (Eq. (3)), where $I_3$ is the Rényi tripartite information and $a$ the convention parameter of the genuine multi-entropy, note that "this reduction is not a consequence of the Coxeter structure", and ask in §6 "Whether there is a similar counting argument for $n = 3$ case is an open question." The verbatim definitions are in the literature note.

Issue #13575 reads Eq. (3) as a theorem about every normalized pure qubit stabilizer state and every assignment of its qubits to four labelled parties (parties may be empty), for every real $a$, together with positivity of every replica partition function $Z_3^{(\mathtt q)}$ so that every logarithm is defined.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/StabilizerMultiEntropyCollapse.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPrime multiplicities in Bouras's continued fractionsEvery occurring prime other than five has exactly two indices in Bouras A363347.Proved

The questionBouras, OEIS A363347, COMMENTS, Conjecture 3:

Why it mattersThe results determine unbounded occurrence counts, rather than checking a finite prefix.

The ideaFactorial divisibility and elementary residue exclusions constrain a surviving prime denominator.

Our contributionEvery occurring prime other than five has exactly two indices in Bouras A363347.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime multiplicities in Bouras's continued fractions

M. Bouras, OEIS A363347, COMMENTS, Conjecture 3:

Conjecture 3: Except for 5, the primes all appear exactly twice.

M. Bouras, OEIS A363102, COMMENTS, Conjecture 2:

Conjecture 2: All prime numbers appear either twice (same as A356247 and A357127) or three times.

For $n\ge3$, let $K(n,x)$ be the reciprocal of the right-folded fraction $2-3/(3-4/(\cdots- n/x))$. Define $a_4(n)=\operatorname{den}(K(n,-4))$ and $a_2(n)=\operatorname{den}(K(n,-2))$, using the positive reduced rational denominator. The first claim counts exactly two indices for every occurring prime $p\ne5$; the second counts exactly two or three for every occurring prime $p$. This dossier anchors A363347 Conjecture 3; the sibling settlement has its own dossier. Neither statement asserts that every prime occurs. This is the binding reading in preregistrations #18626 and #18627, both tier 1 external OEIS problems.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Continued fraction, terminal value $-4$ | BourasA363347.a, using BourasA356360.K and nested | Literal reciprocal and right fold; Rat.den | | Continued fraction, terminal value $-2$ | BourasA363102.a, using the same K | Literal reciprocal and right fold; Rat.den | | Prime values | Nat.Prime and an existential occurrence guard | Includes precisely the primes taken by the sequence | | Except for $5$ | BourasA363347.claim has the guard $p\ne5$ | No exclusion is added to the second claim | | Indices start at three | Both occurrence sets include $3\le n$ | Matches the OEIS offsets | | Exactly twice | BourasA363347.claim uses Set.ncard = 2 | Its proof constructs a finite pair | | Twice or three times | BourasA363102.claim uses the disjunction of cardinalities two and three | Its proof constructs a pair or triple |

The natural polynomial BourasPolynomials.U is $U(n)=n^2-2$, with truncated natural subtraction. At $n\ge3$ it agrees with the integer polynomial. The source's rational fold expressions are represented by the existing frozen BourasA356360.nested and K declarations in this module. The public BourasA372763OnceRefutation.nested and K owners are textually equal; reusing them here would require changing or removing those frozen declarations, so both copies remain. No duplicate-retention authorization is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BourasA363347PrimeMultiplicity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2007Unique minimum of the nearest-neighbor QES chain potentialThe nearest-neighbor QES chain potential has a unique minimum at its site configuration.Proved

The questionRodríguez, arXiv:0704.3046v1, §3, printed p. 13, after Eqs. (31)–(32), state:

Why it mattersThis supplies the uniqueness premise identified by the source as the first limitation of the usual freezing-trick argument.

The ideaGaussian decay supplies a maximum on the closed monotone chamber.

Our contributionThe nearest-neighbor QES chain potential has a unique minimum at its site configuration.

Verification details
Release source recordProved in Lean

Open problem

Unique minimum of the nearest-neighbor QES chain potential

A. Enciso, F. Finkel, A. González-López and M. A. Rodríguez, arXiv:0704.3046v1, §3, printed p. 13, after Eqs. (31)–(32), state:

The first one is the requirement that ξ be the unique minimum of the potential U in the domain C. Although our numerical calculations suggest that this is indeed the case, we have not been able to provide a rigorous proof of this fact.

The chamber is $C={x\in\mathbb R^N:x_1<\cdots<x_N}$ from Eq. (10). With cyclic indices $x_0=x_N$ and $x_{N+1}=x_1$, Eq. (31) is

$$ U(x)=\sum_i x_i^2+\sum_i\frac{2}{(x_i-x_{i+1})^2} +\sum_i\frac{2}{(x_i-x_{i-1})(x_i-x_{i+1})}. $$

The sites obey Eq. (6), $\xi_i=1/(\xi_i-\xi_{i-1})+1/(\xi_i-\xi_{i+1})$. Issue #11857 fixes the settlement for every natural $N\geq3$, with existence of an increasing site configuration and equality only there.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/NearestNeighborFreezingUniqueMinimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 3/4 · 10% weight

    First preprint year: 2007. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2011The three-state root-fidelity Holevo boundRoot-fidelity matrix entropy bounds the Holevo quantity of every three-state ensemble.Proved

The questionRoga and K. Życzkowski, Matrices of fidelities for ensembles of quantum states and the Holevo quantity, arXiv:1104.2271v1, page 10, Conjecture 1:

Why it mattersIt establishes the unrestricted three-state mixed-state bound.

The ideaTwo polar gauges make the first two cross products positive.

Our contributionRoot-fidelity matrix entropy bounds the Holevo quantity of every three-state ensemble.

Verification details
Release source recordProved in Lean

arXiv

The three-state root-fidelity Holevo bound

M. Fannes, F. de Melo, W. Roga and K. Życzkowski, Matrices of fidelities for ensembles of quantum states and the Holevo quantity, arXiv:1104.2271v1, page 10, Conjecture 1:

For any ensemble E₃ = {(pᵢ, ϱᵢ)} with K = 3 three states of arbitrary dimension d the corresponding Holevo quantity χ(E₃) is bounded from above by the entropy of the matrix C√F(E₃) defined in (16): χ(E₃) ≤ S(C√F(E₃)).

Here $S(X)=-\operatorname{Re}\operatorname{tr}(X\log X)$, $\chi=S(\sum_i p_i\rho_i)-\sum_i p_iS(\rho_i)$, and $C_{ij}=\sqrt{p_ip_j}\operatorname{tr}|\sqrt{\rho_i}\sqrt{\rho_j}|$. The logarithm is natural. The matrices are positive semidefinite of trace one; weights are nonnegative and sum to one.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Three states of arbitrary dimension | claim quantifies d : ℕ, three matrices indexed by Fin 3 on Fin d | No invertibility assumption; dimension zero has no trace-one state | | Density matrices | Each matrix is positive semidefinite and has trace one | Singular matrices included | | Ensemble probabilities | Each weight is nonnegative, and their sum is one | Zero weights included | | Fidelity and matrix (16) | The literal trace-norm/square-root expression in rootFidelityMatrix | The trace norm is the existing FiniteTraceDistance.traceNorm | | Holevo quantity and entropy | holevo; the literal trace-log expression in claim | Ordinary complex matrices, with the L2 operator norm for functional calculus | | Upper bound | FannesRootFidelityHolevo.result | The full expanded claim |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Divergence/FannesRootFidelityHolevo.result

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Quantum informationQuestion resolvedFirst preprint 2004Quantum cube-root scalar curvature is strictly Schur-increasingQuantum cube-root scalar curvature is strictly Schur-increasing.Proved

The questionGibilisco and Isola, On the monotonicity of scalar curvature in classical and quantum information geometry, arXiv:math-ph/0407007v2, Conjecture 4.2, printed page 11:

Why it mattersIt identifies a quantum curvature quantity that strictly increases under spectral mixing.

The ideaThe normal derivative and the coordinate Gauss equation express curvature through the Hermitian Jordan operator.

Our contributionQuantum cube-root scalar curvature is strictly Schur-increasing.

Verification details
Release source recordProved in Lean

Open problem

Quantum cube-root scalar curvature is strictly Schur-increasing

Gibilisco and Isola, On the monotonicity of scalar curvature in classical and quantum information geometry, arXiv:math-ph/0407007v2, Conjecture 4.2, printed page 11:

Suppose $n\geq2$. For the function $\mathrm{Scal}_p(\cdot):\mathcal D^1_n\to\mathbb R$ one has the following properties: if $p\in(1,2)$ then $\mathrm{Scal}_p$ is strictly Schur-decreasing; if $p\in(2,+\infty]$ then $\mathrm{Scal}_p$ is strictly Schur-increasing.

This settlement proves the quantum $p=3$ slice, equivalently $\alpha=1/3$, in every dimension $n\ge2$. The faithful density states are positive definite complex Hermitian matrices of trace one. The metric is the Euclidean pull-back of $A\mapsto3A^{1/3}$, with real trace inner product. Scalar curvature is defined by coordinate Christoffel symbols and Ricci contraction, rather than by the spectral formula proved below. The source's order calls $x$ more mixed than $y$ when the decreasingly sorted prefix sums of $x$ are at most those of $y$, with equal total sums.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every dimension $n\ge2$ | claim quantifies natural n with hypothesis 2 ≤ n. | Exact. | | Faithful complex density states | densityStates requires PosDef and trace one for both matrices; positive definiteness supplies Hermiticity. | Exact. | | Cube-root pull-back geometry | HermitianCubeDerivative.alphaEmbed3, HermitianDensityCoordinates.rho, and metric give the literal derivative pull-back. scal3 contracts coordinate curvature at chartPoint. | Exact; the fixed trace-one chart is proved equivalent to the Hermitian trace-one space. | | Eigenvalue majorization | eig selects the Hermitian eigenvalue vector; PositiveSimplexGeometryAndMixing.moreMixed is the existing decreasing-sort owner. | Exact and independent of permutation order. | | Schur increase | result concludes scal3 B ≤ scal3 A when eig A is more mixed than eig B. | Exact source convention. | | Strictness outside permutations | The second conjunct concludes scal3 B < scal3 A unless the eigenvalue vectors differ by a permutation. | Exact. | | Other values of $p$ and the decreasing clause | No corresponding conclusion in claim. | Open; outside this slice. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/AlphaGeometry/QuantumCubeRootScalarCurvatureMonotone.result

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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2011The discriminant clears average mixing denominatorsFor every finite simple graph, the discriminant of the rational minimal polynomial of its adjacency matrix times its average mixing matrix has integer entries.Proved

The questionGodsil, Average mixing of continuous quantum walks, arXiv:1103.2578v3, section 11, Question 1 (PDF page 20), asks:

Why it mattersIt reuses idempotent and avgMixing from AverageMixingTraceMaximum; the transition-integral interpretation is represented by the source's Lemma 1.1 spectral formula.

The ideaThe rational minimal polynomial of the symmetric integer adjacency matrix is monic, squarefree and has integer coefficients.

Our contributionFor every finite simple graph, the discriminant of the rational minimal polynomial of its adjacency matrix times its average mixing matrix has integer entries.

Verification details
Release source recordProved in Lean

Open problem

The discriminant clears average mixing denominators

C. Godsil, Average mixing of continuous quantum walks, arXiv:1103.2578v3, section 11, Question 1 (PDF page 20), asks:

Is it true that if $D$ is the discriminant of the minimal polynomial of $X$, then $D\widehat M_X$ is an integer matrix?

For a finite simple graph $X$ with adjacency matrix $A$, the minimal polynomial is taken over $\mathbb Q$. The source's Lemma 1.1 identifies $\widehat M_X=\sum_\theta E_\theta\circ E_\theta$, with one orthogonal spectral projector for each distinct eigenvalue. Lemma 3.1 establishes only the factor $D^2$. Issue #12339 preregisters the exact all-graphs assertion, with no connectedness or simple-spectrum restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/AverageMixingDiscriminantIntegrality.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2011. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture provedFirst preprint 2025Lichtenfelz–Modin–Preston Conjecture 1: four Zeitlin six-j identitiesRacah finite sums and the Zeitlin six-j identities.Proved

The questionConjecture 1 of Lichtenfelz, Modin and Preston, arXiv:2508.09833v1, section 2.2, asserts all four identities (2.10)–(2.13).

The ideaAn exact four-spin WZ identity produces the finite Jacobi action, and endpoint WZ normalization fixes the orthogonality constant.

Our contributionRacah finite sums and the Zeitlin six-j identities.

Verification details
Release source recordProved in Lean

Open problem

Lichtenfelz–Modin–Preston Conjecture 1: four Zeitlin six-j identities

Conjecture 1 of Lichtenfelz, Modin and Preston, arXiv:2508.09833v1, section 2.2, asserts all four identities (2.10)–(2.13). The complete quotation and the two Wigner-symbol conventions are in the literature note lichtenfelz2026zeitlin. The domain is every integer N ≥ 2, with 1 ≤ j,l < N; only (2.10) assumes j ≠ l. The sums traverse i = 1,...,N−1, λ(i) = i(i+1), and H(j) is the j-th harmonic number. The preregistration is issue #11604.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/ZeitlinSixJ/SumRules.result

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Quantum informationQuestion resolvedFirst preprint 2026The Tsirelson bound of the d-outcome Bell functional of Perito et al.Every finite unitary strategy satisfies the Perito Bell upper bound.Proved

The questionThe verbatim statements are in the literature note.

The ideaThe proof uses spectral estimates.

Our contributionEvery finite unitary strategy satisfies the Perito Bell upper bound.

Verification details
Release source recordProved in Lean

Open problem

The Tsirelson bound of the d-outcome Bell functional of Perito et al.

I. Perito, R. D'Avino, M. Jung, P. Mironowicz, A. Acín and R. Augusiak, Bell inequalities tailored to optimal global randomness certification, arXiv:2606.21362v3, §III.2, define for $d$-outcome measurements in unitary notation ($A_x=\sum_a\omega^a\Pi_{a|x}$, $\omega=e^{2\pi i/d}$) $\mathcal I_d=\frac12\sum_{x<2}\sum_{y<d}\omega^{xy}A_x\otimes B_y+\text{h.c.}$ and state:

Conjecture 1. The Tsirelson bound of $\mathcal{I}_d$ is given by Eq. (tsir_bound) [$2/\sin(\pi/(2d))$] and is attained by taking $A_0=Z$, $A_1=X$; the state $\ket{\psi_d} = \frac{1}{\sqrt d} \sum_k \ket{kk}$; and $B_y$ defined by Eq. (optbob).

The verbatim statements are in the literature note. Issue #14106 reads it for every $d\ge2$, every pair of finite dimensions, every density matrix and every projective strategy ($A_x$, $B_y$ unitary with $A_x^d=B_y^d=I$), together with the validity and the value of the stated strategy.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/PeritoTsirelson.result

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Number theoryQuestion resolvedFirst preprint 2026Unique integral circulant energy maximum at order $p^2q^3$For distinct odd primes p and q, one divisor set uniquely maximises integral circulant energy at order p squared q cubed, including q = 3.Proved

The questionRoldán, Graph Energy Maximisation for Integral Circulant Graphs of Order n = p^2q^3, arXiv:2604.09491v1, Section 1, Conjecture 1.3 (Universality):

Why it mattersA uniform proof for all distinct odd primes closes this exceptional family and Conjecture 1.3 as stated.

The ideaFourier diagonalization gives eigenvalues as Ramanujan sums.

Our contributionFor distinct odd primes p and q, one divisor set uniquely maximises integral circulant energy at order p squared q cubed, including q = 3.

Verification details
Release source recordProved in Lean

Open problem

Unique integral circulant energy maximum at order $p^2q^3$

D. G. Roldán, Graph Energy Maximisation for Integral Circulant Graphs of Order n = p^2q^3, arXiv:2604.09491v1, Section 1, Conjecture 1.3 (Universality):

For every pair of distinct odd primes (p, q), the set D* = {1, p^2, pq, q^2, p^2q^2, pq^3} is the unique maximiser of E(p^2q^3, D) over all nonempty D ⊆ div(p^2q^3) ∖ {p^2q^3}.

The graph has vertices $\mathbb Z_n$ and adjacency condition $\gcd(a-b,n)\in D$. Its energy is the sum of the absolute values of its adjacency eigenvalues, with multiplicity. Source quotations and locators are in the literature note. Preregistration: #19144.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Vertices $\mathbb Z_n$, modular difference, gcd adjacency | adjacency n D, on Fin n | Literal preregistered expression; subtraction is modulo $n$. | | Energy sums $|\lambda_i|$ with multiplicity | energy n D, using the multiset roots of the real characteristic polynomial | The proof identifies all roots through Fourier diagonalization of the real symmetric adjacency. | | Distinct odd primes $p,q$ | claim quantifies $p,q$ with both primality, inequality and oddness hypotheses | No ordering of $p,q$ and no restriction $q\ge5$. | | $D^={1,p^2,pq,q^2,p^2q^2,pq^3}$ | Dstar p q | Literal six-term finite set. | | Allowed, nonempty proper-divisor sets | First conjunct of claim and the guards on its quantified D | The first conjunct proves admissibility of $D^$; $1\in D^*$ gives nonemptiness. | | Unique maximizer | Strict energy inequality for every different allowed nonempty D | Full proposition, proved by result : claim. |

No additional premise restricts the published conjecture. The spectral argument proves the comparison even when the competing divisor set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/RoldanCirculantEnergyP2Q3.result

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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2017The sharp Hoggar-SIC sum negativity is seven eighthsThe sum negativity of the Hoggar-SIC Q-minus representation in dimension eight has global maximum 7/8 over all density matrices.Proved

The questionFuchs, Negativity Bounds for Weyl–Heisenberg Quasiprobability Representations, arXiv:1703.08272v2, Section 6, page 17, write:

Why it mattersThus the numerical value in Section 6 is the global maximum, including mixed states.

The ideaTransport to complex matrices gives the SIC overlaps and positive-semidefinite trace-one projectors.

Our contributionThe sum negativity of the Hoggar-SIC Q-minus representation in dimension eight has global maximum 7/8 over all density matrices.

Verification details
Release source recordProved in Lean

Open problem

The sharp Hoggar-SIC sum negativity is seven eighths

John B. DeBrota and Christopher A. Fuchs, Negativity Bounds for Weyl–Heisenberg Quasiprobability Representations, arXiv:1703.08272v2, Section 6, page 17, write:

Although dimension $5$ was the last in which we were able to explicitly calculate the sum negativity for the SIC Q-reps by exhaustive combinatorial searching, we suspect that we have found the correct sum negativity for ${Q_j^-}$ constructed with the Hoggar SIC in dimension $8$. Rather than calculating the eigenvalues of every partial sum matrix (since this is infeasible for $2^{64}$, $8\times8$ matrices), we used a numerical local maximization procedure and around $10^6$ random pure state seeds. The overall maximum value we found, $7/8$, occurred frequently in our data and is significantly larger than all of the smaller local maxima. Of course, we could still be falling short of the global maximum value if it occurs at very hard to access positions. The states whose quasiprobability representations achieve the sum negativity of $7/8$ consist of $28$ copies of value $-1/32$ and $36$ copies of value $5/96$.

The precise question, preregistered in issue #11512, is whether $7/8$ is the attained greatest value of sum negativity over all density matrices. With the three-qubit Pauli orbit of $v=(-1+2i,1,1,1,1,1,1,1)^T$, write $\Pi_j=u_ju_j^*/12$, $Q_j^-=3\Pi_j-I/4$ and $q_\rho(j)=\operatorname{Re}\operatorname{Tr}(\rho Q_j^-)/8$. Equations (8)–(11) define $N^1(\rho)=\sum_j(|q_\rho(j)|-q_\rho(j))/2$ and maximize it over positive-semidefinite complex $8\times8$ matrices of trace one.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/HoggarSicSumNegativity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2014The original Jaco directed path-count equivalenceActual shortest directed paths in the finite Jaco truncations.Proved

The questionIts proof placeholder is “(Conjectured)”; the final list explicitly asks “[Open problem: Prove Proposition 4.4]”.

Why it mattersThe source asks whether adjacent repetition of the infinite outdegrees exactly matches adjacent repetition of actual shortest-directed-path cardinalities.

The ideaEvery forward path stays between its endpoints.

Our contributionActual shortest directed paths in the finite Jaco truncations.

Scope & next questionAnalogous count laws for other Jaco functions or further source questions remain open.

Verification details
Release source recordProved in Lean

arXiv

The original Jaco directed path-count equivalence

Johan Kok, Paul Fisher, Bettina Wilkens, Mokhwetha Mabula and Vivian Mukungunugwa pose Proposition 4.4 in Section 4 of https://arxiv.org/abs/1404.1714v1. Its proof placeholder is “(Conjectured)”; the final list explicitly asks “[Open problem: Prove Proposition 4.4]”.

For every natural $k\geq7$ and every Fibonacci bracket index $i$ with $F_i<k\leq F_{i+1}$, the complete question is

$$ \bigl(d^+(v_{k-1})\ne d^+(v_k)\ \land\ d^+(v_k)\ne d^+(v_{k+1})\bigr) \quad\Longleftrightarrow\quad \bigl(\psi(v_{k-1})\ne\psi(v_k)\ \land\ \psi(v_k)\ne\psi(v_{k+1})\bigr). $$

The vertices have positive labels. An original directed arc goes from $v_h$ to $v_j$ exactly when $h<j$ and $j\leq2h-d^-(v_h)$. The degree is the infinite outdegree. Each $\psi(v_n)$ counts distinct shortest directed edge sequences from $v_1$ to $v_n$ in that endpoint's own induced finite truncation $J_n(1)$. Length counts edges. Both adjacent comparisons, the iff, the lower bound and all bracket indices are part of the claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/JacoDirectedPathCount.result

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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2019A product decomposition of the generalized CHSH groupsFor every n at least two, the presented generalized CHSH group is a semidirect product of C(2n) x C2^(n-2) by Cn, with an explicit cyclic action.Proved

The questionCui, Mehta, Mousavi and Nezhadi, A generalization of CHSH and the algebraic structure of optimal strategies, Quantum 4, 346 (2020), arXiv:1911.01593v5, Section 1.4 (Further work), item 2, ask:

Why it mattersIt supplies a uniform structural answer to the source's further-work question.

The ideaThe two composites fix the generators, yielding the multiplicative equivalence.

Our contributionFor every n at least two, the presented generalized CHSH group is a semidirect product of C(2n) x C2^(n-2) by Cn, with an explicit cyclic action.

Verification details
Release source recordProved in Lean

Open problem

A product decomposition of the generalized CHSH groups

Cui, Mehta, Mousavi and Nezhadi, A generalization of CHSH and the algebraic structure of optimal strategies, Quantum 4, 346 (2020), arXiv:1911.01593v5, Section 1.4 (Further work), item 2, ask:

In Section 5, we give efficient explicit presentations for G_n and its multiplication table. Can we go further and characterize these groups in terms of direct and semidirect products of small well-known groups? The first few cases are as follows

For every natural $n\ge2$, the answer is

$$G_n\cong\left(C_{2n}\times C_2^{,n-2}\right)\rtimes_{\alpha_n}C_n.$$

The group $G_n$ has the source's Section 5 presentation on $P_0,P_1,J$: the relators are $P_0^n,P_1^n,J^n,[J,P_0],[J,P_1]$ and $J^i(P_0^iP_1^{-i})^2$ for $1\le i\le\lfloor n/2\rfloor$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The Section 5 presentation | P0, P1, J, rels, G | The relators are words equal to the identity; the negative exponent uses the integer cast of the natural index. | | Every $n\ge2$ | claim quantifies over natural n with 2 ≤ n | No further assumption on $n$. | | Small cyclic factors | L is Multiplicative (ZMod (2*n)) times the functions from Fin (n-2) to Multiplicative (ZMod 2) | The independent involutions have zero-based indices. | | An explicit semidirect product | alpha and the MulEquiv in claim | The action is the binding action in the definition comment on #16204. | | A product characterization | result : claim | An actual multiplicative equivalence, rather than an order comparison. |

The binding comment fixes cyclicHom, s, t, liftL, ts, u, e, step, stepAut and alpha. Here t n i represents the source's $t_{i+1}$, and u n k uses one-based indices. In particular, $u_{n-1}=s^n\prod_{i=1}^{n-2}t_i$, while $u_k=1$ outside the indicated range. For $n=2$, $u_1=s^2$. The action below two is trivial; the claim has no conclusion for those values.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/GeneralizedCHSHGroupDecomposition.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2015First-parameter coefficient signs of the normalized incomplete beta functionStrict coefficient signs for real shifts of the first incomplete beta parameter.Proved

The questionKarp, Normalized incomplete beta function: log-concavity in parameters and other properties, §3, Conjecture 2:

Why it mattersCoefficient signs constrain each term of the normalized series separately, strengthening the corresponding pointwise Wright inequalities.

Our contributionStrict coefficient signs for real shifts of the first incomplete beta parameter.

Verification details
Release source recordProved in Lean

Open problem

First-parameter coefficient signs of the normalized incomplete beta function

Karp, Normalized incomplete beta function: log-concavity in parameters and other properties, §3, Conjecture 2:

Conjecture 2 The coefficients ψ_k are negative for all a, α, β > 0 and x ∈ (0, 1) if 0 < b < 1 and positive if b > 1, so that a → x^{−a}(1 − x)^{−b} I_x(a, b) is strictly coefficient-wise Wright log-convex on (0, ∞) for α, β > 0, x ∈ (0, 1) and 0 < b < 1; and a → x^{−a}(1 − x)^{−b} I_x(a, b) is strictly coefficient-wise Wright log-concave on (0, ∞) for α, β > 0, x ∈ (0, 1) and b > 1.

Equations (1) and (6) give

$$c_j(t,b)=\frac{\Gamma(t+b+j)}{\Gamma(b)\Gamma(t+j+1)},\qquad \psi_n=\sum_{k=0}^n\bigl[c_k(a+\alpha,b)c_{n-k}(a+\beta,b) -c_k(a,b)c_{n-k}(a+\alpha+\beta,b)\bigr],$$

$$\psi_{a,b,\alpha,\beta}(x) =x^{2a+\alpha+\beta}(1-x)^{2b}\sum_{n\ge0}\psi_nx^n.$$

Formal reading

Unless otherwise indicated, Lean names have prefix D5.S3.Analytic.IncompleteBetaShape.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $c_j(t,b)$ in (1) | D5.S3.Analytic.IncompleteBeta.coeff | The same Gamma quotient; no new coefficient definition. | | $\psi_n$ in (6) | Defs.psi | Exactly the displayed finite convolution difference, with natural index $n$. | | $S_n(p,q;b)$ | Mixture.convolution | $\sum_{k=0}^n c_k(p,b)c_{n-k}(q,b)$. | | Uniform shifted beta mixture | Mixture.mixtureDensity | The average of Beta$(p+k+1,q+n-k+1)$ densities for $0\le k\le n$. | | Sliding kernel | Kernel.F, Kernel.H | The uniform affine average and its beta average used below. | | Conjecture 2 | Defs.claim, Result.result | Both strict signs for all real $a,b,\alpha,\beta>0$ and all natural $n$. | | $x\in(0,1)$ | No additional coefficient hypothesis | The coefficients do not depend on $x$, so this quantifier is vacuous for them. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/IncompleteBetaShape/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedFirst preprint 2020Lin–Miao–Yu–Sheng Conjecture 5.11: second-norm uniquenessTwo nonisomorphic connected graphs on six vertices and seven edges have the same second Ky Fan norm of their LI-matrices.Refuted

The questionZhen Lin, Lianying Miao, Guanglong Yu and Han Sheng, On the Ky Fan k-norm of the LI-matrix of graphs, Electronic Journal of Linear Algebra 38 (2022), 680–696, Conjecture 5.11 (page 694), states:

Why it mattersThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

The ideaIt uses the existing D5.S3.Combinatorics.Graph.KyFanGap.kyFan definition rather than defining another norm.

Our contributionTwo nonisomorphic connected graphs on six vertices and seven edges have the same second Ky Fan norm of their LI-matrices.

Verification details
Release source recordRefuted in Lean

Open problem

Lin–Miao–Yu–Sheng Conjecture 5.11: second-norm uniqueness

Zhen Lin, Lianying Miao, Guanglong Yu and Han Sheng, On the Ky Fan k-norm of the LI-matrix of graphs, Electronic Journal of Linear Algebra 38 (2022), 680–696, Conjecture 5.11 (page 694), states:

Among all connected graphs with n and m edges n ≤ m ≤ 2n−3, the Gₘ,ₙ is the unique graph with maximal value of ‖LI(G)‖F₂ and ‖LI(G)‖F₃.

Here $n$ is the number of vertices, $m$ the number of unordered edges, $L(G)=D(G)-A(G)$, $LI(G)=L(G)-(2m/n)I_n$, and the Ky Fan $k$-norm is the sum of the $k$ largest singular values, with multiplicity. The assertion is read as a separate maximum and equality characterization for each of $k=2$ and $k=3$. The interpretation involving a unique simultaneous maximizer is outside the claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite simple undirected graphs of order $n$ | SimpleGraph (Fin n) | Every finite labelled graph of that order is represented. | | Connected, with $m$ edges | G.Connected, Nat.card G.edgeSet = m | Mathlib connectivity and unordered edges. | | $n\le m\le2n-3$ | The two range hypotheses in claim | m + 3 ≤ 2 * n avoids truncated subtraction; together the hypotheses force $n\ge3$. | | $LI(G)=L(G)-(2m/n)I_n$ | LI | Literal source-specific shift of Mathlib's Laplacian. | | $G_{m,n}$ in Figure 2.1 | Gmn n m | Vertex $0$ is universal; vertex $1$ meets $2,\ldots,m-n+2$. Argument order is $n,m$. | | Separate assertions for $F_2$ and $F_3$ | claim quantifies over $k=2$ or $k=3$ | Both the upper bound and the equality iff isomorphism are retained. | | Refutation | result | The public theorem proves the negation of that whole claim using its $k=2$ equality clause. |

Section 2 prints $2n-m+3$ pendant edges. That count conflicts with the stated order, size and figure; the figure's construction has $2n-m-3$ pendant vertices. The definition uses the figure and order/size convention fixed in #21801. There is no additional mathematical hypothesis in the public result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/KyFanGap/LinMiaoYuShengSecondNormUniquenessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Cube polynomial of the associated Mersenne graphsThe cube polynomial of the associated Mersenne graphs is determined by a cleared-denominator generating function.Proved

The questionWei and Yang, Associated Mersenne graphs, arXiv:2407.08237v1, Section 6, Question 6.3 asks:

The ideaThe construction and uniqueness are supplied by cubemapimage, inducedcubeiff, and downCubeunique.

Our contributionThe cube polynomial of the associated Mersenne graphs is determined by a cleared-denominator generating function.

Verification details
Release source recordProved in Lean

Open problem

Cube polynomial of the associated Mersenne graphs

Wei and Yang, Associated Mersenne graphs, arXiv:2407.08237v1, Section 6, Question 6.3 asks:

What is cube polynomial for Associated Mersenne graph $\mathcal{M}_n$?

The associated Mersenne graph has the circular run constrained Boolean words as vertices, with Hamming distance one as adjacency. Its cube polynomial is $C_n(x)=\sum_k c_n(k)x^k$, where $c_n(k)$ counts induced copies of $Q_k$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hamming/AssociatedMersenneCubePolynomial.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2015Svinin's Discrete-Polynomial IdentitySvinin's discrete-polynomial convolution identity holds over every commutative ring.Proved

The questionSvinin, On integrals for some class of ordinary difference equations admitting a Lax pair representation, arXiv:1505.06394v2, Conjecture 1, Eq. (48), asks whether

Why it mattersIt establishes the prerequisite that the source uses for subsequent relations between difference-equation families and first integrals in the same specialization.

The ideaPartition weighted monotone and strictly monotone tuples according to whether their first coordinate is zero.

Our contributionSvinin's discrete-polynomial convolution identity holds over every commutative ring.

Verification details
Release source recordProved in Lean

Open problem

svinin 2015 discrete polynomial identity

A. K. Svinin, On integrals for some class of ordinary difference equations admitting a Lax pair representation, arXiv:1505.06394v2, Conjecture 1, Eq. (48), asks whether

$$ \langle S,T\rangle_s^r(i)=\sum_{j=0}^{r-2}(-1)^j u(i+r-j-2)u(i+s+j)S_{s+j+2}^{r-j-2}(i)T_{s-r+j}^{j}(i+r-j) $$

is an identity for every $r\ge1$, $s\ge r-1$ and $i\in\mathbb Z$ in the specialization $(h,n)=(1,1)$. The source says: “Unfortunately, by now we can prove this conjecture only for r = 1, 2, 3.” The settlement is Proved, under preregistration #19287. The coefficient field is an arbitrary function $u:\mathbb Z\to R$ into any commutative ring $R$.

Statement echo

| Source clause | Lean counterpart | Correspondence | | --- | --- | --- | | Eq. (4): strictly increasing tuples $0\le\lambda_1<\cdots<\lambda_r\le s-1$ | T | Maps Fin r → Fin s.toNat, filtered by StrictMono; factor $a$ has offset $a$. | | Section 2.3: weakly increasing tuples with descending offsets | S | The Monotone filter and integer offset $r-1-a$ reproduce the factor pattern. The printed first-offset $k$ is read as $r$. | | Section 3: $(h,n)=(1,1)$ | T, S | The lattice increments in the two literal sums are one. | | Section 3.4: alternating binary operation | bracket | The sum has indices $j=0,\ldots,r$ and exactly the source's degree, subscript and lattice shifts. | | Conjecture 1, Eq. (48), all degrees | claim, result | All rings, functions, degrees $r\ge1$, subscripts $s\ge r-1$ and lattice points are quantified. | | Degree-zero and nonpositive-subscript conventions | T, S | Empty products are one; positive-degree maps into Fin 0 do not exist. Natural subtraction is truncated; lattice arithmetic uses integers. |

No nonvanishing, field-inverse, characteristic or integral-domain hypothesis is added. For $r=1$ the right-hand side is empty. The Library note Library/RecurrenceComposition/svinin2015discrete.md gives the verbatim source expressions and locator.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/SvininDiscretePolynomialIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025The original ari coproduct preserves every right-factor level filtrationThe ordinary ari coproduct preserves every right-factor level filtrationProved

The questionAnnika Burmester and Ulf Kühn, On post-Lie structures for free Lie algebras, arXiv:2504.19661v1, Conjecture 4.29, state:

Why it mattersThe assertion controls how the original ari coproduct distributes letters away from a chosen generator in its right tensor factor.

The ideaIt uses only the valid nonzero-multiplicity implication from Remark 4.3, including preservation of zero positions.

Our contributionThe ordinary ari coproduct preserves every right-factor level filtration

Verification details
Release source recordProved in Lean

Open problem

The original ari coproduct preserves every right-factor level filtration

Annika Burmester and Ulf Kühn, On post-Lie structures for free Lie algebras, arXiv:2504.19661v1, Conjecture 4.29, state:

For k ≥ 0 we define the k-level of a word v_{k_1}⋯v_{k_d} by d − #{j | k_j=k}, and extend this to an increasing filtration on Q⟨V⟩. For w∈Q⟨V⟩, write as usual Δ_a(w)=w_(1)⊗w_(2). Then, we always have k-level(w_(2)) ≤ k-level(w).

Let $H=\mathbb Q\langle v_0,v_1,\ldots\rangle$ be the algebra of finite ordinary rational word polynomials, including its unit. For a word $w$, let $\lambda_k(w)$ count letters different from $v_k$ and put

$$ F_r^k=\operatorname{span}_{\mathbb Q} {w:\lambda_k(w)\le r}. $$

The precise filtration assertion is, for every $k,r\in\mathbb N$ and every $P\in F_r^k$,

$$ \Delta_a(P)\in\operatorname{range} \bigl(\operatorname{id}H\otimes\iota_r^k: H\otimes{\mathbb Q}F_r^k\longrightarrow H\otimes_{\mathbb Q}H\bigr), $$

where $\iota_r^k$ is the natural submodule inclusion. All tensor products are ordinary algebraic tensor products. A Sweedler expression is read through this subspace assertion, rather than through a chosen tensor decomposition. The resolution is Proved by D5/S1/Words/PostLie/AriCoproductFiltration.result, with claim : Prop and result : claim.

| Source clause | Formal counterpart | | --- | --- | | Original rational word algebra on all nonnegative indices | AriWordOperations.H = RationalWordPolynomial Nat, with Word = FreeMonoid Nat and the existing ordinary monoid-algebra basis | | Every natural $k$, including zero | The unrestricted k : Nat in claim; the generator $v_0$ and mixed-zero multiplicities retain their source conventions | | Increasing level filtration at every $r$ | level k w = w.countP' (· ≠ k) and filtration k r, the rational span of all words of level at most $r$ | | Every finite rational polynomial in that span | P : H with membership in filtration k r; no length, weight, homogeneity, coefficient-sign or cancellation restriction | | Original ari product and dual coproduct | letterAction, primitiveUnshuffle, star, transposeProperty, and the proved uniquely characterized delta | | Full ordinary pairing | pairing, tensorPairing, delta_transpose and delta_basis_coefficient, including empty words and both unit tensor terms | | Membership in the actual right-factor tensor image | rightFiltrationImage k r = LinearMap.range (TensorProduct.map id (filtration k r).subtype) and result |

The source operations and locators are documented in the source note. The claim is the entire target preregistered in #23147.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/PostLie/AriCoproductFiltration.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

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Quantum informationCounterexample foundFirst preprint 2007Extended local permutations do not suffice for block-shift formA three-qubit root vector has an above-diagonal support entry after every extended local permutation, excluding contiguous block-shift form.Refuted

The questionSchulte-Herbrüggen, Relative C-numerical ranges for applications in quantum control and quantum information, Linear and Multilinear Algebra 56 (2008), 27–51, arXiv:math-ph/0702005v1, Conjecture 3.6, pp. 26–27:

Why it mattersThe counterexample separates these two transformation classes.

The ideaSome supported row lies above its column in Kronecker order.

Our contributionA three-qubit root vector has an above-diagonal support entry after every extended local permutation, excluding contiguous block-shift form.

Scope & next questionDetermining a smallest or natural sufficient signed-permutation subgroup remains open, including the proposed class of all signed permutations commuting with the local torus action.

Verification details
Release source recordRefuted in Lean

Open problem

Extended local permutations do not suffice for block-shift form

G. Dirr, U. Helmke, M. Kleinsteuber and T. Schulte-Herbrüggen, Relative C-numerical ranges for applications in quantum control and quantum information, Linear and Multilinear Algebra 56 (2008), 27–51, arXiv:math-ph/0702005v1, Conjecture 3.6, pp. 26–27:

Every element of $E(t_{\mathrm{loc}})$ is similar to a block-shift matrix via an element of $\Pi^{\mathrm{ex}}{\mathrm{loc}} := \Pi{\mathrm{loc}}\cdot\Pi_{\mathrm{out}}$.

The definitions and source quotations are in the literature note. Issue #14776 fixes DHKS-1 and the literal Lean conventions. The universal claim ranges over every $n\ge2$ and every complex matrix in the union of the nonzero root spaces of the local diagonal torus. The tensor basis has its first factor as its most significant bit.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/LocalPermutationBlockShiftRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2025Concatenation can increase the cHaar moment normThe second cHaar moment at system and environment dimension two has larger squared Hilbert-Schmidt norm after five concatenations than after four.Refuted

The questionCerezo, Moments of Quantum Channel Ensembles, PRX Quantum 7, 033050 (2026), DOI 10.1103/grdp-jph9, arXiv:2511.12700v2, Appendix B, Conjecture 1, asks whether the cHaar moment norm is non-increasing with concatenations:

Why it mattersCHaarConcatenationNonmonotone.result refutes the finite-dimensional concatenation assertion.

The ideaThe exact values in instanceN4 and instanceN5 contradict claim.

Our contributionThe second cHaar moment at system and environment dimension two has larger squared Hilbert-Schmidt norm after five concatenations than after four.

Verification details
Release source recordRefuted in Lean

Open problem

Concatenation can increase the cHaar moment norm

M. Duschenes, D. García-Martín, Z. Holmes and M. Cerezo, Moments of Quantum Channel Ensembles, PRX Quantum 7, 033050 (2026), DOI 10.1103/grdp-jph9, arXiv:2511.12700v2, Appendix B, Conjecture 1, asks whether the cHaar moment norm is non-increasing with concatenations:

$$|\widehat T_{\mathrm{cHaar}}^{(t)k'}|{HS}^2\leq|\widehat T{\mathrm{cHaar}}^{(t)k}|_{HS}^2\quad(k'\geq k).$$

The verbal source clause is: “Further, it is conjectured that the cHaar norm is monotonically non-increasing with environment dimension and concatenations,”. Only the concatenation clause is refuted. Preregistration #15553 specifies the conventions and witness. The Library note contains the source definitions, norm convention and locators.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Haar unitary on system and environment, with fixed pure ancilla; partial trace over environment. | cHaarChannel d e U X. | Reuses PartialTraceMutualInformation.partialTraceRight; its sum is exactly the preregistered sum. Matrix.single 0 0 1 is the pure coordinate-zero ancilla. | | The order-$t$ ensemble moment of the channel tensor power. | momentOp d e t. | All tensor factors use the same Haar sample, for every $t$. | | Hilbert–Schmidt norm, squared. | hsNormSq, used in claim. | The definition is the double sum of Complex.normSq specified by #15553. Private hsNormSqCorrespondence equates this sum with the frozen RHLinalg.frobSq used by normReindex, normNormal and result. | | All positive dimensions, moment orders and concatenation counts, with $k'\geq k$. | claim. | Positive $d,t,k$ and nonzero $e$; positive counts suffice for the witness. | | Refutation at $d=e=t=2$, $k=4$, $k'=5$. | result : ¬ claim. | Literal universal claim negation; exact instance values are instanceN4, instanceN5. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/RandomCircuits/CHaarConcatenationNonmonotone.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Discrete structuresConjecture provedFirst preprint 2021Sun Conjecture 4.2(i) absolute-difference permanent congruencesBoth absolute-difference permanent congruences in Sun Conjecture 4.2(i).Proved

The questionThe source permanent is the finite sum over permutations of products of row-indexed entries.

Why it mattersThe result settles the two explicitly numbered congruences in the source's open-conjecture section and gives a uniform prime-order mechanism for both absolute-difference shifts.

The ideaClearing the factor 2 gives the two integer divisibilities.

Our contributionBoth absolute-difference permanent congruences in Sun Conjecture 4.2(i).

Verification details
Release source recordProved in Lean

Open problem

Sun Conjecture 4.2(i) absolute-difference permanent congruences

Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723v7, Conjecture 4.2(i), equations (4.3) and (4.4), asks whether for every odd prime $p$ the permanents of the matrices with entries $|j-k|$ and $|j-k+1|$ are congruent to $-1/2$ and $1/2$ modulo $p$, respectively. The integer encoding is $A_s(j,k)=|((j.val:\mathbb Z)+1)-((k.val:\mathbb Z)+1)+s|$ for $j,k : \operatorname{Fin} p$ and $s=0,1$.

Statement echo

The source permanent is the finite sum over permutations of products of row-indexed entries. The Lean claim clears the invertible denominator 2 and states both divisibilities for the same universally quantified odd prime.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/SunAbsoluteDifferenceCongruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2021A finite set supplies one state entangled for all fixed tensor partitions in every frameEvery composite dimension admits a finite set with one state entangled for all fixed tensor partitions in each frame.Proved

The questionScarani, Absolutely entangled sets of pure states for bipartitions and multipartitions, Physical Review A 104, 032414 (2021), DOI 10.1103/PhysRevA.104.032414; arXiv:2106.16216v3, Section III, immediately after the Corollary (main.tex, line 509):

Why it mattersA positive answer therefore supplies a common witness of nonseparability across all the specified factorizations in each frame, rather than merely a witness for each partition separately.

The ideaIts product-state ingredients are productStatedensity, requiring every local matrix to be positive semidefinite and of trace one, and productStatecolumnminor, which gives a zero corner minor for every column of a product of arbitrary local matrices.

Our contributionEvery composite dimension admits a finite set with one state entangled for all fixed tensor partitions in each frame.

Scope & next questionMinimal size, a quantitative noise margin, and a formal relation to stronger multipartite-entanglement predicates remain open here. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

A finite set supplies one state entangled for all fixed tensor partitions in every frame

B. Yu, P. Jayachandran, A. Burchardt, Y. Cai, N. Brunner and V. Scarani, Absolutely entangled sets of pure states for bipartitions and multipartitions, Physical Review A 104, 032414 (2021), DOI 10.1103/PhysRevA.104.032414; arXiv:2106.16216v3, Section III, immediately after the Corollary (main.tex, line 509):

This result guarantees that, given any $U$ and any partition $P$, there exists at least one entangled state. An interesting open question is whether one can find an AES, such that, for every $U$, there exists a state that is entangled according to all partitions. This would arguably be the strongest form of genuine multipartite entanglement.

The quotation is from the v3 source member main.tex, md5 e11d4dc540598e9ef271fc11bad389a0. Definition 3 is at lines 310–317, and the preceding Corollary at lines 499–508. The definitions and source locator are also in the literature note.

Issue #22238 fixes the reading. A partition $p=(d_1,\ldots,d_k)$ is an ordered factorization with $k\ge2$, every $d_i\ge2$, and $\prod_i d_i=d$. Its tensor coordinates are fixed by partitionEquiv, the mixed-radix identification of local index tuples with the standard basis of $\mathbb C^d$. The matrix productState p τ has entries equal to the products of the corresponding local entries. Separable p ρ means membership in the real convex hull of such products with every local matrix positive semidefinite and of trace one.

The formal assertion is: for every $d,a,b\in\mathbb N$ with $a,b\ge2$ and $d=ab$, there is a finite set $S$ of positive semidefinite $d\times d$ complex matrices of trace one such that

$\forall U\in SU(d),\ \exists\rho\in S,\ \forall p\in\mathrm{Partition}(d),\ \neg\mathrm{Separable}(p,U\rho U^\dagger).$

The member may depend on $U$ and must serve every partition simultaneously. This fixed-coordinate assertion is proved. Purity is used in the construction; it is not an additional predicate in the type of claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/AbsolutelyEntangledAllPartitions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedSource 2017Ziętak's strict spectral approximant convergence conjectureSchatten-p approximants need not converge to the strict spectral approximant.Refuted

The questionZiętak, “From the strict Chebyshev approximant of a vector to the strict spectral approximant of a matrix,” Banach Center Publications 112 (2017), Conjecture 1.1, asks whether Schatten-p approximant residuals converge to the residual of a strict spectral approximant for every complex matrix subspace.

Why it mattersZiętak's conjecture is a named published open problem.

The ideaHermitian projection reduces a complex minimizer to the real comparison.

Our contributionSchatten-p approximants need not converge to the strict spectral approximant.

Verification details
Release source recordRefuted in Lean

Open problem

Ziętak's strict spectral approximant convergence conjecture

K. Ziętak, “From the strict Chebyshev approximant of a vector to the strict spectral approximant of a matrix,” Banach Center Publications 112 (2017), Conjecture 1.1, asks whether Schatten-p approximant residuals converge to the residual of a strict spectral approximant for every complex matrix subspace.

The source statement is:

Let M be a linear subspace of Cᵐˣⁿ and let Bₚ ∈ M be the cₚ-approximant to A ∈ Cᵐˣⁿ. As p tends to ∞, the matrix Rₚ = A − Bₚ converges to A − B⁽ˢᵗ⁾, where B⁽ˢᵗ⁾ is the strict spectral approximant to A.

Statement echo

The Lean claim quantifies over all natural dimensions, complex matrix subspaces, targets, strict spectral approximants, and real-parameter functions of approximants. Its hypotheses require strict spectral minimality and a Schatten-$p$ minimizing approximant for every $p>1$; its conclusion is convergence of residual matrices. The module's result is the literal negation ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/NormCompression/StrictSpectralPolyaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2017.

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    Year recorded in the source bibliography: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2021No randomisation-consistent time reversal on the full channel setRandomisation and complete positivity obstruct reversal of every unitary channel.Proved

The questionChiribella, Aurell and Życzkowski, “Symmetries of quantum evolutions”, arXiv:2101.04962v5, Section V.B, conjecture that the no-go theorem for time reversal continues to hold for symmetries of the full set of quantum channels:

Why it mattersSection V.B, PDF page 7, identifies the extra hypothesis in the prior channel no-go theorem of Chiribella–Liu (Ref. [56]):

The ideaIts negative diagonal entry contradicts positivity in QubitNoGo.inverseunitarynogo.

Our contributionRandomisation and complete positivity obstruct reversal of every unitary channel.

Verification details
Release source recordProved in Lean

Open problem

No randomisation-consistent time reversal on the full channel set

Chiribella, Aurell and Życzkowski, “Symmetries of quantum evolutions”, arXiv:2101.04962v5, Section V.B, conjecture that the no-go theorem for time reversal continues to hold for symmetries of the full set of quantum channels:

We conjecture that this assumption can be lifted, and that the no-go theorem to time reversal applies also to symmetries of the set of quantum channels.

For every natural dimension $d \ge 2$, IsChannelSymmetry S is the predicate saying that S is a bijection of the completely positive trace-preserving maps on $d\times d$ complex matrices and preserves every binary randomisation. The claim requires that no such S sends every unitary channel Ad_U to Ad_{U†}, and that none sends every Ad_U to Ad_{Uᵀ}.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/ChannelTimeReversalNoGo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2014The fully entangled fraction is at most one minus half the steering-ellipsoid centreProved

The questionRudolph, “Quantum correlations of two-qubit states with one maximally mixed marginal”, Physical Review A 90, 024302 (2014), arXiv:1404.3951v2, Section V, Conjecture 2:

Verification details
Release source recordProved in Lean

Open problem

The fully entangled fraction is at most one minus half the steering-ellipsoid centre

A. Milne, D. Jennings, S. Jevtic and T. Rudolph, “Quantum correlations of two-qubit states with one maximally mixed marginal”, Physical Review A 90, 024302 (2014), arXiv:1404.3951v2, Section V, Conjecture 2:

Let $\rho$ be a general two-qubit state with $\mathcal{E}$ centred at $\vec c$. The fully entangled fraction is tightly bounded as $f(\rho)\leq 1-c/2$.

Here $f(\rho)=\max_{|\phi\rangle}\langle\phi|\rho|\phi\rangle$ over maximally entangled $|\phi\rangle$ (Section III), $\mathcal E$ is Alice's steering ellipsoid, and $c=|\vec c|$. The centre is given in the companion paper S. Jevtic, M. Pusey, D. Jennings and T. Rudolph, Physical Review Letters 113, 020402 (2014), arXiv:1303.4724v2: $\vec c=(\boldsymbol a-T\boldsymbol b)/(1-b^2)$, and $\vec c=\boldsymbol a$ when $b=1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SteeringEllipsoidFullyEntangledFraction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2014. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2020Hyperbolic bounds for the Kilbas–Saigo functionHyperbolic bounds for the Kilbas-Saigo function on the negative half-line.Proved

The questionBoudabsa and Simon, Some properties of the Kilbas–Saigo function, arXiv:2012.05666v1, §4.1, Conjecture 4.5 (Conjecture 4 in the journal):

Why it mattersThe bounds compare an entire special function with two explicit rational functions throughout the nonnegative argument range.

The ideaThe ratio criterion gives absolute convergence of every signed derivative series at every real argument, and uniform local bounds justify differentiation.

Our contributionHyperbolic bounds for the Kilbas-Saigo function on the negative half-line.

Verification details
Release source recordProved in Lean

Open problem

Hyperbolic bounds for the Kilbas–Saigo function

Boudabsa and Simon, Some properties of the Kilbas–Saigo function, arXiv:2012.05666v1, §4.1, Conjecture 4.5 (Conjecture 4 in the journal):

For every α ∈ (0, 1], m > 0, l > m − 1/α and x ≥ 0, one has 1/(1 + [Γ(1+α(l−m))/Γ(1+α(l−m+1))] x) ≤ E_{α,m,l}(−x) ≤ 1/(1 + [Γ(1+αl)/Γ(1+α(1+l))] x).

Here

$$E_{\alpha,m,l}(z)=\sum_{n=0}^{\infty}c_nz^n,\qquad c_n=\prod_{k=0}^{n-1} \frac{\Gamma(1+\alpha(km+l))}{\Gamma(1+\alpha(km+l+1))},\quad c_0=1.$$

The source records the cases $l=m-1$ and $m=1$ as known.

Formal reading

All Lean names below have prefix D5.S3.Analytic.KilbasSaigo.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $c_n$ | Defs.ksCoeff | The product over $0\le k<n$, including the empty product at $n=0$. | | $E_{\alpha,m,l}(z)$ | Defs.kilbasSaigo | The real sum of $c_nz^n$; Series.summable_kilbasSaigo proves convergence for every real $z$. | | $a=1+\alpha(l-m)$, $b=\alpha m$, $\phi(s)=\Gamma(a+bs+\alpha)/\Gamma(a+bs)$ | Series.phi | The hypotheses imply $a,b>0$ and positivity of every Gamma argument used. | | $f(x)=E_{\alpha,m,l}(-x)$ | Series.g at order zero | Series.kilbasSaigo_neg_eq_g0 identifies the two series. | | $g_k=(-1)^kf^{(k)}$ | Series.g | Series.hasDerivAt_g gives $g_k'=-g_{k+1}$ for every natural $k$. | | Both bounds and their quantifiers | Defs.claim, Result.result | All real $\alpha,m,l,x$ with exactly the five displayed hypotheses; both inequalities are non-strict. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KilbasSaigo/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2020Cachin Rényi information combining convexity and concavityHirche Conjecture IV.18 holds with threshold five quarters, including the Shannon order and all entropy endpoints.Proved

The questionHirche, “Rényi Bounds on Information Combining”, ISIT 2020, arXiv:2004.14408v1, §IV-D, Conjecture IV.18, Eq. (88):

Why it mattersThis provides the scalar convexity and concavity inputs for binary Rényi information combining.

The ideaContinuity gives concavity on the closed interval.

Our contributionHirche Conjecture IV.18 holds with threshold five quarters, including the Shannon order and all entropy endpoints.

Verification details
Release source recordProved in Lean

Open problem

Cachin Rényi information combining convexity and concavity

The target is the whole of C. Hirche, “Rényi Bounds on Information Combining”, ISIT 2020, arXiv:2004.14408v1, §IV-D, Conjecture IV.18, Eq. (88):

There exists a value $1 < \hat\alpha < 1.3863$, such that $h_\alpha(h_\alpha^{-1}(x) \ast h_\alpha^{-1}(y))$ is convex for $0 < \alpha < \hat\alpha$ and concave for $\alpha \ge 2$.

The formal claim supplies the witness $\hat\alpha=5/4$ and quantifies over every fixed $y\in[0,\log2]$, with convexity or concavity in $x$ on the whole closed entropy interval. Put

$$ h_\alpha(p)= \begin{cases} -p\log p-(1-p)\log(1-p),&\alpha=1,\ \log(p^\alpha+(1-p)^\alpha)/(1-\alpha),&\alpha\ne1, \end{cases} \qquad p\star q=p+q-2pq. $$

The inverse branch has probability domain $[0,1/2]$. Logarithms are natural, all orders are real, and the Shannon value is Mathlib's Real.binEntropy.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/RenyiCombining/HircheCachinCombiningConcavity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2021Kim–Stanton Conjecture 8.3: integer polynomial denominatorsInteger polynomial denominators of contiguous basic hypergeometric coefficients.Proved

The questionKim and Stanton, Three families of q-Lommel polynomials, arXiv:2105.10096v3, Section 8, Conjecture 8.3.

The ideaPole onset and the reciprocal coefficient induction give the floor envelope.

Our contributionInteger polynomial denominators of contiguous basic hypergeometric coefficients.

Verification details
Release source recordProved in Lean

Open problem

Kim–Stanton Conjecture 8.3: integer polynomial denominators

Kim and Stanton, Three families of q-Lommel polynomials, arXiv:2105.10096v3, Section 8, Conjecture 8.3.

Let $R=\mathbb Z[a,b,c,q]$ and $K=\operatorname{Frac}(R)$. The formal series is $F(A,B;C;q,z)=\sum_{r\ge0}(A;q)_r(B;q)_r z^r/((C;q)_r(q;q)r)$, where $(u;q)r=\prod{j=0}^{r-1}(1-uq^j)$. Put $D_n=\prod{k=0}^n(1-cq^k)^{\lfloor(n+1)/(k+1)\rfloor}$.

Define $\gamma_n=z^n$. The assertion is $\forall n\in\mathbb N;\exists P_n\in R:;\gamma_n/(1-c)=P_n/D_n$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The standard basic hypergeometric series | KimStantonDenominatorCore.F and qShiftedFactorial | Both lower products, including $(q;q)_r$, are present. | | The displayed contiguous quotient | KimStantonSimultaneousDenominator.gamma | The upper and lower parameter shifts are literal. | | Every natural coefficient index | KimStantonSimultaneousDenominator.claim | The quantifier includes zero. | | An integer polynomial in all four variables | The existential MvPolynomial (Fin 4) ℤ binder | The variables are ordered $a,b,c,q$. | | The prescribed floor exponent product | KimStantonDenominatorCore.D | Natural division is floor division. | | The claimed identity | KimStantonSimultaneousDenominator.result | No additional hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MotzkinHankel/KimStantonSimultaneousDenominator.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2021Kim–Stanton Conjecture 8.4: integer polynomial denominatorsInteger polynomial denominators of contiguous basic hypergeometric coefficients.Proved

The questionKim and Stanton, Three families of q-Lommel polynomials, arXiv:2105.10096v3, Section 8, Conjecture 8.4.

The ideaPole onset and the reciprocal coefficient induction give the floor envelope.

Our contributionInteger polynomial denominators of contiguous basic hypergeometric coefficients.

Verification details
Release source recordProved in Lean

Open problem

Kim–Stanton Conjecture 8.4: integer polynomial denominators

Kim and Stanton, Three families of q-Lommel polynomials, arXiv:2105.10096v3, Section 8, Conjecture 8.4.

Let $R=\mathbb Z[a,b,c,q]$ and $K=\operatorname{Frac}(R)$. The formal series is $F(A,B;C;q,z)=\sum_{r\ge0}(A;q)_r(B;q)_r z^r/((C;q)_r(q;q)r)$, where $(u;q)r=\prod{j=0}^{r-1}(1-uq^j)$. Put $D_n=\prod{k=0}^n(1-cq^k)^{\lfloor(n+1)/(k+1)\rfloor}$.

Define $\gamma_n=z^n$. The assertion is $\forall n\in\mathbb N;\exists P_n\in R:;\gamma_n/(1-c)=P_n/D_n$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The standard basic hypergeometric series | KimStantonDenominatorCore.F and qShiftedFactorial | Both lower products, including $(q;q)_r$, are present. | | The displayed contiguous quotient | KimStantonSingleDenominator.gamma | The upper and lower parameter shifts are literal. | | Every natural coefficient index | KimStantonSingleDenominator.claim | The quantifier includes zero. | | An integer polynomial in all four variables | The existential MvPolynomial (Fin 4) ℤ binder | The variables are ordered $a,b,c,q$. | | The prescribed floor exponent product | KimStantonDenominatorCore.D | Natural division is floor division. | | The claimed identity | KimStantonSingleDenominator.result | No additional hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MotzkinHankel/KimStantonSingleDenominator.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Discrete structuresQuestion resolvedPosed in 2013Parity of the A222013 factorial product sum coefficientsThe coefficients of OEIS A222013 are odd exactly one below a power of two.Proved

The questionHanna created OEIS A222013 on Feb 04 2013; the parity conjecture is dated Dec 06 2024.

Why it mattersThis is a first-tier OEIS conjecture from a 2013 entry, with the conjecture dated 2024.

The ideaterm r F is r! times X^r times F^{r(r+1)/2} times invOfUnit of the product of (1 + kXF^k) for k from 1 through r, with unit argument 1.

Our contributionThe coefficients of OEIS A222013 are odd exactly one below a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A222013 factorial product sum coefficients

Paul D. Hanna created OEIS A222013 on Feb 04 2013; the parity conjecture is dated Dec 06 2024. These quotations are copied verbatim from Library/Recurrence/hanna2024a222013.md.

NAME:

G.f. satisfies: A(x) = Sum_{n>=0} n! * x^n * A(x)^(n*(n+1)/2) / Product_{k=1..n} (1 + k*x*A(x)^k).

COMMENT:

Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0. - Paul D. Hanna, Dec 06 2024

The infinite sum is formalised with the exact finite window at each coefficient, as stated in Route and quoted in Lean syntax in Evidence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/FactorialProductSumCatalanParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresRecurrence provedPosed in 2020Mathar's D-finite recurrence for OEIS A308815Mathar's D-finite recurrence for the alternating Catalan numbers.Proved

The questionConjecture: D-finite with recurrence: n(2n+1)a(n) -32(9n-4)(n-1)a(n-1) +1024(3n^2-23n+29)a(n-2) +65536(n-2)(2n-5)a(n-3)=0. - R.

Why it mattersFarkas, Moschetti, Naranjo and Pirola's Theorem 1.1, page 1 of arXiv:1906.10406v2, identifies this signed sum with alternating Catalan numbers and odd-cover counts for general curves of genus at least three.

The ideaThe imported PowerSeries.catalanSeriessqmulXaddone supplies its quadratic equation.

Our contributionMathar's D-finite recurrence for the alternating Catalan numbers.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's D-finite recurrence for OEIS A308815

OEIS A308815, FORMULA section, states:

Conjecture: D-finite with recurrence: n*(2*n+1)*a(n) -32*(9*n-4)*(n-1)*a(n-1) +1024*(3*n^2-23*n+29)*a(n-2) +65536*(n-2)*(2*n-5)*a(n-3)=0. - R. J. Mathar, Jan 27 2020

Its PARI definition is:

a(n) = 16^n*sum(i=0, n, (-2)^i*binomial(n, i)*C(2*n-i));

Here $C_k$ is the ordinary Catalan number. Put

$$a_n=16^n\sum_{i=0}^n(-2)^i\binom ni C_{2n-i}.$$

For every natural $n\ge3$, the conjecture asserts

$$n(2n+1)a_n-32(9n-4)(n-1)a_{n-1} +1024(3n^2-23n+29)a_{n-2}+65536(n-2)(2n-5)a_{n-3}=0.$$

The external tier-1 claim is preregistered in https://github.com/the-omega-institute/trureturing/issues/15713. The signed PARI formula is the sequence convention; the NAME line's unsigned $2^i$ conflicts with the displayed data.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Natural sequence index and finite sum $0\le i\le n$ | a308815, Finset.range (n+1) | Exact natural indices; integer summands | | $16^n(-2)^i\binom ni C_{2n-i}$ | a308815, Mathlib Nat.choose and catalan | Literal signed formula; natural subtraction at $2n-i$ | | Every $n\ge3$ | claim | No extra hypothesis | | Four recurrence coefficients and lags | claim | Integer coefficient arithmetic; natural lagged arguments | | The recurrence equals zero | result : claim | Whole assertion proved |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/A308815DFiniteRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Mathar's D-finite recurrence for the alternating Catalan numbers.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2020.

    Read source ↗
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

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    Dated conjecture statement: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsQuestion resolvedFirst preprint 2015Iterated moment minor inequalitiesSuccessive moment minors for positive probability laws.Proved

The questionSimić, Moment inequalities of the second and third order, §2:

Why it mattersBoth inequalities refine the nonnegativity of moment minors by controlling cross terms from independently chosen parameter pairs.

The ideaAt zero and one use f0(x) = x−1−log x and f1(x) = x log x−x+1.

Our contributionSuccessive moment minors for positive probability laws.

Verification details
Release source recordProved in Lean

Open problem

Iterated moment minor inequalities

Simić, Moment inequalities of the second and third order, §2:

Conjecture 1 The inequality φ(r₁,s₁;u₁,v₁)φ(r₂,s₂;u₂,v₂) ≥ [(ξ((r₁+r₂)/2,(s₁+s₂)/2) − ξ((r₁+s₂)/2,(s₁+r₂)/2))(ξ((u₁+u₂)/2,(v₁+v₂)/2) − ξ((u₁+v₂)/2,(v₁+u₂)/2)) − (ξ((r₁+u₂)/2,(s₁+v₂)/2) − ξ((r₁+v₂)/2,(s₁+u₂)/2))(ξ((u₁+r₂)/2,(v₁+s₂)/2) − ξ((u₁+s₂)/2,(v₁+r₂)/2))]² holds for arbitrary r_i, s_i, u_i, v_i ∈ I, i ∈ {1, 2}?

With σ_a(x) := ξ(x, x+a) and θ_a(x,y) := σ_a(x)σ_a(y) − σ_a²((x+y)/2):

Conjecture 2 Is it true that θ_a(x,y)θ_a(z,v) ≥ [θ_a((x+z)/2,(y+v)/2) − θ_a((x+v)/2,(y+z)/2)]²?

For a positive random variable X with mean m, the definitions are

$$\lambda_s=\begin{cases} \log m-\mathbb E\log X,&s=0,\ \mathbb E(X\log X)-m\log m,&s=1,\ \dfrac{\mathbb E X^s-m^s}{s(s-1)},&s\ne0,1, \end{cases}\qquad \xi(s,t)=\lambda_s\lambda_t-\lambda_{(s+t)/2}^2,$$

$$\varphi(s,t;u,v)=\xi(s,t)\xi(u,v) -\left[\xi!\left(\frac{s+u}{2},\frac{t+v}{2}\right) -\xi!\left(\frac{s+v}{2},\frac{t+u}{2}\right)\right]^2.$$

Formal reading

All Lean names below have prefix D5.S3.Analytic.MomentMinors.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Positive probability law of X | Defs.Admissible μ I | A Borel probability measure on ℝ, positive almost everywhere, with integrable identity. | | Moments exist for every s ∈ I | Defs.Admissible.integrable_rpow | Integrability of each real power; log X at 0 ∈ I and X log X at 1 ∈ I are separate conditions. | | Interval I | I.OrdConnected | Any order-connected real set, with all needed midpoints still in I. | | λ_s, including s = 0,1 | Defs.lam, Defs.mean | Exactly the displayed three-case definition, with m the integral of the identity. | | ξ, φ | Defs.xi, Defs.phi | The displayed second and iterated second minors. | | σ_a, θ_a | Defs.sigma, Defs.theta | The displaced pair and its midpoint second minor. | | Conjecture 1 | Defs.claim1 | All eight endpoint parameters independently range over I, in the quoted index order. | | Conjecture 2 | Defs.claim2 | Every real a,x,y,z,v such that x,y,z,v and x+a,y+a,z+a,v+a belong to I. No sign restriction on a. | | Both assertions | Defs.claim, Result.result | The conjunction, for every interval and every admissible law. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MomentMinors/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresRecurrence provedFirst preprint 2014Thanatipanonda's shifted binomial power linearizationEvery positive power of choose(n+dk,k) has an integral expansion in choose(n+dj,j), with coefficients generated by one integer recursion independent of the power index.Proved

The questionThanatipanonda, Beyond Zudilin's Conjectured q-analog of Schmidt's problem, arXiv:1403.4962v1, Section 4, Conjecture 4.1, p. 6:

Why it mattersFor every d,k it supplies one integer-valued function T(j,i), chosen before the power index r, with the required initial row, recursion, support and identity for every natural argument n.

The ideaThe binomial product identity is used locally:

Our contributionEvery positive power of choose(n+dk,k) has an integral expansion in choose(n+dj,j), with coefficients generated by one integer recursion independent of the power index.

Verification details
Release source recordProved in Lean

Open problem

Thanatipanonda's shifted binomial power linearization

T. Thanatipanonda, Beyond Zudilin's Conjectured q-analog of Schmidt's problem, arXiv:1403.4962v1, Section 4, Conjecture 4.1, p. 6:

For any integers d, k ≥ 0 and r ≥ 1, there exist integers $a^{(r)}{d,k,j}$ such that $\binom{n+dk}{k}^r = \sum_j a^{(r)}{d,k,j}\binom{n+dj}{j}$ for all n = 0, 1, 2, .... Moreover $a^{(r)}{d,k,j}$ can be defined as following: $a^{(1)}{d,k,k} = 1$, $a^{(1)}{d,k,j} = 0$ for j ≠ k and $a^{(r+1)}{d,k,j} = \sum_i S_d(k, j, i)a^{(r)}_{d,k,i}$, where, $S_d(k, j, i)$ are integers, independent of r, for all d, k, j, i.

The formal statement uses natural numbers for every nonnegative source parameter and casts binomial values to integers before taking powers. It includes both sentences and proves finite support, at j≤rk for the power coefficients and at j≤i+k for the structure constants.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BinomialBases/ShiftedBinomialPowerLinearization.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    Every positive power of choose(n+dk,k) has an integral expansion in choose(n+dj,j), with coefficients generated by one integer recursion independent of the power index.

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  • Historical context

    First preprint year: 2014.

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Quantum informationCounterexample foundFirst preprint 2023Singlet padding refutes the universal precession GME thresholdA spin-1/2 singlet tensored with the spin-3/2 endpoint state scores 3/4 at K = 3 while remaining separable across the pair–rest bipartition.Refuted

The questionKhoi-Nguyen Huynh-Vu, Lin Htoo Zaw and Valerio Scarani, Certification of genuine multipartite entanglement in spin ensembles with measurements of total angular momentum, Phys.

Why it mattersThe source anchor is Library/QuantumStates/huynhvu2024universal.md.

The ideaFinite Hermitian spectral calculus respects this intertwining.

Our contributionA spin-1/2 singlet tensored with the spin-3/2 endpoint state scores 3/4 at K = 3 while remaining separable across the pair–rest bipartition.

Verification details
Release source recordRefuted in Lean

Open problem

Singlet padding refutes the universal precession GME threshold

Khoi-Nguyen Huynh-Vu, Lin Htoo Zaw and Valerio Scarani, Certification of genuine multipartite entanglement in spin ensembles with measurements of total angular momentum, Phys. Rev. A 109, 042402, arXiv:2311.00806v2, Conjecture 3, PDF p. 9:

Consider a spin ensemble. Perform the precession protocol with odd $K \geq 3$ on the total angular momentum of the system. If the score $P_K > \mathbf{P}_K^{\mathrm{conj}}$ is obtained, then the spin ensemble is GME.

Here $J_k=\cos(2\pi k/K)J_x+\sin(2\pi k/K)J_y$ is the sum of the particle observables, $Q_K=K^{-1}\sum_k\operatorname{pos}(J_k)$ and $P_K=\operatorname{tr}(\rho Q_K)$. The spectral weight is $(1+\operatorname{sgn}m)/2$, including one half at zero. Separability across a cut means a normalized finite convex mixture of products of density matrices. GME excludes convex mixtures over all nontrivial cuts. Preregistration: #13586.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PrecessionUniversalThresholdRefutation.result

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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryConjecture refutedFirst preprint 2025A square discriminant refutes the odd clause of Levens Conjecture 6.4The square discriminant of the degree twenty-four binomial truncation refutes the odd-index Galois-group conjecture at n = 49.Refuted

The questionOwen Levens, Polynomials Arising from Sorted Binomial Coefficients, arXiv:2511.03082v1, Section 6.2, Conjecture 6.4 states:

Why it mattersA strict cardinality bound on the actual splitting-field automorphism group rules it out.

The ideaIts nonzeroness follows from the distinct roots in signeqoneofvandermondesq, which forces every induced permutation to have sign one.

Our contributionThe square discriminant of the degree twenty-four binomial truncation refutes the odd-index Galois-group conjecture at n = 49.

Verification details
Release source recordRefuted in Lean

Open problem

A square discriminant refutes the odd clause of Levens Conjecture 6.4

Owen Levens, Polynomials Arising from Sorted Binomial Coefficients, arXiv:2511.03082v1, Section 6.2, Conjecture 6.4 states:

P_n(z) has full Galois group for even n. For n = 2m + 1, the Galois group of P_n(z) is the mth hyperoctahedral group.

The source defines the Pascalian numbers by $\langle n,k\rangle=\binom n{\lfloor(n-k)/2\rfloor}$ for $0\le k\le n$ and $P_n(z)=\sum_{k=0}^n\langle n,k\rangle z^{n-k}$. The target is the second sentence, with $m\ge1$. The counterexample is $m=24$, $n=49$. Preregistration: #19007.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Pascalian coefficient and full polynomial | P over rational polynomials, summing every index from zero through n | Natural division is the stated floor; the exponent uses natural subtraction in the source range. | | n = 2m + 1 | claim substitutes 2 * m + 1 into P | Every natural m with 1 ≤ m is included; 24 is included. | | mth hyperoctahedral group | hyperoctahedral m, the centralizer of signFlip m on Fin m × Bool | The two Boolean values represent the signs. These are precisely signed permutations. | | The Galois group is that group | Existence of an abstract multiplicative equivalence from Polynomial.Gal | This weakest isomorphism reading is refuted, hence so is the permutation-group reading. | | Refutation of the universal odd sentence | result : Not claim | No extra mathematical hypothesis. The even sentence is outside claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Galois/LevensPascalian.result

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Discrete structuresSharp bound provedFirst preprint 2026Chen–He–Lai–Li Conjecture 6.1: the local average at a stemThe average order of dominating sets containing a stem satisfies the sharp bound (4n+1)/6.Proved

The questionTingyun Chen, Weihua He, Hong-Jian Lai and Jianping Li, “On the local average order of dominating sets”, arXiv:2610.11446v1, Conjecture 6.1:

Why it mattersThe global bound of Beaton and Cameron is 2n/3.

The ideaLet L be the l leaf neighbours of v, and H the induced graph after removing v and L, of order h=n−l−1.

Our contributionThe average order of dominating sets containing a stem satisfies the sharp bound (4n+1)/6.

Verification details
Release source recordProved in Lean

Open problem

Chen–He–Lai–Li Conjecture 6.1: the local average at a stem

Tingyun Chen, Weihua He, Hong-Jian Lai and Jianping Li, “On the local average order of dominating sets”, arXiv:2610.11446v1, Conjecture 6.1:

Let G be a graph of order n ≥ 2 without isolated vertices, and let v be an l-stem of G, where l ≥ 1. Then avd_v(G) ≤ (4n + 1)/6, with equality if and only if l = 1 and G is star-like.

A dominating set contains or neighbours every vertex. The local average is the mean cardinality over the dominating sets containing v. A leaf has degree one; an l-stem has exactly l leaf neighbours. Literally following Iain Beaton and Ben Cameron, “A Tight Upper Bound on the Average Order of Dominating Sets of a Graph”, arXiv:2208.10475v2, a graph is star-like when every vertex is a leaf or a stem with one or two leaf neighbours.

The closed Lean claim quantifies over V : Type, its finite and decidable equality structures, a finite simple graph with decidable adjacency, v and l. It retains the order, positive-degree and leaf-count hypotheses and both the upper bound and the equivalence characterizing equality. All averages are exact rational numbers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/LocalDominatingStemBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The average order of dominating sets containing a stem satisfies the sharp bound (4n+1)/6.

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  • Historical context

    First preprint year: 2026.

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Discrete structuresQuestion resolvedFirst preprint 2026Every vertex an identity Seidel switchA finite nonempty simple graph has every vertex as an identity Seidel switch if and only if it has one vertex.Proved

The questionGervacio, On identity Seidel switches, arXiv:2601.04530v1, Section 6, Problem 6.1, page 13:

Why it mattersThe source definitions and locators are in Library/GraphInvariants/gervacio2026identityseidel.md.

The ideaPartition the edges into incident and nonincident edges.

Our contributionA finite nonempty simple graph has every vertex as an identity Seidel switch if and only if it has one vertex.

Verification details
Release source recordProved in Lean

arXiv

Every vertex an identity Seidel switch

S. V. Gervacio, On identity Seidel switches, arXiv:2601.04530v1, Section 6, Problem 6.1, page 13:

Characterize graphs $G$ for which every vertex is a vertex-ISS. Lemma 4.4 gives a necessary condition in terms of the minimum and maximum degree; can this be strengthened to a complete characterization?

The source's graphs are simple graphs on a nonempty finite vertex set. Switching across $S$ complements exactly the pairs with one endpoint in $S$. An identity Seidel switch (ISS) means $S(G)\cong G$; a vertex-ISS uses $S={v}$. The complete answer is $|V(G)|=1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SeidelSwitching/VertexIdentitySwitches.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedPosed in 2026Coefficient congruence for OEIS A396807Every positive-index coefficient of OEIS A396807 is congruent to one modulo ten.Proved

The questionNAME "G.f. satisfies A(x) = x + A^5(x)A^6(x)." where A^n(x) denotes the n-th iteration (compositional power) of A(x).

Why it mattersThe KPI is open problems resolved: the target is the congruence for every positive index, without a finite cutoff.

The ideaFor zero-constant series over any commutative ring, the transformation A ↦ X + A∘5 · A∘6 improves coefficient agreement by one degree.

Our contributionEvery positive-index coefficient of OEIS A396807 is congruent to one modulo ten.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficient congruence for OEIS A396807

NAME "G.f. satisfies A(x) = x + A^5(x)*A^6(x)." where A^n(x) denotes the n-th iteration (compositional power) of A(x).

COMMENT "Conjecture: a(n) = 1 (mod 10) for n >= 1." — Paul D. Hanna, Jun 16 2026.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CompositionalIterateCongruence.coefficient_congruence

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Discrete structuresConjecture provedFirst preprint 2025Zhang–Tian Conjecture 5.1: Steiner diameter and anchored radius of treesTerminal-cut averaging proves Zhang–Tian Conjecture 5.1 for every finite tree.Proved

The questionQingnan Zhang and Yingzhi Tian, On the Steiner k-diameter and Steiner (k, k′)-radius of trees, arXiv:2511.22492v1, p. 14, Conjecture 5.1, states:

Why it mattersIt proves the conjecture for all finite trees and all source parameters.

The ideasteinerDistcut identifies the defining distance with the number of edges whose deletion separates two terminals.

Our contributionTerminal-cut averaging proves Zhang–Tian Conjecture 5.1 for every finite tree.

Verification details
Release source recordProved in Lean

Open problem

Zhang–Tian Conjecture 5.1: Steiner diameter and anchored radius of trees

Qingnan Zhang and Yingzhi Tian, On the Steiner k-diameter and Steiner (k, k′)-radius of trees, arXiv:2511.22492v1, p. 14, Conjecture 5.1, states:

Let n, k, k′ be integers with n ≥ k ≥ 3 and k > k′ ≥ 1 and let T be a tree with order n. Then Sd_k(T) ≤ k/(k − k′) Sr_{k,k′}(T) − k′(k′ − 1)/(k − k′).

Here $d_T(S)$ is the minimum edge count among connected subgraphs containing $S$; $\varepsilon_k(v;T)$ maximizes $d_T(S)$ over $k$-sets containing $v$; $Sd_k(T)$ maximizes these vertex eccentricities. For an anchor set $B$, $\varepsilon_{k,r}(B;T)$ maximizes $d_T(S)$ over $k$-sets containing $B$, and $Sr_{k,r}(T)$ minimizes this quantity over $r$-sets $B$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A finite simple tree of order $n$ | V : Type, Fintype V, DecidableEq V, T : SimpleGraph V, T.IsTree; $n$ is Fintype.card V | Exact; decidable equality is classical bookkeeping on a finite carrier. | | $n\ge k\ge3$ | k ≤ Fintype.card V, 3 ≤ k | Exact. | | $k>k'\ge1$ | r < k, 1 ≤ r | Exact; $r$ denotes $k'$. | | Minimum connected-subgraph edge count | steinerDist | Literal source definition, using a natural infimum. | | Vertex eccentricity and diameter | steinerEcc, steinerDiam | Literal nested maxima, as natural suprema. | | Anchored eccentricity and radius | steinerSetEcc, steinerRadius | Literal extrema over terminal sets of the specified cardinality. | | The rational inequality | claim, proved by result | All distances and parameters are cast into $\mathbb Q$ before subtraction and division. | | Auxiliary cut representation | Separates, cutEdges, steinerDist_cut | Repository-derived objects characterize the source's literal distance; they do not replace its definition. | | Arbitrary terminal $k$-set $A$ and anchor $r$-set $B$ | pointwise_bound | Strengthening; no disjointness, hull containment or diameter-maximization hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Eccentric/ZhangTianSteinerTreeRadius.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026The right-special rotor word is the reversed prefix after the initial zeroFor every irrational slope greater than one, the unique right-special rotor factor of each positive length is the reversed prefix after the initial zero.Proved

The questionKrishnaswami and Anirudh Rameshan, Symbol sequences from three-rotor coincidences and their word-complexity, arXiv:2606.25766v1 [nlin.CD] (2026), §5.3.1, “Conjecture on right-special word”, state:

Why it mattersThe source uses rotor-coincidence sequences to study the word statistics of three-rotor dynamics.

The idea3. Uniqueness survives encoding. A nonempty right-special rotor factor cannot contain the unique zero and must end in 2: every other symbol has forced successor 2.

Our contributionFor every irrational slope greater than one, the unique right-special rotor factor of each positive length is the reversed prefix after the initial zero.

Verification details
Release source recordProved in Lean

Open problem

The right-special rotor word is the reversed prefix after the initial zero

Govind S. Krishnaswami and Anirudh Rameshan, Symbol sequences from three-rotor coincidences and their word-complexity, arXiv:2606.25766v1 [nlin.CD] (2026), §5.3.1, “Conjecture on right-special word”, state:

we conjecture that the $(n-1)$-word that branches (in two ways) coincides with the unique $n$-word starting with 0 read backwards (excluding the 0).

This is a verbatim excerpt from line 701 of the downloaded v1 TeX 3rotor-symb-dyn.tex, md5 5d78bd4b19257909d542bbe5b1469202. The surrounding scope is UHE quasiperiodic trajectories starting at the origin with irrational slope $\sigma>1$ (line 663). Discussion question (1), line 917, explicitly asks for a settlement. The source specifies its encoding in §5.3.4, line 835:

Given a Sturmian sequence $s_1 s_2 s_3 \cdots$, the corresponding rotor sequence is $02 s_1 2 s_2 2 s_3 2 \cdots$.

Issue #20821 fixes the formal reading. For $m\ge1$, define $s_m=3$ exactly when $m=\lfloor k\sigma\rfloor_{\mathbb N}+k$ for some natural $k\ge1$, and $s_m=1$ otherwise. Here $\lfloor\cdot\rfloor_{\mathbb N}$ is the natural-valued floor used by cutting. Define $R_0=0$, $R_{2j-1}=2$ and $R_{2j}=s_j$ for $j\ge1$, as in rotor.

A finite list $w$ of natural symbols is a factor when $w_t=R_{i+t}$ for some natural starting position $i$ and every $t<|w|$. It is right-special when $wa$ and $wb$ are factors for two distinct natural symbols $a\ne b$; the occurrences may start at different positions. These are IsFactor and IsRightSpecial.

The claim is: for every irrational real $\sigma>1$, every natural $n\ge2$, and every finite list $w$ of length $n-1$,

$w\text{ is right-special in }R\quad\Longleftrightarrow\quad w=\operatorname{reverse}(R_1R_2\cdots R_{n-1}).$

This reading is proved. The literature note gives the source quotations and locators.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Mechanical/RotorCoincidenceRightSpecial.result

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Discrete structuresConjecture provedFirst preprint 2023Gallardo–Rahavandrainy Conjecture 3.11The binary-polynomial Collatz length of (M²+M+1)ⁿ equals n+1 for n≥2.Proved

The questionThe total definition of oddPart also accepts zero.

The ideaThe six public prerequisites Dynamics.nonzeroofodd, Dynamics.nonzeroeqone, Dynamics.degreeq, Dynamics.qnezero, Dynamics.eqoneofdegreezero and Dynamics.oddPartfactor are consumed by GallardoRahavandrainyCollatzLengthBound.

Our contributionThe binary-polynomial Collatz length of (M²+M+1)ⁿ equals n+1 for n≥2.

Verification details
Release source recordProved in Lean

Open problem

Gallardo–Rahavandrainy Conjecture 3.11

Gallardo and Rahavandrainy, On Collatz Conjecture for binary polynomials, arXiv:2308.01181v2, Conjecture 3.11, p. 15, state: “If $A=(M^2+M+1)^n$ with $n\ge2$, then $\ell_A$ equals $n+1$.” Here $M=x^2+x+1$ in $\mathbb F_2[x]$. The source calls a polynomial odd when it is coprime to $x(x+1)$, removes all powers of these two linear factors to obtain its odd part, and iterates the odd part of $1+MA$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Binary polynomials and $M=x^2+x+1$ | Polynomial (ZMod 2) and M | Literal coefficient ring and polynomial. | | Remove $x^{val_x(S)}(x+1)^{val_{x+1}(S)}$ | oddPart | rootMultiplicity at 0 and 1; X-C 1=X+1 in characteristic two. | | The odd subsequence $A_1,A_3,\ldots$ | oddSeq A k | Source index $2k+1$; its initial term is the odd part of $A$. | | Length counts the initial odd term and terminal 1 | collatzLength | First hitting index of 1, plus one. | | Every natural $n\ge2$ | claim | The quantified natural exponent has hypothesis $2\le n$. | | $\ell_{(M^2+M+1)^n}=n+1$ | result : claim | Proved, with no extra hypothesis. |

The total definition of oddPart also accepts zero. The proof only transports nonzero polynomials with the required oddness; this total extension does not weaken or change the stated conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/GallardoRahavandrainyBinaryCollatz.result

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Quantum informationConjecture provedFirst preprint 2024Conjecture 3 of Tavakoli–Morelli: the Schmidt-number bound for incomplete sets of MUBsIncomplete families of mutually unbiased bases obey the Schmidt-number correlation trace-norm bound of Tavakoli and Morelli's Conjecture 3.Proved

The questionMorelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys.

Why it mattersFor r = 1, (54) is the separability bound 1 + (m − 1)/d.

The idea1. Shared lemma. TavakoliMorelliIncompleteEAM.schmidttraceNormleofmoment: two families of nonnegative multiples of pure projectors whose second moments satisfy Σa (tr H Ea)² ≤ α tr H² + β (tr H)² for every Hermitian H give ‖table‖tr ≤ α + βr for every state of Schmidt number at most r.

Our contributionIncomplete families of mutually unbiased bases obey the Schmidt-number correlation trace-norm bound of Tavakoli and Morelli's Conjecture 3.

Verification details
Release source recordProved in Lean

Open problem

Conjecture 3 of Tavakoli–Morelli: the Schmidt-number bound for incomplete sets of MUBs

A. Tavakoli and S. Morelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys. Rev. A 110, 062417 (2024), bound the trace norm of the table of joint outcome probabilities of local measurements for states of Schmidt number at most r.

  • Incomplete MUBs. Each side measures in m mutually unbiased bases of C^d, with m < d + 1; Q_m is the md × md table of the joint probabilities tr[ρ (|g^l_a⟩⟨g^l_a| ⊗ |h^k_b⟩⟨h^k_b|)] of the md local projectors.
  • Conjecture 3 (§V B, p. 7): "For any state of Schmidt number r it holds that ‖Q_m‖_tr ≤ 1 + (m − 1)r/d. (54)" The introduction leaves deciding the conjectures as an open problem.

The verbatim statements are in the literature note. Issue #17745 fixes the reading:

  • for all d ≥ 1, 1 ≤ m ≤ d, 1 ≤ r ≤ d, any two families of m mutually unbiased bases of C^d chosen independently, and every state of Schmidt number at most r, the bound (54) holds;
  • a state of Schmidt number at most r is a finite convex combination of pure states whose coefficient matrices have rank at most r.

This conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TavakoliMorelliIncompleteMUB.result

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Discrete structuresQuestion resolvedFirst preprint 2022Molinari's honeycomb trapezium determinantThe determinant of a honeycomb trapezium's Hückel matrix equals the determinant of Molinari's reduced Pascal matrix, for arbitrary boundary weights in any commutative ring.Proved

The questionMolinari's Conjecture 2 in Graphene nanocones and Pascal matrices (arXiv:2206.14428v2, section 5, page 9) asks whether the determinant of the Hückel matrix of a honeycomb trapezium with rows of lengths 2k+1, 2k+3, ..., 2n+1 equals the determinant of the displayed reduced Pascal matrix of size n+1-k.

Why it mattersThe paper proves palindromy, degree, and square-coefficient properties and reports only finite hand reductions for the determinant identity.

The ideaEvery step is carried out in an arbitrary commutative ring.

Our contributionThe determinant of a honeycomb trapezium's Hückel matrix equals the determinant of Molinari's reduced Pascal matrix, for arbitrary boundary weights in any commutative ring.

Verification details
Release source recordProved in Lean

Open problem

Molinari's honeycomb trapezium determinant

Molinari's Conjecture 2 in Graphene nanocones and Pascal matrices (arXiv:2206.14428v2, section 5, page 9) asks whether the determinant of the Hückel matrix of a honeycomb trapezium with rows of lengths 2k+1, 2k+3, ..., 2n+1 equals the determinant of the displayed reduced Pascal matrix of size n+1-k. The weights are arbitrary boundary parameters.

The formal statement is

claim := ∀ (R : Type) [CommRing R] (k n : ℕ), k ≤ n → ∀ (x y : ℕ → R),
  (huckel k n x y).det = (reducedPascal k n x y).det
result : claim
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/HoneycombTrapeziumHuckelDeterminant.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedFirst preprint 2026A disconnected counterexample to the token-radius conjectureThe graph consisting of one edge and two isolated vertices has Laplacian spectral radius 2, as does its two-token graph, but it is not a star.Refuted

The questionZhang, The Algebraic Connectivity and Laplacian Spectral Radius of Token Graphs, arXiv:2610.00500v1, section 1, page 4, state:

Why it mattersThe equality would characterize stars using the spectra of all token graphs in the specified range.

The ideaIts two-token vertices are 01, 02, 03, 12, 13, 23, and its two edges are 02–12 and 03–13.

Our contributionThe graph consisting of one edge and two isolated vertices has Laplacian spectral radius 2, as does its two-token graph, but it is not a star.

Verification details
Release source recordRefuted in Lean

Open problem

A disconnected counterexample to the token-radius conjecture

X. Song, C. Dalfó, M. À. Fiol and S. Zhang, The Algebraic Connectivity and Laplacian Spectral Radius of Token Graphs, arXiv:2610.00500v1, section 1, page 4, state:

Conjecture 1.1. Let G be a graph of order n(≥ 4), the equality ρ(F_k(G)) = ρ(G) holds for all k with 2 ≤ k ≤ ⌊n/2⌋ if and only if G ≅ S_n.

Here the token vertices are k-subsets, adjacency means that their symmetric difference is an edge of G, and ρ is the largest Laplacian eigenvalue. The literal statement has no connectedness hypothesis. Issue #12457 preregisters this reading and the counterexample K₂ ⊔ 2K₁.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/TokenGraphLaplacianRadiusRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2024A qubit dephasing midpoint refutes convexity of Wasserstein complexityA qubit dephasing midpoint refutes convexity of Wasserstein complexity.Refuted

The questionVirosztek, On the metric property of quantum Wasserstein divergences, Phys.

Why it mattersIts supporting module supplies coupling optimization, unitary covariance and the exact self cost.

The ideaA diagonal state's self cost is attained by the vectorized square-root coupling; a principal-minor bound proves optimality.

Our contributionA qubit dephasing midpoint refutes convexity of Wasserstein complexity.

Verification details
Release source recordRefuted in Lean

Open problem

A qubit dephasing midpoint refutes convexity of Wasserstein complexity

G. Bunth, J. Pitrik, T. Titkos and D. Virosztek, On the metric property of quantum Wasserstein divergences, Phys. Rev. A 110 (2024), 022211, arXiv:2402.13150v4, Section 5.1, asks:

We note that whether the Wasserstein complexity (37) is convex or not is an open question.

For a finite Hermitian observable family $\mathcal A$, the complexity is $C_W(\Phi)=\max_\rho d_{\mathcal A}(\rho,\Phi(\rho))$. The density matrices are positive semidefinite and have trace one. The couplings have marginals $\omega$ and $\rho^T$, and the quadratic cost and divergence are those of equations (eq:2-Wass-quant-def) and (eq:mod-qw-dist-def). The literature note gives the definitions and locators. Preregistration #15149 specifies the universal convexity statement, its conventions and the qubit counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/WassersteinComplexityNonconvex.result

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Quantum informationQuestion resolvedFirst preprint 2019The complete graph maximizes the trace of the average mixing matrixFor every n and every connected graph G on n vertices, the trace of the average mixing matrix of the adjacency quantum walk on G is at most that of the complete graph Kn.Proved

The questionChris Godsil, Krystal Guo and Mariia Sobchuk, Diagonal entries of the average mixing matrix, arXiv:1910.02039v1, Section 9 (Conjecture 9.1 of Australas.

The ideaThe proof uses spectral estimates.

Our contributionFor every n and every connected graph G on n vertices, the trace of the average mixing matrix of the adjacency quantum walk on G is at most that of the complete graph Kn.

Verification details
Release source recordProved in Lean

Open problem

The complete graph maximizes the trace of the average mixing matrix

Chris Godsil, Krystal Guo and Mariia Sobchuk, Diagonal entries of the average mixing matrix, arXiv:1910.02039v1, Section 9 (Conjecture 9.1 of Australas. J. Combin. 86(3) (2023) 373–386), write:

Based on the computations summarized in Table 2 and on Corollary 6.3, we also make the following conjecture. The complete graph on $n$ vertices attains the maximum trace with respect to the $\widehat{M}_{A}$ for all $n$.

For a graph $X$ with adjacency matrix $A=\sum_\theta\theta E_\theta$ over its distinct eigenvalues, $E_\theta$ the orthogonal projection onto the $\theta$-eigenspace, the average mixing matrix of the continuous quantum walk is $\widehat M_A(X)=\sum_\theta E_\theta\circ E_\theta$ (Schur product), and $\operatorname{tr}\widehat M_A(K_n)=(n^2-2n+2)/n$.

The precise question, preregistered in issue #11821, is whether $\operatorname{tr}\widehat M_A(X)\le\operatorname{tr}\widehat M_A(K_n)$ for every $n$ and every connected graph $X$ on $n$ vertices. Over all graphs the statement is false: the empty graph has $\widehat M_A=I$ and trace $n$. The paper's Laplacian results on the maximum trace are stated for connected graphs, and A. Mohan, C. Tamon, Y. Xu and H. Zhan (arXiv:2608.20739, 2026) restate the conjecture over connected graphs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/AverageMixingTraceMaximum.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsConjecture refutedFirst preprint 2023Refutation of Nikolov–Savov Conjecture 3.7 at k = 4Nikolov and Savov's Conjecture 3.7 fails at k = 4: the polynomial x^2(x+y+z)+xy is a degree-three renewal minorant strictly above Q4 at interior points of the probability simplex.Refuted

The questionSavov, Properties and conjectures regarding discrete renewal sequences, arXiv:2307.00545v2, §3, Conjecture 3.7, states:

The ideaAt k = 4 write x = p₁, y = p₂, z = p₃ and w = 1 − x − y − z.

Our contributionNikolov and Savov's Conjecture 3.7 fails at k = 4: the polynomial x^2(x+y+z)+xy is a degree-three renewal minorant strictly above Q4 at interior points of the probability simplex.

Scope & next questionRefuting maximality of Q₄ does not settle existence of another maximal minorant. - Open: the separate assertion that Qk is largest in 𝒜̂k is not used and remains unsettled. - Proved boundary: source results that rely specifically on the universal maximality of Qk cannot use that clause at k = 4 without an additional hypothesis or replacement minorant.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Nikolov–Savov Conjecture 3.7 at k = 4

N. Nikolov and M. Savov, Properties and conjectures regarding discrete renewal sequences, arXiv:2307.00545v2, §3, Conjecture 3.7, states:

For any k ≥ 3, Q_k is a maximal element in 𝒜_k and it is largest in 𝒜̂_k.

The formal claim settles the first clause:

claim := ∀ k ≥ 3, Q_k ∈ 𝒜_k ∧ ∀ P ∈ 𝒜_k, Q_k ≺ P → P = Q_k.

Here A_k is the simplex of nonnegative first k−1 masses with sum at most one, u is the renewal sequence of Eq. (2.2), Q_k is the product of partial sums in Eq. (2.5), and 𝒜_k is the degree-bounded class of polynomial lower bounds in Eq. (2.7). The hatted class is not used.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/RandomWalks/RenewalMinorantMaximalityRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture provedFirst preprint 2024Alternative q-Charlier Fibonacci positivity: Petrov–Scott Conjecture 6.3Alternative q-Charlier Fibonacci positivity at rho = q = 1/4.Proved

The questionLeonid Petrov and Jeanne Scott, Random Fibonacci words via clone Schur functions, Forum of Mathematics, Sigma 14 (2026), e15, DOI 10.1017/fms.2025.10165, arXiv:2412.21126v2, Conjecture 6.3, p. 31:

Why it mattersThe conjecture asks for an actual positive specialization, rather than positivity of finitely many words.

The ideaThe proof uses induction.

Our contributionAlternative q-Charlier Fibonacci positivity at rho = q = 1/4.

Verification details
Release source recordProved in Lean

Open problem

Alternative q-Charlier Fibonacci positivity: Petrov–Scott Conjecture 6.3

Leonid Petrov and Jeanne Scott, Random Fibonacci words via clone Schur functions, Forum of Mathematics, Sigma 14 (2026), e15, DOI 10.1017/fms.2025.10165, arXiv:2412.21126v2, Conjecture 6.3, p. 31:

There exist values of ρ and q for which the following specialization (which we call the alternative q-Charlier) is Fibonacci positive:

$$ x_k=\frac{q^{k-1}(1+\rho q^{k-2}+\rho q^{k-1}-\rho q^{2k-2})}{(1+\rho q^{2k-3})(1+\rho q^{2k-1})},\qquad y_k=\frac{\rho q^{3k-2}(1-q^k)(1+\rho q^{k-1})}{(1+\rho q^{2k})(1+\rho q^{2k-1})^2(1+\rho q^{2k-2})},\qquad k\ge1. $$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (2.10): first tridiagonal determinant, including the empty determinant | A | Every matrix entry is literal; size is any natural number. | | Equation (2.10): second determinant, first row $y_1,x_1y_2$, then diagonal $x_3,x_4,\ldots$ | B | B k has size $k+1$. | | Definition 2.5: leading ones, then a two and a shifted remaining word | sAux, s | Shift is the inline function $k\mapsto u(k+\lvert w\rvert)$; the weight is List.sum. The extension to other natural letters does not alter the predicate on Fibonacci words. | | Definition 4.1: strict positivity for every word over ${1,2}$ | FibonacciPositive | Includes the empty word and arbitrary weight; no finite cutoff. | | Conjecture 6.3: the displayed specializations for every $k\ge1$ | xAQC, yAQC | Integer powers retain negative exponents at $k=1$. Values at index zero are outside the source domain and unused by the word determinants. | | There exist values of $\rho,q$ | claim, result | Both range over $\mathbb R$ without extra assumptions; the proof chooses $\rho=q=1/4$. |

Every public definition has its full body in the Blueprint. No source clause is dropped. The private stronger induction applies to the extension on all natural-letter lists; the public result retains exactly the source's Fibonacci-word restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/YoungFibonacci/PetrovScottAltQCharlierPositivity.result

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Discrete structuresQuestion resolvedFirst preprint 2023Max-minus-min homomesy for rectangle interval-closed rowmotionThe max-minus-min statistic is zero-sum on every literal interval-closed rectangle rowmotion orbit.Proved

The questionElder, Lafrenière, McNicholas, Striker and Welch, Toggling, rowmotion, and homomesy on interval-closed sets, arXiv:2307.08520v2, Conjecture 4.9, states that for every product of two finite chains the number of maximal elements minus the number of minimal elements is 0-mesic under rowmotion on interval-closed sets.

Why it mattersThe result gives the exact all-rectangle resolution requested by the original conjecture.

The ideaThese facts yield an integer coboundary for maxMinusMin.

Our contributionThe max-minus-min statistic is zero-sum on every literal interval-closed rectangle rowmotion orbit.

Scope & next questionThe follow-up paper's separate signed-cardinality result, its explicit orbit descriptions for [2] × [n], and any broader poset or higher-product statement are outside this result.

Verification details
Release source recordProved in Lean

Open problem

Max-minus-min homomesy for rectangle interval-closed rowmotion

Elder, Lafrenière, McNicholas, Striker and Welch, Toggling, rowmotion, and homomesy on interval-closed sets, arXiv:2307.08520v2, Conjecture 4.9, states that for every product of two finite chains the number of maximal elements minus the number of minimal elements is 0-mesic under rowmotion on interval-closed sets. Lafrenière, Lewis, McNicholas, Striker and Welch, Interval-closed set rowmotion and homomesy on products of two chains, arXiv:2505.04000v1, restates this target as Conjecture 4.2.

The exact formal target keeps all source quantifiers: positive m,n, every N and equivalence e : Fin N ≃ Fin m × Fin n satisfying the reverse linear-extension condition, and every order-convex initial set I, including the empty set. literalOrbit e I is the finite set of distinct states reached by iterating the literal successive toggle trace. maxMinusMin is the integer cardinality difference of global maxima and minima in the actual state. The conclusion is the sum of this statistic over the actual distinct orbit states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/RectangleRowmotionHomomesy.result

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Quantum informationQuestion resolvedFirst preprint 2022Bernards–Gühne hybrid classical boundThe sharp Bernards--Gühne classical bound holds for two cells of arbitrary sizes k,m >= 2, including the conjectured range m > 2.Proved

The questionGühne, Bell inequalities for nonlocality depth, arXiv:2205.04250v2, Phys.

Why it mattersSection II explicitly allows signaling inside a cell, so these responses may depend on its entire setting string.

The ideaEncode setting 1 by false and setting 2 by true.

Our contributionThe sharp Bernards--Gühne classical bound holds for two cells of arbitrary sizes k,m >= 2, including the conjectured range m > 2.

Verification details
Release source recordProved in Lean

Open problem

Bernards–Gühne hybrid classical bound

F. Bernards and O. Gühne, Bell inequalities for nonlocality depth, arXiv:2205.04250v2, Phys. Rev. A 107, 022412 (2023), Section III.F, Eq. (28), define

$$F_n=\sum_{\ell=1}^n(-1)^{1+\lceil\ell/2\rceil}\ell (1\ldots1,2\ldots2)\le n2^{n-2}.$$

The bracket contains $n-\ell$ settings 1 and $\ell$ settings 2, and sums distinct party permutations. Section V states:

For (k, m) models with m > 2, we have a conjecture for the classical bound. Proving this bound or finding a counterexample remains an open problem.

Issue #12316 preregisters this Tier 1 question, its source, full quantifiers, interpretation and literature check. The encoded claim is the sharp bound for every $k,m\ge2$ with $n=k+m$, including $m>2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/HybridMerminDepthBound.result

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Discrete structuresQuestion resolvedFirst preprint 2023Complete homogeneous symmetric norms of pathsComplete homogeneous symmetric functions of finite path spectra.Proved

Why it mattersThe two expressions of the later Bouthat–Chávez–Sheng Remark 4.3 are settled separately in 18566; its one-term simplification is refuted in 18567.

The ideaThe path determinant satisfies the constant-weight continuant recurrence.

Our contributionComplete homogeneous symmetric functions of finite path spectra.

Verification details
Release source recordProved in Lean

Open problem

Complete homogeneous symmetric norms of paths

Chávez et al., Extremal polynomial norms of graphs, arXiv:2311.04689v3, Section 5, Problem 1: “Compute ‖P_n‖_d for d ≥ 4.” The surrounding norm definition requires even $d$.

For $d=2m$, set $L=2m+n-1$ and interpret $\binom Lr$ as zero outside $0\le r\le L$. The answer is

$$h_{2m}(\lambda(P_n))=\sum_{k=0}^{\lfloor m/(n+1)\rfloor}\left[\binom L{m-k(n+1)}-2\binom L{m-k(n+1)-1}+\binom L{m-k(n+1)-2}\right].$$

The norm is the real $2m$-th root of this expression, for every $n\ge1,m\ge2$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Path adjacency | Literal matrix with entry one exactly at consecutive natural indices | pathAdjHermitian identifies a Hermitian path matrix. | | Complete homogeneous symmetric polynomial | chs, using Mathlib hsymm at Hermitian eigenvalues | Exact source object; eigenvalue ordering is immaterial. | | Norm from Theorem 1.4 | chsNorm, real power with reciprocal degree | Exact real-root convention. | | Every even degree at least four | claim: all natural $n,m$, $1\le n$, $2\le m$ | $d=2m$. | | Compute the value | Both equalities in claim, proved by result | Gives the power and the norm; no source clause is dropped. | | Zero-extended binomial | Frozen IntegerIndexBinomial.binom | Its value is zero outside the natural binomial range. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PathCHS/PathCHSNormClosedForm.result

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Quantum informationQuestion resolvedFirst preprint 2024Simple-spectrum Gibbs-stochastic matrices for every positive Gibbs vectorEvery strictly positive Gibbs vector admits a column-stochastic matrix with simple complex spectrum.Proved

The questionF. vom Ende, Almost All Quantum Channels Are Diagonalizable, arXiv:2403.19643v3 [quant-ph], Open Systems & Information Dynamics 31(3), 2450012 (2024), DOI 10.1142/S1230161224500124, leaves a construction question in the Outlook.

Why it mattersIn vom Ende (2024), arXiv:2403.19643v3, Lemma lemmahartfiel (TeX lines 170–181) relates existence of one simple-spectrum element to density of such elements in an analytically path-connected class.

The ideaAll named helpers below are in GibbsStochasticSimpleSpectrum unless qualified otherwise.

Our contributionEvery strictly positive Gibbs vector admits a column-stochastic matrix with simple complex spectrum.

Verification details
Release source recordProved in Lean

Open problem

Simple-spectrum Gibbs-stochastic matrices for every positive Gibbs vector

F. vom Ende, Almost All Quantum Channels Are Diagonalizable, arXiv:2403.19643v3 [quant-ph], Open Systems & Information Dynamics 31(3), 2450012 (2024), DOI 10.1142/S1230161224500124, leaves a construction question in the Outlook.

The v3 TeX source channel_diag_approx.tex, line 511, states:

for the case of thermal operations, for example, one would have to first show that almost all Gibbs-stochastic matrices have only simple eigenvalues---regardless of the exact Gibbs state

Its footnote, lines 512–513, reads:

For those interested: The last point is what makes this non-trivial. While the $\beta$-permutations \cite{Alhambra19,Mazurek19,PolytopeDegen22} are candidates for Gibbs-stochastic matrices with distinct eigenvalues many of them are only valid in certain parameter regimes, i.e.~only if the eigenvalues of the Gibbs state are interrelated in a specific way. We leave a general construction that is valid for all Gibbs states as an open problem.

These quotations are from the downloaded v3 source, md5 18bc93c300903edbf34bf142ad00519d; the source's definition of a simple eigenvalue as one of algebraic multiplicity one is at lines 123–125. See also the literature note.

Issue #21152 fixes the positive-Gibbs-vector reading: for every $n\in\mathbb N$ and every $d:\mathrm{Fin},n\to\mathbb R$ with $d_i>0$ for all $i$, there exists a real $n\times n$ matrix $P$ with nonnegative entries, every column summing to one, $Pd=d$, and no repeated roots in its complex characteristic polynomial. This existence statement is proved. Normalization of $d$ is not required; $n=0$ is included. The source's density consequence is separate from the formal claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/GibbsStochasticSimpleSpectrum.result

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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2026Every positive-uniform adjacent class inclusion is properFor every positive k, the finite graphs represented by two k-uniform words form a proper subclass of those represented by two (k+1)-uniform words.Proved

The questionAdamson, Dietz, Fleischmann, Huch and Sacher, 2-word-π-representable Graphs, arXiv:2605.27183v1, Conjecture 31, PDF p. 15, states:

Why it mattersThe earlier Dietz–Fleischmann–Huch–Sacher four-author contribution in IFIG Report 2501, printed pp. 21–24 (PDF pp. 29–32), states the same assertion as Conjecture 3.2 on printed p. 23.

The ideaThe known all-positive-k nonuniversality component is proved inside the full consumer.

Our contributionFor every positive k, the finite graphs represented by two k-uniform words form a proper subclass of those represented by two (k+1)-uniform words.

Scope & next questionAn assumed nonuniversality theorem, a fixed-k result, or vertex extension alone does not settle either the full family or this named problem.

Verification details
Release source recordProved in Lean

arXiv

Every positive-uniform adjacent class inclusion is proper

Adamson, Dietz, Fleischmann, Huch and Sacher, 2-word-π-representable Graphs, arXiv:2605.27183v1, Conjecture 31, PDF p. 15, states:

For each k ∈ ℕ, the inclusion 𝒢_k ⊆ 𝒢_(k+1) is proper.

The source convention starts ℕ at 1. For a finite simple graph G on V, membership requires actual words w,v over V, each containing every vertex exactly k times. For every distinct a,b, G.Adj(a,b) holds if and only if their ordered two-letter List projections in w and v are equal. Positive counts give both full alphabets. Single-word alternation and unions of word representation languages are different models.

The exact full target has two clauses for every positive k: inclusion for every finite graph, and existence of a finite graph on one and the same carrier with a (k+1)-uniform representation and no k-uniform representation. An assumed nonuniversality theorem, a fixed-k result, or vertex extension alone does not settle either the full family or this named problem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/GraphRepresentation/UniformHierarchy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026An identity ancilla defeats maximal measurement coherence monotonicityMaximal spectral-distance measurement coherence can increase under a CP unital map.Refuted

The questionTable I marks the corresponding entry under any fuzzifying operations as open.

The ideaThe proof uses spectral estimates and convexity.

Our contributionMaximal spectral-distance measurement coherence can increase under a CP unital map.

Scope & next questionThe diamond-distance and dimension-preserving questions remain open.

Verification details
Release source recordRefuted in Lean

arXiv

baek jo nha 2026 maximal coherence monotonicity

Baek, Jo and Nha, Maximal coherence of quantum measurement and the resource theory of sharpness, arXiv:2607.05847v1, asks whether $C_D^{\max}$ remains monotone under general CP unital maps for non-dichotomic POVMs. Table I marks the corresponding entry under any fuzzifying operations as open.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/MaximalCoherenceAncillaNonmonotone.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026All-ring coordinates for unrestricted face-magic hypercubesAll-ring coordinates classify unrestricted face-magic hypercubes.Proved

The questionManjul Bhargava's arXiv:2609.26762v1, Section 6.2 asks: What is the structure of the R-module Un(R) of face magic hypercubes of dimension n over R when the characteristic is even and positive? The source concerns unrestricted entries and one common sum on every two-coordinate face.

Why it mattersIt keeps the full cube, every dimension, and an existential common face constant in the carrier, rather than adjoining that constant as stored data.

The ideaA parity calculation proves that every face has sum C.

Our contributionAll-ring coordinates classify unrestricted face-magic hypercubes.

Verification details
Release source recordProved in Lean

arXiv

All-ring coordinates for unrestricted face-magic hypercubes

Manjul Bhargava's arXiv:2609.26762v1, Section 6.2 asks: What is the structure of the R-module U_n(R) of face magic hypercubes of dimension n over R when the characteristic is even and positive? The source concerns unrestricted entries and one common sum on every two-coordinate face. The source question is recorded in preregistration #19560.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Hypercube/FaceMagicModule.face_magic_module_structure

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The Gorenstein pairs for alternating adjacent-sum polytopesFor s at least two, the alternating adjacent-sum polytope in dimension 2r is Gorenstein exactly when s = 3 and r = 1.Proved

The questionXinru Jiang, Suzhen Wen and Yueming Zhong, Alternating adjacent-sum polytopes: transfer matrices and Ehrhart series, arXiv:2607.14887v1, §4, Question Q1 (q-gorenstein), ask:

Why it mattersThe source note is Library/Geometry/jiangwenzhong2026alternating.md.

Our contributionFor s at least two, the alternating adjacent-sum polytope in dimension 2r is Gorenstein exactly when s = 3 and r = 1.

Scope & next questionQ2 on unimodality and real-rootedness remains open; this module does not settle it.

Verification details
Release source recordProved in Lean

arXiv

The Gorenstein pairs for alternating adjacent-sum polytopes

Xinru Jiang, Suzhen Wen and Yueming Zhong, Alternating adjacent-sum polytopes: transfer matrices and Ehrhart series, arXiv:2607.14887v1, §4, Question Q1 (q-gorenstein), ask:

For $s=3$, direct computation (Propositions 3.22 and 3.23) shows that only $d=2$ is Gorenstein in even dimensions $d\le6$; we conjecture this extends to all $d\ge4$. Characterize all $(s,r)$ with $s\ge2$ for which $\mathcal P_{2r}^{(s)}$ is Gorenstein.

Their set is $\mathcal P_d^{(s)}={x\in\mathbb R_{\ge0}^d: x_i+x_{i+1}\le s+\delta_i,\ 1\le i\le d-1}$, where $\delta_i=0$ for odd $i$ and $\delta_i=1$ for even $i$. The Gorenstein characterization used here is the source's interior-lattice translation condition: some integer $q\ge1$ and $c\in\mathbb Z^d$ satisfy $\operatorname{int}((n+q)P)\cap\mathbb Z^d =c+(nP\cap\mathbb Z^d)$ for every integer $n\ge0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/AlternatingAdjacentSumGorenstein.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Further frontiersCounterexample foundFirst preprint 2026Question 1.10: regular Toeplitz and strictly ergodic strengtheningThe regular Toeplitz and strictly ergodic branch is refuted by one base.Refuted

The questionThe cited v1 asks whether a given minimal/strictly ergodic (Toeplitz) subshift admits a permutative sliding block code of length ℓ≥2 whose entire inverse image is minimal/strictly ergodic.

Why it mattersThe original finite extension has the actual fibre A^(ℓ−1).

The ideaThe original factorial-word base and UniversalRefutation.result remain valid.

Our contributionThe regular Toeplitz and strictly ergodic branch is refuted by one base.

Verification details
Release source recordRefuted in Lean

Open problem

Question 1.10: regular Toeplitz and strictly ergodic strengthening

The cited v1 asks whether a given minimal/strictly ergodic (Toeplitz) subshift admits a permutative sliding block code of length ℓ≥2 whose entire inverse image is minimal/strictly ergodic. This dossier includes the delivered general minimal refutation and its strengthening to a regular Toeplitz, strictly ergodic base. Both concern the same numbered external question.

The original UniversalRefutation.claim quantifies over all finite discrete alphabets A with at least two letters and all nonempty minimal two-sided subshifts. Its conclusion asks for some k≥0 and h:A^(k+2)→A, bijective in both endpoints with all remaining letters fixed, whose whole inverse image is minimal. UniversalRefutation.result : Not claim is the general-branch refutation retained as a preliminary result.

The strengthened BalancedRegularRefutation.claim already restricts the input to binary subshifts that are nonempty and minimal, have exactly one invariant probability measure, and are generated by one aperiodic Toeplitz point with a genuine essential period structure. The structure has positive nonempty essential periods, strict growth, divisibility into the next period, and coverage of every signed coordinate; its actual normalized periodic skeleton densities tend to one. The conclusion still allows every length k+2 and every bipermutative local rule, and asks for minimality of the whole inverse image. BalancedRegularRefutation.result : ¬ claim is its literal negation. Since strict ergodicity implies minimality, it also rules out a strictly ergodic whole lift. A binary counterexample suffices for the original finite-alphabet question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Dynamics/Symbolic/PermutativeLifts/BalancedRegularRefutation.result

D5/S1/Dynamics/Symbolic/PermutativeLifts/UniversalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Mazorchuk–Xu: every first-column deficit has a counting polynomialThe top weak-coordinate patterns are exactly a choice of position values and a permutation of the complementary cutoff values.Proved

The questionThere is no eventual threshold, fixed-deficit bound, nonempty-row or representation hypothesis.

Why it mattersPropositions 11–14 already settle deficits zero through three; those cases are prior work.

The ideaSort the anchor neighbors and their complement.

Our contributionThe top weak-coordinate patterns are exactly a choice of position values and a permutation of the complementary cutoff values.

Verification details
Release source recordProved in Lean

Open problem

Mazorchuk–Xu: every first-column deficit has a counting polynomial

Volodymyr Mazorchuk and Shuoyang Xu, On the number of cokernel-closed additive subcategories for uniformly oriented A_n quivers, arXiv:2610.07999v2, Section 5.7, conjecture:

It seems reasonable to conjecture that $r(n,n+1-i)$ is a polynomial in $n$ of degree $2i$ and leading coefficient $\frac{1}{i!}$.

The preceding sentence specifies “any fixed number of missing points.” This dossier anchors the whole conjecture: for every natural $i$, one rational polynomial $P_i$ has exact degree $2i$, leading coefficient $1/i!$, and satisfies $P_i(n)=r(n,n+1-i)$ for every natural $n$ with $i\le n+1$.

Section 3, before Section 3.1, defines $T(n)={(a,b):0\le a\le b\le n}$ and its subsets satisfying Condition (star). Three selected points $(a,b),(c,d),(e,f)$, allowing repetitions, with $a\le c,e\le b\le d,f$ force $(\max(c,e),\max(d,f))$ and, if $b<\min(d,f)$, also $(b+1,\min(d,f))$. Section 5.1 defines $r(n,k)$ as the number of these subsets with exactly $k$ selected first-column points. Empty subsets and rows are permitted.

Source correspondence

| Source clause | Lean counterpart | | --- | --- | | Literal triangular subsets | triangle, Star, family, count in MazorchukXuClosure | | Repeated selected points and both forced conclusions | All three membership premises in Star; star_iff_rules proves equivalence with the ordered-edge rules | | Exact first-column statistic | firstColumnCard; count_eq_edgeFamily_card preserves its value | | Every fixed deficit | Outer ∀ i : ℕ, followed by ∃ p : Polynomial ℚ before ∀ n | | Every defined rank | ∀ n : ℕ, i ≤ n+1 → p.eval (n : ℚ) = (count n (n+1-i) : ℚ) | | Exact degree and leading coefficient | p.natDegree = 2*i and p.leadingCoeff = (i.factorial : ℚ)⁻¹ | | Boundary cases | No exclusions of i=0, n=0, i=n+1, zero or tied coordinates, or empty rows |

There is no eventual threshold, fixed-deficit bound, nonempty-row or representation hypothesis. The source does not require integer coefficients. The canonical embedding of the rational polynomial into the reals preserves evaluation, degree and leading coefficient. The separate recurrence convention at rank $-1$ is outside this domain.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BinomialBases/MazorchukXuTopPatterns.mazorchukXu_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedBala's Fibonacci-power strong divisibility conjectureFibonacci minus one at powers of any positive odd base is a strong divisibility sequence.Proved

The questionPeter Bala's COMMENTS entry of Dec 05 2022 states:

Why it mattersSubtracting one destroys the ordinary Fibonacci strong-divisibility identity at arbitrary indices.

The ideaFor the natural representative a of x, p^e divides a(a+1).

Our contributionFibonacci minus one at powers of any positive odd base is a strong divisibility sequence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's Fibonacci-power strong divisibility conjecture

OEIS A000071 has a(t)=F_t-1, with Fibonacci initial values F_0=0, F_1=1. Peter Bala's COMMENTS entry of Dec 05 2022 states:

Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)).

The exact formal statement is:

theorem result (k n m : ℕ) (hk : Odd k) (hn : 0 < n) (hm : 0 < m) :
    Nat.gcd (Nat.fib (k ^ n) - 1) (Nat.fib (k ^ m) - 1) =
      Nat.fib (k ^ Nat.gcd n m) - 1

Oddness of a natural k supplies its positivity. The source's k=1 and every pair of positive indices are included; there is no cutoff, changed offset, or conditional prime restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/BalaFibonacciPowerStrongDivisibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThree tetrahelix coordinate generating functionsThe ordered Boerdijk-Coxeter tetrahelix has Kagey's three exact rational coordinate series.Proved

The questionThe ordered initial vertices are (-1,-1,-1), (-1,1,1), (1,-1,1), (1,1,-1).

Why it mattersTheir common recurrence makes them one connected problem, not three independent resolutions.

The ideaThe D5 result constructs vertices from the ordered initial tetrahedron by the order-four reflection recurrence.

Our contributionThe ordered Boerdijk-Coxeter tetrahelix has Kagey's three exact rational coordinate series.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Three tetrahelix coordinate generating functions

OEIS A400216, A400217 and A400218, Peter Kagey (2026), conjecture the coordinate generating functions of one Boerdijk-Coxeter tetrahelix. The ordered initial vertices are (-1,-1,-1), (-1,1,1), (1,-1,1), (1,1,-1). At each step the oldest vertex is reflected across the face through the next three. The coordinates are scaled by 1 through index 3 and by 3^(n-3) thereafter. In the order x, y, z, the proposed series are

  • (54t^6-84t^5-66t^4+23t^3+9t^2+t-1)/((3t-1)^2(9t^2+4t+1));
  • (-54t^6+84t^5-90t^4+15t^3+3t^2+3t-1)/((3t-1)^2(9t^2+4t+1));
  • (18t^5+26t^4-24t^3-5t^2+1)/(27t^3+3t^2-t-1).

The denominators are units in rational formal power series, with respective constant coefficients 1, 1 and -1. Preregistration is issue #10182.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/BoerdijkCoxeterGeneratingFunctions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedSpezia's Pell circulant permanent-to-determinant ratioThe permanent divided by the signed determinant of the Pell circulant tends to one.Proved

The questionStefano Spezia's OEIS A392131, revision 16, FORMULA states:

Why it mattersThe conjecture compares the unsigned sum over permutation products with its signed counterpart for a dense, sequence-defined matrix.

The ideaThe Pell recurrence and companion positivity are reused from PellCompanionGcd.

Our contributionThe permanent divided by the signed determinant of the Pell circulant tends to one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Spezia's Pell circulant permanent-to-determinant ratio

Stefano Spezia's OEIS A392131, revision 16, FORMULA states:

Conjecture: lim_{n->oo} A392132(n)/a(n) = 1.

OEIS A392132, revision 10, FORMULA states the same conjecture as:

Conjecture: lim_{n->oo} a(n)/A392131(n) = 1.

In A392131, $a(n)$ is the signed determinant; in A392132, $a(n)$ is the permanent. These are two records of one assertion. The source is OEIS-only, submitted in 2025 and published in 2026, with no source DOI or arXiv identifier. Issue #21886 fixes the conventions and the settlement criterion. The source quotations and locators are in the Library note.

Statement echo

Let $P_0=0$, $P_1=1$ and $P_{k+2}=2P_{k+1}+P_k$. With indices in $\operatorname{Fin}(n)$, the matrix is $C_n(i,j)=P_{n-1-\operatorname{val}(i-j)}$, with natural subtraction truncated at zero and modular subtraction in $i-j$. Its first row is $(P_{n-1},P_0,\ldots,P_{n-2})$, the initial right rotation specified by the source COMMENT and four-by-four EXAMPLE. The formalization reuses the existing Pell sequence D5.S1.Recurrence.PellCompanionGcd.P and Mathlib's Matrix.circulant, Matrix.det and Matrix.permanent.

| Source clause | Formal counterpart | | --- | --- | | A392131 COMMENT: first column reversed Pell vector, first row initially rotated right | C, with the literal binding vector and Mathlib's modular circulant indexing | | A392131 determinant, including the source empty value | A392131 n = Matrix.det (C n); the determinant retains its sign | | A392132 permanent of the same matrix | A392132 n = Matrix.permanent (C n) | | The ratio tends to one | claim: every real $\varepsilon>0$ admits $N\ge2$ such that every $n\ge N$ has $|A392132(n)/A392131(n)-1|<\varepsilon$ | | No additional matrix-family hypothesis | result : claim has no hypotheses |

The threshold condition $N\ge2$ specifies an equivalent tail of the limit; it imposes no extra condition on the sequence. Both empty values are one. The zero determinant at $n=1$ does not affect a tail assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/SpeziaA392131PellCirculantRatio.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryEquivalence provedSeven and five versus binary carryless multiplicationSeven and five define equivalent ordinary versus binary carryless multiplier conditions.Proved

The questionDigits are indexed from zero at the least significant bit and have value zero or one.

Why it mattersThe source compares sets defined by different pairs of ordinary and carryless multipliers.

The ideaIt contains the pair of initial zero-carry, zero-mask states.

Our contributionSeven and five define equivalent ordinary versus binary carryless multiplier conditions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Seven and five versus binary carryless multiplication

OEIS A115770 defines the nonnegative integers $k$ satisfying $7k=\operatorname{A048720}(11,k)$ and conjectures that they are exactly the integers satisfying $5k=\operatorname{A048720}(13,k)$. The exact assertion is

$$ \forall k\in\mathbb{N},\qquad 7k=C(11,k)\ \Longleftrightarrow\ 5k=C(13,k). $$

The same $k$ occurs on both sides, $\mathbb{N}$ includes zero, and there is no upper bound on $k$. Multiplication outside $C$ is ordinary multiplication. The operation $C$ is binary coefficient convolution over $\mathrm{GF}(2)$, as defined by OEIS A048720:

$$ \operatorname{bit}(C(u,k),t) =\left(\sum_{i=0}^{t} \operatorname{bit}(u,i)\operatorname{bit}(k,t-i)\right)\bmod 2. $$

Digits are indexed from zero at the least significant bit and have value zero or one. The row-five/row-seven conjecture in A115872 expresses this same question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/CarrylessSevenFiveEquivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Seven and five define equivalent ordinary versus binary carryless multiplier conditions.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's full logarithmic arithmetic-subsequence closureIntegral logarithmic series remain integral under every positive affine extraction.Proved

The questionFor integers m>0, b>=0, Sum{n>=0} L(mn+b) log(1+x)^n/n! is an integer series whenever Sum{n>=0} L(n)log(1+x)^n/n! is an integer series.

Why it mattersArithmetic extraction is a distinct closure operation from pointwise powers.

The ideaOriginal reconstruction supplies a:N->Z with L(N)=sum(k<=N) ak qk(N).

Our contributionIntegral logarithmic series remain integral under every positive affine extraction.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's full logarithmic arithmetic-subsequence closure

For integers m>0, b>=0, Sum_{n>=0} L(m*n+b) * log(1+x)^n/n! is an integer series whenever Sum_{n>=0} L(n)*log(1+x)^n/n! is an integer series.

The source is the CONJECTURE comment in OEIS A167540. An integer series has integer ordinary coefficients. The exact formal statement concerns every characteristic-zero field K, arbitrary L:N->K, every positive natural m and arbitrary natural b, at every ordinary degree including zero. The only mathematical premise is integrality of the entire input logarithmic transform. Integral values of L are derived, with arbitrary L(0). There is no sign, normalization, support, degree, dilation, shift or analytic restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaLogarithmicAffineClosure.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedOEIS A302975 divisor-count power-ratio denominator squareEvery reduced denominator in OEIS A302975 is a square.Proved

The questiontau(n) = the number of the divisors of n (A000005).

Why it mattersKrizek's 2018 OEIS conjecture asks whether every reduced denominator in this divisor-count power ratio is a square.

The ideaFor every prime p, vp(D n) = max(t vp(n) - n vp(t), 0).

Our contributionEvery reduced denominator in OEIS A302975 is a square.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A302975 divisor-count power-ratio denominator square

OEIS A302975, NAME (%N, verbatim):

a(n) = denominator of tau(n)^n / n^tau(n)

COMMENTS (%C, verbatim):

tau(n) = the number of the divisors of n (A000005).

Conjecture: all terms are squares.

The formal claim is the literal statement ∀ n ≥ 1, IsSquare (D n), where D is the reduced denominator of the displayed rational ratio. The definition is totalized with D 0 = 1; that value is not a source claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/KrizekDivisorCountPowerRatioDenominatorSquare.krizek_a302975

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    Source 1
Discrete structuresQuestion resolvedOEIS A381190 trapezohedral domination seriesThe original A381190 rational generating function for all n>=3.Proved

The question-2x^3(4x^5 + 8x^4 + 4x^3 - 9x^2 - 8x - 3)/(x^3 + x^2 - 1)^2.

Why it mattersThis is the full revision-16 OEIS conjecture.

The ideaApex classification then shows that every admissible connected subset has exactly one apex and the opposite rim label.

Our contributionThe original A381190 rational generating function for all n>=3.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A381190 trapezohedral domination series

OEIS A381190 is named “Number of connected minimal dominating sets in the n-trapezohedral graph.” Its conjectured formula is

-2*x^3*(4*x^5 + 8*x^4 + 4*x^3 - 9*x^2 - 8*x - 3)/(x^3 + x^2 - 1)^2.

The resolved statement is the literal ordinary formal-power-series identity for every integer n >= 3. For each n, the graph has distinct labelled vertices alpha, beta, a_0,...,a_(n-1), and b_0,...,b_(n-1), the alternating rim cycle, alpha adjacent to every a_i, beta adjacent to every b_i, and no other edges. The coefficient counts actual subsets S whose closed neighbourhood is the full graph, every proper subset of S fails to dominate, and the induced graph on S is connected. Minimality ranges over all proper subsets independently of their connectivity; no quotient by rotations, reflections, or other automorphisms is taken. Coefficients below order three are zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/TrapezohedralDominationSeries.actual_series

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    Source 1
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
Further frontiersQuestion resolvedSource 2026The orthogonal absolute-product minimum is attained at oneEvery nonsingular real square matrix has orthogonal absolute-product minimum exactly one, attained by a Gram-Schmidt basis.Proved

The questionUser3 (MathOverflow user 600125), A nontrivial lower bound for the orthogonally transformed absolute product, question 515549, revision 1, GUID CB719581-BCE3-4573-931E-15327002F861, asks:

Why it mattersThe source distinguishes the entrywise absolute-product spectral radius from a norm or a matrix absolute value.

The ideaThe proof uses spectral estimates.

Our contributionEvery nonsingular real square matrix has orthogonal absolute-product minimum exactly one, attained by a Gram-Schmidt basis.

Verification details
Release source recordProved in Lean

Open problem

The orthogonal absolute-product minimum is attained at one

User3 (MathOverflow user 600125), A nontrivial lower bound for the orthogonally transformed absolute product, question 515549, revision 1, GUID CB719581-BCE3-4573-931E-15327002F861, asks:

Let $P\in\mathbb{R}^{K\times K}$ be nonsingular and define $$M(Q)=|PQ|,|Q^\top P^{-1}|,\qquad Q\in O(K).$$ $|M∣$ means the matrix obtained by taking the absolute value of each entry. Since $$|PQ(Q^\top P^{-1})|=I\le |PQ|,|Q^\top P^{-1}|=M(Q),$$ one obtains the trivial bound $\rho(M(Q))\ge 1$. $\rho(M(Q))$ is the spectral radius of the matrix $M$. The question is whether there exists a nontrivial quantity $\beta(P)$, depending only on $P$, such that $$\rho(M(Q))\ge\beta(P)>1\qquad\text{for all }Q\in O(K),$$ for an appropriate class of nonsingular matrices $P$. Equivalently, can the optimization problem $$\inf_{Q\in O(K)}\rho!\left(|PQ|,|Q^\top P^{-1}|\right)$$ be bounded below, or even characterized explicitly, in terms of intrinsic properties of $P$?

The quotation is attributed to User3, MathOverflow, under CC BY-SA 4.0. Its source identity and license are recorded in the Library note.

The single resolution anchor is the exact optimization characterization: for every natural $K>0$ and every $P\in\mathbb R^{K\times K}$ with $\det P\ne0$,

$$ \left(\forall Q\in O(K),\quad 1\le\rho!\left(|PQ|,|Q^\top P^{-1}|\right)\right) \quad\land\quad \left(\exists Q\in O(K),\quad \rho!\left(|PQ|,|Q^\top P^{-1}|\right)=1\right). $$

Thus the minimum and infimum are both one, with actual attainment. In particular, no $\beta(P)>1$ can bound the objective below for every orthogonal $Q$, for any nonsingular $P$ in positive dimension. This answers the characterization question and negates the proposed strict improvement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Eigenstructure/OrthogonalAbsoluteProductMinimum.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
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Quantum informationQuestion resolvedFirst preprint 2022Exact quantum bound of the Bernards–Gühne nonlocality-depth familyFor every n >= 3, the Bernards--Gühne Bell functional has exact quantum maximum sqrt(2) n 2^(n-2), over all finite local dimensions, binary observables and density matrices.Proved

The questionGühne, Bell inequalities for nonlocality depth, arXiv:2205.04250v2, Phys.

Why it mattersThe source's two-cell classical bound has a separate owner, HybridMerminDepthBound.result; it is a different maximum.

The ideaIts conclusion holds for all natural phase shifts and arbitrary finite local carriers.

Our contributionFor every n >= 3, the Bernards--Gühne Bell functional has exact quantum maximum sqrt(2) n 2^(n-2), over all finite local dimensions, binary observables and density matrices.

Verification details
Release source recordProved in Lean

Open problem

Exact quantum bound of the Bernards–Gühne nonlocality-depth family

F. Bernards and O. Gühne, Bell inequalities for nonlocality depth, arXiv:2205.04250v2, Phys. Rev. A 107, 022412, Section III.F, Eqs. (28)–(29), define the permutation-symmetric Bell functional

$$F_n=\sum_{\ell=1}^n(-1)^{1+\lceil\ell/2\rceil}\ell (1\ldots1,2\ldots2).$$

The bracket sums distinct party permutations, with $n-\ell$ settings 1 and $\ell$ settings 2. The source states:

In the following, we discuss numerical findings concerning the Bell inequalities F_n for n ≤ 8 parties. We find, that the quantum bound of the inequalities is F_n = √2 n 2^{n−2} (29) up to numerical precision, which is achieved by choosing A₁ = B₁ = … = σ_x and A₂ = B₂ = … = σ_z

For every integer $n\ge3$, the maximum over all finite local dimensions, binary observables and density matrices is exactly $Q_n=\sqrt2,n2^{n-2}$. This is a uniform extension and proof of the numerical finding; the source does not state a general-$n$ conjecture. The preregistration is #16626.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Every party number $n\ge3$ | claim, quantified n and lower bound | All such natural numbers | | Finite local dimensions | d, local carriers Fin (d j) | Arbitrary finite dimensions, including empty carriers | | Two binary measurements at each party | A j s, literal Hermitian and shift conditions | Hermitian matrices with both shifts positive semidefinite; general binary POVMs | | Full correlators and distinct party permutations | Entrywise product in bellOp, Boolean setting strings | Entrywise tensor product; Hamming weight supplies the coefficient and multiplicity | | Arbitrary states | Positive semidefinite ρ with trace one in claim | Literal body of the existing GHZMeasureBiseparableBound.IsDensity; no duplicate predicate | | Exact quantum maximum | IsGreatest in claim; result | Universal upper bound and attained value | | Source's X/Z optimum | X/Y phase-adjusted GHZ realization in result | Equivalent local measurement plane; the explicit X/Z conjugation is a prose identification | | Shift and involution hypotheses of the auxiliary estimate | Upper.unweighted_bound | Auxiliary inputs discharged by dilation and party decomposition inside result; no extra assumption in claim |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/NonlocalityDepthQuantumBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2023Complete homogeneous symmetric norms of regular graphsExtrema of complete homogeneous symmetric norms over connected regular graphs.Proved

The questionChávez et al., Extremal polynomial norms of graphs, arXiv:2311.04689v3, Section 8, Problem 3: “Can one find the extrema of the CHS norms over other families of graphs such as regular, unicyclic and complete multipartite graphs?”

Why it mattersThe two endpoint graphs belong to the connected regular family for every admissible order.

The ideaIts nonnegative terms imply nonnegativity of all adjacency CHS coefficients and monotonicity under domination of closed-walk counts.

Our contributionExtrema of complete homogeneous symmetric norms over connected regular graphs.

Verification details
Release source recordProved in Lean

Open problem

Complete homogeneous symmetric norms of regular graphs

Chávez et al., Extremal polynomial norms of graphs, arXiv:2311.04689v3, Section 8, Problem 3: “Can one find the extrema of the CHS norms over other families of graphs such as regular, unicyclic and complete multipartite graphs?”

For even degree $d=2m\ge2$, Theorem 1.4 defines

$$|G|{2m}=h{2m}(\lambda_1,\ldots,\lambda_n)^{1/(2m)},$$

where $\lambda_1,\ldots,\lambda_n$ are the real adjacency eigenvalues. Among connected regular simple graphs of order $n\ge3$, the cycle $C_n$ attains the minimum and the complete graph $K_n$ attains the maximum:

$$|C_n|{2m}\le|G|{2m}\le|K_n|_{2m}.$$

Here connected means that every two vertices are joined by a walk. Section 2 of the source uses the word connected for graphs without isolated vertices; under that reading the minimum changes for even orders, since a perfect matching is regular without isolated vertices and has smaller coefficients, for example $h_2(2K_2)=2<4=h_2(C_4)$. The formal statement uses ordinary connectedness. In fact the lower coefficient bound $h_{2m}(C_n)\le h_{2m}(G)$ holds for every $r$-regular simple graph of order $n\ge3$ with $r\ge2$, without connectedness. This dossier concerns the regular family. The unicyclic branch is not addressed; the complete multipartite branch is settled separately.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Simple graph of order n | SimpleGraph (Fin n) | Literal finite simple graph. | | Adjacency eigenvalues and complete homogeneous polynomial | Defs.chs | MvPolynomial.hsymm evaluated at the real Hermitian adjacency eigenvalues. | | Positive even-degree norm | Defs.chsNorm, 2 * m, 1 ≤ m | Real power with exponent one divided by the degree. | | Connected regular family | G.Connected and ∃ r, G.IsRegularOfDegree r | Usual graph connectedness; no isolated-vertex convention is substituted. | | Every n at least three | 3 ≤ n in Defs.claim | Excludes the smaller orders. | | Minimum and maximum with attainment | Result.result : Defs.claim | Both inequalities and connected regular endpoint graphs. | | Lower bound without connectedness | Private Result.chs_cycle_le_regular | Every regular degree at least two, at every even coefficient degree. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/RegularCHS/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2022Reduced Denominators for Hankel Determinants of Boundary-Weighted Motzkin PathsCigler's boundary-weighted Motzkin Hankel series has the specified denominator and a numerator of exact degree binom(m+1,3) + 1 for every positive shift m.Proved

The questionJohann Cigler, Some remarks and conjectures about Hankel determinants of polynomials which are related to Motzkin paths, arXiv:2204.09910v4, Section 2, Conjecture 2.1, equation (2.3).

The ideaAn orthogonal-polynomial (Christoffel-type) formula writes dm(n,t,s) as an m × m determinant of coefficients of the polynomials pr(y) = Fr(y − t) + (t − s)F{r−1}(y − t), confluent at the zero node.

Our contributionCigler's boundary-weighted Motzkin Hankel series has the specified denominator and a numerator of exact degree binom(m+1,3) + 1 for every positive shift m.

Verification details
Release source recordProved in Lean

Open problem

Reduced Denominators for Hankel Determinants of Boundary-Weighted Motzkin Paths

Johann Cigler, Some remarks and conjectures about Hankel determinants of polynomials which are related to Motzkin paths, arXiv:2204.09910v4, Section 2, Conjecture 2.1, equation (2.3). M_{n,k}(t,s) is the weighted number of Motzkin paths from (0,0) to (n,k) that never go below the axis, with up and down steps of weight 1, horizontal steps of weight s at height 0 and t above; d_m(n,t,s) = det(M_{m+i+j,0}(t,s)){0≤i,j<n}. With L_0 = 2, L_1 = t, L_r = tL{r−1} − L_{r−2}, A_{0,0}(x,t) = 1 − x and A_{0,r}(x,t) = 1 − L_r(t)x + x² for r > 0, the conjecture states

Σ_{n≥0} d_m(n,t,s) xⁿ = R_m(x,t,s) / ∏_{j=0}^{⌊m/2⌋} A_{0,m−2j}(x,t)^{1+j(m−j)},

with R_m an integer polynomial in x, s, t of x-degree binom(m+1,3) + 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MotzkinHankel/CiglerMotzkinHankel.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Reduced Denominators for Hankel Determinants of Motzkin-Triangle ColumnsAt every positive shift, each Motzkin-triangle column has Cigler's specified Hankel-series denominator and a numerator of the exact stated degree.Proved

The questionJohann Cigler, Some remarks and conjectures about Hankel determinants of polynomials which are related to Motzkin paths, arXiv:2204.09910v4, Section 1, Conjecture 1.3, equation (1.30).

The ideaSpecializing the boundary weight s to t in the proof of Conjecture 2.1 of the same paper (CiglerMotzkinHankel), a Christoffel-type formula writes dm^{(k)}(n,t) as a determinant with k fixed Chebyshev nodes and m confluent zero nodes, with sign (−1)^{binom(k+1,2)}.

Our contributionAt every positive shift, each Motzkin-triangle column has Cigler's specified Hankel-series denominator and a numerator of the exact stated degree.

Verification details
Release source recordProved in Lean

Open problem

Reduced Denominators for Hankel Determinants of Motzkin-Triangle Columns

Johann Cigler, Some remarks and conjectures about Hankel determinants of polynomials which are related to Motzkin paths, arXiv:2204.09910v4, Section 1, Conjecture 1.3, equation (1.30). M_{n,k}(t) is the weighted number of Motzkin paths from (0,0) to (n,k) that never go below the axis, with up and down steps of weight 1 and horizontal steps of weight t; d_m^{(k)}(n,t) = det(M_{m+i+j,k}(t)){0≤i,j<n} and D_m^{(k)}(x,t) = Σ_n d_m^{(k)}(n,t) xⁿ. With e = (−1)^{binom(k+1,2)}, A{k,0} = 1 − e x^{k+1} and A_{k,r} = 1 − e L_r(t) x^{k+1} + x^{2(k+1)} for r > 0 (equation (1.28)), the conjecture states

D_m^{(k)}(x,t) = r_m^{(k)}(x,t) / ∏_{j=0}^{⌊m/2⌋} A_{k,(k+1)(m−2j)}(x,t)^{1+j(m−j)},

with deg_x r_m^{(k)} = binom(m+1,3) + k (binom(m,1) + binom(m,2) + binom(m,3)).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MotzkinHankel/CiglerMotzkinColumn.result

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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2022The exact linear-span dimension of binary bond-two uniform matrix product statesThe span of binary uniform matrix product states of bond dimension two has an exact parity-dependent quartic dimension.Proved

The questionSeynnaeve, The Linear Span of Uniform Matrix Product States, SIGMA 18 (2022), 099, DOI 10.3842/SIGMA.2022.099; arXiv:2204.10363v3, Section 3.2, PDF page 9 (sigma22-099.tex, lines 540–547):

Why it mattersThe source studies which linear equations vanish on uniform matrix product states, before addressing their higher-degree defining equations.

The ideaRank-nullity and the annihilator dimension formula give the equality.

Our contributionThe span of binary uniform matrix product states of bond dimension two has an exact parity-dependent quartic dimension.

Verification details
Release source recordProved in Lean

Open problem

The exact linear-span dimension of binary bond-two uniform matrix product states

C. De Lazzari, H. J. Motwani and T. Seynnaeve, The Linear Span of Uniform Matrix Product States, SIGMA 18 (2022), 099, DOI 10.3842/SIGMA.2022.099; arXiv:2204.10363v3, Section 3.2, PDF page 9 (sigma22-099.tex, lines 540–547):

Conjecture 3.8 (Corollary of Conjecture 3.7).

$\dim \langle \mathrm{uMPS}(2,2,d)\rangle = \begin{cases}\frac{1}{192}\big(d^4-4d^2+192d+192\big) & \text{for }d\text{ even,}\ \frac{1}{192}\big(d^4-10d^2+192d+201\big) & \text{for }d\text{ odd.}\end{cases}$

For two complex $2\times2$ matrices $X,Y$, let $A_0=X$, $A_1=Y$, and let the tensor $\phi_d(X,Y)$ have coefficient $\operatorname{tr}(A_{i_1}\cdots A_{i_d})$ at the binary word $(i_1,\ldots,i_d)$. The source defines $\mathrm{uMPS}(2,2,d)$ as the closure of the image of $\phi_d$ (the parametrization and following paragraph; TeX lines 215–225). Its linear span is taken over $\mathbb C$.

Issue #23810 fixes the reading for every positive integer $d$. The Lean claim uses UniformMPSExpectedDimensionRefutation.uMPS 2 2 d, the image of this ordered trace parametrization, and asserts the displayed equality after casting the natural-number dimension to $\mathbb Q$. The image/closure bridge is explained below. This reading of Conjecture 3.8 is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/UniformMPSTwoByTwoLinearSpan.result

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    Source 1
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Discrete structuresQuestion resolvedSource 2023Ducci Modular Cycle FactorizationEvery cycle of the modular multiplicative Ducci game factors through one constant and two generator cycles.Proved

The questionFellman and Klyve, INTEGERS 23 (2023), Article A86, Section 7, state:

Why it mattersThe proof does not use the cocomposite hypothesis, so it covers coprime cycles as well and reproves the authors' own Theorem 13 by an argument that never inverts an entry.

The ideaLet u lie on a cycle, step^[L] u = u, and write Q = u0 u1 u2 u3.

Our contributionEvery cycle of the modular multiplicative Ducci game factors through one constant and two generator cycles.

Verification details
Release source recordProved in Lean

Open problem

Ducci Modular Cycle Factorization

Fellman and Klyve, INTEGERS 23 (2023), Article #A86, Section 7, state:

Conjecture 1. Every cocomposite cycle is a product of a constant cycle and two generating cycles.

and add: "This conjecture has been verified for all moduli up to 161. One possible direction of proving this conjecture is to show that all cocomposite cycles are a product of a cocomposite constant cycle and a coprime cycle, but we have been unsuccessful on this front."

The game replaces the difference map of the classical Ducci game by [a b c d] ↦ [ab bc cd da] on (Z_n)^4. A 4-tuple is in a cycle when iterating returns it; by their Lemma 2 the entries of such a tuple are then either all units or all non-units, and the cycle is called coprime or cocomposite accordingly. A cycle is constant when its tuples have four equal entries. A generator 4-tuple is [1 x 1 x^{-1}] with x a unit lying on a cycle, and a cycle containing one is a generator cycle. Products are taken entry by entry.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/DucciModularCycleFactorization.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2023.

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2022Distance Seidel energy after deleting an edgeDeleting any edge of a complete bipartite graph with both parts of size at least two strictly increases its distance Seidel energy for the matrix J - I - 2D.Proved

The questionChithra, Distance Seidel matrix of a connected graph, arXiv:2210.05940v2, page 13, Problem 5.1:

Why it mattersThe source is recorded in Library/GraphInvariants/harithachithra2022distanceseidel.md.

The ideaThe proof uses spectral estimates.

Our contributionDeleting any edge of a complete bipartite graph with both parts of size at least two strictly increases its distance Seidel energy for the matrix J - I - 2D.

Verification details
Release source recordProved in Lean

Open problem

Distance Seidel energy after deleting an edge

T. Haritha and A. V. Chithra, Distance Seidel matrix of a connected graph, arXiv:2210.05940v2, page 13, Problem 5.1:

Let G = K_{a,b}, with a, b > 1. Is E_{D^S}(K_{a,b} − e) > E_{D^S}(K_{a,b}), for any edge e of K_{a,b}?

The answer is affirmative for the source's matrix equation $D^S(G)=J-I-2D$, for every $a,b\ge2$ and every edge. The source's casewise expression instead gives diagonal one. The universal statement here concerns the zero-diagonal matrix equation; the diagonal-one alternative has only the finite evidence below.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $D^S(G)=J-I-2D$ | distSeidel | Literal entrywise expression with shortest-path distance and zero diagonal. | | $E_{D^S}(G)=\sum_r|\partial_r^S|$ | energy | Absolute real characteristic roots, counted with multiplicity; the matrices used by the theorem are real symmetric. | | $G=K_{a,b}$ | completeBipartiteGraph (Fin a) (Fin b) | Carrier Fin a ⊕ Fin b; graph constructions are Mathlib's. | | $a,b>1$ | Both natural-number hypotheses in claim | Exactly $2\le a$ and $2\le b$. | | Any edge $e$ | Universal quantifier over edgeSet in claim | Unordered edges; no particular edge is selected in the public conclusion. | | Delete $e$ and increase energy strictly | deleteEdges and strict inequality in claim, proved by result | The singleton edge set is removed; all source clauses are retained. | | Finite decidable vertex carrier | The instance parameters of distSeidel and energy | Supplies finite matrices and sums, without an additional graph restriction. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/KyFanGap/HarithaChithraDistanceSeidelEnergy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Generic logarithmic Lie-length of su(n)Generic Lie-length in the traceless skew-Hermitian matrix Lie algebra grows logarithmically for every fixed generator count at least two.Proved

The questionYifan Jia and Angela Capel, A generic quantum Wielandt's inequality, arXiv:2301.08241v3, Quantum 8 (2024) 1331, Conjecture 6, p. 19:

Why it mattersThe source identifies logarithmic-length spanning words as its missing ingredient.

The ideaGive the root colour zero and give its two children the other two colours recursively modulo three.

Our contributionGeneric Lie-length in the traceless skew-Hermitian matrix Lie algebra grows logarithmically for every fixed generator count at least two.

Verification details
Release source recordProved in Lean

Open problem

Generic logarithmic Lie-length of su(n)

Yifan Jia and Angela Capel, A generic quantum Wielandt's inequality, arXiv:2301.08241v3, Quantum 8 (2024) 1331, Conjecture 6, p. 19:

Let $S$ be a random Lie-generating set of $\mathfrak{su}(n)$, then $\mathtt{Lie}\ell(S)=\Theta(\log n)$ almost surely.

Definition 3, p. 17, counts generator letters in right-nested brackets. It takes $U_1=\mathrm{span}(S)$, $U_{k+1}=\mathrm{span}[S,U_k]$, and the least positive $L$ for which the cumulative layers span the Lie algebra. The number $g\ge2$ of generators is fixed independently of $n$. The random law is Lebesgue measure in real linear coordinates on $\mathfrak{su}(n)^g$; the almost-everywhere assertion also applies to probability laws absolutely continuous with respect to this measure (by the definition of absolute continuity).

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | $\mathfrak{su}(n)$ | GenericLieLengthFiltration.suSet | Real submodule with $X^*=-X$ and $\operatorname{tr}X=0$. | | Right-nested brackets and cumulative layers | layer, lieLength, rightWordEval, rightSpan | layer at zero is bottom; at one it is the real span of the generators. | | Fixed generator count | GenericLieLength.claim | For every $g\ge2$, positive constants precede the dimension quantifier. | | Dimensions | claim | Every $n:\mathbb N$ with $2\le n$; dimension and Lie-length are cast to $\mathbb R$ in the estimates. | | Random, almost surely | suCoordinateMap, suPairMeasure, claim | Measure.map of coordinate Lebesgue measure; a null exceptional set at each dimension. | | $\Theta(\log n)$ | GenericLieLength.result | $c_g=1/\log(g+1)$ and $C=32/\log 2$ satisfy both inequalities. | | Lie-generating | GenericLieLength.ae_span_top_of_witness with the explicit spanning pair | Generation follows almost everywhere; it is not an extra hypothesis that narrows the random law. |

The layer recurrence advances from $k+1$ to $k+2$, so the successor rule is asserted for $1\le k$. Matrix units are Mathlib's Matrix.single. The dimension quantifier and real casts are as stated in the table. These conventions avoid applying a positive-layer recurrence to the zero layer.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/GenericLieLength.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2023Coefficientwise log-concavity and log-convexity of Catalan-power seriesBoth coefficient signs of Karp and Zhang's Conjecture 4.2.Proved

The questionDmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 8, Conjecture 4.2:

Why it mattersThe index law makes the unweighted transform log-neutral.

The ideaTheir convolution index law and its derivative at zero give

Our contributionBoth coefficient signs of Karp and Zhang's Conjecture 4.2.

Verification details
Release source recordProved in Lean

Open problem

Coefficientwise log-concavity and log-convexity of Catalan-power series

Dmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 8, Conjecture 4.2:

Assume that ${h_k}_{k=0}^{\infty}$ is log-concave (log-convex ) and independent of $\mu$. Then the function $\mu\to h(\mu;x)$ defined in (15) is coefficient-wise log-concave (log-convex ).

Formula (15) defines

$$ h(t;x)=\sum_{k\ge0}h_k a_k(t)x^k,\qquad a_k(t)=\frac{(t)_{2k}}{(t+1)_k k!}. $$

Definition 1.1, page 2, starts with a nonnegative nontrivial real sequence. The log-concave class has $h_k^2\ge h_{k-1}h_{k+1}$ for every positive integer $k$ and interval support, with the precise printed zero-support disjunction. The log-convex clause reads:

If $f_k^2\le f_{k-1}f_{k+1}$ for all $k\in\mathbb N$, the sequence is called log-convex (this inequality implies that a nontrivial sequence is strictly positive).

Here $\mathbb N={1,2,\ldots}$. The literal inequalities permit the exception supported on ${0}$; no extra strict-positivity assumption is imposed. For $\mu,\alpha,\beta\ge0$, set

$$ \delta_m=\sum_{k=0}^{m}h_kh_{m-k} \left[a_k(\mu+\alpha)a_{m-k}(\mu+\beta) -a_k(\mu)a_{m-k}(\mu+\alpha+\beta)\right]. $$

The assertion is $\delta_m\ge0$ for log-concave $h$ and $\delta_m\le0$ for log-convex $h$, for every natural degree $m$ and every nonnegative real base parameter and shifts. It concerns finite formal coefficients, without convergence or growth assumptions.

All Lean names in the table have prefix D5.S3.Analytic.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Formula (15) | KarpZhangCatalanPower.Coeff.a | Literal rising-factorial quotient, including $k!$. | | Definition 1.1, log-concave class | KarpZhangReciprocal.Result.IsLogConcave | Existing predicate, including nontriviality and exact interval support. | | Definition 1.1, log-convex class | KarpZhangCatalanPower.Result.IsLogConvex | Nonnegativity, nontriviality and adjacent inequalities at positive indices; no support restriction. | | Turanian coefficient | KarpZhangCatalanPower.Result.delta | Exactly the finite Cauchy sum above. | | Conjecture 4.2 | KarpZhangCatalanPower.Result.claim | Both implications for the same $h,\mu,\alpha,\beta,m$. | | Settlement | KarpZhangCatalanPower.Result.result | A proof of exactly claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KarpZhangCatalanPower/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2023Coefficientwise log-concavity of the two Karp--Zhang seriesBoth coefficient assertions of Karp and Zhang's Conjecture 5.2.Proved

The questionDmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 12, Conjecture 5.2:

Why it mattersThe conjecture asks for preservation of coefficientwise log-concavity by two different parameter-dependent transforms of an arbitrary log-concave sequence.

The ideaThe second derivative of the logarithm of their sum is the score variance minus the expected negative logarithmic curvature.

Our contributionBoth coefficient assertions of Karp and Zhang's Conjecture 5.2.

Verification details
Release source recordProved in Lean

Open problem

Coefficientwise log-concavity of the two Karp--Zhang series

Dmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 12, Conjecture 5.2:

Suppose that ${y_k}_{k=0}^{\infty}$ is a log-concave sequence. Then the functions $\mu\to\widetilde y(\mu;x)$ and $\mu\to\widehat y(\mu;x)$ are coefficient-wise log-concave on $\mu>0$.

The two series are

$$ \widetilde y(\mu;x)=\sum_{k=0}^{\infty}y_k\frac{(2\mu)_k}{(\mu)k k!}x^k, \qquad \widehat y(\mu;x)=\sum{k=0}^{\infty}y_k\frac{\Gamma(\mu+k)}{\Gamma(2\mu+k)k!}x^k. $$

Here $(a)_k$ is the rising factorial. Definition 1.1, page 2:

Let ${f_k}{k=0}^{\infty}$ be a non-negative nontrivial real sequence. We call ${f_k}{k=0}^{\infty}$ log-concave (or Pólya frequency two – $PF_2$, or doubly positive) if $f_k^2\ge f_{k-1}f_{k+1}$ for each $k\in\mathbb N$, and ${f_k}{k=0}^{\infty}$ is supported on an interval of integers (i.e., $f_N=0$ implies either $f{N+i}=0$ for all $i\in\mathbb N_0$ or $f_{N-i}=0$ for $i=0,\ldots,N$).

The Lean predicate retains nonnegativity, nontriviality, adjacent inequalities for positive natural indices, and the printed zero-support disjunction. Natural subtraction is used only with the stated bounds.

For either series $f$, coefficientwise log-concavity means that every coefficient of

$$ f(\mu+\alpha;x)f(\mu+\beta;x)-f(\mu;x)f(\mu+\alpha+\beta;x) $$

is nonnegative for every $\mu>0$ and every real $\alpha,\beta\ge0$. This is an assertion about finite Cauchy sums in each natural degree, so it requires no convergence or growth assumption on $y$.

Formal correspondence. All Lean names below have prefix D5.S3.Analytic.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Definition 1.1 | KarpZhangReciprocal.Result.IsLogConcave | All four sequence conditions, including nontriviality and the exact zero convention. | | Reciprocal-series coefficient before multiplication by $y_k$ | KarpZhangReciprocal.Coeff.c | $(2\mu)_k/((\mu)_k k!)$. | | Reciprocal Turanian coefficient | KarpZhangReciprocal.Result.delta | The degree-$m$ finite Cauchy coefficient with both factorials retained. | | Gamma-series coefficient before multiplication by $y_k$ | KarpZhangGamma.Series.gammaCoeff | $\Gamma(\mu+k)/(\Gamma(2\mu+k)k!)$. | | Gamma Turanian coefficient | KarpZhangGamma.Series.gammaDelta | The degree-$m$ finite Cauchy coefficient of the gamma-series Turanian. | | Conjecture 5.2 | KarpZhangConjecture.Result.claim | Both coefficient inequalities for the same $y,\mu,\alpha,\beta,m$, universally quantified. | | Settlement | KarpZhangConjecture.Result.result | A proof of exactly claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KarpZhangConjecture/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

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    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2024Koutschan–Krattenthaler–Schlosser Conjecture 23, equation (10.12)The binomial determinant identities follow from a common column multiplier and triangular determinant reductions.Proved

The questionKoutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”.

Why it mattersThe determinant families share a rational column transformation.

The ideaThe multiplier has constant coefficient one, so its lower triangular coefficient matrix preserves determinants.

Our contributionThe binomial determinant identities follow from a common column multiplier and triangular determinant reductions.

Verification details
Release source recordProved in Lean

Open problem

Koutschan–Krattenthaler–Schlosser Conjecture 23, equation (10.12)

Koutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”. Equation (10.12) is

$G_{2,1,2,0}(n)=2G_{7,4,7,-1}(n-1)$.

The preregistration is #19228. The Library note records the source definition and all three equations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (1.2): falling factorial divided by $p!$ for $p\ge0$, zero for $p<0$ | ShiftedG.binom | Rational Ring.choose on the nonnegative branch, integer lower argument. | | Section 10 determinant | ShiftedG.G | Integer parameters, rational entries, integer powers of 2 and zero-based Fin n indices. | | “for all n ≥ 3” | ShiftedG.claim | Natural size and the original lower bound. | | Every consecutive equality of (10.12) | ShiftedG.claim and ShiftedG.result | Original coefficients, signs and size shifts; a chain is a conjunction of adjacent equalities. |

Natural subtraction is truncated at zero; within the source range it agrees with integer subtraction. The rational/integer coercions express the same source entries. There are no additional assumptions on the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KoutschanKrattenthalerSchlosserShiftedG.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Cubic Grover walks cannot have twice an odd periodNo connected 3-regular graph is 2l-periodic for an odd multiple l of 3.Proved

The questionYoshino, Regular graphs to induce even periodic Grover walks, arXiv:2307.13227v1, Discrete Mathematics 348 (2025) 114345, ask after Theorem 4.10:

Why it mattersThe obstruction acts on the exact arc evolution matrix; it requires neither a spectral classification nor a restricted family of graphs.

The ideaConnectedness provides a vertex, and degree three provides an incident arc.

Our contributionNo connected 3-regular graph is 2l-periodic for an odd multiple l of 3.

Verification details
Release source recordProved in Lean

Open problem

Cubic Grover walks cannot have twice an odd period

S. Kubota, H. Sekido and K. Yoshino, Regular graphs to induce even periodic Grover walks, arXiv:2307.13227v1, Discrete Mathematics 348 (2025) 114345, ask after Theorem 4.10:

Question 4.11. Let l be an odd integer that is a multiple of 3. Do 2l-periodic 3-regular graphs exist?

Graphs have the scope of §2.1: finite, simple, connected and undirected. An arc is an ordered adjacent pair, represented by Mathlib's G.Dart; its origin, terminus and reversal are a.fst, a.snd and a.symm. Section 2.2 defines the complex Grover matrix by

$$ U_{a,b}=\begin{cases} 2/\deg_G(t(b))-1 & a=b^{-1},\ 2/\deg_G(t(b)) & t(b)=o(a),\ a\ne b^{-1},\ 0 & t(b)\ne o(a). \end{cases} $$

The period is the least positive integer $\tau$ with $U^\tau=I$. The precise claim in issue #12338 uses $l\in\mathbb N$, Odd l, 3 ∣ l, a finite vertex type, connectedness, degree three at every vertex, and the negation of period $2l$. Oddness already excludes zero; the natural-number formulation preserves the source's positive periods.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/CubicGroverTwiceOddPeriod.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023A closed form for the generalized Stirling numbers of boson orderingThe generalized Stirling numbers with parameters (-1, 2; r), which govern boson operator orderings, have the closed form conjectured by Maier for every integer r, once the first binomial coefficient is read as the generalized binomial coefficient.Proved

The questionMaier uses the Hsu–Shiue generalized Stirling numbers to express orderings of boson creation and annihilation operators.

Why it mattersThe numbers Ŝ{n,k}(−1, 2; r) enter the ordering identities of §6 of the source.

The ideaExtend the sum to all integers j; outside the stated range the factor C(n + 1, 2j + r − 1) vanishes.

Our contributionThe generalized Stirling numbers with parameters (-1, 2; r), which govern boson operator orderings, have the closed form conjectured by Maier for every integer r, once the first binomial coefficient is read as the generalized binomial coefficient.

Verification details
Release source recordProved in Lean

Open problem

A closed form for the generalized Stirling numbers of boson ordering

Maier uses the Hsu–Shiue generalized Stirling numbers to express orderings of boson creation and annihilation operators. With (y)^{n, α} = y(y − α) ⋯ (y − (n − 1)α), the rescaled numbers Ŝ_{n,k}(α, β; r), 0 ≤ k ≤ n, are defined by (βx + r)^{n, α} = Σ_k Ŝ_{n,k}(α, β; r) C(x, k), equivalently (Theorem 4.1) Ŝ_{n,k} = Σ_{x=0}^{k} (−1)^{k−x} C(k, x) (βx + r)^{n, α}. Conjecture 5.3 of arXiv:2308.10332v4 states, as found heuristically:

For all r ∈ ℤ, Ŝ_{n,k}(−1, 2; r) = Σ_{j=⌊(2−r)/2⌋}^{⌊(n+2−r)/2⌋} C(n − j, n − k) n! C(n + 1, 2j + r − 1).

The source remarks that the upper argument n − j may be negative. Issue #10170 fixes the readings: n, k ∈ ℕ with k ≤ n, r ∈ ℤ; Ŝ is the finite sum of Theorem 4.1 with the rising factorial; C(n − j, n − k) is the generalized binomial coefficient Ring.choose; the sum runs over integers j between the stated floors.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/FockSpace/BosonOrderingStirlingClosedForm.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Martinez's mirror-step gambler's ruin limitThe mirror-step ruin probability has Martinez's conjectured limit.Proved

The questionMartinez, A symbolic computational approach to the generalized gambler's ruin problem in one and two dimensions, arXiv:2412.07667v1, Conjecture 5.1, page 16:

Why it mattersThe proof concerns the literal boundary recurrence and its limit.

The ideafinitesolution verifies the two endpoints and the three-destination recurrence. harmonicnonpos chooses an attained maximum on the finite interval.

Our contributionThe mirror-step ruin probability has Martinez's conjectured limit.

Verification details
Release source recordProved in Lean

Open problem

Martinez's mirror-step gambler's ruin limit

L. Martinez, A symbolic computational approach to the generalized gambler's ruin problem in one and two dimensions, arXiv:2412.07667v1, Conjecture 5.1, page 16:

Consider the generalization of the gambler’s ruin problem when we add a mirror step. If the particle starts at $x=1$, then $$\lim_{N\to\infty} f_N^{(p)}(1)=\frac{\sqrt{(p+1)(1-3p+4p^2)}-(1-2p)(p+1)}{2p(p+1)}$$ whenever we restrict the particle moves by either moving from $x$ to $x-1$ with probability $q_1$, or from $x$ to $x+1$ with probability $p$, or from $x$ to $N-x$ with probability $q_2$ where $q_1=q_2=\frac{1-p}{2}$.

Here $0<p<1$, and $f_N^{(p)}(x)$ is the probability of absorption at $N$. The source note records the versioned locator. The statement and conventions are preregistered in #19080.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every real $0<p<1$ | p : ℝ, 0 < p, p < 1 in claim | Full open interval | | States $0,\ldots,N$ with absorbing endpoints | f 0 = 0, f N = 1 in IsWinProbability | Values outside the interval are unrestricted | | Left step of weight $(1-p)/2$ | First term in IsWinProbability | Argument $x-1$ | | Right step of weight $p$ | Second term in IsWinProbability | Argument $x+1$ | | Mirror step of weight $(1-p)/2$ | Third term in IsWinProbability | Argument $N-x$ | | First-step equation at every interior state | ∀ x, 0 < x → x < N → … | Coincident destinations add their weights | | Starts at $x=1$, $N\to\infty$ | Filter.Tendsto (fun N => f N 1) Filter.atTop … | Natural-number limit | | The displayed radical | limitValue p | Literal numerator, denominator and positive square root | | Winning-probability family | ∀ f, satisfying the recurrence for every $N≥2$ | Uniqueness identifies every such family |

The universal family quantifier introduces no extra restriction on the source's solution: existence follows from the finite formula, and harmonic_unique proves uniqueness on the interval. The lower bound $N≥2$ ensures the starting point is interior and removes only finitely many indices from the limit. Natural subtraction is truncated, and at interior points agrees with integer subtraction. Conjecture 5.1 assigns $p$ to the right step; the symmetric example in Section 4 assigns $p$ to the mirror step. The claim uses Conjecture 5.1's assignment.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Probability/MartinezMirrorGamblersRuin.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsConjecture provedFirst preprint 2024Mathar's Conjecture 1 on two non-bonding dominoesTwo dominoes on the r × c board are non-bonding when every square of one is at L1 distance at least 2 from every square of the other, so that they share at most a corner point: hard dimers with nearest-neighbour exclusion.Proved

The questionMathar (arXiv:2404.18806, 2024) counts placements of dominoes on an r × c board that are non-bonding: every square of one is at L1 distance at least 2 from every square of another.

The ideaEvery domino is horizontal or vertical and is determined by its anchor square.

Our contributionTwo dominoes on the r × c board are non-bonding when every square of one is at L1 distance at least 2 from every square of the other, so that they share at most a corner point: hard dimers with nearest-neighbour exclusion.

Verification details
Release source recordProved in Lean

arXiv

Mathar's Conjecture 1 on two non-bonding dominoes

R. J. Mathar (arXiv:2404.18806, 2024) counts placements of dominoes on an r × c board that are non-bonding: every square of one is at L1 distance at least 2 from every square of another. They are hard dimers with nearest-neighbour exclusion. His Conjecture 1 (equation (21)) reads:

D(r,c,2) = 2c²r² − 2(cr² + c²r) + ½(r² + c²) − 22cr + (59/2)(c + r) − 30; r, c ≥ 3.

Issue #11068 fixes the reading. A domino is a set of two board squares at L1 distance 1, and D(r, c, 2) counts the sets of two pairwise non-bonding dominoes. The identity holds in ℚ for all r, c ≥ 3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/HardCore/NonBondingDominoPairs.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Gibonacci power-window gcd normalizationInteger seed content scales the gcd of all shifted power windows by its exponent.Proved

The questionSection 6.1, Question 6.2, PDF p. 14 of arXiv:2204.12626v2 asks:

Why it mattersThe paper gives the corresponding seed-content scaling for first and second powers and asks whether it persists for every higher power.

The ideaThe set-gcd universal property, cancellation by a positive scalar and nonnegative divisibility antisymmetry give scaling of the whole-family gcd.

Our contributionInteger seed content scales the gcd of all shifted power windows by its exponent.

Verification details
Release source recordProved in Lean

Open problem

Gibonacci power-window gcd normalization

Section 6.1, Question 6.2, PDF p. 14 of arXiv:2204.12626v2 asks:

Fix $G_0, G_1 \in \mathbb{Z}$ and set $d := \gcd(G_0,G_1)$. Consider the two Gibonacci sequences $(G_n){n\geq 0}$ and $(G_n^\prime){n\geq 0}$, where $(G_n^\prime)_{n=0}^\infty$ is generated by the relatively prime initial conditions $G_0^\prime = \frac{G_0}{d}$ and $G_1^\prime = \frac{G_1}{d}$. Then the following identities holds:

$\mathcal G_{G_0,G_1}(k)=d\cdot\mathcal G_{G'0,G'1}(k)$; $\mathcal G^2{G_0,G_1}(k)=d^2\cdot\mathcal G^2{G'_0,G'_1}(k)$.

The first formula holds from Guyer-Mbirika [4, Theorem 19], while the second formula holds from Theorem 3.6 in this current paper. Will it be the case that $\mathcal{G}{G_0, G_1}^n!(k) = d^n \cdot \mathcal{G}{G_0^\prime, G_1^\prime}^n!(k)$ for all $n \geq 3$?

For every signed integer seed pair $(a,b)≠(0,0)$, every natural $k\geq1$ and every natural $e\geq3$, let $G(0)=a$, $G(1)=b$ and $G(n+2)=G(n+1)+G(n)$. Define the nonnegative gcd of the entire shift family by

$$C_e(a,b;k)=\gcd_{t\in\mathbb N}\left(\sum_{i=1}^{k}G(t+i)^e\right).$$

With $d=\gcd(a,b)>0$, the target is

$$C_e(a,b;k)=d^e C_e(a/d,b/d;k).$$

The quotients are exact integers. Both gcds satisfy their divisibility iff for every integer divisor, including zero and negative divisors. Individual zero seeds, zero windows and odd exponents remain in scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/GibonacciPowerGcdNormalization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Geometry & algebraQuestion resolvedFirst preprint 2022Generalized doubly nonnegative cone inclusionThe generalized doubly nonnegative cone inclusion for orthant and Lorentz products.Proved

The questionThe cited article, arXiv:2204.12119v4, Section 5.1, Conjecture 5.1, asks for the inclusion of the NN-type generalized doubly nonnegative cone in the BD-type cone for products of a nonnegative orthant and second-order cones.

Why it mattersThe result compares the two specified approximations for the full orthant and Lorentz product family.

The ideaA nonnegative real homogeneous quadratic polynomial is a sum of squares of linear forms: diagonalize its positive semidefinite symmetric coefficient matrix and absorb the square roots of its nonnegative eigenvalues.

Our contributionThe generalized doubly nonnegative cone inclusion for orthant and Lorentz products.

Verification details
Release source recordProved in Lean

Open problem

Generalized doubly nonnegative cone inclusion

The cited article, arXiv:2204.12119v4, Section 5.1, Conjecture 5.1, asks for the inclusion of the NN-type generalized doubly nonnegative cone in the BD-type cone for products of a nonnegative orthant and second-order cones. The statement reads verbatim: “If K is a direct product of a nonnegative orthant and second-order cones, DNN^NN(K) ⊆ DNN^BD(K) holds.” The article notes that the case of a single second-order cone follows from its Proposition 4.12 and that the general inclusion relationship is unknown.

Formal reading

Let $p,r\ge0$ and $k_j\ge1$ for $1\le j\le r$, and put

$$K=\mathbb R_+^p\times\prod_{j=1}^r {(t,z)\in\mathbb R\times\mathbb R^{k_j}:|z|_2\le t}.$$

Let $q$ square each orthant coordinate and send each Lorentz block $(t,z)$ to $(t^2+|z|2^2,2tz)$. Write $\Sigma{n,4}$ for the convex hull of squares of homogeneous quadratic polynomials. With the standard coordinate pairing and $n=p+\sum_j(k_j+1)$, define

$$\mathcal K_{\mathrm{NN},0}(K) ={A\in\mathbb S^n:q(x)^\mathsf{T}Aq(x)\in\Sigma_{n,4}}.$$

Every $X\in\mathbb S^n$ satisfying $\operatorname{tr}(XA)\ge0$ for every $A\in\mathcal K_{\mathrm{NN},0}(K)$ is positive semidefinite and satisfies $Xs\in K$ for every extreme ray generator $s$ of $K^*$. Thus $\mathrm{DNN}^{\mathrm{NN}}(K)\subseteq \mathrm{DNN}^{\mathrm{BD}}(K)$.

| Source object | Lean counterpart | Correspondence | | --- | --- | --- | | Orthant and Lorentz product | Defs.Idx, Defs.cone | Orthant coordinates and a time coordinate plus $k_j$ spatial coordinates per block. | | Jordan square | Defs.jsq | Homogeneous quadratic polynomials with the displayed block formula. | | Quartic SOS cone | Defs.sos4 | Real convex hull of homogeneous quadratic squares. | | Zeroth approximation | Defs.quartic, Defs.knn0 | Symmetric matrices with SOS quartic pullback. | | NN-type dual | Defs.dnnNN | Symmetric trace dual of exactly this approximation. | | BD-type cone | Defs.dualCone, Defs.extremeRays, Defs.dnnBD | Positive semidefiniteness and preservation of every dual extreme ray generator. | | All product dimensions | Defs.claim, Result.result | All $p,r$ and all positive spatial dimensions; no factor is omitted. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/ConicGDNN/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2023Kurkov's binomial identity for binary-weight productsBinary-weight products satisfy Kurkov's binomial identity at every natural index.Proved

The questiona(0) = 1, and for n > 1, a(n) = (1 + A000120(n))a(floor(n/2)); also a(n) = A000005(A283477(n)).

Why it mattersSuccessive binary shifts define a product whose factors depend on digit weight.

The ideaThis derivation is a local have in each product induction.

Our contributionBinary-weight products satisfy Kurkov's binomial identity at every natural index.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's binomial identity for binary-weight products

OEIS A284005, NAME (%N, verbatim):

a(0) = 1, and for n > 1, a(n) = (1 + A000120(n))*a(floor(n/2)); also a(n) = A000005(A283477(n)).

The settled FORMULA (%F, verbatim):

Conjecture: a(2^m*(2n+1)) = Sum_{k=0..m+1} binomial(m+1, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1. - Mikhail Kurkov, Apr 24 2023

AUTHOR (%A, verbatim):

Antti Karttunen, Mar 18 2017

Here wt(n)=A000120(n) counts ones in the binary expansion, represented by (Nat.digits 2 n).sum. The NAME specifies the recurrence for n>1; the entry's data fixes a(1)=2. Accordingly, the formal definition is a(0)=1 and a(n+1)=(1+wt(n+1))*a((n+1)/2) for every natural n, where / is natural floor division. This reproduces the initial terms 1,2,4,6,8,12,18,24,16,24,36,48,54,72,96,120.

The exact Lean statement is:

theorem result (m n : ℕ) :
    a (2 ^ m * (2 * n + 1)) =
      ∑ k ∈ Finset.range (m + 2), Nat.choose (m + 1) k * a (2 ^ k * n)

Both quantifiers include zero. The sum runs from k=0 through k=m+1, and every quantity is a natural number. Only the quoted binomial conjecture is asserted: the 2019 bit-flip recursion, the 2023 mod-2 binomial-transform conjecture for A329369, and the representation A000005(A283477(n)) are not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/KurkovWeightProductBinomial.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    Dated conjecture statement: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresRecurrence provedSource 2024The Alternating Stirling Weighted Sum of Even-Index Fibonacci and Lucas NumbersThe alternating factorial-Stirling weighted sum of the even-index terms of a Fibonacci recurrence equals the signed weighted sum of the preceding Stirling row.Proved

The questionParisse, Integers 24 (2024), Article A70, Section 4, states two conjectures.

Why it mattersNeither identity depends on the initial values of the sequence, so the recorded proof establishes the common statement for an arbitrary integer sequence satisfying u{n+2} = un + u{n+1} and reads off the two conjectures as the Fibonacci and Lucas instances.

The ideaThis is the classical Fubini polynomial, also called the ordered Bell polynomial; Qn(1) is the number of ordered partitions of an n-element set.

Our contributionThe alternating factorial-Stirling weighted sum of the even-index terms of a Fibonacci recurrence equals the signed weighted sum of the preceding Stirling row.

Verification details
Release source recordProved in Lean

Open problem

The Alternating Stirling Weighted Sum of Even-Index Fibonacci and Lucas Numbers

Parisse, Integers 24 (2024), Article A70, Section 4, states two conjectures. The first, after Equation (4.7):

Conjecture 1. For all ℓ ∈ N_0, we have

∑_{m=0}^{ℓ} (-1)^m m! S(ℓ+1, m+1) F_{2(m+1)} = (-1)^ℓ ∑_{m=0}^{ℓ} m! S(ℓ, m) F_{m+2} .

The second, after Equation (4.9):

Conjecture 2. For all ℓ ∈ N_0, we have

∑_{m=0}^{ℓ} (-1)^m m! S(ℓ+1, m+1) L_{2(m+1)} = (-1)^ℓ ∑_{m=0}^{ℓ} m! S(ℓ, m) L_{m+2} .

Here S(j, m) is the Stirling number of the second kind, F the Fibonacci numbers with F_1 = F_2 = 1, and L the Lucas numbers with L_0 = 2, L_1 = 1. Both are the constant term of the article's own formula for the weighted sums ∑_{k=0}^{n} k^ℓ F_k and ∑_{k=0}^{n} k^ℓ L_k, rewritten through Equation (3.14), c_{ℓ,m}(0) = (-1)^m m! S(ℓ+1, m+1).

The article's remarks fix the reading of the stacked bracket symbols. The right-hand side of Conjecture 1 is named there as A000557, whose recorded formula is a(n) = ∑_{k=0..n} k! Stirling2(n, k) Fibonacci(k+2); the right-hand side of Conjecture 2, times -1, is named as A263968, whose recorded formula is a(n) = (-1)^{n+1} ∑_{k=0..n} k! Lucas(k+2) Stirling2(n, k). Both readings are therefore Stirling numbers of the second kind, not binomial coefficients.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ParisseStirlingFibonacciAlternatingSum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The alternating factorial-Stirling weighted sum of the even-index terms of a Fibonacci recurrence equals the signed weighted sum of the preceding Stirling row.

    Read source ↗
  • Historical context

    Year in the primary source attribution: 2024.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2026The overpartition threshold identityA finite identity relates the full threshold statistic on native overpartitions to total and nonoverlined part counts.Proved

The questionBhoria, Eyyunni and Santra, On the number of missing integers in partitions, arXiv:2604.12557v1, Section 4, Question 3, asks:

Why it mattersThe ordinary-partition identity in Theorem 2.9 motivates this overpartition analogue.

The ideaThe support-corner equivalence cornerEquiv proves that native Ferrers conjugation preserves the support cardinality.

Our contributionA finite identity relates the full threshold statistic on native overpartitions to total and nonoverlined part counts.

Verification details
Release source recordProved in Lean

Open problem

The overpartition threshold identity

Bhoria, Eyyunni and Santra, On the number of missing integers in partitions, arXiv:2604.12557v1, Section 4, Question 3, asks:

Is there an overpartition analogue of Theorem 2.9, namely, a concise identity for the number of integers less than or equal to the largest part that appear less than k times in all overpartitions of n.

Let $\mathcal O_n$ contain every partition of $n$ together with an arbitrary subset of its distinct positive part sizes. A chosen size has its first occurrence overlined. Write $m_r(o)$ for its total multiplicity, $S(o)$ for the chosen subset, and $L(o)$ for its largest part, with $L(o)=0$ for the empty partition. Define

$$ U_k(n)=\sum_{o\in\mathcal O_n}#{r:1\le r\le L(o),\ m_r(o)<k}, \qquad T(n)=\sum_{o\in\mathcal O_n}#\operatorname{parts}(o), $$

$$ N_j(n)=\sum_{o\in\mathcal O_n}\bigl(m_j(o)-\mathbf1_{j\in S(o)}\bigr). $$

Thus $U_k(n)$ includes absent sizes, $T(n)$ includes every occurrence, and $N_j(n)$ counts actual nonoverlined occurrences. The answer is the finite identity

$$ \boxed{U_k(n)=T(n)-2\sum_{j=k}^{n}(-1)^{j-k}N_j(n)} \qquad(k>0,\ n\ge0). $$

The equality is in the integers. Every legal marking pattern is retained, including marking every distinct size. For $k>n$ the sum is empty. The case $n=0$ is a harmless extension supplied here, not a quotation from the paper. The source asks for an analogue; it does not propose the displayed formula as a conjecture.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | All overpartitions of $n$ | Overpartition n | A native Nat.Partition n and any subset of its support. | | Integers at most the largest part | largest, U | The closed positive interval $[1,L(o)]$. | | Appearing fewer than $k$ times | multiplicity, U | Total multiplicity includes the overlined occurrence; absent sizes have multiplicity zero. | | A concise identity for every threshold | result | Every natural $k>0$ and every natural $n$, with no bounded enumeration or extra restriction. | | Counts on the right side | T, N | All occurrences and actual nonoverlined occurrences, respectively. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Partitions/OverpartitionThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The whole finite-cyclic SGB extensionThe complete finite-cyclic extension of the Section 5 SGB assertions is false at order one.Refuted

The questionDas–Erfanian–Nath, Certain topological indices and spectral properties of SGB-graphs of finite cyclic groups, arXiv:2602.07587v1, closes Section 5 with an unlabelled invitation to extend its preceding complete list to any cyclic group.

Why it mattersSubgroupGenerating.CyclicExtensionRefutation.result proves the negation of this entire universal.

The ideaThe source graph is defined directly on arbitrary finite group carriers, so it requires no identification with a divisors-only model or a labelled cyclic representative.

Our contributionThe complete finite-cyclic extension of the Section 5 SGB assertions is false at order one.

Verification details
Release source recordRefuted in Lean

Open problem

The whole finite-cyclic SGB extension

Das–Erfanian–Nath, Certain topological indices and spectral properties of SGB-graphs of finite cyclic groups, arXiv:2602.07587v1, closes Section 5 with an unlabelled invitation to extend its preceding complete list to any cyclic group. The source's ambient groups are finite. The list was established there for orders $p^n,pq,p^2q,p^2q^2$, with distinct primes $p,q$ and $n\geq1$; those families are known source results.

For every finite cyclic group $G$, including order one, let $B(G)$ have vertex type $(G\times G)\sqcup\operatorname{Subgroup}(G)$. All ordered and repeated pairs, the identity pair, and all subgroups including the trivial one remain present. Cross adjacency is exactly $\langle a,b\rangle=H$; same-part vertices are never adjacent.

The whole extension asserts the conjunction of these ten clauses:

  1. The adjacency spectrum is not integral.
  2. The Laplacian spectrum is integral.
  3. The signless Laplacian spectrum is integral.
  4. The common-neighbour spectrum is integral.
  5. $E(B(G))<|V(B(G))|$.
  6. $E(B(G))\leq E(K_{|V(B(G))|})$.
  7. $LE(B(G))\leq LE(K_{|V(B(G))|})$.
  8. $LE^+(B(G))\leq LE^+(K_{|V(B(G))|})$.
  9. $E_{CN}(B(G))\leq E_{CN}(K_{|V(B(G))|})$.
  10. $E(B(G))\leq LE(B(G))$.

Here $A$ is literal adjacency, $D$ is diagonal degree, $L=D-A$, $Q=D+A$, and $CN$ counts common neighbours off the diagonal and is zero on it. Spectra are actual symmetric-matrix spectra with algebraic multiplicity. Integral means every real eigenvalue is the real image of an integer. Energies sum absolute eigenvalues; only $LE,LE^+$ subtract the whole graph's $2m/N$. Each benchmark uses its respective energy of an actual same-carrier complete graph. Thus no separate formula or indexing transport is assumed for a benchmark.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SubgroupGenerating/CyclicExtensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Daugherty et al., Latin tableaux Conjecture 6.3For every integer k at least five, let lambda=(k,k-1,...,5,4,3,3).Proved

The questionFor every integer k at least five, let lambda=(k,k-1,...,5,4,3,3).

Why it mattersThe source lists this family as a separate open problem and states that for k at least six its Latin property is not supplied by the preceding results.

The ideaSource.sourcerowsymbols proves from the full conjugate type that a row of length m contains exactly integers 1 through m.

Our contributionFor every integer k at least five, let lambda=(k,k-1,...,5,4,3,3).

Scope & next questionConjectures 6.1 and 6.2 and the general wide-partition conjecture remain outside this result; no consequence about their truth is claimed.

Verification details
Release source recordProved in Lean

arXiv

Daugherty et al., Latin tableaux Conjecture 6.3

For every integer k at least five, let lambda=(k,k-1,...,5,4,3,3). Conjecture 6.3 of arXiv:2610.29291v1 states LT(lambda)=9*4^(k-3). The source count is the number of all fillings of the fixed left-aligned cells by actual integer symbols, with no repetition in any row or column, and type lambda-prime. It is not the reduced count. Both bottom rows remain distinct positions. There is no upper cutoff or additional wide-partition hypothesis.

The exact Lean statement is Counting.conjecture63_integer: for k : Int with 5 <= k, Source.LT k.toNat = 9 * 4^(k.toNat - 3). On this domain conversion to Nat is exact. Source.shape k.toNat is precisely the source diagram. Source.SourceLatin imposes the conjugate multiplicity equation for every integer, including symbols outside the positive row intervals.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Latin/Tableaux/Counting.conjecture63_integer

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The spatiotemporal Born rule for the LvN distribution has exactly two casesA universal sequential Born representation exists exactly for a maximally mixed state or a discard-and-prepare channel.Proved

The questionWu, The spatiotemporal Born rule is quasiprobabilistic, arXiv:2507.16919v2; published as Quasiprobabilistic nature of the spatiotemporal Born rule, Phys.

Why it mattersA fixed bipartite operator gives a trace pairing additive in each measurement argument.

The ideaOtherwise two distinct eigenvalues provide a comparison vector for every eigenspace.

Our contributionA universal sequential Born representation exists exactly for a maximally mixed state or a discard-and-prepare channel.

Scope & next questionStability under approximate Born representations and extensions to infinite dimensions remain open here. - Both branches are needed.

Verification details
Release source recordProved in Lean

Open problem

The spatiotemporal Born rule for the LvN distribution has exactly two cases

J. Fullwood, Z. Ma and Z. Wu, The spatiotemporal Born rule is quasiprobabilistic, arXiv:2507.16919v2; published as Quasiprobabilistic nature of the spatiotemporal Born rule, Phys. Rev. A 113, 032204 (2026). Section VI, final paragraph (PRA_rev.tex, line 577), asks the following. The source macro \E denotes $\mathcal E$.

It is then natural to ask whether $\rho$ being maximally mixed or $\E$ being a discard-and-prepare channel is also a necessary condition for the existence of the spatiotemporal Born rule for the LvN distribution. While it is straightforward to show that this is indeed the case when $A$ consists of a single qubit, for systems of arbitrary dimension we leave it as an open question.

Issue #21996 fixes the statement for all positive finite input and output dimensions. For every $d,e\ge1$, every density matrix $\rho\in M_d(\mathbb C)$ and every CPTP complex linear map $\mathcal E:M_d(\mathbb C)\to M_e(\mathbb C)$,

$$ \begin{aligned} &\exists W\in M_{de}(\mathbb C);\forall P,Q\text{ orthogonal projections},\quad \operatorname{tr}[W(P\otimes Q)] =\operatorname{tr}[\mathcal E(P\rho P)Q] \ &\qquad\Longleftrightarrow\quad \rho=I_d/d\quad\text{or}\quad \exists\sigma\in M_e(\mathbb C),\quad \sigma\succeq0,\quad\operatorname{tr}\sigma=1,\quad \forall X\in M_d(\mathbb C),\quad\mathcal E(X)=\operatorname{tr}(X)\sigma. \end{aligned} $$

Here $P$ acts on the input and $Q$ on the output; an orthogonal projection is a self-adjoint idempotent. A density matrix is positive semidefinite with trace one. Complete positivity means positivity of every finite amplification, and trace preservation is required on every matrix. The same $W$ must work for all projections, without a positivity, Hermiticity or normalization assumption on $W$. This equivalence is proved.

The source defines the LvN distribution by $\mathbf P(i,j)=\operatorname{tr}[\mathcal E(P_i\rho P_i)Q_j]$ in Section II (PRA_rev.tex, lines 247–250), and its universal Born representation in Section VI, lines 534–537. The literature note supplies the corresponding source locators.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/SpatiotemporalBornRuleDichotomy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresEquivalence provedSource 2025Binary automatic apwenian sequences have finite level generatorsBinary automatic apwenian sequences arise as the parity of successive whole levels of one finite sum-equivalent morphism.Proved

The questionYing-Jun Guo and Guo-Niu Han, On a family of automatic apwenian sequences, Discrete Mathematics 348(5), 114399 (2025), Section 3, Conjecture 1, printed page 6 of the author manuscript:

Why it mattersIt permits several odd letters and places no fixed-point or prolongability requirement on the generated sequence.

The ideaAdjoin a fresh zero-loop state z to the source DFAO.

Our contributionBinary automatic apwenian sequences arise as the parity of successive whole levels of one finite sum-equivalent morphism.

Scope & next questionIt makes no claim about central-bottleneck status, later citation impact or global priority.

Verification details
Release source recordProved in Lean

Open problem

Binary automatic apwenian sequences have finite level generators

Ying-Jun Guo and Guo-Niu Han, On a family of automatic apwenian sequences, Discrete Mathematics 348(5), 114399 (2025), Section 3, Conjecture 1, printed page 6 of the author manuscript:

Let {u(n)}ₙ≥₀ be a 2-automatic apwenian sequence over {0,1}. Then there exist a finite alphabet Σ ⊂ N and a sequence σ over A(Σ) such that {u(n)}ₙ≥₀ = ϕ(σ) (mod 2).

The full domain consists of every binary sequence computed by a finite DFAO on canonical binary expansions in most-significant-first order, with u(0) = 1 and u(n) ≡ u(2n+1) + u(2n+2) modulo two for every nonnegative n. The alphabet Σ is a finite set of natural numbers containing one. Every member of A(Σ) takes each letter of Σ to exactly two letters of Σ whose sum is congruent to the original letter modulo two. Set X₀ = [1] and Xⱼ₊₁ = σⱼ(Xⱼ). The generated sequence ϕ(σ) is the concatenation X₀X₁X₂⋯, and the conclusion is equality of its parity with u at every nonnegative index.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Apwenian/GuoHanGeneration.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Binary automatic apwenian sequences arise as the parity of successive whole levels of one finite sum-equivalent morphism.

    Read source ↗
  • Historical context

    Year in the primary source attribution: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture refutedSource 2026Self-conjugate even polynomial Conjecture 2Every odd power of X^4+X^2 from the third onward obstructs the proposed sparse-power characterization.Refuted

The questionThe sole target is Section 2, Conjecture 2 of the versioned source identified by the literature note, PDF page 4 immediately after Theorem 9:

Why it mattersTranslation-invariant polynomial algebra gives a direct way to test the proposed sparse characterization.

The ideaFor every odd natural r at least three, Frobenius gives U(X+1)=U(X), hence P(r)(X+1)=P(r)(X).

Our contributionEvery odd power of X^4+X^2 from the third onward obstructs the proposed sparse-power characterization.

Verification details
Release source recordRefuted in Lean

arXiv

Self-conjugate even polynomial Conjecture 2

The sole target is Section 2, Conjecture 2 of the versioned source identified by the literature note, PDF page 4 immediately after Theorem 9:

Let p(x) be an even polynomial over GF(2). Then p(x) is self-conjugate if and only if p(x)=x^{2^{k_1}}+x^{2^{k_2}}+...+x^{2^{k_m}}, where m is odd if x^0 is a term and m is even otherwise.

The source defines self-conjugacy by f(x)=f(x+1), as equality of formal polynomials. For each of the six explicit conventional readings e of even, let Domain(e,p) mean, respectively: every supported exponent is even; the natural degree is even; the support cardinality is even; the constant coefficient is zero; X divides p; or p(-X)=p(X). For each such e the complete published assertion is [ \forall p\in\mathbb F_2[X],\quad \operatorname{Domain}(e,p)\Longrightarrow \left[p(X+1)=p(X)\ \Longleftrightarrow
\exists m\in\mathbb N,\ k:\operatorname{Fin}(m)\to\mathbb N, \ p=\sum_{i\in\operatorname{Fin}(m)}X^{2^{k(i)}} \ \land\ \operatorname{SourceParity}(p,m)\right], ] where SourceParity(p,m) is the conjunction of the implications from nonzero constant coefficient to odd m and from zero constant coefficient to even m. The indices may repeat; m=0 and k(i)=0 remain allowed. The optional-constant variant separately adds an arbitrary c in the same field as C(c) to the sum. It is robustness of the witnesses and does not repair the literal display.

The Lean claim is the existential disjunction, over these six readings, of the complete published iff or its optional-constant counterpart. Thus its literal negation refutes every complete published iff and every separately labelled robustness counterpart. There is no auxiliary family condition in the claim whose failure could stand in for a source refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/SelfConjugateEvenPolynomialRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresConjecture refutedSource 2026Nonuniqueness of the fractional concavity transitionAn admissible polynomial forcing produces two distinct zeros of the second derivative.Refuted

The questionThe source explicitly states that uniqueness remains open.

Why it mattersIts precise conclusion is the negation of the supplied conjecture.

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionAn admissible polynomial forcing produces two distinct zeros of the second derivative.

Scope & next questionThe source explicitly states that uniqueness remains open.

Verification details
Release source recordRefuted in Lean

Open problem

li morgan siegrist 2026 inflection points conjecture 6 1

Li, Morgan and Siegrist, Concavity structure of fractional differential equations and zeros of Mittag–Leffler functions, Conjecture 6.1, asks whether every solution under Theorem 5.2 has exactly one sign-changing zero of its second derivative in $(0,1)$. The source explicitly states that uniqueness remains open. The verbatim source and literature screen are in #18772, and the source locators are in the Library note.

Statement echo

| Source clause | Formal counterpart | Interpretation | | --- | --- | --- | | Definition 1.1, the left-sided Riemann–Liouville fractional integral | rlIntegral | Lower endpoint zero, Real.Gamma, real powers and the interval integral; positive order and positive argument give the source interpretation. | | Equation (7), the solution and its boundary constant | solution | The source expression with the boundary constant substituted. Equivalence with the fractional Sobolev weak solution is supplied by the paper. | | $1<\alpha<2$ | Two order hypotheses of claim | All real orders in this interval. | | $f\in C([0,1])\cap C^1((0,1])$ | First two conjuncts of Hypotheses | ContinuousOn on the closed interval and ContDiffOn on the half-open interval. | | $f(1)=0$, $f\geq0$, $f\not\equiv0$ | Next three conjuncts of Hypotheses | Endpoint equality, pointwise nonnegativity and one nonzero point. | | Weighted right derivative limit zero | Last conjunct of Hypotheses | deriv f, right-sided neighborhood filter and exponent $(3-\alpha)/(2-\alpha)$. | | Exactly one zero in $(0,1)$ | Existence and uniqueness conjuncts of claim | The derivative is iteratedDeriv 2; every interior zero equals the selected zero. | | Changes sign at that zero | Last conjunct of claim | Negative product at every sufficiently close pair on opposite sides. |

There are no additional hypotheses on the public conclusion. Lean's integral and real powers are totalized; the displayed source interpretation uses their stated domain conditions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/LiMorganSiegristInflectionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresConjecture provedHanna's A365574 Lagrange parity conjectureTriangular inversion determines Hanna's integral series and dyadic odd support.Proved

The questionA(x) satisfying [x^(n-1)] (1 + (n+1)xA(x))^n / A(x)^n = n(n+2)^(n-2) for n > 1.

The ideaThe rational images are canonicalR and canonicalA.

Our contributionTriangular inversion determines Hanna's integral series and dyadic odd support.

Scope & next questionThat derivation, Conjecture 2 modulo 16, and any analytic convergence assertion remain open here.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A365574 Lagrange parity conjecture

Paul D. Hanna's OEIS A365574 states:

Expansion of g.f. A(x) satisfying [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) for n > 1.

Conjecture 1: a(k) is odd iff k = 2^n - 2 for n >= 1.

The literal binding conventions and source screen are in #18829. All series are formal; no analytic convergence is assumed.

Statement echo

All unqualified names below belong to D5.S3.Combinatorics.PowerSeries.HannaLagrangeParity.

| Source clause or binding convention | Lean counterpart | Fidelity | | --- | --- | --- | | Normalization and every diagonal identity for $n>1$ | IsGF | Rational formal series, $A(0)=1$, and the literal coefficient expression are retained. | | Odd exactly at $k=2^n-2$, $n\ge1$ | claim, result | Every natural $k$, including zero, has an integer coefficient witness and the biconditional. | | Existence and integrality | claim, result | These rule out vacuity and supply the integer meaning of oddness; they add no hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PowerSeries/HannaLagrangeParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedZero-Prepended First Sums and Odd PartsZero-prepended first sums of increasing positive lists correspond to odd-part partitions.Proved

The questionConjecture: Reversed integer partitions of this type (with weakly increasing 0-based partial alternating sums) are counted by A000009.

Why it mattersThis is a first-tier 2026 OEIS comment conjecture.

The ideaReuse FirstSumsPartitionCharacterization.firstSums, the existing adjacent sums of a list.

Our contributionZero-prepended first sums of increasing positive lists correspond to odd-part partitions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Zero-Prepended First Sums and Odd Parts

OEIS A392694, revision 7 (January 25, 2026), states:

Conjecture: Reversed integer partitions of this type (with weakly increasing 0-based partial alternating sums) are counted by A000009.

A reversed partition is a weakly increasing positive integer list y. The alternating recurrence is s0=0, sj=yj-s(j-1), and the condition is that 0,s1,...,sk is weakly increasing. A000009(n) counts distinct-part partitions of n. The claim has no shift.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/ZeroPrependedFirstSumsOddParts.card_zeroPrependedFirstSums_eq_distincts

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresRecurrence provedFirst preprint 2025Thue-Morse reduced abelian complexity: the odd-length recurrenceThe all-start reduced abelian complexity of the Thue-Morse word obeys the odd-index recurrence and equals three at every power of two plus one.Proved

The questionThis dossier deliberately anchors only the equality rho^{ab,red}t(2n+1) = rho^{ab,red}t(n+1) for every nonnegative integer n, proposed in Section 3 (Conclusion), page 15 of arXiv:2509.16034v1 and printed page 14 of the verified journal version.

Why it mattersIt provides the exact odd recurrence needed for this external proposition, making the all-start interpretation explicit rather than depending on a finite sampled prefix.

The ideaThe formal thueMorse is zero-indexed binary digit parity; factor length start ranges over every natural start, runCompress uses List.destutter, and R length counts the resulting reduced Parikh classes.

Our contributionThe all-start reduced abelian complexity of the Thue-Morse word obeys the odd-index recurrence and equals three at every power of two plus one.

Scope & next questionThe full recursion and four further questions in the source paragraph, namely the nonzero sign of rho(4n+2)-rho(4n), a recursion for rho(4n), equation (11), and non-k-automaticity of sequence (10), are deliberately out of scope.

Verification details
Release source recordProved in Lean

Combinatorics on words

Thue-Morse reduced abelian complexity: the odd-length recurrence

The number of reduced abelian classes at length 2n+1 equals the count at length n+1, for every nonnegative n.

This dossier deliberately anchors only the equality rho^{ab,red}_t(2n+1) = rho^{ab,red}_t(n+1) for every nonnegative integer n, proposed in Section 3 (Conclusion), page 15 of arXiv:2509.16034v1 and printed page 14 of the verified journal version. The shared Library note records the journal citation. The source sentence is:

Although it appears that rho^{ab,red}_t(2n+1) = rho^{ab,red}_t(n+1) for nonnegative integers n, the problem of determining a full recursion for rho^{ab,red}_t(n) seems to be challenging.

red(w) collapses every maximal constant run; reduced abelian complexity counts all length-n factors up to rearrangement of their reduced words with equal reduced length. The full recursion and four further questions in the source paragraph, namely the nonzero sign of rho(4n+2)-rho(4n), a recursion for rho(4n), equation (11), and non-k-automaticity of sequence (10), are deliberately out of scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Complexity/ThueMorseReducedAbelianOdd.reducedAbelianComplexity_odd

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The all-start reduced abelian complexity of the Thue-Morse word obeys the odd-index recurrence and equals three at every power of two plus one.

    Read source ↗
  • Historical context

    First preprint year: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2025Wang Conjecture 3.26: finite algebraic constant does not bound the relative constantsOne infinite commutative periodic semigroup over Q has algebraic Davenport constant at most one and unbounded relative constants in its actual Green quotient.Refuted

The questionGuoqing Wang, Semigroup rings and algebraically independent sequences with respect to idempotents in commutative semigroups, arXiv:2509.18724v1, Conjecture 3.26, states:

Why it mattersThe algebraic constant measures cancellation of products, while relative irreducibility can also detect independent semilattice coordinates.

The ideaThe carrier is one fixed infinite commutative monoid

Our contributionOne infinite commutative periodic semigroup over Q has algebraic Davenport constant at most one and unbounded relative constants in its actual Green quotient.

Verification details
Release source recordRefuted in Lean

arXiv

Wang Conjecture 3.26: finite algebraic constant does not bound the relative constants

Guoqing Wang, Semigroup rings and algebraically independent sequences with respect to idempotents in commutative semigroups, arXiv:2509.18724v1, Conjecture 3.26, states:

Let S be a commutative periodic semigroup, and let R be a commutative unitary ring. If d(S,R) < infinity then sup {D_bar(a)(S/H): a in S and <a> is not a group} < infinity.

Conjecture 5.3 is a separate finite-semigroup assertion and is not a second settlement of this question.

| Source clause | Formal counterpart | | --- | --- | | every commutative semigroup, including infinite and nonunital ones | S : Type, [AddCommSemigroup S] | | periodic | Periodic S: each native AddSubsemigroup.closure {a} is finite | | positive cyclic generation | cyclic a, without adjoining a zeroth multiple | | actual cyclic idempotent | IsCyclicIdempotent and cyclicIdempotent; pointwise existence uses finite cyclic closure | | every commutative unitary ring | R : Type, [CommRing R]; no field hypothesis | | ordinary semigroup algebra | upstream AddMonoidAlgebra R S, retaining all labels | | all independently nonzero positional coefficients | GoodSequence: every c : N -> R, with c i != 0 for each occupied position | | finite algebraic constant | algebraicD S R < top, supremum of finite good lengths | | actual principal ideals | principalIdeal a = {a} union {a+c : c in S} | | Green quotient | native quotient of greenCon S, whose relation is equality of those ideals | | cyclic nongroup, with its own possible identity | negation of CyclicIsGroup a, quantifying identity and inverses inside cyclic a | | relative constant at the canonical image | relativeD ((greenCon S).toQuotient a) | | relative irreducibility | no shorter sublist, including empty, has the same sum | | one uniform bound over all filtered elements | nongroupSup S < top | | whole implication and exact negation | closed claim : Prop; designated result : Not claim |

All supremums are in ENat. Infinity means unbounded finite lengths; no infinite product or finite truncation substitutes for the source assertion. Nonempty products require no algebra unit. Length zero contributes only zero to the length supremum. The empty sequence sums only to an existing two-sided identity. In a monoid this is exactly the usual list sum, so the empty proper subsequence is included in the counterexample's irreducibility. Lists enumerate the source's unordered sequences and their submultisets; commutativity makes enumeration irrelevant.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Algebra/WangConjecture326Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Positivity of the Two-Color Partition Series D'(2,2,n)The two-color partition series D-prime at parameters 2 and 2 has nonnegative coefficients.Proved

The questionAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 3, Conjecture 3.

The ideaTheorem 4 of the paper and Theorem 1 for k = m = 2 (a finite Heine-type q-binomial identity) express the series through q²/(1 − q)² minus a Lambert series and the product (q², q⁴; q²)∞ / (q; q²)∞².

Our contributionThe two-color partition series D-prime at parameters 2 and 2 has nonnegative coefficients.

Verification details
Release source recordProved in Lean

Open problem

Positivity of the Two-Color Partition Series D'(2,2,n)

George E. Andrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 3, Conjecture 3. With Σ_n D'(k,m,n) qⁿ = Σ_{j≥0} q^{m(2j+2)} (q^{2j+4}, q^{2j+2+2k}; q²)_∞ / (q^{2j+3}; q²)_∞² (equation (1.3)), the conjecture states that D'(2,2,n) ≥ 0 for every n ≥ 0. Equivalently, among the two-color partitions of Definition 2 with k = m = 2, those with an odd number of even parts above the smallest part never outnumber those with an even number.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TwoColorPartition/AndrewsElBachraouiThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraEquivalence provedFirst preprint 2026A partition strictly between group isomorphism and simultaneous power equivalenceA common refinement strictly between isomorphism and simultaneous power equivalence on all finite groups.Proved

The questionProblem 6.2 of Fernandes–Qureshi–Reis–Ribas, arXiv:2609.20516v1, asks for a partition P on all finite groups with P₁<P<P₂, P₂<P<P₅ or P₅<P<P₆.

Why it mattersThe first interval asks whether simultaneous power maps forget group information that can be restored partially without recovering isomorphism.

The ideaDefine R(G,H) to be simultaneous power equivalence and agreement on abelianity.

Our contributionA common refinement strictly between isomorphism and simultaneous power equivalence on all finite groups.

Verification details
Release source recordProved in Lean

Open problem

A partition strictly between group isomorphism and simultaneous power equivalence

Problem 6.2 of Fernandes–Qureshi–Reis–Ribas, arXiv:2609.20516v1, asks for a partition P on all finite groups with P₁<P<P₂, P₂<P<P₅ or P₅<P<P₆. The exact wording and definitions are in the literature note. Here P₁ is group isomorphism and P₂ requires one bijection preserving all integer powers simultaneously. The strict order is finer-to-coarser.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/PowerInvariantRefinement.finite_source_endpoint

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    A common refinement strictly between isomorphism and simultaneous power equivalence on all finite groups.

    Read source ↗
  • Historical context

    First preprint year: 2026.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSource 2025Purely Automatic Apwenian Sequences over an Alphabet with One Odd LetterPurely automatic sequences over a finite alphabet whose only odd letter is one are apwenian exactly when their parity is period doubling.Proved

The questionYing-Jun Guo and Guo-Niu Han, On a family of automatic apwenian sequences, Discrete Mathematics 348 (2025) 114399, Conjecture 2.

The ideaSince 1 is the only odd letter, b(n) = a(n) mod 2 equals 1 exactly when a(n) = 1, so every parity-one position carries a copied block of the iterated morphism.

Our contributionPurely automatic sequences over a finite alphabet whose only odd letter is one are apwenian exactly when their parity is period doubling.

Verification details
Release source recordProved in Lean

Open problem

Purely Automatic Apwenian Sequences over an Alphabet with One Odd Letter

Ying-Jun Guo and Guo-Niu Han, On a family of automatic apwenian sequences, Discrete Mathematics 348 (2025) 114399, Conjecture 2. A sequence a over ℕ is apwenian if a(0) = 1 and a(n) ≡ a(2n+1) + a(2n+2) (mod 2) for every n. Let Σ be a finite set of nonnegative integers whose only odd element is 1, and let a be a fixed point of a p-uniform morphism on Σ with p ≥ 2. The conjecture states that a is apwenian if and only if a(n) ≡ P(n) (mod 2) for every n, where P is the period-doubling sequence, the fixed point of 1 ↦ 10, 0 ↦ 11.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Apwenian/GuoHan.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2010Collinear triples in the four-point gridCollinear triples in every finite Cartesian power of the four-point grid have an exact count.Proved

The question%N A178294 Number of collinear point triples in a 4 X 4 X 4 X... n-dimensional cubic grid.

Why it mattersThe target is the full count for every dimension, with the OEIS object represented as unordered three-element subsets of the four-point Cartesian grid.

The ideaThe proof uses a bijective argument.

Our contributionCollinear triples in every finite Cartesian power of the four-point grid have an exact count.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Collinear triples in the four-point grid

OEIS A178294 records:

%N A178294 Number of collinear point triples in a 4 X 4 X 4 X... n-dimensional cubic grid.

%A A178294 R. H. Hardin, suggested by R. J. Mathar in the Sequence Fans Mailing List, May 24 2010

%F A178294 Conjecture: a(n) = 8^n/2-3*4^n/2+6^n. a(n) = +18*a(n-1) -104*a(n-2) +192*a(n-3). G.f.: 4*x*(-1+7*x)/((6*x-1)*(8*x-1)*(4*x-1)). - R. J. Mathar, May 24 2010

The recurrence and generating function on the FORMULA line are consequences of the displayed closed form, rather than separate targets here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Lattices/FourGridCollinearTriples.mathar_collinear_triples

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 3/4 · 10% weight

    Dated conjecture statement: 2010. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedTrimmed Alternating Sums of Ordinary PartitionsDistinct trimmed alternating sums characterize strict partition tails and count strict partitions of n+1.Proved

The questionFor an ordinary partition q of n, list the positive parts in weakly decreasing order.

Why it mattersThe previous worker audited all 42 source xrefs and found the exact shifted statement qualified by "we appear to have" on A392703.

The ideaThe structural theorem proves that trimmed sums are distinct exactly when q.tail is strictly decreasing.

Our contributionDistinct trimmed alternating sums characterize strict partition tails and count strict partitions of n+1.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Trimmed Alternating Sums of Ordinary Partitions

For an ordinary partition q of n, list the positive parts in weakly decreasing order. Set s_0=0 and s_j=q_j-s_(j-1) in the integers. Count partitions for which s_1,...,s_k are pairwise distinct, excluding s_0. The corrected conjecture asserts that this count is the number of partitions of n+1 into distinct positive parts, for every n including zero. The unshifted A000009(n) sentence on A392698 is incorrect; the source note records the correction and the equivalent conjectural A392703 comment.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/TrimmedAlternatingPartitions.card_trimmedSums_eq_distincts

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Shapiro Conjecture 1(iii): identity words on finite treesLeaf intervals characterize identity words on every finite tree.Proved

The questionBoris Shapiro, Transposition Cayley Spectra and Subgraph Counts, arXiv:2610.14373v1, Section 4.1, page 14, Conjecture 1(iii), states:

Why it mattersThe theorem replaces the bounded verification of the local criterion by a uniform proof for all finite trees.

The ideaShapiroProbe.multiplicitytwo uses acyclicity: an edge used exactly once would transport a vertex across the cut left by deleting that edge.

Our contributionLeaf intervals characterize identity words on every finite tree.

Verification details
Release source recordProved in Lean

arXiv

Shapiro Conjecture 1(iii): identity words on finite trees

Boris Shapiro, Transposition Cayley Spectra and Subgraph Counts, arXiv:2610.14373v1, Section 4.1, page 14, Conjecture 1(iii), states:

For a tree T, a word of length 2|E(T)| in the edges of T using every edge has product 1 if and only if every edge occurs exactly twice and, for every vertex x of T, the subword formed by the edges incident to x is a non-crossing double occurrence word.

Each edge acts by the transposition of its endpoints. Multiplication uses $(\sigma\rho)(v)=\sigma(\rho(v))$. Alternating occurrences of distinct letters are crossings; nested pairs are allowed. The one-vertex tree and empty word are included.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every finite tree | n : ℕ, T : SimpleGraph (Fin n), T.IsTree in claim | Relabeling of $[n]$; no bound on $n$ | | A word in the edges | w : List (Sym2 (Fin n)); every member belongs to T.edgeFinset | No nonedge letters | | Length twice the edge count | w.length = 2 * T.edgeFinset.card | Exact equality | | Uses every edge | Every member of T.edgeFinset belongs to w | Onto edge support | | Product is identity | (w.map edgeSwap).prod = 1 | Existing endpoint transposition; literal list product | | Every edge occurs twice | w.count a = 2 for each tree edge | Exact multiplicity | | Every vertex's incident subword is noncrossing | NoncrossingDoubleOccurrence of the order-preserving membership filter | Existing Crossing predicate on the position pairs | | If and only if | Both directions of result : claim | No dropped clause |

NoncrossingDoubleOccurrence displays the source convention literally: every occurring letter has two copies and distinct letters have no crossing position pairs. claim retains the preregistered body. The proofs reuse SupportAndEdgeWordReflection.edgeSwap and NoncrossingNonnestingGraphRecurrence.Crossing.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SupportForestMoments/ShapiroTreeIdentityWords.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedSource 2025Wu's k-gonal-pyramidal complement formulaWu's formula enumerates every positive integer outside a k-gonal-pyramidal sequence for k at least nine.Proved

The questionChai Wah Wu, Algorithms for Complementary Sequences, INTEGERS 25 (2025), A95, printed page 12, Conjecture 1, states:

Why it mattersThis is an exact enumeration theorem, not an asymptotic estimate: it connects a geometrically defined integer sequence to the rank of every positive integer in its complement.

The ideaLet H=nthRoot(3,(6n)/(k-2)) and prove that H is exactly the printed real-floor cube root.

Our contributionWu's formula enumerates every positive integer outside a k-gonal-pyramidal sequence for k at least nine.

Verification details
Release source recordProved in Lean

Open problem

Wu's k-gonal-pyramidal complement formula

Chai Wah Wu, Algorithms for Complementary Sequences, INTEGERS 25 (2025), A95, printed page 12, Conjecture 1, states:

For k >= 9, (6) holds for all n >= 1.

For positive m, let P_k(m)=m(m+1)(m(k-2)-(k-5))/6, and let the complementary sequence enumerate the positive integers outside {P_k(m) | m>=1} in increasing order. Equation (6), on printed page 11, sets h=floor((6n/(k-2))^(1/3)) and gives n+h+1 when 6n is at least (k-2)h^3+3(k-1)h^2+(2k-1)h+6; if that test fails, it gives n+h-1 when 6n is at most h(h-1)(h(k-2)+k+1), and gives n+h otherwise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/WuPyramidalComplement.wu_conjecture_one

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedChen and Chen's step-m Eulerian parity positivityFor every integer step at most minus three, two-step propagation makes each positive even row strictly positive.Proved

The questionFor each m=-3,-4,-5, let T map pairs of natural numbers to integers, with T(n,1)=T(n,n)=1 for n>=1 and

Why it mattersThe finite verification through row 400 in the source does not prove the unbounded claim.

The ideaConstruct a total solution recursively in the row index.

Our contributionFor every integer step at most minus three, two-step propagation makes each positive even row strictly positive.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Chen and Chen's step-m Eulerian parity positivity

For each m=-3,-4,-5, let T map pairs of natural numbers to integers, with T(n,1)=T(n,n)=1 for n>=1 and

T(n,k)=(m*(n-k)+1)*T(n-1,k-1)+(m*(k-1)+1)*T(n-1,k)

for 1<k<n. The conjectures in OEIS A392664, A392665 and A392666 assert T(n,k)>=0 for every positive even n and every 1<=k<=n. The exact Lean claim conjoins, for each parameter, existence of a solution and this assertion for every solution. Row zero and entries outside the triangle are unconstrained, and no uniqueness of the total extension is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/ChenChenParityPositivity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationConjecture provedFirst preprint 2012Beccaria–Hagendorf endpoint-density conjecture for the open staggered M1 chainJordan-Wigner fermions, hard-core compression and the M1 boundary sum rule.Proved

The questionBeccaria and Hagendorf, arXiv:1206.4194v2, §3.2.3, printed page 13:

Why it mattersThe boundary sum rule relates two local occupation probabilities for arbitrary chain length.

The ideaThe hard-core subtype uses the existing admissible-word predicate AdmissibleCount.Adm.

Our contributionJordan-Wigner fermions, hard-core compression and the M1 boundary sum rule.

Scope & next questionThe source's deduction of last-site thermodynamic expressions from its conjectured first-site expressions therefore has an established finite-size algebraic relation; the analytic and first-site conjectures remain open.

Verification details
Release source recordProved in Lean

Open problem

Beccaria–Hagendorf endpoint-density conjecture for the open staggered M1 chain

Beccaria and Hagendorf, arXiv:1206.4194v2, §3.2.3, printed page 13:

As similar pattern is found by probing if a particle is present on the last site. The data is consistent with $\rho_n^{(N)}(y)=y^{-2}\rho_n^{(1)}(y)$ for finite $n\leq8$. We conjecture this to hold for arbitrary system sizes.

The chain has $N=3n$, $n\geq1$, nearest-neighbour exclusion, the two inaccessible empty boundary sites, Jordan–Wigner fermionic signs, and staggering II $(y,y,1)$. The preregistration is issue #12308. The quantified formal claim covers every real $y\ne0$ and every nonzero zero-energy state, so it contains the source's ground-state assertion without an existence or uniqueness premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/StaggeredM1EndpointDensity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryStructural characterizationPosed in 2013Products of distinct Mersenne primes satisfy exactly the sigma-tau rational-power relationKrizek's sigma-tau rational-power condition characterizes products of distinct Mersenne primes.Proved

The questionNumbers that are a product of distinct Mersenne primes (elements of A000668).

Why it mattersThe conjecture gives a structural characterization of every positive integer whose divisor sum and divisor count are related by positive rational powers.

The ideaEquality of positive powers gives equality of the prime supports of sigma(n) and tau(n).

Our contributionKrizek's sigma-tau rational-power condition characterizes products of distinct Mersenne primes.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Products of distinct Mersenne primes satisfy exactly the sigma-tau rational-power relation

OEIS A046528, NAME (%N, verbatim):

Numbers that are a product of distinct Mersenne primes (elements of A000668).

The known COMMENT (%C, verbatim) states:

n is a product of distinct Mersenne primes iff sigma(n) is a power of 2: see exercise in Sivaramakrishnan, or Shallit.

Jaroslav Krizek's COMMENT (%C, verbatim) states:

Supersequence of A051281 (numbers n such that sigma(n) is a power of tau(n)). Conjecture: numbers n such that sigma(n) = tau(n)^(a/b), where a, b are integers >= 1. Example: sigma(93) = 128 = tau(93)^(7/2) = 4^(7/2). - Jaroslav Krizek, May 04 2013

The literal proved statement is ∀ n : ℕ, 0 < n → (isMersenneProduct n ↔ ratPow n), where isMersenneProduct n means ∃ S : Finset ℕ, n = ∏ p ∈ S, p ∧ ∀ p ∈ S, p.Prime ∧ ∃ k : ℕ, 0 < k ∧ p + 1 = 2 ^ k, and ratPow n means ∃ a b : ℕ, 0 < a ∧ 0 < b ∧ (σ 1 n) ^ b = (σ 0 n) ^ a. Here σ 1 is the divisor-sum function and σ 0 is the divisor-count function tau. Since both bases are positive integers for positive n, this integer-power equality is equivalent to Krizek's rational-exponent form.

No statement about A051281 beyond the displayed equivalence is claimed. The infinitude or distribution of Mersenne primes and novelty of the known sigma-equals-a-power-of-two classification are also NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Mersenne/KrizekSigmaTauRationalPowerMersenne.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Krizek's sigma-tau rational-power condition characterizes products of distinct Mersenne primes.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2013.

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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Further frontiersQuestion resolvedPosed in 2014The differences of A049473 occupy two shifted Beatty sequencesRounded multiples of one over sqrt two have binary differences at the two shifted Beatty position sequences.Proved

The questionClark Kimberling's COMMENT (%C, verbatim, including the complete sentence and attribution):

Why it mattersThe difference clause identifies both the range and every position of the adjacent differences of the nearest-integer sequence A049473.

The ideaRewrite real rounding as the floor of n/sqrt(2) + 1/2, and bound its adjacent increment between zero and one.

Our contributionRounded multiples of one over sqrt two have binary differences at the two shifted Beatty position sequences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The differences of A049473 occupy two shifted Beatty sequences

OEIS A049473, NAME (%N, verbatim):

Nearest integer to n/sqrt(2).

Clark Kimberling's COMMENT (%C, verbatim, including the complete sentence and attribution):

Let s(n) = zeta(3) - Sum_{k=1..n} 1/k^3. Conjecture: for n >=1, s(a(n)) < 1/n^2 < s(a(n)-1), and the difference sequence of A049473 consists solely of 0's and 1, in positions given by the nonhomogeneous Beatty sequences A001954 and A001953, respectively. - Clark Kimberling, Oct 05 2014

OEIS A001953, NAME (%N, verbatim):

a(n) = floor((n + 1/2) * sqrt(2)).

OEIS A001954, NAME (%N, verbatim):

a(n) = floor((n+1/2)*(2+sqrt(2))); winning positions in the 2-Wythoff game.

Writing a(n) = round(n/sqrt(2)), lower(k) = floor((k+1/2)*sqrt(2)), and upper(k) = floor((k+1/2)*(2+sqrt(2))), the literal proved clause says that every a(n+1)-a(n) is zero or one, that it is one exactly when n = lower(k) for some natural k, and that it is zero exactly when n = upper(k) for some natural k.

The first clause of the same comment, the zeta(3) tail squeeze s(a(n)) < 1/n^2 < s(a(n)-1), is NOT claimed. No other property of A001953 or A001954 is claimed. The general complementarity of nonhomogeneous Beatty sequences is classical 2-Wythoff theory and is cited as a literature prerequisite, not claimed as a new result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Deficit/Beatty/KimberlingRoundedHalfRootTwoDifference.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2014. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2026The k-positive region of the diagonal-perturbed Tomiyama mapsFor every 2 ≤ k ≤ d, the k-positive maps of the diagonal-perturbed Tomiyama family on d × d matrices are exactly those whose parameters lie in the quadrilateral spanned by the identity, the two extremal completely positive maps and the k-positive Tomiyama map.Proved

The questionThe verbatim statements are in the literature note.

The ideaThe four half-planes cut out the quadrilateral.

Our contributionFor every 2 ≤ k ≤ d, the k-positive maps of the diagonal-perturbed Tomiyama family on d × d matrices are exactly those whose parameters lie in the quadrilateral spanned by the identity, the two extremal completely positive maps and the k-positive Tomiyama map.

Verification details
Release source recordProved in Lean

Open problem

The k-positive region of the diagonal-perturbed Tomiyama maps

A. Bera, B. Bhattacharya and D. Chruściński, Tomiyama-type maps with a diagonal perturbation, arXiv:2604.18600v1, study $\Phi_{\alpha,\beta}=(1-\alpha-\beta),\mathrm{id}+\alpha,\tau_0+\beta,\Delta$ on $M_d$, where $\tau_0(X)=\mathrm{tr}(X),I/d$ and $\Delta$ keeps the diagonal. Conjecture 2.4 states that the $k$-positive maps of the family form the quadrilateral $\mathcal P_k=\mathrm{conv}{\Psi_0,\Psi_1,\Psi_2,\mathcal T_k}$ with parameter points $(0,0)$, $(0,\tfrac d{d-1})$, $(\tfrac d{d-1},-\tfrac1{d-1})$ and $(\tfrac{kd}{kd-1},0)$, and §4 states that the full proof is missing. The verbatim statements are in the literature note.

Issue #13894 reads the conjecture for $2\le k\le d$, with $k$-positivity of $\Phi$ meaning that $\mathrm{id}_k\otimes\Phi$ preserves positive semidefiniteness, and the convex hull taken in the $(\alpha,\beta)$ plane.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/TomiyamaDiagonalKPositivity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresRecurrence provedFirst preprint 2025The whole positive coefficient anti-recurrence conjectureEvery positive integral anti-recurrence has a unique complementary source and a finite automaton for its exact discrepancy.Proved

The questionRobbert Fokkink and Gandhar Joshi, Anti-Recurrence Sequences, arXiv:2506.13337v1, Conjecture 4; journal version, INTEGERS 26, A24, DOI 10.5281/zenodo.18305140, Conjecture 2:

Why it mattersThe theorem constructs a source for every such vector, proves its uniqueness, and supplies a finite DFAO with integer output computing Aₙ − (kτ+1)n on the standard base-τ digits read most significant first.

The ideaA finite-history complement enumeration constructs successive A terms.

Our contributionEvery positive integral anti-recurrence has a unique complementary source and a finite automaton for its exact discrepancy.

Verification details
Release source recordProved in Lean

Open problem

The whole positive coefficient anti-recurrence conjecture

Robbert Fokkink and Gandhar Joshi, Anti-Recurrence Sequences, arXiv:2506.13337v1, Conjecture 4; journal version, INTEGERS 26, A24, DOI 10.5281/zenodo.18305140, Conjecture 2:

Let a = (a₁,…,aₖ) be positive and integral of dimension k > 1. Let Aₙ be the anti-recurrence sequence generated by the linear form f(x) = a·x. Then Aₙ − κn is τ-automatic for κ = kτ+1 and τ the trace of the linear form.

Here τ = Σ aⱼ. The strictly increasing positive sequences A and B partition the positive integers and satisfy Aₙ = Σⱼ₌₁ᵏ aⱼ Bₖ₍ₙ₋₁₎₊ⱼ for every n ≥ 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/AntiRecurrenceAutomaticity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Every positive integral anti-recurrence has a unique complementary source and a finite automaton for its exact discrepancy.

    Read source ↗
  • Historical context

    First preprint year: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Analysis & dynamicsConjecture provedFirst preprint 2025Garcia–Volčič Conjecture 4.5Proved

The questionVolčič, “A noncommutative generalization of Hunter's positivity theorem”, arXiv:2503.12376v2, Conjecture 4.5, states:

The ideaEven and odd degrees use separate contractions.

Verification details
Release source recordProved in Lean

Open problem

Garcia–Volčič Conjecture 4.5

S. R. Garcia and J. Volčič, “A noncommutative generalization of Hunter's positivity theorem”, arXiv:2503.12376v2, Conjecture 4.5, states:

Let $n,d\ge 2$. For all tuples of hermitian operators $X_1,\dots,X_n$ on a Hilbert space, $\ker(H_{2d}(X_1,\dots,X_n)-\mu_{n,d}(X_1^{2d}+\cdots+X_n^{2d}))=\ker X_1\cap\cdots\cap\ker X_n$.

The settled assertion is exactly the bounded complex Hilbert-space statement preregistered in #13735. Here $H_k(X)$ is the ordered word sum, with each word $w$ weighted by $\prod_i m_i(w)!/k!$. This is the reciprocal cardinality of the word's abelianization fibre. The constant $\mu_{n,d}$ retains the source's $n=1$, odd-degree and even-degree branches.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Hunter/NCHunterKernelRigidity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedSource 2007Degrees of the order-eleven quasi-Fibonacci polynomialsEvery order-eleven quasi-Fibonacci polynomial in the five families has full degree from index five onward.Proved

The questionWituła and Słota define the five polynomial sequences by system (3.12) on printed page 5:

Why it mattersThe source proves only that the maximum of the five degrees is n.

The ideaLet ln be the vector of coefficients of X^n in (An,Bn,Cn,Dn,En).

Our contributionEvery order-eleven quasi-Fibonacci polynomial in the five families has full degree from index five onward.

Verification details
Release source recordProved in Lean

Open problem

Degrees of the order-eleven quasi-Fibonacci polynomials

Wituła and Słota define the five polynomial sequences by system (3.12) on printed page 5:

A_{n+1}(δ) = A_n(δ) + 2δB_n(δ) - δE_n(δ),

B_{n+1}(δ) = δA_n(δ) + B_n(δ) + δC_n(δ) - δE_n(δ),

C_{n+1}(δ) = δB_n(δ) + C_n(δ) + δD_n(δ) - δE_n(δ),

D_{n+1}(δ) = δC_n(δ) + D_n(δ),

E_{n+1}(δ) = δD_n(δ) + (1 - δ)E_n(δ),

with the initial values printed immediately below it:

A_0(δ) = 1, B_0(δ) = C_0(δ) = D_0(δ) = E_0(δ) = 0.

The sentence on printed page 19 is:

Problem. Is it true that deg A_n(∆) = deg B_n(∆) = deg C_n(∆) = deg D_n(∆) = deg E_n(∆) = n for every n = 5, 6, . . .?

Section 6 introduces ∆ := 1/δ and writes the same polynomial family as A_n(∆),...,E_n(∆). The formal module names the indeterminate X. It reads degree in WithBot ℕ, rather than natDegree, takes the printed threshold literally as 5 ≤ n, and defines the recurrence over the coefficient ring ℤ.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/WitulaSlotaQuasiFibonacciOrderElevenDegree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2007.

    Read source ↗
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    A released resolution of an explicitly recorded external question.

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    The delivered argument combines documented mathematical mechanisms.

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    The proved statement applies to the stated general family, not just a checked finite range.

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  • Historical context · 3/4 · 10% weight

    Year recorded in the source bibliography: 2007. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresRecurrence provedHardin–Barker five-column clockwise-edge-difference formulasExact five-column array counts, parity, recurrence and formal generating series.Proved

The questionHardin's OEIS A209723, revision 11, NAME states:

Why it mattersThe source formulas count the literal arrays at every positive height.

The ideaHere the branching indicator has six supported starts, the branching degree sums to eight, and the deterministic indicator has sixteen supported starts.

Our contributionExact five-column array counts, parity, recurrence and formal generating series.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hardin–Barker five-column clockwise-edge-difference formulas

R. H. Hardin's OEIS A209723, revision 11, NAME states:

1/4 the number of (n+1) X 5 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

The FORMULA field states the empirical recurrence and Colin Barker's conjectures:

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).

G.f.: x*(6 + x - 11x^2) / ((1 - x)(1 - 2*x^2)).

a(n) = 3*2^(n/2 - 1) + 4 for n even.

a(n) = 2^((n + 1)/2) + 4 for n odd.

The offset is $1,1$. Entries are labelled; no symmetry quotient is taken. For a square with top row $a,b$ and bottom row $c,d$, the differences are the signed integers $b-a,d-b,c-d,a-c$. Let $a(n)$ be one quarter of the number of valid $(n+1)$-by-$5$ arrays in the rationals.

Statement echo

| Source clause | Lean counterpart | Domain and fidelity | | --- | --- | --- | | Distinct clockwise differences | DistinctClockwise | Four integer differences, pairwise unequal; entries have type Fin 3. | | Every consecutive square | IsSourceArray | Every Fin n row gap and Fin 4 column gap is tested. | | One quarter of the array count | a | Nat.card of the entire labelled-array subtype, coerced to rationals and divided by four. | | Positive even formula | First conjunct of claim | $a(2(m+1))=3\cdot2^m+4$ for every natural $m$. | | Positive odd formula | Second conjunct of claim | $a(2m+1)=2^{m+1}+4$ for every natural $m$. | | Recurrence | Third conjunct of claim | $a(m+4)=a(m+3)+2a(m+2)-2a(m+1)$ for every natural $m$. | | Ordinary generating function | IsSourceSeries, fourth and fifth conjuncts of claim | Existence and the identity for every rational formal series with zero constant coefficient and coefficient $a(n)$ for $n\ge1$. |

The natural-index substitutions exactly cover the source domains. The extra series-existence conjunct prevents a vacuous universal assertion. The generating identity is cross-multiplied by $(1-X)(1-2X^2)$, whose constant coefficient is one. The auxiliary value $a(0)$ is outside the source formulas and does not supply the series constant term.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Pressure/HardinBarkerA209723EdgeDifferences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Exact five-column array counts, parity, recurrence and formal generating series.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Arbitrary even-order convexity of split covariance intersectionThe split covariance intersection covariance has nonnegative derivatives of every positive even order for both its log determinant and its trace, with semidefinite input covariances and positive definite pair sums.Proved

The questionHao Li, “Conjecture About Arbitrary Even-Order Convexity of w-Optimization of the Split CIF”, arXiv:2606.27260v1, Section 2.2, equations (7a) and (7b), states:

Why it mattersFor real symmetric matrices A, B, C, D of any finite dimension, assume each is positive semidefinite and A+B and C+D are positive definite.

The ideaFor positive definite inputs, form the affine three-block pencil

Our contributionThe split covariance intersection covariance has nonnegative derivatives of every positive even order for both its log determinant and its trace, with semidefinite input covariances and positive definite pair sums.

Verification details
Release source recordProved in Lean

arXiv

Arbitrary even-order convexity of split covariance intersection

Hao Li, “Conjecture About Arbitrary Even-Order Convexity of w-Optimization of the Split CIF”, arXiv:2606.27260v1, Section 2.2, equations (7a) and (7b), states:

The proposed conjecture is that the w-optimization problem has arbitrary even-order convexity, more specifically, for $k \in {1,2,3,\cdots}$ we always have

$$\frac{d^{2k}}{dw^{2k}}\ln\det(P(w))\geq 0, \qquad \frac{d^{2k}}{dw^{2k}}\operatorname{tr}(P(w))\geq 0.$$

Issue #11592 preregisters the source quotations, quantified statement, Tier 1 classification and literature-search boundary. The reading retains both inequalities for every positive even order and every interior weight, with the semidefinite hypotheses of Section 2.1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/SplitCovarianceIntersectionEvenConvexity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedMathar's first recurrence for OEIS A152661The source permutation count in OEIS A152661 satisfies Mathar's first recurrence at every natural index at least four.Proved

The questionIts first conjectural FORMULA, attributed to R.

Why it mattersThe recurrence relates three consecutive values of the source permutation count at every admissible index.

The ideaRepresent a permutation by a bijection of Fin n, giving its value in position i+1 the natural label p(i).val+1.

Our contributionThe source permutation count in OEIS A152661 satisfies Mathar's first recurrence at every natural index at least four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's first recurrence for OEIS A152661

OEIS A152661, revision 23, defines a(n), for n >= 2, as the number of permutations of {1,...,n} whose first two entries have the same parity. Its first conjectural FORMULA, attributed to R. J. Mathar, states

$$ (3-n)a(n)+2(n-2)a(n-1)+(n-1)(n-2)(n-3)a(n-2)=0. $$

The assertion here is this exact equality in the integers for every natural n >= 4. Each sequence value is the actual permutation cardinality, cast to the integers. The domain ensures that n, n-1 and n-2 all lie in the source domain. The adjacent conjecture a(n)=2*A077613(n-1) is excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/MatharSameParityRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The source permutation count in OEIS A152661 satisfies Mathar's first recurrence at every natural index at least four.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedMarcus's A344083 nonpolygonal Pythagorean triple conjectureThe three-four-five triple is the unique positive Pythagorean triple whose sides all belong to A090467.Proved

The questionMichel Marcus's COMMENT in OEIS A344083 states:

Why it mattersThe result resolves the full unbounded uniqueness claim, including the existence of the named example, rather than checking a finite range of Pythagorean triples.

The ideaSquares modulo three are zero or one, so the Pythagorean equation forces at least one leg to be divisible by three.

Our contributionThe three-four-five triple is the unique positive Pythagorean triple whose sides all belong to A090467.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Marcus's A344083 nonpolygonal Pythagorean triple conjecture

Michel Marcus's COMMENT in OEIS A344083 states:

Conjecture: there are no other Pythagorean triples that give this minimum. In other words, it is the only triple with 3 A090467 terms.

OEIS A090467 consists of the natural numbers that are not polygonal numbers of order greater than two. In the formal statement, nonpolygonal n means that no k > 2 and m > 2 satisfy n = m + (k - 2) * Nat.choose m 2.

The conjecture asserts both that (3,4,5) has three nonpolygonal sides and that every positive Pythagorean triple with three nonpolygonal sides has unordered legs {3,4} and hypotenuse 5.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/MarcusPythagoreanNonpolygonalTriple.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedDivisibility of the A363560 coefficientsEvery positive coefficient index outside the class one modulo seven in OEIS A363560 has coefficient divisible by three.Proved

The questionHanna, Aug 12 2023, records the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2023a363560.md:

Why it mattersThis is the first-tier recent OEIS conjecture selected in the implementation brief.

The ideaThe exact polynomial factorization is t + t² + t⁹ = (t² + t + 1)·P(t), where P(t) = t − t³ + t⁴ − t⁶ + t⁷.

Our contributionEvery positive coefficient index outside the class one modulo seven in OEIS A363560 has coefficient divisible by three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A363560 coefficients

OEIS A363560, Paul D. Hanna, Aug 12 2023, records the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2023a363560.md:

Expansion of g.f. A(x) satisfying A(x)^3 = 1 + x*(A(x) + A(x)^2 + A(x)^9).

Conjecture: a(n) == 0 (mod 3) for n > 0 except when n == 1 (mod 7).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CubicNinthPowerSubstitutionModThree.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedPosed in 2025The A378252 running-sum residue conjectureA running-sum residue recurrence is a doubling in disguise, so it vanishes exactly when its modulus is a power of two.Proved

The questionFix integers with the start value coprime to the modulus.

Why it mattersDirect inspection on September 9, 2026 found the statement still labelled Conjecture, at revision 16 with no later revision supplying a proof.

The ideaThe recurrence appears to accumulate history but does not.

Our contributionA running-sum residue recurrence is a doubling in disguise, so it vanishes exactly when its modulus is a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A378252 running-sum residue conjecture

Fix integers with the start value coprime to the modulus. Let the first term be the start value reduced modulo the modulus, and each later term be the start value plus every earlier term, again reduced. Lechoslaw Ratajczak conjectured on October 13, 2025 that some term is zero exactly when the modulus is a power of two, the exponent zero being allowed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/RatajczakDoublingZero.exists_zero_iff_modulus_pow_two

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    A running-sum residue recurrence is a doubling in disguise, so it vanishes exactly when its modulus is a power of two.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2025.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHowroyd's full Andrasfai mutual-visibility generating functionThe full original generating identity for all labelled mutual-visibility subsets of the Andrasfai graphs.Proved

The questionConjectured g.f.: x(4 - 19x + 33x^2 - 40x^3 + 8x^4)/(1 - 10x + 29x^2 - 32x^3 + 8x^4).

Why it mattersThis is a named whole-family enumeration problem.

The ideaOriginal-host shortest paths reduce to common-neighbor obstructions.

Our contributionThe full original generating identity for all labelled mutual-visibility subsets of the Andrasfai graphs.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Howroyd's full Andrasfai mutual-visibility generating function

OEIS A391632, revision 27, attributes this assertion to Andrew Howroyd:

Conjectured g.f.: x*(4 - 19x + 33x^2 - 40x^3 + 8x^4)/(1 - 10x + 29x^2 - 32x^3 + 8x^4).

The entry's name is "Number of mutual-visibility sets in the n-Andrásfai graph" and its offset is 1,1. For every integer $n\ge1$, let $G_n$ be the simple undirected Cayley graph on the labelled vertices $\mathbb Z/(3n-1)\mathbb Z$, with connection set ${1,4,\ldots,3n-2}$. A subset $S$ is mutual-visible when every distinct pair $u,v\in S$ admits a shortest $u$-$v$ path in the original host $G_n$ whose internal vertices avoid $S$. Different pairs may choose different paths. Let $a(n)$ count every such labelled subset, including the empty set and singletons. The entire question is the identity in $\mathbb Q[[X]]$

$$ \sum_{n\ge1}a(n)X^n =\frac{X(4-19X+33X^2-40X^3+8X^4)}{1-10X+29X^2-32X^3+8X^4}. $$

The series has constant coefficient zero; no graph of index zero is introduced. There is no quotient by graph symmetries, maximality, independence or induced-host restriction. The quotient denotes multiplication by the formal inverse of a series with constant coefficient one, without an analytic convergence premise.

Statement correspondence

Put $m=n-1$ and $N=3m+2=3n-1$. Andrasfai.host m uses ZMod N, with adjacency determined by the canonical representative of $v-u$ being one modulo three. Andrasfai.count m is the cardinality of the original Configuration (host m) subtype of actual finite vertex subsets. AndrasfaiActual.generatingSeries is defined directly as $X\operatorname{mk}(m\mapsto(\operatorname{count}(m):\mathbb Q))$. AndrasfaiActual.result has no hypotheses and states the displayed identity for that series.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/MutualVisibility/AndrasfaiActual.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedHanna's power-of-two parity conjecture for OEIS A393173The normalized inverse-diagonal equation has integral coefficients, odd exactly at powers of two in positive degrees.Proved

The questionA(x) satisfies [x^n] 1/(A(x) - n^2x) = 0 for n >= 1.

Why it mattersThe normalized rational inverse-diagonal solution exists, its positive coefficients are integers, and their odd support is exactly the powers of two.

The ideaThus the correction that adds the offending inverse coefficient is a strict coefficient contraction.

Our contributionThe normalized inverse-diagonal equation has integral coefficients, odd exactly at powers of two in positive degrees.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's power-of-two parity conjecture for OEIS A393173

Paul D. Hanna's OEIS A393173 states:

G.f. A(x) satisfies [x^n] 1/(A(x) - n^2*x) = 0 for n >= 1.

Conjecture: for n > 0, a(n) is odd iff n is a power of 2.

The offset-zero data specify $a(0)=1$. Preregistration #18824 classifies the published conjecture as tier 1. The source is documented in the Library note.

Statement echo

Names below belong to D5.S1.Recurrence.Invariants.HannaA393173PowerTwoParity.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Offset zero, $a(0)=1$ | IsGF constant coefficient | Literal normalization. | | $x^n^{-1}=0$ for $n\ge1$ | IsGF inverse coefficient, natural index cast into $\mathbb Q$ | Literal source expression. | | Positive indices | claim, $n>0$ | Zero is excluded. | | $a(n)$ is odd iff $n=2^k$ | claim, integer coefficient witness and both directions of the equivalence | Exponent $k$ is natural. | | Existence and integrality implicit in the sequence | claim, an existing solution and an integer witness for every coefficient of every solution | Explicit strengthened clauses from the preregistration; these prevent vacuity. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaA393173PowerTwoParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe seventh compositional iterate of A396798 modulo eightThe seventh compositional iterate of A396798 has coefficient period 7,1,3,5 modulo eight from degree two.Proved

The questionHanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)A^5(x)." Its formula field says:

The ideaReduce the fixed integer source to F over ZMod8 and set U=I3(F).

Our contributionThe seventh compositional iterate of A396798 has coefficient period 7,1,3,5 modulo eight from degree two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The seventh compositional iterate of A396798 modulo eight

Paul D. Hanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)*A^5(x)." Its formula field says:

G.f. A(x) = Sum_{n>=1} a(n)*x^n has the following properties, where A^n(x) denotes the n-th iteration of A(x).

The seventh comment states exactly:

Conjecture: [x^n] A^7(x) == [7,1,3,5] repeating (mod 8) for n > 1.

Let I0(A)=X and I(k+1)(A)=Ik(A) composed with A. For the unique zero-constant integer ordinary series satisfying A=X+I4(A)*I5(A), the result is:

forall n : Nat, 1 < n ->
  (8 : Int) ∣ PowerSeries.coeff n
    (CompositionalIterateCongruence.iterate
      IterateProductFourFiveEighthModEight.generatingSeries 7) -
    (if (n - 2) % 4 = 0 then (7 : Int)
     else if (n - 2) % 4 = 1 then 1
     else if (n - 2) % 4 = 2 then 3 else 5)

Every natural n>1 is included, with no upper bound or extra premise. Residues 0,1,2,3 of (n-2)%4 select 7,1,3,5, starting at n=2. A^7 is compositional iteration; products are ordinary multiplication, without factorial normalization or convergence premises. The existing source is D5/S1/Recurrence/Residue/IterateProductFourFiveEighthModEight.generatingSeries, declaration identity sha256:ec9535ab3a9b0a1e907e9e393ae6d5cccfb485ff6209d712252354e4035bdfbc.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductFourFiveSeventhModEight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe integer source and dyadic parity of A397592Integer existence and uniqueness for the literal source of A397592, together with its dyadic-neighbor parity assertion for every source solution.Proved

The questionHanna's OEIS A397592 (July 10, 2026), revision 18, has NAME:

Why it mattersThis first-tier 2026 external named conjecture is the questionanswered in 8006, created 2026-09-15T03:15:36Z before probes.

The ideaMultiply the rational NAME sum in row m=n+1 by m^n.

Our contributionInteger existence and uniqueness for the literal source of A397592, together with its dyadic-neighbor parity assertion for every source solution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The integer source and dyadic parity of A397592

Paul D. Hanna's OEIS A397592 (July 10, 2026), revision 18, has NAME:

G.f. A(x) such that the sum of the first n coefficients in A(x/n)^n equals 2^(n-1) for n >= 1.

Its sole conjecture is:

Conjecture: for n > 3, a(n) is odd iff n is of the form 2^k-1 or 2^k+1 (k > 1).

The FORMULA states A(x)=Sum_{n>=0} a(n)*x^n; the original offset is 0 and a(0)=1. For A in Z[[X]], A_Q is its coefficientwise image in Q[[X]], namely A.map (Int.castRingHom Rat). The exact source and single result are:

def SourceA397592 (A : PowerSeries Int) : Prop :=
  PowerSeries.constantCoeff A = 1 ∧
    ∀ m : Nat, 0 < m →
      (Finset.range m).sum (fun j => PowerSeries.coeff j
        ((PowerSeries.rescale ((m : Rat)⁻¹)
          (A.map (Int.castRingHom Rat))) ^ m)) = (2 : Rat) ^ (m - 1)

theorem result :
  (∃! A : PowerSeries Int, SourceA397592 A) ∧
  (∀ A : PowerSeries Int, SourceA397592 A → ∀ n : Nat, 3 < n →
    (Odd (PowerSeries.coeff n A) ↔
      ∃ k : Nat, 1 < k ∧ (n = 2 ^ k - 1 ∨ n = 2 ^ k + 1)))

These are ordinary formal powers, not compositional iterates; there is no convergence or factorial normalization. The sum includes j=0,...,m-1 for every natural m>0. The separate integer-existence-and-uniqueness conjunct excludes an unproved integrality premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LinearRows/DoubledLinearExponentDyadicSupport.result

Frozen module record

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Number theoryQuestion resolvedPosed in 2026A399755: generalized Pell classes and the least square partnerThe least square partner and integral-unit classes of the norm-(k^2+1) equation.Proved

The questionThe single external named target is Chai Wah Wu's A399755 conjecture, attributed in the entry to September 10, 2026:

Why it mattersIts parameter is 6j and its conclusion is a norm and coordinate identity.

The ideaThe proof uses elementary integer algebra and descent.

Our contributionThe least square partner and integral-unit classes of the norm-(k^2+1) equation.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A399755: generalized Pell classes and the least square partner

The single external named target is Chai Wah Wu's A399755 conjecture, attributed in the entry to September 10, 2026:

Conjecture: A399491(k) < 4k^3+3k if and only if k is a term of this sequence.

The A399755 name is:

Numbers k such that the generalized Pell equation x^2-(k^2+1)*(y^2+1) = 0 has more than one fundamental solution in (x, y).

The companion A399491 name is:

The smallest integer m > n such that (n^2+1)*(m^2+1) is a perfect square.

For every natural k, define D = Discriminant(k) = (k : Int)^2 + 1. All generalized Pell coordinates and all unit coordinates below are integers. The exact definitions and sole target are:

Discriminant(k : Nat) : Int = (k : Int)^2 + 1
Sol(k : Nat, x y : Int) := x^2 - Discriminant(k)*y^2 = Discriminant(k)
SameClass(k : Nat, x y r s : Int) :=
  exists u v : Int,
    u^2 - Discriminant(k)*v^2 = 1 AND
    x = r*u + Discriminant(k)*s*v AND y = r*v + s*u
MultipleClasses(k : Nat) :=
  exists x y r s : Int,
    Sol(k,x,y) AND Sol(k,r,s) AND NOT SameClass(k,x,y,r,s)
NextSquarePartner(k m : Nat) :=
  k < m AND IsSquare((k^2+1)*(m^2+1)) AND
  forall n : Nat,
    k < n -> IsSquare((k^2+1)*(n^2+1)) -> m <= n
result : forall k : Nat, 0 < k ->
  exists m : Nat, NextSquarePartner(k,m) AND
    (m < 4*k^3+3*k IFF MultipleClasses(k))

IsSquare here is the predicate on natural numbers. The existential result includes totality and the full minimum property, not only a bound on an assumed partner. Membership refers to k itself, not to the kth value listed in A399755. The source convention selects one fundamental representative per generalized solution class; it is distinct from the unique fundamental positive solution of the norm-one equation.

SameClass is Robertson's integral norm-one relation, including the unit -1 and hence simultaneous negation of both coordinates. It does not impose positive coordinates, primitivity, or squarefree D, and it does not use fractional units of a maximal order. Any representative-based formulation must be proved equivalent to this predicate on arbitrary integral pairs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/PellClassSuccessor.result

Frozen module record

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    Source 1
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    Source 1
Quantum informationConjecture refutedFirst preprint 2004Refutation of Braunstein–Ghosh–Severini Conjecture 6.7Refutation of the Braunstein–Ghosh–Severini star maximum conjecture.Refuted

The questionBraunstein, Ghosh and Severini, The Laplacian of a graph as a density matrix: a basic combinatorial approach to separability of mixed states, Annals of Combinatorics 10 (2006), Conjecture 6.7, states:

The ideaA local projective measurement isolates the non-root blocks.

Our contributionRefutation of the Braunstein–Ghosh–Severini star maximum conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Braunstein–Ghosh–Severini Conjecture 6.7

Braunstein, Ghosh and Severini, The Laplacian of a graph as a density matrix: a basic combinatorial approach to separability of mixed states, Annals of Combinatorics 10 (2006), Conjecture 6.7, states:

Let $\mathcal G_n^c$ be the set of all connected graphs on $n$ vertices. Let $G\in\mathcal G_n^c$ with $|V|=pq$. Then $$\max_{G\in\mathcal G_n^c}E_F(\sigma(G))=E_F(\sigma(K_{1,n-1})).$$

Here $\sigma(G)=L(G)/d_G$ is the graph Laplacian density matrix and $E_F$ is the convex-roof entanglement of formation over finite pure-state ensembles.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GraphDensity/BraunsteinGhoshSeveriniStarRefutation.result

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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2004. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Further frontiersCounterexample foundFirst preprint 2025The negative-index root-amplitude attainment clause is refutedMane's negative-index root-amplitude attainment fails at k = 3 and n = -15.Refuted

The questionMane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, Section 6, Conjecture 6.6, states:

Why it mattersThe refutation addresses the attainment sentence; it does not negate the separate upper-bound inequality.

The ideaIts fixed-parameter exact certificate has utility kind=certified-instance, with refutes pointing to claim.

Our contributionMane's negative-index root-amplitude attainment fails at k = 3 and n = -15.

Verification details
Release source recordRefuted in Lean

arXiv

The negative-index root-amplitude attainment clause is refuted

S. R. Mane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, Section 6, Conjecture 6.6, states:

For fixed k ≥ 3 and n < 0, the upper bound is ζ_{n,k} ≤ ⌊|n|/k⌋. The bound is attained whenever r_{n,k} = 1 (equivalently n = −sk, where s ≥ 1), i.e. it is a tight bound.

The recurrence (1.2)–(1.3) and initial values are

$$ \mathcal F_{n,k}(x)=\sum_{j=1}^{k}x^{k-j}\mathcal F_{n-j,k}(x),\qquad \mathcal F_{1,k}=1,\qquad \mathcal F_{n,k}=0\quad (-(k-2)\le n\le0). $$

The source statistic is the maximum of $|x|^k$ over nonzero complex roots; it is zero when there are none, including vanishing polynomials. The Lean encoding uses a k m for $\mathcal F_{1-m,k}$, F k n := a k (1-n).toNat, and the real supremum of the nonzero-root amplitude image. Its empty and unbounded totalizations give zero. The guards in claim are $k\ge3$ and $s\ge1$, with integer index $-sk$ and real-valued right side $s$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/NegativeIndexTribonacciAttainmentRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

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Discrete structuresQuestion resolvedFirst preprint 2026Rivin's cycle-geodesic midpoint formulaThe midpoint formula holds with a uniform quadratic error and even-dimensional vanishing.Proved

The questionIgor Rivin, “Permanents of matrix ensembles: computation, distribution, and geometry”, arXiv:2602.10141v3, Section 8.4, Open Problem 2, states:

Why it mattersComplex absolute value is encoded by the complex norm; the displayed scale uses natural subtraction and division only on positive odd dimensions.

The ideaGaudinPermanent.gaudinpermanent identifies the reciprocal-difference Cauchy permanent with a Gaudin determinant.

Our contributionThe midpoint formula holds with a uniform quadratic error and even-dimensional vanishing.

Verification details
Release source recordProved in Lean

arXiv

Rivin's cycle-geodesic midpoint formula

Igor Rivin, “Permanents of matrix ensembles: computation, distribution, and geometry”, arXiv:2602.10141v3, Section 8.4, Open Problem 2, states:

Prove the midpoint formula $\operatorname{perm}(\gamma(\frac12))=(-1)^{(n-1)/2}\cdot 2e^{-n}(1+\frac{1}{3n}+O(n^{-2}))$.

The literal Section 4.1 matrix is

$$\gamma_n(t)_{jl}=\frac{e^{2\pi it}-1}{n(e^{2\pi i(l-j+t)/n}-1)},\qquad j,l\in{0,\ldots,n-1}.$$

Observation 5 also states the even-dimensional zero clause. Dimension is positive: $n\ge1$. The empty permanent at $n=0$ is $1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/CycleGeodesic/CycleGeodesicMidpoint.result2

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2026Rivin's cycle-geodesic universal functionThe permanent rate tends to the closed cotangent expression, with the odd midpoint limit one.Proved

The questionIgor Rivin, “Permanents of matrix ensembles: computation, distribution, and geometry”, arXiv:2602.10141v3, Section 8.4, Open Problem 3, states:

The ideaGaudinPermanent.gaudinpermanent identifies the reciprocal-difference Cauchy permanent with a Gaudin determinant.

Our contributionThe permanent rate tends to the closed cotangent expression, with the odd midpoint limit one.

Verification details
Release source recordProved in Lean

arXiv

Rivin's cycle-geodesic universal function

Igor Rivin, “Permanents of matrix ensembles: computation, distribution, and geometry”, arXiv:2602.10141v3, Section 8.4, Open Problem 3, states:

Find a closed form for the universal function $f(t)$.

Observation 4 defines the finite-dimensional quantity as $-n^{-1}\log|\operatorname{perm}\gamma_n(t)|$ for the literal matrix

$$\gamma_n(t)_{jl}=\frac{e^{2\pi it}-1}{n(e^{2\pi i(l-j+t)/n}-1)},\qquad j,l\in{0,\ldots,n-1}.$$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/CycleGeodesic/CycleGeodesicScaling.result3

Frozen module record

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    Source 1
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2025Finite-Fock beam-splitter states need not be extreme Wigner-positive statesFinite-Fock beam-splitter states need not be extreme Wigner-positive states.Refuted

The questionChabaud, Extreme non-negative Wigner functions, arXiv:2512.14831v3, Section 6, printed page 15, ask:

The ideaThe module uses the source's Laguerre expansion, complex Wigner function, positive semidefinite trace-one matrices and midpoint definition of extremality.

Our contributionFinite-Fock beam-splitter states need not be extreme Wigner-positive states.

Verification details
Release source recordRefuted in Lean

arXiv

Finite-Fock beam-splitter states need not be extreme Wigner-positive states

Z. Van Herstraeten, J. Davis, N. C. Dias, J. N. Prata, N. J. Cerf and U. Chabaud, Extreme non-negative Wigner functions, arXiv:2512.14831v3, Section 6, printed page 15, ask:

This leads us to formulate the following open problem: for any two Fock-bounded pure states $\ket\psi,\ket\phi$ is the beamsplitter state $\hat\sigma(\psi,\phi)$ an extreme WPS?

The source is the version-3 paper. The Tier-1 statement, finite-matrix conventions and literature reading are in preregistration #14447. The answer is Refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/FockAttenuator/FiniteFockBeamSplitterNonExtremality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

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Quantum informationQuestion resolvedFirst preprint 2026Disorder-free localization on even hypercubesFor every even dimension d at least four, the all-plus spin-coupled Grover operator has a nonzero fixed vector with nonconstant position weights.Proved

The questionIts equivalent clause asks for a nonzero fixed point with nonuniform position distribution.

Why it mattersCorollary 4.4 of the paper converts that clause to disorder-free localization.

The ideaThe proof uses spectral estimates.

Our contributionFor every even dimension d at least four, the all-plus spin-coupled Grover operator has a nonzero fixed vector with nonconstant position weights.

Verification details
Release source recordProved in Lean

arXiv

Disorder-free localization on even hypercubes

Wang, arXiv:2609.07267v1, Conjecture 4.1 asks whether every even $d\ge4$ has disorder-free localization for the spin-coupled Grover walk with $\phi_\sigma=|\sigma|\pi/d$ and all spins $+1$. Its equivalent clause asks for a nonzero fixed point with nonuniform position distribution. The source identity and locators are in Library/QuantumChannels/wang2026hypercubelocalization.md.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/SpinCoupledEvenHypercubeGroverFixedPoint.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

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    The exact recorded statement is proved.

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Quantum informationQuestion resolvedFirst preprint 2026Maximum purity in the 4-by-4 absolutely PPT sectorThe literal 4-by-4 APPT purity maximum is 2/27 and is attained by the explicit diagonal spectrum (1/6,1/18,...,1/18).Proved

The questionAhiable, Kothakonda and Winter, arXiv:2608.03390v2, Conjecture 6.7, page 29:

Why it mattersAn attaining state alone supplies a lower bound.

The ideaThe corner uses the three largest and six smallest ordered eigenvalues.

Our contributionThe literal 4-by-4 APPT purity maximum is 2/27 and is attained by the explicit diagonal spectrum (1/6,1/18,...,1/18).

Verification details
Release source recordProved in Lean

Open problem

Maximum purity in the 4-by-4 absolutely PPT sector

Ahiable, Kothakonda and Winter, arXiv:2608.03390v2, Conjecture 6.7, page 29:

Let $\mathcal P_{m,n}\subseteq\mathrm{APPT}{m,n}$ be the inscribed absolute PPT polytope with $2\leq m\leq n,;n>2$. Then $$\max{\lambda\in\mathrm{APPT}{m,n}}\sum{i=1}^{mn}\lambda_i^2 =\max_{\lambda\in\mathcal P_{m,n}}\sum_{i=1}^{mn}\lambda_i^2$$ and occurs at the spectra given by Eq. (44).

The exact target here is its $m=n=4$ sector. A density matrix is a complex matrix $\rho$ on $\mathbb C^4\otimes\mathbb C^4$ with $\rho\succeq0$ and $\operatorname{Tr}\rho=1$. Literal APPT requires $(U\rho U^\dagger)^\Gamma\succeq0$ for every global unitary $U$, where $\Gamma$ transposes the second tensor factor.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsolutePPT/QuadbitMaximumPurity.result

Frozen module record

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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2025Infinitely many non-star trees admit Laplacian peak state transferThere are arbitrarily large non-star trees with Laplacian peak state transfer.Proved

The questionGabriel Coutinho, Krystal Guo and Vincent Schmeits, Peak state transfer in continuous quantum walks, arXiv:2505.11986v4, Section 7, Open Problem 7.1 (p. 21), ask:

Why it mattersPreregistration 14801 names this settlement LPST-1.

The ideaBoth the root and a hub have degree at least two, excluding a star.

Our contributionThere are arbitrarily large non-star trees with Laplacian peak state transfer.

Verification details
Release source recordProved in Lean

Open problem

Infinitely many non-star trees admit Laplacian peak state transfer

Gabriel Coutinho, Krystal Guo and Vincent Schmeits, Peak state transfer in continuous quantum walks, arXiv:2505.11986v4, Section 7, Open Problem 7.1 (p. 21), ask:

Determine whether infinitely many such trees, not isomorphic to the star graph, admit Laplacian peak state transfer.

Figure 10 (p. 22) includes a root with three hubs and two leaves per hub. For the real Laplacian $L=D-A=\sum_\theta\theta E_\theta$, Section 3 defines $U(t)=e^{itL}$ and $B(L)=\sum_\theta|E_\theta|$, summing over distinct eigenvalues and taking absolute values entrywise. Peak transfer between distinct vertices $r,h$ means $|U(\tau){h,r}|=B(L){h,r}$ at some real time. The source does not require that value to equal one.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/LaplacianPeakTransferTrees.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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Geometry & algebraQuestion resolvedFirst preprint 2004Logarithmic scalar curvature is strictly Schur-increasingStrict Schur increase of logarithmic simplex scalar curvature.Proved

The questionGibilisco and Isola, On the monotonicity of scalar curvature in classical and quantum information geometry, arXiv:math-ph/0407007v2, Conjecture 4.1, printed page 11:

Why it mattersThe endpoint gives an intrinsic curvature quantity that increases under mixing for every positive normalized vector of dimension at least three.

The ideaThe coordinate logarithmic metric is diagonal plus rank one.

Our contributionStrict Schur increase of logarithmic simplex scalar curvature.

Verification details
Release source recordProved in Lean

Open problem

Logarithmic scalar curvature is strictly Schur-increasing

Gibilisco and Isola, On the monotonicity of scalar curvature in classical and quantum information geometry, arXiv:math-ph/0407007v2, Conjecture 4.1, printed page 11:

Suppose $n>2$. For the function $\mathrm{Scal}_p(\cdot):\mathcal P^1_n\to\mathbb R$ one has the following properties: if $p\in(1,2)$ then $\mathrm{Scal}_p$ is strictly Schur-decreasing; if $p\in(2,+\infty]$ then $\mathrm{Scal}_p$ is strictly Schur-increasing.

This settlement proves the $p=\infty$ endpoint only. The positive simplex is $\mathcal P^1_n={x\in\mathbb R^n: x_i>0,\ \sum_i x_i=1}$. The source says $x$ is more mixed than $y$ when the decreasingly sorted initial sums of $x$ are at most those of $y$, with equal total sums. Strict Schur increase means weak curvature increase for all such pairs and strict increase unless $x$ is a permutation of $y$.

Definitions 4.3–4.4 define the $\alpha$-geometry as the Euclidean pull-back under $A_p(x)=p x^{1/p}$ for finite nonzero $p$ and $A_\infty(x)=\log x$, and $\mathrm{Scal}_p$ as its scalar curvature. The literal logarithmic pull-back, Christoffel symbols, Ricci contraction and decreasing-sort majorization are the conventions of #15145.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/AlphaGeometry/LogScalarCurvatureMonotone.result

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Discrete structuresQuestion resolvedFirst preprint 2025First longest Thue-Morse progressions at power-of-two differencesFirst starts of longest Thue-Morse progressions at power-of-two differences.Proved

The questionJoshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, Theoretical Computer Science 1050 (2025), 115391, arXiv:2501.05830v2, Section 3.2.2, Conjecture 3.8:

Why it mattersThis 2025 specialist conjecture concerns three unbounded families.

The ideaBinary block parity and complementary residues express progression letters through carries or borrows.

Our contributionFirst starts of longest Thue-Morse progressions at power-of-two differences.

Verification details
Release source recordProved in Lean

Open problem

First longest Thue-Morse progressions at power-of-two differences

Joshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, Theoretical Computer Science 1050 (2025), 115391, arXiv:2501.05830v2, Section 3.2.2, Conjecture 3.8:

We have i(2^n+1)=3·2^(2n)-2^n-1, i(2^(2n)-1)=3·2^(4n)-2^(2n)+1, i(2^(2n+1)-1)=2^(2n+1)-1.

The display omits ranges. The contextual interpretation, from the preceding maximum-length formulas, is n>=2 in the first equation and n>=1 in the other two. Equivalently, for every natural e>=2, with q=2^e, i(q+1)=3q^2-q-1, i(q-1)=3q^2-q+1 for even e, and i(q-1)=q-1 for odd e. This range is an explicitly disclosed interpretation, not a printed quantifier. The previously known i(3)=45 excludes the plus-family extension to e=1; it is not a new refutation.

The word is the zero-indexed Thue-Morse word 0110100110010110..., with t(0)=0, t(2u)=t(u) and t(2u+1)=1-t(u). Definitions 2.2, 2.3 and 2.5 make A(d) the global maximum length over every nonnegative start and either letter, and i(d) the least start attaining it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/ThueMorseMapFirstStart.result

Frozen module record

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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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Discrete structuresQuestion resolvedFirst preprint 2025First longest progressions at Fibonacci differencesLocate the first longest monochromatic progression at every Fibonacci difference.Proved

The questionJoshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, Theoretical Computer Science 1050 (2025), 115391, arXiv:2501.05830v2, Conjecture 3.19:

Why it mattersThe mechanical bridge identifies the actual substitution word with the golden rotation at index j+1.

The ideaWrite r=1/tau, eps=r^m, and use the open true-letter interval (r^2,1).

Our contributionLocate the first longest monochromatic progression at every Fibonacci difference.

Verification details
Release source recordProved in Lean

Open problem

First longest progressions at Fibonacci differences

Joshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, Theoretical Computer Science 1050 (2025), 115391, arXiv:2501.05830v2, Conjecture 3.19:

i(F_(2n+1)) = F_(2n+3) - 2, and i(F_(2n)) = F_(4n) - 1.

The quantifier is every natural n >= 1, as independently clarified in the author's A364648 comment. Here F_0=0, F_1=1. The word is the zero-indexed fixed point 010010100100101... of 0 -> 01, 1 -> 0. For a positive difference d, A(d) is the global maximum of positive lengths over every nonnegative start and both symbols; i(d) is the least start attaining that maximum, again allowing either symbol. Definitions 2.2, 2.3 and 2.5 supply these conventions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/FibonacciMapFirstStart.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

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Discrete structuresQuestion resolvedFirst preprint 2026The Zero Forcing Number of P(n,3) Is Eight for Every n At Least ThirteenEight vertices are necessary and sufficient to zero-force P(n,3) for every n at least thirteen.Proved

The questionArnav Krishnan, A correction to the Zero Forcing Number of the Generalized Petersen Graphs P(n,3), arXiv:2607.19412v1, Conjecture 5:

Why it mattersThe lower-bound argument transfers a statement about forcing sets into a vertex-isoperimetric inequality for P(n,3), and the isoperimetric inequality is reduced from arbitrary n to finitely many column patterns by deleting empty columns.

The ideaUpper bound: the eight consecutive outer vertices u0, …, u7 force P(n,3) for every n ≥ 9.

Our contributionEight vertices are necessary and sufficient to zero-force P(n,3) for every n at least thirteen.

Verification details
Release source recordProved in Lean

Open problem

The Zero Forcing Number of P(n,3) Is Eight for Every n At Least Thirteen

Arnav Krishnan, A correction to the Zero Forcing Number of the Generalized Petersen Graphs P(n,3), arXiv:2607.19412v1, Conjecture 5:

Z(P(n,3)) = 8 for every n ≥ 13.

The paper proves the upper bound for n ≥ 9 (Proposition 4) and the lower bound for 13 ≤ n ≤ 20 by exhaustive search, and states that a lower-bound proof valid for all n is missing.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GeneralizedPetersen/ZeroForcingThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

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    Source 1
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Discrete structuresQuestion resolvedThe third compositional iterate of A396798 modulo eightThe third compositional iterate of A396798 has coefficient period 3,1,7,5 modulo eight from degree two.Proved

The questionHanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)A^5(x)." Its formula field says that A^k denotes the k-th iteration.

Why it mattersThe public fifth and eighth results on that same source reduce the third-iterate question to a compositional inverse identity over ZMod8.

The ideaReduce the fixed integer source to F over ZMod8.

Our contributionThe third compositional iterate of A396798 has coefficient period 3,1,7,5 modulo eight from degree two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The third compositional iterate of A396798 modulo eight

Paul D. Hanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)*A^5(x)." Its formula field says that A^k denotes the k-th iteration. The third comment states:

Conjecture: [x^n] A^3(x) == [3,1,7,5] repeating (mod 8) for n > 1.

Let I0(A)=X and I(k+1)(A)=Ik(A) composed with A. For the unique zero-constant integer series satisfying A=X+I4(A)*I5(A), the target is forall n:Nat, 1<n -> 8 divides coeff n (I3(A)) - r(n), where r(n)=if (n-2)%4=0 then 3 else if (n-2)%4=1 then 1 else if (n-2)%4=2 then 7 else 5. The period starts at n=2. Multiplication is ordinary series multiplication, with no factorial normalization or convergence assumption. The source is D5/S1/Recurrence/Residue/IterateProductFourFiveEighthModEight.generatingSeries.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductFourFiveThirdModEight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 3/4 · 30% weight

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    Source 1
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    Source 1
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Number theoryQuestion resolvedFirst preprint 2026Rosenzweig–Stanfill even residue vanishingThe Rosenzweig-Stanfill residue bracket vanishes for every even index.Proved

The questionBart Rosenzweig and Jonathan Stanfill, On the fundamental solutions of two nonlocal parabolic equations related to logarithmic Laplacians, arXiv:2606.04225v1, Open Problem 1.6(iv), printed page 4:

Why it mattersThe even-index cancellation is part of the paper's proposed description of the zero and nonzero residues of K₂(t,x,x).

The ideaLet C(s) be the ordinary series of the primary c-array.

Our contributionThe Rosenzweig-Stanfill residue bracket vanishes for every even index.

Verification details
Release source recordProved in Lean

Open problem

Rosenzweig–Stanfill even residue vanishing

Bart Rosenzweig and Jonathan Stanfill, On the fundamental solutions of two nonlocal parabolic equations related to logarithmic Laplacians, arXiv:2606.04225v1, Open Problem 1.6(iv), printed page 4:

Consider the notation of Theorem 1.5. Then the following are conjectured to be true:

(iv) For every α ∈ (0, π), (1.13) is equal to zero whenever m ≥ 0 is even.

The Lean proposition is ∀ m : ℕ, Even m → ∀ α : ℝ, 0 < α → α < Real.pi → paperBracket m α = 0. It includes m = 0 and α = π/2. The bracket uses the literal finite sums in (1.13), the primary recursive c-array in (1.14), its separate Bernoulli formula at π/2, and the partial ordinary Bell profile sum in (1.20)–(1.21). The convention is B₁ = −1/2. Natural subtraction and integer division are used only in indices; coefficient arithmetic takes place in ℂ. The external factor 2 exp(γ_E m)/π is nonzero, so the zero bracket gives the zero value of (1.13).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/LogLaplacianEvenResidueVanishing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

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    Source 1
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    Source 1
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    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2013The alternating vortex polygon has exactly $N/2-2$ distinct real eigenvalue pairsThe alternating unit polygon has exactly m minus two distinct pairs of nonzero real eigenvalues.Proved

The questionKevrekidis, Few-particle vortex cluster equilibria in Bose–Einstein condensates: existence and stability, J.

Why it mattersThe conjecture specifies how many distinct real eigenvalue pairs occur as the alternating vortex polygon grows.

Our contributionThe alternating unit polygon has exactly m minus two distinct pairs of nonzero real eigenvalues.

Verification details
Release source recordProved in Lean

Open problem

The alternating vortex polygon has exactly $N/2-2$ distinct real eigenvalue pairs

A. M. Barry and P. G. Kevrekidis, Few-particle vortex cluster equilibria in Bose–Einstein condensates: existence and stability, J. Phys. A: Math. Theor. 46, 445001 (2013), arXiv:1306.1545v2, Section 4.1, Conjecture 4.4 (BEC_VFP6.tex, lines 630–631):

For $N\geq 6$, the $N$-gon has $\frac{N}{2}-2$ distinct pairs of real eigenvalues.

The setting is the precessing point-vortex system of Section 2, Eqs. (1)–(2), with $b=2$:

$$\dot x_j=-S_j\Omega y_j-\sum_{k\ne j}S_k\frac{y_j-y_k}{(x_j-x_k)^2+(y_j-y_k)^2},\qquad \dot y_j=S_j\Omega x_j+\sum_{k\ne j}S_k\frac{x_j-x_k}{(x_j-x_k)^2+(y_j-y_k)^2}.$$

Section 4.1 specifies the charges:

We will assume hereafter that the vortices are of alternating charge, namely $S_j=(-1)^{j+1}$, $j=1,2, \dots,N$.

This is BEC_VFP6.tex, lines 506–507. The polygon has vertices $z_j=e^{2\pi i j/N}$, $1\le j\le N$ (line 515); the even polygon is an equilibrium at $\Omega=1/2$ (Lemma 4.2, lines 524–532). Thus the conjecture concerns even $N\ge6$ in this equilibrium family.

Issue #26188 fixes the reading. For every natural $m\ge3$, let $J_{2m}$ be the real Cartesian Jacobian of this field at the unit regular $2m$-gon. Every real root of its characteristic polynomial is accompanied by its negative, and the set of nonzero real roots has exactly $2(m-2)$ elements. Pairs count distinct nonzero values, without algebraic multiplicity. This statement is proved.

The source and encoding are described in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/Vortices/AlternatingPolygonSpectrum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

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    Source 1
  • Historical context · 2/4 · 10% weight

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    Source 1 · Source 2
Further frontiersQuestion resolvedFirst preprint 2019Informationally complete pairs are exactly complementary pairs with forest fiber graphsTwo finite real random variables determine every probability weight exactly when their joint value map is injective and their bipartite fiber graph is a forest.Proved

The questionGudder, Finite-Dimensional Convex Effect Algebras (2019), arXiv:1912.05110v1, Section 5, ends with this question.

Why it mattersGudder (2019), arXiv:1912.05110v1, Section 5, line 433, asks about pairs of individually incomplete observables:

The ideaNo probability hypothesis is needed for this lemma.

Our contributionTwo finite real random variables determine every probability weight exactly when their joint value map is injective and their bipartite fiber graph is a forest.

Scope & next questionCharacterizations in those settings and quantitative recovery from approximate distributions remain open here.

Verification details
Release source recordProved in Lean

Open problem

Informationally complete pairs are exactly complementary pairs with forest fiber graphs

S. Gudder, Finite-Dimensional Convex Effect Algebras (2019), arXiv:1912.05110v1, Section 5, ends with this question. The following is the verbatim TeX text at line 588 of finitedimensionalCEA.tex (md5 9dfe03bd484881c9db3b8b737095b005):

We conclude from Example~5 that complementarity is a necessary but not a sufficient condition for IC. It is an open problem to find a simple characterization of IC for a pair $\brac{f,g}\subseteq\rscript (X)$.

The source macros denote ${f,g}$ and $\mathcal R(X)$, respectively. Here $X={1,\ldots,n}$ with $n\ge2$, $\mathcal R(X)$ consists of all maps $X\to\mathbb R$, and a probability weight $\mu$ has $\mu(x)\ge0$ and $\sum_x\mu(x)=1$. Its distribution under $f$ is $\Phi_{f,\mu}(\Delta)=\sum_{x:,f(x)\in\Delta}\mu(x)$ for every $\Delta\subseteq\mathbb R$ (TeX lines 435–439). The pair is informationally complete (IC) when, for every two probability weights $\mu,\nu$, equality of their distributions under both $f$ and $g$ implies $\mu=\nu$ (line 474).

Let $\Gamma(f,g)$ have vertices in two tagged copies of $\mathbb R$, with a left value $a$ adjacent to a right value $b$ exactly when some $x\in X$ satisfies $f(x)=a$ and $g(x)=b$. Issue #23501 fixes the reading:

${f,g}$ is IC if and only if $x\mapsto(f(x),g(x))$ is injective and $\Gamma(f,g)$ is acyclic.

Injectivity is precisely the source's complementarity condition: every nonempty intersection of an $f$-level set and a $g$-level set is a singleton (lines 507–508). The characterization is proved, for every finite domain, without the restriction $n\ge2$. The definitions and source locators are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Observer/Tomography/InformationallyCompletePairForest.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

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    Source 1
  • Historical context · 2/4 · 10% weight

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Quantum informationConjecture provedFirst preprint 2020Hanaoka–Konno Conjecture 4.2: self-similarity of the Riesz walkFinite Riesz moments, orthogonal Laurent polynomials and scale-four projection identities.Proved

The questionKonno, “Return probability and self-similarity of the Riesz walk,” Quantum Information and Computation 21 (2021), 409–422, arXiv:2010.04518v3, DOI 10.26421/QIC21.5-6-5, state Conjecture 4.2:

Why it mattersThe conjecture connects the return law of a quantum walk with the fourth-power self-similarity of the Riesz measure.

The ideaThe Riesz moments stabilize in finite products, giving a positive form on Laurent polynomials.

Our contributionFinite Riesz moments, orthogonal Laurent polynomials and scale-four projection identities.

Verification details
Release source recordProved in Lean

Open problem

Hanaoka–Konno Conjecture 4.2: self-similarity of the Riesz walk

R. Hanaoka and N. Konno, “Return probability and self-similarity of the Riesz walk,” Quantum Information and Computation 21 (2021), 409–422, arXiv:2010.04518v3, DOI 10.26421/QIC21.5-6-5, state Conjecture 4.2:

Let Z_> = {1,2,3,...}. Put T_0(n)={0} and T_k(n)=((k-1)/n,k/n] for 1 <= k <= n. For the Riesz walk with initial state Psi_0=[1,0]^T delta_0, P(X_2t/(2t) in T_k(t)) = P(X_8t/(8t) in T_k(t)) for every t in Z_>.

Issue #15758 preregistered the exact natural-number reading. The delivered declaration is D5/S3/Quantum/Walks/RieszWalkSelfSimilarity.result : claim, and its module state has statement identity sha256:4b9935aefe3ec7033ed45b54d5dec573a17d5471a67ae35f412951ac228f6269.

Statement echo

The source's t >= 1 is the binder t : Nat with 1 <= t; its 0 <= k <= t is k : Nat with k <= t. The source's normalized probability is prob (2t) (Tset t k) and prob (8t) (Tset t k); Tset gives the singleton at k=0 and the left-open, right-closed rational interval for 1 <= k <= t. The conclusion is the equality in claim; result proves exactly that proposition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Walks/RieszWalkSelfSimilarity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

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    Source 1
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    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedFirst preprint 2013Mourad–Abbas Conjecture 5.1 is falseAn infinite family of positive-trace block matrices refutes Mourad–Abbas Conjecture 5.1.Refuted

The questionBassam Mourad and Hassan Abbas, On the symmetric doubly stochastic matrices that are determined by their spectra, arXiv:1310.1273v1, printed page 10, Conjecture 5.1:

Why it mattersThe positive-trace restriction does not eliminate reducible spectrally determined matrices.

The ideaThe quotient and its fibers assemble a single permutation, proving DS.

Our contributionAn infinite family of positive-trace block matrices refutes Mourad–Abbas Conjecture 5.1.

Verification details
Release source recordRefuted in Lean

Open problem

Mourad–Abbas Conjecture 5.1 is false

Bassam Mourad and Hassan Abbas, On the symmetric doubly stochastic matrices that are determined by their spectra, arXiv:1310.1273v1, printed page 10, Conjecture 5.1:

For $0<a\leq n$, the only elements of $\Delta_n^s(a)$ that are DS in $\Delta_n^s$ are points on the line segments $[I_n,C_n]$, $[I_n,P]$ and $[C_n,P]$ where $P$ is a vertex of $\Delta_n^s$.

Here $\Delta_n^s$ is the full polytope of real symmetric nonnegative matrices with every row and column sum equal to one, and $\Delta_n^s(a)$ is its trace-$a$ slice. A matrix is DS if every similar member of that full polytope is permutationally similar to it. Similarity means conjugation by an arbitrary invertible real matrix. All segments include their endpoints.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Real matrices of order $n\geq2$ | Matrix (Fin n) (Fin n) ℝ, order binder in claim | Unrestricted real entries; order bound makes $C_n$ meaningful. | | Symmetric doubly stochastic matrices | DeltaSymm, using doublyStochastic and Matrix.IsSymm | Full polytope, with nonnegativity and both stochastic sums. | | Trace $a$, $0<a\leq n$ | traceSlice and the real bounds in claim | No restriction to the constructed family. | | $C_n$ | C | Zero diagonal, off-diagonal $1/(n-1)$, with real subtraction. | | Permutation similarity | IsPermutationallySimilar | One permutation with $B=P^TAP$. | | Every similar symmetric doubly stochastic $B$ | IsDS, using IsConj | All real $B$ of the same order, with no orthogonality or irreducibility assumption. | | Only points of the three segment types | Conclusion of claim, using segment ℝ | Necessary condition; no converse is added. | | $P$ is a vertex of $\Delta_n^s$ | (DeltaSymm n).extremePoints ℝ | Extreme point of the full polytope, rather than the trace slice. |

The source conventions use existing Matrix, stochasticity, symmetry, conjugation, permutation-matrix, segment and extreme-point owners directly. The six public definitions retain their source expressions; family constructions and proof helpers are private.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/MouradAbbasC51Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2019Entropy saturation of a Lüders MIC channel forces SIC existenceSaturation of the Haar-averaged output entropy of a Lüders MIC channel forces a SIC.Proved

The questionStacey, Lüders channels and the existence of symmetric-informationally-complete measurements, Phys.

Why it mattersThe source interprets the average output entropy as the typical sharpening of expectations associated with a Lüders measurement.

The ideaThe integrated entropy gap is a sum of two nonnegative terms, so both vanish.

Our contributionSaturation of the Haar-averaged output entropy of a Lüders MIC channel forces a SIC.

Verification details
Release source recordProved in Lean

Open problem

Entropy saturation of a Lüders MIC channel forces SIC existence

J. B. DeBrota and B. C. Stacey, Lüders channels and the existence of symmetric-informationally-complete measurements, Phys. Rev. A 100, 062327 (2019), arXiv:1907.10999v2, Section III, “Entropic Optimality”, Conjecture 1 (lmc.tex, lines 417–420):

Equality is achievable in the statement of Theorem \ref{avgentropybnd} only if a SIC exists in dimension $d$.

The reference is to Theorem 2 (lines 385–390): every Lüders MIC channel has Haar-averaged pure-input output entropy at most $B_d=\log(d+1)-\frac{2}{d+1}\log2$, with equality achievable if a SIC exists in dimension $d$.

A MIC is a family $E=(E_i)_{i=1}^{d^2}$ of positive semidefinite complex $d\times d$ matrices, linearly independent over $\mathbb C$, with $\sum_iE_i=I$. Its Lüders channel is $\mathcal E_E(X)=\sum_i\sqrt{E_i},X\sqrt{E_i}$, using the positive square roots. A SIC is a rank-one equiangular MIC. These are the definitions in the source, lines 80–90 and 104–114, also recorded in the literature note.

Issue #25617 fixes the statement for every positive natural dimension $d$: if a MIC $E$ has Haar-averaged pure-input output entropy $B_d$, then there exists a SIC in dimension $d$. This is the proposition claim, proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/LudersEntropySaturationSIC.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Source 1
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Analysis & dynamicsQuestion resolvedSource 2020Derivative signs of the power-weighted hyperbolic cotangent remainderDerivative signs of the power-weighted hyperbolic cotangent remainder.Proved

The questionZhu, Completely monotonic integer degrees for a class of special functions, Conjecture 6.4:

Why it mattersThe two consecutive derivative orders locate the loss of positivity for a power-weighted Bernoulli remainder throughout the full integer parameter range.

The ideaThe cotangent partial fractions at an imaginary argument, followed by finite geometric division and the even zeta values, give

Our contributionDerivative signs of the power-weighted hyperbolic cotangent remainder.

Verification details
Release source recordProved in Lean

Open problem

Derivative signs of the power-weighted hyperbolic cotangent remainder

Zhu, Completely monotonic integer degrees for a class of special functions, Conjecture 6.4:

Let m ≥ 1, n ≥ 2, and pₙ(t) be defined as (2.2). Then ((−1)ⁿ tᵐ pₙ(t))^{(m+2n−2)} > 0 holds for all t ∈ (0, ∞) and ((−1)ⁿ tᵐ pₙ(t))^{(m+2n−1)} > 0 is not true for all t ∈ (0, ∞).

Equation (2.2) is

$$p_n(t)=\frac{\coth t}{t} -\sum_{k=0}^{n}\frac{2^{2k}B_{2k}}{(2k)!}t^{2k-2}.$$

Put $P(t)=(-1)^nt^mp_n(t)$ and $M=m+2n$. The first clause is (6.12), and the failure of universal strict positivity in the second clause is (6.13).

Formal reading

All Lean names below have prefix D5.S3.Analytic.CothBernoulliDegree.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $p_n(t)$ in (2.2) | Defs.p | The finite Bernoulli sum includes $k=0$ and $k=n$; its powers of $t$ are integer powers. | | $P(t)=(-1)^nt^mp_n(t)$ | Defs.P | The sign and the natural power $t^m$ are retained. | | Integers $m\ge1$, $n\ge2$ | Natural m, n with 1 ≤ m, 2 ≤ n | The domains coincide on the stated positive integer ranges. | | $P^{(M-2)}(t)>0$ for all $t>0$ | First conjunct of Defs.claim | The derivative is iteratedDeriv (m + 2 * n - 2) (P m n). | | $P^{(M-1)}(t)>0$ is not true for all $t>0$ | Second conjunct of Defs.claim | The universal strict positivity assertion is negated, with the same real argument domain. | | Both clauses | Result.result | Exactly Defs.claim, with no additional hypothesis. |

The value assigned by real division and integer powers at zero is immaterial: PartialFraction.hasSum_iteratedDeriv_P uses neighborhood agreement at each positive argument to transfer all iterated derivatives.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/CothBernoulliDegree/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2020.

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Quantum informationEquivalence provedFirst preprint 2022PPT-squared and IPT-squared equivalenceThe complex PPT-squared conjecture is equivalent to the IPT-squared conjectures over the complex and real fields.Proved

The questionRahaman, Positive maps and entanglement in real Hilbert spaces, arXiv:2207.02510v2, Ann.

Why it mattersIt does not prove the PPT-squared conjecture itself.

The ideaRestrict the lift to real matrices and complexify it back; the restriction is real IPT.

Our contributionThe complex PPT-squared conjecture is equivalent to the IPT-squared conjectures over the complex and real fields.

Scope & next questionIt is a numerical check, not a universal proof. - Open.

Verification details
Release source recordProved in Lean

Open problem

PPT-squared and IPT-squared equivalence

G. Chiribella, K. R. Davidson, V. I. Paulsen and M. Rahaman, Positive maps and entanglement in real Hilbert spaces, arXiv:2207.02510v2, Ann. Henri Poincaré 24 (2023), DOI 10.1007/s00023-023-01325-x, Section 9.2, asks:

We do not know if these conjectures are equivalent.

The source defines PPT-squared over complex matrices and IPT-squared over real and complex matrices. The source implications are $\mathrm{PPT}^2(\mathbb C) \Rightarrow \mathrm{IPT}^2(\mathbb C) \Rightarrow \mathrm{IPT}^2(\mathbb R)$.

Statement echo

The source conventions are represented by the public definitions SEP, IsEB, ptFirst, IsPPT, IsIPT, PPT2C, IPT2 and claim in D5/S3/Quantum/QuantumChannels/IptPptSquaredEquivalence.lean. Amplification and complete positivity use MatrixMap.kron and MatrixMap.IsCompletelyPositive from FiniteKrausChannel; the consumed kron_id_apply proves the source coordinate formula. The input-first Choi expression is MatrixMap.choi_matrix from FiniteKrausRepresentation, with submatrix Prod.swap Prod.swap; its entries are exactly the source formula. Thus the claim remains $(\mathrm{PPT2C}\leftrightarrow\mathrm{IPT2}(\mathbb C))\land(\mathrm{IPT2}(\mathbb C)\leftrightarrow\mathrm{IPT2}(\mathbb R))$. The settling declaration is result : claim. The matrix indices use Fin, scalars are generic over [RCLike 𝕜], and the flag identification is (finCongr (Nat.mul_comm 2 n)).trans finProdFinEquiv.symm.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/IptPptSquaredEquivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The complex PPT-squared conjecture is equivalent to the IPT-squared conjectures over the complex and real fields.

    Read source ↗
  • Historical context

    First preprint year: 2022.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Equal Sorting Classes and Outputs of the Stacks Avoiding 312, 31-2 and 3-12The stacks avoiding 312, 31-2 and 3-12 sort the same permutations and agree on their outputs.Proved

The questionZhao, arXiv:2410.17057v1, Conjecture 5.1 (Section 5, printed page 20):

The ideaAvoidance of the classical pattern and of the two vincular patterns coincide on stacks that are maintained by the machines, so the three machines run identically until the first push that one of them refuses and another accepts.

Our contributionThe stacks avoiding 312, 31-2 and 3-12 sort the same permutations and agree on their outputs.

Scope & next questionThe source lists the enumeration of Sortn(SC{312}) as left open by Cerbai, Claesson and Ferrari; the tuple-label recurrence does not settle it, and the agreement of the first nine terms with OEIS A202062 is not proved for all n.

Verification details
Release source recordProved in Lean

Open problem

Equal Sorting Classes and Outputs of the Stacks Avoiding 312, 31-2 and 3-12

Zhao, arXiv:2410.17057v1, Conjecture 5.1 (Section 5, printed page 20):

The sorting classes of SC_{312} and SC_{31̲2} are identical. That is, Sort_n(SC_{312}) = Sort_n(SC_{31̲2}) = Sort_n(SC_{31̲2̲}). Furthermore, for τ ∈ Sort_n(SC_{312}), we have SC_{312}(τ) = SC_{31̲2}(τ) = SC_{31̲2̲}(τ).

SC_σ is the right-greedy stack map whose stack, read from top to bottom, must avoid σ; σ is the classical pattern 312, the vincular pattern in which the entries playing 3 and 1 are adjacent, or the vincular pattern in which the entries playing 1 and 2 are adjacent. Sort_n(SC_σ) is the set of permutations of length n whose image avoids 231.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/VincularStack/VincularStackThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedHanna's diagonal ghost-power integrality conjectureDiagonal self-powers preserve integrality of formal exponential ghost series.Proved

The questionHanna's second comment in OEIS A156216, internal revision 3, reads:

Why it mattersThe source note is Library/RecurrenceComposition/hanna2009a156216.md.

The ideaThe reused PrimitiveEulerLedger.uniqueprimitiveeulerledger gives its integer coefficients and stable finite products.

Our contributionDiagonal self-powers preserve integrality of formal exponential ghost series.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's diagonal ghost-power integrality conjecture

Paul D. Hanna's second comment in OEIS A156216, internal revision #3, reads:

More generally, if exp( Sum_{n>=1} C(n) * x^n/n ) equals a power series in x with integer coefficients, then exp( Sum_{n>=1} C(n)^n * x^n/n ) also equals a power series in x with integer coefficients (conjecture).

For a characteristic-zero field $K$ and $C:\mathbb N\to K$, put $H_C=\sum_{n\ge1}C_nX^n/n$ and $F_C=\exp(H_C)$, interpreted formally. The assertion is $F_C\in\mathbb Z[[X]]\Rightarrow F_{(C_n^n)}\in\mathbb Z[[X]]$, with the rings of integer series understood through the canonical embedding into $K[[X]]$.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | The sequence $C(n)$ | claim quantifies $K$, its field and characteristic-zero structures, and $C:\mathbb N\to K$ | Every binding field and sequence | | $\sum_{n\ge1}C_nX^n/n$ | logarithmicInput has coefficient zero at zero and $C_n/(n:K)$ otherwise | Entry $C_0$ is unused | | The formal exponential | exponentialSum substitutes the logarithmic input into PowerSeries.exp K | Ordinary coefficients; no analytic condition | | Integer input coefficients | The antecedent is the existing HannaLogarithmicPowerClosure.IntegralSeries | Canonical integer image | | $C(n)^n$ | The consequent uses the sequence $n\mapsto C_n^n$ | Index-dependent natural power | | Integer output coefficients | The consequent uses the same IntegralSeries | No positivity or nonzero restriction |

The existing HannaDiagonalDerivativeParity.step a 0 is used in the private differential-equation proofs; the public claim retains the literal formal exponential. No source clause is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Witt/HannaDiagonalGhostPowerIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The linear spectral condition (31) and absolute separabilityThe linear spectral condition in Eq. (31) implies absolute separability.Proved

The questionWinter, The geometry of absolute separability and other convex matrix properties from spectrum, arXiv:2608.03390v2, Section 7 (page 39):

The ideaThe proof uses spectral estimates.

Our contributionThe linear spectral condition in Eq. (31) implies absolute separability.

Verification details
Release source recordProved in Lean

Open problem

The linear spectral condition (31) and absolute separability

J. Ahiable, N. B. T. Kothakonda and A. Winter, The geometry of absolute separability and other convex matrix properties from spectrum, arXiv:2608.03390v2, Section 7 (page 39):

Thus, another matter of interest is to resolve Conjecture 6.7 in the positive, and to show that the linear condition in Eq. (31) is sufficient for absolute separability: because then the maximum purity over all three sets must coincide.

Theorem 6.2 (page 23) proves that a state on $\mathcal M_m\otimes\mathcal M_n$ whose decreasingly ordered eigenvalues satisfy $2\lambda_{mn}+\sum_{k=1}^{m-1}\lambda_{mn-k}\ge\sum_{k=1}^{m-1}\lambda_k$ (31) is absolutely PPT. A state is absolutely separable when every $U\rho U^\dagger$, $U\in\mathcal U(mn)$, is separable. The verbatim texts are in the literature note. Issue #14958 preregisters the reading ($2\le m\le n$, the repository's finite-sum separable cone) and the route.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsoluteSeparability/AhiableKothakondaWinterEq31.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Eventual positivity of the complete D-prime familyAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Conjecture 5: “If k>m, then sum of D'(k,m,n) q^n is eventually positive.” The source defines positive to mean that all coefficients are nonnegative.Proved

The questionAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Conjecture 5: “If k>m, then sum of D'(k,m,n) q^n is eventually positive.” The source defines positive to mean that all coefficients are nonnegative.

Why it mattersThe assertion includes both coefficient parities, the smallest pair (2,1), the m = 1 family and the m = k−1 family.

The ideaTwo applications of the existing nilpotent q-binomial identity prove the full source transformation modulo X^{n+1}.

Our contributionAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Conjecture 5: “If k>m, then sum of D'(k,m,n) q^n is eventually positive.” The source defines positive to mean that all coefficients are nonnegative.

Verification details
Release source recordProved in Lean

Open problem

Eventual positivity of the complete D-prime family

George E. Andrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Conjecture 5: “If k>m, then sum of D'(k,m,n) q^n is eventually positive.” The source defines positive to mean that all coefficients are nonnegative. Its parameters k and m are positive integers.

The exact assertion is: for every natural k and m with 1 ≤ m < k, there exists a natural N depending on k and m, such that D'(k,m,n) ≥ 0 for every natural n ≥ N. Equation (1.3) defines this coefficient by

Σ_n D'(k,m,n) q^n = Σ_{j≥0} q^{m(2j+2)} (q^{2j+4};q²)_∞ (q^{2j+2+2k};q²)_∞ / (q^{2j+3};q²)_∞².

The original AndrewsElBachraouiDefs.dCoeff uses precisely these factors and degree n, without a coefficient shift. For m ≥ 1, omitted summands and product factors have degree above n. The finite truncation therefore retains the source coefficient exactly.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TwoColorPartition/AndrewsElBachraouiFive.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBala's A193237 congruences modulo five and sevenBala's unbounded modulo-five and modulo-seven coefficient congruences.Proved

The questionThe existence and universal-integrality clauses make the source's normalized generating-function interpretation explicit.

Why it mattersD5.S1.Recurrence.Residue.BalaA193237ModFiveSeven.result proves the full claim.

The ideaThe shared BalaA193237ModFiveSeven.resolve combines integral-root construction, triangular support, Frobenius and rational uniqueness.

Our contributionBala's unbounded modulo-five and modulo-seven coefficient congruences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A193237 congruences modulo five and seven

Peter Bala's FORMULA line in OEIS A193237 states: “Conjectures: a(5n+4) == 0 (mod 5) and a(7n+5) == 0 (mod 7) (both checked up to n = 200).” The source is preregistered in #22112. The companion claim is #22113.

Write $T_j=j(j+1)/2$ and $F(x)=\sum_{j\ge0}(2j+1)3^j(-1)^{T_j}x^{T_j}$. The question concerns the constant-one reciprocal cube root $A(x)=F(x)^{-1/3}$ and every natural index, without a finite cutoff.

Statement echo

| Source clause or convention | Formal counterpart | Scope | | --- | --- | --- | | The displayed source series | BalaA193237ModFiveSeven.F | Its degree-$m$ coefficient sums the literal source weights over $j<m+1$ with $T_j=m$. The truncation is exact because $T_j=m$ implies $j\le m$. | | The constant-one reciprocal cube root | BalaA193237ModFiveSeven.IsGF | Rational power series with constant coefficient one and $A^3F=1$. | | A generating function exists | First conjunct of claim | An explicit integral inverse gives a rational normalized solution; the universal clause is nonvacuous. | | Coefficients of the sequence | Integer sequence quantified in claim | Every coefficient of every normalized rational solution equals the cast of its corresponding integer value. | | Modulo five | First divisibility in claim | $5\mid a(5n+4)$ for every $n\in\mathbb N$. | | Modulo seven | Second divisibility in claim | $7\mid a(7n+5)$ for the same universally quantified $n$. |

The existence and universal-integrality clauses make the source's normalized generating-function interpretation explicit. The source-specific definitions are literal; the shared proof uses BalaA193237ModFiveSeven.triangular with the source weights.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/BalaA193237ModFiveSeven.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedBala's OEIS A363411 and A363412 valuation conjecturesInteger quadratic products and Bala's valuation conjectures.Proved

Why it mattersThese four statements are explicit OEIS conjectures.

The ideaThe type-one theorem extends finite residue checks with strong induction.

Our contributionInteger quadratic products and Bala's valuation conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's OEIS A363411 and A363412 valuation conjectures

P. Bala's OEIS A363411 asks whether the real coordinate of $$P_n=\prod_{k=0}^{n}(1+k\sqrt{-3})$$ satisfies $v_5(a(n+2))=A079998(n)$, where $a(n)=\operatorname{Re}(P_n)$ and A079998 is the characteristic function of the multiples of five. The companion A363411 type-one statement lists the primes at most $101$ that divide no real term. The companion A363412 statements concern $b(n)=\operatorname{Im}(P_n)/\sqrt3$: $3\nmid b(3n+1)$ and $v_3(b(3n))=v_3(b(3n-1))$ for $n\ge1$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $P_n=\prod_{k=0}^{n}(1+k\sqrt{-3})$ in $\mathbb Z[\sqrt{-3}]$ | D5.S3.Arith.Congruence.BalaA363411.P | Literal finite product over Finset.range (n + 1) | | A363411 real coordinate | BalaA363411.a | Zsqrtd.re of P | | A363412 imaginary coordinate divided by $\sqrt3$ | BalaA363411.b | Zsqrtd.im coordinate | | $v_5(a(n+2))=1$ when $5\mid n$, and $0$ otherwise | BalaA363411.claim and result | Exact all-$n$ quantifier | | Type-one primes through $101$ | BalaA363411TypeOne.claim and result | Exact finite prime bound and all-term quantifier | | $3\nmid b(3n+1)$ | BalaA363412.claim and result | Exact all-$n$ quantifier | | $v_3(b(3n))=v_3(b(3n-1))$ for $n\ge1$ | BalaA363412Valuation.claim and result | Exact positive-index condition |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BalaA363411.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Equal third and unequal second principal specializations for a caterpillar familyFor every positive j, the source caterpillars (5,j,j,4,3) and (4,j,j,5,3) have different second and identical third ordinary chromatic symmetric principal specializations.Proved

The questionAram Bingham, Kyle Celano, Patricia Commins, Rosa Orellana, Jianping Pan, and Mary Claire Simone, Principal specializations of the t-chromatic symmetric function, arXiv:2610.31785v1, Remark 3.11, first question:

Why it mattersThe source reports verification through j = 7.

The ideaRestrict a proper coloring to the spine and its leaves.

Our contributionFor every positive j, the source caterpillars (5,j,j,4,3) and (4,j,j,5,3) have different second and identical third ordinary chromatic symmetric principal specializations.

Verification details
Release source recordProved in Lean

Open problem

Equal third and unequal second principal specializations for a caterpillar family

Aram Bingham, Kyle Celano, Patricia Commins, Rosa Orellana, Jianping Pan, and Mary Claire Simone, Principal specializations of the t-chromatic symmetric function, arXiv:2610.31785v1, Remark 3.11, first question:

Do (5,j,j,4,3) and (4,j,j,5,3) exhibit this phenomenon for all j?

The single target is the conjunction, for every positive integer j, that the actual caterpillars of Definition 2.1 associated with these two compositions have unequal ordinary second principal specializations and equal ordinary third principal specializations. Equality and inequality are those of formal integer polynomials in X. Equation (2.1) uses colors 0 through k−1 and sums over all proper colorings; every color need not occur. The coefficient inclusion into the source's complex polynomial ring is injective. The t refinement is outside this question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PrincipalSpecialization/BinghamCaterpillarPs3Family.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Caldwell and coauthors' super Catalan action-graph identityFor every nonnegative n, the weighted sum of the path-table columns of the recursively constructed action graph equals S(0, n + 1).Proved

The questionTebbe, Catalan Number Sequences and Generalized Action Graphs, Ball State Undergraduate Mathematics Exchange 18(1), Fall 2024, 88–106, Conjecture 5; arXiv:2507.22719v1, Conjecture 5.6:

Why it mattersThe graph encoding retains distinct vertices as child-list occurrences; repeated equal subtrees retain multiplicity.

The ideaDeleting the last edge yields pathsgrowsucc; summing over all starting vertices yields totalPathsgrowzero and totalPathsgrowsucc.

Our contributionFor every nonnegative n, the weighted sum of the path-table columns of the recursively constructed action graph equals S(0, n + 1).

Verification details
Release source recordProved in Lean

Open problem

Caldwell and coauthors' super Catalan action-graph identity

D. Caldwell, A. Cochran, N. Glisson, B. Jennings, K. McDicken, L. Proctor, S. Klanderman and A. Tebbe, Catalan Number Sequences and Generalized Action Graphs, Ball State Undergraduate Mathematics Exchange 18(1), Fall 2024, 88–106, Conjecture 5; arXiv:2507.22719v1, Conjecture 5.6:

The subsequent super Catalan number can be computed from the n-table of its previous action graph via S(0, n + 1) = Σ_{ℓ=0}^{n} (2/2^ℓ) Σ_{v=0}^{n} Kℓ,v,n.

For every $n\ge0$, with precisely the graphs of Definition 5.2 and the path counts of Definition 5.4, the named assertion is

$$S(0,n+1)=\sum_{\ell=0}^{n}\frac{2}{2^\ell}\sum_{v=0}^{n}K_{\ell,v,n}.$$

Here Definition 5.1 is used literally: $S(m,n)=(2m)!(2n)!/(m!n!(m+n)!)$, in $\mathbb Q$. Preregistration: #14721, CALD-1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SuperCatalanActionGraphRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The diameter of the inverse-closed consecutive 3-cycle Schreier coset graphFor 2 <= L <= N and N > 3, on binary words with L zeros and N - L ones, where a move rotates three consecutive letters one place in either direction, every word can be turned into every other in at most the ceiling of L(N - L)/2 moves, and turning the sorted word 0...01...1 into its reversal needs that many, as conjectured for k = 3 in Conjecture 16 of CayleyPy-4.Proved

The questionCayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph SN / (SL × S{N−L}) whose vertices are the binary words with L zeros and N − L ones and whose generators are the consecutive k-cycles (i, i + 1, …, i + k − 1), 0 ≤ i ≤ N − k, for N > k.

Why it mattersThe paper proposes these diameters as bivariate quasipolynomials in N and N − L, with leading term L(N − L)/(k − 1).

The ideaThe inversion count of a word is the number of positions p < q holding a one at p and a zero at q.

Our contributionFor 2 <= L <= N and N > 3, on binary words with L zeros and N - L ones, where a move rotates three consecutive letters one place in either direction, every word can be turned into every other in at most the ceiling of L(N - L)/2 moves, and turning the sorted word 0...01...1 into its reversal needs that many, as conjectured for k = 3 in Conjecture 16 of CayleyPy-4.

Verification details
Release source recordProved in Lean

Open problem

The diameter of the inverse-closed consecutive 3-cycle Schreier coset graph

CayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph S_N / (S_L × S_{N−L}) whose vertices are the binary words with L zeros and N − L ones and whose generators are the consecutive k-cycles (i, i + 1, …, i + k − 1), 0 ≤ i ≤ N − k, for N > k. In the inverse-closed case a move rotates k consecutive letters one place in either direction. The k = 3 clause of Conjecture 16 states:

For $k=3$ from $L\ge 2$(?): $D(L, N) = \left\lceil \frac{L (N - L)}{2} \right\rceil$

Issue #10230 fixes the readings: words are List Bool with false for 0; "diameter" is read both as CayleyPy computes it, the largest distance from the central state [0]^L + [1]^{N−L}, and as the largest distance between two vertices; the range is 2 ≤ L ≤ N and N > 3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ShrunkenGrassmannianThreeCycleDiameter.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedFirst preprint 2025Coulter–Do Conjecture 5.8The literal differential operator and native matching operator use exactly one occurrence coefficient on every partition.Proved

The questionXavier Coulter and Norman Do, arXiv:2506.04002v1, Conjecture 5.8 (con:b-laplace, original TeX lines 2144–2149):

Why it mattersThe sum of the original deformed Jucys–Murphy operators is defined without any commutation assumption.

The ideaThe candidate uses the original partner maps, actual transposition action and output-matching charges.

Our contributionThe literal differential operator and native matching operator use exactly one occurrence coefficient on every partition.

Verification details
Release source recordProved in Lean

Open problem

Coulter–Do Conjecture 5.8

Xavier Coulter and Norman Do, arXiv:2506.04002v1, Conjecture 5.8 (con:b-laplace, original TeX lines 2144–2149):

For all $\lambda \vdash k$, $\mathrm{ch}^{(b)}((\mathcal J_1+\cdots+\mathcal J_k)\cdot\mathbf p_\lambda) =D(b)\cdot\mathrm{ch}^{(b)}(\mathbf p_\lambda)$.

The original source uses the macros \J and \p for the two symbols above. Here every positive integer k and every partition lambda of k is in scope, the scalar field is C(b), and b is an indeterminate. The class vector has coefficient one at precisely the matchings of coset type lambda. Membership of the operator image in the class-constant space is part of the assertion, since the characteristic map has that space as its domain. Explicitly, one coefficient function c on all partitions of k must satisfy both

  1. S_X(indicator lambda)(p) = c(ct(p)) for every actual matching p;
  2. sum_mu (c(mu)/A_mu) p_mu = D_X((A_lambda)^(-1) p_lambda).

The normalization is literally A_mu=(1+b)^(number of parts of mu) product_r r^(m_r(mu)) m_r(mu)!. The differential operator acts on C(b)[p_1,p_2,...], with all positive derivative indices and the original ordered cut and join sums.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PerfectMatchings/DeformedJucysMurphyOccurrence.conjecture58

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Analysis & dynamicsQuestion resolvedFirst preprint 2025The distinct upper degrees of deterministic simplicial complexes shift by dimensionThe distinct upper degrees in positive dimension are zero together with the vertex degrees at the dimension-shifted generation.Proved

The questionKrapivsky, Deterministic simplicial complexes, arXiv:2507.07402v1; J.

Why it mattersThe source studies local structure through upper degrees, alongside simplex counts and Hodge Laplacian spectra.

The ideaUpper degree is the number of singleton members of the link.

Our contributionThe distinct upper degrees in positive dimension are zero together with the vertex degrees at the dimension-shifted generation.

Verification details
Release source recordProved in Lean

Open problem

The distinct upper degrees of deterministic simplicial complexes shift by dimension

S. N. Dorogovtsev and P. L. Krapivsky, Deterministic simplicial complexes, arXiv:2507.07402v1; J. Stat. Mech. (2025) 123404, Section II.3, equation (34). The source TeX DSC-7.tex, lines 505–510, states:

Another useful general relation

$K^{(d)}(n) = 0 \cup K^{(0)}(n-d)$

is conjecturally valid for all $n \geq d + 1$ and $d \geq 1$.

Here $K^{(d)}(n)$ lists the distinct upper degrees in ascending order, without multiplicities (TeX lines 415–425); the $0$ on the right denotes adjoining the value zero. The growth rule and upper degrees are also described in the literature note.

Issue #20715 fixes the precise reading. Let $V$ be a type with decidable equality, let $v_0\in V$, and let $v:\mathbb N\to\operatorname{Finset}(V)\to V$ be injective as a function of the pair $(s,\sigma)$, with $v(s,\sigma)\ne v_0$ for every $s,\sigma$. Define the finite families of nonempty faces by

$C_0={{v_0}}$,

$C_{s+1}=C_s\cup\bigcup_{\sigma\in C_s}{{v(s+1,\sigma)}\cup\beta:\beta\subseteq\sigma}$.

Every existing nonempty face receives its own fresh vertex, all subsets $\beta$ including the empty subset are used, and earlier faces persist. The labeling hypotheses ensure that a new vertex differs from every earlier vertex and from the other vertices added in the same step.

For a finite family $C$ and a finite vertex set $\tau$, define

$\operatorname{upperDegree}(C,\tau)=#{\rho\in C:\tau\subseteq\rho,\ |\rho|=|\tau|+1}$,

$\operatorname{degreeSet}(C,d)={k\in\mathbb N:\exists\tau\in C,\ |\tau|=d+1,\ \operatorname{upperDegree}(C,\tau)=k}$.

Write $K^d(n)=\operatorname{degreeSet}(C_n,d)$. The claim is that, for every such $V,v_0,v$ and all natural numbers $d,n$ with $d\geq1$ and $n\geq d+1$,

$K^d(n)={0}\cup K^0(n-d)$.

Subtraction is natural subtraction; the hypotheses imply $n-d\geq1$. Equality of these sets expresses equality of the source's lists of distinct values. This claim is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/DeterministicSimplicialDegreeShift.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025The Katz cut-off limit for path graphsThe cut-off values αn converge to 1/√5 from above.Proved

The questionEvans, Francis and Jones, Katz similarity index comparisons, arXiv:2509.02874v1, Conjecture 4:

Why it mattersThe cut-off separates the Katz and resistance orderings of pairs of path vertices.

The ideaThe positive continued-fraction recursion in Monotone proves Monotone.RABstrictMonoOn and Monotone.Rstrictsucc.

Our contributionThe cut-off values αn converge to 1/√5 from above.

Verification details
Release source recordProved in Lean

Open problem

The Katz cut-off limit for path graphs

Evans, Francis and Jones, Katz similarity index comparisons, arXiv:2509.02874v1, Conjecture 4:

The cut-off values α_n converge to 1/√5 from above.

The source defines the Katz matrix by $K_{P_n}(\alpha)=(I-\alpha A_n)^{-1}-I$ and the cut-off values as roots of $p_{n,1}(\alpha)=\min_i K_{i,i+1}-\max_i K_{i,i+2}$, with each extremum restricted to existing vertex pairs. Theorem 19 locates a root between $1/\sqrt5$ and $1/2$ for every $n\ge6$. The preregistration is #20122.

Statement echo

| Source clause or convention | Lean counterpart | Scope | | --- | --- | --- | | The path $P_n$ and its adjacency matrix | SimpleGraph.pathGraph n, in Reduction.katz | Vertices are Fin n, indexed from zero. | | $(I-\alpha A_n)^{-1}-I$ and its entries | Reduction.katz, Reduction.katzEntry | The matrix inverse is Mathlib's inverse. Out-of-range natural indices are extended by zero and excluded from both extrema. | | The minimum for separation one and maximum for separation two | Reduction.p1 | The ranges are Fin (n-1) and Fin (n-2). | | Theorem 19's existence | First conjunct of Result.claim | Every $n\ge6$ has a root in $(1/\sqrt5,1/2)$. | | Convergence from above | Second conjunct of Result.claim, proved by Result.result | Every sequence choosing roots in $(0,1/2)$ for $n\ge6$ lies above $1/\sqrt5$ there and tends to it. The first six values are unrestricted. |

All abbreviated declaration names in this dossier have prefix D5.S1.Recurrence.KatzCutoff..

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/KatzCutoff/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The five-cycle graph state is strongly 1-resistantThe five-cycle graph state is genuinely multipartite entangled, remains genuinely multipartite entangled after any one-qubit loss, and becomes fully separable after any two-qubit loss.Proved

The questionHan, Zhang and Zhang, arXiv:2606.08561v1, page 8, Discussion, ask:

The ideaFor the initial density matrix ρ₅ use the witness W = I/2 − ρ₅.

Our contributionThe five-cycle graph state is genuinely multipartite entangled, remains genuinely multipartite entangled after any one-qubit loss, and becomes fully separable after any two-qubit loss.

Verification details
Release source recordProved in Lean

arXiv

The five-cycle graph state is strongly 1-resistant

Han, Zhang and Zhang, arXiv:2606.08561v1, page 8, Discussion, ask:

Several open problems remain. First, do C₅ and C₆ give strongly m-resistant graph states for m = 1 and m = 2, respectively? Here, “strong” means genuine multipartite entanglement rather than mere entanglement.

This dossier concerns the C₅, m = 1 clause. The five-cycle graph-state amplitudes are the cyclic controlled-Z phase on five qubits, divided by √32. Definition 2 of arXiv:2505.06567v1 requires initial genuine multipartite entanglement (GME), GME after every one-qubit loss, and full separability after every two-qubit loss. The formal claim is IsStrongResistant 1 (cycleGraphState 5).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/CycleFiveStrongOneResistance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Discrete structuresQuestion resolvedHanna's three congruences for A365771Base-five digit descent proves Hanna's first residue and the shared classification.Proved

The questionConjectures: given A033042 is the sums of distinct powers of 5, then (1) a(5A033042(n)) == 4 (mod 5) for n > 0, (2) a(5A033042(n) + 1) == 2 (mod 5) for n > 0, (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5A033042(k) for some k >= 0.

Why it mattersThe first congruence is one part of a complete base-five description of this Catalan-binomial sequence.

The ideaThe resulting public residueclassification gives result after appending a zero digit.

Our contributionBase-five digit descent proves Hanna's first residue and the shared classification.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's three congruences for A365771

OEIS A365771, revision 23, defines $a(n)=\binom{2n+1}{n}/(2n+1)\cdot\binom{3n-1}{n}$ for $n\geq0$ and states:

Conjectures: given A033042 is the sums of distinct powers of 5, then (1) a(5*A033042(n)) == 4 (mod 5) for n > 0, (2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0, (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.

The problem is preregistered in #18656. The three settlements are Proved. For every positive sum $m$ of distinct powers of five, $a(5m)\equiv4\pmod5$ and $a(5m+1)\equiv2\pmod5$. For every positive $n$ outside $n=5m$ or $n-1=5m$ with $m$ a sum of distinct powers of five (including zero), $a(n)\equiv0\pmod5$. The companion statements are preregistered in #18657 and #18658.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | The formula for $a(n)$, including $n=0$ | HannaA365771.a | Literal natural-number binomial quotient and product; subtraction is truncated. | | “Sums of distinct powers of 5.” | HannaA365771.InA033042 | Every digit of Nat.digits 5 m is at most one, including the empty digit list at zero. | | “for n > 0” in the enumeration | The positive-value hypothesis in HannaA365771.claim | A033042 starts at zero, so the positive enumeration indices select exactly its positive values. | | Residue four at five times the selected value | HannaA365771.result | Proves the full claim, with no extra hypothesis. |

Unqualified formal names in this dossier belong to D5.S1.Recurrence.Residue.HannaA365771. The companion settling declarations are D5.S1.Recurrence.Residue.HannaA365771Two.result and D5.S1.Recurrence.Residue.HannaA365771Zero.result; their claims retain the positive-$m$ restriction for (2) and the negated existential, including $m=0$, for (3). Natural division agrees with the displayed quotient because the first factor is an integer Catalan number.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA365771.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026Perfect homogeneous Wronskian extensions exist exactly at order oneA perfect homogeneous one-generator polynomial extension exists exactly at differential order one.Proved

The questionThe following is the verbatim TeX paragraph at line 1259 of FinDim220526v2.tex (md5 777135f176c9db701f21e5554216fb33):

Why it mattersThe source's research problem concerns perfect finite-dimensional polynomial algebras with the complete Wronskian as bracket (TeX lines 525–527).

The ideaRepetitions give equal columns and zero determinant; injective selections differ from one omission bracket by a column-permutation sign.

Our contributionA perfect homogeneous one-generator polynomial extension exists exactly at differential order one.

Verification details
Release source recordProved in Lean

Open problem

Perfect homogeneous Wronskian extensions exist exactly at order one

Markuss G. Ķēniņš and Arthemy V. Kiselev, Explicit class of finite-dimensional polynomial algebras with Wronskians over $\mathbb R^d$ as $N$-ary Lie brackets: beyond $\mathfrak{sl}(2)$, arXiv:2605.27305v2 (2026), subsection (Im)perfect N-ary SH-Lie polynomial algebras, Open problem open:perfect-algebras. The following is the verbatim TeX paragraph at line 1259 of FinDim220526v2.tex (md5 777135f176c9db701f21e5554216fb33):

Let the base dimension $d\geqslant 2$ and differential order $k\geqslant 1$ be arbitrary. Put the complete generalised Wronskian $W_d^k=\Bone\wedge \partial_\bx\wedges \partial^k_{\bx...\bx}$ as the $N$\nobreakdash-ary bracket, $N={d+k\choose d}$, on the $N$\nobreakdash-ary SH-Lie polynomial algebra $\Bbbk[x^1,...,x^d]$. Does there exist an homogeneous polynomial $p(\bx)\in \Bbbk[\bx]$ of degree $k+1$ such that the $N$\nobreakdash-ary SH-Lie polynomial subalgebra $\EuA=\Bbbk_k[\bx]\oplus\langle p(\bx)\rangle$ is \emph{perfect} w.r.t\ the Wronskian, i.e.\ $\Span_\Bbbk W_d^k(\EuA,...,\EuA)=\EuA$?

The source macros are retained in the quotation: \Bone is the identity operator, \bx denotes $\boldsymbol x$, \wedges denotes $\wedge\cdots\wedge$, \EuA denotes $\mathcal A$, and \Span denotes linear span. The source takes $\Bbbk=\mathbb R$ or $\mathbb C$ and writes $\Bbbk_k[\boldsymbol x]$ for polynomials of degree at most $k$ (TeX lines 538–542 and 613–623).

Issue #22115 fixes the statement over a characteristic-zero field $K$. Put

$I={\alpha\in\mathbb N^d:|\alpha|\le k},\qquad W(f)=\det(\partial^\alpha f_\beta)_{\alpha,\beta\in I},$

$A_p=K_{\le k}[\boldsymbol x]+Kp,\qquad D_p=\operatorname{span}_K{W(f):f:I\to A_p}.$

Here $\partial^\alpha$ is ordinary iterated partial differentiation, without divided-power normalization. The settled statement is: for every characteristic-zero field $K$, every $d\ge2$ and every $k\ge1$,

$\bigl(\exists p\in K[\boldsymbol x],\ p\ne0,\ p\text{ homogeneous of degree }k+1,\ D_p=A_p\bigr)\ \Longleftrightarrow\ k=1.$

This classification is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/WronskianPerfectLonelyAlgebra.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026The simple evolution-algebra automorphism maximumThe full automorphism maximum for simple complex evolution algebras.Proved

The questionKaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, Conjecture 4, section 2.3, page 28:

Why it mattersSimplicity adds strong connectivity to the invertible structure map.

Our contributionThe full automorphism maximum for simple complex evolution algebras.

Verification details
Release source recordProved in Lean

Open problem

The simple evolution-algebra automorphism maximum

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, Conjecture 4, section 2.3, page 28:

Over ℂ, the maximal order of the automorphism group of a simple n-dimensional evolution algebra is max{n!, n(2ⁿ−1)}.

The dimension is positive. The source's natural-basis condition is zero mixed products. The product is complex bilinear and commutative, without associativity or a multiplicative unit. Automorphisms are the entire group of complex linear multiplication-preserving equivalences.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Every positive dimension | claim quantifies over $n\in\mathbb N$ with $1\le n$ | | Every complex vector space of dimension $n$ | Arbitrary carrier, complex module, finite-dimensional instance and finrank equality | | Evolution product and natural basis | Bilinear product, arbitrary finite basis, zero mixed products and commutativity | | Simple algebra | IsSimple: $A^2\ne0$ and every complex-linear two-sided ideal is zero or the whole space | | Full finite automorphism group and upper bound | Finite (Aut μ) together with the bound on Nat.card | | Maximum is attained | A product on $\mathbb C^n$ with the dimension, commutativity, natural basis, simplicity, finiteness and exact cardinality |

The simplicity definition is literal: “We recall that an algebra A is simple if A² ≠ 0 and 0 is the only proper ideal.” This sentence is from the survey's cited arXiv:1601.01832v2, section 4, page 18. All two-sided complex-linear ideals are quantified, including those not spanned by natural-basis vectors. The coordinate representative in the attainment clause does not restrict the universal clause. Finite prevents the natural cardinality of an infinite group from satisfying a bound through its value zero. The extra dimension equality in the attaining clause records that the chosen representative has the required dimension.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/SimpleEvolutionAutomorphismMaximum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Strict containment for graph quantum magic squares on the five-cycleThe C5 graph quantum magic squares strictly contain the matrix convex hull of graph quantum permutation matrices.Proved

The questionLa Piana, Graph Quantum Magic Squares and Free Spectrahedra, arXiv:2512.08797v3, Question 1.3 asks:

Why it mattersThis settlement supplies a size-two graph quantum magic square outside the entire matrix convex hull, including compressions from arbitrary finite internal sizes.

The ideaProducts of entries in the first two rows construct these families; orthogonality and the cycle recurrence extend the identity to every row. matrixconvexdihedral preserves this form under normalized blockwise compressions.

Our contributionThe C5 graph quantum magic squares strictly contain the matrix convex hull of graph quantum permutation matrices.

Verification details
Release source recordProved in Lean

Open problem

Strict containment for graph quantum magic squares on the five-cycle

F. La Piana, Graph Quantum Magic Squares and Free Spectrahedra, arXiv:2512.08797v3, Question 1.3 asks:

Given a graph Γ, does the matrix convex hull of its quantum permutation matrices P^(Γ) generate all graph quantum magic squares? That is, mconv(P^(Γ)) = M^(Γ)?

Section 4.1 names $C_5$ and the Petersen graph among the candidates for strict containment. The settlement is Proved for $C_5$, under the complete conventions preregistered in #19303:

$\operatorname{mconv}(\mathcal P^{(C_5)})_s\subseteq\mathcal M^{(C_5)}_s$ at every internal size, and some $X\in\mathcal M^{(C_5)}_2$ lies outside that hull.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Definition 2.1: positive semidefinite blocks and identity row/column sums | IsQMS | All external and internal natural-number sizes; complex matrices | | Definition 3.1: adjacency-block commutation | IsGraphQMS | Every pair of vertices, both adjacency-filtered sums | | Definition 3.2: graph magic squares | M | Set of blocks satisfying IsGraphQMS | | Definition 3.2: idempotent, Hermitian graph magic squares | P | Both projection equations for every block | | Definition 2.3: normalized finite blockwise compressions | IsMatrixConvex | All finite families, varying input sizes and all output sizes | | Smallest containing matrix convex family | mconv | Intersection of every containing matrix convex family | | Section 4.1, named candidate $C_5$ | claim, result | All-level containment and existential separation |

The source uses positive sizes and a Hermitian ambient family. The preregistered encoding uses arbitrary complex block families and includes level zero; IsQMS itself enforces Hermitian blocks through positivity. The size-zero matrices are unique. Fin indices, the matrix identity 1, and SimpleGraph.cycleGraph 5 implement the source's indices, identity and graph. The strictness statement is existential in the internal size, with a size-two witness; it does not assert strictness at every size.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/LaPianaGraphQuantumMagicSquares.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2025Exact rational attainment of finite-group subgroup conjugacy ratiosEvery rational strictly between zero and one is the subgroup-conjugacy ratio of an actual finite group.Proved

The questionMihai-Silviu Lazorec and Marius Tărnăuceanu, On the number of conjugacy classes of subgroups of a finite group, arXiv:2508.14267v2, Section 4, Problem 4.1:

Why it mattersExact attainment distinguishes a single finite realization from a limit of realizations.

The ideaTrivial intersection makes projection injective.

Our contributionEvery rational strictly between zero and one is the subgroup-conjugacy ratio of an actual finite group.

Verification details
Release source recordProved in Lean

Open problem

Exact rational attainment of finite-group subgroup conjugacy ratios

Mihai-Silviu Lazorec and Marius Tărnăuceanu, On the number of conjugacy classes of subgroups of a finite group, arXiv:2508.14267v2, Section 4, Problem 4.1:

Let a,b be integers such that $1\leq a<b$. In Section 2 we proved that any ratio of the form $\frac{a}{a+1}$ can be achieved by computing $d'$ for specific non-nilpotent supersolvable groups having one conjugacy class of non-normal subgroups. We also showed that each ratio of the form $\frac{a}{b}$ can be achieved as the limit of $(d'(G_n)){n\geq1}$ where $(G_n){n\geq1}$ is a sequence of nilpotent groups formed as direct products of some modular p-groups. Hence, we pose the following question: Is it true that there is a group G such that $d'(G)=\frac{a}{b}$?

Section 1 specifies that all groups are finite, defines $L(G)$ as the subgroup lattice and $k'(G)$ as the number of conjugacy classes of subgroups, and sets $d'(G)=k'(G)/|L(G)|$. Bottom and top belong to $L(G)$. Conjugacy means inner conjugation by elements of $G$.

The exact settling statement is

∀ a b : ℤ, 1 ≤ a → a < b → ∃ W : FiniteGroup,
  b * (classCount W.Carrier : ℤ) = a * (latticeCount W.Carrier : ℤ)

The source clauses correspond as follows.

| Source clause | Formal meaning | | --- | --- | | Integers $a,b$ | Universal a b : ℤ, with no coprimality or reduced-fraction condition | | $1\leq a<b$ | Exactly 1 ≤ a and a < b | | A finite group $G$ | FiniteGroup contains a carrier, its Group dictionary and its Finite proof; groups are nonempty | | The whole lattice $L(G)$ | Subgroup W.Carrier, including bottom and top, counted by Nat.card | | $k'(G)$ | Nat.card of the orbit quotient of that lattice for ConjAct W.Carrier; the action is actual inner conjugation | | $d'(G)=a/b$ | Exact integer equation b * classCount = a * latticeCount; the finite lattice and its quotient are both nonempty and have positive counts | | Every allowed pair | No bounded range, structural restriction, supplied count formula or family-selection hypothesis in the final theorem |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/RationalSubgroupConjugacyAttainment.rational_attainment

Frozen module record

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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedSource 2026Strict monotonicity of Wallis gamma remaindersStrict monotonicity of three Wallis gamma remainders.Proved

The questionDo the functions μ(x)=−x²−{(3/x)[ln(√x Γ(x+1/2)/Γ(x+1))]^{−1}+24}^{−1}, θ(x)=(23040x⁵/31){3x/(24x²+1)+ln(√x Γ(x+1/2)/Γ(x+1))}, χ(x)=(358400x⁷/257){(120x²+31)/(96x(10x²+3))+ln(√x Γ(x+1/2)/Γ(x+1))} satisfy the property that μ(x) and χ(x) are strictly increasing, while θ(x) is strictly decreasing for all x>0?

Why it mattersStrict monotonicity determines the direction of improvement of three normalized rational approximations to a Wallis-type Gamma quotient.

The ideaobeys the same half-step equation and tends to zero.

Our contributionStrict monotonicity of three Wallis gamma remainders.

Verification details
Release source recordProved in Lean

Open problem

Strict monotonicity of Wallis gamma remainders

Section 4, Open Question of the source identified in D5/L/Asymptotics/mahmoudmortici2026wallisratio asks:

Do the functions μ(x)=−x²−{(3/x)[ln(√x Γ(x+1/2)/Γ(x+1))]^{−1}+24}^{−1}, θ(x)=(23040x⁵/31){3x/(24x²+1)+ln(√x Γ(x+1/2)/Γ(x+1))}, χ(x)=(358400x⁷/257){(120x²+31)/(96x(10x²+3))+ln(√x Γ(x+1/2)/Γ(x+1))} satisfy the property that μ(x) and χ(x) are strictly increasing, while θ(x) is strictly decreasing for all x>0?

Formal reading

All Lean names below have prefix D5.S3.Analytic.WallisRemainder.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $B(x)=\sqrt{x},\Gamma(x+1/2)/\Gamma(x+1)$ | Defs.wallisB | The exact real Gamma quotient. | | $\mu(x)$ | Defs.mu | The displayed nested reciprocal, with exactly the constants 3 and 24. | | $\theta(x)$ | Defs.theta | The displayed difference multiplied by $23040x^5/31$. | | $\chi(x)$ | Defs.chi | The displayed difference multiplied by $358400x^7/257$. | | Implicit reciprocal conditions | First conjunct of Defs.claim | For every $x>0$, $\ln B(x)\ne0$ and $(3/x)[\ln B(x)]^{-1}+24\ne0$. | | Strict monotonicity | Remaining conjuncts of Defs.claim | For every real $0<x<y$, $\mu(x)<\mu(y)$, $\theta(y)<\theta(x)$ and $\chi(x)<\chi(y)$. | | Complete assertion | Result.result : Defs.claim | Unconditional theorem with no representation assumption. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/WallisRemainder/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

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Number theoryQuestion resolvedFirst preprint 2025Mane's negative-index factorizationMane's negative-index factorization holds for all k >= 2 and 2 <= s <= k+1.Proved

The questionMane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, p. 19, Section 6, Conjecture 6.4, states:

Why it mattersPreregistration 14688 selects MANE-2, the full factorization identity.

The ideaInformation-escape registration is paused under CLAUDE.md §3.9.

Our contributionMane's negative-index factorization holds for all k >= 2 and 2 <= s <= k+1.

Verification details
Release source recordProved in Lean

arXiv

Mane's negative-index factorization

S. R. Mane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, p. 19, Section 6, Conjecture 6.4, states:

For k ≥ 2 and n = −sk, where s ∈ [2, k + 1] (the pattern fails for s ≥ k + 2), 𝓕_{−sk,k}(x) = −x(x^k + 1)^{s−2}(x^k + s). (6.4) Hence 𝓕_{−sk,k}(x) has roots x^k = −s.

The recurrence (1.2)–(1.3) is

$$ \mathcal F_{n,k}(x)=\sum_{j=1}^k x^{k-j}\mathcal F_{n-j,k}(x),\qquad \mathcal F_{1,k}=1,\qquad \mathcal F_{n,k}=0\quad (-(k-2)\le n\le0). $$

The frozen definitions a and F in D5/S3/Zeros/NegativeIndexTribonacciAttainmentRefutation.lean encode $a_k(m)=\mathcal F_{1-m,k}$ and F k n = a k (1-n).toNat. The claim is an identity in $\mathbb Z[X]$ with natural $k,s$, $k\ge2$ and $2\le s\le k+1$; the index is the integer $-sk$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/NegativeIndexFibonacciFactorization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedMathar's recurrence for the surviving walks A026023The number a(n) of walks s(0), ..., s(n) on the nonnegative integers with steps of size 1 and s(0) = 3 satisfies (n + 4)(n - 1) a(n) + (n - 1)(n + 1) a(n - 1) - 2(n + 1)(2n + 1) a(n - 2) - 4(n - 1)(n + 1) a(n - 3) = 0 for every n at least 3, as conjectured by R.Proved

The questionZiff, a(n)/2^n is the probability that a random walker started at x = 4 has not been adsorbed at x = 0 by time n.

The ideaSplitting on the first step, W(n + 1, x) = W(n, x + 1) + [x ≥ 1] W(n, x − 1) for the number W(n, x) of walks of length n from x.

Our contributionThe number a(n) of walks s(0), ..., s(n) on the nonnegative integers with steps of size 1 and s(0) = 3 satisfies (n + 4)(n - 1) a(n) + (n - 1)(n + 1) a(n - 1) - 2(n + 1)(2n + 1) a(n - 2) - 4(n - 1)(n + 1) a(n - 3) = 0 for every n at least 3, as conjectured by R.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's recurrence for the surviving walks A026023

OEIS A026023 counts the sequences s(0), …, s(n) of nonnegative integers with |s(i) − s(i − 1)| = 1 and s(0) = 3; by a comment of R. M. Ziff, a(n)/2^n is the probability that a random walker started at x = 4 has not been adsorbed at x = 0 by time n. Its formula field records:

Conjecture: (n+4)*(n-1)*a(n) +(n-1)*(n+1)*a(n-1) -2*(n+1)*(2*n+1)*a(n-2) -4*(n-1)*(n+1)*a(n-3)=0. - R. J. Mathar, Sep 29 2012

Issue #10761 fixes the reading: a is the entry's name read literally, the recurrence is asserted for every n ≥ 3 and read in ℤ.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/RandomWalks/SurvivingWalkRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedA004123 totient-period conjectureWeighted Stirling transforms have totient periods, proving A064618 and A004123.Proved

The questionNumber of generalized weak orders on n points.

Why it mattersThis is a first-tier OEIS conjecture recorded in 2022.

The ideaIts live proof path has three steps. Twindow collapses the sum to the fixed range k < m: terms with k ≥ m vanish because m ∣ k!, and terms with k > n vanish because S(n,k) = 0. Texponentialsum rewrites that window by inclusion-exclusion as a fixed integer combination of jⁿ. powaddtotient supplies the prime-power split: for each prime power q = p^e dividing m, a unit base uses Euler's theorem together with φ(q) ∣ φ(m); a non-unit base has both powers vanishing once the exponent reaches e.

Our contributionWeighted Stirling transforms have totient periods, proving A064618 and A004123.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A004123 totient-period conjecture

OEIS A004123 NAME:

Number of generalized weak orders on n points.

OEIS A004123 COMMENT (Peter Bala, Jul 08 2022):

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

The finite formula for this entry is a(n) = T (fun k => (2 : ℤ)^k) (n - 1); its proved onset is n ≥ m + 1 for every positive modulus m. The entry has offset 1, so the transform index is n - 1.

The companion entry is recorded separately with its own Library note and dossier:

A064618 NAME (offset 0):

Stirling transform of (n!)^2.

A064618 COMMENT (Peter Bala, Jan 15 2018):

Conjecture: for fixed k = 1,2,..., the sequence a(n) (mod k) is eventually periodic with the exact period dividing phi(k), where phi(k) is the Euler totient function A000010.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/StirlingTransformTotientPeriod.bala_conjecture_a004123

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedA064618 totient-period conjectureWeighted Stirling transforms have totient periods, proving A064618 and A004123.Proved

The questionConjecture: for fixed k = 1,2,..., the sequence a(n) (mod k) is eventually periodic with the exact period dividing phi(k), where phi(k) is the Euler totient function A000010.

Why it mattersThis is a first-tier OEIS conjecture recorded in 2018.

The ideaIts live proof path has three steps. Twindow collapses the sum to the fixed range k < m: terms with k ≥ m vanish because m ∣ k!, and terms with k > n vanish because S(n,k) = 0. Texponentialsum rewrites that window by inclusion-exclusion as a fixed integer combination of jⁿ. powaddtotient supplies the prime-power split: for each prime power q = p^e dividing m, a unit base uses Euler's theorem together with φ(q) ∣ φ(m); a non-unit base has both powers vanishing once the exponent reaches e.

Our contributionWeighted Stirling transforms have totient periods, proving A064618 and A004123.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A064618 totient-period conjecture

OEIS A064618 NAME:

Stirling transform of (n!)^2.

OEIS A064618 COMMENT (Peter Bala, Jan 15 2018):

Conjecture: for fixed k = 1,2,..., the sequence a(n) (mod k) is eventually periodic with the exact period dividing phi(k), where phi(k) is the Euler totient function A000010.

The finite formula for this entry is a(n) = T (fun k => (k.factorial : ℤ)) n; its proved onset is n ≥ m for every positive modulus m. The entry has offset 0.

The companion entry is recorded separately with its own Library note and dossier:

A004123 NAME (offset 1):

Number of generalized weak orders on n points.

A004123 COMMENT (Peter Bala, Jul 08 2022):

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/StirlingTransformTotientPeriod.bala_conjecture_a064618

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedA Gray-code closed form for the A163617 binary recurrenceYanev's Gray-code formula for the nonnegative half of the binary recurrence.Proved

The questiona(2n) = 2a(n), a(2n + 1) = 2a(n) + 2 + (-1)^n, for all n in Z.

Why it mattersThis first-tier OEIS conjecture was recorded in 2016.

The ideaThe invariant a n = gray n + corr n is then carried through the custom recurrence from the initial value a 0 = 0.

Our contributionYanev's Gray-code formula for the nonnegative half of the binary recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A Gray-code closed form for the A163617 binary recurrence

OEIS A163617, %N (verbatim):

a(2*n) = 2*a(n), a(2*n + 1) = 2*a(n) + 2 + (-1)^n, for all n in Z.

FORMULA (verbatim; Velin Yanev, Dec 17 2016):

Conjecture: a(n) = A003188(n) + (6*n + 1 - (-1)^n)/4.

OEIS A003188, %N (verbatim), followed by its %F XOR formula:

Decimal equivalent of Gray code for n.

a(n) = n XOR floor(n/2), where XOR is the binary exclusive OR operator.

Only the natural-number half, initialized by a(0) = 0, is formalized. The negative-index half is OEIS A163618 and is not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/GrayCodeBinaryRecurrenceClosedForm.yanev_a163617

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Yanev's Gray-code formula for the nonnegative half of the binary recurrence.

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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA266489: clause C1, index divisibility on the frozen objectNormalized vanishing diagonals inherit index-power divisibility from their exponents.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^n ) = 0 for n>1. COMMENT: CONJECTURES: COMMENT: (C1) n divides a(n) for n>=1: A268293(n) = a(n)/n. COMMENT: (C2) a(n) == 0 (mod 2) for n>=2. AUTHOR: Paul D.

Why it mattersThis is a first-tier OEIS conjecture attributed to Paul D.

The ideaThe module constructs a normalized integer power series by a triangular coefficient update.

Our contributionNormalized vanishing diagonals inherit index-power divisibility from their exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A266489: clause C1, index divisibility on the frozen object

Quoted directly from the seat-local file oeis-A266489.src (seats/opB-diagidx2/oeis-A266489.src in the supplied workspace):

NAME: G.f. A(x) satisfies: [x^n] A( x/A(x)^n ) = 0 for n>1. COMMENT: CONJECTURES: COMMENT: (C1) n divides a(n) for n>=1: A268293(n) = a(n)/n. COMMENT: (C2) a(n) == 0 (mod 2) for n>=2. AUTHOR: Paul D. Hanna, Feb 07 2016

The target is only clause (C1) for this entry. The formal object has a(0)=a(1)=1 and exponent e(n)=n. The formal vanishing condition is imposed for n>1, as in the quoted NAME.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalVanishingIndexDivisibility.hanna_conjecture_a266489

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA292394: square-index divisibilityNormalized vanishing diagonals inherit index-power divisibility from their exponents.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^(n^2) ) = 0 for n>1. COMMENT: a(n) is divisible by n^2 for n>=1 (conjecture): A292395(n) = a(n)/n^2. AUTHOR: Paul D.

Why it mattersThis is a first-tier OEIS conjecture attributed to Paul D.

The ideaThe module constructs a normalized integer power series by a triangular coefficient update.

Our contributionNormalized vanishing diagonals inherit index-power divisibility from their exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A292394: square-index divisibility

Quoted directly from the seat-local file oeis-A292394.src (seats/opB-diagidx2/oeis-A292394.src in the supplied workspace):

NAME: G.f. A(x) satisfies: [x^n] A( x/A(x)^(n^2) ) = 0 for n>1. COMMENT: a(n) is divisible by n^2 for n>=1 (conjecture): A292395(n) = a(n)/n^2. AUTHOR: Paul D. Hanna, Sep 15 2017

The target is the quoted divisibility conjecture for this entry. The formal object has a(0)=a(1)=1 and exponent e(n)=n^2. The formal vanishing condition is imposed for n>1, as in the quoted NAME.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalVanishingIndexDivisibility.hanna_conjecture_a292394

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA300732: index divisibilityNormalized vanishing diagonals inherit index-power divisibility from their exponents.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^(2n) ) = 0 for n>=1. COMMENT: Conjecture: n divides a(n) for n>=1. AUTHOR: Paul D.

Why it mattersThis is a first-tier OEIS conjecture attributed to Paul D.

The ideaThe module constructs a normalized integer power series by a triangular coefficient update.

Our contributionNormalized vanishing diagonals inherit index-power divisibility from their exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A300732: index divisibility

Quoted directly from the seat-local file oeis-A300732.src (seats/opB-diagidx2/oeis-A300732.src in the supplied workspace):

NAME: G.f. A(x) satisfies: [x^n] A( x/A(x)^(2*n) ) = 0 for n>=1. COMMENT: Conjecture: n divides a(n) for n>=1. AUTHOR: Paul D. Hanna, Mar 11 2018

The target is the quoted divisibility conjecture for this entry. The formal object has a(0)=a(1)=1 and exponent e(n)=2*n. The printed degree-one boundary requires the correction disclosed under ASSUMED-UNVERIFIED.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalVanishingIndexDivisibility.hanna_conjecture_a300732

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA300733: index divisibilityNormalized vanishing diagonals inherit index-power divisibility from their exponents.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^(3n) ) = 0 for n>=1. COMMENT: Conjecture: n divides a(n) for n>=1. AUTHOR: Paul D.

Why it mattersThis is a first-tier OEIS conjecture attributed to Paul D.

The ideaThe module constructs a normalized integer power series by a triangular coefficient update.

Our contributionNormalized vanishing diagonals inherit index-power divisibility from their exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A300733: index divisibility

Quoted directly from the seat-local file oeis-A300733.src (seats/opB-diagidx2/oeis-A300733.src in the supplied workspace):

NAME: G.f. A(x) satisfies: [x^n] A( x/A(x)^(3*n) ) = 0 for n>=1. COMMENT: Conjecture: n divides a(n) for n>=1. AUTHOR: Paul D. Hanna, Mar 11 2018

The target is the quoted divisibility conjecture for this entry. The formal object has a(0)=a(1)=1 and exponent e(n)=3*n. The printed degree-one boundary requires the correction disclosed under ASSUMED-UNVERIFIED.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalVanishingIndexDivisibility.hanna_conjecture_a300733

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA300734: square-index divisibilityNormalized vanishing diagonals inherit index-power divisibility from their exponents.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^(2n^2) ) = 0 for n>1. COMMENT: Conjecture: n^2 divides a(n) for n>=1. AUTHOR: Paul D.

Why it mattersThis is a first-tier OEIS conjecture attributed to Paul D.

The ideaThe module constructs a normalized integer power series by a triangular coefficient update.

Our contributionNormalized vanishing diagonals inherit index-power divisibility from their exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A300734: square-index divisibility

Quoted directly from the seat-local file oeis-A300734.src (seats/opB-diagidx2/oeis-A300734.src in the supplied workspace):

NAME: G.f. A(x) satisfies: [x^n] A( x/A(x)^(2*n^2) ) = 0 for n>1. COMMENT: Conjecture: n^2 divides a(n) for n>=1. AUTHOR: Paul D. Hanna, Mar 11 2018

The target is the quoted divisibility conjecture for this entry. The formal object has a(0)=a(1)=1 and exponent e(n)=2*n^2. The formal vanishing condition is imposed for n>1, as in the quoted NAME.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalVanishingIndexDivisibility.hanna_conjecture_a300734

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedBala's OEIS A305550 totient-period conjectureIntegral exponential substitutions have totient periods, including OEIS A305550.Proved

The questionThe sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

Why it mattersThis is a first-tier OEIS conjecture from the 2022 entry/comment.

The ideaThe binomial theorem expands each power of exp X − 1; powers of exp are rescaled exponentials whose degree-n coefficients are jⁿ/n!; the imported Stirling inclusion-exclusion identity gives k!·S(n,k).

Our contributionIntegral exponential substitutions have totient periods, including OEIS A305550.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's OEIS A305550 totient-period conjecture

OEIS A305550 NAME (entry by Ilya Gutkovskiy, Jun 15 2018):

Expansion of e.g.f. Product_{k>=1} (1 + (exp(x) - 1)^k).

COMMENT (Peter Bala, Jul 08 2022):

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

In the same comment:

More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/IntegralEgfTotientPeriod.bala_conjecture_a305550

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedBala's general integral e.g.f. substitution conjectureIntegral exponential substitutions have totient periods, including OEIS A305550.Proved

The questionThe sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

Why it mattersThis is a first-tier OEIS conjecture from the 2022 entry/comment.

The ideaThe binomial theorem expands each power of exp X − 1; powers of exp are rescaled exponentials whose degree-n coefficients are jⁿ/n!; the imported Stirling inclusion-exclusion identity gives k!·S(n,k).

Our contributionIntegral exponential substitutions have totient periods, including OEIS A305550.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's general integral e.g.f. substitution conjecture

OEIS A305550 NAME (entry by Ilya Gutkovskiy, Jun 15 2018):

Expansion of e.g.f. Product_{k>=1} (1 + (exp(x) - 1)^k).

COMMENT (Peter Bala, Jul 08 2022):

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k).

In the same comment:

More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series.

The broader sentence appears verbatim on both A305550 and A004123. These are two source locations of ONE conjecture settled once, not two resolutions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/IntegralEgfTotientPeriod.bala_conjecture_egf_general

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedHanna's A379204 evenness conjectureThe bilateral equation of OEIS A379204 implies Hanna's parity and square-shift conjectures.Proved

The questionHanna, Dec 20 2024) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2024a379204.md:

Why it mattersThis is a first-tier OEIS conjecture from 2024.

The ideaThe bilateral NAME is read after multiplication by xA.

Our contributionThe bilateral equation of OEIS A379204 implies Hanna's parity and square-shift conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A379204 evenness conjecture

OEIS A379204 (Paul D. Hanna, Dec 20 2024) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2024a379204.md:

G.f. A(x) satisfies 1/x = Sum_{n=-oo..+oo} A(x)^n * (A(x)^n + 4)^(n+1).

Conjecture: a(n) is even for n > 1.

The evenness conjecture is this dossier's sole resolution claim, anchored by hanna_conjecture. The entry also states:

Conjecture: a(n) == 2 (mod 4) iff n = (k-1)^2 + 1 for some k > 1.

The companion hanna_conjecture_mod_four proves that second conjecture. The NAME is formalized after multiplication by x*A, using exact finite coefficient windows; the source-to-formal interpretation is bounded as described below.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/BilateralQuarticPowerThetaModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A379764 theta-four coefficient conjecturesHanna's A379764 theta-four coefficient conjectures.Proved

The questionA(x) satisfies Sum{n=-oo..+oo} (4^nA(x)^n - 4A(x))^(n+1) = 2theta4(x) = 2Sum{n=-oo..+oo} (-x)^(n^2).

Why it mattersThis is an externally stated OEIS conjecture (tier 1).

The ideaTheir exponents increase with the index; pairorder and tailwindow justify each finite coefficient window.

Our contributionHanna's A379764 theta-four coefficient conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A379764 theta-four coefficient conjectures

Paul D. Hanna's OEIS A379764 states:

G.f. A(x) satisfies Sum_{n=-oo..+oo} (4^n*A(x)^n - 4*A(x))^(n+1) = 2*theta_4(x) = 2*Sum_{n=-oo..+oo} (-x)^(n^2).

Conjecture: for n > 0, a(n) is odd iff n is a square.

Conjecture: for n > 0, a(n) == (-1)^(n-1) (mod 4) iff n is a square, otherwise a(n) is divisible by 4 if n is nonsquare.

The statement source and binding formal conventions are in #18830. The normalization is $A(0)=0$ and $a(1)=1$. Integer powers of the summand are taken in rational Laurent series. The bilateral equation means equality at coefficient $N$ for every symmetric window $[-K,K]$ with $N+1\le K$; no analytic convergence assertion is made.

Statement echo

All names below belong to D5.S1.Recurrence.Bilateral.HannaThetaFourModFour.

| Source or binding convention | Lean counterpart | Fidelity | | --- | --- | --- | | The sum defining $\theta_4$ | theta4 | Its coefficient at $N$ sums $(-1)^N$ at the integer roots of $j^2=N$ in $[-N,N]$. | | The Laurent summand with integer index | term | The source expression is retained literally after HahnSeries.ofPowerSeries embeds $A$. | | Normalization and every coefficient window | IsGF | Both normalization equalities and all $N,K$ with $N+1\le K$ are required. | | Every positive index; odd coefficient iff square | claim, result | Every normalized rational solution has an integer coefficient witness and the odd-square biconditional. | | Signed residue iff square; divisibility otherwise | claim, result | The same coefficient witness satisfies the second biconditional and nonsquare divisibility. | | Existence of the normalized series | First conjunct of claim | This strengthens the source convention by ruling out a vacuous universal assertion. | | Rational-to-integer coefficient correspondence | Integer witness in claim | This proves integrality at every positive coefficient; it adds no hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/HannaThetaFourModFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedSigned coefficient modes of physicists Hermite polynomialsThe smallest exponent with maximum signed physicists Hermite coefficient changes by one or minus three.Proved

The questionIts full comment continues with finer patterns, spacing and square-root growth.

Why it mattersAlternating signs restrict a positive signed maximum to exponents congruent to n modulo four.

The ideaIts true indices form an initial segment: the left side increases with r and the right side decreases.

Our contributionThe smallest exponent with maximum signed physicists Hermite coefficient changes by one or minus three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a381423 hermite differences

OEIS A381423 revision 8, Mike Sheppard, has the independent conjecture clause: “Differences are either 1 or -3”. Its full comment continues with finer patterns, spacing and square-root growth. The independent target is preregistered at #25634.

For every natural n, including zero, let a(n) be the smallest j in zero through n maximizing the signed coefficient of $x^j$ in the physicists Hermite polynomial $H_n$. The exact assertion is

$$ (a(n+1):\mathbb Z)-(a(n):\mathbb Z)=1\quad\text{or}\quad-3. $$

The source's Mathematica first PositionLargest and PARI upward scan establish the tie convention. Its n=5 example selects exponent 1, distinguishing signed order from absolute coefficient order.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Physicists H_n | physicistsHermite n, the literal DLMF 18.5.13 finite factorial expansion over ℚ. | | Coefficient at exponent j | polynomial_coeff proves extraction equals signedCoeff n j throughout zero through n. | | Signed maximum | IsSmallestMax compares rational coefficients directly, with no absolute value. | | Smallest tied exponent | The last conjunct of IsSmallestMax; the proof chooses the largest positive-term index. | | Every n, including zero | Unrestricted ∀ n : ℕ in both conjuncts of claim. | | Integer difference | Both values of a are cast to ℤ before subtraction. | | No uniqueness assumption | No assumption is present; all tied maximizers enter the selector predicate. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Polynomial/HermiteSignedMode.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedThe complete A381737 Conjecture 2 recurrenceThe fifth-lag recurrence for the complete increasing enumeration of physicists Hermite absolute-tie orders.Proved

The questionOrders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Why it mattersThe original forecasts and clock remain those immutable source records.

The ideaHermiteAbsoluteTieOrders.result supplies positivity, strict increase and exhaustive unique positive indexing for the original predicate, with

Our contributionThe fifth-lag recurrence for the complete increasing enumeration of physicists Hermite absolute-tie orders.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The complete A381737 Conjecture 2 recurrence

OEIS A381737 revision 28, by Mike Sheppard (2025), names the sequence:

Orders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Its Formula field states:

Conjecture 2: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).

The offset is 1,1. Let $H_k$ be the physicists Hermite polynomial represented by HermiteSignedMode.physicistsHermite over the rationals. For every natural order $k$ and exponent $j$, put

$$ M(k,j)\iff j\le k\ \land\ \forall i\le k,\quad |[x^i]H_k|\le |[x^j]H_k|. $$

Let $S$ consist of exactly the natural orders $k$ for which there are natural exponents $u<v$ such that, for every natural $j$, $M(k,j)$ holds if and only if $j=u$ or $j=v$. Thus exactly two distinct exponents attain the absolute maximum of the entire row $0,\ldots,k$, including parity-zero coefficients.

The complete assertion is that there exists $g:\mathbb Z\to\mathbb N$ satisfying all of the following:

  • For every integer $n\ge1$, $g(n)>0$.
  • For all integers $1\le n<m$, $g(n)<g(m)$.
  • For every natural order $k$, including zero, $k\in S$ if and only if there is exactly one integer $n\ge1$ with $k=g(n)$.
  • For every integer $n\ge6$,

$$ (g(n):\mathbb Z)=(g(n-1):\mathbb Z)+2(g(n-2):\mathbb Z) -2(g(n-3):\mathbb Z)-(g(n-4):\mathbb Z)+(g(n-5):\mathbb Z). $$

Every sequence value is cast separately to the integers before subtraction. The positive-index restriction of $g$ is the actual increasing exhaustive one-based enumeration of $S$. Its values at nonpositive indices impose no restriction on this event. There is no finite cutoff, eventual qualification, selected subfamily, signed maximum, probabilists normalization or recurrence-defined replacement sequence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Polynomial/HermiteTieRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The fifth-lag recurrence for the complete increasing enumeration of physicists Hermite absolute-tie orders.

    Read source ↗
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Discrete structuresQuestion resolvedPosed in 2025Eventual periodicity of OEIS A383165Every scaled logarithm column is eventually periodic modulo every positive integer.Proved

The questionThe entry-specific Library note Library/ArithSums/bala2026a383165.md records the NAME and conjecture verbatim as follows:

Why it mattersThis is a first-tier OEIS conjecture: entry year 2025, conjecture year 2026, as recorded in its own Library note.

The ideaHere the substitution is exp 2x − 1, and the outer function log(1 + y/2)^r / r! is not integral: its coefficients carry powers of two and factors of the coefficient index j in denominators.

Our contributionEvery scaled logarithm column is eventually periodic modulo every positive integer.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Eventual periodicity of OEIS A383165

The entry-specific Library note Library/ArithSums/bala2026a383165.md records the NAME and conjecture verbatim as follows:

Expansion of e.g.f. log(1 + (exp(2*x) - 1)/2)^2 / 2.

Conjecture: the sequence {a(n)} reduced modulo a positive integer k is eventually periodic.

The note attributes the entry to Seiichi Manyama (April 18, 2025) and the conjecture to Peter Bala (February 17, 2026). This dossier claims only this entry's eventual-periodicity conjecture, not a divisibility clause of a negative-power diagonal family. The claimed conclusion is existence of an onset and a positive period, without a specified or minimal period.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/ScaledLogColumnPeriodicity.bala_conjecture_a383165

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPosed in 2025Eventual periodicity of OEIS A383166Every scaled logarithm column is eventually periodic modulo every positive integer.Proved

The questionThe entry-specific Library note Library/ArithSums/bala2026a383166.md records the NAME and conjecture verbatim as follows:

Why it mattersThis is a first-tier OEIS conjecture: entry year 2025, conjecture year 2026, as recorded in its own Library note.

The ideaHere the substitution is exp 2x − 1, and the outer function log(1 + y/2)^r / r! is not integral: its coefficients carry powers of two and factors of the coefficient index j in denominators.

Our contributionEvery scaled logarithm column is eventually periodic modulo every positive integer.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Eventual periodicity of OEIS A383166

The entry-specific Library note Library/ArithSums/bala2026a383166.md records the NAME and conjecture verbatim as follows:

Expansion of e.g.f. log(1 + (exp(2*x) - 1)/2)^3 / 6.

Conjecture: the sequence {a(n)} reduced modulo a positive integer k is eventually periodic.

The note attributes the entry to Seiichi Manyama (April 18, 2025) and the conjecture to Peter Bala (February 17, 2026). This dossier claims only this entry's eventual-periodicity conjecture, not a divisibility clause of a negative-power diagonal family. The claimed conclusion is existence of an onset and a positive period, without a specified or minimal period.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/ScaledLogColumnPeriodicity.bala_conjecture_a383166

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedParity conjecture C.1 of OEIS A384267The absolute reciprocal-square series satisfies conjecture C.1 of OEIS A384267.Proved

The questionThe NAME and COMMENT below are verbatim excerpts of the claim in Library/Recurrence/hanna2025a384267.md.

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaA is the stabilized fixed point of Phi(F) = absSeries (1 + X invOfUnit (F^2) 1).

Our contributionThe absolute reciprocal-square series satisfies conjecture C.1 of OEIS A384267.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity conjecture C.1 of OEIS A384267

OEIS A384267, Paul D. Hanna, Jun 19 2025. The NAME and COMMENT below are verbatim excerpts of the claim in Library/Recurrence/hanna2025a384267.md.

NAME:

a(n) = abs([x^n] 1 + x/A(x)^2), where A(x) is the g.f.

COMMENT:

(C.1) a(n) == binomial(3*n-1,n)/(3*n-1) (mod 2) (cf. A006013).

This dossier anchors exactly C.1 for n >= 1. The entry lists three conjectures; C.2 (sign pattern) and C.3 (asymptotic ratios) are untouched. The companion a_zero gives a(0) = 1. Here abs means coefficientwise absolute value of the series 1 + x*A(x)^(-2), represented by the frozen absSeries applied to 1 + X * invOfUnit (A ^ 2) 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/AbsoluteReciprocalSquareShiftParity.hanna_conjecture

Frozen module record

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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe fourth compositional iterate of A396798 modulo eightEvery coefficient above degree two in the fourth compositional iterate of A396798 is divisible by eight.Proved

The questionHanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)A^5(x)." Its formula field says that A^k denotes the k-th iteration.

Why it mattersThe eighth-iterate module supplies the common source definition.

The ideaThe common source definition takes coefficient n from H^n+1, where H(F)=X+I4(F)I5(F).

Our contributionEvery coefficient above degree two in the fourth compositional iterate of A396798 is divisible by eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The fourth compositional iterate of A396798 modulo eight

Paul D. Hanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)*A^5(x)." Its formula field says that A^k denotes the k-th iteration. The fourth comment states:

Conjecture: [x^n] A^4(x) == 0 (mod 8) for n > 2.

Let I0(A)=X and I(k+1)(A)=Ik(A) composed with A. For the unique zero-constant integer series satisfying A=X+I4(A)*I5(A), prove forall n:Nat, 2<n -> (8:Int) divides coeff n (I4(A)). Multiplication is ordinary multiplication of formal series. There is no factorial normalization or analytic convergence hypothesis.

The source is the existing D5/S1/Recurrence/Residue/IterateProductFourFiveEighthModEight.generatingSeries. The fourth-iterate theorem uses precisely this definition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductFourFiveFourthModEight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A396803 iterated-exponential coefficientsEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.Proved

The questionHanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

Why it mattersThe candidate comes from the repository's own triage notes Library/Words/oeis2026triage0909.md and oeis2026triage0911b.md.

The ideaCoefficients are integral EGF coefficients, eCoeff f n = n! coeff n f.

Our contributionEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A396803 iterated-exponential coefficients

Paul D. Hanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

%N A396803 E.g.f. satisfies A(x) = x*exp( A^3(x) ).
%C A396803 Conjecture: a(n) is odd iff n is odd for n >= 1.
%C A396803 Conjecture: a(n) == [1,2,0] repeating (mod 3) for n >= 1.
%A A396803 Paul D. Hanna, Jun 08 2026

In this entry A^3(x) denotes the 3-fold COMPOSITIONAL ITERATE of A, not the 3-th power; the entry's formula field states that convention explicitly. The offset is 1 and the first terms are 1, 2, 21, 472, 17165, 885696, 60160723.

This dossier concerns the parity conjecture. The modulo-three conjecture is proved separately by IterateExponentialModThree.result; see the modulo-three dossier. It is not a theorem of the parity module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialParity.parity_iterate_three

Frozen module record

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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A396804: the fourth-iterate EGF modulo fourEvery positive-index EGF coefficient in OEIS A396804 is congruent to its index modulo four.Proved

The questionHanna's OEIS A396804 (2026-06-09) defines the exponential generating function by A(x)=x exp(A^4(x)), explicitly specifying that superscripts mean compositional iteration.

Why it mattersThis is a first-tier recent OEIS conjecture with an unbounded arithmetic target.

The ideaDefine integer composition by the chain-rule recurrence C(f,g)(0)=f(0) and C(f,g)(n+1)=Σ(i=0..n) binomial(n,i) C(shift f,g)(i) g(n-i+1).

Our contributionEvery positive-index EGF coefficient in OEIS A396804 is congruent to its index modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A396804: the fourth-iterate EGF modulo four

Paul D. Hanna's OEIS A396804 (2026-06-09) defines the exponential generating function by A(x)=x exp(A^4(x)), explicitly specifying that superscripts mean compositional iteration. Its independent modulo-four conjecture is: a(n) == [1,2,3,0] repeating (mod 4) for n >= 1.

Precisely: A is the unique rational formal power series with constant term zero and A=X*(exp Rat).subst(iterateComp A 4), where the zeroth iterate is X. The independently established natural sequence satisfies (a(n):Rat)=n!*[x^n]A; the target is Nat.ModEq 4 (a n) n for every n≥1. The old, refuted modulo-three pattern is outside this dossier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/QuarticEGFModFour.mod_four

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A396805 iterated-exponential coefficientsEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.Proved

The questionHanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

Why it mattersThe candidate comes from the repository's own triage notes Library/Words/oeis2026triage0909.md and oeis2026triage0911b.md.

The ideaCoefficients are integral EGF coefficients, eCoeff f n = n! coeff n f.

Our contributionEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A396805 iterated-exponential coefficients

Paul D. Hanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

%N A396805 E.g.f. satisfies A(x) = x*exp( A^5(x) ).
%C A396805 Conjecture: a(n) is odd iff n is odd for n >= 1.
%C A396805 Conjecture: a(n) == [0,1,0] repeating (mod 3) for n >= 3.
%C A396805 Conjecture: a(n) == [1,2,3,4,0] repeating (mod 5) for n >= 1.
%A A396805 Paul D. Hanna, Jun 09 2026

In this entry A^5(x) denotes the 5-fold COMPOSITIONAL ITERATE of A, not the 5-th power; the entry's formula field states that convention explicitly. The offset is 1 and the first terms are 1, 2, 33, 1264, 80505, 7365456, 893006317.

This dossier settles the parity conjecture. The separate modulo-three dossier binds IterateExponentialFiveModThree.result, which proves the pattern from n=3. The modulo-five assertion has the non-kernel counterexamples recorded under Triage; the parity theorem does not formalize that refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialParity.parity_iterate_five

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    Source 1
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Discrete structuresQuestion resolvedParity of the A396806 iterated-exponential coefficientsEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.Proved

The questionHanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

Why it mattersThe candidate comes from the repository's own triage notes Library/Words/oeis2026triage0909.md and oeis2026triage0911b.md.

The ideaCoefficients are integral EGF coefficients, eCoeff f n = n! coeff n f.

Our contributionEvery compositional iterate count gives fixed-point EGF coefficients odd exactly at positive odd indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A396806 iterated-exponential coefficients

Paul D. Hanna's NAME and every conjecture COMMENT, quoted verbatim from the entry snapshot:

%N A396806 E.g.f. satisfies A(x) = x*exp( A^6(x) ).
%C A396806 Conjecture: a(n) is odd iff n is odd for n >= 1.
%C A396806 Conjecture: a(n) == [1,2,0] repeating (mod 3) for n >= 1.
%C A396806 Conjecture: a(n) == [1,2,3,4,5,0] repeating (mod 6) for n >= 1.
%A A396806 Paul D. Hanna, Jun 09 2026

In this entry A^6(x) denotes the 6-fold COMPOSITIONAL ITERATE of A, not the 6-th power; the entry's formula field states that convention explicitly. The offset is 1 and the first terms are 1, 2, 39, 1804, 139625, 15563526, 2301954109.

This dossier settles the parity conjecture through parity_iterate_six. The independent result IterateExponentialModSix.result settles the modulo-six assertion and consequently the modulo-three assertion; see Problems/oeis-a396806-iterate-exponential-mod-six.md. These residue conclusions are not additional statements of the parity module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialParity.parity_iterate_six

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Discrete structuresQuestion resolvedParity of the A397242 coefficientsThe coefficients of OEIS A397242 are odd exactly one below a power of two.Proved

The questionHanna, Jun 19 2026, gives the following NAME and first COMMENT, quoted from Library/Recurrence/hanna2026a397242b.md:

Why it mattersThis is a first-tier OEIS conjecture from 2026.

The ideaThe escape content is the integral normalization d(n) = a(n)/n for n >= 1, with d(1) = 1.

Our contributionThe coefficients of OEIS A397242 are odd exactly one below a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A397242 coefficients

OEIS A397242, Paul D. Hanna, Jun 19 2026, gives the following NAME and first COMMENT, quoted from Library/Recurrence/hanna2026a397242b.md:

O.g.f. satisfies A(x) = exp( x + Sum_{n>=2} (n^2-1) * a(n)*x^n / n^2 ).

Conjecture: a(n) is odd iff n+1 is a power of 2.

The formal reading is A(0)=1 and X*A' = M*A, where M = X + sum_{n>=2} (n^2-1) a(n) X^n / n. The integral normalization d(n) = a(n)/n for positive n realizes these weights. The entry's two other comments conjecture classifications modulo 3; neither is resolved here. This dossier makes one resolution claim, on hanna_conjecture alone.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ExponentialSquareWeightCatalanParity.hanna_conjecture

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Discrete structuresConjecture provedThe residue-one conjecture of OEIS A397242The two modulo-three support conjectures of OEIS A397242 hold in both directions.Proved

The questionO.g.f. satisfies A(x) = exp( x + Sum{n>=2} (n^2-1) a(n)x^n / n^2 ).

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaSetting T = xV and F = xU gives T = x + T³, F = T + T², and F = x + x²A over ZMod(3).

Our contributionThe two modulo-three support conjectures of OEIS A397242 hold in both directions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The residue-one conjecture of OEIS A397242

OEIS A397242 (Paul D. Hanna, Jun 19 2026), NAME:

O.g.f. satisfies A(x) = exp( x + Sum_{n>=2} (n^2-1) * a(n)*x^n / n^2 ).

COMMENT (the sole resolution target of this dossier):

Conjecture: for n > 0, a(n) == 1 (mod 3) iff n+2 is of the form 3^k or 2*3^k.

The entry's parity COMMENT was already proved and frozen here as D5/S1/Recurrence/Residue/ExponentialSquareWeightCatalanParity.hanna_conjecture. It is NOT re-claimed. This module imports that frozen module and reuses its sequence rather than redefining it. The other mod-three COMMENT has its own dossier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ExponentialSquareWeightTernarySupport.hanna_conjecture_one

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    Source 1
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    Source 1
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Discrete structuresConjecture provedThe residue-two conjecture of OEIS A397242The two modulo-three support conjectures of OEIS A397242 hold in both directions.Proved

The questionO.g.f. satisfies A(x) = exp( x + Sum{n>=2} (n^2-1) a(n)x^n / n^2 ).

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaSetting T = xV and F = xU gives T = x + T³, F = T + T², and F = x + x²A over ZMod(3).

Our contributionThe two modulo-three support conjectures of OEIS A397242 hold in both directions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The residue-two conjecture of OEIS A397242

OEIS A397242 (Paul D. Hanna, Jun 19 2026), NAME:

O.g.f. satisfies A(x) = exp( x + Sum_{n>=2} (n^2-1) * a(n)*x^n / n^2 ).

COMMENT (the sole resolution target of this dossier):

Conjecture: for n > 0, a(n) == 2 (mod 3) iff n+2 is a sum of 2 distinct powers of 3 (A038464).

The entry's parity COMMENT was already proved and frozen here as D5/S1/Recurrence/Residue/ExponentialSquareWeightCatalanParity.hanna_conjecture. It is NOT re-claimed. This module imports that frozen module and reuses its sequence rather than redefining it. The other mod-three COMMENT has its own dossier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ExponentialSquareWeightTernarySupport.hanna_conjecture_two

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    Source 1
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    Source 1
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Number theoryQuestion resolvedFirst preprint 2025Optimality of the Ordered Zeckendorf Long Game StrategyComplete ordered Long Game Strategy optimality with every permitted switch choice.Proved

The questionA state is an ordered list of Fibonacci numbers.

Why it matters- Erasing order maps a game state to RawDigits.

The ideaThe weighted split-only subproblem is now proved in SplitStabilization.splitprefixpromotion, splitphasepromotion, and greedysplitoptimality.

Our contributionComplete ordered Long Game Strategy optimality with every permitted switch choice.

Verification details
Release source recordProved in Lean

Open problem

Optimality of the Ordered Zeckendorf Long Game Strategy

A state is an ordered list of Fibonacci numbers. The game starts with n copies of F_1. Legal adjacent moves are (F_i, F_{i+1}) -> F_{i+2}; (F_1, F_1) -> F_2; (F_i, F_i) -> (F_{i-2}, F_{i+1}) for i > 2; (F_2, F_2) -> (F_1, F_3); and switching an inversion (F_i, F_j) -> (F_j, F_i) when i > j.

The Long Game Strategy uses the priority: all switch moves in any order, combine adjacent ones from the left, split from the right, then merge from the left.

Conjecture 1.7, quoted from arXiv:2508.20222v2:

“The LGS has the longest game length.”

Status statement, quoted from the same version:

“The longest-game strategy described in Conjecture 1.7 is supported by empirical simulations. However, a rigorous proof establishing its optimality remains an open problem.”

Formal statement, with the five moves represented exactly:

∀ n > 0, ∃ g, CompleteLGSRun n g
∀ n > 0, ∀ g, CompleteLGSRun n g ->
  ∀ h, LegalTerminalRun n h -> h.length ≤ g.length

Every permitted switch ordering is retained in the proved universal comparison. Priorities restart after every move, including a switch.

The paper says the conjecture is backed by exhaustive simulations rather than a proof. In its broader exact game-tree analysis it also states:

“Due to the combinatorial explosion in the number of game states and legal move sequences, a full resolution of the winner for n>25 was computationally infeasible.”

That second quote concerns winner computation, not directly Conjecture 1.7, but it identifies the same state-space obstacle. For length, the paper proves only upper and lower asymptotics and a structural lemma about repetitions under LGS.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Carry/OrderedGame/Completion.result

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Discrete structuresConjecture provedFirst preprint 2026Raney Conjecture 7.1: zero-run record familiesFor every admissible Raney sequence, nonzero residues occur arbitrarily far out and every actual maximal zero-run length lies in finitely many affine prime-power families.Proved

The questionEu, Huang, and Kao, arXiv:2609.25742v1, Conjecture 7.1, state:

Why it mattersThe problem asks for a uniform description of record zero-run lengths for the entire Raney family, extending beyond the paper's (3,2) focus and the known Catalan (2,1) case.

The ideaMaximalBlockEvolution formalizes literal morphism powers, support stabilization, actual maximal intervals including first=0, and boundary-safe uniform versions of BKS Lemmas 11 and 12. MaximalBlockDescent uses one stabilizing power throughout, builds actual predecessor/image steps with literal context words, descends strictly in the right endpoint, and reaches finite families of early roots, bounded late root words, and bounded contexts.

Our contributionFor every admissible Raney sequence, nonzero residues occur arbitrarily far out and every actual maximal zero-run length lies in finitely many affine prime-power families.

Verification details
Release source recordProved in Lean

arXiv

Raney Conjecture 7.1: zero-run record families

Eu, Huang, and Kao, arXiv:2609.25742v1, Conjecture 7.1, state:

Fix k, r and a prime p not dividing kr. The left-to-right record values of the zero-run sequence of R_(k,r)(n) mod p belong to a finite union of families of the form (a p^m + b)/c, where a,b,c depend only on k,r,p.

The source quantifies over positive k,r and prime p with p not dividing k*r. Its Raney numbers are

R_(k,r)(n) = r * binomial(k*n+r,n) / (k*n+r).

The source's left-to-right records allow ties and come from maximal zero runs in the full infinite sequence. The formal target was preregistered in issue #9544 with these parameters, the literal natural quotient, tied records, and the source-to-formal implication. It does not ask for a converse saying that every term of every family occurs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Raney/ZeroRunFamilies.raney_zero_run_coefficient_families

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Quantum informationQuestion resolvedFirst preprint 2026The constraint polynomials of the two-photon Rabi model at and beyond the critical couplingThe constraint polynomials of the two-photon asymmetric quantum Rabi model factor into linear terms y + 2n(2n + 2rho - 1) at the critical coupling x = 1, and for x > 1 and nonnegative bias all their coefficients are positive, because the tridiagonal matrix behind them is positive definite.Proved

The questionReyes-Bustos and Wakayama define, for the two-photon asymmetric quantum Rabi model with parity ρ ∈ {0, 1}, bias ε, x = (2g)^2 and y = Δ^2, the polynomials

Why it mattersx = 1 is the critical coupling g = 1/2 where the two-photon model acquires continuous spectrum, and the absence of positive roots for x > 1 matches the absence of Juddian points past the critical coupling.

The ideaAt x = 1 the bias cancels from the recursion.

Our contributionThe constraint polynomials of the two-photon asymmetric quantum Rabi model factor into linear terms y + 2n(2n + 2rho - 1) at the critical coupling x = 1, and for x > 1 and nonnegative bias all their coefficients are positive, because the tridiagonal matrix behind them is positive definite.

Verification details
Release source recordProved in Lean

Open problem

The constraint polynomials of the two-photon Rabi model at and beyond the critical coupling

Reyes-Bustos and Wakayama define, for the two-photon asymmetric quantum Rabi model with parity ρ ∈ {0, 1}, bias ε, x = (2g)^2 and y = Δ^2, the polynomials

P_0 = 1, P_1 = y + 2x(4N + 2ρ + 2ε − 1) − 4(1 + ε), P_k = (y + 2xk(4N + 2ρ − 2k + 2ε + 1) − 4k(k + ε)) P_{k−1} − 4k(k − 1)(2(N − k + 1) + ρ)(2(N − k + 1) + ρ − 1) x P_{k−2},

(Definition 4.1 of arXiv:2609.00750v1) and state Conjecture 4.5:

We have P_N^{(N,ρ,ε)}(1, y) = ∏_{n=1}^{N} (y + 2n(2n + 2ρ − 1)). Moreover, for x > 1, the polynomial P_N^{(N,ρ,ε)}(x, y) has positive coefficients, and therefore, no positive roots for y.

Issue #10148 fixes the readings: P_k is a real polynomial in y with real parameters x and ε; the first statement is taken for every real ε; the second statement is taken for ε ≥ 0 and means that the coefficients of y^0, …, y^N are all positive and that P_N^{(N,ρ,ε)}(x, y) ≠ 0 for every y > 0. The source derives its constraint condition for ε ≥ 0 and writes P_N^{(N,ρ,±ε)} for the two signs, while the conjecture writes +ε only.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/TwoPhotonRabiConstraintPolynomials.result

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    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Zeros of Binary Digit Hankel Determinants at Twice a Root of UnityBartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, Conjecture 5.7.Proved

The questionBartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, Conjecture 5.7.

The ideaThe determinant recursions of the paper (Theorem 4.2 in scalar polynomial form, extended to the degenerate case by polynomial identity) relate H and an auxiliary bordered determinant at n to their values at a smaller index, with explicit fixed bases depending on the binary phase of n.

Our contributionBartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, Conjecture 5.7.

Verification details
Release source recordProved in Lean

Open problem

Zeros of Binary Digit Hankel Determinants at Twice a Root of Unity

Bartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, Conjecture 5.7. For u = Σ_j ε_j 2^j with binary digits ε_j put S(u,t) = Σ_j ε_j t^j and H(n,t) = det[S(i+j,t)]{0≤i,j<n}. For d ≥ 2 and ℓ ≥ d + 1 write ℓ = qd + r with q ≥ 1 and 1 ≤ r ≤ d, let z{d,ℓ} = 2^r (2^{qd} − 1)/(2^d − 1) − 2 for r < d and z_{d,ℓ} = 2^d (2^{qd} − 1)/(2^d − 1) − 1 for r = d, and let A_d be the set of positive integers in ⋃_{ℓ≥d+1} ⋃_{s odd} (2^ℓ s − z_{d,ℓ} − 1, 2^ℓ s + z_{d,ℓ} + 1]. The conjecture states that for every primitive d-th root of unity ζ and every n ≥ 2, H(n, 2ζ) = 0 if and only if n ∈ A_d.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/DigitHankel/CyclotomicDigitHankel.result

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Quantum informationQuestion resolvedFirst preprint 2026Token-1 spectral contraction for every finite positive-velocity chainToken-1 damping strictly contracts the spectrum and every centered trajectory.Proved

The questionFekete, Pass the Bucket: Efficient, Robust, Local Load Balancing for Teams of Heterogeneous Robots, arXiv:2608.27085v1, Section IV.B.1, page 4:

Why it mattersThat product bound alone allows individual eigenvalues of modulus at least one.

The ideaThe two adjacent terms have the decomposition

Our contributionToken-1 damping strictly contracts the spectrum and every centered trajectory.

Verification details
Release source recordProved in Lean

Open problem

Token-1 spectral contraction for every finite positive-velocity chain

T. Wallner, D. Krupke, A. Schmidt and S. P. Fekete, Pass the Bucket: Efficient, Robust, Local Load Balancing for Teams of Heterogeneous Robots, arXiv:2608.27085v1, Section IV.B.1, page 4:

Conjecture 1 (Spectral contraction): The damped transfer map Mα is repeatedly applied and its spectral radius satisfies ρ(Mα) < 1, such that u → 0 for 0 < α < 1.

The statement concerns the paper's centered linear map in its rotational regime. There are $n=m+1$ robots, $m\ge1$ pair coordinates and positive velocities $v_1,\ldots,v_n$. The same damping parameter $\alpha\in(0,1)$ is used at every step. The delivered claim contains both the strict bound for every complex spectral value and convergence $M_\alpha^k u\to0$ for every real initial vector.

The binding source identifiers $B_\alpha$ and $M_\alpha$ are spelled Balpha and Malpha in Lean, as authorized by #14768.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/TridiagonalSweeps/TokenDampedSweepContraction.result

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  • Mathematical advance · 3/4 · 15% weight

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Quantum informationCounterexample foundFirst preprint 2010Log-negativity entangling capacity depends on the input resource for CPTP mapsA two-qutrit CPTP map separates assisted and separable-input log-negativity capacities.Refuted

The questionCampbell, Optimal entangling capacity of dynamical processes, Phys.

Why it mattersResource independence asks whether prior entanglement can improve the maximum increase of the same entanglement measure.

The ideaFor arbitrary local ancillas, positivity of conditional matrices follows through the existing IptPptSquaredEquivalence.weightedpsd theorem.

Our contributionA two-qutrit CPTP map separates assisted and separable-input log-negativity capacities.

Scope & next questionThe unitary clause and the source's two-qutrit unitary conjecture remain open.

Verification details
Release source recordRefuted in Lean

Open problem

Log-negativity entangling capacity depends on the input resource for CPTP maps

E. T. Campbell, Optimal entangling capacity of dynamical processes, Phys. Rev. A 82, 042314 (2010), arXiv:1007.1445v2, Section VII, p. 7, asks:

The most obvious open question is whether log-negativity EC is resource independent for all unitaries, or indeed all CPTP maps, or whether this is a singular feature of 2-qubit gates.

The all-CPTP clause is refuted. The unitary clause and the source's two-qutrit unitary conjecture remain open. The statement and source scope are fixed in #17047; the source quotations are in the Campbell note.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Every bipartite CPTP map | claim quantifies local dimensions and finite Kraus families with completeness | Finite-dimensional square input/output spaces; the Choi–Kraus representation supplies this interpretation. | | Free local ancillary systems | extend, and the suprema in capSep and capAssisted | Every pair of natural-number ancillary dimensions; the cut is $AA':BB'$. | | Eq. (1), separable inputs | capSep | Density matrices that are finite convex mixtures of products of density matrices. | | Eq. (2), increase of entanglement | capAssisted | All density inputs, subtracting the input's own log-negativity. | | Eqs. (4)–(5), logarithmic rescaling | Real.logb 2 of traceNorm (partialTranspose ...) | Base two; the partial transpose acts on $BB'$. | | Resource independence | Equality of the two capacities in claim | Suprema are in EReal, preserving unbounded suprema. | | A negative answer to the all-CPTP clause | result | A fixed two-qutrit channel with 18 Kraus operators and $t=1$. |

The generalized IptPptSquaredEquivalence.IsSeparable retains the finite-convex-mixture condition. Its local factors are positive semidefinite with trace one, the body of the reused IsDensity predicate. The auxiliary density condition in capSep is consistent with those mixtures; it does not exclude a source separable state.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/LogNegativityCapacityResourceDependence.result

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Discrete structuresRecurrence provedPosed in 2014The A226857 Fibonacci sum-product recurrence is provedThe Fibonacci sum-product sequence satisfies Barker's three-and-six-step recurrence.Proved

The questionNumbers that are both the sum of two Fibonacci numbers and the product of two Fibonacci numbers.

Why it mattersThe entry records a 2014 recurrence conjecture for the increasing sequence of numbers with both Fibonacci-sum and Fibonacci-product representations.

The ideaThe product gap invariant proves, by two-step double induction, the strict interval in the product indices.

Our contributionThe Fibonacci sum-product sequence satisfies Barker's three-and-six-step recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A226857 Fibonacci sum-product recurrence is proved

OEIS A226857, NAME (%N, verbatim):

Numbers that are both the sum of two Fibonacci numbers and the product of two Fibonacci numbers.

FORMULA conjecture line (%F, verbatim):

Conjecture: a(n) = a(n-3)+a(n-6) for n>12. - Colin Barker, Nov 09 2014

Empirical generating function (%F, verbatim):

Empirical g.f.: -x^2*(x^10 +x^9 +x^8 +2x^7 +3x^6 +3x^5 +3x^4 +3x^3 +3x^2 +2*x +1) / (x^6 +x^3 -1). - Colin Barker, Nov 09 2014

COMMENT (%C, verbatim):

All Fibonacci numbers are in the sequence. The only prime numbers in this sequence are prime Fibonacci numbers.

AUTHOR (%A, verbatim):

Alonso del Arte, Jun 19 2013

The first %S terms are 0,1,2,3,4,5,6,8,9,10,13,15,16,21,24,26,34,39,42,55,63,68,89,102,110.

The literal proved statement is ∀ n : ℕ, 12 < n → a n = a (n - 3) + a (n - 6), where mem x := (∃ i j : ℕ, x = Nat.fib i + Nat.fib j) ∧ (∃ r s : ℕ, x = Nat.fib r * Nat.fib s) and a n := Nat.nth mem (n - 1). Nat.nth enumerates members from zero, so the one-based OEIS index is represented by the subtraction n - 1 in natural numbers. Equal summands and equal factors are allowed. The empirical generating-function line and the %C corollaries are NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/BarkerFibonacciSumProductRecurrence.barker_a226857

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The Fibonacci sum-product sequence satisfies Barker's three-and-six-step recurrence.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2014.

    Read source ↗
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    The argument includes a structural formulation; this alone does not establish a new concept.

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    Dated conjecture statement: 2014. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2013The temporal Hardy maximum 1/4 is attained with spin measurements for every spinFor every positive integer n, spin-n/2 measurements attain temporal Hardy success probability one quarter.Proved

The questionGhosh, Persistence of Hardy's nonlocality in time, arXiv:1302.5296v1, Appendix B:

Why it mattersFor spatial Hardy nonlocality, spin systems of higher dimension were found to behave more classically.

The ideaFix n ≥ 1, s = n/2 and r = √(2^{1/n} − 1), so (1 + r²)ⁿ = 2.

Our contributionFor every positive integer n, spin-n/2 measurements attain temporal Hardy success probability one quarter.

Verification details
Release source recordProved in Lean

arXiv

The temporal Hardy maximum 1/4 is attained with spin measurements for every spin

S. K. Choudhary, S. K. Goyal, T. Konrad and S. Ghosh, Persistence of Hardy's nonlocality in time, arXiv:1302.5296v1, Appendix B:

We conjecture that the maximum of the success probability appearing in the temporal version of Hardy's argument is $25%$ for any spin-$s$ system.

The conclusion of the source states the same conjecture for spin measurements. The source proves that the success probability is at most 1/4 for every system and every projective measurement, and attains it with spin components for s = 1 and s = 3/2. The experiment and the four probabilities are quoted in the literature note.

Issue #19643 fixes the reading. For every spin s = n/2 with n ≥ 1, some state and four spin-component observables u·S, with Lüders spectral projectors for chosen outcomes, satisfy the three zero conditions and have success probability 1/4. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/TemporalHardySpinAttainment.result

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Quantum informationQuestion resolvedFirst preprint 2023The worst-case CNOT count for reflection-symmetric diagonal unitariesEvery positive even number of wires admits a reflection-symmetric diagonal unitary forcing the exponential CNOT lower bound.Proved

The questionMatsushita, Optimized Synthesis of Circuits for Diagonal Unitary Matrices with Reflection Symmetry, J.

Why it mattersHuang, Kosugi, Nishi and Matsushita (2024, arXiv:2310.06676v2) use reflection symmetry to reduce circuit resources for diagonal unitaries occurring in first-quantized Hamiltonian simulation.

The idea2. Circuit normal form and return. Wire labels start at the identity rows, and a CNOT replaces its target label by target plus control.

Our contributionEvery positive even number of wires admits a reflection-symmetric diagonal unitary forcing the exponential CNOT lower bound.

Scope & next questionThe optimal depth question in the source's Conclusion (lines 847–853) remains open here: a gate-count lower bound does not determine how many disjoint gates can run simultaneously.

Verification details
Release source recordProved in Lean

Open problem

The worst-case CNOT count for reflection-symmetric diagonal unitaries

X. Huang, T. Kosugi, H. Nishi and Y.-i. Matsushita, Optimized Synthesis of Circuits for Diagonal Unitary Matrices with Reflection Symmetry, J. Phys. Soc. Jpn. 93, 054002 (2024), arXiv:2310.06676v2, state in Sect. 6, Conclusion (main_arXiv2.tex, line 847):

Although we conjecture that the number of entanglement gates is at best $2^{2n-1}+2n-2$ in the ${\mathrm{CNOT}, R_z}$-based circuit for the worst scenario of diagonal unitary matrices that satisfy Eq. \eqref{intro:eq-theta}, […]

The quoted TeX label denotes Eq. (1): for $n\ge1$, the real phases of a $2^{2n}\times2^{2n}$ diagonal unitary satisfy $\theta_{j,j'}=\theta_{2^n-1-j,,2^n-1-j'}$ for every $0\le j,j'<2^n$ (main_arXiv2.tex, lines 207–227). Equivalently, writing $m=2n$ and $x\in{0,1}^m$, $\theta(\bar x)=\theta(x)$, where $\bar x$ complements every bit. The source specifies ancilla-free ${\mathrm{CNOT},R_z}$ circuits at lines 194 and 1225 and realization up to a global phase at lines 429–445.

Issue #24527 fixes the reading: for every $n\ge1$ there is such a phase function $\theta$ for which every finite circuit realizing $\operatorname{diag}(e^{\mathrm i\theta(x)})$ up to a global phase has at least $2^{2n-1}+2n-2$ CNOT gates. This lower-bound statement is proved. The source's matching upper bound from Algorithm 1 is separate from claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/ReflectionDiagonalCNOTLowerBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024Wei–Liu CCZ magic-capacity thresholdThe noisy CCZ gate preserves magic with a reference at depolarizing strength 1/2.Refuted

The questionFuchuan Wei and Zi-Wen Liu, Noise robustness and threshold of many-body quantum magic, arXiv:2410.21215v1, Section VII:

The ideaDiagonal restriction of a six-qubit stabilizer normal form is either zero or a scalar multiple of a three-qubit normal form.

Our contributionThe noisy CCZ gate preserves magic with a reference at depolarizing strength 1/2.

Verification details
Release source recordRefuted in Lean

arXiv

Wei–Liu CCZ magic-capacity threshold

Fuchuan Wei and Zi-Wen Liu, Noise robustness and threshold of many-body quantum magic, arXiv:2410.21215v1, Section VII:

We conjecture that the magic capacity threshold for $\mathrm{CC}Z$ under local depolarizing noise is $1/3$.

The capacity maximizes robustness of magic over pure six-qubit stabilizer inputs, with $\mathrm{CC}Z$ and independent single-qubit depolarizing noise acting on the three system qubits $A$ and the three reference qubits $B$ untouched. Robustness is faithful: it equals one exactly on the real convex hull $\mathrm{STAB}_6$ of pure stabilizer density matrices. The single-qubit channel is $\rho\mapsto(1-\lambda)\rho+\lambda\operatorname{Tr}(\rho)I_2/2$.

The literal claim has two conjuncts: every pure stabilizer output belongs to STAB 6 for $\lambda\in[1/3,1]$, and a pure stabilizer input with output outside STAB 6 exists for every $\lambda\in[0,1/3)$. Noise follows the gate. The result contradicts the first conjunct at $\lambda=1/2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CCZMagicCapacityThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2005No single polynomial inequality characterizes separability in every dimensionAn affine family of qubit–qutrit density operators excludes a single polynomial inequality characterizing separability, even when the polynomial may depend on the local dimensions.Refuted

The questionWerner, Some Open Problems in Quantum Information Theory, arXiv:quant-ph/0504166v1, Problem 3, "Polynomial entanglement invariants", the third closely related problem:

Why it mattersThe main question of the problem, the completeness of polynomial invariants, is answered in the source by the Onishchik–Vinberg theorem.

Our contributionAn affine family of qubit–qutrit density operators excludes a single polynomial inequality characterizing separability, even when the polynomial may depend on the local dimensions.

Verification details
Release source recordRefuted in Lean

arXiv

No single polynomial inequality characterizes separability in every dimension

O. Krüger and R. F. Werner, Some Open Problems in Quantum Information Theory, arXiv:quant-ph/0504166v1, Problem 3, "Polynomial entanglement invariants", the third closely related problem:

Decide whether the set of separable states can be described in terms of a polynomial invariant $f$, such that $f(\rho)\ge 0$ is equivalent to separability. There are many weaker versions of this statement, which may be of interest. For example, we might merely ask for a sufficient or a necessary separability criterion, and we might allow $f$ to depend on the dimensions.

Here a polynomial invariant is a real function on bipartite density operators that is constant on local-unitary orbits $\rho\mapsto(U_A\otimes U_B)\rho(U_A\otimes U_B)^*$ and is computed by a polynomial in the matrix elements of $\rho$. The definitions are quoted in the literature note.

Issue #20112 fixes the reading: for all local dimensions $d_A,d_B\ge2$ there is a real polynomial $f$ in the real and imaginary parts of the entries, invariant under local unitaries on density operators, with $f(\rho)\ge0$ iff $\rho$ is separable for every density operator $\rho$. This dimension-dependent version, the weakest the source names for a characterization, is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SeparabilitySinglePolynomialRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2005. Historical context is capped; this does not establish sustained unsuccessful research.

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    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Alexander HolevoThe entanglement of formation has a supporting affine functional at every pure stateIn every finite-dimensional bipartite system, each pure state admits a Hermitian supporting functional for the entanglement of formation.Proved

The questionShirokov, Log-Sobolev inequality, von Neumann entropy and Entanglement of Formation, arXiv:2609.12667v3, Introduction:

Why it mattersA supporting affine functional at ρ is equivalent to a one-sided Lipschitz lower bound for EF at ρ.

The ideaLet R be the marginal of Pψ, c = Σi ri^{−1} over its positive eigenvalues ri (the module's choice; the smaller c = 1/rmin also works) and K = 2bc, and put W = −(log₀ R) ⊗ I − K(I − Pψ).

Our contributionIn every finite-dimensional bipartite system, each pure state admits a Hermitian supporting functional for the entanglement of formation.

Verification details
Release source recordProved in Lean

arXiv

The entanglement of formation has a supporting affine functional at every pure state

A. S. Holevo and M. E. Shirokov, Log-Sobolev inequality, von Neumann entropy and Entanglement of Formation, arXiv:2609.12667v3, Introduction:

It is somewhat surprising that the above problem is nontrivial even in the case when the state $\rho$ is pure. We and GPT-5.6 have found neither a general proof nor counterexamples which would confirm or refute the hypothesis of the existence of a supporting affine functional for the EoF at any pure bipartite state.

Section 3.1 restates the finite-dimensional question. The verbatim passages are in the literature note.

Issue #19308 fixes the reading: for all finite a, b and every unit ψ ∈ ℂ^a ⊗ ℂ^b, there is a Hermitian W with Tr(W P_ψ) = E_F(P_ψ) and Tr(Wσ) ≤ E_F(σ) for every density matrix σ. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/PureState/PureStateFormationSupportingFunctional.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    First preprint year: 2026.

    Read source ↗
  • Scholarly context

    Alexander Holevo is a leading contributor to the mathematical foundations of quantum information, including quantum channels and information bounds.

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    Alexander Holevo is a leading contributor to the mathematical foundations of quantum information, including quantum channels and information bounds.

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Quantum informationQuestion resolvedFirst preprint 2014CHSH nonlocality is tightly bounded by the steering-ellipsoid centreEvery two-qubit density matrix with steering-ellipsoid centre of length c has CHSH nonlocality at most max(2 sqrt(2(1 - c)), 2).Proved

The questionRudolph, Quantum correlations of two-qubit states with one maximally mixed marginal, Phys.

Why it mattersThe paper proves the bound only for canonical states, whose Bob marginal is maximally mixed (Theorem 1). §V explains why that does not extend: CHSH nonlocality does not transform simply under the local filtering that makes a state canonical.

The idea1. CHSH to a plane. For unit b, b′ write b ± b′ = r e₁, s e₂ with e₁ ⊥ e₂ and r² + s² = 4.

Our contributionEvery two-qubit density matrix with steering-ellipsoid centre of length c has CHSH nonlocality at most max(2 sqrt(2(1 - c)), 2).

Verification details
Release source recordProved in Lean

Open problem

CHSH nonlocality is tightly bounded by the steering-ellipsoid centre

A. Milne, D. Jennings, S. Jevtic and T. Rudolph, Quantum correlations of two-qubit states with one maximally mixed marginal, Phys. Rev. A 90, 024302 (2014), arXiv:1404.3951v2, define the CHSH nonlocality in §II as β(ρ) = max_B |tr(ρB)|, with B = α·σ ⊗ (β + β′)·σ + α′·σ ⊗ (β − β′)·σ and the maximum over all directions α, α′, β, β′. §V, Conjecture 1:

Let ρ be a general two-qubit state with E centred at c. The CHSH nonlocality is tightly bounded as β(ρ) ≤ max[2√(2(1 − c)), 2].

Here E is Alice's steering ellipsoid and c = |c⃗| is the length of its centre. The verbatim statements and the centre convention are in the literature note. The repository settled Conjecture 2 of the same paper in the fully-entangled-fraction dossier.

Issue #16558 fixes the reading:

  • β is the supremum of |tr(ρB)| over four unit vectors of ℝ³;
  • the centre is the companion module's centre: (a − Tb)/(1 − |b|²) for |b| ≠ 1 and a for |b| = 1.

The conjecture is proved in this form. "Tightly bounded" is proved as a supremum at every centre length t ∈ [0, 1]: for every ε > 0 some state with c = t has β > max(2√(2(1 − t)), 2) − ε.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SteeringEllipsoidCHSH.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

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Quantum informationCounterexample foundFirst preprint 2021Local versus joint Gaussian thermometry for unequal-frequency modesThe local-versus-global Gaussian thermometry conjecture in Eq. (47) of Cenni, Lami, Acín and Mehboudi is refuted by two unequal-frequency modes.Refuted

The questionMehboudi, “Thermometry of Gaussian quantum systems using Gaussian measurements”, arXiv:2110.02098v4, §3.2, Eq. (47), ask whether the optimal joint Gaussian Fisher information equals the optimal local Gaussian Fisher information for every product thermal state:

Why it mattersFor m independent thermal modes at a common temperature T > 0 and frequencies ωk > 0, set νk = coth(ωk/(2T)), νk′ = (ωk/(2T²))(νk²−1), σ = ⊕k νk I₂ and ∂Tσ = ⊕k νk′ I₂.

The ideaUse m=2, T=1 and a=artanh(1/2)>0, with frequencies (2a,4a), equivalently (ln 3,2 ln 3).

Our contributionThe local-versus-global Gaussian thermometry conjecture in Eq. (47) of Cenni, Lami, Acín and Mehboudi is refuted by two unequal-frequency modes.

Verification details
Release source recordRefuted in Lean

arXiv

Local versus joint Gaussian thermometry for unequal-frequency modes

M. F. B. Cenni, L. Lami, A. Acín and M. Mehboudi, “Thermometry of Gaussian quantum systems using Gaussian measurements”, arXiv:2110.02098v4, §3.2, Eq. (47), ask whether the optimal joint Gaussian Fisher information equals the optimal local Gaussian Fisher information for every product thermal state:

"${\cal F}^{\rm C}(\oplus_k {\bm \sigma}{k}; {\bm \sigma}^M{\max}) \overset{?}{=} {\cal F}^{\rm C}(\oplus_k {\bm \sigma}{k}; \oplus_k{\bm \sigma}{k,\max}^M)$"

Based on these observations we conjecture this is generally true, however, a rigorous proof is missing currently.

Issue #11666 preregisters this Tier 1 conjecture, the quantified version 2 and the two-mode counterexample. The formal claim compares each physical joint value with the supremum over all physical block-diagonal measurements.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/GaussianThermometryJointMeasurementRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2020Chen–Kato–Brandão completely contractive CMI data processingA dephasing-plus-shift channel has only scalar correctable observables on the whole input space, yet preserves positive conditional mutual information for a singular input state.Refuted

The questionBrandão, Matrix Product Density Operators: when do they have a local parent Hamiltonian?, arXiv:2010.14682v3, Section III.C, Conjecture III.1:

Why it mattersIts channel has only scalar correctable observables on the whole input space, yet preserves a bit encoded in two selected preparations.

The ideaThe diagonal Kraus products force every commuting observable to be diagonal.

Our contributionA dephasing-plus-shift channel has only scalar correctable observables on the whole input space, yet preserves positive conditional mutual information for a singular input state.

Scope & next questionThis stronger noise hypothesis does not settle the scalar whole-space correctable-algebra conjecture. - Open — nearest surviving multi-step statement.

Verification details
Release source recordRefuted in Lean

arXiv

Chen–Kato–Brandão completely contractive CMI data processing

Chi-Fang Chen, Kohtaro Kato and Fernando G. S. L. Brandão, Matrix Product Density Operators: when do they have a local parent Hamiltonian?, arXiv:2010.14682v3, Section III.C, Conjecture III.1:

For any channel $\mathcal E:C\rightarrow C'$ with trivial correctable algebra $\mathcal A(\mathcal E)=\mathbb C I$, there exists a constant $\eta<1$ such that for any tripartite system $ABC$ and any state $\rho_{ABC}$, it holds that $I(A:C'|B){\mathcal E(\rho)}\le\eta I(A:C|B)\rho$.

Equation (22) defines the whole-space algebra as $\mathcal A(\mathcal E)=\mathrm{Alg}{O:[O,K_i^\dagger K_j]=0\ \forall i,j}$. Equation (1) defines $I(A:C|B)=S(AB)+S(BC)-S(B)-S(ABC)$, with von Neumann entropy in bits. The Lean expression uses the same eigenvalue entropy in nats; all terms are multiplied by the common positive factor $\log 2$. Singular density matrices are part of the source's domain.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ChenKatoBrandaoCMIRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture provedFirst preprint 2020Period-doubling PPL-difference: Conjecture 17Proved

The questionFrid, Laborde and Peltomäki, On prefix palindromic length of automatic words, Theoretical Computer Science 891, 13–23, arXiv:2009.02934v2, Section 5.1, printed page 13, Conjecture 17:

The ideaThe literal substitution is related to valuation parity.

Scope & next questionNon-2-automaticity excludes the relevant uniform automatic representation, and does not settle general morphicity.

Verification details
Release source recordProved in Lean

Open problem

Period-doubling PPL-difference: Conjecture 17

Frid, Laborde and Peltomäki, On prefix palindromic length of automatic words, Theoretical Computer Science 891, 13–23, arXiv:2009.02934v2, Section 5.1, printed page 13, Conjecture 17:

The sequence $d_{pd}$ of the period-doubling word $\mathbf{u}{pd}$ is not $2$-automatic, and so the prefix palindromic length $\operatorname{PPL}{pd}(n)$ of $\mathbf{u}_{pd}$ is not $2$-regular.

The word is the fixed point beginning with a of a → ab, b → aa. Write P(n) for the minimum number of nonempty palindrome factors of its length-n prefix, P(0)=0, and d_pd(n)=P(n+1)−P(n) for n≥0. C17 asks whether the set of sequences n ↦ d_pd(2^e n+r), over all natural e and r<2^e, is infinite. No finite-prefix restriction is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Palindromes/PeriodDoubling/PrefixPalindromicLengthNotAutomatic.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is proved.

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Analysis & dynamicsConjecture refutedFirst preprint 2015The Karp–Prilepkina Meijer zero-count conjectureA two-component Gamma quotient violates the proposed comparison of interior zero counts.Refuted

The questionKarp and Prilepkina, Completely monotonic gamma ratio and infinitely divisible H-function of Fox, arXiv:1501.05388v2, Section 2, Conjecture 1:

Why it mattersThe conjecture compares oscillation of an inverse Mellin transform with a finite power sum built from the same parameter vectors.

The ideaThe partial fractions follow by polynomial interpolation at the distinct denominator roots.

Our contributionA two-component Gamma quotient violates the proposed comparison of interior zero counts.

Verification details
Release source recordRefuted in Lean

Open problem

The Karp–Prilepkina Meijer zero-count conjecture

Karp and Prilepkina, Completely monotonic gamma ratio and infinitely divisible H-function of Fox, arXiv:1501.05388v2, Section 2, Conjecture 1:

If aᵢ ≥ 0, i = 1, …, p and Σᵢ₌₁ᵖ (bᵢ − aᵢ) > 0 then # zeros of G^{p,0}_{p,p}(x | b/a) on (0, 1) ≤ # zeros of v(t) on (0, 1).

Here $p\ge1$, $a_i,b_i$ are real,

$$v(t)=\sum_{k=1}^p(t^{a_k}-t^{b_k}),\qquad \psi=\sum_{i=1}^p(b_i-a_i),$$

and the contour-integral definition is

$$G^{p,0}{p,p}!\left(x;\middle|;\begin{matrix}b\a\end{matrix}\right) =\frac{1}{2\pi i}\int_L \frac{\prod{i=1}^p\Gamma(a_i+s)} {\prod_{i=1}^p\Gamma(b_i+s)}x^{-s},ds.$$

The contour separates the numerator poles in the usual Mellin–Barnes representation. The zero counts below count distinct zeros.

Formal reading

All Lean names below have prefix D5.S3.Analytic.MeijerZeroCount.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $R(s)=\prod_i\Gamma(a_i+s)/\prod_i\Gamma(b_i+s)$ | Defs.gammaQuot | Complex Gamma quotient of the real parameter vectors. | | $G^{p,0}_{p,p}(x\mid b/a)$ | Defs.meijerG | mellinInv 1 (gammaQuot a b) x, with normalization $1/(2\pi)$ in the real integral. | | $v(t)=\sum_k(t^{a_k}-t^{b_k})$ | Defs.control | Real powers of real arguments; the comparison uses $0<t<1$. | | $p\ge1$, $a_i\ge0$, $\psi>0$ | Defs.claim | The formal claim restricts to $\psi>1$. It quantifies over natural $p$; the case $p=0$ cannot satisfy $\psi>1$. | | Number of zeros on $(0,1)$ | Set.encard of each displayed zero set | Extended natural cardinality, including infinite zero sets. | | $a=(n+c,n+c)$, $b=(c,2n+d+c)$ | Result.family | All natural $n,d\ge2$ and real $c>0$; $p=2$ and $\psi=d$. | | Refutation of the comparison | Result.result | The unconditional theorem has type ¬ Defs.claim. |

The formal claim restricts to $\psi>1$, where the contour may be taken as the vertical line $\Re s=1$. Since $a_i\ge0$, this line is to the right of every numerator Gamma pole. The Gamma quotient decays as $O(|\Im s|^{-\psi})$ on this line, so the integral is absolutely convergent. A counterexample in this restricted range is also a counterexample to the source statement with $\psi>0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MeijerZeroCount/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2021Mahalingam–Pandoh maximum number of scattered palindromesThe sharp maximum of distinct scattered palindromes and its alphabet sizes.Proved

The questionPandoh, Counting scattered palindromes in a finite word, arXiv:2108.02729v1, §5, Conjecture 5.6, states:

Why it mattersThe supporting inequality concerns general monotone discretely convex functions and disjoint endpoint pairs.

The ideaPartition the palindromes by their outer letter.

Our contributionThe sharp maximum of distinct scattered palindromes and its alphabet sizes.

Verification details
Release source recordProved in Lean

arXiv

Mahalingam–Pandoh maximum number of scattered palindromes

K. Mahalingam and P. Pandoh, Counting scattered palindromes in a finite word, arXiv:2108.02729v1, §5, Conjecture 5.6, states:

The maximum number of scattered palindromic subwords in a word of length $n$ is $3(2^{\lceil n/2\rceil-1})-2$ if $n$ is odd, $2^{n/2+1}-2$ if $n$ is even, and this bound is achieved only when $q=\lceil n/2\rceil$ if $n$ is odd, $\frac n2$ and $\frac n2+1$ if $n$ is even.

$SP(w)$ counts distinct nonempty palindromic scattered subwords, rather than their embeddings. A scattered subword is a subsequence; $q$ counts the distinct letters in $w$. Write $M(n)$ for the displayed maximum.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Scattered containment and palindromes (§2) | spal uses List.sublists.toFinset and excludes the empty word while requiring reversal equality. | Every word over a decidable alphabet; duplicate embeddings count once. | | $SP(w)=|SPAL(w)|$ and $q=|Alph(w)|$ (§2) | sp and alphabetSize. | Cardinalities of the literal source objects. | | The two parity cases of $M(n)$ | maxSP. | Natural division $(n+1)/2$ represents the ceiling; subtraction is truncated natural subtraction. | | Maximum over all length-$n$ words | First conjunct of claim; result. | Universal bound over every decidable alphabet, including the empty word. | | Attainment of the maximum | Second conjunct of claim; attainment; result. | Every $n\ge1$ and every permitted $q$; witnesses are natural-number words. | | Only the specified alphabet sizes | permittedQ, third conjunct of claim; result. | Every extremal nonempty word; the even alternatives form a disjunction. |

The positive-length assumption applies to attainment and necessity. The universal bound includes $n=0$ as an additional proved boundary case. Decidable equality enables finite-set cardinalities; it does not restrict a finite word's letter pattern. Natural-number letters provide existence witnesses and impose no restriction on the universal clauses.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Palindromes/ScatteredPalindromeMaximum.result

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Number theoryQuestion resolvedFirst preprint 2019Minimal rational faces of the fan of interlace certsMinimal interlace fan faces containing nonzero rational points.Proved

The questionRobins, Dragging the roots of a polynomial to the unit circle, arXiv:1908.03208v3, Section 10, Problem 3:

Why it mattersIt identifies the rational structure of the fan and answers the existence question throughout every minimal rational face.

The ideaA rational polynomial taking a rational value at one primitive root takes the same value at every root of that order, by cyclotomic minimal-polynomial divisibility.

Our contributionMinimal interlace fan faces containing nonzero rational points.

Verification details
Release source recordProved in Lean

arXiv

Minimal rational faces of the fan of interlace certs

A. Mandel and S. Robins, Dragging the roots of a polynomial to the unit circle, arXiv:1908.03208v3, Section 10, Problem 3:

What are the minimal faces of the FOIC that contain rational points? Most rays clearly do not (except in darga 6). On the other hand, Section 9.2 gives several examples of faces that do. Are there interlace rational polynomials in those minimal faces?

For $n\ge2$, write $J={0,\ldots,\lfloor n/2\rfloor}$ and $O_d={j\in J:n/\gcd(n,j)=d}$ for $d\mid n$. For nonempty $S\subseteq J$, let $F_S$ be the closed face where every index in $S$ attains the maximum of the interlace functionals of equation (12). The classification concerns nonzero rational points; zero belongs to every closed face.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Darga and sigma coordinates | n ≥ 2, x : Fin (n / 2) → ℝ | Coordinate k represents sigma index k+1. | | Interlace certs and fan faces | I, certIdx, interlaceNumber, face | Equations (12)–(15), Proposition 5.5, and the Interlace Formula. | | Rational points | IsRationalPt, HasRatPt | Nonzero rational coordinates. | | Minimal rational faces | Second conjunct of claim | Maximal cert sets are exactly the complements of individual order classes. | | Interlace rational points in those faces | Third conjunct of claim | Every nonzero rational point has strict inequality on the omitted class and rational maximum. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/InterlaceFan/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresRecurrence provedPosed in 2019Barker's recurrence for crossing- and nesting-free labeled graphsBarker's recurrence counts labeled simple graphs whose edges neither cross nor nest.Proved

The questionNumber of labeled n-vertex simple graphs without crossing or nesting edges.

Why it mattersThis 2019 OEIS conjecture asks for an exact unbounded recurrence for a literal class of labeled simple graphs.

The ideaFor an avoiding graph G, let A(G) be the vertices c such that every edge whose right endpoint is greater than c is incident with c.

Our contributionBarker's recurrence counts labeled simple graphs whose edges neither cross nor nest.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barker's recurrence for crossing- and nesting-free labeled graphs

OEIS A326244, NAME:

Number of labeled n-vertex simple graphs without crossing or nesting edges.

COMMENTS:

Two edges {a,b}, {c,d} are crossing if a < c < b < d or c < a < d < b, and nesting if a < c < d < b or c < a < b < d.

FORMULA:

Conjectures from Colin Barker, Jun 28 2019: (Start)

G.f.: (1 - x)(1 - 4x) / (1 - 6x + 8x^2 - 4*x^3).

a(n) = 6*a(n-1) - 8*a(n-2) + 4*a(n-3) for n>2.

(End)

The entry line is _Gus Wiseman_, Jun 20 2019. The generating-function line is the same third-order recurrence together with initial values 1, 1, 2; no separate generating-function identity is claimed here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/NoncrossingNonnestingGraphRecurrence.barker_a326244

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    Barker's recurrence counts labeled simple graphs whose edges neither cross nor nest.

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    Dated conjecture statement: 2019.

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Analysis & dynamicsQuestion resolvedFirst preprint 2021The supremum of the elliptic remainder parameter exponentThe supremum of the elliptic remainder exponentProved

The questionBao, Wang and Qiu, Monotonicity properties of the generalized elliptic integrals with respect to the parameter, arXiv:2112.13993v1, Open problem 3.5:

Why it mattersThe exponent controls the upper parameter threshold for the generalized elliptic remainder in Theorem 3.1(2).

The ideaThe endpoint coefficient has the finite-product expression

Our contributionThe supremum of the elliptic remainder exponent

Verification details
Release source recordProved in Lean

Open problem

The supremum of the elliptic remainder parameter exponent

Bao, Wang and Qiu, Monotonicity properties of the generalized elliptic integrals with respect to the parameter, arXiv:2112.13993v1, Open problem 3.5:

(2) What is the value of λ̄₁₀ in Lemma 3.2 (4) for n ∈ ℕ?

Lemma 3.2(4) states:

Let λ̄₁₀ = sup_{a∈(0,1/2]} {λ₁₀(a)} for n ∈ ℕ₀. Then λ̄₁₀ = 2 if n = 0.

Equation (3.2), specialized to $c=1$, gives

$$g_{6,k}(a)=a(a)_k(1-a)_k,\qquad (a)k=\prod{j=0}^{k-1}(a+j).$$

Write $S_n(a)=\sum_{k=0}^n g_{6,k}(a)/((k+1)(k!)^2)$. Equation (3.6) is

$$\lambda_{10}(a)=\frac{a}{1-a} \frac{\sin(\pi a)+\pi(1-a)\cos(\pi a) -\pi(1-a)^2\sum_{k=0}^n\frac{g'{6,k}(a)}{(k+1)(k!)^2}} {\sin(\pi a)-\pi(1-a)\sum{k=0}^n\frac{g_{6,k}(a)}{(k+1)(k!)^2}}.$$

Formal reading

All Lean names below have prefix D5.S3.Analytic.EllipticTailExponent.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $g_{6,k}(a)$ | Defs.g6 | Exactly $a(a)k(1-a)k$, with empty products equal to one. | | $S_n(a)$ | Defs.S | The finite sum through index $n$, inclusive; deriv (S n) is its real derivative. | | $\lambda{10}(a)$ in (3.6) | Defs.lambda10 n a | The same trigonometric quotient, with truncation order explicit. | | $a\in(0,1/2]$ | Set.Ioc 0 (1 / 2) | The left endpoint is excluded and the right endpoint included. | | $\bar\lambda{10}=2$ for all $n\in\mathbb N_0$ | Defs.claim, Result.result | IsLUB (lambda10 n '' Set.Ioc 0 (1 / 2)) 2 for every n : ℕ, including zero. | | Positive denominator and strict upper bound | Result.denominator_pos, Result.lambda10_lt_two | Both hold at every point of the stated interval. | | Limit at zero from the right | Result.tendsto_lambda10 | Tendsto (lambda10 n) (𝓝[>] 0) (𝓝 2) for each natural $n$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/EllipticTailExponent/Result.result

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Quantum informationQuestion resolvedSource 2016Bhatia–Jain: Stolarsky mean matrices in class APositive profile curvature implies conditional negativity and the class A property for Stolarsky mean matrices.Proved

The questionJain, Mean matrices and conditional negativity, Electronic Journal of Linear Algebra 29 (2016), 206–222, section 2.9, defines the Stolarsky means by

Why it mattersThe real parameter range cannot be reduced to integer power factorizations.

The ideaThis holds also at zero by continuity of both sides.

Our contributionPositive profile curvature implies conditional negativity and the class A property for Stolarsky mean matrices.

Verification details
Release source recordProved in Lean

Open problem

Bhatia–Jain: Stolarsky mean matrices in class A

R. Bhatia and T. Jain, Mean matrices and conditional negativity, Electronic Journal of Linear Algebra 29 (2016), 206–222, section 2.9, defines the Stolarsky means by

Sγ (a, b) = ((aγ − bγ)/(γ(a − b)))^{1/(γ−1)}, −∞ < γ < ∞

with their continuous extensions at equal inputs and at parameters zero and one. Class A means positive entries and exactly one positive eigenvalue, counted with multiplicity. The cited source states:

This family does not yield readily to our analysis. We have some evidence for the following: Conjecture 1. Let λ1, λ2, . . . , λn be positive numbers. The matrix [Sγ(λi, λj)] is in the class A if γ ≥ 2 but this is not always so when 0 ≤ γ < 2.

The exact formal target quantifies over every real parameter at least two and every nonempty positive finite family, allowing repeated nodes. For each parameter from zero inclusive to two exclusive it asks for some positive finite family outside class A. The source proves parameters three and four in Theorems 2.10 and 2.11 and relates parameters zero, one half, and two to earlier sections.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/StolarskyMean/Result.result

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Quantum informationQuestion resolvedFirst preprint 2019The lower boundary of the entropy region of qutrit channels and their complements is the conjectured curveComplementary qutrit channels with a qutrit environment have a sharp lower boundary for their map entropies, attained by a one-parameter family and its complementary channels.Proved

The questionBraun and K. Życzkowski, Trade-off relations for operation entropy of complementary quantum channels, Int.

Why it mattersThe trade-off between the map entropies quantifies how the information leaking to the environment constrains the information kept by the system (§III).

The ideaLet P = VV†/3, a state on B ⊗ E with marginals ρB and ρE.

Our contributionComplementary qutrit channels with a qutrit environment have a sharp lower boundary for their map entropies, attained by a one-parameter family and its complementary channels.

Verification details
Release source recordProved in Lean

Open problem

The lower boundary of the entropy region of qutrit channels and their complements is the conjectured curve

J. Czartowski, D. Braun and K. Życzkowski, Trade-off relations for operation entropy of complementary quantum channels, Int. J. Quantum Inf. 17, 1950046 (2019), arXiv:1908.03492v1, study the pairs (S_map(Φ), S_map(Φ̃)) of the map entropy of a channel and of its complementary channel.

  • The map entropy. S_map(Φ) is the von Neumann entropy of the normalized Choi state. By Proposition 1, S_map(Φ) = S(Φ̃(I/N)) and S_map(Φ̃) = S(Φ(I/N)).
  • The conjecture. Conjecture 1 (page 5): "For qutrit channels with qutrit environment the lower boundary of the allowed set A_3 is given by the parametric curve p(a) = (−a log a − (1 − a) log(1 − a), log 3/3 − (a + 1/3) log(a + 1/3) − (a − 1/3) log(a − 1/3)) (20) with a ∈ (0, 1/3), together with reflection through line S = S̃".

The verbatim statements are in the literature note.

Issue #17038 fixes the reading:

  • The last term of Eq. (20) is read as −(1/3 − a) log(1/3 − a): the printed term is not real for a < 1/3, and the source's attaining channels give the corrected one.
  • Put h(a) = H(1 − a, a) and g(a) = H(1/3 + a, 1/3, 1/3 − a). The lower boundary means two things for a ∈ [0, 1/3]:
    • for every isometry V : ℂ³ → ℂ³ ⊗ ℂ³, S_map(Φ) ≤ h(a) implies S_map(Φ̃) ≥ g(a), and the same with Φ and Φ̃ exchanged;
    • the curve and its reflection are attained.

The conjecture is proved in this reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/QutritComplementaryEntropyFrontier.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2018The sharp weighted Bessel K tail boundSquared-order convexity and an exact integral identity give the sharp weighted Bessel tail bound.Proved

The questionGaunt, Inequalities for some integrals involving modified Bessel functions, Proc.

The ideaDifferentiation under the integral and the fundamental theorem give

Our contributionSquared-order convexity and an exact integral identity give the sharp weighted Bessel tail bound.

Verification details
Release source recordProved in Lean

Open problem

The sharp weighted Bessel K tail bound

R. E. Gaunt, Inequalities for some integrals involving modified Bessel functions, Proc. Amer. Math. Soc. 147 (2019), 2937–2951, DOI 10.1090/proc/14433, arXiv:1806.00524v2, Conjecture 2.4, manuscript page 7. The source states:

“Conjecture 2.4. Let n < 1. For ν ≥ 1 − n + √(2(1 − n)), define α_{ν,n} := ν + n − √((ν + n)² − 2ν − 1). Then, for all x > 0, ∫_x^∞ K_{ν+n}(t)/t^ν dt > K_{ν+n−α_{ν,n}}(x)/x^ν, ν ≥ 1 − n + √(2(1 − n)). (2.17) We have equality in (2.17) when n = 1, and this equality is valid for all ν ∈ ℝ. Let α′{ν,n} := ν + n + √((ν + n)² − 2ν − 1), so that ν + n − α′{ν,n} = −(ν + n − α_{ν,n}) and thus K_{ν+n−α′{ν,n}}(x) = K{ν+n−α_{ν,n}}(x). Then, the index ν + n − α_{ν,n} (and equivalently ν + n − α′{ν,n}) in inequality (2.17) is best possible, in the sense that, for any β ∉ {α{ν,n}, α′{ν,n}}, either K{ν+n−β}(x)/x^ν < K_{ν+n−α_{ν,n}}(x)/x^ν for all x > 0, or there exists a y > 0 such that ∫_y^∞ K_{ν+n}(t)/t^ν dt < K_{ν+n−β}(y)/y^ν.”

The quotation is the source statement. The Lean correspondence and its literal definitions are given below.

Formal reading

Lean names have prefix D5.S3.Analytic.GauntBesselTail.Result.

| Object | Lean declaration | Correspondence | | --- | --- | --- | | Bessel K | D5.S3.Analytic.GammaGammaNBU.Result.besselK | The existing hyperbolic integral, DLMF 10.32.9, on positive arguments. | | Two shifts | alpha, alphaPrime | Literal source square-root formulas. | | Weighted upper tail | tail | Real-power weight on t>x>0. | | Entire conjecture | claim | The conjunction of the strict bound, its either/or optimality, and the unrestricted real-order equality. | | Settlement | result | Exactly claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/GauntBesselTail/Result.result

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Geometry & algebraQuestion resolvedFirst preprint 2015The closed form of the higher-antibracket coefficientsFor every n >= 2, the coefficients c1^n, ..., cn^n that express the higher Koszul bracket Phi^(n+1) through the operators rho1, ..., rhon are given by the closed formula conjectured by M.Proved

The questionRicciardi, Universal Lie Formulas for Higher Antibrackets, SIGMA 12 (2016), 053, arXiv:1509.09032v3, Conjecture 2.4:

Why it mattersHigher Koszul brackets (higher antibrackets) measure the failure of an operator to be a derivation and govern the BV formalism.

Our contributionFor every n >= 2, the coefficients c1^n, ..., cn^n that express the higher Koszul bracket Phi^(n+1) through the operators rho1, ..., rhon are given by the closed formula conjectured by M.

Verification details
Release source recordProved in Lean

Open problem

The closed form of the higher-antibracket coefficients

M. Manetti and G. Ricciardi, Universal Lie Formulas for Higher Antibrackets, SIGMA 12 (2016), 053, arXiv:1509.09032v3, Conjecture 2.4:

For every $n\ge 2$ the coefficients $c^n_i$ of Theorem~{\rm \ref{thm.standardform}} are given by the formula

$c_i^n=\frac{\displaystyle (-1)^{n} \prod_{j=2}^{i}\dfrac{n(n-1)-(j-1)(j-2)}{2}}{\displaystyle \sum_{h=2}^n h\left(\prod_{j=2}^{h}\dfrac{n(n-1)-(j-1)(j-2)}{2}\right)\left(\prod_{j=h}^{n-1}\dfrac{(1-j)(j+2)}{2}\right)}$,

where every empty product is intended to be equal to $1$. Moreover $(-1)^{n}c^n_i>0$ for every $n\ge i\ge 1$.

The coefficients of Theorem 2.3 are the unique rationals with $\Phi^{n+1}_f=(c_1^n\rho_1^n+c_2^n\rho_1^{n-2}\rho_2+\dots+c_n^n\rho_n)f$ for every linear operator $f$ of every graded commutative algebra. The paper proves Theorem 2.3 from Theorem 6.4, the same identity $\Phi^{n+1}=(c_1^n\rho_1^n+\dots+c_n^n\rho_n)\Phi^1$ for the operators $\Phi^{m,i}$ and $\rho_k$ on $\mathbb{Q}[x]$, through the $\mathfrak g$-module morphism of Lemma 6.3; uniqueness in Theorem 2.3 identifies the two sequences. The authors verified the formula for $n\le12$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/HigherAntibracketCoefficients.result

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2015Nikiforov: asymptotic graph Ky Fan gapsInfinitely many graph Ky Fan prefixes have strict asymptotic gaps.Proved

The questionNikiforov, Extrema of graph eigenvalues, arXiv:1502.00359v1, §4, Conjecture 4.7.

Why it mattersThe stronger target is a positive gap for every nonsquare positive k, uniformly over every order n≥k and every symmetric zero-one matrix, including matrices with nonzero diagonal.

The ideaCenter A by B=2A-J, normalize T=B/n, and choose a signed orthonormal family attaining the Ky Fan sum.

Our contributionInfinitely many graph Ky Fan prefixes have strict asymptotic gaps.

Verification details
Release source recordProved in Lean

arXiv

Nikiforov: asymptotic graph Ky Fan gaps

The source is V. Nikiforov, Extrema of graph eigenvalues, arXiv:1502.00359v1, §4, Conjecture 4.7. Its definition is:

ξk (n) = max λ∗1 (G) + · · · + λ∗k (G)

The maximum is over graphs of order n. Starred eigenvalues denote singular values, equivalently the absolute eigenvalues of the real symmetric adjacency matrix, in decreasing order. The text preceding the conjecture states:

Theorem 4.3 does not shed any light on the case when k is not an exact square, so we suggest the following concrete conjecture.

Conjecture 4.7 states:

There exist infinitely many integers k such that lim_{n→∞} ξk (n)/n < (1+√k)/2.

The repeated statement is Conjecture 3.20 of Beyond graph energy: norms of graphs and matrices, arXiv:1510.02850v2, §3.3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/KyFanGap/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2018Every incoherent channel with coherence-breaking index at least two is amended by one unitaryEvery incoherent trace-preserving Kraus channel of finite coherence-breaking index at least two admits one unitary conjugation repeated between its successive applications.Proved

The questionWang, Amending Coherence-Breaking Channels via Unitary Operations (Int.

Why it mattersCoherence-breaking channels destroy all coherence; amendment asks whether interleaved unitary control can keep some coherence through repeated uses of a noisy incoherent channel.

The ideaLet O be the projection onto off-diagonal matrices, and D = 1 − O.

Our contributionEvery incoherent trace-preserving Kraus channel of finite coherence-breaking index at least two admits one unitary conjugation repeated between its successive applications.

Verification details
Release source recordProved in Lean

Open problem

Every incoherent channel with coherence-breaking index at least two is amended by one unitary

L.-M. Yang, B. Chen, T. Li, S.-M. Fei and Z.-X. Wang, Amending Coherence-Breaking Channels via Unitary Operations (Int. J. Theor. Phys. 57, 3776 (2018); arXiv:1810.09642), study incoherent channels.

  • Definitions. An incoherent channel is coherence-breaking if it maps every state to an incoherent state. The coherence-breaking index n(Φ) is the least k with Φ^k coherence-breaking.
  • Results. For qubit channels with n(Φ) = 1 and 2 they find unitaries Λα that make Λα ◦ Φ and Φ ◦ Λα ◦ Φ not coherence-breaking.
  • The conjecture. Section 5 states: "for a d-dimensional quantum channel Φ with n(Φ) ≥ 2, one may conjecture that the channel Φ can also be amended by Φ ◦ Λα ◦ Φ ◦ · · · ◦ Φ ◦ Λα ◦ Φ [n Φ] with proper unitary operations Λα."
  • The later preprint. MiS Preprint 7/2019 asserts the qudit statement in its conclusions without a proof.

The verbatim statements are in the literature note.

Issue #18385 fixes the reading: for every incoherent trace-preserving Kraus channel with n(Φ) = n ≥ 2, some unitary U makes Φ ◦ (Λ_U ◦ Φ)^{n−1} not coherence-breaking, with the same U in every slot. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/CoherenceBreakingAmendment.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2012Schur-channel diamond-distance equalityFor every Schur channel, the diamond distance to all mixed-unitary channels equals the distance to Schur mixed-unitary channels.Proved

The questionNengkun Yu, Runyao Duan and Quanhua Xu, Bounds on the distance between a unital quantum channel and the convex hull of unitary channels, IEEE Transactions on Information Theory 63(2), 1299–1310, Section V, Theorem 3, Equation (8), ask:

Why it mattersAn arbitrary mixed-unitary competitor can be replaced by a diagonal-unitary mixture without increasing its distance from the target Schur channel.

The ideaRetain the second copy while the input channel acts on the first.

Our contributionFor every Schur channel, the diamond distance to all mixed-unitary channels equals the distance to Schur mixed-unitary channels.

Verification details
Release source recordProved in Lean

Open problem

Schur-channel diamond-distance equality

Nengkun Yu, Runyao Duan and Quanhua Xu, Bounds on the distance between a unital quantum channel and the convex hull of unitary channels, IEEE Transactions on Information Theory 63(2), 1299–1310, Section V, Theorem 3, Equation (8), ask:

It seems quite likely that in Eq. (8) the first inequality should be strict and the second one should be an equality. However, this is still an unsettled issue.

The equality in the second inequality is $D(\Phi,\operatorname{Conv}(U_d))=D(\Phi,\Lambda_d)$ for every Schur channel $\Phi$, where $\Lambda_d=S_d\cap\operatorname{Conv}(U_d)$. The distance is the unhalved diamond distance. The literature note contains the definitions and source quotations. Preregistration #16745 specifies the full quantified statement.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Every dimension and every channel | claim quantifies natural d and QuantumChannel (Fin d) (Fin d) | Includes dimension zero; complete positivity and trace preservation are carried by the channel type. | | Schur action $\Phi(X)=C\circ X$ on every matrix | IsSchurChannel | Literal existence of a matrix giving Hadamard action; no replacement by a recovery certificate. | | Convex hull of unitary channels | MixedUnitary | Finite families, nonnegative real weights of sum one, and unitary conjugation on every complex matrix. | | $\Lambda_d=S_d\cap\operatorname{Conv}(U_d)$ | SchurMixedUnitary | Literal intersection; diagonal mixtures are obtained in the proof. | | Infimum of diamond distances | distToSet | Real sInf of the image of the unhalved diamondDistance; no attained minimum is assumed. | | Equality in the second inequality of (8) | result : claim | Proved with the Schur hypothesis as the only additional hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/SchurChannelMixedUnitaryDistance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationConjecture refutedFirst preprint 2023Han–Liu's twirling phase-retrievability conjectureThree characters of C3 with multiplicity two contradict the proposed twirling phase index formula.Refuted

The questionLiu, Zero-Error Correctibility and Phase Retrievability for Twirling Channels, arXiv:2311.16349v1 (27 November 2023), published in Reports on Mathematical Physics 93(1) (2024), 87–102, DOI 10.1016/S0034-4877(24)00012-0, state the final conjecture of Section 5:

The ideaCharacter orthogonality identifies the C3 twirl with three-block pinching.

Our contributionThree characters of C3 with multiplicity two contradict the proposed twirling phase index formula.

Verification details
Release source recordRefuted in Lean

Open problem

Han–Liu's twirling phase-retrievability conjecture

D. Han and K. Liu, Zero-Error Correctibility and Phase Retrievability for Twirling Channels, arXiv:2311.16349v1 (27 November 2023), published in Reports on Mathematical Physics 93(1) (2024), 87–102, DOI 10.1016/S0034-4877(24)00012-0, state the final conjecture of Section 5:

Let $\pi = m_1\pi_1\oplus\dots\oplus m_d\pi_d$ be a unitary representation of $G$ such that $\pi_i$ and $\pi_j$ are inequivalent irreducible representations for all $i\ne j$ and $\mathcal I_k\le d<\mathcal I_{k+1}$. Then $pr(\Phi_\pi)=\max{\beta(\Phi_\pi),k}$.

Issue #15229 preregistered the source, the finite-group specialization, the route, the readings and the settlement criteria. The frozen Lean declaration D5/S3/Quantum/QuantumChannels/TwirlingPhaseRetrievabilityCounterexample.result refutes the universal statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/TwirlingPhaseRetrievabilityCounterexample.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Further frontiersConjecture refutedFirst preprint 2023Herrera's function-approximation conjectureA positive-Fourier defect dilation and a three-point certificate refute Herrera's operator-monotone function-approximation conjecture.Refuted

The questionHerrera, On the commutator modulus of continuity for operator monotone functions, arXiv:2311.17448v1, Conjecture 2 (numbered 1.10 in print), states:

Why it mattersThe positive-Fourier approximation problem supplies upper estimates for commutator moduli.

The ideaNo moment assumption or interchange of two integrals is required.

Our contributionA positive-Fourier defect dilation and a three-point certificate refute Herrera's operator-monotone function-approximation conjecture.

Scope & next questionConjectures 1 and 3 concern commutator inequalities and remain open here.

Verification details
Release source recordRefuted in Lean

arXiv

Herrera's function-approximation conjecture

D. Herrera, On the commutator modulus of continuity for operator monotone functions, arXiv:2311.17448v1, Conjecture 2 (numbered 1.10 in print), states:

Let $f$ be a non-negative function that is operator monotone on $[0, \infty)$. Then $E_{f}(c) \leq Cf(c)$ holds for all $c > 0$ when $C = 1$.

The source defines

$$ E_f(c)=\inf_{g'\in\widehat{\mathcal M}+} \left(\sup{x\ge0}(f-g)(x)-\inf_{x\ge0}(f-g)(x)+cg'(0)\right), $$

where $\widehat{\mathcal M}_+$ consists of Fourier transforms of finite positive Borel measures on $\mathbb R$. Section 5, equation (28), requires $g\in C^1(\mathbb R)$. The conjecture is refuted by $f(x)=x/(1+x)$ and $c=1$. Preregistration: #15279.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | “$g$ is $C^1(\mathbb R)$ and $g'$ is the Fourier transform of a finite positive Borel measure” | IsAdmissible | ContDiff and a finite positive measure, with the complex exponential sign convention. | | The displayed oscillation plus $cg'(0)$ | approxCost | Supremum and infimum on the subtype of nonnegative reals. | | Infimum over admissible $g$ | approxError | Real infimum over finite oscillation costs. | | “a non-negative function” | The nonnegativity premise of claim | Every nonnegative real argument. | | “operator monotone on $[0,\infty)$” | OperatorMonotoneOn | Loewner order for positive semidefinite complex matrices at every finite dimension. | | “for all $c>0$ when $C=1$” | The positive-$c$ conclusion of claim | The constant is exactly one. | | Failure of the conjecture | result : ¬ claim | The witness satisfies all premises; the lower bound exceeds $f(1)$. |

approxError is a real infimum; Lean's real empty-set infimum convention is zero. The refutation instance has a nonempty finite cost set, witnessed by $g=0$, and a uniform finite lower bound. Discarding infinite-cost candidates preserves the source infimum in this domain. No extended-real equivalence is claimed for targets without a finite-cost approximant. Matrix monotonicity at all finite dimensions is the classical Löwner formulation used here; no separate infinite-dimensional transport theorem is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Fourier/PositiveFourierApproximationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Dicke optimality for detector-sensor identificationBalanced Dicke states maximize the success of equal-prior detector-sensor identification.Proved

The questionHillery, Optimizing Initial State of Detector Sensors in Quantum Sensor Networks, arXiv:2306.17401v6, ACM Transactions on Quantum Computing 5(2), Article 10, DOI 10.1145/3655028, Section 5, Conjecture 1, asks whether a state with uniform coefficient-squares on a minimizing weight partition is an optimal initial state.

Why it mattersThe settlement supplies a globally optimal pure probe on the stated angle range, including arbitrary coefficient phases.

The ideaThe public surface consists of fire, isoSuccess, rCount, lCount, T, claim and the single settling theorem result.

Our contributionBalanced Dicke states maximize the success of equal-prior detector-sensor identification.

Scope & next questionThe full-angle combination, ancillary-probe reduction and mixed-probe reduction remain open here.

Verification details
Release source recordProved in Lean

Open problem

Dicke optimality for detector-sensor identification

C. Zhan, H. Gupta and M. Hillery, Optimizing Initial State of Detector Sensors in Quantum Sensor Networks, arXiv:2306.17401v6, ACM Transactions on Quantum Computing 5(2), Article 10, DOI 10.1145/3655028, Section 5, Conjecture 1, asks whether a state with uniform coefficient-squares on a minimizing weight partition is an optimal initial state. The Library note contains the source quotation and locators.

There are $n\ge3$ qubit sensors, one firing event with equal priors $1/n$, and a unitary with eigenvalues $e^{i\theta}$ and $e^{-i\theta}$. Write $q=\lceil n/2\rceil$ and $T=\tfrac12\arccos(-(q-1)/q)$. The conjecture concerns $\theta\in(0,T]\cup[\pi-T,\pi)$. For a tensor eigenbasis word $x$, its weight is the number of plus eigenvectors. For two distinct sensors, $R_k$ and $L_k$ count weight-$k$ words whose selected bits agree or differ. Every minimizing $l$ of $(R_k+\cos(2\theta)L_k)/(R_k+L_k)$, over $0\le k\le n$, is allowed. The proposed state has coefficient-squares $1/\binom nl$ on weight $l$ and zero elsewhere, with arbitrary phases.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Qubit tensor eigenbasis | Fin n → Bool, true for the plus eigenvector | A coordinate representation of all pure input states | | One firing event at sensor $i$ | fire | The local eigenphase multiplies each coefficient | | Equal priors and global POVM | isoSuccess; the frozen TriangleIndependentSourceNoise.LocalPOVM | Positive semidefinite effects sum to the identity; no locality restriction | | $n\ge3$ | n, hn in claim | Includes every permitted sensor number | | Both angle intervals and threshold endpoints | T, the disjunction in claim | Radians replace degrees without changing the intervals | | Every minimizing partition | l, l ≤ n, IsMinOn on Set.Iic n | The selected distinct sensor indices have values zero and one | | Uniform coefficient-squares, arbitrary phases | φ, the pointwise Complex.normSq hypothesis | The weight-cardinality expression is inlined literally | | Any normalized competing pure state | ψ, the sum of norm-squares equals one | No symmetry assumption on the competitor | | Optimal initial state | For every M, an attaining M' with the success comparison | result : claim proves the universal comparison |

The natural division in T gives $\lceil n/2\rceil$. Fin-index proof fields witness index bounds and impose no additional hypothesis. The existing LocalPOVM supplies the stated POVM definition directly. The weight expression is inlined rather than duplicated beside the existing I-level DynamicCoolingParallelTieUniqueness definition; the settling module has generality G. No source clause is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/DetectorSensorDickeOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
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    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst sums of decreasing partitionsA decreasing partition is not a first-sums list exactly when it has a part below four.Proved

The questionNAME "Number of integer partitions of n that are not the first sums of any composition with all parts > 1."

Why it mattersThe KPI is open problems resolved, and the target is the universal characterization and its partition counting corollary, with no finite search bound.

The ideaNecessity: if q = firstSums x and every part of x is at least two, each adjacent sum is at least four (mingefourofisFirstSums).

Our contributionA decreasing partition is not a first-sums list exactly when it has a part below four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

First sums of decreasing partitions

OEIS A391620 (Gus Wiseman, Dec 30 2025) states:

NAME "Number of integer partitions of n that are not the first sums of any composition with all parts > 1."

COMMENT "Conjecture: These are integer partitions of n whose least part is < 4."

The NAME line is copied verbatim from Library/Words/wiseman2025a391620.md. The reading used by Lean writes a partition q as a weakly decreasing, nonempty list of positive parts. It is the first sums of a composition x = (x_1, ..., x_{r+1}), with every x_j >= 2, exactly when q_i = x_i + x_{i+1} for each i. Equality preserves this order; it is not equality after arbitrarily permuting the adjacent sums. The conjecture is equivalent to (not first sums) iff min q < 4. For a nonempty partition, the latter condition is equivalent to the existence of a part below four.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/FirstSumsPartitionCharacterization.wiseman_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedFirst preprint 2016Lin's matrix majorization conjecture fails at exponent fourLin's majorization conjecture fails for positive semidefinite three-by-three matrices at exponent four.Refuted

The questionThe source clause is “Let A and B be n × n positive semidefinite matrices.

Why it mattersThe source presents the majorization statement as a strengthening of the determinantal question and cites an integer-power comparison.

The ideaEqual traces and the two-entry prefix condition contradict majorization.

Our contributionLin's majorization conjecture fails for positive semidefinite three-by-three matrices at exponent four.

Verification details
Release source recordRefuted in Lean

Open problem

Lin's matrix majorization conjecture fails at exponent four

Lin, On a determinantal inequality arising from diffusion tensor imaging, Communications in Contemporary Mathematics 19 (2017), Conjecture 4.3, asks whether for positive semidefinite matrices $A,B$ and every real $p>0$,

$$ \lambda(A^2+|BA|^p) \succ \lambda(A^2+|AB|^p). $$

Here $|X|=(X^*X)^{1/2}$ and the eigenvalue lists are in descending order. The Lean claim uses the literal carrier Matrix (Fin n) (Fin n) ℂ, the continuous functional calculus expression for $|X|^p$, and the source's prefix-sum definition of majorization.

Statement echo

The source clause is “Let A and B be n × n positive semidefinite matrices. Then λ(A² + |BA|^p) ≻ λ(A² + |AB|^p) for all p > 0.” The declaration D5/S3/Quantum/Matrix/LinDTIMajorizationRefutation.claim binds $n,A,B,p$, the two positive-semidefinite hypotheses, and the two ordered eigenvalue lists with exactly those operations.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/LinDTIMajorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Number theoryCounterexample foundFirst preprint 2017A self-dual irreducible extreme thermal process can factor on proper subspacesA five-level self-dual irreducible extreme thermal process factors into two extreme processes acting on fewer levels, refuting the non-self-duality conjecture.Refuted

The questionHorodecki, Decomposability and convex structure of thermal processes, New J.

Why it mattersThe problem concerns whether detailed balance obstructs building an extreme thermal process from extreme operations that each leave some energy level untouched.

The idea2. Extremality. Nonnegativity forces both terms of any strict convex decomposition to vanish wherever the decomposed matrix vanishes.

Our contributionA five-level self-dual irreducible extreme thermal process factors into two extreme processes acting on fewer levels, refuting the non-self-duality conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

A self-dual irreducible extreme thermal process can factor on proper subspaces

P. Mazurek and M. Horodecki, Decomposability and convex structure of thermal processes, New J. Phys. 20, 053040 (2018), arXiv:1707.06869v2, Section IV.B, “Detailed balance symmetry”, Conjecture 1:

If an extremal TP $C$ for $d$ dimensional space is decomposable into a sequence of extremal TPs $A$, $B$, each acting non-trivially on at most $d-1$ dimensional space: $C=AB$, and $C$ is not a direct sum of extremal TPs from lower dimensional subspaces, then $C$ is not self-dual with respect to the conjugation associated with the operator scalar product.

This quotation is from manuscript.tex, lines 425–427, read from the arXiv v2 source, md5 69e7001dd049721e95968ea93ad550c2. The source's matrix characterization of thermal processes is at line 159, its discussion of proper-subspace factors at lines 208–211, and its conjugation at lines 415–420. These passages are also located in the literature note.

Issue #20157 fixes the following precise reading. For every $d\ge2$ and every positive, pairwise distinct weight vector $g\in\mathbb R^d$ with $g_i\le g_0$ for every $i$, let

$\mathrm{TP}(g)={T\in\mathbb R^{d\times d}:T_{ij}\ge0\text{ for all }i,j,\ \sum_iT_{ij}=1\text{ for all }j,\ Tg=g}$.

For all extreme points $A,B,C$ of this set, suppose $C=AB$, each of $A$ and $B$ has a diagonal entry equal to one, and there is no nonempty proper set of levels whose two cross-blocks in $C$ are zero. The claim concludes that $C$ is not self-dual: $C_{ij}g_j\ne C_{ji}g_i$ for some $i,j$. The module calls these predicates thermalProcesses, actsOnFewerLevels, irreducible and selfDual, and calls the universal implication claim.

The universal claim is refuted. Its additional restriction to distinct weights makes it weaker than the source's unrestricted conjecture, so refuting this restriction also refutes Conjecture 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/ThermalProcessSelfDualFactorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2016Binomial and kappa states do not exhaust truncated entropy minimizersThe loss-dephasing channel of Memarzadeh and Mancini has a qutrit input with strictly smaller output entropy than every binomial or kappa candidate at the same input energy.Refuted

The questionLaleh Memarzadeh and Stefano Mancini, Minimum output entropy of a non-Gaussian quantum channel, arXiv:1605.04525v1, Phys.

Why it mattersThe witness is a qutrit input of energy 3/2 outside both candidate families; its output entropy is strictly smaller than that of every candidate at the same channel parameters and energy.

The ideaDefine the channel using the source's Kraus operators Aj Pk.

Our contributionThe loss-dephasing channel of Memarzadeh and Mancini has a qutrit input with strictly smaller output entropy than every binomial or kappa candidate at the same input energy.

Verification details
Release source recordRefuted in Lean

Open problem

Binomial and kappa states do not exhaust truncated entropy minimizers

Laleh Memarzadeh and Stefano Mancini, Minimum output entropy of a non-Gaussian quantum channel, arXiv:1605.04525v1, Phys. Rev. A 94, 022341 (2016), section IV, Conjecture 1, state:

In a truncated Hilbert space of dimension K+1, the minimal output entropy of the quantum channel (3) is achieved either by binomial states of Eq.(14) or by states |κ_α⟩ of Eq. (11), depending on the values of ε and t.

Issue #11680 specifies the fixed-energy reading: for every K ≥ 1, N ∈ [0,K], ε ∈ [0,1] and t ≥ 0, some binomial or κ candidate minimizes output entropy among positive semidefinite trace-one inputs whose number expectation is N. Binomial candidates satisfy 1 ≤ M ≤ K, μ ∈ [0,1] and Mμ=N; κ candidates allow every real phase α.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/TruncatedLossDephasingOptimizerRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2013Conditional subentropy and guessing min-entropyThe simplex-integral conditional subentropy of a classical-quantum state is bounded by guessing min-entropy.Proved

The questionDatta, Dorlas, Jozsa and Benatti, Properties of subentropy, arXiv:1310.1312v1, §2.5.1, Conjecture 2:

Why it mattersPreregistration 18776 identifies this as a tier-1 explicit published conjecture.

The ideaRadial decomposition of the exponential law and its Gamma moments express the normalized simplex integral as a Gaussian quadratic expectation.

Our contributionThe simplex-integral conditional subentropy of a classical-quantum state is bounded by guessing min-entropy.

Scope & next questionThis module makes no claim about infinite-dimensional states or a complete equality classification.

Verification details
Release source recordProved in Lean

Open problem

Conditional subentropy and guessing min-entropy

Datta, Dorlas, Jozsa and Benatti, Properties of subentropy, arXiv:1310.1312v1, §2.5.1, Conjecture 2:

For a c-q state ρ_XB, the conditional subentropy Q(X|B) satisfies the following bound: Q(X|B) ≤ H_min(X|B) = − ln p_guess(X|B).

For every finite family of positive semidefinite complex matrices $A_x$ of total trace one, the joint state is $\bigoplus_x A_x$ and its marginal is $\sum_x A_x$. The claim uses equation (4), with normalized simplex measure from equation (3), for subentropy. Natural logarithms are used. Zero blocks, zero eigenvalues and repeated eigenvalues remain in scope.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (2), $C_n=1/2+\cdots+1/n$ | C | Inclusive sum from 2 to n. | | Equation (3), normalized simplex measure | simplexCoords, subentropySpectrum | Nonnegative first n−1 coordinates of sum at most one; factorial times Lebesgue measure. | | Equation (4), $Q=n\int\eta(\lambda\cdot x),dx-C_n$ | subentropySpectrum, subentropy | Real.negMulLog, the existing simplex chart, and all Hermitian eigenvalues. | | Finite c-q state | claim | Arbitrary finite X and B; positive semidefinite blocks, total trace one. | | Equation (13), $Q(X\mid B)=Q(\rho_{XB})-Q(\rho_B)$ | claim | Block diagonal state and summed marginal. | | All POVMs; guessing min-entropy | pGuess, result | Supremum of the exact finite POVM scores; negative natural logarithm. | | Typeclass and Hermitian binders | claim | Finite indexing and matrix well-formedness; Hermiticity follows from positivity. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/ConditionalSubentropyGuessingBound.result

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Quantum informationCounterexample foundFirst preprint 2018Entropy maximisers need not have balanced degreesAn unbalanced three-edge multigraph maximises orientation-count entropy.Refuted

The questionPelekis, Le Matematiche 74(1), pp. 119–130, Conjecture 1.3, p. 122: “Let the positive integers m,n,r be such that m ≥ n > r ≥ 2.

Why it mattersThe conjecture proposes degree balance as a necessary condition for entropy maximality.

The ideaLogarithm monotonicity proves maximality; vertices zero and two have degree difference two, contradicting claim.

Our contributionAn unbalanced three-edge multigraph maximises orientation-count entropy.

Verification details
Release source recordRefuted in Lean

Open problem

Entropy maximisers need not have balanced degrees

Pelekis, Le Matematiche 74(1), pp. 119–130, Conjecture 1.3, p. 122: “Let the positive integers m,n,r be such that m ≥ n > r ≥ 2. A hypergraph H = (V,ℰ) from D_{n,m,r} for which it holds H(X_H) ≥ H(X_G), for all G ∈ D_{n,m,r} is such that |deg_H(v₁) − deg_H(v₂)| ≤ 1, for all v₁,v₂ ∈ V.” Here $D_{n,m,r}$ includes repeated edges and isolated vertices. Each edge independently selects a uniformly random vertex, and $X$ counts the vertices selected at least once. Entropy uses logarithms of base two and $0\log 0=0$. The title's zero-in-degree count is $n-X$; that bijection preserves entropy.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $m\ge n>r\ge2$ | claim quantifies over n m r, with n ≤ m, r < n, 2 ≤ r | All parameters are positive under these hypotheses. | | $r$-uniform multi-hypergraphs | E : Fin m → Finset (Fin n) and every edge has cardinality r | Ordered edges retain multiplicities; isolated vertices remain. | | Uniform independent orientations | The literal orientation filter inside entropyX r E, with denominator $(r:\mathbb R)^m$ | Mathlib Fintype.piFinset supplies the public correspondence. | | $X$ is the number with nonzero in-degree | X E o counts the image of o | The edge argument labels the family; the value depends on o. | | Shannon entropy | entropyX r E sums $-p\log_2p$ over $k=0,\ldots,n$ | r is explicit; zero-probability outcomes are included. | | Entropy at least every competitor | Every uniform F satisfies entropyX r F ≤ entropyX r E | Full competitor class, with no sampling restriction. | | Absolute degree difference at most one | Every ordered pair satisfies the integer difference bound | Both orders give the two signs of the absolute-value bound. | | Refutation | result : ¬ claim | Unconditional negation of the complete conjecture. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Hypergraphs/OrientationEntropyBalance.result

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Quantum informationQuestion resolvedFirst preprint 2021Tightness of the 0-fidelity lower bound and the worst-case processesThe product SIC states test an n-qubit channel through their survival probabilities.Proved

The questionA description of the worst-case process is an open problem.

Why it mattersThe 0-fidelity is estimated from product states only.

Our contributionThe product SIC states test an n-qubit channel through their survival probabilities.

Verification details
Release source recordProved in Lean

Open problem

Tightness of the 0-fidelity lower bound and the worst-case processes

K. Mayer, A short note on the 0-fidelity, arXiv:2109.09629v1, proves for every $n$-qubit process $\mathcal E$ that $1-\frac32(1-F_0)\le F\le F_0$, where $F$ is the process fidelity and $F_0$ the 0-fidelity on tensor products of single-qubit SIC-POVM states, and writes:

This leads us to conjecture that the lower bound in Th.~1 is tight for all $n$.

A description of the worst-case process is an open problem.

The verbatim statements are in the literature note. Issue #14119 reads the two statements for every $n\ge1$, every choice of single-qubit SIC-POVMs and every finite Kraus family: the lower bound holds with equality exactly when every Kraus operator lies in the span of the identity and the single-qubit Pauli operators, and every value $f\in[\frac13,1]$ of $F_0$ is attained with equality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/MayerZeroFidelityTightness.result

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Quantum informationCounterexample foundFirst preprint 2019Measurement can extract more than half the ergotropy of a non-stationary stateA non-stationary pure state in dimension four has metrotropy strictly greater than half its ergotropy.Refuted

The questionCampisi, Maximal energy extraction via quantum measurement, J.

Why it mattersA positive answer would make unitary driving at least twice as effective as any rank-one measurement for extracting energy from every finite quantum system, extending the stationary bound of the source to states with coherence in the energy basis.

The ideaThe module defines measured, metrotropy, ergotropy and claim as above.

Our contributionA non-stationary pure state in dimension four has metrotropy strictly greater than half its ergotropy.

Verification details
Release source recordRefuted in Lean

Open problem

Measurement can extract more than half the ergotropy of a non-stationary state

A. Solfanelli, L. Buffoni, A. Cuccoli and M. Campisi, Maximal energy extraction via quantum measurement, J. Stat. Mech. (2019) 094003, arXiv:1905.10262v2, compare the metrotropy $\mathcal M$, the maximal energy extractable by a non-selective rank-one projective measurement, with the ergotropy $\mathcal W$, the maximal energy extractable by a unitary. For stationary states they prove $\mathcal M\le\mathcal W/2$; for non-stationary states they prove $\mathcal M\le\mathcal W$. Section 3.1 of the source TeX Metrotropy02.tex, lines 277–308, ends:

Whether the relation $\mathcal M \leq \mathcal W/2$ holds in the case $[H,\rho]\neq 0$ remains an open question.

The repository proposition claim reads the question for finite dimensions. For every $d$ and all $d\times d$ complex matrices $h,\rho$ with $h$ Hermitian, $\rho$ a density matrix and $h\rho\ne\rho h$, [ \operatorname{metrotropy}(h,\rho)\le\operatorname{ergotropy}(h,\rho)/2 . ] Here measured u ρ is the non-selective measurement $\sum_k\pi_k\rho\pi_k$ with $\pi_k$ the projector onto the $k$th column of a unitary $u$; metrotropy h ρ is $\operatorname{Re}\operatorname{Tr}(h\rho)$ minus the infimum of the energies of measured u ρ over all unitaries $u$; ergotropy h ρ is $\operatorname{Re}\operatorname{Tr}(h\rho)$ minus the infimum of the energies of $v\rho v^\dagger$ over all unitaries $v$.

The Lean declaration is result : \lnot claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Thermal/MetrotropyHalfErgotropyRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2018Strong subadditivity of bosonic Gaussian Rényi entropy for $0<\alpha\le2$, $\alpha\ne1$Bosonic Gaussian Renyi entropy satisfies strong subadditivity for every order from zero to two, with zero and one excluded.Proved

The questionEliëns, On the Strong Subadditivity of the Rényi entropies for bosonic and fermionic Gaussian states, arXiv:1810.07070v2 (Phys.

Why it mattersThe source studies when Rényi entropy combinations can quantify correlations.

The ideaAll generic finite matrix index types below have decidable equality.

Our contributionBosonic Gaussian Renyi entropy satisfies strong subadditivity for every order from zero to two, with zero and one excluded.

Verification details
Release source recordProved in Lean

Open problem

Strong subadditivity of bosonic Gaussian Rényi entropy for $0<\alpha\le2$, $\alpha\ne1$

G. Camilo, G. T. Landi and S. Eliëns, On the Strong Subadditivity of the Rényi entropies for bosonic and fermionic Gaussian states, arXiv:1810.07070v2 (Phys. Rev. B 99, 045155, 2019), Section IV.A, RenyiGaussian_v2.tex, line 463:

It provides clear evidence that SSA is violated for all $\alpha>2$, while no violation is found for $0\le\alpha\le2$. We conjecture that this result holds true in general.

Section V, line 532, requests:

An obvious continuation of this work is to prove the conjectured domains of validity of the SSA for the R'enyi-$\alpha$ entropies.

For $n$ modes, put $\Omega=\bigoplus_{j=1}^n\begin{pmatrix}0&1\-1&0\end{pmatrix}$. The covariance formulation takes a real symmetric $2n\times2n$ matrix $V$ with $V+i\Omega\succeq0$. For a mode set $X$, retain both quadratures to obtain $V_X$ and $\Omega_X$. The module defines

$$s_\alpha(\nu)=\frac{\log\left[((\nu+1)/2)^\alpha-((\nu-1)/2)^\alpha\right]}{\alpha-1},\qquad S_\alpha(V_X)=\frac12\sum_{\lambda\in\operatorname{spec}(V_X^{1/2}(i\Omega_X)V_X^{1/2})}s_\alpha(|\lambda|),$$

with eigenvalues counted with multiplicity. This is the covariance entropy formula of Eq. renyi_bosonic_1 (TeX lines 216–220), using both signs of the symplectic spectrum. The empty sum is zero.

Issue #26322 fixes the statement: for every $n$, every such physical $V$, every pairwise disjoint mode sets $A,B,C$, and every $0<\alpha\le2$ with $\alpha\ne1$,

$$S_\alpha(V_{A\cup B})+S_\alpha(V_{B\cup C})\ge S_\alpha(V_{A\cup B\cup C})+S_\alpha(V_B).$$

This statement is proved. The settlement excludes order zero, order one, the existence of violations at every order above two, fermions and non-Gaussian states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Thermal/GaussianRenyiStrongSubadditivity.result

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Quantum informationQuestion resolvedFirst preprint 2018The inverse Gram matrix of an orthocross MIC is half-integralFor every dimension d and every orthonormal basis of C^d, the Gram matrix G of the orthocross MIC is invertible and every entry of G^(-1) is an integer or a half-integer.Proved

The questionQuantum Inf. 19 (2021) 2040005) study minimal informationally complete measurements (MICs) and their Gram matrices G{ij} = tr(Ei Ej).

Why it mattersThe inverse Gram matrix of a MIC is the matrix that turns the outcome probabilities of the MIC back into the state, so its entries control the probabilistic representation of quantum theory that the MIC provides.

The idea1. Ω is the identity plus a positive semidefinite matrix, so it is positive definite, (Ω^{−1/2})² = Ω⁻¹ and G{αβ} = tr(Πα Ω⁻¹ Πβ Ω⁻¹); a unitary change of basis leaves G unchanged.

Our contributionFor every dimension d and every orthonormal basis of C^d, the Gram matrix G of the orthocross MIC is invertible and every entry of G^(-1) is an integer or a half-integer.

Verification details
Release source recordProved in Lean

Open problem

The inverse Gram matrix of an orthocross MIC is half-integral

J. B. DeBrota, C. A. Fuchs and B. C. Stacey (arXiv:1812.08762, quant-ph; Int. J. Quantum Inf. 19 (2021) 2040005) study minimal informationally complete measurements (MICs) and their Gram matrices G_{ij} = tr(E_i E_j). For an orthonormal basis {|j⟩} of ℂ^d, the orthocross MIC has elements E_α = Ω^{−1/2} Π_α Ω^{−1/2}, where Π_α runs over the d² projectors onto |j⟩, (|j⟩ + |k⟩)/√2 and (|j⟩ + i|k⟩)/√2 (j < k) and Ω = Σ_α Π_α. The paper states:

Conjecture. For any orthocross MIC, the entries in G^{−1} are integers or half-integers.

Issue #11385 fixes the reading: every dimension d and every orthonormal basis (the columns of a unitary U), Ω^{−1/2} the positive square root of Ω⁻¹, and "integers or half-integers" as 2 (G⁻¹)_{αβ} ∈ ℤ, with G invertible.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/OrthocrossGramHalfInteger.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2016The inverse of a critical Carleman Jordan chain is an increasing-path sumThe inverse of each normalized critical Carleman Jordan chain is an increasing-path sum.Proved

The questionHuillet, Some combinatorial aspects of discrete non-linear population dynamics, Chaos, Solitons & Fractals 93, 71–79 (2016), arXiv:1610.05919v1, §2.2, state the following conjecture (snlmatrixRR.tex, lines 449–452; the displayed formula is rendered inline):

Why it mattersThe inverse coefficients supply the increasing-path weights in the paper's critical-iteration representation immediately following the conjecture (lines 456–466).

The ideaThe proof uses induction.

Our contributionThe inverse of each normalized critical Carleman Jordan chain is an increasing-path sum.

Verification details
Release source recordProved in Lean

Open problem

The inverse of a critical Carleman Jordan chain is an increasing-path sum

N. Grosjean and T. Huillet, Some combinatorial aspects of discrete non-linear population dynamics, Chaos, Solitons & Fractals 93, 71–79 (2016), arXiv:1610.05919v1, §2.2, state the following conjecture (snlmatrixRR.tex, lines 449–452; the displayed formula is rendered inline):

Then, we conjecture that $W^{-1}(k,k^{\prime })=\sigma (k-1,k^{\prime })$, $k^{\prime }\geq k\geq 1$.

The paper defines $\phi(x)=\sum_{r\geq1}\alpha_rx^r$ and $P(k,k')=[x^{k'}]\phi(x)^k$ in Eq. (2) (lines 126–141). In the critical case $\alpha_1=1$, its normalized columns satisfy $P\mathbf w_1=\mathbf w_1$, $P\mathbf w_k=\mathbf w_{k-1}+\mathbf w_k$ for $k>1$, $\mathbf w_1=e_1$, $\mathbf w_k(1)=0$ for $k>1$, and $\mathbf w_k(r)=0$ for $r>k$ (lines 403–428). Its recursion divides by $P(r,r+1)=r\alpha_2$ (lines 420–436), so this construction requires $\alpha_2\ne0$.

For $m\geq1$, $\sigma(m,j)$ is the sum of $\prod_{\ell=1}^mP(i_\ell,i_{\ell+1})$ over $1=i_1<\cdots<i_{m+1}=j$; it is zero when $j\leq m$, and $\sigma(0,j)=\delta_{1,j}$ (lines 442–447). The definitions and quotation are also recorded in the literature note.

Issue #24105 fixes the finite reading: for every real coefficient sequence with $\alpha_0=0$, $\alpha_1=1$, $\alpha_2\ne0$, and every size $N\geq1$, a normalized upper triangular Jordan-chain matrix exists for the principal truncation, and every such matrix is invertible with the stated inverse entries for $1\leq k\leq k'\leq N$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CarlemanJordanChainInverse.result

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Quantum informationQuestion resolvedSource 2017Jain: multiplicity two for absolute-cosine Hadamard powersExactly double nonzero eigenvalues for the odd absolute cosine power matrix.Proved

The questionTanvi Jain, Hadamard powers of some positive matrices, Linear Algebra and its Applications 528 (2017), 147–158, states in the final unnumbered paragraph of section 3, after the proof of Theorem 3.3:

Why it mattersReflection pairs the nonconstant Fourier modes of an odd real symmetric circulant matrix.

The ideaA generalized Descartes bound for real exponential sums, the exact trigonometric coefficient variation counts, and the forced zeros at 0,2,...,2j-2 give

Our contributionExactly double nonzero eigenvalues for the odd absolute cosine power matrix.

Verification details
Release source recordProved in Lean

Open problem

Jain: multiplicity two for absolute-cosine Hadamard powers

Tanvi Jain, Hadamard powers of some positive matrices, Linear Algebra and its Applications 528 (2017), 147–158, states in the final unnumbered paragraph of section 3, after the proof of Theorem 3.3:

We conjecture that the multiplicity of every nonzero eigenvalue of |C|∘r∘ (except the top eigenvalue) equals two.

For every natural m at least one, put n=2m+1 and C_ab=cos((a-b)pi/n), with a,b from zero through n-1. The conjecture concerns the matrix A(r) whose entries are |C_ab|^r for every real r at least zero. Every nonzero characteristic-polynomial root other than the top Fourier value must have algebraic multiplicity exactly two. The formal definition Defs.claim retains all these quantifiers and both exceptions.

The sentence is attributed to the publisher's indexed full text. The journal PDF was inaccessible, so its page number is unverified. The first version of arXiv:1803.06803 omits this sentence; the journal article is the source of the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/JainAbsCosine/Result.result

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Discrete structuresQuestion resolvedPosed in 2019The Knight-Dialler RecurrenceThe knight-dialler counts satisfy the conjectured fourth-order recurrence, because the residual of that recurrence on the all-ones vector sits entirely on the one key a knight can never leave, and one further step annihilates it.Proved

The questionNumber of length-n phone numbers that can be dialed by a chess knight on a 0-9 keypad that starts on any number and takes n-1 steps

The ideaWrite u n for the vector whose entry at digit d counts the dialable sequences of n steps that start at d.

Our contributionThe knight-dialler counts satisfy the conjectured fourth-order recurrence, because the residual of that recurrence on the all-ones vector sits entirely on the one key a knight can never leave, and one further step annihilates it.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Knight-Dialler Recurrence

OEIS A327692 is defined by

Number of length-n phone numbers that can be dialed by a chess knight on a 0-9 keypad that starts on any number and takes n-1 steps

and carries, at revision #41 of Apr 22 2024, a formula line still labelled a conjecture:

Conjectures from Colin Barker, Oct 01 2019: (Start) G.f.: 2*x*(5 + 10*x - 7*x^2 - 8*x^3 + 2*x^4) / (1 - 6*x^2 + 4*x^4). a(n) = 6*a(n-2) - 4*a(n-4) for n>6. (End)

A later comment by Francesca Arici, dated Apr 17 2024, adds that the recurrence also holds at n = 6 and sketches a route through the eigenvalues of the adjacency matrix, ending at "reduces to checking an algebraic condition on the nonzero eigenvalues". The condition is not checked there and the formula line was not reclassified.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/KnightDiallerRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedPosed in 2012Mathar's recurrence for the zero-quantum transition pairs A108958The number a(n) of unordered pairs of distinct binary words of length n with the same number of 1's (the zero-quantum transitions of n spins 1/2) satisfies n(n - 2)a(n) + 2(-3n^2 + 7n - 3)a(n - 1) + 4(n - 1)(2n - 3)a(n - 2) = 0 for every n at least 2, as conjectured by R.Proved

The questionConjecture: n(n-2)a(n) +2(-3n^2+7n-3)a(n-1) +4(n-1)(2n-3) a(n-2)=0. - R.

The idea1. 2 a(n) = Σk C(n,k)^2 − Σk C(n,k) = C(2n, n) − 2^n.

Our contributionThe number a(n) of unordered pairs of distinct binary words of length n with the same number of 1's (the zero-quantum transitions of n spins 1/2) satisfies n(n - 2)a(n) + 2(-3n^2 + 7n - 3)a(n - 1) + 4(n - 1)(2n - 3)a(n - 2) = 0 for every n at least 2, as conjectured by R.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's recurrence for the zero-quantum transition pairs A108958

OEIS A108958 counts the unordered pairs of distinct binary words of length n with the same number of 1's, a(n) = Σ_{k=0}^{n} C(C(n,k), 2); for n spins 1/2 these are the nontrivial zero-quantum transitions. Its formula field records:

Conjecture: n*(n-2)*a(n) +2*(-3*n^2+7*n-3)*a(n-1) +4*(n-1)*(2*n-3) *a(n-2)=0. - R. J. Mathar, Apr 04 2012

Issue #10598 fixes the reading: a is defined by the entry's first formula line (also at n = 0, value 0), and the recurrence is asserted for every n ≥ 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ZeroQuantumPairsRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresRecurrence provedPosed in 2012Mathar's recurrence for OEIS A113409The binomial sum defining OEIS A113409 satisfies Mathar's five-term recurrence at every index at least four.Proved

The questiona(n) = Sum{k=0..floor(n/2)} C(n-k, k)C(k, floor(k/2)).

Why it mattersThe recurrence gives an unbounded exact relation among five consecutive values of the sequence defined by the source sum.

The ideaWrite b(k) = C(k, floor(k/2)) and extend the triangular transform to Tb(n) = sum{k=0..n} C(n-k,k)b(k); the added terms vanish.

Our contributionThe binomial sum defining OEIS A113409 satisfies Mathar's five-term recurrence at every index at least four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's recurrence for OEIS A113409

OEIS A113409 is named "A transform of the central binomial coefficients A001405." Its FORMULA field defines the sequence literally:

a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*C(k, floor(k/2)).

The conjecture in the same field is:

Conjecture: (n+2)*a(n)-2*(n+1)*a(n-1) +(n-4)*a(n-2) +2*a(n-3) +4*(2-n)*a(n-4)=0. - R. J. Mathar, Nov 07 2012

Preregistration issue #13334 fixes the nonnegative-index domain: every natural n >= 4. The binomial sum and its indices are natural numbers; the displayed recurrence is evaluated in the integers. Its statement is claim, and D5/S1/Recurrence/MatharCentralBinomialTransformRecurrence.result proves it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MatharCentralBinomialTransformRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The binomial sum defining OEIS A113409 satisfies Mathar's five-term recurrence at every index at least four.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2012.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2020The Miró-Roig--Tran Strict-Sign ConjectureThe literal Miró-Roig--Tran alternating integer coefficient is strictly negative for every natural n at least two.Proved

The questionMiró-Roig and Tran, immediately after the proof of Proposition 3.12 in arXiv 2001.06143v1, conjecture that for every integer n>=2,

Why it mattersThe registration fixes the source, quantifiers, and strict inequality; it does not establish worldwide novelty or priority.

The ideaFor natural m,q and real x, define the single normalized finite positive-part family

Our contributionThe literal Miró-Roig--Tran alternating integer coefficient is strictly negative for every natural n at least two.

Verification details
Release source recordProved in Lean

Open problem

The Miró-Roig--Tran Strict-Sign Conjecture

Miró-Roig and Tran, immediately after the proof of Proposition 3.12 in arXiv 2001.06143v1, conjecture that for every integer n>=2,

sum_{k=0}^n (-1)^k binom(2n+2,k)
  * (2n^2 - 1 - (2n-1)k)^(2n-1) < 0.

The sum, affine subtraction, and power in the formal owner are literal integer operations. Proposition 3.12(c) has a separate n>=4 premise for a preceding weak-Lefschetz implication; the conjectural display following the proof explicitly returns to every n>=2. The paper's finite computation through n=400, empirical monotonicity suggestion, and WLP consequence are not the formal target.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Splines/MiroRoigTranStrictSign.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryRecurrence provedPosed in 2017Yanev's position identity for binary reversalBinary reversal is determined by the lexicographic position recurrence after removing the leading bit.Proved

The question0,1,1,3,1,5,3,7,1,9,5,13,3,11,7,15,1,17,9,25,5,21,13,29,3,19,11,27,7,

Why it mattersThe conjecture connects two independent views of a binary word: reversal of its digits and its lexicographic rank when the least significant digit is read first.

The ideaDefine w by Heinz's even and odd recurrence and rev by Stephan's even and odd recurrence, both through Nat.binaryRec.

Our contributionBinary reversal is determined by the lexicographic position recurrence after removing the leading bit.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Yanev's position identity for binary reversal

OEIS A030101, OFFSET (%O, verbatim):

0,4

OEIS A030101, initial sequence line (%S, verbatim):

0,1,1,3,1,5,3,7,1,9,5,13,3,11,7,15,1,17,9,25,5,21,13,29,3,19,11,27,7,

OEIS A030101, NAME (%N, verbatim):

a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

Ralf Stephan's A030101 recurrence (%F, verbatim):

a(n) = 0, a(2n) = a(n), a(2n+1) = a(n) + 2^(floor(log_2(n)) + 1). For n > 0, a(n) = 2*A030109(n) - 1. - Ralf Stephan, Sep 15 2003

Velin Yanev's A030101 conjecture (%F, verbatim):

Conjecture: a(n) = 2*w(n) - 2*w(A053645(n)) - 1 for n > 0, where w = A264596. - Velin Yanev, Sep 12 2017

OEIS A030101, AUTHOR (%A, verbatim):

David W. Wilson

OEIS A264596, NAME (%N, verbatim):

Let S_n be the list of the first n nonnegative numbers written in binary, with least significant bits on the left, and sorted into lexicographic order; a(n) = position of n in S_n, starting indexing at 0.

Heinz's A264596 recurrence (%F, verbatim):

a(2n) = a(n), a(2n+1) = a(n) + n+1, a(0) = 0. - Alois P. Heinz, Nov 19 2015

OEIS A053645, NAME (%N, verbatim):

Distance to largest power of 2 less than or equal to n; write n in binary, change the first digit to zero, and convert back to decimal.

OEIS A053645, defining formula (%F, verbatim):

a(n) = n - 2^A000523(n).

The formal definitions are:

def w : ℕ → ℕ :=
  Nat.binaryRec 0 (fun bit n previous => if bit then previous + n + 1 else previous)

def stripTop (n : ℕ) : ℕ := n - 2 ^ Nat.log 2 n

def rev : ℕ → ℕ :=
  Nat.binaryRec 0 fun bit n previous =>
    if bit then if n = 0 then 1 else previous + 2 ^ (Nat.log 2 n + 1) else previous

The proved statement is:

theorem result (n : ℕ) (hn : 1 ≤ n) :
    rev n + 2 * w (stripTop n) + 1 = 2 * w n

The source writes the result with subtraction on the right. Lean natural subtraction truncates, so the theorem moves the two subtracted quantities to the additive side. It proves the source equality together with the non-negativity needed for the source subtraction. The printed recurrence begins a(n) = 0, rather than a(0) = 0; its initial sequence value and the rest of the recurrence identify the intended zero boundary used by rev.

Not claimed: A264596 contains a separate Yanev conjecture dated Sep 12 2017. That conjecture is outside this dossier and is not asserted or resolved here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/YanevBinaryReversalPositionIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Binary reversal is determined by the lexicographic position recurrence after removing the leading bit.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2017.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedPosed in 2017Yanev's maximum prime exponent recurrenceThe maximum prime exponent drops by one when a positive integer greater than one is divided by its radical.Proved

The questionMaximum exponent in the prime factorization of n.

Why it mattersYanev's conjecture characterizes maximum prime multiplicity by repeatedly removing one copy of every prime divisor.

The ideaThe empty prime-factor set at one gives the base value.

Our contributionThe maximum prime exponent drops by one when a positive integer greater than one is divided by its radical.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Yanev's maximum prime exponent recurrence

OEIS A051903, NAME (%N, verbatim):

Maximum exponent in the prime factorization of n.

Velin Yanev's FORMULA conjecture (%F, verbatim):

Conjecture: a(n) = a(A003557(n)) + 1. This relation together with a(1) = 0 defines the sequence. - Velin Yanev, Sep 02 2017

AUTHOR (%A, verbatim):

Labos Elemer, Dec 16 1999

With a n := n.primeFactors.sup n.factorization and the frozen D5/S1/Deficit/AlmostAdditivity.primeRadical, the literal theorem is

a 1 = 0 ∧ (∀ n : ℕ, 1 < n → a n = a (n / primeRadical n) + 1) ∧
  ∀ b : ℕ → ℕ, b 1 = 0 → (∀ n : ℕ, 1 < n → b n = b (n / primeRadical n) + 1) →
    ∀ n : ℕ, 0 < n → b n = a n

Here primeRadical is A007947, and A003557(n) = n / primeRadical(n) on the positive sequence domain. The quotient uses natural-number division. The empty supremum gives a 0 = 0, an out-of-range extension; no recurrence at zero or one is claimed.

The third conjunct formalizes the %F line's second sentence: the relation together with a(1) = 0 determines the sequence on positive indices. Every natural-valued function b with the same base value and recurrence agrees with a at every positive index; its value at zero is unrestricted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/YanevMaxExponentRadicalRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedPosed in 2018Barker's and Hasler's conjectures for the rows of Rule 201The elementary cellular automaton with Wolfram rule 201, started from a single ON cell at the origin and updated at every cell of the integers, has row n read on the cells -n, ..., n equal to 2 4^n - (2 [n odd] + 5 [n > 0]) 2^(n-1) - 1 in base 2, as conjectured by M.Proved

The questiona(n) = 5a(n-1)-20a(n-3)+16a(n-4) for n>4. G.f.: (1-5x+21x^2+14x^3-40x^4) / ((1-x)(1-2x)(1+2x)(1-4x)).

Why it mattersRule 201 is the local update rule of the Floquet-PXP cellular automaton (Wilkinson, Klobas, Prosen and Garrahan 2020).

The ideaHence the window of row n ≥ 1 in base b is Σ{j < 2n+1} b^j − b^{n+1} − b^{n−1} − [n odd] b^n.

Our contributionThe elementary cellular automaton with Wolfram rule 201, started from a single ON cell at the origin and updated at every cell of the integers, has row n read on the cells -n, ..., n equal to 2 4^n - (2 [n odd] + 5 [n > 0]) 2^(n-1) - 1 in base 2, as conjectured by M.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barker's and Hasler's conjectures for the rows of Rule 201

OEIS A267681 and A267680 read row n of the Rule 201 elementary cellular automaton, started from a single ON cell, on the cells -n, …, n: in base 2 (A267681) and as a string of decimal digits (A267680). Their formula fields record

a(n) = 5*a(n-1)-20*a(n-3)+16*a(n-4) for n>4. G.f.: (1-5*x+21*x^2+14*x^3-40*x^4) / ((1-x)*(1-2*x)*(1+2*x)*(1-4*x)).

Conjecture: a(n) = 2*4^n - (n%2*2 + [n]*5)*2^(n-1) - 1, where [n] = 1 iff n > 0; n%2 = 1 iff n is odd. - M. F. Hasler, Jul 28 2018

for A267681 (the first two by Colin Barker, Jan 19 2016), and

a(n) = 101*a(n-1)-10100*a(n-3)+10000*a(n-4) for n>4. G.f.: (1-101*x+10101*x^2+89910*x^3-101000*x^4) / ((1-x)*(1-10*x)*(1+10*x)*(1-100*x)).

for A267680 (Colin Barker, Jan 19 2016 and Apr 20 2019). Issue #11102 fixes the reading: every cell of the integers is updated at every step, the recurrences are asserted for n > 4, and each generating function is stated as the product of the series with its denominator, which has constant term 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/Rule201Rows.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2016A277829 and A278770: squared-Catalan Hankel limitsBoth Kotesovec squared-Catalan Hankel conjectures have logarithmic rate two times log two.Proved

The questionThese are exactly Kotesovec's November 29, 2016 A277829 conjecture and November 28, 2016 A278770 conjecture.

Why it mattersThe two OEIS sequences differ only by the adjacent Hankel shift, while their published conjectures assert the same exact logarithmic rate.

The ideaThe supporting module defines the scaled beta Catalan moment law and proves the literal identity integral (4x)^m = Cm.

Our contributionBoth Kotesovec squared-Catalan Hankel conjectures have logarithmic rate two times log two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A277829 and A278770: squared-Catalan Hankel limits

Let

C_m = binomial(2*m,m)/(m+1),
D_r(n) = det[(C_(i+j+r))^2] over i,j in Fin n.

The empty determinant is one. Prove both conjuncts

lim_(n->infinity) log(D_1(n))/n^2 = 2*log(2),
lim_(n->infinity) log(D_2(n))/n^2 = 2*log(2).

These are exactly Kotesovec's November 29, 2016 A277829 conjecture and November 28, 2016 A278770 conjecture. Their one-based programs use C_(i+j-1)^2 and C_(i+j)^2, respectively, so the zero-based shifts are exactly one and two. The source assertions were preregistered in https://github.com/the-omega-institute/trureturing/issues/9530 before the proof probes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Constants/Moments/CatalanSquareHankelLimits.catalan_square_hankel_log_limits

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryConjecture provedPosed in 2020Barry's A331474 Hankel generating-function conjectureThe actual determinant sequence from the literal A331473 moments has Barry's conjectured rational generating function.Proved

The questionsn = Sum{k=0..n} (-1)^(n-k) binomial(2k+2,k).

Why it mattersThe displayed coefficients begin 1, 3, 2, -50, -43, 535, 487, -4983, -4654, 43174 and have alternating-sign blocks incompatible with a positive-real Gram argument.

The ideaDifferentiate the Catalan series to obtain the literal unsigned moments, then derive two period-two tails and their continuant error identity.

Our contributionThe actual determinant sequence from the literal A331473 moments has Barry's conjectured rational generating function.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barry's A331474 Hankel generating-function conjecture

OEIS A331473 defines, at offset zero,

s_n = Sum_{k=0..n} (-1)^(n-k) * binomial(2*k+2,k).

OEIS A331474 is its literal Hankel transform

H_n = det(s_(i+j))_(0<=i,j<=n)

and Paul Barry's January 17, 2020 FORMULA line conjectures the full identity

Sum_{n>=0} H_n*x^n = (1 + 3*x + 16*x^2 - 8*x^3 + 36*x^4 - 12*x^5 + x^6 - x^7) / (1 + 7*x^2 + x^4)^2.

The resolved statement is D5/S3/Constants/Moments/A331474HankelGeneratingFunction.a331474_hankel_generating_function. It is an equality for PowerSeries.mk H, so the left side is the actual determinant sequence, not a recurrence-defined surrogate. The denominator has constant coefficient 1 and the formal right side uses its unit inverse.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Constants/Moments/A331474HankelGeneratingFunction.a331474_hankel_generating_function

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedSource 2021Quasi-Injective Order SeparationEvery order has a function that is quasi-injective just below it and fails at it.Proved

Why it mattersThe orders form a decreasing chain of conditions, and the article's own examples do not separate two consecutive links: τ and the Jordan totients are quasi-injective of every order, and for σs only order two is established.

The ideaGive an odd number 2t + 1 with t ≥ 1 the level t mod (m − 1), and define

Our contributionEvery order has a function that is quasi-injective just below it and fails at it.

Verification details
Release source recordProved in Lean

Open problem

Quasi-Injective Order Separation

Pongsriiam, Journal of Integer Sequences 24 (2021), Article 21.10.1, Definition 1, reads: "We call a function f : N → C a quasi-injective function if for all a, b ∈ N, the condition f(an) = f(bn) for all n ∈ N implies a = b. In addition, if f : N → N and ℓ ∈ N, then we say that f is quasi-injective of order ℓ if f, f^(2), f^(3), …, f^(ℓ) are quasi-injective, that is, for any a, b, k ∈ N with 1 ≤ k ≤ ℓ, if f^(k)(an) = f^(k)(bn) for all n ∈ N, then a = b." Section 5 asks:

Question 17. For each m ≥ 2, is there a function f : N → N such that f is quasi-injective of order m − 1 but not of order m?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/QuasiInjectiveOrderSeparation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2021.

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Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedPosed in 2024Kurkov's A003319 double-array conjectureThe actual indecomposable-permutation count is the common exact left border of Kurkov's two arrays.Proved

The questionThe source audit for this resolution read the canonical OEIS A003319 entry and these immutable OEIS-data revisions:

Why it mattersThe conjecture identifies two independently specified triangular-array borders with the actual cardinality of a permutation class for every index, not merely with another recurrence-defined proxy.

The ideaFor each positive n, choose the least positive prefix stabilized by a permutation.

Our contributionThe actual indecomposable-permutation count is the common exact left border of Kurkov's two arrays.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's A003319 double-array conjecture

The source audit for this resolution read the canonical OEIS A003319 entry and these immutable OEIS-data revisions:

  • A003319 at commit dacce77176c40534c4309db03464e7f4c08d83dd;
  • A370380 at commit a32b213583dbdd91a31cc0975e203471b3afe0fb;
  • A370381 at commit ed2582877dea2190aa163abe737afe7d42179fae.

Kurkov's April 26, 2024 conjecture is quoted there verbatim as:

Conjecture: a(n) = A370380(n-2, 0) = A370381(n-2, 0) for n > 1 with a(0) = a(1) = 1. - Mikhail Kurkov, Apr 26 2024

The sequence a(n) counts permutations of {1,...,n} that preserve no proper nonempty initial interval. The formal subtype uses permutations of Fin n; translating labels by one gives the same condition. Its literal formal count is the finite cardinality c(n) = Fintype.card (IndecomposablePerm n), rather than a sequence defined by either recurrence.

For natural row and column indices, the two source arrays are formalized literally as

U(0,k) = 1
U(m+1,k) = (k+2) U(m,k+1) + Sum_{j=0..k} U(m,j)

V(0,k) = 1
V(m+1,k) = Sum_{j=0..k+1} binomial(k+2,j+1) V(m,j).

The exact frozen theorem is

c 0 = 1 ∧ c 1 = 1 ∧
  ∀ n : ℕ, 2 ≤ n → c n = U (n - 2) 0 ∧ c n = V (n - 2) 0
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/IndecomposablePermutationDoubleArray.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Historical context · 1/4 · 10% weight

    Dated conjecture statement: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2022Braun–Bruegge Conjecture 5.1Braun–Bruegge's symmetric edge-polytope facet-sum conjectureProved

The questionBruegge, Facets of Symmetric Edge Polytopes for Graphs with Few Edges, arXiv:2201.13303v4, Conjecture 5.1 (Conjecture 3.15 in v1), asks for two endpoint maxima and two monotone transfers of the facet sum from Proposition 4.4.

The ideaReflection combines the two signs of the distance, with multiplicity one at zero and two elsewhere.

Our contributionBraun–Bruegge's symmetric edge-polytope facet-sum conjecture

Verification details
Release source recordProved in Lean

arXiv

Braun–Bruegge Conjecture 5.1

B. Braun and K. Bruegge, Facets of Symmetric Edge Polytopes for Graphs with Few Edges, arXiv:2201.13303v4, Conjecture 5.1 (Conjecture 3.15 in v1), asks for two endpoint maxima and two monotone transfers of the facet sum from Proposition 4.4. The verbatim statement is recorded in #19695 and the cited Library note.

For sorted positive natural numbers $a\ge b\ge c$ of common parity, the literal source specialization is

$$ F(a,b,c)=\sum_{j=0}^{c}\binom cj\binom b{(b-c)/2+j}\binom a{(a-c)/2+j}. $$

The natural divisions are exact on this domain. Outside it, the Lean definition remains a total natural-valued expression; the conjecture asserts nothing there.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Proposition 4.4, three entries, ordered and of common parity | F, Admissible | The sum and the five domain conjuncts are literal binding conventions. | | $n=2k$, $k\ge2$, sum $n+1$, maximum at $(n-1,1,1)$ | result.1 | Every admissible triple with that sum is bounded by the endpoint. | | $n=2k-1$, $k\ge3$, sum $n+1$, maximum at $(n-3,2,2)$ | result.2.1 | Natural subtraction has the source value under the stated bound on $k$. | | Transfer from the third coordinate | result.2.2.1 | $c\ge3$ makes $(a+2,b,c-2)$ positive and sorted. | | Transfer from the second coordinate, preserving order | result.2.2.2 | The additional condition $c\le b-2$ is exactly preservation of the lower order inequality. |

The endpoint triples themselves satisfy the domain and sum conditions. There are no additional hypotheses on the four assertions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BinomialBases/BraunBrueggeFacetSumConjecture.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
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    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Number theoryQuestion resolvedFirst preprint 2023Arithmetic complexity of singleton partitionsThe arithmetic complexity of singleton partitions has a positive normalized lower limit.Proved

The questionAlexander Bors, Daniel Panario and Qiang Wang, Functional graphs of generalized cyclotomic mappings of finite fields, arXiv:2304.00181v1, Question 6.3.1 (arXiv page 189; published memoir page 224):

Why it mattersQuestion 6.3.1 asks for a matching positive uniform lower order.

Our contributionThe arithmetic complexity of singleton partitions has a positive normalized lower limit.

Verification details
Release source recordProved in Lean

Open problem

Arithmetic complexity of singleton partitions

Alexander Bors, Daniel Panario and Qiang Wang, Functional graphs of generalized cyclotomic mappings of finite fields, arXiv:2304.00181v1, Question 6.3.1 (arXiv page 189; published memoir page 224):

Is it true that lim inf_{m→∞} AC(T_m)/(log² m/log log m) > 0?

Here $T_m$ is the partition of $\mathbb Z/m\mathbb Z$ into singletons and $\mathrm{AC}(T_m)$ is the least number of congruences with positive moduli dividing $m$ whose truth values separate all residues. The source is documented in the Library note. The mathematical preregistration is #19381.

Statement echo

| Source clause | Formal counterpart | Correspondence | | --- | --- | --- | | $x\equiv b\pmod a$, $a\ge1$, $a\mid m$ | MCongruence and MCongruence.Holds | Positive natural modulus, integer residue, and Int.ModEq; positive natural and integer moduli describe the same tests. | | Standard representatives $0,\ldots,m-1$ | Natural representatives with bounds in SpansTrivial | Equality of all test truth values forces equality of representatives. | | $K$ congruences span the singleton partition | Fin K family in SpansTrivial | Each possible sign system is a truth vector; each nonempty block has one representative. | | Smallest admissible $K\in\mathbb N_0$ | acTrivial | The natural infimum of separating-family sizes; the set is nonempty for $m>0$. | | $\liminf_{m\to\infty}\mathrm{AC}(T_m)/(\log^2m/\log\log m)>0$ | claim, proved by result | Natural index at infinity, real natural logarithms and real division, with the ratio embedded in EReal. |

The result has no hypotheses. Values at $m=0,1$ use the total Lean operations; changing a finite initial segment does not change the lower limit at infinity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BorsPanarioWangArithmeticComplexity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2024Necessary and sufficient conditions for the vertical slice resultVertical slices admit criteria in terms of two consecutive traces, integer companion powers under a unit discriminant, and half the companion period when four is nonzero.Proved

The questionWrite Vn = Vn(p, q) for the companion Lucas sequence and piV(m) for its least period modulo m.

Why it mattersThe entry point is the paper's organising statistic and every criterion it proves is phrased through it.

The ideaThe paper's m = 41 numbers are the pairs Vj + V{piV/2 - j}, and V{-k} = Vk turns that into Vn + V{n + piV/2} = 0 for all n, which refers to no entry point.

Our contributionVertical slices admit criteria in terms of two consecutive traces, integer companion powers under a unit discriminant, and half the companion period when four is nonzero.

Verification details
Release source recordProved in Lean

arXiv

Necessary and sufficient conditions for the vertical slice result

Write V_n = V_n(p, q) for the companion Lucas sequence and pi_V(m) for its least period modulo m. Drawing one fundamental period around a circle, Fiebig, Mbirika and Spilker call it the vertical slice result when a line through the circle meets two terms that are additive inverses of one another. Their Remark 4.12 records that it holds in many cases where the entry point e_V(m) does not exist at all, and their Theorem 4.11 proves it for q = 1 and order statistic omega_V(m) = 4, where the axis passes through the two zeros. Question 5.5 of arXiv:2408.14632v2 asks:

Can we find necessary and sufficient conditions for the "vertical slice result" to hold for (V_n(p,q)) modulo m with q = 1 and omega_V != 4?

The sentence quoted from Remark 4.12 is looser than the property the paper actually tracks. Modulo 28 with p = 7 and q = 1 the period is 2, 7, 19, 14, 23, 7, 26, 7, 23, 14, 19, 7, and the chord joining positions 0 and 6 does join two additive inverses, 2 + 26 = 28; the paper nonetheless lists m = 28 as a case where the result fails. So the property is that every pair reflected in one axis cancels, not that some chord joins a cancelling pair.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LucasVerticalSlice.verticalSlice_iff_two_traces

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Historical context · 1/4 · 10% weight

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    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2023The most classical orthonormal basis of spin 3/2Every orthonormal basis of spin 3/2 has mean one-qubit anticoherence at least 1/9, attained by an explicit classical basis.Proved

The questionBurchardt and K. Życzkowski, Orthonormal bases of extreme quantumness, Quantum 8, 1234 (2024), arXiv:2306.00532v3, Conjecture 1:

Why it mattersA complete orthogonal measurement on a spin cannot consist of coherent states. ℬ₁ measures how far the best such measurement must stay from classical, and the source found its minimum for N = 4 only numerically, by imposing the symmetries of a random-walk optimum.

The idea2. Determinant one. Rescaling one column by a phase leaves every pi unchanged, so U ∈ SU(4).

Our contributionEvery orthonormal basis of spin 3/2 has mean one-qubit anticoherence at least 1/9, attained by an explicit classical basis.

Verification details
Release source recordProved in Lean

Open problem

The most classical orthonormal basis of spin 3/2

M. Rudziński, A. Burchardt and K. Życzkowski, Orthonormal bases of extreme quantumness, Quantum 8, 1234 (2024), arXiv:2306.00532v3, Conjecture 1:

The matrix $U_4^{c}$ given in (21), for which $\mathcal{B}_1=\frac{1}{9}$, leads to the most spin-coherent (classical) basis in $\mathcal{H}_4$.

Here $\mathcal{B}_1(U)$ is the mean, over the four basis vectors given by the columns of $U$, of the one-qubit anticoherence $\mathcal{A}_1(\psi) = 2[1 - \operatorname{tr}\rho_1(F\psi)^2]$, where $F$ sends a spin-3/2 state to the symmetric state of three qubits and $\rho_1$ is its one-qubit reduced state (Eqs. (8) and (12)–(15)). The verbatim statements are in the literature note.

Issue #18510 fixes the reading: 1/9 ≤ ℬ₁(U) for every U ∈ U(4), and the printed $U_4^c$ is unitary with ℬ₁ = 1/9. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/SpinThreeHalvesClassicalBasis.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

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Quantum informationCounterexample foundFirst preprint 2020Pure programs need more dimension than mixed covariant programsAn irreducible representation of the dihedral group of order eight admits a mixed qubit program, while pure qubit programs fail.Refuted

The questionGschwendtner, Bluhm and Winter, Programmability of covariant quantum channels, Quantum 5, 488, DOI 10.22331/q-2021-06-29-488, arXiv:2012.00717v2, Section 5, ask:

Why it mattersThe source's abelian-commutant construction already attains the mixed lower bound with pure programs.

The ideaThe complete Kraus family turns the output for a fixed basis input and pure program into the Gram matrix of a linear program-to-environment matrix map.

Our contributionAn irreducible representation of the dihedral group of order eight admits a mixed qubit program, while pure qubit programs fail.

Verification details
Release source recordRefuted in Lean

Open problem

Pure programs need more dimension than mixed covariant programs

Gschwendtner, Bluhm and Winter, Programmability of covariant quantum channels, Quantum 5, 488, DOI 10.22331/q-2021-06-29-488, arXiv:2012.00717v2, Section 5, ask:

We leave it as an open question whether the upper bounds can be improved to match the lower ones also for pure program states.

The setting is a compact group, an irreducible unitary input representation $U$, and an arbitrary unitary output representation $V$. A single exact processor must implement every $UV$-covariant channel. The mixed optimum is $d_c=\sum_k n_k$, where the commutant of $\bar U\otimes V$ is $\bigoplus_k\mathbb1_{b_k}\otimes\mathcal B(\mathbb C^{n_k})$. The affirmative assertion is that pure programs attain this same dimension. Issue #15518 fixes the source question and its quantifiers.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Compact group $G$ | claim quantifies over a group with Fintype | A finite group is compact in its discrete topology; this restriction suffices for a counterexample. | | Unitary $U,V$ | MonoidHom into Matrix.unitaryGroup | Every group element acts by its coerced complex matrix. | | Irreducible input | Frozen NiceErrorBasisNonNormalStabilizer.IrredProj | Its Euclidean-space invariant subspaces are transported by WithLp.linearEquiv to coordinate functions. | | Completely positive, trace-preserving channel | Supporting IsChannel | A finite complete Kraus family and the frozen MatrixMap.of_kraus action. | | $UV$ covariance | IsCovariant | The literal two-sided conjugation equality, for every group element and input matrix. | | One fixed processor for every channel | MixedProgrammable, PureProgrammable | The processor is existentially quantified before the channel. | | Mixed density program | Frozen GHZMeasureBiseparableBound.IsDensity | The literal positive-semidefinite, trace-one predicate on Matrix; DensityState uses a different CStarMatrix subtype carrier. | | Pure density program | Frozen PureStateHandshake.rankOneDensity and normalized vector | Its body is vecMulVec with the starred vector; normalization is the conjugate dot product equal to one. | | Tensor input | Mathlib Matrix.kronecker | The product-index matrix has the source tensor expression. | | Match the mixed lower bound | claim | Every mixed program dimension also admits pure programs. The source's Theorems 25 and 31 identify the mixed optimum with $d_c$. | | Negative answer | Settling result | A closed negation of claim, with $G=\mathrm{DihedralGroup}\ 4$, $d_1=d_2=d_P=2$. |

The arbitrary finite input index types in IsChannel include the processor's product indices. Equality on all matrices expresses exact channel equality. These conventions add no restriction that could exclude the source's counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/CovariantPureProgramDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresCounterexample foundSource 2026The full zero-sum prefix-count characterizationAn alternating positive integer sequence has C-finite zero-sum counts for every finite integer digit set while its expanding characteristic roots have a ratio which is not a root of unity.Refuted

The questionThe target is the entire displayed iff of MathOverflow 516232 revision 2.

Why it mattersCayley–Hamilton makes observations of powers of one finite matrix satisfy a fixed recurrence.

The ideaBounded carries give finite integer transfer matrices.

Our contributionAn alternating positive integer sequence has C-finite zero-sum counts for every finite integer digit set while its expanding characteristic roots have a ratio which is not a root of unity.

Scope & next questionA replacement classification, extensions to other alternating sequences, and consequences for any other source conjectures remain open; no such extension is asserted here.

Verification details
Release source recordRefuted in Lean

Open problem

The full zero-sum prefix-count characterization

For every positive integer sequence satisfying a constant integer recurrence, let $P$ be the characteristic polynomial of its minimal recurrence. For every finite $D\subset\mathbb Z$, count the ordered original-coordinate assignments $S_n(D)=#{d\in D^n:\sum d_i a_i=0}$. The assertion is that all these count sequences are C-finite if and only if $P$ has no unit-circle roots, all its expanding roots are simple, and all ratios of two expanding roots are roots of unity. The target is the entire displayed iff of MathOverflow 516232 revision 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MoZeroSumCountRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2026.

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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresRecurrence provedKagey's order-eight recurrence for binary null polynomials modulo threeBinary integer polynomials of each positive degree that vanish modulo three at every integer satisfy Kagey's order-eight counting recurrence.Proved

The questionPeter Kagey's OEIS A329479, revision 25, final FORMULA line, states:

Why it mattersPreregistration 21714 identifies the final conjectured recurrence as the named external problem.

The ideaEvaluation at zero forces the constant coefficient to be zero.

Our contributionBinary integer polynomials of each positive degree that vanish modulo three at every integer satisfy Kagey's order-eight counting recurrence.

Scope & next questionThe two product formulas involving A024493 and A024495 are source context; independent formal settlements of those entries and minimality of order eight remain open here.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kagey's order-eight recurrence for binary null polynomials modulo three

Peter Kagey's OEIS A329479, revision 25, final FORMULA line, states:

Conjectured recurrence: a(n) = 2a(n-1) + 2a(n-2) - 5a(n-3) - 2a(n-4) + 10a(n-5) - 4a(n-6) - 4a(n-7) + 8a(n-8).

The source NAME is: “Number of degree n polynomials f with all nonzero coefficients equal to 1 such that f(k) is divisible by 3 for all integers k.” Thus $a(n)$ counts distinct integer polynomials of degree exactly $n$, with leading coefficient one, every coefficient in ${0,1}$, and $3\mid f(k)$ for every integer $k$. The source offset is one. The conjecture is interpreted for every natural $n\ge9$, using integer arithmetic for the counts and natural subtraction for the indices.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Degree $n$ integer polynomials | a: subtype of Polynomial Int with natDegree = n | Exact degree; distinct polynomials remain distinct | | All nonzero coefficients equal one | a: every coefficient is zero or one, with Monic | For the positive degrees used here, monicness expresses the forced leading one | | Divisible by three for all integers | a: divisibility of every integer evaluation | Full universal condition | | Eight-term recurrence | claim; result proves it | All eight coefficients and signs retained | | Offset one | Guard $9\le n$ | Every preceding index is at least one |

The source COMMENT describes strings of length $n$, while the NAME and six degree-seven examples use polynomial degree $n$. The formal count follows the NAME and examples, so the coefficient string has length $n+1$. No recurrence-defined sequence or quotient by polynomial functions replaces this count.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BinaryNullPolynomials/KageyA329479Recurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    Binary integer polynomials of each positive degree that vanish modulo three at every integer satisfy Kagey's order-eight counting recurrence.

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Number theoryQuestion resolvedPalindrome-square reversal infinitude in A133901Brockhaus and Seidov's sequence of palindromes with reversed-square witnesses is infinite.Proved

The questionNumbers in A128921 whose square is not a palindrome.

Why it mattersBrockhaus and Seidov recorded this unboundedness conjecture in 2007.

The ideaGiven a bound B, put j = B + 5, t = 10^j, and

Our contributionBrockhaus and Seidov's sequence of palindromes with reversed-square witnesses is infinite.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Palindrome-square reversal infinitude in A133901

OEIS A133901, NAME (verbatim):

Numbers in A128921 whose square is not a palindrome.

OEIS A128921, NAME (verbatim):

Palindromes m such that reverse of m^2 is also a square.

OEIS A133901, COMMENT (verbatim):

Conjecture: Sequence is infinite.

Thus membership means a decimal palindrome p whose square is not a palindrome and whose reversed square is a perfect square. In the formal module, rev10 is the numeric value of the reversed decimal string; it is distinct from the digit-list reversal D in D5/S3/Arith/SumInConcatenation.lean.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/PalindromeSquareReversalInfinitude.brockhaus_seidov_a133901

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Conceptual contribution · 2/4 · 20% weight

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    The delivered argument documents a mathematical mechanism.

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    The exact recorded statement is proved.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025Ta Problem 11.6: three invariants do not determine the countTwo finite nonabelian groups have equal order, center order and class number, but different complete counts of good involutions.Refuted

The questionLực Ta, Good involutions of conjugation subquandles, arXiv:2505.08090v5, Section 11, Problem 11.6, asks:

Why it mattersThe requested determination would force any two finite nonabelian groups with identical (order, center order, class number) to have identical good-map counts.

The ideaBoth are nonabelian groups with invariant triple (64,4,22).

Our contributionTwo finite nonabelian groups have equal order, center order and class number, but different complete counts of good involutions.

Scope & next questionNo claim about knot classification, all finite groups' counts, or counts modulo isomorphism is substituted for the original.

Verification details
Release source recordRefuted in Lean

Open problem

Ta Problem 11.6: three invariants do not determine the count

Lực Ta, Good involutions of conjugation subquandles, arXiv:2505.08090v5, Section 11, Problem 11.6, asks:

Let G be a finite nonabelian group of order n. Is the number of good involutions of Conj G completely determined by n, the order of the center Z(G), and the class number k(G) of G? If so, is there an exact formula to compute this number?

Here s_x(y)=xyx^-1. A good involution is a function rho with rho squared equal to the identity, commuting with every s_x, and satisfying s_(rho(x))=s_x^-1. The count is of all such functions on the whole group, without quotienting by quandle isomorphism. This is separate from 11.7's question about a bound for conjugation subquandles.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/Quandles/TaThreeInvariantRefutation.result

Frozen module record

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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2024The MUB and SIC correlation trace norms agree up to the factor KSIC and complete-MUB correlation matrices have proportional rectangular trace norms.Proved

The questionMorelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys.

Why it mattersResults 1 and 2 of the source give Schmidt-number criteria ‖P‖tr ≤ (1 + r)/K and ‖Q‖tr ≤ 1 + r.

The idea1. SIC frame. The d² SIC projectors are linearly independent, from their Gram matrix (d δij + 1)/(d + 1), so they form a basis of the operator space.

Our contributionSIC and complete-MUB correlation matrices have proportional rectangular trace norms.

Verification details
Release source recordProved in Lean

Open problem

The MUB and SIC correlation trace norms agree up to the factor K

A. Tavakoli and S. Morelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys. Rev. A 110, 062417 (2024), Conjecture 1:

Provided that both a SIC and a complete set of MUBs exist in dimension $d_A$ and $d_B$, the criteria in Result 1 and 2 are equivalent. It then holds that $\left|\mathbf{Q}\right|\text{tr}=K\left|\mathbf{P}\right |\text{tr}$.

The SIC and MUB definitions, the correlation matrices P and Q and the constant K = √(d_A(d_A+1)) √(d_B(d_B+1)) are quoted in the literature note.

Issue #19503 fixes the reading. For all d_A, d_B ≥ 2, every density matrix ρ, every pair of local SICs and every pair of complete local MUB families, ‖Q‖_tr = K ‖P‖_tr. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/SicMubCorrelationTraceNormEquality.result

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Quantum informationConjecture provedFirst preprint 2024Conjecture 2 of Tavakoli–Morelli: the Schmidt-number bound for equiangular measurementsEquiangular measurements obey the Schmidt-number correlation trace-norm bound of Tavakoli and Morelli's Conjecture 2.Proved

The questionMorelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys.

Why it mattersFor r = 1, (52) is the separability bound, and for larger r it detects states of Schmidt number greater than r with n < d² outcomes per side.

The idea1. Pure states as a Gram product. For a unit vector ψ with coefficient matrix Ψ = U D Vᴴ (singular value decomposition, D = diag(λ)), put S = D^{1/2}.

Our contributionEquiangular measurements obey the Schmidt-number correlation trace-norm bound of Tavakoli and Morelli's Conjecture 2.

Verification details
Release source recordProved in Lean

Open problem

Conjecture 2 of Tavakoli–Morelli: the Schmidt-number bound for equiangular measurements

A. Tavakoli and S. Morelli, Enhanced Schmidt number criteria based on correlation trace norms, arXiv:2402.09972v2, Phys. Rev. A 110, 062417 (2024), bound the trace norm of the table of joint outcome probabilities of local measurements for states of Schmidt number at most r.

  • Equiangular measurement (EAM). n > d unit vectors ψ_a in C^d with Σ_a |ψ_a⟩⟨ψ_a| = (n/d)𝟙 (Eq. (49)) and |⟨ψ_a|ψ_a′⟩|² = (n − d)/(d(n − 1)) for a ≠ a′ (Eq. (50)); its effects are E_a = (d/n)|ψ_a⟩⟨ψ_a|.
  • The table. For Alice's EAM {E^A_a} and Bob's {E^B_b}, (P_n)_ab = tr[ρ (E^A_a ⊗ E^B_b)].
  • Conjecture 2 (§V A, p. 7): "For any state of Schmidt number r it holds that ‖P_n‖_tr ≤ [d(d − 1) + r(n − d)]/[n(n − 1)]. (52)" The introduction leaves deciding the conjectures as an open problem.

The verbatim statements are in the literature note. Issue #17745 fixes the reading:

  • for all d ≥ 1, 1 ≤ r ≤ d, n > d, any two n-element EAMs on C^d chosen independently, and every state of Schmidt number at most r, the bound (52) holds;
  • a state of Schmidt number at most r is a finite convex combination of pure states whose coefficient matrices have rank at most r.

This conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TavakoliMorelliIncompleteEAM.result

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Discrete structuresQuestion resolvedFirst preprint 2024All-odd randomized subtraction games do not convergeAll-odd finite subtraction games have nonconvergent winning probabilities.Proved

The questionNicolás Capitelli and Francisco Somma, “On sequences arising from randomizing subtraction games”, arXiv:2405.19593v1, Section 3, p. 7, state:

Why it mattersThe source proves convergence when the gcd-one subtraction set contains an even element.

The ideaEvery term in the finite window tends to zero, contradicting its positive constant value.

Our contributionAll-odd finite subtraction games have nonconvergent winning probabilities.

Verification details
Release source recordProved in Lean

Open problem

All-odd randomized subtraction games do not converge

Nicolás Capitelli and Francisco Somma, “On sequences arising from randomizing subtraction games”, arXiv:2405.19593v1, Section 3, p. 7, state:

However, we do not show that an all-odd subtraction set necesarily do not converge (although we conjeture it) nor we provide the exact limits of the even and odd subsequences for the (conjectured) non-convergent case.

The section assumes a finite nonempty subtraction set of positive integers with gcd one. A player selects uniformly among legal subtractions, and loses when none is available. Writing $A_n={s\in S:0<s\le n}$, the next player's winning probability is $a_n=0$ if $A_n$ is empty and otherwise $a_n=1-|A_n|^{-1}\sum_{s\in A_n}a_{n-s}$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite subtraction set | S : Finset ℕ in claim | Finite sets, without an ordering or multiplicities. | | Nonempty set | S.Nonempty | Preserved. | | All subtractions odd and positive | ∀ s ∈ S, Odd s; positivity follows for natural numbers | Preserved. | | Section 3 assumes gcd one | S.gcd id = 1 | Preserved in the public claim. | | Uniform legal subtraction; no move loses | legalMoves, winProb | Literal preregistered recursion, including real division and truncated natural subtraction. | | Does not converge | Absence of a real $L$ with Tendsto (winProb S) atTop (𝓝 L) | Preserved. |

The public result proves claim. No source clause is dropped. The private theorem nonconvergence proves the stronger statement without the gcd assumption.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Games/CapitelliSommaAllOddNonconvergence.result

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Analysis & dynamicsCounterexample foundFirst preprint 2022Abdesselam's Problem 3: the PGG inequalities at arbitrary positive exponentsThe padded general Ginibre inequalities for stable determinantal polynomials fail at exponent 1/4.Refuted

The questionAbdelmalek Abdesselam, “Non-Abelian correlation inequalities and stable determinantal polynomials”, arXiv:2207.07603v2, Problem 3, printed p. 14:

Why it mattersTheorem 2.3 proves the PGG inequalities for positive half-integer exponents.

The ideaUse η = 1/4, n = 3, q = 2, L = 0 and the zero parity map.

Our contributionThe padded general Ginibre inequalities for stable determinantal polynomials fail at exponent 1/4.

Verification details
Release source recordRefuted in Lean

Open problem

Abdesselam's Problem 3: the PGG inequalities at arbitrary positive exponents

Abdelmalek Abdesselam, “Non-Abelian correlation inequalities and stable determinantal polynomials”, arXiv:2207.07603v2, Problem 3, printed p. 14:

In the light of investigations of spin models with non-integer number of components N, as in [8], it would be interesting to see if Thm. 2.3 still holds for P^{−η} where η is any positive real number instead of being restricted to half integers.

The question concerns all of Theorem 2.3 (p. 10). For real symmetric positive semidefinite q by q matrices A_j with positive definite sum, let P(x) = det(Σ_j x_j A_j). Given a parity homomorphism ρ from ℤⁿ to (ℤ/2ℤ)ᴸ, an even multi-index lies in its kernel. For every m ≥ 0, natural m by n matrix V with even 1_m V, signs ε_i in {−1,1}, and positive even natural padding u, the proposed extension requires

$$ \sum_{\substack{\alpha,\beta\in\mathbb N^m\\alpha+\beta=\mathbf 1_m}} \mathbf 1{\alpha V\text{ even}},\varepsilon^\beta P(u+\alpha V)^{-\eta}P(u+\beta V)^{-\eta}\geq 0 \quad\text{for every real }\eta>0. $$

The Lean claim retains these quantifiers and premises, including L = 0. Row-vector multiplication uses Mathlib Matrix.vecMul. The single public settling theorem result : ¬ claim answers this question negatively. Problem 3's subsequent question about characterizing distributions with bilinear positivity is a separate open boundary.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/PaddedGinibreNonHalfIntegerRefutation.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2024Question 6 of Alofi and Dukes: do the stochastically recurrent states of K_{m,n} dominate?In the stochastic sandpile model on the complete bipartite graph K{m,n}, m at least 2 and n at least 1, with the sink in the first part, the stochastically recurrent states number at most half of the n^(m-1) m^n stable configurations, and exactly half if and only if m = 2.Proved

The questionIn the stochastic sandpile model of Chan, Marckert and Selig, a stable configuration c on a graph with sink is stochastically recurrent when some orientation O satisfies inO(v) ≥ d(v) − c(v) at every non-sink vertex.

The ideaEvery orientation directs each of the mn edges into exactly one vertex, so the in-degrees sum to mn; summing inO(v) + c(v) ≥ d(v) over the non-sink vertices gives at least 2mn − n − mn = n(m − 1) grains.

Our contributionIn the stochastic sandpile model on the complete bipartite graph K{m,n}, m at least 2 and n at least 1, with the sink in the first part, the stochastically recurrent states number at most half of the n^(m-1) m^n stable configurations, and exactly half if and only if m = 2.

Verification details
Release source recordProved in Lean

Open problem

Question 6 of Alofi and Dukes: do the stochastically recurrent states of K_{m,n} dominate?

In the stochastic sandpile model of Chan, Marckert and Selig, a stable configuration c on a graph with sink is stochastically recurrent when some orientation O satisfies in_O(v) ≥ d(v) − c(v) at every non-sink vertex. For the complete bipartite graph K_{m,n} with the sink in the first part, Alofi and Dukes (arXiv:2411.02667, Discrete Math. 348 (2025) 114323) record n^{m−1} m^{n−1} ≤ |Sto(K_{m,n})| ≤ n^{m−1} m^n and ask:

Question 6. Can it be determined whether or not the number of stochastically recurrent states dominates the set of stable states? I.e. can it be decided |Sto(K_{m,n})| ≶ n^{m−1} m^n / 2 ?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Sandpiles/StochasticSandpileBipartiteHalf.result

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Discrete structuresRecurrence provedFirst preprint 2023Ferrers Boards with Prescribed Ordered Level NumbersThe Gaussian recurrence counts literal fixed-slot source boards, with the published top-down order restored.Proved

The questionKenneth Barrese, Weighted File Placements on Singleton Boards, arXiv:2312.10855v1, Section 2.2, asks:

Why it mattersThe joint conclusion uses BarreseOrderedLevels.source2023count for the fixed rectangle and BarreseSourceNormalization.lOperatorcount for the actual native all-width operator fiber.

The ideaLet G(N,k) be the finite Gaussian polynomial, defined by G(N,0)=1, G(0,k+1)=0 and G(N+1,k+1)=G(N,k+1)+q^(N-k)G(N,k).

Our contributionThe Gaussian recurrence counts literal fixed-slot source boards, with the published top-down order restored.

Scope & next questionIf C is a singleton board, then perhaps counting the number of Ferrers boards B with l(B) = C would be a good start to this problem, but this too remains open.

Verification details
Release source recordProved in Lean

Open problem

Ferrers Boards with Prescribed Ordered Level Numbers

Kenneth Barrese, Weighted File Placements on Singleton Boards, arXiv:2312.10855v1, Section 2.2, asks:

This suggests the open question, given an ordered set of level numbers: (l_1, l_2, . . . , l_n), how many unique Ferrers boards have those specific level numbers.

Section 5 repeats this question. For every positive m and n, count every weakly increasing n-tuple of column heights between zero and mn in the fixed n-column, mn-row rectangle, with a prescribed top-down vector l of n nonnegative level totals. A level has m rows; its total counts actual occupied cells. Leading zero columns and empty top levels are retained. The input board need not be singleton, and no permutation of l is allowed.

The same prescribed-level counting problem has an earlier all-width formulation in Kenneth Barrese, Nicholas Loehr, Jeffrey Remmel and Bruce E. Sagan, Bijections on m-level Rook Placements, arXiv:1508.05944v1, Section 6.2:

If C is a singleton board, then perhaps counting the number of Ferrers boards B with l(B) = C would be a good start to this problem, but this too remains open.

For every positive m and every nonempty positive singleton target C, count all normalized Ferrers boards B of arbitrary width with l(B)=C. Definition 4 reverses the nonempty bottom-up level totals. Here normalization uses positive weakly increasing column-height lists: Section 4.1 aligns lower-right cells and treats leading zero padding as another notation for the same board. Arbitrary horizontal translations as distinct unbounded absolute-grid subsets are a different carrier. The bibliographic header identifies the later 2023 question; the original 2015 formulation and its governing conventions are identified explicitly here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Partitions/BarreseOrderedLevels.source2023_count

D5/S3/Combinatorics/Partitions/BarreseSourceNormalization.lOperator_count

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    The Gaussian recurrence counts literal fixed-slot source boards, with the published top-down order restored.

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    First preprint year: 2023.

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Quantum informationCounterexample foundFirst preprint 2023Pretty good state transfer in double blow-ups of pathsThe double blow-up of P11 admits pretty good state transfer between the two copies of source vertex 4, refuting Conjecture 1 of Bhattacharjya, Monterde and Pal.Refuted

The questionBhattacharjya, Monterde and Pal, Quantum walks on blow-up graphs, arXiv:2308.13887v2, J.

The ideaThe settling module uses the imported TopEigenspace.sinemodeeigenvector and EnergyEigenstateStationarity.expmulVecofeigenvector to obtain the sine expansion of the path propagator.

Our contributionThe double blow-up of P11 admits pretty good state transfer between the two copies of source vertex 4, refuting Conjecture 1 of Bhattacharjya, Monterde and Pal.

Scope & next questionThe complete parameter classification requested in Section 12 remains open here. 4.

Verification details
Release source recordRefuted in Lean

Open problem

Pretty good state transfer in double blow-ups of paths

Bhattacharjya, Monterde and Pal, Quantum walks on blow-up graphs, arXiv:2308.13887v2, J. Phys. A 57 (2024) 335303, Section 8, Conjecture 1:

Let n = 2^t r − 1, where t ≥ 2 and r is an odd prime number. If u is a multiple of 2^{t−1}, then PGST does not occur between (0, u) and (1, u) in the double blow-up of P_n.

Section 12 asks whether this conjecture is true and, if not, for which parameters pretty good state transfer occurs. The source uses $U(s)=\exp(isA)$. PGST means the existence of a sequence of real times whose transition-entry moduli converge to one. Source vertex $u=j+1$ is represented by $j:\mathrm{Fin}(n)$; the blow-up carrier is $\mathrm{Fin}(2)\times\mathrm{Fin}(n)$ and adjacency ignores the copy coordinate. Issue #14649 preregisters BMP-1, the literal definitions, the tier-1 classification, the witness and the literature readings.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PathDoubleBlowUpPGSTRefutation.result

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Number theoryQuestion resolvedFirst preprint 2024Bonin's shadow Markoff positivity regionThe nonnegative integer shadow Markoff tree has exactly Bonin's quadrilateral of initial points.Proved

The questionNathan Bonin, Conditions of positivity on a shadow Markoff tree, J.

Why it mattersThe source's Theorem 1 gives some convex polygon and supplies numerical evidence for the proposed quadrilateral.

The ideaThe dual identity changes the rational mutation to a polynomial mutation and preserves integrality.

Our contributionThe nonnegative integer shadow Markoff tree has exactly Bonin's quadrilateral of initial points.

Verification details
Release source recordProved in Lean

Open problem

Bonin's shadow Markoff positivity region

Nathan Bonin, Conditions of positivity on a shadow Markoff tree, J. Integer Sequences 27 (2024), Article 24.8.3, arXiv:2403.07668v2, Conjecture 1, asks for the quadrilateral of projective initial shadows that give an integer positive shadow Markoff tree. The source and its literal quotations are recorded in the Library note and preregistration #17060.

The precise assertion uses the rooted tree of Sections 2 and 5: start at $(1,\alpha,1,\beta,1,\gamma)$, include the root, apply the mandatory right and left mutations, then every finite word in the two moves. For $(\alpha,\beta,\gamma)\in\mathbb N^3\setminus{(0,0,0)}$, every node has all six entries in $\mathbb N$ if and only if

$$\alpha+\beta\leq 2\gamma,\qquad 2\alpha\leq\beta+\gamma.$$

Equivalently, $\gamma>0$ and $(\alpha/\gamma,\beta/\gamma)$ belongs to the rational convex hull of $(0,0),(1/2,0),(1,1),(0,2)$. Natural numbers include zero, as in Equation (8) and the boundary vertices. The result does not quantify over independent root permutations.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | The two directions of Section 5 | Move | Left mutates the second pair; right the first | | Mutation formulas (5), rational division | step | Both real and shadow coordinates are literal source expressions | | Root and mandatory right, left moves, followed by every finite word | path, universal word and node quantifiers in claim | List.scanl includes the initial node | | All six coordinates in $\mathbb N$, Equation (8) | IsNatNode | Six natural witnesses cast to $\mathbb Q$ | | Nonzero initial natural shadow triple | The three natural binders and nonzero hypothesis in claim | Excludes the triple with no projective point | | Four projective vertices in Conjecture 1 | quadrilateral | Rational convex hull in the chart $\gamma=1$ | | Tree positivity exactly in that region | result : claim | Both implications, including the boundary | | Positive chart denominator | The $0<\gamma$ conjunct of claim | Follows from the inequalities and nonzero triple; no extra restriction on the region |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/BoninShadowMarkoffPositivity.result

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Discrete structuresQuestion resolvedFirst preprint 2023Complete homogeneous symmetric norms of complete multipartite graphsThe complete multipartite branch of the cited article's CHS norm problem.Proved

The questionChávez et al., Extremal polynomial norms of graphs, arXiv:2311.04689v3, Section 8, Problem 3: “Can one find the extrema of the CHS norms over other families of graphs such as regular, unicyclic and complete multipartite graphs?”

Why it mattersThe result addresses the complete multipartite family named in Section 8, Problem 3; it does not address regular or unicyclic graphs.

The ideaThe determinant identity for the complete multipartite adjacency pencil is

Our contributionThe complete multipartite branch of the cited article's CHS norm problem.

Scope & next questionThe regular and unicyclic branches remain open.

Verification details
Release source recordProved in Lean

Open problem

Complete homogeneous symmetric norms of complete multipartite graphs

Chávez et al., Extremal polynomial norms of graphs, arXiv:2311.04689v3, Section 8, Problem 3: “Can one find the extrema of the CHS norms over other families of graphs such as regular, unicyclic and complete multipartite graphs?”

For a complete $r$-partite graph with positive part sizes $a_1,\ldots,a_r$ and total order $n$, the cited article defines, for even $d=2m$, the CHS norm by

$$|G|{2m}=h{2m}(\lambda_1,\ldots,\lambda_n)^{1/(2m)},$$

where the $\lambda_i$ are the adjacency eigenvalues. This dossier settles the complete multipartite branch only, for each fixed number of parts. The regular and unicyclic branches remain open.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Complete multipartite graph | Vtx a and adj a | Literal vertices and adjacency matrix for the part sizes. | | Complete homogeneous symmetric polynomial | chs | Evaluation of MvPolynomial.hsymm at the real adjacency eigenvalues. | | Even-degree norm | chsNorm | Real power with exponent 1 / d, specialized to d = 2 * m. | | Minimum part multiset | minParts r n | One part of size n-r+1 and r-1 singleton parts. | | Maximum part multiset | balancedParts r n | Parts differ by at most one, with the prescribed quotient and remainder. | | Unique extrema | claim and Result.result | Both inequalities and both equality classifications up to permutation. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/MultipartiteCHS/Result.result

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Quantum informationQuestion resolvedFirst preprint 2024The Cobucci–Tavakoli ten-setting GHZ witness in odd dimensionsThe ten-setting GHZ witness bounds fidelity and genuine multipartite Schmidt dimension in every odd local dimension.Proved

The questionTavakoli, “Detecting the dimensionality of genuine multi-particle entanglement”, Science Advances 10, eadq4467, DOI 10.1126/sciadv.adq4467, arXiv:2402.06234v4, Supplementary Material V, printed page 22, states:

Why it mattersTen measurement settings certify genuine three-party Schmidt dimension.

The ideacharacterorthogonality, constrainedcharactersum, and witnessentry express the entries through additive characters and the two coordinate differences. countlethree bounds the multiplicity of a nonzero frequency fibre among the nine labelled pairs.

Our contributionThe ten-setting GHZ witness bounds fidelity and genuine multipartite Schmidt dimension in every odd local dimension.

Scope & next questionEven-dimensional analogues remain open here. - Computed — finite experiment.

Verification details
Release source recordProved in Lean

Open problem

The Cobucci–Tavakoli ten-setting GHZ witness in odd dimensions

G. Cobucci and A. Tavakoli, “Detecting the dimensionality of genuine multi-particle entanglement”, Science Advances 10, eadq4467, DOI 10.1126/sciadv.adq4467, arXiv:2402.06234v4, Supplementary Material V, printed page 22, states:

We have shown this explicitly for odd d ≤ 17 and conjecture it to hold for odd d in general.

For the nine labelled triples in (V1), the three-party witness operator $\mathcal W_d$ in (V2), and every odd natural local dimension $d$, the conjectured operator comparison is

$$ \mathcal W_d\preceq9|\mathrm{ghz}{3,d}\rangle \langle\mathrm{ghz}{3,d}|+3\mathbf1. \tag{V5} $$

Consequently, every density state $\rho$ satisfies

$$ F_{\mathrm{ghz}}(\rho)\geq\frac{W_d(\rho)-3}{9}, \qquad W_d(\rho)\leq3+\frac{9D_{\mathrm{GME}}(\rho)}{d}. \tag{V4,V3} $$

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | All odd local dimensions | claim, binder d and hypothesis Odd d | All odd natural dimensions, including composite dimensions and 1 | | Nine triples (V1) | tuples | A list retains all nine labelled multiplicities | | Witness (V2) | WindowRegister.windowRoot, basisVec, PureStateHandshake.rankOneDensity, witness, GHZMeasureBiseparableBound.expect | The source exponential phase, coordinate vectors, rank-one projectors and labelled witness sum | | GHZ vector, Eq. (5) and fidelity | ghz, RHLinalg.hermForm | The normalized three-party vector and real expectation | | Schmidt rank and Schmidt number | cutRank, SchmidtNumberLE | Literal coefficient matrices indexed by Fin d and Fin d × Fin d, and finite convex decompositions across each cut | | Definition 1, Eq. (2) | GMEDecomposable, gmeDim | Mixtures across all three cuts; the natural infimum is attained | | (V5) | witness_quadratic_bound, witness_density_bound | Quadratic-form and density-state comparisons | | (V4) and (V3) | result : claim | Both inequalities for every density state |

Except for the named phase, projector, real trace expectation and quadratic-form owners, names in the table belong to D5.S3.Quantum.Entanglement.GHZ.OddGhzTenSettingWitness. The state carrier is the canonical FiniteStateChannel.DensityState. The witness expressions follow the formulas fixed by preregistration #16991.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GHZ/OddGhzTenSettingWitness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedFirst preprint 2023Ferenczi–Zamboni Conjecture 1 on longest clustering wordsA longer perfectly clustering Arnoux–Rauzy factor refutes the Ferenczi–Zamboni conjecture.Refuted

The questionZamboni, Clustering and Arnoux–Rauzy words, arXiv:2305.18986v1, Advances in Applied Mathematics 153 (2024) 102621, DOI 10.1016/j.aam.2023.102621, state Conjecture 1:

The ideaThe displayed reverse-order blocks prove perfect clustering.

Our contributionA longer perfectly clustering Arnoux–Rauzy factor refutes the Ferenczi–Zamboni conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Ferenczi–Zamboni Conjecture 1 on longest clustering words

S. Ferenczi and L. Q. Zamboni, Clustering and Arnoux–Rauzy words, arXiv:2305.18986v1, Advances in Applied Mathematics 153 (2024) 102621, DOI 10.1016/j.aam.2023.102621, state Conjecture 1:

In a given AR language $\Lambda$ satisfying $(Rabc)$, the word in Proposition [14] is the longest clustering word, or, if this fails, the longest clustering word conjugate to a standard AR word.

The formal statement uses the source's three-letter Arnoux–Rauzy blocks, languages, standard words, conjugacy, Burrows–Wheeler transform and clustering definitions. It records the second alternative of the conjecture as claim.

Statement echo

The alphabet is $\mathrm{Fin},3$, with $a=0$, $b=1$, and $c=2$; words are lists. The three AR rules are the pointwise update that fixes the selected block and appends it to each other block. arBlocks uses Nat.rec and Function.update for that recurrence. InfOften is the universal occurrence condition expressed using Filter.Frequently at Filter.atTop, equivalent to $\forall x,N,\ \exists k\ge N,\ d(k)=x$. The boundary $\mu$ is Nat.findGreatest of the positions $j$ with $d(j-1)\ne d(\mu_y-1)$, bounded by $\mu_y$; the value is $0$ when no such positive position exists. Stage blocks are enumerated by List.ofFn and sorted by length. For BWT, the non-strict permutation-rank order is the negation of reverse List.lex; List.insertionSort sorts the rotations and List.getLast? extracts their last letters. The quantified conditions in claim are exactly the weaker standard-conjugacy alternative of Conjecture 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/ArnouxRauzyLongestClusteringRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedFirst preprint 2024O'Bryant's seven-point positive-real sum-product cellA seven-element set of positive reals with thirteen products has at least twenty-four sums.Proved

The questionKevin O'Bryant, Visualizing the Sum-Product Conjecture, arXiv:2411.08139v2, §5.2:

Why it mattersAt equality, positivity forces a geometric progression, so a specific corner of the sum-product spectrum becomes a problem about collisions between sums of powers of one real parameter.

The ideaFor B={b₀<⋯<bₘ₋₁} of positive reals, the chain

Our contributionA seven-element set of positive reals with thirteen products has at least twenty-four sums.

Verification details
Release source recordProved in Lean

Open problem

O'Bryant's seven-point positive-real sum-product cell

Kevin O'Bryant, Visualizing the Sum-Product Conjecture, arXiv:2411.08139v2, §5.2:

We believe that (23, 13) is not in SPP_{R+}(7).

For a domain X, the spectrum is

$$ \operatorname{SPP}_X(n) = {(|A+A|,|AA|): A\subseteq X,\ |A|=n}. $$

Here R+ denotes the strictly positive reals, and A+A and AA denote the pointwise sumset and product set. The exact assertion is that every finite set A of positive reals with seven elements satisfies

$$ \neg\bigl(|A+A|=23\ \land\ |AA|=13\bigr). $$

The Lean claim retains these quantifiers and hypotheses. Its theorem result : claim follows from the stronger statement

$$ |AA|=13\quad\Longrightarrow\quad |A+A|\geq24. $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SumProductSevenPointCell.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is proved.

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Number theoryQuestion resolvedFirst preprint 2023Question 3 on outgoing digit-sum arcsThe attainable positive digit sums of multiples form an additive semigroup whose gcd determines its eventual support and density.Proved

The questionPhunphayap, Noppakaew and Pongsriiam, Directed graphs of integers with arcs determined by an arithmetic function, arXiv:2303.13017v1, Question 3 (Question 3.3 in Utilitas Mathematica 124, DOI 10.61091/um124-06), ask:

Why it mattersThe source note is Library/Digit/phunphayap2023digitsumarcs.md.

The ideaConcatenating digit blocks gives sumconcat and additive closure outadd.

Our contributionThe attainable positive digit sums of multiples form an additive semigroup whose gcd determines its eventual support and density.

Scope & next questionIt introduces no claim about the optimal threshold.

Verification details
Release source recordProved in Lean

Open problem

Question 3 on outgoing digit-sum arcs

Phunphayap, Noppakaew and Pongsriiam, Directed graphs of integers with arcs determined by an arithmetic function, arXiv:2303.13017v1, Question 3 (Question 3.3 in Utilitas Mathematica 124, DOI 10.61091/um124-06), ask:

We obtain in Theorem 2 that Out(s_b, n) contains every multiple of s_b(n). Does Out(s_b, n) contain an infinite number of positive integers that are not divisible by s_b(n)? When d = 1 and s_b(n) ≥ 2, the answer is yes. Can we determine a simple necessary and sufficient condition for the infinitely many elements in A ∩ Out(s_b, n) where A is the set of positive integers that are not divisible by s_b(n)? In that case, does Out(s_b, n) has a natural density?

For $b\ge2$ and $n\ge1$, write $s=s_b(n)$ and $d=\gcd(b-1,n)$. The answer is that the difference set is infinite exactly when $d<s$, and $\operatorname{Out}(s_b,n)$ has natural density $1/d$. It contains every sufficiently large positive multiple of $d$. Preregistration: #19415.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Base $b\ge2$ and positive $n$ | claim binders b n : ℕ, hypotheses 2 ≤ b, 1 ≤ n | Full parameter range. | | Positive arc from $n$ to $u$, witnessed by positive $N$ with $n\mid N$ and $s_b(N)=u$ | Out | Both positivity clauses, divisibility and the complete digit-list sum are literal. | | Infinitely many elements of $A\cap\operatorname{Out}$ | First conjunct of claim | Set difference by the multiples of $(\operatorname{Nat.digits}\ b\ n).sum$. | | Necessary and sufficient condition | First conjunct of result | Biconditional with $\gcd(b-1,n)<s_b(n)$. | | Natural density | Second conjunct of result | Real limit of the count in $[1,x]$, divided by $x$, equal to $1/d$. | | Cofiniteness among positive multiples of $d$ | Private out_cofinite, consumed by result | An existential threshold; no numerical threshold is asserted. |

Out and claim inline the preregistered digit-sum and density expressions. Proofs use MaxDigitSumBase.digitSum and the shifted AsymptoticDensity.HasDensity; density_count_shift and density_as_limit connect the interval conventions. No additional public mathematical statement or strengthened hypothesis is used.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Density/PhunphayapNoppakaewPongsriiamDigitSumArcs.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023In-between-state optimality for Bayesian transmissivity sensingA vacuum–two-photon superposition has strictly smaller Bayesian transmissivity error than every phased in-between state at mean photon number one half, for an equiprobable two-point prior.Refuted

The questionGagatsos, “Bayesian minimum mean square error for transmissivity sensing”, arXiv:2304.05539v1, Section V, p. 4, state:

Why it mattersThe quantified claim asserts that, for every admissible prior, positive mean energy and finite normalized competitor at that energy, some phase of the in-between state has no larger finite-POVM MMSE.

The ideaInside result, the operator variance of any finite measurement is expressed as a sum of positive semidefinite sandwiches.

Our contributionA vacuum–two-photon superposition has strictly smaller Bayesian transmissivity error than every phased in-between state at mean photon number one half, for an equiprobable two-point prior.

Verification details
Release source recordRefuted in Lean

Open problem

In-between-state optimality for Bayesian transmissivity sensing

B. Zhou, B. A. Bash, S. Guha and C. N. Gagatsos, “Bayesian minimum mean square error for transmissivity sensing”, arXiv:2304.05539v1, Section V, p. 4, state:

For the two-point prior PDF (15), we verify that for integer n̄ the optimal state is the Fock state with the same photon number. For real n̄, our numerical results support the conjecture that the optimal state has the form of the state (27), up to a phase, i.e., |Φ′ₙ̄⟩ = e^{iφn̂}|Φₙ̄⟩.

Issue #11653 preregisters this Tier 1 conjecture, the quantified reading and the witness. The two-point prior is $P(\tau)=q\delta(\tau-\tau_0)+(1-q)\delta(\tau-\tau_1)$, with $q,\tau_0,\tau_1\in[0,1]$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/TransmissivityTwoPointProbeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresExact formula provedFirst preprint 2022The extremal matching number of almost regular triangle-free graphsThe least matching number of an almost regular, triangle-free, factor-critical graph has an exact formula for every odd degree at least twenty-one.Proved

The questionYıldız, Maximum size of a triangle-free graph with bounded maximum degree and matching number, arXiv:2207.02271v1, Conjecture 6.5:

Why it mattersThe two odd residue classes have uniform extremal constructions, while the lower bound rules out smaller orders.

The ideaFactor-criticality identifies the order as twice the matching number plus one.

Our contributionThe least matching number of an almost regular, triangle-free, factor-critical graph has an exact formula for every odd degree at least twenty-one.

Verification details
Release source recordProved in Lean

Open problem

The extremal matching number of almost regular triangle-free graphs

M. Ahanjideh, T. Ekim and M. A. Yıldız, Maximum size of a triangle-free graph with bounded maximum degree and matching number, arXiv:2207.02271v1, Conjecture 6.5:

For d ≥ 21 and odd, we have Z(d) = ⌊5(d + 1)/4⌋.

The journal locator is DOI 10.1007/s10878-024-01123-z. The quotations refer to version 1. The source defines $Z(d)$, for $d\ge2$, as the least matching number of a triangle-free factor-critical graph that is regular of degree $d$ when $d$ is even, and almost regular of degree $d$ when $d$ is odd.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | “if d(w) = d − 1 for some w ∈ V(G) and d(v) = d for all v ∈ V(G) − w, then G is said to be almost d-regular.” | FactorCriticalMatching.IsAlmostRegular | One deficient vertex and every other degree are specified literally. | | “A graph G is said to be factor-critical if G \ v has a perfect matching for all v ∈ V(G).” | FactorCriticalMatching.IsFactorCritical | Perfect subgraph matchings in each induced vertex complement. | | “For any d ≥ 2, let Z(d) be the smallest natural number n such that there exists a d-regular (if d is even) or almost d-regular (if d is odd) triangle-free and factor-critical graph G with ν(G) = n.” | AhanjidehEkimYildizZ.Z | Natural infimum of exactly those matching-number witnesses. | | “For d ≥ 21 and odd, we have Z(d) = ⌊5(d + 1)/4⌋.” | AhanjidehEkimYildizZ.claim and .result | Every odd natural degree at least 21; natural division is the floor. |

Finite simple graphs have carrier Fin n; the matching number is the existing DimerVirialFourthDifferenceRefutation.matchingNumber. The formal definition of $Z$ is total on natural numbers and admits an empty carrier. The proved range has a deficient vertex and is nonempty. No statement about the paper's convention at even or small degrees is inferred from this total extension.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ComponentPacking/AhanjidehEkimYildizZ.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The least matching number of an almost regular, triangle-free, factor-critical graph has an exact formula for every odd degree at least twenty-one.

    Read source ↗
  • Historical context

    First preprint year: 2022.

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Quantum informationQuestion resolvedFirst preprint 2024Without shared entanglement the odd-prime communication task succeeds with probability at most 2/(d + 1)For every odd prime dimension, unentangled quantum messages have sharp average success probability 2/(d + 1), including finite mixtures with shared classical randomness.Proved

The questionTavakoli, Scalable entanglement certification via quantum communication, PRX Quantum 5, 020319 (2024), arXiv:2401.00796v2, certify entanglement through a symmetric prepare-and-measure task. - The task. For an odd prime d, Alice and Bob receive x, y ∈ {0, …, d − 1}² and send d-dimensional messages; Charlie receives z ∈ {0, …, d}, may measure the two messages jointly, and outputs c.

Why it mattersRoch i Carceller and Tavakoli (arXiv:2410.15388, Phys.

The idea1. Character expansion. With wz = az(x) + bz(y), the indicator of c = wz is d⁻¹ Σk χ(k(wz − c)).

Our contributionFor every odd prime dimension, unentangled quantum messages have sharp average success probability 2/(d + 1), including finite mixtures with shared classical randomness.

Verification details
Release source recordProved in Lean

Open problem

Without shared entanglement the odd-prime communication task succeeds with probability at most 2/(d + 1)

P. Bakhshinezhad, M. Mehboudi, C. Roch i Carceller and A. Tavakoli, Scalable entanglement certification via quantum communication, PRX Quantum 5, 020319 (2024), arXiv:2401.00796v2, certify entanglement through a symmetric prepare-and-measure task.

  • The task. For an odd prime d, Alice and Bob receive x, y ∈ {0, …, d − 1}² and send d-dimensional messages; Charlie receives z ∈ {0, …, d}, may measure the two messages jointly, and outputs c. He wins when c = w_z, with w_z = x₁ + y₁ − 2z(x₀ − y₀) for z ≠ d and w_d = x₀ − y₀ mod d (Eq. (27)), and R_d is the average success probability (Eq. (28)).
  • Without shared entanglement. p(c | x, y, z) = tr(α_x ⊗ β_y M_c|z) (Eq. (19)); sending x₀ and y₀ achieves R_d = 2/(d + 1).
  • The conjecture. After see-saw searches at d = 3 and d = 5: "Conjecture 1. For every odd prime d, the largest average success probability achievable in a quantum model without shared entanglement is R_d = 2/(d + 1). (29)" With shared entanglement and product measurements R_d = 1, so the bound would certify every state with a non-trivial maximally entangled fraction.

The verbatim statements are in the literature note.

Issue #16770 fixes the reading: messages are d × d density matrices, Charlie's measurement for each z is any d-outcome POVM on the joint d²-dimensional space, and shared classical randomness is a finite mixture of strategies with one common index. The conjecture is proved for every odd prime d.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/UnentangledOddPrimeCommunication.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2022Fixed-point-free permutation solutions of the n-simplex equationFor n > 2, a permutation without fixed points gives a solution of the n-simplex equation on every set exactly when n is even, and then the product of the transpositions of adjacent pairs of indices is such a solution.Proved

The questionLeshkov, Set-theoretical solutions of simplex equations, arXiv:2206.08906v1, Section 4.2 (Question 4.22 of Siberian Adv.

Our contributionFor n > 2, a permutation without fixed points gives a solution of the n-simplex equation on every set exactly when n is even, and then the product of the transpositions of adjacent pairs of indices is such a solution.

Verification details
Release source recordProved in Lean

Open problem

Fixed-point-free permutation solutions of the n-simplex equation

V. Bardakov, B. Chuzinov, I. Emel'yanenkov, M. Ivanov, T. Kozlovskaya and V. Leshkov, Set-theoretical solutions of simplex equations, arXiv:2206.08906v1, Section 4.2 (Question 4.22 of Siberian Adv. Math. 34(1) (2024) 1–40, and of its Russian original in Мат. труды 27(1) (2024)), ask:

  1. We know that the permutation $P_{12}$ is a solution of the YBE. For which $n > 2$ there are non-identity permutations without fixed points that gives solutions of the \SE?

The $n$-simplex equation $R_{\overline 1}\cdots R_{\overline{n+1}} = R_{\overline{n+1}}\cdots R_{\overline 1}$ acts on $X^N$, $N=n(n+1)/2$, with the multi-indices given by the rows of the matrix $MI_n$; a permutation $s$ gives the simple map $(x_1,\dots,x_n)\mapsto(x_{s(1)},\dots,x_{s(n)})$. The precise question, preregistered in issue #11874, asks for the set of $n>2$ for which some permutation without fixed points gives a solution on every set.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/VertexModels/SimplexFixedPointFreePermutations.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedSource 2023Inversion Sequences Avoiding {010, 100, 102, 210} and {011, 201, 210}The two pattern-avoidance classes in Class 152 are equinumerous at every length.Proved

The questionDavid Callan and Toufik Mansour, Inversion Sequences Avoiding Quadruple Length-3 Patterns, Integers 23 (2023), A78, Section 1, Conjecture 1:

The ideaIn the left class a value that occurs after a larger value occurs only once, and every adjacent descent starts at the global maximum; the sequence is cut at its first adjacent descent into a weakly increasing prefix and a suffix of interleaved maxima and singleton values.

Our contributionThe two pattern-avoidance classes in Class 152 are equinumerous at every length.

Verification details
Release source recordProved in Lean

Open problem

Inversion Sequences Avoiding {010, 100, 102, 210} and {011, 201, 210}

David Callan and Toufik Mansour, Inversion Sequences Avoiding Quadruple Length-3 Patterns, Integers 23 (2023), #A78, Section 1, Conjecture 1:

Class 152: {010, 100, 102, 210} ∼_I {011, 201, 210}

Here ∼_I means that, for every n, equally many inversion sequences of length n avoid each set. The authors checked the conjecture up to n = 13 and list the comparison as "still open" in Table 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/InversionSeq/InversionSeqClass152.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2023.

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Quantum informationQuestion resolvedFirst preprint 2021Classical Markov monogamy for every finite alphabet and every permutationThe diagonal sum of mutual informations dominates every permuted sum along a finite classical Markov chain.Proved

The questionModi, Quantum Markov monogamy inequalities, Phys.

Why it mattersConjecture 1 constrains how mutual information can be distributed among pairs across a channel and its pre-processing and post-processing stages.

The ideaThat lemma requires a finite, nonnegative joint mass of total mass one.

Our contributionThe diagonal sum of mutual informations dominates every permuted sum along a finite classical Markov chain.

Verification details
Release source recordProved in Lean

Open problem

Classical Markov monogamy for every finite alphabet and every permutation

M. Capela, L. C. Céleri, R. Chaves and K. Modi, Quantum Markov monogamy inequalities, Phys. Rev. A 106, 022218 (2022), arXiv:2108.11533v4, Section II, Conjecture 1 (PRA-v1.tex, lines 153–159):

Consider the Markovian process $X_n \rightarrow \cdots \rightarrow X_1 \rightarrow Y_1 \rightarrow \cdots \rightarrow Y_n$. The variables $X_1$ and $Y_1$ are to be interpreted as input and output of a given channel, respectively. The variables $X_i$ and $Y_i$, with $i=2,\dots,n$, are interpreted as pre-processed and post-processed variables, respectively. Then for any bijective function $f \colon {1,\dots,n} \rightarrow {1,\dots,n}$, it holds that

$\sum_{i=1}^{n} I(X_i:Y_i) \geq \sum_{i=1}^{n} I(X_i:Y_{f(i)}),$

where $I(X:Y)$ denotes the mutual information of random variables $X$ and $Y$.

The source reports at line 161:

The conjecture above has been checked for $n$ up to 4, that is, Markov chains with 8 random variables

Here $I(X:Y)=H(X)+H(Y)-H(X,Y)$, with Shannon entropy. The definitions and source locators are in the literature note.

Issue #22399 fixes the finite-alphabet reading. The module proves it for every $n\in\mathbb N$, including $n=0$ with empty sums: choose finite alphabets $\alpha_t$, a nonnegative initial mass $p:\alpha_0\to\mathbb R$ with $\sum_xp(x)=1$, and matrices $k_t:\alpha_t\to\alpha_{t+1}\to\mathbb R$ satisfying $k_t(x,y)\ge0$ and $\sum_y k_t(x,y)=1$ whenever $t+1<2n$. For every permutation $f$ of $\mathrm{Fin},n$, the sum paired by $f$ is at most the diagonal sum.

Precisely, composedTransition is the identity at equal positions and the product of successive matrices for ordered positions $s\le t$; write it as $K_{s,t}$. The definitions mixture and mixtureJoint give $p_s(x)=\sum_zp(z)K_{0,s}(z,x)$ and $q_{s,t}(x,y)=p_s(x)K_{s,t}(x,y)$. For $i\in\mathrm{Fin},n$, position $n-1-i.val$ represents $X_{i.val+1}$ and position $n+i.val$ represents $Y_{i.val+1}$. Each summand in claim is mutualInformation of the corresponding $q_{s,t}$, defined as its KL divergence from the product of its marginals, in nats.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/MarkovMonogamyPermutationInequality.result

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Quantum informationQuestion resolvedFirst preprint 2022The bound L_k(M) ≤ k L(M) is not tight for k ≥ 3Averaging signs improves the prepare-and-measure bound uniformly for every k at least three.Proved

The questionVértesi, Certification of qubits in the prepare-and-measure scenario with large input alphabet and connections with the Grothendieck constant, arXiv:2211.17185v2 (Sci.

The ideaThe proof uses convexity.

Our contributionAveraging signs improves the prepare-and-measure bound uniformly for every k at least three.

Verification details
Release source recordProved in Lean

Open problem

The bound L_k(M) ≤ k L(M) is not tight for k ≥ 3

P. Diviánszky, I. Márton, E. Bene and T. Vértesi, Certification of qubits in the prepare-and-measure scenario with large input alphabet and connections with the Grothendieck constant, arXiv:2211.17185v2 (Sci. Rep. 13, 13200, 2023), Methods, after Lemma III.4:

It is an open question whether Lemma~\ref{LkkL} is tight or not.

Hence, for $k=2$ the upper bound in Lemma~\ref{LkkL} is tight. We conjecture that the bound is not tight for greater values of $k$.

Here, for a real $n\times m$ matrix $M$:

  • $L(M)=\max_{v\in{\pm1}^n}|vM|_1$, the classical bound of the correlation expression $M$;
  • $L_k(M)=\max_W\sum_i|(W^{\mathsf T}M)_i|_1$ over the $n\times k$ matrices $W$ whose rows are standard basis vectors, which groups the inputs into $k$ classes;
  • Lemma III.4 states $L_k(M)\le kL(M)$.

The definitions are quoted in the literature note.

Issue #20049 fixes the reading: for every $k\ge3$ there is $c_k<k$ with $L_k(M)\le c_kL(M)$ for all matrices. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Optimization/PrepareMeasureLkBound.result

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Discrete structuresQuestion resolvedFirst preprint 2024The single exceptional modulus for equality of even Lucas periodsThe two even-parameter open period questions are settled: the matrix and companion periods agree exactly outside q = 1, modulus 4, and 4 dividing p.Proved

The questionQuestion 5.4 of arXiv:2408.14632v2, as printed:

Why it mattersThis is the sharp form of the previous question.

The ideaWork with the companion matrix M = !![p, -q; 1, 0] over ZMod m; piU is its order in the unit group and piV the least period of its trace.

Our contributionThe two even-parameter open period questions are settled: the matrix and companion periods agree exactly outside q = 1, modulus 4, and 4 dividing p.

Verification details
Release source recordProved in Lean

arXiv

The single exceptional modulus for equality of even Lucas periods

Question 5.4 of arXiv:2408.14632v2, as printed:

... when p = 0 (mod 4), then it appears that the corollary holds for all even values m > 2 except for the single value of m = 4. And in that case, we have e_V(4) = 1 but pi_U(4) = 4 != 2 = pi_V(4) ... However, when p = 2 (mod 4) and m > 2 is even, the existence of e_V(m) appears to always guarantee that pi_U(m) = pi_V(m). Can we prove that the corollary holds for all even values p and m with m > 2, except in the singular case when m = 4 and p = 0 (mod 4)?

The excerpt is taken from a subsection whose standing hypothesis is q = 1, so the exceptional case has three conditions, not two: q = 1, m = 4, and 4 dividing p. That matters because with q = -1 the same modulus and the same divisibility give equal periods, which is what the sibling entry fiebig-mbirika-spilker-ballot-extended-even records. The theorem below carries all three conditions in its exceptional clause.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LucasEvenPeriods.even_lucas_periods

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Quantum informationQuestion resolvedFirst preprint 2022Quantum max-flow equals min-cut in the symmetric bridge regionsQuantumMaxFlowMinCut supplies the width-three bridge max-flow proof.Proved

The questionThe literature note gives the verbatim source clauses and locators. 14652 records the complete target and literature scope.

Why it mattersQuantumMaxFlowMinCut.result proves both equalities at width three, for every positive natural dimension quadruple in the two symmetric regions.

The ideaThe construction uses rational three-slice matrices.

Our contributionQuantumMaxFlowMinCut supplies the width-three bridge max-flow proof.

Verification details
Release source recordProved in Lean

Open problem

Quantum max-flow equals min-cut in the symmetric bridge regions

F. Gesmundo, V. Lysikov and V. Steffan, Quantum max-flow in the bridge graph, arXiv:2212.09794v2, Conjecture 1.4 (also Conjecture 3.14), state: “Let $(a,b),(a',b')\in U_w\cup V_w\cup W_w$. Then” $\operatorname{QMaxFlow}=\operatorname{QMinCut}=\min{ab',a'b}$. Conjecture 3.15 is the width-three case. The literature note gives the verbatim source clauses and locators. #14652 records the complete target and literature scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/BridgeGraph/QuantumMaxFlowMinCut.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Discrete structuresConjecture provedFirst preprint 2024Finite Buffered Hosts for Ordinary AmoebasA finite layered supertree preserves every initial tree, realizes all arm paths simultaneously for every copy, and is confining under total-edge minimization.Proved

The questionVladimir Gurvich, Matjaž Krnc and Mikhail Vyalyi, Growing Trees and Amoebas’ Replications, arXiv:2401.07484v1/v2 and DOI 10.1007/s00025-025-02421-6, state Conjecture 1 in Section 2.1 of arXiv v1 and on page 11 of the published PDF: if an amoeba A=(H,m) has maximum degree greater than its number of roots k by more than two, then A is totally confined.

Why it mattersThe result addresses the complete named conjecture, including its universal quantifiers over copies, initial hosts, arms and total-edge minimizers.

The ideaThe formalization uses a finite layered tree.

Our contributionA finite layered supertree preserves every initial tree, realizes all arm paths simultaneously for every copy, and is confining under total-edge minimization.

Verification details
Release source recordProved in Lean

Open problem

gurvich krnc vyalyi conjecture 1 total amoeba confinement

Vladimir Gurvich, Matjaž Krnc and Mikhail Vyalyi, Growing Trees and Amoebas’ Replications, arXiv:2401.07484v1/v2 and DOI 10.1007/s00025-025-02421-6, state Conjecture 1 in Section 2.1 of arXiv v1 and on page 11 of the published PDF: if an amoeba A=(H,m) has maximum degree greater than its number of roots k by more than two, then A is totally confined.

The formal target keeps the complete ordinary finite-host reading: every finite simple tree H, every multiplicity function m, every extension length l, every finite initial tree T containing H, every multiplicity-preserving copy and transported multiplicity placement, every simultaneous family of distinct arms, and every minimum growth ordered by total host edges. Length zero, zero multiplicities and k=0 remain included. Total confinement requires a finite host containing T in which every such copy is dead.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Walks/AmoebaHost.total_confinement

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2022Johnson–Wakefield Question 6.19: nonvanishing in rank threeNo finite rank-three nonmodular matroid has zero J-Möbius evaluation at minus one.Proved

The questionJohn Johnson and Max Wakefield, A non-associative incidence near-ring with a generalized Möbius function, Ars Mathematica Contemporanea 24, P2.03, Question 6.19, ask:

Why it mattersThis settlement does not claim an independent escape witness.

The ideaThe namespace below is D5.S3.Combinatorics.Matroids.JohnsonWakefieldRankThreeMoebius.

Our contributionNo finite rank-three nonmodular matroid has zero J-Möbius evaluation at minus one.

Verification details
Release source recordProved in Lean

Open problem

Johnson–Wakefield Question 6.19: nonvanishing in rank three

John Johnson and Max Wakefield, A non-associative incidence near-ring with a generalized Möbius function, Ars Mathematica Contemporanea 24, #P2.03, Question 6.19, ask:

Is there a rank 3 non-modular connected matroid $M$ such that $\mathcal M(M,-1)=0$?

The answer for finite matroids is negative. The precise source definitions and finite-carrier conventions are fixed by #19762. The Library source is Library/FiniteGeometry/johnsonwakefield2024incidencenearring.md.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Lattice of flats $L(M)$, ordered by inclusion | Flats M, the subtype of Matroid.IsFlat | Literal binding convention | | Rank and top rank | rk M F, M.eRank | Extended rank converted to a natural number; finite carrier | | Definition 5.1, $x\le a\le y\le b\le z$, with diagonal value one and all other flag values zero | IsJFunction | The rectangle equation, rather than the product formula, defines the public predicate | | Definition 6.2, the partial-flag sum with exponent $3r(M)-r(x)-r(y)-r(z)$ | MoebiusPoly | All sums are finite; exponent subtraction is in $\mathbb N$ | | Rank three, connected, nonmodular | claim hypotheses M.eRank = 3, Connected M, and ¬ Modular M | Connectedness is retained literally | | No such vanishing example | result : claim | Every solution of the defining rectangle equation has nonzero evaluation | | Finite matroid setting | Finite α with M : Matroid α | No infinite-carrier assertion |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Matroids/JohnsonWakefieldRankThreeMoebius.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2023Rectangular-torus rotation descentA decorated plane lift fixed by a translated transposing right action descends to the square torus whose side is the gcd of the source periods.Proved

The questionPeter Kagey and William Keehn, Counting tilings of the n x m grid, cylinder, and torus, arXiv:2311.13072v1, Section 5.1, Conjecture 5.1.1 (the journal alias is JIS 27 (2024), 24.6.1, Section 6.1, Conjecture 42), ask whether a plane tiling represented by an n x m torus and fixed by a translated right action r, r^3, rf, or r^3f is equivalent to a gcd(n,m) x gcd(n,m) torus tiling.

Why it mattersThe source models a finite torus as a repeating decorated plane.

The ideaFor a source translation representative u and center c, the decorated graph map is (z,d) -> (Ah(z+u)+c, d·h).

Our contributionA decorated plane lift fixed by a translated transposing right action descends to the square torus whose side is the gcd of the source periods.

Verification details
Release source recordProved in Lean

arXiv

Rectangular-torus rotation descent

Peter Kagey and William Keehn, Counting tilings of the n x m grid, cylinder, and torus, arXiv:2311.13072v1, Section 5.1, Conjecture 5.1.1 (the journal alias is JIS 27 (2024), 24.6.1, Section 6.1, Conjecture 42), ask whether a plane tiling represented by an n x m torus and fixed by a translated right action r, r^3, rf, or r^3f is equivalent to a gcd(n,m) x gcd(n,m) torus tiling.

The complete reading quantifies over every positive n,m, every applicable subgroup R <= D8, every actual right R-set T of decorated tile designs, every f : ZMod n -> ZMod m -> T with literal periodic lift F : Z x Z -> T, every h in {r,r^3,rf,r^3f} belonging to R, every allowed translation representative u before h, and every integral affine center. It asks for a genuine H on the full product ZMod gcd(n,m) x ZMod gcd(n,m) whose literal plane lift is pointwise identical to F. Equal and unequal dimensions, unit dimensions, nonprimitive periods, all quotient representatives, and nontrivial design actions are included. No minimality claim and no D8 action on an unequal finite rectangle is part of the problem.

The operational source maps are r : (x,y) -> (-y,x), r^3 : (x,y) -> (y,-x), rf : (x,y) -> (y,x), and r^3f : (x,y) -> (-y,-x), with labels d -> d·h. The source's inverse conjugation coordinate map is kept distinct from these operational maps.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TorusTilings/KageyKeehnRectangularTorus.rectangular_torus_gcd_descent

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2020Normalization of weak shuffle productsMammez's weak shuffle normalization conjecture holds for every alphabet.Proved

The questionMammez, Weak stuffle algebras, International Electronic Journal of Algebra 32 (2022), 1–45, DOI 10.24330/ieja.1060709, arXiv:2003.09142v3, Conjecture 3.10 (arXiv v3 Conjecture 14) states: “Proposition [3.9] is still true for any countable alphabet.” Proposition 3.9 asserts that a weak shuffle product over a characteristic-zero field is algebra-isomorphic to its normalized product; its proof covers alphabets with two or three letters.

Why it mattersPreregistration 15815 supplies the verbatim source, the characteristic-zero hypotheses, the explicit normalization conventions and the bounded literature screen.

The ideaThe proof establishes barrier laws from coefficient comparisons in commutativity and associativity.

Our contributionMammez's weak shuffle normalization conjecture holds for every alphabet.

Verification details
Release source recordProved in Lean

Open problem

Normalization of weak shuffle products

C. Mammez, Weak stuffle algebras, International Electronic Journal of Algebra 32 (2022), 1–45, DOI 10.24330/ieja.1060709, arXiv:2003.09142v3, Conjecture 3.10 (arXiv v3 Conjecture 14) states: “Proposition [3.9] is still true for any countable alphabet.” Proposition 3.9 asserts that a weak shuffle product over a characteristic-zero field is algebra-isomorphic to its normalized product; its proof covers alphabets with two or three letters.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | A field of characteristic 0 | K, Field K, CharZero K in claim | Literal algebraic hypotheses | | Noncommutative polynomials on an alphabet | KX X K, finitely supported coefficients on List X | Every alphabet, including empty and uncountable ones | | Empty word is the unit and the two-branch weak shuffle recursion | wsWord, with the first branch removing the left head and the second removing the right head | Literal recursion; the contraction coefficient is zero | | Bilinear extension; associative and commutative product | wsMul and IsWeakShuffle | Both quantified identities occur in the predicate | | Exceptional off-diagonal coefficient becomes one, its reverse becomes zero | norm1 | Both ordered tests are explicit | | Other off-diagonal coefficients are retained and diagonal coefficients are unchanged | norm1, norm2 | The reverse exceptional test expresses the first bullet's forced reverse value | | An algebra isomorphism between the two products | claim, result | A linear equivalence preserves the unit and intertwines multiplication | | Any countable alphabet | X : Type in claim | Stronger: no countability hypothesis |

DecidableEq X is the implementation interface for finite-support words. Classical choice supplies it for every alphabet, so it does not restrict the mathematical scope. The implicit Zero K of KX and classical coefficient tests likewise do not change the definitions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/WeakShuffleNormalization.result

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Quantum informationConjecture provedFirst preprint 2023Novaes's strong-absorption cumulant conjectureStrongAbsorptionCumulants for the strong-absorption cumulant expansion.Proved

The questionNovaes, Time delay statistics for chaotic cavities with absorption, arXiv:2305.18235v1, Section 5, p. 7, states “we conjecture that, for n > 1,” followed by Eq. (58):

Why it mattersThe conclusion determines two coefficients for every finite channel number and cumulant order, with an eventual absolute remainder estimate.

The ideaThe normalized inverse-rate law is a polynomial perturbation of two Laguerre weights.

Our contributionStrongAbsorptionCumulants for the strong-absorption cumulant expansion.

Verification details
Release source recordProved in Lean

Open problem

Novaes's strong-absorption cumulant conjecture

M. Novaes, Time delay statistics for chaotic cavities with absorption, arXiv:2305.18235v1, Section 5, p. 7, states “we conjecture that, for n > 1,” followed by Eq. (58):

$$ (-1)^n k_n(\gamma) =\frac{(n-1)!}{M^{2n-2}\gamma^{2n}} -\frac{(2n-1)n(n-1)!}{M^{2n-2}\gamma^{2n+1}} +O(\gamma^{-2n-2}). $$

Here $M\geq1$ is an integer channel number and $n\geq2$ is an integer cumulant order. The symmetry class is $\beta=2$, with perfect coupling and uniform absorption. The journal publication is Journal of Statistical Physics 190, 182 (2023), DOI 10.1007/s10955-023-03183-7; its Eq. (65) is the same conjecture.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Every positive integer channel number | claim quantifies every natural M with 1 ≤ M. | | Every cumulant order greater than one | Every natural n with 2 ≤ n. | | Grabsch's inverse-rate law at $\beta=2$, $N=M$ | Foundations.weight is the literal joint density below; Foundations.law divides its measure by total mass. | | Wigner time delay in dwell-time units | Foundations.tauW is $M^{-1}\sum_i(\gamma+x_i)^{-1}$. | | The $n$th cumulant | Foundations.kappa is the $n$th derivative at zero of the observable's cumulant generating function. | | Both coefficients of Eq. (58), including $(-1)^n$ | The literal unsimplified factorial, channel and absorption powers in claim. | | $O(\gamma^{-2n-2})$ as $\gamma\to\infty$ | There exist real $C,\gamma_0$ with $\gamma_0>0$, and the absolute error is at most $C/\gamma^{2n+2}$ for every $\gamma\geq\gamma_0$. Constants may depend on $(M,n)$. |

Grabsch, arXiv:1909.01002v2, Eqs. (3)–(4), supplies the model law. Writing the inverse delay rates as $\Gamma_i=\gamma+x_i$, its unnormalized weight on $x_i\geq0$ and $0\leq y_j\leq\gamma$ is

$$ e^{-M(\sum_i x_i+\sum_j y_j)}\Delta(x)^2\Delta(y)^2 \prod_{i,j}(\gamma+x_i-y_j). $$

The five conventions weight, law, tauW, kappa, claim are literal to #16825. The probability normalization and bounded observable are proved for this law. Identification of this law with the physical random-matrix model is the cited mathematical-model premise, rather than a theorem about a physical cavity's dynamics.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Transport/StrongAbsorptionCumulants.result

Frozen module record

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Discrete structuresRecurrence provedPosed in 2024The EGF coefficients of A338193 and Kurkov's recurrenceThe EGF coefficients in OEIS A338193 equal Kurkov's two-index recurrence.Proved

The questionThe October 26, 2024 conjecture by Mikhail Kurkov in OEIS A338193 asserts that a(n)=f(0,n-1) for every positive n.

Why it mattersThe equality connects the integral equation to a recurrence entirely in natural numbers.

The ideaInduction first on degree and then on row index proves F(j)=F(0)R^j.

Our contributionThe EGF coefficients in OEIS A338193 equal Kurkov's two-index recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The EGF coefficients of A338193 and Kurkov's recurrence

The October 26, 2024 conjecture by Mikhail Kurkov in OEIS A338193 asserts that a(n)=f(0,n-1) for every positive n. Here a(n)=n! times coefficient n of the rational formal series satisfying A(0)=1 and A=1+Integral ((x/A)' / (x/A^2)') dx, with integral constant zero. The independent natural-valued recurrence has f(j,0)=1, f(0,m+1)=f(0,m)+(m+1)f(1,m), and f(j+1,m+1)=f(j,m+1)+(m+1)(f(j+1,m)+f(j+2,m)).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/SchroderIntegralEGF.egf_coeff_eq_f

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About this highlight
  • Contribution

    The EGF coefficients in OEIS A338193 equal Kurkov's two-index recurrence.

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  • Historical context

    Dated conjecture statement: 2024.

    Read source ↗
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Quantum informationQuestion resolvedFirst preprint 2023For two qubits the SVD entanglement entropy is at least the smaller of the two entanglement entropiesFor two unit two-qubit states with nonzero reduced transition, SVD entanglement entropy is at least the smaller input entanglement entropy.Proved

The questionIt lies between 0 and log dA. - The open question. "In Figure 8, we provide numerical evidence suggesting that a non-trivial lower bound of min{S(ρ¹A), S(ρ²A)} might hold for qubits, fails in general for qudits … Rigorous proofs of these observations for qubits … are still lacking" (§3.3).

Why it mattersThe SVD entropy measures entanglement of a transition between two states.

The idea5. Entropy. The entropy of a 2 × 2 density matrix is an increasing function of its determinant p(1 − p), so S(Q) ≥ min{S(ρψ), S(ρφ)}.

Our contributionFor two unit two-qubit states with nonzero reduced transition, SVD entanglement entropy is at least the smaller input entanglement entropy.

Verification details
Release source recordProved in Lean

Open problem

For two qubits the SVD entanglement entropy is at least the smaller of the two entanglement entropies

A. J. Parzygnat, T. Takayanagi, Y. Taki and Z. Wei, SVD entanglement entropy, JHEP 12 (2023) 123, arXiv:2307.06531v2, define an entropy for a pair of pure states.

  • The SVD entropy. For |ψ¹⟩, |ψ²⟩ in H_A ⊗ H_B, the reduced transition matrix is τ_A = Tr_B(|ψ¹⟩⟨ψ²|)/⟨ψ²|ψ¹⟩, the SVD density matrix is ρ^{1|2}_A = √(τ_A†τ_A)/Tr √(τ_A†τ_A), and the SVD entanglement entropy is its von Neumann entropy (§2). It lies between 0 and log d_A.
  • The open question. "In Figure 8, we provide numerical evidence suggesting that a non-trivial lower bound of min{S(ρ¹_A), S(ρ²_A)} might hold for qubits, fails in general for qudits … Rigorous proofs of these observations for qubits … are still lacking" (§3.3). The authors prove it only when the second state is a local unitary image of the first (Eq. (3.37)), "a restricted version of S(ρ^{1|2}_B) ?≥ min{S(ρ¹_B), S(ρ²_B)} (3.38)".

The verbatim statements are in the literature note.

Issue #16821 fixes the reading: two-qubit pure states as coefficient matrices of Frobenius norm 1, the unnormalized transition Tr_B|ψ⟩⟨φ| (the factor 1/⟨φ|ψ⟩ cancels in the SVD density matrix, so orthogonal pairs with nonzero transition are included), and natural-logarithm entropies. The bound is proved for every such pair.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/TwoQubitSVDEntropyLowerBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedSource 2023Ramane–Kitturmath's regular graph energy conjectureThe adjacency energy of a regular graph cannot equal 2k√2 for a positive integer k.Proved

The questionKitturmath, On the conjecture of Sombor energy of a graph, Examples and Counterexamples 3 (2023), 100115, Conjecture 3.11, page 8:

Why it mattersThe source note is Library/GraphInvariants/ramanekitturmath2023regularenergy.md.

The ideaThe rational adjacency characteristic polynomial splits in a real normal extension.

Our contributionThe adjacency energy of a regular graph cannot equal 2k√2 for a positive integer k.

Verification details
Release source recordProved in Lean

Open problem

Ramane–Kitturmath's regular graph energy conjecture

H. S. Ramane and D. V. Kitturmath, On the conjecture of Sombor energy of a graph, Examples and Counterexamples 3 (2023), 100115, Conjecture 3.11, page 8:

There is no regular graph with adjacency energy 2k√2, where k∈Z⁺.

The quotation is the search seat's source reading recorded in preregistration #18578; independent verification of the publisher text and the scope of the corrigendum, DOI 10.1016/j.exco.2024.100135, is ASSUMED-UNVERIFIED.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Regular graph | A finite type V, G : SimpleGraph V, and G.IsRegularOfDegree r for r : ℕ | Includes disconnected graphs, empty vertex types and degree zero. | | Adjacency energy | Imported graphEnergy G MatrixKind.adjacency | Sum of absolute adjacency eigenvalues, with multiplicities and center zero. | | $k\in\mathbb Z^+$ | k : ℕ with 0 < k | Exactly the positive integers. | | There is no such graph | Universal inequality in claim; result : claim | Positive settlement. | | Finite enumeration and decidability instances | Anonymous Fintype, DecidableEq and DecidableRel instances | Represent finite simple graphs without adding a mathematical restriction. |

The only public definition is claim. It states the source proposition through the definitions explicitly fixed in #18578; it introduces no replacement notion of energy or regularity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/RegularGraphEnergyQuadraticExclusion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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    Year recorded in the source bibliography: 2023.

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Quantum informationQuestion resolvedFirst preprint 2023The noisy qutrit Werner–Holevo channel is a GKSL exponential only at zero noise on $[0,2/3]$On the interval from zero to two thirds, the noisy qutrit Werner-Holevo channel is a GKSL exponential exactly at zero noise.Proved

The questionShayan Roofeh and Vahid Karimipour, The noisy Werner-Holevo channel and its properties, Phys.

Why it mattersThe question distinguishes a noise channel specified by its input-output action from a finite-time exponential of a time-independent Lindblad generator.

The ideaThese are symmetricProjection, antisymmetricProjection, symmetricTrace and antisymmetricTrace.

Our contributionOn the interval from zero to two thirds, the noisy qutrit Werner-Holevo channel is a GKSL exponential exactly at zero noise.

Verification details
Release source recordProved in Lean

Open problem

The noisy qutrit Werner–Holevo channel is a GKSL exponential only at zero noise on $[0,2/3]$

Shayan Roofeh and Vahid Karimipour, The noisy Werner-Holevo channel and its properties, Phys. Rev. A 109, 052620 (2024), arXiv:2310.15353v7, subsection “Spectral properties of the channel,” ask:

The question remains whether, within the range of $0\leq x\leq \frac{2}{3}$, the channel can be described as the exponential of Lindbladian dynamics. We will explore this question in our future work.

This is verbatim from ModifiedLSChannel-12.tex, line 249, in the downloaded v7 source, md5 bcff52a86196a2e1a0054a34c48a8117. The channel formula is at line 153; its spectral multipliers and determinant are at lines 237–247. The literature note records these locators.

The module defines, for every real $x$, the complex-linear map on $M_3(\mathbb C)$

$\Lambda_x(X)=(1-x)X-\frac{x}{2}X^{\mathsf T}+\frac{x}{2}\operatorname{tr}(X)I_3.$

This is wernerHolevo; the source uses the same formula with $0\le x\le1$. IsGKSLGenerator G means that there exist $m\in\mathbb N$, a Hermitian $H\in M_3(\mathbb C)$ and matrices $V_k\in M_3(\mathbb C)$ indexed by $k\in\mathrm{Fin},m$ such that, for every $X\in M_3(\mathbb C)$,

$G(X)=-i(HX-XH)+\sum_k\left(V_kXV_k^\dagger-\frac12(V_k^\dagger V_kX+XV_k^\dagger V_k)\right).$

The empty jump family is allowed. generatorExp G is the ordinary exponential of the $9\times9$ complex matrix of $G$ in the matrix-unit basis, transported back to a linear map. Equality is equality on the entire complex matrix space.

Issue #21274 fixes the statement: for every $x\in[0,2/3]$, there exists such a $G$ with $\operatorname{generatorExp}(G)=\Lambda_x$ if and only if $x=0$. This statement is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/NoisyWernerHolevoLindbladEmbedding.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2024Explicit flat-spectrum probe probabilitiesExplicit positive probabilities solve the flat-spectrum system for every copy number.Proved

The questionVolkoff, Quantum programmable reflections, arXiv:2411.03648v2, Section 4.2.1, Conjecture 4.6:

Why it mattersA flat spectrum makes the source's probe-state entropy and program-dimension argument explicit.

The ideaExtracting the common square root in Racah's formula gives a rational binomial convolution.

Our contributionExplicit positive probabilities solve the flat-spectrum system for every copy number.

Verification details
Release source recordProved in Lean

Open problem

Explicit flat-spectrum probe probabilities

E. Schoute, D. Grinko, Y. Subaşı and T. Volkoff, Quantum programmable reflections, arXiv:2411.03648v2, Section 4.2.1, Conjecture 4.6:

The probabilities $q_j$ for $j\in{(n\bmod 2)/2,\ldots,n/2}$ in Eq. (181), which define the probe state $\Phi_n$, could be chosen such that $\lVert|v_J\rangle\rVert^2=1/\binom{n+2}{2}$ for every $J\in{n\bmod 2,\ldots,n-2,n}$.

Equation (183) is the equivalent system

$$ \frac{2J+1}{\binom{n+2}{2}} =\sum_{j\ge J/2}\frac{q_j}{2j+1} \left|\sum_{m=-j}^{j}C^{J0}_{jm,j-m}\right|^2. $$

Issue #17082 fixes the doubled-integer Racah convention and this system as the binding statement. The settlement is Proved by D5.S3.Quantum.Measurement.SGVFlatSpectrum.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every copy number $n\ge1$ | claim quantifies natural $n$ with $1\le n$ | Exact | | $j=n/2-k$, $0\le k\le\lfloor n/2\rfloor$ | The finite range in claim | Exact reindexing | | Probabilities | Nonnegativity and the sum equal to one in claim | Exact | | Real Condon–Shortley coefficients | S uses the frozen doubled-spin CG owner; equal_spin_coefficient proves its source Racah expression on the admissible summation inputs | Exact binding convention | | Coherent sum over magnetic indices | S sums $i=0,\ldots,2j$, with $m=j-i$ | Exact | | $J\le n$ of parity $n$ | The bound and remainder equality in claim | Exact | | Terms with $2j\ge J$ | The conditional inside the finite sum | Exact | | Flat spectrum (182) | The equality (183) in claim; source Eq. (181) divides by $2J+1$ | Source algebraic equivalence |

Integer division inside Racah's formula is Euclidean division; factorials use Nat.factorial after Int.toNat. Natural subtractions are truncated, and the summation guards preserve their intended ranges. The coefficient owner DampedSpinKernelPositivityRefutation.CG and this module have generality I. S calls the owner in doubled-integer argument order; the Scribe displays the source Racah expression. Inside equal_spin_coefficient, the square-root product and summation-range correspondence establish equality on every admissible summation input.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/SGVFlatSpectrum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Normal limits for Speyer g-polynomials of growing uniform-matroid rankGrowing uniform-matroid ranks satisfy both the central and local normal limit theorems.Proved

The questionRong Zhang and James Jing Yu Zhao, Analytic properties of Speyer’s g-polynomial of uniform matroids, arXiv:2408.15506v1, Conjecture 4.8:

Why it mattersThe underlying probability estimates also apply to polynomial arrays whose normalized law is a one-step shift of a Bernoulli product and whose variance diverges.

The ideaNormalizing these linear factors gives the shifted Bernoulli factorization in gbernoullifactorization.

Our contributionGrowing uniform-matroid ranks satisfy both the central and local normal limit theorems.

Verification details
Release source recordProved in Lean

Open problem

Normal limits for Speyer g-polynomials of growing uniform-matroid rank

Rong Zhang and James Jing Yu Zhao, Analytic properties of Speyer’s g-polynomial of uniform matroids, arXiv:2408.15506v1, Conjecture 4.8:

If $d=d(n)$ is a real function of $n$ such that $d(n)\to+\infty$ as $n\to+\infty$, then the coefficient of Speyer’s g-polynomial $g_{n,d(n)}(t)$, namely $S_i(n,d(n))$, is asymptotically normal by local and central limit theorems.

The source defines $g_{n,d}$ for integer ranks and uses the standing restriction $1\le d\le\lfloor n/2\rfloor$ after its rank-complement symmetry. The formal claim quantifies $d:\mathbb N\to\mathbb N$, imposes that restriction eventually, and assumes $d(n)\to\infty$. It does not extend the polynomial to arbitrary real-valued rank functions. Both limits are uniform in every real threshold and use the source's derivative-ratio mean and variance.

Statement echo

| Source clause or convention | Lean counterpart | Scope | | --- | --- | --- | | Proposition 1.1 factorial sum | SpeyerUniformNormality.g | Natural subtraction occurs before casting, on the prescribed finite summation range. | | Integer matroid rank under the standing half-rank restriction | d : ℕ → ℕ and eventual 1 ≤ d n ∧ d n ≤ n / 2 | The source's wording “real function” is interpreted within its integer rank domain; eventual validity allows arbitrary finitely many initial values. | | Rank tends to positive infinity | Tendsto d atTop atTop | No monotonicity or growth-rate hypothesis is added. | | Theorem 4.2 mean and variance | BernoulliTriangularLimits.mu, sigmaSq, sigma | The same derivative ratios and nonnegative square root. | | Coefficient probability | BernoulliTriangularLimits.prob | Coefficient divided by evaluation at one, and zero at negative integer indices. | | Standard normal distribution and density | StatLean.HypothesisTesting.stdNormalCDF and stdNormalPDF | Reused Gaussian owners inside CLT and LLT; claim is their conjunction for the Speyer polynomial sequence. Their integral and density correspondences are established inside the consuming proofs. | | Central limit theorem, uniform over real thresholds | BernoulliTriangularLimits.CLT | The epsilon form uses the finite coefficient support. | | Local limit theorem, with integer floor | BernoulliTriangularLimits.LLT | The epsilon form retains the literal integer floor. | | Both limits | SpeyerUniformNormality.result : claim | The conclusion is their conjunction for every admissible divergent rank sequence. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Matroids/SpeyerUniformNormality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2026Class 69 Mesh Patterns Are Not Equidistributed on InvolutionsThe four Class 69 length-2 mesh patterns fail equidistribution on involutions at length three.Refuted

The questionSun, On mesh patterns of short length: Equidistribution and enumeration, arXiv:2606.14367v1, Concluding remarks, Conjecture 1:

Why it mattersAt length three the avoidance counts for R0 and R2 are respectively two and one.

The ideaRepresent permutations by Equiv.Perm (Fin n) and involutions by the existing equation s s = 1.

Our contributionThe four Class 69 length-2 mesh patterns fail equidistribution on involutions at length three.

Scope & next questionFinite enumeration does not resolve this unbounded question. - Computed: the two avoidance sequences above have no matches in the specified OEIS queries.

Verification details
Release source recordRefuted in Lean

Open problem

Class 69 Mesh Patterns Are Not Equidistributed on Involutions

Q. Fang, S. Fu, S. Kitaev, H. Li, X. Su and Z. Sun, On mesh patterns of short length: Equidistribution and enumeration, arXiv:2606.14367v1, Concluding remarks, Conjecture 1:

The patterns in the set {\pattern{scale=0.5}{2}{1/1,2/2}{1/2,1/1,2/1,0/0}, \pattern{scale=0.5}{2}{1/1,2/2}{2/2,0/1,1/1,1/0}, \pattern{scale=0.5}{2}{1/1,2/2}{0/2,1/1,2/1,1/0}, \pattern{scale=0.5}{2}{1/1,2/2}{1/2,0/1,1/1,2/0}} are equidistributed on involutions. (The first two patterns, as well as the last two patterns, are trivially equidistributed via the composition of reverse and complement.)

The macro shades the unit box with lower-left corner (x,y) for each x/y in its fourth argument; the dots are at (1,1) and (2,2). Let R0, R1, R2, R3 denote those shaded-cell sets in that order. Equidistribution requires equal numbers of involutions having exactly k occurrences, for every length n, count k, and pair of pattern indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MeshPattern/Class69InvolutionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The uniform single-left-multiplier Toeplitz factorization assertionNo fixed number of ordinary Toeplitz factors suffices after a free invertible left multiplier.Refuted

The questionIgnacio García-Marco, Irene Márquez-Corbella and Daniel Seco, On the minimum number of Toeplitz factors of a matrix, arXiv:2506.16432v1, Section 5.2, Conjecture 5.1, original archive Toeplitz-GMS.tex, lines 1170–1175 (super-conjecture), states:

Why it mattersF asks whether a free invertible change of rows leaves only a bounded Toeplitz multiplication cost, uniformly in dimension, even for singular matrices.

The ideaChoose an r by r complex matrix X whose r² coordinates are algebraically independent over the rationals, and set

Our contributionNo fixed number of ordinary Toeplitz factors suffices after a free invertible left multiplier.

Verification details
Release source recordRefuted in Lean

Open problem

The uniform single-left-multiplier Toeplitz factorization assertion

Ignacio García-Marco, Irene Márquez-Corbella and Daniel Seco, On the minimum number of Toeplitz factors of a matrix, arXiv:2506.16432v1, Section 5.2, Conjecture 5.1, original archive Toeplitz-GMS.tex, lines 1170–1175 (super-conjecture), states:

There exists a constant C > 0 such that every matrix can be expressed as a product of an invertible matrix and C Toeplitz matrices. In particular, Toep_n ≤ Toep_n^Inv + C.

The source's introduction, equation (1), defines ordinary complex Toeplitz matrices by equality of adjacent southeast diagonal entries. Section 4 explicitly writes the left-multiplier constructions M = P T_1 T_2 and M = P_0 rref(M). The ambient matrices are complex and square, including singular matrices. The version was submitted 2025-06-19; the inspected arXiv history lists v1 only and no journal reference. The literature note gives the original definitions and neighboring questions.

The target F is the complete first assertion:

$$ \exists C\in\mathbb N,\quad 0<C\ \land
\forall n\in\mathbb N,\quad 0<n\ \Longrightarrow
\forall M\in\mathbb C^{n\times n},\quad \exists P\in\mathbb C^{n\times n},\ \exists T:\operatorname{Fin}(C)\to\mathbb C^{n\times n},\quad \operatorname{IsUnit}(P)\ \land
(\forall k\in\operatorname{Fin}(C),\ \operatorname{Toeplitz}(T(k)))\ \land
M=P,\operatorname{List.prod}(\operatorname{List.ofFn}(T)). $$

For every i,j : Fin n and proofs hi : i.val + 1 < n, hj : j.val + 1 < n, the Toeplitz condition is T i j = T ⟨i.val + 1, hi⟩ ⟨j.val + 1, hj⟩. This also defines the empty-size predicate; F uses positive sizes only. IsUnit P is actual matrix invertibility. The multiplier is on the left. List.ofFn orders the factors by their finite indices, and List.prod uses prod([])=1, prod(A::L)=A*prod(L), with ordinary finite-sum matrix multiplication. The factor count is independent of both n and M. Exact C and at most C are equivalent by padding with the ordinary Toeplitz identity.

The result negates this entire statement. The separately printed numerical assertion J, Toep_n ≤ Toep_n^Inv + C, remains unsettled. No converse J ⇒ F, negation of J, or unbounded global numerical gap is established. Conjectures 5.2 and 5.3 are separate and excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/Toeplitz/FactorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Gazeau Problem 4.3: no normal Hilbert realizationThe ordinary Gazeau block action on the full Schwartz core has no closed densely defined self-adjoint, skew-adjoint, or normal realization under any complete positive Hilbert metric on the fixed complex vector space.Refuted

The questionJean-Pierre Gazeau, Bi-Oriented Interlaced Matrix Multiplication and an Operator Factorisation of the Quantum Harmonic Oscillator, arXiv:2609.28511v1, Problem 4.3, asks:

Why it mattersAn arbitrary algebraic complex-linear equivalence to a complete complex Hilbert space represents the metric; continuity relative to the standard L2 norm and equivalence of norms are not assumed.

The ideaThe existing Gaussian Schwartz construction supplies the actual real Gaussian.

Our contributionThe ordinary Gazeau block action on the full Schwartz core has no closed densely defined self-adjoint, skew-adjoint, or normal realization under any complete positive Hilbert metric on the fixed complex vector space.

Verification details
Release source recordRefuted in Lean

arXiv

Gazeau Problem 4.3: no normal Hilbert realization

Jean-Pierre Gazeau, Bi-Oriented Interlaced Matrix Multiplication and an Operator Factorisation of the Quantum Harmonic Oscillator, arXiv:2609.28511v1, Problem 4.3, asks:

Is D essentially self-adjoint, skew-adjoint, or normal with respect to a natural Hilbert-space structure on L2(R) direct sum L2(R)?

The ordinary action on the full complex Schwartz core is $D_0(f,h)=(xf+i h',-i f'+xh)$. The fixed vector space is the product of the usual complex Lebesgue L2 spaces, including their almost-everywhere equivalence classes. A candidate positive Hermitian metric must make that entire vector space a complete Hilbert space.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Analysis/GazeauNormalRealization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2026The ordinary nilpotency spectrum of evolution algebrasAn ordinary nilpotency index outside the proposed binary spectrum.Refuted

The questionKaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.3, Conjecture 9 (Spectrum of nilpotency indices), state:

Why it mattersThe source links binary expansions to degree doubling in natural squares.

The ideaSet all mixed products to zero and specify the natural squares, in order, by

Our contributionAn ordinary nilpotency index outside the proposed binary spectrum.

Verification details
Release source recordRefuted in Lean

Open problem

The ordinary nilpotency spectrum of evolution algebras

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.3, Conjecture 9 (Spectrum of nilpotency indices), state:

Let $\Bbbk$ be an infinite field. An $n$-dimensional nilpotent evolution $\Bbbk$-algebra of nilpotency index $k+1$ exists if and only if either $k=2^a$ with $0\leq a\leq n-1$, or $k$ has at least two non-zero binary digits and $\lfloor\log_2 k\rfloor+s_2(k)\leq n-1$, where $s_2(k)$ is the number of non-zero binary digits of $k$.

The whole assertion quantifies over every infinite field $K$, each integer $n\geq0$ and each integer $k\geq1$. The algebra is a vector space of dimension $n$ with a $K$-bilinear product and a natural basis satisfying $e_i e_j=0$ for distinct indices. No associativity, power-associativity, unit, nondegeneracy, perfectness, nonzero-square or condition on bases of lower-power subspaces is imposed. The source's section 1.1 defines the ordinary lower powers by

$$ A^1=A,\qquad A^m=\sum_{i=1}^{m-1}A^iA^{m-i}\quad(m\geq2), $$

where a product of subspaces is the linear span of all pair products and the sum is a sum of subspaces. Thus every positive split and every parenthesization is included. Index $k+1$ means $A^{k+1}=0$ and $A^k\ne0$. The logarithm is its integer floor on positive $k$; all bounds, including $n-1$ at zero dimension, are integer bounds. Neither $k=0$ nor a zeroth power convention can settle this assertion. Both directions and both arithmetic alternatives belong to this one target.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/NilpotencySpectrumRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2026A qutrit obstruction to strong CP-filter equivalenceA unital measure-and-prepare channel on three-dimensional complex matrices has diagonal transpose range and noncommuting original range.Refuted

The questionMárquez González, Feasibility Ordering of Entanglement-Source Placement for Qubit Channels, arXiv:2609.18803v1, printed PDF p. 5, Section VI, Eq. (47), asks:

Why it mattersA counterexample at this dimension refutes the universal-dimensional assertion.

The ideaIts Choi matrix has a positive semidefinite decomposition; spectral decomposition supplies a finite Kraus family.

Our contributionA unital measure-and-prepare channel on three-dimensional complex matrices has diagonal transpose range and noncommuting original range.

Scope & next questionThe present noncommuting-range obstruction does not settle that class. - Open: strong CP-filter equivalence for other dimension-three channel classes beyond the explicit measure-and-prepare counterexample. - Open: sufficient hypotheses for strong equivalence in higher dimensions and the range of dimensions admitting this obstruction.

Verification details
Release source recordRefuted in Lean

Open problem

A qutrit obstruction to strong CP-filter equivalence

Samuel A. Márquez González, Feasibility Ordering of Entanglement-Source Placement for Qubit Channels, arXiv:2609.18803v1, printed PDF p. 5, Section VI, Eq. (47), asks:

This separates the higher-dimensional problem into two levels. The strong question is whether every unital qudit channel $\Upsilon$ is CP-filter equivalent to its transpose, i.e., whether there exist invertible CP maps $\mathcal L$ and $\mathcal R$, with CP inverses, such that $\Upsilon^T=\mathcal L\circ\Upsilon\circ\mathcal R$.

Issue #13180 preregisters the universal question for every dimension $d\geq2$ and the refutation at $d=3$. A channel is completely positive and trace preserving, with unitality $F(I)=I$. Both filters have two-sided completely positive inverses; neither filter must preserve trace. The channel transpose is the source's Eqs. (15)–(16): a finite Kraus family $K_i$ gives $F^T(X)=\sum_i K_i^T X\overline{K_i}$ in the fixed basis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/CPFilterTransposeRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationQuestion resolvedFirst preprint 2025Quantum twins in three finite classical graphsAdjacency determines algebra homomorphisms for finite twin-free graphs.Proved

The questionBochniak, Chełstowski, Kasprzak and Sołtan, “Quantum Mycielskians: symmetries, twin vertices and distinguishing labelings”, arXiv:2502.10521v2, Question 3.17, p. 13:

Why it mattersIts public general lemma, algHomeqofadjacencyeq, treats every finite twin-free simple graph and every complex target algebra.

The ideaThe public algHomeqofadjacencyeq proves equality of the homomorphisms.

Our contributionAdjacency determines algebra homomorphisms for finite twin-free graphs.

Verification details
Release source recordProved in Lean

Open problem

Quantum twins in three finite classical graphs

Bochniak, Chełstowski, Kasprzak and Sołtan, “Quantum Mycielskians: symmetries, twin vertices and distinguishing labelings”, arXiv:2502.10521v2, Question 3.17, p. 13:

Do the above-mentioned graphs possess quantum twin vertices?

The graphs are $K_2\mathbin\square K_5$, $C_4\mathbin\square C_3$ and the tesseract $Q_4$. Definition 3.8 requires two distinct actions of a compact quantum group whose compositions with the adjacency operator agree. Issue #19959 fixes the stronger algebraic claim: for each graph and every unital complex algebra $B$, no two distinct unital complex algebra homomorphisms $\mathbb C^V\to B$ agree after adjacency.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Classical function algebra and adjacency | adjacencyOperator, with function carrier $V\to\mathbb C$ and Matrix.toLin' of SimpleGraph.adjMatrix | Literal preregistered expression | | Two different actions agreeing after adjacency | HasTwinsIn retains distinctness and agreement, allowing arbitrary AlgHom targets | A strengthening: excluding all such homomorphisms excludes actions | | $K_2\mathbin\square K_5$ | prism | Mathlib complete graphs and Cartesian product | | $C_4\mathbin\square C_3$ | cycleProduct | Mathlib cycle graphs and Cartesian product | | Tesseract | The literal Hamming-distance-one graph on Fin 4 → Bool in claim | The proof uses Hypercube.hypercube through the private equality tesseract_eq_hypercube; no duplicate graph definition | | Negative answer for the three graphs | claim, proved by result | Every complex algebra $B$, without commutativity or finite-dimensionality assumptions | | General twin-free hypothesis | algHom_eq_of_adjacency_eq assumes injectivity of adjacency rows | A hypothesis of the general lemma, discharged for all three graphs in result; not an extra settlement premise |

The compact quantum group structure and the action-to-homomorphism passage are mathematical interpretation, not additional Lean declarations in this module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/BochniakChelstowskiKasprzakSoltanQuantumTwins.result

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  • Mathematical advance · 2/4 · 15% weight

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Discrete structuresCounterexample foundFirst preprint 2026The theta representation set: Question (iv), first convexity clauseThe five-cycle has a nonconvex trace-two orthonormal representation set.Refuted

The questionMarcel K Carli Silva, Gabriel Coutinho, Thiago Oliveira and Levent Tunçel, On some k-fold generalizations of Lovász theta and their sandwich theorems, arXiv:2609.06312v1, §7, Open questions, item (iv), ask:

Why it mattersThis is a recent explicit external question, the first research tier of the preregistration, rather than a repository-created assertion.

The ideaEach edge joins orthogonal unit vectors, and the common diagonal matrix is a PSD contraction of trace two.

Our contributionThe five-cycle has a nonconvex trace-two orthonormal representation set.

Scope & next questionChanging exact trace or the saturated coordinates removes hypotheses used by this proof; the outcome for those changed problems remains open here.

Verification details
Release source recordRefuted in Lean

arXiv

The theta representation set: Question (iv), first convexity clause

Marcel K Carli Silva, Gabriel Coutinho, Thiago Oliveira and Levent Tunçel, On some k-fold generalizations of Lovász theta and their sandwich theorems, arXiv:2609.06312v1, §7, Open questions, item (iv), ask:

Is the orthonormal representation set $\texttt{M}_k(G)$ convex? If not, can one characterize its closed convex lower hull, i.e., the convex corner whose support function is $\varphi_k(G,\cdot)$?

This dossier identifies only the first convexity clause of that version and item. The closed convex lower hull characterization is a separate, unresolved question. The preregistration is issue #15881.

In §5.3, the proof of Theorem 16 defines, for a finite ordinary simple graph $G=(V,E)$ and $n=|V|$,

$$ \texttt{M}k(G)=\left{x\in\mathbb R+^V:\begin{array}{l} \exists M\in\mathbb S_+^V,\quad I-M\in\mathbb S_+^V,\quad \operatorname{tr}(M)=k,\ \exists u:V\to\mathbb R^V,\quad \forall i\in V,\ \langle u_i,u_i\rangle=1,\ \forall i,j\in V,\ ij\in E\Rightarrow\langle u_i,u_j\rangle=0,\ \forall i\in V,\ x_i=u_i^{\mathsf T}Mu_i \end{array}\right}. $$

Here $\mathbb S_+^V$ consists of real symmetric positive semidefinite $V\times V$ matrices. The source's order $M\preccurlyeq I$ means $I-M\succeq0$, and its matrix inner product is $\langle A,B\rangle =\operatorname{tr}(AB)$, so $\langle I,M\rangle=k$ is exact trace equality. One common matrix and one common vector family realize every coordinate. There is no rank or projector restriction. The ambient real dimension is exactly $|V|$. In the source's convention, $u$ is an orthonormal representation of $\overline G$: orthogonality is imposed on edges of $G$, with no condition on its nonedges. Vectors at nonadjacent vertices may coincide; the family need not be a mutually orthogonal basis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Orthogonality/ThetaRepresentationNonconvex.result

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Discrete structuresCounterexample foundFirst preprint 2025Even-hypergraph inverse-positivity refutationSix four-element edges on eight vertices give inverse coefficient minus fourteen.Refuted

The questionSection 8.1 of arXiv:2507.20847v1, PDF p. 23, states: “Let 𝒢 be a simple hypergraph.

Why it mattersEven edge cardinality is proposed as a complete characterization of inverse positivity.

The ideaSix distinct four-element sets have even cardinality at least two and satisfy the inclusion clause of simplicity.

Our contributionSix four-element edges on eight vertices give inverse coefficient minus fourteen.

Verification details
Release source recordRefuted in Lean

Open problem

Even-hypergraph inverse-positivity refutation

Section 8.1 of arXiv:2507.20847v1, PDF p. 23, states: “Let 𝒢 be a simple hypergraph. Then we have I(𝒢,−x)⁻¹ ≥ 0 if and only if all edges of 𝒢 must have even number of elements.” Quotations preserve the prose verbatim; mathematical glyphs and whitespace are normalized.

Section 3.1, p. 7, begins: “Let n be a positive integer.” Definition 2(1), p. 7, says: “A hypergraph 𝒢 is called simple if for any e,f ∈ ℰ, we have |e| ≥ 2 and e ⊆ f implies e = f.” Definition 2(5), p. 7, says: “A subset I ⊆ 𝒱 is called independent if no edge of 𝒢 is entirely contained in I, i.e., e ⊈ I for all e ∈ ℰ.”

Definition 3, p. 7, includes every independent subset, including the empty set, in the ordinary independence polynomial. The coefficient ring is rational (p. 1) and positivity is coefficientwise (p. 2). Relabel [n] as Fin n. Here indicator S is one on S and zero elsewhere, and signedIndependence E is the sum, over all independent S ⊆ Fin n, of the monomial with exponent indicator S and coefficient (-1 : ℚ)^S.card. The precise claim is

∀ n : ℕ, 0 < n → ∀ E : Finset (Finset (Fin n)),
  ((∀ e ∈ E, 2 ≤ e.card) ∧
    (∀ e ∈ E, ∀ f ∈ E, e ⊆ f → e = f)) →
  ((∀ m : Fin n →₀ ℕ,
      0 ≤ MvPowerSeries.coeff m (signedIndependence E)⁻¹) ↔
    ∀ e ∈ E, Even e.card)

The inverse is in MvPowerSeries (Fin n) ℚ and all multiindices are quantified. There is no marked-vertex parameter. The scope is the full Section 8.1 equivalence, also introduced by Conjecture 1 on p. 5.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Hypergraphs/EvenInversePositivityRefutation.result

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Discrete structuresCounterexample foundSource 2026Strictly positive coefficients are not necessary for generalized action graphsThe constant sequence one admits generalized action graphs at every stage: the directed paths from zero to n.Refuted

The questionTebbe, Conditions for Building Generalized Action Graphs From Sequences, The PUMP Journal of Undergraduate Research 9 (2026), 158–179, DOI 10.46787/pump.v9i.6304, Section 4, page 178, asks:

Why it mattersTheorem 1.1 supplies sufficient sequence conditions for constructing the whole graph family.

The ideaThe full formal graph predicate expresses the source clauses as follows:

Our contributionThe constant sequence one admits generalized action graphs at every stage: the directed paths from zero to n.

Verification details
Release source recordRefuted in Lean

Open problem

Strictly positive coefficients are not necessary for generalized action graphs

S. Klanderman, K. McDicken and A. Tebbe, Conditions for Building Generalized Action Graphs From Sequences, The PUMP Journal of Undergraduate Research 9 (2026), 158–179, DOI 10.46787/pump.v9i.6304, Section 4, page 178, asks:

Another open question is whether the hypotheses of Theorem 1.1 are in fact necessary conditions for generalized action graphs to exist.

This dossier anchors exactly that question, with the strict positivity in Theorem 1.1. The theorem assumes positive integers $s_n$ for every $n\ge0$ and $z_i$ for every $i\ge1$, with $s_0=1$ and

$$ s_n=\sum_{i=1}^{n}z_i s_{n-i}\qquad(n\ge1). $$

The necessity assertion quantifies every positive integer sequence and the existence of one entire family of generalized action graphs, and concludes the normalization and existence of coefficients satisfying all these conditions. The value $z_0$ is irrelevant and unrestricted. The earlier arXiv:2507.22861v1, Section 6, is an alias of this question: its Definition 4.1 and Theorem 2.1 correspond to journal Definition 1.1 and Theorem 1.1. It is not a second target. Preregistration is #23113.

Definition 1.1, pages 158 and 172, gives the full source model:

The sequence Gn of generalized action graphs for a particular sequence sn of positive integers is a sequence of directed, labeled, rooted trees such that: Axiom 1 Define G0 as the graph with s0 vertices labeled 0 and no edges. Construct Gn from Gn−1 by adding sn new vertices, which are each labeled n. Axiom 2 For vertex v in Gn, the subtree of Gn with root v is isomorphic to some Gk such that k ≤ n. Axiom 3 All leaves in the graph Gn have label n.

The paragraph following the repeated definition on page 172 specifies:

Note that subtree of G with root v here means the subgraph of G consisting of all descendants of v and edges between them in G.

Section 3.1, page 172, specifies the label convention:

Note that when we consider isomorphism in generalized action graphs, the labels on the vertices may be shifted, but the underlying structure of the graphs are the same.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ActionGraphs/PositiveNecessity.result

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    Year in the primary source attribution: 2026.

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Number theoryQuestion resolvedEquivalent ordinary and binary carryless multiplier conditionsTwo ordinary versus binary carryless multiplier equations define the same natural numbers.Proved

The questionThe binary digits are indexed from the least significant bit.

Why it mattersThe source proposes equality of two sets specified by different ordinary and carryless multiplier pairs.

The ideaSplitting an equality into its low bit and remaining bits proves this transition rule.

Our contributionTwo ordinary versus binary carryless multiplier equations define the same natural numbers.

Scope & next questionA proposed binary-run characterization and the older neighboring 17 -> 9 question remain open in this dossier; neither is a consequence claimed by this resolution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Equivalent ordinary and binary carryless multiplier conditions

OEIS A115801 defines the nonnegative integers $k$ satisfying $9k = \operatorname{A048720}(25,k)$ and conjectures that the same integers satisfy $15k = \operatorname{A048720}(19,k)$. The exact assertion is

$$ \forall k\in\mathbb{N},\qquad 9k=C(25,k)\ \Longleftrightarrow\ 15k=C(19,k). $$

Here $\mathbb{N}$ includes zero, and the multiplications $9k$ and $15k$ are ordinary natural-number multiplication. The binary digits are indexed from the least significant bit. The operation $C$ is genuine coefficient convolution over $\mathrm{GF}(2)$:

$$ \operatorname{bit}(C(u,k),t)

\left(\sum_{i=0}^{t} \operatorname{bit}(u,i)\operatorname{bit}(k,t-i)\right)\bmod 2. $$

This is the binary polynomial multiplication table OEIS A048720, rather than ordinary multiplication.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/CarrylessMultiplierEquivalence.result

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Discrete structuresRecurrence provedFirst preprint 2026Shao–Zhao cycle Italian-domination recurrenceThe total number of Italian dominating functions on the labelled cycle Cn, for n at least three, satisfies a constant-coefficient recurrence of order five.Proved

The questionZhao, Counting Weight-k Italian Dominating Sets on Trees and Cycles, arXiv:2610.00108v1, §9 (verbatim):

Why it mattersThe source leaves the transfer-matrix recurrence open.

The ideaThe Italian condition is exactly the admissibility condition on each transition of the nine-state transfer matrix T.

Our contributionThe total number of Italian dominating functions on the labelled cycle Cn, for n at least three, satisfies a constant-coefficient recurrence of order five.

Scope & next questionSeveral directions remain open for future work: … • Deriving a complete linear recurrence for the total count Σk dI(Cn, k) using the transfer matrix formulation (the observed limiting ratio ≈ 2.7843 is the dominant eigenvalue of the corresponding transfer matrix).

Verification details
Release source recordProved in Lean

Open problem

Shao–Zhao cycle Italian-domination recurrence

P. Shao and C. Zhao, Counting Weight-k Italian Dominating Sets on Trees and Cycles, arXiv:2610.00108v1, §9 (verbatim):

Several directions remain open for future work: … • Deriving a complete linear recurrence for the total count Σ_k d_I(C_n, k) using the transfer matrix formulation (the observed limiting ratio ≈ 2.7843 is the dominant eigenvalue of the corresponding transfer matrix).

The preregistered issue is #12435. The settled statement is that, for every natural n ≥ 3, the total count a(n) satisfies a(n+5) = 2a(n+4) + 2a(n+3) + a(n+2) − a(n+1) − a(n), with a(3), …, a(7) = 23, 60, 167, 467, 1297.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/CycleItalianDominationRecurrence.result

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  • Contribution

    The total number of Italian dominating functions on the labelled cycle Cn, for n at least three, satisfies a constant-coefficient recurrence of order five.

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  • Historical context

    First preprint year: 2026.

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Quantum informationQuestion resolvedFirst preprint 2026Zhang–Li rank-one Hadamard product variational problemZhang–Li Problem 3 for matrices of rank at most one.Proved

The questionLei-Hong Zhang and Ren-Cang Li, On a Conjecture Related to Eigenvalue Perturbations, arXiv:2609.25552v1, §1 Problem 3 and §5:

Why it mattersThe question compares arbitrary changes of basis with discrete permutation matchings in a norm expression associated with eigenvalue perturbations.

The ideaThis identity supplies expansion for every positive scale.

Our contributionZhang–Li Problem 3 for matrices of rank at most one.

Scope & next questionThe rank-at-least-two question in Zhang–Li §5 remains OPEN for general unitarily invariant norms.

Verification details
Release source recordProved in Lean

Open problem

Zhang–Li rank-one Hadamard product variational problem

Lei-Hong Zhang and Ren-Cang Li, On a Conjecture Related to Eigenvalue Perturbations, arXiv:2609.25552v1, §1 Problem 3 and §5:

Problem 3: Relate min_{W nonsingular} ‖W∘G‖_ui ‖W^{-1}∘G^T‖ui to min{P permutation} ‖P∘G‖_ui^2.

In fact, for rank-one G, Romeo and Tilli [23, Theorem 1.2] proved (1.1) with c1=1 and the unitary restriction (2.3) with c2=1. They explicitly left the corresponding nonsingular version of Problem 3 unresolved; see [23, Remark 3.1].

For every natural $n$, every complex $n\times n$ matrix $G$ with $\operatorname{rank}(G)\le1$, and every unitarily invariant norm $N$, the precise conclusion is

$$ \inf_{W\in\mathrm{GL}n(\mathbb C)} N(W\circ G)N(W^{-1}\circ G^{\mathsf T}) =\min{P\text{ permutation}}N(P\circ G)^2. $$

A permutation minimizing $N(P\circ G)$ attains the infimum. The transpose of $G$ is plain transpose. The zero matrix, zero weights, and $n=0$ are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Schatten/ZhangLiRankOneProblemThree.result

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Discrete structuresQuestion resolvedFirst preprint 2026Connected graphs with ordinary domination root minus fourA rooted-component decomposition gives a general weighted domination identity.Proved

The questionSaeid Alikhani and Max Griswold, On the Integer Domination Root Conjecture, arXiv:2608.00109v1, Section 4 item 3 asks:

Why it mattersThe source's Theorem 2.1 already supplies a graph G33 with 33 vertices and 36 edges whose domination polynomial has a factor x+4.

The ideaThe general Decomposition interface permits arbitrary dimensions n,k,m, dependent branch sizes, a graph, vertex labels, roots, incidence data and an integer weight x.

Our contributionA rooted-component decomposition gives a general weighted domination identity.

Verification details
Release source recordProved in Lean

Open problem

Connected graphs with ordinary domination root minus four

Saeid Alikhani and Max Griswold, On the Integer Domination Root Conjecture, arXiv:2608.00109v1, Section 4 item 3 asks:

Does there exist an infinite family of graphs G satisfying D(G, -4) = 0?

The source uses finite simple undirected connected graphs. For ordinary domination, a subset S dominates when each vertex belongs to S or has a neighbor in S. The polynomial counts each such subset exactly once:

$$D(G,x)=\sum_{S\text{ dominates }G}x^{|S|}.$$

The exact result is: for every natural N there exist n>N and a simple graph G on Fin n such that G is connected and D(G,-4)=0. Unbounded orders ensure infinitely many isomorphism classes. Connectedness applies to G; the dominating subsets need not themselves be connected. There is no additional degree, planarity, cyclomatic-number, or root-multiplicity hypothesis.

The declaration is D5/S3/Combinatorics/Graph/Domination/MinusFourReassembly.result. Its ordinary dominating subsets are exactly DominatingSetAverage.domSets; the summands are integer powers of the integer minus four. Both existential witnesses follow the universal N, and the inequality is strict.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/MinusFourReassembly.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

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  • Mathematical advance · 3/4 · 15% weight

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    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2009Monotonicity and exact ranges of the shifted Kummer ratioStrict monotonicity and exact ranges of Kummer ratios with shifted upper parameters.Proved

The questionKarp and Sitnik, Log-convexity and log-concavity of hypergeometric-like functions, arXiv:0902.3073v1, the unnumbered Conjecture after Theorem 4:

Why it mattersThe conclusion determines the full variation of the ratio in the real argument, including the sharp Gamma endpoints in both parameter regimes.

The ideaThe comparison uses positivity on the closed positive ray, continuity of the solution and its first derivative there, and two derivatives satisfying the equation on the positive ray.

Our contributionStrict monotonicity and exact ranges of Kummer ratios with shifted upper parameters.

Verification details
Release source recordProved in Lean

Open problem

Monotonicity and exact ranges of the shifted Kummer ratio

Karp and Sitnik, Log-convexity and log-concavity of hypergeometric-like functions, arXiv:0902.3073v1, the unnumbered Conjecture after Theorem 4:

Conjecture. The ratio in the middle of (10) is monotone decreasing from (0, ∞) onto (1, A), where A is the left-hand side of (10); the ratio in the middle of (11) is monotone increasing from (−∞, 0) onto (B, 1), where B is the left-hand side of (11).

Write $M(s;c;x)={}_1F_1(s;c;x)$ and

$$R(x)=\frac{M(b+\delta;c;x)M(a;c;x)} {M(a+\delta;c;x)M(b;c;x)}.$$

Theorem 4 assumes $\delta>0$ throughout. Its inequality (10), for $b>a>0$ and $c,x>0$, is

$$\frac{\Gamma(a+\delta)\Gamma(b)} {\Gamma(b+\delta)\Gamma(a)}<R(x)<1.$$

Its inequality (11), for $a<b<c-\delta$ and $c>0>x$, is

$$\frac{\Gamma(c-a-\delta)\Gamma(c-b)} {\Gamma(c-b-\delta)\Gamma(c-a)}<R(x)<1.$$

The printed “onto (1, A)” names the endpoints in the order $x\to0^+$, $x\to+\infty$. The formal image is the open interval $(A,1)$. On the negative ray the image is $(B,1)$, with the displayed Gamma quotient as $B$.

Formal reading

All Lean names below have prefix D5.S3.Analytic.KummerRatio.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $(s)_n/((c)_n n!)$ | Defs.kummerCoeff | Ascending Pochhammer evaluations divided by the lower Pochhammer evaluation and factorial. | | $M(s;c;x)={}_1F_1(s;c;x)$ | Defs.kummer | The infinite real series with these coefficients. | | Middle ratio in (10) and (11) | Defs.ratio | Exactly $M(b+\delta;c;x)M(a;c;x)/(M(a+\delta;c;x)M(b;c;x))$. | | Positive-ray clause | Defs.claim, Result.result | Every $c,\delta>0$ and $0<a<b$; strict decrease and image $(A,1)$. | | Negative-ray clause | Defs.claim, Result.result | Every $c,\delta>0$ and $a<b<c-\delta$; strict increase and image $(B,1)$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KummerRatio/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 3/4 · 10% weight

    First preprint year: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedStrict First Sums and Gapfree Odd PartitionsStrict zero-prepended first sums correspond to gapfree odd partitions of the same weight.Proved

The questionConjecture: Reversed integer partitions of this type (with strictly increasing 0-based partial alternating sums) are counted by A053251.

Why it mattersThis is a first-tier 2026 OEIS comment conjecture.

The ideaUse the existing firstSums definition and the existing weak correspondence.

Our contributionStrict zero-prepended first sums correspond to gapfree odd partitions of the same weight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Strict First Sums and Gapfree Odd Partitions

OEIS A392707, revision 10, states:

Conjecture: Reversed integer partitions of this type (with strictly increasing 0-based partial alternating sums) are counted by A053251.

The formal target here is the self-contained combinatorial bridge: for every n, the number of partitions with a strictly increasing positive preimage s under firstSums(0::s) equals the number of partitions whose support is exactly {1,3,...,2k-1} for some k. The two classes include the empty partition. The further equality with mock theta coefficients is not formalized here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/StrictFirstSumsGapfreeOddParts.card_strictFirstSums_eq_gapfreeOdd

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedWiseman's nonchoosable partition identityThe two partition predicates give equal counts.Proved

The questionThe source expressions are retained in the two count definitions.

Why it mattersThe source is documented in Library/PermutationPatterns/oeis2025wisemanpartitions.md.

The ideaChoosableInitialIntervals.choosableInitialcongrperm transports choosability between the descending and ascending multiset sorts. ChoosableInitialIntervals.choosableInitialiffsuperdiagonal applies to the ascending sort.

Our contributionThe two partition predicates give equal counts.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Wiseman's nonchoosable partition identity

OEIS A388712 states: “Triangle read by rows where T(n,k) is the number of integer partitions of n into k parts without choosable initial intervals.” Its conjecture is: “Also the number of non-superdiagonal reversed partitions of n into k parts, where a sequence (y_1, ..., y_k) is superdiagonal iff y_i >= i for all i = 1..k.” The question is preregistered in #18601.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Partitions of $n$ into $k$ positive parts | Nat.Partition n, filtered by p.parts.card = k | | Initial intervals admit distinct positive representatives | The imported ChoosableInitial predicate on descending parts | | Without choosable initial intervals | Negation in WisemanA388712.T | | Reversed partition, weakly increasing | p.parts.sort (· ≤ ·) | | $y_i\ge i$ with one-based indices | The imported Superdiagonal predicate uses zero-based index plus one | | Non-superdiagonal, with the same number of parts | Negation and the same cardinality filter in WisemanA388712.nonSuperdiagonal | | Equality for every $n,k$ | WisemanA388712.claim, proved by WisemanA388712.result |

All declaration names above have namespace prefix D5.S1.Words.Compositions.. The source expressions are retained in the two count definitions. The existing predicates are reused without new aliases. Classical finite cardinalities do not restrict the quantified natural numbers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/WisemanA388712.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2023Advantage-quantum measurements are not quantum n-simulableNo binary advantage-quantum measurement admits an exact quantum simulation on any finite number of copies.Proved

The questionArai, The Diversity of Entanglement Structures with Self-Duality and Non-Orthogonal State Discrimination in General Probabilistic Theories, Ph.D. thesis, arXiv:2302.12361v1, Section 5.2 (Open Problems), Theme B:

Why it mattersThe thesis studies entanglement structures beyond standard quantum theory and asks whether their measurements can be reproduced inside quantum theory by acting on several copies of the state.

The ideaThe proof uses spectral estimates.

Our contributionNo binary advantage-quantum measurement admits an exact quantum simulation on any finite number of copies.

Verification details
Release source recordProved in Lean

arXiv

Advantage-quantum measurements are not quantum n-simulable

H. Arai, The Diversity of Entanglement Structures with Self-Duality and Non-Orthogonal State Discrimination in General Probabilistic Theories, Ph.D. thesis, arXiv:2302.12361v1, Section 5.2 (Open Problems), Theme B:

Similarly, it is considered that AQ measurement might be non-simulable because AQ measurement also has an extraordinary performance. This is an open problem.

Here:

  • DOVM: a two-outcome measurement ${M_1,M_2}$ on $\mathcal H_A\otimes\mathcal H_B$ whose effects are block positive (nonnegative on product vectors) and sum to $I$.
  • AQ: one effect satisfies $\lambda_1(M_i)<0$ and $1+\lambda_1(M_i)<\lambda_d(M_i)<1$.
  • Domain: the states $\rho$ with $\mathrm{tr},\rho M_i\ge0$ for both $i$.
  • Quantum $n$-simulable: some POVM ${N_i}$ on $(\mathcal H_A\otimes\mathcal H_B)^{\otimes n}$ has $\mathrm{tr},\rho^{\otimes n}N_i=\mathrm{tr},\rho M_i$ on the domain.

The definitions are quoted in the literature note.

Issue #20035 fixes the reading: no AQ DOVM is quantum $n$-simulable for any $n$, with the simulation identity required only on trace-one domain states, which makes the statement stronger than the literal domain. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/AdvantageQuantumNonSimulability.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Four MUBs detect a PPT state in dimension eightFour mutually unbiased bases in dimension eight give a negative expectation on a positive-partial-transpose density matrix for the witness of Bae, Bera, Chruściński, Hiesmayr and McNulty.Refuted

The questionMcNulty, How many mutually unbiased bases are needed to detect bound entangled states?, Journal of Physics A 55 (2022), 505303, arXiv:2108.01109v2, Section VIII, Conjecture 1:

Why it mattersThis is the source's measurement-resource question for its own witness.

The ideaCongruence, finite sums, submatrices and nonnegative scaling produce rhodensity and rhoppt.

Our contributionFour mutually unbiased bases in dimension eight give a negative expectation on a positive-partial-transpose density matrix for the witness of Bae, Bera, Chruściński, Hiesmayr and McNulty.

Verification details
Release source recordRefuted in Lean

Open problem

Four MUBs detect a PPT state in dimension eight

J. Bae, A. Bera, D. Chruściński, B. C. Hiesmayr and D. McNulty, How many mutually unbiased bases are needed to detect bound entangled states?, Journal of Physics A 55 (2022), 505303, arXiv:2108.01109v2, Section VIII, Conjecture 1:

Conjecture 1. If d = 2^r and r > 1, the minimal number of MUBs required for bound entanglement detection equals m = d/2 + 1 = 2^{r−1} + 1.

The preceding sentence specifies the witness $W^\Gamma(\mathcal M_m,s)$. Equation (4) defines $W(\mathcal M_m,s)=((d+m-1)/d)I-B(\mathcal M_m,s)$; partial transposition conjugates the second factor of each equal-column tensor projector. The computational term pairs $\ell$ with $\ell+s$ modulo $d$. The source's Eqs. (2)–(3) establish nonnegative expectation on separable states. This source result supplies the interpretation of negative expectation as entanglement detection; this module does not formalize separability or the separable-state bound.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $d=2^r$, $r>1$ | claim quantifies over natural r with 2 ≤ r and dimension Fin (2 ^ r) | The same integer range | | Fewer than $2^{r-1}+1$ MUBs cannot detect bound entanglement | Every natural m below the threshold in claim | Necessity half of the minimum equality | | Pairwise mutually unbiased orthonormal bases | IsMUBFamily: unitary matrices and all distinct-basis squared overlaps equal to $1/d$ | Columns are the basis vectors | | Canonical basis $B_0$ | IsMUBFamily sets basis zero to the identity whenever 0 < m | Source convention; no nonempty-family assumption is added | | Source witness and modular shift | witnessGamma with every s : Fin d | Literal Eq. (4) after partial transposition; column projectors use Matrix.vecMulVec directly | | PPT density matrix with negative expectation | DetectsBoundEntanglement uses the existing IsDensity and partialTransposeB, and the real trace is strictly negative | Literal binding convention in #16454 | | Refutation of the universal necessity statement | result : ¬ claim, using $r=3$, $m=4$, $s=1$ | One counterexample refutes the equality; optimality of four is not asserted |

The four public definitions have complete defining formulas in the Blueprint. Natural subtraction in $r-1$ is harmless under $2\le r$. The predicates are total at zero dimension and zero family size, but the refutation uses dimension eight and four bases. The computational basis convention is from the source, and positivity, normalization and strict negativity all remain requirements.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/FourMUBBoundEntanglementRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2024Problem 5 of Brešar, Hedžet and Herrman: m(C_n × P_m, 2)In 2-neighbour bootstrap percolation on the direct product of the cycle Cn and the path Pm, the least size of a percolating set is n for every n at least 3 and m at least 1.Proved

The questionIn r-neighbour bootstrap percolation a vertex becomes infected once it has at least r infected neighbours, and m(G, r) is the least size of a nonempty initial set that eventually infects every vertex.

The ideaThe layer {(a, 0)} percolates: (a, b + 1) has the two distinct neighbours (a ± 1, b) in layer b.

Our contributionIn 2-neighbour bootstrap percolation on the direct product of the cycle Cn and the path Pm, the least size of a percolating set is n for every n at least 3 and m at least 1.

Verification details
Release source recordProved in Lean

Open problem

Problem 5 of Brešar, Hedžet and Herrman: m(C_n × P_m, 2)

In r-neighbour bootstrap percolation a vertex becomes infected once it has at least r infected neighbours, and m(G, r) is the least size of a nonempty initial set that eventually infects every vertex. For the direct product C_n × P_m of a cycle and a path, Brešar, Hedžet and Herrman (arXiv:2403.10957, Discuss. Math. Graph Theory 46 (2026)) ask:

Problem 5. Determine m(C_n × P_m, 2).

Their Proposition 3 gives m(C_n × P_m, 2) ≤ n, and they note that equality appears to hold for small examples.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Percolation/DirectProductCyclePathBootstrap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2024Divisibility of connected encoder equivalence classesThe commuting order-three rotations of the matched input-pivot pairs act freely on each connected class of normalized encoders.Proved

The questionKhesin and Li, Equivalence Classes of Quantum Error-Correcting Codes, arXiv:2406.12083v1, Conjecture 7.1, state:

Why it mattersThe conjecture predicts a divisibility law in the authors' encoder-class tabulations.

The ideaApply local complementation first at an input and then at its matched pivot.

Our contributionThe commuting order-three rotations of the matched input-pivot pairs act freely on each connected class of normalized encoders.

Verification details
Release source recordProved in Lean

Open problem

Divisibility of connected encoder equivalence classes

Khesin and Li, Equivalence Classes of Quantum Error-Correcting Codes, arXiv:2406.12083v1, Conjecture 7.1, state:

For positive integers n > k, the equivalence classes of [n,k] codes with a single connected component have sizes divisible by 3^k.

The proposition concerns the labelled normalized encoders counted in Section 7. Every normalized member of the class is required to be connected, as in the figures' restriction to classes in which every graph is prime. Completeness of the operations as a characterization of physical code equivalence is a separate question, Conjecture 6.3.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive integers $n>k$ | Natural parameters in claim, with 0 < k and k < n | Both positivity and strict inequality are retained. | | Encoder graph with $k$ inputs and $n$ outputs | SimpleGraph (Fin k ⊕ Fin n) | Inputs and outputs have distinct labels. | | Operations 1–5 of Conjecture 6.3 | Step | Local complementation, separate input/output permutations, modulo-two row addition and input-edge removal are explicit constructors. | | Removed output Clifford decorations | Decoration-free carrier | Operation 6 has no graph action; this is the source's normalized convention. | | Claim 6.4 normalized adjacency matrices | Normalized | No input-input or pivot-pivot edges; the first $k$ output columns form the identity block. | | One operation followed by renormalization | NormalizedStep, using Step and Relation.ReflTransGen RenormStep | Normalization is required at both endpoints; an empty renormalization sequence is allowed. | | Equivalence classes of distinct normalized encoders | Relation.EqvGen NormalizedStep and classSize | The subtype is counted by Nat.card; no quotient by graph isomorphism is taken. | | A single connected component | The hypothesis that every normalized class member is Connected | This is the preregistered interpretation of the source's connected-class restriction. | | Size divisible by $3^k$ | result proves claim | No additional conclusion is assumed. |

The source's XOR expression for local complementation is written literally in Step.complement. The consumed private correspondence lc_source identifies it with the existing graph operation GraphStateMonogamyForbiddenSubgraph.lc used in the proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/KhesinLiEquivalenceClasses.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2022The C*-algebraic Schoenberg Conjecture is falseA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.Refuted

The questionMahesh Krishna, "C-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2.

Why it mattersThe same cubic matrix polynomial settles this conjecture and the two linked conjectures in the de Bruin-Sharma dossier (krishna-2022-cstar-de-bruin-sharma.md) and the Kushel-Tyaglov dossier (krishna-2022-cstar-kushel-tyaglov.md).

The ideaThe first inequality has right side minus left side D2.1^(1)=-2t^2 E11/9.

Our contributionA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

The C*-algebraic Schoenberg Conjecture is false

K. Mahesh Krishna, "C*-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2. The source text is quoted verbatim:

**Conjecture 2.1 (C*-algebraic Schoenberg Conjecture).** Let $\mathcal{A}$ be a C*-algebra. Let $d\in \mathbb{N}\setminus{1}$, $P(z) \coloneqq (z-a_1)(z-a_2)\cdots (z-a_d)$ be a polynomial over $\mathcal{A}$ with $a_1, a_2, \dots, a_d \in \mathcal{A}$. If $P'$ can be written as $P'(z)= d(z-b_1)(z-b_2)\cdots (z-b_{d-1})$ on $\mathcal{A}$ with $b_1, b_2, \dots, b_{d-1} \in \mathcal{A}$, then $$\sum_{k=1}^{d-1}b_kb_k^*\leq \frac{1}{d^2}\left[\sum_{j=1}^{d}a_j\right]\left[\sum_{j=1}^{d}a_j\right]^*+ \frac{d-2}{d}\sum_{j=1}^{d}a_ja_j^*$$ and $$\sum_{k=1}^{d-1}b_k^*b_k\leq \frac{1}{d^2}\left[\sum_{j=1}^{d}a_j\right]^*\left[\sum_{j=1}^{d}a_j\right]+ \frac{d-2}{d}\sum_{j=1}^{d}a_j^*a_j.$$

The ordered derivative is defined in the source immediately before the conjecture. The encoding uses zero-based Fin indices, complex scalar coefficients, star, and the C*-order. It restricts the source to unital C*-algebras in Type with PartialOrder and StarOrderedRing, and d >= 2; this subclass is enough for a counterexample to the source's universal assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarSchoenberg.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture provedFirst preprint 2024Louvet symmetric absolutely-PPT spectrum conjectureThe conjectured symmetric qudit spectrum is SAPPT at and above its threshold.Proved

Why it mattersThe paper leaves this explicitly stated qudit conjecture after proving the qubit case.

The ideaThe permutation symmetrizer has a positive partial-transpose floor on every site bipartition.

Our contributionThe conjectured symmetric qudit spectrum is SAPPT at and above its threshold.

Verification details
Release source recordProved in Lean

Open problem

Louvet symmetric absolutely-PPT spectrum conjecture

J. Louvet, E. Serrano-Ensástiga, T. Bastin and J. Martin, Nonequivalence between absolute separability and positive partial transposition in the symmetric subspace, arXiv:2411.16461v2 and Physical Review A 111, 042418 (2025), state Conjecture 1: every symmetric $N$-qudit state with

$$D = \binom{N+d-1}{d-1}$$

non-zero eigenvalues of the form

$$1-(D-1)p/D,\quad p/D,\ldots,p/D,$$

where $p \in [p_{\min},1]$ and

$$p_{\min}=1/(1+2[D\binom{N}{\lfloor N/2\rfloor}]^{-1}),$$

is symmetric absolutely PPT. The formal claim uses Matrix.PosSemidef, the symmetric-column predicate, the full-space characteristic polynomial with $d^N-D$ zero roots, and positivity of every site partial transpose.

Statement echo

| source clause | Lean counterpart | fidelity | | --- | --- | --- | | symmetric $N$-qudit state | IsSymmetricOperator ρ together with ρ.PosSemidef | exact quantified predicate | | $D=\binom{N+d-1}{d-1}$ | D N d | literal definition | | $p_{\min}=1/(1+2[D\binom{N}{\lfloor N/2\rfloor}]^{-1})$ | pmin N d | literal definition with Nat.div | | prescribed non-zero spectrum and $d^N-D$ zero roots | the ρ.charpoly equality in claim | exact characteristic-polynomial factors | | $p\in[p_{\min},1]$ | pmin N d ≤ p and p ≤ 1 | exact interval hypotheses | | symmetric absolutely PPT | IsSAPPT ρ | exact quantified conclusion |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsoluteSeparability/LouvetSymmetricAbsolutelyPPT.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2023Non-backtracking Laplacian gap and minimum degreeThe non-backtracking transition spectrum has gap at most the reciprocal minimum-degree excess.Proved

The questionZucal, There is no going back: Properties of the non-backtracking Laplacian, arXiv:2303.00373v2, Section 5, Conjecture 5.3:

The ideaThe proof uses spectral estimates.

Our contributionThe non-backtracking transition spectrum has gap at most the reciprocal minimum-degree excess.

Verification details
Release source recordProved in Lean

Open problem

Non-backtracking Laplacian gap and minimum degree

R. Mulas, D. Zhang and G. Zucal, There is no going back: Properties of the non-backtracking Laplacian, arXiv:2303.00373v2, Section 5, Conjecture 5.3:

If $G$ is not the cycle graph, then $\varepsilon\leq\frac{1}{\delta-1}$.

The source assumes a finite simple graph of minimum degree $\delta\ge2$. Its transition matrix is $P=\mathcal D^{-1}\mathcal A$, and its non-backtracking Laplacian is $\mathcal L=I-P$. The gap from $1$ is $\varepsilon=\min_{\lambda\in\sigma(P)}|\lambda|$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/NonBacktracking/LaplacianGapMinDegree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024A strict gap between measured and Uhlmann average fidelitiesThree qubit states have measured average pairwise fidelity above the Uhlmann average.Proved

The questionWilde, Multivariate fidelities, arXiv:2404.16101v3, Remark 5.45, page 34, ask:

Why it mattersThis external named question has no digestion atom.

The idea3. scalareffectbound and effectlowerbound scale this estimate to arbitrary positive effects. uhlmanndiagonal identifies the classical measured fidelity. povmpairsumbound sums the estimates over any finite POVM: the scalar coefficients sum to one and both Bloch coefficients sum to zero.

Our contributionThree qubit states have measured average pairwise fidelity above the Uhlmann average.

Verification details
Release source recordProved in Lean

Open problem

A strict gap between measured and Uhlmann average fidelities

T. Nuradha, H. K. Mishra, F. Leditzky and M. M. Wilde, Multivariate fidelities, arXiv:2404.16101v3, Remark 5.45, page 34, ask:

Furthermore, it is interesting to determine whether there exists a tuple of states for which F_M(ρ_1, …, ρ_r) > F_U(ρ_1, …, ρ_r)—we leave this as an open question.

The answer is affirmative. For the three qubit states $R=I/2$, $A=(I+Z/2)/2=\operatorname{diag}(3/4,1/4)$ and $B=(I+X/2)/2$, the proof gives $F_U(R,A,B)<24/25\le F_M(R,A,B)$. Here $F(\rho,\sigma)=|\sqrt\rho\sqrt\sigma|_1$, $F_U$ is the average over unordered pairs, and $F_M$ takes one infimum over all finite POVMs of every outcome size.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Proposition 3.1: root fidelity | uhlmann | The existing trace norm applied to the product of positive square roots. | | Equation (5.13): average over unordered pairs | FU | Real normalization $2/(r(r-1))$ and Finset.Ioi. | | Definition 5.40: the measurement channel | The two diagonal trace expressions in FM | The literal measurement expression is inlined; no second channel definition. | | Infimum over finite POVMs of every size | The dependent-sum index in FM | A single real infimum over size and POVM together; empty outcome sizes do not supply artificial zero costs. | | Existence of a tuple with strict inequality | claim, result | Natural dimensions and tuple sizes and positive trace-one matrix states; witness dimension two and tuple size three. |

The public definition bodies are the binding expressions of #19068, with the measurement action expanded. No hypothesis is added to the existential question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/NMLWMeasuredFidelitySeparation.result

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedFirst preprint 2022Minimum Seidel energy of chain graphsEvery connected chain graph has at least the Seidel energy of each member of Gamma(n,k), with equality precisely for the members of that family.Proved

The questionSantanu Mandal, Ranjit Mehatari and Kinkar Chandra Das, On the spectrum and energy of Seidel matrix for chain graphs, arXiv:2205.00310v1, Conjecture 4.14:

Why it mattersThe source and literal conventions are recorded in 21681 and Library/GraphInvariants/mandalmehataridas2022seidelchain.md.

The ideaSign switching merges the first 0-run with the last 1-run.

Our contributionEvery connected chain graph has at least the Seidel energy of each member of Gamma(n,k), with equality precisely for the members of that family.

Verification details
Release source recordProved in Lean

Open problem

Minimum Seidel energy of chain graphs

Santanu Mandal, Ranjit Mehatari and Kinkar Chandra Das, On the spectrum and energy of Seidel matrix for chain graphs, arXiv:2205.00310v1, Conjecture 4.14:

Let $G$ be a chain graph of order $n$ with the binary string $b=0^{s_1}1^{t_1}0^{s_2}\ldots0^{s_k}1^{t_k}$ such that $\sum_i(s_i+t_i)=n$, $s_i\ge1$ and $t_i\ge1$. Then $SE(G)\ge SE(H)$, where $H\in\Gamma_{n,k}$. Moreover, the equality holds if and only if $G\in\Gamma_{n,k}$.

Here $\Gamma_{n,k}$ has binary word $0^{s_1}(10)^{k-1}1^{t_k}$ with positive endpoints, each at most $n-2k+1$, and $s_1+t_k=n-2k+2$. Seidel matrices are $J-I-2A$ and energy sums the absolute eigenvalues with multiplicity. The theorem quantifies over every comparator $H$.

Statement echo

The local names below belong to D5.S3.Combinatorics.Graph.KyFanGap.MandalMehatariDasSeidelMinimum; the five graph conventions lie in its MandalMehatariDasSeidel namespace.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive alternating runs $0^{s_1}1^{t_1}\cdots0^{s_k}1^{t_k}$ | binaryString, positive hypotheses in claim | All run lengths and their order are retained. | | A later 1 is adjacent precisely to earlier 0-vertices | chainGraph | The symmetric adjacency relation has no loops. | | Order $n$ | Sum hypothesis in claim | The vertices are the positions of the binary list; its length is the sum of the runs. | | $S=J-I-2A$ | seidel | Literal all-ones matrix, identity and scalar adjacency matrix. | | $SE$ is the sum of absolute eigenvalues | energy, energy_eigenvalues | Real characteristic roots with multiplicity; every graph matrix used is real symmetric. | | Every $H\in\Gamma_{n,k}$ | Universal list h and Gamma n k h in claim | No representative comparator is substituted. | | Minimum and equality precisely in $\Gamma_{n,k}$ | Both conclusions of claim, proved by MandalMehatariDasSeidel.result | Full inequality and equivalence, including $k=1$. | | $k\ge1$ and $2k\le n$ | Explicit guards in claim | These follow from nonempty positive runs and introduce no source restriction. | | Finite ordered vertices | Fin and the finite type instances | These encode the finite graph and its construction order. |

Standalone Gamma outside $1\le k$ and $2k\le n$ is outside the source interpretation: natural subtraction truncates. The settling proposition has both guards. The literal binary and Seidel conventions are retained; the proofs also use shared spectral owners.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/KyFanGap/MandalMehatariDasSeidelMinimum.result

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2015Optimal quantum states depend on the Rényi ordersA qutrit basis projector is Pareto optimal for the Shannon dual pair (1,1), but ceases to be optimal for the dual pair (3/5,3).Refuted

The questionKais Abdelkhalek, René Schwonnek, Hans Maassen, Fabian Furrer, Jörg Duhme, Philippe Raynal, Berthold-Georg Englert and Reinhard F.

Why it mattersOptimality at the first pair is proved against every density state, including mixed competitors.

The ideaOnly the first projector weakly dominates the first projector at (1,1).

Our contributionA qutrit basis projector is Pareto optimal for the Shannon dual pair (1,1), but ceases to be optimal for the dual pair (3/5,3).

Verification details
Release source recordRefuted in Lean

Open problem

Optimal quantum states depend on the Rényi orders

Kais Abdelkhalek, René Schwonnek, Hans Maassen, Fabian Furrer, Jörg Duhme, Philippe Raynal, Berthold-Georg Englert and Reinhard F. Werner, “Optimality of entropic uncertainty relations”, arXiv:1509.00398v1, section V.E, Conjecture V.8 (Independence of the optimal states of $(\alpha,\beta)$), state:

If $\rho$ is an optimal state for any unitary operator and any $\alpha,\beta>\frac12$ satisfying the duality relation (2), then $\rho$ is also an optimal state for all other dual pairs.

The relation is $1/\alpha+1/\beta=2$. Section II defines $f(\rho)=(H_\alpha(p_X^\rho),H_\beta(p_Y^\rho))$ and $\rho\sqsubseteq\sigma$ by both entropy coordinates being at most those of $\sigma$. Its definition of optimality is:

We call a state $\rho$ optimal if $\rho'\sqsubseteq\rho$ implies $\rho\sqsubseteq\rho'$, and hence $f(\rho)=f(\rho')$.

The formal claim quantifies over all natural dimensions, complex unitaries, complex positive trace-one density matrices, and both finite real dual pairs with every order strictly above $1/2$. The source explicitly excludes ${1/2,\infty}$. Issue #11782 preregisters this reading and the refutation before any Lean probe.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/RenyiOptimalStateDependenceRefutation.result

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Discrete structuresCounterexample foundPosed in 2014Refutation of Detlefs's A000040 Fibonacci-Fermat prime characterizationThe composite 219781 refutes Detlefs's Fibonacci-Fermat prime characterization.Refuted

The questionGary Detlefs's FORMULA of May 25, 2014 (verbatim):

Why it mattersThe May 25 formula identifies the prime numbers exactly with five and the non-five values satisfying both a Fibonacci residue test and a Fermat residue test.

The ideaA ZMod 219781 computation by reducemodchar, followed by the natural-cast congruence, establishes 2^219780 mod 219781 = 1.

Our contributionThe composite 219781 refutes Detlefs's Fibonacci-Fermat prime characterization.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Detlefs's A000040 Fibonacci-Fermat prime characterization

OEIS A000040, NAME (verbatim):

The prime numbers.

Gary Detlefs's FORMULA of May 25, 2014 (verbatim):

Conjecture: Sequence = {5 and n <> 5| ( Fibonacci(n) mod n = 1 or Fibonacci(n) mod n = n - 1) and 2^(n-1) mod n = 1}. - Gary Detlefs, May 25 2014

His FORMULA of May 28, 2014 (verbatim):

Conjecture: Sequence = {5 and n <> 5| ( Fibonacci(n) mod n = 1 or Fibonacci(n) mod n = n - 1) and 2^(3n) mod 3n = 8}. - Gary Detlefs, May 28 2014

For natural numbers, define fibTest(n) := (Fibonacci(n) mod n = 1) or (Fibonacci(n) mod n = n-1), define fermatTest(n) := 2^(n-1) mod n = 1, and define inDetlefsSet(n) := (n=5) or (n!=5 and fibTest(n) and fermatTest(n)). The literal refuted statement is claim := for every natural n, 0<n implies (Prime(n) iff inDetlefsSet(n)).

Only the May 25 characterization is claimed and refuted here. The May 28 variant is disclosed but not claimed, although 219781 also satisfies its congruence. Detlefs's September 10, 2010 Wilson-type characterizations in the same FORMULA block are not claimed or assessed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/DetlefsFibonacciFermatPrimeCharacterizationRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2023The minimum of the long-time algebra OTOCFor a block structure with sectors of type (nJ, dJ), J = 1, ..., r, and D = sumJ nJ dJ, the non-resonant long-time average of the algebra OTOC over all eigenbases of the D-dimensional Hilbert space is at least 1 - (sumJ dJ + sumJ nJ - r) / D, and the product basis attains this value.Proved

The questionZanardi, Long-time Quantum Scrambling and Generalized Tensor Product Structures, Phys.

The ideaThe proof uses spectral estimates.

Our contributionFor a block structure with sectors of type (nJ, dJ), J = 1, ..., r, and D = sumJ nJ dJ, the non-resonant long-time average of the algebra OTOC over all eigenbases of the D-dimensional Hilbert space is at least 1 - (sumJ dJ + sumJ nJ - r) / D, and the product basis attains this value.

Verification details
Release source recordProved in Lean

Open problem

The minimum of the long-time algebra OTOC

F. Andreadakis, E. Dallas, P. Zanardi, Long-time Quantum Scrambling and Generalized Tensor Product Structures, Phys. Rev. A 109, 052424 (2024), arXiv:2312.13386v2, the Conjecture of Section IV:

Let $H$ be an NRC Hamiltonian and $\mathcal{A}f$ be the equivalence class of isomorphic algebras of the form $\mathcal{A}\cong \oplus{J=1}^{d_\mathcal{Z}} \mathds{1}{n_J} \otimes \mathcal{L}(\mathbb{C}^{d_J} )$. Then, the algebra corresponding to a gTPS (\cref{structure_H}) for which the Hamiltonian eigenstates have the form ${\ket{\psi_J}\otimes\ket{\phi_J} , \vert , J=1,\dots d\mathcal{Z}; , \psi_J = 1,\dots, n_J; , \phi_J = 1,\dots , d_J }$ minimizes the $\mathcal{A}$-OTOC LTA over the family of isomorphic algebras. The minimum value is $\overline{G_{\mathcal{A}}}^{NRC}{min}=1-\frac{1}{d} \left( \sum_J d_J + \sum_J n_J - d\mathcal{Z} \right).$

The long-time average under the non-resonance condition (NRC) is Eq. (6) (Proposition 1 (ii)): for the eigenbasis ${\phi_k}$ of $H$ and $\Pi_k=|\phi_k\rangle\langle\phi_k|$, $$\overline{G}^{NRC}=1-\frac1d\sum_{\mathcal X\in{\mathcal A,\mathcal A'}} \Big(\operatorname{Tr}(R^{(0),\mathcal X}R^{(1),\mathcal X'})-\tfrac12 \operatorname{Tr}(R^{(0),\mathcal X}DR^{(1),\mathcal X'}D)\Big),$$ with $R^{(0),\mathcal X}{lk}=|\mathbb P{\mathcal X}(|\phi_k\rangle \langle\phi_l|)|2^2$, $R^{(1),\mathcal X}{kl}=\langle\mathbb P_{\mathcal X}(\Pi_k),\mathbb P_{\mathcal X}(\Pi_l)\rangle$, $\mathbb P_{\mathcal A}=\oplus_J\frac{\mathbb 1_{n_J}}{n_J}\otimes \operatorname{Tr}{n_J}$ and $\mathbb P{\mathcal A'}=\oplus_J \operatorname{Tr}{d_J}\otimes\frac{\mathbb 1{d_J}}{d_J}$. The paper proves the minimum for Hamiltonians that are block diagonal in the sectors, and for $d_\mathcal Z=1$ over all NRC Hamiltonians (Proposition 2); its Section IV A reports numerical evidence up to $d=40$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/AlgebraOTOCLongTimeMinimum.result

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Discrete structuresQuestion resolvedFirst preprint 2024Greatest Pell indices dividing initial partial sumsPell partial sums have greatest dividing indices in four residue classes.Proved

The questionByrapuram et al., Fibonacci Partial Sums Tricks, arXiv:2409.01296v1, Section 4.5, states verbatim:

Why it mattersIt is a preregistration locator rather than proof evidence.

The ideaReuse the Pell and companion definitions P, Q and their coupled recurrence, same-index coprimality and companion oddness.

Our contributionPell partial sums have greatest dividing indices in four residue classes.

Verification details
Release source recordProved in Lean

Open problem

Greatest Pell indices dividing initial partial sums

Byrapuram et al., Fibonacci Partial Sums Tricks, arXiv:2409.01296v1, Section 4.5, states verbatim:

Conjecture 18. For the Pell sequence, we have m^P_{4k+1}=m^P_{4k+2}=1, m^P_{4k-1}=2k, and m^P_{4k}=2k+1.

Here P(0)=0, P(1)=1, P(n+2)=2P(n+1)+P(n), and S(n)=P(1)+...+P(n). For positive n, m_n^P is the greatest positive j satisfying P(j)|S(n). The first two equalities hold for every natural k; the latter two hold for every k>=1. The formal predicate is IsGreatest {j | 1<=j and P(j)|S(n)} m, so divisibility, positivity of the selected index, and maximality are all part of each clause.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/PellPartialSumMaxIndex.result

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Discrete structuresExact formula provedFirst preprint 2023Shifted Hankel Determinants of Odd Catalan PowersThe shifted Hankel determinant of an odd Catalan power has Cigler's closed form, including both upper boundary shifts.Proved

The questionJohann Cigler, Some experimental observations about Hankel determinants of convolution powers of Catalan numbers, arXiv:2308.07642v2, Section 2.2, Conjecture 11: with C{r,j} = r/(2j+r) · binom(2j+r, j) for j ≥ 0, C{r,j} = 0 for j < 0, and D{r,s}(N) = det(C{r,i+j+s}){0 ≤ i,j < N}, D{r,s}(0) = 1, for every k ≥ 1, 0 ≤ m ≤ k+1 and n ≥ 0, D{2k+1,m−k+1}((2k+1)n+k) = (−1)^{kn+binom(k,2)} (2k+1)^m (n+1)^m.

The ideaThe coefficients C{2k+1,t−k} are the moments ℒ(X^t pk) of a monic orthogonal family pj for the Jacobi operator with boundary weight 1 and interior weight 2, proved through the Pascal recurrence of the coefficient array.

Our contributionThe shifted Hankel determinant of an odd Catalan power has Cigler's closed form, including both upper boundary shifts.

Verification details
Release source recordProved in Lean

Open problem

Shifted Hankel Determinants of Odd Catalan Powers

Johann Cigler, Some experimental observations about Hankel determinants of convolution powers of Catalan numbers, arXiv:2308.07642v2, Section 2.2, Conjecture 11: with C_{r,j} = r/(2j+r) · binom(2j+r, j) for j ≥ 0, C_{r,j} = 0 for j < 0, and D_{r,s}(N) = det(C_{r,i+j+s}){0 ≤ i,j < N}, D{r,s}(0) = 1, for every k ≥ 1, 0 ≤ m ≤ k+1 and n ≥ 0, D_{2k+1,m−k+1}((2k+1)n+k) = (−1)^{kn+binom(k,2)} (2k+1)^m (n+1)^m. The shift m − k + 1 may be negative.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CatalanPowerHankel/CiglerEleven.result

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About this highlight
  • Contribution

    The shifted Hankel determinant of an odd Catalan power has Cigler's closed form, including both upper boundary shifts.

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    First preprint year: 2023.

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Discrete structuresQuestion resolvedFirst preprint 2024Hankel Determinants of Backward Shifts of Partial Theta CoefficientsEvery normalized backward-shifted partial theta Hankel quotient is a monic integer polynomial with the conjectured degree and values at zero and one.Proved

The questionJohann Cigler, Hankel determinants of backward shifts of the coefficients of a partial theta function, arXiv:2407.05768v2, Section 1, the Conjecture after equation (2): with a(s, q) = q^{binom(s,2)} for s ≥ 0 and a(s, q) = 0 for s < 0, D{−m,N}(q) = det(a(−m + i + j, q)){0≤i,j<N}, and r{m,n}(q) defined by D{−m,n+m+1}(q) = (−1)^{binom(m+1,2)} r{m,n}(q) q^{m binom(n,2)} D{0,n+1}(q),

The ideaD{0,n+1}(q) is a Vandermonde determinant in q⁰, q¹, …, qⁿ times a monomial, hence a monomial times ∏{d=1}^{n} (q^d − 1)^{n+1−d}, which is nonzero.

Our contributionEvery normalized backward-shifted partial theta Hankel quotient is a monic integer polynomial with the conjectured degree and values at zero and one.

Verification details
Release source recordProved in Lean

Open problem

Hankel Determinants of Backward Shifts of Partial Theta Coefficients

Johann Cigler, Hankel determinants of backward shifts of the coefficients of a partial theta function, arXiv:2407.05768v2, Section 1, the Conjecture after equation (2): with a(s, q) = q^{binom(s,2)} for s ≥ 0 and a(s, q) = 0 for s < 0, D_{−m,N}(q) = det(a(−m + i + j, q)){0≤i,j<N}, and r{m,n}(q) defined by D_{−m,n+m+1}(q) = (−1)^{binom(m+1,2)} r_{m,n}(q) q^{m binom(n,2)} D_{0,n+1}(q),

The functions r_{m,n}(q) are monic polynomials with integer coefficients with deg r_{m,n}(q) = mn(n + m + 2)/2 which satisfy r_{m,n}(1) = 1 and r_{m,n}(0) = (−1)^{mn}.

The paper proves the cases m = 1 and m = 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PartialTheta/PartialThetaHankel.result

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Quantum informationQuestion resolvedFirst preprint 2024Vacuum and one-photon Wigner-majorization chainsVacuum and one-photon mixtures form two oppositely ordered chains meeting at one half.Proved

The questionDe Boer, Di Giulio, Keski-Vakkuri and Tonni, Continuous majorization in quantum phase space for Wigner-positive states and proposals for Wigner-negative states, arXiv:2412.19698v3, Section V.A.1, ask for an analytic understanding of ordered subsets of the vacuum/one-photon mixture family.

Why it mattersIts cancellation theorem treats continuous profiles with an everywhere derivative, an integrable derivative, one nonnegative peak position, the corresponding derivative signs, and zero endpoint values at zero and infinity.

The ideaThe proof uses convexity.

Our contributionVacuum and one-photon mixtures form two oppositely ordered chains meeting at one half.

Verification details
Release source recordProved in Lean

Open problem

Vacuum and one-photon Wigner-majorization chains

De Boer, Di Giulio, Keski-Vakkuri and Tonni, Continuous majorization in quantum phase space for Wigner-positive states and proposals for Wigner-negative states, arXiv:2412.19698v3, Section V.A.1, ask for an analytic understanding of ordered subsets of the vacuum/one-photon mixture family. The v1 question is: “It would be interesting to identify subsets of W̃_u in which all elements exhibit a Wigner majorization relationship with one another. We leave this question for future works.” The v3 conjecture concerns the branch above one half. Preregistration #18571 specifies DBKT-1: both closed intervals form chains under Proposal 1.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Fock Wigner function, Eq. (99) | fockWigner n x p, with the reused genLaguerre 0 n | Literal expression over real phase-space coordinates | | Mixture, Eq. (100) | mixtureWigner u x p | $(1-u)W_0+uW_1$ | | Positive-part functional, Eq. (97) | levelFunctional W t | Product Lebesgue integral on $\mathbb R^2$ of $\max(|W|-t,0)$ | | Proposal 1, Eq. (98) | WignerMajorizes W₁ W₂ | Every real threshold $t\geq0$ | | DBKT-1(A) | First conjunct of claim and result | $1/2\leq u\leq v\leq1\Rightarrow\widetilde W_v\succ_W\widetilde W_u$ | | DBKT-1(B) | Second conjunct of claim and result | $0\leq u\leq v\leq1/2\Rightarrow\widetilde W_u\succ_W\widetilde W_v$ |

Definitions of the functions are meaningful for every real parameter; the theorem imposes the source's physical parameter intervals. Endpoints and all nonnegative thresholds are included. No clause is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/FockAttenuator/VacuumOnePhotonWignerMajorization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding 2413 and 2431, or 2431 and 3241Two Fishburn avoidance classes have the same binomial-Catalan enumeration.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaA permutation in either class decomposes uniquely as a direct sum of indecomposable components, each in the same class, so the generating function is F = 1/(1 − Find).

Our contributionTwo Fishburn avoidance classes have the same binomial-Catalan enumeration.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding 2413 and 2431, or 2431 and 3241

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.13. For all n ≥ 1, |Fn(2413, 2431)| = |Fn(2431, 3241)| = Σ_{k=1}^{n} C(n−1, k−1) C_{n−k}.

Here C_m is the m-th Catalan number. A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) is the set of Fishburn permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/FishburnTenThirteen/FishburnTenThirteen.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Discrete structuresConjecture provedFirst preprint 2024Koutschan–Krattenthaler–Schlosser Conjecture 23, equation (10.11)The binomial determinant identities follow from a common column multiplier and triangular determinant reductions.Proved

The questionKoutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”.

Why it mattersThe determinant families share a rational column transformation.

The ideaThe multiplier has constant coefficient one, so its lower triangular coefficient matrix preserves determinants.

Our contributionThe binomial determinant identities follow from a common column multiplier and triangular determinant reductions.

Verification details
Release source recordProved in Lean

Open problem

Koutschan–Krattenthaler–Schlosser Conjecture 23, equation (10.11)

Koutschan, Krattenthaler and Schlosser, arXiv:2401.08481v1, Conjecture 23, states: “Moreover, the following identities are conjectured to hold for all n ≥ 3:”. Equation (10.11) is

$G_{1,1,0,-2}(n)=-\frac1{49}G_{2,3,0,-6}(n)=-\frac27G_{6,4,5,-3}(n-1)=-\frac4{5577}G_{7,6,5,-7}(n-1)$.

The preregistration is #19237. The Library note records the source definition and all three equations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (1.2): falling factorial divided by $p!$ for $p\ge0$, zero for $p<0$ | ShiftedG.binom | Rational Ring.choose on the nonnegative branch, integer lower argument. | | Section 10 determinant | ShiftedG.G | Integer parameters, rational entries, integer powers of 2 and zero-based Fin n indices. | | “for all n ≥ 3” | TenEleven.claim | Natural size and the original lower bound. | | Every consecutive equality of (10.11) | TenEleven.claim and TenEleven.result | Original coefficients, signs and size shifts; a chain is a conjunction of adjacent equalities. |

Natural subtraction is truncated at zero; within the source range it agrees with integer subtraction. The rational/integer coercions express the same source entries. There are no additional assumptions on the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KoutschanKrattenthalerSchlosserTenEleven.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Number theoryQuestion resolvedPosed in 2022The Marcus and Bala formulas for the power of three in F(4n)The highest power of three dividing F(4n) satisfies the Marcus and Bala formulas.Proved

The questionHighest power of 3 dividing F(4n) where F(k) is the k-th Fibonacci number.

Why it mattersThe two formulas connect Fibonacci divisibility with a divisor-sum identity for every positive natural index.

The ideaUse the Fibonacci gcd identity and the divisors of 12 to prove, for every positive k, that 9 divides F(k) exactly when 12 divides k.

Our contributionThe highest power of three dividing F(4n) satisfies the Marcus and Bala formulas.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Marcus and Bala formulas for the power of three in F(4n)

OEIS A074724, NAME (%N, verbatim):

Highest power of 3 dividing F(4n) where F(k) is the k-th Fibonacci number.

Marcus's FORMULA (%F, verbatim):

a(n) = 3^A051064(n) (conjectured). - Michel Marcus, May 17 2022

Bala's FORMULA (%F, verbatim):

Conjecture: a(n) = (sigma(3*n) - sigma(n))/(sigma(3*n) - 3*sigma(n)), where sigma(n) = A000203(n). Equivalently, a(n) = A088838(n) - A074724(n). - Peter Bala, Jun 10 2022

The AUTHOR line (%A, verbatim) is:

Benoit Cloitre, Sep 04 2002

With a n = 3 ^ padicValNat 3 (Nat.fib (4 * n)), the exact claim is forall n : Nat, 0 < n -> (a n = 3 ^ (padicValNat 3 n + 1) and a n * (sigma 1 (3 * n) - 3 * sigma 1 n) = sigma 1 (3 * n) - sigma 1 n). Here A051064(n) is v_3(3n) = v_3(n)+1. Bala's denominator is multiplied out in natural numbers; its positivity is proved within the proof. The separate "Equivalently" remark involving A088838 is not part of this claim. The positive-index guard excludes the totalized Lean value at zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CloitreFibFourThreeAdicValuationSigma.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    Source 1
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    Source 1
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Discrete structuresQuestion resolvedPosed in 2022Werner's all-column formula for the affine polynomial pairThe affine polynomial pair has Werner's all-column binomial formula.Proved

The questionConjecture: T(n,k) = binomial(n-1,2k+1) + binomial(n,2k+1). - Knud Werner, Jan 11 2022

Why it mattersThe identity gives every coefficient of an unbounded family specified by a coupled affine polynomial recursion.

The ideaUse the actual pair recursion with stage m corresponding to source index m+1.

Our contributionThe affine polynomial pair has Werner's all-column binomial formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Werner's all-column formula for the affine polynomial pair

OEIS A208509 revision 28, FORMULA, attributes this conjecture to Knud Werner:

Conjecture: T(n,k) = binomial(n-1,2*k+1) + binomial(n,2*k+1). - Knud Werner, Jan 11 2022

Clark Kimberling authored the sequence entry. Its companion A208508 revision 10 supplies the jointly generated first polynomial. The complete defining laws are u(1,X)=v(1,X)=1, u(n,X)=u(n-1,X)+X*v(n-1,X), and v(n,X)=u(n-1,X)+v(n-1,X)+1 for every n>=2, over Z[X]. For every natural n>=1 and every natural k>=0, T(n,k) is the coefficient of X^k in this actual v(n,X), including all zero coefficients beyond its degree. The formula's two natural binomial coefficients are cast to integers. No assertion at n=0 is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/AffinePolynomialPairBinomialCoefficients.werner_a208509

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  • Mathematical advance · 3/4 · 15% weight

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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2023Original cyclic-morphism string-attractor minimumOriginal cyclic morphism prefixes have the unrestricted attractor minimum of Conjecture 42.Proved

The questionFor k >= 2, let c be a length-k vector of natural coefficients with c(0) >= 1 and c(k-1) >= 1.

Why it mattersThe existing word-power API reads repeated literal blocks modulo their original length.

The ideaNesting and U(n) >= n+1 establish fixed-point-prefix semantics.

Our contributionOriginal cyclic morphism prefixes have the unrestricted attractor minimum of Conjecture 42.

Verification details
Release source recordProved in Lean

arXiv

Original cyclic-morphism string-attractor minimum

For k >= 2, let c be a length-k vector of natural coefficients with c(0) >= 1 and c(k-1) >= 1. Set d to c with the last entry decreased by one and assume d is lexicographically at least every cyclic rotation, allowing equality. The original substitution sends a < k-1 to c(a) zeros followed by a+1, and sends k-1 to c(k-1) zeros.

Let u be its fixed point starting with zero and U(n) the length of the n-th iterate of that singleton. Conjecture 42 states that for every m >= 1, the prefix of length m has minimum attractor size i+1 when 0 <= i <= k-2 and U(i) <= m < U(i+1), and size k when m >= U(k-1). The minimum ranges over every position subset; each nonempty factor needs an equal occurrence wholly inside the same prefix crossing a selected original position.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Attractors/CyclicMorphismAttractorMinimum.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Analysis & dynamicsQuestion resolvedFirst preprint 2022The sharp upper bound on skewness minus kurtosis of a unidirectional unicyclic currentFor every finite cycle size at least two and every positive real rate vector, S − K is at most B = 3(107 + 51√17)/2048; B is the least upper bound of all these actual values.Proved

The questionPtaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys.

Why it mattersThe difference constrains the joint higher current cumulants of one positive unidirectional cycle.

The ideaThe definitions normalizedRates, p, q, r and the theorem normalizedmoments are the original suppliers in UnicyclicSkewnessPlusKurtosisLowerBound.

Our contributionFor every finite cycle size at least two and every positive real rate vector, S − K is at most B = 3(107 + 51√17)/2048; B is the least upper bound of all these actual values.

Verification details
Release source recordProved in Lean

Open problem

The sharp upper bound on skewness minus kurtosis of a unidirectional unicyclic current

K. Ptaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys. Rev. E 106, 024119 (2022), arXiv:2204.08421v4, Section III E 1, main.tex line 432, gives

$$ \max(\mathcal S-\mathcal K)=B=\frac{3(107+51\sqrt{17})}{2048}, \qquad A=\frac{5-\sqrt{17}}4. $$

Line 424 specifies the asymptotic family: the exceptional clockwise rate is $ak$, the other rates are $k$, and $a=A/N$ as $N\to\infty$. Appendix line 792 states that the extremal-topology assertion was numerically conjectured. Thus the analytic endpoint on the proposed topology is source input; its universal validity for arbitrary positive rate vectors needs proof.

Source sign discrepancy. The detailed display at line 432 and Figure 6 concern $\mathcal S-\mathcal K$, but the summary at line 133 prints $\mathcal K-\mathcal S\in[-0.15,0.465]$. The two signs conflict. This problem follows the detailed display and Figure 6; it does not identify the reversed-sign summary with that claim.

The exact reading is preregistered in #25765 and its unchanged endpoint comment. For every natural $N\ge2$ and every $k:\mathrm{Fin},N\to\mathbb R$ with $k_i>0$ for all $i$, put

$$ R_j=\sum_{i:\mathrm{Fin},N}k_i^{-j},\qquad \mathcal S(k)=\frac{3R_2^2}{R_1^4}-\frac{2R_3}{R_1^3},\qquad \mathcal K(k)=\frac{15R_2^2}{R_1^4}-\frac{20R_3}{R_1^3} +\frac{6R_4}{R_1^2R_2}. $$

Define the set of actual admissible values

$$ V={s\in\mathbb R\mid \exists N\in\mathbb N,\ N\ge2\ \land
\exists k:\mathrm{Fin},N\to\mathbb R,\ (\forall i,\ k_i>0)\ \land
s=\mathcal S(k)-\mathcal K(k)}. $$

The proved claim is the conjunction of $\mathcal S(k)-\mathcal K(k)\le B$ for every such $N,k$ and $\operatorname{IsLUB}(V,B)$. Both ratios use the same vector, repeated rates retain their multiplicities, the coordinates are arbitrary positive reals, and $N$ is unbounded. The claim includes $N=2$ and assumes no extremal topology. Sharpness concerns this full actual-value set, not a finite prefix or a relaxation to independently chosen moments.

The source's $\mathcal S$ and $\mathcal K$ are third-to-first and fourth-to-second scaled current-cumulant ratios (main.tex lines 53–54), not standardized central moments. The inverse-rate formulas follow by substitution in the source's winding-number and waiting-time cumulant expressions (Appendix lines 772–775 and 787–790). Their original Lean definitions are UnicyclicSkewnessMinimum.skewness and UnicyclicKurtosisLowerBound.kurtosis. See the source note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/UnicyclicSkewnessMinusKurtosisUpperBound.result

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Discrete structuresQuestion resolvedPosed in 2023Richman's congruence for the deranged-matching numbersFor the deranged-matching numbers a(n), which are also the central moments of the chi-squared distribution with one degree of freedom, the signed values (-1)^n a(n) are periodic with period q modulo every odd q.Proved

The questionConjecture: if m == n (mod q) for q odd, then (-1)^ma(m) == (-1)^na(n) (mod q). - Harry Richman, Aug 29 2023

Why it mattersThe entry reads a(n) as the number of deranged matchings of 2n people and as the n-th central moment E[(X² − 1)^n] of the chi-squared distribution with one degree of freedom, X a standard normal variable; its links include Costa, Dobrescu and Fox, Chiral Abelian gauge theories with few fermions (arXiv:2001.11991).

The idea1. b(n) = (−1)^n a(n) satisfies b(n + 2) = 2(n + 1)(b(n) − b(n + 1)), b(0) = 1, b(1) = 0.

Our contributionFor the deranged-matching numbers a(n), which are also the central moments of the chi-squared distribution with one degree of freedom, the signed values (-1)^n a(n) are periodic with period q modulo every odd q.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Richman's congruence for the deranged-matching numbers

OEIS A053871 is the sequence a(n) = 2(n − 1)(a(n − 1) + a(n − 2)) with a(0) = 1, a(1) = 0. Its FORMULA section states

Conjecture: if m == n (mod q) for q odd, then (-1)^m*a(m) == (-1)^n*a(n) (mod q). - Harry Richman, Aug 29 2023

Issue #10354 fixes the reading: a is the sequence of the recurrence in the entry's name, q is a positive odd integer, and the congruence is in the integers. The statement settled is that for every odd q and all m ≡ n (mod q), (−1)^m a(m) ≡ (−1)^n a(n) (mod q).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/DerangedMatchingCongruence.result

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Analysis & dynamicsBijective proofFirst preprint 2026The Yang–Baxter cellular automata over F_{2^n} have period dividing 2^nThe cellular automaton thus constructed over a finite field of order 2^n has a period that is a divisor of the order of the field.Proved

The questionGeom. 29, 30 (2026)) build a cellular automaton over a finite field F from the R-matrix R(x, y) = (y + f(x + y), x − f(x + y)).

Why it mattersThe paper's numerical tables for orders 8 and 16 show fundamental periods 1, 2, 4, 8 and 16 and no others, and the authors single out this strict periodicity as a feature of the integrable automata: an R-matrix violating the Yang–Baxter equation gives other periods.

The ideaOnly the additive group is used, and x + x = 0.

Our contributionThe cellular automaton thus constructed over a finite field of order 2^n has a period that is a divisor of the order of the field.

Verification details
Release source recordProved in Lean

Open problem

The Yang–Baxter cellular automata over F_{2^n} have period dividing 2^n

A. Araoka and T. Tokihiro (arXiv:2602.17148, nlin.SI; Math. Phys. Anal. Geom. 29, 30 (2026)) build a cellular automaton over a finite field F from the R-matrix R(x, y) = (y + f(x + y), x − f(x + y)). In characteristic 2 the Yang–Baxter equation for R reduces to f(x) + f(x + f(y)) = f(x + f(y + f(x))). The automaton runs R along a row of N cells, R(x_i, y_{i−1}) = (x_i', y_i) with y_0 = b, and the helical boundary condition feeds y_N back as the next boundary value. The paper states:

Conjecture. The cellular automaton thus constructed over a finite field of order 2^n has a period that is a divisor of the order of the field.

and proves it for orders 4 and 8. Issue #11405 fixes the reading: every finite field of characteristic 2, every bijective solution f of the reduced equation, every N, and "period divides q" as step^q = id on the pair of cell values and boundary value.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/YangBaxterAutomatonPeriod.result

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About this highlight
  • Contribution

    For every finite field F of characteristic 2 and every bijection f of F satisfying f(x) + f(x + f(y)) = f(x + f(y + f(x))), the cellular automaton that runs the R-matrix R(x, y) = (y + f(x + y), x - f(x + y)) along a row of N cells with the helical boundary condition returns to its initial state after |F| steps, for every N.

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  • Historical context

    First preprint year: 2026.

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Analysis & dynamicsQuestion resolvedFirst preprint 2026Lee's reduction operators are invertibleEvery reduction operator is a unit for all species counts, word lengths, valid block positions, and parameters in the closed unit cube.Proved

The questionEunghyun Lee, Integrability of multispecies long-range swap models with species-dependent interpolation, arXiv:2604.12136v1, Remark 3.1 after Equation (28), source label 1122am331, conjectures:

Why it mattersThe source's Proposition 2.2, “Integrability under invertibility”, uses these operators to reduce the model to two-particle interactions.

The ideaThe general attained-maximum principle forces every zero-boundary harmonic chain to vanish.

Our contributionEvery reduction operator is a unit for all species counts, word lengths, valid block positions, and parameters in the closed unit cube.

Scope & next questionCommand /usr/bin/time -l python3 /tmp/op-lee/exactcheck.py exited 0.

Verification details
Release source recordProved in Lean

Open problem

lee 2026 reduction operator invertibility

Eunghyun Lee, Integrability of multispecies long-range swap models with species-dependent interpolation, arXiv:2604.12136v1, Remark 3.1 after Equation (28), source label 1122am331, conjectures:

These results lead us to conjecture that $\mathfrak A_k$ is invertible for all parameters $\mu_i \in [0,1]$.

LEE-A is preregistered in #14343. For every $N\ge1$, $n\ge2$, $j\ge1$, $k\ge0$ with $j+k+1\le n$ and $\mu\in[0,1]^N$, the literal recursion $\mathfrak A_0=I$, $\mathfrak A_k=I-\mathcal B_{k+1}\mathfrak A_{k-1}^{-1}\mathcal B'_k$ has invertible matrices on words $\operatorname{Fin}(n)\to\operatorname{Fin}(N)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/LongRangeSwap/LeeReductionInvertibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2025The all-order increasing-function endpoint-jet assertionThe original all-order endpoint-jet assertion, with its monotone input class and one simultaneous smooth witness.Proved

The questionBurke, Maleeha Haris and Madhavendra, Repeated integrals of increasing functions, arXiv:2512.02151v1, Conjecture 1.2, p. 2:

Why it mattersThe conjecture asks whether every attainable finite endpoint jet admits one simultaneous smooth realization, with the specified strict interior sign and full endpoint conditions, at every order.

The ideaNonconstancy gives positive mass; continuity gives atomlessness.

Our contributionThe original all-order endpoint-jet assertion, with its monotone input class and one simultaneous smooth witness.

Verification details
Release source recordProved in Lean

Open problem

The all-order increasing-function endpoint-jet assertion

Maxim R. Burke, Maleeha Haris and Madhavendra, Repeated integrals of increasing functions, arXiv:2512.02151v1, Conjecture 1.2, p. 2:

$(P_n)$ is true for all nonnegative integers $n$.

For each $n\geq 0$, let $W_n\subseteq\mathbb R^{n+1}$ consist of vectors $b=(b_0,\ldots,b_n)$ for which some $f\in C^n[0,1]$ has nondecreasing, nonconstant $D^n f$ and satisfies $D^j f(0)=0$ and $D^j f(1)=b_j$ for every $0\leq j\leq n$. Here $D^0f=f$ and endpoint derivatives are one-sided. Nondecreasing does not mean strictly increasing, and absolute continuity is not an input hypothesis.

The assertion $(P_n)$ requires both openness of $W_n$ and, for every $b\in W_n$, one function $F\in C^\infty[0,1]$ satisfying all of the following:

  • $D^jF(0)=0$ and $D^jF(1)=b_j$ for every $0\leq j\leq n$;
  • $D^{n+1}F(x)>0$ for every $0<x<1$;
  • $D^{n+1}F(0)=D^{n+1}F(1)=1$;
  • $D^jF(0)=D^jF(1)=0$ for every $j>n+1$.

The sole problem anchor is this original Conjecture 1.2, including $n=0$. The follow-up arXiv:2512.23949v1 calls the same interval property $Q_n$ and retains its all-order assertion as Conjecture 2.5.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Interpolation/OriginalIncreasingFunctionJets.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Zero-magic-gap three-dimensional subspaces in every power-of-two dimensionFor every qudit dimension divisible by four, and so for every power of two at least four, some three-dimensional subspace has zero average stabilizer entropy gap.Proved

The questionThe verbatim definitions and statements are in the literature note.

Why it mattersThe ASE gap measures the extra nonstabilizerness that an encoding into a larger system costs on average.

Our contributionFor every qudit dimension divisible by four, and so for every power of two at least four, some three-dimensional subspace has zero average stabilizer entropy gap.

Verification details
Release source recordProved in Lean

Open problem

Zero-magic-gap three-dimensional subspaces in every power-of-two dimension

S. Cepollaro, G. Cuffaro, M. B. Weiss, S. Cusumano, A. Hamma and S. Lloyd, Stabilizer Entropy of Subspaces, arXiv:2512.23013v1, define the average stabilizer entropy gap of an isometry $\mathcal E:\mathbb C^{d_S}\to\mathbb C^{d_B}$ as the Haar average over pure states of the difference between the linear stabilizer entropy of $\mathcal E(\psi)$ for the single-qudit Weyl–Heisenberg group of $\mathbb C^{d_B}$ and that of $\psi$ for the group of $\mathbb C^{d_S}$, and write it in closed form as $\Delta M(\mathcal E)=d_S,\mathrm{tr}(Q_S\mathcal A_4^S)-d_B,\mathrm{tr}(Q_B\mathcal E^{\otimes4}(\mathcal A_4^S))$. In §V.A they report from numerical optimization that "when $d_B$ is a power of 2, a zero ASE subspace of dimension three is always achievable", and in their conclusion that "it is unclear why" this happens. The verbatim definitions and statements are in the literature note.

Issue #13822 reads a three-dimensional subspace as the image of an isometry $V:\mathbb C^3\to\mathbb C^{d_B}$ and the Haar fourth moment in the paper's closed form $\binom{6}{4}^{-1}\frac1{24}\sum_{\sigma\in S_4}T_\sigma$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/CepollaroPowerOfTwoZeroMagicGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The minimum of purity plus time-reversal overlapFor every bipartition of N qubits into a part A of k qubits and its non-empty complement, the minimum over pure states of the purity of the reduced state plus its overlap with the time-reversed reduced state is 2^(k-N) when 2k > N and 2^(1-k) when 2k <= N.Proved

The questionA 113 (2026) 012415) bound the sector lengths of multiqubit states through the reduced states ρA of a pure state and their time-reversed partners ρ̃ = σy^{⊗k} ρ^ σy^{⊗k}, with the overlap Rρ = Tr(ρ ρ̃).

Why it mattersThe paper uses Tr(ρA²) + R{ρA} to derive sector-length inequalities, and the conjecture identifies the exact minimum of this quantity.

The ideaWrite d = 2^k, r = 2^{N−k}, M the d × r matrix of amplitudes, so that ρA = M Mᴴ with Tr ρA = 1, and Φ = Y M̄ with Y = σy^{⊗k}.

Our contributionFor every bipartition of N qubits into a part A of k qubits and its non-empty complement, the minimum over pure states of the purity of the reduced state plus its overlap with the time-reversed reduced state is 2^(k-N) when 2k > N and 2^(1-k) when 2k <= N.

Verification details
Release source recordProved in Lean

Open problem

The minimum of purity plus time-reversal overlap

E. Serrano-Ensástiga, O. Giraud and J. Martin (arXiv:2507.12680, quant-ph; Phys. Rev. A 113 (2026) 012415) bound the sector lengths of multiqubit states through the reduced states ρ_A of a pure state and their time-reversed partners ρ̃ = σ_y^{⊗k} ρ^* σ_y^{⊗k}, with the overlap R_ρ = Tr(ρ ρ̃). Their Proposition 1 gives Tr(ρ_A²) + R_{ρ_A} ≥ 2^{−min(k, N−k)}, and they state:

Let $A|\bar{A}$ be a bipartition of an $N$-qubit system into $k=|A|$ and $N-k=|\bar{A}|$ qubits. Then, the minimum of $\mathrm{Tr}(\rho_A^2) + R_{\rho_A}$ taken over all pure states $|\psi\rangle$ of the $N$-qubit system, is given by $\min_{|\psi\rangle} \big(\mathrm{Tr}(\rho_A^2) + R_{\rho_A}\big) = 2^{k-N}$ if $2k > N$, and $2^{-k + 1}$ if $2k \leq N$.

Issue #11730 fixes the reading: A is any set of k qubits with 1 ≤ k ≤ N − 1, a pure state is a unit vector of amplitudes on the 2^N configurations, ρ_A is the partial trace over the other qubits, and the minimum means a lower bound for every unit vector together with a unit vector that attains it. The result proves the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PurityTimeReversalOverlapMinimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Degree recovery from the ordinary Loopy polynomial with loopsEquality of ordinary Loopy polynomials determines the complete actual degree multiset.Proved

The questionProblem 11.4 of kirillov2026loopy asks exactly:

Why it mattersDefinition 1.1 and Proposition 2.1 supply the ordinary Loopy recursion and choice independence.

The ideaThe endpoint keeps two graph inductions live inside its proof and applies them to both inputs.

Our contributionEquality of ordinary Loopy polynomials determines the complete actual degree multiset.

Verification details
Release source recordProved in Lean

arXiv

Degree recovery from the ordinary Loopy polynomial with loops

Problem 11.4 of kirillov2026loopy asks exactly:

Does L_G determine the degree sequence of an arbitrary graph G, loops included?

In repository coordinates, the complete target quantifies independently over finite vertex sets V W, edge-occurrence lists E F, and accumulated-loop functions ell eta. From Valid V E ell, Valid W F eta, and loopy V E ell = loopy W F eta, it concludes degreeMultiset V E ell = degreeMultiset W F eta. It assumes no equality of orders. Lists preserve parallel occurrences; both endpoints of a pending loop are counted, accumulated loops contribute twice, and mapping all vertices in the carrier includes isolates. No connectedness, nonemptiness or simplicity hypothesis is present.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Combinatorics/LoopyDegreeSequence.loopy_determines_degree_multiset

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryConjecture provedPosed in 2025The A146557 collinear triple divisibility conjectureTranslation orbits prove the square-divisibility conjecture in OEIS A146557.Proved

The questionPeter Bala's July 24, 2025 comment conjectures that n squared divides a(n) when three does not divide n.

Why it mattersDirect inspection of the entry on September 8, 2026 found the statement still labelled Conjecture.

The ideaTranslation preserves the coordinate restrictions and the determinant.

Our contributionTranslation orbits prove the square-divisibility conjecture in OEIS A146557.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A146557 collinear triple divisibility conjecture

Peter Bala's July 24, 2025 comment conjectures that n squared divides a(n) when three does not divide n. Here a(n) counts unordered collinear triples in (ZMod n) squared whose first coordinates, and second coordinates, are pairwise distinct. Collinearity means the difference determinant is zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/CollinearTripleTranslationOrbits.square_dvd_card_collinear_triples

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2024Reverse level order labeling gives one nonzero entry in every nonfinal column tailA labeling that decreases with distance from a tree's root gives exactly one nonzero entry in every column tail before the last column.Proved

The questionMišta Jr., Minimal criteria for continuous-variable genuine multipartite entanglement, arXiv:2401.04376v3; Quantum 9, 1837 (2025), Section 4.1.3, “Generic tree”, state the conjecture in Quantum-accepted.tex, lines 837–839:

Why it mattersLeskovjanová and Mišta Jr. (2025, arXiv:2401.04376v3), Section 4.1, represent a minimal set of two-mode marginal covariance matrices by a tree.

The ideaThe adjacency hypothesis is essential to this equivalence.

Our contributionA labeling that decreases with distance from a tree's root gives exactly one nonzero entry in every column tail before the last column.

Verification details
Release source recordProved in Lean

Open problem

Reverse level order labeling gives one nonzero entry in every nonfinal column tail

O. Leskovjanová and L. Mišta Jr., Minimal criteria for continuous-variable genuine multipartite entanglement, arXiv:2401.04376v3; Quantum 9, 1837 (2025), Section 4.1.3, “Generic tree”, state the conjecture in Quantum-accepted.tex, lines 837–839:

We checked by a direct calculation that for all configurations of up to $N=10$ the established reverse level order labeling leads to an adjacency matrix, where each vector $A_{i+1:N,i}$ contains only one non-zero component and therefore the respective quadrature combinations (\ref{uxpi}) contain at most two non-zero terms. This leads us to a conjecture that this statement holds generally for an arbitrary $N$-vertex tree.

The source construction, lines 828–835, roots the tree at its center vertex or at an endpoint of its central edge, assigns label $N$ to the root, and assigns decreasing labels level by level, from left to right within each level. The paper assumes $N\ge3$ for its entanglement application (lines 429–431). The source quotations and bibliographic entry are also in the literature note. The checked TeX member has MD5 b1e8a86c3fefec6858c4ac36ca46f9fd.

Issue #21778 fixes the graph statement with an arbitrary root and arbitrary order within each level. In the module, for every $n\in\mathbb N$, every simple graph $g$ on $\operatorname{Fin} n$ with decidable adjacency and $g.\mathrm{IsTree}$, every root $r\in\operatorname{Fin} n$, and every bijection $\ell:\operatorname{Fin} n\simeq\operatorname{Fin} n$ satisfying

$$ \operatorname{dist}_g(r,u)<\operatorname{dist}_g(r,v) \ \Longrightarrow\ \ell(v)<\ell(u), $$

define $A^\ell_{ji}$ to be the natural-valued adjacency entry of $g$ at $\ell^{-1}(j),\ell^{-1}(i)$: it is $1$ for adjacent vertices and $0$ otherwise. Then claim states

$$ \forall i\in\operatorname{Fin} n,\quad i.\mathrm{val}+1<n\ \Longrightarrow \exists! j\in\operatorname{Fin} n,\quad i<j\ \land\ A^\ell_{ji}\ne0. $$

The domain preorder in StrictAnti is lifted along $\operatorname{dist}_g(r,\cdot)$; its codomain has the usual label order. Thus equal levels impose no label comparison. Lean's labels $0,\ldots,n-1$ correspond to the source's $1,\ldots,N$. The graph statement is proved by result : claim; the Cholesky and quadrature implications are outside that declaration.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/ReverseLevelOrderTreeLabeling.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedMarcus's tent-map cycle length and the signed suborder of twoMarcus's tent-map cycle length equals the signed suborder of two.Proved

The questionLeast number m > 0 such that 2^m == +-1 (mod 2n + 1).

Why it mattersThe tent map folds doubling into an interval, and its even numerator representatives turn this signed modular return into an ordinary periodic orbit.

The ideaFor arbitrary N>=2, induction on m shows that (tent N)^[m] 2 lies in [0,N], is even, and equals either sign of 2^(m+1) in ZMod N.

Our contributionMarcus's tent-map cycle length equals the signed suborder of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Marcus's tent-map cycle length and the signed suborder of two

OEIS A003558, NAME (%N, verbatim):

Least number m > 0 such that 2^m == +-1 (mod 2n + 1).

The settled COMMENT sentence (%C, verbatim):

It appears that a(n) is the length of the cycle starting at 2/(2n+1) for the map x->1 - abs(2x-1). - Michel Marcus, Jul 16 2025

AUTHOR (%A, verbatim):

N. J. A. Sloane

The offset is %O 0,3. The claim is read for every natural n>0, with N=2n+1: the least positive return time of 2/N under the tent map is the least positive m such that 2^m is congruent to 1 or -1 modulo N. The residue -1 is written 2n in natural congruence notation.

The exact Lean statement is:

theorem result (n : ℕ) (hn : 0 < n) :
    IsLeast {m : ℕ | 0 < m ∧ (2 ^ m ≡ 1 [MOD 2 * n + 1] ∨
      2 ^ m ≡ 2 * n [MOD 2 * n + 1])}
      (Function.minimalPeriod (tent (2 * n + 1)) 2)

Here tent N k := if 2*k <= N then 2*k else 2*N - 2*k is a natural numerator map. For N>0 and k<=N, its branches divided by N are 2k/N and (2N-2k)/N, exactly the two branches of the rational tent map. Since k<=N, the second natural subtraction is not truncated. The orbit invariant supplies this bound for every iterate from 2. Division by the fixed positive N is injective, so numerator and rational return times coincide. This rational interpretation is an algebraic explanation; the kernel theorem is the displayed numerator statement.

Only the quoted Marcus sentence is settled. A003558's other comments about Kaprekar cycles, x^2-2 iteration, and the Kappraff-Adamson base conjecture are not claimed. The hypothesis 0<n excludes n=0, where N=1 and 2/N is outside [0,1].

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MarcusTentMapSuborder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThree-column permutation arrays with compatible row ordersPositive-length three-column permutation arrays have count 2+(n+1)(n+2)(n+3)/6.Proved

The questionFor each positive integer n, count the actual n by 3 arrays whose rows are permutations of 1,2,3, whose adjacent-descent counts are nondecreasing along the original row indices, and whose rows are lexicographically nonincreasing.

Why it mattersHardin in 2013, and the displayed formula is attributed to Colin Barker in 2018.

The ideaIn increasing lexicographic order the six rows are 123,132,213,231,312,321, with descent counts 0,1,1,1,1,2.

Our contributionPositive-length three-column permutation arrays have count 2+(n+1)(n+2)(n+3)/6.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Three-column permutation arrays with compatible row orders

For each positive integer n, count the actual n by 3 arrays whose rows are permutations of 1,2,3, whose adjacent-descent counts are nondecreasing along the original row indices, and whose rows are lexicographically nonincreasing. Rows may repeat. In the source carrier a row is Equiv.Perm (Fin 3) and the source symbol in column j is (r j).val + 1, so the carrier is exactly the six permutations of 1,2,3. OEIS A222001 gives the formula 2+(n+1)*(n+2)*(n+3)/6 with offset 1,1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PermutationArrays/DescentLexCount.count_arrays

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The component count is the graph count plus loop-free strange componentsQuantum graph components equal graph components plus loopless isolated strange edges.Proved

The questionNechita, Quantum Graph Theory by Example, arXiv:2603.23651v3 [math.OA, cross-listed math-ph, quant-ph] (Quantum 10, 2228, 2026), leave the component-count question in the conclusion.

Why it mattersThe settlement supplies that formula in the formal model and separates the exceptional two-vertex, loop-free case from all other connected components.

The idea2. Commutant reduction. The columns of the realigned projector give a Kraus representation of adjacency A C d.

Our contributionQuantum graph components equal graph components plus loopless isolated strange edges.

Verification details
Release source recordProved in Lean

Open problem

The component count is the graph count plus loop-free strange components

G. L. Spitzer and I. Nechita, Quantum Graph Theory by Example, arXiv:2603.23651v3 [math.OA, cross-listed math-ph, quant-ph] (Quantum 10, 2228, 2026), leave the component-count question in the conclusion. The source text states in content/conclusion.tex, line 8:

It would be interesting to determine the exact number of connected components of $X_{A, \ccdot, C}$ in terms of $\mathfrak{G}(A, C)$, and in particular to understand the role played by isolated strange edges with phase $\pi$.

The paper defines the strange graph $\mathfrak{G}(A,C)$ from the nonzero two-vertex blocks of a quantum graph of the form $X_{A,B,C}$. The formal module fixes the $B=\operatorname{diag}A$ reading through the following declarations. For $A,C : \operatorname{Mat} n$ and $d : \operatorname{Fin} n \to \mathbb C$, realignedProjector A C d is the matrix with diagonal-coordinate entries $d$ and off-diagonal blocks determined by $A$ and $C$, and adjacency A C d is the linear map obtained from that realigned matrix. HasComponents G k means that there are $k$ nonzero star projections summing to the identity whose cross terms $P_s G(P_t Z)$ vanish for every $s \ne t$ and every $Z$. underlyingGraph A C has an edge exactly when the corresponding off-diagonal entry of $A$ or $C$ in either direction is nonzero. IsLoopFreeStrangeComponent A C d Q says that a connected component $Q$ has support ${i,j}$ for distinct vertices, is adjacent, has $A_{ij}=1/2$, and has $d_i=d_j=0$. The two counts are graphComponentCount A C and loopFreeStrangeComponentCount A C d, the latter counting such components.

The Lean claim quantifies over every natural $n$, every $A,C : \operatorname{Mat} n$, and every $d : \operatorname{Fin} n \to \mathbb C$. Under $A_{ii}=d_i$, $C_{ii}=d_i$, Hermiticity and idempotence of realignedProjector A C d, and Hermiticity of offDiag A, it asserts that the set of $k$ for which HasComponents (adjacency A C d) k holds has greatest element graphComponentCount A C + loopFreeStrangeComponentCount A C d.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/QuantumGraphStrangeComponentCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2017Non-isotropic mixed-unitary qubit channels do not commute with marginal pinchingEvery non-isotropic mixed-unitary qubit channel fails to commute with marginal pinching on a two-qubit density matrix with nondegenerate input and output marginals.Proved

The questionThat is, qubit mixed-unitary channels that are not isotropic all fail to satisfy the condition.

Why it mattersThe source uses commutation with the discord-destroying map as a sufficient condition for monotonicity of relative-entropy diagonal discord.

The ideaNon-isotropy therefore excludes a scalar Gram matrix.

Our contributionEvery non-isotropic mixed-unitary qubit channel fails to commute with marginal pinching on a two-qubit density matrix with nondegenerate input and output marginals.

Verification details
Release source recordProved in Lean

Open problem

Non-isotropic mixed-unitary qubit channels do not commute with marginal pinching

Z.-W. Liu, R. Takagi and S. Lloyd, Diagonal quantum discord, J. Phys. A: Math. Theor. 52, 135301 (2019), arXiv:1708.09076v2, Section 3.2, “$\pi$-commuting local operations and monotonicity”, conjecture following the probabilistic-Hadamard example:

We conjecture (which is not important for our current purpose) that ${\bf ISO} = X_A(\pi_A)$. That is, qubit mixed-unitary channels that are not isotropic all fail to satisfy the condition.

This quotation is line 463 of the source archive's revised_jphysa_clean.tex (md5 3ae0def434187ebe1875d87961e23048). The relevant definitions are at lines 261, 325–329 and 347; line 339 assumes the input marginal nondegenerate. See also the literature note.

The formal definitions use the following conventions:

  • IsMixedUnitary means a finite mixture $\Phi(a)=\sum_j p_j U_j a U_j^\dagger$ on every complex $2\times2$ matrix, with real $p_j\ge0$, $\sum_jp_j=1$, and unitary $U_j$.
  • IsIsotropic means that, for some fixed $\gamma\in[0,1]$ and unitary $U$, either $\Phi(a)=(1-\gamma)UaU^\dagger+\gamma\operatorname{tr}(a)I/2$ for every $a$, or $\Phi(a)=(1-\gamma)Ua^{\mathsf T}U^\dagger+\gamma\operatorname{tr}(a)I/2$ for every $a$. Transpose gives the complex linear extension of the antiunitary action on Hermitian states.
  • localMap applies $\Phi$ to each first-qubit block; for these linear maps it is $\Phi\otimes\mathrm{id}_B$. The marginal $\rho_A=\operatorname{tr}_B\rho$ is partialTraceRight.
  • IsNondegenerate means Hermitian with two distinct eigenvalues. For a Hermitian marginal, pinchA is $\pi_A(\rho)=\sum_i(P_i\otimes I)\rho(P_i\otimes I)$, where $P_i$ are its rank-one spectral projectors; its spectral basis is recomputed on the argument to which pinching is applied.

Issue #22787 fixes the reading to the quotation's second sentence: for every mixed-unitary qubit map $\Phi$ that is not isotropic, there is a positive semidefinite two-qubit matrix $\rho$ of trace one such that $\rho_A$ and $\Phi(\rho_A)$ are nondegenerate and $\pi_A((\Phi\otimes\mathrm{id}_B)(\rho))\ne(\Phi\otimes\mathrm{id}_B)(\pi_A(\rho))$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/MixedUnitaryPinchingCommutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2019Godrèche–Luck moving-average record boundThe record probability of two successive iid summands is at most 3/(2n) for every atomless real probability law and n at least 3.Proved

The questionLuck, Records for the moving average of a time series, arXiv:1907.07598v2, J.

Why it mattersThe source presents the inequality as a data-supported absolute upper bound.

The ideaOutside the null tie set, the record event is the union of the maximum-at-the-terminal-edge event and the event that the terminal edge is the runner-up while the next cyclic edge is maximal.

Our contributionThe record probability of two successive iid summands is at most 3/(2n) for every atomless real probability law and n at least 3.

Verification details
Release source recordProved in Lean

Open problem

Godrèche–Luck moving-average record bound

C. Godrèche and J.-M. Luck, Records for the moving average of a time series, arXiv:1907.07598v2, J. Stat. Mech. (2020) 023201, DOI 10.1088/1742-5468/ab5d08, state after equation (6.33) that the data suggest $3/(2n)$ is an absolute upper bound for the record probability $Q_n$.

Statement echo

The source's two-term moving average is $Y_n=X_{n-1}+X_n$, with records counted from the first complete sum $Y_2$. The formal recordProb event is the strict comparison $Y_n>\max(Y_2,\ldots,Y_{n-1})$ in the preregistered zero-based coordinates. The theorem movingAverage_record_bound proves the bound for every atomless probability measure on $\mathbb R$ and every $n\ge3$; result specializes it to the density-form claim. Its Continuous f hypothesis is the density restriction fixed by preregistration #15864; discontinuous densities in the paper are covered by the public atomless theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Records/MovingAverageRecordBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2026The Fourier LDOI LOCC lower bound is not tightJohnston and Russo ask whether their lower bound for the uniform Fourier LDOI ensemble is tight.Refuted

The questionJohnston and Russo, “Distinguishability of locally diagonal orthogonally invariant quantum states,” Example 17, arXiv:2604.12808v1, ask:

Why it mattersThe source constructs a PPT measurement attaining the upper bound.

The ideaClassical communication enters through history-dependent child trees.

Our contributionJohnston and Russo ask whether their lower bound for the uniform Fourier LDOI ensemble is tight.

Scope & next questionThe source's broader Conclusion question, equality of LOCC and PPT for every uniform orthonormal LDOI ensemble outside the stated balanced-amplitude class, remains open.

Verification details
Release source recordRefuted in Lean

arXiv

The Fourier LDOI LOCC lower bound is not tight

Johnston and Russo, “Distinguishability of locally diagonal orthogonally invariant quantum states,” Example 17, arXiv:2604.12808v1, ask:

Whether the lower bound in Equation (61) is tight—that is, whether opt_LOCC(E) = 1/2 − (n − 2)/(2n²)—remains an open question.

The uniform ensemble contains $n^2$ states, with prior $1/n^2$ each. Its diagonal vectors are $\phi_{ii}=\sum_k U_{ik}|kk\rangle$, where $U$ is the normalized Fourier matrix. For $i<j$, its two ordered labels are $\phi_{ij}=(|ij\rangle+|ji\rangle)/\sqrt2$ and $\phi_{ji}=(|ij\rangle-|ji\rangle)/\sqrt2$. The question is the equality for every $n\ge3$. Issue #14672 fixes the source conventions and the literal equality claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/FourierLDOILOCCRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedFirst preprint 2026Sahbi's Conjecture 4.2: bipartite two-independence and upper star formationA bipartite graph on ten vertices has a minimal two-star-forming set of size six, but every two-independent set has size at most five.Refuted

The questionRafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 4.2:

Why it mattersThe paper proves equality for complete bipartite graphs and chain graphs.

The ideaTake vertices aᵢ = i and bⱼ = 5+j for 0 ≤ i,j < 5.

Our contributionA bipartite graph on ten vertices has a minimal two-star-forming set of size six, but every two-independent set has size at most five.

Scope & next questionThese are finite computations, not a proof for all grids; their execution was not repeated here. - [open] Conjecture 8.4 on grids remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Sahbi's Conjecture 4.2: bipartite two-independence and upper star formation

Rafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 4.2:

For every bipartite graph G, β₂(G) = SF₂(G).

For a finite simple graph, a set I is k-independent when every vertex of G[I] has degree less than k; βₖ(G) is the maximum such cardinality. Definition 2.1 calls S k-star-forming if, for each v outside S, G[S ∪ {v}] contains a copy of K₁,ₖ containing v. The copy need not be induced. A star-forming set is minimal if no proper subset is star-forming; SFₖ(G) is the maximum cardinality among these minimal sets.

The Lean claim quantifies over all n and all decidable simple graphs on Fin n, assumes Colorable 2, and asserts beta G 2 = SF G 2. The theorem result : ¬ claim refutes this statement. Fin n represents every finite vertex type up to bijection, and all the defining adjacency, subset and cardinality conditions are invariant under such a relabelling.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/StarFormingDissociationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Cao–Zhang–Shi–Zhao MABK positivity for the two remaining blocksThe five-term MABK self-testing expression is nonnegative throughout the real cube for every n >= 6 and either of the two small index blocks h = 1, 2.Proved

The questionShen Cao, Xingjian Zhang, Fei Shi and Qi Zhao, Size-Independent Robustness in Multipartite Bell Self-Testing, arXiv:2608.30851v1, Supplemental Material, “Efficient Verification of Optimal Lower Bound”, state:

Why it mattersThe operator reductions and extractability statements are separate, unformalized obligations.

The ideaThe proof uses induction and factorization.

Our contributionThe five-term MABK self-testing expression is nonnegative throughout the real cube for every n >= 6 and either of the two small index blocks h = 1, 2.

Verification details
Release source recordProved in Lean

Open problem

Cao–Zhang–Shi–Zhao MABK positivity for the two remaining blocks

Shen Cao, Xingjian Zhang, Fei Shi and Qi Zhao, Size-Independent Robustness in Multipartite Bell Self-Testing, arXiv:2608.30851v1, Supplemental Material, “Efficient Verification of Optimal Lower Bound”, state:

For $n \le 5$ this bound is already established analytically [20, 23]; for $n \geq 6$ the two remaining cases ($h = 1, 2$) are an open conjecture supported by the numerical evidence above.

The conjecture is that the five-term expression $\lambda_{\vec a}$ on PDF p. 37 is nonnegative at every point of $[0,\kappa]^n$, where $\kappa=1-1/\sqrt2$, for every $n\ge6$ and $h\in{1,2}$. The source has complementary index sets $T$ and $J$, of cardinalities $h$ and $d=n-h$. Its convention $h\le d$ follows from these hypotheses. Issue #11561 preregisters the literal expression, full quantifiers, Tier 1 classification and literature check.

The definition lambdaA preserves all five terms, their coefficients, both complementary products in each paired term, and the real square roots. The bijection from Fin n to the source indices is $i\mapsto i+1$. The definitions kappa and claim express the endpoint and complete conjecture. The theorem result : claim proves it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/MabkSelfTestingPositivity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2025Early state exclusion in seven-site chains with symmetric spectrumEscobar, Garcia and Minenkova (arXiv:2507.18767, Section 4) conjecture that a 7 x 7 Jacobi matrix with symmetric spectrum realizing perfect state transfer does not have early state exclusion if and only if its positive eigenvalues are integer multiples of the smallest positive eigenvalue, and they prove two one-parameter families of spectra.Proved

The questionMinenkova, Early State Exclusion in 7-Qubit Spin Chains, arXiv:2507.18767v1, Section 4.

Why it mattersEarly state exclusion was introduced (arXiv:2411.06047) as a time before the first perfect transfer at which the sender site is empty, so that a new state could be injected early.

The ideaThe proof uses spectral estimates and convexity.

Our contributionEscobar, Garcia and Minenkova (arXiv:2507.18767, Section 4) conjecture that a 7 x 7 Jacobi matrix with symmetric spectrum realizing perfect state transfer does not have early state exclusion if and only if its positive eigenvalues are integer multiples of the smallest positive eigenvalue, and they prove two one-parameter families of spectra.

Verification details
Release source recordProved in Lean

Open problem

Early state exclusion in seven-site chains with symmetric spectrum

M. G. Escobar, V. Garcia and A. Minenkova, Early State Exclusion in 7-Qubit Spin Chains, arXiv:2507.18767v1, Section 4. A Jacobi matrix is a real symmetric tridiagonal matrix with positive off-diagonal entries (Eq. (1.1): "for $a_i \in \R$ and $b_i > 0$"). The source defines (quoted with its displayed equations set inline and without \emph and \label markup):

A Jacobi matrix $J$ realizes \textbf{perfect state transfer} (PST) between the end-vertices of the weighted path in Figure \ref{fig:nearestneighbor} at time $T > 0$ if $e^{-iJT}\e_0 = e^{i\phi}\e_N$ for some phase $\phi \in \R$.

We consider a special case where $J$ has symmetric spectrum in section 3, meaning that for ${\lambda_k}{k = 0}^N$, $\lambda_j = -\lambda{N - j}$ for all $1 \leq j \leq N - 1$.

Let $J$ be a Jacobi matrix realizing PST at earliest time $T > 0$. If there exists some $\tau \in (0, T)$ such that $\langle e^{-iJ\tau}\e_0, \e_0 \rangle = 0$, then we say that $J$ exhibits \textbf{early state exclusion} (ESE) at time $\tau$.

and conjectures:

Let $J$ be a $7 \times 7$ Jacobi matrix with symmetric spectrum realizing PST. Then $J$ does not have ESE if and only if its positive eigenvalues are integer multiples of the smallest positive eigenvalue.

Issue #14487 fixes the reading. "Symmetric spectrum" is read as closure of the spectrum under negation, the definition of arXiv:2606.04353 ("if $-\lambda \in \sigma(J)$ whenever $\lambda \in \sigma(J)$") and the sense in which the source uses it for ${0, \pm x, \pm y, \pm z}$. The matrix is pathHamiltonian r q with r : Fin 6 → ℝ positive and q : Fin 7 → ℝ arbitrary; perfect state transfer at time t is HasPST J t 0 (Fin.last 6), the modulus-one form of the source's condition; the eigenvalues are Mathlib's (pathHamiltonian_isHermitian r q).eigenvalues, and the spectrum is symmetric when the negative of every eigenvalue is an eigenvalue; T is an earliest time when 0 < T, there is transfer at T and at no time in (0, T); early state exclusion is a τ ∈ (0, T) with hamiltonianPropagator J τ 0 0 = 0; the divisibility condition asks for a positive eigenvalue that is the least positive one and of which every positive eigenvalue is an integer multiple. The statement is universally quantified over r, q and T.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/SevenSiteEarlyStateExclusion.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026First occurrences in Kimberling's arrayThe first natural with v occurrences in the golden power array is the Lucas number of index four v minus five.Proved

The questionAlex Ashburn, arXiv:2610.05776v1, Section 5, Conjecture 15:

Why it mattersThe source proves that each positive occurrence count is attained infinitely many times.

The ideaConsecutive rows are disjoint: the Fibonacci inverse-power identity makes an integer weighted average lie strictly between N and N+1 if the same N belongs to both rows.

Our contributionThe first natural with v occurrences in the golden power array is the Lucas number of index four v minus five.

Verification details
Release source recordProved in Lean

Open problem

First occurrences in Kimberling's array

Alex Ashburn, arXiv:2610.05776v1, Section 5, Conjecture 15:

For every v ≥ 2, the least N with a(N) = v is L_{4v−5}.

Put φ = (1 + √5)/2 and R_i = {⌊kφ^i⌋ : k ≥ 1}, for positive natural row indices i. The occurrence count a(N) is the number of rows containing N. The Lucas sequence has L₀ = 2, L₁ = 1 and L_{n+2} = L_{n+1} + L_n.

The Lean claim says, for every natural v ≥ 2, both a(L_{4v−5}) = v and that every positive N < L_{4v−5} has a(N) ≠ v. The count uses the finite index interval 1 ≤ i ≤ N; the proved row-index bound shows that this loses no containing row.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/KimberlingArrayFirstOccurrence.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024No β-invertible S₃ symmetry in the ℤ_N × ℤ_N SymTFT when 3 ∣ NWhen 3 divides N, no copy of the symmetric group S3 inside the anyon permutation symmetries of the ZN x ZN SymTFT has all of its non-identity elements beta-invertible, as conjectured by D.-C.Proved

The questionZhang (arXiv:2406.12151, JHEP 11 (2025) 081) study G-ality defects of 2d QFTs through the SymTFT of the finite Abelian symmetry A.

Why it mattersTogether with the paper's even case it shows that a β-invertible S₃ can exist only when every prime factor of N is at least 5, in line with the paper's tables, where N = 5 is the smallest admissible value.

The ideaReduction modulo 3 maps ρ to a homomorphism into the matrices over ZMod 3 that preserve Q, with invertible β at every g ≠ 1: every vector over ZMod 3 lifts, and the determinant of β reduces to a unit.

Our contributionWhen 3 divides N, no copy of the symmetric group S3 inside the anyon permutation symmetries of the ZN x ZN SymTFT has all of its non-identity elements beta-invertible, as conjectured by D.-C.

Verification details
Release source recordProved in Lean

Open problem

No β-invertible S₃ symmetry in the ℤ_N × ℤ_N SymTFT when 3 ∣ N

D.-C. Lu, Z. Sun and Z. Zhang (arXiv:2406.12151, JHEP 11 (2025) 081) study G-ality defects of 2d QFTs through the SymTFT of the finite Abelian symmetry A. The anyons of the SymTFT are the pairs (a, â) ∈ A ⊕ Â, and an anyon permutation symmetry is a block matrix U = (α β; γ δ) with β : Â → A that preserves the self-statistics and the braiding; U is β-invertible when β is invertible. A G-ality extension needs a subgroup isomorphic to G whose non-identity elements are all β-invertible. For A = ℤ_N × ℤ_N and G = S₃ the paper states (§1.1):

We find that for any N containing a factor of 2 or 3, there are no β-invertible S_3 symmetries in the SymTFT, thus there cannot be S_3-ality defects for these N. We are able to prove this analytically when N is even, and we conjecture this is true generally for N being a multiple of 3 based on our numerical evidence.

Issue #11228 fixes the reading: Â is identified with ℤ_N × ℤ_N through the dot product, the symmetries are the invertible 4 × 4 matrices over ZMod N preserving Q(a, â) = a·â, β is the upper-right 2 × 2 block, and a β-invertible S₃ is a homomorphism ρ : S₃ → GL₄(ZMod N) into these symmetries with det β(ρ g) a unit for every g ≠ 1. Only this group-theoretic statement is settled; the physical consequence rests on the Etingof–Nikshych–Ostrik classification of G-extensions and is not formalized.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.result

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Discrete structuresQuestion resolvedSource 2026Affine support of the odd cyclic sign-matrix traceCyclic overlap constraints bound the constant-derivative space and assemble the complete odd-length affine-support theorem.Proved

The questionRicardorr (MathOverflow user 49534; real-name attribution unverified) asks in question 512655, revision 1, GUID E06A6743-2C15-41CD-9390-04D1862ABA02:

Why it mattersThe question connects an ordered matrix trace to the affine geometry of a Boolean Fourier support.

The ideaProve the odd-cycle inverse and change variables.

Our contributionCyclic overlap constraints bound the constant-derivative space and assemble the complete odd-length affine-support theorem.

Verification details
Release source recordProved in Lean

Open problem

Affine support of the odd cyclic sign-matrix trace

Ricardorr (MathOverflow user 49534; real-name attribution unverified) asks in question 512655, revision 1, GUID E06A6743-2C15-41CD-9390-04D1862ABA02:

For every odd $n$ and every nonzero $b$, does the affine hull of $\mathrm{Supp}(b)$ have dimension at least $n-1$? Equivalently: is $\mathrm{Supp}(b)$ contained in no affine subspace of codimension $\ge 2$?

The original revision is attributed under CC BY-SA 4.0. The title's hyperplane wording is a locator; the target is the dimension lower bound in the body.

Precisely, let $n$ be any positive odd integer, $b\in\mathbb F_2^n$ nonzero, and $a\in\mathbb F_2^n$. All indices are modulo $n$. Put $g_k=a_{k-1}$ and $h_k=a_k+a_{k-2}+b_k$ in $\mathbb F_2$, and define the integer matrix

$$ A_k(a)=\begin{pmatrix}1&1\(-1)^{h_k}&(-1)^{h_k+g_k}\end{pmatrix}. $$

Equivalently, its entry indexed by $s,t\in{0,1}$ is $(-1)^{h_ks+g_kst}$. With the literal ordered integer product, set

$$ W_b(a)=\operatorname{Tr}(A_0(a)A_1(a)\cdots A_{n-1}(a)),\qquad \mathrm{Supp}(b)={a\in\mathbb F_2^n:W_b(a)\ne0}. $$

The assertion is $\dim_{\mathbb F_2}\operatorname{aff}(\mathrm{Supp}(b))\ge n-1$. Length one, coincident cyclic indices, every nonzero mask, and every tested word including zero and all-one are included. There is no length or weight cutoff, additional nonempty-support premise, or equality assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CyclicBoolean/CyclicSignTrace/RankAssembly.affine_trace_support_dimension

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    Year recorded in the source bibliography: 2026.

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Discrete structuresConjecture provedFirst preprint 2026Sahbi Conjecture 6.4For every Boolean cube of dimension at least two, the largest number of colors in a sub-quorum coloring is exactly one parity class, namely two to the power n-1.Proved

The questionFor every integer n>=2, prove that the sub-quorum chromatic number of the Boolean cube is

Why it mattersSahbi's source proves the equality through dimension six and poses the all-dimensional statement as Conjecture 6.4.

The ideaFor an arbitrary admissible coloring, divide the colors into those with an internal colored edge and those without.

Our contributionFor every Boolean cube of dimension at least two, the largest number of colors in a sub-quorum coloring is exactly one parity class, namely two to the power n-1.

Further research on sub-quorum colorings of graphs — see more details here: arXiv.

Verification details
Release source recordProved in Lean

Open problem

Sahbi Conjecture 6.4

For every integer n>=2, prove that the sub-quorum chromatic number of the Boolean cube is

psi_sq(Q_n)=2^(n-1).

Definition 2.2 permits a partial coloring: it is an onto map from a colored support S to a positive set of k colors. At every v in S, the center and the same-color colored neighbors must comprise at least half of the center and all colored neighbors. The center is counted once and uncolored neighbors are excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/HypercubeSubQuorum.subQuorumChromaticNumber_hypercube

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Discrete structuresQuestion resolvedFirst preprint 2026Tyagi's sharper Sprague–Grundy bound for Divisor NimFor every nonempty board of positive heaps, the Grundy value is at most twice the smallest heap.Proved

The questionSatyam Tyagi, “Removing m from m-pile Divisor Nim”, arXiv:2610.06925v1, §5, printed p. 23, states:

Why it mattersThe source proves g(P) ≤ 2hmin(P) + ⌊log₂ hmin(P)⌋, independently of the number and sizes of the other heaps.

The ideaLet k be the minimum 2-adic valuation and let r count heaps at that valuation.

Our contributionFor every nonempty board of positive heaps, the Grundy value is at most twice the smallest heap.

Verification details
Release source recordProved in Lean

Open problem

Tyagi's sharper Sprague–Grundy bound for Divisor Nim

Satyam Tyagi, “Removing m from m-pile Divisor Nim”, arXiv:2610.06925v1, §5, printed p. 23, states:

For every nonempty position P, g(P) ≤ 2hmin(P).

A position is a finite multiset of positive natural numbers. A move chooses one occurrence of a heap h and a positive integer d ≤ h, requires d to divide every other heap, and replaces h by h − d, omitting a zero remainder. The removal need not divide h. The Grundy value is the mex of follower values.

The Lean proposition claim states that every positive nonempty multiset P and every member m that is at most every member of P satisfy g(P) ≤ 2m. The theorem result : claim preserves these quantifiers and the literal move rule, including repeated heaps and singletons.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Games/DivisorNimBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2022Aliabadi–Zerbib Question 2.7 has a negative answerA sparse paving integer matroid need not be matched to itself.Refuted

The questionZerbib, “Matchings in matroids over abelian groups”, arXiv:2202.07719v3, published in Journal of Algebraic Combinatorics (2024), Question 2.7, ask:

Why it mattersThe resulting matroid and its dual remain paving.

The ideaThe four-slot escape audit for result is unfinished under 20096; its source realization, enrolled readout, variation/sensitivity evidence and continuation record are missing.

Our contributionA sparse paving integer matroid need not be matched to itself.

Verification details
Release source recordRefuted in Lean

Open problem

Aliabadi–Zerbib Question 2.7 has a negative answer

M. Aliabadi and S. Zerbib, “Matchings in matroids over abelian groups”, arXiv:2202.07719v3, published in Journal of Algebraic Combinatorics (2024), Question 2.7, ask:

Let $G$ be an abelian torsion-free group. Let $M$ be a matroid over $G$ and $N$ be a sparse paving matroid over $G$, both of the same rank $n$, so that $|E(M)| = |E(N)| - 1$ or $|E(M)| = |E(N)|$. If $0 \notin E(N)$, is $M$ matched to $N$?

The source note records the DOI and arXiv locator. The preregistration is #19681. The answer is Refuted by the settling declaration D5/S3/Combinatorics/Matroids/AliabadiZerbibMatroidMatching.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Abelian torsion-free group | G, [AddCommGroup G], [IsAddTorsionFree G] in claim | The group is arbitrary within the stated class | | Finite loopless matroids | M.Finite, N.Finite, M.Loopless, N.Loopless | The source standing convention is explicit in the proposition | | Same positive rank | 0 < M.eRank, M.eRank = N.eRank | Rank is Mathlib's extended natural rank | | Sparse paving target | IsSparsePaving N | Both the matroid and its dual are paving | | Ground-size alternatives | M.E.ncard + 1 = N.E.ncard ∨ M.E.ncard = N.E.ncard | Both alternatives are retained | | Exclusion of zero | (0 : G) ∉ N.E | The source's zero condition is literal | | Matching of ordered bases | BasesMatched and IsMatchedTo | Injective Fin n enumerations and Mathlib bases are used | | Negative answer | result : ¬ claim | The proposition is refuted by an exact integer witness |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Matroids/AliabadiZerbibMatroidMatching.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2024The steering substitution need not preserve powersAn admissible five-dimensional projective family fails steering power preservation.Refuted

The questionPhys. 27, 064501 (2025), arXiv:2410.20477v2, Appendix “Robust self-testing”, Conjecture 1:

Why it mattersThe substitution transfers the source's reference Heisenberg–Weyl operators into operators formed from Bob's unknown measurements.

The ideaAll witness objects below are the module's fixed five-dimensional objects.

Our contributionAn admissible five-dimensional projective family fails steering power preservation.

Scope & next questionThe module does not settle the source's assertion for all admissible qutrit phases, nor classify primes beyond five. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

The steering substitution need not preserve powers

A. C. Orthey Jr and R. Augusiak, Certifying classes of $d$-outcome measurements with quantum steering, New J. Phys. 27, 064501 (2025), arXiv:2410.20477v2, Appendix “Robust self-testing”, Conjecture 1:

For any prime $d$, we have that $\widetilde{B}_k^{(n)}=\widetilde{B}_k^n$, for every $n\in{1,\ldots,d-1}$. Note that this statement is true for $d=3$.

This quotation is from the downloaded v2 source main2.tex, line 1272, MD5 b3c5b40a34f721352981dcf58659087e. Its projective setting is explicit at line 1263:

For simplicity, we will assume that the state is pure and the measurements performed by Bob are projective.

Write $\omega=e^{2\pi\mathrm i/d}$ and let $\oplus$ denote addition modulo $d$. The coefficients and substitution of lines 204–206 and 246–249 are

$\gamma_{yk}^{(n)}=\frac1d\omega^{kn(n-1)/2}\sum_{l=0}^{d-1}e^{-\mathrm i\sum_{m=1}^n\phi_{l\oplus m,y}}\omega^{knl}$,

$\widetilde B_k^{(n)}=\sum_y\gamma_{yk}^{(n)}B_y^n$, with $\widetilde B_k=\widetilde B_k^{(1)}$ in the projective setting of lines 293–305.

Issue #21069 fixes the following reading. For every prime $d$, every real phase array $\phi$ satisfying $\sum_j\phi_{j,y}=0$ for every $y$ and

$\sum_k\overline{\gamma_{y'k}^{(n)}}\gamma_{yk}^{(n)}=\delta_{yy'}$

for every $1\le n\le d-1$ and every $y,y'$, every family of unitary $d\times d$ complex matrices $B_y$ with $B_y^d=I$, every $k\in{0,\ldots,d-1}$ and every $1\le n\le d-1$, the identity $\widetilde B_k^{(n)}=(\widetilde B_k^{(1)})^n$ holds. This is claim; result refutes it. The definitions and source locators are also given in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/SteeringPowerPreservationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2018Three two-strand boxes refute Jones–Wenzl generation at a root of unityThree two-strand Jones-Wenzl projectors at q = exp(pi i/6) give a corner not generated by its projected boundary generators.Refuted

The questionPeltola, Generators, projectors, and the Jones-Wenzl algebra (2018), arXiv:1811.12364v2, Section 1, Conjecture 1.2, TeX source WJalgebragen-sec1-intro.tex, lines 544–546, with macros expanded and references resolved:

Why it mattersThe distinction between existence of the individual projector boxes and generation of their whole corner determines whether the source's generator description extends to arbitrary admissible root-of-unity parameters.

The ideaAll the representation certificates below concern the fixed parameter and matrices defined in this module.

Our contributionThree two-strand Jones-Wenzl projectors at q = exp(pi i/6) give a corner not generated by its projected boundary generators.

Scope & next questionThe source's active references to Conjecture 1.2 discuss the proposed extension and its special cases (introduction, lines 539–556); no other named theorem is identified there as conditional on the conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Three two-strand boxes refute Jones–Wenzl generation at a root of unity

S. M. Flores and E. Peltola, Generators, projectors, and the Jones-Wenzl algebra (2018), arXiv:1811.12364v2, Section 1, Conjecture 1.2, TeX source WJalgebra_gen-sec1-intro.tex, lines 544–546, with macros expanded and references resolved:

Item 1 in theorem 1.1 holds whenever $\max \varsigma < \bar{\mathfrak p}(q)$.

Here $q\in\mathbb C^\times$, $\nu=-q-q^{-1}$ and $\varsigma=(s_1,\ldots,s_d)$ has positive entries, with total size $n_\varsigma=\sum_i s_i$. The source also permits the exceptional multiindex $(0)$. The bound $\bar{\mathfrak p}(q)$ is infinite at $q=\pm1$ and otherwise is the order of $q^2$, interpreted as infinite when that order is infinite. The composite Jones–Wenzl projector $P_\varsigma$ places projector boxes of sizes $s_i$ side by side, and $\mathrm{JW}\varsigma(\nu)=P\varsigma\mathrm{TL}{n\varsigma}(\nu)P_\varsigma$.

Item 1 of Theorem 1.1, at lines 502–516, states that, assuming $n_\varsigma<\bar{\mathfrak p}(q)$, the unit $P_\varsigma$ and the projected boundary generators $U_i^{\mathrm{JW}}=P_\varsigma U_{s_1+\cdots+s_i}^{\mathrm{TL}}P_\varsigma$, $1\le i<d$, form a minimal generating set. The conjecture replaces the total-size hypothesis by the individual-box hypothesis.

Issue #25910 fixes the generation assertion claim: for every nonzero complex $q$, every list $\varsigma$ of positive sizes individually below $\bar{\mathfrak p}(q)$, and every $T\in\mathrm{TL}{n\varsigma}(-q-q^{-1})$, the sandwich $P_\varsigma TP_\varsigma$ belongs to the ambient unital complex subalgebra generated by $P_\varsigma$ and the $U_i^{\mathrm{JW}}$. This assertion is refuted, which suffices to refute the conjecture independently of minimality. The definitions and source locator are in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/JonesWenzlGenerationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2015Zhang–Singh–Pati: strong subadditivity of subentropyA Bell pair and an independent maximally mixed qubit violate strong subadditivity of subentropy by exactly one twelfth.Refuted

The questionThe question attributed to arXiv:1510.08859v2, §7.1 is recorded in zhang2017subentropy.

Why it mattersSubentropy is a spectral information quantity whose subadditivity properties are distinct from those of von Neumann entropy.

The ideaPartial fractions at positive distinct eigenvalues give the spectral formula in datta2014subentropy; continuity is needed at the counterexample's repeated and zero eigenvalues.

Our contributionA Bell pair and an independent maximally mixed qubit violate strong subadditivity of subentropy by exactly one twelfth.

Verification details
Release source recordRefuted in Lean

Open problem

Zhang–Singh–Pati: strong subadditivity of subentropy

The question attributed to arXiv:1510.08859v2, §7.1 is recorded in zhang2017subentropy.

A similarly question can be asked for the strong subadditivity of subentropy: Q(ρ ABC) + Q(ρ B) ⩽ Q(ρ AB) + Q(ρ BC).

For every finite-dimensional tripartite density operator $\rho_{ABC}$, let $\rho_{AB}=\operatorname{Tr}C\rho{ABC}$, $\rho_{BC}=\operatorname{Tr}A\rho{ABC}$ and $\rho_B=\operatorname{Tr}{AC}\rho{ABC}$. The proposed assertion is

$$ Q(\rho_{ABC})+Q(\rho_B) \le Q(\rho_{AB})+Q(\rho_{BC}). $$

The formal statement quantifies over arbitrary finite coordinate types and the repository's DensityState; changing the parentheses of a product coordinate type only relabels matrix entries. The original assertion does not impose commutation, full rank, separability or purity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Subentropy/StrongSubadditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2021Asgarli–Yip two-dimensional pseudo-Paley cliquesThe two-dimensional cliques containing one in quartic pseudo-Paley graphs are exactly the odd-generator planes.Proved

The questionThe formal convention is GaloisField p 4, with ZMod p as the prime field, cyclotomicClass, connection, and PP as defined in the settling module.

Why it mattersThe source leaves Conjecture 5.9 open after a finite verification.

The ideaEqual classes give the cross-product equation for two affine parameters.

Our contributionThe two-dimensional cliques containing one in quartic pseudo-Paley graphs are exactly the odd-generator planes.

Verification details
Release source recordProved in Lean

Open problem

Asgarli–Yip two-dimensional pseudo-Paley cliques

Conjecture 5.9 of S. Asgarli and C. H. Yip, “The subspace structure of maximum cliques in pseudo-Paley graphs from unions of cyclotomic classes”, arXiv:2110.07176v5 (2024), asks for the classification of two-dimensional cliques containing $1$ in $PP(p^4,p+1,I)$:

Let $V$ be a $2$-dimensional subspace in $mathbb F_{p^4}$, such that $1\in V$. Then $V$ is a clique in $PP(p^4,p+1,I)$ for some $I$ if and only if $V=\mathbb F_p\oplus a\mathbb F_p$, where $a=g^{(p+1)k}$ and $k$ is odd.

The formal convention is GaloisField p 4, with ZMod p as the prime field, cyclotomicClass, connection, and PP as defined in the settling module. The quantified claim ranges over odd primes, primitive roots, two-dimensional subspaces containing $1$, and index sets contained in Finset.range (p+1) with cardinality $(p+1)/2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PseudoPaley/TwoDimensionalCliques.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2018Multipartite partial tightness of the entangled-measurement boundComputational and Fourier bases in disjoint sectors attain every multipartite partial bound with a multiple of d separable outcomes.Proved

The questionBrunner, Semi-device-independent characterisation of multipartite entangled states and measurements, arXiv:1805.00377v1 (Phys.

Why it mattersResult 2 turns an observed score into a lower bound on the number of entangled measurement operators.

Our contributionComputational and Fourier bases in disjoint sectors attain every multipartite partial bound with a multiple of d separable outcomes.

Verification details
Release source recordProved in Lean

Open problem

Multipartite partial tightness of the entangled-measurement bound

A. Tavakoli, A. A. Abbott, M.-O. Renou, N. Gisin and N. Brunner, Semi-device-independent characterisation of multipartite entangled states and measurements, arXiv:1805.00377v1 (Phys. Rev. A 98, 052333, 2018), Appendix C, SDI_manuscript.tex, line 497:

We leave it as an open question whether a similar partial tightness result holds in the more general $n$ partite case.

The source text at lines 474–496 concerns attainment of Eq. (10) for two parties with exactly $k=md$ fully separable measurement operators, for $m=0,\ldots,d$. Result 2, lines 265–272, bounds the average winning probability by $\mathcal A_{n,d}\le(d^n-k+k/d)/d^n$ whenever at least $k$ of the $d^n$ measurement operators are fully separable, for any shared state and local transformations in the source's scenario.

The game has uniformly distributed inputs $\mathbf x,\mathbf y\in(\mathbb Z_d)^n$ and winning output $C_1=\sum_jx_j$, $C_j=y_j-y_1$ for $j\ge2$, all modulo $d$; $\mathcal A_{n,d}=d^{-2n}\sum_{\mathbf x,\mathbf y}P(C(\mathbf x,\mathbf y)\mid\mathbf x,\mathbf y)$ (lines 137–159). The definitions and source locator are also in the literature note.

Issue #23843 fixes the affirmative reading: for every natural $n,d\ge2$ and $0\le m\le d^{n-1}$, there exist a density matrix on $(\mathbb C^d)^{\otimes n}$, local unitary encodings and a projective measurement with exactly $dm$ fully separable outcome operators and score $(d^n-dm+m)/d^n$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/MultipartiteEntangledMeasurementPartialTightness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundSource 2026Andoyo Question 5.2: infinitude of exact prime fibersThe exact odd-prime fiber of the attained balance -1 for seeds (0,3) is {3}.Refuted

The questionAndoyo, (a,b)-Fibonacci-Legendre Cordial Graphs and k-Pisano-Legendre Primes, Question 5.2, asks:

Why it mattersIts infinitude requirement concerns the exact unbounded fiber, rather than the number of observed occurrences in a prime sample.

The ideaFinite invertible recurrence-state dynamics gives a positive period.

Our contributionThe exact odd-prime fiber of the attained balance -1 for seeds (0,3) is {3}.

Verification details
Release source recordRefuted in Lean

Open problem

Andoyo Question 5.2: infinitude of exact prime fibers

Jason D. Andoyo, (a,b)-Fibonacci-Legendre Cordial Graphs and k-Pisano-Legendre Primes, Question 5.2, asks:

Suppose that $a$ and $b$ are integers with $(a,b)\ne(0,0)$. For every $k\in\chi(a,b)$, is $|\vartheta_{(a,b)}(k)|=\aleph_0$?

The exact locator is the versioned preprint, Question 5.2. The journal text has DOI 10.62072/acm.2026.09011. The existing Library citation andoyo2026pisanolegendrecollision identifies this same article and is reused.

Statement echo: $F_0=a$, $F_1=b$, $F_{n+2}=F_{n+1}+F_n$; $\pi_p(a,b)$ is the least positive scalar period, valid at every index including zero. For $j\in{-1,0,1}$, $N_j$ counts indices $0\le n<\pi_p(a,b)$ with Legendre symbol $j$, with multiplicity. Set $K=N_1-N_{-1}-N_0$ in the integers, $\chi(a,b)={K(p,a,b):p\text{ odd prime}}$, and $\vartheta_{(a,b)}(k)={p:p\text{ odd prime},K(p,a,b)=k}$. The Lean claim asserts that every such attained fiber is infinite, equivalently countably infinite because it is a subset of the natural numbers. Seeds are signed and may be nonprimitive; primes dividing both are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/PisanoLegendre/AndoyoPrimeFiberRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Equation (11) for the reduced abelian complexity of the Thue-Morse wordTwo Thue-Morse letters decide the even-length gap in reduced abelian complexity.Proved

The questionWrite rho(n) for the reduced abelian complexity of the Thue-Morse word: the number of classes of length-n factors, where two factors are equivalent when collapsing each maximal constant run to one letter leaves words of equal length that are rearrangements of one another.

Why it mattersThat module already supplies the all-start factor definitions, the run-collapsed Parikh vector, the code that decides equivalence, and the odd recurrence.

The ideaLet alternations n s count the letter changes across n edges in the length-n+1 factor at start s, and let minAlternations n and maxAlternations n be its extremes over all starts.

Our contributionTwo Thue-Morse letters decide the even-length gap in reduced abelian complexity.

Scope & next questionThis dossier's resolution concerns only the displayed absolute difference.

Verification details
Release source recordProved in Lean

Open problem

Equation (11) for the reduced abelian complexity of the Thue-Morse word

Write rho(n) for the reduced abelian complexity of the Thue-Morse word: the number of classes of length-n factors, where two factors are equivalent when collapsing each maximal constant run to one letter leaves words of equal length that are rearrangements of one another. Campbell, Currie and Rampersad display

|rho(4n+2) - rho(4n)| = 0  if t(n+1) = t(3n+1),  and 1 otherwise,

as equation (11) in the Conclusion of arXiv:2509.16034v1, with the sentence "It appears that (11) holds", and state a few lines later that they leave proving it as an open problem. The verified journal version retains equation (11) and this question in Section 3, printed page 14; its citation is recorded in Library/Words/campbell2025reduced.md. The paper's letter indices start at one, so the zero-indexed reading is t(n) = t(3n). Factor lengths do not shift.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Complexity/ThueMorseReducedAbelianEven.reducedAbelianComplexity_even_difference

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Symmetric Order and No Connections Do Not Force Full Palindromic ComplexityA four-letter path language has symmetric orders and no connection, but only three palindromes of length three.Refuted

The questionZamboni, Clustering, order conditions, and languages of interval exchanges, arXiv:2507.17370v2, Section 4, Conjecture 1, page 14:

The ideaThe departure order is numerical and the arrival order is its reverse.

Our contributionA four-letter path language has symmetric orders and no connection, but only three palindromes of length three.

Verification details
Release source recordRefuted in Lean

arXiv

Symmetric Order and No Connections Do Not Force Full Palindromic Complexity

Sébastien Ferenczi and Luca Q. Zamboni, Clustering, order conditions, and languages of interval exchanges, arXiv:2507.17370v2, Section 4, Conjecture 1, page 14:

A language on $k$ letters satisfying a symmetric order condition and with no connection has palindromic complexity $1$ for $n$ even, $k$ for $n$ odd.

The source language is factorial and two-sided extendable. A symmetric pair has arrival order opposite to departure order. Connections are the consecutive-letter, missing-cross-edge configurations of Definition 7. Recurrence and closure under reversal are not hypotheses of this conjecture. Preregistration: #15331.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/SymmetricOrderPalindromicComplexityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryConjecture provedSource 2025Fried's Fibonacci Shift ConjectureNo Fibonacci number equals the n-th Fibonacci number plus 2n plus one times the next one.Proved

The questionFried, Journal of Integer Sequences 28 (2025), Article 25.4.3, Section 6, states:

The ideaThe recurrence gives X n = F(n) + (2n + 1)F(n + 1).

Our contributionNo Fibonacci number equals the n-th Fibonacci number plus 2n plus one times the next one.

Verification details
Release source recordProved in Lean

Open problem

Fried's Fibonacci Shift Conjecture

Fried, Journal of Integer Sequences 28 (2025), Article 25.4.3, Section 6, states:

Nevertheless, we were not able to show that the two sets { nF(n/2 + 3) − (n − 1)F(n/2 + 2) : n ≥ 1 is even }, { F((n+1)/2 + 2) : n ≥ 1 is odd } are disjoint, or, equivalently, that for every n ∈ N, the number F(n + 2) + 2nF(n + 1) is not a Fibonacci number. We conjecture that this is so.

Here F(1) = F(2) = 1. Both sets are indexed by n ≥ 1, so the N of the equivalent form starts at one; n = 0 gives F(2) = 1, which is a Fibonacci number, and is outside the range.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FriedFibonacciShiftNonFibonacci.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument combines documented mathematical mechanisms.

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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryConjecture refutedSource 2005Graham–O'Bryant Conjecture 5.2: ordinary residue-set refutationGraham and O'Bryant's inverse-sine conditions admit noncanonical positive residue sets.Refuted

The questionOn printed page 302/PDF page 20, Graham and O'Bryant write:

Why it mattersThe source proposes the inverse-sine assertion as an auxiliary step toward restricting counterexamples to Fraenkel's tiling conjecture.

The ideaFor every n>=3, take q=2ⁿ−1 and pᵢ=2ⁱ except for the final entry pₙ₋₁=2ⁿ⁻¹−1.

Our contributionGraham and O'Bryant's inverse-sine conditions admit noncanonical positive residue sets.

Scope & next questionThe construction is not a Beatty partition, does not refute the main Fraenkel conjecture, and does not settle the proposed modulus bound for genuine Fraenkel counterexamples.

Verification details
Release source recordRefuted in Lean

Open problem

Graham–O'Bryant Conjecture 5.2: ordinary residue-set refutation

On printed page 302/PDF page 20, Graham and O'Bryant write:

Suppose that p₁, …, pₙ are distinct and relatively prime to q > (7/4)ⁿ, with ∑ pₖ ≤ q, and for each k ∈ [n], 2/sin(π/q) ≤ ∑ᵢ₌₁ⁿ 1/|sin(π pₖ p̄ᵢ/q)|. Then q = 2ⁿ − 1 and {p₁, …, pₙ} ≡ {1, 2, …, 2ⁿ⁻¹} (mod q).

The quantifiers range over every positive integer n and q and every n-tuple of distinct positive integer representatives coprime to q. The last set contains all n powers 2⁰ through 2ⁿ⁻¹. The barred quantity is a modular multiplicative inverse, and every complete row includes its diagonal. The conclusion compares ordinary residue sets, without independent signs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/GrahamObryantInverseSineRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2005.

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    The delivered argument combines documented mathematical mechanisms.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Year recorded in the source bibliography: 2005. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedHanna's Half-Iterate Conjecture Modulo ThreeThe normalized half-iterate of X + 4 X cubed has integral coefficients supported modulo three on powers of three.Proved

The questionNo nonvanishing assertion at powers of three is imposed.

Why it mattersThe congruence requires a common integral half-iterate and its reduction modulo three.

The ideaAn even correction cancels the next residual coefficient.

Our contributionThe normalized half-iterate of X + 4 X cubed has integral coefficients supported modulo three on powers of three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

hanna a107099 half iterate mod three

Paul D. Hanna's OEIS A107099 comment states: "Coefficients [x^n] A(x) = 0 (mod 3) except at n = 3^k (conjecture)." The Name states: "G.f. satisfies A(A(x)) = x + 4*x^3, where A(x) = Sum_{n>=0} a(n)x^(2n+1)." The displayed expansion begins with $x$, fixing the linear coefficient to one.

The exact binding claim of #18858 includes existence and integrality: there exists an odd $A\in\mathbb Q[[x]]$ with linear coefficient one and $A\circ A=x+4x^3$; for every such $A$ and every $n\in\mathbb N$, its coefficient equals an integer $m$, and $3\mid m$ whenever $n$ is not a power of three. No nonvanishing assertion at powers of three is imposed.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Odd generating function | IsGF: every even coefficient vanishes | The literal preregistered condition, including degree zero. | | Expansion begins with $x$ | IsGF: linear coefficient one | The source normalization. | | $A(A(x))=x+4x^3$ | IsGF: formal self-substitution equation | Substitution is Mathlib formal power-series substitution. | | Coefficients vanish modulo three outside $3^k$ | claim: every natural degree has an integer coefficient representative and the off-power implication | The rational coefficient is compared with the cast of that integer. | | The entry defines a generating function | claim: an IsGF solution exists | Prevents vacuity. Universal solution quantification introduces no additional hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaHalfIterateCubicModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedHanna's A323689 Gaussian coefficient paritySymmetric coefficient pairs leave exactly the odd diagonal contribution.Proved

The questionThe source is OEIS-only, with no identified DOI or arXiv id.

Why it mattersA symmetric integer array and fourth-root phases reduce parity to diagonal terms of a finite symmetric support.

Our contributionSymmetric coefficient pairs leave exactly the odd diagonal contribution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A323689 Gaussian coefficient parity

Paul D. Hanna's OEIS A323689 defines $\sum_{m\ge0}(x^m+i)^m/(1+ix^m)^{m+1}$, with constant coefficient replaced by $1$. Its COMMENTS conjecture says odd coefficients occur only at $N=r^2$; its FORMULA also asserts oddness at every such index. The preregistration is #21417. The source is OEIS-only, with no identified DOI or arXiv id.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Literal rational summand in NAME and FORMULA | HannaGaussianSquareParity.summand | Same powers, imaginary unit and formal inverse; no leading factor. | | Locally finite coefficient series | HannaGaussianSquareParity.A | Degree zero is $1$; at positive degree $N$, the sum has $1\le m\le N$. | | Real coefficients and integer parity | HannaGaussianSquareParity.claim | Every natural degree has an ordinary integer representative embedded in the complex numbers. | | Off-support conjecture and on-support FORMULA | HannaGaussianSquareParity.result | Oddness is equivalent to $N=r^2$, including degree zero. |

The settling result has no hypotheses. The public definitions retain the binding source conventions. The auxiliary recursive array and rational kernel are used within the proofs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaGaussianSquareParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedHanna's A324300 Gaussian coefficient paritySymmetric coefficient pairs leave exactly the odd diagonal contribution.Proved

The questionThe source is OEIS-only, with no identified DOI or arXiv id.

Why it mattersA symmetric integer array and fourth-root phases reduce parity to diagonal terms of a finite symmetric support.

Our contributionSymmetric coefficient pairs leave exactly the odd diagonal contribution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A324300 Gaussian coefficient parity

Paul D. Hanna's OEIS A324300 defines $\sum_{m\ge0}x^m(x^{m+1}+i)^m/(1+ix^{m+1})^{m+1}$. Its COMMENTS conjecture says odd coefficients occur only at $N=k(k+2)$; its FORMULA also asserts oddness at every such index. The preregistration is #21418. The source is OEIS-only, with no identified DOI or arXiv id.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Literal rational summand in NAME and FORMULA | HannaGaussianShiftedSquareParity.summand | Same powers, imaginary unit, formal inverse and leading factor $X^m$. | | Locally finite coefficient series | HannaGaussianShiftedSquareParity.A | Degree $N$ sums $0\le m\le N$; no constant replacement. | | Real coefficients and integer parity | HannaGaussianShiftedSquareParity.claim | Every natural degree has an ordinary integer representative embedded in the complex numbers. | | Off-support conjecture and on-support FORMULA | HannaGaussianShiftedSquareParity.result | Oddness is equivalent to $N=k(k+2)$, including degree zero. |

The settling result has no hypotheses. The public definitions retain the binding source conventions. The auxiliary recursive array and rational kernel are used within the proofs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaGaussianShiftedSquareParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Kauffman–Mukherjee–Vojtěchovský Problem 5.24Finite conjugation racks answer Problem 5.22 negatively and Problem 5.24 positively.Refuted

The questionPrinted page 21 of arXiv:2504.09368v1, section 5.5, states:

Why it mattersThe source presents this as one of several open problems about finite connected racks.

The ideaThe subset {(1 2 3),(1 3 2),(4 5 6),(4 6 5)}, indexed by {40,45,2,4}, is an inclusion-maximal R-clique of size 4.

Our contributionFinite conjugation racks answer Problem 5.22 negatively and Problem 5.24 positively.

Verification details
Release source recordRefuted in Lean

Open problem

Kauffman–Mukherjee–Vojtěchovský Problem 5.24

Printed page 21 of arXiv:2504.09368v1, section 5.5, states:

Problem 5.24. Does there exist a finite connected rack Q and a maximal R-clique C of Q such that |C| does not divide |Q|?

The printed question is existential. The preregistered claim24 is its negation: every inclusion-maximal R-clique in every finite connected rack has cardinality dividing the carrier cardinality. The result refutes this universal divisibility assertion; it answers the printed existence question Yes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KauffmanMukherjeeVojtechovskyRCliqueRefutations.result24

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Kauffman–Mukherjee–Vojtěchovský Problem 5.22Finite conjugation racks answer Problem 5.22 negatively and Problem 5.24 positively.Refuted

The questionPrinted page 21 of arXiv:2504.09368v1, section 5.5, states:

Why it mattersThe source presents this as one of several open problems about finite connected racks.

The ideaOn points 0 through 7, let C₇ contain the seven translations x ↦ x XOR a for a = 1,…,7.

Our contributionFinite conjugation racks answer Problem 5.22 negatively and Problem 5.24 positively.

Verification details
Release source recordRefuted in Lean

Open problem

Kauffman–Mukherjee–Vojtěchovský Problem 5.22

Printed page 21 of arXiv:2504.09368v1, section 5.5, states:

Problem 5.22. Do all maximal R-cliques in a finite connected rack have the same size?

The affirmative reading says that any two inclusion-maximal R-cliques C and D have equal cardinalities. The formal claim22 quantifies over every Q : Type with Fintype Q and DecidableEq Q, every binary operation mul, and both cliques, with IsRack mul and Connected mul as premises.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KauffmanMukherjeeVojtechovskyRCliqueRefutations.result22

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026The single-pass butterfly fails the relative exterior-square two-design boundA minor of every single-pass stride-doubling unitary butterfly vanishes.Refuted

The questionIordanis Kerenidis, Scalable Quantum Machine Learning: Trainability, Expressivity and Efficiency, arXiv:2607.24014v2, Conjecture 20 (quantumv2.tex, conj:lambda2design), asserts:

Why it mattersThe lower relative bound alone forces ε ≥ 1 in every dimension 2^K with K ≥ 2, contradicting an O(1/n) error.

The ideaThe phase at mode j is exp(i (sumi φi/2 − φj)), retaining the global vacuum phase.

Our contributionA minor of every single-pass stride-doubling unitary butterfly vanishes.

Verification details
Release source recordRefuted in Lean

Open problem

The single-pass butterfly fails the relative exterior-square two-design bound

Iordanis Kerenidis, Scalable Quantum Machine Learning: Trainability, Expressivity and Efficiency, arXiv:2607.24014v2, Conjecture 20 (quantum_v2.tex, conj:lambda2_design), asserts:

The unitary butterfly W_n at uniformly random parameters is an ε-approximate unitary 2-design on U(n) in the antisymmetric 2-particle representation Λ² ℂ^n, in the relative (multiplicative) sense, with ε = O(1/n).

The displayed assertion is (1−ε) Φ₂^(Λ²,Haar) ⪯ Φ₂^(Λ²,W_n) ⪯ (1+ε) Φ₂^(Λ²,Haar) as completely positive maps on End(Λ²ℂ^n ⊗ Λ²ℂ^n). The Library note quotes the complete conjecture, including its separate fixed-point-space sentence, and the literal circuit definition. Issue #11543 preregisters the quantified reading: there are a real c and natural N₀ such that every n=2^K ≥ N₀ admits 0 ≤ ε ≤ c/n satisfying both inequalities. The circuit has one pass through increasing strides, with the full phase layer before each disjoint RBS layer.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ButterflyLambdaTwoRelativeDesignRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The minimum number of arcs requiring nSSPReal centralizers and directed cycle spaces constrain exact nSSP supports.Proved

The questionSara Koljančić and Polona Oblak, Combinatorial aspects of the nonsymmetric strong spectral property for graphs, Problem 1.8, printed page 759:

Why it mattersThe assertion determines the sharp support-size threshold for universal real nSSP at every positive order.

The ideaThe lower-bound argument embeds an n-dimensional real space in the centralizer, using the classical polynomial-module structure theorem.

Our contributionReal centralizers and directed cycle spaces constrain exact nSSP supports.

Verification details
Release source recordProved in Lean

Open problem

The minimum number of arcs requiring nSSP

Sara Koljančić and Polona Oblak, Combinatorial aspects of the nonsymmetric strong spectral property for graphs, Problem 1.8, printed page 759:

Is the minimum number of arcs in a directed graph G on n vertices that requires the nSSP equal to 2n−1?

The source and preregistration are linked from issue #25108. The Library note records source versions, attribution and licensing. The endpoint-looped double path is the source's Theorem 5.5 and Remark 2.10(1), not a new upper construction.

| Source clause | Exact Lean meaning | | --- | --- | | n vertices | Fin n, for every natural n with 1 ≤ n | | Directed graph, loops allowed | Finset (Fin n × Fin n); each ordered pair appears at most once | | Exact matrix support | A i j ≠ 0 if and only if (i,j) belongs to E | | nSSP | Every real X with A i j * X i j = 0 and A * X.transpose − X.transpose * A = 0 is zero | | Requires nSSP | Every real A of that exact support has nSSP | | Minimum equals 2n−1 | All requiring E have at least 2n−1 arcs, and at the same n some requiring E has exactly 2n−1 arcs |

These definitions occur in D5.S3.Combinatorics.SpectralGraph.KoljancicOblakMinimumArcs. MinimumArcs is the entire two-clause assertion; result states it directly. There is no additional identity-Hadamard, trace, sign, connectivity, symmetry, eigenvalue, genericity or finite-order restriction in that assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/KoljancicOblakMinimumArcs.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Even double paths with a nonempty loop set allow nSSPExact-support witnesses for every positive even double path and every nonempty loop set.Proved

The questionKoljančić–Oblak Question 5.11 asks whether every positive even order n and every nonempty loop set L on the double path admit a real nSSP matrix with exactly that pattern.

Why it mattersAn exact-support witness for every nonempty loop set requires both a one-loop construction at any vertex and a support enlargement that preserves injectivity.

The ideaChoose a single loop in L, construct a symmetric weighted path with that loop, and enlarge the exact support after establishing nSSP.

Our contributionExact-support witnesses for every positive even double path and every nonempty loop set.

Scope & next questionThe theorem does not settle odd orders or the empty-loop-set question.

Verification details
Release source recordProved in Lean

Open problem

Even double paths with a nonempty loop set allow nSSP

Koljančić–Oblak Question 5.11 asks whether every positive even order n and every nonempty loop set L on the double path admit a real nSSP matrix with exactly that pattern. Both orientations of every consecutive path edge occur, and only the loops at vertices in L are added. The final witness has no sign, symmetry, trace, spectrum or genericity restriction. The real test matrix X is unrestricted; the defining equations are entrywise A times X equal to zero and A times transpose X minus transpose X times A equal to zero.

The exact Lean proposition is D5.S3.Combinatorics.SpectralGraph.KoljancicOblakEvenPath.claim. Unit #25129, preregistration comment 6554378006, owns this event. The companion D5/S3/Combinatorics/SpectralGraph/KoljancicOblakEvenPathNSSP.result has this exact proposition as its type. Both mathematical hosts are frozen. The result proves the full event stated here.

The source is Sara Koljančić and Polona Oblak, Combinatorial aspects of the non-symmetric Strong Spectral Property for graphs, arXiv:2604.21552v1, Question 5.11, with Definitions 1.1, 1.3 and 1.5 fixing nSSP, allowing a pattern and double-path loops. The later Electronic Journal of Linear Algebra publication, volume 42, pages 755–779, DOI 10.13001/ela.2026.10523, states Question 5.11 on PDF page 22, printed page 776. The linked Library note owns the source and license correspondence.

For every natural n with 0 < n and Even n, and every nonempty finite set L of vertices in Fin n, there exists a real matrix A indexed by Fin n whose nonzero entries are exactly the two orientations of consecutive path edges and the diagonal entries selected by L. For every real matrix X on the same indices, A ∘ X = 0 together with A * Xᵀ − Xᵀ * A = 0 implies X = 0. Here ∘ is the entrywise product. This includes n = 2 and singleton endpoint loops; no restriction on X or extra premise on A is added to the event.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/KoljancicOblakEvenPathNSSP.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2025Independent extreme-point tests in polyhedral Banach spacesA complete real Banach plane refutes the assertion that any independent extreme-point basis detects scalar isometries in every polyhedral space.Refuted

The questionManna, Mandal, Paul and Sain, An improvement of the Blanco-Koldobsky-Turnšek characterization of isometries, arXiv:2511.20345v1, final Open Question, S2.p45.1:

Why it mattersThe distinction between a spanning test set and the geometry of its normal cones determines whether local tests force a global isometry.

The ideaAlgebraic coordinates give dimension two; finite-dimensional completeness and continuity give a Banach space and an actual bounded endomorphism.

Our contributionA complete real Banach plane refutes the assertion that any independent extreme-point basis detects scalar isometries in every polyhedral space.

Verification details
Release source recordRefuted in Lean

arXiv

Independent extreme-point tests in polyhedral Banach spaces

Manna, Mandal, Paul and Sain, An improvement of the Blanco-Koldobsky-Turnšek characterization of isometries, arXiv:2511.20345v1, final Open Question, S2.p45.1:

Open Question: In an n-dimensional polyhedral Banach space, does the preservation of Birkhoff-James orthogonality by a bounded linear operator at any n linearly independent extreme points necessarily imply that the operator is a scalar multiple of an isometry?

This dossier anchors only this question. The standing field is real or complex; local preservation means every direction and every base-field scalar. The test tuple is chosen independently, rather than requiring preservation simultaneously at every independent tuple. The conclusion permits any one common scalar, including zero, and a linear isometry of the same carrier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BanachGeometry/PolyhedralBJRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedSource 2026Level Sequences Avoiding 101 and 102The level sequences avoiding 101 and 102 are counted by the Catalan numbers.Proved

The questionToufik Mansour, Wilf Classes for Level Sequences Avoiding Patterns of Length Three, Mathematics 14(11), 1983 (2026), Section 4:

The ideaA word avoids 101 and 102 exactly when every descent is permanent: after wi > wj with i < j, every later entry is smaller than wi.

Our contributionThe level sequences avoiding 101 and 102 are counted by the Catalan numbers.

Verification details
Release source recordProved in Lean

Open problem

Level Sequences Avoiding 101 and 102

Toufik Mansour, Wilf Classes for Level Sequences Avoiding Patterns of Length Three, Mathematics 14(11), 1983 (2026), Section 4:

Problem 1. We have that F_{101,102}(x) = (1 − √(1 − 4x))/(2x) − 1.

A level sequence of length n is a sequence w_1 ⋯ w_n of nonnegative integers with w_1 = 0 and w_i ≤ 1 + lev(w_1 ⋯ w_{i−1}), where lev counts adjacent equal entries; the problem asks for the number of level sequences avoiding the patterns 101 and 102.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/LevelSequence/LevelSequenceCatalan.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2026.

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedFirst preprint 2026Nathanson Problem 9: existence of a nontrivial proper spectrumEvery infinite group admits a decreasing sequence with a nontrivial proper product intersection spectrum.Proved

The questionNathanson, Problems and results on intersections of product sets and sumsets in semigroups, arXiv:2604.04781v2, Section 4, Problem 9, asks:

Why it mattersThere is one selected unit and one Proved OpenProblemResolutionClaim.

The ideaIf the cube range is finite, finite pigeonhole gives an infinite common-cube fibre.

Our contributionEvery infinite group admits a decreasing sequence with a nontrivial proper product intersection spectrum.

Verification details
Release source recordProved in Lean

Open problem

Nathanson Problem 9: existence of a nontrivial proper spectrum

Melvyn B. Nathanson, Problems and results on intersections of product sets and sumsets in semigroups, arXiv:2604.04781v2, Section 4, Problem 9, asks:

Let G be an infinite torsion group. Describe the product intersection sets H_Q(G). Must there exist X in H_N^*(G) with X != {1} and X != N?

This dossier anchors exactly the final existence question. The preceding classification request is outside its selected unit. The source uses N = {1,2,3,...}, ordered h-fold products with repetitions, and H_N^*(G) for the spectra of asymptotically strictly decreasing sequences of subsets of the same group G. Empty common intersections are allowed.

Source correspondence

| Source clause | Canonical counterpart | | --- | --- | | Every infinite torsion group | result quantifies over {G : Type u} [Group G] [Infinite G]; it proves the statement for all infinite groups, so every torsion instance is included. | | Arbitrary cardinality and group law | There is no Countable G, CommGroup G, bounded-exponent or subgroup assumption. | | Positive-indexed subsets of G | A : PNat → Set G; q+1 is positive successor. | | Adjacent inclusion and infinitely many strict changes | ASD A is (∀ q, A (q+1) ⊆ A q) ∧ {q | A q ≠ A (q+1)}.Infinite. | | Arbitrarily late strict changes | Infinitude of that subset of positive naturals implies ∀ q₀, ∃ q ≥ q₀, A (q+1) ≠ A q; each constructed family is in fact strictly decreasing at every index. | | Ordered h-fold products, repetitions allowed | Mathlib pointwise set multiplication and natural powers; both factor orders and all triple support and sign patterns are covered. | | Core A and full positive spectrum | spectrum A = {h : Nat | 0 < h ∧ (⋂ q, A q)^h = ⋂ q, (A q)^h}. | | X in H_N^*(G) | Take X = spectrum A; ASD A supplies the defining membership. | | X differs from {1} and from N | Both inequalities in result concern the complete set spectrum A, with exponent zero excluded. |

The exact torsion-specialized, arbitrarily-late formulation is an application of this single theorem and receives no additional retained theorem or resolution claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GroupWords/NathansonProblem9.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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Discrete structuresRecurrence provedPosed in 2002Kurkov's highest-bit recurrence for inverse Gray codeThe inverse Gray-code sequence OEIS A006068 satisfies Kurkov's highest-bit recurrence.Proved

The questionConjecture: a(n) = a(A053645(A063946(n))) + A053644(n) for n > 0 with a(0) = 0. - Mikhail Kurkov, Sep 09 2023

Why it mattersThe result connects this recurrence to Hanna's independent XOR-prefix definition and proves the inverse property for every natural input.

The ideaDefine a by Hanna's recursion with termination on the natural input.

Our contributionThe inverse Gray-code sequence OEIS A006068 satisfies Kurkov's highest-bit recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's highest-bit recurrence for inverse Gray code

OEIS A006068, NAME (%N, verbatim):

a(n) is Gray-coded into n.

Mikhail Kurkov's conjecture (%F, verbatim):

Conjecture: a(n) = a(A053645(A063946(n))) + A053644(n) for n > 0 with a(0) = 0. - Mikhail Kurkov, Sep 09 2023

The AUTHOR line (%A, verbatim) is:

N. J. A. Sloane

Hanna's unconjectured XOR-prefix formula (%F, June 4, 2002) is:

a(n) = n XOR [n/2] XOR [n/4] XOR [n/8] ... XOR [n/2^m] where m = [log(n)/log(2)] (for n>0) and [x] is integer floor of x.

His inverse formula (%F, January 18, 2012) is:

a(n) XOR [a(n)/2] = n.

The exact formal result is:

(∀ n : ℕ, gray (a n) = n) ∧ a 0 = 0 ∧
∀ n : ℕ, 0 < n →
  a n = a (complementSecondBit n - msb (complementSecondBit n)) + msb n

Here gray is imported from the frozen D5/S1/Digit/Admissibility/GrayCodeBinaryRecurrenceClosedForm. The definition of a is Hanna's XOR-prefix formula written as the recursion n XOR a(n/2), terminating at zero. For positive inputs, msb n := 2 ^ Nat.log 2 n is A053644, and subtraction of this value is A053645. The totalized formula gives msb 0 = 1, while A053644(0) is 0; every msb argument in the positive recurrence is positive.

The definition of complementSecondBit is A063946, fixing zero and one. Its two arithmetic %F cases, proved and used inside result, are:

If 2*2^k <= n < 3*2^k then a(n) = n + 2^k; if 3*2^k <= n < 4*2^k then a(n) = n - 2^k

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/KurkovGrayInverseRecurrence.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    The inverse Gray-code sequence OEIS A006068 satisfies Kurkov's highest-bit recurrence.

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    Dated conjecture statement: 2002.

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Number theoryQuestion resolvedBala's A347854 fractional-factorial integralityProved

The questionThe integrality conjecture quantifies over every natural index, including zero.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347854 fractional-factorial integrality

Peter Bala, OEIS A347854, states:

a(n) = (6*n)!/((3*n)!*(2*n)!) * (n/2)!/(3*n/2)!.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347854BalaHalfIndexIntegral.a | Literal expression; natural factorials are cast to the reals. | | Fractional factorials use Gamma | OeisA347854BalaHalfIndexIntegral.ffact | $x! = \Gamma(1+x)$; the natural-index factorials are Nat.factorial. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347854BalaHalfIndexIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347854BalaHalfIndexIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347854BalaHalfIndexIntegral.result

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Number theoryQuestion resolvedBala's A347855 fractional-factorial integralityProved

The questionThe integrality conjecture quantifies over every natural index, including zero.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347855 fractional-factorial integrality

Peter Bala, OEIS A347855, states:

a(n) = (4*n)!/((2*n)!*(n)!) * (n/3)!/(4*n/3)!.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347855BalaThirdIndexIntegral.a | Literal expression; natural factorials are cast to the reals. | | Fractional factorials use Gamma | OeisA347854BalaHalfIndexIntegral.ffact | $x! = \Gamma(1+x)$; the natural-index factorials are Nat.factorial. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347855BalaThirdIndexIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347855BalaThirdIndexIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347855BalaThirdIndexIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedThe literal A361552 equation and ordinary partition parityThe literal pentagonal unit equation and Hanna's ordinary-partition parity assertion.Proved

The questionConjecture: a(n)/2 == A000041(n-1) (mod 2) for n >= 1; that is, a(n) is even, and a(n)/2 has the same parity as the number of partitions of n-1, for n >= 1.

Why it mattersThis is a first-tier externally published all-index conjecture from 2023.

The ideaThe original equation is equivalent to A⁻¹ = 1-2X+nonzeroSum A.

Our contributionThe literal pentagonal unit equation and Hanna's ordinary-partition parity assertion.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The literal A361552 equation and ordinary partition parity

Paul D. Hanna's OEIS A361552 revision #14 states:

Conjecture: a(n)/2 == A000041(n-1) (mod 2) for n >= 1; that is, a(n) is even, and a(n)/2 has the same parity as the number of partitions of n-1, for n >= 1.

The original definition is:

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following. (1) 2*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

The source, attribution, immutable locator and classical suppliers are in Library/Recurrence/hanna2023a361552.md. The preregistered exact question is https://github.com/the-omega-institute/trureturing/issues/16442.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/HannaPentagonalPartitionParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedThe A363104 bilateral generating function modulo fourThe integer solution of the A363104 bilateral equation is theta3 modulo four.Proved

The questionHanna, May 21 2023) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2023a363104.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2023 entry.

The ideaRead the bilateral NAME formally using bilateralTerm A n for integer n.

Our contributionThe integer solution of the A363104 bilateral equation is theta3 modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A363104 bilateral generating function modulo four

OEIS A363104 (Paul D. Hanna, May 21 2023) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2023a363104.md:

Expansion of g.f. A(x) satisfying 4 = Sum_{n=-oo..+oo} (-x)^n * (4*A(x) + x^(n-1))^(n+1).

Conjecture: g.f. A(x) == theta_3(x) (mod 4); a(n) == 2 (mod 4) iff n is a nonzero square and a(n) == 0 (mod 4) iff n is nonsquare.

The NAME is formalised coefficientwise with exact finite windows after the two Laurent terms cancel. mod_four_identity proves the generating-function clause; hanna_conjecture proves both coefficient biconditionals. This dossier has one resolution claim, attached only to hanna_conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/BilateralQuarterShiftThetaModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedParity of the A375439 coefficientsThe odd coefficients of OEIS A375439 occur exactly at powers of three and twice powers of three.Proved

The questionHanna, Aug 21 2024) supplies the following statements, copied from the claim in Library/Arith/hanna2024a375439.md:

Why it mattersThis is a first-tier recent OEIS conjecture selected in the implementation brief.

The ideaFrobenius modulo three gives cubecongrsubstthree: for any integer formal series B, B^3 and B(x^3) agree coefficientwise modulo three.

Our contributionThe odd coefficients of OEIS A375439 occur exactly at powers of three and twice powers of three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A375439 coefficients

OEIS A375439 (Paul D. Hanna, Aug 21 2024) supplies the following statements, copied from the claim in Library/Arith/hanna2024a375439.md:

NAME "Expansion of g.f. A(x) satisfying A(x) = x + x^2 + (2*A(x)^3 + A(x^3))/3."

COMMENT "Conjecture: a(n) is odd iff n is in A038754, which consists of numbers of the form 3^k and 2*3^k."

The sequence starts at index one; the integer formal generating series is normalized by its zero constant coefficient.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CubicTernarySubstitutionParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedCoefficients of the A378580 quotient theta composition modulo fourThe positive coefficients of OEIS A378580 are two modulo four exactly at square degrees.Proved

The questionHanna, Jan 08 2025) gives the following NAME and COMMENT, quoted verbatim from Library/ArithSums/hanna2025a378580.md.

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaThe escape content is inverseagreement: for power series with constant coefficient 1, the inverse-difference identity f⁻¹ − g⁻¹ = f⁻¹ (g − f) g⁻¹ shows that divisibility of f − g by X^d survives inversion, so X/f and X/g agree one degree further. quotienttriangular uses that to compare the quotient substitution arguments and isolate the degree-n outer difference.

Our contributionThe positive coefficients of OEIS A378580 are two modulo four exactly at square degrees.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficients of the A378580 quotient theta composition modulo four

OEIS A378580 (Paul D. Hanna, Jan 08 2025) gives the following NAME and COMMENT, quoted verbatim from Library/ArithSums/hanna2025a378580.md.

NAME:

G.f. A(x) satisfies: A(x/A(x)) = theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4) iff n is square, else a(n) is divisible by 4 if n is nonsquare.

Here A has integer coefficients and constant coefficient one; x/A(x) means X times its formal unit inverse. The Lean construction and uniqueness theorem identify the sequence defined by this quotient equation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/QuotientThetaCompositionModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedDyadic denominators of the A381670 compositional-square seriesEvery reduced denominator of the normalized compositional-square series is a power of two.Proved

The questionThe function A(x) = x+(1/2)x^2-(1/16)x^4... = Sum{k >= 0} x^kA381669(k)/a(k) satisfies the functional equation: x(A(x)+1) = A(A(x)).

Why it mattersThis is a first-tier OEIS conjecture from the 2025 entry.

The ideaThe delivered escape content is the integral rescaling F(x)=A(4x)/4.

Our contributionEvery reduced denominator of the normalized compositional-square series is a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dyadic denominators of the A381670 compositional-square series

OEIS A381670 (Thomas Scheuerle, Mar 03 2025), NAME, as quoted in Library/ArithSums/scheuerle2025a381670.md:

The function A(x) = x+(1/2)*x^2-(1/16)*x^4... = Sum_{k >= 0} x^k*A381669(k)/a(k) satisfies the functional equation: x*(A(x)+1) = A(A(x)).

The COMMENT that is the target, from the same Library note:

Conjecture: All terms are powers of two.

The brief separately supplied the purported comment:

All terms appear to be odd.

That sentence does not occur in the supplied Library note. It cannot describe these denominators: the displayed coefficient of x^2 has denominator 2. Its attribution is unverified, and it is not a second resolution claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/CompositionalSquareDyadicDenominators.scheuerle_conjecture

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    The delivered argument combines documented mathematical mechanisms.

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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedChoosable Initial Intervals and Reversed PartitionsDistinct representatives of increasing initial intervals exist exactly above the diagonal.Proved

The questionIts comment conjectures that the same number counts superdiagonal reversed partitions: the parts in increasing order satisfy yi >= i at every one-based position.

Why it mattersThis is a first-tier recent OEIS comment conjecture.

The ideaFor a weakly increasing list, restrict a representative map to its first i+1 positions.

Our contributionDistinct representatives of increasing initial intervals exist exactly above the diagonal.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Choosable Initial Intervals and Reversed Partitions

OEIS A388711 counts partitions of n into k positive parts whose initial intervals admit distinct representatives. Its comment conjectures that the same number counts superdiagonal reversed partitions: the parts in increasing order satisfy y_i >= i at every one-based position.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/ChoosableInitialIntervals.card_choosable_eq_superdiagonal

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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedDivisibility of the A396795 coefficientsEvery coefficient above degree one in OEIS A396795 is divisible by five.Proved

The questionHanna, Jun 06 2026, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2026a396795.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaThe escape content is iterateproductlift, general in the natural modulus m and in both natural iterate indices i, j.

Our contributionEvery coefficient above degree one in OEIS A396795 is divisible by five.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A396795 coefficients

OEIS A396795, Paul D. Hanna, Jun 06 2026, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2026a396795.md:

G.f. A(x) satisfies A^2(x) * A^3(x) = x^2 + 25*x^3, where A^n(x) is the n-th iteration of A(x).

Conjecture: a(n) == 0 (mod 5) for n > 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductTwentyFiveModFive.hanna_conjecture

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    The exact recorded statement is proved.

    Source 1
Quantum informationConjecture provedSource 2025Pronko's anti-adjoint expansion conjecture for the periodic Fredkin chainThe total spin operators and the anti-adjoint action are defined on the spin-word basis of the periodic Fredkin chain.Proved

The questionPronko fixes the one-site Pauli matrices in section 2.1, printed p. 3:

Why it mattersThe operators Σ^± commute with the periodic Fredkin Hamiltonian (Theorem 2 of the paper), while the total spin operators do not; the theorem expresses the nonlocal symmetry generators as explicit finite expressions in S^±.

The ideaFor spin words y, x let b count sites with (yi, xi) = (↑, ↓) and c count sites with (↓, ↑).

Our contributionThe total spin operators and the anti-adjoint action are defined on the spin-word basis of the periodic Fredkin chain.

Verification details
Release source recordProved in Lean

Open problem

Pronko's anti-adjoint expansion conjecture for the periodic Fredkin chain

Pronko fixes the one-site Pauli matrices in section 2.1, printed p. 3:

The basis in End(ℂ²) is provided by the Pauli matrices σ⁺ = [[0, 1], [0, 0]], σ⁻ = [[0, 0], [1, 0]], σᶻ = [[1, 0], [0, −1]].

Section 2.1, printed p. 3, normalizes the total spin operators in an unnumbered display:

S^± = Σ_{j=1}^N σ_j^±, S^z = ½ Σ_{j=1}^N σ_j^z.

Theorem 2, equation (3.1), printed pp. 6-7, defines

Σ^± = Σ_{r₁,...,r_N ∈ {−1,0,1}, r₁+⋯+r_N = ±1} σ₁^{r₁} ⋯ σ_N^{r_N}, where σ_i^0 = 1 and σ_i^{±1} = σ_i^±.

Conjecture 2, printed p. 7, is:

For the operators Σ^± there exists the representation Σ^± = Σ_{k=1}^{⌈N/2⌉} γ_k (ãd S^± ãd S^∓)^{k−1} S^±, where γ_k are some coefficients and ãd denotes the anti-adjoint action, (ãd a) b ≡ {a, b} = ab + ba.

The paper gives Σ^± = −¼ S^± + ⅛ {S^±, {S^∓, S^±}} for N = 3 and the coefficients for N ≤ 10 in Table 1, printed p. 8.

Issue #9982 fixes the readings: (i) Σ^± is (3.1) literally; (ii) S^± has the section 2.1 normalization; (iii) (ãd S^± ãd S^∓)^{k−1} S^± is the map X ↦ {S^±, {S^∓, X}} iterated k − 1 times on S^±; (iv) one family of complex coefficients serves both signs; (v) every natural number N.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PronkoFredkinAntiAdjointExpansion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2025.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedSource 2025Separable states outside every local-unitary stabilizer frameA rank-two mixture of two nonorthogonal product states remains outside the full convex stabilizer hull under every local unitary.Proved

The questionQian and Wang, Quantum non-local nonstabilizerness, arXiv:2502.06393v4, Appendix F, conjecture that some separable multi-qubit states cannot be transformed into STAB using only local unitary transformations.

Why it mattersThe source defines separable states as convex hulls of pure product states and STAB as the convex hull of pure stabilizer states.

The ideaThe settling theorem D5.S3.Quantum.Information.SeparableStateLocalUnitaryStabilizerObstruction.result uses the family theorem on its live proof path.

Our contributionA rank-two mixture of two nonorthogonal product states remains outside the full convex stabilizer hull under every local unitary.

Verification details
Release source recordProved in Lean

Open problem

qian wang 2025 separable states stab local unitaries

Qian and Wang, Quantum non-local nonstabilizerness, arXiv:2502.06393v4, Appendix F, conjecture that some separable multi-qubit states cannot be transformed into STAB using only local unitary transformations. Their statement is that they are unaware of a proof that every separable state can be so transformed and conjecture that this is false.

The settled two-qubit claim is the preregistered QW-F statement from GitHub issue #14115: for real parameters $0 < p < 1$, $c > 0$, $s > 0$, $c^2+s^2=1$, and $c^2 \ne 1/2$, let $b=c|0\rangle+s|1\rangle$ and $$ \rho=p|00\rangle\langle00|+(1-p)|bb\rangle\langle bb|. $$ Then no $U_A\otimes U_B$ with single-qubit local unitaries sends $\rho$ into the full convex stabilizer hull STAB.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SeparableStateLocalUnitaryStabilizerObstruction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2025.

    Read source ↗
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    Source 1
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Number theoryConjecture refutedFirst preprint 2026Conjecture 3.1: APN functions and large binary flatsAn almost perfect nonlinear map on a field with sixteen elements sends a three-dimensional binary flat onto another three-dimensional binary flat.Refuted

The questionRodríguez-Aldama–Šehović–Pasalic–Kudin, arXiv:2609.22394v2, §3.1:

Why it mattersThe conjecture connects differential nonlinearity to the destruction of large affine subspaces.

The ideaLet W be the binary vector space with four coordinates, labeled (a₀,a₁,a₂,a₃) by a₀+2a₁+4a₂+8a₃.

Our contributionAn almost perfect nonlinear map on a field with sixteen elements sends a three-dimensional binary flat onto another three-dimensional binary flat.

Verification details
Release source recordRefuted in Lean

Open problem

Conjecture 3.1: APN functions and large binary flats

Rodríguez-Aldama–Šehović–Pasalic–Kudin, arXiv:2609.22394v2, §3.1:

Definition 3.1. We say that a mapping F : F_{2^n} → F_{2^n} breaks a flat A of F_{2^n} if F(A) is not a flat of F_{2^n}. If F breaks all k-dimensional flats, F is said to be k-breaking.

Conjecture 3.1. Any APN function F over F_{2^n} is k-breaking for ⌊n/2⌋ + 1 ≤ k ≤ n − 1.

A k-flat is a coset a+V of a k-dimensional binary subspace. APN means that F(x+a)+F(x)=b has at most two solutions for every nonzero a and every b. The Lean claim : Prop quantifies over all finite fields K of characteristic two and cardinality 2^n, every such APN function, every permitted k, every offset and every submodule of binary dimension k. Its conclusion excludes an image equal to a coset of a subspace of any dimension. Characteristic two provides the canonical ZMod 2-algebra and module. A decidable equality instance enables finite fiber counting and places no restriction on the mathematical fields.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/APNBreakingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026Rosenzweig–Stanfill Open Problem 1.3The Bell polynomial values in Rosenzweig and Stanfill's Open Problem 1.3 are nonzero at every positive natural index.Proved

The questionRosenzweig and Stanfill state in Open Problem 1.3 (arXiv:2606.04225v1, page 2):

The ideaThe proof works over the 2-adic valuation on ℚ.

Our contributionThe Bell polynomial values in Rosenzweig and Stanfill's Open Problem 1.3 are nonzero at every positive natural index.

Scope & next questionThe bridge from this result to the paper's separate Open Problem 1.6(i), and any analytic consequence requiring the paper's Theorem 1.2 and residue formula (1.7), remains open in this repository.

Verification details
Release source recordProved in Lean

Open problem

Rosenzweig–Stanfill Open Problem 1.3

Rosenzweig and Stanfill state in Open Problem 1.3 (arXiv:2606.04225v1, page 2):

Show that $p_{2m,S_1}(\tfrac{1}{2}-m)\neq0$ for all $m\in\mathbb{N}$ where the sequence $S_1$ satisfies \eqref{Eq:S1}.

Their Definition 1.1 gives the Bell-polynomial expression

Given a sequence of numbers, $S$, indexed over a set $J\supseteq \mathbb{N}$, we define $p_{j,S}(t):=\sum_{k=0}^j \frac{(-1)^k}{k!} \widehat{B}{j,k}(s_1,\dots,s{j-k+1})t^k=\frac{(-1)^j}{j!}\det\mathcal{N}_{j,S}(t),\quad j\in\mathbb{N}_0,$

and equation (1.8) gives

$s_k^{(1)}=(1-2^{1-k})\frac{2 B_{k}}{k}=-\frac{2}{k}B_k(1/2),\quad k\in\mathbb{N}.$

The Lean encoding uses m : ℕ with 1 ≤ m for the source's positive naturals and uses the first Bell-polynomial expression.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/LogLaplacianBellNonvanishing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Shah–Kiselev Conjecture 3.1: the late growing ratioThe Shah–Kiselev permutation ratio tends to one and exceeds one at every index at least two.Proved

The questionFor σ ∈ SN define the partial sums Tk := Σ{i=N−k+1}^{N} (σ(i) − 1 − p), k = 1, …, N − 1. … We denote by Φp the set of all permutations σ ∈ SN with σ(1) = 1 and Tk ⩾ 0 for all k = 1, …, N − 1.

Why it mattersThe factorial normalization counts circular orderings of the residual alphabet.

The ideaFor a finite integer alphabet S with distinct letters and sum zero, let L(S) count orderings with nonnegative prefix sums and P(S) count orderings with positive nonempty proper prefix sums.

Our contributionThe Shah–Kiselev permutation ratio tends to one and exceeds one at every index at least two.

Scope & next questionThis theorem does not settle that separate alphabet and normalization.

Verification details
Release source recordProved in Lean

Open problem

Shah–Kiselev Conjecture 3.1: the late growing ratio

Kian C. Shah and Arthemy V. Kiselev, arXiv:2610.08636v1, §3:

Definition 3.1. For σ ∈ S_N define the partial sums T_k := Σ_{i=N−k+1}^{N} (σ(i) − 1 − p), k = 1, …, N − 1. … We denote by Φ_p the set of all permutations σ ∈ S_N with σ(1) = 1 and T_k ⩾ 0 for all k = 1, …, N − 1.

Here N = 2p. The conjecture states:

Conjecture 3.1. As p → +∞, the ratio |Φ_p|/(2p − 2)! converges from above to 1 + 0.

The Lean claim uses the existing zero-based PhiTailEncoding.phi and asserts convergence to 1 together with ∀ n : ℕ, 2 ≤ n → (2*n-2).factorial < (phi n).card. At p = 1 the ratio equals 1. “From above” means the stated strict inequality, without an additional monotonicity assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/LateGrowingRatioLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Mathematical advance · 2/4 · 15% weight

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Discrete structuresConjecture provedFirst preprint 2026Sun Conjecture 5.2: both lowercase strict Turan clausesThe first complete closed-domain clause of Sun's Conjecture 5.2 holds at every positive index.Proved

The questionSun, arXiv:2608.13192v1, Conjecture 5.2 asks for both assertions, for every integer n >= 1:

Why it mattersThe new source target tests an all-index strict determinant property for two different literal three-term recurrences over negative real half-lines.

The ideaIn the formal result the order of the two real neighbor factors is reversed from the source's v display by commutativity.

Our contributionThe first complete closed-domain clause of Sun's Conjecture 5.2 holds at every positive index.

Scope & next questionA proof of only one clause, a finite-index check, or a non-strict statement does not settle this source question.

Verification details
Release source recordProved in Lean

Open problem

Sun Conjecture 5.2: both lowercase strict Turan clauses

Sun, arXiv:2608.13192v1, Conjecture 5.2 asks for both assertions, for every integer n >= 1:

$$ g_n(x)^2>g_{n-1}(x)g_{n+1}(x)\quad\text{for every real }x\leq-1, $$

and

$$ v_n(x)^2>v_{n+1}(x)v_{n-1}(x)\quad\text{for every real }x\leq-1/8. $$

The source's literal initial values and multiplied quadratic/cubic recurrences are reproduced in Library/Recurrence/sun2026generalizations.md. Both endpoints are included. A proof of only one clause, a finite-index check, or a non-strict statement does not settle this source question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Sun/GStrict.g_strict_turan

D5/S1/Recurrence/Sun/VStrict.v_strict_turan

Frozen module record

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Further frontiersQuestion resolvedFirst preprint 2025Tang–Zhang: the sharp order −2 Schoenberg inequalityTang–Zhang Conjecture 6.3 holds for every finite nonempty family of nonzero complex roots, with equality exactly for collinearity through the origin.Proved

The questionZhang, "Sharp Schoenberg type inequalities and the de Bruin–Sharma problem", arXiv:2508.10341v3 (5 Dec 2025), §6, p. 31.

Why it mattersIt is an explicit external conjecture in the first problem tier.

The ideaWrite bj=1/zj, D=diag(bj), J for the n by n all-ones matrix, and B=D(I+J)=D+DJ.

Our contributionTang–Zhang Conjecture 6.3 holds for every finite nonempty family of nonzero complex roots, with equality exactly for collinearity through the origin.

Scope & next questionWhether the same similarity yields sharp bounds for order −2m, including a derivation or refinement of Section 7's order −4 inequality, remains open here.

Verification details
Release source recordProved in Lean

Open problem

Tang–Zhang: the sharp order −2 Schoenberg inequality

Q. Tang, T. Zhang, "Sharp Schoenberg type inequalities and the de Bruin–Sharma problem", arXiv:2508.10341v3 (5 Dec 2025), §6, p. 31.

The identity (5.2) leads us to the following conjectural refinements of Theorem 6.1 in the case of order −2.

Conjecture 6.3. Let P(z) = z ∏_{j=1}^n (z − z_j), z_j ∈ C \ {0}, and let w_1, …, w_n denote the zeros of P′(z). Then Σ_{k=1}^n 1/|w_k|² ≤ 3 Σ_{j=1}^n 1/|z_j|² + |Σ_{j=1}^n 1/z_j|². (6.9) Moreover, equality in (6.9) holds whenever z_1, …, z_n all lie on the same line passing through the origin.

In the case n = 2 we can verify Conjecture 6.3 by an explicit computation. (Proposition 6.4.)

Theorem 6.1 there proves the weaker right side (n+3) Σ 1/|z_j|². The exact public definitions and result are:

noncomputable def P {n : ℕ} (z : Fin n → ℂ) : Polynomial ℂ :=
  Polynomial.X * ∏ j, (Polynomial.X - Polynomial.C (z j))

def claim : Prop := ∀ (n : ℕ) (hn : 1 ≤ n) (z : Fin n → ℂ), (∀ j, z j ≠ 0) →
    (((Polynomial.derivative (P z)).roots.map (fun w => ‖w‖⁻¹ ^ 2)).sum ≤
      3 * ∑ j, ‖z j‖⁻¹ ^ 2 + ‖∑ j, (z j)⁻¹‖ ^ 2) ∧
    ((((Polynomial.derivative (P z)).roots.map (fun w => ‖w‖⁻¹ ^ 2)).sum =
      3 * ∑ j, ‖z j‖⁻¹ ^ 2 + ‖∑ j, (z j)⁻¹‖ ^ 2) ↔
        ∀ j, ∃ r : ℝ, z j = (r : ℂ) * z ⟨0, hn⟩)

theorem result : claim

Here n≥1, all z_j are nonzero complex numbers, and roots of the derivative are counted with multiplicity. No simplicity, distinctness, reality or common argument hypothesis is imposed on the nonzero roots. Since P′(0)=∏_j(−z_j)≠0, no critical point is zero. The final equivalence says that all z_j are real multiples of the nonzero first root, which is exactly collinearity on a line through the origin.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/CriticalPoints/TangZhangOrderMinusTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Global-coalition graph universalityThe edgeless two-vertex graph has no global coalition realization, because every partition block has an actual coalition partner.Refuted

The questionAnholcer, Besharati, Mojdeh and Samadzadeh, Global coalitions in graphs, arXiv:2509.15386v4, Section 7 Problem 58, PDF p. 29, asks whether every graph is realisable as a global coalition graph.

Why it mattersEvery vertex of every global coalition graph has a distinct adjacent partner because the partition's partners field supplies an actual block forming a global coalition.

The ideaThe source note is Library/FiniteGeometry/anholcer2025globalcoalitions.md, with the v4 arXiv URL, v1 alias, TeX/PDF locators and comparison literature.

Our contributionThe edgeless two-vertex graph has no global coalition realization, because every partition block has an actual coalition partner.

Verification details
Release source recordRefuted in Lean

arXiv

Global-coalition graph universality

Anholcer, Besharati, Mojdeh and Samadzadeh, Global coalitions in graphs, arXiv:2509.15386v4, Section 7 Problem 58, PDF p. 29, asks whether every graph is realisable as a global coalition graph. The earlier alias is Besharati, Mojdeh and Samadzadeh, Global coalition sets in graphs, arXiv:2509.15386v1, Section 6 Problem 4, PDF p. 19. Both aliases identify one target.

Original question and definitions

The following is the complete question and its conventions as recorded in preregistration #23084:

The current complete question asks whether every finite simple undirected graph H of order at least two is isomorphic to the global coalition graph of some finite simple undirected host G of order at least two and some actual gc-partition of G. A gc-partition need not maximize the number of blocks. There is no connectedness, isolate-free, positive-edge-count or independent-block restriction.

For the same vertex set, S dominates G precisely when every vertex is in S or adjacent to a member of S. Global domination means domination of both G and its genuine loopless complement. Two actual subsets A,B form a global coalition precisely when they are disjoint, neither is individually globally dominating, and their union globally dominates. The partition consists of nonempty pairwise-disjoint actual subsets covering all host vertices; every block has another distinct block in that same partition forming such a coalition. There is no singleton-dominating exception. The global coalition graph has these actual blocks as vertices and adjacency exactly the full global-coalition relation. Realization means genuine graph isomorphism.

Current v4 places the order>=2 convention after Definition 2. V1 has no corresponding global lower-order restriction; its unrestricted full question is preserved separately, not silently narrowed. The proposed two-vertex witness obeys both versions and they count as one target. V3 withdrew v2 because it had been uploaded without consulting all authors and contained incorrect results; current v4 is restored and retains the question.

The formal claim is the literal closed assertion

def claim : Prop :=
  ∀ (V : Type) [Finite V], 2 ≤ Nat.card V → ∀ H : SimpleGraph V,
    ∃ (W : Type) (_ : Finite W) (G : SimpleGraph W), 2 ≤ Nat.card W ∧
      ∃ π : GCPartition G, Nonempty (globalCoalitionGraph G π ≃g H)

The model retains actual finite Type-valued target and host types, native simple graphs and native loopless complements, genuine Setoid.IsPartition nonempty/disjoint-cover structure, and a distinct actual coalition partner for every actual block. The v1 unrestricted-order alias is a source boundary, not a second theorem or credit.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/GlobalCoalitionUniversality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2025The peakon transfer matrices expand in the Hamiltonians h^(k) with explicit universal coefficientsThe Camassa–Holm peakon transfer matrices have a partition expansion with explicit universal rational coefficients satisfying the values of Conjecture 5.2.Proved

The questionRagoucy, Quadratic Poisson brackets for the Camassa–Holm peakons, arXiv:2512.11066v3, J.

Why it mattersThe expansion expresses every transfer matrix through the single family of Hamiltonians h^(k) of Proposition 5.1.

The ideaThe proof works with formal power series in t over ℂ.

Our contributionThe Camassa–Holm peakon transfer matrices have a partition expansion with explicit universal rational coefficients satisfying the values of Conjecture 5.2.

Verification details
Release source recordProved in Lean

Open problem

The peakon transfer matrices expand in the Hamiltonians h^(k) with explicit universal coefficients

J. Avan, L. Frappat and E. Ragoucy, Quadratic Poisson brackets for the Camassa–Holm peakons, arXiv:2512.11066v3, J. Phys. A 59, 245202 (2026), study the two-parameter deformation of the Camassa–Holm peakon Lax matrix.

  • Its matrices are A, T and S of Eqs. (2.1)–(2.3), and its Lax matrix is L(z) = S + Λ(z)T + ξ(z)A with ξ² = Λ² + γΛ + 1 (5.18).
  • The transfer matrices are τ_m(z) = tr L(z)^m.
  • Proposition 5.1 gives their degree-one term in Λ, h^(m) = m tr((T + γA/2)(S + A)^{m−1}) (5.17).

Conjecture 5.2 asserts that τ_m(z) = Σ_j Λ^j Σ a_j(k_1, …, k_j) h^(k_1)⋯h^(k_j), summed over partitions k_1 ≥ ⋯ ≥ k_j of m (5.20), with a_1(m) = 1 and the values a_m(1, …, 1) = 1 and a_{j+1}(m − j, 1, …, 1) = m/(m − j) for 0 ≤ j < m − 1 (5.21). It leaves "the other coefficients … to be determined", and the conclusion calls for a proof. The verbatim statements are in the literature note.

Issue #16323 fixes the reading: for every N and m ≥ 1, there are coefficients independent of N and of all parameters with these values such that (5.20) holds. It also preregisters the explicit coefficients a_j(k) = m (j − 1)! / (∏_i k_i · ∏_r e_r!), where e_r counts the parts equal to r.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/Solitons/CamassaHolmPeakonTransferExpansion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Badhani–Das noisy-unitary conditional min-entropyWhite-noise mixtures of arbitrary bipartite unitaries with qubit conditioning attain the Choi-state upper bound on conditional min-entropy.Proved

The questionDas, Conditional channel entropy sets fundamental limits on thermodynamic quantum information processing, arXiv:2604.01217v2, Section VI.1, conjecture after Proposition 10:

Why it mattersThe source note is Library/Entropy/badhani2026conditionalchannel.md.

The ideaThe normalized reduced Choi matrix has two maximally mixed qubit marginals.

Our contributionWhite-noise mixtures of arbitrary bipartite unitaries with qubit conditioning attain the Choi-state upper bound on conditional min-entropy.

Verification details
Release source recordProved in Lean

Open problem

Badhani–Das noisy-unitary conditional min-entropy

H. Badhani and S. Das, Conditional channel entropy sets fundamental limits on thermodynamic quantum information processing, arXiv:2604.01217v2, Section VI.1, conjecture after Proposition 10:

We conjecture that Proposition 10 can be extended to all unitaries with added white noise. We leave this as an open problem.

Proposition 10 assumes $|B'|=|B|=2$. The statement here retains that hypothesis and takes $A'=A=\mathbb C^d$, for every positive natural $d$. For every unitary $U$ on $\mathbb C^d\otimes\mathbb C^2$ and every $p\in[0,1]$, put $N_p(Z)=p\operatorname{tr}(Z)I/(2d)+(1-p)UZU^\dagger$. Then

$$ S_\infty[A\mid B]{N_p}=S\infty(R_AA\mid R_BB)_{\Phi^{N_p}}-\log_2 d. $$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Arbitrary positive input/output dimension on A | The natural dimension and positivity binder in claim | No upper bound on $d$. | | Qubit input/output conditioning | Matrix.unitaryGroup (Fin d × Fin 2) ℂ | Both B systems have dimension two. | | Arbitrary unitary channel | The universally quantified unitary in claim | No selected matrix or spectrum. | | Added white noise | noisyUnitary | Literal replacement-plus-unitary expression. | | $p\in[0,1]$ | The two real inequalities in claim | Both endpoints included. | | Choi operator/state | choiN, choiState | Index order $(R_A,A),(R_B,B)$, normalization $1/(2d)$. | | State/channel conditional entropy | condMinEntropy, channelCondMinEntropy | Density states and quantum channels range over the existing repository carriers. The channel formula inlines its matrix-unit Choi expression. | | Subtract $\log|A'|$ | The base-two logarithm term in claim | $|A'|=d$. | | Universal equality | result | Full claim, without additional hypotheses. |

The public definitions retain the expressions prescribed in #19060. Dmax uses the real logarithm followed by coercion to EReal; at zero input it gives zero rather than an extended logarithm of zero. The settlement uses trace-one states and positive lower bounds on feasible costs, so that zero-input extension is outside its entropy-scaling argument.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/BadhaniDasNoisyUnitary.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Positive polygonal couplings realize every discrete Fourier transformThe polygonal coupling model uses real symmetric circulant matrices with zero diagonal.Proved

The questionThe verbatim statements are in the literature note.

The ideaThe proof uses spectral estimates.

Our contributionThe polygonal coupling model uses real symmetric circulant matrices with zero diagonal.

Verification details
Release source recordProved in Lean

arXiv

Positive polygonal couplings realize every discrete Fourier transform

Barriga et al., N-dimensional discrete Fourier transform via bosonic Hamiltonian, arXiv:2609.05644v1, §II.B, model an array of $N$ waveguides on a regular polygon by a real circulant coupling matrix with zero diagonal and $C_l=C_{N-l}$, evolve it by $U=e^{-i\mathcal C}$, and ask for diagonal phase shifters with $\Phi^{\rm out}U\Phi^{\rm in}=F_N$, $F_N=(\omega^{jk})/\sqrt N$, $\omega=e^{-2\pi i/N}$. They find analytic solutions for $N=4,5,6$ and numerical ones up to $N=31$, and write:

Whether the polygonal model can always yield solutions to obtain $F_N$ from Eq.~\eqref{eq:Coupling_polygon} or not is beyond our theoretical efforts.

The verbatim statements are in the literature note. Issue #13954 reads the question for every $N\ge1$, with exact equality (no column relabelling) and, in addition, strictly positive couplings.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PolygonalFourierCouplings.result

Frozen module record

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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The original simplex coverage optimizerThe q-ary simplex code minimizes actual uniform physical iid expected full-span retrieval time among every full-rank generator of the exact simplex length.Proved

The questionBertuzzo–Ravagnani–Yaakobi, arXiv:2603.06489v1, Section 3, Conjecture 3.2, asks whether the q-ary simplex code solves Problem B.

Why it mattersThis decides the source's complete optimizer assertion at the simplex parameters.

The ideaThe source's Theorem 3.1 already computes the simplex expectation.

Our contributionThe q-ary simplex code minimizes actual uniform physical iid expected full-span retrieval time among every full-rank generator of the exact simplex length.

Verification details
Release source recordProved in Lean

Open problem

The original simplex coverage optimizer

Bertuzzo–Ravagnani–Yaakobi, arXiv:2603.06489v1, Section 3, Conjecture 3.2, asks whether the q-ary simplex code solves Problem B. The predecessor is arXiv:2507.20639v1, Section III, its unnumbered simplex-optimizer paragraph. Issue https://github.com/the-omega-institute/trureturing/issues/11799 preregisters the complete assertion before its mathematical probes.

For every finite field K of cardinality q and every k at least two, set n=(q^k-1)/(q-1). For every rank-k k-by-n physical generator G and every actual simplex generator S with exactly one nonzero representative of each projective line, prove E[T_S] <= E[T_G]. T is the original minimum time at which the sampled columns span K^k; each physical Fin n position is drawn independently, uniformly and with replacement. Zero, repeated and scalar-parallel competitor positions remain in the alphabet without conditioning or renormalization.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Resource/SimplexCoverageOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025A single-formula feasibility witness for the qubit simplex-and-interval inclusionThe six real symmetric matrices of Appendix A.1 are positive semidefinite over the complex numbers and satisfy all four affine constraints throughout the closed interval.Proved

The questionThe verbatim statements are in the literature note.

The idea- the affine equations hold identically because Eq. (47) solves them;

Our contributionThe six real symmetric matrices of Appendix A.1 are positive semidefinite over the complex numbers and satisfy all four affine constraints throughout the closed interval.

Verification details
Release source recordProved in Lean

Open problem

A single-formula feasibility witness for the qubit simplex-and-interval inclusion

A. Bluhm, E. Evert, I. Klep, V. Magron and I. Nechita, Inclusion constants for free spectrahedra with applications to quantum incompatibility, arXiv:2512.17706v1, Appendix A.1, conjecture that the explicit tuple with $C_1=(\tfrac1{\sqrt3}-\tfrac12)\begin{pmatrix}1&1\1&1\end{pmatrix}$, the displayed $C_2(\theta)$ and $C_3,\dots,C_6$ from Eq. (47) is feasible for the SDP (37) at $X(\theta)$, $\gamma=4/(1+\sqrt3)$, for every $\theta\in[0,\pi/2]$. The verbatim statements are in the literature note.

Issue #14338 reads "feasible" as membership in $\mathfrak D_{\gamma,X(\theta)}$: all six matrices positive semidefinite in $M_2(\mathbb C)$ and the four affine equations of (37).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/FreeSpectrahedronFeasibilityWitness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The exact compatibility degree of four dichotomic qubit measurementsFourQubitCompatibilityDegree: exact analytic statements for four-qubit white-noise compatibility.Proved

The questionBluhm, Evert, Klep, Magron and Nechita, Inclusion constants for free spectrahedra with applications to quantum incompatibility, arXiv:2512.17706v1, Conjecture 6.10, printed page 44:

Why it mattersThe minimum is attained by the trine-plus-perpendicular tuple.

The ideaIn FourQubitCompatibilityDegree, their associated tuple has a twelve-outcome endpoint parent and a matching dual witness.

Our contributionFourQubitCompatibilityDegree: exact analytic statements for four-qubit white-noise compatibility.

Verification details
Release source recordProved in Lean

Open problem

The exact compatibility degree of four dichotomic qubit measurements

Bluhm, Evert, Klep, Magron and Nechita, Inclusion constants for free spectrahedra with applications to quantum incompatibility, arXiv:2512.17706v1, Conjecture 6.10, printed page 44:

We conjecture that $s_{\mathbb C}(2,4)=2/\sqrt{13}$.

Definitions 2.13, 2.15 and 2.16 specify one joint POVM with the original effects as marginals, white noise $E_i(s)=sE_i+(1-s)I/2$, the maximum feasible $s\in[0,1]$, and the minimum of these degrees over all four-tuples. Issue #14687 preregisters this literal statement and its literature boundary.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/FourQubitCompatibilityDegree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Strongly polynomial graph families with constant depthStrongly polynomial graph families have constant depth one and growing degree.Proved

The questionTristram Bogart and Santiago Jiménez-Salazar, Strongly regular graphs, strongly polynomial sequences, and two-level polynomials, arXiv:2609.32077v1, Question 3.19, p. 14:

Why it mattersThe question separates the growth of the degree of a homomorphism polynomial from the number of its leading coefficients that vary polynomially with the source parameter.

The ideaThe first two coefficient functions give depth one.

Our contributionStrongly polynomial graph families have constant depth one and growing degree.

Verification details
Release source recordProved in Lean

Open problem

Strongly polynomial graph families with constant depth

Tristram Bogart and Santiago Jiménez-Salazar, Strongly regular graphs, strongly polynomial sequences, and two-level polynomials, arXiv:2609.32077v1, Question 3.19, p. 14:

More dramatically, do there exist strongly polynomial families $G_q$, $H_n$ such that the depth function of $\hom(G_q,H_n)$ is constant while the degree function is a polynomial of positive degree?

The binding assertion in #21296 reads depth as maximal uniform integer constant depth. Definition 1.2 is downward closed in depth and does not select a pointwise greatest function. The assertion requires one integer $d\geq-1$ for which the coefficient condition holds at every nonnegative parameter, and excludes every larger integer constant. Pointwise greatest depth and maximality on every tail are separate assertions.

Statement echo

All Lean names below belong to D5.S3.Combinatorics.ExtremalDensity.BogartJimenezSalazarConstantDepthHom unless an owner is stated.

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Definition 1.1: numerical on every integer | Numerical | Integer inputs and integer outputs, cast to rational numbers; negative inputs included. | | Definition 1.2: eventually nonnegative degree and the set $S$ | TwoLevel | Numerical degree, eventual nonnegativity and guards $0\leq g(q)$. | | Definition 1.2(a): degree and descending coefficients | TwoLevel | Nonzero $f_q$, degree $g(q)$, and polynomial coefficient functions; coefficients beyond the degree are zero. | | Definition 1.2(b): extension outside $S$ | TwoLevel, constantDepth | Zero polynomial and depth $-1$ outside $S$. | | Definition 3.2: every fixed finite graph and every natural parameter | StronglyPolynomial | Fin-labelled finite simple graphs and the existing ThreeClassHomomorphismLimit.homCount; all adjacency-preserving maps. | | Question 3.19: existence of both graph families and their counting polynomials | claim, proved by result | Independent strong polynomiality of both families and equality for every $q,n\in\mathbb N$. | | Positive degree and constant depth | claim | Positive natural degree of $g$, a finite integer depth at least $-1$, and failure of every larger uniform integer depth. |

The finiteness encoding by Fin is a labelling choice; no injectivity, connectivity or positive-parameter assumption is added. Rational polynomial coefficients suffice for polynomial integer-count functions. The eventual-nonnegative and guarded coefficient bodies are inlined literally, without duplicate aliases; Polynomial.reverse is used inside the private proof through reverse_coeff.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ExtremalDensity/BogartJimenezSalazarConstantDepthHom.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresStructural characterizationFirst preprint 2026Half letter frequency characterizes balanced bordersHalf letter frequency characterizes balanced borders of nonempty binary words.Proved

The ideaThe first-hit decomposition and overlap bijection prove the cleared-denominator counting identity (ForbiddenWordCounting.forbiddenwordcountingidentity).

Our contributionHalf letter frequency characterizes balanced borders of nonempty binary words.

Verification details
Release source recordProved in Lean

arXiv

Half letter frequency characterizes balanced borders

Miklós Bóna, Balázs Maga and Jacob Richey, Letter frequency in shifts of finite type with one forbidden word, arXiv:2606.06655v2, Section 6, Conjecture 6.1: “A word $w$ has $\rho^w = 1/2$ if and only if $w$ has balanced borders. The same holds with $\rho^w$ replaced by $q^w$.”

Only the first sentence is settled. For every nonempty binary word $w$, let $\Omega_m^w$ be the length-$m$ binary words without $w$ as a factor and let

$$ \rho_m^w=\frac{\sum_{u\in\Omega_m^w}|u|_1}{m|\Omega_m^w|}. $$

A border is a nonempty prefix that is also a suffix, including $w$ itself. Balanced borders means $2|b|_1=|b|$ for every border $b$. Since the source defines $\rho^w$ as the limit if it exists, the literal settlement is $\rho_m^w\to1/2$ if and only if every border is balanced; no independent convergence assumption is added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Forbidden/BonaMagaRicheyHalfFrequency.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Half letter frequency characterizes balanced borders of nonempty binary words.

    Read source ↗
  • Historical context

    First preprint year: 2026.

    Read source ↗
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Christiansen's fermionic correlational boundCompleted fermionic operators and their correlational estimate.Proved

Why it mattersIts Hilbert-space assumptions are NormedAddCommGroup, InnerProductSpace ℂ, CompleteSpace and TopologicalSpace.SeparableSpace; no finite-dimensional assumption is present.

The ideaThe particle carrier is the physically antisymmetric subspace of the actual completed Hilbert tensor power.

Our contributionCompleted fermionic operators and their correlational estimate.

Verification details
Release source recordProved in Lean

arXiv

Christiansen's fermionic correlational bound

Martin Ravn Christiansen, A Correlational Bound for Eigenvalues of Fermionic 2-Body Operators, arXiv:2505.21167v1, Section 1.1, Conjecture 3, asks for a constant $C>0$, independent of the normalized canonical pair state $\Phi=\sum_k\lambda_k u_k\wedge v_k$, such that every even $N\ge2$ with $N\lambda_{\max}^2\le1$ and every normalized $N$-particle state satisfy

$$ \langle\Phi,\gamma_2^\Psi\Phi\rangle\le N\left(1-\frac{N-2}{2}\sum_k\lambda_k^4+C(N\lambda_{\max}^2)^2\right). $$

The source note quotes the conjecture and the RDM definition. Preregistration #14135 fixes CHR-3 with $C=5/8$ over every separable complex Hilbert space, including infinite-dimensional spaces. The source conventions are $\lambda_k\ge0$, $\sum_k\lambda_k^2=1$, mutual orthonormality of the pair modes, $u\wedge v=(u\otimes v-v\otimes u)/\sqrt2$, and RDM trace $N(N-1)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/FermionicCorrelationalBound/SourceCorrelationalBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Expansions of Narayana-Weighted Dyck Paths in an Even StripCigler's two expansions express the Narayana-weighted Dyck path sums in an even strip through bounded Dyck skeletons and binomial coefficients.Proved

The questionJohann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 4, Conjecture 3 (Conjecture 2 in v1): for m ≥ 1 and n ≥ 0,

The ideaRemove the first and last steps of a path of semilength n + 1 in the strip of height 2m and pair the remaining 2n steps.

Our contributionCigler's two expansions express the Narayana-weighted Dyck path sums in an even strip through bounded Dyck skeletons and binomial coefficients.

Verification details
Release source recordProved in Lean

Open problem

Expansions of Narayana-Weighted Dyck Paths in an Even Strip

Johann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 4, Conjecture 3 (Conjecture 2 in v1): for m ≥ 1 and n ≥ 0,

C_{n+1}^{(2m)}(t) = Σ_{j=0}^{⌊n/2⌋} C_j^{(m−1)} binom(n, 2j) t^j (1 + t)^{n−2j},

c_{n+1}^{(2m)}(t) = Σ_{j=0}^{⌊n/2⌋} (−1)^j C_j^{(m−1)} binom(⌊n/2⌋, j) t^j (1 + t)^{n−2j}.

Here C_n^{(h)}(t) and c_n^{(h)}(t) are the weighted counts of Dyck paths of semilength n in the strip 0 ≤ y ≤ h, a down-step to height k having weight τ_k with τ = (1, t, 1, t, …) and τ = (1, t, −1, −t, …) respectively, and C_j^{(m−1)} is the number of Dyck paths of semilength j in the strip of height m − 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/NarayanaStrip/CiglerStripExpansion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Signed Dyck Paths in the Strip of Height 4k − 2 Count Walks on the 4k-CycleSigned Dyck paths in the strip of height 4k minus two count walks on the cycle with 4k vertices.Proved

The questionJohann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 3, Conjecture 1, equation (74) (absent from v1): for k ≥ 1, with Uk the adjacency matrix of the cycle on 4k vertices,

The ideaPairing the steps of a Dyck path in the strip of height 4k − 2 gives a weight-preserving bijection with two-coloured Motzkin paths in the strip of height 2k − 1 (the bijection of CiglerStripExpansionPairing).

Our contributionSigned Dyck paths in the strip of height 4k minus two count walks on the cycle with 4k vertices.

Verification details
Release source recordProved in Lean

Open problem

Signed Dyck Paths in the Strip of Height 4k − 2 Count Walks on the 4k-Cycle

Johann Cigler, Some sequences and number triangles which are related to Narayana polynomials and to q-Narayana polynomials for q=-1, arXiv:2608.03363v2, Section 3, Conjecture 1, equation (74) (absent from v1): for k ≥ 1, with U_k the adjacency matrix of the cycle on 4k vertices,

c_{2n+1}^{(4k−2)} = U_k^{2n+1}(0, 1), c_{2n+2}^{(4k−2)} = U_k^{2n+2}(0, 0) = 2 c_{2n+1}^{(4k−2)}.

Here c_r^{(H)} = c_r^{(H)}(1) is the weighted count of Dyck paths of semilength r in the strip 0 ≤ y ≤ H, a down-step to height j having weight (−1)^{⌊j/2⌋}, that is the q = −1 weights (1, t, −1, −t, …) at t = 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/NarayanaStrip/CiglerCycleWalk.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Crown order-polytope f-vector log-concavityEvery positive crown has a log-concave full geometric face vector.Proved

The questionConjecture 3.7 on PDF page 12 of the pinned v3 source states:

Why it mattersIt motivates the polynomial route; it is not a proof dependency or an exact solution to this problem.

The ideaMap source vertex j to zero-based coordinate j-1 : Fin (2n). crownRelation puts the even coordinates below their two cyclic neighbors. crownGeometricFaceCount counts nonempty real exposed faces by the dimension of their affine-span directions, independently of any enumeration.

Our contributionEvery positive crown has a log-concave full geometric face vector.

Verification details
Release source recordProved in Lean

Open problem

Crown order-polytope f-vector log-concavity

Conjecture 3.7 on PDF page 12 of the pinned v3 source states:

For any positive integer n, the entries of the f-vector of O(C₂ₙ) form a log-concave sequence.

The calligraphic symbols are transcribed here as O and C. The crown has source cover relations 1<2>3<...>2n-1<2n>1. Its order polytope consists of real coordinates between zero and one respecting those comparisons. Its full f-vector includes the empty face and the whole 2n-dimensional polytope. Precisely, for all n,k : Nat with 0<n, 0<k, and k<2*n+1, the assertion is F(n,k-1)*F(n,k+1) <= F(n,k)^2, where F(n,0)=1 and F(n,k)=f_(k-1) for k>0. The full vector has 2n+2 entries.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/CrownOrderPolytopeLogConcavity.crownGeometricFVector_log_concave

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Wilf Classes of Rotation Avoidance for Patterns of Length FourRotation avoidance has exactly eight Wilf classes for patterns of length four.Proved

The questionÖmer Eğecioğlu, Collier Gaiser and Mei Yin, Pattern avoidance in permutations and their rotations, arXiv:2607.20750v1, Section 6, Open Question 6.1: for k ≥ 4, are the Wilf classes of Sn^{(k)}(q), the permutations of [n] whose first k rotations all avoid q, over the patterns q ∈ S4 exactly the eight complement–reverse orbits?

The ideaComplement and reverse preserve a(n,k;q), which gives the direction from orbits to equal counts.

Our contributionRotation avoidance has exactly eight Wilf classes for patterns of length four.

Verification details
Release source recordProved in Lean

Open problem

Wilf Classes of Rotation Avoidance for Patterns of Length Four

Ömer Eğecioğlu, Collier Gaiser and Mei Yin, Pattern avoidance in permutations and their rotations, arXiv:2607.20750v1, Section 6, Open Question 6.1: for k ≥ 4, are the Wilf classes of S_n^{(k)}(q), the permutations of [n] whose first k rotations all avoid q, over the patterns q ∈ S_4 exactly the eight complement–reverse orbits? Writing a(n,k;q) for the number of such permutations, the statement is: for every k ≥ 4 and q, s ∈ S_4, a(n,k;q) = a(n,k;s) for every n ≥ k if and only if s lies in the orbit of q under complement and reverse.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/RotationAvoidance/RotationAvoidance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraConjecture provedFirst preprint 2026Fan Conjecture 5.9: moving three-point circle capacityThe exact measure-theoretic definitions in Fan's moving three-point circle capacity conjecture.Proved

The questionConjecture 5.9 of Qiuling Fan, Riesz capacity ratios with negative exponents, arXiv:2609.11186v1, printed page 19, states:

Why it mattersThe moving point is constrained by a fixed circle and a fixed longest chord; this is not the unconstrained three-point optimizer or a capacity-ratio problem.

The ideaFor a moving angle t, set x=(phi-t)/2 and y=(phi+t)/2.

Our contributionThe exact measure-theoretic definitions in Fan's moving three-point circle capacity conjecture.

Verification details
Release source recordProved in Lean

Open problem

Fan Conjecture 5.9: moving three-point circle capacity

Conjecture 5.9 of Qiuling Fan, Riesz capacity ratios with negative exponents, arXiv:2609.11186v1, printed page 19, states:

Suppose r ≥ 2. Let A, B, C be three points on the unit circle. Fix B and C such that they are symmetric with respect to the x-axis, having polar angles φ and −φ, respectively, where φ ∈ (π/2, 2π/3]. Move A along the circle with polar angle ψ ∈ [0, 2π − 3φ], so that BC remains the longest side of the triangle △ABC. Then the capacity of the three-point set {A, B, C} is maximized when ψ = 2π − 3φ, i.e., when AC = BC.

The formal target keeps that original domain exactly. For every real r >= 2, pi/2 < phi <= 2*pi/3, and 0 <= psi <= 2*pi-3*phi, it compares the set of actual complex unit-circle points at angles psi, phi, and -phi with the isosceles endpoint. Capacity is Fan's negative-exponent Riesz capacity: the 1/r power of the supremum of the positive r-power distance energy over all probability measures on the actual point subtype. The target does not restrict r to integers, impose positive singleton masses, use a finite grid, or claim strictness or uniqueness.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/Distances/FanCircleCapacity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Feldman's Conjecture 9.3: balanced zigzag choicesBoth parity equivalences and finite Laurent coefficients give closed formulas for Feldman's actual balance-only count and the complete three-clause Conjecture 9.3.Proved

The questionFeldman, The Missing Zigzag: Cycles of Semitone Trichords and a Conservation Law in Equal Temperament, arXiv:2609.26114v1, Section 5, equation (2), Conjecture 9.3.

Why it mattersThe endpoint in FeldmanConjectureNineThree.feldmanconjectureninethree carries all three clauses in a single theorem.

The ideaWeightedPaths counts the positive-sector path charges in a finite Laurent polynomial.

Our contributionBoth parity equivalences and finite Laurent coefficients give closed formulas for Feldman's actual balance-only count and the complete three-clause Conjecture 9.3.

Verification details
Release source recordProved in Lean

arXiv

Feldman's Conjecture 9.3: balanced zigzag choices

David V. Feldman, The Missing Zigzag: Cycles of Semitone Trichords and a Conservation Law in Equal Temperament, arXiv:2609.26114v1, Section 5, equation (2), Conjecture 9.3. For each integer t>=2, set n=3t. For every k=2,...,n-2, choose one labelled directed form in ZMod n:

| Label | From | To | | --- | --- | --- | | I | 1 | k-1 | | II | k | 1-k | | III | -1 | k | | IV | k-1 | -k | | V | -k | 1 | | VI | 1-k | -1 |

At k=2, exactly II, III, IV, and V are permitted. Labels remain different choices when their endpoint pairs coincide. Balance means integer outdegree minus indegree zero at each nonzero residue; no closedness, connectivity, Eulerian-circuit, or cycle condition is added. Let a(t) be the number of these actual balanced labelled choices. The full conjecture is

a(2)=4; for every r>=2, a(2r)=4a(2r-1); and for every r>=1, (2r)a(2r+1)=6(2r-1)a(2r).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Zigzag/FeldmanConjectureNineThree.feldman_conjecture_nine_three

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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Analysis & dynamicsConjecture provedFirst preprint 2026Generalized Glasby–Paseman Conjecture 1.1(d)The finite maximum of the actual powered-sum ratio has its exact asymptotic constant.Proved

The questionConjecture 1.1(d) asks, for every fixed positive real a and positive integer l, whether

Why it mattersIt does not by itself identify the maximum's Gaussian prefactor.

The ideaIts definition maximumValue takes the finite maximum over the image of range(m+1) under poweredRatio, so it includes every allowed index and does not choose a unique peak.

Our contributionThe finite maximum of the actual powered-sum ratio has its exact asymptotic constant.

Verification details
Release source recordProved in Lean

arXiv

Generalized Glasby–Paseman Conjecture 1.1(d)

The version-1 source, D5/L/Analytic/byun2026unimodality, page 2, defines the sequence (1.4) by

R_{m,l,a}(r)= \frac{\sum_{i=0}^{r}(\binom mi a^i)^l} {\sum_{i=0}^{r}(\binom ri a^i)^l},\qquad 0\le r\le m.

Conjecture 1.1(d) asks, for every fixed positive real a and positive integer l, whether

\lim_{m\to\infty}\frac{\max_{r\in\{0,\ldots,m\}}R_{m,l,a}(r)}{A_m}=1, \qquad A_m=\frac{\sqrt l}{\sqrt{2\pi m}} \frac{\sqrt{1+2a}(1+a)a^{(l-2)/2}}{(1+a)^l-1} \left(\frac{1+2a}{1+a}\right)^{(m+1/2)l}.

The displayed constant transcribes clause (d); source footnote 1 specifies the ratio-to-one meaning. Both weighted products are raised to l before summation. The denominator is a complete row-r powered sum, not (1+a)^(rl). All integer r with 0<=r<=m are included, and ties are allowed. The exponent (l-2)/2 uses real subtraction and equals -1/2 at l=1. The quantifiers fix a and l before taking m to infinity; uniformity in varying a or l is not asserted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/AnalyticClosure/BinomialPoweredRatioMaximum.maximum_asymptotic

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
Quantum informationQuestion resolvedFirst preprint 2025GNP arXiv v4 Question 8.20: Dicke extension conesStar and complete graph Bose symmetric fully PPT extensions of unnormalized Dicke diagonal operators.Proved

The questionGulati, Nechita and Park, Positive maps and extendibility hierarchies from copositive matrices, arXiv:2509.15201v4, Question 8.20, asks:

Why it mattersIt equates the full star and complete-graph existence predicates, retaining the same matrix witness in each implication.

The ideaThe two-particle antisymmetric projector annihilates every Dicke vector.

Our contributionStar and complete graph Bose symmetric fully PPT extensions of unnormalized Dicke diagonal operators.

Verification details
Release source recordProved in Lean

Open problem

GNP arXiv v4 Question 8.20: Dicke extension cones

Gulati, Nechita and Park, Positive maps and extendibility hierarchies from copositive matrices, arXiv:2509.15201v4, Question 8.20, asks:

For mixtures of Dicke states, are the following two sets equivalent?

$$ \mathsf{PPTBExt}{n}^{(\mathcal K_r)} =\mathsf{PPTBExt}{n}^{(r-1)},\qquad r\ge2. $$

The target is exactly this versioned question. For every integer $n\ge1$, $r\ge2$ and every real upper-triangular family $Y_{ij}\ge0$, put

$$ D_{ii}=e_i\otimes e_i,\qquad D_{ij}=\frac{e_i\otimes e_j+e_j\otimes e_i}{\sqrt2}\quad(i<j),\qquad \rho=\sum_{1\le i\le j\le n}Y_{ij}|D_{ij}\rangle\langle D_{ij}|. $$

The star predicate is existence of $E\succeq0$ on $(\mathbb C^n)^{\otimes r}$ with $(I\otimes\Pi_{r-1})E(I\otimes\Pi_{r-1})=E$, every root-leaf marginal $\rho$, and $E^{T_S}\succeq0$ for every nonempty proper register subset $S$. The complete-graph predicate is existence of $F\succeq0$ on the same tensor power with $\Pi_rF\Pi_r=F$, every two-register marginal $\rho$, and $F^{T_S}\succeq0$ for every such $S$. Here $\Pi_k$ is the orthogonal projector onto $\operatorname{Sym}^k(\mathbb C^n)$, marginals are genuine unnormalized partial traces, and transposes use the standard tensor basis. The assertion is equivalence of these predicates for every $n,r,Y$, including all-zero weights, without trace normalization or invertibility.

The Library note holds the source identity and bounded source checks. The original preregistration fixes this event before execution; its tier is the first-tier short complete-proof unit.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SymmetricExtensions/DickeExtension.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Gupta's uniform residual degree boundThe exact residual count satisfies Gupta's total degree bound.Proved

The questionChakshu Gupta, One construction for the Miura-ori flip-graph degree sequence, arXiv:2607.05567v3, Conjecture 8.3 (Degree bound), states: “For d ≥ 5 and every split (a,b) with a+b=d and a≤b, the remainder R(a,b) has total degree at most d−3.” Lemma 8.2 identifies one bivariate polynomial on all m,n≥2d−1.

Why it mattersThe source had already established polynomiality and the separable degree bound; its introduction leaves the nonseparable case open for d≥9.

The ideaLegal orthogonal deletions commute; overlapping and separated same-axis deletions satisfy the one-step diamond. MiuraCanonicalJoint.jointreducedcoreunique joins arbitrary deletion choices and proves equality of the dependent pairs containing both dimensions and the entire integer function.

Our contributionThe exact residual count satisfies Gupta's total degree bound.

Verification details
Release source recordProved in Lean

Open problem

Gupta's uniform residual degree bound

Chakshu Gupta, One construction for the Miura-ori flip-graph degree sequence, arXiv:2607.05567v3, Conjecture 8.3 (Degree bound), states: “For d ≥ 5 and every split (a,b) with a+b=d and a≤b, the remainder R_(a,b) has total degree at most d−3.” Lemma 8.2 identifies one bivariate polynomial on all m,n≥2d−1. Section 7, Lemma 7.3 identifies the remainder with the configurations whose minima do not all lie on one common boundary side. The source's splits have positive a and b.

The original question is preregistered in issue #22211, preregistration.

Statement echo

| Original clause | Formal counterpart | | --- | --- | | Every d≥5, positive a+b=d, a≤b | The complete telescope of MiuraResidualCount.claim | | One rational bivariate polynomial before all dimensions | ∃ p : MvPolynomial (Fin 2) ℚ precedes ∀ m n; the witness is MiuraPolynomialCoreSum.corePolynomial a b | | Total degree at most d−3 | p.totalDegree ≤ d - 3 | | Every m,n≥2d−1 | Both universal dimension binders and their original bounds | | Every normalized integer unit-edge height function counted once | Residual on the complete function Fin m × Fin n → ℤ, normalized at the origin; residualCount is the cardinality of that proved finite set | | Exactly a strict minima and b strict maxima | Separate exact cardinalities of minima and maxima, testing every existing physical neighbour | | No single common physical side contains all minima | The negation of ∃ sd, AllOnSide ... sd; corners belong to both incident sides |

No symmetry quotient, identification of orientations, assumed fiber cardinality or replacement counting object enters the claim. Zero-based Fin coordinates are sent to the source's one-based physical coordinates by adding one to each coordinate. All subtractions in the formal bounds are natural-number subtractions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/MiuraResolution.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Periodicity of the Shift n + 2 Hankel Determinants of q-Metallic NumbersFor every n at least two, the shift n+2 Hankel determinants of the q-metallic series have signed period 2n(n+1) and values in {-2,-1,0,1,2}.Proved

The questionGuo-Niu Han and Emmanuel Pedon, Hankel continued fractions and Hankel determinants for q-deformed metallic numbers, arXiv:2502.05993v2, Section 1, Conjecture E, part 1.

The ideaThe quadratic equation determines Φn over ℤ, and concrete quadratic transitions between successive shifted series, uniform in n ≥ 2, replace the Hankel continued fraction of the shift n + 1.

Our contributionFor every n at least two, the shift n+2 Hankel determinants of the q-metallic series have signed period 2n(n+1) and values in {-2,-1,0,1,2}.

Verification details
Release source recordProved in Lean

Open problem

Periodicity of the Shift n + 2 Hankel Determinants of q-Metallic Numbers

Guo-Niu Han and Emmanuel Pedon, Hankel continued fractions and Hankel determinants for q-deformed metallic numbers, arXiv:2502.05993v2, Section 1, Conjecture E, part 1. Let Φ_n(q) = Σ_i f_i qⁱ be the power series with constant term 1 satisfying q Φ² + ((1 + qⁿ)(1 − q) − q[n]q) Φ = 1, where [n]q = 1 + q + ⋯ + q^{n−1}, and let Δ_j^{(ℓ)} = det(f{ℓ+a+b}){0≤a,b<j}. The conjecture states that for n ≥ 2 the sequence Δ^{(n+2)} satisfies Δ_{j+2n(n+1)}^{(n+2)} = (−1)ⁿ Δ_j^{(n+2)} for every j and takes only the values −2, −1, 0, 1, 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MetallicHankel/MetallicHankel.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Tight penalty coefficients for the QUBO reformulations of max k-cutFor max k-cut with real edge weights, every optimal solution of the one-hot QUBO reformulation is an optimal k-cut as soon as each penalty cv exceeds max(d+v / k, -d-v / 2), and every optimal solution of the reduced R-QUBO reformulation is one as soon as cv exceeds d+v - d-v, where d+v and d-v are the sums of the positive and of the negative weights at v.Proved

The questionHarkness et al. (arXiv:2511.01108, quant-ph) study the penalty (QUBO) reformulations of max k-cut used with QAOA, quantum annealers and Ising machines.

Why it mattersThe penalty coefficients control the energy landscape seen by QAOA and by annealers, and the paper's experiments use the tightest valid values.

The ideaLet tv be the number of parts of v in a maximiser x̂, and V₂ the set of vertices with tv ≥ 2.

Our contributionFor max k-cut with real edge weights, every optimal solution of the one-hot QUBO reformulation is an optimal k-cut as soon as each penalty cv exceeds max(d+v / k, -d-v / 2), and every optimal solution of the reduced R-QUBO reformulation is one as soon as cv exceeds d+v - d-v, where d+v and d-v are the sums of the positive and of the negative weights at v.

Verification details
Release source recordProved in Lean

Open problem

Tight penalty coefficients for the QUBO reformulations of max k-cut

A. Harkness et al. (arXiv:2511.01108, quant-ph) study the penalty (QUBO) reformulations of max k-cut used with QAOA, quantum annealers and Ising machines. With edge weights of either sign, d⁺_v and d⁻_v the sums of the positive and of the negative weights at v, the one-hot QUBO objective is q(x) = Σ_{uv∈E} w_uv (1 − Σ_j x_uj x_vj) − Σ_v c_v (Σ_j x_vj − 1)², and the reduced R-QUBO objective on k − 1 columns is q̄(x) = Σ_{uv∈E} w_uv (1 − Σ_j x_uj x_vj − (1 − Σ_j x_uj)(1 − Σ_j x_vj)) − Σ_v c_v Σ_{i<j} x_vi x_vj. The paper proves that every maximiser is an optimal k-cut when c_v > max(d⁺_v/k, −(3/2) d⁻_v) (Theorem 1), respectively c_v > d⁺_v − 2 d⁻_v (Theorem 2), and conjectures:

Conjecture 1. Let G(V, E) be a graph with edge weights w_uv for all {u,v} ∈ E. Let x̂ ∈ {0,1}^{n×k} be an optimal solution of the QUBO formulation with c_v > max{d⁺_v/k, −d⁻_v/2} for every vertex v ∈ V. Then, x̂ is an optimal solution of the max k-cut problem.

Conjecture 2. Let G(V,E) be a graph with edge weights w_uv for all {u,v} ∈ E. Let x̂ ∈ {0,1}^{n×(k−1)} be an optimal solution for the R-QUBO model with c_v > d⁺_v − d⁻_v for every vertex v ∈ V. Then, x̂ is an optimal solution of the max k-cut problem.

Issue #11239 fixes the reading. Vertices are Fin n and parts Fin k. The weights are a symmetric real function, with weight 0 on non-edges, and the edges are the pairs u < v. Matrices are Boolean, with entries read as 0 or 1 by the frozen bit. "An optimal solution of the max k-cut problem" means feasible for the BQO formulation (one-hot rows) and maximal for the BQO objective among feasible points; for Conjecture 2 the same holds for the R-BQO formulation (rows with at most one entry). The claim follows the paper's scope k ≥ 3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/MaxKCutQuboPenalty.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025The Hemmer–Straub–Westrem Ev character identityThe alternating Ev sum over even three-row characters equals the Ev sum of the two-row character for every partition of positive size.Proved

The questionSection 6 asks for the extension from single part sizes to all partitions.

The ideaThe factor identity minusfactor and the symmetric homogeneous coefficient detector coefficientdetector give leftcoefficient.

Our contributionThe alternating Ev sum over even three-row characters equals the Ev sum of the two-row character for every partition of positive size.

Verification details
Release source recordProved in Lean

Open problem

The Hemmer–Straub–Westrem Ev character identity

Hemmer, Straub and Westrem, New identities in the character table of symmetric groups involving Riordan numbers, arXiv:2509.02796v1, Conjecture 2.3, equation (2.4), asks whether, for every partition $\lambda\vdash n$,

$$ \sum_{\widetilde\lambda\in\operatorname{Ev}(\lambda)} \sum_{\mu\in R_3(2n)}(-1)^{\ell(\widetilde\lambda)} \chi^\mu_{\widetilde\lambda} =\sum_{\widetilde\lambda\in\operatorname{Ev}(\lambda)} \chi^{(n,n)}_{\widetilde\lambda}. $$

The statement here has $n\geq1$. Each occurrence of a part $a$ independently contributes either $(2a)$ or $(a,a)$; coincident cycle types retain multiplicity. The set $R_3(2n)$ consists of decreasing triples of even natural numbers with sum $2n$, including zero padding. Section 6 asks for the extension from single part sizes to all partitions. The journal version is Electronic Journal of Combinatorics 33(1) (2026), P1.57, DOI 10.37236/14401. Preregistration: #17766.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Every partition of positive size | claim: every natural n, 1 ≤ n, and lam : Nat.Partition n | Positive parts and their sum follow from the partition type. | | Power-sum product | powerSumProd | Product of Mathlib's three-variable MvPolynomial.psum, with repeated parts retained. | | Frobenius character formula | vandermonde3 and frobChar | The coefficient at the partition plus $(2,1,0)$ of the Vandermonde times the power-sum product. | | Independent doubling choices | evTypes | Both recursion branches and the empty-list case retain multiplicity. | | Even partitions with at most three parts | R3 | Fixed-sum triples, decreasing order, and all entries even. | | Sign and two-row character | claim | The sign is the integer $(-1)^{\operatorname{length}(\tau)}$; $(n,n)$ is padded by zero. | | Equality (2.4) | result : claim | No single-part-size or sorting hypothesis. |

The identification of frobChar with the ordinary irreducible character is the classical Frobenius formula, attested through the source's Lemma 3.1 and Stanley's Corollary 7.17.4. This representation-theoretic identification is a literature input, not a Lean construction of representations or Specht modules. The public definitions retain the preregistered coefficient conventions, with the polynomial-first coefficient API.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Partitions/HemmerStraubWestrem.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2025A null direct summand in a tensor power of a singular semigroup moduleA tensor power of every singular finite-semigroup representation contains a one-dimensional null summand; every indecomposable-summand count is asymptotic to the tensor dimension.Proved

The questionDavid He and Daniel Tubbenhauer, Tensor powers of representations of (diagram) monoids, arXiv:2508.04054v1, Conjecture 4.2, page 17:

Why it mattersIt proves the entire claim, including count-sequence existence and the limit for every internal indecomposable decomposition.

The ideaTensor ranks are bounded by individual-rank powers.

Our contributionA tensor power of every singular finite-semigroup representation contains a one-dimensional null summand; every indecomposable-summand count is asymptotic to the tensor dimension.

Verification details
Release source recordProved in Lean

Open problem

A null direct summand in a tensor power of a singular semigroup module

David He and Daniel Tubbenhauer, Tensor powers of representations of (diagram) monoids, arXiv:2508.04054v1, Conjecture 4.2, page 17:

Let V be a kS-module on which no element acts invertibly, then there is some m ≥ 1 such that Z is a summand of V^{⊗m}, and consequently a(n)=(dim V)ⁿ.

The field is algebraically closed, the semigroup is nonempty and finite with no two-sided identity, and the representation is finite dimensional and nonunital. The one-dimensional null representation has zero action. The source defines $a(n)$ as an asymptotic equivalent of the indecomposable-summand count $b(n)$, with multiplicity. Thus the final clause means $b(n)/(\dim V)^n\to1$, rather than exact equality at finite exponents. Preregistration: #22531.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Algebraically closed field of arbitrary characteristic | claim: field and algebraic-closure instances, with no characteristic restriction | | Nonempty finite semigroup without identity | claim: semigroup, finiteness and nonemptiness instances, and the negated two-sided-identity existence hypothesis | | Finite-dimensional nonunital module | claim: finite-dimensional vector space and a multiplicative map into endomorphisms | | Every element acts noninvertibly | claim: nonunit hypothesis for every action | | Some positive tensor power contains the null representation as an actual direct summand | HasNullSummand and the first conjunct of claim; complementary subspace is invariant | | Indecomposable summands counted with multiplicity | IsIndecomposableSummand, IsSummandCount, internal direct sum indexed by Fin b | | The asymptotic equivalent is the tensor dimension | Count-sequence existence and the ratio limit for every such sequence, the second and third conjuncts of claim |

Nonemptiness and singularity exclude the zero space; there is no additional nonzero-space hypothesis. The public definitions retain the literal preregistered expressions. The tensor action uses the upstream PiTensorProduct.map construction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/TensorNullSummand.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The minimum output overlap of orthogonal C–Q inputsOrthogonal pure inputs to a classical–quantum channel have average output overlap at least the minimum over distinct basis inputs.Proved

The questionA pure state is a unit vector, and a quantum state is positive semidefinite with trace one.

Why it mattersIt compares an optimization over orthonormal frames with the finite set of computational basis choices.

Our contributionOrthogonal pure inputs to a classical–quantum channel have average output overlap at least the minimum over distinct basis inputs.

Verification details
Release source recordProved in Lean

Open problem

The minimum output overlap of orthogonal C–Q inputs

Hutton, Mehta and Vukovic, Deciding whether a C–Q channel preserves a bit is QCMA-complete, arXiv:2508.10664v1, Conjecture 1.5, page 3, ask whether, for quantum states $\sigma_1,\ldots,\sigma_n\in M_d(\mathbb C)$ and $k\le n$ pairwise orthogonal pure states $u_1,\ldots,u_k\in\mathbb C^n$,

$$ \frac{1}{k(k-1)}\sum_{r\ne s}\operatorname{Tr}!\left(\Bigl(\sum_i |(u_r)i|^2\sigma_i\Bigr)\Bigl(\sum_j |(u_s)j|^2\sigma_j\Bigr)\right) \ge \min{i_1,\ldots,i_k\text{ distinct}}\frac{1}{k(k-1)}\sum{r\ne s}\operatorname{Tr}(\sigma_{i_r}\sigma_{i_s}). $$

The denominator requires $2\le k$. A pure state is a unit vector, and a quantum state is positive semidefinite with trace one. Definition 1.3 specifies the channel on a pure input by $\Phi(|u\rangle\langle u|)=\sum_i|u_i|^2\sigma_i$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Arbitrary finite dimensions $n,d$ | Natural-number parameters of claim | All dimensions; empty impossible state/frame cases are retained | | Arbitrary quantum states $\sigma_i$ | CoPRelativeQuantumnessRefutation.IsDensity: positive semidefinite and trace one | Literal conjunction supplied directly by the existing public owner | | $k\le n$ orthogonal pure inputs | k ≤ n, IsOrthonormalFamily u | Coordinate inner products are the Kronecker delta, including unit norms | | Defined average over unequal ordered pairs | averageOutputOverlap | Real reciprocal $1/(k(k-1))$, zero diagonal, real part of each trace | | Minimum over $k$ distinct indices | minDistinctOverlap | Infimum over injective Fin k → Fin n; finite and nonempty under $k\le n$ | | Lower bound | result : claim | Same inequality direction | | Denominator defined | 2 ≤ k | Makes the source normalization positive |

No source clause is dropped. The five local public definitions and the directly reused public CoPRelativeQuantumnessRefutation.IsDensity retain the literal conventions of #20116.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/HuttonMehtaVukovicOutputOverlap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026Kania–Kąkol Problem 8.2: the full Talagrand construction under CHCH alone gives one literal Talagrand realization with the full signed and topological endpoints.Proved

The questionKania and Kąkol, Talagrand compacta, 2DCP, and pointwise quotients, arXiv:2607.06808v2, revised 10 July 2026, PDF p.19, ask:

Why it mattersTheorem A (p.2) assumes diamond on a stationary co-stationary subset of omega-one.

The ideaThe well-founded coherent ordinal history constructs all actual stages from CH.

Our contributionCH alone gives one literal Talagrand realization with the full signed and topological endpoints.

Verification details
Release source recordProved in Lean

Open problem

Kania–Kąkol Problem 8.2: the full Talagrand construction under CH

Kania and Kąkol, Talagrand compacta, 2DCP, and pointwise quotients, arXiv:2607.06808v2, revised 10 July 2026, PDF p.19, ask:

Problem 8.2. Can the prediction principle in Theorem A be weakened from ♢(S) to CH?

Exact statement correspondence

There is one mathematical input, hCH : Cardinal.continuum = ℵ₁, and one literal carrier K := (SignedOrdinalRealization.finalPrefix hCH).LimitSpace. The public theorem is TalagrandTopologicalConsequences.actual_CH_Talagrand_scheme_and_endpoints. All clauses below occur together in its statement and are consumed by its proof.

| Original requirement | Lean statement and derivation | | --- | --- | | Actual closed nonempty coordinate compacta and onto restrictions | The first conjunction retains all H.carrier, H.map and H.projection readbacks and surjectivity. | | Initial omega cube | Every carrier through omega is the whole cube; the final restriction onto the omega cube is onto. | | Closed-cover successor split for omega ≤ alpha < omega-one | Closed Y,Z cover H.Space a; H.carrier (nextStage a) is literally the range of Boolean insertion from splitCarrier Y Z. | | Proper limit and final thread equations | Each proper limit above omega is exactly its proper-slice limit carrier; the final carrier is exactly all compatible restrictions. | | Countable-stage and final topology | Every proper stage is compact Hausdorff, zero-dimensional and metrizable. K is nonempty compact Hausdorff and zero-dimensional, with its actual limiting cone. | | Whole original enumerations | U, B and nu are onto; B and V are the original whole births. Every signed measure has regular Jordan parts, variation mass one, support concentration and positive mass at least one half. | | Original A–F and every signed selection | Every proper cutoff has the original P.Laws D, including every-choice signed F, with every original-label measure readback. No final-stage F is postulated. | | G and cofinal bookkeeping | One eta gives the same support launches and signed source measures. The fixed G clopen is the new-coordinate-true cylinder. Its fixed increasing column sequence works for arbitrary complete signed norm-one lifts, with infinitely many evaluations at least one sixth and infinitely many exact zeros. Every row/label pair is treated cofinally above any base. | | Theorem A closed-copy prohibition | For every closed A and continuous f : A → K with disjoint range, the range is finite. Every two disjoint homeomorphic subspaces, one closed, are consequently finite. This is stronger than excluding disjoint nonmetrizable closed copies. | | No 2DCP | No sequence of nonempty compact T(n) has two disjoint copies of T(n+1) in T(n). No nesting or infinitude premise is added. | | Grothendieck | Every sequence in the full real continuous dual converging pointwise to FInf converges under every continuous real bidual test. There is no input norm bound. | | Entire signed dual unit ball | Every continuous map from all of StoneCech ℕ into WeakDual ℝ C(K,ℝ) bounded by one is noninjective, including principal and free points. | | Point-space consequences | Every convergent sequence is eventually its actual limit; no continuous StoneCech injection exists; K is infinite, not first countable and not metrizable. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Topology/InverseLimit/TalagrandTopologicalConsequences.actual_CH_Talagrand_scheme_and_endpoints

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2025Kok's unique maximum-degree values and infinite realizing familyThe positive maximum-degree values attained at exactly one recursive Jaco graph order are the complete source sequence, realized by one explicit infinite indexed family.Proved

The questionJohan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500v1, Section 2.2, Conjecture 2.3, printed page 5 (primary locator), states:

Why it mattersThe existing all-order degree theorem for the same recursive graph makes the multiplicity of each positive degree value accessible.

The ideaUse the existing source graph JacoDomPathDiameterRefutation.jaco and JacoMaximumDegree.result directly.

Our contributionThe positive maximum-degree values attained at exactly one recursive Jaco graph order are the complete source sequence, realized by one explicit infinite indexed family.

Verification details
Release source recordProved in Lean

Open problem

Kok's unique maximum-degree values and infinite realizing family

Johan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500v1, Section 2.2, Conjecture 2.3, printed page 5 (primary locator), states:

For $t = 1,2,3,\dots$ the values given by sequence A003622 i.e. the Wythoff compound sequence AA: $a(t) = \lfloor t \times (\frac{1+\sqrt{5}}{2})^2\rfloor -1$ are the unique maximum degrees of corresponding linear Jaco graphs.

Furthermore, it implies that sequence A003622 is $p$-graphical where $p(G) = \Delta(G)_{unique}$ and $\mathcal{F} \in {G:G=J_n(x), n = 1,2,3,\dots}$.

The preceding paragraph says the maximum-degree values that appear once across graph orders are intended and explicitly excludes $\Delta(J_1(x))=0$. This is numerical value uniqueness across orders, not a singleton set of maximizing vertices within one graph; Conjecture 2.5 concerns the latter.

Retain the original recursive graph with $f(x)=x$. For every positive $i$, let $d_i=#{j:1\le j<i,\ 2j-d_j\ge i}$. For every positive $n$, $J_n$ is the simple undirected graph on ${1,\ldots,n}$ with an edge ${i,j}$ exactly when $1\le i<j\le n$ and $2i-d_i\ge j$. Write $\Delta_n$ for its ordinary maximum vertex degree, $\phi=(1+\sqrt5)/2$, and $a(t)=\lfloor t\phi^2\rfloor-1$. For an integer $k$, define

$$ U(k)\iff k\ge1\ \land\ \exists n\ge2,\quad \Delta_n=k\ \land\ \forall m\ge2,\ (\Delta_m=k\Rightarrow m=n). $$

The complete assertion, with natural graph orders and term indices, is

$$ \left(\forall k\in\mathbb Z,\ k\ge1\Rightarrow [U(k)\iff\exists t\ge1,\ k=a(t)]\right) \ \land
\left(\exists r:\mathbb N_{>0}\to\mathbb N_{>0},\ \forall t\ge1,
r(t)\ge2\ \land\ a(t)\ge1\ \land\ \Delta_{r(t)}=a(t) \ \land\ \forall m\ge1,\ [\Delta_m=a(t)\Rightarrow m=r(t)]\right). $$

There is one whole infinite indexed family $G_t=J_{r(t)}$, not only separate finite prefixes. Every positive graph order remains in the reference class, including $m=1$ in the realizing clause. The Lean witness is a function on all natural numbers whose assertions are restricted to positive indices. It additionally identifies $r$ with realizingOrder; no value at zero is claimed as a source sequence term.

The source conventions are explicit. Read the printed family/element notation memberwise: each realizing graph belongs to the class of source graphs. The introduction's existential realizing-family definition permits selection of corresponding orders; neither exhaustion of all orders nor $r(t)=t$ is required. Extend that termwise definition to the positive infinite index set printed in Conjecture 2.3. On $D={J_n:n\ge1,\ U(\Delta_n)}$, interpret the qualified parameter as $p_D(G)=\Delta(G)$. No value is assigned outside $D$. Read “are the unique maximum degrees” as the full iff classification, retaining both inclusions. These conventions normalize the source's notation; they are not an author-issued correction or a certificate of uniquely intended semantics.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/JacoUniqueMaximumValues.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Skew-Merged Permutations Are Exactly Those With One or Two Rook PlacementsA permutation grid has exactly one or two complete rook placements if and only if its permutation is skew-merged.Proved

The questionJoel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Section 3.3:

The ideaRook placements of a permutation grid are the perfect matchings of its word graph; every permutation grid has at least one.

Our contributionA permutation grid has exactly one or two complete rook placements if and only if its permutation is skew-merged.

Verification details
Release source recordProved in Lean

Open problem

Skew-Merged Permutations Are Exactly Those With One or Two Rook Placements

Joel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Section 3.3:

Conjecture 3.9. A permutation w ∈ S_n is skew-merged if and only if Grid(w) admits exactly one or two rook placements.

Grid(w) is the n × n board whose black squares are (i, w(i)); a rook placement is a set of white squares meeting every across word and every down word exactly once. A permutation is skew-merged when its entries split into an increasing and a decreasing subsequence. The paper proves the forward direction (Proposition 3.8) and characterizes exactly one placement (Theorem 3.7).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CrosswordGrid/SkewMergedRook.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2025A Finite Andrews–Gordon Companion IdentityLi and Uncu's finite Andrews-Gordon companion identity holds for every k at least five.Proved

The ideaA deletion–insertion bijection on weighted lattice paths with bounded height and a boundary state gives a recurrence for the multiple-sum side, with the doubled boundary shift and the primed convention.

Our contributionLi and Uncu's finite Andrews-Gordon companion identity holds for every k at least five.

Verification details
Release source recordProved in Lean

Open problem

A Finite Andrews–Gordon Companion Identity

Runqiao Li and Ali K. Uncu, A MacMahon Analysis View of Cylindric Partitions, arXiv:2501.19272v1, Section 1, Conjecture 1.3, equation (1.5): for integers n ≥ 0, k ≥ 5 and k > i ≥ 1, Σ_{n₁≥⋯≥n_{k−1}≥n_k=0} q^{n₁²+⋯+n_{k−1}²+n_i+⋯+n_{k−1}} ∏_{j=1}^{k−1} [2n − 2Σ_{ℓ<j} n_ℓ − n_j − n_{j+1} − 2α_{ij} ; n_j − n_{j+1}]′q equals Σ{r∈ℤ} (−1)^r q^{r((2k+1)r+2k−2i+1)/2} [2n ; n − (2k+1)r/2 + (2k−2i+1)((−1)^r − 1)/4]q, where α{ij} = max(j − i + 1, 0), the Gaussian binomial vanishes unless its lower index lies between zero and the upper index, and the primed one equals 1 when the lower index is zero, even for a negative upper index.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CylindricPartition/LiUncu.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Quantum informationQuestion resolvedFirst preprint 2025The averaged Zeitlin Ricci curvature tends to minus half of the harmonic number minus oneAveraged Ricci curvature of the Zeitlin metric.Proved

The questionPreston, Ricci curvature for hydrodynamics on the sphere, arXiv:2508.09833v1; Commun.

Why it mattersConjecture 2 fixes the exact limiting value behind that interpretation.

Our contributionAveraged Ricci curvature of the Zeitlin metric.

Verification details
Release source recordProved in Lean

Open problem

The averaged Zeitlin Ricci curvature tends to minus half of the harmonic number minus one

L. Lichtenfelz, K. Modin and S. C. Preston, Ricci curvature for hydrodynamics on the sphere, arXiv:2508.09833v1; Commun. Math. Phys. 407, 37 (2026), Section 2.2, Conjecture 2: "For each fixed $\ell > 1$, the averaged Ricci curvature $\tilde r_\ell(N)$ of the Zeitlin metric on $SU(N)$ becomes negative for sufficiently large $N$, and in the limit $\tilde r_\ell(N)\to-(H_\ell-1)/2$, as $N\to\infty$." Here $\tilde r_\ell(N)=(r_\ell(N)^+-r_\ell(N)^-)/(N^2-1)$ with $r_\ell(N)^\pm$ the finite sums of squared Wigner 6j symbols ${\ell,k,k';,s,s,s}$, $s=(N-1)/2$, over $1\le k,k'<N$ with $k+k'+\ell$ odd, in eq. (2.4). The verbatim statements are in the literature note.

Issue #14664 reads both clauses for every natural $\ell\ge2$: the limit along $N\to\infty$ and eventual negativity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/ZeitlinSixJ/RicciLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    Source 1
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Discrete structuresQuestion resolvedThe first columns of A098052 and A098530In every dimension d, the d-dimensional partitions of n that extend in exactly d ways to a partition of n + 1, and the d-dimensional partitions of n + 1 that contain exactly one partition of n, are the boxes, so both are counted by the number of ordered factorizations of n, respectively n + 1, into d factors; for d = 4 these are the first columns of A098052 and A098530 on solid partitions.Proved

The questionT(n,k) counts the solid partitions of n that can be extended to a solid partition of n+1 in exactly (k+3) ways.

Why it mattersSolid partitions are the four-dimensional analogue of plane partitions.

The idea1. box v is a lower set with ∏ vi cells, and it determines v; so the boxes of n cells correspond to the ordered factorizations of n.

Our contributionIn every dimension d, the d-dimensional partitions of n that extend in exactly d ways to a partition of n + 1, and the d-dimensional partitions of n + 1 that contain exactly one partition of n, are the boxes, so both are counted by the number of ordered factorizations of n, respectively n + 1, into d factors; for d = 4 these are the first columns of A098052 and A098530 on solid partitions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The first columns of A098052 and A098530

OEIS A098052 (Wouter Meeussen, 2004) counts the solid partitions of n by their number of extensions:

T(n,k) counts the solid partitions of n that can be extended to a solid partition of n+1 in exactly (k+3) ways. Equivalently, the number of solid partitions of n that have exactly k+3 partitions of n+1 majoring them.

and its COMMENTS state

First column is conjectured to be A007426 = tau_4(n).

The twin entry A098530 (Meeussen, 2004) counts the solid partitions of n + 1 by the number of solid partitions of n they just cover and states

First column conjectured to be the (beheaded) A007426.

A007426 is tau_4(n), the number of ordered factorizations n = rstu. Issue #10330 fixes the readings, and issue #10404 registers the statement in every dimension d: a d-dimensional partition of n is read through its Ferrers diagram, a lower set of n cells of ℕ^d (for d = 4 the solid partitions of A000293); for the entries, a solid partition of n is its four-dimensional Ferrers diagram, a lower set of n cells of ℕ⁴ in the coordinatewise order; an extension of I is a solid partition of n + 1 containing I, and a shrinking of J is a solid partition of n contained in J. The statement proved is that for every d ≥ 1 and n ≥ 1 exactly τ_d(n) partitions of n have d extensions and exactly τ_d(n + 1) partitions of n + 1 have one shrinking, τ_d being the number of ordered factorizations into d factors; d = 4 settles both conjectures. The same statement at d = 3 is the first column of A098529 (stated there without proof), and A007426 repeats the d = 4 observation ("Appears to equal the number of solid partitions of n that can be extended in exactly 4 ways").

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SolidPartitionFirstColumn.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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Further frontiersQuestion resolvedFirst preprint 2026Metya–Mukherjee's exact independent-convolution L1 minimumThe full L1 deviation of independent finite convolution from uniform has an attained exact minimum.Proved

The questionNilava Metya and Satyaki Mukherjee, Approximate Uniformity in Finite Convolution Models, arXiv:2609.38243v1, §4.3, Conjecture 4.7, states:

Why it mattersThe source separates this L1 objective from its squared-L2, relative-entropy and line-Wasserstein objectives.

The ideaA zero actual output coefficient immediately gives one deviation equal to t.

Our contributionThe full L1 deviation of independent finite convolution from uniform has an attained exact minimum.

Verification details
Release source recordProved in Lean

Open problem

Metya–Mukherjee's exact independent-convolution L1 minimum

Nilava Metya and Satyaki Mukherjee, Approximate Uniformity in Finite Convolution Models, arXiv:2609.38243v1, §4.3, Conjecture 4.7, states:

The optimal value of (21) is (2n−1)⁻¹ for n≥3.

The source declares CC BY-SA 4.0. The quotation and mathematical source correspondence here are attributed to that version under the same license. The proof below is a repository formalization of the sharp bound.

Equation (21) minimizes the full L1 norm of the deviation from uniform over ordinary convolutions of two independent probability vectors. For every natural n≥3 and real closed-simplex vectors p,q on Fin n, put

r_k=\sum_{i,j<n,\ i+j=k}p_iq_j,\qquad t=\frac1{2n-1}.

The exact target is that t is the least member of the set of all \sum_{k=0}^{2n-2}|r_k-t| obtained from these legal pairs. Both factors range over the whole real closed simplex. They may be unequal, asymmetric, irrational or have zero entries. Addition is ordinary natural addition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/TotalVariation/IndependentConvolutionL1Minimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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Quantum informationQuestion resolvedFirst preprint 2025Integral adjacency support gives transfer through uniform clone matchingsIntegral adjacency support gives perfect state transfer at pi/2 after adding the same perfect matching in every clone set.Proved

The questionKirkland, Laplacian quantum walks on blow-up graphs (2025), arXiv:2504.11585v1, DOI 10.1016/j.laa.2025.10.001, ask in Section 11, Future work (main.tex, line 814):

Why it mattersIntegral adjacency support is required only at the vertex whose clones are to transfer; the graph need not have integral spectrum at every vertex.

The ideaInvolutivity and the absence of fixed points are not hypotheses of this lemma.

Our contributionIntegral adjacency support gives perfect state transfer at pi/2 after adding the same perfect matching in every clone set.

Verification details
Release source recordProved in Lean

Open problem

Integral adjacency support gives transfer through uniform clone matchings

H. Monterde, H. Pal and S. Kirkland, Laplacian quantum walks on blow-up graphs (2025), arXiv:2504.11585v1, DOI 10.1016/j.laa.2025.10.001, ask in Section 11, Future work (main.tex, line 814):

Lastly, we ask: does Theorem \ref{pt1} hold for the adjacency case? That is, if $n\equiv 0$ (mod 4), will the addition of an appropriate matching in $\up{n}G$ induce adjacency LPST?

Their Laplacian Theorem pt1, Section 10, Edge perturbation (main.tex, lines 666–668), states:

Let $n\equiv 0$ (mod 4) and $v$ be a vertex in $G$ with $\sigma_v(G)\subset\Zl.$ Then the addition of a matching in $T_v$ results in LPST in $\up{n}G$ at $\frac{\pi}{2}$ between the end vertices of every edge inserted. Moreover, the resulting graph is periodic at all vertices in $T_v$ that are not incident to the newly added edges with period $\frac{\pi}{2}.$

These quotations retain the source macros: \up{n}G denotes the $n$-fold blow-up and \Zl denotes the integers. In the source, $\sigma_v(G)$ is Laplacian eigenvalue support. The quotations were checked against the downloaded v1 TeX member main.tex, md5 b21c68f0d57c75bbf8eb7530da0c4d04; the definitions are discussed in the literature note.

Issue #22707 fixes the reading settled here. Let $G$ be a finite simple graph with adjacency matrix $A_G$, let $v$ have integral adjacency eigenvalue support, and let $4\mid n$. Choose a fixed-point-free involution $\rho$ of the $n$ clone indices and insert its perfect matching in every clone set. With vertices indexed by $(v,j)$, the resulting matrix is $H=A_G\otimes J_n+I_m\otimes P_\rho$. For every clone index $j$, the settlement proves adjacency perfect state transfer at $\pi/2$ between $(v,j)$ and $(v,\rho j)$. This answers the appropriate-matching reading; it does not assert the literal variant with a matching inserted only in $T_v$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/BlowUpMatchingAdjacencyTransfer.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Discrete structuresQuestion resolvedPosed in 2003The A005590 zero set at multiples of three is characterizedThe zeros of A005590 at multiples of three are exactly the Fibbinary indices.Proved

The questiona(0) = 0, a(1) = 1, a(2n) = a(n), a(2n+1) = a(n+1) - a(n).

Why it mattersIdentifying its zeros with A003714 connects the signed recurrence to the standard binary language with no adjacent one bits.

The ideaStrong induction proves the adjacent-pair invariant I(x,y) = x(x-y): (r(n),r(n+1)) ≠ (0,0) and 0 ≤ r(n) (r(n)-r(n+1)).

Our contributionThe zeros of A005590 at multiples of three are exactly the Fibbinary indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A005590 zero set at multiples of three is characterized

OEIS A005590, NAME (%N, verbatim):

a(0) = 0, a(1) = 1, a(2n) = a(n), a(2n+1) = a(n+1) - a(n).

Ralf Stephan's FORMULA conjecture (%F, verbatim):

Conjecture: a(3n)=0 iff n in A003714. - Ralf Stephan, May 02 2003

OEIS A003714, NAME (%N, verbatim):

Fibbinary numbers: if n = F(i1) + F(i2) + ... + F(ik) is the Zeckendorf representation of n (i.e., write n in Fibonacci number system) then a(n) = 2^(i1 - 2) + 2^(i2 - 2) + ... + 2^(ik - 2). Also numbers whose binary representation contains no two adjacent 1's.

The literal proved statement is ∀ n, r(3n)=0 ↔ No11 n, rendered in Lean as ∀ n : ℕ, r (3 * n) = 0 ↔ No11 n, where r : ℕ → ℤ satisfies r(0)=0, r(1)=1, r(2n)=r(n), and r(2n+1)=r(n+1)-r(n), and where No11(n) := ∀ j : ℕ, (n >>> j) % 4 ≠ 3. The growth rate attributed to Reznick and the partial recurrences recorded by Chai Wah Wu are NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/StephanA005590ZeroSetCharacterization.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture provedPosed in 2026The A038867 no-square conjectureCubes five apart differ by two modulo three, and no square does.Proved

The questionTake the difference between two cubes whose bases are five apart.

Why it mattersDirect inspection on September 9, 2026 found the statement still labelled Conjecture, with the page carrying only the polynomial expansion and the numerical verification.

The ideaThe difference expands to five times a quadratic whose constant term is twenty-five and whose other terms carry a factor of three.

Our contributionCubes five apart differ by two modulo three, and no square does.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A038867 no-square conjecture

Take the difference between two cubes whose bases are five apart. Klaus Purath conjectured on May 15, 2026 that no such difference is a perfect square, noting verification through one hundred thousand indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/CubeGapFiveNoSquare.cube_gap_five_ne_sq

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    Source 1
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Discrete structuresConjecture provedPosed in 2004Stephan's A054096 second-column conjectureThe second column of Kimberling's row-sum triangle is A006183 shifted right.Proved

The questionRalf Stephan's target COMMENT (%C, verbatim):

Why it mattersRalf Stephan stated the conjecture in January 2004.

The ideaThe cross-row induction proves that an entry in one row is the difference between the first column and the next column of the following row.

Our contributionThe second column of Kimberling's row-sum triangle is A006183 shifted right.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Stephan's A054096 second-column conjecture

OEIS A054096, NAME (%N, verbatim):

T(n,2), array T as in A054090.

Ralf Stephan's target COMMENT (%C, verbatim):

Conjecture: sequence is A006183 shifted right. - Ralf Stephan, Jan 15 2004

OEIS A054090, NAME (%N, verbatim):

Triangular array generated by its row sums: T(n,0) = 1 for n >= 0, T(n,1) = r(n-1), T(n,k) = T(n,k-1) - (-1)^k * r(n-k) for k = 2, 3, ..., n, n >= 2, r(h) = sum of the numbers in row h of T.

OEIS A054090, FORMULA (%F, verbatim):

T(n, k) = T(n, k-1) - (-1)^k * Sum_{j=0..n-k} T(n-k, j), with T(n, 0) = 1, and T(n, 1) = Sum_{j=0..n-1} T(n-1, j).

OEIS A006183, NAME (%N, verbatim):

a(n) = (n+1)*a(n-1) + (2-n)*a(n-2).

The literal proved statement is

forall T B, IsA054090Triangle T -> IsShiftedA006183 B -> forall n, 2 <= n -> T n 2 = B (n - 1).

A054096 has offset 2 and A006183 has offset 1, and their %S and %T data lines are term-for-term identical. Thus "shifted right" means A054096(n) = A006183(n-1) for n >= 2. Writing B(j) = a(j-1) turns A006183's recurrence into B(j) = j*B(j-1) + (3-j)*B(j-2), exactly the recurrence encoded by IsShiftedA006183.

Not claimed here are other columns of A054090; the row-sum sequence A054091; the generating-function form of A006183, including Gladkovskii's 2013 continued fraction %F, or its equivalence to the recurrence; or any particular term of A054096.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/StephanTriangleSecondColumn.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Dated conjecture statement: 2004. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresRecurrence provedPosed in 2003Quet's A079278 reduced-denominator recurrenceThe reduced denominators of Quet's rational iteration satisfy his recurrence.Proved

The question%N A079278 Define a rational sequence {b(n)} as b(1) = 1, b(n) = b(n-1) + 1/(1 + 1/b(n-1)) for n > 1; a(n) is the denominator of b(n). %F A079278 Conjecture (Quet): a(m+1) = a(m)^2 + a(m)^3 / a(m-1)^2 - a(m)a(m-1)^2 for m >= 2. %D A079278 Suggested by Leroy Quet, Feb 14 2003. %A A079278 N.

Why it mattersThe DATE line says Suggested by Leroy Quet, Feb 14 2003, and the AUTHOR line says N.

The ideaWriting a reduced rational as p/q, direct field algebra gives

Our contributionThe reduced denominators of Quet's rational iteration satisfy his recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Quet's A079278 reduced-denominator recurrence

OEIS A079278 states (verbatim):

%N A079278 Define a rational sequence {b(n)} as b(1) = 1, b(n) = b(n-1) + 1/(1 + 1/b(n-1)) for n > 1; a(n) is the denominator of b(n). %F A079278 Conjecture (Quet): a(m+1) = a(m)^2 + a(m)^3 / a(m-1)^2 - a(m)*a(m-1)^2 for m >= 2. %D A079278 Suggested by Leroy Quet, Feb 14 2003. %A A079278 N. J. A. Sloane, Feb 16 2003 %O A079278 1,2

The literal proved statement is

theorem result (m : ℕ) (hm : 2 ≤ m) :
    (b (m - 1)).den ^ 2 ∣ (b m).den ^ 3 ∧
      (b (m + 1)).den = (b m).den ^ 2 + (b m).den ^ 3 / (b (m - 1)).den ^ 2 -
        (b m).den * (b (m - 1)).den ^ 2

The public definition b is the rational iteration in %N, with only the harmless total extension b 0 = 0. The public theorem result states both conjuncts directly with Lean's reduced-rational denominator projection (b k).den. Consequently the public declarations alone identify a(k) with the denominator of b(k) and entail Quet's %F; no private bridge theorem is needed by an external reader to connect a separately named integer sequence.

Here / and subtraction are natural-number operations. The divisibility conjunct proves that the quotient in Quet's formula is exact, rather than silently interpreting an inexact quotient by floor division. The proof also establishes the inequality required for the natural-number subtraction.

Not claimed here are a recurrence for the numerator sequence A079269; any result about A355615, A080581, or A080582; exhaustive literature coverage; priority for the proof; or that finite computations replace the kernel proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/QuetRationalIterationDenominator.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The reduced denominators of Quet's rational iteration satisfy his recurrence.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2003.

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    A released resolution of an explicitly recorded external question.

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    Dated conjecture statement: 2003. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedPrime periods of the A122399 Stirling power-factorial sumBala's A122399 conjecture: p - 1 is a period modulo every prime p.Proved

The questionThe following NAME and COMMENT text is copied verbatim from the claim in Library/Recurrence/bala2022a122399.md.

Why it mattersThis is a first-tier OEIS conjecture dated 2022 in an entry created in 2006.

The ideaa n is the NAME verbatim over natural numbers: sum{k <= n} k^n k! Nat.stirlingSecond n k. stirling2inclusionexclusion proves the integer identity k! S(n,k) = sum{j <= k} (-1)^(k-j) C(k,j) j^n by induction on the factorial-weighted Stirling recurrence.

Our contributionBala's A122399 conjecture: p - 1 is a period modulo every prime p.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime periods of the A122399 Stirling power-factorial sum

OEIS A122399, entry by Vladeta Jovovic, Aug 31 2006; NAME attribution supplied by the orchestrator: Paul D. Hanna, May 31 2022. The following NAME and COMMENT text is copied verbatim from the claim in Library/Recurrence/bala2022a122399.md.

NAME:

a(n) = Sum_{k=0..n} k^n * k! * Stirling2(n,k).

COMMENT (Peter Bala, May 31 2022):

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1.

The theorem proves that p - 1 is a period for every prime p and every n >= 1. It does not assert that this period is minimal.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/StirlingPowerFactorialPrimePeriod.bala_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresQuestion resolvedA266489: the mod-two congruence clauseOne general theorem proves a congruence clause for each of two negative-power diagonal sequences.Proved

The questionA(x) satisfies: [x^n] A( x/A(x)^n ) = 0 for n>1.

Why it mattersThis is a first-tier OEIS conjecture from the 2016 entry.

The ideaThe escape witness is the general theorem: for every prime p, the unique normalized integer solution of [x^n] A(x / A^e) = 0 for n > 1, with e = (p−1)(n−1)+1 and a 0 = a 1 = 1, satisfies p ∣ a n for n ≥ 2.

Our contributionOne general theorem proves a congruence clause for each of two negative-power diagonal sequences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A266489: the mod-two congruence clause

From Library/ArithSums/hanna2016a266489.md, OEIS A266489, Paul D. Hanna, Feb 07 2016:

NAME:

G.f. A(x) satisfies: [x^n] A( x/A(x)^n ) = 0 for n>1.

COMMENT:

(C2) a(n) == 0 (mod 2) for n>=2.

The normalization is a(0) = a(1) = 1. Only this congruence clause is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/NegativePowerDiagonalModPrime.hanna_conjecture_a266489

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresConjecture provedHanna's prime substitution integrality conjectureEvery prime exponent has a normalized integer power series satisfying Hanna's substitution equation.Proved

The questionMore generally, for prime p there exists an integer series G(x) that satisfies: G(x) = G( x^p + pxG(x)^p )^(1/p) with G(0)=0, G'(0)=1 (conjecture).

The ideaThe triangular first-difference formula makes the next coefficient correction exact.

Our contributionEvery prime exponent has a normalized integer power series satisfying Hanna's substitution equation.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's prime substitution integrality conjecture

More generally, for prime p there exists an integer series G(x) that satisfies: G(x) = G( x^p + p*x*G(x)^p )^(1/p) with G(0)=0, G'(0)=1 (conjecture).

For every prime $p$, there is $G\in\mathbb Z[[x]]$ with $G(0)=0$, $[x]G=1$, and $G^p=G(x^p+pxG^p)$. The entries A271931, A271932 and A271933 specialize to $p=5,7,11$. The powered equation expresses the normalized root convention.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | every prime exponent | claim quantifies over natural p with Nat.Prime | exact registered convention | | integer series | the existential series has coefficient ring integers | exact registered convention | | G(0)=0 and G'(0)=1 | constant coefficient zero and coefficient one equal to one | exact registered convention | | root equation | the pth-power substitution equation in claim | exact registered convention |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaPrimeSubstitutionIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363558 power-of-two coefficient formulaHanna's A363558 power-of-two coefficient formula.Proved

The questionThe binding coefficient convention and the exact claim are in 18659.

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363558 power-of-two coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363558 power-of-two coefficient formula

Paul D. Hanna's OEIS A363558 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (2 + x^n)^(2*n).

a(2^n) = 4^(2^n) for n > 0 (conjecture).

The binding coefficient convention and the exact claim are in #18659. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (2 + x^n)^(2*n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(2^n) = 4^(2^n) for n > 0 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo.claim and D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363558PowerOfTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363559 power-of-two coefficient formulaHanna's A363559 power-of-two coefficient formula.Proved

The questionThe binding coefficient convention and the exact claim are in 18661.

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363559 power-of-two coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363559 power-of-two coefficient formula

Paul D. Hanna's OEIS A363559 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (3 + x^n)^(2*n).

a(2^n) = 9^(2^n) for n > 0 (conjecture).

The binding coefficient convention and the exact claim are in #18661. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (3 + x^n)^(2*n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(2^n) = 9^(2^n) for n > 0 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363559PowerOfTwo.claim and D5.S1.Recurrence.HannaBilateral.OeisA363559PowerOfTwo.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363559PowerOfTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363561 power-of-two coefficient formulaHanna's A363561 power-of-two coefficient formula.Proved

The questionG.f.: Sum{n=-oo..+oo} x^n (sqrt(2) + x^n)^(2n).

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363561 power-of-two coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363561 power-of-two coefficient formula

Paul D. Hanna's OEIS A363561 states:

G.f.: Sum_{n=-oo..+oo} x^n * (sqrt(2) + x^n)^(2*n).

a(2^n) = 2^(2^n) for n > 0 (conjecture).

The binding coefficient convention and the exact claim are in #18663. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | G.f.: Sum_{n=-oo..+oo} x^n * (sqrt(2) + x^n)^(2*n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(2^n) = 2^(2^n) for n > 0 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363561PowerOfTwo.claim and D5.S1.Recurrence.HannaBilateral.OeisA363561PowerOfTwo.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363561PowerOfTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363569 power-of-two coefficient formulaHanna's A363569 power-of-two coefficient formula.Proved

The questionA(x) = Sum{n=-oo..+oo} x^n (i + x^n)^(2n), where i^2 = -1.

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363569 power-of-two coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363569 power-of-two coefficient formula

Paul D. Hanna's OEIS A363569 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (i + x^n)^(2*n), where i^2 = -1.

a(2^n) = 1 for n > 0 (conjecture).

The binding coefficient convention and the exact claim are in #18665. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (i + x^n)^(2*n), where i^2 = -1. | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(2^n) = 1 for n > 0 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363569PowerOfTwo.claim and D5.S1.Recurrence.HannaBilateral.OeisA363569PowerOfTwo.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363569PowerOfTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSquare parity of the unit-reversion series A373312The normalized integer solution of Hanna's A373312 equation has odd coefficients exactly at positive Mersenne indices.Proved

The questionThe following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2024a373312.md.

Why it mattersThe KPI is open problems resolved, and the target is the unbounded parity assertion.

The ideaStabilized iterates of a contracting equation for B construct the integer series without coefficient division. generatingequation proves the equation with the unit-denominator innerSeries, and generatingunique proves uniqueness by degree induction.

Our contributionThe normalized integer solution of Hanna's A373312 equation has odd coefficients exactly at positive Mersenne indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Square parity of the unit-reversion series A373312

OEIS A373312, Paul D. Hanna, Jun 25 2024. The following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2024a373312.md.

NAME:

Expansion of g.f. A(x) satisfying A(x)^2 = A( x*A(x)/(1 - A(x))^2 ).

COMMENT:

Conjecture: a(n) == 1 (mod 2) iff n = 2^k - 1 for k >= 1.

The formalization constructs the integer solution normalized by A(0)=0 and a(1)=1 and proves the parity assertion for every positive index.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/UnitReversionSquareParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedStrip-three ternary Catalan parity for A378578The coefficients of OEIS A378578 have ternary Catalan parity.Proved

The questionHanna, Jan 03 2025) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a378578.md.

Why it mattersThis is a first-tier OEIS conjecture from the 2025 entry.

The ideastrip3 m = m / 3 ^ padicValInt 3 m removes every factor of 3 (and maps zero to zero). strip3modtwo proves its image in ZMod 2 equals that of m, because the removed power of 3 is odd. strip3Series F applies this operation coefficientwise.

Our contributionThe coefficients of OEIS A378578 have ternary Catalan parity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Strip-three ternary Catalan parity for A378578

OEIS A378578 (Paul D. Hanna, Jan 03 2025) gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a378578.md.

NAME:

G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.

COMMENT:

Conjecture: a(n) == binomial(3*n,n)/(2*n+1) (mod 2) for n >= 0.

The quotient is exact: these are the ternary Catalan numbers. The Lean statement uses natural-number division; ternary_catalan_div records the exact quotient identity for n > 0, with n = 0 handled directly.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/StripThreeTernaryCatalanParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsConjecture provedThe complete A381737 Conjecture 3 generating functionThe ordinary rational generating function of the complete increasing enumeration of physicists Hermite absolute-tie orders.Proved

The questionOrders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Why it mattersThe C2 recurrence has the separate fulfilled scope in oeis-a381737-hermite-tie-recurrence.

The ideaHermiteAbsoluteTieOrders.result supplies the full-row predicate, positivity, strict increase and exhaustive unique positive indexing for

Our contributionThe ordinary rational generating function of the complete increasing enumeration of physicists Hermite absolute-tie orders.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The complete A381737 Conjecture 3 generating function

OEIS A381737 revision 28, by Mike Sheppard, names the sequence:

Orders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Its Formula field states:

Conjecture 3: G.f.: (8x + 5x^2 + 5*x^3 - x^4 - x^5) / ((1 - x)^3 * (1 + x)^2).

The offset is 1,1. Let $H_k$ denote the physicists Hermite polynomial over $\mathbb Q$, represented by HermiteSignedMode.physicistsHermite. For natural orders $k$ and exponents $j$, define

$$ M(k,j)\iff j\le k\ \land\ \forall i\le k,\quad |[x^i]H_k|\le |[x^j]H_k|. $$

Write $T(k)$ when there exist natural exponents $u<v$ such that, for every natural $j$, $M(k,j)$ holds if and only if $j=u$ or $j=v$. This is precisely HermiteAbsoluteTieOrders.HasExactlyTwo k: exactly two distinct positions maximize the absolute coefficient over the full row $0,\ldots,k$, including parity-zero coefficients.

The entire assertion is that there exists a total $a:\mathbb N\to\mathbb N$ satisfying all of the following:

  • $a(n)>0$ for every natural $n\ge1$.
  • $a(n)<a(m)$ whenever $1\le n<m$.
  • For every natural order $k$, including zero, $T(k)\iff\exists! n\in\mathbb N,\ 1\le n\land a(n)=k$.
  • There exists $S\in\mathbb Q[[X]]$ with $[X^0]S=0$ and $[X^n]S=(a(n):\mathbb Q)$ for every $n\ge1$, such that

$$ P=8X+5X^2+5X^3-X^4-X^5,\qquad D=(1-X)^3(1+X)^2,\qquad [X^0]D=1,\qquad S=PD^{-1}. $$

The inverse is the formal power-series unit inverse: the constant coefficient of $D$ is one. The zero coefficient of $S$ is padding, not an assertion about $a(0)$. This is a formal series identity at every positive coefficient, with no analytic convergence assertion, finite prefix, selected subfamily or recurrence-defined substitute.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Polynomial/HermiteTieGeneratingFunction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedHanna's A382124 congruenceThe A382124 and A382125 coefficients agree at the Frobenius indices modulo three.Proved

Why it mattersThe exact rational formal series carry integer witnesses for the source's congruence statements.

The ideaThe other primes use the Dold congruences of the divisor sum and their product and index-scaling closure.

Our contributionThe A382124 and A382125 coefficients agree at the Frobenius indices modulo three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A382124 congruence

OEIS A382124 states: "Conjecture: a(n) == A382125(3*n) (mod 3) for n >= 0."

Write $A(x)=\exp(\sum_{n\ge1}\sigma(n)\sigma(2n)x^n/(3n))$ and $B(x)=\exp(\sum_{n\ge1}\sigma(n)\sigma(2n)x^n/n)$, with $\sigma(n)=\sum_{d\mid n}d$. The public rational-series definitions are exactly the preregistered expressions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every index $n\ge0$ | claim quantifies over every natural index | Exact | | A382124 generating function | HannaSigmaProductExpModThree.A | Exact rational exponential substitution | | A382125 generating function | HannaSigmaProductExpModThree.B | Exact rational exponential substitution | | $a(n)\equiv b(3n)\pmod3$ | claim: coeff n A = u, coeff (3*n) B = v, u ≡ v [ZMOD 3] | Exact | | Integer coefficients | ∃ u v : ℤ and the two rational coefficient equalities | Witnesses make congruence well typed; no extra hypothesis |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Robin/HannaSigmaProductExpModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A383377 coefficientsEvery coefficient of OEIS A383377 above index one is even.Proved

The questionThe following quotations are verbatim from Library/Arith/hanna2025a383377.md.

Why it mattersThe target module proves the assertion for every natural index greater than one.

The ideaThe integer series A is built by stabilised iteration (approximation) of the coefficientwise map Phi: at each degree N its coefficient is the sum, for n from zero through N, of coeff N (X ^ n absSeries (invOfUnit F 1 ^ n)).

Our contributionEvery coefficient of OEIS A383377 above index one is even.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A383377 coefficients

Paul D. Hanna, OEIS A383377, May 15 2025. The following quotations are verbatim from Library/Arith/hanna2025a383377.md.

NAME:

G.f. satisfies A(x) = Sum_{n>=0} x^n * abs(1/A(x)^n), where abs(F(x)) equals the series expansion formed by the unsigned coefficients in F(x).

COMMENT:

Conjecture: a(n) is even for n > 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/AbsoluteReciprocalGeometricParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedAlternating dyadic reversion modulo four for A389537The coefficients of OEIS A389537 satisfy Hanna's mod-four classification.Proved

The questionHanna, OEIS A389537, Nov 03 2025, gives the following NAME and COMMENT.

Why it mattersThis is a first-tier recent OEIS conjecture selected by the search seat.

The ideaCoefficient cancellation proves inverseequation, R + R(X^2) = X.

Our contributionThe coefficients of OEIS A389537 satisfy Hanna's mod-four classification.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Alternating dyadic reversion modulo four for A389537

Paul D. Hanna, OEIS A389537, Nov 03 2025, gives the following NAME and COMMENT. These quotations are copied verbatim from the claim in Library/Arith/hanna2025a389537.md.

NAME:

G.f. A(x) satisfies A( A(x) - x ) = A(x)^2.

COMMENT:

Conjecture: for n > 2, a(n) == 2 (mod 4) when n = 3*2^k for k >= 0 otherwise a(n) is divisible by 4.

The formal target uses the normalized integral formal power series with A(0) = 0 and a(1) = 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/AlternatingDyadicReversionModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedCoefficients of A393856 are one modulo fiveScaled reversion proves Hanna's coefficient congruences in OEIS A393856 and A393857.Proved

The questionThe following NAME and COMMENT are quoted verbatim from the claim in Library/Arith/hanna2026a393856.md.

Why it mattersKPI = open problems resolved: the target is the assertion for every positive index, not a finite prefix.

The ideaDefine the inner series integrally as x - x sum{n >= 1} q^(n-1) a(q,n) x^n; no division is performed in the integers.

Our contributionScaled reversion proves Hanna's coefficient congruences in OEIS A393856 and A393857.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficients of A393856 are one modulo five

OEIS A393856, Paul D. Hanna, Mar 11 2026. The following NAME and COMMENT are quoted verbatim from the claim in Library/Arith/hanna2026a393856.md.

NAME:

G.f. A(x) satisfies: A( x - x*A(4*x)/4 ) = x.

COMMENT:

Conjecture: a(n) == 1 (mod 5) for n >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/ScaledReversionCongruence.hanna_a393856

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedCoefficients of A393857 are one modulo sixScaled reversion proves Hanna's coefficient congruences in OEIS A393856 and A393857.Proved

The questionThe following NAME and COMMENT are quoted verbatim from the claim in Library/Arith/hanna2026a393857.md.

Why it mattersKPI = open problems resolved: the target is the assertion for every positive index.

The ideaDefine the inner series integrally as x - x sum{n >= 1} q^(n-1) a(q,n) x^n; no division is performed in the integers.

Our contributionScaled reversion proves Hanna's coefficient congruences in OEIS A393856 and A393857.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficients of A393857 are one modulo six

OEIS A393857, Paul D. Hanna, Mar 11 2026. The following NAME and COMMENT are quoted verbatim from the claim in Library/Arith/hanna2026a393857.md.

NAME:

G.f. A(x) satisfies A( x - x*A(5*x)/5 ) = x.

COMMENT:

Conjecture: a(n) == 1 (mod 6) for n >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/ScaledReversionCongruence.hanna_a393857

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPrime divisibility of the A393867 logarithmic-derivative termsThe nth prime divides the nth logarithmic-derivative term of OEIS A393866 for n greater than one.Proved

The questionHanna's OEIS entries A393866 and A393867, Mar 29 2026, specify:

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaprime n = Nat.nth Nat.Prime (n - 1) is one-based, so prime 1 = 2; ltprime proves n < prime n for 1 <= n.

Our contributionThe nth prime divides the nth logarithmic-derivative term of OEIS A393866 for n greater than one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime divisibility of the A393867 logarithmic-derivative terms

Paul D. Hanna's OEIS entries A393866 and A393867, Mar 29 2026, specify:

A393866 NAME (quoted verbatim in Library/Recurrence/hanna2026a393867.md):

G.f. A(x) satisfies [x^n] A(x)^prime(n) = prime(n) * [x^(n-1)] A(x)^prime(n) for n >= 1.

A393867 NAME (quoted verbatim in the same note):

Logarithmic derivative of A393866.

With constant coefficient 1 for A, this means a(n) = [x^(n-1)] A'(x)/A(x), with offset 1. This dossier concerns A393867. Its target COMMENT (quoted verbatim in the same note) is:

Conjecture: a(n) is divisible by prime(n) for n > 1.

The second A393867 comment, "Conjecture: all terms are odd.", is addressed separately in Problems/oeis-a393867-all-terms-odd.md by the new companion module PrimePowerShiftLogDerivativeOdd; the original module is unchanged. A393866 FORMULA (2), "[x^n] A'(x)/A(x) == 0 (mod prime(n)) for n > 1 (conjecture - see A393867)", is refuted as printed by the companion theorem; it is not a second resolution claim in this dossier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/PrimePowerShiftLogDerivative.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA395833: the mod-three congruence clauseOne general theorem proves a congruence clause for each of two negative-power diagonal sequences.Proved

The questionA(x) satisfies [x^n] A( x/A(x)^(2n-1) ) = 0 for n > 1.

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaThe escape witness is the general theorem: for every prime p, the unique normalized integer solution of [x^n] A(x / A^e) = 0 for n > 1, with e = (p−1)(n−1)+1 and a 0 = a 1 = 1, satisfies p ∣ a n for n ≥ 2.

Our contributionOne general theorem proves a congruence clause for each of two negative-power diagonal sequences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A395833: the mod-three congruence clause

From Library/ArithSums/hanna2026a395833.md, OEIS A395833, Paul D. Hanna, May 07 2026:

NAME:

G.f. A(x) satisfies [x^n] A( x/A(x)^(2*n-1) ) = 0 for n > 1.

COMMENT:

Conjecture: a(n) == 0 (mod 3) for n >= 2.

The normalization is a(0) = a(1) = 1. Only this congruence clause is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/NegativePowerDiagonalModPrime.hanna_conjecture_a395833

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A396659 universal edge-collision ratioFor every n at least two and every fixed starting vertex v on the complete directed graph on n labeled vertices, including loops, count length-k walks that visit every vertex, have distinct first k-1 directed edges, and repeat one of these edges at the final step.Proved

The questionFor every n at least two and every fixed starting vertex v on the complete directed graph on n labeled vertices, including loops, count length-k walks that visit every vertex, have distinct first k-1 directed edges, and repeat one of these edges at the final step.

Why it mattersAt either of these two lengths the distinct prefix uses all directed edges or all but one.

The ideaDeleting the final repeated edge gives T(n,nn+1)=nE.

Our contributionFor every n at least two and every fixed starting vertex v on the complete directed graph on n labeled vertices, including loops, count length-k walks that visit every vertex, have distinct first k-1 directed edges, and repeat one of these edges at the final step.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A396659 universal edge-collision ratio

For every n at least two and every fixed starting vertex v on the complete directed graph on n labeled vertices, including loops, count length-k walks that visit every vertex, have distinct first k-1 directed edges, and repeat one of these edges at the final step. With T(n,k) denoting this count, prove or refute

$$\frac{T(n,n^2)}{T(n,n^2+1)}=\frac{2n-1}{n}.$$

The denominator must be positive. The sequences retain a distinguished start and are not identified under rotation. The source is OEIS A396659 revision 16.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Walks/CollisionPrefix.collision_ratio

D5/S3/Combinatorics/Graph/Walks/CollisionPrefix.long_count_positive

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    The dossier explains a connection to a broader mathematical question.

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    The delivered argument documents a mathematical mechanism.

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  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedCoefficients of the A396797 three-four iterate product series modulo sixEvery positive-index coefficient of OEIS A396797 is congruent to one modulo six.Proved

The questionThe following NAME and COMMENT quotes are supplied by the orchestrator and recorded in Library/Arith/hanna2026a396797.md:

Why it mattersThis is a first-tier recent OEIS conjecture, selected in the lane brief under the recent-small-conjecture route.

The ideaStabilized successive approximations construct the integer coefficient sequence a and generatingSeries; generatingequation proves its zero constant term and defining functional equation.

Our contributionEvery positive-index coefficient of OEIS A396797 is congruent to one modulo six.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficients of the A396797 three-four iterate product series modulo six

OEIS A396797, Paul D. Hanna, Jun 15 2026. The following NAME and COMMENT quotes are supplied by the orchestrator and recorded in Library/Arith/hanna2026a396797.md:

NAME:

G.f. satisfies A(x) = x + A^3(x)*A^4(x).

Here A^n(x) denotes the n-th compositional iteration, not an ordinary power.

COMMENT:

Conjecture: a(n) = 1 (mod 6) for n >= 1.

范围与诚实边界

The resolution concerns the first conjecture for every positive index of the unique zero-constant integer formal power series satisfying the NAME equation. The second comment, asserting [x^n] A^k(x) ≡ k^(n−1) (mod 6) for all k, is not resolved here. Existence and uniqueness identify the constructed series with the defining equation; no coefficient congruence is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/ThreeFourIterateProductModSix.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe fifth compositional iterate of A396798 modulo eightThe fifth compositional iterate of A396798 has coefficient period 5,1,1,5 modulo eight from degree two.Proved

The questionHanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)A^5(x)." Its formula field says that A^k denotes the k-th iteration.

Why it mattersThe fourth and eighth public results on that same source provide the high-degree coefficients needed to turn the fifth-iterate question into linear identities of formal series.

The ideaThe fixed source definition and degree contraction give its zero constant coefficient, fixed-point equation and identity.

Our contributionThe fifth compositional iterate of A396798 has coefficient period 5,1,1,5 modulo eight from degree two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The fifth compositional iterate of A396798 modulo eight

Paul D. Hanna's OEIS A396798, revision 12, defines the ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)*A^5(x)." Its formula field says that A^k denotes the k-th iteration. The fifth comment states:

Conjecture: [x^n] A^5(x) == [5,1,1,5] repeating (mod 8) for n > 1.

Let I0(A)=X and I(k+1)(A)=Ik(A) composed with A. For the unique zero-constant integer series satisfying A=X+I4(A)*I5(A), the exact target is forall n:Nat, 1<n -> 8 divides coeff n (I5(A)) - r(n), where r(n)=if (n-2)%4=0 or (n-2)%4=3 then 5 else 1. The four-term period begins at n=2. Multiplication is ordinary series multiplication, with no factorial normalization or convergence assumption. The source is the existing D5/S1/Recurrence/Residue/IterateProductFourFiveEighthModEight.generatingSeries.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductFourFiveFifthModEight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The Pan–Skandera–Wang Map Is Bruhat-Increasing for Every nThe Pan-Skandera-Wang map raises every source permutation in the strong Bruhat order.Proved

The questionSihong Pan, Mark Skandera and Jiayuan Wang, Permanental Inequalities and Unit Interval Orders, EPTCS 445 (2026), arXiv:2606.13162v1, Section 7:

Why it mattersBy Theorem 7.1 of the paper (Drake, Gerrish, Skandera), each comparison w ≤ fn(w) makes the corresponding difference of monomials totally nonnegative, so the bijection gives the permanental inequality at h = ⌊n/2⌋ for all totally nonnegative matrices.

The ideaWrite Rx(p, q) = {j ≤ p : xj ≥ q} and, for 0 ≤ d ≤ min(p, n − p), S(p, d) = [p − d] ∪ {p − d + 2, p − d + 4, …, p + d}, a set of p positions.

Our contributionThe Pan-Skandera-Wang map raises every source permutation in the strong Bruhat order.

Verification details
Release source recordProved in Lean

Open problem

The Pan–Skandera–Wang Map Is Bruhat-Increasing for Every n

Sihong Pan, Mark Skandera and Jiayuan Wang, Permanental Inequalities and Unit Interval Orders, EPTCS 445 (2026), arXiv:2606.13162v1, Section 7:

Conjecture 7.8. For all n and all w ∈ A_n we have w ≤ f_n(w) in the Bruhat order. A proof of Conjecture 7.8 would extend Theorem 5.1 to all totally nonnegative matrices.

Here A_n is the set of permutations whose first ⌊n/2⌋ entries permute {1, …, ⌊n/2⌋}, and f_n is defined by Algorithm 7.5 from f_4 through the insertion maps ins_p, inss_q and the reverse-complement conjugation. Theorem 7.7 of the paper states the case n ≤ 13, with proof omitted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PanSkanderaWangBruhat.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Rather's book domination-root Conjecture 5.4The literal arXiv v1/v2 all-positive-order book domination-root conjecture fails at the four-cycle.Refuted

The questionBilal Ahmad Rather, arXiv:2604.08998v1, Root geometry of domination polynomials for friendship and book graphs, and v2, On roots of domination polynomials for friendship and book graphs, Section 5, printed page 12, Conjecture 5.4:

Why it mattersThe conjecture concerns finite real-root configurations on the Euclidean real line.

The ideaIts proof shape is honestly bind-only, with no escape witness claimed.

Our contributionThe literal arXiv v1/v2 all-positive-order book domination-root conjecture fails at the four-cycle.

Verification details
Release source recordRefuted in Lean

arXiv

Rather's book domination-root Conjecture 5.4

Bilal Ahmad Rather, arXiv:2604.08998v1, Root geometry of domination polynomials for friendship and book graphs, and v2, On roots of domination polynomials for friendship and book graphs, Section 5, printed page 12, Conjecture 5.4:

The numerical data suggest the following sharper parity-dependent statement. (i) If n is even, then D(B_n,x) has exactly four real roots counting multiplicity: 0 with multiplicity 2, one simple root in (-infinity,-2), and one simple root in (-1/2,0). (ii) If n is odd, then D(B_n,x) has exactly four real roots counting multiplicity: 0 with multiplicity 2 and two simple roots in (-1/2,0).

Section 2, printed page 3, defines the domain and graph:

For n >= 1, the book graph B_n is formed by identifying a common edge in n copies of C_4, see Figure 1(b).

The source uses finite simple undirected graphs and ordinary domination: $S$ dominates $G$ when every vertex is in $S$ or has a neighbor in $S$, equivalently $N[S]=V(G)$. Its polynomial counts each dominating subset once:

$$D(G,X)=\sum_{S\text{ dominates }G}X^{|S|}.$$

The complete closed claim quantifies over every natural $n\geq1$. For even $n$ there are distinct nonzero $a,b\in\mathbb R$ with $a<-2$ and $-1/2<b<0$; for odd $n$ there are distinct nonzero $a,b\in(-1/2,0)$. In each branch, $0,a,b$ are actual roots, their exact multiplicities are $2,1,1$, and for every real $r$,

$$D(B_n,r)=0\quad\Longleftrightarrow\quad r=0\ \lor\ r=a\ \lor\ r=b.$$

The delivered conclusion is the literal negation of this entire universal claim. It retains both parity branches and all intervals, multiplicities and exhaustiveness conditions. The legal odd value $n=1$ is the witness; restricting the source to $n\geq2$ gives a different, unsettled assertion. Proposition 5.3 has a local $n\geq2$ hypothesis, which is not a hypothesis of Conjecture 5.4. Printed page 11 explicitly identifies $B_1=C_4$.

Statement correspondence

| Source object or clause | Lean counterpart | | --- | --- | | Every shared-edge book $B_n$, $n\geq1$ | claim quantifies n : Nat with 1 <= n; bookGraph n has vertices Fin (2*n+2). | | One shared edge and otherwise disjoint pages | Vertices 0,1 are shared; page i : Fin n is 0--1--(2*i.val+2)--(2*i.val+3)--0; pageForward and its reverse give exactly adjacency. | | Ordinary dominating subsets | Existing DominatingSetAverage.domSets and mem_domSets, using existing SimpleGraph.IsDominating. | | Count each subset by cardinality | dominationPolynomial sums Polynomial.X ^ S.card over those actual subsets, with real coefficients. | | Even and odd intervals | Both implications in claim, including the real endpoint -(1/2 : Real). | | Exact multiplicities and no further real root | FourRealRoots, including nonzero/distinct roots, Polynomial.IsRoot, Polynomial.rootMultiplicity values 2,1,1 and the universal root iff. | | Negation of the whole statement | Closed claim : Prop; result : Not claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/RatherBookRootRefutation.result

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Discrete structuresQuestion resolvedFirst preprint 2026Nonvanishing Indices of the Binary Digit Hankel Determinants at t = −2The nonvanishing indices of the binary digit Hankel determinants at t = -2 are exactly the triples around ceil(2^(k+2)/3).Proved

The questionBartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, the case d = 2 of Conjecture 5.7, stated in the sentence following it.

The ideaA binary-carry conjugation of determinant one reduces the Hankel matrix to a sparse block form; paired elimination gives reflection recurrences that relate H at the triple around nk to the triple around n{k−1}, through an auxiliary bordered determinant.

Our contributionThe nonvanishing indices of the binary digit Hankel determinants at t = -2 are exactly the triples around ceil(2^(k+2)/3).

Verification details
Release source recordProved in Lean

Open problem

Nonvanishing Indices of the Binary Digit Hankel Determinants at t = −2

Bartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, the case d = 2 of Conjecture 5.7, stated in the sentence following it. For u = Σ_j ε_j 2^j with binary digits ε_j put S(u,t) = Σ_j ε_j t^j and H(n,t) = det[S(i+j,t)]_{0≤i,j<n}; let n_k = ⌈2^{k+2}/3⌉, so that n_k = 2, 3, 6, 11, 22, 43, …. The statement is that for every n ≥ 2, H(n,−2) ≠ 0 if and only if n ∈ {n_k − 1, n_k, n_k + 1} for some k ≥ 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/DigitHankel/BinaryDigitHankel.result

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Quantum informationQuestion resolvedFirst preprint 2025Endpoint homomorphisms of everywhere overlapping flat pairsEndpoint and negative tropical homomorphisms fail monotonicity on everywhere overlapping flat pairs.Proved

The questionVerhagen, Tomamichel and Haapasalo, Conditions for Large-Sample Majorization of Pairs of Flat States in Terms of α-z Relative Entropies, arXiv:2507.07520v3, Remark 15, p. 21, ask:

Why it mattersThe question concerns channels whose input and output remain in the restricted flat-pair domain.

The ideaThe proof applies D5.S3.Quantum.Foundation.FiniteKrausChannel.finitekrausquantumchannel to the displayed complete Kraus families.

Our contributionEndpoint and negative tropical homomorphisms fail monotonicity on everywhere overlapping flat pairs.

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Release source recordProved in Lean

Open problem

Endpoint homomorphisms of everywhere overlapping flat pairs

Verhagen, Tomamichel and Haapasalo, Conditions for Large-Sample Majorization of Pairs of Flat States in Terms of α-z Relative Entropies, arXiv:2507.07520v3, Remark 15, p. 21, ask:

Namely, it is still an open question whether the homomorphisms for 𝕂 = ℝ₊ or ℝ₊ᵒᵖ that have the same form as (84), but with parameters α = 0, z < 1, or α = 1, z < 1, are monotone or not. The same question of monotonicity can also be asked for the homomorphisms for 𝕂 = 𝕋ℝ₊ that are similar to (85), but with z < 0.

We suspect that these additional homomorphisms are in fact not monotone either, since they correspond to parameters α, z for which the α-z relative entropy is known not to satisfy DPI for general pairs of quantum states (with some conditions on the support).

The settled claim excludes $z=0$, which the source treats as the degenerate trace maps. It asserts failure of both real target orders for every $\alpha\in{0,1}$ and every real $z<1$, $z\ne0$, and failure of the ordinary tropical order for every real $z<0$.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Eq. (31): finite direct sums of two rank-one block matrices; vectors may vanish | FlatPair, FlatPair.A, FlatPair.B | Arbitrary finite block counts and dimensions; sigma indices are written directly | | Eqs. (33)–(34): nonzero overlap or both block vectors zero | FlatPair.IsEverywhereOverlapping | Every block, including zero blocks | | Eq. (37): weights and normalized overlaps | FlatPair.p, FlatPair.q, FlatPair.overlapSq | Unnormalized vectors, real self-dot-product weights | | One channel acts on both matrices | Majorizes | Existing completely positive trace-preserving QuantumChannel | | Eqs. (84)–(85) | Phi, PhiT | Zero blocks contribute zero; the maximum starts at zero | | Both real target orders | IsMonotone, IsMonotoneOp | The same channel preorder and everywhere-overlapping domain | | Endpoint suspicion, excluding trace maps | First conjunct of claim, proved by result | All real parameters $\alpha\in{0,1}$, $z<1$, $z\ne0$ | | Negative tropical suspicion | Second conjunct of claim, proved by result | Every real $z<0$ in the ordinary tropical order |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/FlatPairEndpointHomomorphisms.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2025Gurland's original logarithmic remainderThe original Gurland logarithmic remainder tends to zero for every positive real pair.Proved

The questionHalina Wiśniewska, Some inequalities for Gurland's ratio of the gamma functions, arXiv:2512.07028v1, section 2, Open Problem 1:

Why it mattersThe closed proposition claim is precisely the all-positive limit above.

The ideaIn the real Euler logarithmic sequence at 1+x, 1+y and 1+a, the factorial and log(N) contributions cancel.

Our contributionThe original Gurland logarithmic remainder tends to zero for every positive real pair.

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Release source recordProved in Lean

arXiv

Gurland's original logarithmic remainder

Halina Wiśniewska, Some inequalities for Gurland's ratio of the gamma functions, arXiv:2512.07028v1, section 2, Open Problem 1:

In Corollary 1 we showed that R_n^m(x,y) tends to zero under a sufficient condition. Does R_n^m(x,y) tend to zero hold for every x,y > 0?

For positive real x,y put a=(x+y)/2 and d=(x-y)/2. In equations (2)–(4),

$$ G_* = \frac{\Gamma(1+x)\Gamma(1+y)}{\Gamma(1+a)^2},\qquad H(s,b)=\sum_{j=0}^{\infty}\frac1{(j+b)^s}, $$ $$ S_m=\sum_{k=1}^{m-1}\frac{d^{2k}H(2k,1+a)}{k},\qquad R_m=\log G_*-S_m. $$

The expansion order m tends to infinity through integers, with m >= 2 in the source. Its n is already summed out. Every positive real pair, including x=y and positive integer Gamma arguments, is in scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/SeriesInequalities/GurlandLogRemainder.result

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Analysis & dynamicsQuestion resolvedSource 2026Complete monotonicity of the Xu–Yang Mills remaindersYang, Higher order monotonicity involving the Mills ratio with applications, Journal of Inequalities and Applications 2026:98, DOI 10.1186/s13660-026-03505-9.Proved

The questionYang, Higher order monotonicity involving the Mills ratio with applications, Journal of Inequalities and Applications 2026:98, DOI 10.1186/s13660-026-03505-9.

The ideaAll orders of differentiability follow from the same domination.

Our contributionYang, Higher order monotonicity involving the Mills ratio with applications, Journal of Inequalities and Applications 2026:98, DOI 10.1186/s13660-026-03505-9.

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Release source recordProved in Lean

Open problem

Complete monotonicity of the Xu–Yang Mills remainders

H.-Z. Xu and Z.-H. Yang, Higher order monotonicity involving the Mills ratio with applications, Journal of Inequalities and Applications 2026:98, DOI 10.1186/s13660-026-03505-9.

Mills-ratio definition, printed p. 2:

The Mills ratio [3] is defined by

$$R\equiv R(x)=\frac{1-\Phi(x)}{\phi(x)} =e^{x^2/2}\int_x^\infty e^{-t^2/2},dt,\quad x\ge0,$$

which satisfies the differential equation of first order:

$$R'=xR-1.\qquad(1.1)$$

Here $\phi(x)=e^{-x^2/2}/\sqrt{2\pi}$ and $\Phi(x)=\int_{-\infty}^x\phi(t),dt$ are the standard Gaussian density and cumulative distribution function used in the preceding source sentences.

The bound family in (1.5), printed p. 2, is

$$\varphi_{p,q,r}(x)=\frac{p}{\sqrt{x^2+q}+rx}, \qquad p,q>0,\quad r\ge0.$$

Equation (1.8), printed p. 3, with its complete surrounding statement:

Note that all bounds of the form $\varphi_{p,q,r}(x)$ can be derived from the inequality

$$Y_{a,b}(x):=(x^2+a)R^2(x)+bxR(x)<(>)c\qquad(1.8)$$

for all $x>0$, $a,b\in\mathbb R$, where

$$c=\lim_{x\to0}Y_{a,b}(x)=\frac{a\pi}{2} \quad\text{or}\quad c=\lim_{x\to\infty}Y_{a,b}(x)=b+1.$$

Equation (3.7), printed p. 6, gives the coefficients following the asymptotic expansion in Theorem 3.1:

where

$$u_n\equiv u_n(a,b)=(b+1)\frac{2^n\Gamma(n+1/2)}{\Gamma(1/2)} +(n-a)(n-1)!\sum_{k=0}^{n-1} \frac{2^k\Gamma(k+1/2)}{\Gamma(1/2)k!}.\qquad(3.7)$$

In particular, $u_1=b+2-a$, $u_2=3b+7-2a$, $u_3=15b+36-7a$, and $u_4=105b-36a+249$.

Conjecture 7.1, printed p. 27, complete statement:

Moreover, inspired by Theorem 4.2, we propose the following conjecture.

Conjecture 7.1 Let

$$D_n(x,a)=Y_{a,a-2}(x)-\left(a-1+\sum_{k=1}^{n} \frac{(-1)^ku_k}{x^{2k}}\right),$$

where $u_k$ is defined by (3.7). Then $x\mapsto(-1)^nD_n(x,-1)$ is CM on $(0,\infty)$.

Formal reading

The namespace is D5.S3.Analytic.XuYangMillsRemainder.Result. The definitions millsRatio, Y, u and D retain the displayed Gaussian integral, polynomial expression, Gamma coefficients and finite remainder. Here $\Gamma$ is Euler's Gamma function, $k!$ is the factorial of the nonnegative integer $k$, and all integrals use Lebesgue measure. CompletelyMonotoneOn f means $f$ is $C^\infty$ on $(0,\infty)$ and every signed derivative $(-1)^m f^{(m)}(x)$ is nonnegative there. claim quantifies over every integer $n\ge2$; result states claim. The source does not display a lower bound for $n$ in Conjecture 7.1. The restriction $n\ge2$ is necessary: $n=1$ fails, as explained in Triage.

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Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/XuYangMillsRemainder/Result.result

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  • Historical context

    Year recorded in the source bibliography: 2026.

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Number theoryConjecture provedFirst preprint 2026Xu-Zhao Conjecture 1.3The original coefficients have the full all-composition eventual signs and strict admissible binomial excess.Proved

Why it mattersThe scalar-series, disk, boundary, slit and local Banks results connect the strict-largest-index source to an analytic reciprocal.

The ideaThe disk, admissible-boundary, slit, local Banks and zero-free collar results supply one actual normalized continuation.

Our contributionThe original coefficients have the full all-composition eventual signs and strict admissible binomial excess.

Verification details
Release source recordProved in Lean

Open problem

Xu-Zhao Conjecture 1.3

For every nonempty positive composition $k=(k_1,\ldots,k_d)$ and integer $\ell\ge1$, put $F_k(z)=z^{-d}\operatorname{Li}_k(z)$ with its removable value at zero, and $C_n^{j;k;\ell}=z^n^{-j}F_k(z)^{-\ell}$ as in source equations (7)-(8). Conjecture 1.3 asks for $C_n^{0;k;\ell}<0$ eventually. For every fixed integer $j\ge1$ it also asks for $C_n^{j;k;\ell}>0$ eventually and, if $k_1>1$, for $C_n^{j;k;\ell}>\binom{n+j-1}{n}/\zeta(k)^\ell$ eventually. The threshold may depend on $k,\ell,j$; the strict inequality and all quantifiers are retained.

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Module present in current Truth release; resolution evidence is pinned below.

D5/S3/AnalyticClosure/Polylogarithm/FiniteSlitAdmissibleSmoothing.full_result

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Quantum informationQuestion resolvedFirst preprint 2026All qubit tuples maximizing the imaginary Bargmann invariantFor every n at least three, the unit qubit tuples with imaginary Bargmann invariant tau n form one unitary orbit of a latitude polygon, allowing an independent phase at each vertex.Proved

The questionLin Zhang and Bing Xie, A Survey of Bargmann Invariants: Geometric Foundations and Applications, arXiv:2601.01858v2, p. 26, ask:

Why it mattersThe maximum value alone does not identify all tuples attaining it.

The ideaA positive imaginary Bargmann value makes every cyclic overlap nonzero.

Our contributionFor every n at least three, the unit qubit tuples with imaginary Bargmann invariant tau n form one unitary orbit of a latitude polygon, allowing an independent phase at each vertex.

Verification details
Release source recordProved in Lean

Open problem

All qubit tuples maximizing the imaginary Bargmann invariant

Lin Zhang and Bing Xie, A Survey of Bargmann Invariants: Geometric Foundations and Applications, arXiv:2601.01858v2, p. 26, ask:

The quantity τ_n predicts the upper bound for the imaginary part of any nth-order Bargmann invariant. This immediately raises the question: what is the complete set of n-tuples Ψ = (|ψ_1⟩, …, |ψ_n⟩) in ℂ² that achieve Im Tr(ψ_1 … ψ_n) = τ_n? We leave this characterization, along with the operational interpretation of the condition z ∈ 𝓑_{n+1} ∖ 𝓑_n, as interesting open problems for future work.

The first question is Proved with the exact conventions in preregistration #19224. For every $n\ge3$, a tuple of unit vectors in $\mathbb C^2$ attains

$$\tau_n=\frac{\cos^n(\pi/n)}{\cos^{n-1}(\pi/[2(n-1)])}$$

if and only if $\psi_j=\eta_jUv_j$ for one common $2\times2$ unitary $U$, independent phases $|\eta_j|=1$, and

$$v_j=(\sqrt{1-t_n},\sqrt{t_n},e^{2\pi ij/n}),\qquad t_n=\frac{1-\tan(\pi/[2(n-1)])/\tan(\pi/n)}2.$$

The Library note gives Definition 3.1, Corollary 5.5, Eq. (5.15), and the source question.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every $n\ge3$ | claim universally quantifies n with 3 ≤ n | Exact | | Unit vectors in $\mathbb C^2$ | Ψ : Fin n → Fin 2 → ℂ, with star (Ψ j) ⬝ᵥ Ψ j = 1 for every j | Exact; the wave-function condition is inlined | | Pure state $|\psi_j\rangle\langle\psi_j|$ and ordered trace | bargmann uses Matrix.vecMulVec, List.ofFn, List.prod, and Matrix.trace | Exact; the projector expression is inlined | | Maximal imaginary value | tau is the source's cosine-power quotient | Exact; natural subtraction occurs in the exponent, real subtraction in the angle | | Complete set | result : claim proves both directions of the equality/orbit equivalence | Exact | | Explicit candidate answer | maximizer is the displayed latitude vector; one U ∈ Matrix.unitaryGroup (Fin 2) ℂ and phases η | The answer proposed in #19224, with no added hypothesis |

No clause is dropped or weakened. The source's second question is outside this equivalence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/BargmannImaginaryMaximizers.result

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Analysis & dynamicsCounterexample foundFirst preprint 2009Ayyer–Mallick maximal degeneracy of the annihilation-process generatorAt L = 5, alpha = 1/3 and beta = 2/5, the annihilation-process generator has an eigenspace of dimension at least two at eigenvalue -2, refuting the Ayyer–Mallick maximal-degeneracy conjecture.Refuted

The questionMallick, Exact results for an asymmetric annihilation process with open boundaries, J.

Why it mattersIt refutes the geometric-multiplicity conjecture using positive rates and a real eigenvalue.

The ideaCasting transfers the equalities to the real matrix.

Our contributionAt L = 5, alpha = 1/3 and beta = 2/5, the annihilation-process generator has an eigenspace of dimension at least two at eigenvalue -2, refuting the Ayyer–Mallick maximal-degeneracy conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Ayyer–Mallick maximal degeneracy of the annihilation-process generator

A. Ayyer and K. Mallick, Exact results for an asymmetric annihilation process with open boundaries, J. Phys. A 43, 045003 (2010), arXiv:0910.0693v2, after Eq. (79), p. 14:

We also conjecture that the Markov matrices are maximally undiagonalizable in the sense that each eigenvalue seems to have exactly one eigenvector independent of the number of times the eigenvalue appears (this implies that the Jordan blocks are of the maximum possible size). In other words, the degeneracy seems to be so strong that the Markov matrix M_L which is of size 2^L has only 2L eigenvectors.

A. Ayyer and V. Strehl, The spectrum of an asymmetric annihilation process, arXiv:1008.2134v1, FPSAC 2010, DMTCS Proc. AN, 461–472, DOI 10.46298/dmtcs.2884, Conjecture 1, p. 466:

The matrix M_L is maximally degenerate in the sense that it has exactly 2L eigenvectors.

The model has $L$ Boolean sites, bulk transitions $10\to01$ and $11\to00$ at rate $1$, a symmetric left-site flip at rate $\alpha$, and right-site removal at rate $\beta$. This is the papers' specialization $\lambda=1$. The matrix entry $M_L(t,s)$ is the rate from $s$ to $t$ off the diagonal; each diagonal entry is the negative sum of the other entries in its column.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | $L$ sites, empty or occupied | Fin L → Bool | true denotes a particle; index $0$ is site $1$. | | $10\to01$, $11\to00$, rate $1$ | bulkMove, bulkCount | Clear the occupied site and toggle its right neighbour; sum over all bonds. | | Left-site $0\leftrightarrow1$, rate $\alpha$ | leftCount, offRate | Both directions have the same rate at $\lambda=1$. | | Right-site $1\to0$, rate $\beta$ | rightCount, offRate | Removal requires an occupied last site. | | Target-row/source-column matrix, columns sum to zero | genMat | Diagonal is the negative sum of all off-diagonal outgoing rates. | | One independent eigenvector at each eigenvalue | claim | Every $L\ge1$, positive real $\alpha,\beta$, and real $\mu$ has $\dim\ker(M_L-\mu I)\le1$. | | Negative answer | result : ¬ claim | $L=5$, $\alpha=1/3$, $\beta=2/5$, $\mu=-2$. |

The source assertion includes eigenvalues without restricting their scalar field. A two-dimensional real eigenspace already refutes it. Quantifying over all real $\mu$ also includes numbers outside the spectrum, whose kernels have dimension zero. The definitions at $L=0$ are total extensions; claim requires $L\ge1$. The extension of bulkMove to coinciding indices is never used by a bulk transition. None of these conventions weakens the counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/ParticleProcesses/AnnihilationMaximalDegeneracyRefutation.result

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Quantum informationCounterexample foundFirst preprint 2025The pure-state scale does not confine purely dissipative spectra to the unit diskA traceless diagonal jump on three complex coordinates gives an eigenvalue outside the disk determined by the pure-state scale.Refuted

The questionTarnowski and K. Życzkowski, Spectral delineation of Markov generators: classical vs quantum, arXiv:2504.07903v1 (2025), Section “Beyond complete positivity”, Conjecture 3.

Why it mattersThe source seeks quantum analogues of the spectral enclosure for classical Markov generators.

The ideaEvery value is at most one, providing boundedness above.

Our contributionA traceless diagonal jump on three complex coordinates gives an eigenvalue outside the disk determined by the pure-state scale.

Scope & next questionConjecture 3 and the universal optimality assertion immediately following it (line 1051) cannot stand under the stated purely dissipative scope: the proposed scale is not even sufficient.

Verification details
Release source recordRefuted in Lean

Open problem

The pure-state scale does not confine purely dissipative spectra to the unit disk

D. Chruściński, S. Denisov, W. Tarnowski and K. Życzkowski, Spectral delineation of Markov generators: classical vs quantum, arXiv:2504.07903v1 (2025), Section “Beyond complete positivity”, Conjecture 3. The source is bounding_6.tex, lines 1044–1050, md5 31cb6e7a93cfb4dac99f83105a784d57. The wording below is verbatim; the source's bra–ket macros are expanded for display:

If $\mathcal{L}$ is a generator of positive trace-preserving dynamics and

$\nu_{\text{q},3} := \max_{|\psi|=1} | \langle \psi|\mathcal{L}(|\psi\rangle\langle\psi|)|\psi\rangle|$,

then the spectrum of $\frac{1}{ \nu_{\text{q},3}} \mathcal{L}$ is confined in the unit disk $\mathbb{D}(-1,1)$.

The disk is closed, with centre $-1$ and radius $1$. The paper sets its purely dissipative scope at lines 227 and 702. Its GKLS formula is at lines 205–209; the canonical traceless, Hilbert–Schmidt orthonormal jumps appear at lines 794–799. The source and publication identifiers are collected in the literature note.

Issue #20152 fixes the restricted reading: for every $N,m\in\mathbb N$, every finite family of traceless complex $N\times N$ jumps $L_k$ and nonnegative real rates $\gamma_k$, put $\mathcal L=\sum_{k\in\mathrm{Fin},m}\gamma_k\mathcal D_{L_k}$, where $\mathcal D_L(\rho)=L\rho L^\dagger-\tfrac12(L^\dagger L\rho+\rho L^\dagger L)$. Define $\nu$ as the supremum of $|\langle\psi|\mathcal L(|\psi\rangle\langle\psi|)|\psi\rangle|$ over $\sum_i|\psi_i|^2=1$. If $\nu>0$, every nonzero eigenmatrix $X$ with $\mathcal L(X)=\lambda X$ would satisfy $|1+\lambda/\nu|\le1$. This is claim; result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/DissipatorSpectralDiskRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2026The Thue–Morse prefix 01101001 has quantum Černý complexity 3, not 2Lee, Lee and Kjos-Hanssen (arXiv:2609.40154) define the quantum Cerny complexity qc(w) of a binary word w as the least dimension d for which two quantum channels on d x d density matrices and a start state make w the unique shortest synchronizing word, and conjecture that the Thue-Morse prefix 01101001 has qc = 2.Refuted

Our contributionLee, Lee and Kjos-Hanssen (arXiv:2609.40154) define the quantum Cerny complexity qc(w) of a binary word w as the least dimension d for which two quantum channels on d x d density matrices and a start state make w the unique shortest synchronizing word, and conjecture that the Thue-Morse prefix 01101001 has qc = 2.

Verification details
Release source recordRefuted in Lean

arXiv

The Thue–Morse prefix 01101001 has quantum Černý complexity 3, not 2

Lee, Lee and Kjos-Hanssen, Quantum Černý complexity of binary words, arXiv:2609.40154v1, define the quantum Černý complexity $\mathrm{qc}(w)$ as the least $d$ for which two quantum channels on $d\times d$ density matrices and a start state make $w$ the unique shortest synchronizing word (Definition 1.1). Open problem 1 reads

(Thue--Morse prefix.) We conjecture $\qc(01101001)=2$; the bound $\qc(01101001)\le3$ follows from Theorem~\ref{thm:kmp}.

Issue #13185 fixes the reading: channels and density matrices are the frozen QuantumChannel (Fin d) (Fin d) and DensityState (Fin d), letters act left to right, the reachable set is the set of images of the start state, and qc w = sInf {d | 1 ≤ d ∧ some instance of dimension d has w as its unique shortest synchronizing word}.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/QuantumCernyThueMorseRefutation.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2025A stabilizer measurement exceeds every maximally entangled oneA pure three-qubit ancilla separates the information dimensions of two stabilizer measurements: two Bell pairs and local Z measurements give at least 32, whereas every maximally entangled stabilizer measurement gives at most 31.Refuted

The questionInnocenti, The non-stabilizerness cost of quantum state estimation, arXiv:2510.00157v2, Section IV, Conjecture 1 (PDF page 8), state:

Why it mattersAt n=m=3, one pure ancilla has only 31 nonzero Pauli expectations.

The ideaTake ψ=(1−i,1+i,−2,−2,2,2,2+2i,2−2i)/6 in computational order 000,…,111.

Our contributionA pure three-qubit ancilla separates the information dimensions of two stabilizer measurements: two Bell pairs and local Z measurements give at least 32, whereas every maximally entangled stabilizer measurement gives at most 31.

Verification details
Release source recordRefuted in Lean

Open problem

A stabilizer measurement exceeds every maximally entangled one

G. Lo Monaco, S. Lorenzo, A. Ferraro, M. Paternostro, G. M. Palma and L. Innocenti, The non-stabilizerness cost of quantum state estimation, arXiv:2510.00157v2, Section IV, Conjecture 1 (PDF page 8), state:

For any pure |ψ⟩, the maximal value of s_μ can be obtained with maximally entangled S.

For n data qubits and m ancillas, a stabilizer measurement is an orthonormal common-eigenvector basis of a maximal Abelian Pauli subgroup excluding −I. Its effective effects are μ_b = (I⊗⟨ψ|)|Φ_b⟩⟨Φ_b|(I⊗|ψ⟩), and s_μ is the complex dimension of their linear span. The source defines 2p = n+m−dim(S_n)−dim(S_m), where S_n and S_m are the data-supported and ancilla-supported subgroups. Maximal entanglement means p=min(n,m). The formal claim retains all n,m and all pure normalized ancillas and compares every stabilizer basis for the same ancilla.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/StabilizerPovmMaximalEntanglementRefutation.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2026DomiRank entropy need not decrease with competitionDomiRank entropy can increase with competition on a connected nonregular simple graph.Refuted

The questionYingying Zhang and Chengye Zhao, Theoretical Analysis of DomiRank Centrality: Automorphism, Entropy, and Graph Transformations, arXiv:2610.00107v1, Section 4.1, Remark 4.6, identify the unconditional assertion

Why it mattersPreregistration 13267 classifies this explicit single-paper question as Tier 1.

The ideaAll coordinates are strictly between zero and one.

Our contributionDomiRank entropy can increase with competition on a connected nonregular simple graph.

Scope & next questionThe separate star-minimum Conjecture 4.7 remains outside this result.

Verification details
Release source recordRefuted in Lean

Open problem

DomiRank entropy need not decrease with competition

Yingying Zhang and Chengye Zhao, Theoretical Analysis of DomiRank Centrality: Automorphism, Entropy, and Graph Transformations, arXiv:2610.00107v1, Section 4.1, Remark 4.6, identify the unconditional assertion

the DomiRank entropy of a connected non-regular graph decreases monotonically with σ

as an open problem. The claim quantifies over every finite connected nonregular unweighted undirected simple graph and every pair $0<\sigma<\tau<-1/\lambda_{\min}(A)$. It asserts $H_G(\tau)\le H_G(\sigma)$, with the actual inverse equilibrium $\Gamma(s)=s(I+sA)^{-1}A\mathbf1$, positive clipping, normalization, and Shannon entropy in bits. There is no raw-positivity or extra nonzero-normalizer premise. Conjecture 4.7 is excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/DomiRankEntropyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
Number theoryQuestion resolvedFirst preprint 2026Andoyo's equal Pisano sets questionThe distinct integer seed pairs (0,1) and (1,1) have equal complete Pisano sets.Proved

The questionAre there values of a₁, a₂, b₁, and b₂ for which χ(a₁,b₁)=χ(a₂,b₂)?

Why it mattersTheir complete Pisano sets agree over all odd primes, with the source's least positive scalar periods and all three Legendre index counts.

The ideaEvery positive scalar period of f is one of g.

Our contributionThe distinct integer seed pairs (0,1) and (1,1) have equal complete Pisano sets.

Verification details
Release source recordProved in Lean

Open problem

Andoyo's equal Pisano sets question

Section 5, Question 5.3 asks:

Are there values of a₁, a₂, b₁, and b₂ for which χ(a₁,b₁)=χ(a₂,b₂)?

The preceding paragraph requires a₁,b₁,a₂,b₂∈ℤ, distinct ordered pairs, and neither pair equal to (0,0). Definitions 2.6 and 2.7 specify the least positive scalar period at every nonnegative index, the three index counts for Legendre values 1,−1,0, and their integer difference over all odd primes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/AndoyoPisanoSetCollision.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Equatorial multiplicativity of the qubit Wigner distanceUsing the compact minimum bridge in WignerDistanceMinimum, equatorial qubit states obey exact multiplicativity of one plus the Wigner distance.Proved

The questionSoumyojyoti Dutta and Tushar, A Phase-Space Geometric Measure of Magic in Qubit Systems, arXiv:2603.20792v3, printed p. 8, Conjecture 5.6:

Why it mattersThe two-qubit free set includes entangled stabilizers: it is defined through four-element commuting subgroups of the matrix unitary group with a unique common normalized eigenray.

The ideaEvery single-qubit Wigner vector has at most one negative coordinate.

Our contributionUsing the compact minimum bridge in WignerDistanceMinimum, equatorial qubit states obey exact multiplicativity of one plus the Wigner distance.

Verification details
Release source recordProved in Lean

Open problem

Equatorial multiplicativity of the qubit Wigner distance

Soumyojyoti Dutta and Tushar, A Phase-Space Geometric Measure of Magic in Qubit Systems, arXiv:2603.20792v3, printed p. 8, Conjecture 5.6:

For $\langle Z\rangle_\rho=\langle Z\rangle_\sigma=0$: $C(\rho\otimes\sigma)=C(\rho)+C(\sigma)+C(\rho)C(\sigma)$.

The quantities use the fixed Wootters product frame, the convex hull of Wigner functions of actual Pauli stabilizer states, and the L1 distance. Issue #11541 preregisters the fully quantified density-matrix statement, Tier 1 classification and literature reading before the Lean probe.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/QubitWignerDistanceTensorRules.resultEquatorial

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Self-tensor superadditivity on the nonpositive Bloch-product branchUsing the compact minimum bridge in WignerDistanceMinimum, equatorial qubit states obey exact multiplicativity of one plus the Wigner distance.Proved

The questionSoumyojyoti Dutta and Tushar, A Phase-Space Geometric Measure of Magic in Qubit Systems, arXiv:2603.20792v3, printed p. 8, Conjecture 5.7:

Why it mattersThe two-qubit free set includes entangled stabilizers: it is defined through four-element commuting subgroups of the matrix unitary group with a unique common normalized eigenray.

The ideaEvery single-qubit Wigner vector has at most one negative coordinate.

Our contributionUsing the compact minimum bridge in WignerDistanceMinimum, equatorial qubit states obey exact multiplicativity of one plus the Wigner distance.

Verification details
Release source recordProved in Lean

Open problem

Self-tensor superadditivity on the nonpositive Bloch-product branch

Soumyojyoti Dutta and Tushar, A Phase-Space Geometric Measure of Magic in Qubit Systems, arXiv:2603.20792v3, printed p. 8, Conjecture 5.7:

For any qubit state $\rho$ with $s(\rho)\leq0$: $C(\rho\otimes\rho)\geq2C(\rho)$.

The quantities use the fixed Wootters product frame, the convex hull of Wigner functions of actual Pauli stabilizer states, and the L1 distance. Issue #11541 preregisters the fully quantified density-matrix statement, Tier 1 classification and literature reading before the Lean probe.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/QubitWignerDistanceTensorRules.resultSelfTensor

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture provedFirst preprint 2025The forbidden-subgraph conjecture for MMI in graph states holdsEvery graph state that violates monogamy of mutual information is carried by local complementations to a graph with an induced four-star, and that graph is a generalized star for the partition that places the three leaves of the star in their own parts and every other vertex in the centre.Proved

The questionFuentes, Keeler, Munizzi and Pollack (arXiv:2511.19585v1) conjecture:

The ideaLocal complementation at vertices of a vertex set commutes with restriction to that set, so an induced four-star, or a relabelling of a fixed graph, reached on part of the vertices lifts to the whole graph.

Our contributionEvery graph state that violates monogamy of mutual information is carried by local complementations to a graph with an induced four-star, and that graph is a generalized star for the partition that places the three leaves of the star in their own parts and every other vertex in the centre.

Verification details
Release source recordProved in Lean

Open problem

The forbidden-subgraph conjecture for MMI in graph states holds

Fuentes, Keeler, Munizzi and Pollack (arXiv:2511.19585v1) conjecture:

Conjecture 1 (Forbidden-Subgraph). Any graph state that violates MMI has a graph representation that is $LC$-equivalent to a graph $H$ containing a $K_4$ subgraph. $H$ is $\mathcal{G}$ with respect to some partition.

Issue #10534 fixes the readings from the paper: the entropy of a set of qubits A is the rank over Z_2 of the adjacency block with rows in A and columns outside A; an MMI violation is a triple of pairwise disjoint sets I, J, K with S_IJ + S_IK + S_JK < S_I + S_J + S_K + S_IJK; LC equivalence is a finite sequence of local complementations; K_4 is the four-vertex star K_{1,3}, read as an induced subgraph, which is the stronger reading; and a generalized star with centre C and parts V^1, …, V^k has nonempty, pairwise disjoint parts that with C partition the vertices, no edge between different parts, and an edge from every part to C.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GraphStateMonogamyForbiddenSubgraph.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2026Ghosh–Ramachandran's connected-set Fekete area questionA degree-four Fekete polynomial on the connected interval [−20,20] has larger unit-lemniscate area than a monic competitor with zeros in the same interval.Refuted

The questionRamachandran, A note on the Erdős minimal area problem, arXiv:2604.03036v3, immediately after Proposition 3.6:

The ideaSumming disc areas gives the exact rational bounds

Our contributionA degree-four Fekete polynomial on the connected interval [−20,20] has larger unit-lemniscate area than a monic competitor with zeros in the same interval.

Verification details
Release source recordRefuted in Lean

Open problem

Ghosh–Ramachandran's connected-set Fekete area question

S. Ghosh and K. Ramachandran, A note on the Erdős minimal area problem, arXiv:2604.03036v3, immediately after Proposition 3.6:

While the above proposition proves that Fekete polynomials are not area minimizers in general, it would be of interest to know if they are minimzers when the set $K$ is connected. We leave this as an open question.

In Section 1, $P_n(K)$ consists of monic degree-$n$ polynomials whose zeros all lie in the compact set $K\subseteq\mathbb C$. Write $\Lambda_p={z:|p(z)|<1}$ and $A_n(K)=\inf_{p\in P_n(K)}m(\Lambda_p)$, with $m$ planar Lebesgue measure. A Fekete tuple $a=(a_1,\ldots,a_n)$ maximizes $\prod_{i<j}|a_i-a_j|$ on $K^n$, and defines $F_a(z)=\prod_i(z-a_i)$.

The Lean claim says that for every compact connected $K$, every $n\geq1$, every Fekete tuple $a$ in $K$, and every $p\in P_n(K)$, $m(\Lambda_{F_a})\leq m(\Lambda_p)$. Thus result : ¬ claim refutes exact area minimization in the stated connected-set setting.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Lemniscates/FeketeLemniscateAreaRefutation.result

Frozen module record

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Analysis & dynamicsCounterexample foundFirst preprint 2007Refutation of the sum-condition Luke upper boundAn alternating lower bracket contradicts the sum-condition Luke upper bound.Refuted

The questionSitnik, Inequalities and monotonicity of ratios for generalized hypergeometric function, J.

Why it mattersThe sum condition permits an individual denominator parameter smaller than its numerator partner, destroying the first-order comparison required by the upper expression.

The ideaGeometric comparison gives summability and monotonicity.

Our contributionAn alternating lower bracket contradicts the sum-condition Luke upper bound.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the sum-condition Luke upper bound

D. Karp and S. M. Sitnik, Inequalities and monotonicity of ratios for generalized hypergeometric function, J. Approx. Theory 161 (2009), 337–352, DOI 10.1016/j.jat.2008.10.002, arXiv:math/0703084v2, Conjecture 2, printed p. 14, equation (38):

For 0 < σ ≤ min(a₁, a₂, . . . , a_q), ∑ᵢ₌₁ᵠ (bᵢ − aᵢ) > 0 and x > 0:

$\displaystyle {}{q+1}F_q(\sigma,(a_q);(b_q);-x)<\frac{1}{\left[1+x\prod{i=1}^q\frac{\Gamma(a_i)\Gamma(b_i-\sigma)}{\Gamma(b_i)\Gamma(a_i-\sigma)}\right]^\sigma}.\qquad (38)$

The preceding sentence reads:

The asymptotic formula (16) and numerical experiments suggest the following

Formula (16), printed p. 5, is preceded by:

It follows that for σ < min(a₁, a₂, . . . , a_q) we will have

$$ {}_{q+1}F_q(\sigma,(a_q);(b_q);-z) =\frac{\prod_i\Gamma(a_i-\sigma)\Gamma(b_i)} {\prod_i\Gamma(a_i)\Gamma(b_i-\sigma)}z^{-\sigma} +o(z^{-\sigma}),\qquad z\to\infty. $$

The quotation transcribes the extracted display, which places a $\sigma$ next to the Gamma product. Write $K=\prod_{i=1}^q\Gamma(a_i)\Gamma(b_i-\sigma)/(\Gamma(b_i)\Gamma(a_i-\sigma))$. Separately, the asymptotically precise reading via (16) is $1/(1+xK^{1/\sigma})^\sigma$: the inner exponent $1/\sigma$ makes its leading asymptotic coefficient agree with (16). The counterexample has $\sigma=1$, where both readings coincide.

Formal reading

Lean names below have prefix D5.S3.Analytic.KarpSitnikLuke.UpperResult, except where their frozen owner is specified.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Rising-factorial ratio | D5.S3.Analytic.HypergeometricShiftTuran.TwoBlockResult.pochRatio | Quotient of the products for a and b. | | Summand of (1) | D5.S3.Analytic.KarpSitnikLuke.Result.hyperTerm | Extra upper parameter σ, factorial denominator and zⁿ. | | Series value | D5.S3.Analytic.KarpSitnikLuke.Result.hyper | Real infinite sum of these summands. | | Gamma product in (38) | gammaProduct | Exactly K defined above, with finite index type Fin q. | | Restricted Conjecture 2 | claim | Positive parameters, positive sum(b−a), 0<σ≤aᵢ, 0<x<1, summability, K>0 and the strict upper bound. | | Refutation | result | Exactly ¬ claim. |

Each additional condition weakens the universal assertion: $x<1$ restricts to the power-series disk, summability restricts to convergent series, and $K>0$ restricts the parameter set. The latter, with $x>0$, makes the real-power base positive. Refuting the weaker assertion refutes the conjecture. The $q=0$ positive-sum assumption is impossible.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KarpSitnikLuke/UpperResult.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Geometry & algebraConjecture provedFirst preprint 2026Kaygorodov–Martín González–Viruel Conjecture 5Semisimple derivation algebras are sums of orthogonal algebras.Proved

The questionThe source's Section 1.1 defines an evolution algebra by a natural basis whose distinct basis vectors multiply to zero.

The ideaThe annihilator obstruction shows that semisimplicity forces nonzero natural squares.

Our contributionSemisimple derivation algebras are sums of orthogonal algebras.

Verification details
Release source recordProved in Lean

Open problem

Kaygorodov–Martín González–Viruel Conjecture 5

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, state Conjecture 5: if an evolution algebra is finite-dimensional over an algebraically closed field of characteristic zero and its derivation algebra is semisimple, then the derivation algebra is a finite direct sum of orthogonal Lie algebras $\mathfrak{so}_{m_i}$ with $m_i\geq 3$. The empty sum is allowed.

The source's Section 1.1 defines an evolution algebra by a natural basis whose distinct basis vectors multiply to zero. Proposition 14 realizes the orthogonal summands; its converse realization assertion is not separately formalized here.

Statement echo

The public claim in D5/S3/QuadraticForms/EvolutionAlgebras/SemisimpleDerivationClassification.lean preserves the source's field, algebraic-closedness, characteristic-zero, finite-dimensionality, natural-basis, semisimplicity and finite direct-sum clauses. The public result proves that literal proposition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/SemisimpleDerivationClassification.result

Frozen module record

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    The exact recorded statement is proved.

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Number theoryConjecture provedKotěšovec's first A368959 conjectureEvery positive modulus with nonzero complete monomial sums at every degree at least two is powerful.Proved

The questionVáclav Kotěšovec, OEIS A368959, COMMENTS; preregistration 19241.

Why it mattersA prime occurring exactly once in a modulus forces cancellation at that prime degree.

The ideaChinese remaindering factors the complete additive-character power sum for coprime moduli.

Our contributionEvery positive modulus with nonzero complete monomial sums at every degree at least two is powerful.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kotěšovec's first A368959 conjecture

Václav Kotěšovec, OEIS A368959, COMMENTS; preregistration #19241.

Conjecture 1: This is a subset of A001694.

The NAME is “Numbers m such that Sum_{k=0..m-1} exp(2Pii*k^s/m) <> 0 for all s>=2.” A001694 denotes powerful numbers. A338325 has NAME “Biquadratefree powerful numbers: numbers whose exponents in their prime factorization are either 2 or 3.” Membership uses positive natural numbers.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The complete sum over $0\le k<m$ | KotesovecConjectureOne.monomialSum | Literal summand and range; natural exponentiation precedes complex casting. | | Every degree $s\ge2$ and positive $m$ | KotesovecConjectureOne.InA368959 | Positivity, universal natural degree and nonvanishing are all explicit. | | Prime exponents two or three | KotesovecConjectureOne.InA338325 | Every prime factor of the positive natural number is covered. | | A368959 is a subset of A001694 | claimOne, proved by result | Universal implication to the frozen PowerfulNumber.Powerful predicate; no added hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CharacterSums/KotesovecConjectureOne.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

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Quantum informationConjecture provedSource 2025Pronko's non-cyclic Fredkin eigenstate annihilation conjectureThe periodic Fredkin Hamiltonian and Pronko's nonlocal operators are represented on the spin-word basis.Proved

The questionPronko fixes the one-site basis in section 2.1, printed p. 3:

Why it mattersIt retains the simultaneous Hamiltonian eigenvector hypothesis but proves the stronger fact used by the conjecture: every C-eigenvector with eigenvalue different from one is annihilated by both operators.

The ideaThe exact entry description is Sigma M ε y x = [weight y − weight x = ε]: at each (y,x), at most one Kronecker monomial of (3.1) has a nonzero value, and it lies in the constrained sum exactly when the displayed weight difference equals ε; otherwise no term of (3.1) contributes.

Our contributionThe periodic Fredkin Hamiltonian and Pronko's nonlocal operators are represented on the spin-word basis.

Verification details
Release source recordProved in Lean

Open problem

Pronko's non-cyclic Fredkin eigenstate annihilation conjecture

Pronko fixes the one-site basis in section 2.1, printed p. 3:

We denote basis vectors in ℂ² by |↑⟩ = (1, 0)ᵀ, |↓⟩ = (0, 1)ᵀ.

The basis in End(ℂ²) is provided by the Pauli matrices σ⁺ = [[0, 1], [0, 0]], σ⁻ = [[0, 0], [1, 0]], σᶻ = [[1, 0], [0, −1]].

We will also use projectors onto the spin-up and spin-down states: n↑ = ½(1 + σᶻ), n↓ = ½(1 − σᶻ).

The same page defines the adjacent exchange and antisymmetrizer:

Consider two adjacent copies of ℂ², say, the jth and (j + 1)th. Important objects are the permutation operator P_{j,j+1}, which exchanges these copies, and the projector Π_{j,j+1} onto the spin-0 state, which is essentially the antisymmetrizer, Π_{j,j+1} = ½(1 − P_{j,j+1}).

It then prints the Hamiltonian density and periodic Hamiltonian:

F_{j,j+1,j+2} = n↑_j Π_{j+1,j+2} + Π_{j,j+1} n↓_{j+2}. (2.2)

The periodic boundary conditions mean that the Hamiltonian is the cyclic sum, H = Σ_{j=1}^{N} F_{j,j+1,j+2}. (2.3)

where identification of sites by modulus N is assumed.

Across printed pp. 3-4 the source gives the ordered product and its action:

The operator C = P_{1,2} ⋯ P_{N−2,N−1} P_{N−1,N} is the cyclic shift operator C, acting on sites as i ↦ i + 1.

Theorem 2, equation (3.1), printed pp. 6-7, states:

The operators Σ^± = Σ_{r₁,...,r_N ∈ {−1,0,1}, r₁+⋯+r_N = ±1} σ₁^{r₁} ⋯ σ_N^{r_N}, (3.1) where σ_i^0 = 1 and σ_i^{±1} = σ_i^±.

Conjecture 1, printed p. 7, is:

The operators Σ^± annihilate all non-cyclic invariant (C ≠ 1) eigenstates of the Hamiltonian.

Issue #9359 fixes four readings: (i) an eigenstate is a nonzero H-eigenvector that is also a C-eigenvector; (ii) non-cyclic means its C-eigenvalue c is not one; (iii) (3.1) is represented literally, with Fin 3 values 0,1,2 encoding exponents -1,0,1; and (iv) sites are modulo N, N >= 3, over complex scalars.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PronkoFredkinNonCyclicAnnihilation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2025.

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    Source 1
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  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

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Analysis & dynamicsCounterexample foundFirst preprint 2026Štampach–Waclawek's alternative Birman weight expansionThe Birman weight expansion conjecture fails at order two and p = 11/10.Refuted

The questionWaclawek, Optimal discrete p-Hardy–Rellich–Birman inequalities, arXiv:2605.25238v1, Conjecture 5.6(iii), p. 28:

The ideaInteger-power comparisons give rational enclosures of the required real roots.

Our contributionThe Birman weight expansion conjecture fails at order two and p = 11/10.

Verification details
Release source recordRefuted in Lean

arXiv

Štampach–Waclawek's alternative Birman weight expansion

F. Štampach and J. Waclawek, Optimal discrete p-Hardy–Rellich–Birman inequalities, arXiv:2605.25238v1, Conjecture 5.6(iii), p. 28:

Let $\ell\in\mathbb N$ and $p>1$, and let $\tilde{\rho}^{(\ell,p)}$ be defined by (2.17) and (2.16).

For all $n\geq\ell$, the terms $\tilde{\rho}_{n}^{(\ell,p)}$ admit a power series expansion in negative powers of $n$ with entirely non-negative coefficients; cf. (2.15).

Here $\mathbb N$ means positive integers. The convention is $\tilde{\mathfrak g}n=n^{1-1/p}\prod{j=1}^{\ell-1}(n-j)$ for $n\geq0$, with zero extension to negative integers; $(\nabla u)n=u_n-u{n-1}$ and $(\mathrm{div},u)n=u{n+1}-u_n$; $\nu^{\langle a\rangle}=\nu|\nu|^{a-1}$, with $0^{\langle a\rangle}=0$; and $\tilde\rho_n=(-1)^\ell[\mathrm{div}^{\ell}(m\mapsto(\nabla^{\ell}\tilde{\mathfrak g})_m^{\langle p-1\rangle})]_n/\tilde{\mathfrak g}_n^{p-1}$, with signed power taken before the divergence iterate.

The Lean claim asks for one non-negative coefficient sequence $c:\mathbb N_0\to\mathbb R$ whose series converges to $n^{\ell p}\tilde\rho_n$ at every $n\geq\ell$. It allows any non-negative constant coefficient. The source's (2.15) normalization implies this claim by $c_0=((1/q)_\ell)^p$ and $c_k=c_0 A_k$ for $k\geq1$, where $1/q=1-1/p>0$ and the rising factorial is positive. Thus negating the weaker claim refutes part (iii).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/SeriesInequalities/StampachWaclawekExpansionRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

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Quantum informationQuestion resolvedFirst preprint 2026Conditional hockey-stick contraction coefficients reduce to orthogonal states and do not depend on γConditional hockey-stick ratio sets equal their orthogonal subsets and are independent of gamma at least one.Proved

The questionFor γ = 1 the coefficients η^cTr and η^ΔTr are optimized by orthogonal states (Proposition 8). - The open problem. "When considering the hockey-stick divergence, can the optimization in the contraction coefficient be restricted to orthogonal states, similar to Proposition 8?

Why it mattersFor the trace distance, the restriction to orthogonal states turns the conditional coefficients into tractable optimizations (Proposition 8).

The idea1. Jordan parts. For an admissible pair (ρ, σ) at γ, write ρ − γσ = P − Q with P, Q ⪰ 0 and PQ = 0, and put p = Tr P, q = Tr Q.

Our contributionConditional hockey-stick ratio sets equal their orthogonal subsets and are independent of gamma at least one.

Verification details
Release source recordProved in Lean

arXiv

Conditional hockey-stick contraction coefficients reduce to orthogonal states and do not depend on γ

C. Hirche, I. George, T. Nuradha and M. M. Wilde, Conditional contraction coefficients and their applications to quantum networks, arXiv:2608.27171v1, study contraction coefficients with reference systems.

  • The coefficient. η^Δ_D(N) := sup over |R| and ρ_RA ≠ σ_RA of [D((id ⊗ N)(ρ_RA)‖(id ⊗ N)(σ_RA)) − D(ρ_R‖σ_R)] / [D(ρ_RA‖σ_RA) − D(ρ_R‖σ_R)] (Eq. (2.14)), over pairs with a strictly positive finite denominator (Remark 2).
  • The divergence. E_γ(ρ‖σ) := Tr(ρ − γσ)₊ (Eq. (5.1)). For γ = 1 the coefficients η^c_Tr and η^Δ_Tr are optimized by orthogonal states (Proposition 8).
  • The open problem. "When considering the hockey-stick divergence, can the optimization in the contraction coefficient be restricted to orthogonal states, similar to Proposition 8? This would directly imply that the associated contraction coefficient is monotone in γ" (§9, p. 72).

The verbatim statements are in the literature note.

Issue #17637 fixes the reading for the conditional coefficient η^Δ_{E_γ}, γ ≥ 1:

  • S_γ(N) is the set of ratios over every finite reference system R and all states ρ, σ on R ⊗ A with a positive denominator;
  • S^⊥_γ(N) is its subset from pairs with ρσ = 0;
  • (i), the question asked: S_γ(N) = S^⊥_γ(N);
  • (ii), a strengthening: S_γ(N) = S_δ(N) for all γ, δ ≥ 1.

Both are proved in this reading. The complete coefficient η^c_{E_γ} is not addressed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Divergence/ConditionalHockeyStickOrthogonalReduction.result

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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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Discrete structuresSharp bound provedFirst preprint 2025Lee–Liu integer {2}-domination of three-row gridsThe integer {2}-domination number of the three-row Cartesian grid is ⌊3n/2⌋ + 1 for every positive n.Proved

The questionJia-Ying Lee and Chia-An Liu, The integer {2}-domination number of grids, arXiv:2502.00134v1, Conjecture 2.5:

Why it mattersTheorem 2.4 supplies the alternating-column upper bound.

The ideaThis is feasible and has the required weight.

Our contributionThe integer {2}-domination number of the three-row Cartesian grid is ⌊3n/2⌋ + 1 for every positive n.

Verification details
Release source recordProved in Lean

Open problem

Lee–Liu integer {2}-domination of three-row grids

Jia-Ying Lee and Chia-An Liu, The integer {2}-domination number of grids, arXiv:2502.00134v1, Conjecture 2.5:

For positive integer n, we have γ{2}(G3,n) = ⌊3n/2⌋ + 1.

The settlement is Proved for every positive width. Its preregistration is #17651. Here $G_{m,n}=P_m\mathbin{\square}P_n$. An integer {2}-dominating function assigns each vertex a value in ${0,1,2}$ and has sum at least $2$ on each closed neighbourhood; $\gamma_{{2}}$ is the minimum total weight.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Cartesian product of paths $P_m\mathbin{\square}P_n$ | grid m n, using SimpleGraph.pathGraph and SimpleGraph.boxProd | Vertices in Fin m × Fin n represent the source's one-based labels. | | Values in ${0,1,2}$ | IsTwoDominating m n f bounds every natural value by $2$ | Exactly the abstract and Section 4 domain. | | Sum on each closed neighbourhood is at least $2$ | The second conjunct of IsTwoDominating adds the vertex value to its open-neighbour sum | Each closed-neighbourhood vertex is counted once. | | Minimum total weight | gamma2 m n is the natural infimum of attained feasible weights | The constant function $2$ is feasible, so the weight set is nonempty and its infimum is its minimum. | | Every positive integer $n$ | claim quantifies over every natural $n$ with $1\le n$ | No upper cutoff or additional hypothesis. | | $\lfloor3n/2\rfloor+1$ | Natural division in claim | Natural division equals the floor for these nonnegative integers. | | Equality for three rows | result : claim | Both bounds, including width $1$, are proved. |

Section 1 also allows arbitrary nonnegative integer values. Truncating each value at $2$ preserves feasibility: a neighbourhood containing a value at least $2$ still has sum at least $2$, and otherwise its values are unchanged. Weight cannot increase. Thus the two minimum-weight formulations coincide. This correspondence is a prose argument; the Lean predicate uses the bounded domain. The general row parameter in the definitions names the graph family and adds no hypothesis to the three-row conclusion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/LeeLiuThreeRowIntegerDomination.result

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  • Contribution

    A potential on pairs of three-entry columns gives the lower bound, and alternating columns attain it.

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  • Historical context

    First preprint year: 2025.

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Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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Discrete structuresCounterexample foundFirst preprint 2024A general obstruction to the pinnacle middle-order latticeFor every n at least six, an ordered pinnacle class contains two permutations without a common upper bound in the middle order.Refuted

The questionSagan, Log-concavity and log-convexity via distributive lattices, Order 43 (2026), Article 21, DOI 10.1007/s11083-026-09735-2, arXiv:2408.02782v2, Section 8, printed pages 34–35, Question 8.6 and its preceding definition:

Why it mattersThe ordered pinnacle condition constrains positions, while the middle order constrains inversion counts.

The ideaTheir relevant inversion counts are ιn(An)=n−4 and ι{n−1}(Bn)=n−2.

Our contributionFor every n at least six, an ordered pinnacle class contains two permutations without a common upper bound in the middle order.

Verification details
Release source recordRefuted in Lean

Open problem

A general obstruction to the pinnacle middle-order lattice

J. Liang and B. E. Sagan, Log-concavity and log-convexity via distributive lattices, Order 43 (2026), Article 21, DOI 10.1007/s11083-026-09735-2, arXiv:2408.02782v2, Section 8, printed pages 34–35, Question 8.6 and its preceding definition:

Let σ be a permutation of distinct positive integers. Let Pin_n(σ) = {π ∈ S_n | the pinnacles of π are the elements of σ in that order}. … Computation of examples raise the following questions. Question 8.6. For all n and σ: (a) Is the restriction of the middle order to Pin_n(σ) a distributive lattice? (b) Is the sequence (pin_n(σ)) log-concave?

Section 6, “Permutations”, printed pages 17–18, defines inversions and inversion tops:

Given π = π1 π2 . . . πn ∈ Sn in one-line notation, its set of inversion (values) is Inv π = {(πi, πj) | i < j and πi > πj}.

We also say that πi is an inversion top if (πi, πj) ∈ Inv π for some πj.

On printed page 18 the inversion coordinate and middle order are specified separately:

where ιi = # of inversions in which i is an inversion top.

We can now use the bijection just given to define the middle order (Sn, ⪯) by π ⪯ σ if and only if ι(π) ≤ ι(σ) where the order on inversion tables is component-wise.

Pinnacles are the values π_j with π_{j−1}<π_j>π_{j+1}.

Formal reading

Lean names have prefix D5.S0.Certificates.Combinatorics.LiangSaganPinnacle.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Permutation word | Equiv.Perm (Fin n) | Increasing positions; Fin value k represents k+1. | | Ordered pinnacle sequence | pinnacleWord | Values at internal strict local maxima, from left to right. | | Inversion-top table | iota | Number of smaller values to the right of a value. | | Middle order | middleLe | Componentwise comparison of iota. | | Ordered pinnacle class | Pin | Fibre of pinnacleWord. | | Universal distributive-lattice assertion | claim | A distributive-lattice structure with exactly this order on every fibre. | | Refutation | result | Exactly ¬ claim. |

The source uses distinct positive integers. Repeated values cannot occur in a pinnacle word. A source sequence containing a value outside [n] has empty fibre; the counterexample uses only distinct in-range values.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/LiangSaganPinnacle/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

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  • Mathematical advance · 3/4 · 15% weight

    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
Number theoryQuestion resolvedSource 2011Merca's lifted residue sum at prime orderMerca's lifted residue sum over an even multiplicative-order cycle for a prime modulus.Proved

The questionMerca defines the remainder operation on printed page 2:

Why it mattersConjecture 2 asks for an exact identity over the full multiplicative-order cycle.

The ideaThe half-order power satisfies a^s = -1 in ZMod m: its square is one, while minimality of the order rules out a^s = 1.

Our contributionMerca's lifted residue sum over an even multiplicative-order cycle for a prime modulus.

Verification details
Release source recordProved in Lean

Open problem

Merca's lifted residue sum at prime order

Merca defines the remainder operation on printed page 2:

When m is integer and n is a positive integer the quotient of m divided by n is ⌊m/n⌋ and the value m mod n = m − n⌊m/n⌋ is the remainder (or residue) of the division.

The multiplicative-order convention appears on printed page 17:

For every positive integer m and every integer a relatively prime to m, we denote by ord_m(a) the multiplicative order of a modulo m, i.e., the smallest positive integer n such that a^n ≡ 1 (mod m), namely ord_m(a) = min {n ∈ N∗ | a^n ≡ 1 (mod m)}

The observation leading to Conjecture 2 appears on printed page 23:

Using Maple to determine the value of some sums as Σ_{i=1}^{ord_m(a)} ((2a^i + m) mod 2m), necessary to determine the arithmetic mean (31) for round function, we notice another interesting identity.

The conjecture is stated on the same printed page:

Conjecture 2. Let a and m be relatively prime positive integers. If m is prime and ord_m(a) is even then Σ_{i=1}^{ord_m(a)} ((2a^i + m) mod 2m) = m · ord_m(a).

The printed display brackets the summand as ((2a^i + m) mod 2m): the remainder is taken of the whole quantity 2a^i + m. Formula (39) on the same page encloses each residue summand as (a^i mod m) in the same typography.

With the printed bracketing, the proved statement is: for every positive natural number a and every prime natural number m coprime to a, if the multiplicative order of a modulo m is even, then

Sum_{i=1}^{ord_m(a)} ((2*a^i + m) mod (2*m)) = m*ord_m(a).
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/MercaLiftedResidueSumPrimeOrder.result

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  • Historical context

    Year recorded in the source bibliography: 2011.

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    An explicitly identified research question in the original source.

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    Year recorded in the source bibliography: 2011. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsConjecture provedThe complete A381737 Conjecture 1 enumerationThe exact increasing enumeration of physicists Hermite orders with exactly two largest absolute coefficients.Proved

The questionOrders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Why it mattersThe neighboring absolute-mode result supplies coefficient maximality, while the tie classification and enumeration identify every qualifying order.

The ideaThe coefficient extraction theorem transfers this statement to every exponent in the full polynomial row; parity zeros cannot maximize a positive mode weight.

Our contributionThe exact increasing enumeration of physicists Hermite orders with exactly two largest absolute coefficients.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The complete A381737 Conjecture 1 enumeration

OEIS A381737, revision 28, by Mike Sheppard (2025), names the sequence:

Orders k of Hermite polynomials whose maximal coefficient in absolute value appears twice.

Its Formula field states:

Conjecture 1: a(n) = 2*n*(n + 2) + (n + 1)*(-1)^(n+1).

The offset is 1,1. Let $H_k$ be the physicists Hermite polynomial, with the original coefficients represented by HermiteSignedMode.physicistsHermite over the rationals. For natural $k$ and $j$, write

$$ M(k,j)\iff j\le k\ \land\ \forall i\le k,\quad |[x^i]H_k|\le |[x^j]H_k|. $$

Exactly two maxima means that there exist $u<v$ such that, for every natural $j$, $M(k,j)$ holds if and only if $j=u$ or $j=v$. The entire row from zero through $k$ is compared, including zero coefficients of the other parity. For every positive natural index, define the integer value

$$ a(n)=2n(n+2)+(n+1)(-1)^{n+1}. $$

The complete assertion is that $a(n)>0$ for every $n\ge1$, that $a(n)<a(m)$ whenever $1\le n<m$, and that for every natural order $k$, including zero, exactly two maxima occur if and only if there is exactly one $n\ge1$ with $(k:\mathbb Z)=a(n)$. The additions and products in the formula are integer operations; $n+1$ in the exponent is natural. There is no finite cutoff or asymptotic qualification.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Polynomial/HermiteAbsoluteTieOrders.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2020Second-family singlet-like data exist in every dimensionCyclic translations supply the orthogonal matrices, difference labels, and sum-indexed shield vectors required for the second family in every dimension.Proved

The questionVértesi, Bound entangled singlet-like states for quantum metrology, Phys.

Why it mattersSupplying the data in every dimension closes the second family's construction gap.

The ideaThe proof uses factorization.

Our contributionCyclic translations supply the orthogonal matrices, difference labels, and sum-indexed shield vectors required for the second family in every dimension.

Scope & next questionConjecture 1, the optimality assertion at TeX line 221, still requires an upper bound over all PPT states and remains open here.

Verification details
Release source recordProved in Lean

Open problem

Second-family singlet-like data exist in every dimension

K. F. Pál, G. Tóth, E. Bene and T. Vértesi, Bound entangled singlet-like states for quantum metrology, Phys. Rev. Research 3, 023101 (2021), arXiv:2002.12409v4, section “Second family of states and its quantum Fisher information,” state:

For this family of states we have constructed analytical examples only for $d=3$ and $2^n.$ However, we believe that such states exist for all $d$.

This quotation is from line 674 of the v4 TeX source metroboundent53.tex. Definition 2 and its accompanying requirements are at lines 556–640; the source's analytic constructions are in the appendix “Special properties of orthogonal matrices,” lines 1196–1308. See the literature note.

Issue #25511 fixes the question as existence of the data required by Definition 2 for every natural $d\ge2$. In the module, claim asserts that there exist real $d\times d$ matrices $Q^k$, real vectors $\bar q_m\in\mathbb R^d$, labels $\mu_{ij}\in\mathrm{Fin},d$, real coefficients $a_{ij}$ and complex vectors $s_m\in\mathbb C^{d\times d}$ satisfying all six conditions below. All indices range over $\mathrm{Fin},d$; the coefficient $a_{ij}$ is the source's $S_{ij}$.

Write $N_m=#{(i,j):a_{ij}\ne0,\ \mu_{ij}=m}$, the module's multiplicity, and define the module's tVec by $t_m(i,j)=a_{ij}/\sqrt d$ when $a_{ij}\ne0$ and $\mu_{ij}=m$, and $t_m(i,j)=0$ otherwise. The conditions in IsSecondFamilyData are:

  1. $(Q^k)^{\mathsf T}Q^k=I_d$ for every $k$.
  2. $\sum_k\bar q_m(k)\bar q_n(k)=\delta_{mn}$ for every $m,n$.
  3. $Q^k_{ij}=a_{ij}\bar q_{\mu_{ij}}(k)$ for every $i,j,k$.
  4. $a_{ij}\ne0$ implies $a_{ij}^2=d/N_{\mu_{ij}}$ for every $i,j$.
  5. $\sum_{i,j}\overline{s_m(i,j)}s_n(i,j)=\delta_{mn}$ for every $m,n$.
  6. $(\sum_m s_ms_m^\dagger)^\Gamma=\sum_m t_mt_m^\dagger$, where $M^\Gamma_{(i,j),(i',j')}=M_{(i,j'),(i',j)}$.

The last operation is AbsolutePPT.Bipartite.partialTranspose. The fixed $|01\rangle_{AB}$ factor of the source's shield vectors is suppressed in these coordinates. The theorem is result : claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/BoundEntanglement/SecondFamilyAllDimensions.result

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  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryConjecture provedPosed in 2018Ordowski's binomial and power-sum least-prime conjectureOrdowski's binomial and power-sum conditions define the same least prime in OEIS A133907.Proved

The questionLeast prime number p such that binomial(n+p, p) mod p = 1.

Why it mattersThe independent question is Ordowski's published least-prime characterization for every positive natural index, registered in issue 8099.

The ideaFor every prime p, instantiate Choose.choosemodEqchoosemodmulchoosedivnat to obtain C(n+p,p) ≡ floor(n/p) + 1 (mod p).

Our contributionOrdowski's binomial and power-sum conditions define the same least prime in OEIS A133907.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Ordowski's binomial and power-sum least-prime conjecture

OEIS A133907, NAME (%N, verbatim):

Least prime number p such that binomial(n+p, p) mod p = 1.

Ordowski's COMMENT (%C, verbatim):

Conjecture: a(n) is the smallest prime p such that Sum_{k=1..n} k^(p-1) == n (mod p). Thus a(n) >= A317358(n). - Thomas Ordowski, Jul 29 2018

The offset is %O 1,1. The AUTHOR line (%A, verbatim) is:

Hieronymus Fischer, Oct 20 2007

The related unsigned COMMENT (%C, verbatim) is:

Also the least prime number p such that p divides floor(n/p) or p > n.

Define a n = sInf {p : Nat | p.Prime ∧ (n+p).choose p ≡ 1 [MOD p]}. For each natural n, let S(n) = {p : Nat | p.Prime ∧ (∑ k ∈ Finset.Icc 1 n, k^(p-1)) ≡ n [MOD p]}. The full-quantifier reading is: for every n : Nat, if 0 < n, then a n ∈ S(n) and, for every q : Nat, q ∈ S(n) implies a n ≤ q. Thus membership includes primality and the sum congruence, and leastness compares against every prime satisfying that congruence. The exact Lean type is:

theorem result (n : ℕ) (_hn : 0 < n) :
    IsLeast {p : ℕ | p.Prime ∧
      (∑ k ∈ Finset.Icc 1 n, k ^ (p - 1)) ≡ n [MOD p]} (a n)

The sum includes both endpoints, division is natural-number floor division, and p - 1 is natural subtraction. The unused hypothesis _hn : 0 < n retains the OEIS domain; no strengthened premise is introduced. The defining prime set is nonempty, so the natural infimum is an attained least element. Only the Conjecture: sentence is settled. The clause Thus a(n) >= A317358(n) is a consequence, not claimed separately, and the unsigned comment receives no separate public theorem or resolution claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/OrdowskiBinomialPowerSumLeastPrime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2022Refutation of the dual-frame cross-Gramian Welch boundA spanning family of three nonzero vectors in the real plane and its dual have cross-Gramian coherence 3/19, below the proposed Welch bound 1/3.Refuted

The questionKaczanowski, Cross-Frame Potential, arXiv:2205.05613v3, Section 4.2, Conjecture 41, Eq. (21), page 17, states:

The ideaThe proof verifies both reconstruction equations for every vector and uses them to prove both spanning conditions.

Our contributionA spanning family of three nonzero vectors in the real plane and its dual have cross-Gramian coherence 3/19, below the proposed Welch bound 1/3.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the dual-frame cross-Gramian Welch bound

R. Aceska and M. Kaczanowski, Cross-Frame Potential, arXiv:2205.05613v3, Section 4.2, Conjecture 41, Eq. (21), page 17, states:

Let F be a frame for Fⁿ, and let G be one of its dual frames. Then $\mu(\mathrm{Gr}(F,G))\ge\sqrt{\frac{nk-n^2}{k^2(k-1)}}$.

The source permits real or complex Hilbert spaces and assumes $k\ge n$. It imposes no equal-diagonal, tightness or equal-norm condition. A real counterexample therefore refutes its universal statement. The encoded claim quantifies all natural dimensions and family sizes, both real vector families, n ≤ k, 2 ≤ k, and the frame and dual-frame predicates. The condition 2 ≤ k ensures a nonempty off-diagonal maximum; the witness has n = 2 and k = 3, so this restriction does not remove the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/Designs/CrossFrameWelchBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2026All-direction maximal clique partitions of mixed-alphabet Hamming graphsEvery mixed-alphabet Hamming graph in at least four dimensions admits a partition into maximal full coordinate lines using every direction.Proved

The questionAhmed and Kunjwal, Characterizing unitaries via quasi-process functions, arXiv:2610.00579v1, Conjecture V.1 in Section V.2, ask for genuine multipartiteness of maximal-clique partitions whenever the number of coordinates exceeds three.

Why it mattersThe graph partition is the source's combinatorial formulation of using each party in mutually exclusive controlled-circuit constructions.

The ideaA binary selector chooses a coordinate at each vertex and keeps that choice when the chosen coordinate is flipped.

Our contributionEvery mixed-alphabet Hamming graph in at least four dimensions admits a partition into maximal full coordinate lines using every direction.

Verification details
Release source recordProved in Lean

Open problem

All-direction maximal clique partitions of mixed-alphabet Hamming graphs

Ahmed and Kunjwal, Characterizing unitaries via quasi-process functions, arXiv:2610.00579v1, Conjecture V.1 in Section V.2, ask for genuine multipartiteness of maximal-clique partitions whenever the number of coordinates exceeds three. The surrounding definitions mean that every coordinate direction must occur. Alphabet sizes are independent and at least two. Definition V.1 defines adjacency by disagreement at exactly one coordinate; maximal means inclusion-maximal, not maximum cardinality.

The exact formal statement is: for every natural n with 4 ≤ n, every d : Fin n → ℕ with ∀ i, 2 ≤ d i, and V = (i : Fin n) → Fin (d i), there is a set P : Set (Set V) such that

  • Every S ∈ P is nonempty, is a full coordinate line, and satisfies Maximal (graph d).IsClique S.
  • For every x : V, there is exactly one S such that S ∈ P ∧ x ∈ S.
  • For every i : Fin n, some x : V has line i x ∈ P.

There is no equal-alphabet assumption, upper dimension bound, finite search bound, or assumed binary matching supplier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hamming/HammingMultipartitePartition.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026An infinite twin-free theta family separating the Laplacian equalitiesAn infinite twin-free theta separating family.Proved

The questionAlfuraidan, Khan and Sauras-Altuzarra, Metric dimension and maximum eigenvalue multiplicity of graph Laplacians, arXiv:2610.26875v1, Problem 30, p. 27, ask:

Why it mattersThe source's finite examples do not establish an infinite family.

The ideaExact distance formulas give a two-landmark resolving set, and a lower-bound argument excludes a single landmark.

Our contributionAn infinite twin-free theta separating family.

Verification details
Release source recordProved in Lean

Open problem

An infinite twin-free theta family separating the Laplacian equalities

Alfuraidan, Khan and Sauras-Altuzarra, Metric dimension and maximum eigenvalue multiplicity of graph Laplacians, arXiv:2610.26875v1, Problem 30, p. 27, ask:

Do there exist infinite families of connected twin-free graphs satisfying exactly one of the two equalities mm_L(G)=dim(G), mm_Q(G)=dim(G)? The computational census shows that finite twin-free examples occur in both directions.

The result answers this existence question affirmatively. Here $\dim(G)$ is the minimum size of a resolving set for actual shortest-path distances, $\operatorname{mm}_L$ and $\operatorname{mm}_Q$ maximize real eigenspace dimension over every real eigenvalue, and twin-freeness excludes both true and false twins. The signless Laplacian is $D(G)+A(G)$.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Finite simple graphs | SimpleGraph (Fin (orders i)) in claim | | Connected | the connectivity conjunct for every natural index | | Twin-free, both kinds | TwinFree, defined using neighborhoods with the other vertex removed | | Metric dimension | minimum cardinality of an injective distance representation | | Maximum multiplicity | maximumMultiplicity, using actual real eigenspaces and all real eigenvalues | | Exactly one equality | disjunction of one equality and the other inequality | | Infinite family | all-natural indexing and pairwise nonisomorphism |

The concrete witness has $G_i=\Theta_{3,3,6(i+1)+4}$ and order $6(i+1)+9$. The source permits either equality direction; the construction uniformly supplies the Laplacian direction. No source clause is dropped. The fixed finite carrier only labels the graph; pairwise nonisomorphism expresses infinitude independently of labels.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/TwinFreeThetaSeparation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Halving partitions admit no finite integral decimation familyThe exact halving-partition count admits no finite integral family closed under both decimations.Refuted

The questionAlekseyev, Halving Partitions and Hyperbinary Expansions, arXiv:2611.01243v1, Definition 1 and section 6, asks:

Why it mattersThe question asks whether finitely many integer sequence coordinates suffice for both binary decimations.

The ideaThe proof classifies every actual source list at total 2^m+1, for m at least two.

Our contributionThe exact halving-partition count admits no finite integral family closed under both decimations.

Verification details
Release source recordRefuted in Lean

arXiv

Halving partitions admit no finite integral decimation family

Max A. Alekseyev, Halving Partitions and Hyperbinary Expansions, arXiv:2611.01243v1, Definition 1 and section 6, asks:

This raises the question of whether a(n) is 2-regular: does there exist a finite family of integer sequences containing a(n) such that, for each sequence b(n) in the family, both b(2n) and b(2n+1) are integer linear combinations of sequences in the family [16]?

This dossier anchors exactly that existence question. The negative answer is D5/S1/Recurrence/Partitions/HalvingPartitionNonregularity.result. The resolution kind Refuted refers to the proposed existence of such a family.

Statement echo

For every natural total n, the source carrier consists of every finite ordered list of positive integers summing to n. With virtual predecessor p0=n, each successive part belongs to the value set consisting of the lower and upper integer halves of its predecessor. Equal halves give one value. Repeated parts, arbitrary finite lengths and repeated terminal ones are retained. At zero the empty list is the sole object.

The count a(n) is the cardinality of that independently defined carrier. Its integer cast admits no positive finite dimension d, vector sequence F and fixed integer matrices M0,M1 satisfying, for every n including zero,

$$ F(n)_0=a(n),\qquad F(2n)=M_0F(n),\qquad F(2n+1)=M_1F(n). $$

The same theorem proves both directions of the matrix equivalence with finite indexed families for every integer sequence, including arbitrary finite index types with a distinguished member. Duplicate and dependent constituent sequences are allowed. The distinguished member makes the family nonempty; no dimension bound, basis assumption or invertibility condition enters the statement. The result also identifies the actual-list count with the filtered Composition cardinality for every n and proves a(0)=1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Partitions/HalvingPartitionNonregularity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
Analysis & dynamicsQuestion resolvedThe toppling count of the n × 1 torus sandpileOn the n x 1 torus, fill every cell except c[0,0] with 4 grains and topple cells holding at least 4 grains, the grains sent to c[0,0] being lost.Proved

The questionFill each cell except c[0,0] with 4 grains of sand.

Why it mattersThe abelian sandpile model is the standard example of self-organized criticality.

The ideaA toppling of (i, 0) removes 4 grains and returns 2 through the two self-neighbours, so every cell other than c[0,0] holds 4 − 2v(i) + v(i − 1) + v(i + 1) grains, and c[0,0] is never toppled.

Our contributionOn the n x 1 torus, fill every cell except c[0,0] with 4 grains and topple cells holding at least 4 grains, the grains sent to c[0,0] being lost.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The toppling count of the n × 1 torus sandpile

OEIS A293452 (Joerg Arndt, 2017) counts the iterations of the sandpile of A249872 on the n × k torus:

Let the lattice be c[i,j], 0 <= i,j < n. Fill each cell except c[0,0] with 4 grains of sand. Until all c[i,j] < 4, do the following: Find a c[i,j] >= 4. (According to Knuth, it does not matter which cell is chosen, the result will be the same.) Decrement the chosen cell by 4 and increment its 4 neighbors by 1. c[0,0] is never increased, sand grains placed here are lost.

and its FORMULA section states

Conjecture: T(n,1) = A023855(n).

with A023855 a(n) = 1*(n) + 2*(n-1) + ... + (n+1-k)*k, k = floor((n+1)/2).

Issue #10122 fixes the readings. The printed identity is false at every n (T(1,1) = 0 but A023855(1) = 1; T(2,1) = 1 but A023855(2) = 2): the column T(n,1) = 0, 1, 2, 7, 10, 22, 28, 50, 60, 95 of the entry is A023855 shifted by one place, so the statement settled is the corrected identity T(n,1) = A023855(n − 1) for n ≥ 2. On the n × 1 torus the four neighbours of (i, 0) are (i ± 1, 0) and the cell itself twice. The number of iterations is read, following Knuth's remark, as the length of every legal toppling sequence that ends in a final state, and the existence of such a sequence is part of the statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Sandpiles/TorusColumnToppling.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2025Divisibility of the rectangular sandpile limitThe direct limit of the rectangular sandpile groups with Shkolnikov's signed reflected tiling maps is divisible: every element has a multiple-root at a later rectangular stage.Proved

The questionMikhail Shkolnikov, Algebraic Limits of Sandpiles, arXiv:2506.09912v3, Section 3, states that the image of the direct limit of rectangular sandpile groups may be a rational subtorus or a divisible group.

Why it mattersDivisibility supplies a structural property of the direct limit that is stronger than the previously known existence of elements of every finite order.

The ideaA finite-denominator lift of the first row lies in a finite subgroup, so the injective transfer orbit is periodic.

Our contributionThe direct limit of the rectangular sandpile groups with Shkolnikov's signed reflected tiling maps is divisible: every element has a multiple-root at a later rectangular stage.

Verification details
Release source recordProved in Lean

Open problem

arxiv 2506 09912 shkolnikov sandpile limit divisible

Mikhail Shkolnikov, Algebraic Limits of Sandpiles, arXiv:2506.09912v3, Section 3, states that the image of the direct limit of rectangular sandpile groups may be a rational subtorus or a divisible group. The settled question is the divisibility alternative: for every positive rectangle, every circle-valued sandpile function on it, and every positive integer $k$, a later rectangular stage contains a sandpile function whose $k$-fold sum equals the signed reflected image of the original function.

Statement echo

The Lean declaration D5.S3.StatisticalMechanics.Sandpiles.ShkolnikovSandpileLimitDivisible.claim preserves the six binding conventions from issue #20011: the carrier is AddCircle (1 : ℝ); InRect is the open integer rectangle; IsSandpile is zero extension plus the four-neighbour equation; unfold is the parity-reflected coordinate; embed is the signed reflected zero extension; and the quantifiers require positive $p,q,k$, a sandpile $f$, and a later $g$ with the exact equality. The theorem result proves this proposition without weakening its quantifiers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Sandpiles/ShkolnikovSandpileLimitDivisible.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBala's A193236 congruences modulo five and sevenBala's unbounded modulo-five and modulo-seven coefficient congruences.Proved

The questionThe existence and universal-integrality clauses make the source's normalized generating-function interpretation explicit.

Why it mattersD5.S1.Recurrence.Residue.BalaA193236ModFiveSeven.result proves the full claim.

The ideaThe shared BalaA193237ModFiveSeven.resolve combines integral-root construction, triangular support, Frobenius and rational uniqueness.

Our contributionBala's unbounded modulo-five and modulo-seven coefficient congruences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A193236 congruences modulo five and seven

Peter Bala's FORMULA line in OEIS A193236 states: “Conjectures: a(5n+4) == 0 (mod 5) and a(7n+5) == 0 (mod 7) (both checked up to n = 200).” The source is preregistered in #22113. The companion claim is #22112.

Write $T_j=j(j+1)/2$ and $F(x)=\sum_{j\ge0}(-3)^j(2j+1)x^{T_j}$. The question concerns the constant-one reciprocal cube root $A(x)=F(x)^{-1/3}$ and every natural index, without a finite cutoff.

Statement echo

| Source clause or convention | Formal counterpart | Scope | | --- | --- | --- | | The displayed source series | BalaA193236ModFiveSeven.F | Its degree-$m$ coefficient sums the literal source weights over $j<m+1$ with $T_j=m$. The truncation is exact because $T_j=m$ implies $j\le m$. | | The constant-one reciprocal cube root | BalaA193236ModFiveSeven.IsGF | Rational power series with constant coefficient one and $A^3F=1$. | | A generating function exists | First conjunct of claim | An explicit integral inverse gives a rational normalized solution; the universal clause is nonvacuous. | | Coefficients of the sequence | Integer sequence quantified in claim | Every coefficient of every normalized rational solution equals the cast of its corresponding integer value. | | Modulo five | First divisibility in claim | $5\mid a(5n+4)$ for every $n\in\mathbb N$. | | Modulo seven | Second divisibility in claim | $7\mid a(7n+5)$ for the same universally quantified $n$. |

The existence and universal-integrality clauses make the source's normalized generating-function interpretation explicit. The source-specific definitions are literal; the shared proof uses BalaA193237ModFiveSeven.triangular with the source weights.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/BalaA193236ModFiveSeven.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedSource 2026A three-dimensional minimal idempotent subspaceA five-dimensional complex evolution algebra contains a three-dimensional idempotent subspace with no nonzero proper idempotent subspace.Proved

The questionBarei, with Muse Spark via meta.ai and review by Nicolás Jaramillo Torres, states in Remark 3.5:

Why it mattersThe result separates minimal idempotent subspaces from idempotent lines and from subspaces possessing natural bases.

The ideaThese products span V: uv supplies v, uw then supplies w, and u² supplies u.

Our contributionA five-dimensional complex evolution algebra contains a three-dimensional idempotent subspace with no nonzero proper idempotent subspace.

Verification details
Release source recordProved in Lean

Open problem

A three-dimensional minimal idempotent subspace

Barei, with Muse Spark via meta.ai and review by Nicolás Jaramillo Torres, states in Remark 3.5:

Given a finite-dimensional evolution algebra E, we are yet to find idempotent subspaces of dimension greater than two which do not properly contain a nontrivial idempotent subspace.

The reading fixed in #12879 is an existence question over ℂ. An evolution algebra is a vector space with a bilinear, possibly nonassociative multiplication and a natural basis whose mixed products vanish. For every complex linear subspace U, let $U^2=\operatorname{Span}_{\mathbb C}{xy:x,y\in U}$. Idempotent means $U^2=U$, and nontrivial means nonzero. V is not required to be an evolution algebra in its own right. Minimality ranges over all complex linear subspaces, not only coordinate subspaces or subspaces defined over ℚ.

The precise positive answer is: there exist such an E and V with $\dim_{\mathbb C}V=3$, $V^2=V$, and $U\le V$, $U\ne0$, $U^2=U$ implying $U=V$ for every U. The displayed remark is an unresolved search, not an asserted universal nonexistence theorem. Barei's Example 2.1 is literature and is excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/BareiRemarkThreeFive.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Every upset in a neighbourhood convex geometry is convexLet V be any finite nonempty vertex set and G any simple undirected graph on V.Proved

The questionDaniela Bubboloni and José Cáceres, The neighbourhood convexity, arXiv:2608.25912v1, “Conclusions and future lines of work,” paragraph beginning “By exploring the connection with the neighbourhood preorder P(G)”:

Why it mattersLemma 29(iii) already supplies convex principal upsets, and Lemma 29(iv) supplies the converse implication from convex sets to upsets.

The ideaSymmetric incidence gives antitonicity, extensivity of double polarity and the triple-polarity identity.

Our contributionLet V be any finite nonempty vertex set and G any simple undirected graph on V.

Verification details
Release source recordProved in Lean

Open problem

Every upset in a neighbourhood convex geometry is convex

Daniela Bubboloni and José Cáceres, The neighbourhood convexity, arXiv:2608.25912v1, “Conclusions and future lines of work,” paragraph beginning “By exploring the connection with the neighbourhood preorder P(G)”:

We conjecture that, given a convex geometry (G,n), any upset in P(G) is an n-convex set, having checked it computationally for graphs up to 9 vertices.

For every finite nonempty set $V$ and simple undirected graph $G=(V,E)$, put $B(x)={x}\cup{y:xy\in E}$ and $N(X)=\bigcap_{x\in X}B(x)$, with $N(\varnothing)=V$. Let $C={N(Y):Y\subseteq V}\cup{\varnothing}$, let $h(\varnothing)=\varnothing$ and $h(X)=N(N(X))$ for nonempty $X$, and put $\operatorname{ex}(K)={x\in K:K\setminus{x}\in C}$. Assume $h(\operatorname{ex}(K))=K$ for every $K\in C$. Then, for every $U\subseteq V$,

$$ \left(\forall x\in U,\ \forall y\in V,\ B(x)\subseteq B(y)\Rightarrow y\in U\right) \Rightarrow U\in C. $$

The source anchors are Sections 2.1 and 2.2, Definitions 3, 5, 9 and 28, and Proposition 12. The quantifiers include all positive graph orders, disconnected graphs, universal vertices, equal closed neighbourhoods and the empty upset. No connected, star-free or twin-free hypothesis is substituted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/BubboloniCaceresNeighbourhoodUpsets.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSource 2025Costa–Catarino bidimensional Fibonacci matricesBidimensional Fibonacci matrices have nonzero determinant at every pair of natural indices.Proved

The questionCosta and Catarino, On Matrices with Bidimensional Fibonacci Numbers, Annales Mathematicae Silesianae 40(1), 59–76 (2026), DOI 10.2478/amsil-2025-0014, state an unnumbered conjecture at the end of §4 (p. 73): every order-two bidimensional Fibonacci matrix has nonzero determinant for all nonnegative indices.

Why it mattersThe conjecture concerns the determinant of the matrices generated by the two-coordinate Fibonacci recurrence.

The ideaThe closed form sourceClosedForm satisfies both guarded recurrences and the four corner conditions.

Our contributionBidimensional Fibonacci matrices have nonzero determinant at every pair of natural indices.

Verification details
Release source recordProved in Lean

Open problem

Costa–Catarino bidimensional Fibonacci matrices

Costa and Catarino, On Matrices with Bidimensional Fibonacci Numbers, Annales Mathematicae Silesianae 40(1), 59–76 (2026), DOI 10.2478/amsil-2025-0014, state an unnumbered conjecture at the end of §4 (p. 73): every order-two bidimensional Fibonacci matrix has nonzero determinant for all nonnegative indices. The source definitions are Definition 2.1 (p. 63), Lemma 2.5 (p. 64), and Definition 3.1 (p. 67). Preregistration and the verbatim statement are in issue #22448.

Statement echo

The Lean declaration D5.S1.Recurrence.CostaCatarinoBidimensionalFibonacci.claim keeps the source recurrences with guards $2 \le m$ and $2 \le n$, all four complex corner values, and the matrix with both negative entries. The definition sourceClosedForm separately retains the literal closed form. The claim adds an existence conjunct so the universal determinant assertion cannot be vacuous, and quantifies over every source sequence, every $m$, and every $n$; axes and the diagonal are included.

Proof record

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D5/S1/Recurrence/CostaCatarinoBidimensionalFibonacci.result

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    Year recorded in the source bibliography: 2025.

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Discrete structuresQuestion resolvedSource 2026The first q = 2 Schreier-multiset recurrenceThe first q=2 Schreier-multiset counting sequence obeys its conjectured recurrence.Proved

The questionChu, Geng, King, Miller, Tresch, and Vasseur define on printed page 19:

Why it mattersThe source prints thirteen terms and labels the recurrence conjectural.

The ideaThe live escape witness (W) is the bridge identity

Our contributionThe first q=2 Schreier-multiset counting sequence obeys its conjectured recurrence.

Verification details
Release source recordProved in Lean

Open problem

The first q = 2 Schreier-multiset recurrence

Chu, Geng, King, Miller, Tresch, and Vasseur define on printed page 19:

A^{(s)}_{p,q,n} := {F ⊂ {1, …, 1, …, n − 1, …, n − 1, n} : n ∈ F and q min F ≥ p|F|}

In the printed display, an underbrace labelled s spans the copies of each entry from 1 through n − 1: there are s copies of every one of those entries, followed by one copy of n. The page then says:

Below are the data and conjectured recurrences we gather:

Its first item is:

(|A^{(2)}{1,2,n}|)^∞{n=1}: 1, 2, 4, 9, 19, 41, 88, 189, 406, 872, 1873, 4023, 8641, . . . with a_n = a_{n−1} + 2a_{n−2} + a_{n−3};

The formal reading makes three decisions. First, s = 2 means two copies of each of 1,...,n−1 and one copy of n. Second, (p,q) = (1,2) means 2 · min F ≥ |F|, with min F = F.toFinset.min' on the witness n ∈ F and |F| = Multiset.card. Third, the recurrence starts at 4 ≤ n, the least index for which all four sequence indices are at least one; at the boundary, a_4 = 9 = a_3 + 2a_2 + a_1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/ChuSchreierMultisetRecurrenceQTwo.result

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    Year recorded in the source bibliography: 2026.

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Discrete structuresQuestion resolvedFirst preprint 2026The L n-omino instance sequenceThe minimum size of a polyomino containing N translated L n-ominoes is the Nth term of S(n, 1, 2).Proved

The questionWilliams, Polyomino Density, arXiv:2608.29231v1, Section 6.3, p. 28:

Why it mattersThe source's Theorem 4.9 supplies a closed form for the instance minimum.

The ideaPut d=n-2 and Hd(K)=sum{0 <= i < K} floor(i/d).

Our contributionThe minimum size of a polyomino containing N translated L n-ominoes is the Nth term of S(n, 1, 2).

Verification details
Release source recordProved in Lean

Open problem

The L n-omino instance sequence

D. M. Condon, E. B. Dugan, L. M. Goldman and E. R. Williams, Polyomino Density, arXiv:2608.29231v1, Section 6.3, p. 28:

We believe S(5, 1, 2) is the same as the instance sequence for [L pentomino], and we suspect that the instance sequence for the L n-omino is S(n, 1, 2) in general.

The bracketed label represents the inline L pentomino diagram. For every integer n >= 3, the L n-omino has n-1 left-aligned bottom-row cells and one top-row cell. Instances are translated copies, as in Section 2.1. For every N >= 1, its instance minimum is the least cardinality of a nonempty edge-connected finite board containing at least N instances. The sequence S(n,1,2) starts at n, then increases by one at position k if k appears before position k, and by two otherwise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Polyomino/LPolyominoSkippedSequence.result

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Discrete structuresConjecture provedDement's odd-index square conjecture for OEIS A113249Every odd-index term in Dement's fourth-order integer family is a square.Proved

The questionConjecture: a(m, 2n+1) is a perfect square for all m, n.

The ideaThe public result applies IsSquare.sq to the resulting square expression.

Our contributionEvery odd-index term in Dement's fourth-order integer family is a square.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dement's odd-index square conjecture for OEIS A113249

The Name field of OEIS A113249 defines the integer family by $a_0=-1$, $a_1=4$, $a_2=-13+6(m-1)+3(m-1)^2$, $a_3=(m^2-8)^2$, and $a_{n+4}=m^4a_n+4m^2a_{n+1}-4a_{n+3}$. Its Comments field states:

Conjecture: a(m, 2*n+1) is a perfect square for all m, n.

The parameter and index range over natural numbers including zero, and the values are integers. Subtraction in $m-1$ is integer subtraction. Preregistration #19316 fixes these conventions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Four stated initial values | a at indices 0, 1, 2, 3 | Literal integer expressions, including all casts of the natural parameter. | | Fourth-order recurrence | a at index n+4 | The source recurrence with all indices shifted by four. | | All family parameters and odd indices | claim | Both quantifiers range over the naturals, including zero. | | Perfect square | Mathlib IsSquare over the integers | Existence of an integer whose product with itself is the term. | | Entire conjecture | result | No additional hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/OEISA113249OddIndexSquares.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Weak-four metric dimension of complete graph direct squaresExact weak-four metric dimension of complete graph direct squaresProved

The questionYero, The weak k-metric dimension of the direct product of complete graphs, arXiv:2605.22307v1, Section 3, paragraph S3.p17.1:

Why it mattersThe source gives exact formulas for the other weak parameters and leaves the weak-four case open.

The ideaView the landmarks as edges of a bipartite incidence graph on n row labels and n column labels.

Our contributionExact weak-four metric dimension of complete graph direct squares

Verification details
Release source recordProved in Lean

Open problem

Weak-four metric dimension of complete graph direct squares

Mohammad Farhan, Dorota Kuziak and Ismael G. Yero, The weak k-metric dimension of the direct product of complete graphs, arXiv:2605.22307v1, Section 3, paragraph S3.p17.1:

The result above clearly leads to a question concerning the exact value of wdim4(Kn × Kn) when n ≥ 4.

Section 6, first bullet, asks for the remaining exact formula at k = 4 and whether the upper bound in Theorem 3.3 is the exact value. This is one question over all integer orders at least four.

Exact value and semantics

For every integer $n\ge4$,

$$ \operatorname{wdim}4(K_n\times K_n) =2n+\left\lceil\frac{2n}{11}\right\rceil+\mathbf 1{n=5}. $$

The vertex set is $[n]\times[n]$. Two vertices are adjacent exactly when both coordinates differ. This is the direct product in the source. For $n\ge3$ the actual shortest-walk distance is zero at equal vertices, one when both coordinates differ, and two otherwise. In the last case, a third row and a third column give a common neighbor. The graph is connected.

A finite landmark set $S$ is weak four-resolving if

$$ \sum_{w\in S}|d(x,w)-d(y,w)|\ge4 \quad\text{for every distinct }x,y. $$

The introductory source definition omits the word “distinct.” Proposition 2.2 explicitly restricts to distinct pairs and supplies the coordinate identities used by the paper. Self-pairs have separation zero and cannot be intended when the paper discusses positive k. The formal predicate retains this explicit distinct-pair intent.

The dimension is the least cardinality of such a set. The formal construction proves that its feasible family is nonempty for every order in the question; the universal lower bound and an equality-size set prove attainment. Natural number division represents the ceiling as $(2n+10)/11$.

At order nine the answer is twenty. The source upper bound is twenty-one there, so Theorem 3.3 is not always sharp. The answer covers every order; this strict inequality alone is not the whole result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/DirectSquareWeakFour/Exact.result

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Quantum informationQuestion resolvedFirst preprint 2026Fukai–Yoshida–Katsura dissipative free-fermion root expectationFor a finite claw-free, even-hole-free graph with a simplicial clique, real couplings, a positive dissipation rate and no common root of the two independence polynomials, every root of the dissipative polynomial lies in the open upper half-plane.Proved

The questionKatsura, Dissipative free fermions in disguise, arXiv:2603.22163v1, printed p. 4, after Eq. (17):

Why it mattersIt supplies the root sign used in the paper's dissipative-mode interpretation.

The ideaFor a finite induced vertex domain U and a clique K contained in U, independent configurations meet K in at most one vertex.

Our contributionFor a finite claw-free, even-hole-free graph with a simplicial clique, real couplings, a positive dissipation rate and no common root of the two independence polynomials, every root of the dissipative polynomial lies in the open upper half-plane.

Verification details
Release source recordProved in Lean

Open problem

Fukai–Yoshida–Katsura dissipative free-fermion root expectation

K. Fukai, H. Yoshida and H. Katsura, Dissipative free fermions in disguise, arXiv:2603.22163v1, printed p. 4, after Eq. (17):

We expect Im ũ_k > 0 to hold for general ECF graphs, as observed numerically in the boundary-driven Fendley model, so that ε̃_k ≡ 1/ũ_k satisfies Im ε̃_k < 0.

The roots ũ_k are those of the plus polynomial in Eq. (16):

$$ \widetilde P_G^+(u)=P_G(u^2)+i\gamma u P_{G\setminus K_s}(u^2). $$

Equation (7), printed p. 3, defines the actual independent-set sum:

$$ P_G(x)=\sum_{S\in\mathcal S_G}(-x)^{|S|}\prod_{j\in S} b_j^2. $$

The couplings are real, γ > 0 is the dissipation rate, and ECF means even-hole-free and claw-free. The simplicial clique K_s satisfies the source's closed-neighborhood condition, printed p. 3:

Equivalently, denoting the closed neighborhood of j by Γ[j] ≡ {j} ∪ {ℓ ∈ V(G) | A_jℓ = 1}, K_s is simplicial if and only if Γ[j] \ K_s is a clique for all j ∈ K_s.

The standing exclusion after Eq. (26), printed p. 7, is retained:

Equation (26) shows that N_k = 0 if P_G and P_{G\K_s} share the root u_k². Throughout this work, we exclude this nongeneric case [80].

The quantified statement ranges over every finite graph on Fin n, every real coupling vector, every simplicial clique, every positive γ, and every complex root. The shared-root exclusion ranges over all complex x. Neither nonzero individual couplings nor simple roots are assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/DissipativeFreeFermionRootLocation.result

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Number theoryQuestion resolvedSource 2025Closed Form for the Greedy Fibonacci-Average SequenceThe greedy sequence whose running averages are Fibonacci takes the conjectured closed form from the tenth term onwards.Proved

The questionFried, Journal of Integer Sequences 28 (2025), Article 25.4.3, Section 6, writes:

Why it mattersBoth of Barker's conjectures follow: the odd-index one is the second line read at n = 2k+1, and the order-eight recurrence holds because each branch, as a function of the half index, is a linear combination of F(j+c) and jF(j+c) and so satisfies the square of the Fibonacci recurrence.

The ideaWrite Xj = F(j+2) + 2jF(j+1), which is the even branch of the displayed formula, since 2jF(j+3) − (2j−1)F(j+2) = 2j(F(j+3) − F(j+2)) + F(j+2) = F(j+2) + 2jF(j+1).

Our contributionThe greedy sequence whose running averages are Fibonacci takes the conjectured closed form from the tenth term onwards.

Verification details
Release source recordProved in Lean

Open problem

Closed Form for the Greedy Fibonacci-Average Sequence

Fried, Journal of Integer Sequences 28 (2025), Article 25.4.3, Section 6, writes:

As mentioned earlier, we observed the statement of the previous theorem while working on a conjecture stated in A248982, which is defined to be the sequence of distinct least positive numbers such that the average of the first n terms is a Fibonacci number. Let (a_n)_{n≥1} be this sequence. Refining the conjecture stated in A248982 regarding a closed-form formula for (a_n), it seems that, for n ≥ 10, we have

a_n = nF(n/2 + 3) − (n − 1)F(n/2 + 2),  if n is even;
a_n = F((n+1)/2 + 2),                    otherwise.

A proof of this likely proceeds along similar lines as the previous theorem.

Here F(1) = F(2) = 1. The sequence entry A248982 carries two further standing conjectures, both recorded by Colin Barker on 2014-10-19 and both still labelled Conjecture: that a(n) = 2a(n−2) + a(n−4) − 2a(n−6) − a(n−8) for n > 17, and that a(2n+1) = F(n+3) for n > 4. An Empirical g.f. equivalent to the first is recorded alongside them.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/GreedyFibonacciAverageClosedForm.result

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    Year in the primary source attribution: 2025.

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Quantum informationQuestion resolvedFirst preprint 2025The binary kite construction works exactly when $n$ is not one modulo threeBinary kite forms are doubly perfect exactly off one residue modulo three.Proved

The questionGoedicke, Thirty-six officers, artisanally entangled (2025), arXiv:2504.15401v2, subsection “Construction of doubly perfect functions”, Remarks after Theorem 5 (artisanal.tex, line 2316):

Why it mattersTheir Theorem 2 turns doubly perfect functions into complex Hadamard matrices proportional to two-unitaries (statement at lines 505–526; proof at lines 2144–2202).

Our contributionBinary kite forms are doubly perfect exactly off one residue modulo three.

Verification details
Release source recordProved in Lean

Open problem

The binary kite construction works exactly when $n$ is not one modulo three

D. Gross and P. Goedicke, Thirty-six officers, artisanally entangled (2025), arXiv:2504.15401v2, subsection “Construction of doubly perfect functions”, Remarks after Theorem 5 (artisanal.tex, line 2316):

As a concrete example, computer experiments indicate that the matrices with ``kite-shaped support'', $A_{ij}=B_{ij}=1$ iff $i+j\leq n+1$, give a solution for all $n$ not congruent to $1$ modulo $3$.

The quotation is from the downloaded v2 TeX source artisanal.tex, MD5 29b677f0287fbbb335cd79eb4f5f33d2. Its surrounding paragraph (lines 2304–2316) takes $A,B$ to be nonsingular symmetric binary matrices with $AB-I$ also nonsingular, and uses $N=\operatorname{diag}(A,B)$ as the coefficient matrix of a quadratic exponential. See the literature note.

For $n\in\mathbb N$, let $K_n\in M_n(\mathbb F_2)$ have entry $1$ exactly when $i+j\le n+1$, with indices $1,\ldots,n$, and entry $0$ otherwise; at $n=0$ this is the empty matrix. Put $N_n=\operatorname{diag}(K_n,K_n)$ and $J_n=\begin{pmatrix}0&I_n\-I_n&0\end{pmatrix}$.

Issue #20969 fixes the following reading, including the converse to the source's observation: for every $n\ge1$,

  • $K_n$ is symmetric and has trivial right kernel;
  • $N_n$ has trivial right kernel;
  • $K_n^2-I_n$ has trivial right kernel if and only if $n\not\equiv1\pmod3$;
  • $N_n+J_n$ has trivial right kernel if and only if $n\not\equiv1\pmod3$.

This matrix statement is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/KiteShapedDoublyPerfectForm.result

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Discrete structuresQuestion resolvedHanna's mixed trigonometric and hyperbolic congruenceEvery positive-index factorial-normalized coefficient of Hanna's reciprocal is an integer congruent to three modulo ten.Proved

The questionHanna's OEIS A227051, revision 10, first FORMULA states:

Why it mattersThe result establishes the conjectured last digit for all positive indices of the source sequence.

The ideaStrong induction proves integrality and uniqueness of its inverse over any commutative ring.

Our contributionEvery positive-index factorial-normalized coefficient of Hanna's reciprocal is an integer congruent to three modulo ten.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's mixed trigonometric and hyperbolic congruence

Paul D. Hanna's OEIS A227051, revision 10, first FORMULA states:

a(n) == 3 (mod 10) for n>0 (conjecture).

Its NAME is “E.g.f.: 1/(cos(x) - sin(x)*sinh(x)), omitting the zero-valued coefficients of odd powers of x.” The source offset is 0,2. Preregistration: #21582.

Statement echo

Over $\mathbb Q[[x]]$, let $D(x)=\cos x-\sin x\sinh x$, $A(x)=D(x)^{-1}$ and $a_n=(2n)![x^{2n}]A(x)$. The five public definitions sinhSeries, denominator, A, a and claim retain these literal formal Taylor expressions. The factorial normalization uses the existing IntegralEGFComposition.eCoeff owner. The result states that for every natural $n\ge1$ there is an integer $z$ with $a_n=z$ in $\mathbb Q$ and $z\equiv3\pmod {10}$. The natural-number guard is exactly the source's positive-index condition; the existential integer makes integrality explicit. There is no finite cutoff or convergence hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaMixedTrigHyperbolicModTen.result

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Discrete structuresQuestion resolvedHanna's integral composition equation modulo tenThe rational exponential generating series of A277403 has integral coefficients divisible by ten above index two.Proved

The questionHanna's sole COMMENTS conjecture in OEIS A277403 states:

Why it mattersIt proves the entire published conjecture with the binding source interpretation.

The ideaRing-homomorphism transport (seqmap) and sequnique identify it with the integral solution's reduction.

Our contributionThe rational exponential generating series of A277403 has integral coefficients divisible by ten above index two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

h277403 div10

Paul D. Hanna's sole COMMENTS conjecture in OEIS A277403 states:

a(n) is divisible by 10 for n>2 (conjecture).

The NAME field states:

E.g.f. satisfies: A(x - Integral A(x) dx) = x + Integral A(x) dx.

The FORMULA field introduces the primitive:

Let G(x) = Integral A(x) dx, then e.g.f. A(x) also satisfies:

The source is revision #21, with offset one and example beginning $A(x)=x+2x^2/2!+10x^3/3!+90x^4/4!+\cdots$. There is no source DOI, arXiv identifier or page number. The preregistration is #21746.

Statement echo

All declarations below belong to D5.S1.Recurrence.Residue.HannaIntegralCompositionDivTen.

| Source clause or interpretation | Lean counterpart | Scope | | --- | --- | --- | | Rational formal EGF | IsSolution parameters A, G | Every pair of rational formal power series | | Example has zero constant term | First conjunct of IsSolution | $A(0)=0$ | | Zero-constant formal integral | Second and third conjuncts of IsSolution | $G(0)=0$ and $G'=A$ | | Displayed defining equation | Fourth conjunct of IsSolution | $A(x-G(x))=x+G(x)$ | | Exponential coefficients, offset one | First conclusion of claim | $n![x^n]A\in\mathbb Z$ for every $n\ge1$, concluded rather than assumed | | Sole COMMENTS conjecture | Second conclusion of claim, proved by result | $n![x^n]A\in10\mathbb Z$ for every $n>2$ |

The existing IntegralEGFComposition.eCoeff supplies the literal factorial-scaled coefficient expression. The existing QuarticEGFFixedPoint.encode constructs rational series from coefficient sequences. Neither definition is duplicated. No integrality or desired congruence is assumed in IsSolution.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaIntegralCompositionDivTen.result

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    The proved statement applies to the stated general family, not just a checked finite range.

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Number theoryQuestion resolvedHanna's prime-power correction formulaEvery bounded integral square-weight correction equals p minus one at p to any positive power.Proved

The questionHanna, OEIS A384819, approved revision 14, NAME:

Why it mattersIt resolves the source's prime-power assertion rather than extending a finite table.

The ideaThis is the necessary integrality direction in the reused integralinputledger, specialized by correctionghost.

Our contributionEvery bounded integral square-weight correction equals p minus one at p to any positive power.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's prime-power correction formula

Paul D. Hanna, OEIS A384819, approved revision 14, NAME:

Nonnegative numbers a(n) < n for n >= 1 such that exp( Sum_{n>=1} (n^2 - a(n))*x^n/n ) is a power series with integral coefficients.

The sole COMMENTS conjecture is:

Conjecture: a(p^n) = p - 1 when p is prime for n >= 1.

All series are formal over the rationals. The offset is 1,3; the unused value $a(0)$ is unconstrained. The conclusion quantifies over every integer sequence satisfying the defining condition, every natural prime $p$, and every natural exponent $r\ge1$.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Nonnegative numbers $a(n)<n$ for $n\ge1$ | IsCorrection: integer-valued sequence and both positive-index bounds | Literal | | Integral coefficients of the specified formal exponential | IsCorrection: literal PowerSeries.mk, rational division, PowerSeries.subst, PowerSeries.exp, and reused IntegralSeries | Literal; the exponent's constant coefficient is zero | | $p$ prime | claim: Nat.Prime p | Includes 2 | | $n\ge1$ | claim: positive natural exponent $r$ | Bound variable renamed | | $a(p^n)=p-1$ | result proves claim | Unbounded in both parameters | | Every solution; natural index zero | Universal sequence binder and an unused $a(0)$ | No choice of a canonical solution and no added restriction |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/HannaPrimePowerCorrection.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A274479 coefficient congruenceThe normalized generating equation has an integral solution, and every positive coefficient of every normalized solution is one modulo three.Proved

The questionHanna, OEIS A274479, Name and Comments (unpaginated), as quoted in the source note.

Why it mattersThe published entry explicitly states this coefficient congruence as a conjecture.

The ideaThe integral construction first solves the normalized reversion equation.

Our contributionThe normalized generating equation has an integral solution, and every positive coefficient of every normalized solution is one modulo three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A274479 coefficient congruence

Paul D. Hanna, OEIS A274479, Name and Comments (unpaginated), as quoted in the source note.

G.f. satisfies: A(x)^2 = A( x^2/(1 - 2*x - 4*x^2) ).

a(n) = 1 (mod 3) for n>=1 (conjecture).

For $A(x)=x+\sum_{n\ge2}a(n)x^n$, the equation is $A(x)^2=A!\left(\frac{x^2}{1-2x-4x^2}\right)$. The resolution is Proved: a normalized rational power-series solution exists, and every normalized solution has integral coefficients $a(n)\equiv1\pmod3$ for all $n\ge1$. Preregistration: #18850. The other Hanna problem is separately preregistered and has its own resolution claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The displayed generating equation | HannaGeometricSquareSubstitutionModThree.IsGF | Literal preregistered equation over rational power series, with the denominator inverse interpreted in that ring. | | The example begins $x+x^2+\cdots$ | Constant coefficient zero and coefficient one at $X$ in IsGF | The normalization is explicit. | | $a(n)$ for $n\ge1$ | coeff n A, with the hypothesis 1 ≤ n | Every positive natural index. | | Integral coefficients congruent to one modulo three | Integer witness equal to coeff n A and Int.ModEq 3 m 1 in claim | Coercion into the rationals is explicit. | | A generating function is defined by the equation | Existence conjunct in claim | Excludes a vacuous universal statement. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaGeometricSquareSubstitutionModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's bilateral theta congruence, OEIS A380060Hanna's A380060 equation has an integral solution with residue two exactly at positive squares.Proved

The questionHanna's OEIS A380060 defines the generating function by:

Why it mattersThe residue pattern identifies exactly which positive coefficients have 2-adic valuation one.

The ideaIterating from one therefore stabilizes every coefficient and constructs an integer fixed point.

Our contributionHanna's A380060 equation has an integral solution with residue two exactly at positive squares.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's bilateral theta congruence, OEIS A380060

Paul D. Hanna's OEIS A380060 defines the generating function by:

G.f. A(x) satisfies A(x) = Sum_{n=-oo..+oo} x^(n^2) * (A(x) - x^n)^n/(1 - x^n*A(x))^n.

The Comments state:

The term x^(n^2) * (A - x^n)^n/(1 - x^n*A)^n is invariant under sign reversal of n.

Conjecture: for n > 0, a(n) == 2 (mod 4) iff n is a square, and a(n) is divisible by 4 when n is nonsquare.

Formula (2) states:

A(x) = 1 + 2*Sum_{n>=1} x^(n^2) * (A(x) - x^n)^n/(1 - x^n*A(x))^n.

The source is OEIS A380060, with its citation in the Library note. Preregistration: #18872.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Bilateral generating equation | IsGF uses the source's formula (2) over rational power series | The zero term is one and opposite indices coincide; no separate Lean Laurent-series folding theorem is asserted. | | Infinite sum | Degree $m$ is read from indices $1\le n\le m$ | Each summand contains $X^{n^2}$, so omitted indices contribute zero. | | Coefficients $a(n)$ | coeff n A | Every rational solution of IsGF is quantified. | | Positive index | Natural $n$ with $0<n$ | Degree zero is excluded from the residue assertion. | | Integer coefficients | An integer $z$ with coeff n A = (z : ℚ) | Integrality is a conclusion. | | Residue two iff square | $z\equiv2\pmod4\iff\operatorname{IsSquare}(n)$ | Both directions are included. | | Divisible by four otherwise | $\neg\operatorname{IsSquare}(n)\Rightarrow4\mid z$ | Integer divisibility. | | A generating function exists | The existence conjunct of claim | Additional existence, rather than an extra hypothesis restricting the conjecture. |

The public definitions IsGF and claim are the literal conventions in #18872. The reused Agree and squareSupport definitions retain their original bodies. No source clause is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaBilateralThetaModFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's derivative-substitution congruences for A232694Both nominated congruences hold for every factorial-normalized coefficient of OEIS A232694.Proved

The questionThe source equation and its nominated congruences are:

Why it mattersThe preregistration 21485 supplies the source conventions and literature screen for A232694.

The ideaThe shared machinery constructs the unique integral coefficient recurrence, transports coefficient congruences from an integral approximate solution, and cancels integral unit powers.

Our contributionBoth nominated congruences hold for every factorial-normalized coefficient of OEIS A232694.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's derivative-substitution congruences for A232694

Paul D. Hanna's OEIS A232694 COMMENTS conjectures concern integer exponential coefficients $a_n=n![x^n]A$ of rational formal power series with $A(0)=1$. The source equation and its nominated congruences are:

%N A232694 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^4) with A(0)=1. %C A232694 CONJECTURES. %C A232694 a(n) == 0 (mod 3) for n>=3. %C A232694 a(n) == 1 (mod 8) for n>=0.

| Record | Formal equation | Nominated congruences | Settling declaration | | --- | --- | --- | --- | | A232694 | $A'(x)=A(x(A'(x))^4)$ | $3\mid a_n$ for $n\ge3$; $8\mid a_n-1$ for every $n\ge0$ | HannaA232694Congruences.result |

The settling declaration belongs to D5.S1.Recurrence.Residue. It universally quantifies over rational formal solutions and concludes all-index integrality as well as the two displayed congruences. The source note records the OEIS record and its literal source text.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A'(x)=A(x(A'(x))^4)$ and $A(0)=1$ | IsSolution in HannaA232694Congruences | The formal derivative, substitution and constant coefficient are literal. | | $a_n=n![x^n]A$ is integral | The existential a : ℕ → ℤ with eCoeff A n = (a n : ℚ) | Integrality is a conclusion at every index. | | $3\mid a_n$ for $n\ge3$ | The first divisibility conjunct in claim | The threshold and modulus are literal. | | $8\mid a_n-1$ for all $n\ge0$ | The second divisibility conjunct in claim | The all-index residue is literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA232694Congruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's derivative-substitution congruences for A232695Both nominated congruences hold for every factorial-normalized coefficient of OEIS A232695.Proved

The questionThe source equation and its nominated congruences are:

Why it mattersThe preregistration 21486 supplies the source conventions and literature screen for A232695.

The ideaThe shared machinery constructs the unique integral coefficient recurrence, transports coefficient congruences from an integral approximate solution, and cancels integral unit powers.

Our contributionBoth nominated congruences hold for every factorial-normalized coefficient of OEIS A232695.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's derivative-substitution congruences for A232695

Paul D. Hanna's OEIS A232695 COMMENTS conjectures concern integer exponential coefficients $a_n=n![x^n]A$ of rational formal power series with $A(0)=1$. The source equation and its nominated congruences are:

%N A232695 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^5) with A(0)=1. %C A232695 CONJECTURES. %C A232695 a(n) == 1 (mod 5) for n>=0. %C A232695 a(n) == 0 (mod 2) for n>=4.

| Record | Formal equation | Nominated congruences | Settling declaration | | --- | --- | --- | --- | | A232695 | $A'(x)=A(x(A'(x))^5)$ | $5\mid a_n-1$ for every $n\ge0$; $2\mid a_n$ for $n\ge4$ | HannaA232695Congruences.result |

The settling declaration belongs to D5.S1.Recurrence.Residue. It universally quantifies over rational formal solutions and concludes all-index integrality as well as the two displayed congruences. The source note records the OEIS record and its literal source text.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A'(x)=A(x(A'(x))^5)$ and $A(0)=1$ | IsSolution in HannaA232695Congruences | The formal derivative, substitution and constant coefficient are literal. | | $a_n=n![x^n]A$ is integral | The existential a : ℕ → ℤ with eCoeff A n = (a n : ℚ) | Integrality is a conclusion at every index. | | $5\mid a_n-1$ for all $n\ge0$ | The first divisibility conjunct in claim | The all-index residue is literal. | | $2\mid a_n$ for $n\ge4$ | The second divisibility conjunct in claim | The threshold and modulus are literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA232695Congruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's derivative-substitution congruences for A232696Both nominated congruences hold for every factorial-normalized coefficient of OEIS A232696.Proved

The questionThe source equation and its nominated congruences are:

Why it mattersThe preregistration 21487 supplies the source conventions and literature screen for A232696.

The ideaThe shared machinery constructs the unique integral coefficient recurrence, transports coefficient congruences from an integral approximate solution, and cancels integral unit powers.

Our contributionBoth nominated congruences hold for every factorial-normalized coefficient of OEIS A232696.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's derivative-substitution congruences for A232696

Paul D. Hanna's OEIS A232696 COMMENTS conjectures concern integer exponential coefficients $a_n=n![x^n]A$ of rational formal power series with $A(0)=1$. The source equation and its nominated congruences are:

%N A232696 E.g.f. A(x) satisfies: A'(x) = A(x/A'(x)) with A(0)=1. %C A232696 CONJECTURES. %C A232696 a(n) == 0 (mod 2) for n>=4. %C A232696 a(n) == 0 (mod 2^2) for n>=10.

| Record | Formal equation | Nominated congruences | Settling declaration | | --- | --- | --- | --- | | A232696 | $A'(x)=A(x/A'(x))$ | $2\mid a_n$ for $n\ge4$; $4\mid a_n$ for $n\ge10$ | HannaA232696Congruences.result |

The settling declaration belongs to D5.S1.Recurrence.Residue. It universally quantifies over rational formal solutions and concludes all-index integrality as well as the two displayed congruences. The source note records the OEIS record and its literal source text.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A'(x)=A(x/A'(x))$ and $A(0)=1$ | IsSolution in HannaA232696Congruences | The formal derivative, substitution and constant coefficient are literal. | | $a_n=n![x^n]A$ is integral | The existential a : ℕ → ℤ with eCoeff A n = (a n : ℚ) | Integrality is a conclusion at every index. | | $2\mid a_n$ for $n\ge4$ | The first divisibility conjunct in claim | The threshold and modulus are literal. | | $4\mid a_n$ for $n\ge10$ | The second divisibility conjunct in claim | The threshold and modulus are literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA232696Congruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedSource 2026The all-dimensional Hypercube InequalityThe determinant-defined Hypercube Inequality holds in every positive dimension.Proved

The questionLet d>=1, let xv>=0 for every vertex v of {0,1}^d, and let Ad be the integer matrix whose column at v is (v,1).

Why it mattersAverkov, von Dichter and Soprunov state this as the Hypercube Inequality in Section 4.4, Equation (16), after reducing a proposed geometric log-submodularity inequality to it.

The ideaIt contains only the two source definitions and the theorem result.

Our contributionThe determinant-defined Hypercube Inequality holds in every positive dimension.

Verification details
Release source recordProved in Lean

arXiv

The all-dimensional Hypercube Inequality

Let d>=1, let x_v>=0 for every vertex v of {0,1}^d, and let A_d be the integer matrix whose column at v is (v,1). For each unordered (d+1)-vertex subset S, let A_S be the square submatrix of those actual columns. Prove

(sum_{|S|=d+1} |det A_S| * product_{v in S} x_v)
  * (sum_v x_v)^(d-1)
<= product_{i in Fin d}
     (sum_{v_i=0} x_v) * (sum_{v_i=1} x_v).

Each subset occurs once and there is no factorial factor. The formal definitions are augmentedCube and hypercubeWeight; result proves the displayed assertion for every positive natural dimension, including zero weights, zero facets and singular determinant data.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/FiniteGeometry/HypercubeInequality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Ishida's cyclic stable-marriage arc-lemma upper boundFor n >= 1, every complete matching in the cyclic Hollow-Shell profile has at most floor((n - 1)^2 / 4) blocking pairs.Proved

The questionYoshiteru Ishida, A Quadratic Lower Bound for the Shield Number of the Stable Marriage Problem: Rearrangement, Extremal Construction, and Biclique Realizability, arXiv:2609.17418v1, Section 7.1, p. 8:

Why it mattersTheorem 5.3 of the source supplies a matching attaining the target.

The ideaPut U = {i : i≤μ(i)} and W = {i : μ(i)<i}, with cardinalities k and q.

Our contributionFor n >= 1, every complete matching in the cyclic Hollow-Shell profile has at most floor((n - 1)^2 / 4) blocking pairs.

Verification details
Release source recordProved in Lean

Open problem

Ishida's cyclic stable-marriage arc-lemma upper bound

Yoshiteru Ishida, A Quadratic Lower Bound for the Shield Number of the Stable Marriage Problem: Rearrangement, Extremal Construction, and Biclique Realizability, arXiv:2609.17418v1, Section 7.1, p. 8:

This statement does not assert that µ∗ is maximum-blocking; that is exactly the open arc-lemma upper-bound problem.

Conclusion, p. 10:

It does not claim the exact Shield-Core equality for general n, a universal Hall-feasible biclique theorem, or the general upper bound β(C_n) ≤ ⌊(n − 1)²/4⌋.

The cyclic profile (1) on labels 0 through n−1 is r_M(g,h) = (h−g) mod n + 1 and r_W(h,g) = n − (h−g) mod n. Smaller ranks are preferred. A complete matching is a bijection μ from men to women; (g,h) blocks it when both r_M(g,h) < r_M(g,μ(g)) and r_W(h,g) < r_W(h,μ⁻¹(h)). Write B(μ) for the number of such ordered pairs and β(C_n) for its maximum over matchings.

The quantified target is: for every natural n≥1 and every permutation μ of Fin n, B(μ) ≤ (n−1)²/4, where the division is natural division, hence the floor.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/CyclicStableMarriageBlockingBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedKurkov's weighted-ballot formula for OEIS A107667Kurkov's weighted-ballot formula for the A107667 triangle.Proved

The questionMikhail Kurkov's FORMULA conjecture in Paul D.

Why it mattersThe recurrence has a parameter-sensitive first-return convolution.

The ideaNo separate path-set bijection is formalized.

Our contributionKurkov's weighted-ballot formula for the A107667 triangle.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's weighted-ballot formula for OEIS A107667

Mikhail Kurkov's FORMULA conjecture in Paul D. Hanna's OEIS A107667, approved revision 25 asserts $T(n+m,m)=(n+m+1)P(n+1,n+1,m)$ for all $n,m\in\mathbb N$. The preregistration is #21639. The source is OEIS-only, with no claim-source DOI or arXiv identifier identified.

The source NAME defines a lower triangular matrix by $T=D+\operatorname{SHIFT_LEFT}(T^2)$ with diagonal $1,2,3,\ldots$. Thus $T(a,b)=0$ for $a<b$, $T(a,a)=a+1$, and $T(a,b)=\sum_{j=b+1}^{a}T(a,j)T(j,b+1)$ for $b<a$. Kurkov's three-argument table satisfies $P(r,1,m)=1$ for $r\ge1$, $P(r,k,m)=P(r-1,k,m)+(r+m)P(r,k-1,m)$ for $1<k<r$, and $P(r,r,m)=(r+m)P(r,r-1,m)$ for $r\ge2$. Values outside $r\ge k\ge1$ are unrestricted. This table is distinct from the source's earlier two-argument conjugating matrix also called $P$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Lower triangular $T=D+\operatorname{SHIFT_LEFT}(T^2)$, diagonal $1,2,3,\ldots$ | IsSourceTriangle | All three literal recurrence clauses, with zero-based indices and a finite interval sum. | | Base, interior and diagonal rules for three-argument $P$ | IsPTable | Base includes $P(1,1,m)=1$; diagonal starts at row $2$. | | The two-parameter identity | claim, result | Every natural-valued pair of tables satisfying those rules, every $n,m\ge0$, and the full displayed equality. |

The additional table binders and recurrence hypotheses express the source definitions; no path formula or first-return identity is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KurkovA107667WeightedBallot.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2025Finite generation of one-parameter polynomial Vassiliev algebrasEvery family of rational univariate polynomials has a finite subfamily generating the same algebra.Proved

The questionSleptsov, Algebraic structures of Vassiliev invariants for knot families, arXiv:2508.02385v2, Nuclear Physics B 1022 (2026), 117268, conjecture in the abstract (main.tex, line 165):

Why it mattersThe source's primary invariants generate the descended algebra: every restricted invariant is a polynomial in them.

The ideaNatural degree zero includes the zero polynomial.

Our contributionEvery family of rational univariate polynomials has a finite subfamily generating the same algebra.

Verification details
Release source recordProved in Lean

Open problem

Finite generation of one-parameter polynomial Vassiliev algebras

E. Lanina and A. Sleptsov, Algebraic structures of Vassiliev invariants for knot families, arXiv:2508.02385v2, Nuclear Physics B 1022 (2026), 117268, conjecture in the abstract (main.tex, line 165):

We conjecture that such 1-parametric algebra of Vassiliev invariants is always finitely generated, while in the case of more parameters, we provide example of the knot family with infinite number of generators.

Section 4.1, “Generic one-parametric family” (main.tex, line 1166), asks:

Can this force such a 1-parametric algebra of Vassiliev invariants to be infinitely generated? Examples and reasonings provide the intuition that a 1-parametric algebra of polynomial Vassiliev invariants always has only a finite number of generators. Still, we currently lack strict mathematical proof.

The source restricts attention to rational-valued invariants (main.tex, line 350), defines primary invariants as generators of the algebra restricted to a family (line 356), and represents the one-parameter restrictions by polynomials with rational coefficients (lines 1114–1117). These passages are recorded in the literature note.

Issue #22008 fixes the algebraic reading: for every index type $\iota:\mathrm{Type}$ and every family $p:\iota\to\mathbb Q[X]$, there is a finite set $s$ of indices such that

$\mathbb Q[p_i\mid i\in s]=\mathbb Q[p_i\mid i:\iota]$.

Here $\mathbb Q[S]$ denotes the smallest rational subalgebra containing $S$. Thus the generators can be chosen from the original family of restricted invariants. This is the statement of claim, proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Topology/KnotFamilyVassilievFiniteGeneration.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The largest Bloch weight in the HKNN stateThe block Bloch state has strictly largest HKNN weightProved

The questionZimeng Li and Ning Wu, Exact momentum-space analysis of small spin-1/2 J1-J2 rings, arXiv:2604.23149v1, conjecture after Eq. (47), p. 12:

Why it mattersThe conjecture identifies the configuration with all down spins consecutive as the largest component of the HKNN singlet in the momentum basis.

The ideaCrossing pairings correspond to bijections from down sites to up sites.

Our contributionThe block Bloch state has strictly largest HKNN weight

Verification details
Release source recordProved in Lean

Open problem

The largest Bloch weight in the HKNN state

Zimeng Li and Ning Wu, Exact momentum-space analysis of small spin-1/2 J1-J2 rings, arXiv:2604.23149v1, conjecture after Eq. (47), p. 12:

Thus, we conjecture that this property holds for arbitrary even number N, i.e., the Bloch state |ξ1,1,...,1(−π)⟩ (there are N/2 − 1 1’s) should have the largest weight in |ψHKNN(N)⟩.

Eq. (6), p. 4, defines the unnormalized HKNN vector by summing the products of singlets over every pair partition, with each pair's smaller endpoint first and no permutation sign. The singlet is [i,j] = |↑⟩i|↓⟩j − |↓⟩i|↑⟩j. The normalized Bloch states are the nonzero translation sums of computational-basis configurations, divided by their Hilbert norms; Eq. (7) gives the N=6 examples.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/HKNNBlockBlochLargestWeight.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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Discrete structuresRecurrence provedMathar and Kimberling's self-convolution recurrenceThe central self-convolution of the four-diamondized Pascal row satisfies the six-term recurrence in OEIS A027182.Proved

The questionThe proposition asserts this recurrence; it does not assert a separate formal D-finiteness predicate.

Why it mattersNo escape witness or independent supporting module is claimed.

The ideaFour instances of Nat.succmulcentralBinomsucc give hrecurrence:

Our contributionThe central self-convolution of the four-diamondized Pascal row satisfies the six-term recurrence in OEIS A027182.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar and Kimberling's self-convolution recurrence

OEIS A027182 revision 8 states Mathar's FORMULA conjecture for the self-convolution of the rows of Kimberling's A027170 revision 32. The verbatim source is in the source note, and the quantified statement is preregistered in #23023.

For $n,k\in\mathbb N$, define $t(n,k)=\binom{n+2}{k+1}+\binom{n+1}{k}+\binom{n+1}{k+1}+\binom nk$, $T(n,k)=t(n,k)-t(0,0)+1$, with natural subtraction, and $g_n(X)=\sum_{k=0}^n T(n,k)X^k\in\mathbb N[X]$. Put $a_n=[X^n]g_n(X)^2$. The settling theorem proves, for every $n\ge6$,

$$ 2(n+2)a_n+(-17n-23)a_{n-1}+(47n+36)a_{n-2} +2(-22n-17)a_{n-3}+4(-n+20)a_{n-4} +8(2n-9)a_{n-5}+24(12n-47)=0. $$

The coefficient arithmetic is in $\mathbb Z$ and sequence indices use natural subtraction. The proposition asserts this recurrence; it does not assert a separate formal D-finiteness predicate.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Four binomial terms defining $t$ in A027170 | t | Same terms, order and natural carrier. | | $T(n,k)=t(n,k)-t(0,0)+1$ | T | Literal natural subtraction, with the source row domain $0\le k\le n$. | | Row generating polynomial over $k=0,\ldots,n$ | g | Literal finite sum, including both endpoints. | | Central coefficient of the squared row polynomial | a | Coefficient $n$ of $g_n^2$. | | Six sequence terms and affine forcing term in the FORMULA field | claim, result | Every coefficient and lag is preserved, over all natural $n\ge6$. | | Published offset $1$ | Total a with $a_0=1$ | The extension is explicit; no extra hypothesis restricts the published range. | | No additional source clause | Integer casts and total definitions | Coefficient arithmetic is integer arithmetic; totality outside the source row domain does not weaken the quantified claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MatharA027182SelfConvolution.result

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  • Contribution

    The central self-convolution of the four-diamondized Pascal row satisfies the six-term recurrence in OEIS A027182.

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Discrete structuresRecurrence provedMathar's OEIS A174193 order-four recurrenceMathar's order-four recurrence holds for the specified algebraic power series.Proved

Why it mattersThe source still labels the recurrence a conjecture.

The ideaTwo explicit polynomial certificates cancel the square-root terms.

Our contributionMathar's order-four recurrence holds for the specified algebraic power series.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's OEIS A174193 order-four recurrence

R. J. Mathar's unnumbered FORMULA entry for OEIS A174193 gives the algebraic series [ \frac{1-8x+13x^2+24x^3-48x^4-(1-6x+x^2+34x^3-26x^4-4x^5)\sqrt{1-4x}}{2x^2(1-x)(1-4x)^2} ] and conjectures the displayed order-four recurrence for its coefficients. The formal statement uses rational power series, a branch $s$ with constant coefficient one and $s^2=1-4X$, and the cleared equation $DA=P-Bs$. It asserts existence of such a source series and the recurrence for every such series and every natural $n\ge4$.

Statement echo

The Lean claim contains the existence conjunct, the universal source-series predicate, the guard $4\le n$, natural lagged indices $n-1$ through $n-4$, and the five rational polynomial coefficient terms exactly. IsSourceSeries contains $\exists s$, $s(0)=1$, $s^2=p$, and $DA=P-Bs$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/MatharA174193OrderFourRecurrence.result

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    Mathar's order-four recurrence holds for the specified algebraic power series.

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Number theoryQuestion resolvedSource 2026Uniform prime-square approximation for the full local momentThe exact infinite local moment has a uniform prime-square approximation for every sufficiently large real moment parameter.Proved

The questionThe source is MathOverflow question 516091, revision 1, by unkown, GUID 5D544532-802E-47EE-9240-92FCF941D961.

Why it mattersThe retained prime-square term can grow on the moving window.

The ideaThus the real tsum has its intended series value and the logarithm is taken on a positive argument.

Our contributionThe exact infinite local moment has a uniform prime-square approximation for every sufficiently large real moment parameter.

Verification details
Release source recordProved in Lean

Open problem

Uniform prime-square approximation for the full local moment

The source is MathOverflow question 516091, revision 1, by unkown, GUID 5D544532-802E-47EE-9240-92FCF941D961. The attributed source note gives the original definitions, question and bounded literature comparison. Issue #18820 specifies the quantified target.

All local Lean names below belong to D5.S3.Analytic.SeriesInequalities.PrimeSquareMomentApproximation.

| Source clause | Lean counterpart | Domain and endpoints | | --- | --- | --- | | $e_p(0)=1$ | factor p 0 | Index zero is present | | Rational $e_p(a)$ for integer $a\ge1$ | Nonzero branch of factor | Every positive natural exponent | | Real $s>0$ | Real.rpow in localMoment | The final real threshold is chosen with $s\ge1$ | | Full infinite sum over $a\ge0$ | ∑' a : ℕ, (factor p a)^s / p^a | No truncation or finite proxy | | $c=\log(7/6)/2$, $P=e^{cs}$ | c, Real.exp (c*s) | $P^2=\exp(2cs)$ | | Sufficiently small fixed $\delta>0$ | 0 < δ, δ < 1 - c3/c | c3 = log(5/4)/3; positive nonempty interval | | Every fixed $B>2$ | ∀ B : ℝ, 2 < B → | Threshold may depend on $δ,B,ε$ | | Uniform $o(1)$ as real $s\to\infty$ | ∀ ε>0, ∃ S : ℝ, ∀ s≥S, ∀ p, ... | One threshold precedes both $s$ and $p$ | | Prime $p\equiv1\pmod4$ | p : ℕ, p.Prime, p % 4 = 1 | No change of prime or congruence domain | | $P^{1-δ}<p\le BP$ | exp ((1-δ)*c*s) < p, p ≤ B*exp(c*s) | Strict lower, closed upper | | Logarithmic error $<ε$ | originalClaim, result : originalClaim | Absolute difference of the two original logarithms |

For positive $P$, its real power $P^{1-δ}$ equals the lower exponential cutoff by the positive-base real-power definition. The two expressions for $P^2$ also agree by exponential multiplication. No integer restriction on $s$ is introduced.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/SeriesInequalities/PrimeSquareMomentApproximation.result

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  • Historical context

    Year recorded in the source bibliography: 2026.

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Quantum informationExact optimum establishedFirst preprint 2025Optimal unambiguous discrimination without repetition for equal-overlap statesIndependent local measurements attain the global unambiguous optimum for sequences of distinct equal-overlap states.Proved

The questionTheir conjecture appears in the TeX source revised02.tex, lines 970–984, subsection “Conjecture for multipartite systems”.

Why it mattersThe 2026 paper studies whether nonlocality without entanglement depends on the discrimination measure: local operations and classical communication (LOCC) can attain the global optimum for unambiguous discrimination even when they are suboptimal for minimum-error discrimination.

The ideaIts hypothesis includes positivity of every effect, completeness and zero probability of every erroneous output.

Our contributionIndependent local measurements attain the global unambiguous optimum for sequences of distinct equal-overlap states.

Verification details
Release source recordProved in Lean

Open problem

Optimal unambiguous discrimination without repetition for equal-overlap states

S. Murshid, T. Gupta, V. Russo and S. Bandyopadhyay, Quantum nonlocality without entanglement and state discrimination measures, arXiv:2506.20560v3; Quantum 10, 2174 (2026), consider the uniform ensemble $\mathcal E_{N,k}$ of ordered products of $k$ distinct members of $N$ normalized states $\psi_i\in\mathbb C^N$ with $\langle\psi_i,\psi_j\rangle=s\in(0,1)$ for $i\ne j$.

Their conjecture appears in the TeX source revised_02.tex, lines 970–984, subsection “Conjecture for multipartite systems”. The quotations retain the source macro \E, meaning $\mathcal E$, and its equation reference; the displays are written inline.

For the family of ensembles of multipartite pure product states $\E_{N,k}$ given by Equation \eqref{eq:gensemble} the following holds for all $N \geq 3$ and $2 \leq k \leq N$

$p_{{L}}^{me}\left(\E{N,k}\right) <p_{{G}}^{me}\left(\E{N,k}\right),$

$p_{{L}}^{ud}\left(\E{N,k}\right) =p_{{G}}^{ud}\left(\E{N,k}\right).$

Moreover, the optimal unambiguous discrimination probability of $\E_{N,k}$ is given by

$p_{{G}}^{ud}\left(\E{N,k}\right) = \begin{cases} (1-s)^k, & \text{ when } k < N \ (1-s)^{N-1}(1+\left(N-1\right)s), & \text{ when } k=N. \end{cases}$

T. Gupta, S. Murshid and S. Bandyopadhyay, Unambiguous discrimination of sequences of quantum states, arXiv:2402.06365v2; Phys. Rev. A 109, 052222 (2024), pose the earlier no-repetition question in GMBjournal-revision-02.tex, lines 883–903:

In our scenario, as well as in more general ones (which we could not solve), repetitions of states are allowed in a sequence. The third problem considers the situation where it is not. Restrict $k<N$ and by $S'_{N,k}$ denote the set of sequences of length $k$, where no element is repeated.

Now assume that the inner products of the states $\left|\psi_{i}\right\rangle$ are equal and positive, say, $s>0$. Then $S_{N,k}^{\prime}\subset S_{N,k}$, where $S_{N,k}$ is the set of sequences considered in this paper. Under these restricted conditions, numerical experiments ($N=3$, $k=3$) still suggest that the optimal probability for distinguishing between the elements of $S_{N,k}^{\prime}$ unambiguously once again obeys $(1-s)^{k}$. However, for arbitrary values of both $N$ and $k$, whether one could benefit from collective measurements in this scenario is an interesting problem to consider in the future.

The source definitions and locators are also recorded in the 2026 literature note and the 2024 literature note.

Issue #21233 fixes the settlement as the unambiguous clauses of the conjecture, together with the 2024 question for $2\le k<N$ under these hypotheses. These are proved. The minimum-error clause of the conjecture is not settled by this result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/PureState/NoRepetitionSequenceUnambiguousDiscrimination.result

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  • Contribution

    Independent local measurements attain the global unambiguous optimum for sequences of distinct equal-overlap states.

    Read source ↗
  • Historical context

    First preprint year: 2025.

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Analysis & dynamicsQuestion resolvedFirst preprint 2026The Rule 84 center column modulo threeThe canonical polynomial lift of Rule 84 from a single seed has center column 1 at time zero and the repeating block (1, 2, 2) at every positive time modulo three.Proved

The questionTigran Nersissian, Diagonal Bases and Diagonal Periods of Elementary Cellular Automata, arXiv:2609.25078v1, Remark 6, page 12, states:

Why it mattersNonzero center values give the unit condition needed by the source's all-window diagonal-basis universality criterion.

The ideaThe half-plane x < 0 remains zero by induction, since (0+0+00)(1+c) = 0.

Our contributionThe canonical polynomial lift of Rule 84 from a single seed has center column 1 at time zero and the repeating block (1, 2, 2) at every positive time modulo three.

Verification details
Release source recordProved in Lean

Open problem

The Rule 84 center column modulo three

Tigran Nersissian, Diagonal Bases and Diagonal Periods of Elementary Cellular Automata, arXiv:2609.25078v1, Remark 6, page 12, states:

(The outstanding Rule 84 case) The polynomial of Rule 84 is (a + b + ab)(1 + c). Modulo three, its center column begins 1, 1, 2, 2, 1, 2, 2, … and follows the repeating block (1, 2, 2) after the first entry for the tested times 0 ≤ t < 2048. All center values are also units modulo 9 and 27 on that range, as Corollary 2 predicts. An all-time proof of the observed pattern is not supplied here. If Rule 84 is universal modulo three, it is universal at every power of three; three could not be the sole exceptional modulus.

Section 19, page 42, states:

The classification reduces the remaining single-seed universality question to the eight rules in E at odd primes. The observed Rule 84 pattern modulo three requires an invariant or an all-time recurrence proof. A finite nonzero prefix is insufficient.

This dossier anchors exactly that all-time pattern. Definition 1, page 4, fixes the canonical positive-coefficient polynomial lift and the synchronous integer-lattice orbit. For the modulus 3 and the single seed at zero, let A 0 x = if x = 0 then 1 else 0 and A (t+1) x = (a+b+a*b)*(1+c), where a = A t (x-1), b = A t x and c = A t (x+1) in ZMod 3. The statement is A 0 0 = 1 ∧ ∀ s : ℕ, A (3*s+1) 0 = 1 ∧ A (3*s+2) 0 = 2 ∧ A (3*s+3) 0 = 2. It includes every natural time, with no time cutoff or extra hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/Rule84ModThreeCenterColumn.result

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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresQuestion resolvedFirst preprint 2026Nikandish Problem 4.2The exact clique number for all nonzero binary subspaces in every positive dimension.Proved

The questionSection 4 of the cited source asks: "Determine the exact value of omega(On) for general n." For every integer n >= 1, the vertices are all nonzero linear subspaces of F2^n.

Why it mattersThe target concerns the whole subspace lattice, including degenerate subspaces with nonzero intersections, and cannot be replaced by a graph on lines or nondegenerate subspaces.

The idea(orthogonalityGraph n).cliqueNum = max n (nonzeroSubspaceCount (n/2) + n%2).

Our contributionThe exact clique number for all nonzero binary subspaces in every positive dimension.

Verification details
Release source recordProved in Lean

Open problem

Nikandish Problem 4.2

Section 4 of the cited source asks: "Determine the exact value of omega(O_n*) for general n." For every integer n >= 1, the vertices are all nonzero linear subspaces of F_2^n. Distinct vertices are adjacent when every vector in one is orthogonal to every vector in the other under the standard symmetric, nonalternating dot form.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Orthogonality/NikandishClique.result

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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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Discrete structuresQuestion resolvedBala's cubic recurrence and its growing three-adic congruenceBala's cubic recurrence satisfies the conjectured power-of-three congruence.Proved

The questionThe settled FORMULA line (%F, verbatim), inside Peter Bala's Nov 15 2022 (Start) ... (End) block:

Why it mattersThe conjecture gives an increasing power of 3 shared by consecutive terms of a cubic recurrence.

The ideaFor the step, a(n+1)+2=(a(n)-1)^2(a(n)+2), and the induction hypothesis implies 3 ∣ a(n)+2, hence 3 ∣ a(n)-1.

Our contributionBala's cubic recurrence satisfies the conjectured power-of-three congruence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's cubic recurrence and its growing three-adic congruence

OEIS A002000, NAME (%N, verbatim):

a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

The settled FORMULA line (%F, verbatim), inside Peter Bala's Nov 15 2022 (Start) ... (End) block:

Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.

AUTHOR (%A, verbatim):

N. J. A. Sloane

The offset is %O 0,1. Both n and r range over the natural numbers, including zero, and the only hypothesis is r <= n. The sequence has values in the integers and is defined by the NAME recurrence. Congruence means that 3^(n+r+2) divides the integer difference a(n+1)-a(n).

The exact Lean statement is:

theorem result (n r : ℕ) (hr : r ≤ n) :
    (3 : ℤ) ^ (n + r + 2) ∣ a (n + 1) - a n

Only the Conjecture line is settled, for every n and every r <= n. The block's Lucas representation, 3-adic limit, and product formula are not claimed. In particular, a is not defined through Lucas numbers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/BalaCubicThreeAdicCongruence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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Number theoryQuestion resolvedStern values invert the next Farey-tree numeratorStern's diatomic values invert the next Farey-tree numerator modulo its denominator.Proved

The questionStern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2n) = a(n), a(2n+1) = a(n) + a(n+1).

Why it mattersThe Stern scalar recurrence and the level-by-level mediant construction define their objects independently.

The ideaProve the row-coordinate invariant fareyRow m k = (stern k, stern (2^m+k)) for k <= 2^m by induction over all levels, separating copied even positions from newly inserted odd mediants.

Our contributionStern's diatomic values invert the next Farey-tree numerator modulo its denominator.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Stern values invert the next Farey-tree numerator

OEIS A002487, NAME (%N, verbatim):

Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

Gary W. Adamson's December 18, 2023 COMMENT (%C, verbatim):

It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1). For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7.

The A002487 AUTHOR line (%A, verbatim):

N. J. A. Sloane

OEIS A007305, NAME (%N, verbatim):

Numerators of Farey (or Stern-Brocot) tree fractions.

OEIS A007306, NAME (%N, verbatim):

Denominators of Farey tree fractions (i.e., the Stern-Brocot subtree in the range [0,1]).

The listing has 0/1 and 1/1 at indices 0 and 1. Each later level lists its new mediants in increasing order in [0,1]. For r = 2^m + j, 1 <= j <= 2^m, entry r is fareyRow (m+1) (2*(j-1)+1). Equivalently, for r >= 2, take m = Nat.log 2 (r-1) and k = r-1-2^m, then select row m+1, position 2*k+1. The first and second coordinates are fareyNum and fareyDen.

The literal Lean statement is:

theorem result (n : ℕ) (hn : 0 < n) :
    stern n * fareyNum (n + 1) % fareyDen (n + 1) = 1 % fareyDen (n + 1)

Only Adamson's sentence is settled, for positive natural indices. The Yurramendi frequency conjecture, Torres's conjectures and other Farey-tree properties are not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/PrimeForms/SternBrocot/AdamsonSternFareyInverse.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedWythoff column residue pairsThe first two Wythoff columns attain all residue pairs modulo every positive integer.Proved

The questionClark Kimberling's 4 June 2025 comment on OEIS A035513 states:

Why it mattersThis is a tier-one named external conjecture.

The ideaThe irrationality of nphi and the quadratic identity phi^2=phi+1 give floor(Aphi)=A+n-1 and floor(Aphi^2)=2A+n-1 for every positive row.

Our contributionThe first two Wythoff columns attain all residue pairs modulo every positive integer.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Wythoff column residue pairs

Clark Kimberling's 4 June 2025 comment on OEIS A035513 states:

Conjecture: If m >= 2, then {(T(n,1), T(n,2)) mod m} has cardinality m^2.

Here n>=1 ranges over positive row indices and phi=(1+sqrt(5))/2. The first two columns are T(n,1)=floor(floor(n*phi)*phi) and T(n,2)=floor(floor(n*phi)*phi^2). For every integer modulus m>=2, their residue image is the full product (Z/mZ)^2. Equivalently, for every a,b modulo m there is a positive row with those two residues. The formal statement also includes m=1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Wythoff/ColumnResiduePairs.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedLeast-two-element subset countThe least-two-element subset count agrees with OEIS A077866 after shifting the index by three.Proved

The questionKimberling's 2022 OEIS A077866 comment asks whether the number b(N) of subsets of {1,...,N} with more than one element, whose least two elements sum to the maximum, satisfies b(0)=b(1)=b(2)=0 and b(n+3)=A077866(n) for every n>=0.

Why it mattersThis is a tier-one named external conjecture.

The ideaFor a counted set, write its least two elements as 0<a<b.

Our contributionThe least-two-element subset count agrees with OEIS A077866 after shifting the index by three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Least-two-element subset count

Kimberling's 2022 OEIS A077866 comment asks whether the number b(N) of subsets of {1,...,N} with more than one element, whose least two elements sum to the maximum, satisfies b(0)=b(1)=b(2)=0 and b(n+3)=A077866(n) for every n>=0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/KimberlingLeastTwoSubsetCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedPerry's A094802 least factorial index conjectureFor n >= 5, the prefix lcm divides the factorial of the largest prime <= n, which is the least factorial index.Proved

The question%N A094802 a(n) = smallest k such that all of 1 through n divides k!. %C A094802 It is conjectured that after n=4 the sequence is prime for n prime or the previous prime for n not prime. %A A094802 Jon Perry, Jun 11 2004 %O A094802 1,2

Why it mattersRevision 1 (September 2004) already contains the conjecture, and revision 10 (2024-04-01) still marks it as a conjecture without a settlement comment.

The ideaThe lower bound observes that a prime p at most n divides lcmUpto n; for any admissible k it therefore divides k!, and Nat.Prime.dvdfactorial forces p <= k.

Our contributionFor n >= 5, the prefix lcm divides the factorial of the largest prime <= n, which is the least factorial index.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Perry's A094802 least factorial index conjecture

The OEIS entry states, verbatim:

%N A094802 a(n) = smallest k such that all of 1 through n divides k!. %C A094802 It is conjectured that after n=4 the sequence is prime for n prime or the previous prime for n not prime. %A A094802 Jon Perry, Jun 11 2004 %O A094802 1,2

The literal proved statement is: for all natural n and p, if 5 <= n, Nat.Prime p, p <= n, and every prime q at most n satisfies q <= p, then IsLeast {k : Nat | Nat.lcmUpto n | k !} p. For a prime n, p is n; otherwise it is the previous prime. Thus p is precisely the largest prime at most n. The entry's example a(6)=5 as 5!=120 which is divisible by 1,2,3,4,5 and 6. agrees. The bound is necessary: a(4)=4 is not prime. Mathlib defines Nat.lcmUpto n as (Finset.Icc 1 n).lcm id, so the %N divisibility clause is precisely Nat.lcmUpto n | k !.

Not claimed: a general formula for n at most four beyond the computed values, any general theorem about the Kempner function away from this prefix-lcm input, or an asymptotic estimate for lcmUpto.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/PerryLeastFactorialDivisibleByPrefixLcm.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedRatajczak's A110491 T-transform determinant conjectureRatajczak's literal T-transform determinant equals the independently specified A110491 sequence in every positive matrix order.Proved

The questionLechoslaw Ratajczak's August 4, 2021 COMMENT on OEIS A110491 states:

Why it mattersThe conjecture identifies a determinant transform of one independently specified OEIS sequence with another sequence at all orders.

The ideaSubtract every predecessor row from its successor.

Our contributionRatajczak's literal T-transform determinant equals the independently specified A110491 sequence in every positive matrix order.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Ratajczak's A110491 T-transform determinant conjecture

Lechoslaw Ratajczak's August 4, 2021 COMMENT on OEIS A110491 states:

Conjecture: {a(n-1), n>=1} is the T-transform of A093178, where T maps a sequence {b(n), n>=1} to the sequence {c(n)} defined by c(n) = det(M_n), where M_n is the n X n matrix with elements M_n(i,j) = b(2*j) for i>j and M_n(i,j) = b(i+j-1) for i<=j.

Here A093178 is b(r)=1 when r is even and b(r)=r when r is odd. The matrix indices i,j in the comment are one-based. Independently, define the integer sequence a by a(0)=1, a(1)=2, and

a(m+2)=2*a(m+1)+4*(m+1)*m*a(m).

For zero-based finite indices set i1=i+1 and j1=j+1, and define the order-n integer matrix by the literal source branches

M_n(i,j)=b(2*j1) if i1>j1, and M_n(i,j)=b(i1+j1-1) if i1<=j1.

The resolved statement is the unbounded assertion

theorem result (m : ℕ) :
    (sourceMatrix (m + 1)).det = sourceA m

for every natural m. This is exactly the source conjecture at every positive order, not a finite prefix, a surrogate matrix, a definition of a by determinants, or a consequence of an assumed generating function.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/RatajczakTTransformDeterminant.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A112302: Meijer's Lerch recurrence limitThe Lerch convolution source, its summability and unique existence, and its limit at the Somos constant.Proved

The questionMeijer's June 27, 2016 COMMENTS paragraph in OEIS A112302 states the exact external problem:

The ideaAll auxiliary constructions and facts in the proof are local to result; its only top-level mathematical declarations are the three definitions and the single theorem above.

Our contributionThe Lerch convolution source, its summability and unique existence, and its limit at the Somos constant.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A112302: Meijer's Lerch recurrence limit

Johannes W. Meijer's June 27, 2016 COMMENTS paragraph in OEIS A112302 states the exact external problem:

With Phi(z, p, q) the Lerch transcendent, define LP(n) = (1/n) * sum(Phi(1/2, n-k, 1) * LP(k), k=0..n-1), with LP(0) = 1. Conjecture: Lim_{n -> infinity} LP(n) = A112302.

The proof retains all three registered definitions and the complete four-conjunct theorem:

noncomputable def LerchKernel (s : ℕ) : ℝ :=
  ∑' j : ℕ, (1 / 2 : ℝ)^j / ((j : ℝ) + 1)^s

noncomputable def SomosConstant : ℝ :=
  Real.exp (∑' j : ℕ, Real.log ((j : ℝ) + 1) / (2 : ℝ)^(j + 1))

def LPSource (a : ℕ → ℝ) : Prop :=
  a 0 = 1 ∧ ∀ n : ℕ, 0 < n →
    a n = (1 / (n : ℝ)) *
      ∑ k : Fin n, LerchKernel (n - k.val) * a k.val

theorem result :
  (∀ s : ℕ, 0 < s →
    Summable (fun j : ℕ => (1 / 2 : ℝ)^j / ((j : ℝ) + 1)^s)) ∧
  Summable (fun j : ℕ => Real.log ((j : ℝ) + 1) / (2 : ℝ)^(j + 1)) ∧
  (∃! a : ℕ → ℝ, LPSource a) ∧
  (∀ a : ℕ → ℝ, LPSource a →
    Filter.Tendsto a Filter.atTop (nhds SomosConstant))

A112302 denotes the real Somos constant through its decimal expansion. The kernel is Phi(1/2,s,1)=2 Li_s(1/2). For k in Fin n, n-k is positive; the recurrence divides by n only under n>0. Both summability clauses and unique source existence are part of the result, not assumed premises.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LerchSomosLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedCubic divisibility of generalized harmonic numerators along Adamchuk's progressionEvery exponent in Adamchuk's progression gives a generalized harmonic numerator divisible by 37 cubed.Proved

The questionNumbers n such that 37^3 divides the numerator of generalized harmonic number H(36,n) = Sum[ 1/k^n, {k,1,36} ].

Why it mattersA universal proof settles every term of that progression rather than extending the finite list of verified exponents.

The ideaLet L=36!, bj=L/j, and N(n)=Sumj bj^n, so that H(n)=N(n)/L^n.

Our contributionEvery exponent in Adamchuk's progression gives a generalized harmonic numerator divisible by 37 cubed.

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Integer sequences (OEIS)

Cubic divisibility of generalized harmonic numerators along Adamchuk's progression

OEIS A116184, NAME (%N, verbatim):

Numbers n such that 37^3 divides the numerator of generalized harmonic number H(36,n) = Sum[ 1/k^n, {k,1,36} ].

COMMENT (%C, verbatim):

Conjecture: All terms of the arithmetic progression 3+36k belong to a(n).

AUTHOR (%A, verbatim):

Alexander Adamchuk, Apr 08 2007

The first terms (%S) are:

3, 37, 39, 73, 75, 111, 147, 148, 183, 185, …

The literal claim is ∀ k ∈ ℕ, 37³ ∣ num(H(36, 3+36k)), where Rat.num is the reduced numerator. The same OEIS comment also discusses primes of the form 37m-1 and cases of divisibility by 37^4; neither assertion is claimed here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/AdamchukGeneralizedHarmonicThirtySevenCubeProgression.adamchuk_a116184

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A193020 minus-four eventual identityA triangular set of three unary gaps gives the exact distinct support of a double-tail self-shuffle.Proved

The questionLayman and the OEIS contributors, states the separate assertion a(2^k-4)=k(k-1)/2 with its sufficiently-large lower-bound caveat.

Why it mattersAn exhaustive description of the unbounded family's distinct support separates word cardinality from the number of labelled interleaving schedules.

The ideaThe ordinary binary expansion of 2^k-4 consists of m ones followed by two zeros.

Our contributionA triangular set of three unary gaps gives the exact distinct support of a double-tail self-shuffle.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A193020 minus-four eventual identity

OEIS A193020, revision 13, attributed to John W. Layman and the OEIS contributors, states the separate assertion a(2^k-4)=k*(k-1)/2 with its sufficiently-large lower-bound caveat. For positive n, a(n) counts distinct complete order-preserving self-shuffle output words of the ordinary binary expansion of n. Different schedules producing the same word count once.

The exact selected assertion is

$$\exists K\in\mathbb N,\quad 3\le K\ \land
\forall k\in\mathbb N,\quad K\le k\ \Longrightarrow
a(2^k-4)=\frac{k(k-1)}2.$$

The OEIS entry includes a(0). The selected event uses positive inputs throughout its tail; it does not impose a source convention at zero. This dossier anchors only the minus-four question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/DoubleTailShuffleSupport.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A222014 factorial square-exponent sumThe coefficients of OEIS A222014 are odd exactly one below a power of two.Proved

The questionA(x) = Sum{n>=0} n! x^n A(x)^(n^2) / Product{k=1..n} (1 + kxA(x)^n).

Why it mattersThe requested first-tier theorem covers every natural index, including zero.

The ideaparameterizedTerm e d r F is the product of r!, X^r, F^(e r), and the unit inverse of the product of 1 + (k+1)XF^(d r k) over k < r.

Our contributionThe coefficients of OEIS A222014 are odd exactly one below a power of two.

Verification details
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Integer sequences (OEIS)

Parity of the A222014 factorial square-exponent sum

OEIS A222014 defines its generating series by

A(x) = Sum_{n>=0} n! * x^n * A(x)^(n^2) / Product_{k=1..n} (1 + k*x*A(x)^n).

Paul D. Hanna's comment of Dec 06 2024 states, verbatim:

Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

The infinite sum is formalized coefficientwise by its exact finite window.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/FactorialSquareExponentSumParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedCommon positive threshold minima for reciprocal-sigma sumsReciprocal sigma sums and the first crossing of every positive integer threshold.Proved

The question— Jaroslav Krizek, OEIS A265714, revision 23.

Why it mattersThe two predicates ask respectively for a first floor level and a first threshold crossing of the same exact divisor sum.

The ideaPrime divisor terms give a lower bound by sums of 1/(p+1).

Our contributionReciprocal sigma sums and the first crossing of every positive integer threshold.

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Integer sequences (OEIS)

Common positive threshold minima for reciprocal-sigma sums

a(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

Conjecture: a(n) is also the smallest number k such that Sum_{d|k} 1/sigma(d) >= n.

— Jaroslav Krizek, OEIS A265714, revision 23.

This dossier anchors exactly the common-minimum conjecture. For every natural r>0, let g(k) be the exact rational sum of 1/sigma(d) over all positive divisors d of the positive natural k, where sigma(d) is the sum of all positive divisors of d. The sets {k>0 : floor(g(k))=r} and {k>0 : g(k)>=r} must both be nonempty and have the SAME least element. The floor is ordinary floor, r=1 is included, and neither r nor k has a finite upper bound. There is no additional existence or integrality premise. The source's separate integrality question and proposed numerical minima are outside this single assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/ReciprocalSigmaThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBarker's recurrence for no-leaf subgraphs of the 3 X n gridBarker's recurrence and Kagey's modulo-ten congruence for no-leaf subgraphs of the three-by-n grid.Proved

The questionNumber of no-leaf subgraphs of the 3 X n grid.

Why it mattersThe order-four recurrence was recorded as a conjecture in 2018.

The ideaThe proof constructs a bijection between edge-subgraphs and paths of column masks.

Our contributionBarker's recurrence and Kagey's modulo-ten congruence for no-leaf subgraphs of the three-by-n grid.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barker's recurrence for no-leaf subgraphs of the 3 X n grid

OEIS A301976, NAME (verbatim from Library/Words/kagey2018a301976.md):

Number of no-leaf subgraphs of the 3 X n grid.

FORMULA (verbatim; attributed there to Colin Barker, Mar 30 2018):

a(n) = 12*a(n-1) - 6*a(n-2) - 20*a(n-3) - 5*a(n-4) for n>4.

The literal object is the set of spanning edge-subgraphs of the 3 X n grid graph, with all 3n vertices retained, in which every vertex has degree different from one. Degree zero is allowed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/GridNoLeafSubgraphRecurrence.barker_a301976

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPrime periods of the A338040 weighted Stirling power sumBala's A338040 conjecture: p - 1 is a period modulo every prime p.Proved

The questionThe supplied attribution also identifies Paul D.

Why it mattersThis is a first-tier OEIS conjecture: the entry dates to 2020 and Bala's conjecture to 2022.

The ideaa n is the natural-valued finite sum Σ{k ≤ n} 4^k · k^n · k! · Nat.stirlingSecond n k, the entry's finite FORMULA and the formal reading of its e.g.f. Σ 4^j (e^{jx} - 1)^j. aeqwindow reduces a(n) modulo p to k < p: Stirling numbers vanish above the diagonal, and p divides k! for k >= p.

Our contributionBala's A338040 conjecture: p - 1 is a period modulo every prime p.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime periods of the A338040 weighted Stirling power sum

OEIS A338040, entry by Vaclav Kotesovec, Oct 08 2020, has the following NAME, quoted verbatim from Library/Recurrence/bala2022a338040.md:

E.g.f.: Sum_{j>=0} 4^j * (exp(j*x) - 1)^j.

The supplied attribution also identifies Paul D. Hanna, May 31 2022. The COMMENT by Peter Bala, May 31 2022, states:

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1.

The theorem proves periodicity with period p - 1 for every prime p and n >= 1. It does not assert minimality of the period. The e.g.f. identity is read through the entry's finite FORMULA quoted in Evidence; that identity is not separately formalized.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/WeightedStirlingPowerPrimePeriod.bala_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedDivisibility modulo three of the A373308 coefficientsThe coefficients of OEIS A373308 at indices 3n+1 and 3n+2 are divisible by three.Proved

The questionHanna, Jun 20 2024, gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2024a373308.md.

Why it mattersThe target is the divisibility assertion for every natural index, with KPI = open problems resolved.

The ideaThe infinite product in the NAME, Π{n≥0}(1 − x^{2^n})³, is represented by its defining functional equation A = (1 - X)^3 A.subst (X^2) with A(0) = 1.

Our contributionThe coefficients of OEIS A373308 at indices 3n+1 and 3n+2 are divisible by three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility modulo three of the A373308 coefficients

OEIS A373308, Paul D. Hanna, Jun 20 2024, gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2024a373308.md.

NAME:

Expansion of Product_{n>=0} (1 - x^(2^n))^3.

COMMENT:

Conjecture: a(3*n+1) and a(3*n+2) are divisible by 3 for n >= 0.

The entry's separate Thue-Morse congruence for a(3n) and zero-index characterisation conjectures are context only; neither is claimed here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CubedBinaryProductModThree.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedAlternating row sums of the Fibonacci harmonic triangleSchulte's alternating row sums of the Fibonacci harmonic triangle have the conjectured closed form.Proved

The questionConjecture: Alt. row sums of the harmonic triangle are Fibonacci(n-2) / Fibonacci(n+1), where Fibonacci(-1) = 1.

Why it mattersThis is a first-tier recent OEIS conjecture selected in the implementation brief.

The ideaMoving the diagonal replaces its old denominator and adds the next diagonal term, giving, for n >= 2, S(n+1) - S(n) = 2(-1)^n / (F(n+1)F(n+2)).

Our contributionSchulte's alternating row sums of the Fibonacci harmonic triangle have the conjectured closed form.

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Integer sequences (OEIS)

Alternating row sums of the Fibonacci harmonic triangle

OEIS A378277 records the following conjecture:

Conjecture: Alt. row sums of the harmonic triangle are Fibonacci(n-2) / Fibonacci(n+1), where Fibonacci(-1) = 1.

Werner Schulte, Nov 21 2024.

Its FORMULA gives the denominators:

T(n, k) = Fibonacci(n) * Fibonacci(n+1) if k = n, and Fibonacci(k) * Fibonacci(k+2) if 1 <= k < n.

The harmonic triangle has entries 1 / T(n, k) for 1 <= k <= n. Write F(n) = Nat.fib n and S(n) = sum_{k=1}^n (-1)^(k-1) / T(n, k), with the first term positive.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/HarmonicTriangleAlternatingRowSums.schulte_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A378575 shifted-iterate fixed-point congruenceFor r and m at least two with m dividing r-1, the shifted compositional fixed point has all positive coefficients equal to one modulo m.Proved

The questionHanna, Dec 01 2024, states the following NAME and COMMENT, reproduced verbatim from Library/RecurrenceComposition/hanna2024a378575.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2024 entry.

The ideaThe module is parametrised by the iterate count r.

Our contributionFor r and m at least two with m dividing r-1, the shifted compositional fixed point has all positive coefficients equal to one modulo m.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A378575 shifted-iterate fixed-point congruence

OEIS A378575, Paul D. Hanna, Dec 01 2024, states the following NAME and COMMENT, reproduced verbatim from Library/RecurrenceComposition/hanna2024a378575.md:

G.f. A(x) satisfies A(x) = x + x*A(A(A(A(A(x))))), so that this sequence shifts left under the 5th self-COMPOSE.

Conjecture: a(n) == 1 (mod 4) for n >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/ShiftedIterateFixedPointCongruence.hanna_conjecture_five

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A378576 shifted-iterate fixed-point congruenceFor r and m at least two with m dividing r-1, the shifted compositional fixed point has all positive coefficients equal to one modulo m.Proved

The questionHanna, Dec 01 2024, states the following NAME and COMMENT, reproduced verbatim from Library/RecurrenceComposition/hanna2024a378576.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2024 entry.

The ideaThe module is parametrised by the iterate count r.

Our contributionFor r and m at least two with m dividing r-1, the shifted compositional fixed point has all positive coefficients equal to one modulo m.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A378576 shifted-iterate fixed-point congruence

OEIS A378576, Paul D. Hanna, Dec 01 2024, states the following NAME and COMMENT, reproduced verbatim from Library/RecurrenceComposition/hanna2024a378576.md:

G.f. A(x) satisfies A(x) = x + x*A(A(A(A(A(A(x)))))), so that this sequence shifts left under the 6th self-COMPOSE.

Conjecture: a(n) == 1 (mod 5) for n >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/ShiftedIterateFixedPointCongruence.hanna_conjecture_six

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A380066 bilateral product congruenceHanna's A380066 bilateral product congruence.Proved

The questionA(x) satisfies 1/2 = Sum{n=-oo..+oo} x^nA(x)^n (A(x)^n + x)^(n-1) (A(x) + x^n)^(n-1).

Why it mattersThe source is an explicitly stated, externally published conjecture (tier 1).

The ideaThe induction in UnitExponent.approximationstable stabilizes the existing approximations; generatingfixed and generatingequation establish the normalized integral solution.

Our contributionHanna's A380066 bilateral product congruence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A380066 bilateral product congruence

Paul D. Hanna's OEIS A380066 states:

G.f. A(x) satisfies 1/2 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + x)^(n-1) * (A(x) + x^n)^(n-1).

Conjecture: for n > 0, a(n) == 2 (mod 4).

The offset is zero and the listed coefficients begin $1,6,62,826,13562$. The binding convention in #18805 uses integer power series with $A(0)=1$. Each coefficient $N$ is read through every symmetric integer window $[-K,K]$ with $N\le K$ in rational Laurent series. This is a formal coefficient statement, with no analytic convergence claim.

Statement echo

All new names belong to D5.S1.Recurrence.Bilateral.BilateralProductModFourUnitExponent.

| Source clause or binding convention | Lean counterpart | Fidelity | | --- | --- | --- | | The displayed summand, with every integer index | Frozen BilateralProductModFour.laurentTerm at parameter one | The source expression is reused literally. | | Offset zero and constant coefficient one | HannaA380066.IsGF, first conjunct | The integer-series convention and normalization are explicit. | | The bilateral equation | HannaA380066.IsGF, every N, K and window hypothesis | Every coefficient and every radius at least its index are required. | | Every positive index has residue two modulo four | HannaA380066.claim, universal conjunct; result | Every normalized integral solution is covered. | | Existence | HannaA380066.claim, existential conjunct; result | This rules out a vacuous universal assertion and adds no hypothesis. |

The two public definitions retain the literal binding conventions. The proof uses the existing family's owners internally. There are no additional source assumptions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/BilateralProductModFourUnitExponent.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A392095 column and A088713 source seriesThe antidiagonal natural array and the normalized rational source series are uniquely determined, and their first-column coefficients agree after a one-step shift.Proved

The questionMikhail Kurkov's A392095 (December 30, 2025) states:

Why it mattersThis is a first-tier named OEIS conjecture from 2025, connecting an antidiagonal recurrence to a separately specified 2003 generating series.

The ideaThe array is constructed by the lexicographic measure (n+k,n); induction first on n+k and then on n proves uniqueness.

Our contributionThe antidiagonal natural array and the normalized rational source series are uniquely determined, and their first-column coefficients agree after a one-step shift.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A392095 column and A088713 source series

Mikhail Kurkov's A392095 (December 30, 2025) states:

Array read by downward antidiagonals: A(n,k) = A(n-1,k+1) + Sum_{j=0..k} A(n-1,j)*A(k-j,0) with A(0,k) = 1.

Conjecture: A(n,0) = A088713(n+1).

Paul D. Hanna's A088713 (October 12, 2003) states:

G.f. A(x) satisfies A(x/A(x)) = 1/(1-x).

Both sequences start at index zero. A088713 has constant coefficient one. The exact sources, author lines, immutable locators and licenses are in Library/Recurrence/kurkov2025a392095.md and Library/Recurrence/hanna2003a088713.md.

The full assertion concerns a total array T : N -> N -> N with T(0,k)=1 for every k and T(n+1,k)=T(n,k+1)+sum(j=0..k) T(n,j)T(k-j,0) for every n,k. There is a unique such T. Independently there is a unique F in Q[[X]] with F(0)=1 and F(XF^(-1))=(1-X)^(-1). There is a unique b : N -> N whose rational coefficient series satisfies that equation; b(0)=1, [X^m]F=b(m) for every m, and T(n,0)=b(n+1) for every n, including zero. Inverses here are multiplicative unit inverses, and the inner substitution series has zero constant coefficient. No extra source equation is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/AntidiagonalArraySourceSeries.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe greedy-brick all-index identityThe first brick widths in every positive row satisfy the greedy-brick composition identity.Proved

The questionConjecture: when n = a(m) for some m, we have a(n) = n(n+3)/2 - m.

Why it mattersThis first-tier externally named conjecture asks for an identity at every positive index.

The ideaThe literal capacity trajectory has a least birth for every positive row.

Our contributionThe first brick widths in every positive row satisfy the greedy-brick composition identity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The greedy-brick all-index identity

OEIS A395531, revision 30, Aleksei Udovenko:

Conjecture: when n = a(m) for some m, we have a(n) = n(n+3)/2 - m. Verified experimentally up to m=1172, n=688491.

Rows are numbered from one. The value a(m) is the width of the first brick in row m under the literal highest-supported-row placement rule. The exact assertion is a(a(m)) = a(m)(a(m)+3)/2-m for every integer m >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GreedyBrick/OriginalIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A395833: diagonal-exponent self-divisibilityAffine vanishing-diagonal exponents divide their own integer coefficients.Proved

The questionThe Library note quotes the NAME and divisibility conjecture verbatim:

Why it mattersThis is a first-tier OEIS conjecture from 2026 (the supplied AUTHOR line is May 07 2026).

The ideaExpanding the vanishing diagonal writes the nth coefficient as a negative sum of smaller coefficients times inverse-power coefficients c.

Our contributionAffine vanishing-diagonal exponents divide their own integer coefficients.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A395833: diagonal-exponent self-divisibility

The Library note quotes the NAME and divisibility conjecture verbatim:

G.f. A(x) satisfies [x^n] A( x/A(x)^(2*n-1) ) = 0 for n > 1. Conjecture: (2*n-1) divides a(n) for n >= 1.

The supplied local OEIS source identifies the entry and its separate conjecture:

%N A395833 G.f. A(x) satisfies [x^n] A( x/A(x)^(2*n-1) ) = 0 for n > 1.
%C A395833 Conjecture: (2*n-1) divides a(n) for n >= 1.
%C A395833 Conjecture: a(n) == 0 (mod 3) for n >= 2.
%A A395833 _Paul D. Hanna_, May 07 2026

The problem resolved here is the divisibility clause only: (2*n-1) divides a(n) for every n >= 1. The separate modulo-three clause is outside this claim. The sequence is the unique normalized integer series with constant and linear coefficients one satisfying the displayed vanishing-diagonal equation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalExponentSelfDivisibility.exponent_self_divisibility

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA396805 coefficients modulo three from index threeConjecture: a(n) == [0,1,0] repeating (mod 3) for n >= 3.Proved

The questionHanna's OEIS A396805, dated June 9, 2026, states:

Why it mattersThe existing parityiteratefive settles parity.

The ideaWrite M(f,g)(n)=Sum{i=0}^n binomial(n,i)f(i)g(n-i) for multiplication in integral EGF coordinates.

Our contributionConjecture: a(n) == [0,1,0] repeating (mod 3) for n >= 3.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A396805 coefficients modulo three from index three

Paul D. Hanna's OEIS A396805, dated June 9, 2026, states:

E.g.f. satisfies A(x) = x*exp( A^5(x) ). Conjecture: a(n) == [0,1,0] repeating (mod 3) for n >= 3.

The formula field defines A^5 as the fifth compositional iterate and specifies A(x)=Sum_{n>=1} a(n)x^n/n!. For the unique zero-constant rational formal solution A=Xexp(A^{[5]}), the exact conclusion is forall n : Nat, 3 <= n -> Nat.ModEq 3 (IterateExponentialParity.aK 5 n) (if n % 3 = 1 then 1 else 0). The residues 0,1,0 begin at n=3. The source value a(2)=2 excludes extension of this formula to all positive indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialFiveModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA396806 coefficients modulo sixThe positive-index coefficients of OEIS A396806 agree with their indices modulo six.Proved

The questionHanna's OEIS A396806 entry, dated June 9, 2026, states:

Why it mattersThe existing IterateExponentialParity module settles the parity comment and identifies the coefficient sequence.

The ideaIn integral EGF coordinates over ZMod 3, set l(n)=n and T(f)(n)=sumj binomial(n,j)f(j)j^(n-j).

Our contributionThe positive-index coefficients of OEIS A396806 agree with their indices modulo six.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A396806 coefficients modulo six

Paul D. Hanna's OEIS A396806 entry, dated June 9, 2026, states:

E.g.f. satisfies A(x) = x*exp( A^6(x) ). Conjecture: a(n) == [1,2,0] repeating (mod 3) for n >= 1. Conjecture: a(n) == [1,2,3,4,5,0] repeating (mod 6) for n >= 1.

The formula field defines A^6 as the sixth compositional iterate and gives A(x)=Sum_{n>=1} a(n)x^n/n!. For the unique zero-constant rational formal solution A=Xexp(A^{[6]}), the assertion is forall n : Nat, 1 <= n -> Nat.ModEq 6 (IterateExponentialParity.aK 6 n) n. It entails the quoted modulo-three clause because three divides six.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialModSix.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedPalindromic numerators of OEIS A398690Every simplified Verlinde row has a palindromic numerator of its expected degree.Proved

The questionFor every n >= 0, let Fn(z) = sum{q>=0} A(n+3,q) z^q.

Why it mattersChapoton's 2026 entry explicitly labels the all-row palindromicity assertion a conjecture.

The ideaChebyshev factorization gives a product over the sine nodes.

Our contributionEvery simplified Verlinde row has a palindromic numerator of its expected degree.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Palindromic numerators of OEIS A398690

OEIS A398690 defines A(r,q) by its signed finite sine sum. For every n >= 0, let F_n(z) = sum_{q>=0} A(n+3,q) z^q. Its comment says that F_n has denominator (1-z)^(n+1) and conjectures that its numerator is palindromic. The formal result proves that (1-z)^(n+1) F_n(z) is a polynomial of degree exactly n and that its coefficients at j and n-j agree for every 0 <= j <= n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/OeisA398690Palindromic.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedMaximum length of a two-dense divisor blockThe largest two-dense divisor block has ruler-scaled maximum odd count.Proved

The questionFor every positive natural n, split its complete strictly increasing list of positive divisors into maximal contiguous blocks, cutting precisely at b>2a.

Why it mattersThis first-tier recent externally named conjecture has an elementary universal ordinary proof and a separately missing formal membership bridge.

The ideaFor the actual ordered partition, consecutive blocks A,C have 2 last(A)<head(C).

Our contributionThe largest two-dense divisor block has ruler-scaled maximum odd count.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Maximum length of a two-dense divisor block

Conjecture: a(n) = A001511(n) * A400195(n).

— Omar E. Pol, OEIS A400194 revision 18.

For every positive natural n, split its complete strictly increasing list of positive divisors into maximal contiguous blocks, cutting precisely at b>2a. Let M(n) be the maximum block length and O(n) the maximum odd count over the SAME actual block family. The exact assertion is M(n)=(v₂(n)+1)O(n). Equality b=2a and singleton blocks are retained.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/TwoDenseDivisorBlockMaximum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedOEIS A400361: the exact second-order cumulative formulaThe exact second-order summatory formula for the Pillai minus floor-gcd sequence.Proved

The questionWrite P(n)=sum{k=1}^n gcd(n,k), F(n)=sum{k=1}^n gcd(n,floor(n/k)), and a(n)=P(n)-F(n), using signed subtraction.

Why it mattersThe exact source formula isolates the joint divisor and quotient relation in F(n).

The ideaExpand gcd by the sum of totients over common divisors.

Our contributionThe exact second-order summatory formula for the Pillai minus floor-gcd sequence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A400361: the exact second-order cumulative formula

Write P(n)=sum_{k=1}^n gcd(n,k), F(n)=sum_{k=1}^n gcd(n,floor(n/k)), and a(n)=P(n)-F(n), using signed subtraction. The COMMENT conjecture in revision 17 asks, along all positive integers N, for

$$ \sum_{n=1}^{N}a(n)=\frac{3}{\pi^2}N^2\log N+CN^2+o(N^2), $$

with the literal constant

$$ C=\frac{3}{\pi^2}\left(-\frac12+2\gamma- \frac{6\zeta'(2)}{\pi^2}\right)-\frac{\pi^2}{12} -\frac12\sum_{d\ge1}\frac{\varphi(d) (\psi(1+1/d)+\gamma)}{d^2}. $$

The full literal source and attribution are in oeis2026a400361. The preregistration is issue 18775. The target is first tier, an external OEIS conjecture. Its recent source version does not imply novelty; the bounded literature scope and remaining Bu-Xu full-text uncertainty are recorded in the source dossier.

Native statement

D5.S3.ArithSums.A400361SecondTerm.originalClaim : Prop conjoins absolute summability of the literal real coefficient sequence with little-o of the exact remainder against N^2 at the natural atTop filter. The theorem D5.S3.ArithSums.A400361SecondTerm.result : originalClaim has no hypotheses. The definitions preserve the positive source range, integer subtraction, real Gamma logarithmic derivative, real zeta restriction, and every term of the source constant. Normalizing the remainder by N^2 is proved equivalent to the stated little-o. No subsequence or finite proxy is used.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/A400361SecondTerm.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedThe normalized expected range of a knight's random walk is an integerFor a knight's random walk on the infinite chessboard, with each of the eight moves chosen uniformly at every step, the expected number E(n) of distinct squares visited in n steps (the start included) satisfies: E(n) 2^(3n-3) is an integer for every n at least one (OEIS A309221).Proved

The questionSloane, 2019) normalizes the expected number of distinct squares visited by a knight's random walk on an infinite chessboard, A326954/A326955 (the starting square counted):

Why it mattersThe number of distinct sites visited by a random walk, its range, is a basic quantity of lattice random walks in statistical physics.

The ideaThe eight symmetries of the square lattice act on the eight knight moves simply transitively: for every move m there are a signed permutation g of the coordinates and a permutation s of the moves with g(move i) = move(s(i)) and s(0) = m.

Our contributionFor a knight's random walk on the infinite chessboard, with each of the eight moves chosen uniformly at every step, the expected number E(n) of distinct squares visited in n steps (the start included) satisfies: E(n) 2^(3n-3) is an integer for every n at least one (OEIS A309221).

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The normalized expected range of a knight's random walk is an integer

OEIS A309221 (N. J. A. Sloane, 2019) normalizes the expected number of distinct squares visited by a knight's random walk on an infinite chessboard, A326954/A326955 (the starting square counted):

a(0)=1; for n>0, a(n) = (A326954(n)/A326955(n))*2^(3*n-3). (It is only a conjecture that this is always an integer).

Issue #10116 fixes the readings: E(n) is the expected number of distinct squares of ℤ² visited by a walk from the origin whose n steps are chosen independently and uniformly among the eight knight moves, the start included; equivalently E(n) = S(n)/8^n with S(n) the total over all 8^n step sequences; the conjecture asserts that E(n)·2^(3n−3) is an integer for every n ≥ 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/RandomWalks/KnightWalkRangeIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedHigher exterior traces of Spezia's perimeter matrixHigher exterior traces of the Spezia perimeter matrix vanish.Proved

The questionStefano Spezia's OEIS A330700, COMMENTS, revision 23, gives this second line in its Conjectures block:

Why it mattersThis is a universal symbolic statement about a source-defined family; it is not a finite certificate.

The ideaThese are precisely the two branches of the public matrix definition.

Our contributionHigher exterior traces of the Spezia perimeter matrix vanish.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Higher exterior traces of Spezia's perimeter matrix

Stefano Spezia's OEIS A330700, COMMENTS, revision 23, gives this second line in its Conjectures block:

For k > 2, the trace of the k-th exterior power of M(n) is equal to zero.

The same block and OEIS A330287, COMMENTS, revision 31, specify:

The matrix M(n) is defined as M[i,j,n] = i*j if i < 3 or j < 3 and M[i,j,n] = 2*(i + j) - 4 otherwise.

Indices are one-based. For every positive natural $n$, interpret the integer entries in $\mathbb C$ and take every natural $k>2$, including $k>n$. The exterior power is represented by the existing compoundMatrix; its trace is Matrix.trace. The target is exactly this second conjecture, as preregistered in #21999.

Statement echo

| Source clause or convention | Lean counterpart | Fidelity | | --- | --- | --- | | The literal piecewise entry formula | SpeziaPerimeterHigherExteriorTraces.M | Source indices are i.val+1, j.val+1; the comparison is in natural numbers and entries and subtraction are in $\mathbb C$. | | A square matrix of positive order | claim quantifies over natural $n>0$ | Positivity is the explicit preregistered scope. | | Every exterior order $k>2$ | claim quantifies over natural $k>2$ | No upper bound $k\le n$ is added. | | Trace of the exterior power equals zero | Matrix.trace (compoundMatrix (M n) k) equals zero | The existing ordered-subset basis also includes the empty-space case. | | The universal conjecture holds | SpeziaPerimeterHigherExteriorTraces.result | The theorem proves claim without factorization or rank hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/SpeziaPerimeterHigherExteriorTraces.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025Ta Problem 11.7: the good-involution bound need not be strictThe conjugation subquandle {±i, ±j} of the quaternion group Q8 has four good involutions and eight automorphisms.Refuted

The questionLực Ta, Good involutions of conjugation subquandles, arXiv:2505.08090v5, Problem 11.7 (printed page 27), asks:

Why it mattersThe question asks whether the displayed upper bound is always a strict inequality under the two lower bounds on the center and orbit count.

The ideaTake G = QuaternionGroup 2 and X = {QuaternionGroup.a 1, QuaternionGroup.a 3, QuaternionGroup.xa 0, QuaternionGroup.xa 2}, the subquandle {±i, ±j}.

Our contributionThe conjugation subquandle {±i, ±j} of the quaternion group Q8 has four good involutions and eight automorphisms.

Verification details
Release source recordRefuted in Lean

Open problem

Ta Problem 11.7: the good-involution bound need not be strict

Lực Ta, Good involutions of conjugation subquandles, arXiv:2505.08090v5, Problem 11.7 (printed page 27), asks:

In the setting of Corollary 6.12, suppose that |Z(H)|, |X/H| ≥ 2 (so, in particular, Conj X is not connected). Is the upper bound on |Good(Conj X)| in Corollary 6.12 always strict?

Corollary 6.12 gives the bound |Good(Conj X)| <= min(|Aut(Conj X)|, |Z(H)|^(|X/H|)).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/Quandles/TaGoodInvolutionBoundRefutation.result

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Number theoryQuestion resolvedTebni's OEIS A350759 prime congruencesBoth prime congruences for Tebni's OEIS A350759 sequence.Proved

The questionMélika Tebni's OEIS A350759 revision 17 lists these COMMENTS conjectures:

Why it mattersThe source gives a concrete unbounded prime-congruence problem rather than a finite data fit.

The ideaThe proof first obtains the exact rising-factorial expansion of the source EGF.

Our contributionBoth prime congruences for Tebni's OEIS A350759 sequence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Tebni's OEIS A350759 prime congruences

Mélika Tebni's OEIS A350759 revision 17 lists these COMMENTS conjectures:

For all p prime, (a(p) + a(p+1) - 2) == 0 (mod p),
a(p+1) == 1 (mod ((p+1)*p)).

Its FORMULA line is the exponential generating function exp(-1 + (1 + x)*(1 - log(1 + x))). The resolved Lean statement is D5/S3/Arith/Congruence/TebniA350759PrimeCongruences.claim: an integer sequence with that exact EGF exists, and every such sequence satisfies both congruences. The settling declaration is D5/S3/Arith/Congruence/TebniA350759PrimeCongruences.result, which proves claim.

Statement echo

The Lean sourceSeries is the published formal EGF, with PowerSeries.log ℚ interpreted as the formal log(1+X) and PowerSeries.subst taking the inner series first. IsSourceSequence a states that the EGF of the integer sequence a : ℕ → ℤ, with coefficients cast to ℚ, equals sourceSeries. The first conjunct of claim is existence of such an a; its second conjunct quantifies every such a, every natural p, and Nat.Prime p, then asserts Int.ModEq (p : ℤ) (a p + a (p + 1) - 2) 0 and Int.ModEq (((p + 1) * p : ℕ) : ℤ) (a (p + 1)) 1. The natural product is formed before its cast to ℤ, and the prime p=2 is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/TebniA350759PrimeCongruences.result

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Quantum informationQuestion resolvedFirst preprint 2025All odd kinetic frequency momentsThe kinetic contribution obeys the all-order odd-moment formula for every isotropic momentum distribution with finite required moments.Proved

The questionVorberger, Kinetic contribution to the arbitrary order odd frequency moments of the dynamic structure factor, arXiv:2508.17810v1, Sec. 2.3, Eq. (16), state: “Therefore, our conjecture states that the following result holds for the interacting uniform electron gas”, followed by

The ideaDifferent-particle terms commute; same-particle opposite translations cancel.

Our contributionThe kinetic contribution obeys the all-order odd-moment formula for every isotropic momentum distribution with finite required moments.

Verification details
Release source recordProved in Lean

Open problem

All odd kinetic frequency moments

P. Tolias, T. Dornheim and J. Vorberger, Kinetic contribution to the arbitrary order odd frequency moments of the dynamic structure factor, arXiv:2508.17810v1, Sec. 2.3, Eq. (16), state: “Therefore, our conjecture states that the following result holds for the interacting uniform electron gas”, followed by

$$ M_{\mathrm{S,KIN}}^{(2k+1)}(q)= \left(\frac{\hbar q^2}{2m}\right)^{2k+1}\frac{1}{2k+2} \sum_{i=0}^{k}\binom{2k+2}{2i+1} \left(\frac{2}{\hbar q}\right)^{2i}\langle p^{2i}\rangle_0, $$

“where $k$ is an arbitrary non-negative integer.” The paper explicitly checks $k=0,1,2,3$. Its summary also asks whether similar expressions hold for other spin statistics. Issue #13703 fixes the operator representation and averaging conventions. The quotations and locators are in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/KineticMoments/OddMoment.result

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Quantum informationQuestion resolvedFirst preprint 2025Spectral bound for conditionally 2-positive mapsConditionalTwoPositiveSpectralBound for the spectral bound of conditionally two-positive maps.Proved

The questionMuratore-Ginanneschi, Universal Bound on the Eigenvalues of 2-Positive Trace-Preserving Maps, arXiv:2506.02145v1, §3 “Outlook”, page 11, question (10), asks:

The ideaIts Hilbert–Schmidt adjoint is 2-positive and trace preserving, so the formalized Theorem 1 applies.

Our contributionConditionalTwoPositiveSpectralBound for the spectral bound of conditionally two-positive maps.

Verification details
Release source recordProved in Lean

Open problem

Spectral bound for conditionally 2-positive maps

F. vom Ende, D. Chruściński, G. Kimura and P. Muratore-Ginanneschi, Universal Bound on the Eigenvalues of 2-Positive Trace-Preserving Maps, arXiv:2506.02145v1, §3 “Outlook”, page 11, question (10), asks:

Hence one may wonder whether all (generators of) completely positive maps, resp. 2-positive maps satisfy

$$ \mathrm{tr}(L)\le d\min\Re\sigma(L)+(d^2-d)\max\Re\sigma(L). $$

A map $L$ is called conditionally 2-positive if $e^{tL}$ is 2-positive for all $t\ge0$.

With this, (10) becomes really a conjecture about the spectrum of arbitrary conditionally 2-positive maps.

VCKM-10 quantifies every integer $d\ge1$ and every complex-linear endomorphism $L$ of $\mathbb C^{d\times d}$ satisfying this definition. There is no trace-preservation assumption. The trace is the superoperator trace; the spectrum is the set of roots of the characteristic polynomial. The Lean claim also asserts that the superoperator trace is real.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ConditionalTwoPositiveSpectralBound.result

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Discrete structuresQuestion resolvedWiseman A344611: odd-length partition injectionNonnegative reverse-alternating partitions inject into odd-length partitions of the next weight.Proved

The questionGus Wiseman's OEIS A344611 revision 27, FORMULA, states: “Conjecture: a(n) <= A160786(n).

Why it mattersThe named external conjecture and its source and literature screen are recorded in 23757.

The ideaFor odd source length, increase the first part by one.

Our contributionNonnegative reverse-alternating partitions inject into odd-length partitions of the next weight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

wiseman a344611

Gus Wiseman's OEIS A344611 revision 27, FORMULA, states: “Conjecture: a(n) <= A160786(n). The difference is 0, 0, 0, 0, 1, 2, 4, 9, 16, 28, 48, 79, ...”. Its NAME is “Number of integer partitions of 2*n with reverse-alternating sum >= 0.” Its COMMENTS define “The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.” The OFFSET is “0,2”, including the empty partition. A160786 revision 25 gives “a(n) = A027193(2n+1).” A027193 revision 78 counts partitions into an odd number of parts. The electronic sources have no page numbers, DOI or arXiv identifier.

For a decreasing positive partition $\lambda=(\lambda_1,\ldots,\lambda_k)$ of weight $m$, set $R(\lambda)=\sum_{i=1}^k(-1)^{k-i}\lambda_i$, with $R(())=0$. The assertion is

$$ \forall n\in\mathbb N,\quad #{\lambda\vdash2n:R(\lambda)\ge0} \le#{\mu\vdash2n+1: \text{the number of parts of }\mu\text{ is odd}}. $$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive decreasing integer partitions of a specified weight | Nat.Partition, with parts.sort (· ≥ ·) | Positive multiset parts with the weight constraint; decreasing sorting is inlined. | | Reverse signs and the empty sum | reverseAlternatingSum | Existing integer recurrence TrimmedAlternatingPartitions.trimmedSums, last endpoint with default zero; the update is $z\mapsto a-z$. | | Partitions of $2n$ with $R\ge0$ | a344611 | Nat.card of the literal nonnegative-endpoint subtype. | | Odd number of parts at weight $2n+1$ | a160786 | Odd p.parts.card; no parity restriction on individual parts. | | Every $n\ge0$ and non-strict comparison | claim, result | The natural-number quantifier includes zero; no added source hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/WisemanA344611OddLengthInjection.result

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Discrete structuresQuestion resolvedWalks of near-maximal algebraic areaFor k < n, the 2n-step walks on the square lattice with algebraic area n^2 - k number twice the coefficient of x^k in phi(x)/f(-x), and the (2n+1)-step walks with area n^2 + n - k number four times the coefficient of x^k in psi(x^2)/f(-x).Proved

The questionIt appears that T(2n, n^2 - k) = 2 A029552(k) for k < n and T(2n+1, n^2+n - k) = 4 A098613(k) for k < n.

Why it mattersThe enumeration of lattice walks by algebraic area is the combinatorial side of the Hofstadter model of a particle hopping on the square lattice in a perpendicular magnetic field, studied by Ouvry and Polychronakos for closed walks (arXiv:1810.04098, arXiv:2110.09394) and for open walks closed by a straight line (Gan, Ouvry and Polychronakos, arXiv:2206.12428).

The ideaLet r, l, u, d count the right, left, up and down steps of a walk of length L.

Our contributionFor k < n, the 2n-step walks on the square lattice with algebraic area n^2 - k number twice the coefficient of x^k in phi(x)/f(-x), and the (2n+1)-step walks with area n^2 + n - k number four times the coefficient of x^k in psi(x^2)/f(-x).

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Walks of near-maximal algebraic area

OEIS A385672 (Andrei Zabolotskii, 2025) counts the n-step walks on the square lattice by their algebraic area (Sum_{steps right} y) - (Sum_{steps left} y), the integral of y dx. Its FORMULA section states

It appears that T(2*n, n^2 - k) = 2 * A029552(k) for k < n and T(2*n+1, n^2+n - k) = 4 * A098613(k) for k < n.

where A029552 and A098613 are the coefficients of phi(x)/f(-x) and psi(x^2)/f(-x), with generating functions (1 + 2 Σ_{j>0} x^(j^2)) / ∏_{i>0} (1 − x^i) and Σ_{j>0} x^(j^2 − j) / ∏_{i>0} (1 − x^i). Issue #10337 fixes the readings: walks start at the origin, and the coefficients are read through 1 / ∏ (1 − x^i) = Σ p(m) x^m with p the partition numbers, so A029552(k) = p(k) + 2 Σ_{j ≥ 1, j² ≤ k} p(k − j²) and A098613(k) = Σ_{j ≥ 1, j² − j ≤ k} p(k − j² + j).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/LatticeWalkNearMaximalArea.result

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Quantum informationQuestion resolvedFirst preprint 2017The lazy two-causal inequality $J_1$ induces a facet for every $N\ge2$The lazy two-causal inequality J1 induces a facet for every party count at least two.Proved

The questionBranciard, Genuinely Multipartite Noncausality, arXiv:1708.07663v2 (Quantum 1, 39, 2017), Section III.C, ask in genuinemultinoncausal.tex, line 343:

Why it mattersThe source uses two-causal inequalities to distinguish correlations admitting a causal order between two complementary groups from genuinely multipartite noncausal correlations.

The ideaThus polytopespanle places the direction space in homogeneousSpace N, consisting of tables with the same forbidden zeros and row sums zero.

Our contributionThe lazy two-causal inequality J1 induces a facet for every party count at least two.

Scope & next questionFacetness in other input/output scenarios and analogous classifications for other inequalities remain open relative to this settlement. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

The lazy two-causal inequality $J_1$ induces a facet for every $N\ge2$

A. A. Abbott, J. Wechs, F. Costa and C. Branciard, Genuinely Multipartite Noncausality, arXiv:1708.07663v2 (Quantum 1, 39, 2017), Section III.C, ask in genuine_multi_noncausal.tex, line 343:

It is interesting to ask if this inequality is tight (i.e., facet inducing) for all $N$.

Their conjecture is stated at line 344:

By explicitly enumerating the vertices of the 2-causal polytope for $N=4$ (of which there are $136,818,592$) we were able to verify that $J_1(4)\ge 0$ is indeed also a facet, and we conjecture that this is true for all $N$.

Let $U={0,\ldots,N-1}$ and represent binary inputs and outputs by their supports $T,A\subseteq U$. A deterministic lazy two-causal strategy consists of a nonempty proper first group $K\subsetneq U$ and a function $f$ on input supports such that $f(T)\subseteq T$ for every $T$, and $T\cap K=S\cap K$ implies $f(T)\cap K=f(S)\cap K$ for every $T,S$. The complementary group's outputs may depend on the entire input. Its deterministic table is $P_f(T,A)=1$ when $f(T)=A$ and $0$ otherwise. Let $C_N$ be the real convex hull of these tables.

In these coordinates, Eq. (10) is $J_1(p)=\sum_{\varnothing\ne K\subsetneq U}p(K,K)-p(U,U)\ge0$. Laziness makes $p(K,K)$ exactly the marginal event in the source's formula. Definition 2, the deterministic characterization, and the lazy convention occur at TeX lines 146–151, 203–205, and 213–214; Eq. (10) is at lines 333–337. See the literature note.

Issue #23904 fixes the reading: for every natural $N\ge2$, $J_1$ is nonnegative on $C_N$, and $F_N=C_N\cap{p\mid J_1(p)=0}$ has affine dimension one less than $C_N$. This is proved. Precisely, the module uses $d(S)=\dim_{\mathbb R}\operatorname{span}{p-q\mid p,q\in S}$ and asserts $d(F_N)+1=d(C_N)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/LazyTwoCausalJOneFacet.result

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Quantum informationQuestion resolvedFirst preprint 2017The two-qubit binegativity lower bound at fixed negativityFor every two-qubit density matrix, the binegativity is bounded below by the Girard–Gour curve at its negativity.Proved

The questionGour, The binegativity of two qubits, arXiv:1701.02724v3 (2017), Eq. (8):

Why it mattersThe binegativity is a computable quantity that the paper introduces to study entanglement conversion.

The ideaWrite Γ for the partial transpose on the second qubit.

Our contributionFor every two-qubit density matrix, the binegativity is bounded below by the Girard–Gour curve at its negativity.

Verification details
Release source recordProved in Lean

Open problem

The two-qubit binegativity lower bound at fixed negativity

M. W. Girard and G. Gour, The binegativity of two qubits, arXiv:1701.02724v3 (2017), Eq. (8):

Based on numerical evidence, we conjecture that the binegativity is bounded below by $\frac{\nu}{2} \left(1+\frac{c_\nu}{\sqrt{(1-c_\nu)^2+c_\nu^2}}\right) \leq N_2(\sigma)$ where we define $c_\nu=\sqrt{2\nu(\nu+1)}-\nu$.

Here $\nu = N(\sigma) = 2\operatorname{Tr}[(\sigma^\Gamma)-]$ and $N_2(\sigma) = \operatorname{Tr}[(\sigma^\Gamma)-] + 2\operatorname{Tr}[(((\sigma^\Gamma)-)^\Gamma)-]$, with $X_-$ the negative component of a self-adjoint $X$. The verbatim statements are in the literature note.

Issue #18490 fixes the reading: for every two-qubit density matrix σ, with the partial transpose on the second qubit, the inequality holds. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2020A bound on n² times a falling factorialFor every integer k >= 1 and every integer 0 <= n <= k, n^2/k (1/k)^n k!/(k - n)! <= 1, that is, n^2 k(k - 1)...(k - n + 1) <= k^(n + 1).Proved

The questionLopez-Mobilia (arXiv:2004.07168, cond-mat.stat-mech) define an entropy through the number of samples needed to see a repeated microstate.

Why it mattersThe bound gives the corollary h(k') ∈ (1/k', k') for the sum h of eq. (003.21), which the paper uses to estimate its entropy for black holes.

The ideaFor n ≤ 3 the inequality is a polynomial one: 4(k − 1) ≤ k² for n = 2, and k⁴ − 9k(k − 1)(k − 2) = k((k − 3)³ + 9) ≥ 0 for n = 3.

Our contributionFor every integer k >= 1 and every integer 0 <= n <= k, n^2/k (1/k)^n k!/(k - n)! <= 1, that is, n^2 k(k - 1)...(k - n + 1) <= k^(n + 1).

Verification details
Release source recordProved in Lean

arXiv

A bound on n² times a falling factorial

T. A. Malik and R. Lopez-Mobilia (arXiv:2004.07168, cond-mat.stat-mech) define an entropy through the number of samples needed to see a repeated microstate. After eq. (003.21) they state:

Conjecture. (n²/k') (1/k')^n k'!/(k'−n)! ≤ 1 for integer n ∈ [0, k'].

with "We have not found a proof for this conjecture. However, we have analyzed it numerically and it seems to hold." Issue #11361 fixes the reading: for every integer k ≥ 1 and every integer 0 ≤ n ≤ k, equivalently n² · k(k − 1)⋯(k − n + 1) ≤ k^{n+1}.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/FallingFactorialMomentBound.result

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Quantum informationQuestion resolvedFirst preprint 2021A Bernoulli–harmonic–zeta identity for the free scalar on hyperbolic spaceFor every natural number k, the combination of harmonic numbers, Bernoulli numbers and values of the Riemann zeta function at the non-positive integers displayed as eq. (C.12) of T.Proved

The questionSato (arXiv:2101.02399, hep-th; JHEP 05 (2021) 074) compute the free energy of a conformally coupled free scalar on hyperbolic space.

Why it mattersThe identity removes all terms other than ζ'(−j) from ∂s ζ{H^d}(0, 1/2), which the paper uses for the free energy on H^d and its defect C-theorem.

The ideaWith ζ(−j) = (−1)^j B{j+1}/(j+1) and after multiplying by 2^N (k+1), the middle sum is Σ{1≤i≤N−1} bi (1/i + 1/(N−i)) and the zeta sum is Σ{1≤i≤N} C(N, i) 2^i Bi H{i−1}.

Our contributionFor every natural number k, the combination of harmonic numbers, Bernoulli numbers and values of the Riemann zeta function at the non-positive integers displayed as eq. (C.12) of T.

Verification details
Release source recordProved in Lean

Open problem

A Bernoulli–harmonic–zeta identity for the free scalar on hyperbolic space

T. Nishioka and Y. Sato (arXiv:2101.02399, hep-th; JHEP 05 (2021) 074) compute the free energy of a conformally coupled free scalar on hyperbolic space. In the derivative of the spectral zeta function for even dimension they need, for every k ≥ 0,

`−2^{−2k−2}/(k+1) H_{2k+1} − Σ_{m=1}^{k} 2^{−2k−2}(2^{2m}−2)/(k−m+1) · B_{2m}/(2m)

  • Σ_{j=0}^{2k+1} (−1)^j/2^{2k−j} · C(2k+1, j) H_j ζ(−j) + (1 − 2^{−2k−1}) H_{2k+1} B_{2k+2}/(k+1) = 0`

(eq. (C.12) of the published text), and state: "We confirmed (C.12) up to k = 100 numerically. However, we do not know a proof of (C.12)." Issue #11322 fixes the reading: H_j the harmonic numbers, B_n the Bernoulli numbers with B_1 = −1/2, ζ the Riemann zeta function.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result

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Quantum informationCounterexample foundSource 2018Hayajneh–Hayajneh–Kittaneh: mixed-power monotonicity for unitarily invariant normsTwo positive definite rational matrices reverse the trace-norm comparison at powers one half and three quarters, refuting the proposed monotonicity for all unitarily invariant norms.Refuted

The questionKittaneh, On some classical trace inequalities and a new Hilbert–Schmidt norm inequality, Mathematical Inequalities & Applications 21(4) (2018), 1175–1183, DOI 10.7153/mia-2018-21-81, §1, Conjecture 1, printed p. 1177.

Why it mattersConjecture 1 asks whether these special cases extend to every allowed parameter pair and every such norm.

The ideaThe proof uses spectral estimates.

Our contributionTwo positive definite rational matrices reverse the trace-norm comparison at powers one half and three quarters, refuting the proposed monotonicity for all unitarily invariant norms.

Verification details
Release source recordRefuted in Lean

Open problem

Hayajneh–Hayajneh–Kittaneh: mixed-power monotonicity for unitarily invariant norms

M. Hayajneh, S. Hayajneh and F. Kittaneh, On some classical trace inequalities and a new Hilbert–Schmidt norm inequality, Mathematical Inequalities & Applications 21(4) (2018), 1175–1183, DOI 10.7153/mia-2018-21-81, §1, Conjecture 1, printed p. 1177. See hayajneh2018uinorm.

CONJECTURE 1. Let A be a positive semidefinite matrix and B be a Hermitian matrix. Then for 1/2 ⩽ s ⩽ t ⩽ 1 and for every unitarily invariant norm, we have |||A^s B + BA^{1−s}||| ⩽ |||A^t B + BA^{1−t}|||.

The formal statement retains every quantifier: every natural dimension $n$, all $A,B\in M_n(\mathbb C)$ with $A$ positive semidefinite and $B$ Hermitian, all real $1/2\le s\le t\le1$, and every unitarily invariant norm on $M_n(\mathbb C)$. Real powers are interpreted by continuous functional calculus. No lower spectral bound $A\ge I$ is part of the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/HayajnehConjectureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2021Entanglement assistance raises the communication value by at most the output dimensionShared entanglement increases communication value by at most the output dimension.Proved

The questionJunge, The Communication Value of a Quantum Channel, arXiv:2109.11144v1 (IEEE Trans.

Why it mattersThe communication value measures how well a single use of a channel transmits a classical message in a guessing game; the ceiling of its entanglement-assisted version is the zero-error classical cost of simulating the channel with quantum non-signalling resources.

Our contributionShared entanglement increases communication value by at most the output dimension.

Verification details
Release source recordProved in Lean

Open problem

Entanglement assistance raises the communication value by at most the output dimension

E. Chitambar, I. George, B. Doolittle and M. Junge, The Communication Value of a Quantum Channel, arXiv:2109.11144v1 (IEEE Trans. Inf. Theory 69, 1660–1679, 2023), section on the entanglement-assisted communication value:

\noindent\textit{Conjecture}: For any channel $\mc{N}\in\cptp(A\to B)$, $\cv^*(\mc{N})\leq d_B\cdot \cv(\mc{N})$.

Here:

  • cv(N): the supremum, over all finite families of input states $\rho_x$ and output POVMs ${\Pi_y}$, of $\sum_y\max_xP(y|x)$ with $P(y|x)=\mathrm{tr}[\Pi_y\mathcal N(\rho_x)]$.
  • cv*(N): the same supremum when sender and receiver share a state $\varphi^{A'B'}$ of any dimension, the sender applies channels $\mathcal E_x^{A'\to A}$ and the receiver measures $BB'$.

The definitions are quoted in the literature note.

Issue #20033 fixes the reading with the operational definitions of the source. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/EntanglementAssistedCommunicationValue.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2013Maximal mutual correlation dominates the square root of geometric discord for all two-qubit statesFor every two-qubit density matrix, maximal mutual correlation is at least the square root of normalized geometric discord measured on the first qubit.Proved

The questionJakóbczyk, Spontaneous emission can locally create quantum discord out of classical correlations, arXiv:1306.0130v2, Quantum Inf.

Why it mattersThe source uses CM as a measure of total correlations and DG^A as a measure of quantum correlations.

The ideaWrite xi = tr(ρ σi ⊗ I), yj = tr(ρ I ⊗ σj), Tij = tr(ρ σi ⊗ σj) and Q = T − x yᵀ.

Our contributionFor every two-qubit density matrix, maximal mutual correlation is at least the square root of normalized geometric discord measured on the first qubit.

Verification details
Release source recordProved in Lean

Open problem

Maximal mutual correlation dominates the square root of geometric discord for all two-qubit states

M. Gwóźdź and L. Jakóbczyk, Spontaneous emission can locally create quantum discord out of classical correlations, arXiv:1306.0130v2, Quantum Inf. Process. 13 (2014) 171–183, Section II:

C_{M}(\ro)\geq \sqrt{D_{G}^{A}(\ro)}\geq N(\ro).

The second inequality was shown in Ref. \cite{Adesso} whereas the first one can be proved analytically for Bell-diagonal states and numerical simulations suggest that it is valid for all mixed two-qubit states \cite{J}.

D_G^A is the normalized one-sided geometric discord, d/(d − 1) times the least squared Hilbert–Schmidt distance between ρ and its image under a projective measurement of A. C_M is the maximal mutual correlation, the supremum of |ω_ρ(ab) − ω_ρ(a) ω_ρ(b)| over local elements of norm 1. Both definitions are quoted in the literature note.

Issue #19704 fixes the reading: for every two-qubit density matrix ρ, √(D_G^A(ρ)) ≤ C_M(ρ). The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/GeometricDiscordMutualCorrelationBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2016Maximal mutual correlation dominates the trace-norm geometric discord of two qubitsFor every two-qubit density matrix, one-sided trace-norm geometric discord is at most maximal mutual correlation.Proved

The questionJakóbczyk, Maximal mutual correlation: an algebraic measure of correlations in bipartite quantum systems, arXiv:1607.06585v1, abstract:

Why it mattersThe source proposes CM as an algebraic measure of correlations.

The ideaWrite x for the Bloch vector of A, T for the correlation matrix, and Q = T − x yᵀ for the connected correlation matrix.

Our contributionFor every two-qubit density matrix, one-sided trace-norm geometric discord is at most maximal mutual correlation.

Verification details
Release source recordProved in Lean

arXiv

Maximal mutual correlation dominates the trace-norm geometric discord of two qubits

L. Jakóbczyk, Maximal mutual correlation: an algebraic measure of correlations in bipartite quantum systems, arXiv:1607.06585v1, abstract:

In the states containing some quantum correlations, maximal mutual correlation is always greater or equal to geometric measure of quantum discord. This relation is proved to be valid in the class of Bell - diagonal states and it is conjectured that is also true for general two - qubit states.

The quantities are defined as follows:

  • Trace-norm geometric discord: D₁(ρ) = min over projective measurements 𝒫 on A of ‖ρ − 𝒫(ρ)‖₁, where 𝒫(ρ) = Σ_k (P_k ⊗ I) ρ (P_k ⊗ I).
  • Maximal mutual correlation: C_M(ρ) is the supremum of |ω(ab) − ω(a)ω(b)| over local a, b of operator norm 1.

The definitions are quoted in the literature note; C_M is the quantity of the companion settlement D5/S3/Quantum/Information/GeometricDiscordMutualCorrelationBound.

Issue #19706 fixes the reading: for every two-qubit density matrix ρ, D₁(ρ) ≤ C_M(ρ). This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/TraceNormDiscordMutualCorrelationBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationEquivalence provedFirst preprint 2020Compatibility under every Heisenberg channel equals compatibility in every subspaceCompatibility of a finite POVM family under every isometric compression is equivalent to compatibility under every unital completely positive map.Proved

The questionBrunner, Quantum measurement incompatibility in subspaces, arXiv:2010.04048v2 (Phys.

Why it mattersCompressing an incompatible assemblage to a subspace models measuring a system of bounded dimension; the source shows that some incompatibility disappears in every smaller subspace.

Our contributionCompatibility of a finite POVM family under every isometric compression is equivalent to compatibility under every unital completely positive map.

Verification details
Release source recordProved in Lean

Open problem

Compatibility under every Heisenberg channel equals compatibility in every subspace

R. Uola, T. Kraft, S. Designolle, N. Miklin, A. Tavakoli, J.-P. Pellonpää, O. Gühne and N. Brunner, Quantum measurement incompatibility in subspaces, arXiv:2010.04048v2 (Phys. Rev. A 103, 022203, 2021), section on incompressible incompatibility:

It is worth mentioning that one can modify the concept of incompressible incompatibility by demanding that a measurement assemblage becomes compatible under every Heisenberg channel (a channel preserving identity, but not necessarily trace preserving) to a smaller dimensional system. Although we will leave open the question of whether this provides a strict subset of measurement assemblages that are compatible in every subspace, …

Here:

  • Compatible in every subspace of dimension $n$: every compression ${V^\dagger M_{a|x}V}$ by an isometry $V:\mathbb C^n\to\mathbb C^d$ has a parent POVM.
  • Heisenberg channel: a completely positive unital map $\Phi:M_d\to M_n$, applied to each effect.

The definitions are quoted in the literature note.

Issue #20047 fixes the reading: for each $n\le d$ the two conditions are equivalent, so the subset is not strict. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/HeisenbergChannelCompressionEquivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Compatibility of a finite POVM family under every isometric compression is equivalent to compatibility under every unital completely positive map.

    Read source ↗
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    First preprint year: 2020.

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Analysis & dynamicsCounterexample foundFirst preprint 2010A counterexample to the gamma-gamma NBU assertionThe gamma-gamma family does not universally satisfy the NBU property.Refuted

The questionVuorinen, Functional inequalities for modified Bessel functions, Expo.

Why it mattersA product of two exponential random variables is already a gamma-gamma law, and supplies a counterexample to the universal reliability assertion.

The ideaThis proves the product integrability needed for Fubini and yields the displayed survival identity.

Our contributionThe gamma-gamma family does not universally satisfy the NBU property.

Verification details
Release source recordRefuted in Lean

Open problem

A counterexample to the gamma-gamma NBU assertion

Á. Baricz, S. Ponnusamy and M. Vuorinen, Functional inequalities for modified Bessel functions, Expo. Math. 29(4) (2011), 399–414, DOI 10.1016/j.exmath.2011.07.001, arXiv:1009.4814v2, Question 8, printed p. 13:

Question 8. Is it true that the gamma-gamma distribution satisfies the NBU property?

The preceding definition, printed p. 13, reads:

By definition a life distribution satisfies the new-is-better-than-used (NBU) property if u ↦ log F̄(u) is sub-additive, i.e.

$$\overline F(u_1+u_2)\le\overline F(u_1)\overline F(u_2)$$

for all u₁, u₂ > 0.

The density, §4, printed p. 10, is introduced as follows:

The probability density function fₐ,ᵦ,α : (0, ∞) → (0, ∞) of the three parameter gamma-gamma random variable is defined by (see [21])

$$f_{a,b,\alpha}(u)=\frac{2(ab)^{(a+b)/2}u^{(a+b)/2-1}}{\Gamma(a)\Gamma(b)\alpha^{(a+b)/2}}K_{a-b}!\left(2\sqrt{\frac{ab}{\alpha}u}\right),$$

where a, b > 0 are the distribution shaping parameters, Kν stands for the modified Bessel function of the second kind, and α > 0 is the mean of the gamma-gamma random variable.

Formal reading

Lean names have prefix D5.S3.Analytic.GammaGammaNBU.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Modified Bessel function of the second kind | besselK | DLMF 10.32.9: $\int_0^\infty e^{-z\cosh t}\cosh(\nu t),dt$, for real $\nu$ and $z>0$. | | Source density | density | The displayed density with Real.Gamma, real powers and nonnegative square root. | | Survival function | survival | The Lebesgue integral of the density over $(u,\infty)$. | | Universal NBU assertion | claim | Every $a,b,\alpha>0$ and every $u_1,u_2>0$. | | Refutation | result | Exactly $\neg\mathrm{claim}$. |

Only positive arguments occur in the counterexample. The total real functions outside the source domains impose no additional hypotheses and do not change the universally quantified positive-domain assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/GammaGammaNBU/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 1996Fuchs's physical-basis logarithm questionA faithful qutrit pair refutes a physical-basis diagonalization of Fuchs's logarithm argument.Refuted

The questionFuchs, Distinguishability and Accessible Information in Quantum Theory, arXiv:quant-ph/9601020v1, §3.5.3, pp. 77–78, asks after equation (3.244):

Why it mattersA physical change of basis acts on the original three-dimensional Hilbert space; its induced action on matrix space is constrained.

The ideaThe proof expands to existing matrix/CFC identities, finite entry calculations, algebraic normalization and their consumed logical consequences.

Our contributionA faithful qutrit pair refutes a physical-basis diagonalization of Fuchs's logarithm argument.

Verification details
Release source recordRefuted in Lean

arXiv

Fuchs's physical-basis logarithm question

C. A. Fuchs, Distinguishability and Accessible Information in Quantum Theory, arXiv:quant-ph/9601020v1, §3.5.3, pp. 77–78, asks after equation (3.244):

An interesting open question is whether there exists a basis in which to represent the operators ρ̂₀, ρ̂₁, etc., so that the matrix appearing in the logarithm above is diagonal. If so, this would greatly simply the computation of L(t).

The source note identifies the version and equations. Preregistration #18588 fixes the literal Lean conventions. The universal physical-basis assertion is Refuted by D5.S3.Quantum.Foundation.FuchsLogarithmBasisRefutation.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Kronecker sum, equations (3.237), (3.242) | ksum | $I\otimes X+\overline X\otimes I$, with the preregistered column-vectorization order | | Matrix inside the logarithm, equation (3.244) | logArg | $I+t\widetilde\rho_0^{-1/2}\widetilde\Delta\widetilde\rho_0^{-1/2}$; inverse of CFC.sqrt on both sides | | Density operators | d, rho0, rho1, positive-definiteness and trace-one hypotheses in claim | Faithful finite-dimensional states; a counterexample here refutes a universal claim covering these states | | Invertible difference in the source derivation | (rho1-rho0).det ≠ 0 in claim | Preserved, with witness determinant $-1/4608$ | | A basis in which to represent both operators | U ∈ Matrix.unitaryGroup (Fin d) ℂ, and both conjugations by $U^*\rho U$ | A physical basis induces $\overline U\otimes U$ on matrix space | | Source's basis independent of $t$ | ∀ t, 0 < t → t ≤ 1 → ∃ U, … | Permitting a separate basis at each $t$ weakens the positive assertion; its negation at $t=1$ also refutes a single basis for all $t$ | | Diagonal logarithm argument | Matrix.IsDiag in claim | The entire matrix, in the induced basis | | Negative answer | result : ¬ claim | Negation of the whole quantified assertion |

The explicit dimension, faithfulness, trace, determinant and interval binders are the preregistered finite density-state interpretation. The construction and every binder are retained literally. The proof uses the frozen PathSpectralClassical.relativeCongruence sandwich and EntropyProductionCoherenceDeletionIdentity.cfc_diagonal for square roots of diagonal matrices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Foundation/FuchsLogarithmBasisRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    First preprint year: 1996. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2007Refutation of the sum-condition Luke lower boundA convergent hypergeometric series contradicts the sum-condition Luke lower bound.Refuted

The questionSitnik, Inequalities and monotonicity of ratios for generalized hypergeometric function, J.

Why it mattersThe sum condition allows one denominator parameter to be smaller than its numerator partner.

The ideaGeometric comparison proves absolute convergence and monotonicity of the magnitudes.

Our contributionA convergent hypergeometric series contradicts the sum-condition Luke lower bound.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the sum-condition Luke lower bound

D. Karp and S. M. Sitnik, Inequalities and monotonicity of ratios for generalized hypergeometric function, J. Approx. Theory 161 (2009), 337–352, DOI 10.1016/j.jat.2008.10.002, arXiv:math/0703084v2, Conjecture 1, printed p. 14:

Theorem 3 is true for all σ > 0 and ∑ᵢ₌₁ᵠ (bᵢ − aᵢ) > 0.

Theorem 3, printed p. 6, equation (23):

For bₖ > aₖ > 0, k = 1, . . . , q, x > −1 and σ ≥ 1 the inequality

$\displaystyle\frac{1}{(1+x\prod_{i=1}^q(a_i/b_i))^\sigma}<{}_{q+1}F_q(\sigma,(a_q);(b_q);-x)$

holds true.

Formal reading

Lean names below have prefix D5.S3.Analytic.KarpSitnikLuke.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Rising-factorial product ratio | D5.S3.Analytic.HypergeometricShiftTuran.TwoBlockResult.pochRatio | Existing quotient of upper and lower rising-factorial products. | | Summand of (1) | hyperTerm | Extra upper parameter σ, factorial denominator and power zⁿ. | | Function inside the convergence disk | hyper | Real infinite sum of these summands. | | Restricted Conjecture 1 | claim | All positive vectors with positive sum(b−a), σ>0, −1<x<1, positive real-power base and summability. | | Refutation | result | Exactly ¬ claim. |

Restricting to the power-series disk and adding summability and positive base assumptions gives a weaker assertion than the conjecture with analytic continuation on x>−1. Its refutation refutes the conjecture. For q=0 the positive-sum assumption is impossible.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KarpSitnikLuke/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The refutation is supported by a documented parameterized argument or family.

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Quantum informationCounterexample foundFirst preprint 2022A stochastic qutrit channel that is not unitalA completely positive trace-preserving qutrit map has the identity vector as a Choi eigenvector with eigenvalue one half, but does not preserve the identity matrix.Refuted

The questionWallman, Designing Stochastic Channels, arXiv:2201.07156v1, Section II, ask:

Why it mattersThe source supplies a nonunital example at dimension four and every even dimension greater than two, and Appendix A proves unitality at dimension two.

The ideaThe proof uses spectral estimates.

Our contributionA completely positive trace-preserving qutrit map has the identity vector as a Choi eigenvector with eigenvalue one half, but does not preserve the identity matrix.

Verification details
Release source recordRefuted in Lean

arXiv

A stochastic qutrit channel that is not unital

M. Graydon, J. Skanes-Norman and J. J. Wallman, Designing Stochastic Channels, arXiv:2201.07156v1, Section II, ask:

We leave open the question of whether or not all stochastic channels are unital for d = 3.

Eq. (2) defines the normalized Choi state $J(\Phi)=d^{-1}(\Phi\otimes\mathrm{id})(\operatorname{col}(I_d) \operatorname{col}(I_d)^\dagger)$. Eq. (11) requires that $\Phi$ be completely positive and trace preserving and that $J(\Phi)\operatorname{col}(I_d)=\lambda\operatorname{col}(I_d)$ for some real $0<\lambda\leq1$. Unitality means $\Phi(I_d)=I_d$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Complex-linear maps on qutrit matrices | claim quantifies over MatrixMap (Fin 3) (Fin 3) ℂ | Exact dimension and carrier | | CPTP | Frozen CoPRelativeQuantumnessRefutation.IsCPTP within IsStochastic | Complete positivity and trace preservation on every matrix | | Some real $0<\lambda\leq1$ | The existential real parameter and both inequalities in IsStochastic | No endpoint or sign clause dropped | | Normalized Choi state and identity vector | Their preregistered bodies are inlined in IsStochastic | Output-first Choi coordinates; vector entries are one exactly on equal indices | | Every stochastic channel is unital | The implication in claim concludes $\Phi(I_3)=I_3$ | Literal universal question | | Refutation | result has type ¬ claim | A counterexample suffices |

The definition also admits dimension zero as a total Lean expression; the question and its refutation use dimension three. Proofs use the existing ProductUnitaryChoiSpectrum.omega and PassiveMemoryNonclassicalRefutation.choi, with correspondence by unfolding.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/QutritStochasticNonunitalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023A binary-stabilized six-qubit state that is not locally a stabilizer stateDakic (arXiv:2309.09815) that every state stabilized by binary operators and the identity is locally equivalent to a standard stabilizer state.Refuted

The questionA 57, 455301 (2024)) call a state of N qubits stabilized by a stabilizing set 𝒜 when it is, up to a complex factor, the unique common +1 eigenvector of finitely many tensor products of elements of 𝒜.

The ideaLet v have coefficient 1 on the six labels of Hamming weight one, −1 on the six of weight five and 0 elsewhere, and take O1 = (−X) ⊗ X^{⊗5}, O2 = (−Z) ⊗ Z^{⊗5}, O3 = H^{⊗6}, O4 = K^{⊗6} with H = (X+Z)/√2 = A{π/4,0} and K = (X+Y)/√2 = A{π/2,π/4}.

Our contributionDakic (arXiv:2309.09815) that every state stabilized by binary operators and the identity is locally equivalent to a standard stabilizer state.

Verification details
Release source recordRefuted in Lean

Open problem

A binary-stabilized six-qubit state that is not locally a stabilizer state

É. Descamps and B. Dakić (arXiv:2309.09815, quant-ph; J. Phys. A 57, 455301 (2024)) call a state of N qubits stabilized by a stabilizing set 𝒜 when it is, up to a complex factor, the unique common +1 eigenvector of finitely many tensor products of elements of 𝒜. Standard stabilizer states are those stabilized by 𝒫 = {±1, ±i}·{𝟙, X, Y, Z}. After an exhaustive analysis of two and three qubits the paper states:

\begin{conj} All states stabilized by the set $\mathcal A$ composed of binary
operators and the identity {\it i.e.}, $\mathcal{A}=\{A_{\theta,\phi},\1_2\}$,
are locally equivalent to standard stabilizer states (\textit{i.e.},
stabilized by the Pauli group
$\mathcal{P}=\{\pm 1,\pm i\}\cdot\{\1,\sigma_X,\sigma_Y,\sigma_Z\}$).
\end{conj}

Issue #11565 fixes the reading. A_{θ,φ} = cos θ Z + sin θ (cos φ X + sin φ Y), stabilization needs no commutativity, and local equivalence is ψ = (U_1 ⊗ ⋯ ⊗ U_N) φ with U_i ∈ U(2). The formal claim is the conjecture for every N; the result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/BinaryStabilizerLocalInequivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Existing proof account records a substantive content argument.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023Nonnegative Chebyshev expansions of snake polynomialsAn even convex majorant has a degree-six snake with a negative Chebyshev coefficient.Refuted

The questionKryvonos (ed.), Open problems on polynomials, their zero distribution and related questions: 2023, arXiv:2312.07754v1, section “Snake polynomials with positive Chebyshev expansions”, states:

Why it mattersNonnegative Chebyshev expansion is the hypothesis under which Nikolov–Shadrin, arXiv:1210.7795, prove extremality of the snake polynomial in Markov–Duffin–Schaeffer inequalities with a majorant.

The ideaThe maximum of the constant convex function and the positively scaled absolute-value function with an added constant is convex.

Our contributionAn even convex majorant has a degree-six snake with a negative Chebyshev coefficient.

Scope & next questionUP24 Problem 7.1, asking for classes with nonnegative or sign-alternating expansions, remains open.

Verification details
Release source recordRefuted in Lean

arXiv

Nonnegative Chebyshev expansions of snake polynomials

L. Kryvonos (ed.), Open problems on polynomials, their zero distribution and related questions: 2023, arXiv:2312.07754v1, section “Snake polynomials with positive Chebyshev expansions”, states:

Consider a majorant µ ∈ C[−1, 1], µ(x) ≥ 0, x ∈ [−1, 1]. If there exists P ∈ Pn, P ≠ 0, such that −µ(x) ≤ P(x) ≤ µ(x), x ∈ [−1, 1], then there exists a unique (up to orientation) polynomial ωµ ∈ Pn called snake polynomial associated with µ, which oscillates most between ±µ.

The n-th snake polynomial ωµ is uniquely determined by the following properties: a) |ωµ(x)| ≤ µ(x) x ∈ [−1, 1]; b) There exists a set δ* = (τi*)_{i=0}^n, 1 ≥ τ0* > · · · > τn* ≥ −1, such that ωµ(τi*) = (−1)^i µ(τi*), i = 0, . . . , n

Conjecture (G.N.) If µ ≥ 0 is a continuous even convex function in [−1, 1], then the associated with µ snake polynomials have non-negative expansion in the Chebyshev polynomials of the first kind.

The assertion is repeated as Conjecture 7.1 (Geno Nikolov) in Open problems UP24, arXiv:2504.04845, §7. Here $\mathcal P_n$ denotes polynomials of degree at most $n$, and $T_k$ denotes the first-kind Chebyshev polynomial.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Chebyshev/SnakeChebyshevExpansionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2020The exact triplewise joint-measurability threshold of biased trine qubit measurementsProved

The questionKunjwal, Qualitative equivalence between incompatibility and Bell nonlocality, arXiv:2008.10100v3; Phys.

Why it mattersThe paper studies which quantum realizations of a joint measurability structure can violate a Bell inequality.

The ideaUnder the same joint-measurability hypothesis, necessitycertificate proves

Scope & next questionThe settlement supplies the previously conjectural triplewise ingredient; it does not formalize the pairwise criterion. - Neighbouring questions and Bell nonlocality.

Verification details
Release source recordProved in Lean

Open problem

The exact triplewise joint-measurability threshold of biased trine qubit measurements

S. A. Yadavalli, N. Andrejic and R. Kunjwal, Qualitative equivalence between incompatibility and Bell nonlocality, arXiv:2008.10100v3; Phys. Rev. A 110, L060201 (2024), Supplemental Material, Section III, Example 1. Its TeX source qualequiv_supp_arxiv.tex, lines 151–158, states:

Based on our numerical evaluation and visual inspection, we conjecture the exact form of $\eta_1^{(3)}(b)$ to be $$\eta_1^{(3)}(b)=\begin{cases} 1+b,\quad -1\leq b<-\frac{1}{3},\ \frac{2}{3},\quad b\in\left[-\frac{1}{3},\frac{1}{3}\right],\ 1-b,\quad \frac{1}{3}< b\leq 1. \end{cases}$$

The family is defined in lines 112–118, and the threshold is defined in lines 147–150 by triplewise joint measurability if and only if $\eta\leq\eta_1^{(3)}(b)$. For $k\in{1,2,3}$ and $x\in{-1,+1}$, its effects on $\mathbb C^2$ are

$$E_k(x)=\frac12\big((1+xb)I+\eta x,\vec n_k\cdot\vec\sigma\big),\qquad \vec n_k=\cos\frac{2k\pi}{3}\vec e_x+\sin\frac{2k\pi}{3}\vec e_z.$$

The parameter domain is $-1\leq b\leq1$, $0\leq\eta\leq1$ and $|b|\leq1-\eta$. Triplewise joint measurability means the existence of eight positive semidefinite complex $2\times2$ matrices $G_s$, indexed by $s\in{-1,+1}^3$, such that $\sum_sG_s=I$ and $\sum_{s:s_k=x}G_s=E_k(x)$ for every $k$ and $x$.

Issue #22692 fixes the reading: for every real $b,\eta$ in this domain, these three measurements are triplewise jointly measurable if and only if $\eta\leq\eta_1^{(3)}(b)$ for the displayed piecewise function. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/BiasedTrineJointMeasurabilityThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Discrete structuresCounterexample foundPosed in 2011Refutation of Detlefs's A008365 four-residue characterizationThe prime 17 refutes Detlefs's four-residue characterization of 13-rough numbers.Refuted

The question13-rough numbers: positive integers that have no prime factors less than 13.

Why it mattersA single positive 13-rough value outside those classes refutes the universal biconditional.

The ideaThe theorem Nat.Prime.eqoneorselfofdvd then shows that every prime divisor of 17 is 17, establishing isRough13(17).

Our contributionThe prime 17 refutes Detlefs's four-residue characterization of 13-rough numbers.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Detlefs's A008365 four-residue characterization

OEIS A008365, NAME (verbatim):

13-rough numbers: positive integers that have no prime factors less than 13.

Gary Detlefs's COMMENT of December 30, 2011 (verbatim):

Conjecture: Numbers n such that n^24 is congruent to {1,421,631,841} mod 2310. - Gary Detlefs, Dec 30 2011

For natural numbers, define isRough13(n) := for every natural p, Prime(p) implies p divides n implies 13 <= p, and define inResidueSet(n) := n^24 mod 2310 is one of {1,421,631,841}. The literal refuted statement is claim := for every natural n, 0<n implies (isRough13(n) iff inResidueSet(n)).

The five-residue correction {1,421,631,841,1681} is disclosed but not claimed. Its structural reason is that 2310 = 235711: for a value coprime to 2310, the twenty-fourth power is congruent to 1 modulo 235*7, while modulo 11 it equals the fourth power and takes the five quartic residues {1,3,4,5,9}; 1681 is congruent to 9 modulo 11. The converse inclusion from the original four-residue set to 13-rough values is also not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/DetlefsRoughResidueCharacterizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2011. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedPosed in 2002Refutation of the A018800 least-prime prefix suffix-bound conjectureThe value a(1) = 11 refutes Murthy's suffix bound for OEIS A018800.Refuted

The questionConjecture: If a(n) = (n concatenated with k) then k < n. - Amarnath Murthy, May 01 2002

Why it mattersMurthy's 2002 comment states a universal strict bound on every appended suffix.

The ideaThe number 11 belongs to the defining prefix-prime set for n = 1: it is prime and satisfies 10 <= 11 < 20 with m = 1.

Our contributionThe value a(1) = 11 refutes Murthy's suffix bound for OEIS A018800.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A018800 least-prime prefix suffix-bound conjecture

OEIS A018800, NAME (verbatim):

Smallest prime that begins with n.

COMMENTS conjecture line (verbatim; Amarnath Murthy, May 01 2002):

Conjecture: If a(n) = (n concatenated with k) then k < n. - Amarnath Murthy, May 01 2002

The AUTHOR line is:

David W. Wilson

The sequence begins 11, 2, 3, 41, 5, 61, 7, 83, 97, 101, 11, 127, 13, 149, 151, 163, 17, 181, 19, 2003, 211, .... For a natural number n, the formal sequence value is a(n) = sInf {p : Nat | Prime(p) and exists m : Nat, n * 10^m <= p and p < (n + 1) * 10^m}. The interval is the decimal-prefix reading of "begins with n". The literal refuted statement is for all n >= 1, m >= 1, k < 10^m, a(n) = n * 10^m + k -> k < n. The condition m >= 1 discloses the nonempty-suffix reading of "n concatenated with k".

The existence of a(n) in general, the FORMULA line a(n) = prime(A085608(n)), and an equivalence between the interval encoding and a Nat.digits formulation are not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/MurthyLeastPrimePrefixSuffixRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument gives a structural characterization or normal form.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2002. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundRefutation of the A129598 differing-position characterizationThe value 60 refutes Karttunen's A089966 characterization of A129598 differences.Refuted

The questionAntti Karttunen's COMMENT of May 1, 2007 (verbatim):

Why it mattersA single differing position outside A089966 refutes this literal universal characterization.

The ideaThe A129598 value is a(60)=605=300, but g(300)=gcd(300,phi(300))=20, so a(60) is not a preimage of 60 under g.

Our contributionThe value 60 refutes Karttunen's A089966 characterization of A129598 differences.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A129598 differing-position characterization

OEIS A129598, NAME (verbatim):

a(n) = n * A111089(n).

Antti Karttunen's COMMENT of May 1, 2007 (verbatim):

Conjecture: differs from A050399 at the positions given by A089966. E.g., a(15)=75, instead of A050399(15)=225, a(30)=150, instead of A050399(30)=450, a(33)=363, instead of A050399(33)=1089, a(45)=225, instead of A050399(45)=1225.

The NAME of A050399 states (verbatim):

Least k such that n = A009195(k) (= gcd(phi(k), k)).

The NAME of A089966 states (verbatim):

Numbers k such that omega(k) = gpf(k) mod lpf(k), where omega=A001221, gpf=A006530 and lpf=A020639.

For natural numbers, define g(m) := gcd(m, phi(m)). Let greatestPrimeFactor(n) be the last element of the prime-factor list, with default zero, and define a(1):=2 and a(n):=n*greatestPrimeFactor(n) otherwise. Define b(n) to be the least positive m satisfying g(m)=n when such an m exists, and zero otherwise. Finally, define inA089966(n) to mean n=1, or that n is positive and the number of its distinct prime factors equals greatestPrimeFactor(n) mod minFac(n). The literal refuted statement is for every natural n, 1<n implies ((a(n) != b(n)) iff inA089966(n)).

Karttunen's separate Conjecture 2, that a(n) divides A050399(n), is not claimed or assessed. No exact value or minimality statement for b(n) is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KarttunenGcdTotientPositionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2012Bergman's Question 15Distinct nonzero sixth powers of three semigroup elements do not prevent a nonzero complex vector in their contracted span from having zero sixth power.Refuted

The questionBergman, Thoughts on Eggert's Conjecture, Question 15, asks:

Why it mattersDistinct nonzero pure powers need not prevent cancellation among mixed monomials.

The ideaKeep every positive exponent triple of degrees 1 through 5 as a singleton.

Our contributionDistinct nonzero sixth powers of three semigroup elements do not prevent a nonzero complex vector in their contracted span from having zero sixth power.

Verification details
Release source recordRefuted in Lean

Open problem

Bergman's Question 15

Bergman, Thoughts on Eggert's Conjecture, Question 15, asks:

Suppose X is a finite subset of a commutative semigroup S with zero, n a positive integer such that the n-th power map is one-to-one on X and does not take any nonzero element of X to 0, and F a field of characteristic 0. Must every nonzero element of the span FX of X in F_0 S have nonzero n-th power?

The semigroup is associative and commutative with absorbing zero, with no identity or finiteness assumption. The field is arbitrary of characteristic zero. The subset may be empty or contain zero. The algebra is the free vector space on the nonzero semigroup elements, with the semigroup zero sent to the zero vector.

| Source clause | Lean meaning | | --- | --- | | arbitrary commutative semigroup with zero | S : Type, SemigroupWithZero S, IsMulCommutative S | | characteristic-zero field | F : Type, Field F, CharZero F | | finite subset | X : Set S, X.Finite | | positive exponent | index n : ℕ represents exponent n+1 | | positive power without identity | positivePower a 0 = a, followed by repeated multiplication by a | | injective power map on X | Set.InjOn (fun s => positivePower s n) X | | nonannihilation on nonzero X | ∀ s ∈ X, s ≠ 0 → positivePower s n ≠ 0 | | contracted algebra | Contracted F S = {s : S // s ≠ 0} →₀ F | | contracted zero | delta F S 0 = 0 | | linear span | Submodule.span F (delta F S '' X) | | claimed conclusion | every nonzero vector in this span has nonzero positive power |

The utility claim is closed over ordinary small carriers (Type), including arbitrary infinite semigroups and fields. The algebra definitions remain universe-polymorphic. The finite counterexample lies in these universes and therefore also refutes the unrestricted mathematical question; no finite-carrier restriction is imposed on the universal claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Algebra/BergmanQuestion15Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Discrete structuresConjecture refutedFirst preprint 2020A circulant counterexample to Davies–Marigliano Conjecture 4.4A graph on thirty vertices has a cyclic symmetry, at least eight orbits of unordered vertex pairs under its full automorphism group, and at most seven distinct real adjacency eigenvalues.Refuted

The questionIsobel Davies and Orlando Marigliano, Coloured Graphical Models and their Symmetries, arXiv:2012.01905v5, Section 4, Conjecture 4.4, PDF page 11:

The ideaTranslation by one is a cyclic graph symmetry.

Our contributionA graph on thirty vertices has a cyclic symmetry, at least eight orbits of unordered vertex pairs under its full automorphism group, and at most seven distinct real adjacency eigenvalues.

Verification details
Release source recordRefuted in Lean

arXiv

A circulant counterexample to Davies–Marigliano Conjecture 4.4

Isobel Davies and Orlando Marigliano, Coloured Graphical Models and their Symmetries, arXiv:2012.01905v5, Section 4, Conjecture 4.4, PDF page 11:

If $G$ is an uncoloured circulant graph, then the number of orbits of the action of $\operatorname{Aut}(G)$ on the set $V*V$ of unordered pairs of vertices of $G$ coincides with the number of distinct eigenvalues of its adjacency matrix.

The preceding definition says: “By definition, these are graphs that have a symmetry which is a cyclic permutation of its vertices.” Corollary 3.3, page 7, uses the entrywise action of the full automorphism group; its count $\binom{n+1}{2}$ includes diagonal pairs. The source's adjacency matrix is the real symmetric zero-one matrix of the simple undirected graph. Preregistration: #21260.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every uncoloured simple undirected graph of positive order | claim quantifies over every natural number and every SimpleGraph (Fin n); positivity is part of IsCirculant | No connectivity assumption is added | | A symmetry which is a cyclic permutation of all vertices | IsCirculant uses a graph self-isomorphism whose permutation is Equiv.Perm.IsCycleOn on Set.univ | The whole vertex set is used | | Unordered pairs, including repetitions | pairOrbits uses Sym2 (Fin n) | Diagonal pairs remain | | Entrywise action of the full automorphism group | Each orbit is the range of Sym2.map over every graph self-isomorphism; the outer range collects distinct orbit sets | No subgroup substitution | | Number of distinct adjacency eigenvalues | eigenvalueCount counts Module.End.Eigenvalues of the real adjacency endomorphism with Nat.card | Scalar values are counted without multiplicity | | Equality for every circulant | claim; result proves its negation | No source clause is dropped |

The public definitions retain the literal preregistered expressions. Proofs use Mathlib's common-neighbour and circulant-matrix constructions directly, with the finite adjacency correspondence established in intMatrix_adj.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hamming/DaviesMariglianoCirculantOrbits.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Further frontiersConjecture refutedFirst preprint 2020Positive joint weights and all-vertex physical weights in the tree Lee–Yang conjecturePositive joint split weights refute the tree Lee-Yang conjecture.Refuted

The questionRios Fukelman, The planar limit of N=2 superconformal quiver theories, JHEP 08 (2020) 161, arXiv:2006.06379v2, Section 2.2, Conjecture 1:

Why it mattersThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionPositive joint split weights refute the tree Lee-Yang conjecture.

Scope & next questionIt does not settle arbitrary other total multisets, other trees, other tau, the full tree ensemble, Conjecture 2, the all-tau family, or priority.

Verification details
Release source recordRefuted in Lean

Open problem

Positive joint weights and all-vertex physical weights in the tree Lee–Yang conjecture

B. Fiol, J. Martínez-Montoya and A. Rios Fukelman, The planar limit of N=2 superconformal quiver theories, JHEP 08 (2020) 161, arXiv:2006.06379v2, Section 2.2, Conjecture 1:

For any tree with $e$ edges, any fixed positive integers $n_1,\dots,n_e$ and arbitrary $\rho(k_i,n_i)>0$ the polynomials $P(\tau,x,n_1,\dots,n_e)$ have all their $x$ roots on the unit circle.

Equation (2.30) sums over positive edge splits $1\le k_i\le n_i-1$ and all Boolean spin states. The mass at a vertex is the sum of the integers assigned to its incident endpoints. A state contributes $\tau^{e_\pm}x^{\sum_{v\text{ up}}m_v}$, where $e_\pm$ counts opposite-spin edges. The Lee–Yang parameter range is $-1\le\tau\le1$. The source's one-edge case is proved through the Lakatos–Losonczi criterion.

The source sentence at quivers_n_2.tex lines 343–345 makes $\rho$ a distribution on the whole split configuration. Thus Conjecture 1 permits an arbitrary positive joint function; it does not impose edgewise factorization. The physical specialization in (2.31) includes vertex factors $\mathcal V_i$ that depend jointly on the incident splits, as defined at source lines 198–203.

Statement echo

The public definitions are splits, mass, treePoly and claim. The tree condition is the frozen D5.S3.Observer.AgencyHolonomy.NamedTreePhaseTransport.NamedSpanningTree with Set.univ, selecting every edge: its nonloop, distinct and isTree fields state exactly the finite-type tree convention. The subtype of all edges is in bijection with the edge type, so its injectivity and support conditions are the same as the source convention.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/LeeYang/QuiverTreeIsingRefutation.result

D5/S3/Zeros/LeeYang/QuiverTreePhysicalE8.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2017The upper binegativity bound for two qubitsGirard and Gour (arXiv:1701.02724) conjecture that the binegativity of every two-qubit state is at most nu (c + nu)^2 / (2 (c^2 + nu^2)), where nu is the negativity and c the concurrence.Refuted

The questionGour, The binegativity of two qubits, arXiv:1701.02724v3 (2017), Eq. (9):

Why it mattersThe binegativity is computable from the spectral decomposition of the partial transpose, vanishes exactly on separable two-qubit states, and is conjectured by the authors to be an entanglement monotone.

The ideaThe proof uses spectral estimates and convexity.

Our contributionGirard and Gour (arXiv:1701.02724) conjecture that the binegativity of every two-qubit state is at most nu (c + nu)^2 / (2 (c^2 + nu^2)), where nu is the negativity and c the concurrence.

Verification details
Release source recordRefuted in Lean

Open problem

The upper binegativity bound for two qubits

M. W. Girard, G. Gour, The binegativity of two qubits, arXiv:1701.02724v3 (2017), Eq. (9):

In general, based on numerical evidence from randomly generated states, we conjecture that the the binegativity of two-qubit states is bounded by $\nu\frac{(c+\nu)(\nu+1)}{(c+\nu)^2+2c(1-c)}\leq N_2(\sigma)\leq \frac{\nu}{2}\frac{(c+\nu)^2}{c^2+\nu^2}$ for all states $\sigma$ with fixed negativity $N(\sigma)=\nu$ and concurrence $C(\sigma)=c$.

Here $N(\sigma)=2\operatorname{Tr}[(\sigma^\Gamma)-]$ and $N_2(\sigma)=\operatorname{Tr}[(\sigma^\Gamma)-]+2\operatorname{Tr}[ (((\sigma^\Gamma)-)^\Gamma)-]$, with $X_\pm$ the positive and negative components of a self-adjoint $X$ and $C$ the two-qubit concurrence. The target is the upper inequality; it is a preprint conjecture (no journal version is listed).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TwoQubitBinegativityUpperBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2012Refutation of reciprocal Gamma coefficient positivityRefutation of reciprocal Gamma coefficient positivity.Refuted

The questionKarp, Log-concavity for series in reciprocal gamma functions and applications, arXiv:1206.4814v1, Conjecture 1:

Why it mattersThe constant unit sequence satisfies every sequence hypothesis.

The ideaThe formal proof uses Euler reflection and the strict bound

Our contributionRefutation of reciprocal Gamma coefficient positivity.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of reciprocal Gamma coefficient positivity

S. I. Kalmykov and D. B. Karp, Log-concavity for series in reciprocal gamma functions and applications, arXiv:1206.4814v1, Conjecture 1:

Under hypotheses of Theorem 2 the function µ → g(µ, x) is log-concave on (0, ∞) for each fixed x ≥ 0. Moreover, the function x → g(µ + α, x)g(µ + β, x) − g(µ, x)g(µ + α + β, x) has positive power series coefficients for µ ≥ −1 and µ + α ≥ 0, µ + β ≥ 0, where α, β > 0.

Theorem 2's sequence hypotheses are nontriviality, nonnegativity, log-concavity $g_{n-1}g_{n+1}\le g_n^2$ for $n\ge1$, and no internal zeros. Equation (20) is

$$g(\mu,x)=\sum_{n=0}^{\infty}\frac{g_n x^n}{\Gamma(n+\mu)},$$

where reciprocal Gamma is zero at its poles. Only the “Moreover” clause is refuted; the first assertion is not addressed.

Formal reading

All Lean names below have prefix D5.S3.Analytic.ReciprocalGammaTuran.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $1/\Gamma(s)$ | Defs.rgamma | Inverse real Gamma, zero at nonpositive integers. | | Theorem 2's sequence hypotheses | Defs.AdmissibleSeq | Nontrivial, nonnegative, log-concave, without internal zeros. | | Equation (20)'s formal series | Series.formalSeries | Degree $n$ coefficient $g_n/\Gamma(n+\mu)$, with no factorial weight. | | Degree $m$ Turanian coefficient | Defs.turanCoeff, Series.coeff_mul | Exact finite Cauchy sum over $0\le k\le m$. | | Moreover clause | Defs.claim | Every admissible sequence and all real shifts in the printed range, at every natural degree. | | Boundary failure | Boundary.d_two_neg | For every $0<t<1$, degree two is negative at $(-t,t,t)$. | | Interior failure | Interior.interior_failure | Positive shifted parameters and a negative second coefficient. | | Refutation | Result.result | Exactly $\neg,\texttt{Defs.claim}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/ReciprocalGammaTuran/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsConjecture refutedFirst preprint 2016The function clause of the shifted Gamma Turanian conjectureA convergent-series counterexample to the function clause of Conjecture 1.Refuted

The questionKarp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, Conjecture 1, page 6, states verbatim:

Why it mattersA negative Taylor coefficient does not imply a negative value of the function.

Our contributionA convergent-series counterexample to the function clause of Conjecture 1.

Verification details
Release source recordRefuted in Lean

Open problem

The function clause of the shifted Gamma Turanian conjecture

S. I. Kalmykov and D. B. Karp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, Conjecture 1, page 6, states verbatim:

Conjecture 1. All conclusions of Theorem 3 hold for all α, β ≥ 0 if µ ≥ 1.

The relevant branch of Theorem 3, page 5, states verbatim (line breaks joined):

Theorem 3. If p ≤ q and conditions (13) are satisfied, then ∆f (α, β; x) ≥ 0 for x ≥ 0, µ, β ≥ 0 and α ∈ N, where ∆f (α, β; x) is defined in (9) with f from (10). Moreover, if α ≤ β + 1, then δm ≥ 0 for all m ∈ N0.

For positive parameter vectors with $p\le q$, equations (9) and (10) are

$$ c(t)=\frac{\prod_i\Gamma(a_i+t)}{\prod_j\Gamma(b_j+t)},\qquad f(\mu;x)=\sum_{n=0}^\infty c(\mu+n)\frac{x^n}{n!}, $$

$$ \Delta_f(\alpha,\beta;x)=f(\mu+\alpha;x)f(\mu+\beta;x) -f(\mu;x)f(\mu+\alpha+\beta;x). $$

Condition (13) is

$$ \frac{e_{q-j}(\mathbf b)}{e_{p-j}(\mathbf a)} \le\frac{e_{q-j-1}(\mathbf b)}{e_{p-j-1}(\mathbf a)}\quad(0\le j<p). $$

This dossier concerns $\Delta_f\ge0$ for all $x\ge0$, $\mu\ge1$ and $\alpha,\beta\ge0$. It has no extra $\alpha\le\beta+1$ restriction. Section 2, page 3, says (source text p.txt, line 167, line breaks joined):

We will always assume that a₂ is not empty (i.e. p₂ ≥ 1), while all other subvectors are allowed to be empty.

It then says:

In this section we will consider log-convexity of the function f (µ; x) defined in (3) under these assumptions.

Section 3 sets $a_1,b_1$ empty, identifying $a=a_2$ in (10). The first sentence admits a standing-scope reading; the next admits a section-local reading. The source-scope ambiguity is unresolved. The encoded claim imposes $1\le p$. Its nonempty-numerator counterexample refutes both that restricted universal claim and the broader claim allowing $p=0$.

| Source object | Lean name in D5.S3.Analytic.HypergeometricShiftTuran | | --- | --- | | Gamma product ratio | frozen Defs.coeff | | Condition (13) | frozen Defs.cond13, using Defs.esymm | | Equation (10), real infinite sum | FunctionResult.fval | | Equation (9) | FunctionResult.turanian | | Universal function clause | FunctionResult.claim | | Its negation | FunctionResult.result |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/HypergeometricShiftTuran/FunctionResult.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2020The Shapley allocation of a graph energy game can leave its coreComplete bipartite graphs refute Shapley core membership for graph energy games.Refuted

The questionGerardo Arizmendi and Octavio Arizmendi, The graph energy game, arXiv:2012.14114v2, Conjecture 3.10, page 8:

Why it mattersEfficiency then conflicts with the lower payoff bound of the coalition excluding that vertex.

Our contributionComplete bipartite graphs refute Shapley core membership for graph energy games.

Verification details
Release source recordRefuted in Lean

Open problem

The Shapley allocation of a graph energy game can leave its core

Gerardo Arizmendi and Octavio Arizmendi, The graph energy game, arXiv:2012.14114v2, Conjecture 3.10, page 8:

For the energy game, the Shapley value is in the core.

The source defines the energy of a finite simple undirected graph as the sum of absolute adjacency eigenvalues, with multiplicity. Its game has $w_G(S)=\mathcal E(G[S])$. For each vertex $i$ its allocation is

$$ \phi_i(w_G)=\sum_{S\subseteq V\setminus{i}} \frac{|S|!(|V|-|S|-1)!}{|V|!} \bigl(w_G(S\cup{i})-w_G(S)\bigr). $$

The core consists of all $x$ satisfying $\sum_{i\in S}x_i\ge w_G(S)$ for every coalition $S$ and $\sum_{i\in V}x_i=w_G(V)$. The conjecture quantifies over every finite simple graph, without connectedness, degree or bipartiteness assumptions. Preregistration #21951 binds this universal proposition and its expected refutation. The Library note records the versioned source and the literal defining formulas.

Statement echo

All local names below belong to D5.S3.Combinatorics.SpectralGraph.ArizmendiEnergyGameShapleyRefutation.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite simple undirected graph, every size | claim, n : Nat, G : SimpleGraph (Fin n) | Includes size zero; no extra graph hypothesis | | Energy of the induced coalition | worth | Reuses CyclicExtensionRefutation.graphEnergy with MatrixKind.adjacency, whose center is zero | | Every subset of the other players | shapley, powerset of univ.erase i | Includes empty and full predecessor sets | | Factorial coefficient and marginal contribution | shapley | Natural factorials and subtraction, followed by real coercions and real division | | Coalition payoff total | Finite sum in core | Inlined sum, with no duplicate payoff-total definition | | Every coalition inequality and efficiency | The two conjuncts of core | Singleton, empty and grand coalitions included | | Universal core membership | claim | Literal source conclusion | | Refutation | result | Negates the entire universal claim |

The explicit vertex labels $0,1$ in Fin 52 choose a counterexample; they add no hypothesis to the source claim. The private parameter $m\ge50$ restricts a sufficient counterexample family, not the quantified claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/ArizmendiEnergyGameShapleyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2013Floors of reciprocal generalized Fibonacci tailsGeneralized Fibonacci reciprocal tails have the conjectured integer floor.Proved

The questionAmdeberhan, Chen, Moll and Sagan, Generalized Fibonacci polynomials and Fibonomial coefficients, Annals of Combinatorics 18 (2014), 541–562, Conjecture 4.6, first clause:

Why it mattersIt proves the complete first clause, including convergence, with symbolic parameters and stride.

The ideaThe reciprocal limits vanish by linear growth.

Our contributionGeneralized Fibonacci reciprocal tails have the conjectured integer floor.

Scope & next questionThe squared-reciprocal clause and claims about other source results depending on it remain open here.

Verification details
Release source recordProved in Lean

Open problem

Floors of reciprocal generalized Fibonacci tails

Amdeberhan, Chen, Moll and Sagan, Generalized Fibonacci polynomials and Fibonomial coefficients, Annals of Combinatorics 18 (2014), 541–562, Conjecture 4.6, first clause:

If $s>t\geq1$ with $(s,-t)\ne(2,-1)$ and $n,r\in\mathbb P$ then $\left\lfloor\left(\sum_{k=n}^{\infty}1/{rk}{s,-t}\right)^{-1}\right\rfloor ={rn}{s,-t}-{r(n-1)}_{s,-t}-1$.

The source defines $\mathbb P$ as the positive integers and ${0}{s,t}=0$, ${1}{s,t}=1$, ${j+2}{s,t}=s{j+1}{s,t}+t{j}_{s,t}$. The squared-reciprocal clause is a separate question.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Integral coefficients and the three recurrence cases | U | Literal recursive definition over integers and natural indices. | | $s>t\geq1$ | claim's t < s and 1 ≤ t | Same inequalities. | | $(s,-t)\ne(2,-1)$ | The excluded-pair hypothesis | Same pair and signs. | | $n,r\in\mathbb P$ | Natural indices with 1 ≤ n, 1 ≤ r | Exactly positive indices. | | Tail starting at $n$ | Real reciprocal of U s (-t) (r * (n + k)), summed over natural $k$ | Reindexing, with the integer denominator cast to the reals. | | Convergence presupposed by the source | Explicit Summable conjunct | Prevents a divergent tsum from satisfying the identity vacuously. | | Reciprocal, integer floor and difference endpoint | Second conjunct of claim | Natural subtraction at $n-1$ is ordinary subtraction because $n\geq1$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LinearRows/ACMSReciprocalTailFloor.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
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    First preprint year: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2019A Unique Indecomposable Basis Without Intersection ClosureA unique indecomposable basis need not imply intersection closure.Refuted

The questionPranab Basu and Navin Kashyap, The Lattice Structure of Linear Subspace Codes, arXiv:1911.00721v1, Section 6, Conjecture 6.1 (page 23; TeX label UIB):

Why it mattersThe code has a unique indecomposable basis and is not closed under intersection.

The ideaThe literal integer distance equals the weight of the Klein difference; translation preserves it, and the other three code axioms hold.

Our contributionA unique indecomposable basis need not imply intersection closure.

Verification details
Release source recordRefuted in Lean

arXiv

A Unique Indecomposable Basis Without Intersection Closure

Pranab Basu and Navin Kashyap, The Lattice Structure of Linear Subspace Codes, arXiv:1911.00721v1, Section 6, Conjecture 6.1 (page 23; TeX label UIB):

A linear code $\mathcal U$ in $\mathbb P_q(n)$ has a unique indecomposable basis if and only if $\mathcal U$ is closed under intersection.

A linear subspace code contains the zero subspace and carries an abelian, exponent-two group operation $\boxplus$ whose identity is zero and whose translations preserve $d_S(X,Y)=\dim X+\dim Y-2\dim(X\cap Y)$. A nonzero codeword is indecomposable when no two codewords of strictly smaller ambient dimension have it as their $\boxplus$-sum. A basis is an unordered finite set of codewords giving exactly one finite-subset-sum representation of each codeword, with respect to the same operation. Thus the basis belongs to the code's $\mathbb F_2$-space, rather than to the ambient $\mathbb F_q$-space.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SubspaceCodes/BasuKashyapUniqueIndecomposableBasisRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2019Choi–Kiem–Kye maximal bi-rank edge statesProved

The questionKye, Entangled edge states of corank one with positive partial transposes, J.

Why it mattersThe theorem supplies examples in unbounded dimensions, beyond the source's finite numerical verification.

The ideaConstructionReduction gives the literal Section 6 matrices and the first-factor partial transpose.

Scope & next questionTheir necessity, alternative anchor phases, and maximal-rank edge states for other bipartite systems remain open here.

Verification details
Release source recordProved in Lean

Open problem

Choi–Kiem–Kye maximal bi-rank edge states

J. Choi, Y.-H. Kiem and S.-H. Kye, Entangled edge states of corank one with positive partial transposes, J. Math. Phys. 61, 062202 (2020), arXiv:1903.10745v2, Conjecture 6.4 (Library note):

For any $n\ge 3$, the open set of parameters $(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_n)$ which give PPT entangled edge states of bi-rank $(n^2-1,n^2-2n+3)$ is nonempty.

The binding statement in issue #13859 asks for the Section 6 construction in every dimension $n\ge3$, with unit phases, $\alpha_1=\beta_n=1$, $\alpha_i^{-1}\alpha_j\ne\beta_i^{-1}\beta_j$ for $i<j$, and $r$ the largest real root of $\det D_n^{\alpha,\beta}(r)$. Partial transpose acts on the first tensor factor. The operators are unnormalized. Entanglement is expressed through the nonzero edge condition under that convention.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ChoiKiemKye/MaximalBirankEdgeStates.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2019The optimal T-parameter is not monotone in edge densityThe optimal T-parameter need not increase with edge density.Refuted

The questionSarkar, On a Conjecture of Nagy on Extremal Densities, SIAM Journal on Discrete Mathematics 35 (2021), 294–306, DOI 10.1137/19M1296525, arXiv:1910.13465v1, Section 5, p. 14, state verbatim:

Why it mattersA monotone selection would constrain how extremal threshold graphons interpolate with edge density.

The ideaThe supporting module ThreeClassHomomorphismLimit proves the floor-sized three-class homomorphism limit equals the finite weighting sum.

Our contributionThe optimal T-parameter need not increase with edge density.

Scope & next questionWhether every graph without a pendant-star configuration admits a monotone selection remains open.

Verification details
Release source recordRefuted in Lean

Open problem

The optimal T-parameter is not monotone in edge density

A. N. Day and A. Sarkar, On a Conjecture of Nagy on Extremal Densities, SIAM Journal on Discrete Mathematics 35 (2021), 294–306, DOI 10.1137/19M1296525, arXiv:1910.13465v1, Section 5, p. 14, state verbatim:

Conjecture 2. For each graph $G_l$, we have that $q(G_l,\beta)$ is an increasing function of $\beta$.

The source takes a maximiser of $q\mapsto t(G_l,T_n^e(q))$ on $[0,1]$. The public claim uses the weakest reading: for every finite labelled simple graph, there exists a nondecreasing selection of maximisers at all densities in $[0,1]$. Its negation also excludes a strictly increasing selection.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every labelled finite graph $G_l$ | v : Nat, G : SimpleGraph (Fin v) in claim | All finite orders, including zero; labels introduce no restriction. | | Class sizes $\sqrt\beta qn$, $(1-\sqrt{1-\beta(1-q^2)})n$, $(\sqrt{1-\beta(1-q^2)}-\sqrt\beta q)n$ | yW, rW, bW, vertexClass | Yellow and red sizes use natural floors; blue is the remainder. Floors do not change the proved limit. | | Yellow and red cliques; red joined to both other classes; blue independent; no yellow–blue edges | classAllowed, tGraph | Adjacency requires unequal endpoints. | | Adjacency-preserving functions, not necessarily injective | homCount | Literal cardinality of the source homomorphisms. | | Homomorphism density as a limit | tDens | Literal limUnder along natural graph orders; tDens_eq_weightingSum proves the limit formula. | | Parameter attaining the supremum on $[0,1]$ | Membership and IsMaxOn conjuncts in claim | Every chosen parameter is a maximiser. | | Increasing with density | MonotoneOn in claim | Nondecreasing selection is the weaker interpretation. The witness densities lie strictly inside $(0,1)$. | | Fractional independence weightings | independentWeight, admissible, weightingSum | Classes red, yellow and blue have indices $0,1,2$, representing weights $0,1/2,1$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ExtremalDensity/NagyOptimizerMonotonicity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedSource 2018Frid's Fibonacci-prefix palindromic-length conjectureThe exact palindromic length of every Frid prefixProved

The questionFrid, Representations of palindromes in the Fibonacci word, Numeration 2018 (Paris), Conjecture 2, states:

Why it mattersThe conjecture gives an explicit infinite family of prefixes whose palindromic length grows linearly in the numeral parameter.

The ideaCanonical Fibonacci encodings are split into six-bit chunks.

Our contributionThe exact palindromic length of every Frid prefix

Scope & next questionThe exact global limsup constant remains open.

Verification details
Release source recordProved in Lean

Open problem

Frid's Fibonacci-prefix palindromic-length conjecture

A. E. Frid, Representations of palindromes in the Fibonacci word, Numeration 2018 (Paris), Conjecture 2, states:

For every k ≥ 1, the prefix of the Fibonacci word of length (100)²ᵏ⁻¹101 cannot be decomposed as a concatenation of at most 2k palindromes.

The numeral (100)^(2k-1)101 is interpreted in the Fibonacci numeration system with weights G(0)=1 and G(1)=2. PL is the minimum number of nonempty palindrome factors of a finite word.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Palindromes/FridPrefix/FridPrefixPalindromicLength.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2018.

    Read source ↗
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year recorded in the source bibliography: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture provedFirst preprint 2012The local-bound clause of the Grandjean et al. conjecture on (B1)For every number K >= 2 of measurement outcomes, the deterministic expression in (B1) has minimum 6(K-1).Proved

The questionGisin, “Bell inequalities for three systems and arbitrarily many measurement outcomes”, Physical Review A 85, 052113 (2012), arXiv:1204.3829v2, Appendix B, pp. 5–6, gives the arbitrary-output Sliwa seventh inequality (B1).

Why it mattersThe source's fact that deterministic strategies suffice reduces the local-theory minimum to this finite residue expression.

Our contributionFor every number K >= 2 of measurement outcomes, the deterministic expression in (B1) has minimum 6(K-1).

Scope & next questionThe facet-defining property, quantum optimum, and visibility optimum are outside the result.

Verification details
Release source recordProved in Lean

Open problem

The local-bound clause of the Grandjean et al. conjecture on (B1)

B. Grandjean, Y.-C. Liang, J.-D. Bancal, N. Brunner and N. Gisin, “Bell inequalities for three systems and arbitrarily many measurement outcomes”, Physical Review A 85, 052113 (2012), arXiv:1204.3829v2, Appendix B, pp. 5–6, gives the arbitrary-output Sliwa seventh inequality (B1). It states that its local bound is numerically $6(K-1)$, attained by the all-zero deterministic strategy, and conjectures that both the local bound and the facet-defining property hold for every $K$. The settled statement is the local-bound clause:

$$ \forall K\ge 2,\quad \min{J_K(a,b,c,A,B,C):a,b,c,A,B,C\in\mathbb Z/K\mathbb Z}=6(K-1). $$

where $J_K$ is the twelve-term deterministic expression, with each bracket interpreted as the least nonnegative residue.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/SliwaSevenArbitraryOutcomeLocalBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2021First letters of permutations with a fixed inversion numberFor k<n, the mean first letter at inversion number k is M(n+1,k)/M(n,k).Proved

The questionPeter Kagey, “Expected value of letters of permutations with a given number of k-cycles”, arXiv:2112.05281v1, Section 5.2, page 20, Conjecture 5.1 states:

Why it mattersPreregistration 19726 identifies the external tier-1 conjecture and its literal conventions.

The ideaThe order-preserving first-letter decomposition removes exactly the rank of the first letter from the inversion count.

Our contributionFor k<n, the mean first letter at inversion number k is M(n+1,k)/M(n,k).

Verification details
Release source recordProved in Lean

Open problem

First letters of permutations with a fixed inversion number

Peter Kagey, “Expected value of letters of permutations with a given number of k-cycles”, arXiv:2112.05281v1, Section 5.2, page 20, Conjecture 5.1 states:

For fixed k and n > k, the expected value of the first letter of a permutation with a given number of inversions satisfies a rational function in n given by E[π(1) | π ∈ S_n, inv(π) = k] = M(n + 1, k)/M(n, k), where M(n, k), as above, is the number of permutations w ∈ S_n such that inv(w) = k.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Fixed $k$ and $n>k$ | claim universally quantifies both natural parameters and assumes $k<n$. | Exact. | | Permutations in $S_n$ | Equiv.Perm (Fin n) and its finite universal set. | Each permutation occurs once. | | One-based first letter | PSW.word, defined by the existing owner as the list of values plus one; List.headD 0 selects its first entry. | $k<n$ excludes the empty permutation. | | Inversion number $k$ | ListInversions.inv applied to PSW.word. | Strict inverted pairs. | | $M(n,k)$ | M, cardinality of the inversion fiber. | Literal source cardinality. | | Conditional expectation | E, rational first-letter sum divided by rational cardinality. | Literal sum-over-cardinality convention. | | Ratio $M(n+1,k)/M(n,k)$ | The conclusion of claim and result. | Both casts and the strict guard are retained. |

Rational division is total in Lean. The theorem proves the ratio identity directly; its statement does not include a separate positivity assertion for $M(n,k)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/KageyInversionFirstLetter.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundFirst preprint 2019The even Kimoto–Wakayama normalized Apéry congruence is refutedThe normalized even Apéry-like ratio at (p,m,n,s) = (3,5,2,7) fails modulo nine.Refuted

The questionThe compiled proposition claim is the odd-prime restriction of this assertion.

Why it mattersA certified counterexample identifies the missing divisibility mechanism before any proposed repair is treated as a theorem.

The ideanormalizedEven s j is the module’s natural-number recursion over two consecutive rows of rational values.

Our contributionThe normalized even Apéry-like ratio at (p,m,n,s) = (3,5,2,7) fails modulo nine.

Verification details
Release source recordRefuted in Lean

Open problem

The even Kimoto–Wakayama normalized Apéry congruence is refuted

K. Kimoto and M. Wakayama define normalized Apéry-like rational numbers $\widetilde J_k(n)$ by a three-term inhomogeneous recurrence in Section 3.4 of Apéry-like numbers for non-commutative harmonic oscillators and automorphic integrals. The source TeX Eichler20191217.tex, lines 1608–1617, states the even part of Conjecture 3.6 as follows (the source macro \tJ is rendered as $\widetilde J$):

For positive integers $m,n,s$ such that $mp\ge s$ and $mp^{n-1}\ge s$, we have $$\frac{p^{2sn}\widetilde J_{2s+2}(mp^n)}{p^{2s}\widetilde J_{2s+2}(mp)}\equiv\frac{p^{2s(n-1)}\widetilde J_{2s+2}(mp^{n-1})}{p^{2s}\widetilde J_{2s+2}(mp)}\ (\not\equiv0)\pmod{p^n}.$$

The compiled proposition claim is the odd-prime restriction of this assertion. It quantifies over natural $p$ with p.Prime and $p\ne2$, then over positive natural $m,n,s$ satisfying $s\le mp$ and $s\le mp^{n-1}$, and requires both the displayed rational congruence and noncongruence of the right ratio to zero. The declaration result has type $\neg,\texttt{claim}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/KimotoWakayamaEvenCongruenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2017Interior perfect state transfer without endpoint transferA five-vertex weighted path transfers interior states but never transfers endpoint states.Refuted

The questionZhang, Perfect quantum state transfer in weighted paths with potentials (loops) using orthogonal polynomials, arXiv:1708.03283v2 (Linear and Multilinear Algebra 67, 2019), study perfect state transfer (PST) on weighted paths with real edge weights and real diagonal potentials.

Why it mattersThe source's forward implication explains why endpoint PST produces mirror interior transfer, while the conditional converse would let interior transfer certify endpoint transfer when every eigenvector is nonzero at the chosen interior vertex.

The ideaThe three vertex inequalities are discharged by decide, and the interior premise is interiortransfer; the resulting endpoint conclusion contradicts noendpointtransfer.

Our contributionA five-vertex weighted path transfers interior states but never transfers endpoint states.

Scope & next questionA converse under another support or spectral hypothesis, or a converse for a restricted family of paths, is not supplied by this settlement and remains open. - Nearby statements.

Verification details
Release source recordRefuted in Lean

Open problem

Interior perfect state transfer without endpoint transfer

S. Kirkland, D. McLaren, R. Pereira, S. Plosker and X. Zhang, Perfect quantum state transfer in weighted paths with potentials (loops) using orthogonal polynomials, arXiv:1708.03283v2 (Linear and Multilinear Algebra 67, 2019), study perfect state transfer (PST) on weighted paths with real edge weights and real diagonal potentials. Their Proposition prop:internal states (source TeX main.tex, lines 309–312):

For a weighted path with or without potentials, PST between vertices $1$ and $n$ implies PST between vertices $j$ and $n+1-j$, for each $j=2, \ldots, n-1$. If for some $j$ with $2 \le j \le n-1$ there is PST between vertices $j$ and $n+1-j$, and if in addition none of the eigenvectors of the Hamiltonian has a zero entry in the $j$--th position, then the converse holds.

The question immediately following it is (source TeX main.tex, line 315):

It is an open question if the converse holds even if there are eigenvectors with zeros in the jth place.

The formal reading settled here uses the repository's vertices $0,\ldots,m$. For every $m : \mathbb N$ with $3 \le m$, every $r : \mathrm{Fin},m \to \mathbb R$ with $0 < r(k)$ for every $k$, every $q : \mathrm{Fin}(m+1) \to \mathbb R$, and every $j : \mathrm{Fin}(m+1)$ satisfying $1 \le j.val$, $j.val \le m-1$, and $j.val \ne m-j.val$, it asserts

$$ \left(\exists t : \mathbb R,\ 0<t \land \operatorname{HasPST}(\operatorname{pathHamiltonian}(r,q),t,j,\operatorname{Fin.rev}(j))\right) ;\Longrightarrow; \left(\exists s : \mathbb R,\ 0<s \land \operatorname{HasPST}(\operatorname{pathHamiltonian}(r,q),s,0,\operatorname{Fin.last}(m))\right). $$

Here HasPST H t a b is exactly the compiled predicate that the squared modulus of the corresponding matrix-exponential entry is one. The theorem result settles the negation of this universal proposition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/InteriorPathTransferConverseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedPosed in 2012Prime periods of the A220181 alternating-weight Stirling sumInteger-weight Stirling power sums have prime periods, proving Bala's A220181 conjecture.Proved

The questiona(n) = Sum{k=0..n} (-1)^(n-k) k^n k! Stirling2(n,k).

Why it mattersThis is a first-tier OEIS conjecture: the entry dates from 2012 and the conjecture COMMENT from 2022.

The ideaFor A220181, aeqalternating proves a n = (-1)^n aw (fun k => (-1)^k) n from the finite FORMULA. balaconjecture transfers the period: p - 1 is even for odd primes, and the sign is trivial modulo 2.

Our contributionInteger-weight Stirling power sums have prime periods, proving Bala's A220181 conjecture.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime periods of the A220181 alternating-weight Stirling sum

OEIS A220181 gives the following NAME, finite FORMULA, and COMMENT, copied verbatim from Library/Recurrence/bala2022a220181.md:

E.g.f.: Sum_{n>=0} (1 - exp(-n*x))^n.

a(n) = Sum_{k=0..n} (-1)^(n-k) * k^n * k! * Stirling2(n,k).

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [1, 0, 3, 0, 0, 1, 1, 0, 3, 0, 0, 1 ...], with an apparent period of 6. Cf. A122399.

The entry is by Paul D. Hanna, Dec 06 2012; the conjecture COMMENT resolved here is by Peter Bala, Jun 01 2022. The two attributions are recorded separately rather than reassigning the comment to the entry author.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/AlternatingWeightStirlingPrimePeriod.bala_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture provedFirst preprint 2021Pandey–Patra subgraph core conjectureThe entire maximizing set of actual connected-subgraph counts lies in one block.Proved

The questionDinesh Pandey and Kamal Lochan Patra, Some new central parts of connected graphs, arXiv:2110.00738v1, Conjecture 3.25, page 9:

Why it mattersConjecture 3.25 asks whether the vertices maximizing the number of connected subgraphs containing a vertex share one native block.

The ideaSubgraphCoreBlock.subgraphNumber uses the existing countContaining carrier, so it preserves both vertex and edge choices. subgraphCore is the complete maximizer set, and NativeBlock expresses native no-cut connectivity together with maximality under actual subgraph inclusion.

Our contributionThe entire maximizing set of actual connected-subgraph counts lies in one block.

Verification details
Release source recordProved in Lean

arXiv

Pandey–Patra subgraph core conjecture

Dinesh Pandey and Kamal Lochan Patra, Some new central parts of connected graphs, arXiv:2110.00738v1, Conjecture 3.25, page 9:

The subgraph core of a graph G is contained in a block of G.

The native domain is every finite simple connected undirected graph. The subgraph number counts all actual connected subgraphs containing a vertex, including singleton subgraphs and distinct edge choices. The conclusion is one native maximal no-cut block containing the entire set of maximizing vertices. Singleton, bridge-edge, empty-deletion, and unbounded-order cases remain in scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Eccentric/SubgraphCoreBlock.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2021The exact critical concurrence for universal teleportation through qubit channelsNecessity holds for every finite Kraus family, and a non-unital amplitude-damping channel attains sufficiency.Proved

The questionBanerjee, Characterizing qubit channels in the context of quantum teleportation, Phys.

Why it mattersCCritical is the least entanglement of the initial pure state for which some qubit channel on Bob's side can still deliver a state that teleports every input state with fidelity above the classical 2/3.

The ideaPut r = 2a − 1 = √(1 − C²), the length of Alice's Bloch vector, which every channel on Bob and every unitary on Alice preserves.

Our contributionNecessity holds for every finite Kraus family, and a non-unital amplitude-damping channel attains sufficiency.

Verification details
Release source recordProved in Lean

Open problem

The exact critical concurrence for universal teleportation through qubit channels

A. Ghosal, D. Das and S. Banerjee, Characterizing qubit channels in the context of quantum teleportation, Phys. Rev. A 103, 052422 (2021), arXiv:2102.02054v2, send one half of |Ψa⟩ = √a|00⟩ + √(1 − a)|11⟩ (1/2 < a < 1, concurrence C = 2√(a(1 − a))) through a qubit channel Λ and ask when the final state (id ⊗ Λ)(|Ψa⟩⟨Ψa|) is useful for universal quantum teleportation (UQT).

  • The criterion. A two-qubit state is useful for UQT if and only if, in its canonical form under product unitaries, |t₁₁| = |t₂₂| = |t₃₃| > 1/3 (§II.B–§II.C).
  • Unital and non-unital channels. Unital channels need C > 1/2. For the two-Kraus non-unital family of Eq. (50), the final state is useful for UQT exactly when C > √((2/3)(4 + √3)(1 − (4 + √3)/6)) ≈ 0.41.
  • The question. Conjecture 1 states that the final state is useful for UQT for some non-unital channels if and only if C_Critical < C < 1, "where 0 < C_Critical ≤ 0.41. We could not determine the precise value of C_Critical; further investigations are needed."

The verbatim statements are in the literature note.

Issue #16624 fixes the reading: "useful for UQT" means that some product of unitaries brings the state to correlation matrix −tI₃ with t > 1/3, and channels are finite Kraus families. The question settled is the precise value of C_Critical: C_Critical = C_* = √(5 − 2√3)/3 = 0.4131045583…, the exact value of the source's radical. The source's printed bound "C_Critical ≤ 0.41" is the rounding of that radical; taken literally as the rational 41/100 it would be false, so the settlement answers the precise-value question and does not assert the printed inequality. Conjecture 1's further clause that only a strict subset of the non-unital channels succeeds is not part of the formal statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/UniversalTeleportationCriticalConcurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Geometry & algebraQuestion resolvedFirst preprint 2020An extremal geometry forces an exact parameter identityAt Gerzon equality, the parameters satisfy a² = (d + 2)b in every dimension at least two over fields of characteristic different from two.Gerzon equality fixes the equiangular parameters in every dimension at least twoProved

The questionWhat relation must equiangular parameters satisfy when the Gerzon bound is attained?

Why it mattersThe known identity omitted two characteristic classes of dimensions. The result includes them and allows isotropic vectors.

The ideaTrace identities and fourth moments establish the same parameter relation in both scalar-frame branches.

Our contributionAt Gerzon equality, the parameters satisfy a² = (d + 2)b in every dimension at least two over fields of characteristic different from two.

Scope & next questionThe theorem gives a necessary identity at equality; it does not construct extremal configurations in every dimension.

Verification details
Release source recordProved in Lean

Open problem

Gerzon equality fixes the equiangular parameters in every dimension at least two

G. R. W. Greaves, J. W. Iverson, J. Jasper and D. G. Mixon, Frames over finite fields: Equiangular lines in orthogonal geometry, arXiv:2012.13642v2, Example 5.8, page 21, asks:

The authors do not know how to construct an ETF with any of the parameters given in Table 4. In particular, every instance of Gerzon equality that we currently know satisfies the identity a² = (d + 2)b, but this equation does not occur anywhere in Table 4 (where b = 1). Proposition 2.10 implies that a² = (d + 2)b whenever Gerzon’s bound is saturated, d ≢ₚ 0, and d ≢ₚ 1. The authors do not know if these conditions can be removed.

The answer is affirmative for every $d\ge2$. Preregistration: #16342. The Library note records Definition 2.8, Propositions 2.9–2.10 and the source locator.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite field of odd characteristic | F : Type, Field F, Fintype F, ringChar F ≠ 2 in claim | Exact source scope | | Dimension $d\ge2$ | d : ℕ, 2 ≤ d | Explicit boundary of the affirmative answer | | Quadratic space | B : LinearMap.BilinForm F (Fin d → F), B.IsSymm, B.Nondegenerate | Arbitrary symmetric nondegenerate form in coordinates | | $N=\binom{d+1}{2}$ vectors | φ : Fin ((d + 1).choose 2) → Fin d → F | Exact Gerzon cardinality | | $(a,b)$-equiangular | IsEquiangularSystem B φ a b | Every diagonal pairing equals $a$; every distinct squared pairing equals $b$ | | $a^2\ne b$ | a ^ 2 ≠ b | Retained hypothesis of Proposition 2.9 | | Remove both dimension congruences | No assumption that $d$ or $d-1$ is nonzero in $F$ | Neither restriction retained | | Parameter identity | result : claim | $a^2=((d:F)+2)b$ |

The two public definitions are the literal conventions of #16342. Fin n indexes from zero instead of one without changing the pairwise conditions. A choice of basis identifies any $d$-dimensional space with these coordinates; that transport is the coordinate interpretation, not an additional public theorem of this module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/GerzonEqualityParameterIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2015Coefficient-wise Wright log-concavity of normalized incomplete beta functionsStrict positivity of the normalized incomplete beta Turanian coefficients.Proved

The questionKarp, Normalized incomplete beta function: log-concavity in parameters and other properties, §3, Conjecture 1:

Why it mattersThe coefficient assertion strengthens pointwise Wright log-concavity by requiring positivity at every coefficient separately.

The ideaThe first is strictly smaller because both shifts are positive.

Our contributionStrict positivity of the normalized incomplete beta Turanian coefficients.

Verification details
Release source recordProved in Lean

Open problem

Coefficient-wise Wright log-concavity of normalized incomplete beta functions

Karp, Normalized incomplete beta function: log-concavity in parameters and other properties, §3, Conjecture 1:

Conjecture 1 The coefficients φ_k are positive for all a, b, α, β > 0 and x ∈ (0, 1), so that b → x^{−a}(1 − x)^{−b} I_x(a, b) is strictly coefficient-wise Wright log-concave on (0, ∞) for these values of parameters.

Here equations (1), (3) and (4) give

$$c_j(a,b)=\frac{\Gamma(a+b+j)}{\Gamma(b)\Gamma(a+j+1)},\qquad \varphi_n=\sum_{k=0}^n \bigl[c_k(a,b+\alpha)c_{n-k}(a,b+\beta) -c_k(a,b)c_{n-k}(a,b+\alpha+\beta)\bigr].$$

Formal reading

All Lean names below have prefix D5.S3.Analytic.IncompleteBeta.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $c_j(a,b)$ in (1) | coeff in Defs | Exactly the displayed Gamma quotient, with natural index $j$. | | $\varphi_n$ in (3), (4) | phi in Defs | The finite convolution difference over $0\le k\le n$. | | $C_n(s,t)$ | convolution in Moment | $\sum_{k=0}^n c_k(a,s)c_{n-k}(a,t)$, with $a$ explicit. | | The symmetric beta kernel | W in Moment | The finite powered kernel used in the moment representation below. | | Conjecture 1 | claim, Result.result | Every real $a,b,\alpha,\beta>0$ and every natural $n$, with strict positivity. | | $x\in(0,1)$ | No additional coefficient hypothesis | $\varphi_n$ does not depend on $x$, so this source quantifier is vacuous for the coefficients. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/IncompleteBeta/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture refutedFirst preprint 2013Sun's Conjecture 3.8: circular permutations with primitive adjacent productsOver the field of eleven elements, offset two forces the vertices two and nine to share their two cyclic neighbours.Refuted

The questionZhi-Wei Sun, “Some new problems in additive combinatorics”, arXiv:1309.1679, Conjecture 3.8:

Why it mattersThe assertion seeks one circular arrangement satisfying all adjacent primitive-element constraints simultaneously.

The ideaThe residues outside {2,6,7,8} are zero or have fifth or second power one, so none has order ten.

Our contributionOver the field of eleven elements, offset two forces the vertices two and nine to share their two cyclic neighbours.

Verification details
Release source recordRefuted in Lean

Open problem

Sun's Conjecture 3.8: circular permutations with primitive adjacent products

Zhi-Wei Sun, “Some new problems in additive combinatorics”, arXiv:1309.1679, Conjecture 3.8:

Let F_q be a finite field with q > 7 elements, and let a_0 be any element of F_q. Then there is a circular permutation (a_1, …, a_{q−1}) of all the nonzero elements of F_q such that all the q − 1 elements a_0 + a_1a_2, a_0 + a_2a_3, …, a_0 + a_{q−2}a_{q−1}, a_0 + a_{q−1}a_1 are primitive elements of the field F_q.

The Lean definition claim : Prop quantifies over every finite field and every offset. An equivalence from Fin (q - 1) to the nonzero subtype expresses that every nonzero element appears exactly once. finRotate advances cyclically, including the last-to-first edge. IsPrimitiveRoot z (q - 1) means exact multiplicative order q - 1, hence precisely that z generates the multiplicative group, whose order is q - 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SunCircularPrimitivePermutationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2021Non-shortening in the one-wall hypereclectic spin chainPowers of the one-wall hypereclectic Hamiltonian have maximal rank between levels.Proved

The questionStaudacher, Combinatorial Solution of the Eclectic Spin Chain, arXiv:2112.04506v1, JHEP 03 (2022) 028, Appendix A, Eq. (A.4):

Why it mattersMaximal rank excludes unexpected shortening of its chains between levels.

The ideaThe positional cancellation directly uses Finset.sumrangesub'.

Our contributionPowers of the one-wall hypereclectic Hamiltonian have maximal rank between levels.

Scope & next questionIts linear equivalence and intertwining identities remain open formal obligations.

Verification details
Release source recordProved in Lean

Open problem

Non-shortening in the one-wall hypereclectic spin chain

C. Ahn, L. Corcoran and M. Staudacher, Combinatorial Solution of the Eclectic Spin Chain, arXiv:2112.04506v1, JHEP 03 (2022) 028, Appendix A, Eq. (A.4):

“We claim that the rank of A^(k) is always maximal:”

“rank(A^(k)) = min(d_{S−k}, d_S).”

Here $A^{(k)}$ is the matrix of $H^k:W_S^{L,M}\to W_{S-k}^{L,M}$ in the elementary bases, and $d_S=\dim W_S^{L,M}$. The settled statement is the rational encoding of the $K=1$ static sector, for every $1\le M\le L$, every natural level $S$ and every $k\le S$. The rightmost letter is the fixed wall $3$; the other $L-1$ letters consist of $L-M$ ones and $M-1$ twos.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/HypereclecticNonShortening.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresRecurrence provedSource 2017The double-root row of Ballot–Kimberling–Moses Table 5The polynomial tree modulo X squared minus 2 X plus 1 has generation sizes satisfying the Table 5 recurrence with coefficients 3, -3, 1 at every index at least 3.Proved

The questionChristian Ballot, Clark Kimberling and Peter J.

Why it mattersNo claim about the six subsequent restricted families is needed.

The ideaThe declarations generationsuccspec and generationsucceq partition each shell into its top label and disjoint finite intervals. Gsucc counts the intervals, Gsum includes generation zero, and Gdifference yields the consecutive-difference identity used by result.

Our contributionThe polynomial tree modulo X squared minus 2 X plus 1 has generation sizes satisfying the Table 5 recurrence with coefficients 3, -3, 1 at every index at least 3.

Verification details
Release source recordProved in Lean

Open problem

The double-root row of Ballot–Kimberling–Moses Table 5

Christian Ballot, Clark Kimberling and Peter J. C. Moses, Linear Recurrences Originating From Polynomial Trees, The Fibonacci Quarterly 55(5), 15–27, Table 5, gives $q(x)=x^2-2x+1$, initial counts $1,1,2,4,7,11,16,22,29,37,46$, and recurrence coefficients $3,-3,1$. Section 5 states: “Most of the entries in the column headed ‘linear recurrence’ are conjectured as ‘eventual’; i.e., the recurrence applies after some unspecified number of initial terms.” The selected assertion and literal Lean conventions are specified in the problem statement. The Library note records the published PDF quotations and verified locator. A source arXiv identifier has not been located.

Start at $0$ in $\mathbb Z[x]$, applying $p\mapsto(p+1)\bmod q$ or $p\mapsto xp\bmod q$, with monic remainders. Let $S_n$ contain the distinct labels reached within $n$ edges. Put $g(0)={0}$ and $g(n+1)=S_{n+1}\setminus S_n$, and $G_n=|g(n)|$. The assertion is $\exists N\geq3;\forall n\geq N,;G_n=3G_{n-1}-3G_{n-2}+G_{n-3}$, with the recurrence arithmetic over $\mathbb Z$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Table 5 polynomial $x^2-2x+1$ | q | Literal integer polynomial. | | Root $0$; children $p+1$ and $xp$ modulo $q$ | addChild, mulChild, seen | Literal monic-remainder operations and finite recursion. | | Remove repeated labels | seen, generation | Finsets remove same-generation duplicates; set difference removes every earlier label. Parent order does not change the label sets. | | $G_n=|g(n)|$, including the root | G | Natural cardinality of the finite generation. | | Eventual recurrence $3,-3,1$ | claim, result | Existential threshold, every tail index, and integer arithmetic are retained; the proof chooses $N=3$. |

The pair coordinates occur only in private proof definitions; the public sequence retains the source's polynomial operations. Natural predecessor subtraction is used only under $n\geq N\geq3$ in the recurrence. The zero root is a separate case.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PolynomialTrees/BallotKimberlingMosesDoubleRootGeneration.result

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  • Contribution

    The polynomial tree modulo X squared minus 2 X plus 1 has generation sizes satisfying the Table 5 recurrence with coefficients 3, -3, 1 at every index at least 3.

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  • Historical context

    Year recorded in the source bibliography: 2017.

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Number theoryQuestion resolvedPosed in 2016Runs in the Second Binary Digit of the Fibonacci NumbersNo three consecutive Fibonacci numbers carry the same second most significant binary digit.Proved

The questionIts COMMENTS section carries, at revision 34 of Dec 08 2025:

The ideaThe digit becomes arithmetic once the number is bracketed between consecutive powers of two.

Our contributionNo three consecutive Fibonacci numbers carry the same second most significant binary digit.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Runs in the Second Binary Digit of the Fibonacci Numbers

OEIS A272170 is the sequence of second most significant binary digits of the Fibonacci numbers exceeding one. Its COMMENTS section carries, at revision #34 of Dec 08 2025:

It is conjectured that there are no more than two consecutive "0's" or "1's" (tested up to n=10^5). The sequence looks quasiperiodic and its Fourier spectrum seems to have a fractal structure. - Andres Cicuttin, Apr 21 2016

Writing s(m) = m / 2 ^ (size m - 2) % 2 for that digit, the assertion is

∀ n ≥ 3,  ¬ ( s(F n) = s(F (n+1)) ∧ s(F (n+1)) = s(F (n+2)) ).
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FibonacciSecondBitRunLength.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2017No parity Majorana code on at most fourteen modes has distance four and a logical qubitA self-orthogonal binary stabilizer space containing parity on at most fourteen modes has a logical operator of weight below four whenever twice its dimension is less than the number of modes.Proved

The questionHastings, Small Majorana Fermion Codes, Quantum Inf.

Why it mattersThe source's distance-four codes target correction of errors on a single Majorana mode (c.tex, lines 121–124).

The ideaThe proof uses induction.

Our contributionA self-orthogonal binary stabilizer space containing parity on at most fourteen modes has a logical operator of weight below four whenever twice its dimension is less than the number of modes.

Scope & next questionTheir optimality remains open here, as do extensions that remove the parity convention; no classification or uniqueness of sixteen-mode codes is established.

Verification details
Release source recordProved in Lean

Open problem

No parity Majorana code on at most fourteen modes has distance four and a logical qubit

M. B. Hastings, Small Majorana Fermion Codes, Quantum Inf. Comput. 17(13–14), 1191–1205 (2017), arXiv:1703.00612v1, Section II.B, PDF p. 5 (c.tex, line 272):

For $N_{\rm maj}=14$, we performed a numerical search (described in the next section) for a code with $N_{\rm stab}=6$ and did not succeed. Hence, we believe that no such code exists, i.e., we believe that $N_{\rm maj}=16$ is the minimum number of modes to have $d=4,K>0$.

The paper uses $N_{\rm maj}$ for the number of Majorana modes, $N_{\rm stab}$ for the number of independent stabilizers, and $K=N_{\rm maj}/2-N_{\rm stab}$ for the number of logical qubits. Its stabilizer bit strings form a self-orthogonal binary subspace $C$, meaning $x\cdot y=0$ for every $x,y\in C$, including $x=y$ (c.tex, lines 85–105). Logical operators correspond to $C^\perp\setminus C$, and distance is their minimum Hamming weight (lines 108–113). The paper requires the all-ones fermion-parity vector $\mathbf1$ to belong to $C$ (lines 115–118). These definitions and the quotation are also recorded in the literature note.

Issue #23984 fixes the statement: for every natural number $n\le14$ and every binary linear subspace $C\le\mathbb F_2^n$, if $C$ is self-orthogonal, $\mathbf1\in C$, and $2\dim C<n$, then

$\exists v\in C^\perp\setminus C,\quad \operatorname{wt}(v)<4.$

This is claim, and result : claim proves it. The conclusion supplies a logical operator directly; it does not define distance through an infimum. The sixteen-mode existence statement is outside claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/MajoranaFourteenModeDistanceFour.result

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Quantum informationQuestion resolvedFirst preprint 2020Arimoto information combining convexity and concavityHirche's binary Arimoto information combining conjecture holds with threshold three halves on the full closed coordinate range.Proved

The questionHirche, Rényi Bounds on Information Combining, ISIT 2020, arXiv:2004.14408v1, §IV-A, Conjecture IV.6:

Why it mattersThis scalar inequality supplies the separate curvature needed for binary Arimoto conditional Rényi information combining.

The ideaContinuity extends these inequalities to the closed interval.

Our contributionHirche's binary Arimoto information combining conjecture holds with threshold three halves on the full closed coordinate range.

Verification details
Release source recordProved in Lean

Open problem

Arimoto information combining convexity and concavity

C. Hirche, Rényi Bounds on Information Combining, ISIT 2020, arXiv:2004.14408v1, §IV-A, Conjecture IV.6:

There exists a value 1 < α̂ < 1.5783, such that k_α^A(k_α^A⁻¹(x) ∗ k_α^A⁻¹(y)) (47) is convex for 0 < α < 1 and concave for 1 < α ≤ α̂.

Equation (23) defines

$$ K_\alpha^A(X\mid Y)=e^{((1-\alpha)/\alpha)H_\alpha^A(X\mid Y)}. $$

For binary $X\sim{p,1-p}$, $k_\alpha^A(p)=K_\alpha^A(X)$, hence

$$ k_\alpha^A(p)=(p^\alpha+(1-p)^\alpha)^{1/\alpha},\qquad p\star q=p+q-2pq. $$

Write $k_\alpha=k_\alpha^A$ below. Lemma IV.3, cited from Hsu–Asoodeh–Salamatian–Calmon, gives convexity for $\alpha\ge2$; Lemma IV.4 shows neither convexity nor concavity for $\alpha\in(1.58,1.97)$.

The inverse is the restriction to $[0,1/2]$, with range $I_\alpha=[\min(1,\delta_\alpha),\max(1,\delta_\alpha)]$, where $\delta_\alpha=2^{(1-\alpha)/\alpha}$. Orders and exponents are real. For each permitted order and every fixed $y\in I_\alpha$, the claim is the full two-point ConvexOn or ConcaveOn inequality on the closed interval $I_\alpha$. The specified witness is $\hat\alpha=3/2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/RenyiCombining/HircheArimotoCombining.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedPosed in 2017Yanev's general sigma-radical identityYanev's identity expresses every positive divisor-power sum through ordinary divisor sums and the radical.Proved

The questiona(n) = sigman(n): sum of n-th powers of divisors of n.

Why it mattersThe identity expresses a divisor sum of any positive exponent using only ordinary divisor sums and the distinct prime factors of the input.

The ideaOn a prime power, apply the existing sigma and radical formulas.

Our contributionYanev's identity expresses every positive divisor-power sum through ordinary divisor sums and the radical.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Yanev's general sigma-radical identity

OEIS A023887, NAME (%N, verbatim):

a(n) = sigma_n(n): sum of n-th powers of divisors of n.

Velin Yanev's FORMULA conjecture (%F, verbatim):

Conjecture: sigma_m(n) = sigma(n^m * rad(n)^(m-1))/sigma(rad(n)^(m-1)) for n > 0 and m > 0, where sigma = A000203 and rad = A007947. - Velin Yanev, Aug 24 2017

AUTHOR (%A, verbatim):

Olivier Gérard

The exact formal statement, with all variables natural, is forall n m, 0 < n -> 0 < m -> sigma m n * sigma 1 (primeRadical n ^ (m - 1)) = sigma 1 (n ^ m * primeRadical n ^ (m - 1)), where D5/S1/Deficit/AlmostAdditivity.primeRadical n = ∏ p ∈ n.primeFactors, p is the frozen natural radical (A007947). Both sides are positive under these hypotheses. The denominator is also positive, so this multiplied-out form is equivalent to the quoted identity, including its exact natural-number quotient. Subtraction in m - 1 is natural subtraction; the positive-m hypothesis makes it the usual predecessor.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/YanevSigmaRadicalIdentity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedPosed in 2017Schulte's alternating quadratic peak moment for Schröder pathsSchulte's alternating quadratic peak moment for large Schroeder paths.Proved

The questionTriangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.

Why it mattersThe target is the full unbounded identity in the formula field.

The ideaRepresent a path by a U/D/H word with U and D of weight one, H of weight two, total weight 2n, equal U and D counts, and every prefix having at least as many U as D.

Our contributionSchulte's alternating quadratic peak moment for large Schroeder paths.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's alternating quadratic peak moment for Schröder paths

OEIS A060693, NAME:

Triangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.

COMMENT:

T(n,k) is the number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k peaks. Example: T(2,1)=3 because we have UU*DD, U*DH and HU*D, the peaks being shown by *.

FORMULA:

Conjecture: Sum_{k=0..n} (-1)^k*T(n,k)*(n+1-k)^2 = 1+n+n^2. - Werner Schulte, Jan 11 2017

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SchroederPeakAlternatingQuadraticMoment.schulte_a060693

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    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsExact formula provedPosed in 2015Barker's and Zizka's conjectures for the ON cells of Rule 13The elementary cellular automaton with Wolfram rule 13, started from a single ON cell at the origin and updated at every cell of the integers, has n + 1 ON cells among the cells -2n, ..., 2n of row 2n and 3n + 1 ON cells among the cells -(2n + 1), ..., 2n + 1 of row 2n + 1, as conjectured by Ctibor O.Proved

The questionConjectures from Colin Barker, Dec 28 2015 and Apr 14 2019: (Start) a(n) = ((-1)^n(3-2n)+4n+1)/4. a(n) = 2a(n-2)-a(n-4) for n>3. G.f.: (1+x+2x^3) / ((1-x)^2(1+x)^2). (End) Conjecture from Ctibor O.

The ideaBy induction on n, row 2k is ON exactly at the even x with 0 ≤ x ≤ 2k, and row 2k + 1 is OFF exactly at the odd x with -1 ≤ x ≤ 2k + 1: rule 13 sends lcr to 1 exactly when l = 0 and either c = 1 or r = 0.

Our contributionThe elementary cellular automaton with Wolfram rule 13, started from a single ON cell at the origin and updated at every cell of the integers, has n + 1 ON cells among the cells -2n, ..., 2n of row 2n and 3n + 1 ON cells among the cells -(2n + 1), ..., 2n + 1 of row 2n + 1, as conjectured by Ctibor O.

Scope & next questionEffect on the entry's other statements: the five formulas are all the conjectures the entry records, and all five are proved, so none remains open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barker's and Zizka's conjectures for the ON cells of Rule 13

OEIS A266285 counts the ON cells among the cells -n, …, n of row n of the Rule 13 elementary cellular automaton started from a single ON cell. Its formula field records

Conjectures from Colin Barker, Dec 28 2015 and Apr 14 2019: (Start) a(n) = ((-1)^n*(3-2*n)+4*n+1)/4. a(n) = 2*a(n-2)-a(n-4) for n>3. G.f.: (1+x+2*x^3) / ((1-x)^2*(1+x)^2). (End) Conjecture from Ctibor O. Zizka, Mar 11 2025: (Start) a(2*n) = n + 1. a(2*n + 1) = 3*n + 1.(End)

Issue #13151 fixes the reading: every cell of the integers is updated at every step, the closed form is read in the rationals, the recurrence is asserted for n > 3 in the integers, and the generating function is stated as the product of the series with its denominator, which has constant term 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/Rule13OnCellCount.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Zizka for OEIS A266285; the counts therefore satisfy Colin Barker's closed form ((-1)^n (3 - 2n) + 4n + 1) / 4, his recurrence a(n) = 2 a(n - 2) - a(n - 4) for n > 3 and his generating function (1 + x + 2x^3) / ((1 - x)^2 (1 + x)^2).

    Read source ↗
  • Historical context

    Dated conjecture statement: 2015.

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Analysis & dynamicsQuestion resolvedPosed in 2016Barker's conjectures for the ON cells of Rule 41The elementary cellular automaton with Wolfram rule 41, started from a single ON cell at the origin and updated at every cell of the integers, has 1, 8k, 2 and 8k + 3 ON cells among the cells -n, ..., n of row n = 4k, 4k + 1, 4k + 2 and 4k + 3; the counts therefore satisfy Colin Barker's conjectured recurrence a(n) = a(n - 2) + a(n - 4) - a(n - 6) for n > 5 and generating function (1 + x^2 + 3x^3 - 2x^4 + 5x^5) / ((1 - x)^2 (1 + x)^2 (1 + x^2)) for OEIS A266614.Proved

The questionConjectures from Colin Barker, Jan 02 2016 and Apr 18 2019: (Start) a(n) = a(n-2)+a(n-4)-a(n-6) for n>5. G.f.: (1+x^2+3x^3-2x^4+5x^5) / ((1-x)^2(1+x)^2(1+x^2)). (End)

The ideaBy induction on n, the background of row n is OFF for even n and ON for odd n, and row n differs from it exactly at {n}, {n − 2, n − 1, n}, {n − 2, n} or {n − 4, n − 3, n − 1, n} according as n ≡ 0, 1, 2, 3 (mod 4): rule 41 sends lcr to 1 exactly for 000, 011 and 101.

Our contributionThe elementary cellular automaton with Wolfram rule 41, started from a single ON cell at the origin and updated at every cell of the integers, has 1, 8k, 2 and 8k + 3 ON cells among the cells -n, ..., n of row n = 4k, 4k + 1, 4k + 2 and 4k + 3; the counts therefore satisfy Colin Barker's conjectured recurrence a(n) = a(n - 2) + a(n - 4) - a(n - 6) for n > 5 and generating function (1 + x^2 + 3x^3 - 2x^4 + 5x^5) / ((1 - x)^2 (1 + x)^2 (1 + x^2)) for OEIS A266614.

Scope & next questionEffect on the entry's other statements: the two formulas are all the conjectures the entry records, and both are proved, so none remains open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Barker's conjectures for the ON cells of Rule 41

OEIS A266614 counts the ON cells among the cells -n, …, n of row n of the Rule 41 elementary cellular automaton started from a single ON cell. Its formula field records

Conjectures from Colin Barker, Jan 02 2016 and Apr 18 2019: (Start) a(n) = a(n-2)+a(n-4)-a(n-6) for n>5. G.f.: (1+x^2+3*x^3-2*x^4+5*x^5) / ((1-x)^2*(1+x)^2*(1+x^2)). (End)

Issue #13157 fixes the reading: every cell of the integers is updated at every step, the recurrence is asserted for n > 5 in the integers, and the generating function is stated as the product of the series with its denominator, which has constant term 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/Rule41OnCellCount.result

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Discrete structuresConjecture provedFirst preprint 2019Peltomäki's silver-slope abelian period set conjectureProved

The questionPeltomäki, Abelian periods of factors of Sturmian words, Journal of Number Theory 214 (2020), 251–285, arXiv:1905.06138v3, Conjecture on page 283:

Why it mattersThe source's common-factor-language interpretation identifies this language with the factor language of Sturmian words of that slope.

The ideaThe Pell error and determinant identities yield uniform integer approximation exclusions.

Verification details
Release source recordProved in Lean

Open problem

Peltomäki's silver-slope abelian period set conjecture

J. Peltomäki, Abelian periods of factors of Sturmian words, Journal of Number Theory 214 (2020), 251–285, arXiv:1905.06138v3, Conjecture on page 283:

Let $\alpha = [0; \overline{2}]$. The abelian period set of a Sturmian word of slope $\alpha$ is $\mathcal{Q}^+\alpha \cup \mathcal{M}\alpha$.

An abelian period set comprises the minimum abelian periods of all nonempty factors. For $\alpha=\sqrt{2}-1$, let $q_0=1$, $q_1=2$ and $q_{k+1}=2q_k+q_{k-1}$. The asserted set is ${q_k:k\geq0}\cup{2q_k:k\geq0}\cup{q_k+q_{k-1}:k\geq1}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Mechanical/SilverSlopeAbelianPeriods.result

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    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2018The instability count of a clamped Lennard–Jones chainFor a clamped steady Lennard-Jones chain with positive pre-stretch, positive bond lengths, at least one free particle and k ≥ 1 broken bonds, the positive force-Jacobian eigenvalue count lies between k − 1 and k.Proved

The questionDobson, N-break states in a chain of nonlinear oscillators, Phys.

Why it mattersThe conjecture turns a count of broken bonds into a count of unstable spectral directions, independently of chain size and break placement.

The ideaFor a Hermitian matrix and a subspace with this positivity property, RHLinalg.finrankleposIndexofposDefOn bounds the subspace dimension by the positive eigenvalue count.

Our contributionFor a clamped steady Lennard-Jones chain with positive pre-stretch, positive bond lengths, at least one free particle and k ≥ 1 broken bonds, the positive force-Jacobian eigenvalue count lies between k − 1 and k.

Verification details
Release source recordProved in Lean

Open problem

The instability count of a clamped Lennard–Jones chain

A. S. Rodrigues, P. G. Kevrekidis and M. Dobson, N-break states in a chain of nonlinear oscillators, Phys. Rev. E 99, 022201 (2019), arXiv:1802.03477v2, state in the abstract:

Each of these higher ``excited state'' solutions possesses an additional unstable pair of eigenvalues. We thus conjecture that $k$ break solutions will possess at least $k-1$ (and at most $k$) pairs of unstable eigenvalues.

This quotation is from oscillatorsNN_resubmit.tex, lines 55–59, in the v2 source, whose MD5 is f0e3a2d64045c090b53b217d75e86a05. The model and terminology are documented in the literature note.

The module uses $m$ free particles, displacements $u_0,\ldots,u_{m+1}$, and zero-based bond indices $j=0,\ldots,m$. Its definitions are

$\phi(r)=r^{-12}-2r^{-6}$, $r_i=(13/7)^{1/6}$, and $r_j=d+u_{j+1}-u_j$.

The force at free particle $n+1$, indexed by $n=0,\ldots,m-1$, is $F_n=\phi'(r_{n+1})-\phi'(r_n)$. With $c_j=\phi''(r_j)$, the defined matrix $J$ has entries $J_{nn}=-c_n-c_{n+1}$, $J_{n,n+1}=J_{n+1,n}=c_{n+1}$, and zero elsewhere.

Issue #23865 fixes the reading: for every $m\ge1$, $d>0$, and displacement vector with $u_0=u_{m+1}=0$, $F_n=0$ at every free particle, and all $r_j>0$, if exactly $k\ge1$ bonds satisfy $r_j>r_i$, then, for every Hermitian witness $h$ for $J$, the number $U=#{i\in\mathrm{Fin},m:0<h.\mathrm{eigenvalues}(i)}$ satisfies $k-1\le U\le k$. The eigenvalues are counted with multiplicity. This is proved as result : claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/LennardJonesChainBreakInstabilityCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedPosed in 2016Browne's Chebyshev numerator and floor productEvery positive Chebyshev row numerator equals Browne's literal floor product and its radical sign.Proved

The questionAnthony Browne's final FORMULA in OEIS A164660, revision 19, reads:

Why it mattersThe proof is bind-only after expanding its consumed helpers; no nonbinding escape witness or originality claim is asserted.

The ideaPolynomial coefficients and endpoint evaluations give the exact rational row sum, and Rat.numdiveqofcoprime supplies its reduced numerator.

Our contributionEvery positive Chebyshev row numerator equals Browne's literal floor product and its radical sign.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Browne's Chebyshev numerator and floor product

Anthony Browne's final FORMULA in OEIS A164660, revision 19, reads:

a(n+1) = Product_{k=1..n} ( 1-2*(floor(k^n/n)-floor((k^n -1)/n)) ) = (-1)^(A003557(n)) for n>0 (conjecture). - Anthony Browne, May 29 2016

The sequence, introduced by Wolfdieter Lang, defines:

a(n) = numerator(Sum_{m=1..n+1} IT(n,m)), n>=0, with IT(n,m):= A164658(n,m)/A164659(n,m) (coefficient triangle from the indefinite integral Integral_{x} T(n,x), n>=0, in lowest terms).

Let $T_m$ be the first-kind Chebyshev polynomial over $\mathbb Q$ and put $S_m=\sum_{j=1}^{m+1}[X^{j-1}]T_m/j$ and $a(m)=\operatorname{num}(S_m)$, where the numerator is reduced and signed with positive denominator. Let $R(n)=\operatorname{rad}(n)$, including $R(1)=1$. For every positive natural $n$, the theorem proves

$$ a(n+1)=\prod_{k=1}^{n}\left(1-2\left(\left\lfloor\frac{k^n}{n}\right\rfloor-\left\lfloor\frac{k^n-1}{n}\right\rfloor\right)\right)=(-1)^{n/R(n)}. $$

The division and subtraction inside each floor are rational; $n/R(n)$ is natural integer division.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Integrated coefficient triangle and row sum | rowSum | The coefficient of degree $j-1$ is divided by $j$ and summed over the inclusive interval $1\le j\le m+1$. The integrated coefficient is inlined. | | Reduced signed numerator | a | Rat.num retains the positive-denominator convention. | | Literal floor product | floorProduct | Rational subtraction precedes integer floor; the inclusive range is $1\le k\le n$. | | $n>0$, both equalities | claim, result | All positive naturals, including one; no extra hypothesis. | | A003557 exponent | Existing natural radical | $n/\operatorname{rad}(n)$, including the value at one. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/BrowneA164660ChebyshevFloorProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Dated conjecture statement: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedSource 2013Barry's Riordan-Pascal kernel determinantIn this note, we investigate some properties of the two-parameter generalized Pascal triangle M^{(m)}(a, b) given by the Riordan array M^{(m)}(a, b) = ( 1/(1 − ax) , x(1 + bx)/(1 − ax)^m ).Proved

The questionBarry defines the generalized Pascal matrix on printed page 1:

Why it mattersConjecture 32 asks for both an exact matrix identity and its determinant consequence for every truncation size.

The idea(A) Coefficient extraction from Sum{j=0}^{n} Pj(x)Pj(y) gives Sumj T{j,i}T{j,k} in entry (i,k), exactly the corresponding entry of Mn^t Mn.

Our contributionIn this note, we investigate some properties of the two-parameter generalized Pascal triangle M^{(m)}(a, b) given by the Riordan array M^{(m)}(a, b) = ( 1/(1 − ax) , x(1 + bx)/(1 − ax)^m ).

Verification details
Release source recordProved in Lean

Open problem

Barry's Riordan-Pascal kernel determinant

Barry defines the generalized Pascal matrix on printed page 1:

In this note, we investigate some properties of the two-parameter generalized Pascal triangle M^{(m)}(a, b) given by the Riordan array M^{(m)}(a, b) = ( 1/(1 − ax) , x(1 + bx)/(1 − ax)^m ).

The row polynomials are defined on printed page 19:

We can regard the matrix M^{(m)}(a, b) as the coefficient array of a family of polynomials P_n(x; m, a, b). We have P_n(x; m, a, b) = Σ_{k=0}^{n} T^{(m)}_{n,k}(a, b) x^k.

The comparison that introduces the kernel appears later on printed page 19:

It is interesting to compare the matrices defined by the quotient above with those defined by Σ_{j=0}^{n} P_j(x; m, a, b)P_j(y; m, a, b). In the case of monic orthogonal polynomials, such expressions lead to the same matrices. However, M^{(m)}(a, b) is not in general the coefficient array of a family of monic orthogonal polynomials, and hence the two family of matrices will in general differ. In the case m = 2, a = b = 1, we can conjecture the following.

Conjecture 32 is stated on printed page 20:

Conjecture 32. The matrix D̃_n(2, 1, 1) with generating function Σ_{k=0}^{n} P_j(x; 2, 1, 1)P_j(y; 2, 1, 1) is given by M_n^{(2)}(a, b)^t M_n^{(2)}(a, b). We then have |D̃_n(2, 1, 1)| = 1 for n ≥ 0. In the above the notation M_n denotes the matrix formed from the first (n + 1) rows and columns of M.

The page-20 lower summation index is printed as k=0, while its summand uses P_j. The unambiguous comparison display on page 19 uses j=0,...,n, so the formal sum uses that index. The right side is read at (a,b)=(1,1), as fixed by the sentence immediately preceding the conjecture and by its left side. The generic reading fails already at n=1: Dtilde_1 = [[2,1],[1,1]], while M_1(a,b)^t M_1(a,b) = [[1+a^2,a],[a,1]].

The phrase "with generating function" uses the coefficient-array convention from the same section: entry (i,k) is the coefficient of x^i y^k. Section 2 defines a Riordan array entry by T_{n,k} = [x^n] g f^k, with columns indexed from zero. Thus the quantified claim is, for every natural n, Dtilde_n = M_n^t M_n and det Dtilde_n = 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/BarryRiordanPascalKernelDeterminant.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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    Year recorded in the source bibliography: 2013.

    Read source ↗
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Quantum informationQuestion resolvedFirst preprint 2016Four-qubit pair-correlation monogamyThe twenty-seven squared two-body Pauli correlations of a normalized four-qubit vector sum to at most three.Proved

The questionWiseman, Volume monogamy of quantum steering ellipsoids for multi-qubit systems, Phys.

Why it mattersThe conjecture constrains how strongly one qubit can correlate with three other qubits through their two-body spin observables.

The ideaAll helper names in this section belong to FourQubitCorrelationMonogamy unless qualified otherwise.

Our contributionThe twenty-seven squared two-body Pauli correlations of a normalized four-qubit vector sum to at most three.

Verification details
Release source recordProved in Lean

Open problem

Four-qubit pair-correlation monogamy

S. Cheng, A. Milne, M. J. W. Hall and H. M. Wiseman, Volume monogamy of quantum steering ellipsoids for multi-qubit systems, Phys. Rev. A 94, 042105 (2016), arXiv:1606.02390v2, Sec. III C, p. 6, Eq. (36) (finalversion.tex, lines 395–399):

Finally, we conjecture that inequality (35) generalises to

$\operatorname{Tr}[T_{AB}T_{AB}^{\top}]+\operatorname{Tr}[T_{AC}T_{AC}^{\top}]+\operatorname{Tr}[T_{AD}T_{AD}^{\top}]\leq 3.$

for all pure 4-qubit states.

The source defines $T_{jk}=\operatorname{Tr}[\rho_{AB}\sigma_j\otimes\sigma_k]$ for $j,k=1,2,3$ in Sec. II, Eq. (2) (finalversion.tex, lines 129–133). Thus each term in Eq. (36) is the sum of the nine squared real entries of a two-qubit correlation matrix.

Issue #26136 fixes the reading. In the module, a vector $\psi:(\mathrm{Fin},4\to\mathrm{Fin},2)\to\mathbb C$ has its sixteen computational-basis amplitudes indexed by four bits. The qubits are ordered $A,B,C,D$. The indices $X=0,1,2$ denote targets $B,C,D$, and $i,j=0,1,2$ denote Pauli directions $x,y,z$. Let $P_{Xij}$ be wordOp (twoBodyWord X i j): Pauli $i$ on $A$, Pauli $j$ on target $X$, and identities elsewhere. The definition rankOneDensity is the outer product $\psi\psi^{\dagger}$, and correlation is

$$c_{Xij}(\psi)=\operatorname{Re}\operatorname{Tr}[\psi\psi^{\dagger}P_{Xij}].$$

The proposition claim states, exactly,

$$\forall\psi,\quad \sum_q\overline{\psi(q)}\psi(q)=1\quad\Longrightarrow\quad \sum_{X=0}^{2}\sum_{i=0}^{2}\sum_{j=0}^{2}c_{Xij}(\psi)^2\leq3.$$

The module proves result : claim. The correspondence with reduced-state correlation matrices is explained below.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumStates/FourQubitCorrelationMonogamy.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2018Orthocross MIC elements are never orthogonal, and their Gram entries tend to zeroOrthocross Gram entries are weighted squared moduli of inverse-frame pairings.Proved

The questionThe verbatim statements are in the literature note.

Why it mattersThe orthocross MICs, introduced by Caves, Fuchs and Schack for the quantum de Finetti theorem, are the standard reference measurements built from a single basis.

Our contributionOrthocross Gram entries are weighted squared moduli of inverse-frame pairings.

Verification details
Release source recordProved in Lean

Open problem

Orthocross MIC elements are never orthogonal, and their Gram entries tend to zero

J. B. DeBrota, C. A. Fuchs and B. C. Stacey (arXiv:1812.08762v5, quant-ph; Int. J. Quantum Inf. 19 (2021) 2040005) define the orthocross MIC of an orthonormal basis ${|j\rangle}$ of $\mathbb C^d$:

  • $\Pi_\alpha$ runs over the $d^2$ projectors onto $|j\rangle$, $(|j\rangle+|k\rangle)/\sqrt2$ and $(|j\rangle+i|k\rangle)/\sqrt2$ ($j<k$);
  • $\Omega=\sum_\alpha\Pi_\alpha$;
  • $E_\alpha=\Omega^{-1/2}\Pi_\alpha\Omega^{-1/2}$;
  • the Gram matrix is $G_{\alpha\beta}=\operatorname{tr}E_\alpha E_\beta$.

The paper states:

The entries in $G$ for orthocross MICs can become arbitrarily small with increasing $d$, but no two elements of an orthocross MIC can be exactly orthogonal.

The verbatim statements are in the literature note. Issue #14213 reads the conjecture as two clauses, both for every $d$ and every orthonormal basis:

  • (A) every off-diagonal Gram entry is a strictly positive real number;
  • (B) every Gram entry tends to zero uniformly as $d\to\infty$.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/OrthocrossNonorthogonality.result

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Quantum informationQuestion resolvedFirst preprint 2019The exact three-coherence threshold of the Ramsey moment ratio R3The greatest third-moment Ramsey ratio for finite mixtures of unit vectors with at most three nonzero coordinates, over all finite dimensions, integer spectra and unit measurement vectors, is an attained algebraic value.Proved

The questionMintert, Characterisation of multi-level quantum coherence without ideal measurements (Phys.

Why it mattersThe threshold decides when an interference experiment certifies four-coherence.

The ideaWrite N(r) = r⁶ + 30r⁴ + 90r² + 20, D(r) = (r + 2)²(r² + 2)² and P(r) = r⁷ − 13r⁶ + 6r⁵ − 60r⁴ + 30r³ − 30r² + 140r − 20.

Our contributionThe greatest third-moment Ramsey ratio for finite mixtures of unit vectors with at most three nonzero coordinates, over all finite dimensions, integer spectra and unit measurement vectors, is an attained algebraic value.

Verification details
Release source recordProved in Lean

Open problem

The exact three-coherence threshold of the Ramsey moment ratio R3

B. Dive, N. Koukoulekidis, S. Mousafeiris and F. Mintert, Characterisation of multi-level quantum coherence without ideal measurements (Phys. Rev. Research 2, 013220 (2020); arXiv:1901.08599), certify k-coherence from the interference pattern p(t) = ⟨χ|e^{−iHt} ρ e^{iHt}|χ⟩ of a Ramsey-type experiment.

  • The certifier is the moment ratio R3 = M3/M1², with M_q = (1/2π)∫₀^{2π} p(t)^q dt.
  • R3 is convex in ρ.
  • Its maximum over states that are not (k + 1)-coherent is a threshold: a measured R3 above it proves at least (k + 1)-coherence, whatever the Hamiltonian and the measurement vector χ.
  • Table I gives the thresholds 1 (k = 1) and 5/4 (k = 2). For k = 3 it gives only the analytic upper bound 179/96 ≈ 1.86 and the numerical value 1.77, with the remark "In the cases where the exact threshold values are not known, we use numerical methods to approximate them".

The verbatim statements are in the literature note.

Issue #18378 fixes the reading. The k = 3 threshold is the maximum of R3 over:

  • all finite dimensions;
  • integer spectra;
  • unit measurement vectors;
  • finite mixtures of unit vectors with at most three nonzero amplitudes, with M1 > 0.

It is proved to equal N(α)/D(α) ≈ 1.7731757563.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/ThreeCoherenceRamseyThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2021At most binomially many inputs certify failure of d-message simulationAt most binomially many inputs certify that a finite classical channel requires more than d messages.Proved

The questionImmediately after Theorem Thm:main2, the source states (main.tex, line 409):

Why it mattersA bounded certificate makes failure of classical simulation testable on a small subchannel.

The ideaIt produces a q common to all feasible sets by applying channeldecomposition to each row-stochastic encoder and decoder and then mixturecommon.

Our contributionAt most binomially many inputs certify that a finite classical channel requires more than d messages.

Verification details
Release source recordProved in Lean

Open problem

At most binomially many inputs certify failure of d-message simulation

B. Doolittle and E. Chitambar, Certifying the Classical Simulation Cost of a Quantum Channel, arXiv:2102.12543v1 (Physical Review Research 3, 043073, 2021), define a classical channel by its conditional probabilities $P(y|x)$ and define d-simulation by [ P(y|x)=\sum_\lambda q(\lambda)\sum_{m\in[d]}R_\lambda(y|m)T_\lambda(m|x). ] Their signaling polytope $\mathcal C_d^{n\to n'}$ is the set of channels with such a simulation. Immediately after Theorem Thm:main2, the source states (main.tex, line 409):

We conjecture that in general, no more than $\binom{n'}{d}$ inputs are needed to certify that a channel $\mbf{P}\in\mc{P}^{n\to n'}$ has a signaling dimension larger than $d$.

The settled finite formulation uses natural numbers $m,k,d$, a channel p : \mathrm{Fin}\,m \to \mathrm{Fin}\,k \to \mathbb R, the imported predicate IsRowStochastic p, and the finite shared-randomness predicate IsSimulable d p. For $1\le d$ and $d<k$, if $p$ is row stochastic and is not d-simulable, the theorem asserts a nonempty finite input set $i$ with $i.\mathrm{card}\le k.choose\ d$ such that the input restriction $(i : \mathrm{Set}\ _) .\mathrm{domRestrict}\ p$ is not d-simulable.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/SignalingDimensionHellyCertificate.result

Frozen module record

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Quantum informationConjecture provedFirst preprint 2012The Grandjean et al. Mermin-CGLMP facet conjectureThe Mermin-CGLMP inequality has a saturating face of affine codimension one for every K >= 2.Proved

The questionGisin, “Bell inequalities for three systems and arbitrarily many measurement outcomes”, Physical Review A 85, 052113 (2012), arXiv:1204.3829v2, Section II, states:

Why it mattersShared-randomness mixtures of local response functions are exactly convex combinations of deterministic assignments.

The ideaThe general bridge in FacetRigidityBridge.rigidityfacetbridge turns rigidity on saturating generators into affine codimension one.

Our contributionThe Mermin-CGLMP inequality has a saturating face of affine codimension one for every K >= 2.

Verification details
Release source recordProved in Lean

Open problem

The Grandjean et al. Mermin-CGLMP facet conjecture

B. Grandjean, Y.-C. Liang, J.-D. Bancal, N. Brunner and N. Gisin, “Bell inequalities for three systems and arbitrarily many measurement outcomes”, Physical Review A 85, 052113 (2012), arXiv:1204.3829v2, Section II, states:

We conjecture that inequality (1) is indeed facet-defining for all $K\ge2$.

For two settings and $K$ outcomes at each of three parties, its functional is

$$ I_K=\langle[A_2-B_1+C_1]_K\rangle+ \langle[A_1+B_2-C_1]_K\rangle+ \langle[-A_1+B_1+C_2]_K\rangle+ \langle[-A_2-B_2-C_2-1]_K\rangle. $$

Here $[X]_K$ is the least nonnegative residue and $\langle[X]K\rangle=\sum{j=0}^{K-1}jP(X=j\bmod K)$. Let $\mathcal L_K$ be the convex hull of all deterministic full behaviours $p_s(a,b,c\mid x,y,z)=[A_x=a][B_y=b][C_z=c]$. The settled statement is validity $I_K(p)\ge K-1$ on $\mathcal L_K$ and

$$ \dim\operatorname{aff}{p\in\mathcal L_K:I_K(p)=K-1}+1 =\dim\operatorname{aff}\mathcal L_K \qquad(K\ge2). $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/MerminCglmpFacet.result

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Quantum informationQuestion resolvedFirst preprint 2015The qudit CHSH orthogonality graph is (2d − 1)(d − 1)-regular for every prime dFor every prime p, the qudit CHSH projectors are distinct trace-one idempotents and their orthogonality graph has degree (2p-1)(p-1).Proved

The questionVala, Quantum Contextuality with Stabilizer States, Entropy 15 (2013) 2340–2362, arXiv:1501.04342v1, Section 5:

Why it mattersThe graph turns the qudit CHSH inequality into a noncontextuality inequality for stabilizer projectors.

The ideaFor p = 2 the eigenvectors of X and Y are explicit.

Our contributionFor every prime p, the qudit CHSH projectors are distinct trace-one idempotents and their orthogonality graph has degree (2p-1)(p-1).

Verification details
Release source recordProved in Lean

Open problem

The qudit CHSH orthogonality graph is (2d − 1)(d − 1)-regular for every prime d

M. Howard, E. Brennan and J. Vala, Quantum Contextuality with Stabilizer States, Entropy 15 (2013) 2340–2362, arXiv:1501.04342v1, Section 5:

Observe that for all $d \in {2,3,5,7}$ that we have discussed, we see some structure in the graphs $\Gamma_{\textrm{CHSH}}$ obtained from qudit CHSH inequalities. Whereas it is quite easy to see that all graphs will have order $|\Gamma_{\textrm{CHSH}}|=d^3$, we conjecture based on the examples here that $\Gamma_{\textrm{CHSH}}$ is a $(2d-1)(d-1)$-regular graph.

The operators A_j and B_k, the spectral projectors and their rank-one decomposition, and the graph convention are quoted in the literature note.

Issue #19492 fixes the reading:

  • the vertices are the d³ rank-one projectors Π^A_j[a] ⊗ Π^B_k[−jk − a] of the decomposition of the CHSH Bell operator;
  • two vertices are adjacent when the product of their projectors vanishes.

The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/StabilizerCHSHGraphRegularity.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2017The Jacq–Lardizabal mean-hitting-time form is falseA complex reset pair refutes the proposed three-entry formula for gambler's-ruin mean hitting times.Refuted

The questionLardizabal, Open Quantum Random Walks on the Half-Line: The Karlin–McGregor Formula, Path Counting and Foster's Theorem (J.

Why it mattersA counterexample determines that this reduction cannot be used for all complex pairs of transition matrices.

The idea2. wordoperatorappend proves, with no additional hypotheses, that wordOperator l r (a ++ b) = wordOperator l r b wordOperator l r a.

Our contributionA complex reset pair refutes the proposed three-entry formula for gambler's-ruin mean hitting times.

Verification details
Release source recordRefuted in Lean

Open problem

The Jacq–Lardizabal mean-hitting-time form is false

T. S. Jacq and C. F. Lardizabal, Open Quantum Random Walks on the Half-Line: The Karlin–McGregor Formula, Path Counting and Foster's Theorem (J. Stat. Phys. 169, 547–594, 2017; arXiv:1701.08400v3), state the following in Remark 5.3 of the source TeX JL_rev2.tex, lines 1291–1293:

Numerical experiments allows us to recognize a behavior that such OQW has in common with Example 5.1, namely, that the mean hitting time has the form $E_{k,\rho}(\tau)=1+h(M,M-k)\rho_{11}+h(M,k)\rho_{22}+j(M,k)Re(\rho_{12}),$ where $h, j$ are functions of $M$ and $k=1,\dots, M-1$, with $j$ being a function satisfying $j(M,k)=-j(M,M-k)$. We conjecture that this holds for the gambler's ruin problem associated to *every* pair or matrices $L$, $R$ with $L^*L+R^*R=I$ inducing an OQW on vertices ${0,\dots, M}$. A proof of this statement is, up to our knowledge, unknown.

The formal claim fixes two-dimensional complex matrices $l,r$ with $l^*l+r^*r=1$, every natural $m$ with $3\le m$, and functions $h,j:\mathbb N\to\mathbb N\to\mathbb R$. It requires the stated antisymmetry for every $k$ with $1\le k\le m-1$ and, for every such $k$ and every IsDensity matrix $\rho$, requires meanHittingTime to be different from $\top$ and its toReal value to equal $1+h(m,m-k)(\rho\ 0\ 0).re+h(m,k)(\rho\ 1\ 1).re+j(m,k)(\rho\ 0\ 1).re$. Here IsDensity is positive semidefiniteness together with trace one, and meanHittingTime has codomain ℝ≥0∞, the extended nonnegative reals.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Walks/OpenQuantumWalkGamblerHittingTimeRefutation.result

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    The argument establishes an obstruction to a proposed characterization or class of methods.

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Quantum informationQuestion resolvedFirst preprint 2014Realizability of the correlators of boxes violating chained monogamyKłobus, Oszmaniec, Augusiak and Grudka (arXiv:1408.1223, Section 5) conjecture that every vector of the correlators xA^i, yA^i, xB^i, yB^i satisfying their inequalities (ElPrat) is realized by a signaling box whose chained Bell expression plus twice the correlator of B0 and E equals 2M + Delta.Proved

The questionGrudka, Communication strength of correlations violating monogamy relations, Found.

Why it mattersThe communication strength of the violation is the least over violating boxes of the largest capacity of these channels.

Our contributionKłobus, Oszmaniec, Augusiak and Grudka (arXiv:1408.1223, Section 5) conjecture that every vector of the correlators xA^i, yA^i, xB^i, yB^i satisfying their inequalities (ElPrat) is realized by a signaling box whose chained Bell expression plus twice the correlator of B0 and E equals 2M + Delta.

Verification details
Release source recordProved in Lean

Open problem

Realizability of the correlators of boxes violating chained monogamy

W. Kłobus, M. Oszmaniec, R. Augusiak, A. Grudka, Communication strength of correlations violating monogamy relations, Found. Phys. 46 (2016) 620–634, arXiv:1408.1223v2, Section 5:

Although we cannot prove it as in the case $M=2$, we conjecture that all possible values of the correlators in $S_{A\to BE}^i$ and $S_{B\to AE}^i$ that satisfy inequalities (\ref{ElPrat}) can always be realized with some signaling probability distribution $\vec{p}$ for which $R_M(\vec{p})=2M+\Delta$.

A box is a family $p(a,b,e|A_i,B_j)$ of distributions on $a,b,e=\pm1$, $i,j=0,\dots,M-1$, with one setting $E$ for the third party, all one- and three-party expectation values zero, and a common value of $\langle B_0E\rangle_{A_i}$. $R_M=I^M_{AB}+2\langle B_0E\rangle$ with the chained Bell expression $I^M_{AB}=\sum_k(\langle A_kB_k\rangle+\langle A_{k+1}B_k\rangle)$, $A_M=-A_0$, and $\Delta\in[0,2]$. The coordinates are $x_A^i=\langle B_iE\rangle_{A_i}$, $y_A^i=\langle B_iE\rangle_{A_{i+1}}$, $x_B^i=\langle A_iE\rangle_{B_{i-1}}$, $y_B^i=\langle A_iE\rangle_{B_i}$ ($1\le i\le M-1$), $x_B^0=\langle A_0E\rangle_{B_0}$, $y_B^0=\langle A_0E\rangle_{B_{M-1}}$, and (ElPrat) is $$\sum_{i=1}^{M-1}(-1)^{a_i}(x_A^i-y_B^i)+\sum_{i=1}^{M-2}(-1)^{b_i} (x_B^{i+1}-y_A^i)+(-1)^c(y_A^{M-1}+y_B^0)+x_B^1+x_B^0\ge\Delta$$ for all $a_i,b_i,c\in{0,1}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/ChainedMonogamySignalingRealizability.result

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Quantum informationQuestion resolvedFirst preprint 2016Lin: both spectral properties of the map ΨThe map Psi gives both indexed spectral properties in every dimension.Proved

The questionLin, New properties for certain positive semidefinite matrices, arXiv:1609.08231v1, §3, Example 3.6; the journal version is Linear Algebra and its Applications 520 (2017), 32–43.

Why it mattersThese are comparisons at each spectral index, with no commutation, invertibility or rank assumption on the original blocks.

The ideaEntrywise calculation gives Σ Fkl T Fkl=(tr T)I−Tᵀ.

Our contributionThe map Psi gives both indexed spectral properties in every dimension.

Verification details
Release source recordProved in Lean

Open problem

Lin: both spectral properties of the map Ψ

The fixed source is M. Lin, New properties for certain positive semidefinite matrices, arXiv:1609.08231v1, §3, Example 3.6; the journal version is Linear Algebra and its Applications 520 (2017), 32–43. The setup states:

we assume that [A X; X∗ B], where A, X, B are n × n, is positive semidefinite.

Definition 1.1(iii) states:

M is said to have a-property if 2sj (M12 ) ≤ λj (M11 + M22 ), j = 1, . . . , n.

Definition 1.1(iv) states:

M is said to have g-property if sj (M12 ) ≤ λj (M11 ♯ M22 ), j = 1, . . . , n.

Example 3.6 states:

If we consider the map Ψ : X ↦ 2(trX)In − X, then using the approach in [14] we can show that the matrix [Ψ(A) Ψ(X); Ψ(X∗) Ψ(B)] is PPT. Though there are strong numerical evidence suggesting that this block matrix also has g-property, yet we have not even been able to show that it has a-property.

The concluding §5 states:

Besides the challenging problems described in Example 3.5 and Example 3.6, …

For every natural n, complex n-by-n blocks A,X,B with [A X; X* B] positive semidefinite, and every decreasing index j, prove both inequalities

$$ 2s_j(\Psi(X))\le\lambda_j(\Psi(A)+\Psi(B)), \qquad s_j(\Psi(X))\le\lambda_j(\Psi(A)#\Psi(B)), \quad \Psi(T)=2(\operatorname{tr}T)I-T. $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/LinPsiMap/Result.result

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Discrete structuresQuestion resolvedPosed in 2012Mathar's recurrence for the lattice-path returns A068551The sequence a(n) = 4^n - binomial(2n, n), the total number of returns to the axis in all lattice paths with steps (1,1) and (1,-1) from the origin to (2n, 0), satisfies n a(n) + 2(3 - 4n) a(n - 1) + 8(2n - 3) a(n - 2) = 0 for every n at least 2, as conjectured by R.Proved

The questionCritzer it counts the returns to the axis in all ±1 lattice paths (random-walk bridges) from the origin to (2n, 0).

The idea1. u(n) = C(2n, n) satisfies n u(n) = 2(2n − 1) u(n − 1) and v(n) = 4^n satisfies v(n) = 4 v(n − 1).

Our contributionThe sequence a(n) = 4^n - binomial(2n, n), the total number of returns to the axis in all lattice paths with steps (1,1) and (1,-1) from the origin to (2n, 0), satisfies n a(n) + 2(3 - 4n) a(n - 1) + 8(2n - 3) a(n - 2) = 0 for every n at least 2, as conjectured by R.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's recurrence for the lattice-path returns A068551

OEIS A068551 is a(n) = 4^n − C(2n, n); by a comment of G. Critzer it counts the returns to the axis in all ±1 lattice paths (random-walk bridges) from the origin to (2n, 0). Its formula field records:

Conjecture: n*a(n) + 2*(3-4*n)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Apr 01 2012

Issue #10754 fixes the reading: a is the entry's name, the recurrence is asserted for every n ≥ 2 and read in ℤ.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/LatticeReturnsRecurrence.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2018A coherence lower bound on the magic of pure qutritsFor every normalized pure qutrit state, the sum negativity is at least (C/2) sqrt(1 - C/2), where C is its l1 coherence.Proved

The questionPati, Coherence makes quantum systems 'magical', J.

Why it mattersThe sum negativity is a magic monotone for odd dimensions, and incoherent states are stabilizer states.

The idea1. Positive and negative parts. For three reals with positive mass P and negative mass N, the sum of squares is at most P² + N².

Our contributionFor every normalized pure qutrit state, the sum negativity is at least (C/2) sqrt(1 - C/2), where C is its l1 coherence.

Verification details
Release source recordProved in Lean

Open problem

A coherence lower bound on the magic of pure qutrits

C. Mukhopadhyay, S. Sazim and A. K. Pati, Coherence makes quantum systems 'magical', J. Phys. A 51, 414006 (2018), arXiv:1801.04807v3, Eq. (18), stated from numerical simulation:

The following condition on quantum coherence, quantified via the $l_1$-norm, and 'magic', quantified by the sum negativity, holds for qutrit pure states $M_{SN}[|\psi\rangle] \geq \frac{C_{l_{1}}[|\psi\rangle]}{2}\sqrt{1- \frac{C_{l_{1}}[|\psi\rangle]}{2}}$

Here $M_{SN} = \sum_u |W_u| - 1$ for the discrete Wigner function $W$ of the paper's Eq. (12), and $C_{l_1}$ is the sum of the moduli of the off-diagonal entries. The verbatim statements are in the literature note.

Issue #18729 fixes the reading: the bound for every normalized x ∈ ℂ³ with ρ = x x†. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/QutritMagicCoherenceBound.result

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Number theoryStructural characterizationPosed in 2020Schulte's factorial-square primality criterionSchulte's A006472 divisibility condition characterizes prime natural numbers.Proved

The questionConjecture: For n > 1, n divides 2a(n-1) + 4 if and only if n is prime. - Werner Schulte, Oct 04 2020

Why it mattersThe criterion combines an exact quotient of two neighboring factorials with a single divisibility test.

The ideaFor odd prime n, the scale 2^(n-2) is a unit modulo n.

Our contributionSchulte's A006472 divisibility condition characterizes prime natural numbers.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's factorial-square primality criterion

OEIS A006472, NAME (%N, verbatim):

a(n) = n!*(n-1)!/2^(n-1).

The offset is %O 1,3. The settled COMMENT line (%C, verbatim) is:

Conjecture: For n > 1, n divides 2*a(n-1) + 4 if and only if n is prime. - Werner Schulte, Oct 04 2020

The full-quantifier reading is: define, for every natural m, a(m) = m!*(m-1)!/2^(m-1) using natural-number division; then for every natural n with 2 <= n, n divides 2*a(n-1)+4 if and only if n is prime. The denominator divides the factorial product for every m >= 1, so all sequence values used by the criterion are exact quotients.

Only the quoted Conjecture sentence is settled, as a biconditional for every natural n >= 2. No other OEIS comment, formula, or stronger composite divisibility statement is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/SchulteFactorialSquarePrimeCriterion.result

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  • Contribution

    Schulte's A006472 divisibility condition characterizes prime natural numbers.

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  • Historical context

    Dated conjecture statement: 2020.

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Number theoryQuestion resolvedPosed in 2014The A008474 prime-exponent sum iteration reaches fiveEvery prime-exponent sum orbit starting above four reaches five.Proved

The questionIf n = Product (pj^kj) then a(n) = Sum (pj + kj).

Why it mattersThe conjecture asks for a global termination statement about an arithmetic map defined from the complete prime factorization of its input.

The idea(a) Establish coprime additivity of F and the value F(p^e) = p + e for prime p and positive e. (b) Prove F(n) <= n + 1 for every n >= 2. (c) Prove F(n) >= 5 for n >= 5, and one-step descent F(n) < n for composite n > 4 other than six. (d) Prove F(2n) <= F(n) + 3 and the prime two-step descent F(F(p)) < p for primes p >= 11. (e) Verify the four-cycle and close the theorem by strong induction.

Our contributionEvery prime-exponent sum orbit starting above four reaches five.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A008474 prime-exponent sum iteration reaches five

OEIS A008474, NAME (%N, verbatim):

If n = Product (p_j^k_j) then a(n) = Sum (p_j + k_j).

COMMENT (%C, verbatim):

Conjecture: for m > 4, by iterating the map m -> A008474(m) one always reaches 5 [tested up to m = 320000]. - Ivan N. Ianakiev, Nov 10 2014

FORMULA (%F, verbatim):

Additive with a(p^e) = p + e.

AUTHOR (%A, verbatim):

Olivier Gérard

The first terms (%S, verbatim excerpt) are 0,3,4,4,6,7,8,5,5,9,12,8,14,....

For every natural n, define F(n) = Sum (p + factorization(n)(p)) over p in primeFactors(n). The literal proved statement is forall m : Nat, 4 < m -> exists t : Nat, F^[t](m) = 5. Thus "reaches 5" means that some finite iterate equals 5. The values follow the cycle 5 -> 6 -> 7 -> 8 -> 5; four is a fixed point, so m > 4 is sharp; and 2 -> 3 -> 4 never reaches five. The generating function and the identities with A001222 and A008472 are NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/IanakievPrimeExponentSumIterationReachesFive.ianakiev_a008474

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Discrete structuresStructural characterizationPosed in 2021Exact factorial-square divisibility in A096127Exact factorial-square divisibility characterizes prime powers.Proved

The questiona(n) is the largest k such that (n^2)!/(n!)^k is an integer.

Why it mattersThe conjecture asks for an unbounded characterization of the indices where the first universally valid exponent is exact.

The ideaLegendre's formula writes vp(m!) = (m - sp(m))/(p-1), where sp is the base-p digit sum.

Our contributionExact factorial-square divisibility characterizes prime powers.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Exact factorial-square divisibility in A096127

OEIS A096127, NAME (verbatim):

a(n) is the largest k such that (n^2)!/(n!)^k is an integer.

COMMENT (verbatim):

Conjecture: a(n)=n+1 only when n is prime or a power of a prime. [Verified for n=2..5000. - Amiram Eldar, Apr 06 2021]

OEIS A096126 states the same assertion. The A096127 offset is 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/FactorialSquareDivisibilityPrimePower.factorial_square_exact_divisibility_iff_prime_power

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  • Contribution

    Exact factorial-square divisibility characterizes prime powers.

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  • Historical context

    Dated conjecture statement: 2021.

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Number theoryQuestion resolvedPosed in 2015The A112041 prime scalings force the index to be four or divisible by tenFour prime values at successive powers-of-three scalings force k to be four or divisible by ten.Proved

The questionNumbers k such that 1k + 1, 3k + 1, 9k + 1, 27k + 1 are all primes.

Why it mattersDale's comment asserts a divisibility pattern for every term after the first.

The ideaIf k is odd, primality forces k+1=2 and hence k=1, but then 3k+1=4 is composite.

Our contributionFour prime values at successive powers-of-three scalings force k to be four or divisible by ten.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A112041 prime scalings force the index to be four or divisible by ten

OEIS A112041, NAME (%N, verbatim):

Numbers k such that 1k + 1, 3k + 1, 9k + 1, 27k + 1 are all primes.

Harvey P. Dale's COMMENT conjecture (%C, verbatim):

Conjecture: all terms except the first are multiples of 10. - Harvey P. Dale, Mar 26 2015

The literal proved statement is forall k : Nat, 0 < k -> (Nat.Prime (k + 1) and Nat.Prime (3 * k + 1) and Nat.Prime (9 * k + 1) and Nat.Prime (27 * k + 1)) -> (k = 4 or 10 divides k). Infinitude of A112041, a characterization beyond this implication, and all other comments on A112041 are NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/DaleTripleScalingPrimeMultiplesOfTen.result

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Number theoryQuestion resolvedPosed in 2017The A136162 next-prime product has four prime neighbors only at q=3Krizek's next-prime product has four prime neighbors at offsets two and four exactly when the prime input is three.Proved

The questionList of prime quadruplets {p, p+2, p+6, p+8}.

Why it mattersKrizek's comment asks whether multiplying a prime by its successor prime can place all four offsets two and four on primes only in the displayed quadruplet.

The ideaThe prime input q=2 gives Q-2=4, so the four primality assumptions fail; q=3 gives Q=15 and the prime neighbors 11, 13, 17, 19.

Our contributionKrizek's next-prime product has four prime neighbors at offsets two and four exactly when the prime input is three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A136162 next-prime product has four prime neighbors only at q=3

OEIS A136162, NAME (%N, verbatim):

List of prime quadruplets {p, p+2, p+6, p+8}.

Jaroslav Krizek's COMMENT (%C, verbatim, including both sentences and its attribution):

{11, 13, 17, 19} is the only prime quadruplet {p, p+2, p+6, p+8} of the form {Q-4, Q-2, Q+2, Q+4} where Q is a product of a pair of twin primes {q, q+2} (for prime q = 3) because numbers Q-2 and Q+4 are for q>3 composites of the form 3*(12k^2-1) and 3(12k^2+1) respectively (k is an integer). Conjecture: {11, 13, 17, 19} is the only prime quadruplet {p, p+2, p+6, p+8} of the form {q(nextprime(q))-4, q*( nextprime(q))-2, q*( nextprime(q))+2, q*( nextprime(q))+4} where q is a prime (for prime q = 3). - Jaroslav Krizek, Jul 07 2017

Define nextPrime(q) as the least prime greater than or equal to q+1, implemented by Nat.find (Nat.exists_infinite_primes (q + 1)), and define Q(q) = q * nextPrime(q). The literal proved statement is forall q : Nat, q.Prime -> ((Nat.Prime (Q q - 4) and Nat.Prime (Q q - 2) and Nat.Prime (Q q + 2) and Nat.Prime (Q q + 4)) iff q = 3).

The preceding twin-prime-product sentence, any other property or classification of prime quadruplets, and every other A136162 comment are NOT claimed. Natural subtraction is truncated subtraction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/KrizekNextPrimeProductQuadruplet.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryQuestion resolvedPosed in 2017The rational-denominator cube shift in A152020A residue-class gcd identity proves Cicuttin's denominator formula for OEIS A152020.Proved

The questiona(n) = n^2/gcd(n^2, 8). - Andrew Howroyd, Jul 25 2018

Why it mattersThis OEIS conjecture gives a closed expression for the reduced denominator sequence at every index in its offset range.

The ideaApplying this formula to both rational expressions reduces the theorem to gcd(n^2,(n-2)^3) = gcd(n^2,8) for n >= 2.

Our contributionA residue-class gcd identity proves Cicuttin's denominator formula for OEIS A152020.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The rational-denominator cube shift in A152020

OEIS A152020, NAME:

Denominator of 8/(9*n^2) divided by 9.

FORMULA (Andrew Howroyd, Jul 25 2018):

a(n) = n^2/gcd(n^2, 8). - Andrew Howroyd, Jul 25 2018

FORMULA (Andres Cicuttin, Sep 19 2017):

Conjecture: a(n) = denominator((n-2)^3/n^2). - Andres Cicuttin, Sep 19 2017

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/RationalDenominatorCubeShift.cicuttin_a152020

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2018Laguerre(n, 4) has odd numerator and denominatorLaguerre(n,4) has odd reduced numerator and denominator for every natural n.Proved

The questionRobert Israel's COMMENT conjecture (%C, verbatim):

Why it mattersThe two sequences separately record the reduced numerator and denominator of the same rational polynomial value.

The ideaFor every nonconstant summand, combine padicValNatfactorialltofnezero with padicValRat.mul, padicValRat.div, padicValRat.pow, padicValRat.ofnat, and padicValRat.neg to show that its 2-adic valuation is positive.

Our contributionLaguerre(n,4) has odd reduced numerator and denominator for every natural n.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Laguerre(n, 4) has odd numerator and denominator

OEIS A160627, NAME (%N, verbatim):

Numerator of Laguerre(n, 4).

Robert Israel's COMMENT conjecture (%C, verbatim):

Conjecture: all terms are odd. - Robert Israel, Mar 29 2018

OEIS A160628, NAME (%N, verbatim):

Denominator of Laguerre(n, 4).

Robert Israel's COMMENT conjecture (%C, verbatim):

Conjecture: all terms are odd. - Robert Israel, Mar 29 2018

The literal proved statement defines L(n) = sum_{k=0}^n C(n,k)(-4)^k/k! in the rationals and proves forall n : Nat, Odd (L n).num and Odd (L n).den. Here num and den are the numerator and denominator of the reduced rational value. The definition is the explicit binomial sum for the classical Laguerre polynomial at x=4; it does not use a Mathlib Laguerre polynomial API.

No value other than x=4, property beyond reduced numerator and denominator parity, other A160627 or A160628 comment, or priority claim is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/IsraelLaguerreFourParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedPosed in 2017The A208342 coefficient formula for a jointly generated polynomial pairSchulte's binomial sum gives every coefficient in the jointly generated polynomial pair.Proved

The questionTriangle of coefficients of polynomials u(n,x) jointly generated with A208343; see the Formula section.

Why it mattersThe conjecture gives an explicit coefficient formula for every row of a triangle defined only through a coupled polynomial recursion.

The ideaUse simultaneous induction on the recursion stage for the coefficient pair (u,v).

Our contributionSchulte's binomial sum gives every coefficient in the jointly generated polynomial pair.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A208342 coefficient formula for a jointly generated polynomial pair

OEIS A208342, NAME:

Triangle of coefficients of polynomials u(n,x) jointly generated with A208343; see the Formula section.

The defining FORMULA lines are:

u(n,x) = u(n-1,x) + x*v(n-1,x),

v(n,x) = x*u(n-1,x) + x*v(n-1,x),

where u(1,x) = 1, v(1,x) = 1.

With T(n,k) = [x^(k-1)] u(n,x), the conjecture is:

T(n,k) = Sum_{j=0..floor((k-1)/2)} binomial(k-1-j,j)*binomial(n-k+j,j) for k,n>0 and k<=n (conjectured). - Werner Schulte, Mar 07 2017

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/JointPolynomialPairBinomialCoefficients.schulte_a208342

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2021Kurkov's alternating binary run countAlternating overlapping binary one-block counts give Kurkov's A329320 identity.Proved

The questionLet cn(n,k) denote the number of times 11..1 (k 1's) appears in the binary representation of n; a(n) = n - cn(n,1) + cn(n,2) - cn(n,3) + cn(n,4) - ... .

Why it mattersThe left side counts overlapping blocks across all lengths with alternating signs.

The ideaFor a natural m, all of its lowest k bits are one exactly when 2^k divides m+1.

Our contributionAlternating overlapping binary one-block counts give Kurkov's A329320 identity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's alternating binary run count

OEIS A239907, NAME (%N, verbatim):

Let cn(n,k) denote the number of times 11..1 (k 1's) appears in the binary representation of n; a(n) = n - cn(n,1) + cn(n,2) - cn(n,3) + cn(n,4) - ... .

The settled FORMULA (%F, verbatim):

Conjecture: a(n) = n - A329320(n) for n >= 0 (noticed by Sequence Machine). - Mikhail Kurkov, Oct 13 2021

AUTHOR (%A, verbatim):

N. J. A. Sloane, Apr 07 2014

Occurrences overlap. In the binary word 111, the counts for block lengths 1, 2, and 3 are 3, 2, and 1; hence a(7)=7-3+2-1=5, as required by the data. Numbering bits from the least significant end preserves counts of the palindromic block of ones. The index range has length n.size, the binary length, with the empty range at zero.

For every natural n, define cn(n,k) to count starts i<n.size with n.testBit(i+j)=true for every j<k. Define b(n) to count starts i<n.size for which v_2(n div 2^i+1) is odd, where div is natural floor division. For positive n this is the A329320 NAME sum under the supplied characterization 1-A035263(m)=[v_2(m) odd] for positive m. All valuation arguments are positive. At n=0 both ranges are empty, so the identity reads 0=0.

The exact Lean statement, with integer subtraction and integer summands, is:

theorem result (n : ℕ) :
    (n : ℤ) - ∑ k ∈ Finset.Icc 1 n.size, (-1 : ℤ) ^ (k + 1) * (cn n k : ℤ) =
      (n : ℤ) - (b n : ℤ)

Only Kurkov's quoted %F line is settled. The interpretation of A329320 uses the cited valuation characterization; the morphism-to-valuation identity for A035263 is not independently formalized here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/KurkovRunAlternatingCount.result

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    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryEquivalence provedPosed in 2016The A260407 condition holds at every positive multiple of the orderKrizek's A260407 condition at exponent n-1 is equivalent to the condition at every positive multiple of that exponent.Proved

The questionNumbers n such that (n-1)^2+1 divides 2^(n-1)-1.

Why it mattersKrizek's comment asks whether the defining divisibility condition for A260407 already characterizes divisibility at every positive multiple of the exponent.

The ideaFor the converse, specialize the universal statement at k=1 and normalize the exponent.

Our contributionKrizek's A260407 condition at exponent n-1 is equivalent to the condition at every positive multiple of that exponent.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A260407 condition holds at every positive multiple of the order

OEIS A260407, NAME (%N, verbatim):

Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.

Jaroslav Krizek's COMMENT (%C, verbatim):

Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - Jaroslav Krizek, Jun 02 2016

Define modulus(n) = (n-1)^2+1, using natural-number subtraction. The literal proved statement is forall n : Nat, 1 <= n -> (inSequence n iff allMultiples n), where inSequence n means modulus(n) divides 2^(n-1)-1, and allMultiples n means that for every natural k >= 1, modulus(n) divides 2^(k*(n-1))-1. The equality (2^k)^(n-1) = 2^(k*(n-1)) identifies the COMMENT exponent with the formal exponent.

The claim that a(7) = 1382401 is the first composite term, the claim that the Fermat numbers 2^(2^k)+1 for k > 1 form a subsequence, Hasler's further conjecture that those numbers are the intersection with A260406, anything about A260406 or A260072, and every other A260407 comment are NOT claimed. The textbook divisibility fact for arbitrary m and d is also not claimed as a new result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/KrizekAllMultiplesOfTheOrder.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Krizek's A260407 condition at exponent n-1 is equivalent to the condition at every positive multiple of that exponent.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2016.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryQuestion resolvedPosed in 2020Ordowski's A306270 semiprime classificationComposite numbers k such that b^(k(k-1)) == 1 (mod k^2) for every b coprime to k.Proved

The questionComposite numbers k such that b^(k(k-1)) == 1 (mod k^2) for every b coprime to k.

Why it mattersThe independent question is Ordowski's named classification of every semiprime greater than four in A306270.

The ideaConvert the universal congruence into carmichael ((pq)^2) ∣ (pq)(pq-1) through the exponent of the unit group modulo (pq)^2.

Our contributionComposite numbers k such that b^(k(k-1)) == 1 (mod k^2) for every b coprime to k.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Ordowski's A306270 semiprime classification

OEIS A306270, NAME (%N, verbatim):

Composite numbers k such that b^(k(k-1)) == 1 (mod k^2) for every b coprime to k.

The offset is %O 1,1. The explanatory COMMENT (%C, verbatim) is:

These are composites k such that lambda(k^2) divides k(k-1), where lambda is the Carmichael function A002322.

Ordowski's settled COMMENT (%C, verbatim) is:

Conjecture: all semiprimes > 4 in this sequence are in A190275. - Thomas Ordowski, Jul 19 2020

The following COMMENT records the finite verification:

The conjecture was verified up to 1063290841. - Amiram Eldar, Jul 19 2020

OEIS A190275, NAME (%N, verbatim):

Semiprimes of the form p*(p^2 - p + 1).

Define mem k to mean that k is composite, 1 < k, and for every natural b coprime to k, b^(k*(k-1)) ≡ 1 [MOD k^2]. The full-quantifier reading is: for every pair of natural primes p and q with p ≤ q, if 4 < p*q and mem (p*q), then q = p^2-p+1. Thus the semiprime is p*(p^2-p+1), as required for A190275. All subtraction is truncated natural subtraction. The scope wall is only Ordowski's Conjecture sentence for every semiprime k > 4; no assertion about non-semiprimes or the finite verification bound is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/OrdowskiSemiprimeCarmichaelSquareClassification.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Dated conjecture statement: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2019M-resistant qudit states exist for every admissible loss parameterFor every N at least three and every m at most N minus two, a normalized pure state is entangled after every m-particle loss and fully separable after every (m+1)-particle loss.Proved

The questionQuinta, André, Burchardt and Życzkowski, Cut-resistant links and multipartite entanglement resistant to particle loss, Phys.

Why it mattersThe source leaves existence in some local dimension open for the full admissible range.

The ideaThe state is the uniform superposition of the resulting distinct product-basis configurations.

Our contributionFor every N at least three and every m at most N minus two, a normalized pure state is entangled after every m-particle loss and fully separable after every (m+1)-particle loss.

Verification details
Release source recordProved in Lean

Open problem

M-resistant qudit states exist for every admissible loss parameter

Quinta, André, Burchardt and Życzkowski, Cut-resistant links and multipartite entanglement resistant to particle loss, Phys. Rev. A 100, 062329 (2019), arXiv:1908.10390v1, Section V, state:

Conjecture. For any $N$ and $m$, there exists an $m$-resistant $N$-qudit state in some local dimension $d$.

Their Section III definition says that an entangled state remains entangled after tracing out any $m$ subsystems and becomes separable after tracing out any $m+1$ subsystems. This dossier uses finite full separability for the separable predicate and the admissible range $N\geq3$, $m+2\leq N$.

Statement echo

The Lean claim quantifies all natural $N,m$ with $3\leq N$ and $m+2\leq N$, then supplies a natural local dimension $d$, a pure amplitude function on $(\mathrm{Fin},N)\to\mathrm{Fin},d$, unit norm by the finite sum of Complex.normSq, and IsMResistant m ψ. IsMResistant is the conjunction of initial failure of full separability, failure after every loss set of cardinality $m$, and full separability after every loss set of cardinality $m+1$. The pure-state convention is the frozen rankOneDensity declaration.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/MResistantQuditExistence.result

Frozen module record

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Quantum informationConjecture provedFirst preprint 2019Rouleux's affine-determinant conjecture for alternating polynomialsAlternating complex polynomials of total degree at most d in d+1 vector variables are scalar affine determinants; with more variables they vanish.Proved

The questionRouleux, Multimode entanglement for fermions, arXiv:1910.04790v1 (2019), §3, "Affine antisymmetric forms in higher dimensions or of lower degree", states in 3fermiProc.tex, lines 266–269:

Why it mattersThe conjecture asks whether antisymmetry and degree force this algebraic choice, and whether another vector variable is possible at the same degree.

The ideaPermuting the points permutes its columns, giving affineDetalternating d.

Our contributionAlternating complex polynomials of total degree at most d in d+1 vector variables are scalar affine determinants; with more variables they vanish.

Verification details
Release source recordProved in Lean

Open problem

Rouleux's affine-determinant conjecture for alternating polynomials

M. Rouleux, Multimode entanglement for fermions, arXiv:1910.04790v1 (2019), §3, "Affine antisymmetric forms in higher dimensions or of lower degree", states in 3fermiProc.tex, lines 266–269:

We conjecture that: (1) the affine determinant is the only non degenerate antisymmetric tensor $\omega$ on ${\bf C}^d$ of degree $d$, depending on $d+1$ variables, and: (2) there is no non trivial antisymmetric tensor on ${\bf C}^d$ of degree $d$, depending on more than $d+1$ variables.

The preceding discussion specifies "degree being understood as degree of homogeneity" (line 247) and uniqueness "up to a factor" (lines 264–265). Definition 3, lines 251–253, gives the determinant of the differences from the first point. The source and its conventions are described in the literature note.

Issue #23528 fixes the polynomial reading. For natural numbers $d,r$, write $X_{i,j}$ for coordinate $j$ of vector variable $i$, with $0\le i<r$ and $0\le j<d$, and let $f$ be a complex polynomial in these $rd$ coordinates. IsAlternating f means that renaming the vector indices by any permutation $\sigma$, while fixing coordinate indices, multiplies $f$ by $\operatorname{sign}(\sigma)$. No multilinearity is assumed. Write $D_d=\texttt{affineDet}\ d$ for the polynomial whose evaluation is $\det(x_1-x_0,\ldots,x_d-x_0)$.

The four conjuncts of result establish:

  1. For every $d,r$ with $r\ge d+2$, every alternating $f$ of total degree at most $d$ is zero.
  2. For every $d$, $D_d$ is alternating and has total degree at most $d$.
  3. For every $d$, every alternating $f$ in $d+1$ vector variables of total degree at most $d$ is $cD_d$ for some $c\in\mathbb C$.
  4. For every $d\ge2$, the evaluated $D_d$ is non-degenerate: for every family $x_1,\ldots,x_d\in\mathbb C^d$, if $D_d(x_0,x_1,\ldots,x_d)=0$ for every $x_0\in\mathbb C^d$, then the direction of the affine span of $x_1,\ldots,x_d$ has complex dimension at most $d-2$.

This proves the preregistered polynomial reading of the two clauses. The total-degree hypothesis includes homogeneous degree-$d$ polynomials; the non-degeneracy statement has the explicit restriction $d\ge2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Fermionic/AffineDeterminantAntisymmetricForms.result

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2021The remaining correlation domain admits the weight gap exactly when $c<\sigma$On the remaining correlation domain, matrices whose top eigenvectors have smaller weight exist in every sufficiently large dimension exactly when c is less than sigma.Proved

The questionQuestion III.5 in the source TeX, lines 525--527, asks:

Why it mattersThe source relates the mean and spread of off-diagonal correlations to the spectral alignment of the leading eigenvector with the normalized all-ones vector.

The ideaThus wedgeconstruction, assuming h : domainD c σ and hcs : c < σ, supplies one common threshold and a realizing matrix in every dimension beyond it.

Our contributionOn the remaining correlation domain, matrices whose top eigenvectors have smaller weight exist in every sufficiently large dimension exactly when c is less than sigma.

Verification details
Release source recordProved in Lean

Open problem

The remaining correlation domain admits the weight gap exactly when $c<\sigma$

Y. Stepanov, H. Herrmann and T. Guhr, Generic features in the spectral decomposition of correlation matrices, arXiv:2104.08966v1 (Journal of Mathematical Physics 62 (2021), 083505), leave the domain $A\setminus(A_1\cup A_2)$ open. They define $A={(c,\sigma):c>0,\ \sigma>0,\ c^2+\sigma^2<1}$, $A_1={(c,\sigma)\in A:c\geq\frac12\ \text{or}\ \sigma<c}$, and $A_2={(c,\sigma)\in A:\sigma>\sqrt3c\ \text{or}\ c<\sigma<1-c}$. For an orthonormal eigenbasis, $w_j=\langle v_j,\delta_n\rangle^2$ with $\delta_n=(1,\ldots,1)/\sqrt n$, and the source's definition at TeX lines 877--879 compares $w_1$ with the largest weight among the other eigenvectors. Question III.5 in the source TeX, lines 525--527, asks:

Given $(c,\sigma) \in A \setminus (A_1 \cup A_2)$ and $n\geq 2$ sufficiently large, is there an $n\times n$ correlation matrix $C$ with mean correlation $c$ and standard deviation $\sigma$ such that $w_1<w_{\max}$ holds?

The formal statement fixes the remaining domain as domainD and asks, for every point in it, whether there is a threshold after which every dimension has a correlation matrix with the specified meanCorr, stdCorr, and SatisfiesWGap properties.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/CorrelationTopEigenvectorWeight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2018The two-qubit non-selective measurement optimum N/2 + 1/4 is globalFor every pure two-qubit state, the largest negativity after a non-selective projective measurement is half its initial negativity plus one quarter.Proved

The questionPaterek, Entanglement gain in measurements with unknown results (Phys.

Why it mattersAn unread measurement can create entanglement.

The ideaLet ρ = ψψ† and σ = Σj pj Pj with pj = ⟨ej, ρ ej⟩ for an orthonormal basis (ej).

Our contributionFor every pure two-qubit state, the largest negativity after a non-selective projective measurement is half its initial negativity plus one quarter.

Verification details
Release source recordProved in Lean

Open problem

The two-qubit non-selective measurement optimum N/2 + 1/4 is global

M. Zuppardo, R. Ganardi, M. Miller, S. Bandyopadhyay and T. Paterek, Entanglement gain in measurements with unknown results (Phys. Rev. A 99, 042319 (2019); arXiv:1806.05465), study how a non-selective measurement in an orthonormal basis changes entanglement. It maps a pure state |φ⟩ to Σ_j |⟨ψ_j|φ⟩|² |ψ_j⟩⟨ψ_j|.

  • The derived optimum. For two qubits and the measurement bases (28), they derive the largest final negativity: N^max_f − N_i = 1/4 − N_i/2 (Eq. (33)).
  • The open point. "Although we were not able to provide a proof that this is indeed the optimal choice among all possible measurement bases, we have conducted numerical check which supports optimality."

The verbatim statements are in the literature note.

Issue #18399 fixes the reading: for every unit two-qubit ψ and every orthonormal basis, the measured state has negativity at most N(ψψ†)/2 + 1/4, and some basis attains it. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/NonselectiveNegativityGainOptimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    First preprint year: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2026Connectedness of weak Pareto solutions for continuous convex dataContinuous convex objectives on a closed convex strip can have a disconnected weak Pareto solution set.Refuted

The questionVu Trung Hieu, A connectedness result for convex polynomial multi-objective optimization without boundedness, arXiv:2610.01029v1 (2026-10-01), §4.3 “Scope of the hypotheses”, asks:

Why it mattersThe formal counterexample addresses the stated general-convex gap while retaining a closed convex feasible set.

The ideaBoth objectives are continuous on the whole Euclidean space.

Our contributionContinuous convex objectives on a closed convex strip can have a disconnected weak Pareto solution set.

Verification details
Release source recordRefuted in Lean

arXiv

Connectedness of weak Pareto solutions for continuous convex data

Vu Trung Hieu, A connectedness result for convex polynomial multi-objective optimization without boundedness, arXiv:2610.01029v1 (2026-10-01), §4.3 “Scope of the hypotheses”, asks:

whether $\mathrm{Sol}^w(K,f)$ is still connected in the general convex case, with $K$ closed convex and the $f_i$ convex and continuous but not necessarily polynomial

The paper states:

no explicit example of a disconnected weak Pareto solution set is known

and that Theorem 1

shows that no such example can have convex polynomial data.

The theorem in the paper proves connectedness for convex polynomial data without boundedness or a constraint qualification. The question is whether the conclusion survives for general continuous convex objectives.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Convexity/WeakParetoDisconnected.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Geometry & algebraCounterexample foundFirst preprint 2026Kaygorodov–Martín González–Viruel Question 31SL3 refutes universal realization by full evolution-algebra automorphism functors.Refuted

The questionKaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.5, page 24, ask: “Let G be a linear algebraic group over 𝕜.

Why it mattersA natural basis restricts the semisimple derivation algebra to orthogonal factors.

The ideaEvolutionAutomorphismTangentBridge.tangentlieequivalence reconstructs the derivative of a natural realization using dual numbers.

Our contributionSL3 refutes universal realization by full evolution-algebra automorphism functors.

Verification details
Release source recordRefuted in Lean

Open problem

Kaygorodov–Martín González–Viruel Question 31

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.5, page 24, ask: “Let G be a linear algebraic group over 𝕜. Is there a finite-dimensional evolution 𝕜-algebra A with Aut(A) ≅ G(𝕜) as abstract groups? with Aut(A) ≅ G as algebraic groups? with 𝐀𝐮𝐭(A) ≅ G as group schemes?”

The second and third interrogatives have a negative answer. The counterexample is $G=\mathrm{SL}_3$ over $\mathbb C$. The public claim is the affirmative universal statement for polynomial-closed matrix groups and full natural automorphism functors over every commutative unital complex coefficient algebra. It includes coefficient algebras with nilpotents. The abstract-group interrogative is outside this assertion.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Linear algebraic group $G$ over a field. | LinearAlgebraicGroup with a matrix degree, polynomial equations, points, zero-locus characterization and change-of-algebra stability. | The field is specialized to $\mathbb C$, sufficient to refute the universal question. | | Finite-dimensional evolution algebra $A$. | Existential $d\in\mathbb N$ and bilinear product on $\mathrm{Fin}(d)\to\mathbb C$, satisfying IsEvolutionAlgebra. | An arbitrary natural basis is existential; zero dimension and degenerate algebras are included. | | Full automorphism group or group scheme. | automorphisms contains all invertible linear maps preserving multiplication; Realizes gives multiplicative equivalences at every coefficient algebra and naturality for every algebra homomorphism. | No identity or associativity is imposed on multiplication. | | Algebraic groups and group schemes. | The literal functor formulation in claim. | The comparison uses characteristic-zero smoothness; no theorem about arbitrary abstract point groups is asserted. | | Is there such an $A$ for every $G$? | result proves the negation of claim. | The witness group is slGroup 3. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/AlgebraicGroupRealizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2025The CZX parent Lindbladian reaches its conjectured even-ring steady stateThe CZX parent Lindbladian reaches its conjectured even-ring steady state from the uniform state with global bit flip.Proved

The questionCirac, Parent Lindbladians for Matrix Product Density Operators, arXiv:2501.10552v2, Phys.

Why it mattersThe source constructs a parent Lindbladian from the renormalization channels of a CZX matrix product density operator.

The ideaIt states that the channel on a gauged matrix unit is the uniform average over the two middle sites when goodWindow i δ holds and is zero otherwise. windowchannelscommute has hypotheses n : ℕ, hn : 4 ≤ n, i j : Fin n, and an arbitrary matrix a, and gives commutation of the two translated channels. windowidempotent has hypotheses n : ℕ, hn : 4 ≤ n, and i : Fin n, and gives idempotence of the corresponding superoperator.

Our contributionThe CZX parent Lindbladian reaches its conjectured even-ring steady state from the uniform state with global bit flip.

Verification details
Release source recordProved in Lean

Open problem

The CZX parent Lindbladian reaches its conjectured even-ring steady state

Y. Liu, A. Ruiz-de-Alarcón, G. Styliaris, X.-Q. Sun, D. Pérez-García and J. I. Cirac, Parent Lindbladians for Matrix Product Density Operators, arXiv:2501.10552v2, Phys. Rev. Research 8, 013210 (2026), DOI 10.1103/1qyd-59z7, Appendix F, “Evolution,” main.tex lines 5038–5042:

Starting from a trivial state $\rho=\frac{1}{2^N}(\mathbb 1^{\otimes N}+\sigma_x^{\otimes N})$, we numerically verify that: when $N=4,6,8$, the resulting steady state is $\rho_\infty^{(N)}=\frac{1}{2^N}(\mathbb 1_2^{\otimes N}+\frac{1}{2^{N/2-1}}U_{CZX}^{(N)})$. We conjecture that this relation holds for general $N$.

For $n\in\mathbb N$, the module indexes computational-basis configurations by $C_n=\mathrm{Fin}(n)\to\mathrm{Fin}(2)$. Its tensorX n matrix flips every bit, cz n is the diagonal cyclic controlled-$Z$ phase, and czxUnitary n = cz n * tensorX n. The four-site channel is embedded at cyclic windows by windowChannel; parentLindbladian n is the sum of windowSuperoperator n i - 1 over zero-based even starts, and evolution n t a is the matrix exponential of that superoperator acting on the entries of a.

The formal statement is claim: for every n with Even n and $4\le n$, the evolution from $((2:\mathbb C)^n)^{-1},\cdot,(1+\mathrm{tensorX}\ n)$ tends at atTop to $((2:\mathbb C)^n)^{-1},\cdot,(1+(2:\mathbb C)^{1-(n/2:\mathbb N):\mathbb Z},\cdot,\mathrm{czxUnitary}\ n)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/CZXParentLindbladianSteadyState.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationConjecture refutedFirst preprint 2015The Sun–Luo–Ye projection-matrix AM-GM conjectureA rank-one projection family gives a negative quadratic form for the matrix AM-GM difference.Refuted

The questionYe, On the efficiency of random permutation for ADMM and coordinate descent, arXiv:1503.06387v3, Section 3.3, Conjecture 3.1, equation (37), PDF p.20, state:

Why it mattersThe source proposes the conjecture as a route to the best convergence rate of random-permutation coordinate descent for quadratic problems.

Our contributionA rank-one projection family gives a negative quadratic form for the matrix AM-GM difference.

Verification details
Release source recordRefuted in Lean

Open problem

The Sun–Luo–Ye projection-matrix AM-GM conjecture

R. Sun, Z.-Q. Luo, and Y. Ye, On the efficiency of random permutation for ADMM and coordinate descent, arXiv:1503.06387v3, Section 3.3, Conjecture 3.1, equation (37), PDF p.20, state:

We observe that we only need to prove a matrix AM-GM inequality for projection matrices. We conjecture that the following matrix AM-GM inequality holds.

Conjecture 3.1 (matrix AM-GM inequality for projection matrices). Suppose $P_i \in \mathbb R^{N \times N}, i=1,\dots,n$ are projection matrices, then $$\frac{1}{n!}\sum_{\sigma=(\sigma_1,\dots,\sigma_n)\in\Gamma} P_{\sigma_n}P_{\sigma_{n-1}}\cdots P_{\sigma_1}\preceq\left(\frac1n\sum_i P_i\right)^n.$$

Here $\Gamma$ is the set of permutations of ${1,\dots,n}$ and $X\preceq Y$ means that $Y-X$ is positive semidefinite. The source and its coordinate-descent motivation are recorded in the Library note. Preregistration: #15647.

The inspected mathematical source is arXiv v3. The matching journal record is Mathematics of Operations Research 45(1) (2020), 233–271, DOI 10.1287/moor.2019.0990; the journal PDF was not independently inspected. The conjecture number above is the arXiv v3 number.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Real $N\times N$ matrices, indexed by $1,\dots,n$ | claim: dimensions n N, family on Fin n, real matrices on Fin N | Finite reindexing | | Projection matrices | claim: symmetry and idempotence, inlined literally; the private instance proof uses Mathlib IsStarProjection | Orthogonal real projections; no rank restriction in the claim | | $n$ projection matrices | claim: $1\le n$ | The source's reciprocal requires a positive number of factors | | Arithmetic mean, raised to $n$ | meanProj, and its $n$th power in claim | The inverse of the real cast of $n$ scales the sum | | Average over all permutations, with the first entry rightmost | symProd | The real inverse factorial scales the sum of reversed permutation words | | Positive-semidefinite order | claim: the mean power minus the symmetrized product is Matrix.PosSemidef | Exact difference direction | | Refutation of the universal assertion | result: negation of claim | Witness $n=N=3$; no assumptions on the result |

Structured block target and source identities

The separately registered target #17674, with its pre-probe scope notice, restricts (37) to the following complete domain. For every integer $n\ge1$ and positive integers $d_1,\ldots,d_n$, put $N=\sum_i d_i$ and take real $A_i\in\mathbb R^{N\times d_i}$ with $A_i^\top A_i=I_{d_i}$ such that the actual concatenation $A=[A_1,\ldots,A_n]$ is square and nonsingular. Set $P_i=I_N-A_iA_i^\top$, $M=n^{-1}\sum_iP_i$, and $G=(n!)^{-1}\sum_{\sigma\in\Gamma}P_{\sigma_n}\cdots P_{\sigma_1}$. The target is $G\preceq M^n$, with all block permutations uniformly weighted and the number of blocks as the exponent. The six-dimensional witness below refutes this universal restricted assertion.

The block dimensions and normalization come from source Sections 2.1.1 and 2.2, equations (7)–(8) and (23). Claim 4.1, equations (51a)–(51b), PDF p.23, assumes this square nonsingular, block-orthonormal structure and states identities. In the original block coordinates, let $L_\sigma$ have block $(\sigma_i,\sigma_j)$ equal to $A_{\sigma_i}^\top A_{\sigma_j}$ for $j\le i$ and zero otherwise, and put $Q_\sigma=L_\sigma^{-1}$ and $Q=(n!)^{-1}\sum_\sigma Q_\sigma$. Then those identities are

$$ I-AQ_\sigma A^\top=P_{\sigma_n}\cdots P_{\sigma_1},\qquad I-AQA^\top=G. $$

They hold for the witness below; Claim 4.1 is not refuted. Lemma 3, equation (52), and Corollary 4.1, equation (53), prove the weaker bound $G\preceq M$. The target $G\preceq M^n$ is a strengthening on that structured domain and a restriction of unrestricted Conjecture 3.1. The unrestricted conjecture was already refuted by the historical Lean result above before the block probes. No literal published, still-open full-block assertion matching #17674 was established by the inspected pre-probe record. The block result is retained here as exposition, with no new open-problem-resolution admission or literature-priority claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/ProjectionAMGMRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2014Monotonicity of the negativity of randomized graph statesWu, Rossi, Kampermann, Severini, Kwek, Macchiavello and Bruss (arXiv:1403.3828, Section V) ask whether the negativity of a randomized graph state, across any bipartition, increases monotonically with the probability p that each edge is present.Refuted

The questionBruß, Randomized Graph States and their Entanglement Properties, Phys.

Why it mattersThe question asks whether this holds for every graph and every cut.

Our contributionWu, Rossi, Kampermann, Severini, Kwek, Macchiavello and Bruss (arXiv:1403.3828, Section V) ask whether the negativity of a randomized graph state, across any bipartition, increases monotonically with the probability p that each edge is present.

Verification details
Release source recordRefuted in Lean

Open problem

Monotonicity of the negativity of randomized graph states

J.-Y. Wu, M. Rossi, H. Kampermann, S. Severini, L. C. Kwek, C. Macchiavello, D. Bruß, Randomized Graph States and their Entanglement Properties, Phys. Rev. A 89, 052335 (2014), arXiv:1403.3828v3, Section V:

However, even though this conjecture is supported by numerical evidence, it is an open question whether the monotonic behavior of the negativity in terms of the randomness $p$ is a common feature to all RG states.

The randomized graph state of a graph $G$ is $\rho^p_G=\sum_{F\subseteq E(G)}p^{|F|}(1-p)^{|E(G)\setminus F|} |F\rangle\langle F|$, where $|F\rangle=\prod_{ab\in F}\mathrm{CZ}_{ab} |+\rangle^{\otimes n}$, and the negativity across a bipartition $A|B$ is $N(\rho)=(|\rho^{\Gamma_A}|_1-1)/2$, evaluated "with respect to all possible bipartitions of the qubits". The conjecture is that $N$ increases monotonically in $p$; the paper reports it numerically for the complete graphs $K_n$ and the star graphs $S_n$ with $n\le4$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/RandomizedGraphNegativityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2018Positional autocorrelation does not determine finite deletion-channel entropy orderingPositional autocorrelation does not determine the finite entropy order of binary deletion-channel outputs.Refuted

The questionRyan, A Proof of Entropy Minimization for Outputs in Deletion Channels via Hidden Word Statistics, arXiv:1807.11609v1, Section V, state:

Why it mattersThe finite ordering is a separate question from the asymptotic minimization theorem.

The ideaEnumerating all 256 input words gives the embedding histograms below.

Our contributionPositional autocorrelation does not determine the finite entropy order of binary deletion-channel outputs.

Scope & next questionThe domain excludes only zero output length and the no-deletion case, neither of which is needed for the counterexample.

Verification details
Release source recordRefuted in Lean

arXiv

Positional autocorrelation does not determine finite deletion-channel entropy ordering

A. Atashpendar, D. Mestel, A. W. Roscoe and P. Y. A. Ryan, A Proof of Entropy Minimization for Outputs in Deletion Channels via Hidden Word Statistics, arXiv:1807.11609v1, Section V, state:

Beyond establishing this maximization, proving the entropy ordering of x strings determined by κ2(x) for finite n and m represents another open problem.

Section IV-E.3 and Table I specify the direction: larger $\kappa^2$ predicts smaller posterior entropy. The finite claim is false at the source's own parameters $n=8$, $m=5$: for $u=00100$, $v=01110$, $\kappa^2(u)=450>410=\kappa^2(v)$, but $H_8(u)>H_8(v)$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Section III-A.f: the number of masks projecting $y$ to $x$ | embeddings | List.sublists retains index-subsequence multiplicities; filtering and taking the length counts masks. | | Equation (1): $\mu_{n,m}=\binom{n}{m}2^{n-m}$ | D5.S3.Arith.FibonacciAtomic.CommonPrediction.Capacity.pref | Reuses the frozen binomial factor, with natural coefficients and subtraction; claim restricts $m<n$. | | Equation (3): Shannon entropy in bits of $\omega_x(y)/\mu_{n,m}$ | entropyBits | Reuses D5.S3.Entropy.MaxEntropy.shannonEntropy on the posterior law, dividing nats by $\log 2$; incompatible words contribute zero. | | Equation (11): $\sum_{1\le r,s\le m}\binom{r+s-2}{r-1}\binom{2m-r-s}{m-r}[[x_r=x_s]]$ | kappa2 | The literal expression uses zero-based indices and natural coefficients. | | Section V and the decreasing-$\kappa^2$ ordering of Table I | claim | Every positive $m$, every $n>m$, and every pair of binary words of length $m$; strict $\kappa^2$ comparison predicts weak entropy comparison. | | Negation of that ordering | result | A counterexample at $m=5,n=8$ refutes the universal statement. |

The added binders are representations of the source's lengths and strings, not extra hypotheses. Bool encodes binary letters (false is zero); List.ofFn reads increasing positions. The domain excludes only zero output length and the no-deletion case, neither of which is needed for the counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/DeletionChannelEntropyOrdering.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2016Coefficients of the two-block hypergeometric TuranianNegative first Turanian coefficients with a nonempty shifted numerator.Refuted

The questionKarp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, page 6:

Why it mattersA negative coefficient under all of the stated hypotheses refutes the coefficient clause.

The ideaThe empty rising factorial products equal one, so the coefficient reduces to Defs.turanCoeff for the shifted block.

Our contributionNegative first Turanian coefficients with a nonempty shifted numerator.

Verification details
Release source recordRefuted in Lean

Open problem

Coefficients of the two-block hypergeometric Turanian

S. I. Kalmykov and D. B. Karp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, page 6:

Conjecture 2. Suppose p1 ≤ q1 , p2 ≤ q2 and conditions (13) hold for a1 , b1 and a2 , b2 . Then the function µ → f (µ; x) defined in (3) is log-concave on [1, ∞) and the corresponding generalized Turánian ∆f (α, β; x) has nonnegative coefficients at all powers of x.

The standing assumption on page 2 is:

We will always assume that a2 is not empty (i.e. p2 ≥ 1), while all other subvectors are allowed to be empty.

For positive parameter vectors in two blocks, equation (3) defines

$$ f(\mu;x)=\frac{\Gamma(\mathbf a_2+\mu)}{\Gamma(\mathbf b_2+\mu)} ,{}_pF_q(\mathbf a_1,\mathbf a_2+\mu; \mathbf b_1,\mathbf b_2+\mu;x). $$

Here a Gamma value at a vector means the product over its entries. Writing $(a)_n$ for the rising factorial and $c(t)$ for the shifted Gamma product ratio, the Taylor coefficient is

$$ F_n(\mu)= \frac{\prod_i(a_{1,i})n}{\prod_j(b{1,j})n} \frac{c(\mu+n)}{n!},\qquad c(t)=\frac{\prod_i\Gamma(a{2,i}+t)}{\prod_j\Gamma(b_{2,j}+t)}. $$

Equation (2) defines

$$ \Delta_f(\alpha,\beta;x)= f(\mu+\alpha;x)f(\mu+\beta;x) -f(\mu;x)f(\mu+\alpha+\beta;x), $$

whose degree-$m$ coefficient is

$$ \delta_m=\sum_{k=0}^m \bigl[F_k(\mu+\alpha)F_{m-k}(\mu+\beta) -F_k(\mu)F_{m-k}(\mu+\alpha+\beta)\bigr]. $$

For each block separately, condition (13) is

$$ \frac{e_{q-j}(\mathbf b)}{e_{p-j}(\mathbf a)} \le\frac{e_{q-j-1}(\mathbf b)}{e_{p-j-1}(\mathbf a)} \qquad(0\le j<p), $$

with $e_0=1$. The coefficient clause quantifies over $p_1\le q_1$, $1\le p_2\le q_2$, all four positive vectors satisfying their respective chains, every $\mu\ge1$, all $\alpha,\beta\ge0$, and every natural degree $m$.

Formal correspondence. All Lean names below have prefix D5.S3.Analytic.HypergeometricShiftTuran.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Unshifted rising factorial ratio | TwoBlockResult.pochRatio | Products of (ascPochhammer ℝ n).eval at the unshifted parameters. | | Equation (3) Taylor coefficient | TwoBlockResult.seriesCoeff | Rising factorial ratio times Defs.coeff for the shifted block at $\mu+n$, divided by $n!$. | | Equation (2) Cauchy coefficient | TwoBlockResult.twoBlockTuranCoeff | Exact finite sum with the factorials contained in seriesCoeff. | | Shifted Gamma ratio and condition (13) | Defs.coeff, Defs.esymm, Defs.cond13 | Reused definitions, with the ratio chain imposed separately on both blocks. | | Conjecture 2 coefficient clause | TwoBlockResult.claim | All dimensions, positivity assumptions, both chains, nonempty shifted numerator, shifts and coefficient degrees. | | Negative coefficient family | TwoBlockResult.half_parameter_family_neg | Empty unshifted blocks, shifted numerator $(1)$, repeated half denominators, $\mu=1$, $\alpha=\beta=1/2$, and every $q\ge7$. | | Refutation | TwoBlockResult.result | Exactly $\neg,\texttt{TwoBlockResult.claim}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/HypergeometricShiftTuran/TwoBlockResult.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2015Ancilla-free testers with a common normalization state have canonical POVM robustnessAncilla-free testers with a common normalization state have the robustness of their canonical POVMs.Proved

The questionZiman, Incompatible measurements on quantum causal networks, Phys.

Why it mattersFor a common normalization state, the question isolates incompatibility arising from the measurements.

The ideaAll matrix families below have finite coordinate and outcome types; the output type has decidable equality.

Our contributionAncilla-free testers with a common normalization state have the robustness of their canonical POVMs.

Scope & next questionThus it does not settle the mutually unbiased basis conjectures in Section VI.B or Appendix G.

Verification details
Release source recordProved in Lean

Open problem

Ancilla-free testers with a common normalization state have canonical POVM robustness

M. Sedlák, D. Reitzner, G. Chiribella and M. Ziman, Incompatible measurements on quantum causal networks, Phys. Rev. A 93, 052323 (2016), arXiv:1511.00976v2, Section VI.C, ask (verbatim from ti_pra_07.tex, line 807, retaining the source's TeX macros):

It is an open question whether the equality $\Rt({\st A}, {\st B}) = \Rm( \st P, \st Q )$ holds for all ancilla-free testers.

Here $R_{\rm t}$ and $R_{\rm m}$ denote tester and measurement robustness. Section VI restricts attention to testers with the same normalization state (ti_pra_07.tex, lines 758 and 772); that hypothesis is retained. Proposition 3 characterizes ancilla-free testers as $T_a=M_a\otimes\rho$ for an output POVM $M$ and an input density matrix $\rho$ (lines 333–338). Definitions and locators are in the literature note.

Issue #23653 fixes the following finite-dimensional reading. Let $y,x,j,k$ be finite types with decidable equality, indexing output coordinates, input coordinates and the two outcome sets. Let $\rho$ be a complex matrix on $x$ with $\rho\succeq0$ and $\operatorname{Tr}\rho=1$. Let $M=(M_a){a\in j}$ and $N=(N_b){b\in k}$ be complex output POVMs: each effect is positive semidefinite and each family sums to $I_y$. With $r=\operatorname{rank}\rho$, the statement is

$$ R_{\rm t}\bigl((M_a\otimes\rho)_a,(N_b\otimes\rho)b\bigr) =R{\rm m}\bigl((M_a\otimes I_r)_a,(N_b\otimes I_r)_b\bigr). $$

In the module, IsTester means positive semidefinite effects summing to $I_y\otimes\sigma$ for some positive semidefinite $\sigma$ of trace one. TesterCompatibleAt at a real $\lambda$ requires noise testers $E,F$ and a joint tester $C$ with

$$ \sum_b C_{ab}=(1-\lambda)A_a+\lambda E_a, \qquad \sum_a C_{ab}=(1-\lambda)B_b+\lambda F_b. $$

The noises have separate existential witnesses; their normalization states are not prescribed to equal $\rho$, and their effects need not be products. POVMCompatibleAt uses the same marginal equations with noise POVMs and a joint POVM, each positive semidefinite and summing to the identity. testerRobustness and povmRobustness are the real infima of their respective feasible $\lambda\in[0,1]$. The source uses minima; attainment is not part of claim or result. The displayed equality, with these infimum definitions and $I_r$ indexed by Fin ρ.rank, is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/AncillaFreeTesterRobustnessEquality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2015The Zhou–Wu–Chau endpoint lower bound on overall error is falseA rational three-dimensional measurement has overall error below both endpoint quantities.Refuted

The questionChau, State-Independent Error-Disturbance Tradeoff For Measurement Operators, arXiv:1509.03812v1 (2015), published as State-independent error-disturbance trade-off for measurement operators, Physics Letters A 380(22–23), 1918–1924 (2016), DOI 10.1016/j.physleta.2016.03.046, state Conjecture 3 in Section IV.

Why it mattersThe proposed bound would identify the best endpoint cost as a universal lower bound on error plus disturbance evaluated in a common state.

The ideaThe boundedness lemmas justify passing these witnesses to the real suprema in disturbancelower and relaxedlower; finiteness and nonemptiness of the permutation type justify the strict lower bound on its infimum.

Our contributionA rational three-dimensional measurement has overall error below both endpoint quantities.

Verification details
Release source recordRefuted in Lean

Open problem

The Zhou–Wu–Chau endpoint lower bound on overall error is false

S. S. Zhou, S. Wu and H. F. Chau, State-Independent Error-Disturbance Tradeoff For Measurement Operators, arXiv:1509.03812v1 (2015), published as State-independent error-disturbance trade-off for measurement operators, Physics Letters A 380(22–23), 1918–1924 (2016), DOI 10.1016/j.physleta.2016.03.046, state Conjecture 3 in Section IV. The v1 source new_912arxiv.tex, lines 625–638, reads:

For all $d \geq 2$, the following trade-off between error and disturbance holds: $\varepsilon(\mathcal{A},\mathcal{A}') + \eta(\mathcal{A}',\mathcal{B}) \geq f(\mathcal{A},\mathcal{B})$ and $\Delta(\mathcal{A},\mathcal{A}',\mathcal{B}) \geq f(\mathcal{A},\mathcal{B})$, where $f(\mathcal{A},\mathcal{B}) = \min { \varepsilon(\mathcal{A},\mathcal{B}),\eta(\mathcal{A},\mathcal{B}) }$. Moreover, the equalities are both attained either when for all $i$, $\ket{a'_i} = \ket{a_i}$, or when for all $i$, $\ket{a'_i} = \ket{b_i}$.

The source uses rank-one projective measurements associated with orthonormal bases of $\mathbb C^d$. Section II defines the outcome-distance by $D(x,y)=\max_i|x_i-y_i|$ and compares the $\mathcal B$ statistics before and after the nonselective $\mathcal A'$ measurement. Its overall quantity is explicitly

$\Delta(\mathcal{A}, \mathcal{A}', \mathcal{B}) = \max_\rho \left( \varepsilon_{\rho}( \mathcal{A}, \mathcal{A}') + \eta_{\rho}( \mathcal{A}', \mathcal{B})\right).$

This is line 221 of the source. Both terms use one common density matrix. Immediately before the conjecture, lines 616–620 relax the endpoint error by

$\varepsilon(\mathcal{A},\mathcal{B}) = \min_{\tilde{\mathcal B}} \varepsilon(\mathcal{A},\tilde{\mathcal{B}})$,

with the explanation

This relaxed definition of $\varepsilon$ means that we only distinguish between the classical probability distributions resulting from measurements $\mathcal{A}$ and $\mathcal{B}$ with no regard to the labeling of different measurement outcomes.

The definitions and source locator are collected in the literature note.

Issue #23208 fixes the statement: for every natural $d\ge2$ and all three unitary measurement bases $A,A',B$, the minimum of the endpoint error, minimized over outcome permutations, and the disturbance caused by $A$ is at most $\Delta(A,A',B)$. This second inequality is refuted. The first inequality, concerning the sum of two separately maximized quantities, is not settled here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/ErrorDisturbanceEndpointRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

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    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2023Refutation of the fixed-resistance Fiedler invariance conjectureInvariant weights cannot maximize the Fiedler value at fixed resistance on K23.Refuted

The questionCruz Ortega and Menéndez-Conde, Conectividad algebraica máxima para caminos con resistencia total fija, arXiv:2306.08187v1 and Pesquimat 26(2) (2023), Section 5, define

The ideaLeaf permutations and the hub swap act transitively on the edges.

Our contributionInvariant weights cannot maximize the Fiedler value at fixed resistance on K23.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the fixed-resistance Fiedler invariance conjecture

Cruz Ortega and Menéndez-Conde, Conectividad algebraica máxima para caminos con resistencia total fija, arXiv:2306.08187v1 and Pesquimat 26(2) (2023), Section 5, define

$$ \breve a(G)=\max{\lambda_2(G;\omega)\mid R(G;\omega)=R(G)} $$

and conjecture that the maximum is always attained by edge weights invariant under graph automorphisms. Issue #15242 records the literal source statement, the $K_{2,m}$ route and the Lean conventions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/FiedlerResistanceInvariantWeightsRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2022The minimum skewness of a unidirectional unicyclic currentFor every unidirectional cycle with at least two states, the least skewness of positive rates is attained when one rate is the reciprocal of the number of other rates.Proved

The questionPtaszyński, Bounds on skewness and kurtosis of steady state currents, Phys.

Why it mattersA sharp lower bound separates variation of positive rates within a unidirectional cycle from departure from that model: a smaller value cannot come from any rate vector covered by the claim.

Our contributionFor every unidirectional cycle with at least two states, the least skewness of positive rates is attained when one rate is the reciprocal of the number of other rates.

Scope & next questionEquality classification and extension of the moment argument to the other cumulant combinations remain open here.

Verification details
Release source recordProved in Lean

Open problem

The minimum skewness of a unidirectional unicyclic current

K. Ptaszyński, Bounds on skewness and kurtosis of steady state currents, Phys. Rev. E 106, 024119 (2022), arXiv:2204.08421v4, Appendix, main.tex line 792:

As discussed in Sec.~\ref{subsec:unmark}, it was numerically conjectured that the minimum and maximum values of $\mathcal{S}$, $\mathcal{K}$, $\mathcal{S}+\mathcal{K}$, $\mathcal{S}-\mathcal{K}$ and $\mathcal{S} \times \mathcal{K}$ are always obtained for a specific type of network topology, with all rates $k_{ij}$ equal to $k$ apart from one (here $k_{21}$) equal to $ak$, with the parameter $a$ depending on the optimized quantity and the number of states $N$.

The Appendix, main.tex line 804, specifies the skewness minimum:

The minimum value $\min (\mathcal{S})$ is found for $a=1/(N-1)$; it is given by Eq.~\eqref{skewunidet}.

The reading fixed in issue #22468 concerns only the minimum skewness for unidirectional cycles. For every integer $N\ge2$ and positive clockwise rates $k_i$, put

$R_j=\sum_{i=0}^{N-1}k_i^{-j},\qquad \mathcal{S}(k)=3R_2^2/R_1^4-2R_3/R_1^3,\qquad g(N)=-\frac{8-8N+N^2}{16(N-1)^2}.$

The claim is that $g(N)$ is the least value of $\mathcal{S}(k)$ over all such positive rate vectors, including attainment. The source's candidate rates are $(1/(N-1),1,\ldots,1)$, up to an overall positive scale. This minimum claim is proved.

The source defines $\mathcal{S}$ as the third scaled current cumulant divided by the first, rather than the standardized third central moment. Its definition is at main.tex line 53; the Appendix's current and waiting-time cumulant formulas at lines 772–789 give the expression above. The finite-$N$ candidate is displayed in Section III E 1, lines 417–423. See the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/UnicyclicSkewnessMinimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2015Every (n,d)↦1 random access code has a quantum advantage with two measurement settingsFor every n and d at least two, a d-dimensional random access code with two projective measurement settings strictly outperforms every classical strategy with shared randomness.Proved

The questionRai, Quantum advantages in (n,d)↦1 random access codes, arXiv:1510.03045v4, New J.

Why it mattersThe source proves that majority encoding with identity decoding is classically optimal.

The idea1. Classical optimum. A mixture is bounded by its best deterministic strategy.

Our contributionFor every n and d at least two, a d-dimensional random access code with two projective measurement settings strictly outperforms every classical strategy with shared randomness.

Verification details
Release source recordProved in Lean

Open problem

Every (n,d)↦1 random access code has a quantum advantage with two measurement settings

A. Ambainis, D. Kravchenko, S. Sazim, J. Bae and A. Rai, Quantum advantages in (n,d)↦1 random access codes, arXiv:1510.03045v4, New J. Phys. 26 (2024) 123023, Section 5:

We conjecture that the quantum advantages would appear in $(n,d) \mapsto 1$ RACs in general and leave it as an open question. The problem also addresses minimal resources, i.e., two measurement settings, to achieve quantum advantages in information theory.

The scenario, the figure of merit and the classical strategies with shared randomness are quoted in the literature note.

Issue #19414 fixes the reading. For all n, d ≥ 2, some pure-state quantum strategy decoded by two projective measurements has an average success probability strictly above that of every classical strategy with shared randomness. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/RandomAccessCodeTwoSettingAdvantage.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2023The generic quadratic Nambu KHK map has a Cremona decompositionThe homogeneous KHK map of a positive definite quadratic Nambu pair with a simple pencil factors through the cubic Cremona map.Proved

The questionThis is quoted from GGcremonav1clean.tex, lines 3209–3211, Remark ex:celledoni; the same remark says that a complete proof is left for further research.

Why it mattersThe source paper identifies the Euler top as the diagonal special case of these Nambu systems and proves in its Lemma 2.6 that the Euler-top KHK map is a composition of two projectivities and the standard cubic Cremona map.

The ideaThe rank condition is therefore used exactly to obtain pairwise distinct diagonal coefficients.

Our contributionThe homogeneous KHK map of a positive definite quadratic Nambu pair with a simple pencil factors through the cubic Cremona map.

Scope & next questionThe source’s assertion that its invariant net has eight distinct base points, and the implication from that geometric condition to the rank hypothesis, are not formalized here, so the relationship between those two genericity descriptions remains open.

Verification details
Release source recordProved in Lean

Open problem

The generic quadratic Nambu KHK map has a Cremona decomposition

M. Graffeo and G. Gubbiotti study the Kahan–Hirota–Kimura (KHK) discretisation of the quadratic three-dimensional Nambu system $ \dot{x}=\nabla H_1(x)\mathbin{\times}\nabla H_2(x),\qquad H_i(x)=x^{\mathsf T}A_i x. $ In Remark 7.7 of Growth and Integrability of Some Birational Maps in Dimension Three, they write:

“We conjecture that, up to conjugation by a projectivity, the map $\Phi_{h}^{(N)}$ is of type (A)-(i), i.e. it exists an analogue of Lemma 2.6 for the map $\Phi_{h}^{(N)}$.”

This is quoted from GG_cremona_v1_clean.tex, lines 3209–3211, Remark ex:celledoni; the same remark says that a complete proof is left for further research.

The Lean module defines, for $X:\mathrm{Fin},4\to\mathbb C$, the polarised matrix nambuLinearPart, the homogeneous KHK matrix $ K_{A,B,h}(X)=X(3)I_3-2h,\mathrm{nambuLinearPart}(A,B)(X\mathbin{\circ}\mathrm{Fin.castSucc}), $ the homogeneous lift $ \mathrm{kahan}(A,B,h,X) =\mathrm{Fin.snoc}\bigl(\operatorname{adj}(K_{A,B,h}(X)),\mathrm{mulVec}, (X\mathbin{\circ}\mathrm{Fin.castSucc}),\det K_{A,B,h}(X)\bigr), $ and $ \mathrm{cremona}(X)= \bigl(X(1)X(2)X(3),,X(0)X(2)X(3),,X(0)X(1)X(3),,X(0)X(1)X(2)\bigr). $ Its claim states that for every real $3\times3$ matrices $A,B$ with A.PosDef and B.PosDef, with $ \forall t:\mathbb R,\quad 2\le \operatorname{rank}(B-tA), $ and every $h:\mathbb R$ with $0<h$, there are complex matrices $l_1,l_2:\mathrm{Matrix}(\mathrm{Fin},4)(\mathrm{Fin},4)\mathbb C$ such that IsUnit l₁.det, IsUnit l₂.det, and, for every $X:\mathrm{Fin},4\to\mathbb C$, $ \mathrm{kahan}(A,B,h,X) =l_2,\mathrm{mulVec}, \mathrm{cremona}(l_1,\mathrm{mulVec},X). $ The declaration result has type claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/Solitons/NambuKahanCremonaDecomposition.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Source 1
Quantum informationQuestion resolvedFirst preprint 2024Maximal input entropy for binary two-deletion and two-insertion channelsExact binary run counts and input entropy for two deletions and insertions.Proved

The questionSinghvi, Sabary, Bar-Lev and Yaakobi, Conditional Entropies of k-Deletion/Insertion Channels, arXiv:2407.10026v2, Section VII-A, ask:

Why it mattersThe source settles the minimizing sequences for two deletions but leaves the maximizing sequences and values open.

The ideaThis gives the exact joint count distribution and the deletion run functional.

Our contributionExact binary run counts and input entropy for two deletions and insertions.

Verification details
Release source recordProved in Lean

Open problem

Maximal input entropy for binary two-deletion and two-insertion channels

Singhvi, Sabary, Bar-Lev and Yaakobi, Conditional Entropies of k-Deletion/Insertion Channels, arXiv:2407.10026v2, Section VII-A, ask:

While we provide a complete characterization of the sequences that minimize input entropy, the precise identification of the sequences that maximize input entropy for the 2-deletion and 2-insertion channels is still unresolved. Determining these sequences and the associated maximal entropy values remains an important open direction.

Issue #19173 fixes the formal statement. For a received binary word, the posterior on uniformly transmitted inputs is proportional to the number of increasing index sets realizing the received word as a subsequence. Entropy is Shannon entropy in bits. The claim gives the two alternating families and their complements, the exact deletion and large insertion maxima, and the finite insertion cases of received lengths two, three and four.

Statement echo

The Lean claim in D5/S3/Entropy/SinghviMaxInputEntropy.lean inlines the two posterior definitions, quantifies over the same finite binary words, uses the frozen embedding count and Shannon entropy, divides by Real.log 2, and states all seven clauses of the preregistration. The public result : claim is the settling declaration.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/SinghviMaxInputEntropy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026The whole prescribed packing-maximum conjectureRefuted

The questionBlitvić–Kammoun–Steingrímsson, Permutations, Pattern Covers, Posets and Positivity: Part I: Patterns and Posets, arXiv:2610.29846v1, Section 7 item 4, Conjecture 7.1:

Why it mattersThe source's packing/poset correspondence turns this into an extremal linear-extension question.

The ideaBoth patterns are compatible and satisfy the source's increasing and first-value convention.

Verification details
Release source recordRefuted in Lean

Open problem

The whole prescribed packing-maximum conjecture

Blitvić–Kammoun–Steingrímsson, Permutations, Pattern Covers, Posets and Positivity: Part I: Patterns and Posets, arXiv:2610.29846v1, Section 7 item 4, Conjecture 7.1:

For overlap $m\le K/2$, the maximum is obtained when prefix and suffix are interleaved and minimal, that is, for patterns with prefix $135\ldots(2m-1)$ and suffix $246\ldots(2m)$.

For every $K\ge2$, $1\le m<K$, $2m\le K$, $j\ge1$, the assertion requires a prescribed pattern $\rho$ with $C(\pi,m,j)\le C(\rho,m,j)$ for every compatible pattern $\pi$. The middle order of $\rho$ is unrestricted. Here $C$ counts actual permutations of length $K+(j-1)(K-m)$ having all prescribed windows; extra occurrences are permitted. Compatibility means that prefix and suffix have the same relative order. The formal comparison domain does not impose a normalization or identify patterns by symmetry.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/PackingMaximumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Number theoryConjecture refutedFirst preprint 2026Jaffe's unrestricted primitive Frobenius successor-ratio conjectureThe primitive Frobenius successor ratio exceeds five for an admissible pair of integers.Refuted

The questionThe target is exactly Conjecture 7.3 in arXiv:2610.17267v1:

Why it mattersThe source asks for a uniform bound when the second generator advances to its next coprime integer.

The ideaThe integer \(a\) is coprime to 7 and 49, and every integer strictly between them is noncoprime to \(a\).

Our contributionThe primitive Frobenius successor ratio exceeds five for an admissible pair of integers.

Verification details
Release source recordRefuted in Lean

arXiv

Jaffe's unrestricted primitive Frobenius successor-ratio conjecture

The target is exactly Conjecture 7.3 in arXiv:2610.17267v1:

For all a,b >= 3 relatively prime and for c(a,b) the minimum integer satisfying both c(a,b)>b and gcd(c(a,b),a)=1, we have g(a,c(a,b))/g(a,b) <= 5.

Section 3 defines [ R(a,b)={ax+by:x,y\in\mathbb Z_{\geq0},\ \gcd(x,y)=1}, \quad g(a,b)=\max(\mathbb Z\setminus R(a,b)), ] with infinity when a greatest missing integer does not exist. Zero coefficients are allowed. There is no a<b assumption. The successor is the least integer above b coprime to a. The same conjecture is numbered 8.3 in the author's 2023 dissertation; this dossier anchors one assertion, not two. Conjecture 7.2 is a different target.

The Lean definition fullClaim : Prop uses precisely this integer domain, representation predicate, greatest-missing convention, successor and quotient bound. The designated result : Not fullClaim is unconditional.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/JaffeSuccessorRatioRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Pick functions do not preserve embeddable matricesA positive Pick function sends an embeddable stochastic matrix to a matrix with a negative entry.Refuted

The questionTsatsomeros, Powers, Roots, and Positivity, arXiv:2609.06298v1, Table 2, “Behavior of Matrix Classes under Pick Functions and Fractional Powers”, p. 22, row 13 reads verbatim:

Why it mattersRow 13 records preservation by fractional powers.

The ideaIt is therefore embeddable in the sense of Remark 3.28.

Our contributionA positive Pick function sends an embeddable stochastic matrix to a matrix with a negative entry.

Verification details
Release source recordRefuted in Lean

Open problem

Pick functions do not preserve embeddable matrices

C. R. Johnson, S. Mondal, and M. J. Tsatsomeros, Powers, Roots, and Positivity, arXiv:2609.06298v1, Table 2, “Behavior of Matrix Classes under Pick Functions and Fractional Powers”, p. 22, row 13 reads verbatim:

Embeddable Matrix | Open | Yes

The note below the table states verbatim:

The entries marked Open indicate cases for which preservation under Pick functions remains an open question.

Remark 3.28, p. 17, states verbatim:

The embedding problem, introduced in [11], asks whether a given stochastic matrix A can be realized as the transition matrix of a continuous-time Markov process. It was shown in [24] that a nonsingular stochastic matrix A is embeddable if and only if A = e^B for some essentially nonnegative matrix B satisfying Be = 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/EmbeddablePickRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Geometry & algebraQuestion resolvedFirst preprint 2026Aliabadi's finite nonabelian quotient-multiset problemEvery finite nonabelian group has a multiset satisfying the abelianization obstruction which is not a multiset of right quotients of two bijective enumerations.Proved

The questionMohsen Aliabadi, A nonabelian twist on differences of bijections, arXiv:2605.16478v2, Section 7, Problem 7.6, asks:

Why it mattersThe canonical abelianization condition is necessary for every realizable multiset.

The ideaThe equivalence commutatoreqbotiff makes [G,G] nontrivial.

Our contributionEvery finite nonabelian group has a multiset satisfying the abelianization obstruction which is not a multiset of right quotients of two bijective enumerations.

Verification details
Release source recordProved in Lean

Open problem

Aliabadi's finite nonabelian quotient-multiset problem

Mohsen Aliabadi, A nonabelian twist on differences of bijections, arXiv:2605.16478v2, Section 7, Problem 7.6, asks:

Does every finite nonabelian group admit a multiset A of cardinality |G| satisfying the abelianization obstruction but not quotient-realizable?

The following paragraph says that a positive answer would characterize finite abelian groups as precisely those finite groups for which the abelianization obstruction is the only obstruction. The official journal article, DOI 10.61091/um128-18, has the same question, global conventions and relevant definitions. The source note is aliabadi2026nonabeliantwist.

For every finite group G with multiplication not commutative, let n=|G| and let pi:G->G/[G,G] be the canonical abelianization map. There exists a multiset A of elements of G, counted with multiplicity, such that |A|=n and the multiplicity-counted product of pi(a) over A is 1, but there are no bijections b,c:{1,...,n}->G for which A is the multiset of b(i)*c(i)^(-1). Repeated identity entries are allowed. No product-one ordering in G, identity-free condition, group-family, solvability or parity restriction is imposed.

Definition 2.1 supplies exactly this two-bijection meaning of quotient realizability. Proposition 2.3 supplies the necessary abelianization condition: the product is taken after projection, with multiplicities, in the abelian group G/[G,G]. It does not assert an unordered product in G.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/AliabadiQuotientObstruction.exists_obstructed_multiset

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2022The path does not minimize Kemeny's edge derivativeA connected barbell on 23 vertices has a smaller Kemeny edge derivative than the path.Refuted

The questionNix, On the Edge Derivative of the Normalized Laplacian with Applications to Kemeny's Constant, arXiv:2211.01495v2, Conjecture 3.5, printed page 19:

Why it mattersTwo dense cliques joined through a long path concentrate degree mass at widely separated vertices.

The ideaExact rational matrices certify the two instances; their casts to the reals preserve the inverse and voltage identities.

Our contributionA connected barbell on 23 vertices has a smaller Kemeny edge derivative than the path.

Verification details
Release source recordRefuted in Lean

Open problem

The path does not minimize Kemeny's edge derivative

C. Albright, K. P. Hadaway, A. H. Pomerance, J. Jeffries, K. J. Lorenzen and A. K. Nix, On the Edge Derivative of the Normalized Laplacian with Applications to Kemeny's Constant, arXiv:2211.01495v2, Conjecture 3.5, printed page 19:

For a given n, the derivative of Kemeny's constant on simple graphs of order n is smallest on the path on n vertices with respect to the non-edge connecting the ends of the path. The value of this derivative is O(n³).

The minimality sentence is refuted under the literal conventions in preregistration #19527. The Library note records the source definitions. The path's own cubic growth sentence is not negated by this settlement.

Statement echo

All short declaration names below belong to D5.S3.Combinatorics.SpectralGraph.AlbrightKemenyEdgeDerivativeRefutation.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A(t)=A+tB$, with $B$ indicating one unordered pair | paramAdj | The two matrix units have weight one; degrees change with $t$ | | $L=I-D^{-1/2}AD^{-1/2}$ | normLap | $D$ contains the row sums of the perturbed matrix | | $K=\sum_{\lambda_L\ne0}1/\lambda_L$ | kemeny | Nonzero real characteristic roots, counted with multiplicity; the proof identifies them spectrally | | Simple graphs of order $n$ | claim: $G$ on Fin n, $2\le n$ | Connectedness is the domain where the source's random-walk constant is defined | | Edge or non-edge with distinct ends | claim: $a\ne b$ | The witness uses a missing edge, so also refutes a non-edge-only comparison | | Path on $n$ vertices, endpoint perturbation | SimpleGraph.pathGraph n, endpoints $0,n-1$ | Bounds follow from $2\le n$ | | Smallest derivative | claim: two HasDerivAt premises at zero and $d_P\le d_G$ | Derivative existence is required; no default value for a nondifferentiable function | | Universal minimality fails | result | Negates this entire proposition | | The path value is $O(n^3)$ | Outside claim | The refutation makes no contrary assertion about the path value |

The instance class DecidableRel G.Adj supplies computation on the finite graph and imposes no mathematical restriction. Real parameters allow the ordinary two-sided derivative; the proof establishes positive degrees and the inverse update on a neighborhood of zero. No source clause within the minimality proposition is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/AlbrightKemenyEdgeDerivativeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresCounterexample foundFirst preprint 2023Signed cardinality is not 0-mesic on [3] × [12]A literal 73-state orbit on a 3 by 12 rectangle has signed-cardinality sum minus one.Refuted

The questionElder, Lafrenière, McNicholas, Striker and Welch, Toggling, rowmotion, and homomesy on interval-closed sets, arXiv:2307.08520v2, Conjecture 4.12, states:

Why it mattersConjecture 4.12 predicts cancellation of the signed rank statistic on every interval-closed rowmotion orbit in the stated dimensions.

The ideaIt defines the rank-parity point weight and its finite set sum, states the full quantified conjecture, and constructs a row-major reverse extension of Point 3 12.

Our contributionA literal 73-state orbit on a 3 by 12 rectangle has signed-cardinality sum minus one.

Verification details
Release source recordRefuted in Lean

Open problem

Signed cardinality is not 0-mesic on [3] × [12]

Elder, Lafrenière, McNicholas, Striker and Welch, Toggling, rowmotion, and homomesy on interval-closed sets, arXiv:2307.08520v2, Conjecture 4.12, states:

If m = 2 or m = 3, then the signed cardinality statistic is 0-mesic under rowmotion on interval-closed sets of [m]×[n] whenever m + n − 1 is even.

The signed point statistic is 1 on even rank and −1 on odd rank, and the statistic of an interval-closed set is the sum over its members. The follow-up by Lafrenière, Lewis, McNicholas, Striker and Welch, arXiv:2505.04000v1, retains the m = 3 even-n case as an open conjectural statement in Remark 3.31.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/IntervalClosedSignedCardinalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2023Successive new Laplacian layer maxima of token graphsFor the actual connected star on four vertices, the first two new Laplacian layers have maxima 4 and 3.Refuted

The questionFiol, On the spectra and spectral radii of token graphs, arXiv:2310.16929v1, Section 3, Definition 3.1 and Conjecture 3.3; journal version, Boletín de la Sociedad Matemática Mexicana, volume 30, article 11 (2024), DOI 10.1007/s40590-023-00583-3, Section 3:

Why it mattersFull-sector spectral maxima and new-layer maxima are different quantities: inherited eigenvalues remain in the former and are removed, with their multiplicities, from the latter.

The ideaTake the connected star K₁,₃ on four vertices and k=1.

Our contributionFor the actual connected star on four vertices, the first two new Laplacian layers have maxima 4 and 3.

Scope & next questionThe refutation also does not settle the source's conditional equality consequence or any separate full-radius claim.

Verification details
Release source recordRefuted in Lean

Open problem

Successive new Laplacian layer maxima of token graphs

M. A. Reyes, C. Dalfó and M. À. Fiol, On the spectra and spectral radii of token graphs, arXiv:2310.16929v1, Section 3, Definition 3.1 and Conjecture 3.3; journal version, Boletín de la Sociedad Matemática Mexicana, volume 30, article 11 (2024), DOI 10.1007/s40590-023-00583-3, Section 3:

Conjecture 3.3. For any graph G, ρ₁(G) ≤ ρ₂(G) ≤ ⋯ ≤ ρ⌊n/2⌋(G).

Here G is a finite simple unweighted undirected graph of order n. The source uses the ordinary real Laplacian D−A. The vertices of F_j(G) are all j-element subsets of V(G), with an edge precisely when their symmetric difference is an edge of G. F_0(G) is the singleton empty configuration. Writing Sp_j for the full Laplacian spectral multiset, Definition 3.1 sets Δ_j = Sp_j − Sp_(j−1) by multiset subtraction and ρ_j = max Δ_j, for 1 ≤ j ≤ floor(n/2). Thus the conjecture quantifies every 1 ≤ k < floor(n/2) and asserts ρ_k ≤ ρ_(k+1). The source's connected-graph convention is satisfied by the counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/TokenGraph/NewLayerMaximaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresCounterexample foundFirst preprint 2021Shala Question 7.4: weakly but not strongly coset-like atomistic latticesA finite lattice separates weak and strong coset-likeness.Refuted

The questionShala, The Probabilistic Zeta Function of a Finite Lattice, arXiv:2112.13766v6 (published in Rocky Mountain Journal of Mathematics 54 (2024), 1511–1526), asks in Question 7.4: “Does there exist a weakly coset-like atomistic lattice that is not strongly coset-like?

Why it mattersThe source reports weakly coset-like non-strong examples among non-atomistic lattices and asks whether an atomistic example exists.

The ideaThe counterexample carrier has twelve elements: bottom, top, two intermediate elements, and eight singleton atoms.

Our contributionA finite lattice separates weak and strong coset-likeness.

Verification details
Release source recordRefuted in Lean

arXiv

Shala Question 7.4: weakly but not strongly coset-like atomistic lattices

B. Shala, The Probabilistic Zeta Function of a Finite Lattice, arXiv:2112.13766v6 (published in Rocky Mountain Journal of Mathematics 54 (2024), 1511–1526), asks in Question 7.4: “Does there exist a weakly coset-like atomistic lattice that is not strongly coset-like? In particular, is the d-divisible partition lattice weakly coset-like?”

The negation of the first question is formalized as the universal implication that every finite atomistic weakly coset-like lattice is strongly coset-like. The settling theorem refutes that implication and answers the existence question affirmatively; Refuted refers to this universal implication.

Statement echo

Equation (3.1) is represented by P, with Mathlib's incidence-algebra Möbius function and the rational ratio of join-irreducible cardinalities. Definitions 5.1 and 5.2 are represented by StronglyCosetLike and WeaklyCosetLike. The universal first question is claim, and the settling node proves result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Lattices/ShalaCosetLikeAtomisticLattice.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Analysis & dynamicsCounterexample foundFirst preprint 2022Abdesselam's Problem 2: padded general Ginibre inequalities for the XY modelThe XY model violates a padded general Ginibre inequality on a five-cycle.Refuted

The questionAbdelmalek Abdesselam, “Non-Abelian correlation inequalities and stable determinantal polynomials”, arXiv:2207.07603v2, §4, Problem 2, printed p. 14:

Why it mattersGinibre's GG inequalities for the XY model do not answer whether inserting an arbitrary even natural padding preserves the entire PGG collection.

The ideaTake p = 5 and cycle edges (0,1), (1,2), (2,3), (3,4), (0,4).

Our contributionThe XY model violates a padded general Ginibre inequality on a five-cycle.

Scope & next questionThis refutation supplies its negative answer; it does not settle Problems 1 or 4, or change the separate Problem 3 settlement.

Verification details
Release source recordRefuted in Lean

Open problem

Abdesselam's Problem 2: padded general Ginibre inequalities for the XY model

Abdelmalek Abdesselam, “Non-Abelian correlation inequalities and stable determinantal polynomials”, arXiv:2207.07603v2, §4, Problem 2, printed p. 14:

For the XY model, or O(2) model, the GG inequalities were proved by Ginibre [18]. What about the padded generalizations given by the PGG inequalities?

The free O(2) measure is the product of the rotation-invariant probability measures on S¹. Its angle coordinates use normalized Haar measure on Real.Angle, with spin θ = (cos θ, sin θ). The basic observables are all pair inner products σᵢ·σⱼ, i < j. The parity map assigns one bit to each endpoint of each pair; the source's evenness means ρ(u) = 0.

For every p ≥ 1, the proposed assertion requires, for every m ≥ 0, every natural row matrix V, every ε ∈ {−1,1}ᵐ, and every even natural padding u,

$$ \int_{X^2} \mathcal O(x)^u\mathcal O(y)^u \prod_{i=1}^{m}\left[\mathcal O(x)^{V_{i*}} +\varepsilon_i\mathcal O(y)^{V_{i*}}\right] ,d\mu(x),d\mu(y) \ge 0. $$

The Lean claim retains these quantifiers and the literal duplicated integral. Its evenness premise is inline: parity p (fun e => (u e : ℤ)) = 0. The sole settling theorem is result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/XYPaddedGinibreRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedSource 2023The coherence vector and spectrum satisfy the product bound in every positive dimensionEvery majorization upper bound of the coherence set has first component times largest eigenvalue at least one over the dimension.Proved

The questionBosyk, Complementarity between quantum coherence and mixedness: a majorization approach, Communications in Theoretical Physics 75(5), 055102 (2023), DOI 10.1088/1572-9494/acc92f, Section 3.4, journal PDF p. 7:

Why it mattersThe paper studies the tensor product of vectorial measures of coherence and mixedness, retaining their majorization information instead of choosing a scalar measure.

The ideaThe following helper names refer to the canonical module.

Our contributionEvery majorization upper bound of the coherence set has first component times largest eigenvalue at least one over the dimension.

Scope & next questionThe conjectural premise of this consequence is discharged; the tensor-product majorization statement and lattice infimum are not formalized by result. - Neighbouring questions.

Verification details
Release source recordProved in Lean

Open problem

The coherence vector and spectrum satisfy the product bound in every positive dimension

Y. Alvarez, M. Losada, M. Portesi and G. M. Bosyk, Complementarity between quantum coherence and mixedness: a majorization approach, Communications in Theoretical Physics 75(5), 055102 (2023), DOI 10.1088/1572-9494/acc92f, Section 3.4, journal PDF p. 7:

Finally, we state a conjecture for the greatest component of $\nu(\rho)\otimes\lambda(\rho)$ that allows us to obtain a nonoptimal but nontrivial bound for the tensor-product majorization relation, which depends only on the dimension of the Hilbert space.

Conjecture 5. Let $\rho\in\mathcal S(\mathcal H_d)$. Then $\nu_1(\rho)\lambda_1(\rho)\ge\frac1d$. (20)

For $d=2$ the conjecture is clearly valid, whereas for $d=3$ we have numerical evidence of its validity.

Here $\mathcal S(\mathcal H_d)$ is the set of density operators on the $d$-dimensional Hilbert space, with a fixed incoherent basis. Section 2.2, Eqs. (9)–(12), PDF p. 3, defines the pure-state squared-amplitude vector $\mu(|\psi\rangle\langle\psi|)$, its decreasing rearrangement $\mu^\downarrow$, and

$\mathcal U^{\rm psd}(\rho)={\sum_jp_j\mu^\downarrow(|\psi_j\rangle\langle\psi_j|):\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|}$,

where the ensembles are finite, their weights are nonnegative and sum to one, and their vectors have norm one. The source defines $\nu(\rho)=\bigvee\mathcal U^{\rm psd}(\rho)$ in the majorization lattice; Section 2.3, PDF p. 4, defines $\lambda(\rho)$ as the decreasingly ordered spectrum. Thus the subscript $1$ denotes the greatest component in each vector. The definitions are also recorded in the literature note.

Issue #23165 fixes the formal reading: for every $d\ge1$, every positive semidefinite complex $d\times d$ matrix $\rho$ of trace one, and every decreasing, nonnegative vector $\nu$ of sum one that majorizes every member of $\mathcal U^{\rm psd}(\rho)$, one has $\nu_1\lambda_1(\rho)\ge1/d$. This is proved. Applying it to the source's supremum uses that supremum's upper-bound property; the module does not construct the majorization lattice.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CoherenceVectorSpectrumProductBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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    Year in the primary source attribution: 2023.

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Discrete structuresQuestion resolvedFirst preprint 2023Weak Ascent Sequences Avoiding 210 and the Semi-Baxter NumbersWeak ascent sequences avoiding 210 are counted by the semi-Baxter numbers.Proved

The questionRamírez, Pattern Avoidance in Weak Ascent Sequences, arXiv:2309.06518v4 (Discrete Math.

The ideaFor a nonempty 210-avoiding weak ascent sequence put H = 1 + wasc, M = max and D = the largest entry that is the smaller term of some inversion (0 if none).

Our contributionWeak ascent sequences avoiding 210 are counted by the semi-Baxter numbers.

Verification details
Release source recordProved in Lean

Open problem

Weak Ascent Sequences Avoiding 210 and the Semi-Baxter Numbers

Beáta Bényi, Toufik Mansour and José L. Ramírez, Pattern Avoidance in Weak Ascent Sequences, arXiv:2309.06518v4 (Discrete Math. Theor. Comput. Sci. 26:1, 2024), Section 3, Conjecture 3.1:

The sequence w_210(n) coincides with the sequence A117106.

The paper adds that A117106 enumerates permutations avoiding the vincular pattern 2-41-3. Here w_210(n) counts the weak ascent sequences of length n with no indices i < j < k and e_i > e_j > e_k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/WeakAscent/WeakAscentSemiBaxter.result

Frozen module record

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Geometry & algebraCounterexample foundFirst preprint 2023Unequal minima for sumsets with at most two distinct termsOdd restricted sumsets in elementary abelian 2-groups have smaller minima.Refuted

The questionBhanja, Sumsets with a minimum number of distinct terms, Integers 25 (2025), A12, arXiv:2307.03977v2, §4 “Further Remarks”, printed p. 11:

Why it mattersRestricting the number of distinct summands can preserve the integer minimum while changing minima in finite groups.

The ideaA sum with odd length using at most two values is one of those values: exactly one multiplicity is odd.

Our contributionOdd restricted sumsets in elementary abelian 2-groups have smaller minima.

Verification details
Release source recordRefuted in Lean

arXiv

Unequal minima for sumsets with at most two distinct terms

J. Bhanja, Sumsets with a minimum number of distinct terms, Integers 25 (2025), A12, arXiv:2307.03977v2, §4 “Further Remarks”, printed p. 11:

Another sumset that may come immediately to the reader’s mind is the sumset h(≤r) A, which contains those elements of the group G that are the sum of h elements of A with at most r distinct elements. The sumset hA is a special case of h(≤r) A for r = h.

Question 1. Is it always the case that the minimum size of both the sumsets hA and h(≤r) A are the same in any abelian group G, where the minimum runs over the subsets of G of size (say) k?

Formal reading

Names have prefix D5.S3.FiniteGroups.BhanjaRestrictedSumset.Result. hSumset h A is Mathlib's pointwise $h\mathbin{\bullet}A$, with exactly $h$ summands. restrictedSumset h r A is the image of functions $f:\operatorname{Fin}h\to G$ taking values in $A$ whose image has at most $r$ elements, under $f\mapsto\sum_i f(i)$. minSize G k h and minRestrictedSize G k h r are infima in the natural numbers of the corresponding cardinalities over $k$-element finsets. The claim requires a $k$-element subset to exist, so these infima are actual minima. It quantifies over all abelian groups, $h\ge1$, $r\ge2$, and all attainable cardinalities $k$. Allowing $k=0$ adds only a case in which both sumsets are empty for $h\ge1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/BhanjaRestrictedSumset/Result.result

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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Right multiplication maximizes CHSH over the nonlocal-box orbitRight multiplication maximizes CHSH over every parenthesization in the PR-SR-I triangle.Proved

The questionPellegrini, Algebra of Nonlocal Boxes and the Collapse of Communication Complexity, arXiv:2312.00725v3, Conjecture 15 (Dyck paths), ask:

Why it mattersTheorem 13 bounds the tilted orbit under a distillation hypothesis.

The ideaMultiplication on the right by the starting box preserves the second-coordinate order when the first coordinates agree.

Our contributionRight multiplication maximizes CHSH over every parenthesization in the PR-SR-I triangle.

Verification details
Release source recordProved in Lean

arXiv

Right multiplication maximizes CHSH over the nonlocal-box orbit

P. Botteron, A. Broadbent, R. Chhaibi, I. Nechita and C. Pellegrini, Algebra of Nonlocal Boxes and the Collapse of Communication Complexity, arXiv:2312.00725v3, Conjecture 15 (Dyck paths), ask:

We conjecture that the same result actually holds without the tilde, i.e. the right multiplication $P^{\boxtimes k}$ gives the highest CHSH-value of $\mathrm{Orbit}^{(k)}(P)$, as observed numerically.

The source uses the wiring $\mathsf W_{\rm BS}$ and the domain $\widetilde{\mathcal A}={P\in\operatorname{Conv}{\mathrm{PR},\mathrm{SR},\mathrm I}:\operatorname{CHSH}(P)\ge3/4}$. The orbit includes every full binary parenthesization of $k$ copies of $P$. The right product multiplies each fresh copy on the right, so its parentheses are left-nested. Botteron's thesis, arXiv:2510.09457, repeats the question as Conjecture 6.11. The source definitions and their locators are in the Library note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/NonlocalBoxOrbitRightProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Edge irregularity strength of cycle-star graphsCycle-star edge irregularity strength attains both conjectured sharp bounds.Proved

The questionSalma–Nagesh–Narahari, On edge irregularity strength of cycle-star graphs, Proyecciones 43(3), 725–741, arXiv:2405.12263v1, Conjecture 3.1, states:

Why it mattersThe paper establishes the short-cycle cases and asks for the uniform formula at every remaining cycle length.

Our contributionCycle-star edge irregularity strength attains both conjectured sharp bounds.

Verification details
Release source recordProved in Lean

Open problem

Edge irregularity strength of cycle-star graphs

Salma–Nagesh–Narahari, On edge irregularity strength of cycle-star graphs, Proyecciones 43(3), 725–741, arXiv:2405.12263v1, Conjecture 3.1, states:

Let $G=CS_{k,n-k}$ be a cycle-star graph. For $k\geq8$ and $n-k\geq1$, $$es(G):=\begin{cases}\lceil(n+1)/2\rceil&\text{for }k+1\leq n\leq2k-4,\n-k+2&\text{for }n\geq2k-3.\end{cases}$$

The graph consists of a cycle of length $k$ with $n-k$ leaves appended to one vertex. A vertex labeling takes values in ${1,\ldots,K}$ and may repeat labels; different unordered edges must have different endpoint sums. The strength is the least positive feasible bound and includes attainment. The source and binding conventions are in #22868. The source note is Salma–Nagesh–Narahari.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every $k\geq8$ and $n-k\geq1$ | claim quantifies natural $k,n$ with both hypotheses | Exact source domain | | Vertices $v,v_i$, $1\leq i\leq n-1$ | cycleStar has carrier Fin n; zero represents $v$ | Exact indexing | | Leaf edges, both cycle-closing edges, consecutive internal edges | cycleStar uses the three source index ranges inside SimpleGraph.fromRel | Undirected, loopless edge set | | Labels $1$ through $K$, repetitions allowed | IsKLabeling | No vertex-injectivity assumption | | Edge weights are endpoint sums; different edges have different weights | edgeWeight, IsEdgeIrregularLabeling, unordered equality in Sym2 | Reversing orientation preserves the edge | | Minimum feasible label bound | IsEdgeIrregularityStrength uses IsLeast on positive feasible bounds | Attainment and comparison with every feasible bound | | $k+1\leq n\leq2k-4$: $\lceil(n+1)/2\rceil$ | First conjunct of claim, natural division $(n+2)/2$ | Same ceiling | | $n\geq2k-3$: $n-k+2$ | Second conjunct of claim | Exact source value |

No hypothesis assumes a labeling exists. Positivity of feasible bounds is the source's label-bound convention, rather than an additional graph hypothesis. Under the source domain, the two ranges exhaust all parameters.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/EdgeLabeling/CycleStarIrregularity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Nonnesting Permutations Avoiding 1322Nonnesting permutations avoiding 1322 have the asserted binomial enumeration.Proved

The questionSergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1322}: the conjectured number of nonnesting permutations of {1, 1, 2, 2, …, n, n} avoiding 1322 is (1/n) Σ{k=0}^{n-1} C(3n, k) C(2n-k-2, n-1) (OEIS A007297), with the remark

The ideaA nonnesting word is a permutation (its first-occurrence order) together with a Dyck word, and it contains 1322 exactly when some letters a < c < b occur as a, then b, before the first copy of c; hence the first-occurrence order avoids 132 and splits as B A z at its last entry z.

Our contributionNonnesting permutations avoiding 1322 have the asserted binomial enumeration.

Verification details
Release source recordProved in Lean

Open problem

Nonnesting Permutations Avoiding 1322

Sergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1322}: the conjectured number of nonnesting permutations of {1, 1, 2, 2, …, n, n} avoiding 1322 is (1/n) Σ_{k=0}^{n-1} C(3n, k) C(2n-k-2, n-1) (OEIS A007297), with the remark

All the conjectures have been checked for n up to 8.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Nonnesting/NonnestingOneThreeTwoTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2022Ellis–Thanatipanonda Conjecture 1: Family AEvery Family A cut-set has nim-sequence (0, 1)^c (+2).Proved

The questionPaul Ellis and Thotsaporn Aek Thanatipanonda, The Arithmetic-Periodicity of cut for C = {1, 2c}, arXiv:2203.02457v1, Section 7, page 18; Discrete Applied Mathematics 322 (2022), 391–403, DOI 10.1016/j.dam.2022.08.027:

Why it mattersIts only direct D5 owner is D5.S0.Certificates.Games.AbukuEnforceNimberDomination.mex.

The ideaThe proof uses induction.

Our contributionEvery Family A cut-set has nim-sequence (0, 1)^c (+2).

Scope & next questionFamily B, including the source's Conjecture 2, is outside this result, so the combined DDLP conclusion is not claimed.

Verification details
Release source recordProved in Lean

Open problem

ellis thanatipanonda 2022 cut family a conjecture 1

Paul Ellis and Thotsaporn Aek Thanatipanonda, The Arithmetic-Periodicity of cut for C = {1, 2c}, arXiv:2203.02457v1, Section 7, page 18; Discrete Applied Mathematics 322 (2022), 391–403, DOI 10.1016/j.dam.2022.08.027:

Let X to be a non-empty set of even numbers, each of which is at least 4. Let Y to be a non-empty set of odd numbers, each of which is at least 5. Let x = 2c and y be the smallest elements of X and Y, respectively.

Family A: C = {1, 3} ∪ X, or C = {1, 3} ∪ X ∪ Y.

The nim-sequence for all games of cut in this family are all precisely (0, 1)^c (+2).

This is Conjecture 1. Preregistration: #19674. The settling theorem is D5.S0.Certificates.Games.EllisThanatipanondaCutFamilyA.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A move cuts a pile into $d+1$ positive parts, $d\in C$ | G: Fin (d + 1) option, positivity and exact sum | All legal options occur | | Nim-sum of the part values; mex of option values | G: List.ofFn, right fold of natural xor, existing mex | The guard only ensures well-founded recursion and is inactive for Family A options | | Nonempty even $X$, all elements at least $4$ | First two hypotheses of claim | Arbitrary natural sets, including infinite sets | | No extra odd cuts, or nonempty odd $Y$ with elements at least $5$; its least element $y$ | Disjunction in claim | $Y=\varnothing$ gives the first alternative; nonempty natural $Y$ has a least member, which the conclusion does not use | | Smallest even cut $2c$ | $2c=\operatorname{sInf}X$ | Nonempty natural sets have a least member; $c\ge2$ follows | | Sequence $(0,1)^c(+2)$, beginning at $n=1$ | Conclusion of claim, proved by result | $G_C(n)=2\lfloor(n-1)/(2c)\rfloor+((n-1)\bmod2)$ for every $n\ge1$ |

The natural subtraction, integer quotient and remainder are the literal binding conventions. The recursion also defines values outside Family A; guard_value and G_equation identify its legal-option recursion on the positive cut-sets used in the proof. No assertion about an invalid zero cut is needed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/EllisThanatipanondaCutFamilyA.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023Foucart–Liao's Chebyshev-center regularization questionA real Chebyshev center need not minimize any positively weighted quadratic regularization objective.Refuted

The questionLiao, S-Procedure Relaxation: a Case of Exactness Involving Chebyshev Centers, arXiv:2310.09677v1, in Explorations in the Mathematics of Data Science (Springer, 2024), DOI 10.1007/978-3-031-66497-71, §5 “Some Open Questions”, Question 2:

Why it mattersThe positive result under the specific conditions does not settle the unrestricted real case.

Our contributionA real Chebyshev center need not minimize any positively weighted quadratic regularization objective.

Scope & next questionThe example does not settle either restriction.

Verification details
Release source recordRefuted in Lean

Open problem

Foucart–Liao's Chebyshev-center regularization question

S. Foucart and C. Liao, S-Procedure Relaxation: a Case of Exactness Involving Chebyshev Centers, arXiv:2310.09677v1, in Explorations in the Mathematics of Data Science (Springer, 2024), DOI 10.1007/978-3-031-66497-7_1, §5 “Some Open Questions”, Question 2:

Is the Chebyshev center always obtained via regularization?

The authors' gloss is:

From [10], we know that the answer is affirmative under the specific conditions mentioned above, since the Chebyshev center equals the regularizer $f_{\tau_\sharp}$ with parameter $\tau_\sharp\in(0,1)$ satisfying the implicit equation (9). Without the specific conditions, we also know from [8] that the answer is affirmative if one considered the complex setting rather than the real setting.

The section takes $Q=\mathrm{Id}_H$ and, in (25), $\mathcal K={f\in H:|Pf|_H\leq\varepsilon}$ and $\mathcal E={e\in\mathbb R^m:|e|_2\leq\varepsilon}$. The “specific conditions” are that $P$ is an orthogonal projector and $\Lambda\Lambda^*=\mathrm{Id}_m$. Equation (8) defines

$$ f_\tau=\operatorname{argmin}_f \bigl((1-\tau)|Pf|^2+\tau|y-\Lambda f|^2\bigr). $$

The Lean claim quantifies over finite-dimensional real Euclidean spaces, linear $P$ and $\Lambda$, data $y$, and positive bounds $\varepsilon,\eta$. For every nonempty bounded feasible set $C={f:|Pf|\leq\varepsilon,\ |y-\Lambda f|\leq\eta}$, it asks whether every Chebyshev center minimizes this objective for some $\tau\in(0,1)$. The counterexample uses $\varepsilon=\eta=1$, so it also addresses the equal-bound setting of (25).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/OptimalRecovery/ChebyshevCenterRegularization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023Gan–Kim: weak log-majorization monotonicity of Wasserstein power meansTwo positive definite integer matrices reverse the required determinant comparison between powers one half and one for every internal weight, refuting Conjecture 5.7.Refuted

The questionLuyining Gan and Sejong Kim, Weak log-majorization between the geometric and Wasserstein means, arXiv:2301.07934v4, §5, Conjecture 5.7.

Why it mattersThe conjecture removes the extra hypothesis (5.15) from the paper's Theorem 5.6 and proposes an unconditional power-parameter monotonicity law for the Wasserstein mean.

The ideaRiccati uniqueness identifies them with the required inverse geometric means.

Our contributionTwo positive definite integer matrices reverse the required determinant comparison between powers one half and one for every internal weight, refuting Conjecture 5.7.

Verification details
Release source recordRefuted in Lean

Open problem

Gan–Kim: weak log-majorization monotonicity of Wasserstein power means

Luyining Gan and Sejong Kim, Weak log-majorization between the geometric and Wasserstein means, arXiv:2301.07934v4, §5, Conjecture 5.7. See gankim2024wasserstein.

Conjecture 5.7. Let A, B ∈ Pm and t ∈ (0, 1). Then (A^p ⋄t B^p)^{1/p} ց≺w log exp((1 − t) log A + t log B) as p ց 0.

A ⋄t B := (1 − t)² A + t² B + t(1 − t)[A(A^{−1} #B) + (A^{−1} #B)A]

Here $A#B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$ and $\mathbb P_m$ is the cone of complex Hermitian positive definite matrices. Weak log-majorization compares every initial product of decreasingly sorted eigenvalues, including the product of all $m$ eigenvalues.

For $F_t(p)=(A^p\diamond_tB^p)^{1/p}$, the decreasing arrow is the conjunction of the comparisons $F_t(p)\prec_{w\log}F_t(q)$ for all real $0<p<q$ and the right-hand limit $F_t(p)\to\exp((1-t)\log A+t\log B)$ as $p\downarrow0$. The formal statement includes both assertions, every finite dimension, all complex positive definite inputs and every real $0<t<1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/GanKimConjectureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2020Mary Beth RuskaiEvery equal-modulus HMR qutrit channel has a unitary and doubly extreme decompositionEvery equal-modulus Haagerup–Musat–Ruskai qutrit channel is a mixture of a unitary conjugation and a rank-three channel extreme in both the unital and trace-preserving completely positive maps.Proved

The questionRuskai, Extreme Points and Factorizability for New Classes of Unital Quantum Channels, arXiv:2006.03414v4 (Annales Henri Poincaré 22, 3455–3496, 2021), §2.3.1, “Non-extreme cases”, HMR2021J.tex, lines 387–390:

Why it mattersThe paper distinguishes extremality in the intersection UCPT from extremality in either of the larger UCP and CPT sets.

The ideaThese normalizations force the coefficient matrix to be the identity in the respective convex-decomposition arguments.

Our contributionEvery equal-modulus Haagerup–Musat–Ruskai qutrit channel is a mixture of a unitary conjugation and a rank-three channel extreme in both the unital and trace-preserving completely positive maps.

Scope & next questionThe argument supplies neither a classification of the possible unitaries and remainder maps nor an extension to unequal moduli or higher dimensions; those questions remain open here. - Effect on the source.

Verification details
Release source recordProved in Lean

Open problem

Every equal-modulus HMR qutrit channel has a unitary and doubly extreme decomposition

U. Haagerup, M. Musat and M. B. Ruskai, Extreme Points and Factorizability for New Classes of Unital Quantum Channels, arXiv:2006.03414v4 (Annales Henri Poincaré 22, 3455–3496, 2021), §2.3.1, “Non-extreme cases”, HMR2021J.tex, lines 387–390:

It seems reasonable to conjecture that, whenever $|\alpha | = |\beta| = \1rt2$, the channel $\Phi_{\alpha,\beta} $ is a convex combination of a unitary conjugation and a map with Choi-rank $3$ which is extreme in both the UCP and CPT maps. However, one would need a different unitary transformation relating ${ A_k } $ to ${ X_k } $.

The source macro \1rt2 means $1/\sqrt2$ (line 56). For the standard basis of $\mathbb C^3$, the four operators of Theorem thm:UCPT (lines 280–290) are

$A_1=\alpha|e_1\rangle\langle e_1|+|e_2\rangle\langle e_3|$, $A_2=\beta|e_1\rangle\langle e_3|+|e_3\rangle\langle e_2|$,

$A_3=-|e_1\rangle\langle e_2|-\overline\beta|e_3\rangle\langle e_1|$, $A_4=|e_2\rangle\langle e_1|+\overline\alpha|e_3\rangle\langle e_3|$,

and $\Phi_{\alpha,\beta}(\rho)=\tfrac12\sum_{k=1}^4 A_k^*\rho A_k$. Here UCP means unital completely positive, CPT means completely positive and trace preserving, and UCPT means both. The definitions and source locators are in the literature note.

Issue #22743 fixes the statement: for every $\alpha,\beta\in\mathbb C$ with $|\alpha|=|\beta|=1/\sqrt2$, there exist a unitary $U\in M_3(\mathbb C)$, a real $\lambda\in[0,1]$, and a complex linear map $\Psi:M_3(\mathbb C)\to M_3(\mathbb C)$ such that $\Psi$ is completely positive, unital and trace preserving, its Choi matrix has rank $3$, it is separately extreme in the UCP and CPT sets under real convex combinations, and

$\Phi_{\alpha,\beta}(\rho)=\lambda U^*\rho U+(1-\lambda)\Psi(\rho)$

for every complex matrix $\rho\in M_3(\mathbb C)$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/QutritChannelUnitaryExtremeDecomposition.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    First preprint year: 2020.

    Read source ↗
  • Scholarly context

    Mary Beth Ruskai made foundational contributions to quantum entropy inequalities, quantum channels and mathematical quantum information.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Scholarly context · 4/4 · 5% weight

    Mary Beth Ruskai made foundational contributions to quantum entropy inequalities, quantum channels and mathematical quantum information.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023Mixed imitation and best-response dynamics need not equilibrate on general graphsA four-agent coordination network has a persistent mixed imitation and best-response trajectory that never reaches equilibrium.Refuted

The questionRamazi, Equilibration of Coordinating Imitation and Best-Response Dynamics, arXiv:2306.03243v3 [math.DS], IEEE Control Systems Letters 7 (2023) 3078-3083, study finite undirected networks whose agents use either imitation or best response in two-strategy coordination games.

Why it mattersA positive answer to the universal claim would extend equilibration from the source's special network classes to all finite symmetric irreflexive graphs.

Our contributionA four-agent coordination network has a persistent mixed imitation and best-response trajectory that never reaches equilibrium.

Scope & next questionHowever, it remains open whether the dynamics equilibrate under an arbitrary persistent activation sequence.

Verification details
Release source recordRefuted in Lean

Open problem

Mixed imitation and best-response dynamics need not equilibrate on general graphs

N. Hasheminejad and P. Ramazi, Equilibration of Coordinating Imitation and Best-Response Dynamics, arXiv:2306.03243v3 [math.DS], IEEE Control Systems Letters 7 (2023) 3078-3083, study finite undirected networks whose agents use either imitation or best response in two-strategy coordination games. The source defines persistence as every agent becoming active infinitely often. Its abstract asks:

It is therefore, unknown, whether a potential function also exists for mixed populations or if there actually exists a counter example where an activation sequence prevents equilibration.

(main.tex, abstract, l. 71.) In Section 2 it states:

However, it remains open whether the dynamics equilibrate under an arbitrary persistent activation sequence.

(main.tex, l. 225.) The conclusion specifies the general-graph question:

Whether dense trees or general graphs equilibrate under every activation sequence remains an open problem.

(main.tex, l. 668.)

The module's claim is the general-graph statement: for every n : ℕ, every symmetric and irreflexive adjacency relation on Fin n, every imitator assignment and real payoff map satisfying $\max{\mathrm{pay}_i(B,A),\mathrm{pay}_i(A,B)}<\min{\mathrm{pay}_i(A,A),\mathrm{pay}_i(B,B)}$ for every agent, every initial state, and every persistent activation sequence, some trajectory state is an IsEquilibrium. Here false denotes $A$ and true denotes $B$; IsEquilibrium quantifies over every activation sequence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/DecisionValue/MixedImitationBestResponseCycle.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Non-tightness of the antidistinguishability overlap boundFinite-dimensional spectral and overlap bounds for antidistinguishability.Proved

The questionJohnston, Russo and Sikora, “Tight bounds for antidistinguishability and circulant sets of pure quantum states”, Quantum 9, 1622, arXiv:2311.17047v2, Section 6, asks:

Why it mattersIts quantitative theorem gives a uniform guarantee for arbitrary unit-state ensembles, including non-circulant Gram matrices.

The ideaThe proof constructs a POVM from an incoherent positive semidefinite Gram decomposition (AntidistinguishabilityGram.incoherentgramantidistinguishable).

Our contributionFinite-dimensional spectral and overlap bounds for antidistinguishability.

Verification details
Release source recordProved in Lean

Open problem

Non-tightness of the antidistinguishability overlap bound

Johnston, Russo and Sikora, “Tight bounds for antidistinguishability and circulant sets of pure quantum states”, Quantum 9, 1622, arXiv:2311.17047v2, Section 6, asks:

We still do not know whether or not Corollary 5.4 is tight when n ≥ 5.

Corollary 5.4 guarantees antidistinguishability when every distinct-state overlap is at most $c_n=\sqrt{(n-2)/(2(n-1))}$. Issue #15573 specifies the finite-dimensional unit-state convention and the squared-overlap margin. The settlement proves non-tightness for every $n\ge10$.

Statement echo

| Source condition | Formal counterpart | Scope | | --- | --- | --- | | A set of $n$ pure states | $\psi:\operatorname{Fin}(n)\to\mathbb C^d$ | Every finite ambient dimension $d$ | | Normalized states | $|\psi_i|=1$ for every $i$ | No dimension or distinctness restriction | | POVM with zero probability for the corresponding state | IsAntidistinguishable | Positive semidefinite effects, identity sum, zero Born quadratic form | | Tightness of the overlap guarantee | IsTight | Every positive squared-overlap margin admits a non-antidistinguishable family | | Positive source cap | $c_n>0$ for $n\ge10$ | Squaring reparametrizes positive margins: $(c_n+\varepsilon)^2-c_n^2>0$ and $\sqrt{c_n^2+\delta}-c_n>0$ | | Non-tightness | claim and result | Every $n\ge10$; cases $5\le n\le9$ remain open |

The positive margin $\delta_n$ is additional proved information, rather than a hypothesis strengthening the source question. The displayed finite-dimensional POVM definition is literal. The Gram construction is reused as Matrix.gram, and the squared Frobenius norm is reused as RHLinalg.frobSq.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/AntidistinguishabilityOverlapNonTight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2023Coefficientwise log-concavity of the triple rising-factorial seriesTriple rising-factorial series preserve coefficientwise log-concavity.Proved

The questionDmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 8, Conjecture 3.8:

Why it mattersThe conjecture is the remaining multiplier-three case of the source's rising-factorial family.

The ideaThe binomial residue counts obey Pascal recurrence and a six-periodic closed form.

Our contributionTriple rising-factorial series preserve coefficientwise log-concavity.

Verification details
Release source recordProved in Lean

Open problem

Coefficientwise log-concavity of the triple rising-factorial series

Dmitrii Karp and Yi Zhang, Log-concavity and log-convexity of series containing multiple Pochhammer symbols, arXiv:2305.09029v3, page 8, Conjecture 3.8:

Assume that ${f_n}_{n=0}^{\infty}$ is log-concave and independent of $\mu$. Then the function

$$ \mu\to f(\mu;x)=\sum_{n=1}^{\infty} f_n\frac{(\mu)_{3n}}{(3n-1)!}x^n $$

is coefficient-wise log-concave.

Definition 1.1, page 2, uses nonnegative, nontrivial real sequences, $f_k^2\ge f_{k-1}f_{k+1}$ for positive natural $k$, and interval support. In particular, internal zeros are excluded, but zero endpoints and finite supports are allowed. Define

$$ a_0(t)=0,\qquad a_n(t)=\frac{(t)_{3n}}{(3n-1)!}\quad(n\ge1), $$

and, for every natural $m$ and nonnegative real $\mu,\alpha,\beta$,

$$ \delta_m=\sum_{k=0}^{m} f_kf_{m-k} \left[a_k(\mu+\alpha)a_{m-k}(\mu+\beta) -a_k(\mu)a_{m-k}(\mu+\alpha+\beta)\right]. $$

The assertion is $\delta_m\ge0$ for every such sequence, parameter triple and degree. The endpoint summands vanish. These are finite formal Cauchy coefficients, without convergence or growth assumptions.

The coefficient-wise convention on printed page 1 reads:

if the coefficients $\delta_k$ at all powers of $x$ in (2) are non-negative (non-positive) for all $\alpha,\beta\ge0$ we will say that $\mu\to f(\mu;x)$ is coefficient-wise log-concave (log-convex ).

The Lean namespace prefix is D5.S3.Analytic. The coefficient is KarpZhangNinh.Result.Kernel.a, the finite Turanian is KarpZhangNinh.Result.delta, the sequence predicate is the reused KarpZhangReciprocal.Result.IsLogConcave, and the universal proposition is KarpZhangNinh.Result.claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KarpZhangNinh/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationEquivalence provedFirst preprint 2024Bipartite encoder forms in an operation-generated class differ only by permutations and row operationsFor graphs with only input–output edges, equivalence generated by local complementation, the two vertex permutations, binary input-row addition and input-edge removal implies reachability by the two permutations and row additions alone.Proved

The questionLi, Equivalence Classes of Quantum Error-Correcting Codes (2024), arXiv:2406.12083v1, Section 8, Conjecture 8.3 (main.tex, lines 1768–1771):

Why it mattersSection 8 seeks simple representatives of encoder equivalence classes containing bipartite forms.

The ideaLocal complementation uses lcsource and the cut-rank invariance lceq; input permutations reindex the cut, and input-edge removal changes no entry of an output cut.

Our contributionFor graphs with only input–output edges, equivalence generated by local complementation, the two vertex permutations, binary input-row addition and input-edge removal implies reachability by the two permutations and row additions alone.

Scope & next questionExistence of bipartite representatives for other classes, selection of a unique representative, an efficient test, and the Definition 6.2/Conjecture 6.3 bridge remain open here.

Verification details
Release source recordProved in Lean

Open problem

Bipartite encoder forms in an operation-generated class differ only by permutations and row operations

A. B. Khesin and A. M. Li, Equivalence Classes of Quantum Error-Correcting Codes (2024), arXiv:2406.12083v1, Section 8, Conjecture 8.3 (main.tex, lines 1768–1771):

In an equivalence class with bipartite forms, all bipartite forms in the class can be transformed from one to another using only output permutations, input permutations, and row operations on the input-output adjacency matrix.

The source introduces these forms as graphs with “no edges among output nodes” (main.tex, line 1631). Its Section 7 class construction uses the operations of Conjecture 6.3 (main.tex, lines 1557 and 1582–1584). The operations and their locators are recorded in the literature note.

Issue #26139 fixes the operation-generated reading. A graph has vertex set $\mathrm{Fin}(k)\sqcup\mathrm{Fin}(n)$, with inputs on the left and outputs on the right. A bipartite form has only input–output edges: neither input–input nor output–output edges. Equivalence is Relation.EqvGen Step, generated by local complementation, output permutations, input permutations, binary additions of distinct input rows, and input-edge removal. Local Clifford decorations are absent.

For all natural $k,n$ and all such bipartite forms $G,H$, claim says that operation-generated equivalence implies Relation.ReflTransGen BipartiteStep G H: a finite, possibly empty, sequence of output permutations, input permutations and binary additions of distinct input rows. This is proved. There is no positivity, connectedness, full-rank or normalization hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/KhesinLiBipartiteEncoderEquivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    For graphs with only input–output edges, equivalence generated by local complementation, the two vertex permutations, binary input-row addition and input-edge removal implies reachability by the two permutations and row additions alone.

    Read source ↗
  • Historical context

    First preprint year: 2024.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2022The C*-algebraic de Bruin–Sharma Conjecture is falseA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.Refuted

The questionMahesh Krishna, "C-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2.

Why it mattersThe same cubic matrix polynomial settles this conjecture and the two linked conjectures in the Schoenberg dossier (krishna-2022-cstar-schoenberg.md) and the Kushel-Tyaglov dossier (krishna-2022-cstar-kushel-tyaglov.md).

The ideaThe first inequality has right side minus left side D2.3^(1)=2t^4(c^2-7) E11/81.

Our contributionA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

The C*-algebraic de Bruin–Sharma Conjecture is false

K. Mahesh Krishna, "C*-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2. The source text is quoted verbatim:

**Conjecture 2.3 (C*-algebraic de Bruin-Sharma Conjecture).** Let $\mathcal{A}$ be a C*-algebra, $d\in \mathbb{N}\setminus{1}$ and let $P(z) \coloneqq (z-a_1)(z-a_2)\cdots (z-a_d)$ be a polynomial over $\mathcal{A}$ with $a_1, a_2, \dots, a_d \in \mathcal{A}$. Assume that $P'$ can be written as $P'(z)\coloneqq d (z-b_1)\cdots (z-b_{d-1})$ on $\mathcal{A}$ with $b_1, b_2, \dots, b_{d-1} \in \mathcal{A}$. If $\sum_{j=1}^{d}a_j=0,$ then $$\sum_{k=1}^{d-1}(b_kb_k^*)^2\leq \frac{2}{d^2}\left(\sum_{j=1}^{d}a_ja_j^*\right)^2+ \frac{d-4}{d}\sum_{j=1}^{d}(a_ja_j^*)^2$$ and $$\sum_{k=1}^{d-1}(b_k^*b_k)^2\leq \frac{2}{d^2}\left(\sum_{j=1}^{d}a_j^*a_j\right)^2+ \frac{d-4}{d}\sum_{j=1}^{d}(a_j^*a_j)^2.$$

The ordered derivative is defined in the source immediately before the conjecture. The encoding uses zero-based Fin indices, complex scalar coefficients, star, and the C*-order. It restricts the source to unital C*-algebras in Type with PartialOrder and StarOrderedRing, and d >= 2; this subclass is enough for a counterexample to the source's universal assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarDeBruinSharma.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2022The C*-algebraic Kushel–Tyaglov Conjecture is falseA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.Refuted

The questionMahesh Krishna, "C-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2.

Why it mattersThe same cubic matrix polynomial settles this conjecture and the two linked conjectures in the Schoenberg dossier (krishna-2022-cstar-schoenberg.md) and the de Bruin-Sharma dossier (krishna-2022-cstar-de-bruin-sharma.md).

The ideaThe first inequality has right side minus left side D2.4^(1)=-t^4(43c^2+158) E11/81.

Our contributionA cubic polynomial over two by two complex matrices refutes Krishna’s C-algebraic conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

The C*-algebraic Kushel–Tyaglov Conjecture is false

K. Mahesh Krishna, "C*-algebraic Schoenberg Conjecture", arXiv:2206.06653v1 (14 June 2022; single version), Section 2. The source text is quoted verbatim:

**Conjecture 2.4 (C*-algebraic Kushel-Tyaglov Conjecture).** Let $\mathcal{A}$ be a C*-algebra, $n\in \mathbb{N}\setminus{1}$ and let $P(z) \coloneqq (z-a_1)(z-a_2)\cdots (z-a_d)$ be a polynomial over $\mathcal{A}$ with $a_1, a_2, \dots, a_d\in \mathcal{A}$. Assume that $P'$ can be written as $P'(z)\coloneqq d (z-b_1)\cdots (z-b_{d-1})$ on $\mathcal{A}$ with $b_1, b_2, \dots, b_{d-1} \in \mathcal{A}$. Then $$\sum_{k=1}^{d-1}(b_kb_k^*)^2\leq \frac{d-6}{d}\sum_{j=1}^{n}(a_ja_j^*)^2+\frac{1}{d^2}\left(\sum_{j=1}^{d}a_ja_j^*\right)^2+ \frac{1}{d^2}\left[\sum_{j=1}^{d}a_j^2-\frac{1}{d^2}\left(\sum_{k=1}^{d}a_k\right)^2\right] \left[\sum_{j=1}^{d}a_j^2-\frac{1}{d^2}\left(\sum_{k=1}^{d}a_k\right)^2\right]^* +\frac{2}{d}\sum_{j=1}^{d}a_j\left[a_j+\frac{1}{d}\sum_{k=1}^{d}a_k\right]\left[a_j+\frac{1}{d}\sum_{k=1}^{d}a_k\right]^*a_j^*-\frac{4}{d^3}\sum_{j=1}^{d}a_j \left[\sum_{k=1}^{d}a_k \right] \left[\sum_{k=1}^{d}a_k \right]^*a_j^*$$ and $$\sum_{k=1}^{d-1}(b_k^*b_k)^2\leq \frac{d-6}{d}\sum_{j=1}^{d}(a_j^*a_j)^2+\frac{1}{d^2}\left(\sum_{j=1}^{d}a_j^*a_j\right)^2+ \frac{1}{d^2}\left[\sum_{j=1}^{d}a_j^2-\frac{1}{d^2}\left(\sum_{k=1}^{d}a_k\right)^2\right]^* \left[\sum_{j=1}^{d}a_j^2-\frac{1}{d^2}\left(\sum_{k=1}^{d}a_k\right)^2\right] +\frac{2}{d}\sum_{j=1}^{n}a_j^*\left[a_j+\frac{1}{d}\sum_{k=1}^{d}a_k\right]^*\left[a_j+\frac{1}{d}\sum_{k=1}^{d}a_k\right]a_j-\frac{4}{d^3}\sum_{j=1}^{d}a_j^* \left[\sum_{k=1}^{d}a_k \right]^* \left[\sum_{k=1}^{d}a_k \right]a_j.$$

The ordered derivative is defined in the source immediately before the conjecture. In Conjecture 2.4, the encoding reads the printed n as d. Indices become zero-based Fin types and the product order is retained. The encoding uses complex scalar coefficients, star, and the C*-order, and restricts the source to unital C*-algebras in Type with PartialOrder and StarOrderedRing and d >= 2; this subclass is enough for a counterexample to the source's universal assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarKushelTyaglov.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2024The tree–path conjecture for the game cordiality numberA tree on ten vertices forces discrepancy at least three, while Admirable can hold the path on ten vertices to discrepancy at most one.Refuted

The questionKrop, Mittal and Wigal, The Cordiality Game and the Game Cordiality Number, arXiv:2403.18060v1 (2024), published as Graphs and Combinatorics 40 (2024), article 75, state in Section 3, Conjecture 3.2:

Why it mattersThe settling module gives a ten-vertex tree whose game cordiality number is at least three, while an explicit Admirable strategy holds the path value at most one.

The ideaIt proves the minimax correspondence by strong induction.

Our contributionA tree on ten vertices forces discrepancy at least three, while Admirable can hold the path on ten vertices to discrepancy at most one.

Verification details
Release source recordRefuted in Lean

Open problem

The tree–path conjecture for the game cordiality number

Krop, Mittal and Wigal, The Cordiality Game and the Game Cordiality Number, arXiv:2403.18060v1 (2024), published as Graphs and Combinatorics 40 (2024), article 75, state in Section 3, Conjecture 3.2:

For any tree $T$ of order $n$, $c_g(T)\le c_g(P_n)$.

Their Section 1 defines the cordiality game: Admirable labels a selected vertex by $0$, Impish labels one by $1$, edge labels are sums modulo $2$, and the discrepancy is $d=|e_1-e_0|$. Admirable starts in $c_g$ and minimizes the terminal discrepancy while Impish maximizes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Games/CordialityTreePathRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundSource 2023The Fibonacci numbers are not 2-super-accessibleIs the set of Fibonacci numbers 2-super-accessible?Refuted

The questionLandman, Robertson and Robertson, Ramsey properties for integer sequences with restricted gaps, Moscow Journal of Combinatorics and Number Theory 12(3), 181–195, Section 5, question 3, ask:

Why it mattersTheir common coloring refutes 2-super-accessibility. Wesley's note supplies the published modulus-29 parameter.

The ideaThe Fibonacci identity in alphafib and fractional-part calculation in phasefib give

Our contributionIs the set of Fibonacci numbers 2-super-accessible?

Verification details
Release source recordRefuted in Lean

Open problem

The Fibonacci numbers are not 2-super-accessible

Landman, Robertson and Robertson, Ramsey properties for integer sequences with restricted gaps, Moscow Journal of Combinatorics and Number Theory 12(3), 181–195, Section 5, question 3, ask:

Is the set of Fibonacci numbers 2-super-accessible? We know it is 2-accessible but that it is not 2-large.

The source note gives the definitions and locators. The external named question and literal Lean conventions are in #16039.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The Fibonacci set $F={f_n}_{n\ge1}$ | fibSet | Membership is a positive-index Fibonacci value; repeated values do not change a set. | | A $k$-term $S$-ap has positive start and gap in $S$ | HasAP S A k | Every index below $k$ belongs to the color class $A$. | | A positive increasing $S$-diffsequence has successive differences in $S$ | HasDiffSeq S A k | The positivity, strict inequalities, difference memberships and all $k$ color memberships are retained. | | Every $r$-coloring of the positive integers, with literal colors $1,\ldots,r$ | SuperAccessible r S | The total natural-valued function is constrained only on positive inputs. Its value at zero is irrelevant. | | Arbitrarily long color-1 $S$-aps, or arbitrarily long color-$j$ $S$-diffsequences for one $2\le j\le r$ | The disjunction and existential in SuperAccessible | The color $j$ is chosen before the length, exactly as in Definition 14. | | The question for $r=2$ and $S=F$ | claim | The literal definitions above are used. | | A negative answer | result : ¬ claim | One coloring violates both alternatives. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Additive/FibonacciSuperAccessibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2023.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year recorded in the source bibliography: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2024Manabe's restricted amalgamation NimManabe's restricted three-pile amalgamation Nim satisfies Conjecture 1.Proved

The questionHikaru Manabe, An Amalgamation Nim with Restriction, arXiv:2411.14488v1, Section 3, Conjecture 1, page 14:

Why it mattersThe public theorem D5.S0.Certificates.Games.ManabeRestrictedAmalgamationNim.result settles both parts of Conjecture 1 for every position.

The ideaStrong induction on size and mex then give grundyformula, exactformula and result.

Our contributionManabe's restricted three-pile amalgamation Nim satisfies Conjecture 1.

Verification details
Release source recordProved in Lean

Open problem

manabe 2024 restricted amalgamation nim conjecture 1

Hikaru Manabe, An Amalgamation Nim with Restriction, arXiv:2411.14488v1, Section 3, Conjecture 1, page 14:

For n ∈ Z≥0, we have the following (i) and (ii). (i) If the Grundy number of a position (x, y, z) is 2n, then x ⊕ y ⊕ z = 2n or x ⊕ y ⊕ z = 2n + 1. (ii) If the Grundy number of a position (x, y, z) is 2n + 1, then x ⊕ y ⊕ z = 2n or x ⊕ y ⊕ z = 2n + 1.

The game is Definitions 4–5: strictly reduce any one of three natural pile sizes, or merge two piles each at least two, allowing both placements of the sum and zero. Preregistration: #19861.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Three nonnegative pile sizes | Natural arguments of move, grundy and claim | Zero is included | | Definition 5 sets (1)–(3) | First three alternatives of move | All strict reductions | | Definition 5 sets (4)–(6) | Last three alternatives of move | Both piles at least two; both placements | | Grundy value is mex of option values | grundy, with the existing AbukuEnforceNimberDomination.mex | Private move_decreases and grundy_eq remove the well-founded guard on options | | Both even and odd implications, for every natural n | Two conjuncts of claim, proved by result | Neither implication is dropped | | Termination measure in the binding conventions | size: stone total plus number of nonempty piles | An implementation measure, not an extra game hypothesis |

The four public definitions preserve the binding expressions in #19861. The measure and recursion guard add no hypothesis to the universal result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/ManabeRestrictedAmalgamationNim.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Geometry & algebraCounterexample foundFirst preprint 2024Variable-exponent Lebesgue spaces and the Hornich–Hlawka inequalityA two-valued variable exponent gives a Luxemburg norm violating the Hornich–Hlawka inequality.Refuted

The questionNiculescu, A survey of the Hornich-Hlawka inequality, arXiv:2407.03278v1, §3, the paragraph introducing Problem 3 and Problem 3 itself, state verbatim:

Why it mattersVariable exponents permit different power laws on different parts of the measure space.

The ideaThe zero vector has norm zero by the defining infimum; that case is mathematical reasoning rather than a separate theorem in the module.

Our contributionA two-valued variable exponent gives a Luxemburg norm violating the Hornich–Hlawka inequality.

Verification details
Release source recordRefuted in Lean

arXiv

Variable-exponent Lebesgue spaces and the Hornich–Hlawka inequality

D.-Ş. Marinescu and C. P. Niculescu, A survey of the Hornich-Hlawka inequality, arXiv:2407.03278v1, §3, the paragraph introducing Problem 3 and Problem 3 itself, state verbatim:

A natural generalization of Lp spaces is provided by the class of Lp(·) spaces with variable exponent. They differ from the classical Lp spaces in that the exponent p is not constant but a function from the measure space under attention to [1, ∞]. See the monograph of L. Diening, P. Harjulehto, P. Hästö and M. Ruzicka [6] for a thorough presentation. Problem 3. Is every Lp(·)(0, 1) space with a variable exponent p(·) : (0, 1) → [1, 2] a Hornich-Hlawka space?

A Hornich–Hlawka space satisfies

$$ |f+g|+|g+h|+|h+f| \le |f|+|g|+|h|+|f+g+h| $$

for every triple. The norm considered here is the Luxemburg norm. The journal version is A New Look at the Hornich-Hlawka Inequality, Results Math. (2025), doi:10.1007/s00025-025-02464-9; the quotation and locator refer to the arXiv version.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/VariableExponentHornichHlawka.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2021Diophantine exponents of aperiodic episturmian wordsEpisturmian words: DiophantineExponentCriterion.Proved

The questionPeltomäki, Initial nonrepetitive complexity of regular episturmian words and their Diophantine exponents, European Journal of Combinatorics 118 (2024), 103942, DOI 10.1016/j.ejc.2024.103942, arXiv:2103.08351v3, Section 8, Conjecture 8.4:

Why it mattersLiteral palindromic closure constructs the standard word.

The ideacentralJustin supplies the palindromic-prefix morphism identity.

Our contributionEpisturmian words: DiophantineExponentCriterion.

Scope & next questionThe exact critical exponent for bounded partial quotients also remains open here.

Verification details
Release source recordProved in Lean

Open problem

Diophantine exponents of aperiodic episturmian words

J. Peltomäki, Initial nonrepetitive complexity of regular episturmian words and their Diophantine exponents, European Journal of Combinatorics 118 (2024), 103942, DOI 10.1016/j.ejc.2024.103942, arXiv:2103.08351v3, Section 8, Conjecture 8.4:

Let $\mathbf t$ be an aperiodic episturmian word. Then $\mathrm{dio}(\mathbf t)<\infty$ if and only if $\mathbf t$ has bounded partial quotients.

The source proves the regular case in Theorem 6.6. The unrestricted statement is the target of #15858.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | A finite alphabet | Every natural $d$, with letters in Fin d | | A directive word $\Delta$ | Every stream $\mathbb N\to\mathrm{Fin},d$ | | An episturmian word with directive $\Delta$ | Lang t = Lang (cDelta Delta) | | Aperiodic | The literal negation of an eventually periodic natural tail, expanded in claim | | Bounded partial quotients | BoundedPQ Delta, a uniform bound on the distance to the next different directive letter | | Finite Diophantine exponent | dio t < ⊤ | | If and only if | Both implications of DiophantineExponentCriterion.result |

Finite factors use the frozen operator AbelianBorderQuestionDefs.factor at start zero. Its value is the literal list of the first $n$ letters. Directive prepending uses Mathlib's Stream'.cons, and finite powers use Powers.wordPower. No alphabet size, regularity, intercept or recurrence hypothesis is added.

The source's run convention and BoundedPQ agree: if every maximal run has length at most $K$, the next different letter occurs within $K+1$ positions; conversely a witness $B$ bounds each run by $B-1$. An eventually constant directive has neither property. This is a finite run-length argument, proved in prose.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Episturmian/DiophantineExponentCriterion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2023Every classical six-leg perfect tensor factorization has perfect two-wire factorsEvery three-gate factorization of a classical six-leg perfect tensor has perfect two-wire factors.Proved

The questionWanless, Tensor network decompositions for absolutely maximally entangled states, Quantum 8 (2024) 1339, arXiv:2308.07042v3, ask whether the linear factorization result extends to nonlinear classical maps.

Why it mattersFor a classical map, the source's array has rows consisting of the input triple followed by its output triple.

The ideaEvery column selector used in the proof has its injectivity discharged explicitly.

Our contributionEvery three-gate factorization of a classical six-leg perfect tensor has perfect two-wire factors.

Scope & next questionThe source's quantum, nonclassical, and existence questions are outside this result. - Effect on the source.

Verification details
Release source recordProved in Lean

Open problem

Every classical six-leg perfect tensor factorization has perfect two-wire factors

B. Pozsgay and I. M. Wanless, Tensor network decompositions for absolutely maximally entangled states, Quantum 8 (2024) 1339, arXiv:2308.07042v3, ask whether the linear factorization result extends to nonlinear classical maps. In factoringtriunit.tex, the source defines orthogonal columns and orthogonal arrays in lines 891–902, associates a map from three input symbols to three output symbols with an array in lines 929–937, and gives the six-leg factorization $u^{(123)}=\mathcal C^{(23)}\mathcal B^{(13)}\mathcal A^{(12)}$ with classical two-wire functions in lines 945–964. After the linear calculation, lines 1196–1200 state:

In other words, if $G$ represents a linear perfect tensor with 6 legs, and it has a $\triangleright$-decomposition into linear tensors with 4 legs, then those factors will themselves be perfect. It is an open question whether this statement is true in the nonlinear case for $k=6$.

The settled classical statement uses a finite type $x$ and arbitrary maps $a,b,c:x\times x\to x\times x$. The circuit applies $a$ to the first two wires, $b$ to the first output of $a$ and the third wire, and $c$ to the second output of $a$ and the second output of $b$. If its six-column graph is an index-one orthogonal array of strength $3$, the question is whether each four-column gate graph is an index-one orthogonal array of strength $2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Latin/ClassicalPerfectTensorFactorization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2024Convergence of the Shieh-Yang-Yu dotted-pattern machineEvery permutation eventually reaches a fixed point of the 21̇-machine.Proved

The questionMichael Yang, Hansen Shieh, and Ashley Yu, Stack-Sorting with Dotted-Pattern-Avoiding Stacks, arXiv:2411.11914v2, Section 6, printed page 11, state:

Why it mattersThe public statement is ∀ n ≥ 1, ∀ w ~ List.range' 1 n, ∃ t : ℕ, (M^[t + 1]) w = (M^[t]) w, where M w = s (r w).

The ideaA valley is an entry strictly smaller than every preceding entry.

Our contributionEvery permutation eventually reaches a fixed point of the 21̇-machine.

Verification details
Release source recordProved in Lean

arXiv

Convergence of the Shieh-Yang-Yu dotted-pattern machine

Michael Yang, Hansen Shieh, and Ashley Yu, Stack-Sorting with Dotted-Pattern-Avoiding Stacks, arXiv:2411.11914v2, Section 6, printed page 11, state:

Conjecture 6.2. All permutations in Sₙ for all n ≥ 1 are eventually mapped to a fixed point of the 21̇-machine after a finite number of iterations through the machine.

The preceding sentence asks whether this machine has any cycle of length at least two. The machine applies the dotted-pattern map first and West's stack-sorting map second.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/ShiehYangYuMachineConvergence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

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    Source 1
  • Historical context · 1/4 · 10% weight

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    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024The 12̇-Machine Sorts a Central Binomial Number of PermutationsThe 12-dot machine sorts a central binomial number of permutations of each positive size.Proved

The questionMichael Yang, Hansen Shieh and Ashley Yu, Stack-Sorting with Dotted-Pattern-Avoiding Stacks, arXiv:2411.11914v2, Section 6, Conjecture 6.1 (printed page 10):

The ideaWest's criterion is proved from the stack algorithm: s(w) is increasing exactly when w avoids 231.

Our contributionThe 12-dot machine sorts a central binomial number of permutations of each positive size.

Verification details
Release source recordProved in Lean

arXiv

The 12̇-Machine Sorts a Central Binomial Number of Permutations

Michael Yang, Hansen Shieh and Ashley Yu, Stack-Sorting with Dotted-Pattern-Avoiding Stacks, arXiv:2411.11914v2, Section 6, Conjecture 6.1 (printed page 10):

The machine-sortable permutations under the 12̇-machine map in Sₙ are enumerated by (2n−2 choose n−1).

The 12̇-machine is s ∘ s₁₂̇, with s West's stack-sorting map. By the source's Proposition 3.1, s₁₂̇ reverses each peak run, a peak run being a maximal block of consecutive entries that starts at a left-to-right maximum. A permutation is machine-sortable when the machine maps it to the identity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/DottedStack/ShiehYangYuTwelveDot.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is proved.

    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2022Parity-position sequences in the Sobolewski–Ulas recurrenceBartosz Sobolewski and Maciej Ulas, Solutions of certain meta-Fibonacci recurrences, arXiv:2204.04011v1, Question 7.5, pp. 16–17; journal locator: Acta Arithmetica 209 (2023), 269–289, DOI 10.4064/aa220613-19-1.Proved

The questionBartosz Sobolewski and Maciej Ulas, Solutions of certain meta-Fibonacci recurrences, arXiv:2204.04011v1, Question 7.5, pp. 16–17; journal locator: Acta Arithmetica 209 (2023), 269–289, DOI 10.4064/aa220613-19-1.

Why it mattersThere is no digestion atom or coverage claim.

The ideaThe odd and even recurrence identities give the parity table of each aligned sixteen-position block.

Our contributionBartosz Sobolewski and Maciej Ulas, Solutions of certain meta-Fibonacci recurrences, arXiv:2204.04011v1, Question 7.5, pp. 16–17; journal locator: Acta Arithmetica 209 (2023), 269–289, DOI 10.4064/aa220613-19-1.

Verification details
Release source recordProved in Lean

Open problem

Parity-position sequences in the Sobolewski–Ulas recurrence

Bartosz Sobolewski and Maciej Ulas, Solutions of certain meta-Fibonacci recurrences, arXiv:2204.04011v1, Question 7.5, pp. 16–17; journal locator: Acta Arithmetica 209 (2023), 269–289, DOI 10.4064/aa220613-19-1.

Let u_i = (u_i(n)) for i = 0, 1 be the increasing sequence of nonnegative integers such that r(u_i(n)) = i. Is the sequence u_i is k-regular for some k?

Here $h(n)=1$ for $n\leq1$, $h(n)=h(n-h(n-1))+h(n-2)$ for $n>1$, and $r(n)=h(n)\bmod2$. Section 2 requires $k\geq2$ and defines regularity as finite generation of the integer span of the $k$-kernel.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | The recurrence $h=h_{1,1}$ | h | Both initial values are one. Truncated negative arguments map to zero, also of value one. The termination minimum is inactive. | | $r(n)=h(n)\bmod2$ | r | Natural remainder, values zero and one. | | Increasing nonnegative occurrence positions | u | Nat.nth, with zero-based index. Block balance proves both sets infinite. | | Kernel indices $j\geq0$, $0\leq a<k^j$ | kKernel | Literal arbitrary-base subsequences. Binary proofs reuse twoKernel from SparseKernelRank. | | Finite generation over $\mathbb Z$ | IsKRegular | Submodule.span and Submodule.FG over integer-valued sequences. | | Each $i=0,1$, some $k\geq2$ | claim, result | Full parity family without extra hypotheses; the proof supplies $k=2$. |

All unqualified declaration names below belong to D5.S1.Recurrence.MetaFibonacci.SobolewskiUlasParityPositions. The six public definitions reproduce the source conventions. The binary specialization of kKernel agrees definitionally with the imported D5.S1.Words.Palindromes.PeriodDoubling.twoKernel.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MetaFibonacci/SobolewskiUlasParityPositions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresCounterexample foundFirst preprint 2022The vertex position ratio is unbounded on connected graphsA connected graph family has quadratically many vertices on a sphere from one vertex, but only linearly many geodesic rays from another.Refuted

Why it mattersThe source's Lemma 2.3 constructs ratios approaching six.

The ideaThe level function changes by at most one on each edge.

Our contributionA connected graph family has quadratically many vertices on a sphere from one vertex, but only linearly many geodesic rays from another.

Scope & next questionRestricted-class and other-invariant questions remain open here and require their own source and literature checks.

Verification details
Release source recordRefuted in Lean

Open problem

The vertex position ratio is unbounded on connected graphs

Problem 2.4 of M. Thankachy, E. J. Thomas, U. Chandran, J. Tuite, G. Di Stefano and G. Erskine, “On the Vertex Position Number of Graphs”, asks: “Is the ratio $\mathrm{vp}(G)/\mathrm{vp}^-(G)$ bounded for connected graphs?” The source is arXiv:2209.00359v2, page 5, published in Discussiones Mathematicae Graph Theory 44 (2024), 1169–1188, DOI 10.7151/dmgt.2491.

For a fixed vertex $x$, an $x$-position set excludes every other selected vertex from every shortest path from $x$ to a selected vertex. The maximum size is $p_x(G)$; its maximum and minimum over vertices are $\mathrm{vp}(G)$ and $\mathrm{vp}^-(G)$. Endpoints belong to geodesics, so ${x}$ is permitted but a set containing $x$ and another vertex is not. The Lean definitions use the full finite carrier and all its subsets.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PositionSets/VertexPositionRatioRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2022A rotating replacement state refutes purely non-unital eternal non-Markovianity impossibilityA rotating replacement state gives a completely positive trace-preserving qubit dynamics whose unital part is divisible, while its generator fails conditional complete positivity at every positive time.Refuted

The questionSrikanth, On the eternal non-Markovianity of non-unital quantum channels, arXiv:2203.10625v2, International Journal of Quantum Information 22(1), 2350039 (2024), DOI 10.1142/S0219749923500399, Section IV, conjecture:

Why it mattersIt identifies a rotating replacement state as a mechanism by which a divisible isotropic contraction coexists with a generator that fails conditional complete positivity at every positive time.

The idea3. inversepair, generatorcomposition, rationalderiv and phigenerator give the exact inverse and entrywise generator.

Our contributionA rotating replacement state gives a completely positive trace-preserving qubit dynamics whose unital part is divisible, while its generator fails conditional complete positivity at every positive time.

Scope & next questionAlthough we have only studied the qubit case, we conjecture that it is true for non-unital channels in arbitrary finite dimensions.

Verification details
Release source recordRefuted in Lean

Open problem

A rotating replacement state refutes purely non-unital eternal non-Markovianity impossibility

S. Utagi, S. Banerjee and R. Srikanth, On the eternal non-Markovianity of non-unital quantum channels, arXiv:2203.10625v2, International Journal of Quantum Information 22(1), 2350039 (2024), DOI 10.1142/S0219749923500399, Section IV, conjecture:

These diverse examples examined in the Section suggest that a purely non-unital channel (i.e., one lacking any unital component) is impossible. Although we have only studied the qubit case, we conjecture that it is true for non-unital channels in arbitrary finite dimensions.

Section V restates:

However, we conjecture that for a non-unital channel with no such unital component (in the above sense) to be ENM, is impossible.

The source's Definition 1 attributes purely non-unital non-Markovianity to the translation in the Bloch action. The source note gives the locators and quotations. #16599 fixes the complete qubit conventions and the all-positive-times criterion, without a nonzero negative asymptotic bound.

Statement echo

| Source clause or explicit convention | Lean counterpart | Scope | | --- | --- | --- | | Qubit dynamics on all complex matrices | Real-time PhyslibLeaf.MatrixMap (Fin 2) (Fin 2) ℂ in claim | The qubit case suffices to refute an assertion for arbitrary finite dimensions. | | CPTP at every nonnegative time | Existing CoPRelativeQuantumnessRefutation.IsCPTP in claim | Complete positivity and trace preservation are reused literally. | | Dynamics starts at the identity | Identity hypothesis in claim | Required on every matrix, not only states. | | Eqs. (9)–(11): delete the translation | unitalPart | Its complete body is the preregistered subtraction $\Phi_t(X)-\operatorname{Tr}(X)(\Phi_t(I)-I)/2$. | | Definition 1: translation alone supplies non-Markovianity | PurelyNonUnital | Non-unital at every positive time; the unital part has a CPTP intermediate map for every $0\le s\le t$. | | Time-local generator | IsGeneratorAt | Entrywise real-time derivative on every complex matrix. | | Conditional complete positivity | CondCP | The existing output-first MatrixMap.choi_matrix is tested on every trace-free vector. | | Negative generator direction at every $t>0$ | EternallyNonMarkovian | No negative asymptotic bound is asserted. | | Impossibility conjecture | claim, negated by result | Every quantifier and hypothesis above is retained. |

The entrywise derivative and conditional-Choi criterion are the explicit mathematical conventions of #16599. They do not restrict inputs to density matrices or replace the source's unital part by a certificate predicate.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/PurelyNonUnitalEternalNonMarkovianity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedFirst preprint 2024Eventual stability of Zhang's l-complexityEvery positive-multiple orbit reaches permanent linear l-complexity growth.Proved

The questionPengcheng Zhang, The 2-complexity of even positive integers, arXiv:2411.19364v1, Integers 24 (2024), A109, Question 3, page 10:

Why it mattersThe result gives permanent linear growth from a suitable translate, rather than only an observed constant tail in a finite computation.

The ideaAt an addition node both values are at least two, so their sum is at most their product.

Our contributionEvery positive-multiple orbit reaches permanent linear l-complexity growth.

Verification details
Release source recordProved in Lean

Open problem

Eventual stability of Zhang's l-complexity

Pengcheng Zhang, The 2-complexity of even positive integers, arXiv:2411.19364v1, Integers 24 (2024), A109, Question 3, page 10:

Let l > 1 and define A_l := {n ∈ lZ⁺ : ‖l^m · n‖_l = m + ‖n‖_l for all m ≥ 0}.

Can one say anything about the set A_l? For example, is it true that for every n ∈ lZ⁺, there exists m₀ ≥ 0 such that l^{m₀} · n ∈ A_l?

The every-orbit assertion is proved. The broader request to describe $A_l$ and to bound the first stable index remains open here. Preregistration: #19862.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | $l>1$ | Natural parameter l with 1 < l. | | $n\in l\mathbb Z^+$ | Natural parameter n with 0 < n and l ∣ n. | | Minimal number of copies of $l$, with addition, multiplication and parentheses | Public value evaluates the existing KumarPengTranModularComplexity.Expr; public complexity is the natural infimum of attainable Expr.leaves. | | $A_l$ | Public A states positivity, divisibility and the source equality for every natural exponent, including zero. | | Some $m_0\geq0$ with $l^{m_0}n\in A_l$ | claim quantifies over natural m₀; public result : claim proves it. |

The natural evaluation and minimum are the literal conventions of #19862. The private mem_A_iff identifies them with the frozen evaluator and minimum at modulus zero through injectivity of the natural-to-integer cast. The public claim retains the source's natural definitions. No additional premise or finite cutoff is present.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/ZhangLComplexityEventualStability.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Geometry & algebraConjecture provedFirst preprint 2024The original Bajnok–Beatty Conjecture 4.1The original four-clause classification holds for every positive k and s.Proved

The questionBeatty, On the Diameter of Undirected Cayley Graphs of Finite Abelian Groups, arXiv:2406.04045v1, Section 4, Conjecture 4.1:

Why it mattersThe target is the complete original iff under preregistration 25489.

The ideaThe formal endpoint is D5.S3.FiniteGroups.BajnokBeattyAllRegimes.result.

Our contributionThe original four-clause classification holds for every positive k and s.

Verification details
Release source recordProved in Lean

arXiv

The original Bajnok–Beatty Conjecture 4.1

B. Bajnok and W. K. Beatty, On the Diameter of Undirected Cayley Graphs of Finite Abelian Groups, arXiv:2406.04045v1, Section 4, Conjecture 4.1:

Let k and s be positive integers. The group Z2 × Z2k has an s-spanning set of size 2 if, and only if, one of the following holds:

  • s is odd and k ≤ (s² − 1)/2;
  • s is even and k ≤ (s² − s)/2;
  • s ≡ 0 mod 4, k is even, and (s² − s)/2 < k ≤ s²/2; or
  • s ≡ 2 mod 4, k ≡ 2 mod 4, and (s² − s)/2 < k ≤ s²/2.

The signed span consists of $xa+yb$ for integers $x,y$ with $|x|+|y|\le s$, including the zero coefficients. A set of size two is represented by two distinct elements $a,b$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/BajnokBeattyAllRegimes.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Geometry & algebraQuestion resolvedFirst preprint 2022Perfect restricted bases of size s + 1 force cyclicityEvery finite abelian group with a perfect restricted basis of size s + 1 is cyclic.Proved

The questionJust, On Perfect Bases in Finite Abelian Groups, Involve 15 (2022), 525–536, arXiv:2211.13675v1, Question 6:

Why it mattersQuestion 6 asks whether noncyclic groups also occur.

The ideaThis is a fixed-point-free involution, so each such fiber has even cardinality.

Our contributionEvery finite abelian group with a perfect restricted basis of size s + 1 is cyclic.

Verification details
Release source recordProved in Lean

Open problem

Perfect restricted bases of size s + 1 force cyclicity

B. Bajnok, C. Berson and H. A. Just, On Perfect Bases in Finite Abelian Groups, Involve 15 (2022), 525–536, arXiv:2211.13675v1, Question 6:

Are there any noncyclic groups that have a perfect restricted s-basis of size s + 1?

The paper works with finite abelian groups and positive integers $s$. Its definition is:

A is a perfect restricted s-basis of G if each element of G can be written uniquely as the sum of at most s distinct elements of A.

Uniqueness disregards the order of the summands. The empty subset contributes the empty sum $0$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite abelian group $G$ | G : Type, AddCommGroup G, Fintype G | Additive notation; no order or prime-order assumption. | | Positive integer $s$ | s : ℕ, 1 ≤ s | Exact positive range. | | A subset $A$ of size $s+1$ | A : Finset G, A.card = s + 1 | Distinct elements, without an ordering. | | At most $s$ distinct summands | A unique T : Finset G with T ⊆ A, T.card ≤ s and sum equal to the given element | Includes the empty sum. | | No noncyclic example | result : claim, concluding IsAddCyclic G | Universal negative answer to Question 6. | | Decidable equality | DecidableEq G | Computational representation of finsets; every finite group admits it classically. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/BajnokBersonJustPerfectRestrictedBasis.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2024The threshold prior of the G-test is O/(M + O)The G-test has larger Bayesian success probability than the always-unequal test exactly above the prior O/(M+O), where M counts words in a symmetric-group orbit and O counts its G-orbits.Proved

The questionIts completeness is 1 (Eq. (63)). - The Bayesian comparison. With prior {p, 1 − p}, the success probability is Psucc = p Pc + (1 − p) Ps.

Why it mattersThe permutation test (G = Sn) is optimal among tests that use the symmetry of the problem, and Theorem 5 locates the prior below which no test beats the trivial one.

The idea2. Independence of U. ΠG commutes with U^{⊗n}, so Tr[ΠG ρ̃] is the same for every unitary U, hence equal to its Haar average.

Our contributionThe G-test has larger Bayesian success probability than the always-unequal test exactly above the prior O/(M+O), where M counts words in a symmetric-group orbit and O counts its G-orbits.

Verification details
Release source recordProved in Lean

arXiv

The threshold prior of the G-test is O/(M + O)

H. Buhrman, D. Grinko, P. Verduyn Lunel and J. Weggemans, Permutation tests for quantum state identity, arXiv:2405.09626v2, distinguish the twirled unequal-case state ρ̃^µ_≠ (Eqs. (11)–(12)) from the equal-case state ρ^n_= (Eq. (13)).

  • The G-test. For a subgroup G of S_n, the G-test measures Π_= = Π_G = (1/|G|) Σ_{π∈G} ψ(π) (Eqs. (61)–(62)). Its completeness is 1 (Eq. (63)).
  • The Bayesian comparison. With prior {p, 1 − p}, the success probability is P_succ = p P_c + (1 − p) P_s. The trivial test always answers "not equal" and succeeds with probability 1 − p. For G = S_n, Theorem 5 gives the threshold p∗(µ) = 1/(1 + (n choose µ)).
  • The question. §6: "A further open question is to determine, for a given subgroup G and partition µ, the threshold prior p∗(G, µ) below which the trivial test outperforms the G-test, analogous to the threshold p∗(µ) established for the permutation test in Theorem 5."

The verbatim statements are in the literature note.

Issue #17307 fixes the reading:

  • W_µ is the set of words of length n over {1, …, d} with µ_i letters i, of size M = n!/∏_i µ_i!;
  • O is the number of orbits of G on W_µ.

The answer is p∗(G, µ) = O/(M + O):

  • the G-test has strictly larger success probability than the trivial test exactly when p > O/(M + O);
  • the two probabilities are equal exactly when p = O/(M + O);
  • the soundness error is Tr[Π_G ρ̃^µ_≠] = O/M.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/GTestThresholdPrior.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Geometry & algebraConjecture provedFirst preprint 2024Lemos–Moura–Ribas–Silva Conjecture 3.4The consecutive Davenport constant of every finite group equals its order.Proved

Why it mattersThe source note is Library/FiniteGroups/lemos2024consecutivedavenport.md.

The ideaThe intervening nonempty block has product one, giving the upper threshold bound.

Our contributionThe consecutive Davenport constant of every finite group equals its order.

Verification details
Release source recordProved in Lean

Open problem

Lemos–Moura–Ribas–Silva Conjecture 3.4

A. Lemos, A. O. Moura, S. Ribas and A. T. Silva, “A note on weighted consecutive Davenport constant,” arXiv:2404.11312v1, Conjecture 3.4, p. 3, asks whether $C(G)=|G|$ for every finite group $G$. The settlement is Proved under the literal conventions preregistered in #19765.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Ordered sequences over $G$ | List G | The ordered product is List.prod in list order. | | Ordered subsequences $g_i\cdots g_j$ with $1\le i\le j$ | T ≠ [] ∧ T <:+: S | Nonempty contiguous infix is literal. | | Consecutive product-one condition for $A={1}$ | HasConsecutiveProductOne S | T.prod = 1 keeps the unweighted convention. | | Smallest positive threshold | C G | sInf of the positive natural thresholds with the stated property. | | Every finite group has $C(G)=|G|$ | claim and result | result proves the complete quantified proposition. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/LemosMouraRibasSilvaConsecutiveDavenport.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2024n + 1 displaced on-off measurements stay incompatible under pure loss of transmissivity 1/nFor every natural n at least one, some n+1 binary measurements on the full one-mode Fock space remain incompatible under pure loss of transmissivity 1/n.Proved

The questionRahimi-Keshari, Measurement incompatibility under loss, arXiv:2411.05920v3 (Phys.

Why it mattersLoss is the dominant noise in optical quantum communication.

The ideaThis is exact for the vacuum and one-photon inputs of the beam splitter, preserves positivity and sums, and so sends a parent on ℓ² to a parent of the 2 × 2 compressions.

Our contributionFor every natural n at least one, some n+1 binary measurements on the full one-mode Fock space remain incompatible under pure loss of transmissivity 1/n.

Verification details
Release source recordProved in Lean

Open problem

n + 1 displaced on-off measurements stay incompatible under pure loss of transmissivity 1/n

M. Mehboudi, F. Rezaeinia and S. Rahimi-Keshari, Measurement incompatibility under loss, arXiv:2411.05920v3 (Phys. Rev. Lett. 135, 100202 (2025)), model pure loss by a beam splitter of transmissivity τ with vacuum in its second input, E_τ(ρ) = tr₂(U_BS(ρ ⊗ |0⟩⟨0|)U_BS†), and define the dual by tr(M E_τ(ρ)) = tr(ρ E*_τ(M)). Result 2 shows that for τ ≤ 1/n the images under E*_τ of any n measurements are jointly measurable. Result 3 asserts that some n + 1 measurements stay incompatible at τ = 1/n: "We prove this result numerically for 2 ≤ n ≤ 13, and conjuncture it to hold for arbitrary n." The measurements are the displaced on-off detections of Eq. (4), M^k_1 = |μ_k⟩⟨μ_k| and M^k_2 = I − M^k_1, with coherent states of equal modulus separated by the phase 2π/(n + 1). The verbatim statements are in the literature note.

Issue #15985 fixes the reading. For every n ≥ 1 there are n + 1 binary POVMs on the single-mode Fock space ℓ²(ℕ) whose images under E*_{1/n} have no parent POVM. A parent ranges over all 2^{n+1} outcome strings and has exactly the given effects as marginals.

The Lean statement takes E*_τ in the source's own form: N is the image of M when tr(M E_τ(|ψ⟩⟨ψ|)) = ⟨ψ, N ψ⟩ for every ψ. Here E_τ(|ψ⟩⟨ψ|) = tr₂(U_BS(|ψ⟩⟨ψ| ⊗ |0⟩⟨0|)U_BS†) is built from the frozen beam splitter and partial trace, and the trace is the Fock-basis sum. Pure states determine an operator, so this relation determines E*_τ(M). The statement asserts both that such images exist and that no family of them has a parent.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/PureLossIncompatibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2022Every linear N-qubit doily has an odd number of negative linesEvery linear N-qubit doily has an odd number of negative lines.Proved

The questionHolweck, Multi-qubit doilies: enumeration for all ranks and classification for ranks four and five, arXiv:2206.03599v2 (J.

Why it mattersThe paper enumerates linear and quadratic doilies and classifies their observable signatures and negative-line configurations for four and five qubits.

The ideaThe proof uses the multiplication phases of the literal Pauli matrices.

Our contributionEvery linear N-qubit doily has an odd number of negative lines.

Verification details
Release source recordProved in Lean

Open problem

Every linear N-qubit doily has an odd number of negative lines

A. Muller, M. Saniga, A. Giorgetti, H. de Boutray and F. Holweck, Multi-qubit doilies: enumeration for all ranks and classification for ranks four and five, arXiv:2206.03599v2 (J. Comput. Sci. 64, 101853, 2022), §5.4, “Specific behavior of linear doilies”, state:

From our results on three-, four- and five-qubit cases it follows that a linear doily (a) always features an odd number of negative lines, and (b) does not share a perp-set with the distinguished quadric.

We conjecture that Property (a) holds for any number of qubits $N \geq 2$, but we have not yet found of proof of it; we surmise that it has something to do with the fact that a linear doily is ``squeezed'' into a PG$(3,2)$, compared to a quadratic doily that enjoys more degrees of freedom being stretched out in a PG$(4,2)$.

These quotations are from xiv-MSGDH2.tex, lines 2255–2257 and 2259–2263, downloaded from the v2 source, MD5 ba1ee66980eb180ebdcbe7fcd5d8a411. Definitions and source locators are in the literature note.

Issue #20146 fixes the reading: for every $N\ge2$ and every four-dimensional subspace $U\le\mathbb F_2^{2N}$ with nondegenerate restricted symplectic form, an odd number of its isotropic triples ${u,v,u+v}$ have their three canonical, unsigned Pauli observables multiplying to $-I_{2^N}$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/LinearDoilyNegativeLineParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2024Sharp parameter thresholds for the zero-balanced Gauss quotientSharp zero-balanced Gauss quotient classificationsProved

The questionQiu, Ma and Xiang, Turán-type inequalities for Gaussian hypergeometric functions, and Baricz's conjecture, arXiv:2408.15723v1, §6, Conjecture 6.2:

Why it mattersThe result gives the exact parameter range for strict derivative negativity and the exact threshold for the shifted-parameter comparison.

The ideaTermwise parameter differentiation uses locally uniform summable bounds for the coefficient derivatives.

Our contributionSharp zero-balanced Gauss quotient classifications

Verification details
Release source recordProved in Lean

Open problem

Sharp parameter thresholds for the zero-balanced Gauss quotient

Qiu, Ma and Xiang, Turán-type inequalities for Gaussian hypergeometric functions, and Baricz's conjecture, arXiv:2408.15723v1, §6, Conjecture 6.2:

The inequality (1.6) holds for all r ∈ (0, 1) and a ∈ (0, ∞) if and only if b ∈ (0, √(a(a+1))], and the inequality (1.7) holds for all r ∈ (0, 1) if and only if a ∈ [3/7, 1).

Write $F(a,b;c;r)={}_2F_1(a,b;c;r)$. Inequality (1.6) is

$$\frac{\partial}{\partial b} \left[\frac{F(a+1,b+1;a+b+1;r)}{F(a,b;a+b;r)}\right]<0,$$

and inequality (1.7) is

$$\frac{F(a+1,2-a;2;r)}{F(a,1-a;1;r)}

\frac{F(a+1,5/2-a;5/2;r)}{F(a,3/2-a;3/2;r)}.$$

The first classification is read pointwise in the positive parameters: for each $a,b>0$, the derivative is negative at every $r\in(0,1)$ if and only if $b\le\sqrt{a(a+1)}$. The second classification assumes $0<a<1$. This corrects Baricz's 2007 Conjecture A, which the same paper refutes as an unrestricted statement.

Formal reading

All Lean names below have prefix D5.S3.Analytic.GaussZeroBalanced.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $F(a,b;c;x)={}_2F_1(a,b;c;x)$ | Defs.gauss | Mathlib's real ordinaryHypergeometric, with its canonical series. | | Quotient in (1.6) | Defs.quot | Exactly $F(a+1,b+1;a+b+1;x)/F(a,b;a+b;x)$. | | Contiguous quotient $f$ | Defs.f | Exactly $F(a,b;a+b+1;x)/F(a,b;a+b;x)$. | | Inequality (1.7) | Defs.ineq17 | The displayed strict shifted-parameter quotient comparison. | | First clause | Defs.claim, ParamDeriv.clause1 | Every $a,b>0$; negativity of deriv (fun β => quot a β x) b for every $0<x<1$ iff $b\le\sqrt{a(a+1)}$. | | Second clause | Defs.claim, Comparison.clause2 | Every $0<a<1$; (1.7) for every $0<r<1$ iff $3/7\le a$. | | Both clauses | Result.result | The conjunction Defs.claim, without additional hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/GaussZeroBalanced/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2020An isolated qubit ensembleA four-state qubit ensemble with unequal positive priors has a unique minimum-error POVM.Refuted

The questionSpiros Kechrimparis and Joonwoo Bae, Optimal measurement preserving qubit channels, New Journal of Physics 22, 123024, DOI 10.1088/1367-2630/abcc9b, arXiv:2006.16366v2, Conclusions:

Why it mattersIt exhibits an isolated four-state qubit ensemble, answering the published question and refuting the expectation.

The ideaPositive quadratic effect variations give eight first-order stationarity equalities.

Our contributionA four-state qubit ensemble with unequal positive priors has a unique minimum-error POVM.

Scope & next questionThe generalized stationarity theorem allows arbitrary natural dimension and label count; it alone does not settle higher-dimensional isolation or state minimality.

Verification details
Release source recordRefuted in Lean

Open problem

kechrimparis bae 2020 isolated qubit ensembles

Spiros Kechrimparis and Joonwoo Bae, Optimal measurement preserving qubit channels, New Journal of Physics 22, 123024, DOI 10.1088/1367-2630/abcc9b, arXiv:2006.16366v2, Conclusions:

We expect that no isolated ensembles exist but we have not managed to establish the result in general.

Their Definition 1 calls a channel optimal measurement preserving (OMP) when one POVM is optimal for both the original and output ensembles. Isolation means that only the identity CPTP map shares any optimal POVM. Preregistration: #15684.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite qubit ensemble | Natural label count, real priors, complex two-by-two matrices in claim | Literal preregistered convention | | Probabilities of physical states | Strictly positive priors with sum one; frozen IsDensity | No extra hypothesis | | Minimum-error optimum | IsOptimalPOVM: frozen IsPOVM and score comparison with every POVM | The complete defining expression | | One optimum before and after the channel | IsOMP: an existential shared POVM | Does not require all optima to be preserved | | Only the identity preserves an optimum | IsIsolated: every frozen IsCPTP map satisfying IsOMP equals LinearMap.id | The source's trivial OMP set | | No isolated ensembles | Universal claim | Refuted by result : ¬ claim at four states |

The real part of the trace reads the real discrimination score; complete positivity and trace preservation are those of the frozen channel owner. A unique original optimum makes sharing any optimum equivalent to preserving that optimum. The journal title differs from the separate JSAC paper Channel Coding of a Quantum Measurement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/IsolatedQubitEnsemble.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2024The Alcaraz–Ramos lowest-mode conformal prediction failsThe homogeneous N=3 lowest quasienergy cannot have the predicted conformal limit.Refuted

The questionThe Library note gives the source quotations and locators.

Why it mattersThe prediction assigns a conformal arithmetic progression to every fixed low mode.

The ideaThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionThe homogeneous N=3 lowest quasienergy cannot have the predicted conformal limit.

Verification details
Release source recordRefuted in Lean

Open problem

The Alcaraz–Ramos lowest-mode conformal prediction fails

F. C. Alcaraz and L. M. Ramos, Conformally invariant free parafermionic quantum chains with multispin interactions, arXiv:2402.09936v3, Physical Review E 109, 044138, propose the leading prediction $M\varepsilon_i\to\pi A^{(N)}(x_s^{(N)}+i-1)$ for each fixed $i\ge1$. Their conjectured constants for $N=3$ and $M\equiv0\pmod3$ are $A^{(3)}=2\sqrt3$ and $x_s^{(3)}=7/6$. The prediction is Refuted for homogeneous couplings, already at $i=1$ along $M=3m$. Preregistration #19220 fixes the source and Lean conventions; its corrected scaled constant is $7\pi/(3\sqrt3)$. The Library note gives the source quotations and locators. Printed Eqs. (26)–(27), (38), (39), (41) correspond to TeX labels (3.11)–(3.12), (4.7), (4.8), (4.10).

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $P_M=P_{M-1}-z\lambda_M^N P_{M-N}$, with $P_M=1$ for $M\le0$ | poly, specialized to $N=3$, $\lambda_j=1$, with index shift $\operatorname{poly}(n)=P_{n-2}$ | All three initial values and the recurrence are literal | | $P_M$ for nonnegative $M$ | P M = poly (M+2) | Natural-index representation of the source recurrence | | $\varepsilon_i=1/z_i^{1/N}$ over the positive roots | quasienergy M i: roots with multiplicity, positive-root filter, decreasing root order, real exponent $-1/3$ | Increasing quasienergy order; index $i-1$ uses truncated natural subtraction | | Every fixed $i\ge1$ | claim: every natural i with 1 ≤ i | No mode is dropped from the conjecture | | Leading behaviour with $A^{(3)}=2\sqrt3$, $x_s^{(3)}=7/6$ | claim: Tendsto of $3m\varepsilon_i(3m)$ to $\pi(2\sqrt3)(7/6+i-1)$ | The product $3m$ is natural before the real cast; $i$ is cast to real | | $M\equiv0\pmod3$ | The sequence $M=3m$ | Exact nonnegative residue class | | Refutation | result : ¬ claim | The contradiction uses $i=1$ only |

Absent root indices are assigned zero in quasienergy, as fixed by the binding conventions. A positive proposed limit forces the selected quasienergy to be positive eventually; the proof then obtains an actual positive root. This totalization adds no hypothesis to the claim and supplies no spurious counterexample at absent indices. Every binder and conjunct implements a row above; no source clause is weakened or dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/AlcarazRamosConformalTowerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2021Sixfold evil and odious representation countsDyadic coefficient recurrences control the sixfold evil and odious representation counts.Proved

The questionAllouche and Shallit, Additive properties of the evil and odious numbers and similar sequences, arXiv:2112.13627v3, Section 4, Conjecture 12; Funct.

Why it mattersThe published sixfold question concerns every index above the stated thresholds.

Our contributionDyadic coefficient recurrences control the sixfold evil and odious representation counts.

Scope & next questionNo conclusion for these cases is supplied by the sixfold theorem, and whether the same dyadic-norm method settles them remains open.

Verification details
Release source recordProved in Lean

Open problem

Sixfold evil and odious representation counts

Allouche and Shallit, Additive properties of the evil and odious numbers and similar sequences, arXiv:2112.13627v3, Section 4, Conjecture 12; Funct. Approx. Comment. Math. 70(1) (2024), 55–69, DOI 10.7169/facm/2108:

(a) Both $r_6(n)$ and $s_6(n)$ are eventually strictly increasing.

(a) $r_6(n)<r_6(n+1)$ for $n\ge37$.

(b) $s_6(n)<s_6(n+1)$ for $n\ge5$.

The duplicated label (a) is the source's. Equations (2)–(3) count ordered tuples with repetitions allowed. Coordinates range over $\mathbb N={0,1,2,\ldots}$. The Thue–Morse recursion is $t_0=0$, $t_{2n}=t_n$, $t_{2n+1}=1-t_n$; evil and odious coordinates have letters zero and one respectively.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/EvilOdious/SixfoldMonotonicity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2024The Alsing–Birrittella logarithmic-negativity sandwichPositive Hermitian amplitude matrices give an exact trace-norm formula, a logarithmic-negativity sandwich, and equality of the upper bound for real amplitudes.Proved

The questionBirrittella, Numerical Evidence for a Bipartite Pure State Entanglement Witness from Approximate Analytical Diagonalization, arXiv:2404.13725v1, Section IV, p. 11, Equation (22), state:

Why it mattersThis named external problem has no anchor atom.

The ideaThe existing HaarTwoCopyTwirl.replicaSwapcommutes supplies tensor-square commutation.

Our contributionPositive Hermitian amplitude matrices give an exact trace-norm formula, a logarithmic-negativity sandwich, and equality of the upper bound for real amplitudes.

Verification details
Release source recordProved in Lean

Open problem

The Alsing–Birrittella logarithmic-negativity sandwich

P. M. Alsing and R. J. Birrittella, Numerical Evidence for a Bipartite Pure State Entanglement Witness from Approximate Analytical Diagonalization, arXiv:2404.13725v1, Section IV, p. 11, Equation (22), state:

we have numerically found that 𝓛𝓝_a^avg(ρ_ab) ≤ LN_e(ρ_ab) ≤ LN_a(ρ_ab) when the mixed state is written as a pure state decomposition (PSD) ρ_ab = Σ_i p_i |ψ_i⟩⟨ψ_i|, and in addition the quantum amplitude matrix C_i of each pure state component |ψ_i⟩ = Σ_nm C_i,nm |n,m⟩ is uniformly generated (over the Haar measure) as a positive Hermitian matrix, C_i = C_i† ≥ 0.

Further, we find that the second inequality is surprisingly saturated, i.e. LN_e(ρ_ab) = LN_a(ρ_ab) in Eq.(22), if the uniformly randomly generated C_i matrices above are real, i.e. C_i = C_i^T.

The settling theorem proves both inequalities and the real-amplitude equality for every finite ensemble of positive semidefinite complex amplitude matrices, with nonnegative weights summing to one and $\operatorname{tr}(C_iC_i^\dagger)=1$. Different amplitudes need not commute. Zero weights and singular amplitudes are allowed.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite bipartite amplitudes $C_i$ | d K : ℕ, C : Fin K → Matrix (Fin d) (Fin d) ℂ in claim | Every finite dimension and ensemble size; no sampling restriction. | | $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$ | ensembleState | Function.uncurry is the literal amplitude vector; Matrix.vecMulVec forms its outer product, with real weights embedded in $\mathbb C$. | | Probabilities and normalized pure components | Nonnegative weights, sum one, and trace normalization in claim | State and ensemble normalization made explicit. | | Positive Hermitian amplitudes | (C i).PosSemidef | Includes Hermitian symmetry and permits zero eigenvalues. | | Exact logarithmic negativity | exactLogNegativity | Real.logb 2 of the existing trace norm of the existing second-factor partial transpose. | | Equations (11a)–(11b) | pureApproxNegativity, pureApproxLogNegativity | Literal half-sum over ordered unequal pairs and base-two logarithm. | | Equations (17b)–(17c) | mixedApproxNegativity, mixedApproxLogNegativity | Literal mixed entry difference, half-sum and logarithm. | | Average of the pure logarithms | First inequality in claim | Weighted sum of logarithms, with the logarithm inside the ensemble average. | | Upper inequality and real-case equality | Second inequality and guarded equality in claim | For Hermitian matrices, transpose symmetry expresses real entries. |

The public definition bodies are the conventions in #19208. The universal positive-semidefinite class contains the randomly generated source class. No simultaneous diagonalization, invertibility or positive-weight assumption is added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/AlsingBirrittellaLogNegativitySandwich.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024The classical network coherence maximum is $1/4$ for every $n\ge2$The coherence merit of a mixture of two binary product laws has attainable maximum one quarter for every n at least two.Proved

The questionDakić, Quantum coherence in networks, arXiv:2407.11122v1 [quant-ph]; Phys.

Why it mattersThe bound supplies an exact classical threshold for the source's nonlinear witness of coherence in a postselected ring network.

The ideaNo nonnegativity hypothesis is needed for this lemma.

Our contributionThe coherence merit of a mixture of two binary product laws has attainable maximum one quarter for every n at least two.

Scope & next questionThe sharp maximum for exactly three product components is not determined here and remains open in this dossier. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

The classical network coherence maximum is $1/4$ for every $n\ge2$

F. Bibak, F. Del Santo and B. Dakić, Quantum coherence in networks, arXiv:2407.11122v1 [quant-ph]; Phys. Rev. Lett. 133, 230201 (2024), subsection "Generalization of the inequality to $n$ parties", main.tex, lines 277–278:

We obtained classical bound $C_n^{\text{cl}}$ by substituting $p_{\text{cl}}$ (Eq.~\eqref{pncl}) into the expression for $C_n$ and performing extensive numerical optimization. For all tested values of $n \in {2, \ldots, 10}$, we consistently obtained $\text{max}(C_n^{\text{cl}}) = \frac{1}{4}$. Consequently, we conjecture that $\text{max}(C_n^{\text{cl}}) = \frac{1}{4}$ for all $n$.

The quotation preserves the source's words and mathematical content, with inline TeX math delimiters rendered as dollar delimiters and the source line break joined. The source read is main.tex from the v1 source archive, md5 c47930e35162d18bc08d65fb1201100d.

Issue #21125 fixes the reading: for every integer $n\ge2$, maximize over $\gamma\in[0,1]$ and binary probability laws $p_i,q_i$ the merit of

$r(a)=\gamma\prod_{i=0}^{n-1}p_i(a_i)+(1-\gamma)\prod_{i=0}^{n-1}q_i(a_i).$

Each local law is nonnegative and sums to one. This is the mathematical mixture in the source's equation pncl, lines 248–253. Define $E={a:\sum_i a_i\equiv0\pmod2}$ and $O={a:\sum_i a_i\equiv1\pmod2}$. The reading of equation cn, lines 263–271, is

$C_n(r)=\frac12\left|\sum_{\substack{a,b\in E\a\ne b}}r(a)r(b)-\sum_{\substack{a,b\in O\a\ne b}}r(a)r(b)\right|.$

The pairs are ordered, and both strings in each pair have the stated parity. The factor $1/2$ counts each unordered pair once. The conclusion is an attained maximum of $1/4$, as expressed by IsGreatest, rather than an unevidenced supremum. The restriction $n\ge2$ is explicit: at $n=1$, each parity class has only one string and the merit is zero.

The source definitions and quotations are also collected in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/NetworkCoherenceClassicalBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023Model-level ensemble equivalence does not imply measure-level equivalenceA scalar statistic with bounded dyadic weights on binary square matrices separates model-level and measure-level ensemble equivalence.Refuted

The questionGarlaschelli, Description length of canonical and microcanonical models, arXiv:2307.05645v4, Physical Review Research 7, 043057, Section IV.C, following (71):

Why it mattersModel equivalence compares two normalized maximum-likelihood laws averaged over all data.

The ideaFinite binary expansions make the statistic injective on first columns.

Our contributionA scalar statistic with bounded dyadic weights on binary square matrices separates model-level and measure-level ensemble equivalence.

Verification details
Release source recordRefuted in Lean

Open problem

Model-level ensemble equivalence does not imply measure-level equivalence

F. Giuffrida, T. Squartini, P. Grünwald and D. Garlaschelli, Description length of canonical and microcanonical models, arXiv:2307.05645v4, Physical Review Research 7, 043057, Section IV.C, following (71):

i.e., whenever ensemble equivalence holds, the corresponding microcanonical and canonical models are equivalent on the model level. The reverse implication, verified here in the specific case of one-sided local constraints, will be investigated in future work.

Issue #18580 preregisters its refutation. The source and equations are recorded in the Library note.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Binary $n\times m$ matrices, (12) | $\mathrm{Fin}(n)\to\mathrm{Fin}(n)\to\mathrm{Bool}$ | Specialize $m=n$; a counterexample in this subclass refutes the general implication. | | Scalar statistic and fiber cardinality, (15) | omega | Count every binary matrix with the observed statistic. | | Microcanonical uniform law, (16) | pMic | Reciprocal fiber size on the fiber, zero off it. | | Canonical exponential law, (20), and matching, (21) | Exponential expression in IsMatched | Sum over every binary matrix; the parameter is real and finite. | | Matched measure divergence, (22), and equivalence scale, (35) | measureDiv and the conclusion of claim | Matching holds eventually; divergence is divided by $n$. | | Maximized likelihoods, (5), (17) | lCan, lMic | The real supremum includes limiting endpoint likelihoods. | | Normalized maximum likelihood, (8) | nmlMic, nmlCan | The literal likelihood-over-sum expressions. | | Model divergence and its vanishing rate, (69)–(70) | modelDiv and the premise of claim | The normalization scale is $n$, not $n^2$. | | Reverse implication after (71) | result : ¬ claim | Universal statistics and observation sequences, restricted to eventual finite matching. |

The exponential formula is literal in the public definitions. Private proofs use the existing gibbs and meanEnergy constructions; private consumed bridges relate the literal NML quotients to the existing normalized construction. Restricting the source assertion to square matrices, one scalar statistic and eventually matched observations weakens it, so refuting this restriction refutes the general converse.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/ModelLevelEnsembleEquivalenceConverseRefutation.result

Frozen module record

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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Product formulas for periodic Temperley–Lieb seed recurrencesUniform product evaluations of the periodic Temperley–Lieb seed recurrences.Proved

The questionYacine Ikhlef and Alexi Morin-Duchesne, Fusion of irreducible modules in the periodic Temperley–Lieb algebra, Appendix C, Conjectures C.1 and C.2, arXiv:2312.14837v2; SciPost Phys. 17, 132, DOI 10.21468/SciPostPhys.17.5.132.

Why it mattersThese evaluations supply the specialized seed values used in the source's Section 4 discussion of fusion decompositions.

The ideaThe coupled invariant and the terminal induction remain after expanding the same-module helpers.

Our contributionUniform product evaluations of the periodic Temperley–Lieb seed recurrences.

Verification details
Release source recordProved in Lean

Open problem

Product formulas for periodic Temperley–Lieb seed recurrences

Yacine Ikhlef and Alexi Morin-Duchesne, Fusion of irreducible modules in the periodic Temperley–Lieb algebra, Appendix C, Conjectures C.1 and C.2, arXiv:2312.14837v2; SciPost Phys. 17, 132, DOI 10.21468/SciPostPhys.17.5.132. The source note is D5/L/QuantumContext/ikhlefmorinduchesne2024fusion. Preregistration: #18569.

Write $[k]=(q^k-q^{-k})/(q-q^{-1})$, ${k}=q^k+q^{-k}$ and $\alpha_z=z+z^{-1}$. Proposition C.1 defines $X^{(p)}_m$ by (C.3)–(C.4):

$$ X^{(p)}p=1,\qquad X^{(p)}{p+1}=\alpha_x+\frac{\alpha_y\alpha_w}{{p+1}}, $$

$$ X^{(p)}_{k+1}=\left(\alpha_x+ \frac{{p}\alpha_y\alpha_w}{{k}{k+1}}\right)X^{(p)}k+ \frac{({p}^2-{k}^2)(\alpha_y^2-{k}^2)(\alpha_w^2-{k}^2)} {(q-q^{-1})^2[2k-1][2k+1]{k}^2}X^{(p)}{k-1}. $$

Conjecture C.1 states: “For p ⩾ 1, the function X^{(p)}_m(ω, x, y) specialised to y = εq^m evaluates to” (C.5):

$$ X^{(p)}m(w,x,\varepsilon q^m)=(wx)^{p-m} \prod{r=-(m-p-1)/2}^{(m-p-1)/2} (xq^{2r}+\varepsilon w)(\varepsilon q^{2r}+xw). $$

Proposition C.2 defines $\sigma^{(0)}_{0,m}$ by (C.24)–(C.25):

$$ \sigma_0=1,\qquad \sigma_1=\alpha_x+ \frac{\alpha_y\alpha_w}{{1}}, $$

$$ \sigma_{k+1}=\left(\alpha_x+ \frac{2\alpha_y\alpha_w}{{k}{k+1}}\right)\sigma_k- \frac{[k]^2(\alpha_y^2-{k}^2)(\alpha_w^2-{k}^2)} {[2k+1][2k-1]{k}^2}\sigma_{k-1}. $$

Conjecture C.2 states: “The intermediate seed states σ^{(0)}_{0,m}(w, x, y) specialised to y = εq^m evaluate to” (C.26), the same product with $p=0$.

Statement echo

All Lean names in this dossier have prefix D5.S3.Quantum.Algebra.PeriodicTemperleyLiebSeedProduct.

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Complex nonzero $q,x,w$ | claim quantifies complex parameters and requires each nonzero | Exact pointwise domain | | $q$ not a root of unity | Every natural $n\ge1$ has $q^n\ne1$ | All recurrence poles in scope are avoided | | $\varepsilon=\pm1$ | Sign disjunction in claim | Both signs | | $[k]$, ${k}$, $\alpha_z$ | qbr, qcb, al | Literal source expressions | | (C.3)–(C.4) | seedX q x y w p n denotes $X^{(p)}_{p+n}(w,x,y)$ | Three recursion cases, with $k=p+n+1$ | | (C.24)–(C.25) | seedSigma | Independently transcribed three recursion cases | | Symmetric product | seedProduct | $N$ factors, $2r=-N+1+2j$, $0\le j<N$ | | (C.5), every $1\le p\le m$ | First conjunct of claim | $N=m-p$, including $N=0$ | | (C.26), every $m\ge0$ | Second conjunct of claim | $N=m$, including $m=0$ | | Both conjectures | result | Full conjunction; no dropped clause |

The natural subtraction agrees with the source integer difference under $p\le m$. Natural-to-integer casts express the product exponents. Empty products equal one. The recurrence definitions are total on complex arguments; the theorem restricts them to the stated nonzero, non-root domain. No extra hypothesis narrows that domain.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/PeriodicTemperleyLiebSeedProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Geometry & algebraQuestion resolvedSource 2024The full weighted orthogonal-minor inequalityThe weighted two-minor and three-minor inequalityProved

The questionThe source is meler (MathOverflow user 526848; real name unverified), MO469781, revision 1, GUID F53D06CE-3805-43B9-9FEB-43A79F67F7AB, CC BY-SA 4.0, with identical MSE4902019 crosspost.

Why it mattersThis compares weighted squared coordinates of a two-dimensional orthonormal plane with those of its extension by one orthogonal unit vector.

The ideaFor skew pair arrays v and alternating triple arrays b, define lift by z(i)v(j,k)-z(j)v(i,k)+z(k)v(i,j) and lower by summing z(k)b(k,i,j) over k.

Our contributionThe weighted two-minor and three-minor inequality

Verification details
Release source recordProved in Lean

Open problem

The full weighted orthogonal-minor inequality

The source is meler (MathOverflow user 526848; real name unverified), MO469781, revision 1, GUID F53D06CE-3805-43B9-9FEB-43A79F67F7AB, CC BY-SA 4.0, with identical MSE4902019 crosspost. The Library note owns the source citation. The complete external question is:

Fix $n \geq 3$ and take any orthonormal vectors $x,y,z \in \mathbb{R}^n$. Let also $A \in M_n(\mathbb{R})$ be a symmetric matrix with positive entries ($A_{ij} = A_{ji} > 0$). Is the following inequality true? $$\sum_{i<j}A_{ij}\begin{vmatrix}x_i & x_j \newline y_i & y_j \end{vmatrix}^2 \leq \sum_{i<j<k}\max(A_{ij},A_{ik},A_{jk})\begin{vmatrix}x_i & x_j & x_k \newline y_i & y_j & y_k \newline z_i & z_j & z_k \end{vmatrix}^2$$

For ordinary mathematical rendering, the literal \\newline separators in that immutable transcription mean row breaks. Thus the determinants have rows (x,y) and (x,y,z), respectively, and increasing coordinate columns. No statement is changed by that typographic reading.

The source-to-Lean map is exact: n : ℕ covers every original integer n≥3; x y z : EuclideanSpace ℝ (Fin n) are the three real vectors; Orthonormal ℝ ![x,y,z] supplies all unit norms and pairwise orthogonalities; A : Matrix (Fin n) (Fin n) ℝ, ∀ i j, A i j = A j i and ∀ i j, 0 < A i j give precisely the matrix hypotheses. Conditional finite sums select i < j and i < j ∧ j < k. The literal Matrix.det !![...] terms are the displayed two- and three-row determinants, and max (A i j) (max (A i k) (A j k)) is the same three-edge maximum. Zero-based coordinate labels merely relabel the source coordinates. The native theorem quantifies all natural dimensions; below three its orthonormal-triple premise has no instance. No distance, positive-definiteness, nonzero-minor, rationality or finite-dimension-cutoff premise is added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/ProbabilityTransport/WeightedProjectionMinors.weighted_minors

Frozen module record

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  • Historical context

    Year recorded in the source bibliography: 2024.

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Discrete structuresQuestion resolvedPosed in 2023Cardinality-sum-avoiding subset generating functionThe original subset count has the conjectured generating function.Proved

The questionChai Wah Wu's November 21, 2023 conjecture in OEIS A367400 is

Why it mattersThe original count has an exact all-degree formal-series description, including the empty subset and the permitted even midpoint.

The ideaFor positive k, the distinct forbidden pairs are {i,k-i} for 1 <= i <= floor((k-1)/2).

Our contributionThe original subset count has the conjectured generating function.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Cardinality-sum-avoiding subset generating function

For n in N, let a(n) count subsets S of {1,...,n} for which no distinct x,y in S have x+y=|S|. The interval is empty at n=0; a(0)=1. Equal summands are permitted. Chai Wah Wu's November 21, 2023 conjecture in OEIS A367400 is

$$ A(X)=\sum_{n\ge0}a(n)X^n =\frac{1+X^2-X^3}{1-2X+X^2-2X^3+X^4}. $$

The exact retained statement is

(1 - 2*PowerSeries.X + PowerSeries.X^2 - 2*PowerSeries.X^3 +
  PowerSeries.X^4 : PowerSeries ℚ) * PowerSeries.mk (fun n => (a n : ℚ)) =
  1 + PowerSeries.X^2 - PowerSeries.X^3

The definition of a is the literal filtered powerset cardinality, with no recurrence, assumed count identity, or assumed bijection. Preregistration: issue #11453, first-tier external conjecture. The corresponding Library note gives the source and bounded literature scope; neither establishes priority.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CardinalitySumAvoidingSubsets.result

Frozen module record

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Analysis & dynamicsQuestion resolvedFirst preprint 2022The Poncini–Rasmussen central factorial transfer expansionThe eight-vertex transfer matrix equals the central factorial polynomial expansion of its principal Hamiltonian over every algebraically closed field of characteristic zero.Proved

The questionPoncini and Rasmussen, Integrability of planar-algebraic models, arXiv:2206.14462v2, Appendix C, page 47, state Conjecture C.12:

Why it mattersThe source establishes membership in the algebra generated by the Hamiltonian and an inverse-Vandermonde interpolation formula.

The ideaThe scalar module proves centralspectralcoeff by central differences, formal binomial series and recurrence identities, and obtains scalarsettlement by polynomial coefficient comparison and homogenization.

Our contributionThe eight-vertex transfer matrix equals the central factorial polynomial expansion of its principal Hamiltonian over every algebraically closed field of characteristic zero.

Verification details
Release source recordProved in Lean

Open problem

The Poncini–Rasmussen central factorial transfer expansion

Poncini and Rasmussen, Integrability of planar-algebraic models, arXiv:2206.14462v2, Appendix C, page 47, state Conjecture C.12:

For every n ∈ ℕ and all u ∈ F, we have Ť_n(u) = 2 Σ_{k=0}^{n−1} (1 + u²)^{n−k} (2u)^k Σ_{j=0}^{⌊k/2⌋} [n − k − 2 + 2j over 2j]!! (−1)^j m_{k−2j}(h_n)/(k − 2j)!

The domain is $n\ge1$ and an algebraically closed field $F$ of characteristic zero. The source's coefficients and ordered operators are (C.9)–(C.15), the Hamiltonian is (C.40), the central factorial polynomials are (C.44), and the double factorial quotient is (C.65). The source note is D5/L/StatisticalMechanics/poncini2023planar. Preregistration: #19637.

Statement echo

The settling namespace is D5.S3.StatisticalMechanics.VertexModels.PonciniRasmussenTransferExpansion; the scalar namespace is D5.S1.Recurrence.CentralFactorialTransferExpansion.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Algebraically closed characteristic-zero field $F$ | claim quantifies over F, Field F, IsAlgClosed F, CharZero F | Full source domain. | | Every positive length and every spectral parameter | claim quantifies over n, its lower bound, and u | No parameter is specialized. | | Two-dimensional tensor basis | Matrices indexed by functions from Fin n to Fin 2 | Coordinates start at zero. | | Four cases of (C.15) | Rcoef | The same equalities, inequalities and coefficients. | | $K^k_j=\bar K^k_j=\delta_{jk}$ | Kcoef, oneSite | The two boundary operators are identity coefficients on their sites. | | Two-site matrix units, (C.9)–(C.10) | twoSite | Row and column entries use $(k,j)=(x(a),y(a))$ and $(m,l)=(x(b),y(b))$; other coordinates agree. | | Ordered double row, (C.12) | preTrace | Descending first row, first boundary, ascending second row, auxiliary boundary. | | Auxiliary partial trace, (C.13) | transfer | Sum over the same appended auxiliary bit in row and column. | | Adjacent flips, (C.40) | hamiltonian | The bounded next-index proof only supplies the final adjacent site. | | $m_r(x)=\prod_{a=0}^{r-1}(x-(r-1-2a))$ | scalar mPoly | The full product, including $m_0=1$. | | $n_1!!/(n_2!!(n_1-n_2)!!)$, (C.65) | scalar dfBinom | Uses Int.toNat and Nat.doubleFactorial, giving $(-1)!!=0!!=1$. Only $-1$ can occur negatively in the claim. | | Both finite sums and all factors of (C.66) | claim | Natural half-index division and truncated subtraction; the summation bounds ensure the intended nonnegative indices. | | The universal equality | result | Proves the whole claim; its Scribe formula expands it in full. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/VertexModels/PonciniRasmussenTransferExpansion.result

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2021Odd order does not ensure the conjectured minimum of HadamardnessThe proposed odd-order minimum of Hadamardness cannot be attained in order nine.Refuted

The questionNechita, Diagonal unitary and orthogonal symmetries in quantum theory II: Evolution operators, J.

Why it mattersSection 6 connects Hadamardness to the entangling power of real dual orthogonal operators invariant under local diagonal unitary actions (LDUI).

The ideaThe lower-bound clause of claim is not needed for this contradiction.

Our contributionThe proposed odd-order minimum of Hadamardness cannot be attained in order nine.

Verification details
Release source recordRefuted in Lean

Open problem

Odd order does not ensure the conjectured minimum of Hadamardness

S. Singh and I. Nechita, Diagonal unitary and orthogonal symmetries in quantum theory II: Evolution operators, J. Phys. A: Math. Theor. 55 (2022) 255302, arXiv:2112.11123v1, Section 6, main.tex line 848, immediately following Table 1:

We note that for the odd values of $d$ considered above, $\min \mathfrak h(C) = d^3 + d(d-1)$, obtained for matrices $C$ with the property that all the scalar products between different works have absolute value $1$. We conjecture that this is the case for all odd integers $d$.

“Works” is the source's wording. The source defines its measure in Eq. eq:Hada_measure, main.tex lines 780–783. For a real sign matrix, write $G=CC^{\mathsf T}$ and $\mathfrak h(C)=\operatorname{Tr}(G^2)=\sum_{i,j}G_{ij}^2$; conjugate transpose and transpose coincide here. The source and its table are described in the literature note.

Issue #22426 fixes the reading encoded by claim: for every odd natural $d$, there exists an integer $d\times d$ matrix $C$ with every entry equal to $1$ or $-1$, such that $\mathfrak h(C)=v_d$ and every integer sign matrix $B$ of that order satisfies $v_d\leq\mathfrak h(B)$. Here $v_d=d^3+d(d-1)$ is evaluated in the natural numbers, including subtraction, and then embedded in the integers. Thus attainment and the lower bound are both explicit; no infimum convention is used. The conjecture is refuted, using order nine.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/SinghNechitaOddHadamardnessRefutation.result

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    Source 1
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    Year in the primary source attribution: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2024Validity of the five three-outcome PI Bell inequalitiesFive three-outcome PI Bell inequalities hold for every party count.Proved

The questionFadel, Deriving three-outcome permutationally invariant Bell inequalities, arXiv:2406.11792v1, Entropy 26 (2024), 816, Section III.A:

Why it mattersIt proves claim, the normalized finite-hidden-variable formulation specified in 15121. Gnonneg, row3small, row3polynonneg, countBellnonneg, deterministicvalidity, modelnonneg and result have proofshape: content after inlining the same-delivery helpers.

The ideaFor each hidden variable, choose both local responses independently.

Our contributionFive three-outcome PI Bell inequalities hold for every party count.

Verification details
Release source recordProved in Lean

Open problem

Validity of the five three-outcome PI Bell inequalities

A. Aloy, G. Müller-Rigat, J. Tura and M. Fadel, Deriving three-outcome permutationally invariant Bell inequalities, arXiv:2406.11792v1, Entropy 26 (2024), 816, Section III.A:

To give a concrete example, we propose for any $N>3$ the five 3PIBIs shown in Tab. III.

At this point, for each conjectured inequality we have to prove that it is indeed valid for arbitrary number of parties $N$, or at least for all $N$ larger than a minimum number.

There are $N$ parties, inputs $x_i\in{0,1}$ and outcomes $a_i\in{0,1,2}$. The one-party observables sum marginal probabilities; the two-party observables sum products over ordered distinct parties. The five symmetrized observables and all six coefficients of each Table III row are retained literally in P1, P2, Pt0, Pt00, Pt01, Pt10, Pt11, bell and table.

| Row | $(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\beta_c)$ | | --- | --- | | 1 | $(1,1,0,-2,0,0)$ | | 2 | $(1,1,-2,-2,2,0)$ | | 3 | $(-2,1,2,2,0,4)$ | | 4 | $(-6,1,4,4,2,12)$ | | 5 | $(-6,1,4,0,0,24)$ |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ThreeOutcomePermutationallyInvariantBell.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2022The averaged enumerator of B-form codesWe conjecture the form of corresponding averaged enumerator polynomial based on invariance under MacWilliams identity and explicit checks for sufficiently small n and prime pProved

The questionDymarsky (arXiv:2206.14825, hep-th; JHEP 11 (2022) 118) build Narain CFTs from codes over Fp × Fp.

Why it mattersThe averaged enumerator gives the averaged partition function of the code CFT ensemble, which the paper compares with the holographic "U(1) gravity" prediction.

The ideaExchange the sums over B and r; the term r = 0 gives t0^{2c}.

Our contributionWe conjecture the form of corresponding averaged enumerator polynomial based on invariance under MacWilliams identity and explicit checks for sufficiently small n and prime p

Verification details
Release source recordProved in Lean

Open problem

The averaged enumerator of B-form codes

N. Angelinos, D. Chakraborty and A. Dymarsky (arXiv:2206.14825, hep-th; JHEP 11 (2022) 118) build Narain CFTs from codes over F_p × F_p. For prime p the generating matrix is (I | B^T) with B antisymmetric mod p and B_ii = 0; the codewords are (r, B^T r), r ∈ Z_p^c. Averaging the full enumerator polynomial over the p^{c(c−1)/2} such B at x_ab = t_a t_b, t_a = t_{−a}, they state:

We conjecture the form of corresponding averaged enumerator polynomial based on invariance under MacWilliams identity and explicit checks for sufficiently small n and prime p

with eq. (barP): P̄ = t_0^{2c} + (Σ_{k=0}^{p−1} (Σ_{a,b} cos(2πkab/p) t_a t_b)^c − p t_0^c (Σ_a t_a)^c) / p^c.

Issue #11296 fixes the reading: p prime, c ≥ 0, t : Z/p → ℂ even, and in the cosine k, a, b are the representatives 0, …, p − 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/BFormCodeAveragedEnumerator.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2023Half-density modes of the asymmetric K-exclusion processThe half-density K-exclusion mode for small and large positive parameters.Proved

The questionMisra, An exactly solvable asymmetric K-exclusion process, arXiv:2310.03343v2, Section 4.5, page 21:

Why it mattersThe modal comparison distinguishes the two adjacent occupations that share the ground-state contribution.

The ideaThe proof expresses the source weights through finite occupied-slot sums.

Our contributionThe half-density K-exclusion mode for small and large positive parameters.

Verification details
Release source recordProved in Lean

Open problem

Half-density modes of the asymmetric K-exclusion process

A. Ayyer and S. Misra, An exactly solvable asymmetric K-exclusion process, arXiv:2310.03343v2, Section 4.5, page 21:

We focus on when t ≪ 1 and t ≫ 1. From the discussion in Sections 4.2 and 4.3, we know that the most likely number of particles is either k or k + 1. The question we want to address is which one is most likely.

The question is AM-1 of #18570. Write $L=2r\ge4$, $K\ge1$, $0\le k<K$, and $n=(2k+1)r$. There is an $\varepsilon>0$ such that, whenever $t>0$ and either $t<\varepsilon$ or $t>1/\varepsilon$, the unique modal occupation is $k$ if $2k+1<K$, and $k+1$ if $K<2k+1$. If $K=2k+1$, occupations $k$ and $k+1$ tie and strictly exceed every other occupation in these regimes.

Statement echo

Lean names in this dossier belong to D5.S3.Combinatorics.Probability.AyyerMisraHalfDensityMode.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (4), $[k]t=\sum{j<k}t^j$ | qnum | The literal finite sum, including zero at $k=0$. | | Equation (8), $[k]t!=\prod{j<k}[j+1]_t$ | qfact | The literal product, including the empty product. | | Equation (9), the factorial quotient | qbinom | The literal quotient for $i\le k$, zero otherwise. | | Equation (11), steady-state weights | wt | The product uses natural exponent $(a^2+2-3a)/2$ with truncated subtraction; its values at $a=0,1,2,3,4$ are $1,0,0,1,3$. | | Equation (1), configurations | configs | Functions $\operatorname{Fin}L\to\operatorname{Fin}(K+1)$ whose natural values sum to $n$. | | Equation (13), normalized probability | occProb | The selected-occupation weight sum divided by the total configuration weight. | | $L>2$ and half-integer density | claim: r, K, k, hr | $L=2r$, $r\ge2$, $K\ge1$, $k<K$ and $n=(2k+1)r$. Natural types supply nonnegativity. | | A site occupation | claim: s | The first site, the element of $\operatorname{Fin}(2r)$ with value zero. | | Small and large positive $t$ | claim: positive radius and parameter | One radius for both $0<t<\varepsilon$ and $t>1/\varepsilon$. | | Which adjacent occupation is most likely | claim: three conjuncts | Strict comparison with every $b\le K$ outside the stated mode or tied pair. | | Answer to the question | result | The full claim, with no additional hypothesis. |

The public definitions follow the binding source conventions of #18570. They introduce no numerical certificate or experimental premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Probability/AyyerMisraHalfDensityMode.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2023Global ZZ gates admit schedules within the parity boundEvery real symmetric zero-diagonal coupling matrix admits a nonnegative sign schedule within the parity-dependent gate-time bound.Proved

The questionKliesch, Time-optimal multi-qubit gates: Complexity, efficient heuristic and gate-time bounds, arXiv:2307.11160v2 [quant-ph]; Quantum 8, 1279 (2024), Conjecture 16 (conj:gate-time), TeX source Jshaping.tex, lines 919–927:

Our contributionEvery real symmetric zero-diagonal coupling matrix admits a nonnegative sign schedule within the parity-dependent gate-time bound.

Verification details
Release source recordProved in Lean

Open problem

Global ZZ gates admit schedules within the parity bound

P. Baßler, M. Heinrich and M. Kliesch, Time-optimal multi-qubit gates: Complexity, efficient heuristic and gate-time bounds, arXiv:2307.11160v2 [quant-ph]; Quantum 8, 1279 (2024), Conjecture 16 (conj:gate-time), TeX source J_shaping.tex, lines 919–927:

The optimal gate time of $\operatorname{\mathsf{GZZ}}(A)$ with $A \in \mathrm{Sym}_0 (\mathbb{R}^n)$ is tightly upper bounded by

$\vec 1^T \vec \lambda^* \leq |M|{\ell\infty} \cdot \begin{cases} n,, &\text{for odd } n,,\ n-1,, &\text{for even } n,. \end{cases}$

The wording and equation are quoted from the downloaded v2 source (J_shaping.tex, md5 3c751c31a7a40505b6013390b841878a); its macros for the gate, $\mathrm{Sym}_0$, $\mathbb R$ and the norm are expanded for rendering. The source also states, at line 349:

Unfortunately, we were not able to prove this result and state the challenges in \cref{app:proof_problems}.

Here $\mathrm{Sym}0(\mathbb R^n)$ consists of real symmetric matrices with zero diagonal (line 240). The effective coupling is $M=A\oslash J$, with $M{ij}=A_{ij}/J_{ij}$ when $i\ne j$ and $J_{ij}\ne0$, and zero otherwise; the physical interpretation assumes $\operatorname{nz}(A)\subseteq\operatorname{nz}(J)$ (lines 250–263). The source's LP minimizes $T=\sum_m\lambda_m$ over $\lambda_m\ge0$ with

$M_{ij}=\sum_{m\in{-1,+1}^n,\ m_n=+1}\lambda_m m_i m_j\quad(i<j)$.

Only strictly upper triangular entries enter these constraints (lines 265–291 and 355–362). The norm is $d=\max_{i<j}|M_{ij}|$ (line 448).

Issue #21288 fixes the settlement: for every $n\ge2$ and every such $M$, there is a feasible schedule with $T\le Kd$, where $K=n$ for odd $n$ and $K=n-1$ for even $n$. This existence statement is proved. The passage to an optimal LP value and the sharpness argument are distinguished below.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Optimization/GlobalZZGateTimeBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture provedFirst preprint 2023Cowtan–Burton Conjecture D.1: LDPC monic CSS pushoutsBasis-preserving monic pushouts of length-two binary CSS complexes satisfy explicit row and column weight bounds.Proved

The questionAlexander Cowtan and Simon Burton, CSS code surgery as a universal construction, Quantum 8 (2024), 1344, arXiv:2301.13738v6, Appendix D, Conjecture D.1:

Why it mattersLDPC closure supplies a uniform bound for the checks and qubit incidences of code surgery.

The ideaBoth row bounds use the quotient row-support estimate.

Our contributionBasis-preserving monic pushouts of length-two binary CSS complexes satisfy explicit row and column weight bounds.

Verification details
Release source recordProved in Lean

Open problem

Cowtan–Burton Conjecture D.1: LDPC monic CSS pushouts

Alexander Cowtan and Simon Burton, CSS code surgery as a universal construction, Quantum 8 (2024), 1344, arXiv:2301.13738v6, Appendix D, Conjecture D.1:

Let C• ←f• A• →g• D• be a basis-preserving monic span in Ch(MatF2), and let Q• be the pushout chain complex of this monic span. Further, let the monic span be a representative of a family of monic spans which are parameterised by some n ∈ ℕ, and let A•, C• and D• be the Z-type complexes of quantum LDPC codes. Then (Q•, Q•) is also LDPC. Formulating this conjecture properly requires specifying what it means for a monic span to be parameterised.

The settlement is Proved for the proposition in #19488, interpreting parameterisation as an arbitrary natural-indexed family.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Based length-two binary complexes | BasedComplex: bases Fin (dim i), two ZMod 2 matrices and their zero composite | | Basis-preserving chain maps, Definition 3.4 | basisMatrix, BPChainMap, including both commutation identities | | Monic span | BPChainMap.Monic: injectivity of each component matrix map | | Basis-preserving pushout, Definition 5.1 and Lemma 5.4 | glue, PBasis, sumD2, sumD1, pushoutD2, pushoutD1; coefficients are fiber sums of a chosen representative column | | Natural-indexed family | claim quantifies over arbitrary A, C, D, φ, ψ | | Apex and feet are LDPC | The three IsLDPC hypotheses in claim; weights2 and weights1 count all four weights | | Pushout family is LDPC | result : claim, with uniform bound $w_C+w_D$ |

Write $E_i=\operatorname{Fin}(C.\mathrm{dim}(i))\sqcup\operatorname{Fin}(D.\mathrm{dim}(i))$ and $q_i:E_i\to\mathrm{PBasis}(\varphi,\psi,i)$ for the quotient map. Under both monicity hypotheses, D5.S3.Quantum.Information.CssMonicPushoutLdpc.pushoutD2_representative_independent proves, for every $Y:\mathrm{PBasis}(\varphi,\psi,2)$, $y:E_2$ with $q_2(y)=Y$, and $X:\mathrm{PBasis}(\varphi,\psi,1)$,

$$\mathrm{pushoutD2}(\varphi,\psi)(X,Y)=\sum_{x\in E_1,\ q_1(x)=X}\mathrm{sumD2}(C,D)(x,y).$$

D5.S3.Quantum.Information.CssMonicPushoutLdpc.pushoutD1_representative_independent proves the analogous formula for $Y:\mathrm{PBasis}(\varphi,\psi,1)$, $y:E_1$ with $q_1(y)=Y$, and $X:\mathrm{PBasis}(\varphi,\psi,0)$:

$$\mathrm{pushoutD1}(\varphi,\psi)(X,Y)=\sum_{x\in E_0,\ q_0(x)=X}\mathrm{sumD1}(C,D)(x,y).$$

Thus the two matrices represent the induced maps $[y]\mapsto[\partial y]$, independently of the selected representative. The degree-two and degree-one formulas use BPChainMap.comm2 and BPChainMap.comm1, respectively. The identity $\partial^Q_1\partial^Q_2=0$ is checked by the pinned experiment below; it is not delivered as a Lean theorem.

The categorical universal property is proved in prose here, rather than by a Lean categorical theorem. In degree $i$, take the free binary vector space on $\mathrm{PBasis}(\varphi,\psi,i)$. Its maps from the two feet send a basis vector $x$ to $[x]$ and agree on every apex basis vector. Any two linear maps from the feet agreeing on the apex assign equal values to each generating glued pair, so extend uniquely to a linear map on the quotient basis. The representative formulas above identify its differentials with $[y]\mapsto[\partial y]$. If the two maps from the feet are chain maps, that extension commutes with the differential on every quotient generator, hence everywhere by linearity. Uniqueness follows because those generators span. This identifies the construction with the basis-preserving pushout of Lemma 5.4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CssMonicPushoutLdpc.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture provedFirst preprint 2023The complete singular squared-distance Laplacian conjectureThe singular multipartite source Laplacian has exactly the stated positive spectrum and inertia on every original part-size domain.Proved

The questionThe Library note owns the bibliography, equation (3.6), omitted-list locators and source qualifications.

Why it mattersThe singular case requires the actual source matrix, rather than an inverse of singular delta.

The ideaWrite delta as the block matrix with same-part coefficient three and diagonal subtraction four, plus the all-ones matrix.

Our contributionThe singular multipartite source Laplacian has exactly the stated positive spectrum and inertia on every original part-size domain.

Verification details
Release source recordProved in Lean

Open problem

The complete singular squared-distance Laplacian conjecture

The single target is the complete two-clause concluding conjecture identified by D5/L/QuadraticForms/dasmohanty2024inverse. The Library note owns the bibliography, equation (3.6), omitted-list locators and source qualifications. The exact proposition is DasMohantySingularSquaredDistance.completeConjecture; the theorem DasMohantySingularSquaredDistance.result has that proposition as its type. Issue https://github.com/the-omega-institute/trureturing/issues/25619 and immutable preregistration comment 6574192179 identify the external event.

For every finite number t of parts with t at least two, and every positive integer part-size function n on Fin t, let N be the sum of the sizes, s the number of parts of size at least two and h the number of singleton parts. Vertices are pairs consisting of a part and a vertex in that part. The real squared-distance matrix delta has diagonal zero, same-part off-diagonal four and between-part one. Assume delta is singular and the sum of all its cofactors is nonzero. The latter is the sum of all adjugate entries, not a principal cofactor.

The functions phi, psi and theta, their single and double omissions, and all three entry cases of sourceL are exactly the author-defined omitted-list functions and equation (3.6). Empty products are one and empty sums zero. No expanded hatted-Psi display is substituted for the omitted-list definition.

The matrix sourceL is Hermitian. Its entire positive real characteristic-root multiset equals N-t copies of 1/4 together with h-1 copies of one, with algebraic multiplicity and no other positive roots. Its positive, zero and negative indices are N-s-1, one and s, respectively. These are a conjunction on the same actual matrix and domain. All finite part counts, repeated sizes and zero multiplicities are included; no additional matrix identity, genericity, classification or spectral-transfer premise is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/DasMohantySingularSquaredDistance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2022Strong noncommutativity without complete incompatibility in every dimension at least fourIn every complex dimension d ≥ 4, two orthonormal bases can have noncommuting coordinate projectors for every pair of nonempty proper index sets while failing complete incompatibility.Proved

The questionStephan De Bièvre, arXiv:2207.07451v1, §5, conjecture following Proposition 7:

Why it mattersThis settles the quoted conjecture as Proved, while the assertion that (ii) implies (i) fails in that range.

The ideaUse the standard complex orthonormal basis and the columns of the real Householder reflection

Our contributionIn every complex dimension d ≥ 4, two orthonormal bases can have noncommuting coordinate projectors for every pair of nonempty proper index sets while failing complete incompatibility.

Verification details
Release source recordProved in Lean

Open problem

Strong noncommutativity without complete incompatibility in every dimension at least four

Stephan De Bièvre, arXiv:2207.07451v1, §5, conjecture following Proposition 7:

We conjecture it is true that (ii) does not imply (i) in all dimensions $d\geq4$, but we have not produced such examples in other dimensions than $4$ and $6$.

The space is $\mathbb C^d$ and the bases are arbitrary orthonormal bases. Condition (ii) says that every pair of coordinate orthogonal projectors indexed by nonempty proper sets fails to commute. Condition (i), complete incompatibility, says that their images have zero intersection whenever the two index-set cardinalities sum to at most $d$; empty sets are included. Issue #11785 preregisters the exact quantified statement before any Lean.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/NoncommutingNotCompletelyIncompatible.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding 1243 and 2134Mathematical definitions and results for Fishburn permutations and classical pattern avoidance.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaOn both sides every nonempty permutation has two ordinary maximum-insertion children.

Our contributionMathematical definitions and results for Fishburn permutations and classical pattern avoidance.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding 1243 and 2134

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.11. For all n ≥ 0, |Fn(1243, 2134)| = |Sn(123, 3241)|.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenEleven.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding Pairs Counted by Odd-Indexed Fibonacci NumbersMathematical definitions and results for Fishburn permutations and classical pattern avoidance.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaInserting the maximum and deleting it are inverse operations on Fishburn permutations; the active gaps of a permutation are given by exact local tests, and a gap that becomes inactive never recovers.

Our contributionMathematical definitions and results for Fishburn permutations and classical pattern avoidance.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding Pairs Counted by Odd-Indexed Fibonacci Numbers

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.5. For all n ≥ 1, |Fn(1324, 1423)| = |Fn(1324, 3124)| = F_{2n−2}.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenFive.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding Three Patterns from {1324, 2143, 1423, 3124}Three classes of Fishburn permutations avoiding triples of patterns have a common enumeration.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaA Fishburn permutation avoiding 1324 and 1423 has the unique shape D 1 I m J, with D decreasing, I increasing, m the largest entry from 1 on and J decreasing; it is Fishburn exactly when no t in J has t + 1 in I.

Our contributionThree classes of Fishburn permutations avoiding triples of patterns have a common enumeration.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding Three Patterns from {1324, 2143, 1423, 3124}

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.7. For all n ≥ 1, |Fn(1324, 2143, 1423)| = |Fn(1324, 2143, 3124)| = |Fn(1324, 1423, 3124)| = 2^n − n.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) is the set of Fishburn permutations of length n avoiding the classical patterns in B. The source notes that 2^n − n also counts the Grassmannian permutations of length n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/FishburnTenSeven/FishburnTenSeven.result

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Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding 2143 and 3124Mathematical definitions and results for Fishburn permutations and classical pattern avoidance.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaThe active sites of a Fishburn permutation avoiding 2143 and 3124 are characterized completely through its prefix, using layered permutations as an auxiliary class.

Our contributionMathematical definitions and results for Fishburn permutations and classical pattern avoidance.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding 2143 and 3124

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.9. For all n ≥ 1, |Fn(2143, 3124)| = |Sn(231, 4123)|.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenNine.result

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Quantum informationQuestion resolvedFirst preprint 2022Swapping an earliest adjacent pair minimizes dephasing for every number of segmentsEvery complete earliest-pair swapping history minimizes the accumulated dephasing for any finite list of natural ready times.Proved

The questionSchmidt and P. van Loock, Exact rate analysis for quantum repeaters with imperfect memories and entanglement swapping as soon as possible, arXiv:2203.10318v1 (Phys.

Why it mattersThe paper studies linear repeaters with parallel entanglement distribution, deterministic swapping and memory dephasing.

The ideaA complete greedy history (IsGreedyRun) requires such a step at each changing list and ends with length at most one; runCost adds its successive charges.

Our contributionEvery complete earliest-pair swapping history minimizes the accumulated dephasing for any finite list of natural ready times.

Verification details
Release source recordProved in Lean

Open problem

Swapping an earliest adjacent pair minimizes dephasing for every number of segments

L. Kamin, E. Shchukin, F. Schmidt and P. van Loock, Exact rate analysis for quantum repeaters with imperfect memories and entanglement swapping as soon as possible, arXiv:2203.10318v1 (Phys. Rev. Research 5, 023086, 2023), subsection “Optimal swapping scheme”, state in the TeX source skr.tex, lines 1082–1084:

Our conjecture is that the statement is valid for all $n \geqslant 2$, but in this work we consider repeaters with up to eight segments only, and for such $n$ we have verified this statement directly.

The statement is the displayed equality at line 1080:

$\tilde{D}_n(N_1, \ldots, N_n) = D^\star_n(N_1, \ldots, N_n)$.

Here $N_i$ is the positive integer ready time of segment $i$. Swapping adjacent segments adds $|N_i-N_{i+1}|$ and replaces their ready times by $\max(N_i,N_{i+1})$. The source defines $\tilde D_n$ by repeatedly choosing an adjacent pair with least maximum, allowing arbitrary ties (skr.tex, lines 1033–1044). It defines $D^\star_n$ by minimizing the swap cost plus the recursively optimal cost of the shortened list over every adjacent first pair, with $D^\star_2(N_1,N_2)=|N_1-N_2|$ (lines 1049–1065).

Issue #25201 fixes the reading: for every list of $n\ge2$ positive integers and every complete greedy history, its accumulated cost equals $D^\star_n$. This is proved. The formal statement also includes zero ready times and terminal lists.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Optimization/SwapEarliestDephasingOptimality.result

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Geometry & algebraQuestion resolvedFirst preprint 2024The multiplicative Hornich–Hlawka inequality in inner product spacesThe multiplicative Hornich–Hlawka inequality holds in every real or complex inner product space.Proved

The questionNiculescu, A survey of the Hornich-Hlawka inequality, arXiv:2407.03278v1, §6, the paragraph applying Theorem 8 to the logarithm, states:

Why it mattersA Hornich–Hlawka inequality for the logarithmic transform would yield the product inequality by exponentiation.

The ideaTheir sum is the second elementary symmetric function comparison (hpair in scalarinequality).

Our contributionThe multiplicative Hornich–Hlawka inequality holds in every real or complex inner product space.

Verification details
Release source recordProved in Lean

arXiv

The multiplicative Hornich–Hlawka inequality in inner product spaces

D.-Ş. Marinescu, C. P. Niculescu, A survey of the Hornich-Hlawka inequality, arXiv:2407.03278v1, §6, the paragraph applying Theorem 8 to the logarithm, states:

Applying Theorem 8 to the function f (x) = ln(1 + x) one obtains the multiplicative analog of the inequality (2.1), (1 + |x|)(1 + |y|)(1 + |z|)(1 + |x + y + z|) ≥ (1 + |x + y|)(1 + |y + z|)(1 + |z + x|), for all x, y, z ∈ R. No extension of this remark to the case of inner product spaces is known.

The claim replaces absolute values by norms in every real or complex inner product space, with no finite-dimensionality or completeness assumption.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/MultiplicativeHornichHlawka.result

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Quantum informationQuestion resolvedFirst preprint 2023The one-bit bound of the double CGLMP gameFor every d >= 2, two parallel copies of the CGLMPd game have one-bit classical bound 12 when one bit is sent from Alice to Bob, and also when it is sent from Bob to Alice; the truncations to the inputs {00, 01, 11} x {00, 01, 11} and {00, 01, 11} x {00, 11} have local bounds 7 and 4.Proved

The questionVértesi (arXiv:2308.10771, quant-ph; npj Quantum Information 10, 79 (2024)) look for Bell-type inequalities that local hidden variables cannot saturate even with one bit of communication.

Why it mattersQ(CGLMP8^{⊗2}) > 12, so if the one-bit bound is 12 the double CGLMP game gives a four-input Bell-type inequality with one bit of communication that quantum correlations violate; the truncations do the same with three inputs.

The ideaIn one copy, inputs U, V of Alice with bits 0, 1 and W, Z of Bob with bits 0, 1: the four winning conditions would give αU ≤ βZ ≤ αV < βW ≤ αU, so in every rectangle {U, V} × {W, Z} on which Bob answers alike some cell is lost.

Our contributionFor every d >= 2, two parallel copies of the CGLMPd game have one-bit classical bound 12 when one bit is sent from Alice to Bob, and also when it is sent from Bob to Alice; the truncations to the inputs {00, 01, 11} x {00, 01, 11} and {00, 01, 11} x {00, 11} have local bounds 7 and 4.

Verification details
Release source recordProved in Lean

Open problem

The one-bit bound of the double CGLMP game

I. Márton, E. Bene, P. Diviánszky and T. Vértesi (arXiv:2308.10771, quant-ph; npj Quantum Information 10, 79 (2024)) look for Bell-type inequalities that local hidden variables cannot saturate even with one bit of communication. With CGLMP_d = P(A₀ ≥ B₀) + P(A₀ ≤ B₁) + P(A₁ < B₀) + P(A₁ ≥ B₁) and two copies played in parallel, they state:

The one-bit bound L1bit(CGLMP_d^{⊗2}) = 12 in the last column is verified by the branch-and-bound algorithm up to d = 10. We conjecture that this is the exact bound for any d ≥ 2.

For the truncations to the inputs {00, 01, 11} × {00, 01, 11} and {00, 01, 11} × {00, 11} they give "the conjectured local bound" 7 and 4, "which we verified up to d = 20".

Issue #11276 fixes the reading. Local behaviours follow eq. (P_LHV): the hidden variable has an arbitrary probability distribution on ℝ, and the responses are measurable conditional distributions. One-bit behaviours follow eq. (P_LHV1bit), with a measurable bit l(X, λ); both directions of communication are included, as the paper lists the game as bidirectional. Each bound is the greatest value of the Bell expression over its class.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/DoubleCglmpOneBitBound.result

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Quantum informationQuestion resolvedFirst preprint 2023Connected Hamming-weight gate patterns have an inverse-polynomial spectral gapConnected patterns of polynomially many Hamming-weight-preserving gates have an inverse-polynomial spectral gap at fixed weight.Proved

The questionKashefi, Trainability and Expressivity of Hamming-Weight Preserving Quantum Circuits for Machine Learning, Quantum 9, 1745 (2025), DOI 10.22331/q-2025-05-15-1745; arXiv:2309.15547v4, Conjecture conj:spectral-gap, Quantum.tex lines 1980–1985:

Why it mattersFor connected periodic structure ansätze, the source relates gradient variances to powers of these stochastic matrices.

The idea2. Connectivity and Poincaré estimate. Under connectedPattern pat, meaning that every gate has distinct endpoints and the qubit graph is connected, configurationconnected proves that the configuration graph is preconnected: any two configurations are reachable.

Our contributionConnected patterns of polynomially many Hamming-weight-preserving gates have an inverse-polynomial spectral gap at fixed weight.

Verification details
Release source recordProved in Lean

Open problem

Connected Hamming-weight gate patterns have an inverse-polynomial spectral gap

L. Monbroussou, E. Z. Mamon, J. Landman, A. B. Grilo, R. Kukla and E. Kashefi, Trainability and Expressivity of Hamming-Weight Preserving Quantum Circuits for Machine Learning, Quantum 9, 1745 (2025), DOI 10.22331/q-2025-05-15-1745; arXiv:2309.15547v4, Conjecture conj:spectral-gap, Quantum.tex lines 1980–1985:

Let $(U_{0,n}){n\geq2}$ be a sequence of connected patterns of RBS/FBS gates (\cref{def:connected-pattern}), where each $U{0,n}$ is such a pattern over $n$ qubits. If the number $J_n$ of gates in the pattern $U_{0,n}$ satisfies $J_n \in \mathcal{O}\left(\mathrm{poly}(n)\right)$, then for any fixed Hamming weight $k \in \mathcal{O}(1)$, the associated $d_{k,n} \times d_{k,n}$ stochastic matrix $T_n:=T^{(U_{0,n}^k)}$ (\cref{eq:stochastic-matrix-T-def}) satisfies

$\Delta_n \in \Omega\left(1/\mathrm{poly}(n)\right),,$

where $\Delta_n:=1-|\lambda_2(T_n)|$ denotes the spectral gap of the stochastic matrix $T_n$.

The versioned arXiv source archive contains Quantum.tex (2252 lines, md5 057c1aeb9e7f5c3b57445d20f924f4fc). Its line 1975 states: “The actual proof of \cref{conj:spectral-gap} is left for future work.”

Here $X_{n,k}={A\subseteq\mathrm{Fin},n:|A|=k}$ is the weight-$k$ basis, with $N=\binom nk$. A gate on distinct qubits $a,b$ exchanges their labels by $\tau_{ab}$ and has stochastic matrix $P_{ab}=(I+S_{ab})/2$, where $S_{ab}$ is the permutation matrix on $X_{n,k}$ induced by $\tau_{ab}$. A configuration containing both labels or neither is fixed. The pattern matrix is $T=P_J\cdots P_1$, with its first listed gate acting first. The source defines these matrices by squaring the RBS/FBS entries at angle $\pi/4$ (lines 158–192, 1455–1458 and 1486–1489), and its gap by the second-largest eigenvalue modulus (lines 1397–1405).

Issue #22140 fixes the nontrivial sectors $1\le k<n$ and the sum-zero complex eigenvector formulation stated below. This formulation is proved. The source correspondence is described in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/RandomCircuits/HammingWeightPatternSpectralGap.result

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Quantum informationQuestion resolvedFirst preprint 2023Novaes's strong-absorption spectral and Wigner moment conjecturesStrongAbsorptionMoments for strong-absorption moments.Proved

The questionThe journal version is Journal of Statistical Physics 190, 182 (2023), DOI 10.1007/s10955-023-03183-7.

The ideaThe finite-channel law is expanded around the product Laguerre law.

Our contributionStrongAbsorptionMoments for strong-absorption moments.

Verification details
Release source recordProved in Lean

Open problem

Novaes's strong-absorption spectral and Wigner moment conjectures

M. Novaes, Time delay statistics for chaotic cavities with absorption, arXiv:2305.18235v1, Section 5, p. 7, states: “We conjecture that $(1/M)\langle\operatorname{Tr}(Q^n)\rangle$ is given by” Eq. (54) and that $(1/M^n)\langle[\operatorname{Tr}(Q)]^n\rangle$ is given by Eq. (55), for every positive integer $n$:

$$ \frac1{\gamma^n}-\frac n{\gamma^{n+1}}+\frac{n^2}{\gamma^{n+2}} -\frac{(n+1)\bigl(n(5n-2)M^2+n(n+2)\bigr)}{6M^2\gamma^{n+3}} +O(\gamma^{-n-4}), $$

and

$$ \frac1{\gamma^n}-\frac n{\gamma^{n+1}} +\frac{n\bigl((n+1)M^2+(n-1)\bigr)}{2M^2\gamma^{n+2}} +O(\gamma^{-n-3}). $$

Here $M\geq1$ is the channel number, $n\geq1$ is the order, and the model is the finite-channel $\beta=2$ law with uniform absorption. The journal version is Journal of Statistical Physics 190, 182 (2023), DOI 10.1007/s10955-023-03183-7.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Every positive channel number | claim54 and claim55 quantify $M$ with $1\leq M$. | | Every positive integer order | Both claims quantify $n$ with $1\leq n$. | | Spectral observable $(1/M)\operatorname{Tr}(Q^n)$ | traceQPow is $(M:ℝ)^{-1}\sum_i(\gamma+p_1(i))^{-n}$. | | Wigner observable $(1/M^n)[\operatorname{Tr}(Q)]^n$ | tauW M γ p ^ n is the normalized Wigner delay power. | | Expectation under the stated law | Both integrals use law M γ. | | The displayed coefficients and remainder orders | The two absolute remainder inequalities in claim54 and claim55 use the literal coefficients and powers. | | The same external source conjectures both formulas | claim is the conjunction claim54 ∧ claim55. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Transport/StrongAbsorptionMoments.result

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Number theoryConjecture provedPosed in 2024The A290322 admissible residue sum conjectureUnits whose fifth cyclotomic value is again a unit sum to something whose five-adic valuation is one short of the modulus.Proved

The questionFor a modulus at least two, call a residue below it admissible when it is coprime to the modulus and the value of u^4 + u^3 + u^2 + u + 1 at it is coprime to the modulus as well.

Why it mattersDirect inspection on September 9, 2026 found the statement still labelled Conjecture.

The ideaWrite the modulus as a power of five times a factor prime to five, and let the exponent of five be at least one.

Our contributionUnits whose fifth cyclotomic value is again a unit sum to something whose five-adic valuation is one short of the modulus.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A290322 admissible residue sum conjecture

For a modulus at least two, call a residue below it admissible when it is coprime to the modulus and the value of u^4 + u^3 + u^2 + u + 1 at it is coprime to the modulus as well. Robert Israel conjectured on January 23, 2024 that the sum of the admissible residues is not divisible by the modulus whenever five divides the modulus.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CyclotomicFiveResidueSum.residue_sum_ne_zero

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Discrete structuresRecurrence provedPosed in 2024Mathar's linear recurrence for the A372991 factorial-quotient sequenceThe factorial-quotient sequence satisfies Mathar's linear recurrence and is integral.Proved

The questiona(n) = (2n)!/(a(n-1)a(n-2)), where a(0)=1, a(1) = 1.

Why it mattersThis is a first-tier recent OEIS conjecture from 2024.

The ideaDefine the sequence over the rationals, so division does not assume integrality.

Our contributionThe factorial-quotient sequence satisfies Mathar's linear recurrence and is integral.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Mathar's linear recurrence for the A372991 factorial-quotient sequence

OEIS A372991 NAME (Clark Kimberling, Jul 15 2024):

a(n) = (2n)!/(a(n-1)*a(n-2)), where a(0)=1, a(1) = 1.

FORMULA (for n ≥ 3):

Conjecture D-finite with recurrence a(n) -2*n*(2*n-1)*a(n-3)=0. - R. J. Mathar, Jul 16 2024

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/FactorialQuotientRecurrence.mathar_recurrence

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About this highlight
  • Contribution

    The factorial-quotient sequence satisfies Mathar's linear recurrence and is integral.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2024.

    Read source ↗
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Discrete structuresQuestion resolvedPosed in 2023Reversal and the Weighted Sum of a PartitionReversing a partition leaves the divisibility of its weighted sum by the total unchanged, because the weighted sum of a list and of its reverse add to one more than the length times the total.Proved

The questionThe (one-based) weighted sum of a sequence (y1,...,yk) is Sum{i=1..k} iyi.

The ideaFor a finite list y of naturals of length k, write W y for its one-based weighted sum.

Our contributionReversing a partition leaves the divisibility of its weighted sum by the total unchanged, because the weighted sum of a list and of its reverse add to one more than the length times the total.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Reversal and the Weighted Sum of a Partition

OEIS A362559 counts the integer partitions of n whose one-based weighted sum is divisible by n, and fixes the reading in its own comment:

The (one-based) weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i. This is also the sum of partial sums of the reverse.

The next comment line, at revision #18 of Apr 29 2023, is the question:

Conjecture: A partition of n has weighted sum divisible by n iff its reverse has weighted sum divisible by n.

The identical line stands on A362560, which counts the complementary partitions, at revision #6 of the same date. Both are by Gus Wiseman, Apr 24 2023, and both still carry the word Conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PartitionWeightedSumReversal.result

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Quantum informationQuestion resolvedFirst preprint 2024The ideal heralding rate of Fourier photon distillation decreases strictly to 1/4 from n = 5The ideal heralding rate of the photon distillation protocols strictly decreases from n = 5 and converges to one quarter.Proved

The questionRieffel, General protocols for the efficient distillation of indistinguishable photons, Phys.

Why it mattersThe heralding rate sets the resource cost of distillation: about n/hn(0) photons per distilled photon.

The idea1. A finite-interval form. With An = n²∫₀¹ (1 − u)u^{n−1}e^{n(u−1)} du and Tn = 2n!e^{−n}/n^{n−1}, pn = An + (−1)^{n−1}Tn and 0 ≤ Tn ≤ 2n e^{−n}.

Our contributionThe ideal heralding rate of the photon distillation protocols strictly decreases from n = 5 and converges to one quarter.

Verification details
Release source recordProved in Lean

Open problem

The ideal heralding rate of Fourier photon distillation decreases strictly to 1/4 from n = 5

J. Saied, J. Marshall, N. Anand and E. G. Rieffel, General protocols for the efficient distillation of indistinguishable photons, Phys. Rev. Applied 23, 034079 (2025), arXiv:2404.14217v4, distill indistinguishable photons with n-photon protocols built from an n × n unitary whose first row is constant (for example the discrete Fourier transform).

  • The ideal heralding rate. For such a unitary, h_n(0) = (−1/n)^{n−1}(n − 1)! Σ_{t=0}^{n−1} (n − t)(−n)^t/t! for every n > 2 (Theorem III.5, Eq. (III.9)), and lim h_n(0) = 1/4 (Theorem III.5(2)).
  • The conjecture. After a numerical evaluation for every n from 5 to 10⁵ and selected larger n up to 10⁷ (Figure 4): "Conjecture III.7. For n ≥ 5, the sequence h_n(0) monotonically decreases to 1/4."

The verbatim statements are in the literature note.

Issue #16743 fixes the reading: the conjecture is the statement about the sequence of Eq. (III.9), strict decrease for n ≥ 5 together with the limit 1/4. Its correction comment fixes the transcription of Eq. (III.9) ((n − 1)! multiplies the sum). The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/FockSpace/IdealHeraldingRateMonotonicity.result

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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresSharp bound provedFirst preprint 2024Stirling Lucky Displacement SpectrumEvery count of unlucky cars allowed by the bound is attained by a Stirling permutation.Proved

The questionColmenarejo, Dawkins, Elder, Harris, Harry, Kara, Smith and Tenner, Journal of Integer Sequences 27 (2024), Article 24.6.7, study Stirling permutations as parking preference lists.

Why it mattersTheorem 13 and Corollary 45 of the article already supply the two endpoints 2n - 1 and n; the theorem recorded here fills in every count between them and shows the fibres of Problem 49 are all nonempty.

The ideaV(n, j) = (j+1)(j+1) (j+2)(j+2) ⋯ n n j j (j-1)(j-1) ⋯ 1 1 ,

Our contributionEvery count of unlucky cars allowed by the bound is attained by a Stirling permutation.

Verification details
Release source recordProved in Lean

Open problem

Stirling Lucky Displacement Spectrum

Colmenarejo, Dawkins, Elder, Harris, Harry, Kara, Smith and Tenner, Journal of Integer Sequences 27 (2024), Article 24.6.7, study Stirling permutations as parking preference lists. Definition 4 reads: "A permutation of the multiset {1, 1, 2, 2, 3, 3, . . . , n, n} is a Stirling permutation of order n if every value j appearing between the two instances of i satisfies j > i." Definition 39 sets dis(w) = (d(1), …, d(2n)), where d(i) is the spot car i takes minus the spot w(i) it prefers, so d(i) = 0 exactly when car i is lucky.

Corollary 41 reads: "For every w ∈ Q_n, we have 1 ≤ |{i ∈ [2n] : d(i) = 0}| ≤ n", so the number of nonzero entries of dis(w) lies in [n, 2n - 1]. The article then states:

Problem 48. Determine if, for i ∈ [n, 2n − 1], there exists w ∈ Q_n such that dis(w) has exactly i nonzero entries. Equivalently, can we always find a Stirling permutation with i unlucky cars?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/StirlingLuckyDisplacementSpectrum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Every count of unlucky cars allowed by the bound is attained by a Stirling permutation.

    Read source ↗
  • Historical context

    First preprint year: 2024.

    Read source ↗
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    Source 1
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    Source 1
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    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Geometry & algebraCounterexample foundFirst preprint 2026Refutation of the peak-group dichotomyThe constructed finite peak group is neither solvable nor a Frobenius group.Refuted

The questionBahturin, Centrone and Pereira, Quantum regularity of finite dimensional semisimple algebras, arXiv:2608.21936v1, Conjecture 31, asks whether every peak group satisfies exactly one of solvability and the stated Frobenius-complement alternative.

The ideaThe abelian-product theorem transfers peakness.

Our contributionThe constructed finite peak group is neither solvable nor a Frobenius group.

Scope & next questionThe literature reading does not establish which results elsewhere depend on it; this settlement makes no claim about those dependencies. - Follow-up: trivial centre — open.

Verification details
Release source recordRefuted in Lean

Open problem

bahturin centrone pereira 2026 peak group refutation

Bahturin, Centrone and Pereira, Quantum regularity of finite dimensional semisimple algebras, arXiv:2608.21936v1, Conjecture 31, asks whether every peak group satisfies exactly one of solvability and the stated Frobenius-complement alternative. The preregistration is #19254.

The settling declaration is D5.S3.FiniteGroups.BahturinCentronePereiraPeakGroups.BCP31.result. It proves the negation of the literal claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A peak group has a greatest irreducible-character degree for divisibility. | BCP31.IsPeak and BCP31.degreeSet | The degree set is defined by simple finite-dimensional complex representations, and every degree divides the selected degree. | | A Frobenius complement is nontrivial, proper, and has trivial intersection with each conjugate outside it. | BCP31.IsFrobeniusWithComplement | The subgroup-map expression is the literal conjugation condition. | | The complement quotient contains the specified SL(2, 5) × T normal subgroup, with cyclic Sylows and excluded primes. | BCP31.ComplementCondition and BCP31.AlternativeTwo | The quantified finite group $T$ and all three structural conditions are retained. The cyclic-Sylow condition is written directly as a quantification over primes and Sylow subgroups; Mathlib’s IsZGroup states the corresponding property and is not on the proof dependency path. | | Exactly one of solvability and the Frobenius alternative holds for every finite peak group. | BCP31.claim | The result proves the full negation, including the exclusive-or conjunction. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/BahturinCentronePereiraPeakGroups.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Connected progenitor graphs give perfect adaptive fusion strategiesGoodenough, Landahl, Lee, Russo and Thompson (arXiv:2609.02559) conjecture that every [[n, 1, d]] graph code with a connected progenitor graph has a perfect adaptive fusion strategy: the physical fusions can be attempted in an order, with failure axes chosen from the earlier outcomes, so that a single successful physical fusion always gives a successful logical fusion.Proved

The questionA strategy is perfect when one successful fusion suffices.

Why it mattersThe paper characterizes perfect non-adaptive strategies and exhibits a connected five-qubit code with none, while an adaptive strategy exists for it.

Our contributionGoodenough, Landahl, Lee, Russo and Thompson (arXiv:2609.02559) conjecture that every [[n, 1, d]] graph code with a connected progenitor graph has a perfect adaptive fusion strategy: the physical fusions can be attempted in an order, with failure axes chosen from the earlier outcomes, so that a single successful physical fusion always gives a successful logical fusion.

Verification details
Release source recordProved in Lean

arXiv

Connected progenitor graphs give perfect adaptive fusion strategies

Goodenough, Landahl, Lee, Russo and Thompson, Optimal Fusion Strategies for Quantum Computation, arXiv:2609.02559v2, study logical fusions of $[[n,1,d]]$ codes: a failed physical fusion measures both qubits in its failure axis, and failures on a set $W$ cause a logical failure iff the substrings of the failure-axis string on $W$ contain a non-trivial logical operator (Proposition 3.1). A strategy is perfect when one successful fusion suffices. Section 6 reads

So, a natural conjecture and direction of future work is that all $[[n, 1, d]]$ codes with connected progenitor graphs have perfect adaptive strategies.

Issue #13256 fixes the reading. Pauli strings are taken up to phase as $(x,z)$ bits. The stabilizers of the progenitor graph state $G$ with encoding vertex $e$ are the products $\Gamma_U$ of canonical generators. A non-trivial logical operator is the restriction of some $\Gamma_U$ to $V\setminus{e}$ with $\Gamma_U$ not the identity at $e$. An adaptive strategy attempts the physical vertices in a fixed order and chooses each failure axis (X, Y or Z) from the earlier outcomes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/ConnectedProgenitorPerfectAdaptiveFusion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedConstant Blocks and Distinct Run SumsDistinct-sum constant blocks exist exactly when an ordering has distinct maximal run sums.Proved

The questionIts comment conjectures that the same number counts partitions having a permutation with all distinct run-sums.

Why it mattersThis is a first-tier OEIS conjecture by Gus Wiseman, authored March 26, 2025 and created March 28, 2025.

The ideaChoose a valid decomposition with the fewest blocks.

Our contributionDistinct-sum constant blocks exist exactly when an ordering has distinct maximal run sums.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Constant Blocks and Distinct Run Sums

OEIS A382427 counts integer partitions of n that can be partitioned into constant blocks with distinct sums. Its comment conjectures that the same number counts partitions having a permutation with all distinct run-sums. A381717, the complement, explicitly conjectures the pointwise equivalence. Runs are maximal consecutive constant subsequences, as defined by A382876.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/ConstantBlocksDistinctRunSums.card_constantBlocks_eq_distinctRunSums

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Commuting product encoders define exactly the commuting-reflection qubit state classCommuting product unitaries and commuting local reflections encode exactly the same qubit states.Proved

The questionGühne, The Entanglement Content of Quantum Measurement Bases, arXiv:2608.21185v1 (2026), Section VI, asks:

Why it mattersThe paper organizes locally encoded measurement bases by the allowed encoder structure.

The idea1. Local signs. Commutation of the nonzero tensor products makes the corresponding local products proportional.

Our contributionCommuting product unitaries and commuting local reflections encode exactly the same qubit states.

Scope & next questionThis remains open here; result only makes it equivalent to existence of a complete commuting product encoding for every three-qubit pure state.

Verification details
Release source recordProved in Lean

Open problem

Commuting product encoders define exactly the commuting-reflection qubit state class

J. Pauwels and O. Gühne, The Entanglement Content of Quantum Measurement Bases, arXiv:2608.21185v1 (2026), Section VI, asks:

Which other encoding classes define natural state classes, and what do these sets look like?

One of the proposed classes is:

commuting families whose local factors are not Hermitian, the smallest step beyond our theorem

The paragraph states:

For none of these is a characterization known, although $M_4$ now proves that the last class is strictly larger than commuting reflections.

These quotations are from the downloaded source main.tex, lines 1539–1547, md5 217f1dae7e17966b2c9016fa75ef756b. In the last quotation, “the last class” means tables of local Clifford unitaries. The source's local-encoding definition is at lines 328–340, and its encoder-structures definition is at lines 357–368; see also the literature note.

Issue #20897 fixes the following reading. For every $n\ge1$, put $N=2^n$ and let $\psi:{0,1}^n\to\mathbb C$ satisfy $\sum_x|\psi(x)|^2=1$. A commuting product encoding is a family of exactly $N$ operators $U_a=\bigotimes_{j=1}^n U_{a,j}$, with every $U_{a,j}$ a unitary $2\times2$ matrix, such that $U_aU_b=U_bU_a$ and $\langle U_a\psi,U_b\psi\rangle=\delta_{ab}$ for every pair of labels. A commuting-reflection encoding additionally requires $U_{a,j}^\dagger=U_{a,j}$ and $U_{a,j}^2=I$ for every label and site.

The claim is that $\psi$ admits a commuting product encoding if and only if it admits a commuting-reflection encoding. This equality of state classes is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CommutingEncodingReflectionClass.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedInteger existence and dyadic parity for OEIS A397594An integer source series exists for OEIS A397594, and every source series has Hanna's dyadic-neighbor odd support.Proved

The questionA(x) such that the sum of the first n coefficients in A(x/n)^n equals 4^(n-1) for n >= 1.

The ideaThe resulting map preserves constant coefficient one and improves coefficient agreement by one degree, using advancezero and advanceagree.

Our contributionAn integer source series exists for OEIS A397594, and every source series has Hanna's dyadic-neighbor odd support.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Integer existence and dyadic parity for OEIS A397594

Paul D. Hanna's OEIS A397594, revision 15, states:

G.f. A(x) such that the sum of the first n coefficients in A(x/n)^n equals 4^(n-1) for n >= 1.

Its FORMULA gives:

(1) 4^(n-1) = Sum_{k=0..n-1} [x^k] A(x/n)^n for n >= 1. (2) (4*n)^(n-1) = [x^(n-1)] A(x)^n / (1 - n*x) for n >= 1.

The conjecture is:

Conjecture: for n > 3, a(n) is odd iff n is of the form 2^k-1 or 2^k+1 (k > 1).

The series is $A(X)=\sum_{n\geq0}a(n)X^n\in\mathbb Z[[X]]$, with $a(0)=1$. Rational rescaling means coefficientwise mapping to $\mathbb Q[[X]]$ followed by $X\mapsto X/m$. Powers are ordinary formal products. There is no analytic convergence assumption.

Statement echo

| Source or binding convention | Lean counterpart | Scope | | --- | --- | --- | | $a(0)=1$, integer coefficients, offset zero | IsSourceSeries, its constant-coefficient conjunct and PowerSeries ℤ argument | Literal normalization. | | Formula (1), every $m\geq1$, indices $0,\ldots,m-1$ | IsSourceSeries, rational rescaling and Finset.range m | The source's $n,k$ are the binders $m,j$. | | Formula (2) | Local hname_row equivalence in result; the derivation below | The denominator is the constant-one formal inverse. | | $n>3$, odd iff $n=2^k-1$ or $n=2^k+1$, $k>1$ | Universal conjunct of claim, proved by result | Both directions and both strict bounds are retained. | | Integer-source existence and quantification over every source | Existential and universal conjuncts of claim | Required by preregistration #22490; existence excludes vacuity. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LinearRows/QuadrupledLinearExponentDyadicSupport.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2025The Ma–Weigert derivative-sign regions form a descending chainSigned derivative regions of polynomial log-functions form a descending chain with completely monotone intersection.Proved

The questionRourou Ma and Julian Weigert, Complete monotonicity of log-functions, arXiv:2505.04225v2 (Analysis and Mathematical Physics 15, 141, 2025), §4, Conjecture conj:shrinking, infiniteOrderDeriv.tex, lines 154–160 (TeX macros expanded and display mathematics set inline):

Why it mattersThe source studies complete monotonicity in polynomial log-function spaces and gives a finite coefficient-based description using inverse Laplace transforms and nonnegative polynomials.

The ideaNo infimum without an attained minimum is used.

Our contributionSigned derivative regions of polynomial log-functions form a descending chain with completely monotone intersection.

Verification details
Release source recordProved in Lean

Open problem

The Ma–Weigert derivative-sign regions form a descending chain

Rourou Ma and Julian Weigert, Complete monotonicity of log-functions, arXiv:2505.04225v2 (Analysis and Mathematical Physics 15, 141, 2025), §4, Conjecture conj:shrinking, infiniteOrderDeriv.tex, lines 154–160 (TeX macros expanded and display mathematics set inline):

For any $n\in \mathbb N$ the semialgebraic sets $D_k=\left{f\in \mathcal{F}{1,n}\middle|\forall x\in \mathbb R:\left(\frac{-\mathrm d}{\mathrm d x}\right)^kf(x)\geq 0\right}$ form a descending chain under inclusion, i.e. $D_k\supseteq D{k+1}$, which converges to the region $\mathrm{CM}{1,n}=\bigcap{k=1}^\infty D_k$.

The source defines $\mathcal F_{1,n}$ as the smooth functions $f(x)=p(\log x)/x$ on $x>0$, where $p$ is a real polynomial of degree at most $n$ (Intro.tex, lines 20–23). Although the conjecture's display writes $x\in\mathbb R$, the functions' domain and the preceding example's sign condition are $x>0$ (infiniteOrderDeriv.tex, line 150). Issue #21992 fixes this positive-domain reading.

Write $f_p(x)=p(\log x)/x$ and $D_{n,k}={p\in\mathbb R[y]:\operatorname{natDegree}(p)\le n,\ \forall x>0,\ (-1)^k f_p^{(k)}(x)\ge0}$. The module represents regions by their polynomial parameters. Its claim states, for every $n\in\mathbb N_0$:

  • $D_{n,k+1}\subseteq D_{n,k}$ for every natural $k\ge1$;
  • for each real polynomial $p$ with $\operatorname{natDegree}(p)\le n$, membership in every $D_{n,k}$ with $k\ge1$ is equivalent to $(-1)^j f_p^{(j)}(x)\ge0$ for every natural $j\ge0$ and every $x>0$.

Here $f_p^{(j)}$ denotes Lean's iteratedDeriv j (logFunction p). The zero polynomial has natural degree zero. The claim is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/LogFunctionMonotoneChain.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2025The A390088 exception setTwo certificates cover every argument above nine, leaving five exceptions below it.Proved

The questionFor each number, ask whether some base above one writes it with all digits prime.

Why it mattersDirect inspection on September 9, 2026 found it still labelled Conjecture, with the exception set stated verbatim and the sequence marked easy.

The ideaFor an even argument take the base that is half of two less than it, giving the two-digit representation with both digits two; for an odd argument take the base that is half of three less than it, giving digits three and two.

Our contributionTwo certificates cover every argument above nine, leaving five exceptions below it.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A390088 exception set

For each number, ask whether some base above one writes it with all digits prime. Felix Huber conjectured on October 29, 2025 that exactly five numbers admit no such base.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/PrimeDigitBaseClassification.a390088

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedWiseman A387118 total partition identityThe complemented partition predicates give equal totals.Proved

The questionThe formal count a n ranges over Nat.Partition n and filters the descending sort by the negation of ChoosableInitial.

Why it mattersThis is an externally stated OEIS conjecture.

The ideaFor every partition, WisemanA388712.sortedchoosableiffsuperdiagonal gives the equivalence between choosability of the descending parts and superdiagonality of the ascending parts.

Our contributionThe complemented partition predicates give equal totals.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Wiseman A387118 total partition identity

OEIS A387118 asks whether the number of integer partitions of $n$ without choosable initial intervals equals the number of non-superdiagonal reversed integer partitions. The conjecture is preregistered in #18602.

Statement echo

The formal count a n ranges over Nat.Partition n and filters the descending sort by the negation of ChoosableInitial. The comparison count nonSuperdiagonal n filters the ascending sort by the negation of Superdiagonal. The source predicates and sorting convention are imported from D5/S1/Words/Compositions/ChoosableInitialIntervals.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/WisemanA387118.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedWiseman A388713 maximum-refined partition identityThe partition predicates give equal counts with a fixed maximum.Proved

The questionThe formal count T n k filters Nat.Partition n by p.parts.sup = k and superdiagonality of the ascending sort.

Why it mattersThis is an externally stated OEIS conjecture.

The ideaFor every partition, the reverse of WisemanA388712.sortedchoosableiffsuperdiagonal gives superdiagonality of the ascending parts exactly when the descending parts are choosable.

Our contributionThe partition predicates give equal counts with a fixed maximum.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Wiseman A388713 maximum-refined partition identity

OEIS A388713 asks whether the number of superdiagonal reversed partitions of $n$ with maximum $k$ equals the number of integer partitions of $n$ with maximum $k$ and choosable initial intervals. The conjecture is preregistered in #18603.

Statement echo

The formal count T n k filters Nat.Partition n by p.parts.sup = k and superdiagonality of the ascending sort. The comparison count choosable n k uses the same maximum condition and choosability of the descending sort. The source predicates and sorting convention are imported from D5/S1/Words/Compositions/ChoosableInitialIntervals.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Compositions/WisemanA388713.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 1/4 · 20% weight

    The argument includes a structural formulation; this alone does not establish a new concept.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2022Symmetric GHZ resources for complete symmetric Boolean functionsFor 2 ≤ k ≤ n, every symmetric integer lift of C^kn requires a selected size j with k / 2 ≤ j and j ≤ n - k / 2.Proved

The questionGalvão, Quantum advantage in temporally flat measurement-based quantum computation, arXiv:2212.03668v4, Quantum 8, 1312 (2024), Section 4.2, Conjecture 2:

Why it mattersThe source gives numerical support for Conjecture 2 but no all-degree proof.

The ideaTherefore the certificate annihilates the constant and every selected parity layer outside the middle interval.

Our contributionFor 2 ≤ k ≤ n, every symmetric integer lift of C^kn requires a selected size j with k / 2 ≤ j and j ≤ n - k / 2.

Scope & next questionThe corollary's family reduction is a published consequence, but its extension is not separately checked or formalized here; the broader family remains open for this delivery. - Constant phase offsets — proved in prose.

Verification details
Release source recordProved in Lean

Open problem

Symmetric GHZ resources for complete symmetric Boolean functions

M. de Oliveira, L. S. Barbosa and E. F. Galvão, Quantum advantage in temporally flat measurement-based quantum computation, arXiv:2212.03668v4, Quantum 8, 1312 (2024), Section 4.2, Conjecture 2:

The number of qubits in a GHZ state necessary for the deterministic evaluation of a complete symmetric function Cᵏ, of constant degree k, with an n bit input string and a symmetric measurement assignment, scales as Ω(n^{⌊k/2⌋−1}).

The source note quotes Definitions 5, 15 and 16, Equations (70)–(72), and Corollary 1. Issue #16724 specifies the mathematical statement and source correspondence.

Statement echo

All Lean names in this dossier belong to D5.S3.Quantum.Contextuality.SymmetricGHZResourceLowerBound.

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | $n$ Boolean inputs | Fin n → Bool | Every input, including the all-false input | | Parity basis on $S$, Eq. (71) | Literal Odd-cardinality expression in layer | Integer zero or one according to odd subset weight | | All parity bases of each chosen size, Definition 16 | layer, T : Finset ℕ | Each selected size satisfies $1\le i\le n$ | | Equal coefficients within a size, Eq. (72) | a : ℕ → ℝ, ComputesSym | One real coefficient per size; no constant term | | Complete symmetric function, Definition 5 | csf | Degree-$k$ squarefree monomials counted modulo two | | Integer lift congruent modulo two, Definition 15 | ComputesSym | For every input there is an integer with the required real value and parity | | One qubit per selected parity basis | qubits | $\sum_{i\in T}\binom ni$, including selected zero-coefficient layers | | Constant degree and asymptotic lower bound | claim, result | For every $k\ge1$, positive $c$ and threshold $N$ depend only on $k$, uniformly in $n,T,a$ |

There are no additional public hypotheses. The integer exponent in claim is $\lfloor k/2\rfloor-1$, including $-1$ for $k=1$. Selected zero-coefficient layers can only increase the counted resource. The precise conventions are the literal expressions in #16724.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/SymmetricGHZResourceLowerBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2022Factorial Partition Signs: Conjecture 4.4 is FalseFactorial partition signs fail on an unbounded even progression.Refuted

The questionThe settlement is Refuted, specifically for the v1 statement.

Why it mattersThe source note is Library/RecurrenceComposition/gawronulas2022partitions.md.

The ideaThe module uses the literal source definitions.

Our contributionFactorial partition signs fail on an unbounded even progression.

Verification details
Release source recordRefuted in Lean

Open problem

gawron ulas 2022 factorial sign conjecture 4 4

Gawron and Ulas, “Signs behaviour of sums of weighted numbers of partitions”, arXiv:2202.05221v1, Conjecture 4.4, p. 11, asks for two eventual sign inequalities for $U_m={i!:1\le i\le m}$, $m\ge2$. With $c_A(i,n)$ counting unordered partitions of $n$ into exactly $i$ positive parts from $A$, put $S_{A,k}(n)=\sum_{i=0}^n(-1)^i i^k c_A(i,n)$.

The first assertion is $(-1)^{\lfloor(n+1)/2\rfloor}S_{U_m,k}(n)\ge0$ for every sufficiently large $n$ when $0\le k\le m-2$. The second assertion is $(-1)^nS_{U_m,k}(n)\ge0$ for every sufficiently large $n$ when $k\ge m-1$. Each threshold may depend on $m$ and $k$. The settlement is Refuted, specifically for the v1 statement. The preregistration is #19687.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Unordered positive-part partitions with exactly $i$ parts in $A$ | c filters Nat.Partition n by part membership and Multiset.card | Literal count; allowing zero in $A$ adds no partition | | Sum from $0$ through $n$ with weight $(-1)^i i^k$ | S uses Finset.range (n+1) and integer casts | Both endpoints; $0^0=1$ | | $U_m={i!:1\le i\le m}$ | U uses the existential factorial membership condition | Literal set | | Every $m\ge2$, every $k\le m-2$, eventual floor-exponent inequality | First conjunct of claim | Natural division is the floor; no least-period assertion | | Every $m\ge2$, every $k\ge m-1$, eventual alternating inequality | Second conjunct of claim | Independent existential threshold for each pair | | Negation of the whole conjecture | result : ¬ claim | Contradicts the second conjunct at $m=3,k=2$ |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Partitions/GawronUlasFactorialSignConjecture.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Two completely useless channels with a useful concatenation switch branchTwo qubit channels are entanglement breaking before and after each switch branch, while the minus switch branch of their concatenation is not entanglement breaking.Refuted

The questionGhosh, Improvement in quantum communication using quantum switch, Phys.

Why it mattersRelative orientation of two dephasing axes matters for concatenation.

The ideaThe six baseAprepare, baseBprepare, plusAprepare, plusBprepare, minusAprepare and minusBprepare identities give eight-term positive measure-and-prepare representations.

Our contributionTwo qubit channels are entanglement breaking before and after each switch branch, while the minus switch branch of their concatenation is not entanglement breaking.

Scope & next questionUniform identification of the witness and complete uselessness throughout the interval remain open here.

Verification details
Release source recordRefuted in Lean

Open problem

Two completely useless channels with a useful concatenation switch branch

A. Mitra, H. Badhani and S. Ghosh, Improvement in quantum communication using quantum switch, Phys. Scr. 98, 045101 (2023), arXiv:2108.14001v6, Section III.B, Conjecture 1:

Concatenation of two completely useless channels is always completely useless.

The preceding definition is:

A useless channel is completely useless if under both C+ and C− it remains useless.

Here uselessness concerns entanglement transfer and means entanglement breaking. Equation (8) gives the two completely positive switch branches with Kraus operators $(K_iK_j\pm K_jK_i)/2$. These branches need not preserve trace. The conjecture concerns normalized channels, together with entanglement breaking of their original map and both branches.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Qubit Kraus matrices and completeness | IsKrausChannel, complex matrices on Fin 2, and the sum of $K_i^\dagger K_i$ equal to the identity | Literal normalization; no extra hypothesis | | Kraus sum action | Existing MatrixMap.of_kraus K K | The sum of $K_iXK_i^\dagger$; no new forwarding definition | | Equation (8), plus branch | switchPlus | Anticommutator Kraus operators divided by two | | Equation (8), minus branch | switchMinus | Commutator Kraus operators divided by two | | Complete uselessness | CompletelyUseless | All three entanglement-breaking conjuncts, using the existing IsEB | | Concatenation, second after first | composeKraus L K | The pair $(i,j)$ carries $L_jK_i$ | | Every pair of such channels | claim | Every pair of finite index types, both Kraus families, both normalization hypotheses and both complete-uselessness hypotheses | | Refutation | result : ¬ claim | A single pair satisfying all premises and violating the conclusion suffices |

Finite index types are the Kraus-family convention of #16400. Fintype arguments are typeclass arguments, with no extra decidability assumption in the claim. The six public definitions state those conventions; the counterexample construction and all proof helpers are private.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/SwitchUselessConcatenationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryCounterexample foundFirst preprint 2026Moradi–Rampersad–Shallit Problem 1: shifted Fibonacci–Thue–Morse state complexityLiteral refutation of the complete Fibonacci-DFAO Problem 1 statement.Refuted

The questionDelaram Moradi, Narad Rampersad, and Jeffrey Shallit, Complexity of Linear Subsequences of Fibonacci-Automatic Sequences, arXiv:2603.21645v1, Section 4, Problem 1:

Why it mattersThe source's Theorem 14 gives an O(c) upper bound and asks whether a uniform linear lower bound also holds.

The ideaThe raw window retains canonical parity and occupation of the least Fibonacci digit at both endpoints.

Our contributionLiteral refutation of the complete Fibonacci-DFAO Problem 1 statement.

Verification details
Release source recordRefuted in Lean

Open problem

Moradi–Rampersad–Shallit Problem 1: shifted Fibonacci–Thue–Morse state complexity

Delaram Moradi, Narad Rampersad, and Jeffrey Shallit, Complexity of Linear Subsequences of Fibonacci-Automatic Sequences, arXiv:2603.21645v1, Section 4, Problem 1:

Prove the number of states in a minimal automaton generating (t(i+c))ᵢ≥₀ is Θ(c).

This dossier anchors only that complete statement. Put F₀ = 0, F₁ = 1 and Fₙ₊₂ = Fₙ₊₁ + Fₙ. For an MSD-first word x = x₀…xₗ₋₁ over {0,1}, set val(x) = Σⱼ<ₗ xⱼFₗ₊₁₋ⱼ, with val(ε) = 0. Valid words contain no adjacent ones; arbitrary leading-zero padding is valid. The value t(n) is the parity of the number of occupied digits in the canonical Zeckendorf representation of n.

For c ∈ ℕ, s(c) is the attained minimum number of live states of a finite deterministic partial MSD automaton with Boolean state outputs. Every counted state is reachable. The start state is counted and has a zero self-loop; a rejecting sink is omitted. For every binary word x, the output of the partial run is some(t(val(x)+c)) when x is valid and none otherwise. Thus every valid padded word is covered, including ε and all-zero words, which output t(c). Invalid execution is undefined, rather than an output chosen without a constraint. These conventions follow Sections 2.1 and 2.2 of the source.

The full eventual uniform proposition is

$$ \exists a,b\in\mathbb R,\quad a>0\land b>0\land \exists c_0\in\mathbb N,\quad \forall c\in\mathbb N,\quad c\ge c_0\Longrightarrow ac\le s(c)\le bc. $$

The exact Lean definition is D5/S1/Digit/ZeckendorfProblem1Refutation.Problem1. Requiring c₀ ≥ 1 gives the same eventual statement, since the threshold can be replaced by max(c₀,1). The settling theorem D5/S1/Digit/ZeckendorfProblem1Refutation.problem1_refuted : ¬ Problem1 negates the entire proposition with this original machine class.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/ZeckendorfProblem1Refutation.problem1_refuted

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture refutedFirst preprint 2026Dalfó-Fiol-Reyes Conjecture 2.4Conjecture 2.4's literal positive-integer quantifier conflicts with its degree-two graph convention at k equal to one.Refuted

The questionConjecture 2.4 of Dalfó, Fiol, and Reyes, A note on three-quarters circulant digraphs, section 2, quantifies over every integer k >= 1.

Why it mattersThe conjecture proposes one lattice and two-step construction for all k >= 1.

The ideaThreeQuarterCirculantConjecture defines SectorReach with the source's three sign sectors.

Our contributionConjecture 2.4's literal positive-integer quantifier conflicts with its degree-two graph convention at k equal to one.

Verification details
Release source recordRefuted in Lean

Open problem

Dalfó-Fiol-Reyes Conjecture 2.4

Conjecture 2.4 of Dalfó, Fiol, and Reyes, A note on three-quarters circulant digraphs, section 2, quantifies over every integer k >= 1. It specifies N=k^2+4k-2, steps a=1 and b=-k-4, and lattice columns (k+2,1-k) and (2,k). It says these data give a tile of a three-quarters digraph of sector diameter k. The paper defines the sector distance using exactly the three congruences m*a+n*b, -m*a+n*b, and m*a-n*b for m,n >= 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Covering/ThreeQuarterCirculantConjecture24Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2025Pronko's raising and lowering operators of the periodic Motzkin chainOn the periodic Motzkin spin-1 chain with N at least 2 sites, the operators Sigma^+ and Sigma^-, the sums of the products of local powers si^{ri} with r1 + ... + rN equal to 1, respectively -1, commute with the periodic Hamiltonian, send each ground state vm to a nonzero multiple of v(m+1), respectively v(m-1), and annihilate vN, respectively v(-N).Proved

The questionB 1017 (2025) 116963) studies the periodic Motzkin spin-1 chain H^periodic = Σ{i=1}^{N-1} Π{i,i+1} + Π{N,1} with Π = U + D + F and conjectures:

The ideaAn ordered product of operators acting on distinct sites has as entries the products of the local entries.

Our contributionOn the periodic Motzkin spin-1 chain with N at least 2 sites, the operators Sigma^+ and Sigma^-, the sums of the products of local powers si^{ri} with r1 + ... + rN equal to 1, respectively -1, commute with the periodic Hamiltonian, send each ground state vm to a nonzero multiple of v(m+1), respectively v(m-1), and annihilate vN, respectively v(-N).

Verification details
Release source recordProved in Lean

Open problem

Pronko's raising and lowering operators of the periodic Motzkin chain

Pronko (arXiv:2504.00835v3, Nucl. Phys. B 1017 (2025) 116963) studies the periodic Motzkin spin-1 chain H^periodic = Σ_{i=1}^{N-1} Π_{i,i+1} + Π_{N,1} with Π = U + D + F and conjectures:

There exist raising and lowering operators satisfying \eqref{Sigmavec} and \eqref{SpmH}, and they are given by $\Sigma^\pm=\sum_{r_1,\dots,r_N\in{-2,-1,0,1,2},\ r_1+\dots+r_N=\pm 1} s_1^{r_1}\cdots s_N^{r_N}$

where \eqref{Sigmavec} asks Σ^± v_{S^z} = c_±(S^z) v_{S^z ± 1} with c_±(S^z) ≠ 0 for S^z ≠ ±N and Σ^± v_{±N} = 0, and \eqref{SpmH} asks [Σ^±, H^periodic] = 0. Issue #10569 fixes the readings: the ground state v_m is the sum of the basis words of height m (u, f, d count +1, 0, −1), as in Conjecture 1; Π_{i,j} acts with its first factor on site i; N ≥ 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PronkoMotzkinRaisingLowering.result

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Discrete structuresConjecture refutedSource 2025Somer-Krizek uniform-subsequence conjectureA third-order integer recurrence refutes the proposed higher-order uniform-subsequence law.Refuted

The questionSomer and Krizek, Conjecture 4.2, printed page 8, proposes the following.

Why it mattersThe paper proves the corresponding statement for second-order recurrences and asks whether it extends to all higher orders.

The ideaTake k=3, coefficients (0,0,-1), and initial values (0,1,2), so w(n+3)=-w(n) for every n.

Our contributionA third-order integer recurrence refutes the proposed higher-order uniform-subsequence law.

Verification details
Release source recordRefuted in Lean

Open problem

Somer-Krizek uniform-subsequence conjecture

Somer and Krizek, Conjecture 4.2, printed page 8, proposes the following. For every k >= 2, let an integer sequence satisfy

w(n+k) = a1*w(n+k-1) + ... + ak*w(n)

for every n >= 0. Let p be prime, e >= 1, and E > 0. Assume that gcd(ak,p)=1, that the least positive period modulo p^e is exactly p^e E, and that every residue modulo p^e occurs exactly E times in that least period. For every positive g coprime to p and every nonnegative s, put r=E/gcd(g,E). The conjecture asserts that every residue occurs exactly r times among w(s+ng) for 0 <= n < p^e r.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/SomerUniformSubsequenceRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2025.

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Discrete structuresQuestion resolvedFirst preprint 2025Dense Independent Sets of the Square Grid Contain a Full CrossIndependent square-grid sets above density three eighths with correction four divided by the side length contain a full four-neighbour cross.Proved

Why it mattersBy the paper's remark after Problem 29, this determines the odd independence density of the infinite square grid: the paper's lower bound 3/8 is the exact value.

The ideaFor every 3 × 3 window M with no adjacent ones whose centre does not have all four neighbours in the set, 8w(M) + H(L) − H(R) + V(T) − V(B) ≤ 27, where w counts ones, L, R are the left and right 3 × 2 strips, T, B the top and bottom 2 × 3 strips, and H, V are fixed potentials with four nonzero values each (62 admissible windows, 12 with equality).

Our contributionIndependent square-grid sets above density three eighths with correction four divided by the side length contain a full four-neighbour cross.

Verification details
Release source recordProved in Lean

arXiv

Dense Independent Sets of the Square Grid Contain a Full Cross

Y. Caro, M. Petruševski, R. Škrekovski and Zs. Tuza, The odd independence number of graphs, II: Finite and infinite grids and chessboard graphs, arXiv:2510.01897v1, Problem 29: does there exist a sequence ε_n with ε_n → 0 such that any (3/8 + ε_n)n² independent vertices of P_n □ P_n contain all four neighbours of some vertex?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/OddIndependence/OddGrid.result

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Analysis & dynamicsQuestion resolvedFirst preprint 2025The Dzhenzher–Sakbaev random semigroup SLLN holds on every Banach spaceProducts of bounded iid operator exponentials converge almost surely in the strong operator topology.Proved

The questionSakbaev, The Strong Law of Large Numbers for random semigroups on uniformly smooth Banach spaces, arXiv:2507.07658v3 (Advances in Operator Theory 11 (2026), article 4), state Conjecture 2.6 in mains/sllnsmooth.tex, lines 224–229:

Why it mattersConjecture 2.6 asks whether the strong-operator conclusion extends to every Banach space under the same bounded i.i.d. semigroup assumptions.

The ideaIn the nontrivial case it applies deterministicuniform on the full-measure event supplied by stronglawae and the simultaneous bound aealliff.mpr ha; the operator inequality is converted to the required vector inequality using u.leopNorm x.

Our contributionProducts of bounded iid operator exponentials converge almost surely in the strong operator topology.

Verification details
Release source recordProved in Lean

Open problem

The Dzhenzher–Sakbaev random semigroup SLLN holds on every Banach space

S. V. Dzhenzher and V. Zh. Sakbaev, The Strong Law of Large Numbers for random semigroups on uniformly smooth Banach spaces, arXiv:2507.07658v3 (Advances in Operator Theory 11 (2026), article 4), state Conjecture 2.6 in mains/slln_smooth.tex, lines 224–229:

This leads us to the following conjecture. Let $\mathcal{X}$ be a Banach space. Then under the assumptions of Theorem~\ref{t:slln-sot}, $e^{A_1t/n} \ldots e^{A_nt/n}$ converges a.s. in SOT of $\norm{\cdot}_\mathcal{X}$ to $e^{\Expect At}$ uniformly for $t>0$ in any segment.

The formal declaration randomProduct multiplies, in increasing index order, the factors $\operatorname{exp}(((t/n : \mathbb R):\mathbb K)\mathbin{\cdot}a_i(\omega))$ for $i\in\operatorname{Fin}(n)$. The declaration claim quantifies over a scalar type $\mathbb K$ with RCLike, a normed additive commutative group $e$ with a NormedSpace and CompleteSpace structure, a measurable type $\alpha$, a probability measure $\mu$, and a sequence $a : \mathbb N\to\alpha\to(e\toL[\mathbb K]e)$. It assumes AEStronglyMeasurable for every $a_i$, pairwise IndepFun, IdentDistrib (a_i) (a_0) for every $i$, and a real $m$ such that $|a_i(\omega)|\le m$ almost everywhere for every $i$. For every $x:e$ and $t\ge0$, almost every $\omega$ satisfies: for every $\varepsilon>0$, eventually in $n$ and for every $s\in[0,t]$,

$\left|\left(\operatorname{randomProduct}(a,n,s,\omega)-\operatorname{exp}\left(((s:\mathbb R):\mathbb K)\mathbin{\cdot}\int z,,a_0(z),d\mu\right)\right)x\right|<\varepsilon$.

The operator measurable space used by claim is the Borel space on $e\toL[\mathbb K]e$, and its algebra structure is NormedAlgebra.restrictScalars ℝ \mathbb K.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Probability/RandomExponentialProductSLLN.result

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Quantum informationQuestion resolvedFirst preprint 2026The Choi spectrum of unital product-unitary-equivariant mapsFor all positive n₁ and n₂, the Choi matrix of the unital map that acts on the four isotypic components of n₁n₂ × n₁n₂ matrices with weights 1, λ₀₁, λ₁₀ and λ₁₁ has its eigenvalues among four values given by explicit formulas, taken with multiplicities 1, n₁² − 1, n₂² − 1 and (n₁² − 1)(n₂² − 1); coinciding values add their multiplicities.Proved

The questionThe verbatim statements are in the literature note.

Our contributionFor all positive n₁ and n₂, the Choi matrix of the unital map that acts on the four isotypic components of n₁n₂ × n₁n₂ matrices with weights 1, λ₀₁, λ₁₀ and λ₁₁ has its eigenvalues among four values given by explicit formulas, taken with multiplicities 1, n₁² − 1, n₂² − 1 and (n₁² − 1)(n₂² − 1); coinciding values add their multiplicities.

Verification details
Release source recordProved in Lean

Open problem

The Choi spectrum of unital product-unitary-equivariant maps

A. García-Velo and A. Ibort, Schwarz maps with symmetry, arXiv:2601.02282v1, classify the unital, Hermiticity-preserving maps on $M_{n_1}(\mathbb C)\otimes M_{n_2}(\mathbb C)$ that commute with conjugation by $U\otimes V$: they are $\Phi=D_1\otimes D_2+\lambda_{01}D_1\otimes Q_2+\lambda_{10}Q_1\otimes D_2+\lambda_{11}Q_1\otimes Q_2$ with real weights, where $D_i(X)=\mathrm{tr}(X),I/n_i$ and $Q_i=\mathrm{id}-D_i$. Lemma V.8 gives the eigenvalues of the Choi matrix $C_\Phi=\sum_{p,q}E_{pq}\otimes\Phi(E_{pq})$ for $n_1,n_2\in{2,3}$, and Remark V.2 conjectures the same four values, with multiplicities $1$, $n_1^2-1$, $n_2^2-1$ and $(n_1^2-1)(n_2^2-1)$, in every dimension. The verbatim statements are in the literature note.

Issue #14200 reads the conjecture as an identity of characteristic polynomials, which records the eigenvalues with their algebraic multiplicities, for $n_1,n_2\ge1$ and complex weights.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ProductUnitaryChoiSpectrum.result

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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Nilforoushan-Parvaresh Conjecture 8.1Exact k-deck images have the full weighted-Lyndon asymptotic exponent.Proved

The questionConjecture 8.1 of Arman Nilforoushan and Farzad Parvaresh, Improved upper bound on the number of distinct k-decks for any k and alphabet size by counting the independent parameters, arXiv:2609.23106v1, states:

Why it mattersThe source proves the general upper bound but leaves the matching lower bound open outside k=2 and the binary k=3 case.

The ideaThe route has 32 connected owners in nine natural subdomains.

Our contributionExact k-deck images have the full weighted-Lyndon asymptotic exponent.

Verification details
Release source recordProved in Lean

Open problem

Nilforoushan-Parvaresh Conjecture 8.1

Conjecture 8.1 of Arman Nilforoushan and Farzad Parvaresh, Improved upper bound on the number of distinct k-decks for any k and alphabet size by counting the independent parameters, arXiv:2609.23106v1, states:

D_{q,k}(n) = Theta(n^{E_q(k)}) for every fixed q >= 2, k >= 1.

The paper defines D_{q,k}(n) as the number of distinct exact k-decks of length-n words over a q-letter alphabet and defines E_q(k) = sum_{j=1}^k j L_q(j) - 1, where L_q(j) is the number of actual length-j Lyndon words over that alphabet. Preregistration issue #9512 records this complete target, not one of the cases already proved in the source.

The formal target quantifies over an arbitrary finite linearly ordered type A, assumes exactly 2 <= Fintype.card A, takes every natural k with 1 <= k, and proves Real.IsTheta at Filter.atTop for the real-valued cardinality of exactKDeckImage A k n. Its exponent is weightedLyndonExponent A k - 1, with weightedLyndonExponent A k = sum_{i=0}^k i * actualLyndonCount A i and actualLyndonCount A i the cardinality of the subtype of actual Lyndon words of length i. The zero term vanishes, so this is the source exponent. There is no free proxy parameter and no extra hypothesis on n or on k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Complexity/ExactDecks/Asymptotics/PositivePairExactDeckAsymptotics.actual_exactKDeckImage_weightedLyndon_isTheta

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Further frontiersQuestion resolvedFirst preprint 2025Altizio's geometric-sum disk classificationThe geometric-sum bound on the closed disk centered at two holds exactly in degrees five, six and seven.Proved

The questionDavid Altizio, Zeros of Stern polynomials in the complex plane, arXiv:2511.03847v1, Section 4.3, p. 20, Remark 4, states:

Why it mattersPreregistration 20656 selects the degree classification in Remark 4.

The ideais nonnegative for degrees five, six and seven by the certificates below.

Our contributionThe geometric-sum bound on the closed disk centered at two holds exactly in degrees five, six and seven.

Verification details
Release source recordProved in Lean

Open problem

Altizio's geometric-sum disk classification

David Altizio, Zeros of Stern polynomials in the complex plane, arXiv:2511.03847v1, Section 4.3, p. 20, Remark 4, states:

One might suspect that |(z)n| ≥ n|z| more generally whenever z ∈ B. However, numerical evidence suggests that the value min{z∈B} |(z)_n/z| approaches 2π as n → ∞. It seems that this inequality is true only for n = 5, 6, and 7, but we do not have a proof.

Section 2 defines $(z)_n=1+z+\cdots+z^{n-1}$ for positive integers $n$; Section 3 defines $B={z\in\mathbb C:|z-2|\le1}$. The settled statement is

$$ \forall n\in\mathbb N,\quad n>0\Longrightarrow \left[\left(\forall z\in B,\ n|z|\le\left|\sum_{k=0}^{n-1}z^k\right|\right) \Longleftrightarrow n\in{5,6,7}\right]. $$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive integer $n$ | Natural $n$ with $0<n$ in claim | Exact positive range | | $(z)_n=1+z+\cdots+z^{n-1}$ | The literal sum over Finset.range n | Includes its value $n$ at $z=1$ | | $B={|z-2|\le1}$ | Metric.closedBall (2 : ℂ) 1 | The boundary is included | | $|(z)_n|\ge n|z|$ for every $z\in B$ | Natural-to-real cast, complex norms, universal disk quantifier | Same inequality direction | | Only $n=5,6,7$ | Biconditional with the three degree equalities | Sufficiency and necessity | | Suggested $2\pi$ limit | Outside claim | Open |

claim is the only public definition and reproduces the source expression literally. The private proofs use geom_sum_mul for that finite sum, Polynomial.dickson for the boundary recurrence, the frozen coeffL1 for coefficient mass, and the frozen quantitativeSeedRadius for the positive remainder parameter.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/AltizioGeometricSumDisk.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025A polynomial bound for binary-polynomial Collatz lengthEvery nonzero binary polynomial has Collatz length at most two plus the floor of one quarter of its squared degree.Proved

The questionThe natural degree agrees with polynomial degree on the nonzero domain.

Why it mattersThe source's Theorem 1.2 proves termination with an exponential bound.

The ideaA degree-preserving step removes one factor of each linear polynomial, and decreases this minimum by one.

Our contributionEvery nonzero binary polynomial has Collatz length at most two plus the floor of one quarter of its squared degree.

Verification details
Release source recordProved in Lean

Open problem

A polynomial bound for binary-polynomial Collatz length

Gallardo and Rahavandrainy, A variant of Collatz's Conjecture over Binary Polynomials, arXiv:2510.07530v1, §1, state: “However, several computations suggest that here also r_A is bounded by a polynomial in deg(A) (see Section 3).” The question is preregistered in #20129. For every nonzero $A\in\mathbb F_2[x]$, the theorem rA_quadratic proves $r_A\le 2+\lfloor\deg(A)^2/4\rfloor$. The theorem result proves the literal polynomial-bound prediction, using $p(X)=X^2+2\in\mathbb N[X]$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Nonzero binary polynomial $A$ | A : Polynomial (ZMod 2), hA : A ≠ 0 | Same coefficient ring and domain. | | $M_1=x^2+x+1$; removal of all linear factors | Existing M, oddPart | Reused at GallardoRahavandrainyBinaryCollatz. | | $A_0=A$, $A_{2k}=1+M_1A_{2k-1}$ for $k>0$ | evenTerm A k | Literal conditional expression. | | $A_{2k+1}$ | Existing oddSeq A k | The natural index is half the source's odd index, rounded down. | | Least positive $m$ with terminal pair $(x(x+1),1)$ for all $k\ge m$ | rA A | Natural infimum of precisely that set, plus one; the proof constructs a member. | | Bound by a polynomial in degree | claim, result | Existential natural-coefficient polynomial; no extra hypothesis. | | Explicit stronger estimate | rA_quadratic | $r_A\le2+\lfloor\deg(A)^2/4\rfloor$. |

The natural degree agrees with polynomial degree on the nonzero domain. Definitions are total at zero; the theorem excludes zero. The existential polynomial witness is an upper envelope of the sharper floor estimate.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/GallardoRahavandrainyCollatzLengthBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Kumar's remaining rectangular weak two-metric dimensionsFor 4 <= n < m < 2n - 2, the weak two-metric dimension of the rectangular Hamming graph is min(ceil(2(n+m)/3), 2n - 2).Proved

The questionThe rectangular k = 2 values below the threshold m = 2n − 2 remain to be determined.

The ideaConsequently the row and column degree sums must be at least two.

Our contributionFor 4 <= n < m < 2n - 2, the weak two-metric dimension of the rectangular Hamming graph is min(ceil(2(n+m)/3), 2n - 2).

Verification details
Release source recordProved in Lean

Open problem

Kumar's remaining rectangular weak two-metric dimensions

Aryan Kumar, arXiv:2609.32161v1, §7:

The rectangular k = 2 values below the threshold m = 2n − 2 remain to be determined.

The vertices of $K_n\square K_m$ are pairs in $[n]\times[m]$, with $d((i,j),(a,b))=[i\ne a]+[j\ne b]$. For a landmark set $S$, put $\Delta_S(x,y)=\sum_{w\in S}|d(x,w)-d(y,w)|$. A weak $k$-resolving set has $\Delta_S(x,y)\ge k$ for every distinct pair; $\operatorname{wdim}_k$ is its least cardinality.

The exact rectangular answer is

$$ \operatorname{wdim}_2(K_n\square K_m) =\min\left{\left\lceil\frac{2(n+m)}3\right\rceil,,2n-2\right} \qquad(4\le n<m<2n-2). $$

For natural-number division, the ceiling is $(2(n+m)+2)/3$. The Lean claim retains all three size conditions and both upper-bound terms.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/HammingWeakTwoMetricDimension.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025The unrestricted rank-only Jensen converse is falseA signed quadratic norm gives an all-input counterexample to the unrestricted rank-only Jensen converse.Refuted

The questionMátyás Barczy and Zsolt Páles, Jensen convex functions and doubly stochastic matrices, arXiv:2510.03715v1, Remark 2.2(i), ask:

Why it mattersThe question asks whether the dimension of the image of a stochastic matrix alone can force a functional inequality back to midpoint convexity.

The ideaThe rational nonsquare theorem supplies its field structure.

Our contributionA signed quadratic norm gives an all-input counterexample to the unrestricted rank-only Jensen converse.

Verification details
Release source recordRefuted in Lean

arXiv

The unrestricted rank-only Jensen converse is false

Mátyás Barczy and Zsolt Páles, Jensen convex functions and doubly stochastic matrices, arXiv:2510.03715v1, Remark 2.2(i), ask:

However, we do not know whether the assertion in part (i) of Theorem 2.1 holds for every doubly stochastic matrix having rank at least 2. We leave this question as an open problem.

The referenced assertion is Theorem 2.1(i), equation (2.1), with Jensen convexity defined in Definition 1.1. The complete target is the following closed universal assertion:

For every positive integer $n$, every nonempty open real interval $I$ (including unbounded intervals and $I=\mathbb R$), every arbitrary function $f:I\to\mathbb R$, and every real $n\times n$ matrix $P=(p_{ij})$ satisfying

$$ p_{ij}\geq0,\qquad \sum_jp_{ij}=1,\qquad \sum_ip_{ij}=1,\qquad \operatorname{rank}_{\mathbb R}P\geq2, $$

does

$$ f!\left(\frac1n\sum_{j=1}^n x_j\right) \leq\frac1n\sum_{i=1}^n f!\left(\sum_{j=1}^np_{ij}x_j\right) \qquad\text{for every }(x_1,\ldots,x_n)\in I^n $$

imply

$$ f!\left(\frac{u+v}{2}\right)\leq\frac{f(u)+f(v)}2 \qquad\text{for every }u,v\in I? $$

The rank condition replaces the original special column-span premise; it is not conjoined with it. There is no continuity, measurability, boundedness, rational-entry or rational-input assumption. The function is defined on the original interval itself, with no ambient-extension datum. Convexity of the interval puts each mean and each matrix row output in its domain.

The answer is negative. The exact result is the negation of this whole universal assertion:

$$ \neg\Bigl[\forall n\in\mathbb N_{>0};\forall I\subseteq\mathbb R \text{ nonempty open interval};\forall f:I\to\mathbb R; \forall P\in\mathbb R^{n\times n},\quad \bigl(P\text{ doubly stochastic}\ \land\ \operatorname{rank}{\mathbb R}P\geq2\bigr) \Longrightarrow \Bigl(\bigl[\forall x\in I^n,\ f(\tfrac1n\sum_jx_j) \leq\tfrac1n\sum_i f(\sum_jp{ij}x_j)\bigr] \Longrightarrow \bigl[\forall u,v\in I,\ f(\tfrac{u+v}2)\leq\tfrac{f(u)+f(v)}2\bigr]\Bigr)\Bigr]. $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/BarczyPalesJensenConverse.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Further frontiersConjecture refutedFirst preprint 2025Cao-Chen-Miller Conjecture 1: the multidimensional egg-drop boundA correct bounded egg-drop strategy separates all hidden points by fixed-length binary transcripts.Refuted

The questionCao, Chen, and Miller, Egg Drop Problems: They Are All They Are Cracked Up To Be!, arXiv:2511.18330v2, printed page 10, Conjecture 1:

Why it mattersAt d = 4, k = 5, and sides (5,5,5,5), the proposed budget is nine, but there are 625 possible hidden points and only 512 nine-bit words.

The ideaRepresent hidden coordinate xi by xi - 1 in Fin (N i) and physical query coordinates by Fin (N i + 1).

Our contributionA correct bounded egg-drop strategy separates all hidden points by fixed-length binary transcripts.

Verification details
Release source recordRefuted in Lean

Open problem

Cao-Chen-Miller Conjecture 1: the multidimensional egg-drop bound

Cao, Chen, and Miller, Egg Drop Problems: They Are All They Are Cracked Up To Be!, arXiv:2511.18330v2, printed page 10, Conjecture 1:

In a d-dimensional setting with sides N_1, N_2, ..., N_d and k eggs, the minimum number of drops required in the worst-case scenario under the same strategy satisfies

$$P_d(k) \le \left\lceil (k-d+1)(N_1+\cdots+N_d)^{1/(k-d+1)}\right\rceil,$$

for k >= d.

The subscripted names and inequality above transcribe the printed notation. Preregistration issue #8448 writes the positive integer domains explicitly. The hidden point ranges over the product of the integer intervals {1, ..., N_i}. A drop at a returns intact exactly when a_i < x_i for every coordinate; otherwise it breaks one egg. The objective is exact recovery.

The formal claim weakens success of the paper's named strategy to existence of any deterministic adaptive strategy with those outcomes and egg semantics:

forall (d k : Nat) (N : Fin d -> Nat),
  1 <= d -> d <= k -> (forall i, 0 < N i) ->
    exists s : EggStrategy N k (paperBound d k N), Correct s

Refuting this weaker existence assertion refutes the named-strategy assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Observer/Budget/CaoChenMillerEggDropRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Further frontiersCounterexample foundFirst preprint 2026A non-Mersenne exception to the Rule 30 / Rule 22 sign patternRefutes the Rule 30/22 Mersenne sign pattern at row 767.Refuted

The questionMartín-Ruiz, Symmetric Nonlinear Cellular Automata as Algebraic References for Rule 30, arXiv:2604.00165v3, Remark 3, states:

The ideaQuiescence gives a light-cone induction: all sites outside [-m,m] are false.

Our contributionRefutes the Rule 30/22 Mersenne sign pattern at row 767.

Verification details
Release source recordRefuted in Lean

Open problem

A non-Mersenne exception to the Rule 30 / Rule 22 sign pattern

E. Chan-López and A. Martín-Ruiz, Symmetric Nonlinear Cellular Automata as Algebraic References for Rule 30, arXiv:2604.00165v3, Remark 3, states:

A direct computation for m ≤ 256 shows that ϵ(m) ≤ 0 precisely at the Mersenne indices m = 2^k − 1, with ϵ = 0 for k ≤ 3 and ϵ < 0 for 4 ≤ k ≤ 8.

Section 9 asks:

Is the sign pattern of Remark 3 exact for all k?

The quantified sign-pattern interpretation is

claim := ∀ m : ℕ, 1 ≤ m → (eps m ≤ 0 ↔ ∃ k : ℕ, 1 ≤ k ∧ m = 2^k − 1).

Here row g 0 r = decide (r = 0) and row g (m+1) r = g (row g m (r−1)) (row g m r) (row g m (r+1)) on the full integer lattice. The Boolean algebraic normal forms are g30 a b c = a xor b xor c xor (b and c) (Proposition 1) and g22 a b c = a xor b xor c xor (a and b and c) (Eq. (1)). Definition 2 counts the full-row set {r : ℤ | row g m r = true} with Set.ncard. Equation (11) defines eps m as the integer difference of these cardinalities. This interprets the word “precisely” as a biconditional for every positive row; it does not identify that biconditional with the separate strict-negativity assertion at every Mersenne row.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Automata/RuleThirtyTwentyTwoMersenneSignRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Czerwińska–Franco–Merino: radial convexity and matrix monotonicityRadial convexity with respect to the Frobenius norm does not imply matrix monotonicity of the same degree: an explicit rational function separates the two properties in degree three.Refuted

The questionMerino, Towards a Characterization of Matrix Monotonicity via Radial Convexity, arXiv:2610.17393v1, §4, Conjecture 4.10.

Why it mattersThe conjecture proposes that metric growth along ordered intervals characterizes preservation of the matrix order.

The ideaTrace self-adjointness and differentiation of squared Frobenius distances along segments yield both radial inequalities.

Our contributionRadial convexity with respect to the Frobenius norm does not imply matrix monotonicity of the same degree: an explicit rational function separates the two properties in degree three.

Verification details
Release source recordRefuted in Lean

Open problem

Czerwińska–Franco–Merino: radial convexity and matrix monotonicity

M. M. Czerwińska, J. A. Franco and A. Merino, Towards a Characterization of Matrix Monotonicity via Radial Convexity, arXiv:2610.17393v1, §4, Conjecture 4.10. See czerwinska2026radial.

Let f : (0, ∞) → R. Then f is radially convex of order n (with respect to the Frobenius norm) if and only if f is matrix monotone of degree n.

arXiv:2610.17393v1, §4, directly after Conjecture 4.10, verbatim:

By Proposition 3.9, matrix monotonicity of degree n implies radial convexity of order n. Thus, it remains to prove the converse implication. The next theorem establishes the conjecture for n = 2.

The matrices are complex Hermitian positive definite, and the order is Loewner order. Radial convexity requires both endpoint distance inequalities for every ordered triple $A\preceq C\preceq B$. Matrix monotonicity allows either of the two uniform order directions. The universal assertion quantifies over every $n\ge1$ and every real function on the positive half-line.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/CzerwinskaFrancoMerinoCounterexample.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Refutation of the unrestricted strong distance-two period boundThe connected singleton refutes the unrestricted strong distance-two period bound.Refuted

The questionGregory Demo, David Lund and Oleksiy Al-saadi, On the Periodicity of k-distance Graphs, arXiv:2606.25493v1, Section 4, Conjecture 1, printed PDF page 11:

Why it mattersA universal period bound must include every graph admitted by its stated hypotheses.

The ideaIt is connected: every pair of vertices coincides and has a length-zero walk.

Our contributionThe connected singleton refutes the unrestricted strong distance-two period bound.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of the unrestricted strong distance-two period bound

Gregory Demo, David Lund and Oleksiy Al-saadi, On the Periodicity of k-distance Graphs, arXiv:2606.25493v1, Section 4, Conjecture 1, printed PDF page 11:

Conjecture 1. Any connected graph G=(V,E) has ρ²(G)≤|V|−3.

The exact target and integer interpretation are registered in #23003. The source note records the source qualification; the earlier metamour note records the known fixed-point example.

Definitions and statement echo

Let $V$ be a finite vertex set and $G$ a simple undirected graph on $V$. Define $T(G)$ on the same vertices by $u\sim_{T(G)}v\iff\operatorname{edist}_G(u,v)=2$. The extended distance is infinity for unreachable pairs, so $T$ is defined also on disconnected graphs and every iterate remains in its domain. Write $T^0(G)=G$. A positive return is an integer iteration count $t>0$ with an ordinary graph isomorphism $T^t(G)\cong G$. Define $\rho_2(G)=\min{t\in\mathbb N: t>0,\ T^t(G)\cong G}$ when that set is nonempty, and $\rho_2(G)=0$ otherwise. The source starts at $G_1^2=G$, so its return index is $f=t+1$ and its period $f-1=t$.

The full closed assertion is

$$\forall V\text{ finite},\ \forall G\text{ simple on }V,\quad G\text{ connected}\ \Longrightarrow \operatorname{Int.ofNat}(\rho_2(G))\leq \operatorname{Int.ofNat}(|V|)-3.$$

Both casts precede integer subtraction. There is no size lower bound, nontriviality requirement, edge requirement, periodicity hypothesis or restriction on intermediate graphs. Isomorphism need not preserve labels; eventual periodicity is a different assertion.

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Finite simple undirected graph | V : Type, Fintype V, SimpleGraph V in claim | All finite orders admitted by connectedness | | Exact geodesic distance two on the same vertices | distanceTwo | Extended distance also handles disconnected iterates | | Positive isomorphism return to the initial graph | PositiveReturn | Strictly positive count and ordinary graph isomorphism | | Least return, or zero if absent | strongPeriod | Nat.find on precisely that predicate, otherwise zero | | Only connectedness is assumed | claim | No added size hypothesis | | Integer upper bound | claim | Period and cardinality cast before subtracting three | | Negation of the entire assertion | result | Literally Not claim, with no free term or universe parameters |

All declaration names in this table belong to D5.S3.Combinatorics.Graph.Walks.DemoStrongPeriodBoundRefutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Walks/DemoStrongPeriodBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025A qutrit SIC fiducial that is not a Clifford-stabilizer stateThe special Clifford group of a qutrit is finite, and its eigenphase extension has only finitely many one-dimensional fixed spaces.Refuted

The questionGoldstein, Extremizing Measures of Magic on Pure States by Clifford-stabilizer States, arXiv:2512.19657v2, prove that states uniquely stabilized by a subgroup of the finite eigenphase-extended Clifford group extremize several magic measures, and state:

The ideaHence some fiducial is not a Clifford-stabilizer state.

Our contributionThe special Clifford group of a qutrit is finite, and its eigenphase extension has only finitely many one-dimensional fixed spaces.

Verification details
Release source recordRefuted in Lean

Open problem

A qutrit SIC fiducial that is not a Clifford-stabilizer state

M. Erew and M. Goldstein, Extremizing Measures of Magic on Pure States by Clifford-stabilizer States, arXiv:2512.19657v2, prove that states uniquely stabilized by a subgroup of the finite eigenphase-extended Clifford group extremize several magic measures, and state:

Conjecture 1 (Stabilizer nature of SIC fiducials). Every SIC-POVM fiducial state is a Clifford-stabilizer state.

The verbatim definitions are in the literature note. Issue #14173 reads it for every prime $d$ and every normalized single-qudit fiducial.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/ErewGoldsteinSICStabilizerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The extinction set of the wheel W_18 contains 4In the Mukwembi model of HIV infection on the wheel W18 with replacement parameter R = 4, every admissible initial state reaches the all-healthy state by time 25, so 4 belongs to the extinction set of W18 and Conjecture 1 of Espinosa-Garcia et al., which puts that set equal to {3} together with all R >= 17, is false.Refuted

The questionEspinosa-García, Figueroa, Fresán-Figueroa, Maldonado and Sánchez-Solís (arXiv:2608.00340v1) study Mukwembi's graph model of HIV infection.

The ideaAll 2^18 admissible initial states of W18 are simulated at once.

Our contributionIn the Mukwembi model of HIV infection on the wheel W18 with replacement parameter R = 4, every admissible initial state reaches the all-healthy state by time 25, so 4 belongs to the extinction set of W18 and Conjecture 1 of Espinosa-Garcia et al., which puts that set equal to {3} together with all R >= 17, is false.

Verification details
Release source recordRefuted in Lean

Open problem

The extinction set of the wheel W_18 contains 4

Espinosa-García, Figueroa, Fresán-Figueroa, Maldonado and Sánchez-Solís (arXiv:2608.00340v1) study Mukwembi's graph model of HIV infection. States give each vertex the value 0 (healthy), 1 (infected) or 2 (dead). With d_{t,I}(v) the number of infected neighbours of v and a positive replacement parameter R, a healthy vertex becomes infected when d_{t,I}(v) ≥ 1, an infected vertex dies, and a dead vertex is replaced by an infected one when d_{t,I}(v) ≥ R and by a healthy one otherwise. The extinction set 𝓔(G) consists of the R for which every admissible initial state (values in {0, 1}) reaches the all-healthy state. Conjecture 1 states:

The extinction sets of the wheel graphs satisfy $\mathcal{E}(W_n) = {3}\cup{R\in\mathbb{Z^+}:R\geq n-1}$ for every even $n\geq12$, and $\mathcal{E}(W_n) = {4}\cup{R\in\mathbb{Z^+}:R\geq n-1}$ for every odd $n\geq17$.

Issue #10263 fixes the reading: W_n = K₁ ∨ C_{n−1} on Fin n with hub 0 and the cycle 1, 2, …, n − 1, and 𝓔(W_n) as defined, quantified over all 2^n admissible initial states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/WheelHivExtinctionRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Further frontiersCounterexample foundFirst preprint 2008Fisk Question 10: coefficient-Toeplitz minorsA positive upper bivariate polynomial has a non-Hurwitz coefficient-Toeplitz minor.Refuted

The questionSteve Fisk, Questions about determinants and polynomials, arXiv:0808.1850v1, PDF p. 4, Question 10 and matrix (5), asks:

Why it mattersPositive upper bivariate polynomials need not have Hurwitz-stable coefficient-Toeplitz minors.

The ideaThe proof uses factorization.

Our contributionA positive upper bivariate polynomial has a non-Hurwitz coefficient-Toeplitz minor.

Verification details
Release source recordRefuted in Lean

arXiv

Fisk Question 10: coefficient-Toeplitz minors

Steve Fisk, Questions about determinants and polynomials, arXiv:0808.1850v1, PDF p. 4, Question 10 and matrix (5), asks:

Suppose f∈P⁺₂, and write f = ∑ f_i(x)y^i. Form the infinite matrix

Show that the resulting matrix is totally stable. That is, every minor is a stable polynomial.

The matrix has $T_f(r,c)=f_{c-r}$ for $r\le c$ and zero otherwise. The definitions on PDF p. 1 require upper-half-plane nonvanishing and positive coefficients on the support, with Hurwitz stability meaning all roots have nonpositive real part. The related totally stable matrix conjecture is Conjecture 21.84, Section 21.10, printed p. 648 of arXiv:math/0612833v2. Its input class is defined through real substitution, homogeneous positivity and coefficient positivity; no equality of the two source classes is assumed.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Real bivariate polynomial, with outer variable $y$ | Polynomial (Polynomial ℝ) | Literal carrier; coefficient variable is $x$. | | $U_2$: nonzero at every pair in the open upper half plane | Upper | Both complex variables quantified; nested Polynomial.eval₂. | | $P_2^+$: upper and positive coefficients | PositiveUpper | Strict positivity on support; absent coefficients are zero. | | Matrix (5) | toeplitz | Zero-based upper-triangular coefficient shifts. | | Every minor | minor, claim | All natural sizes and strictly increasing row and column maps; no sign reversal. | | All roots in the closed left half plane | HurwitzStable | Roots after mapping coefficients to ℂ. | | Universal assertion is false | result | Negation of the literal claim. |

The size-zero determinant is one. Structural zero minors are admitted by the empty root multiset of the zero polynomial. The exhibited minor is nonzero, so either convention about structural zero stability gives the same refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/FiskToeplitzRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    First preprint year: 2008. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresCounterexample foundFirst preprint 2026Weak-product closure of the edge-complexity equality classK3 times K3 refutes unrestricted weak-product closure of energy equality.Refuted

The questionTian, Edge complexity of graphs, arXiv:2607.15598v1, Section 3, Remark after Corollary 3.10:

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionK3 times K3 refutes unrestricted weak-product closure of energy equality.

Verification details
Release source recordRefuted in Lean

Open problem

Weak-product closure of the edge-complexity equality class

V. Gupta, A. Iosevich, J. Iosevich, B. Song and H. Tian, Edge complexity of graphs, arXiv:2607.15598v1, Section 3, Remark after Corollary 3.10:

Remark. The coprimality hypothesis is not merely a technicality in the present argument: it is what permits the product labeling to be viewed as a cyclic labeling. Proposition 3.7 does not prove closure under arbitrary weak products. Consequently, unrestricted closure of the equality class under the weak product has not been established. Whether such closure holds is a natural open question.

The assertion is that any two finite simple graphs with positive size satisfying $\operatorname{FR}_{\min}(G)=E(G)/\sqrt{2s(G)}$ have a weak product satisfying the same equality. Definition 3.1 gives adjacency precisely when both coordinate pairs are adjacent. The Fourier transform has coefficient $1/N$, the numerator of $\operatorname{FR}$ is entrywise $\ell^1$, and the denominator is Frobenius; $E$ is the sum of the absolute adjacency eigenvalues.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/EdgeComplexityWeakProductRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationConjecture refutedSource 2025The Han–Wang–Yu gluing spectral hull conjecture is falseGluing two stochastic matrices yields an eigenvalue outside their spectral hull.Refuted

The questionYu, PageRank of Gluing Networks and Corresponding Markov Chains, Mathematics 13 (2025), 2080, Conjecture 1, §4.3, page 17:

The ideaThe proof uses spectral estimates and convexity.

Our contributionGluing two stochastic matrices yields an eigenvalue outside their spectral hull.

Scope & next questionAn argument using Conjecture 1 as a universal premise cannot retain that premise; determining which additional hypotheses recover such arguments remains open.

Verification details
Release source recordRefuted in Lean

Open problem

The Han–Wang–Yu gluing spectral hull conjecture is false

X. B. Han, S. Wang and C. Yu, PageRank of Gluing Networks and Corresponding Markov Chains, Mathematics 13 (2025), 2080, Conjecture 1, §4.3, page 17:

Glue Markov chains M_i = (S_i, A_i) to a Markov chain M = (S, A), then all the eigenvalues of A lie in the convex hull of the eigenvalues of all A_i on the complex plane.

The gluing is Definition 8, page 10. The Library note records its four entrywise cases. Preregistration: #19143.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finitely many Markov chains with selected states | RootedChain; families indexed by Fin k | States are Fin (n + 1), selected state 0; the stochastic field is Matrix.colStochastic | | Transition probability from $w$ to $v$ | Matrix entry A v w | Column convention of Definition 8 | | Common root and otherwise disjoint states | Option (Σ i, Fin (M i).n) | none is the common root, some ⟨i,v⟩ corresponds to v.succ | | Definition 8's four transition cases | glue | Root-root weighted sum, private-root weighted entry, root-private unchanged entry, same-chain unchanged entry and cross-chain zero | | $0<\theta_i<1$, $\sum_i\theta_i=1$ | Both antecedents of claim | No weight hypothesis is dropped | | All eigenvalues lie in the input spectral convex hull on the complex plane | claim | Complex spectrum after entrywise algebraMap ℝ ℂ, with convexHull ℝ in $\mathbb C$ | | Refutation | result : ¬ claim | Two identical 3-state chains give an admissible 5-state counterexample |

The state and spectrum expressions are expanded directly rather than introduced as forwarding definitions. Using labelled finite state sets loses no counterexample: these labelled matrices are themselves instances of the source's finite chains. The finite cardinality encoding and column convention introduce no additional hypothesis into the displayed conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/HanWangYuGluingSpectrumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2025.

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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025Hiai's spectral geometric Log-Euclidean comparison failsA positive definite matrix pair refutes Hiai's spectral geometric Log-Euclidean comparison.Refuted

The questionFumio Hiai, Log-majorizations between quasi-geometric type means for matrices, Linear Algebra Appl. 735 (2026), 123–174; arXiv:2510.04691v3.

Why it mattersThe small-power limit of the spectral geometric mean is the Log-Euclidean mean.

The ideaPositivity identifies the square root uniquely.

Our contributionA positive definite matrix pair refutes Hiai's spectral geometric Log-Euclidean comparison.

Verification details
Release source recordRefuted in Lean

Open problem

Hiai's spectral geometric Log-Euclidean comparison fails

Fumio Hiai, Log-majorizations between quasi-geometric type means for matrices, Linear Algebra Appl. 735 (2026), 123–174; arXiv:2510.04691v3.

§2, equation (2.3), verbatim:

Fα(A, B) := (A^{−1}#B)^α A(A^{−1}#B)^α.

§2, equation (2.4), verbatim:

SGα,p(A, B) := Fα(A^p, B^p)^{1/p}.

§2, Log-Euclidean and weighted geometric mean definitions, verbatim:

LEα(A, B) := exp((1 − α) log A + α log B)

A#αB := A^{1/2}(A^{−1/2}BA^{−1/2})^α A^{1/2}

Here # = #_{1/2}.

Problem 3.20, verbatim:

Problem 3.20. The case remaining open is whether SGα,p ≺log LEα for α > 1. Note that this is affirmative if SGα,p ≺log SGα,q if α > 1 and p ≥ q (see Problem 3.12).

The summary table after Problem 3.20 has row “SGα,p ≺log LEα”, with entries “none” for 0 < α < 1 and “?” for α > 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/SpectralGeometricLogEuclideanRefutation.result

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Quantum informationCounterexample foundFirst preprint 2026Same-phase periodic SSH ground states can have positive averaged excess pseudo entropySame-phase half-filled ground states of a periodic three-cell SSH chain have positive averaged excess pseudo-entropy.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersThey report violations only for open chains at small sizes, attributed to edge modes, and present the periodic chain as violation-free.

The idea2. Ground states. For N = 3 and v ∈ {3/8, 5/8}, a rational orthogonal eigenbasis lk of h(v) gives H = Eg + Σk αk Bk† Bk with αk > 0, where Bk is a(lk)† for the negative modes and a(lk) for the positive ones.

Our contributionSame-phase half-filled ground states of a periodic three-cell SSH chain have positive averaged excess pseudo-entropy.

Verification details
Release source recordRefuted in Lean

arXiv

Same-phase periodic SSH ground states can have positive averaged excess pseudo entropy

P. K. Kharel, M. Limbu, N. Khatri, A. Khanal and K. Adhikari, Pseudo entropy and topological phases of matter, arXiv:2606.29311v1, test the conjectured sign of the averaged excess pseudo entropy in the Su–Schrieffer–Heeger model.

  • The quantity. For pure states ψ, φ with ⟨φ|ψ⟩ ≠ 0, T = |ψ⟩⟨φ|/⟨φ|ψ⟩, T_A = Tr_B T and S(T_A) = −Tr(T_A log T_A) (Eqs. (1)–(3)). The averaged excess is ∆S = Re S(T_A) − (S(ρ_A^ψ) + S(ρ_A^φ))/2 (Eq. (21)).
  • The conjecture. "In [25, 26], it was conjectured that the averaged excess entropy is non-positive, ∆S ≤ 0, when the states |ψ⟩ and |φ⟩ are in the same phase" (§IV, p. 5).
  • The assertion. "Under periodic boundary conditions, the conjecture is fully satisfied with no exceptions at all system sizes and across all three phases: topological, trivial, and crossing" (§VI, p. 9), for half-filled ground states of the periodic Hamiltonian of Eq. (15) at w = 1.

The verbatim statements are in the literature note.

Issue #17503 fixes the reading:

  • The Fock space of the 2N modes (1,A), (1,B), …, (N,A), (N,B) carries the Jordan–Wigner annihilation operators in this order.
  • A half-filled ground state is a nonzero N-particle vector of minimal energy in the N-particle sector.
  • A is the first L unit cells, 1 ≤ L < N, and Tr_B is the partial trace over the remaining modes.
  • The entropy of a matrix is −Σ λ log λ over its eigenvalues with multiplicity.
  • The assertion: for all such N, L, all v₁, v₂ in the same phase, and all such ground states, ∆S ≤ 0.

The assertion is refuted in this reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Fermionic/PeriodicSshExcessPseudoEntropyRefutation.result

Frozen module record

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Quantum informationConjecture refutedSource 2026The strengthened Marcus–de Oliveira conjecture has a three-dimensional counterexampleThe determinant of a rational orthogonal conjugate escapes the bounded region enclosed by the exterior boundary of a three-dimensional sigma-point graph.Refuted

The questionKovačec, A conjecture more precise and stronger than the one by Marcus and de Oliveira, DMUC Preprint 26-11, v1, Conjecture 7, p. 5:

Why it mattersThe transposition graph is a smaller geometric object than the convex hull of all permutation products.

Our contributionThe determinant of a rational orthogonal conjugate escapes the bounded region enclosed by the exterior boundary of a three-dimensional sigma-point graph.

Scope & next questionThe original MOC in unrestricted dimension remains open.

Verification details
Release source recordRefuted in Lean

Open problem

The strengthened Marcus–de Oliveira conjecture has a three-dimensional counterexample

A. Kovačec, A conjecture more precise and stronger than the one by Marcus and de Oliveira, DMUC Preprint 26-11, v1, Conjecture 7, p. 5:

Given any two n-multisets of complex numbers A, B such that a Jordan curve J and the bounded component J(A, B) are defined, there holds ∆(A, B) ⊆ closure(J(A, B)).

The source note gives the definitions and URL. Preregistration #17803 specifies the exact witness and the literal Lean conventions. The resolution is Refuted.

Statement echo

| Source clause | Lean counterpart | Fidelity and scope | | --- | --- | --- | | Any two $n$-multisets of complex numbers | claim quantifies over every natural n and a b : Fin n → ℂ | Functions retain multiplicities; no restriction on the spectra is added. | | $z_\sigma=\prod_i(a_i+b_{\sigma(i)})$ | sigmaPoint | The product ranges over all Fin n; permutations are Equiv.Perm (Fin n). | | $\Delta={\det(D_A+UD_BU^*):U\in U(n)}$ | detRange | Matrix.diagonal, Matrix.unitaryGroup, star and determinant give the literal expression. | | Permutations differing by one transposition | graph | Union of the real closed segments for sigma and sigma * Equiv.swap i j, with i ≠ j. | | Every positive prescribed distance admits an unobstructed segment | exterior | All real M > 0, a complex endpoint at distance M, and disjointness from the whole graph. | | Exterior boundary is a Jordan curve | IsJordanCurve applied to the exterior frontier | The entire boundary is the range of a continuous injective map from Circle. | | The bounded component of the complement of the boundary | connectedComponentIn at each point outside that frontier, with Bornology.IsBounded | Every bounded component is quantified. Under Jordan separation this is the source's unique bounded component; no Jordan curve theorem is assumed in Lean. | | $\Delta\subseteq\overline{J(A,B)}$ | The inclusion in claim; result proves its negation | One bounded component containing zero suffices to refute the universal claim. |

The endpoint parameter and boundedness hypothesis identify the source's component; they do not restrict the counterexample's determinant range. No source clause is dropped.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/KovacecStrengthenedMarcusDeOliveiraRefutation.result

Frozen module record

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    Year recorded in the source bibliography: 2026.

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Discrete structuresConjecture refutedFirst preprint 2026Ku's maximum-density conjecture for ordinary hypercubesOrdinary integer-board line avoidance forces a strict upper bound on maximal saturated density.Refuted

The questionKuo-Han Ku, Monochromatic k in a row, arXiv:2606.12880v1, section 5.1.2, states:

Why it mattersThe paper's upper-density formula is suggested by configurations that attain the axis bound for certain parameters.

The ideaProjection onto the first two coordinates is injective, since two points in a projection fiber lie on an axis line.

Our contributionOrdinary integer-board line avoidance forces a strict upper bound on maximal saturated density.

Verification details
Release source recordRefuted in Lean

arXiv

Ku's maximum-density conjecture for ordinary hypercubes

Kuo-Han Ku, Monochromatic k in a row, arXiv:2606.12880v1, section 5.1.2, states:

Conjecture 5.6. For every $k\in\mathbb{N}_{\geq 2}$ and $d\in\mathbb{N}$, $D(k,[k]^d)=1-\frac{2}{k}$.

Definitions 2.2, 2.4 and 2.6 fix the ordinary board $[k]^d={1,\ldots,k}^d$, all full signed integer lines, near-line avoidance, inclusion-maximal saturation, and the maximum saturated density. This universal assertion is refuted at $k=3,d=3$.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Every integer $k\geq2$ and natural dimension $d$ | KuGeometry.claim quantifies over natural k, the hypothesis $2\leq k$, and every natural d. | | Ordinary points in ${1,\ldots,k}^d$ | Board k d is Fin d → Fin k; a label is interpreted as its natural value plus one in the integers. | | Every full line $p+tv$, $0\leq t<k$, $0\ne v\in{-1,0,1}^d$, inside the board | GValid quantifies over every signed nonzero direction and every parameter and coordinate; GOn gives literal integer incidence. | | No $(k-1)$-subset of a line is selected | GSourceConfig; gsource_config_iff identifies it with GConfig for $k\geq2$. | | Saturated means maximal under inclusion | GSaturated: a containing configuration must equal the original one. | | $D$ is the attained maximum of $|S|/k^d$ over saturated configurations | density_maximum_attained supplies a saturated cardinality maximizer; density divides its cardinality by $(k:\mathbb{R})^d$. The denominator is positive for $k\geq2$. | | The claimed equality is $D=1-2/k$ | claim has type Prop; result has type Not claim. | | Refuting parameter pair $(3,3)$ | exact_real_event states density 3 3 < (1 : Real)/3. |

The fixed board has 27 points. source_config_iff and gconfig_33_iff connect the pairwise geometric predicate used in the bound to the general source predicate. Natural dimension zero is allowed by the formal universal quantifier; the refuting positive dimension three also lies in a convention that starts natural dimensions at one. No conclusion depends on dimension zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/KuNearLineRefutation.result

Frozen module record

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Analysis & dynamicsConjecture refutedFirst preprint 2025Refutation of the panchromatic pairing conjectureA six-vertex hypergraph refutes the proposed equality between panchromatic and bipanchromatic extrema.Refuted

The questionLalou, Mbarek, Skender, and Togni, Completely Independent Spanning Trees in Split Graphs: Structural Properties and Complexity, arXiv:2512.15486v2, Section 5, Conjecture 1 (verbatim apart from plain-text typography):

Why it mattersThe published equality has a six-vertex counterexample with three four-element edges and no isolated vertex.

The ideaColoring the core by three distinct colors and all petals by a fourth proves panchromatic feasibility at four.

Our contributionA six-vertex hypergraph refutes the proposed equality between panchromatic and bipanchromatic extrema.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the panchromatic pairing conjecture

Lalou, Mbarek, Skender, and Togni, Completely Independent Spanning Trees in Split Graphs: Structural Properties and Complexity, arXiv:2512.15486v2, Section 5, Conjecture 1 (verbatim apart from plain-text typography):

Every hypergraph H satisfies chi_p^2(H) = chi_p(H) - ceil(alpha_{chi_p(H)}(H)/2).

Here chi_p and chi_p^2 are the maximum feasible panchromatic and bipanchromatic color counts. A bipanchromatic coloring requires every global color class to occur at least twice. The quantity alpha_k is the minimum number of globally unique colors over all panchromatic k-colorings.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/PanchromaticPairingConjectureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2010Refutation of the LeSaulnier-Vijay lower-density conjectureA staged set of positive integers admits a progression-free permutation and has lower density at least four fifteenths, exceeding one quarter.Refuted

The questionLeSaulnier and Vijay, arXiv:1004.1740, Discrete Mathematics 311 (2011), p. 207, define a subset of the positive integers to be 3-free when it admits a permutation containing no three-term arithmetic progression as a subsequence.

Why it mattersAn explicit 3-free set of lower density greater than one quarter refutes the named conjecture.

The ideaThe formal intervals are equivalently \([2Mk,3Mk-1]\) and \([6Mk-2,11Mk-5]\), with inclusive natural endpoints.

Our contributionA staged set of positive integers admits a progression-free permutation and has lower density at least four fifteenths, exceeding one quarter.

Verification details
Release source recordRefuted in Lean

Open problem

lesaulnier vijay 2011 beta3 lower density

LeSaulnier and Vijay, arXiv:1004.1740, Discrete Mathematics 311 (2011), p. 207, define a subset of the positive integers to be 3-free when it admits a permutation containing no three-term arithmetic progression as a subsequence. Its lower density is

\underline d(S)=\liminf_{n\to\infty}\frac{|S\cap[1,n]|}{n}, \qquad \beta(3)=\sup\{\underline d(S):S\text{ is 3-free}\}.

Their Theorem 3 establishes \beta(3)\ge1/4, and they conjecture \beta(3)=1/4. The formal claim is the universal upper bound: for every S\subseteq\mathbb N with 0\notin S, if there is an injective \pi:\mathbb N\to\mathbb N whose range is exactly S and such that

\forall i<j<k,\quad \pi(i)+\pi(k)\ne2\pi(j),

then \underline d(S)\le1/4. The enumeration condition forces infinitude and covers arithmetic progressions in both numerical directions. The target is preregistered in issue #14866.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/LeSaulnierVijayLowerDensityRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2026Five is not the minimal quaternary extension-counterexample lengthThe quaternary quantum extension property already fails at lengths one and three, so length five is not minimal.Refuted

The questionMahmoud, Minimal Counterexamples of the MacWilliams Extension Theorem for Stabilizer Codes, arXiv:2607.26214v1 (2026), §6.2, open problem 2, states:

Why it mattersDistinguishing these requirements is necessary to identify the actual obstruction and its length.

The ideaBoth parameterizations are additive and injective.

Our contributionThe quaternary quantum extension property already fails at lengths one and three, so length five is not minimal.

Verification details
Release source recordRefuted in Lean

Open problem

Five is not the minimal quaternary extension-counterexample length

A. A. Mahmoud, Minimal Counterexamples of the MacWilliams Extension Theorem for Stabilizer Codes, arXiv:2607.26214v1 (2026), §6.2, open problem 2, states:

\textbf{Minimal length for general $q$.} Determine the minimal counterexample length for general $q$. For $q=2$ it is $3=q+1$ (Theorem~\ref{thm:census}); we expect $q+1$ to be minimal in general, in the light of the threshold results for additive isometries in \cite{Dyshko16,DyshkoGeom}, but we claim it only for $q=2$.

This is a verbatim excerpt of MzWi_Ext_Th.tex, line 429, retrieved from the versioned arXiv source; its md5 is ee3d0a4f9ba273634eb55aad8ed7b850. The rest of the item asks separately about full-support minimal lengths. The definitions and source locators are also in the literature note.

The source's Definition 2.3 (TeX lines 237–239) is:

Fix $q$ and $n$. We say the quantum extension property $\mathrm{QEP}(q,n)$ holds if for every pair $C,D\le\A_q^n$ of trace-symplectically self-orthogonal additive codes and every additive bijection $f\colon C\to D$ with $\wt(f(u))=\wt(u)$ for all $u\in C$, the map $f$ extends to a monomial transformation of $\A_q^n$.

Here $A_q=\mathbb F_q^2$, and additive means $\mathbb F_p$-linear, where $p$ is the characteristic (TeX lines 211–217). Definition 2.1 (lines 221–223) specifies a coordinate permutation followed by maps in $\operatorname{Sp}(2,q)$, and explicitly mentions the larger group of all coordinatewise $\mathbb F_p$-linear automorphisms.

Issue #20851 fixes both readings at $q=4$. Write $\mathrm{QEP}(4,n)$ for the specified symplectic coordinate group and $\mathrm{QEP}^{\mathrm{add}}(4,n)$ for the larger additive coordinate group. The formal claim is

$\operatorname{IsLeast}{n\in\mathbb N:1\le n\ \land\ \neg\mathrm{QEP}(4,n)},5\ \lor\ \operatorname{IsLeast}{n\in\mathbb N:1\le n\ \land\ \neg\mathrm{QEP}^{\mathrm{add}}(4,n)},5.$

IsLeast requires both membership in the failure set and being no larger than any member. This disjunction is refuted: neither reading has least positive failure length $5$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/StabilizerExtensionQuaternaryMinimalLength.result

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Discrete structuresConjecture refutedRefutation of the A396844 mod-sixteen conjectureThe normalized series of OEIS A396844 has alternating dyadic residues modulo sixteen.Refuted

The questionHanna, Jul 03 2026, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2026a396844.md:

Why it mattersThis is a first-tier OEIS conjecture from 2026.

The ideaA contracting operator on C has a fixed point constructed by stabilized coefficients. generatingequation gives A(0) = 0, a(1) = 1 and A.subst(XA - 4XA^2) = X^2, exactly the OEIS equation. generatingunique identifies every normalized integral solution with A. modfouridentity gives A = X modulo 4.

Our contributionThe normalized series of OEIS A396844 has alternating dyadic residues modulo sixteen.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A396844 mod-sixteen conjecture

OEIS A396844, Paul D. Hanna, Jul 03 2026, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2026a396844.md:

G.f. A(x) satisfies A( x*A(x) - 4*x*A(x)^2 ) = x^2.

Conjecture: for n > 8, a(n) == 12 (mod 16) iff n is of the form 2*4^k+1 otherwise a(n) is divisible by 16.

The conjecture is refuted at n = 17. The orchestrator reports a(17) = 113101684573952804, congruent to 4 modulo 16. Since 17 is not of the form 2*4^k+1, the "otherwise divisible by 16" clause is false. The supplied numeric-check locator is results/verify-a396844-mod16.out; that file was not available at this worktree path during Stage B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/ScaledQuadraticReversionModSixteen.hanna_conjecture_false

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Quantum informationCounterexample foundFirst preprint 2026Unrestricted qubit generalized-fidelity data processingA fixed bit-flip qubit channel strictly increases the squared generalized Bures--Wasserstein quantity throughout an explicit interval of reference states.Refuted

The questionRajaei's abstract says: "In dimension two, although we do not settle the DPI in full generality,".

Why it mattersThis is the paper's residual dimension-two yes-or-no question, rather than its already refuted dimension-at-least-three statement or classification of all reference bases.

The ideaDirect positive-root certificates for both references and all four sandwiches recover the actual source trace.

Our contributionA fixed bit-flip qubit channel strictly increases the squared generalized Bures--Wasserstein quantity throughout an explicit interval of reference states.

Verification details
Release source recordRefuted in Lean

arXiv

Unrestricted qubit generalized-fidelity data processing

Rajaei's abstract says: "In dimension two, although we do not settle the DPI in full generality,". The exact residual assertion is that every trace-one positive semidefinite $P,Q\in M_2(\mathbb C)$, every positive-definite trace-one $R$, and every completely positive trace-preserving channel $\Phi:M_2(\mathbb C)\to M_2(\mathbb C)$ with positive-definite $\Phi(R)$ obey

$$B_{\Phi(R)}(\Phi(P),\Phi(Q))\le B_R(P,Q).$$

Here $F_R(P,Q)$ is the ordered matrix-root trace in the source, and $B_R(P,Q)=\operatorname{Re}\operatorname{tr}(P+Q)-2\operatorname{Re}F_R(P,Q)$. Singular input states are permitted. Positivity only on three chosen states does not establish complete positivity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/GeneralizedFidelity.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedFirst preprint 2026Song and Lin's parity conjecture for zero transfer on oriented circulant graphsOn the connected oriented circulant graph G(Z30, {5, 6, 9, 20}) the continuous-time quantum walk never moves amplitude between the vertex 0 and the even vertex 2, so the parity restriction conjectured for orders n = 2 (mod 4) fails at n = 30.Refuted

The questionA graph has zero transfer from u to v if U(t){u,v} = 0 for every time t ≥ 0.

Why it mattersZero transfer models the isolation of a vertex from quantum state transfer.

The ideaWrite H = i S, where S is the integer skew-symmetric matrix with entry 1 on arcs and -1 on reversed arcs; then H^k = i^k S^k.

Our contributionOn the connected oriented circulant graph G(Z30, {5, 6, 9, 20}) the continuous-time quantum walk never moves amplitude between the vertex 0 and the even vertex 2, so the parity restriction conjectured for orders n = 2 (mod 4) fails at n = 30.

Verification details
Release source recordRefuted in Lean

Open problem

Song and Lin's parity conjecture for zero transfer on oriented circulant graphs

Song and Lin study the continuous-time quantum walk U(t) = exp(-i t H) on a mixed graph, where the Hermitian adjacency matrix H has entry i on an arc u → v, -i on the reversed arc and 0 elsewhere on an oriented graph. A graph has zero transfer from u to v if U(t)_{u,v} = 0 for every time t ≥ 0. For the oriented circulant graphs G(ℤ_n, C), with arcs a → b for b - a ∈ C, C ⊆ ℤ_n ∖ {0}, C ∩ -C = ∅, and connectedness assumed throughout the section, they state Conjecture 4.1 of arXiv:2608.10643v1:

Let Γ = G(ℤ_n, C) be an oriented circulant graph with n ≡ 2 (mod 4). If zero transfer occurs between vertex v and 0, then v must be odd.

Issue #10142 fixes the readings: vertices are ZMod n; U(t) is the matrix exponential of -(i t) H, taken as the frozen propagator hamiltonianPropagator H t = exp (t • (-i) • H) of D5/S3/Quantum/Dynamics/ProjectionProbabilityFlow; zero transfer between v and 0 is read in both directions at once, which only strengthens the hypothesis; connectedness is that of the underlying undirected graph; "v is odd" means that the representative of v in 0, …, n − 1 is odd.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/OrientedCirculantZeroTransfer.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedFirst preprint 2025Tian–Dokyeesun–Klavžar Conjecture 3.4 refutationInduced vertex deletion can raise the outer general position number by more than the deleted degree.Refuted

The questionConjecture 3.4 in section 3.2 of arXiv:2510.01294v2 states:

Why it mattersThe counterexample has 19 vertices, an actual non-cut vertex of degree two, deletion outer number at least eight, and original outer number at most five.

The ideaReplace two opposite vertices by independent sets A and B of four vertices, preserving their cycle neighbors.

Our contributionInduced vertex deletion can raise the outer general position number by more than the deleted degree.

Verification details
Release source recordRefuted in Lean

Open problem

Tian–Dokyeesun–Klavžar Conjecture 3.4 refutation

Conjecture 3.4 in section 3.2 of arXiv:2510.01294v2 states:

If x is not a cut vertex of a graph G, then gp_o(G-x) <= gp_o(G) + deg_G(x).

The formal claim ranges over all finite simple connected graphs and every non-cut vertex. Deletion is the actual induced graph on vertices other than x. Non-cut means that deletion does not increase the number of connected components. This retains K1, with empty deletion, zero empty outer number, and singleton outer number one.

A pair is Z-positionable when every shortest path between its endpoints has no internal vertex in Z. An outer general position set requires this for every pair with at least one endpoint in Z. The outer number is the maximum cardinality among all such subsets.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/TianDokyeesunKlavzarOuterGeneralPositionRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2026Alai's Staircase for faithful GHZ–Mermin measurement dependenceFaithful GHZ–Mermin measurement dependence: AlaiStaircaseProved

The questionAaron Alai, Exact minimum measurement dependence for faithful local deterministic models of multipartite GHZ–Mermin correlations, arXiv:2608.00124v1, Section IX, Conjecture 1 (Staircase), PDF page 4:

The ideaThe Walsh-square identity for the complete Boolean quadratic form gives the exact spectral magnitude.

Our contributionFaithful GHZ–Mermin measurement dependence: AlaiStaircase

Scope & next questionThe present proof does not settle those questions or richer measurement scenarios.

Verification details
Release source recordProved in Lean

Open problem

Alai's Staircase for faithful GHZ–Mermin measurement dependence

Aaron Alai, Exact minimum measurement dependence for faithful local deterministic models of multipartite GHZ–Mermin correlations, arXiv:2608.00124v1, Section IX, Conjecture 1 (Staircase), PDF page 4:

With $R(n) := 2^{\lfloor (n-1)/2 \rfloor}$ the Mermin violation ratio and $s(n) := (R+1)/(2R)$ the classical satisfiability, $F_{\min}(n) = \frac{R}{2(R+1)} = \frac{1}{4,s(n)}, \qquad F_{\min}\cdot s = \frac{1}{4}.$

Section II specifies $n\ge3$, all even-Y setting strings, and all $4^n$ deterministic X/Y answer tables. At each setting the real density is nonnegative and normalized. A faithful model reproduces the prescribed full correlator and makes every nonempty proper-subset correlator zero. The objective is $F=M/2$, where $M=\max_{s,s'}\sum_\lambda|\rho_s(\lambda)-\rho_{s'}(\lambda)|$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/MerminMeasurementDependence/AlaiStaircase.result

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Number theoryConjecture refutedSource 2025Aliabadi's uncapped subspace-product conjecture is falseThe degree-seven field Q[X]/(X^7-2) refutes Aliabadi's subspace-product lower bound: there are no proper nontrivial intermediate fields, but two five-dimensional subspaces have product span of dimension at most seven, below the proposed lower bound nine.Refuted

The questionMohsen Aliabadi, Conditions for matchability in groups and field extensions II, Discussiones Mathematicae General Algebra and Applications 45 (2025), 135–157, section 5.1, Conjecture 38, printed page 154 / PDF page 20, states:

Why it mattersSection 5.1 seeks a linear counterpart of the group product-set estimate used for sufficient matchability conditions.

The ideaThe selected powers are distinct basis entries.

Our contributionThe degree-seven field Q[X]/(X^7-2) refutes Aliabadi's subspace-product lower bound: there are no proper nontrivial intermediate fields, but two five-dimensional subspaces have product span of dimension at most seven, below the proposed lower bound nine.

Verification details
Release source recordRefuted in Lean

Open problem

Aliabadi's uncapped subspace-product conjecture is false

Mohsen Aliabadi, Conditions for matchability in groups and field extensions II, Discussiones Mathematicae General Algebra and Applications 45 (2025), 135–157, section 5.1, Conjecture 38, printed page 154 / PDF page 20, states:

Let $K\subset L$ be a field extension and $A,B\subset L$ be a $K$-subspace of $L$ of finite positive dimensions. Assume further that for every $a,b\in L$ and every nontrivial proper finite intermediate subfield $K\subset M\subsetneq L$, we have

$$\dim_K(aM\cap A)+\dim_K(bM\cap B)\le [M:K]+1.$$

Then for all $K$-subspaces $S\subset A$ and $T\subset B$, the following holds

$$\dim_K\langle ST\rangle\ge\dim_K S+\dim_K T-1.$$

The same statement is Conjecture 5.1 in arXiv:2309.14664v1, PDF page 14, and v2/v3, PDF page 15. Section 2.2 defines the brackets as the $K$-linear span of the actual product set in $L$. The source note records the primary versions, DOI and locators. The sole target is the whole original universal assertion. Preregistration: #23409.

Statement correspondence

| Source clause | Formal counterpart and witness | | --- | --- | | Arbitrary field extension $K\subset L$ | claim quantifies fields with a K-algebra structure on L. The witness is the actual field AdjoinRoot (X^7-C 2) over $\mathbb Q$, with the canonical injective algebra map and degree seven. | | Finite positive-dimensional $A,B$ | Both finite-dimensional hypotheses and both positive-finrank hypotheses occur in claim; $A=B=\operatorname{powerSpan}(6)$ has dimension six. | | Every $a,b\in L$, including zero | Both unrestricted scalar binders occur in intersectionPremise; translate maps the underlying submodule of $M$ by actual multiplication by the scalar. | | Every finite nontrivial proper intermediate field | IntermediateField K L, ⊥ < M, M < ⊤, and FiniteDimensional K M. Neither endpoint is included. Prime degree proves that no witness field has such an $M$. | | Both intersection dimensions and $[M:K]+1$ | Submodule infima and finranks in intersectionPremise express the exact displayed inequality. No assumed certificate replaces it. | | Every allowed $S\subset A$, $T\subset B$ | claim quantifies all submodules with S ≤ A and T ≤ B, including the zero subspaces. The counterexample has both containments strict, so it also refutes the strict reading of the printed subset symbols. | | Actual $\langle ST\rangle$ | Submodule.span K ((S : Set L) * (T : Set L)), using field pointwise multiplication. | | Uncapped lower bound | finrank K S + finrank K T - 1 ≤ finrank K (span ...); natural subtraction gives the same inequality truth value as integer subtraction, including zero dimensions. | | Whole-claim negation | The sole public theorem result : ¬ claim has no extra mathematical premise. |

The matching definitions in the introduction use equal dimensions and $1\notin B$ when defining matchability. Conjecture 38 states neither matchability nor these conditions as an added premise. The witness nonetheless has $A=B$ and $S=T$, with $1$ absent from all four spaces. It also has $S<A<L$ and $T<B<L$. No assumption that a subspace contain $1$, that the ambient extension be infinite, or that the dimension sum be small is inserted. No ambient cap replaces the source conclusion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Galois/AliabadiSubspaceProductRefutation.result

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    Year in the primary source attribution: 2025.

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Number theoryStructural characterizationBala's A361034 parity conjectureConjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.Proved

The questionConjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

Why it mattersThe retained declarations are unbounded symbolic statements, with no retained finite certificate or enumeration.

The ideaThe imported OeisA347854BalaHalfIndexIntegral.muldivreduce reduces floor terms to quotients and residues, and its factorialproddvdoffloor turns their sum inequality into factorial divisibility.

Our contributionConjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A361034 parity conjecture

Peter Bala's OEIS A361034 defines $a(n)=2520(4n)!/(n!(n+2)!^3)$ and states:

Conjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

The index $n$ is any natural number, including zero. The prerequisite integrality and the exact exponent bound are part of the source's statement in preregistration #18606.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Factorial quotient with coefficient 2520 and shift 2 | BalaA361034Parity.a | Literal source expression, using natural-number division. | | Every natural index, including zero | BalaA361034Parity.claim | Universal quantifier over the naturals. | | Odd exactly when the shifted index is a power of two | BalaA361034Parity.claim | Both directions of the iff; the exponent is natural and at least 1. | | Quotient is an integer | Private BalaA361034Parity.integrality | Denominator divides numerator; its hypotheses are discharged in the settling proof. | | Conjecture holds | BalaA361034Parity.result | Complete claim without extra assumptions. | | Parameters $C,j$ in the shared proof | BalaA361034Parity.quotient_odd | Internal generalization, instantiated with $C=2520$ and $j=2$; no added premise in result. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/BalaA361034Parity.result

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Discrete structuresCounterexample foundFirst preprint 2026The ceiling formula for monochromatic sublattices is falseIn every odd dimension at least eleven, one more member than the chain guarantee is unavoidable.Refuted

The questionBhattacharjee–Mandal–Bhattacharya, arXiv:2610.02833v1, Question 9.3:

Why it mattersThe Erdős 1183 problem page records the trivial bound f(n) ≥ ⌈(n+1)/2⌉, obtained by taking a colour class in a maximal chain.

The ideaSuppose no monochromatic sublattice has more than m members.

Our contributionIn every odd dimension at least eleven, one more member than the chain guarantee is unavoidable.

Verification details
Release source recordRefuted in Lean

Open problem

The ceiling formula for monochromatic sublattices is false

Bhattacharjee–Mandal–Bhattacharya, arXiv:2610.02833v1, Question 9.3:

Is f(n) = O(n)? Is f(n) = ⌈(n + 1)/2⌉ for every n?

Section 1 specifies:

a k-colouring of 2^[n] is a map χ : 2^[n] → [k], and a family ℱ ⊆ 2^[n] is monochromatic if χ is constant on ℱ. … it is a sublattice if it is closed under both union and intersection. … f(n) = min_χ max{|ℒ| : ℒ ⊆ 2^[n] is a monochromatic sublattice}, where the minimum is over all 2-colourings.

Sublattices here are nonempty. The Lean definition uses finite families of subsets of Fin n, constant Bool colour, and closure under both union and intersection. The extremal function is literally a minimum over all colourings of the maximum cardinality of such a family. The universal claim concerns only the second question, using (n + 2) / 2 for the ceiling.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ErdosUlam/SublatticeChainBound.result

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Discrete structuresCounterexample foundFirst preprint 2026The proposed union-closed counting threshold is falseIndependent choices on a level of the Boolean lattice give too many labelled union-closed families for the proposed quasipolynomial threshold.Refuted

The questionBhattacharjee–Mandal–Bhattacharya, arXiv:2610.02833v1, Question 9.2:

Why it mattersThe proposed estimate would combine with Proposition 5.4 to bound the union-closed extremal function at a quasipolynomial scale.

The ideaThere are B subsets of size 5r, by complementation.

Our contributionIndependent choices on a level of the Boolean lattice give too many labelled union-closed families for the proposed quasipolynomial threshold.

Verification details
Release source recordRefuted in Lean

Open problem

The proposed union-closed counting threshold is false

Bhattacharjee–Mandal–Bhattacharya, arXiv:2610.02833v1, Question 9.2:

Is there a constant C such that U(n,N) < 2^{N−1} for all large n and all N ≥ n^{C log₂ log₂ n}?

Proposition 5.4 defines the labelled count and its proposed application:

Let U(n,N) be the number of union-closed families ℱ ⊆ 2^[n] with |ℱ| = N. If U(n,N) < 2^{N−1}, then F(n) < N.

Section 1 defines union-closed by requiring A ∪ B ∈ ℱ whenever A,B ∈ ℱ. The Lean definition counts finite families of subsets of Fin n, with no quotient by ground-set permutations. The claim quantifies over a real constant C, a natural cutoff, every dimension above that cutoff, and every natural N meeting the real-power threshold. Its result is the literal negation of that claim. The logarithms and real power are Mathlib's total functions; the proof uses only dimensions at least 24, where both logarithms have positive arguments and the usual interpretation applies.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ErdosUlam/UnionClosedCountRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026Calderon's reciprocal moments in imaginary quadratic ordersCalderon's reciprocal moments satisfy both congruences for every admissible imaginary quadratic order.Proved

The questionKevin Calderon, Ljunggren–Jacobsthal and Bailey-Type Congruences for Rectangular Gaussian Binomial Coefficients, arXiv:2608.00347v1, Conjecture 6.3, PDF page 19:

Why it mattersThe conclusion supplies the local reciprocal-moment premise used in the source's proposed omega-Ljunggren extension of Gaussian rectangular congruences.

The ideaPrincipal divisibility is exactly membership in the singleton-generated ideals appearing in the public statement.

Our contributionCalderon's reciprocal moments satisfy both congruences for every admissible imaginary quadratic order.

Verification details
Release source recordProved in Lean

Open problem

Calderon's reciprocal moments in imaginary quadratic orders

Kevin Calderon, Ljunggren–Jacobsthal and Bailey-Type Congruences for Rectangular Gaussian Binomial Coefficients, arXiv:2608.00347v1, Conjecture 6.3, PDF page 19:

Let $\omega$ satisfy (6.1), and let $p$ satisfy (6.3). Then, for every $k\geq1$, $H_{1,\omega}(k)\in p^{2k}\mathcal O_{\omega,p}$, $H_{2,\omega}(k)\in p^k\mathcal O_{\omega,p}$.

Equation (6.1) is $\omega^2-T\omega+N=0$, with $T,N\in\mathbb Z$ and $\Delta=T^2-4N<0$. Equation (6.3) requires a rational prime $p>5$, $p\nmid\Delta$, and $(\Delta/p)=-1$. The local order is $\mathcal O_{\omega,p}:=\mathbb Z_p[\omega]$. Its positive representatives are

$$ U_{\omega,k}:={x+y\omega:1\leq x,y\leq p^k,\ p\nmid(x,y)}, \qquad H_{r,\omega}(k):=\sum_{z\in U_{\omega,k}}z^{-r}. $$

The coordinate condition excludes simultaneous divisibility by $p$; both intervals include $p^k$. The Lean ring R is the adjoin-root of $X^2-TX+N$ over PadicInt p, and U is the finite image of exactly this filtered positive rectangle. All its elements are units in the admissible setting. Consequently Ring.inverse in H1 and H2 is the underlying inverse unit value, without adding an assumption or a term to the source's sums.

The fully quantified statement is: for every natural prime $p$, every pair of integers $T,N$ satisfying all four admissibility clauses, and every natural $k\geq1$, both displayed memberships hold. The public claim encodes this statement; result : claim has no additional hypotheses.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/QuadraticOrderMoments/CalderonReciprocalMoments.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2026The uncertainty condition suffices for the mixed discrete GaussianThe covariance uncertainty condition makes the trace-normalized mixed discrete Gaussian a density operator.Proved

The questionNicolae Cotfas, Frame-dependent coherence of Gaussian quantum states, arXiv:2610.05623v1, asks after Eq. (22): “As concern ρ̃σ, we do not know if (11) is sufficient in order to have ρ̃σ ≥ 0.

Why it mattersA trace-normalized mixed discrete-variable Gaussian can serve as a quantum state when it is positive semidefinite.

The ideaA positive diagonal entry excludes zero trace. normalizationisDensity uses GHZMeasureBiseparableBound.IsDensity to complete the density conjunction.

Our contributionThe covariance uncertainty condition makes the trace-normalized mixed discrete Gaussian a density operator.

Verification details
Release source recordProved in Lean

Open problem

The uncertainty condition suffices for the mixed discrete Gaussian

Nicolae Cotfas, Frame-dependent coherence of Gaussian quantum states, arXiv:2610.05623v1, asks after Eq. (22): “As concern ρ̃_σ, we do not know if (11) is sufficient in order to have ρ̃_σ ≥ 0. Based on our numerical investigations, we conjecture that this is true.”

Preregistration #18946 fixes the source formulas and quantifiers. The source note gives the locators and verbatim statements. The settlement is Proved by D5.S3.Quantum.CotfasDVGS.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Eq. (10), continuous Wigner Gaussian | W, real inverse covariance and determinant | Literal formula | | Odd dimension $d=2s+1\ge3$ | Natural $s$ with $1\le s$ in claim | Same range | | Functions on ${-s,\ldots,s}$; addition modulo $d$ | Integer Finset.Icc subtype; ZMod.valMinAbs | Centered modular representative | | Eq. (18), displaced parity | parity, coefficient in row $m$, column $j$ | Same phase and selected column | | Eq. (22), signed double lattice sum | wSum, tsum over $\mathbb Z^2$ | Real division in both coordinates; scalar normalization separated | | Eq. (21), weighted finite parity sum | unnormalized | Both sums use the centered indices | | $c'_\sigma$ is a normalizing constant | rhoTilde, inverse complex trace times unnormalized | The proof establishes a nonzero trace | | Positive Planck parameter | $h\in\mathbb R$, $0<h$ | Universal quantifier | | Real symmetric positive-definite covariance | Real two-by-two Matrix.PosDef | Symmetry follows from this predicate | | Eq. (11), $\sigma+i\begin{pmatrix}0&1\-1&0\end{pmatrix}\succeq0$ | The literal complex matrix in claim | No certificate replaces the condition | | Conjectured positivity | result: claim, including positive semidefiniteness | Exact family | | Density normalization | Trace equals one in claim | Consequence of the source normalization; strengthens positivity |

The public surface comprises W, parity, wSum, unnormalized, rhoTilde, claim and result. All other authored declarations are private and have consumers inside this settlement. The centered index type and density conjunction occur literally in the claim; the proof uses existing covariance and density predicates.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/CotfasDVGS.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026A singular nonzero point of the odd spectral-invariant varietyA nonzero singular point for the odd clause of Conjecture 3.Refuted

The questionSharma, Algebraic Properties of the Ideal of Spectral Invariants for the Discrete Laplacian, arXiv:2602.08304v1, Conjecture 3, p. 11:

Why it mattersThe fifteen-site potential repeats a nonzero five-site solution three times.

The ideaThe shared theorem D5.S3.Combinatorics.SpectralGraph.CyclicTridiagonalTransferTrace.cycleDiagdettransfer identifies the cyclic determinant with the transfer trace minus two over every commutative ring.

Our contributionA nonzero singular point for the odd clause of Conjecture 3.

Verification details
Release source recordRefuted in Lean

Open problem

A singular nonzero point of the odd spectral-invariant variety

M. Faust, L. Friedman, G. O'Malley, R. Ramos and A. Sharma, Algebraic Properties of the Ideal of Spectral Invariants for the Discrete Laplacian, arXiv:2602.08304v1, Conjecture 3, p. 11:

When q is odd, then V(I′) has only regular nonzero points, and when it is an even number greater than 6 there is a single singular point, up to symmetry, of multiplicity 2^{q/2−2}. Moreover, the multiplicity at 𝟎 is 2^{⌊q/2⌋}.

This delivery refutes the odd clause under the Jacobian regularity criterion specified in preregistration #19188. The Library note records the source definitions. Period means a period, without a least-period requirement. The even and origin-multiplicity clauses are not settled here.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $L_V=L_V(1)$, diagonal $V$, cyclic off-diagonal entries one | floquetMatrix | Literal cyclic-adjacency formula; the claim uses sizes at least five | | $D_V(\lambda)=\det(L_V-\lambda I)$ | floquetCharPoly | This sign convention is retained | | $U=(u_1,\ldots,u_m,0,-u_m,\ldots,-u_1)$ | specializedPotential | Zero-based indices; no additional constraint on the coordinates | | $\zeta_U=D_U-D_{\mathbf0}$ | zeta | Full polynomial difference | | $p'_k=(-1)^{n+k}[\lambda^{n-2k}]\zeta_U$ | pPrime | Literal coefficient and sign; only $1\leq k\leq m$ generate the ideal | | A point of $V(I')$ | InVariety | Every generator vanishes | | Regular nonzero points | jacobianAt, IsRegularPoint, u ≠ 0 in claim | Regularity is precisely nonvanishing of the generator Jacobian determinant | | Every odd $q=2m+1>3$ | claim, with $2\leq m$ | No least-period hypothesis | | Refutation | result | Negates this full universal claim |

All names in this table belong to D5.S3.Combinatorics.SpectralGraph.FaustSpectralInvariantsOddRefutation. Section 3, p. 7 refers to multiplicity for the defining ideal rather than its radical. No separate local-algebra equivalence between multiplicity one and Jacobian invertibility is formalized in this delivery.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/FaustSpectralInvariantsOddRefutation.result

Frozen module record

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2025An asymmetric mixed memory outside Gayrard's hierarchyAn asymmetric five-pattern mixed memory for quadratic activation lies outside the composition hierarchy.Refuted

The questionVéronique Gayrard, Mixed memories in Hopfield networks, arXiv:2504.04879v2, Section 1.2.1, Conjecture 1.4, printed p. 6:

Why it mattersThe hierarchy remains a sufficient construction, but cannot classify all mixed memories as defined in the paper.

The ideaAn infinite product of fair Boolean coordinates supplies the patterns; the strong law applied to each bounded coordinate-spin product gives all five almost-sure overlap limits.

Our contributionAn asymmetric five-pattern mixed memory for quadratic activation lies outside the composition hierarchy.

Verification details
Release source recordRefuted in Lean

arXiv

An asymmetric mixed memory outside Gayrard's hierarchy

Véronique Gayrard, Mixed memories in Hopfield networks, arXiv:2504.04879v2, Section 1.2.1, Conjecture 1.4, printed p. 6:

Given any $n\in\mathbb N$ odd, $\xi^{(N)}(m)$ is an $n$-mixed memory of type $F$ if and only if $m\in\mathcal M_{n,F}$.

The source uses a doubly infinite jointly independent family of fair signs, a smooth $F$ with $F'(x)>0$ for $x>0$, fixed odd $n$, a nondecreasing pattern count $M(N)$, and deterministic coefficients in $[-1,1]$. Definition 1.1 requires exactly $n$ nonzero coefficients, binary configurations $\xi_i(m)=\operatorname{sign}(\sum_{\mu=1}^{M}\xi_i^\mu F'(m_\mu) \mathbf1_{m_\mu\ne0})$, almost-sure overlap limits $m_\mu$ on that support, and zero overlap limits on available coordinates outside it. The sign at zero is zero (Section 2.2, printed (2.25)).

The set $\mathcal M_{n,F}$ is defined by allowable ordered compositions: all nonfinal blocks are even and at least two, and the final block is positive and odd. With $\alpha^{(a)}=2^{1-a}\binom{a-1}{\lfloor(a-1)/2\rfloor}$ and $\gamma^{(k)}=\prod_{l=1}^k\alpha^{(n_l)}$, only compositions satisfying $2F'(\gamma^{(k)})>\sum_{r>k}n_r F'(\gamma^{(r)})$ for every nonfinal block are retained. Their block vectors repeat $\gamma^{(k)}$ on $n_k$ coordinates and are padded by zeros. All permutations and all coordinate signs raised to exponent one for odd $F'$ and two otherwise are included.

For every positive block size $a$, the formalization reuses D5.S3.Quantum.Entanglement.PrecessionSpinOneSeparableBound.c, equal to the source's $\alpha^{(a)}=2^{-a+1}\binom{a-1}{\lfloor(a-1)/2\rfloor}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Hopfield/GayrardMixedMemoryRefutation.result

Frozen module record

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    Source 1
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    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Two-parameter greedy three-sumfree membershipGreedy three-sumfree membership has a universal two-parameter periodic formula.Proved

The questionThis dossier deliberately anchors Conjecture 17 of Bosma et al., Journal of Integer Sequences 28 (2025), Article 25.3.8, printed page 18, which is the earlier Conjecture 6 of arXiv:2503.04122v1 with its third-seed bound corrected from z > g+d to z >= g+d.

Why it mattersRecording the external proposition and its Library identity makes that existing formal evidence discoverable in the problem pool.

The ideaIts S is membership in greedyPrefix, which starts at [g+d, g, 1] in reverse order and prepends Nat.find (nextexists s). RestrictedThreeSum requires x < y < w, preserving the paper's distinctness condition.

Our contributionGreedy three-sumfree membership has a universal two-parameter periodic formula.

Scope & next questionConjecture 16 and the paper's other open items are deliberately out of scope.

Verification details
Release source recordProved in Lean

Open problem

Two-parameter greedy three-sumfree membership

This dossier deliberately anchors Conjecture 17 of Bosma et al., Journal of Integer Sequences 28 (2025), Article 25.3.8, printed page 18, which is the earlier Conjecture 6 of arXiv:2503.04122v1 with its third-seed bound corrected from z > g+d to z >= g+d. The caller-supplied reading, 2026-09-07, quotes this single proposition:

Let d >= 2. For every g >= d + 1 the greedy 3-sumfree sequence S_{1,g,g+d} is characterized as follows: z in S_{1,g,g+d} <=> z in {1, g, 2g+d-1, 2g+d} or z >= g+d and z mod 5g+2d in {g+d-2, g+d-1, ..., 2g+d-2}.

Here d, g, and sequence entries are integers, with positive seeds 1 < g < g+d. Every subsequent entry is the least larger positive integer that is not a sum of three distinct earlier entries, as defined on printed page 16. Conjecture 16 and the paper's other open items are deliberately out of scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Sumfree/GreedyThreeSumfreeTwoParameter.conjecture17

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025The tentative Hart–Rowley prefix/suffix equalityGreedy involution prefix and suffix counts can differ in a qualifying Coxeter system.Refuted

The questionRowley, A note on involution prefixes in Coxeter groups, arXiv:2502.00777v1, section 3, printed page 5, immediately before References, write:

Why it mattersThe ancestor decomposition gives a canonical length-additive involution factorization.

The ideaEstablish their Coxeter presentation of type A3, and identify genuine Coxeter length by reduced normal words and a lower bound along every simple-generator step.

Our contributionGreedy involution prefix and suffix counts can differ in a qualifying Coxeter system.

Verification details
Release source recordRefuted in Lean

Open problem

The tentative Hart–Rowley prefix/suffix equality

Sarah B. Hart and Peter J. Rowley, A note on involution prefixes in Coxeter groups, arXiv:2502.00777v1, section 3, printed page 5, immediately before References, write:

We suspect that the involution length of any element always equals its ‘suffix’ involution length, but have not tested this extensively enough to make a firm conjecture.

This is one explicitly tentative, unnumbered published assertion. The quoted text is licensed CC BY 4.0, https://creativecommons.org/licenses/by/4.0/ ; PDF line wrapping and typographic ligatures are normalized. It is not Conjecture 1.2 or 1.3.

The domain is every Coxeter system whose every nonidentity element has a unique maximum-Coxeter-length involution prefix, and every element of such a system. An involution has order exactly two. A prefix u of w satisfies w = uv and length(w) = length(u) + length(v). The unique ancestor maximizes length over all involution prefixes. Remove it and recurse on v, with zero factors at identity. The suffix procedure removes the analogous right factor. The assertion equates these two actual greedy counts. No finiteness, rank, or Coxeter-element restriction is imposed, and neither count minimizes arbitrary factorizations.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/HartRowleyGreedyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Further frontiersConjecture refutedFirst preprint 2025Hoster–Stump Conjecture 1.5: simplicial-poset Chow polynomialsTwo tetrahedra sharing a vertex have a non-real-rooted Chow polynomial.Refuted

The questionElena Hoster and Christian Stump, Chow polynomials of simplicial posets with positive h-vector are real-rooted, arXiv:2508.15538v1, Section 1, Conjecture 1.5, states:

Why it mattersThe face poset of two tetrahedra meeting in one vertex is a finite graded simplicial poset.

The ideaIts maximal Boolean intervals have rank four.

Our contributionTwo tetrahedra sharing a vertex have a non-real-rooted Chow polynomial.

Scope & next questionThe classification of other gluings remains open.

Verification details
Release source recordRefuted in Lean

arXiv

Hoster–Stump Conjecture 1.5: simplicial-poset Chow polynomials

Elena Hoster and Christian Stump, Chow polynomials of simplicial posets with positive h-vector are real-rooted, arXiv:2508.15538v1, Section 1, Conjecture 1.5, states:

Let P be a simplicial poset. Then H_P̂(x), H_P̂*(x) and H^aug_P̂(x) are real-rooted. Moreover, the roots of both H_P̂(x) and H_P̂*(x) interlace the roots of H^aug_P̂(x) = H^aug_P̂*(x).

This dossier anchors that single conjecture. The settlement refutes its first assertion, and therefore the conjunction, without negating its other real-rootedness assertions separately. The definitions are in Section 1, page 1, equations (1.1)–(1.2); the conjecture is on page 3 and on page 4 of the FPSAC 2026 abstract.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/SimplicialPosetChowRefutation.result

Frozen module record

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    Source 1
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Number theoryConjecture provedKotěšovec's second A368959 conjectureEvery biquadratefree powerful number, and each of 243, 972, 1944 and 2401, has nonzero complete monomial sums at every degree at least two.Proved

The questionVáclav Kotěšovec, OEIS A368959, COMMENTS; preregistration 19242.

Why it mattersPrime squares and cubes generate the biquadratefree powerful family through coprime products.

The ideaA missing-preimage argument handles large degrees; odd quadratic sums and the exact small fibers handle the remaining cases.

Our contributionEvery biquadratefree powerful number, and each of 243, 972, 1944 and 2401, has nonzero complete monomial sums at every degree at least two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kotěšovec's second A368959 conjecture

Václav Kotěšovec, OEIS A368959, COMMENTS; preregistration #19242.

Conjecture 2: This sequence contains all the terms from A338325, plus 243, 972, 1944, 2401, ...

The NAME is “Numbers m such that Sum_{k=0..m-1} exp(2Pii*k^s/m) <> 0 for all s>=2.” A001694 denotes powerful numbers. A338325 has NAME “Biquadratefree powerful numbers: numbers whose exponents in their prime factorization are either 2 or 3.” Membership uses positive natural numbers.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The complete sum over $0\le k<m$ | KotesovecConjectureOne.monomialSum | Literal summand and range; natural exponentiation precedes complex casting. | | Every degree $s\ge2$ and positive $m$ | KotesovecConjectureOne.InA368959 | Positivity, universal natural degree and nonvanishing are all explicit. | | Prime exponents two or three | KotesovecConjectureOne.InA338325 | Every prime factor of the positive natural number is covered. | | Every A338325 term is included | First conjunct of claimTwo | Universal family inclusion. | | 243, 972, 1944 and 2401 are included | Four remaining conjuncts of claimTwo | Every named modulus is retained. The ellipsis specifies no further list and is not read as exhaustiveness. | | Entire second conjecture | result : claimTwo | No extra assumptions. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CharacterSums/KotesovecConjectureTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Quantum informationCounterexample foundFirst preprint 2025Partial trace can increase Fisher discordA two-qubit flagged state refutes partial-trace monotonicity of Fisher discord.Refuted

The questionZhang, Fisher discord as a quantifier of quantum complexity, arXiv:2510.15375v1; published as Fisher discord as a quantifier of quantum metrological complexity, Phys.

Why it mattersSeparate monotonicity of two information quantities does not imply monotonicity of their difference.

The ideaThe lifted observable connects only equal flag coordinates.

Our contributionA two-qubit flagged state refutes partial-trace monotonicity of Fisher discord.

Verification details
Release source recordRefuted in Lean

Open problem

Partial trace can increase Fisher discord

H. Li, S. Luo and Y. Zhang, Fisher discord as a quantifier of quantum complexity, arXiv:2510.15375v1; published as Fisher discord as a quantifier of quantum metrological complexity, Phys. Rev. A 114, 032449 (2026).

§VI (Summary), verbatim:

Both the quantum Fisher information based on the symmetric logarithmic derivative and the skew information does not increase under the partial traces, that is, I_F(ρ₁₂, H ⊗ 1) ≥ I_F(ρ₁, H), I_W(ρ₁₂, H ⊗ 1) ≥ I_W(ρ₁, H) … It is natural to consider such an open question: Whether quantum Fisher discord C(ρ₁₂, H ⊗ 1) = I_F(ρ₁₂, H ⊗ 1) − I_W(ρ₁₂, H ⊗ 1) for bipartite states does not increase under the partial trace.

| Source object | Formal reading | |---|---| | Density matrix | A positive semidefinite finite complex matrix of trace one, SLDSpectralEnergy.IsDensity. | | Second-factor partial trace | Defs.partialTrace₂ ρ i j = ∑ b, ρ (i,b) (j,b). | | Local observable | Hermitian H, lifted as Defs.liftLeft H = H ⊗ 1. | | Equation (6) | Defs.IsSLD ρ H L: Hermitian L with $i[\rho,H]=(L\rho+\rho L)/2$; Defs.fisherSLD ρ L is $\tfrac14\operatorname{Re}\operatorname{tr}(\rho L^2)$. | | Equation (7) | Defs.skewWY ρ H is $-\tfrac12\operatorname{Re}\operatorname{tr}[\sqrt\rho,H]^2$, with the positive matrix square root. | | Equation (10) | $C=I_F-I_W$, expressed as the difference of those two real quantities. | | Universal comparison | Defs.claim quantifies over both finite dimensions, every joint density matrix, every Hermitian local observable and every Hermitian SLD solution on both spaces, and requires $C(\operatorname{tr}_2\rho,H)\le C(\rho,H\otimes1)$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/FisherDiscord/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2025Maity–Krishna's mutual abelian-border limit questionBoth binary mutual abelian-border densities have limits, with witnesses 1 and 0.Proved

The questionMaity and Krishna, “Mutually Abelian-Bordered Binary Words”, arXiv:2509.20773v1, Section 5 (Conclusion), question 1, p. 29, ask:

The ideaReverse the first word and interleave it with the complemented second word.

Our contributionBoth binary mutual abelian-border densities have limits, with witnesses 1 and 0.

Verification details
Release source recordProved in Lean

Open problem

Maity–Krishna's mutual abelian-border limit question

Maity and Krishna, “Mutually Abelian-Bordered Binary Words”, arXiv:2509.20773v1, Section 5 (Conclusion), question 1, p. 29, ask:

“Do the limits $\lim_{n\to\infty} \mathcal{M}(n)/2^{2n}$ and $\lim_{n\to\infty} \overline{\mathcal{M}}(n)/2^{2n}$ exist?”

The counts concern ordered pairs of binary words of the same length. An internal abelian border compares a nonempty proper suffix of the first word with a prefix of the second; an external border compares a prefix of the first with a suffix of the second. Abelian equivalence means equal counts of every letter. The two border lengths may differ and the borders may overlap. The first count requires both borders; the second requires neither.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/AbelianBorders/MutualAbelianBorderDensity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2025Cubic graphs are additively rigidEvery finite simple cubic graph has an AR-labeling onto the entire interval from one to its edge count.Proved

The questionAn injective positive-integer edge labeling makes a vertex additively rigid when all subset sums of its incident edge labels are distinct.

Why it mattersThe assertion asks whether cubic incidence constraints can always be satisfied while using every label in the shortest possible positive interval.

The ideaThe orders four and six have explicit labelings.

Our contributionEvery finite simple cubic graph has an AR-labeling onto the entire interval from one to its edge count.

Verification details
Release source recordProved in Lean

Open problem

manattu lakshmanan 2025 cubic ar graphs

Manattu and Lakshmanan S., arXiv:2502.11760, Section 7, state: “whether all cubic graphs are AR-graphs is still an open question.” The exact assertion is preregistered in #14910.

An injective positive-integer edge labeling makes a vertex additively rigid when all subset sums of its incident edge labels are distinct. An AR-labeling of a graph with $m$ edges is a bijection from the edges onto ${1,\ldots,m}$ making every vertex additively rigid. The target includes every finite simple graph of degree three at every vertex, including the empty graph.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/EdgeLabeling/CubicARGraph.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2025Matveeva's literal nonzero-parameter global-boundedness beliefA rational linear solution contradicts the literal nonzero-parameter assertion about global boundedness.Refuted

The questionAnastasia Matveeva, On the integrality of some P-recursive sequences, arXiv:2511.02121v1, Section 4.2 after the general generating-function solution of equation (4.1):

Why it mattersThe counterexample is nonzero and verifies both original global-boundedness clauses, including actual complex summability.

The ideaThus the chosen solution satisfies every premise of claim and is globally bounded, contradicting its conclusion.

Our contributionA rational linear solution contradicts the literal nonzero-parameter assertion about global boundedness.

Verification details
Release source recordRefuted in Lean

arXiv

Matveeva's literal nonzero-parameter global-boundedness belief

Anastasia Matveeva, On the integrality of some P-recursive sequences, arXiv:2511.02121v1, Section 4.2 after the general generating-function solution of equation (4.1):

Supposedly, if e does not vanish in the equation above, S(x) is not globally bounded. However, this question remains open for now.

Section 7 states the parameter condition explicitly:

The case of $b_2 \neq \frac{2a_2b_1-a_1a_2b_0}{a_1}$ is still poorly understood, although we believe it leads to not globally bounded solutions.

The literal assertion quantifies over every rational tuple $(a_1,a_2,b_0,b_1,b_2)$ with $a_1a_2\neq0$ and $b_0\notin{-2,-3,\ldots}$, and every sequence $s:\mathbb N\to\mathbb Q$ with arbitrary rational initial values satisfying

$$ (n+b_0)s_n+(a_1n+b_1)s_{n-1}+(a_2n+b_2)s_{n-2}=0 \qquad\text{for every }n\geq2. $$

It asserts that

$$ E=a_1a_2b_0+a_1b_2-2a_2b_1\neq0 \quad\Longrightarrow\quad s\text{ is not globally bounded}. $$

Definitions 1.4 and 1.5 require both of the following for global boundedness: a single nonzero integer $C$ for which $C^ns_n$ is an integer for every $n\geq1$, and a real $R>0$ for which the full complex series $\sum_{n=0}^{\infty}s_nz^n$ is summable for every complex $z$ with $|z|<R$. The integer witnessing integrality may depend on $n$; there is no integrality restriction on $s_0$.

There is no minimal-order, nonrationality, normalized-initial-values or uncancelled-exponential hypothesis. Requiring an exponential factor to survive simplification of the full generating function would change the assertion. Conjecture 4.1 is a separate assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MatveevaGlobalBoundedness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryCounterexample foundFirst preprint 2025Refutation of both Toda-prime denominator assertionsBoth denominator-based assertions about Toda primes fail.Refuted

The questionStephen McKean, Toda primes, arXiv:2511.19744v1, §5.1, Conjecture 5.6, printed p. 8:

Why it mattersA Bernoulli denominator records predecessor divisibility, whereas Toda primality additionally depends on a prime not dividing the input.

The ideaThe denominator of the prime reciprocal sum divides this product.

Our contributionBoth denominator-based assertions about Toda primes fail.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of both Toda-prime denominator assertions

Stephen McKean, Toda primes, arXiv:2511.19744v1, §5.1, Conjecture 5.6, printed p. 8:

Conjecture 5.6. Let d be a Bernoulli denominator. (i) If F (d) = 4a for some integer a, then T (am) = T (a) whenever D₄ₐₘ = D₄ₐ. (ii) If D₄ₐₘ ≠ D₄ₐ = d, then t(am) ≥ t(a).

Definition 1.1, §1, printed p. 1:

Definition 1.1. Let n be an integer. A Toda prime of n is an odd prime p such that p − 1 | 4n and gcd(p, 4n/(p−1)) = 1. Denote the set of Toda primes of n by T (n) (see Table 1), and let t(n) := |T (n)| (see Figure 1).

Notation 5.5, §5.1, printed p. 8:

Notation 5.5. Given an even integer 2m, let P (2m) denote the set of primes such that p − 1 | 2m, so that D₂ₘ = ∏ₚ∈P(2m) p. Let F (d) = min{2m > 0 : D₂ₘ = d} [PW23].

Formal reading

Lean names have prefix D5.S3.Arith.McKeanTodaPrimes.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $T(n)$ | todaPrimes | Odd primes with predecessor dividing $4n$ and coprime quotient. | | $t(n)$ | t | Cardinality of the finite Toda-prime set. | | $D_k$ | bernDen | Denominator of Mathlib's rational bernoulli k. | | $P(k)$ | denominatorPrimes | Primes whose predecessor divides $k$. | | Conjecture 5.6(i) | clauseI | Positive $a,m$, first even index $4a$, denominator unchanged. | | Conjecture 5.6(ii) | clauseII | Positive $a,m$, denominator changed. | | The full conjecture | claim | Conjunction of the two universal assertions. | | Both negations | refutedClauses | Each component assertion is false. | | Refutation | result | Exactly ¬ claim. |

The condition $F(D_{4a})=4a$ is expressed equivalently as the absence of a smaller positive even $k$ with $D_k=D_{4a}$. The equality at $4a$ is automatic. For positive $n$, the range $p<4n+2$ is exhaustive because $p-1\mid4n$ implies $p-1\le4n$. At $n=0$, the quotient is zero and no prime is coprime to zero, so $T(0)$ is empty. All counterexamples have positive indices. The natural-number restriction suffices to refute the integer conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/McKeanTodaPrimes/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Further frontiersQuestion resolvedFirst preprint 2025Weak mixing and neighborhood shadowing without small periodic setsA weakly mixing compact metric system with neighborhood two-shadowing and without dense nonempty small periodic sets.Proved

The questionJonathan Meddaugh and Elyssa Stephens, Shadowing with small diameter sets of bounded cardinality, arXiv:2509.04312v1, Question 5.6:

Why it mattersOrdinary shadowing and weak mixing yield dense small periodic sets in the source's cited setting.

The ideaLet X consist of binary sequences on the natural numbers such that for every s>=0 and p>=1 some j<=p^2+2 satisfies x(s+jp)=0.

Our contributionA weakly mixing compact metric system with neighborhood two-shadowing and without dense nonempty small periodic sets.

Scope & next questionThe hierarchy Questions 5.1 and 5.2 and language-characterization Questions 5.4 and 5.5 remain outside this result.

Verification details
Release source recordProved in Lean

arXiv

Weak mixing and neighborhood shadowing without small periodic sets

Jonathan Meddaugh and Elyssa Stephens, Shadowing with small diameter sets of bounded cardinality, arXiv:2509.04312v1, Question 5.6:

Is there a weakly mixing system which has the neighborhood N-shadowing property (N>1) that does not have dense small periodic sets?

Use the intended nonempty meaning of a small periodic set: a nonempty compact set B with f^p(B)=B for a positive natural p. The printed definition omits nonemptiness. The phase space is compact metric and the map continuous; the source does not require a homeomorphism.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Dynamics/Shadowing/ProgressionExample.quadratic_progression_example

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    Source 1
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    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Gaussian quotient monotonicity does not force two exceptional signsWeak increase of a Gaussian quotient permits an extra negative coefficient.Refuted

The questionMirzavaziri and Yaqubi, Graph binomial coefficients and a positivity conjecture of Guo and Krattenthaler, arXiv:2610.19249v1, Problem 7.2, p. 16, ask:

Why it mattersThe public surface consists of the three source definitions and this single theorem.

The ideaCancellation and exact polynomial arithmetic verify the first quotient against the literal polynomial smallU.

Our contributionWeak increase of a Gaussian quotient permits an extra negative coefficient.

Verification details
Release source recordRefuted in Lean

Open problem

Gaussian quotient monotonicity does not force two exceptional signs

Mirzavaziri and Yaqubi, Graph binomial coefficients and a positivity conjecture of Guo and Krattenthaler, arXiv:2610.19249v1, Problem 7.2, p. 16, ask:

Characterize the quadruples $(N,k,a,b)$ for which $\genfrac{[}{]}{0pt}{}{N}{k}_q/([a]_q[b]_q)$ is a polynomial whose only negative coefficients are $-1$ at $q^1$ and at $q^{D-1}$. In particular, is this always the case when $\genfrac{[}{]}{0pt}{}{N}{k}_q/[a]_q$ is weakly increasing from $q^1$ to its middle coefficient?

The answer to the final sufficiency question is negative. The full characterization remains open.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Gaussian polynomial and q-integers | Existing LiUncuDefs.gauss and QProductNecessityRefutation.qInteger | The q-Pascal Gaussian and finite geometric sum are reused; q-integers are mapped from natural to integer coefficients. | | The two ordinary quotients are polynomials | claim quantifies integer polynomials $U,P$ with the two exact multiplication identities | These identities state exact division, without rational-function ambiguity. | | Weak increase from $q^1$ to the middle | increasingFromOne | Every adjacent comparison at $2\le s\le\lfloor\deg(U)/2\rfloor$; no comparison from degree zero is imposed. | | Only negative coefficients are $-1$ at $1,D-1$ | twoExceptionSignPattern | Both exceptional coefficients equal $-1$; every other natural index is nonnegative, including the zero tail. Here $D=\deg(P)$. | | Is this always the case? | claim, negated by result | Universal natural indices $N,k,a,b$, with $k\le N$ and positive $a,b$, and all exact polynomial quotients. | | Domain conventions | Natural subtraction and division; $k\le N$, $1\le a,b$ | Subtraction is truncated. These are the preregistered conventions; the numerical screen's narrower conditions are not added as hypotheses. |

The private adjacent_iff_monotoneOn correspondence connects the literal adjacent condition to MonotoneOn on the closed interval. It changes no public definition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GaussianQuotient/MirzavaziriYaqubiMonotonicityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2026The Jordan obstruction for the Jones–Temperley–Lieb endpoint-sector seedsNegami (arXiv:2610.00293, Problem 6.8) asks for a proof of the Jordan obstruction observed numerically for Katz-Long-Moody seeds built from the endpoint sectors of the Jones-Temperley-Lieb path representations.Proved

The questionProblem 6.8 ends: "Also prove the Jordan obstruction for the Jones--Temperley--Lieb endpoint-sector seeds".

The ideaThe proof uses spectral estimates.

Our contributionNegami (arXiv:2610.00293, Problem 6.8) asks for a proof of the Jordan obstruction observed numerically for Katz-Long-Moody seeds built from the endpoint sectors of the Jones-Temperley-Lieb path representations.

Verification details
Release source recordProved in Lean

Open problem

The Jordan obstruction for the Jones–Temperley–Lieb endpoint-sector seeds

H. Negami, Quantum gates from the middle convolution of twisted Burau representations, arXiv:2610.00293v1, builds braid representations for quantum gates by the Katz–Long–Moody construction: a seed representation of $F_n\rtimes B_n\cong B_{1,n}$ on $V$ gives operators $S_i$ on $V^{\oplus n}$, $S_i=s_i^{\oplus n},\mathrm{diag}(I,R_i,I)$ with $R_i=\begin{pmatrix}0&g_i\I&I-g_{i+1}\end{pmatrix}$, and a quotient by $K+L_\lambda$, $K=\bigoplus_j\mathrm{Ker}(g_j-I)$. Remark 6.7(b) tests seeds from the endpoint sectors of the Jones–Temperley–Lieb path representations ($\rho(\sigma_i)=\alpha_\ell I+\alpha_\ell^{-1}E_i$, $\alpha_\ell=ie^{-\pi i/(2\ell)}$) for $n=3$, $\ell\in{4,5,6,7,8,10,12,16}$ and $n=4$, $\ell\in{5,7}$: in sectors with $K=0$ finite-precision rank tests suggest non-trivial Jordan blocks of the ambient $S_i$, which "by themselves … establish neither non-semisimplicity nor unboundedness". Problem 6.8 ends: "Also prove the Jordan obstruction for the Jones--Temperley--Lieb endpoint-sector seeds". The verbatim statements are in the literature note.

Issue #13581 fixes the reading: the obstruction is a Jordan chain of length two of $S_1$ (all $S_i$ are conjugate to it in the braid group) in a sector with $K=0$, with the standard path-model operators $E_i$ (which the source names without writing their entries), and it is proved for every $n\ge3$, $\ell\ge4$ and every endpoint sector of dimension at least two.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/JonesTemperleyLiebLongMoodyJordanObstruction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresConjecture provedPosed in 2025The A084068 Pell gcd conjectureCoprimality and oddness of companion Pell numbers prove the gcd formula in OEIS A084068.Proved

The questionStocke's July 28, 2025 conjecture in OEIS A084068 states A084068(n) = gcd(A001108(n), A001109(n)) for every n at least one.

Why it mattersDefine Pell numbers P by P(0)=0, P(1)=1 and companion Pell numbers Q by Q(0)=Q(1)=1, both with R(n+2)=2R(n+1)+R(n).

The ideaInduction converts the second-order recurrences into P(n+1)=P(n)+Q(n), Q(n+1)=2P(n)+Q(n).

Our contributionCoprimality and oddness of companion Pell numbers prove the gcd formula in OEIS A084068.

Scope & next questionThe other A084068 conjecture concerning A348295 is outside this result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A084068 Pell gcd conjecture

Joseph A. Stocke's July 28, 2025 conjecture in OEIS A084068 states A084068(n) = gcd(A001108(n), A001109(n)) for every n at least one. The entry still labels this formula a conjecture when inspected on September 8, 2026.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/PellCompanionGcd.pell_companion_gcd

Frozen module record

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    Source 1
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Discrete structuresExact formula provedPosed in 2010Closed form of the A107928 harmonic-mean-numerator recurrenceStephan's three-residue closed form for the harmonic-mean-numerator recurrence.Proved

The questiona(n) is the numerator of harmonic mean of a(n-1) and a(n-2).

Why it mattersThe entry has carried Stephan's explicit unbounded conjecture since 2010.

The ideaUse the block invariant (a(3m+1), a(3m+2)) = (38^m, 28^m).

Our contributionStephan's three-residue closed form for the harmonic-mean-numerator recurrence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Closed form of the A107928 harmonic-mean-numerator recurrence

OEIS A107928, NAME (%N, verbatim):

a(n) is the numerator of harmonic mean of a(n-1) and a(n-2).

FORMULA (%F, verbatim):

a(1)=2; a(2)=3; n>=3: a(n) = numerator(2*a(n-1)*a(n-2)/(a(n-2)+a(n-1))).

CONJECTURE (%F, verbatim; Ralf Stephan, Dec 01 2010):

Conjecture: G.f.: x*(2+3*x+12*x^2+8*x^3+8*x^4)/[(1-2*x)*(1+2*x+4*x^2)]. a(3n) = (3/2)*8^n, a(3n+1) = 3*8^n, a(3n+2) = 2*8^n, for n>0. - Ralf Stephan, Dec 01 2010

AUTHOR (%A, verbatim):

%A Zak Seidov, Jun 10 2005

The first terms (%S) are 2, 3, 12, 24, 16, 96, 192, 128, 768, 1536, 1024, 6144, … . The formal claim is ∀ m ≥ 1, a(3m) = 12·8^(m−1) ∧ a(3m+1) = 3·8^m ∧ a(3m+2) = 2·8^m. The Lean definition uses a 0 = 0 as an offset-one sentinel and implements the reduced numerator as Rat.num ... .toNat. Only the three-residue closed form is formalized by elementary block induction; no priority claim is made. The printed generating function with +8x^4 expands to 2, 3, 12, 24, 32, ... and disagrees with the sequence at n = 5 (a(5) = 16), whereas changing the sign to -8x^4 yields the rational generating function x(2+3x+12x^2+8x^3-8x^4)/(1-8x^3) of the proved closed form; the generating-function sentence is neither formalized nor claimed, and the discrepancy is recorded here, not corrected upstream.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/StephanHarmonicMeanNumeratorClosedForm.stephan_a107928

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About this highlight
  • Contribution

    Stephan's three-residue closed form for the harmonic-mean-numerator recurrence.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2010.

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Number theoryCounterexample foundPosed in 2008The A145110 conjectured Redheffer support countsA divisor-pair count refutes Granvik's twenty conjectured Redheffer support values.Refuted

The questionNumber of elements in the Redheffer matrix that contribute to the Moebius function.

Why it mattersThis is an explicit finite conjecture about a union of permutation supports.

The ideawhere τ(d) is the number of positive divisors of d.

Our contributionA divisor-pair count refutes Granvik's twenty conjectured Redheffer support values.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The A145110 conjectured Redheffer support counts

OEIS A145110, NAME (verbatim):

Number of elements in the Redheffer matrix that contribute to the Moebius function.

COMMENT conjecture line (verbatim):

a(11) to a(30) are conjectured to be 11, 23, 13, 18, 19, 25, 17, 29, 19, 30, 26, 26, 23, 45, 27, 30, 32, 39, 29, 48. [From Mats Granvik, Dec 21 2008]

For n≥2, let B_n have rows and columns indexed by 1,…,n. Its entry is 1 exactly when j=1 or i divides j, except that B_n(n,n)=0. A position (i,j) contributes when there is a permutation σ of {1,…,n} with σ(i)=j and B_n(k,σ(k))=1 for every k. Let a(n) be the number of such positions. Thus the conjecture is the conjunction of the twenty displayed equalities, starting with a(11)=11 and ending with a(30)=48.

The entry's array for n=8 is the transpose of this matrix convention and contains 13 positions. At n=1 the literal modified-matrix definition gives 0 against the entry's datum 1; that index is outside the conjectured list.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/GranvikRedhefferSupport.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture provedDequatre's OEIS A343502 divisor-count prime conjectureDequatre's OEIS A343502 conjecture: every member except 2 and 3 is composite.Proved

The questionDequatre's OEIS A343502 is titled “Numbers k such that tau(tau(k)) and tau(k+1) are both prime, where tau is the number of divisors function.” Its comment states: “Conjecture: all members except 2 and 3 are composite.” The entry also asks whether there are finitely many odd terms.

Why it mattersThe conjecture isolates the only prime members visible in the published data and would explain the observed composite tail.

Our contributionDequatre's OEIS A343502 conjecture: every member except 2 and 3 is composite.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dequatre's OEIS A343502 divisor-count prime conjecture

C. H. R. Dequatre's OEIS A343502 is titled “Numbers k such that tau(tau(k)) and tau(k+1) are both prime, where tau is the number of divisors function.” Its comment states: “Conjecture: all members except 2 and 3 are composite.” The entry also asks whether there are finitely many odd terms.

The formal claim is that every natural number $k$ with $1 \le k$, with both divisor-count conditions prime, and with $k \ne 2$ and $k \ne 3$, satisfies $1 < k$ and is not prime.

Statement echo

The membership predicate in D5/S3/Arith/Divisors/OeisA343502DivisorCountPrimes.lean expands tau literally as k.divisors.card, preserving the source convention. The settling declaration is result : claim; its Scribe formula expands the complete claim body.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Divisors/OeisA343502DivisorCountPrimes.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the Catalan-composition series A374570The normalized solution of OEIS A374570 has odd coefficients above degree one exactly at powers of two plus one.Proved

The questionHanna, Jul 11 2024), quoted verbatim from Library/Arith/hanna2024a374570.md:

Why it mattersThe KPI is open problems resolved, not modules, seats, or finite checks.

The ideaConstruct the integer solution A = X B by degree-contracting stabilization of B -> catalanUnit B(X^2 B catalanUnit).

Our contributionThe normalized solution of OEIS A374570 has odd coefficients above degree one exactly at powers of two plus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the Catalan-composition series A374570

OEIS A374570 (Paul D. Hanna, Jul 11 2024), quoted verbatim from Library/Arith/hanna2024a374570.md:

NAME:

Expansion of g.f. A(x) satisfying A(x)^2 = A( A(x)*C(x) ), where C(x) = x + C(x)^2 is the Catalan function (A000108).

COMMENT:

Conjecture: for n > 1, a(n) is odd iff n = 2^k + 1 for k >= 0.

The normalization is A(0) = 0, a(1) = 1, and C(0) = 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CatalanCompositionSquareParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 2/4 · 15% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFibbinary parity of the A374571 coefficientsThe odd coefficients of OEIS A374571 occur exactly at Fibbinary indices.Proved

The questionHanna, Jul 11 2024), quoted verbatim from the claim in Library/Arith/hanna2024a374571.md:

Why it mattersThis is a first-tier recent OEIS conjecture, selected for a universal proof.

The ideaThe integer sequence a is defined by the exact recursion a(0) = 1, a(2n) = a(n), and a(2n+1) = −Σ{j≤n} a(j)·a(n−j).

Our contributionThe odd coefficients of OEIS A374571 occur exactly at Fibbinary indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Fibbinary parity of the A374571 coefficients

OEIS A374571 (Paul D. Hanna, Jul 11 2024), quoted verbatim from the claim in Library/Arith/hanna2024a374571.md:

NAME:

Expansion of g.f. A(x) satisfying A(x) = A(x^2) - x*A(x^2)^2.

COMMENT:

Conjecture: for n > 0, a(n) is odd iff n = A003714(k) for some k > 0, where A003714 lists Fibbinary numbers whose binary representation contains no two adjacent 1's.

The formalization uses the normalization A(0) = 1 and the bitwise predicate Fibbinary n := n &&& (n >>> 1) = 0 for no adjacent one bits.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/FibbinarySquareSubstitutionParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedIntegral Compositional Square Roots and Hanna's Parity ConjectureThe normalized solution of Hanna's system has integer coefficients, odd exactly at powers of two.Proved

The questionHanna's OEIS A378258 Name states: "G.f. satisfies A(x) = B(x) + 5A(x)^2 where B(A(x) - A(x)^2) = x." Its Comments state: "Conjecture: a(n) is odd iff n is a power of 2."

Why it mattersThe parity assertion depends on an integral construction of the generating function.

The ideaDividing the next residual coefficient by two therefore gives an even correction.

Our contributionThe normalized solution of Hanna's system has integer coefficients, odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a378258 hanna square root parity

Paul D. Hanna's OEIS A378258 Name states: "G.f. satisfies A(x) = B(x) + 5*A(x)^2 where B(A(x) - A(x)^2) = x." Its Comments state: "Conjecture: a(n) is odd iff n is a power of 2."

The normalized rational series has constant coefficient zero and linear coefficient one, with offset one. The binding statement of #18825 asserts existence, integrality and dyadic parity for every normalized solution and every $n\ge1$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A=B+5A^2$ | IsGF existentially quantifies the rational series $B$ and includes this equality | Literal source equation. | | $B(A-A^2)=x$ | IsGF includes PowerSeries.subst (A - A ^ 2) B = X | The inner series is the first argument of substitution. | | Offset one and $a(1)=1$ | IsGF requires constant coefficient zero and linear coefficient one | Source normalization. | | A generating function exists | The first conjunct of claim is existence of an IsGF solution | Prevents vacuity. | | Integer sequence and parity conjecture | claim provides an integer representative of each coefficient, with oddness equivalent to $n=2^k$ for some natural $k$ | Every natural degree at least one; no finite cutoff. | | Every normalized solution | Universal quantification over rational $A$ satisfying IsGF | Uniqueness identifies all such solutions; no extra assumption. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaA378258SquareRootParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedDyadic support of the odd coefficients of A380678Hanna's A380678 series has odd coefficients exactly at the indices of A027383.Proved

The questionThe Library note records the month and year but not the exact %A line date.

Why it mattersThis is a first-tier OEIS conjecture from the 2025 entry.

The ideaConstruct A over the integers as the stabilised fixed point of the reversion-type contraction f + X - f.subst (X - f^2 invOfUnit (1-f^2) 1).

Our contributionHanna's A380678 series has odd coefficients exactly at the indices of A027383.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dyadic support of the odd coefficients of A380678

Paul D. Hanna, OEIS A380678, February 2025. The Library note records the month and year but not the exact %A line date. The following NAME and COMMENT are copied verbatim from Library/Recurrence/hanna2025a380678.md.

NAME:

G.f. A(x) satisfies A( x - A(x)^2/(1 - A(x)^2) ) = x.

COMMENT:

Conjecture: for n>= 1, a(n) is odd iff n is a term of A027383, where A027383(2*m) = 3*2^m - 2 and A027383(2*m+1) = 4*2^m - 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/QuadraticSquareReversionDyadicSupportParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFibbinary parity of the A380708 coefficientsThe positive-index coefficients of OEIS A380708 are odd exactly at twice a fibbinary number plus one.Proved

The questionHanna, Feb 08 2025, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a380708.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2025 entry.

The ideaThe integer series A is the stabilized fixed point of Phi(F) = 1 + X (absSeries (invOfUnit F 1)) ^ 2.

Our contributionThe positive-index coefficients of OEIS A380708 are odd exactly at twice a fibbinary number plus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Fibbinary parity of the A380708 coefficients

OEIS A380708, Paul D. Hanna, Feb 08 2025, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a380708.md:

G.f. A(x) satisfies A(x) = 1 + x*abs( 1/A(x) )^2.

Conjecture: for n > 0, a(n) is odd iff n = 2*A003714(k) + 1 for k >= 0, where A003714 are the fibbinary numbers.

Here abs means coefficientwise absolute value of the reciprocal, taken before squaring, as defined in the sibling entries A380709/A380710. A003714 enumerates the nonnegative integers with no two adjacent 1s in their binary expansion. The frozen public predicate is Fibbinary f := f &&& (f >>> 1) = 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/AbsoluteReciprocalSquareFibbinaryParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedAbsolute reciprocal squares and binary parity in A380710The positive-index coefficients of OEIS A380710 are odd exactly at powers of two.Proved

The questionThe following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2025a380710.md.

Why it mattersThe KPI is open problems resolved: the target is the parity assertion for every positive index, together with existence and uniqueness for the exact defining equation.

The ideaThe integer series A is constructed by stabilized iteration of Phi(F) = 1 + X F absSeries (invOfUnit (F ^ 2) 1).

Our contributionThe positive-index coefficients of OEIS A380710 are odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Absolute reciprocal squares and binary parity in A380710

Paul D. Hanna, OEIS A380710, Feb 18 2025. The following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2025a380710.md.

NAME:

G.f. A(x) satisfies A(x) = 1 + x*A(x)*abs( 1/A(x)^2 ).

COMMENT:

Conjecture: for n > 0, a(n) is odd iff n is a power of 2.

Here abs is the series of coefficientwise absolute values, as defined by the entry's formulas and the sibling entry A383377. It is not a pointwise absolute value of an analytic function.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/AbsoluteReciprocalSquareParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Analysis & dynamicsQuestion resolvedAbsolute Hermite coefficient modes: the complete A381524 difference assertionThe smallest exponent of a largest absolute physicists Hermite coefficient changes by one or minus one.Proved

The questionSmallest exponent of x of maximal coefficient (ignoring signs) in Hermite polynomial of order n.

Why it mattersThis is a Tier 1 explicit recent OEIS conjecture under the repository's source classification, not an arXiv assertion or a difficulty measurement.

The ideaThe proof uses the original physicistsHermite, signedCoeff and HermiteSignedMode.polynomialcoeff, whose original statement and proof are preserved with only the consumed extraction made public.

Our contributionThe smallest exponent of a largest absolute physicists Hermite coefficient changes by one or minus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Absolute Hermite coefficient modes: the complete A381524 difference assertion

OEIS A381524, revision 18, by Mike Sheppard (2025), defines:

Smallest exponent of x of maximal coefficient (ignoring signs) in Hermite polynomial of order n.

Its independently printed Comments assertion is:

Conjecture: Differences are either +1 or -1.

The offset is zero. Let $H_n$ be the physicists Hermite polynomial represented by the original HermiteSignedMode.physicistsHermite, and let $b(n)$ be the least $j\in{0,\ldots,n}$ maximizing $|[x^j]H_n|$. The complete event is

$$ \begin{aligned} \forall n\in\mathbb N,\quad &b(n)\le n,\ &\forall k\le n,\quad |[x^k]H_n|\le |[x^{b(n)}]H_n|,\ &\forall k\le n,\quad |[x^k]H_n|=|[x^{b(n)}]H_n|\ \Longrightarrow\ b(n)\le k,\ &b(n+1)-b(n)\in{1,-1}\quad\text{in }\mathbb Z. \end{aligned} $$

Both selector values are cast to integers before subtraction. Every natural order, including zero, every exponent from zero through the order, parity zeros and all ties are included. The source's HermiteH convention and examples fix physicists normalization: $b(5)=3$, and $b(8)=2$ chooses the smaller of the tied exponents two and four. The claim is the one entire Differences assertion with its selector certificate, without a uniqueness premise, finite cutoff or asymptotic qualification.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Polynomial/HermiteAbsoluteMode.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedSigned Catalan cubic substitution modulo threeThe coefficients of OEIS A386666 satisfy Hanna's ternary residue conjecture.Proved

The questionA(x) satisfies: A(x)^2 = A( x^2 + 4A(x)^3 ). Conjecture: a(3n) (mod 3) == 2a(3n+1) (mod 3) == a(3n+2) (mod 3) == 2A039966(n) for n > 0, where A039966(n) equals the number of partitions of n into distinct powers of 3.

The ideaOver ZMod 3, signedCatalanSeries satisfies S + S² = X and the reduced cubic equation.

Our contributionThe coefficients of OEIS A386666 satisfy Hanna's ternary residue conjecture.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Signed Catalan cubic substitution modulo three

OEIS A386666 (Paul D. Hanna, Sep 07 2025) states:

G.f. A(x) satisfies: A(x)^2 = A( x^2 + 4*A(x)^3 ).

Conjecture: a(3*n) (mod 3) == 2*a(3*n+1) (mod 3) == a(3*n+2) (mod 3) == 2*A039966(n) for n > 0, where A039966(n) equals the number of partitions of n into distinct powers of 3.

The conjecture is formalized with distinctPowersOfThree n standing for A039966(n).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SignedCatalanCubicSubstitutionModThree.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe full Laurent rows in A391810 sum to zeroThe nonpositive Laurent coefficients in A391810 have zero full row sum.Proved

The question(2) Sum{k=0..floor((n+1)/2)^2} T(n,k) = 0 for n >= 1 (conjecture).

Why it mattersThis is a Tier 1 explicitly published arithmetic conjecture.

The ideaThe negative outer indices -1 through -n contribute y^j(1-y^j)^(n-j) for j=1 through n.

Our contributionThe nonpositive Laurent coefficients in A391810 have zero full row sum.

Scope & next questionApplicability of this cancellation argument to those neighboring questions remains open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The full Laurent rows in A391810 sum to zero

Paul D. Hanna's OEIS A391810 formula (2) states:

(2) Sum_{k=0..floor((n+1)/2)^2} T(n,k) = 0 for n >= 1 (conjecture).

The defining NAME is:

Irregular triangle, read by rows, where row n lists the coefficients of nonpositive powers of x in the doubly infinite series Sum_{k=-oo..oo} x^k * (1 - x^k)^(n+k) for n >= 1.

Formula (1) identifies the truncated Laurent expansion, including the constant term, with Sum_{k=0..floor((n+1)/2)^2} T(n,k)*x^(-k). All rows with positive integer n are included. Expansion takes place at zero; negative integer exponents mean Laurent-field inverses.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/HannaLaurentRowSums.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A392525 coefficientsThe coefficients of OEIS A392525 are odd exactly at powers of two.Proved

The questionHanna, Jan 15 2026) gives the following NAME, FORMULA (1), and COMMENT.

Why it mattersThe KPI is open problems resolved, not the number of modules or finite coefficient checks.

The ideaInteger existence is proved by exact coefficient corrections and stabilization.

Our contributionThe coefficients of OEIS A392525 are odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A392525 coefficients

OEIS A392525 (Paul D. Hanna, Jan 15 2026) gives the following NAME, FORMULA (1), and COMMENT. The quotations are copied from the claim in Library/Arith/hanna2026a392525.md; the attribution and date were supplied by the orchestrator.

NAME:

G.f. satisfies: A(x) = A( x^3 + 15*x*A(x)^3 )^(1/3), with A(0)=0, A'(0)=1.

FORMULA (1):

(1) A(x)^3 = A( x^3 + 15*x*A(x)^3 ).

COMMENT:

Conjecture: a(n) is odd iff n = 2^k for k >= 0.

范围与诚实边界

The module formalizes FORMULA (1) over integer formal power series with A(0)=0 and A'(0)=1 (constant coefficient zero and linear coefficient one). It does not formalize the cube-root operation in the NAME. The public parity theorem is stated for every n ≥ 1; at n = 0 the normalization gives a(0)=0, and zero is not a power of two. The separate second comment, a(n) ≡ 5 mod 10 iff n = 2^k, k ≥ 1, is not resolved here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CubicFifteenSubstitutionParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThe eighth compositional iterate of A396798 modulo eightEvery coefficient above degree one in the eighth compositional iterate of A396798 is divisible by eight.Proved

The questionHanna's OEIS A396798, revision 12, defines an ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)A^5(x)." Its formula field says that A^k denotes the k-th iteration.

Why it mattersThe existing 3/4 and 5/6 results concern distinct source equations.

The ideaDefine H(F)=X+I4(F)I5(F) and take coefficient n from H^n+1.

Our contributionEvery coefficient above degree one in the eighth compositional iterate of A396798 is divisible by eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The eighth compositional iterate of A396798 modulo eight

Paul D. Hanna's OEIS A396798, revision 12, defines an ordinary generating series by "G.f. satisfies A(x) = x + A^4(x)*A^5(x)." Its formula field says that A^k denotes the k-th iteration. The eighth comment states:

Conjecture: [x^n] A^8(x) == 0 (mod 8) for n > 1.

Let I0(A)=X and I(k+1)(A)=Ik(A) composed with A. For the unique zero-constant integer series satisfying A=X+I4(A)*I5(A), prove forall n:Nat, 1<n -> (8:Int) divides coeff n (I8(A)). The product is ordinary multiplication and the coefficients have no factorial normalization. The source object is constructed, not assumed to exist.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductFourFiveEighthModEight.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFibbinary parity of the A397244 coefficientsThe positive-index coefficients of OEIS A397244 are odd exactly at twice a Fibbinary number plus one.Proved

The questionHanna, gives the following NAME and Fibbinary COMMENT, quoted verbatim from Library/ArithSums/hanna2026a397244.md and matched to oeis-A397244.src supplied by the orchestrator:

Why it mattersThis is a first-tier OEIS conjecture in the recent-entry intake.

The ideaFor n > 1, let P be the strict prefix through degrees j < n.

Our contributionThe positive-index coefficients of OEIS A397244 are odd exactly at twice a Fibbinary number plus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Fibbinary parity of the A397244 coefficients

OEIS A397244, Paul D. Hanna, gives the following NAME and Fibbinary COMMENT, quoted verbatim from Library/ArithSums/hanna2026a397244.md and matched to oeis-A397244.src supplied by the orchestrator:

G.f. A(x) satisfies (2*n) * [x^n] A(x)^(2*n) = (2*n-1) * [x^n] A(x)^(2*n+1) for n > 1.

Conjecture: for n > 0, a(n) is odd iff n = 2*A003714(k) + 1 for some k >= 0, where A003714 lists the Fibbinary numbers.

The initial values are a(0)=a(1)=1. A003714 enumerates the nonnegative integers with no adjacent binary ones; the frozen public predicate is Fibbinary f := f &&& (f >>> 1) = 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/QuadraticPowerDiagonalFibbinaryParity.hanna_conjecture_a397244

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedAlternation modulo three of the A397349 coefficientsSum{n>=1} a(n)x^n/n = log(1+x + Sum{n>=2} 3n/(3n^2 - 1) a(n)x^n ).Proved

The questionHanna's NAME and BOTH conjecture COMMENT lines are quoted verbatim from the orchestrator-supplied oeis-A397349.src:

Why it mattersThis OEIS conjecture is selected from the repository's triage note Library/Words/oeis2026triage0911b.md.

The ideaFor natural indices and integer values, set A k = if k = 1 then 1 else (3k²−1)·s k and B m = if m ≤ 1 then 1 else 3m·s m.

Our contributionSum{n>=1} a(n)x^n/n = log(1+x + Sum{n>=2} 3n/(3n^2 - 1) a(n)x^n ).

Scope & next questionThe separate assertion that a(n) is odd exactly at powers of two remains open and is not claimed as proved.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Alternation modulo three of the A397349 coefficients

Paul D. Hanna's NAME and BOTH conjecture COMMENT lines are quoted verbatim from the orchestrator-supplied oeis-A397349.src:

%N A397349 L.g.f. Sum_{n>=1} a(n)*x^n/n = log(1+x + Sum_{n>=2} 3*n/(3*n^2 - 1) * a(n)*x^n ).
%C A397349 Conjecture: a(n) is odd iff n is a power of 2.
%C A397349 Conjecture: a(n) = [1,2] repeating (mod 3) for n >= 1.

A397349 carries two conjectures. This lane settles ONLY the modulo-three alternation for every positive index. The separate assertion that a(n) is odd exactly at powers of two remains open and is not claimed as proved. The module retains it solely as the unproved definition parity_conjecture : Prop.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/LogarithmicWeightCatalanParity.hanna_conjecture_a397349_mod_three

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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Discrete structuresConjecture refutedFirst preprint 2025Refutation of Pudelko Conjecture 8 at position zeroA uniform count bound refutes the proposed one-quarter limit for Fibonacci minima.Refuted

The questionPudelko, Modular Periodicity of Random Initialized Recurrences, arXiv:2510.24882v5, Conjecture 8, proposes that the probability of a Fibonacci sequence having its absolute minimum at position zero is one quarter.

The ideaFor the bilateral Fibonacci sequence with a0=x and a1=y, the values at indices 1, 2, -1, and -2 are y, x+y, y-x, and 2x-y.

Our contributionA uniform count bound refutes the proposed one-quarter limit for Fibonacci minima.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of Pudelko Conjecture 8 at position zero

Pudelko, Modular Periodicity of Random Initialized Recurrences, arXiv:2510.24882v5, Conjecture 8, proposes that the probability of a Fibonacci sequence having its absolute minimum at position zero is one quarter. Equation (14) gives the bounded-initialization interpretation:

P_N(n) = P(n) + epsilon(N), epsilon(N) -> 0 as N -> infinity.

For integer initial values (x,y) in the inclusive square [-N,N]^2, IsBilateralFibonacci x y a specifies a sequence by a 0 = x, a 1 = y, and a (k + 2) = a (k + 1) + a k for every integer index k. The predicate min0 x y requires the absolute value at position zero to be no greater than the absolute value at every integer index for every sequence satisfying those initial values and the bilateral recurrence. Ties are therefore included. The definition boundedMin0Probability N counts those pairs and divides by (2N+1)^2. The resolved clause, claim, is the quantified convergence assertion

forall epsilon : Rat, 0 < epsilon -> exists N0 : Nat,
  forall N : Nat, N0 <= N -> abs (P_N(0) - 1/4) < epsilon.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/PudelkoFibonacciMinimumLimitRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Tournament solid-metric lower boundsThe source is recorded in Library/ConceptDynamics/hakanenpavan2026solid.md.Proved

The questionThe source is recorded in Library/ConceptDynamics/hakanenpavan2026solid.md.

Why it mattersThe decisive mechanism bounds the number of outside vertices of a solid landmark set by a chain cover of its constrained distance profiles.

The ideaThe literal solid predicate is equivalent to the ordered-pair strict-distance criterion.

Our contributionThe source is recorded in Library/ConceptDynamics/hakanenpavan2026solid.md.

Scope & next questionGeneral sharpness, the arbitrary-order minimum and existence of every intermediate dimension remain open here.

Verification details
Release source recordProved in Lean

arXiv

Tournament solid-metric lower bounds

Hakanen and Pavan, Solid-Resolving Sets on Directed Graphs, arXiv:2610.35041v1, Conjecture 3.4, page 8, asserts that every tournament of orders $6,7,8,9,10,11$ has solid-metric dimension at least $4,5,5,5,5,6$, respectively. The source is recorded in Library/ConceptDynamics/hakanenpavan2026solid.md. Preregistration: #22881.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | All tournaments of order $n$ | Every relation on Fin n satisfying the reused loopless exclusive-or IsTournament | No strong-connectivity or diameter hypothesis | | Directed distance, including unreachable vertices | SolidResolvingTournamentBase.directedDistance, the extended-natural infimum of directed walk lengths | Deleting repeated-vertex cycles identifies the infimum with shortest-path length | | Distance to a nonempty vertex set | SolidResolvingTournamentBase.distanceToSet, the infimum over its subtype | Infinity is retained when every member is unreachable | | Definition 1.1, every nonempty $X\ne{v}$ | SolidResolvingTournamentBase.IsSolidResolving | Literal subset quantifier and inequality | | Minimum solid-set cardinality | SolidResolvingTournamentBase.solidMetricDimension | Natural infimum; full-set solidity gives attainment | | Orders $6,7,8,9,10,11$ and bounds $4,5,5,5,5,6$ | The six conjuncts of SolidResolvingTournament.claim, proved by result | No dropped clause or extra assumption |

The namespaces in this dossier have prefix D5.S3.Combinatorics.Graph.Walks. All auxiliary profile hypotheses are discharged inside the settling proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Walks/SolidResolvingTournament.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedThe second exterior trace of Spezia’s serpentine matrixThe second exterior trace of the serpentine matrix has Spezia’s magnitude.Proved

The questionStefano Spezia, OEIS A322844, COMMENTS revision 68, states:

Why it mattersThe two conjectures ask for uniform identities across every matrix order.

The ideaThe first result therefore proves the absolute-value formula.

Our contributionThe second exterior trace of the serpentine matrix has Spezia’s magnitude.

Scope & next questionThe result changes their status in the characteristic-coefficient formulation; it makes no claim about unrelated Spezia matrix families or a new general rectangular-product identity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The second exterior trace of Spezia’s serpentine matrix

Stefano Spezia, OEIS A322844, COMMENTS revision 68, states:

For n > 1, a(n) is the absolute value of the trace of the 2nd exterior power of an n X n square matrix M(n) defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). Equivalently, a(n) is the absolute value of the coefficient of the term [x^(n-2)] in the characteristic polynomial of the matrix M(n), or the absolute value of the sum of all principal minors of M(n) of size 2.

The matrix definition is shared with A317614 revision 194. With one-based indices $1\leq i,j\leq n$, its integer entries are $M_{ij}=j+n(i-1)$ for odd $i$ and $M_{ij}=ni-j+1$ for even $i$. Put $\varepsilon=n\bmod 2$ and

$$ a(n)=\frac{n^2}{12}\bigl(3(1+n^2)-2(2+n^2)\varepsilon\bigr). $$

Statement echo

| Source clause or convention | Lean counterpart | Fidelity | | --- | --- | --- | | One-based integer serpentine entries | SpeziaSerpentineSecondExteriorTrace.M | Fin n indices are increased by one; entry arithmetic is integer arithmetic. | | Source rational expression for $a(n)$ | SpeziaSerpentineSecondExteriorTrace.a | Rational division and natural remainder; no truncated rational quotient. | | Second exterior trace, characteristic coefficient, principal-minor sum | SpeziaSerpentineSecondExteriorTrace.traceExt | The binding convention uses $(-1)^k[x^{n-k}]\chi_M$ for $k\leq n$ and zero for $k>n$. It does not construct an exterior-power linear map. | | First conjecture, every $n>1$ | SpeziaSerpentineSecondExteriorTrace.claim, result | Integer absolute value, cast to the rationals, equals $a(n)$. | | Second conjecture, every $n$ and $k>2$ | SpeziaSerpentineHigherExteriorTrace.claim, result | Includes $n<k$, including the empty matrix, using the zero exterior-space convention. | | Centered coordinates and two rectangular factors | w, u, U, V in the second-trace module | Auxiliary constructions, not additional hypotheses on either claim. |

All declaration names in this table have prefix D5.S3.Quantum.Matrix.. The source calls the two coefficient/minor descriptions equivalent views of the first conjecture. No additional source conjecture is counted from those equivalent descriptions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/SpeziaSerpentineSecondExteriorTrace.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2025A sharp Weyl–Bloch ℓ1 entanglement criterion for two qutritsThe two-qutrit Weyl--Bloch entrywise one-norm is bounded by 25 on separable states.Proved

The questionA 113, 032440 (2026); arXiv:2508.18393v2, Section VI (p. 10):

Why it mattersThe source characterizes PPT and realignment entanglement for Bell-diagonal qutrit states through the entry-wise 1-norm of their Bloch vectors and asks whether the same norm yields an entanglement criterion for arbitrary bipartite states.

The ideaThe proof uses spectral estimates.

Our contributionThe two-qutrit Weyl--Bloch entrywise one-norm is bounded by 25 on separable states.

Verification details
Release source recordProved in Lean

Open problem

A sharp Weyl–Bloch ℓ1 entanglement criterion for two qutrits

T. C. Sutter, C. Popp and B. C. Hiesmayr, Phys. Rev. A 113, 032440 (2026); arXiv:2508.18393v2, Section VI (p. 10):

Going forward, it is furthermore an open problem whether for general bipartite qudit states, an entanglement criterion based on the entry-wise 1-norm of the Bloch vector, similar to Thm. IV.1, can be derived. … A quick numerical optimization for $d=3$ yields $\Vert\beta\Vert_1 \lesssim 25.9735$ for general pure states $|\psi\rangle\in\mathbb C^3\otimes\mathbb C^3$, whereas separable pure states seem to satisfy $\Vert \beta\Vert_1 \leq 25$.

Here $\beta_{ij,kl}={\rm tr}(\rho,(W_{i,j}\otimes W_{k,l})^\dagger)$ with the Weyl–Heisenberg operators $W_{k,l}=\sum_j\omega^{jk}|j\rangle\langle j+l|$, $\omega=e^{2\pi i/3}$, and $\Vert\beta\Vert_1=\sum|\beta_{ij,kl}|$. The verbatim texts are in the literature note. Issue #14780 preregisters the reading and the proof plan.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/QutritWeylBlochNormSeparableBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is proved.

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Number theoryQuestion resolvedFirst preprint 2025Takemura's fixed-point first appearance degreeTakemura's fixed-point alternating power difference has first degree n - 1 and value n!.Proved

The questionFor n >= 2, let fix(sigma) be the number of fixed points of a permutation sigma of Fin n, and define

Why it mattersThe first appearance degree measures when the alternating projection of a power of the fixed-point statistic first survives.

The ideaExpand fix(sigma)^m as a sum over m-tuples and exchange the tuple and permutation sums.

Our contributionTakemura's fixed-point alternating power difference has first degree n - 1 and value n!.

Verification details
Release source recordProved in Lean

Open problem

Takemura's fixed-point first appearance degree

For n >= 2, let fix(sigma) be the number of fixed points of a permutation sigma of Fin n, and define

APD_m(fix) = sum_{sigma in S_n} sign(sigma) * fix(sigma)^m.

Takemura's Conjecture 1 in arXiv:2512.18169v1 asks to prove that the least positive m for which this sum is nonzero is n - 1:

theorem resultDegree (n : Nat) (hn : 2 <= n) :
    firstAppearanceDegree n fix = n - 1

The source reports numerical verification through n = 10 and leaves the statement unproved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/TakemuraFixedPointAlternatingPowerDifference.resultDegree

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedFirst preprint 2025Takemura's fixed-point first appearance valueTakemura's fixed-point alternating power difference has first degree n - 1 and value n!.Proved

The questionFor n >= 2, let fix(sigma) be the number of fixed points of a permutation sigma of Fin n, and define

Why it mattersThe conjecture gives the exact first nonzero alternating power difference, not only its degree.

The ideaExpand fix(sigma)^(n-1) as a sum over (n - 1)-tuples and exchange the tuple and permutation sums.

Our contributionTakemura's fixed-point alternating power difference has first degree n - 1 and value n!.

Verification details
Release source recordProved in Lean

Open problem

Takemura's fixed-point first appearance value

For n >= 2, let fix(sigma) be the number of fixed points of a permutation sigma of Fin n, and define

APD_m(fix) = sum_{sigma in S_n} sign(sigma) * fix(sigma)^m.

Takemura's Conjecture 2 in arXiv:2512.18169v1 asks to prove the value at degree n - 1:

theorem resultValue (n : Nat) (hn : 2 <= n) :
    apd n fix (n - 1) = (n.factorial : Int)

The source reports numerical verification through n = 10 and leaves the statement unproved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/TakemuraFixedPointAlternatingPowerDifference.resultValue

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Discrete structuresQuestion resolvedTebni's odd-prime congruence for OEIS A364822Integral hyperbolic EGF coefficients have residue two at every odd prime.Proved

The questionTebni's OEIS A364822, revision 25, names the sequence:

Why it mattersThis is an explicitly published OEIS conjecture.

The ideaThe factorial-scaled product formula IntegralEGFComposition.eCoeffmul gives

Our contributionIntegral hyperbolic EGF coefficients have residue two at every odd prime.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Tebni's odd-prime congruence for OEIS A364822

M. Tebni's OEIS A364822, revision 25, names the sequence:

Expansion of e.g.f. cosh(x) / (1 - 2*sinh(x)).

The conjecture resolved here is:

Conjectures: For p prime (p > 2), a(p) == 2 (mod p).

Preregistration: #18643. The companion conjecture is recorded in #18644.

Statement echo

The natural binder $p$ has hypotheses $p$ prime and $2<p$. The conclusion is an integer $z=a(p)$ with $z\equiv2\pmod p$. The rational formal series have coefficients $1/n!$ in even indices for cosh and in odd indices for sinh, and zero in their other indices. The coefficient $a(n)$ is $n!$ times the ordinary coefficient of their formal quotient. The existential integer equals the rational coefficient after casting; thus integrality and the congruence are both conclusions. No finite index bound or extra hypothesis is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/TebniA364822.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedTebni's power-of-two congruence for OEIS A364822Hyperbolic EGF coefficients have residue one at every power of two.Proved

The questionTebni's OEIS A364822, revision 25, names the sequence:

Why it mattersThis is an explicitly published OEIS conjecture.

The ideaThe factorial-scaled product formula IntegralEGFComposition.eCoeffmul gives

Our contributionHyperbolic EGF coefficients have residue one at every power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Tebni's power-of-two congruence for OEIS A364822

M. Tebni's OEIS A364822, revision 25, names the sequence:

Expansion of e.g.f. cosh(x) / (1 - 2*sinh(x)).

The conjecture resolved here is:

For n = 2^m (m natural number), a(n) == 1 (mod n).

Preregistration: #18644. The companion conjecture is recorded in #18643.

Statement echo

The natural binder $m$ ranges over all natural numbers including zero. The conclusion is an integer $z=a(2^m)$ with $z\equiv1\pmod{2^m}$. The rational formal series have coefficients $1/n!$ in even indices for cosh and in odd indices for sinh, and zero in their other indices. The coefficient $a(n)$ is $n!$ times the ordinary coefficient of their formal quotient. The existential integer equals the rational coefficient after casting; thus integrality and the congruence are both conclusions. No finite index bound or extra hypothesis is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/TebniA364822PowerOfTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Proof methods · 2/4 · 15% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Vatter's largest derangement limit below oneBounded decreasing tails yield derangement limits cofinal below one.Refuted

The questionVincent Vatter, "An Assortment of Problems in Permutation Patterns: Unimodality, Equivalence, Derangements, and Sorting", arXiv:2602.16355v2, Section 4, unnumbered paragraph immediately after Question 4.3:

Why it mattersThis is a first-tier question explicitly posed in a 2026 paper.

The ideaFor each natural k, let Ck contain the genuine permutations P(n,a) = (a+1,...,n,a,...,1) for a <= min(k,n).

Our contributionBounded decreasing tails yield derangement limits cofinal below one.

Verification details
Release source recordRefuted in Lean

arXiv

Vatter's largest derangement limit below one

Vincent Vatter, "An Assortment of Problems in Permutation Patterns: Unimodality, Equivalence, Derangements, and Sorting", arXiv:2602.16355v2, Section 4, unnumbered paragraph immediately after Question 4.3:

Is there a largest possible limit strictly less than 1?

Locator: https://arxiv.org/html/2602.16355v2#S4. This dossier concerns only that sentence. It is distinct from the universal convergence question and has no problem number of its own.

A permutation class is a downset of actual permutations under subsequence order-isomorphism. Write C_n for its length-n slice, C_n° for the fixed-point-free members, and r_C(n) = |C_n°| / |C_n|. Let Attained(l) mean that some such class has nonempty slices at every length and r_C(n) -> l as n -> infinity. The affirmative largest-attained-limit proposition is

exists L in R, L < 1 and Attained(L) and
  forall l in R, (l < 1 and Attained(l)) implies l <= L.

The source imposes no growth or finite-basis condition. Nonempty slices are provided by the constructed witnesses, so their ratios do not rely on an empty-denominator convention. Even if a broader convention admits additional limits, any proposed maximum L < 1 is contradicted by the same witness construction with q = L.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/DerangementLimitsCofinal.exists_derangement_limit_between

Frozen module record

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Discrete structuresConjecture refutedFirst preprint 2025Wang's Conjecture 5.3Wang's Conjecture 5.3 fails for a six-element finite commutative monoid over Q: the ordinary algebraic constant is one and the proposed right side is at least two.Refuted

The questionWang, arXiv:2509.18724v1, Conjecture 5.3, asks whether every finite commutative semigroup S and field K satisfying either the splitting condition or the characteristic-p / p-power-exponent condition obey

Why it mattersA finite product semigroup tests whether their local quotient and stabilizer information still predicts nonvanishing in the ordinary semigroup algebra.

The ideaUse all six labels (b,n) in {0,1} x {0,1,2}, with maximum and addition truncated at two.

Our contributionWang's Conjecture 5.3 fails for a six-element finite commutative monoid over Q: the ordinary algebraic constant is one and the proposed right side is at least two.

Verification details
Release source recordRefuted in Lean

arXiv

Wang's Conjecture 5.3

Wang, arXiv:2509.18724v1, Conjecture 5.3, asks whether every finite commutative semigroup S and field K satisfying either the splitting condition or the characteristic-p / p-power-exponent condition obey

$$d(S,K)=\max_{a\in S}{\epsilon_a D_{\bar a}(S/\mathcal H) +d(\Gamma(H_a),K)}.$$

Here epsilon is zero exactly when the positive cyclic subsemigroup is a group with its own identity. The algebra is the ordinary semigroup algebra, the scalar quantifier is universal over individually nonzero scalars, D is the relative irreducible-length constant at the canonical quotient image, and Gamma consists of the actual stabilizer translation maps.

| Source clause | Lean counterpart | | --- | --- | | every finite commutative semigroup | S : Type, AddCommSemigroup S, Finite S in claim | | arbitrary field and both alternatives | Field K, Admissible S K | | ordinary basis and convolution | upstream AddMonoidAlgebra K S and its single | | positive cyclic semigroup and idempotent | positiveMultiple, cyclic, IsCyclicIdempotent, cyclicIdempotent | | period, exponent, splitting | period, semigroupExponent, Splitting | | group property in epsilon | CyclicIsGroup, epsilon | | every independently nonzero coefficient | GoodSequence quantifies all nonzero scalar functions of position | | supremum, including unboundedness | algebraicD : ENat, supremum over every finite list | | actual principal ideals and congruence quotient | Below, Green, greenCon and its upstream canonical quotient | | relative Davenport constant and empty subsequences | SequenceSum, IrreducibleAt, relativeD | | actual stabilizer translations, equality of maps | HClass, Gamma, composition addition | | maximum of the exact summands | rightSide, an ENat supremum over the finite carrier | | whole universal equality and its negation | claim : Prop, result : Not claim |

The source's infinite-length notation is interpreted as unbounded admissible finite lengths, without inventing an infinite product. For a general nonunital semigroup algebra only positive products are used; a bottom length zero does not affect the supremum. The empty sequence sums to an identity only if the semigroup already has one. The model lies in a monoid over Q, where the usual nonzero empty product is available. No zero-ring claim is needed for the field-valued conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Algebra/WangConjecture53Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresCounterexample foundFirst preprint 2026The unrestricted exact Wang–Tian saturation formula is falseThe unrestricted exact eventual formula for the maximum number of arcs in a saturated strict digraph fails at connectivity three and arbitrarily large odd orders.Refuted

The questionThere is no rounding, parity, divisibility restriction or error term.

Why it mattersAn obstruction at arbitrarily large orders defeats the eventual quantifier without computing any extremizer.

The ideaThe left side is even and the right side odd, a contradiction for every threshold.

Our contributionThe unrestricted exact eventual formula for the maximum number of arcs in a saturated strict digraph fails at connectivity three and arbitrarily large odd orders.

Verification details
Release source recordRefuted in Lean

Open problem

The unrestricted exact Wang–Tian saturation formula is false

Qinglin Wang and Yingzhi Tian, Extremal Problems for the Family of k-Strongly Connected Digraphs, arXiv:2605.01269v1, Section 4, Conjecture 1, printed PDF page 13, state: “For sufficiently large $n$, $\operatorname{ex}(n,\mathcal D_k)=\frac32(k-\frac43)(n-k+1)+\binom n2$.” The source is Conjecture 1.

Statement echo

The closed proposition claim quantifies every $k\in\mathbb N$ with $1\le k$, then a threshold $N\in\mathbb N$ depending on $k$, then every $n\in\mathbb N$ with $N\le n$. Its equality is

$$ (\operatorname{ex}(n,\mathcal D_k):\mathbb Q) =\frac32\left((k:\mathbb Q)-\frac43\right) \left((n:\mathbb Q)-(k:\mathbb Q)+1\right) +(\operatorname{Nat.choose}(n,2):\mathbb Q). $$

Every division and subtraction is rational. There is no rounding, parity, divisibility restriction or error term. The local Section 4 reading $k\ge2$ is also contradicted by $k=3$.

Strict digraphs have ordered distinct-endpoint arcs, permit reciprocal arcs, and have no parallel arcs. A subdigraph chooses arbitrary vertex and arc subsets, retaining isolated vertices. Dipaths are vertex-simple directed paths. A separator leaves a nonstrong or singleton remainder. For nonempty records the connectivity is the attained minimum separator size. Positive strong connectivity excludes empty records; the source leaves null connectivity unspecified. This explicit totalization retains the source's Lemma 2.2 consequence that a $k$-strong graph has at least $k+1$ vertices. Saturation means forbidden-free and that every missing ordered arc creates a $k$-strong arbitrary subdigraph. The extremum is the actual attained maximum arc count over a proved nonempty finite family of saturated hosts. The Blueprint explains the invertible relabelling to Fin n; it does not claim a separate compiled transport theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Saturation/WangTianStrongSaturationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationSharp bound provedFirst preprint 2026The exact A|BC constant of the GHZ measureProved

The questionWu, “A measure for genuine tripartite entanglement”, arXiv:2605.02876v3, Section V.C, equation (37), state:

Why it mattersFor a real direction n, the spin is nx X + ny Y + nz Z.

The ideaPut u = EA(a1), u′ = EA(a2) and fjk = EBC(bj,ck).

Verification details
Release source recordProved in Lean

Open problem

The exact A|BC constant of the GHZ measure

S. Wu, K. Zhong and J. Wu, “A measure for genuine tripartite entanglement”, arXiv:2605.02876v3, Section V.C, equation (37), state:

Numerically maximising E_GHZ over all biseparable A|BC states we find the sharp value sup_{ρ ∈ A|BC} E_GHZ(ρ) = ½, (37) attained e.g. by |0⟩_A ⊗ |Φ⁺⟩_BC and coinciding with the product-state value (34); the analytic proof of the exact constant 1/2 remains open. The same 1/2 holds for the B|AC and C|AB partitions by symmetry.

Issue #12719 preregisters the Tier 1 A|BC statement: every product of arbitrary density matrices has E_GHZ at most one half, and some such product attains one half. The specified class is exactly ρ_A tensor ρ_BC, including mixed factors.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GHZMeasureBiseparableBound.result

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  • Contribution

    The GHZ measure has sharp bound one half on arbitrary A|BC product density states.

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    First preprint year: 2026.

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedSource 2026Strong domination under clique gluingThe strong domination number of a clique gluing satisfies Inequality (4) of Alikhani, Ghanbari and Henning for every clique size at least two.Proved

The questionA strong dominating set contains, for every vertex outside it, an adjacent vertex of at least that vertex's degree.

Why it mattersThe cited Library note records the journal locator and source definitions.

The ideaThe proof preserves exterior adjacencies and degrees.

Our contributionThe strong domination number of a clique gluing satisfies Inequality (4) of Alikhani, Ghanbari and Henning for every clique size at least two.

Verification details
Release source recordProved in Lean

Open problem

Strong domination under clique gluing

Saeid Alikhani, Nima Ghanbari and Michael A. Henning, Strong domination number of graphs from primary subgraphs, Quaestiones Mathematicae 49(2), 195–211, Problem 1, p. 202, asks whether, for every $r\ge2$,

$$\gamma_{st}(G)\ge\gamma_{st}(G_1)+\gamma_{st}(G_2)-1-\Sigma_{X,Y}.\tag{4}$$

The finite connected graphs $G_1,G_2$ have order at least $r+1$ and contain enumerated $r$-cliques $A,B$. The graph $G$ identifies corresponding clique vertices. Here $X_i=N_{G_1}(u_i)\setminus A$, $Y_i=N_{G_2}(v_i)\setminus B$, and $\Sigma_{X,Y}=\sum_i|X_i|+\sum_i|Y_i|$. A strong dominating set contains, for every vertex outside it, an adjacent vertex of at least that vertex's degree. The strong domination number is its minimum cardinality.

The settlement is Proved for Inequality (4), under exactly these hypotheses; the preregistration is #19327. The source's request for an optimum using finer parameters is outside this claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every $r\ge2$ | claim quantifies over r with 2 ≤ r | No upper cutoff. | | Two finite simple connected graphs | V1, V2, their finite instances, G1, G2, and connectedness hypotheses | Finite vertex labels and decidable equality add representation, not restrictions on finite graphs. | | Order at least $r+1$ | The two Fintype.card inequalities | Literal order hypotheses. | | Enumerated $r$-cliques | Embeddings A, B from Fin r, with IsClique on their ranges | Zero-based labels correspond to source indices plus one. | | Identify $u_i$ with $v_i$ | embedRight, glue on the sum/subtype vertex type | Edges are the union of the factor edge images; identified clique edges occur once. | | Strong domination | IsStrongDominatingSet, strongDominationNumber | The degree comparison is literal; the natural infimum is attained. | | External-neighbour sum | X, sigmaXY | Neighbours shared by several clique vertices contribute once for each such vertex. | | Inequality (4) | result : claim | Subtraction is in integers; no additional assumptions. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/AlikhaniGhanbariHenningCliqueGluing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2026.

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    An explicitly identified research question in the original source.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Positivity of the Two-Color Partition Series C'(2,4,n)The two-color partition series C-prime at parameters 2 and 4 has nonnegative coefficients.Proved

The questionAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 2, Conjecture 2.

The ideaTheorem 1 of the paper for k = 2, m = 4 and finite partial fractions give an exact formula for C'(2,4,n) in terms of the triangular-pair count of Gauss's identity, a periodic term, and weighted divisor sums.

Our contributionThe two-color partition series C-prime at parameters 2 and 4 has nonnegative coefficients.

Verification details
Release source recordProved in Lean

Open problem

Positivity of the Two-Color Partition Series C'(2,4,n)

George E. Andrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 2, Conjecture 2. With Σ_n C'(k,m,n) qⁿ = Σ_{j≥0} q^{m(2j+1)} (q^{2j+2}, q^{2j+2k}; q²)_∞ / (q^{2j+1}; q²)_∞² (equation (1.2)), the conjecture states that C'(2,4,n) ≥ 0 for every n ≥ 0. In contrast, the paper notes that C'(2,5,n) changes sign.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TwoColorPartition/AndrewsElBachraouiTwo.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The upper-boundary joint instrument is uniqueThe upper-boundary joint instrument of a noisy sharp meter and a depolarizing channel is unique.Proved

The questionToigo, Postponing the choice: advantage of deferred measurements in quantum information processing, arXiv:2502.19871v1 (New J.

Why it mattersThe source quantifies the advantage of deferring the choice of a second measurement by the compatibility region of a noisy sharp meter and a noisy identity channel.

Our contributionThe upper-boundary joint instrument of a noisy sharp meter and a depolarizing channel is unique.

Verification details
Release source recordProved in Lean

Open problem

The upper-boundary joint instrument is unique

C. Carmeli, T. Heinosaari and A. Toigo, Postponing the choice: advantage of deferred measurements in quantum information processing, arXiv:2502.19871v1 (New J. Phys. 27, 094501, 2025), after the proof of the compatibility theorem for a sharp meter and the identity channel:

We conclude that, when the point $(s,t)$ attains the second equality in \eqref{eq:QI_comp_neg}, the instrument $\tilde{\ii}^{(t)}$ is actually the unique $W$-covariant joint instrument for $\Qo_s$ and $I_t$. It is an open question whether in this case there exist other (necessarily not $W$-covariant) joint instruments.

Here:

  • The meter: $\mathsf Q_s(x)=s|\varphi_x\rangle\langle\varphi_x|+(1-s)I/d$.
  • The channel: $I_t=t,\mathrm{id}+(1-t),\mathrm{tr}(\cdot),I/d$.
  • Joint instrument: completely positive maps $J_x$ with $\sum_xJ_x=I_t$ and $\mathrm{tr},J_x(\rho)=\mathrm{tr}(\rho,\mathsf Q_s(x))$.
  • The upper boundary: $d^2s=(d^2-2)(1-t)+2\sqrt{(1-d^2)t^2+(d^2-2)t+1}$, for $t\in[-1/(d^2-1),1]$.

The definitions and the instrument $\tilde{\mathcal J}^{(t)}$ are quoted in the literature note.

Issue #20028 fixes the reading: on the upper boundary, every joint instrument equals $\tilde{\mathcal J}^{(t)}$, for every $d\ge2$ and every orthonormal basis. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/DeferredInstrumentUniqueness.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedSource 2026Centered reduced residue progression maximumThe longest positive-step progression in a short centered squarefree reduced residue system has the conjectured length.Proved

The questionLet n be even and squarefree with at least three distinct prime factors.

Why it mattersThe conjecture gives a closed formula for a maximum over all starts and all positive steps in a source-fixed representative system.

The ideaSquarefreeness and maximality give odd e>=3, prime p>=5, and gcd(d,p)=1.

Our contributionThe longest positive-step progression in a short centered squarefree reduced residue system has the conjectured length.

Verification details
Release source recordProved in Lean

Open problem

Centered reduced residue progression maximum

Let n be even and squarefree with at least three distinct prime factors. Let p be its greatest prime factor and put d=n/p. Under d<2p, determine the longest positive-step arithmetic progression in the fixed centered reduced residue system from -n/2+1 through n/2.

The unnumbered conjecture on printed page 14 of the cited paper states that the maximum length is floor(p-2p/d). The formal target expresses the maximum as an IsGreatest statement, so it requires an attaining progression and a bound for every positive step, together with the exact rational-floor equality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/CenteredReducedResidueProgressions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The minimum average trace distance of a rebit strategy for Elliott's exclusion task is exactly 1/12The minimum average trace distance from the four-input exclusion distribution over all rebit strategies is exactly one twelfth.Proved

The questionElliott, Strict advantage of complex quantum theory in a communication task, arXiv:2505.16523v2, Phys.

Why it mattersThe exact minimum turns the qualitative real–complex separation into a quantitative one.

The idea1. Linear dependence. Real symmetric 2 × 2 matrices form a three-dimensional space, so the four states satisfy Σx wx ρx = 0 for some real w ≠ 0.

Our contributionThe minimum average trace distance from the four-input exclusion distribution over all rebit strategies is exactly one twelfth.

Verification details
Release source recordProved in Lean

Open problem

The minimum average trace distance of a rebit strategy for Elliott's exclusion task is exactly 1/12

T. J. Elliott, Strict advantage of complex quantum theory in a communication task, arXiv:2505.16523v2, Phys. Rev. A 111, 062401 (2025), studies a four-input task with target distribution P(y|x) = (1 − δ_xy)/3 on X = Y = {0, 1, 2, 3}. A complex qubit realizes it exactly, while no real qubit (rebit) strategy does. For the robustness of this separation the source measures the error by d = (1/|X|) Σ_x (1/2) Σ_y |P(y|x) − P̃(y|x)|. It shows that d = 1/12 is achievable with a rebit (Appendix D), and writes: "We therefore only weakly conjecture that d = 1/12 the the minimum distance for a rebit strategy." The verbatim statements are in the literature note.

Issue #16523 fixes the reading. A rebit strategy is:

  • four real symmetric positive-semidefinite 2 × 2 states of trace 1;
  • a four-outcome POVM of real symmetric positive-semidefinite effects summing to the identity.

Mixed states and POVMs cover private randomness and post-processing on either side. The claim is that 1/12 is the least distance attained by such strategies. It is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/RebitExclusionDistance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2007Divergence of the last site of the nearest-neighbor QES chainThe last site of the cyclic nearest-neighbor QES chain exceeds every fixed real bound for all sufficiently large chain lengths.Proved

The questionRodríguez, arXiv:0704.3046v1, §2, printed pp. 9–10, around Eq. (19), ask:

The ideaA maximum-coordinate comparison proves uniqueness of increasing solutions locally inside result.

Our contributionThe last site of the cyclic nearest-neighbor QES chain exceeds every fixed real bound for all sufficiently large chain lengths.

Verification details
Release source recordProved in Lean

Open problem

Divergence of the last site of the nearest-neighbor QES chain

A. Enciso, F. Finkel, A. González-López and M. A. Rodríguez, arXiv:0704.3046v1, §2, printed pp. 9–10, around Eq. (19), ask:

It is also of interest to determine whether the position of the last spin tends to infinity as N → ∞, since according to our interpretation of the chain’s geometry the number 2ξN/π is the radius of the circle on which the spins lie.

The sites are increasing real solutions of Eq. (6),

$$ \xi_i=\frac1{\xi_i-\xi_{i-1}}+\frac1{\xi_i-\xi_{i+1}}, \qquad \xi_0=\xi_N,\quad \xi_{N+1}=\xi_1. $$

Issue #11858 preregisters the fully quantified assertion: for every real $B$ there is a natural $N_0$ such that every $N\geq N_0$, $N\geq3$, and every strictly increasing cyclic site solution satisfy $B<\xi_N$. The source motivates the assertion by Eq. (19), then cautions that the inverse-error-function argument in Eq. (18) is accurate only up to terms of order $1/N$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/NearestNeighborLastSiteDivergence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2007. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2026Equal totals characterize margins of alternating signed square matricesSigned row and column margins admit an alternating signed square matrix exactly when their total sums agree.Proved

The questionZumbrägel, Generalising Latin square orthogonality and Frobenius-König with alternating sign matrices, arXiv:2606.25884v1, Section 8, printed p. 27:

Why it mattersThe total sum must be the same whether all entries are counted by rows or columns.

The ideaLet p,q count positive and negative row margins and p′,q′ the corresponding column margins.

Our contributionSigned row and column margins admit an alternating signed square matrix exactly when their total sums agree.

Verification details
Release source recordProved in Lean

Open problem

Equal totals characterize margins of alternating signed square matrices

A. Ernst, S. Lia, C. O'Brien, J. Sheekey and J. Zumbrägel, Generalising Latin square orthogonality and Frobenius-König with alternating sign matrices, arXiv:2606.25884v1, Section 8, printed p. 27:

Problem 8.3. For which (0, ±1)-vectors R and S of order n does there exist X ∈ W_n with row-sums R and column-sums S?

The same page defines the class:

Section 4 introduces the set W_n consisting of all (0, ±1)-matrices in which the non-zero entries of each row and column alternate in sign, and the sum of each row/column is in {0, ±1}.

For every natural number n and signed integer vectors R,S on Fin n, the exact answer is

$$ \left(\exists X\in W_n:\ \forall i,\ \sum_j X_{ij}=R_i,\quad \forall j,\ \sum_i X_{ij}=S_j\right) \iff \sum_i R_i=\sum_j S_j. $$

Alternation is with respect to the natural row and column order. First and last signs are unrestricted. The zero order and zero lines are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Latin/AlternatingSignMargins.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026Gil–Liang–Odetola–Weiner antidiagonal trafficA uniform strict binomial-ratio bound for antidiagonal grid obstructions at every n at least 496.Proved

The questionFor every n ≥ 496 and every obstruction B on the antidiagonal x+y=n, the maximum of fB is attained at (1,1) and (n-1,n-1).

Why it mattersThis is a first-tier, externally published 2026 conjecture, already selected under issue 8249.

The ideaThe superseded two-neighbor domination is false at (498,241) and is not used; that pair does not refute the arithmetic target.

Our contributionA uniform strict binomial-ratio bound for antidiagonal grid obstructions at every n at least 496.

Verification details
Release source recordProved in Lean

arXiv

Gil–Liang–Odetola–Weiner antidiagonal traffic

Conjecture 7.4 of arXiv:2609.01562v1 states:

For every n ≥ 496 and every obstruction B on the antidiagonal x+y=n, the maximum of f_B is attained at (1,1) and (n-1,n-1).

Here f_B counts monotone north-east paths from (0,0) to (n,n) avoiding B and passing through a grid point other than those two endpoints. The obstruction ranges over all (a,n-a), 0≤a≤n. Existing preregistration: https://github.com/the-omega-institute/trureturing/issues/8249.

The formal arithmetic proposition is exactly

forall n a : Nat, 496 <= n -> 1 <= a -> 2*a < n ->
  n*(n-2*a+1)/(n-a)*choose(n,a)*choose(n-2,a-1)
    /choose(2*n-2,n-1) < 1

The divisions in this expression are rational. The domain includes a=1 and the odd near-central pair (497,248); it is not natural a<n/2. The strict inequality is a sufficient assertion for the verbal conjecture; neither its converse nor uniqueness of grid maximizers is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/GridAntidiagonalTrafficBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Unboundedness of the Large-Shift Hankel Determinants of q-Metallic NumbersFor every n at least one, the shifted Hankel determinants of the q-metallic series are unbounded in absolute value at every shift at least n+3.Proved

The questionGuo-Niu Han and Emmanuel Pedon, Hankel continued fractions and Hankel determinants for q-deformed metallic numbers, arXiv:2502.05993v2, Section 1, Conjecture E, part 2.

The ideaA cofactor identity for the inverse Hankel matrix at the normal indices of the Hankel continued fraction of the shift n + 1, valid across runs of vanishing determinants, together with a transfer computation over the dual numbers uniform in n (and the three-periodic continued fraction for n = 1), gives Δ{Pk−1}^{(n+3)} = (−1)^k 2k for n = 1 and (−1)^{nk}(4n + 2)k for n ≥ 2, where P = 2n(n + 1).

Our contributionFor every n at least one, the shifted Hankel determinants of the q-metallic series are unbounded in absolute value at every shift at least n+3.

Verification details
Release source recordProved in Lean

Open problem

Unboundedness of the Large-Shift Hankel Determinants of q-Metallic Numbers

Guo-Niu Han and Emmanuel Pedon, Hankel continued fractions and Hankel determinants for q-deformed metallic numbers, arXiv:2502.05993v2, Section 1, Conjecture E, part 2. Let Φ_n(q) = Σ_i f_i qⁱ be the power series with constant term 1 satisfying q Φ² + ((1 + qⁿ)(1 − q) − q[n]q) Φ = 1, and let Δ_j^{(ℓ)} = det(f{ℓ+a+b}){0≤a,b<j}. The conjecture states that for every n ≥ 1 and every ℓ ≥ n + 3 the sequence (Δ_j^{(ℓ)}){j≥0} is unbounded. The printed superscript reads n + 2; the varying shift ℓ is intended, as the quantifier ℓ ≥ n + 3 shows.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MetallicHankel/MetallicHankelUnbounded.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe Least Base Maximizing the Digit Sum of n Is ceiling((n+1)/2) for n > 8Past eight, the least base maximizing the digit sum of a number is the first base above half of it, because every base at most half loses more than half of the number to carries while the first base above half keeps all but half.Proved

The questionThe smallest base b < n where the sum of the digits for the number n in the base b is the largest, with 1 < b < n and a(1) = a(2) = 1.

The ideaWrite sb(n) for the base-b digit sum and b0 = floor(n / 2) + 1, which equals ceiling((n + 1) / 2).

Our contributionPast eight, the least base maximizing the digit sum of a number is the first base above half of it, because every base at most half loses more than half of the number to carries while the first base above half keeps all but half.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Least Base Maximizing the Digit Sum of n Is ceiling((n+1)/2) for n > 8

OEIS A394431, by Jean-Marc Rebert, Mar 20 2026:

The smallest base b < n where the sum of the digits for the number n in the base b is the largest, with 1 < b < n and a(1) = a(2) = 1.

The formula line carrying the question, by Sean A. Irvine, Mar 25 2026:

Conjecture: a(n) = ceiling((n+1)/2) for n>8.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/MaxDigitSumBase.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2024Finite Otto congruence quotients are cancellativeFinite semigroups with an Otto congruence have cancellative quotients.Proved

The questionTrevor Jack, Answering Five Open Problems Involving Semigroup Conjugacy, arXiv:2411.16284v1, Section 9, Problem 9.4, p. 14; Semigroup Forum 112 (2026), 450–466, DOI 10.1007/s00233-026-10614-w:

Why it mattersThe source's Problem Four constructs an infinite semigroup with an Otto congruence and a noncancellative quotient.

The ideaThe four public definitions are the literal conventions of 17635.

Our contributionFinite semigroups with an Otto congruence have cancellative quotients.

Verification details
Release source recordProved in Lean

Open problem

Finite Otto congruence quotients are cancellative

Trevor Jack, Answering Five Open Problems Involving Semigroup Conjugacy, arXiv:2411.16284v1, Section 9, Problem 9.4, p. 14; Semigroup Forum 112 (2026), 450–466, DOI 10.1007/s00233-026-10614-w:

Is there a finite semigroup for which ∼o is a congruence and S/∼o is not cancelative?

The answer is negative. Section 2, p. 2 defines the relation by:

∀a, b ∈ S (a ∼o b ⇔ ag = gb and bh = ha for some g, h ∈ S).

Both witnesses lie in $S$; no identity is adjoined.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | finite semigroup | claim quantifies over S : Type with Semigroup S and Finite S | arbitrary finite carrier, with no chosen enumeration | | Otto conjugacy | Otto | exactly the two existential witness equations in Section 2 | | is a congruence | OttoIsCongruence | multiplication compatibility; otto_refl, otto_symm, otto_trans prove equivalence | | quotient is cancelative | QuotientCancelative | both representative cancellation implications | | no such finite semigroup | result : claim | universal negative answer; empty carriers satisfy the cancellation statement vacuously |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/FiniteOttoCongruence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Mobile general position on the seven-cycle and ten-cycle cylindersThe mobile general position number of the seven-cycle and ten-cycle cylinders is four at every height at least five.Proved

The questionKlavžar, Krishnakumar, Kuziak, Shallcross, Tuite and Yero, “Moving through Cartesian products, coronas and joins in general position”, arXiv:2505.00535v2, Discrete Applied Mathematics 379 (2026), 768–780, DOI 10.1016/j.dam.2025.10.041, Section 3, after Proposition 3.7:

Why it mattersThese are the two cylinder families left undetermined by Section 3 of the source.

The ideaThe module combines the cycle and path metrics, classifies the five cycle coordinates in row order, and propagates an invariant of the occupied columns.

Our contributionThe mobile general position number of the seven-cycle and ten-cycle cylinders is four at every height at least five.

Verification details
Release source recordProved in Lean

Open problem

Mobile general position on the seven-cycle and ten-cycle cylinders

Klavžar, Krishnakumar, Kuziak, Shallcross, Tuite and Yero, “Moving through Cartesian products, coronas and joins in general position”, arXiv:2505.00535v2, Discrete Applied Mathematics 379 (2026), 768–780, DOI 10.1016/j.dam.2025.10.041, Section 3, after Proposition 3.7:

Hence, the only unknown values are Mobgp(C₇ □ Pₛ) and Mobgp(C₁₀ □ Pₛ) for s ≥ 5. We conjecture that the answer is four in these cases.

The conclusion is $\operatorname{Mob}{gp}(C_7\square P_s)= \operatorname{Mob}{gp}(C_{10}\square P_s)=4$ for every integer $s\ge5$. It is the tier-1 published conjecture preregistered in issue #17705.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Shortest-path membership | OnGeodesic | Sum of the two distances equals the endpoint distance, on the connected cylinders. | | No three occupied vertices on a common shortest path | IsGeneralPosition | All triples of pairwise distinct occupied vertices are quantified. | | One move along an edge to an unoccupied vertex, preserving general position | LegalMove | Exactly one occupied vertex is erased and the new vertex inserted. | | A sequence of legal moves from the initial set visits every vertex | IsMobile | A finite sequence with the stated initial configuration, every consecutive move legal, and a visit to every vertex. | | Largest cardinality of a mobile general position set | mobGP | Natural-number supremum; the upper bound and a size-four witness establish attainment here. | | Cartesian product of cycle and path | cylinder | Mathlib cycleGraph r and pathGraph s, on Fin r × Fin s; path height counts vertices. | | Both exceptional circumferences, all $s\ge5$, answer four | claim, result | Both equalities, the natural-height quantifier and its guard occur literally. |

Zero-based labels are a relabelling of the source vertices. The pairwise inequalities in IsGeneralPosition express three distinct vertices; the sequence length and configuration indices express the source's finite moves. They impose no extra hypothesis on the cylinder claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/GeneralPosition/CylinderSevenTen.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The q = 2 Singer-quadric qubit codes have minimum distance 2 for every d ≥ 2Kulhandjian and Hanzo (arXiv:2610.02392) build qubit stabilizer codes Q(2,d) from a Singer difference set and the quadric Tr(z^3) = 0 in PG(d,2), prove numerically for d <= 6 that the cross-correlation u = MQ tauH is the all-ones vector, and conjecture (Conjecture 24) that this and the resulting minimum distance 2 hold for every d >= 2.Proved

The questionThe verbatim statements are in the literature note.

Our contributionKulhandjian and Hanzo (arXiv:2610.02392) build qubit stabilizer codes Q(2,d) from a Singer difference set and the quadric Tr(z^3) = 0 in PG(d,2), prove numerically for d <= 6 that the cross-correlation u = MQ tauH is the all-ones vector, and conjecture (Conjecture 24) that this and the resulting minimum distance 2 hold for every d >= 2.

Verification details
Release source recordProved in Lean

Open problem

The q = 2 Singer-quadric qubit codes have minimum distance 2 for every d ≥ 2

M. Kulhandjian and L. Hanzo, Singer-Difference-Set Qudit Stabilizer Codes from Non-Degenerate Quadrics in PG(d,q), arXiv:2610.02392v1, construct for a prime power $q$ and $d\ge2$ a stabilizer code $\mathcal Q(q,d)$ with $H=(A\mid M_QA)$: $A=\mathrm{circ}(\boldsymbol\tau_H)$ is the Singer incidence matrix of $\mathrm{PG}(d,q)$ and $M_Q=\mathrm{circ}(\boldsymbol\tau_Q)$ the quadric circulant. Proposition 23 (Section VII.E) states for $q=2$ and $d\in{2,\dots,6}$ that $u=M_Q\boldsymbol\tau_H\equiv\mathbf 1$, that every $Z_iZ_j$ lies in the centralizer, and, by exhaustive search for $d\le4$, that $Z_0Z_1$ is not a stabilizer, so $d_{\min}=2$. Conjecture 24 asserts the proposition for every $d\ge2$, and the paper's list of open problems asks to settle it analytically. The verbatim statements are in the literature note.

Issue #13470 fixes the reading. At $q=2$: $K=\mathrm{GF}(2^{d+1})$, $n=2^{d+1}-1$, $(\boldsymbol\tau_H)i=1$ iff $\operatorname{Tr}(\alpha^i)=0$, $(\boldsymbol\tau_Q)i=1$ iff $\operatorname{Tr}(\alpha^{3i})=0$, circulants have entries $v{(i-j)\bmod n}$, the centralizer is ${(\mathbf a,\mathbf b):\mathbf aH_x^\top=\mathbf bH_z^\top}$ and the stabilizers are the row space of $H$. claim asserts, for every $d\ge2$ and every primitive $\alpha$ (the paper fixes one): $u=\mathbf 1$; for $i\ne j$, $(e_i+e_j,\mathbf 0)$ is in the centralizer and not a stabilizer; every centralizer element outside the stabilizers has symplectic weight $\ge2$. The witness has weight 2 also as a Hamming weight, so $d{\min}=2$ under either reading of the paper's weight.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SingerQuadricQubitDistanceCeiling.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Every Natural Number Counts the Rook Placements of a Crossword GridAn alternating five-column family realizes every positive count, and a three-by-three grid realizes zero.Proved

The questionJoel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Question 4.2:

Why it mattersThis answers the first part of Question 4.2; the characterization asked in the second part is not addressed.

The ideaFor r ≥ 1 take the grid of side max(5, 2r − 1) whose white cells are the first 2r − 1 cells of column 2 together with, for 0 ≤ i ≤ r − 2, the cells (2i, si), (2i + 1, si), (2i + 2, si), (2i + 1, ti), where (si, ti) = (3, 4) for even i and (1, 0) for odd i.

Our contributionAn alternating five-column family realizes every positive count, and a three-by-three grid realizes zero.

Verification details
Release source recordProved in Lean

Open problem

Every Natural Number Counts the Rook Placements of a Crossword Grid

Joel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Question 4.2:

For which positive integers r does there exist a crossword grid which admits exactly r rook placements? Can the crossword grids which admit exactly r rook placements be characterized?

The paper observes that block-diagonal juxtaposition multiplies counts, so that it suffices to treat prime r, and it conjectures that permutation grids miss infinitely many values.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CrosswordRookCounts.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Every data qubit of the delayed remote-controlled circuit violates the dimension witness S ≤ 4Every Euclidean-unit complex data qubit admits positive preparations and a delayed remote-control dimension witness greater than four.Proved

The questionRead as a prepare-and-measure scenario with four preparations and two binary measurements, a violation of the dimension witness S = E₀₀ − E₀₁ − E₁₀ + E₁₁ − E₂₀ − E₂₁ + E₃₀ + E₃₁ ≤ 4 (Eq. (2)) rules out every classical causal explanation with preshared correlations.

Why it mattersA violation for every input lets a client certify the nonclassicality of the delegated computation whatever data qubit it supplies, without trusting the server's devices.

The idea1. The data branches. u = A d and w = e^{iδ}·B d form an orthonormal basis of the data qubit; the phase of δ removes the relative phase of d.

Our contributionEvery Euclidean-unit complex data qubit admits positive preparations and a delayed remote-control dimension witness greater than four.

Verification details
Release source recordProved in Lean

Open problem

Every data qubit of the delayed remote-controlled circuit violates the dimension witness S ≤ 4

P. Lin, D. Wang, Y. Liu and A. Huang, Nonclassicality of a delayed remote-controlled quantum computing model, arXiv:2606.31103v1, study a delayed 1-U-M remote-controlled quantum computing circuit (§III, Eq. (3), Fig. 3):

  • a Bell pair on the program and remote-control qubits;
  • a controlled-A and a controlled-B gate on the data qubit, with A = diag(1, e^{i(β+δ)}) and B = [[0, −1], [e^{i(β−δ)}, 0]];
  • a Hadamard and a computational-basis measurement on the program qubit;
  • the remote measurements onto |P*_U⟩ = a₀*|0⟩ + a₁*|1⟩, with a₀ = e^{−i(β+δ)/2} cos(γ/2) and a₁ = e^{−i(β−δ)/2} sin(γ/2), whose β and δ must equal those of the gates.

Read as a prepare-and-measure scenario with four preparations and two binary measurements, a violation of the dimension witness S = E₀₀ − E₀₁ − E₁₀ + E₁₁ − E₂₀ − E₂₁ + E₃₀ + E₃₁ ≤ 4 (Eq. (2)) rules out every classical causal explanation with preshared correlations. Simulated annealing found violations at all 900 points of a 30 × 30 grid of initial data qubits, and the authors write: "It is reasonable to conjecture that for any arbitrary initial data qubit |d⟩, there exists a measurement on the data qubit and specific parameter values for β, δ, γ₀ and γ₁, such that, under an appropriate labeling strategy, violating the dimension witness S ≤ 4". The verbatim statements are in the literature note.

Issue #16465 fixes the reading. For every unit data vector d there are β, δ, γ₀, γ₁ and a unit data-measurement vector b such that, with the source's labeling x = 2m + n:

  • every preparation has positive probability;
  • S > 4.

It is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/DelayedRemoteControlDimensionWitness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Joint Symmetry of the Mesh Patterns 123 and 321 with Shading {0,1,2}² ∪ {(3,3)}The mesh patterns 123 and 321 with the common shading R = {0,1,2} squared together with {(3,3)} have a symmetric joint occurrence distribution.Proved

The questionZhang, Joint Equidistributions of Mesh Patterns 123 and 321 with Symmetric and Minus-Antipodal Shadings, arXiv:2501.00357v3, Section 2.4, Conjecture 2 (printed page 9), pair S21: the mesh patterns (123, R) and (321, R) are jointly equidistributed on Sₙ, where R = {0,1,2}² ∪ {(3,3)} as rendered in the source's diagram.

The ideaAn occurrence is determined by its largest value m: it exists exactly when the origin-anchored rectangle Cₘ (positions at most n − m + 3, values at most m) contains exactly three points, including the point of row m and the point of column n − m + 3, and the order type of these three points decides between 123 and 321.

Our contributionThe mesh patterns 123 and 321 with the common shading R = {0,1,2} squared together with {(3,3)} have a symmetric joint occurrence distribution.

Verification details
Release source recordProved in Lean

Open problem

Joint Symmetry of the Mesh Patterns 123 and 321 with Shading {0,1,2}² ∪ {(3,3)}

Shuzhen Lv and Philip B. Zhang, Joint Equidistributions of Mesh Patterns 123 and 321 with Symmetric and Minus-Antipodal Shadings, arXiv:2501.00357v3, Section 2.4, Conjecture 2 (printed page 9), pair S21: the mesh patterns (123, R) and (321, R) are jointly equidistributed on Sₙ, where R = {0,1,2}² ∪ {(3,3)} as rendered in the source's diagram.

An occurrence of (123, R) in π is a triple of positions with increasing values such that no other point of π lies in a shaded cell; (321, R) is the same with decreasing values. Joint equidistribution means that, for all n, k and l, the number of π ∈ Sₙ with k occurrences of (123, R) and l occurrences of (321, R) equals the number with l and k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MeshPattern/MeshPatternS21.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2006The Mišta–Fiurášek qudit partial-estimation trade-offLiteral occupation operators and the exact qudit partial-estimation frontier.Proved

The questionFiurášek, Optimal partial estimation of quantum states from several copies, arXiv:quant-ph/0606112v1, Phys.

Why it mattersThe result gives the exact information–disturbance frontier for all copy counts and qudit dimensions in the source's PSD-seed optimization.

The ideaIn OccupationPartialTransposeSpectral, the occupation indices are partitioned into coupled blocks and isolated singletons. BlockMatrices.weightedblock identifies each coupled restriction as an identity term plus a rank-one term and one distinguished coordinate. GlobalBound.globaloperatorbound combines the exhaustive decomposition with the scalar Rayleigh estimate and monotonic domination by the all-reference block.

Our contributionLiteral occupation operators and the exact qudit partial-estimation frontier.

Verification details
Release source recordProved in Lean

Open problem

The Mišta–Fiurášek qudit partial-estimation trade-off

L. Mišta Jr. and J. Fiurášek, Optimal partial estimation of quantum states from several copies, arXiv:quant-ph/0606112v1, Phys. Rev. A 74, 022316, Introduction, Eq. (2), states:

Moreover, we also conjecture that the optimal fidelity trade-off for an ensemble of N identical pure states of a d-level system has the form

$$ \sqrt{F-\frac1{N+d}}= \sqrt{(d-1)\left(\frac{N+1}{N+d}-G\right)}+ \sqrt{N\left(G-\frac N{N+d}\right)}. $$

Preregistration #16777 fixes the literal covariance-reduced PSD-seed model. The source is documented in the Library note.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $N$ copies of a $d$-level pure state, $N\ge1$, $d\ge2$ | claim quantifies natural N, d with precisely these two inequalities | Every allowed dimension and copy count is retained. | | Symmetric input space tensored with one output qudit | The carrier is the product of the literal occupation subtype ${n:\mathrm{Fin}(d)\to\mathbb N\mid\sum_i n_i=N}$ with Fin d | Occupation carriers occur inline; no alternate certificate space replaces them. | | Eq. (12), the partial transpose of the symmetric projector | OccupationPartialTransposeSpectral.RF, symmetricProjector, occupationEmbedding | The normalized word-sector projector, partial transpose and compression are the definitions. Entry formulas are proved afterward. | | Eq. (10), reference projection and output partial trace | OccupationPartialTransposeSpectral.RG, outputReferenceProjector, reused PartialTraceMutualInformation.partialTraceRight | The matrix product, partial trace and tensor with the output identity are retained. | | Positive seed and Schur normalization | chi.PosSemidef and trace equal to the complex coercion of $\binom{N+d-1}{d-1}$ | No rank restriction is imposed on competitors. | | Output and estimation fidelities | Real parts of trace (chi * RF N d) and trace (chi * RG N d) | Both scores use the literal source operators. | | $N/(N+d-1)\le G\le(N+1)/(N+d)$ | The two bounds on real G in claim | The complete closed interval, including endpoints, is retained. | | Optimal curve, Eq. (2) | IsGreatest of the set of all feasible output scores at fixed G | The displayed squared curve is both attained and an upper bound. Its square-root terms are nonnegative on the stated interval, so it is equivalent to Eq. (2). | | Proof of the conjecture | QuditPartialEstimationTradeoff.result : claim | There are no additional hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/QuditPartialEstimationTradeoff.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 3/4 · 10% weight

    First preprint year: 2006. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2025The non-unital channel conditions imply both network boundsThe three real parameter constraints for the non-unital channel class imply both quadratic and quartic network bounds.Proved

The questionGanguly, Network nonlocality breaking channels, arXiv:2510.26417v1 (2025), Section IV.B, NetworkNonlocality.tex, lines 733–734:

Why it mattersEstablishing the implication from Eq. (6) removes the additional parameter test in Eq. (38) throughout the three-parameter domain used in that section.

The ideaFix a real triple satisfying the three hypotheses.

Our contributionThe three real parameter constraints for the non-unital channel class imply both quadratic and quartic network bounds.

Verification details
Release source recordProved in Lean

Open problem

The non-unital channel conditions imply both network bounds

K. Mukherjee and N. Ganguly, Network nonlocality breaking channels, arXiv:2510.26417v1 (2025), Section IV.B, Network_Nonlocality.tex, lines 733–734:

Comparing Eq.(\ref{resnon1}) with that of the conditions provided by Eq.(\ref{nu3}) over the channel $\mathcal{N}_{NU},$ it is observed that there exist no region in parameter space for which Eq.(\ref{resnon1}) is violated whereas Eq.(\ref{nu3}) is satisfied. However, a formal proof of the same is beyond the scope of present work. We thus provide the following conjecture:

Conjecture 1 reads:

Any non-unital channel from $\mathcal{N}{NU}$( Eqs.(\ref{nu1},\ref{nu2})) acts as a non $n$-locality breaking channel in $\mathbf{N}{lin}.$

The quotations retain the source's TeX reference labels. The corresponding PDF passage is on page 12; nu3 is Eq. (6), resnon1 is Eq. (38), and nu1, nu2 are Eqs. (4), (5). The source TeX has MD5 6bb7806236c7d6b0bcaa73d3eed0bc28.

Issue #20159 fixes the statement to be settled: for every real $t,\lambda_1,\lambda_3$, the three conditions

$|\lambda_1|\leq1$, $|\lambda_3|+|t|\leq1$, and $\lambda_1^2+t^2\leq(1-|\lambda_3|)^2$

imply both

$(|t|+|\lambda_3|)^2+\lambda_1^2\leq1$

and

$2t^2\lambda_1^2+\lambda_1^4+(|t|+|\lambda_3|)^4\leq1$.

This parameter implication is proved. Applying it to channels requires the source's Theorem 3, which is not formalized here. The source passage and parameter correspondence are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/NonUnitalNetworkConditionImplication.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedPosed in 2025The A068012 subset sum doubling conjectureSubset complementation and a three-period symmetry prove the A068012 doubling conjecture.Proved

The questionLet a(n) count subsets of {1,...,n} whose element sum is zero modulo six.

Why it mattersDirect inspection on September 8, 2026 found the exact statement still labelled Conjecture.

The ideaWrite C(m,r) for the count at residue r modulo six.

Our contributionSubset complementation and a three-period symmetry prove the A068012 doubling conjecture.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A068012 subset sum doubling conjecture

Let a(n) count subsets of {1,...,n} whose element sum is zero modulo six. David A. Corneth conjectured on September 13, 2025 that a(n)=2a(n-1) whenever n>2 and three does not divide n-1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/SubsetSumModSixDoubling.subset_sum_mod_six_doubling

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedPosed in 2004Stephan's A091915 maximum even partition-product conjectureThe greatest even product of partition parts triples when the total increases by three.Proved

The question%N A091915 Maximum of even products of partitions of n. %F A091915 For n>6, a(n+3) = 3a(n) (conjectured). - Ralf Stephan, Dec 02 2004 %A A091915 Jon Perry, Feb 12 2004 %O A091915 1,2 %Y A091915 Cf.

Why it mattersRalf Stephan dated the conjecture December 2, 2004.

The ideaThe unconstrained A000792 optimum is 3^k, 23^k, or 43^k, according to the residue class.

Our contributionThe greatest even product of partition parts triples when the total increases by three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Stephan's A091915 maximum even partition-product conjecture

OEIS A091915 states (%N, %F, %A, %O, and %Y, verbatim):

%N A091915 Maximum of even products of partitions of n. %F A091915 For n>6, a(n+3) = 3a(n) (conjectured). - Ralf Stephan, Dec 02 2004 %A A091915 Jon Perry, Feb 12 2004 %O A091915 1,2 %Y A091915 Cf. A000792, A091916.

The literal proved statement is

theorem result (n : ℕ) (hn : 6 < n) :
    ∃ a : ℕ,
      IsGreatest
          {q : ℕ | ∃ p : Nat.Partition n, Even p.parts.prod ∧ p.parts.prod = q} a ∧
        IsGreatest
          {q : ℕ | ∃ p : Nat.Partition (n + 3), Even p.parts.prod ∧ p.parts.prod = q}
          (3 * a)

For n = 1, no partition has an even product, so the feasible set is empty. The existential formulation avoids assigning an arbitrary total-function value there: for every n > 6, it supplies a greatest even product a at total n and proves that 3*a is the greatest even product at n+3. It retains attainment and the upper bound through IsGreatest, rather than asserting only membership.

Not claimed here are any odd-product statement or result about A091916; any generating-function form; any parity result about the number of partitions; an exhaustive literature result or priority claim; or a greatest-even-product value for an empty feasible set.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Partitions/StephanEvenProductPartitionRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 3/4 · 10% weight

    Dated conjecture statement: 2004. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedThe least self-divisor exponent in OEIS A092028The least self-divisor exponent in OEIS A092028 is the least prime factor of n minus one.Proved

The questiona(n) is the smallest m > 1 such that m divides n^m-1

Why it mattersThe equality identifies the least solution of a self-referential divisibility condition for every natural input above two.

The ideaThe number p is prime and p ∣ n − 1, hence n^p ≡ 1 (mod p) and p belongs to the defining set.

Our contributionThe least self-divisor exponent in OEIS A092028 is the least prime factor of n minus one.

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Integer sequences (OEIS)

The least self-divisor exponent in OEIS A092028

OEIS A092028, NAME (%N, verbatim):

a(n) is the smallest m > 1 such that m divides n^m-1

COMMENTS (%C, verbatim; Farideh Firoozbakht, Mar 26 2004):

Conjecture 1: All terms of this sequence are primes.

Conjecture 2: a(n) is the smallest prime factor of n-1 or for n>2, A092028(n) = A020639(n-1).

The literal formal claim is ∀ n > 2, a(n) = minFac(n − 1), where a(n) = sInf {m ∈ ℕ | 1 < m ∧ m ∣ n^m − 1}. In the natural numbers, sInf ∅ = 0; for every n > 2, the defining set is nonempty. Conjecture 1 is a corollary because minFac(n − 1) is prime, and is not formalized as a separate public theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/FiroozbakhtLeastSelfDivisorPowerMinusOne.firoozbakht_a092028

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedLayman's odd-power factorial residue classificationOdd powers of a factorial have Layman's classified residue modulo the corresponding triangular number.Proved

The questionIt appears that f(n)=(n!)^(2k+1) modulo n(n+1)/2 is n if n is one less than an odd prime, else f(n) is 0, for any integer k.

Why it mattersLayman's 2010 comment asks whether the residue classification already stated for n! persists for every odd power of n!.

The ideaWhen n+1 is an odd prime, n is even and the modulus factors as (n/2)(n+1) with coprime factors.

Our contributionOdd powers of a factorial have Layman's classified residue modulo the corresponding triangular number.

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Integer sequences (OEIS)

Layman's odd-power factorial residue classification

OEIS A119690, NAME (%N, verbatim):

n! mod n*(n+1)/2.

John W. Layman's target COMMENT (%C, verbatim):

It appears that f(n)=(n!)^(2k+1) modulo n(n+1)/2 is n if n is one less than an odd prime, else f(n) is 0, for any integer k. See A175567 for related results involving an even power of n!. - John W. Layman, Jul 12 2010

The entry's FORMULA (%F, verbatim):

a(n) = n if n+1 is an odd prime, a(n) = 0 otherwise.

The literal proved statement is

forall n k : Nat, 1 <= n -> (n!)^(2*k+1) mod (n*(n+1)/2) = if Prime(n+1) and Odd(n+1) then n else 0.

The FORMULA is the k = 0 case and is already presented by the entry as a formula, not a conjecture. The new settlement is Layman's generalization to all odd powers 2*k+1; no priority is claimed for the k = 0 case.

Not claimed here are A175567 or any result about even powers, the COMMENT All terms are even, or any other property of A119690.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/LaymanOddPowerFactorialResidue.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A193020 plus-one eventual identityThe endpoint-gap rectangle is exactly the distinct ordinary self-shuffle support.Proved

The questionLayman, states the separate assertion a(2^k+1)=k^2 with its sufficiently-large lower-bound caveat.

Why it mattersCounting the support of the shuffle language makes the distinction between output words and interleaving schedules essential.

The ideaFor k at least one, the ordinary binary numeral of 2^k+1 is one, k-1 zeros, one.

Our contributionThe endpoint-gap rectangle is exactly the distinct ordinary self-shuffle support.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A193020 plus-one eventual identity

OEIS A193020, revision 13, with the 2011 entry attribution to John W. Layman, states the separate assertion a(2^k+1)=k^2 with its sufficiently-large lower-bound caveat. Here a(n), for positive n, counts DISTINCT output words obtained by shuffling the ordinary binary expansion of n with itself. Each shuffle preserves the order within both copies; different schedules producing the same word contribute only once.

The exact selected assertion is

$$\exists K\in\mathbb N,\quad 1\le K\ \land
\forall k\in\mathbb N,\quad K\le k\ \Longrightarrow\ a(2^k+1)=k^2.$$

This dossier anchors only that plus-one question. The other identities in A193020 are separate questions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/EndpointShuffleSupport.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe terminal difference of OEIS A378495For every natural n at least five, the difference T(n,n-2)-T(n,n) for derangements avoiding a cycle length is (n-3)!(n-4)(n-1)/2.Proved

The questionConjecture: T(n,n-2) - T(n,n) = (n-3)!(n-4)(n-1)/2 for n >= 5.

Why it mattersThe source is Peter Kagey's 2024 triangle, with the terminal formula introduced in revision 60.

The ideaThe latter follows because the remaining lengths sum to two and every nontrivial cycle length is at least two.

Our contributionFor every natural n at least five, the difference T(n,n-2)-T(n,n) for derangements avoiding a cycle length is (n-3)!(n-4)(n-1)/2.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The terminal difference of OEIS A378495

The FORMULA field of OEIS A378495, revision 70, states verbatim:

Conjecture: T(n,n-2) - T(n,n) = (n-3)!*(n-4)*(n-1)/2 for n >= 5.

On the source-agreeing columns $2\le k\le n$, $T(n,k)$ counts derangements in $S_n$ with no $k$-cycle; the source explicitly sets $T(n,1)=0$. The triangle is read by rows, with offset 1,5. Both cardinalities are cast to $\mathbb Q$ before ordinary subtraction; the division by two is rational division.

Statement echo

| Source clause | Exact formal counterpart | | --- | --- | | Every natural $n\ge5$ | result (n : Nat) (hn : 5 <= n), with no upper bound | | All permutations in $S_n$ | Equiv.Perm (Fin n), relabelled from zero through $n-1$ | | No fixed point | D n filters by forall x, p x != x | | No $k$-cycle | T n k counts the filter by every cycle factor's support size differing from $k$ | | Cycle-factor interpretation | Consumed noCycle_iff connects this predicate to cycle-type membership | | Columns $n-2$ and $n$ | Exactly T n (n-2) and T n n | | $(n-3)!(n-4)(n-1)/2$ | Natural factorial and natural index subtractions, then rational casts and division | | No assumed counting identity | Only 5 <= n is a hypothesis of result |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Derangements/TerminalDifference.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBinomial parity of the A380709 coefficientsThe coefficients of OEIS A380709 have Hanna's conjectured binomial parity.Proved

The questionThe following NAME and COMMENT are quoted verbatim from the claim in Library/Recurrence/hanna2025a380709.md.

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaInversion, absolute value, and cubing preserve agreement below a degree; the factor X raises that degree, making Phi a degree contraction. generatingequation states A(0) = 1 and A = 1 + X (absSeries (invOfUnit A 1))^3, exactly the NAME with coefficientwise abs taken before cubing. generatingunique covers every integer series B with constant coefficient one satisfying this equation.

Our contributionThe coefficients of OEIS A380709 have Hanna's conjectured binomial parity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Binomial parity of the A380709 coefficients

Paul D. Hanna, OEIS A380709, Feb 09 2025. The following NAME and COMMENT are quoted verbatim from the claim in Library/Recurrence/hanna2025a380709.md.

NAME:

G.f. A(x) satisfies A(x) = 1 + x*abs( 1/A(x) )^3.

COMMENT:

Conjecture: a(n) == binomial(4*n-1, n) (mod 2) for n >= 0 (cf. A263132).

Here abs is the series of coefficientwise absolute values, applied to the reciprocal before cubing, as defined by the entry's formulas and the sibling entries A380710/A383377. The positive-index assertion is hanna_conjecture. At n = 0 the entry reads binomial(-1, 0) = 1; the companion a_zero proves a(0) = 1. This reading does not use truncated natural subtraction to define a generalized negative-upper-index binomial coefficient.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/AbsoluteReciprocalCubeParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A381362 bilateral product coefficient congruenceHanna's bilateral product family has every positive coefficient equal to two modulo four.Proved

The questionA(x) satisfies 1/2 = Sum{n=-oo..+oo} x^nA(x)^n (A(x)^n + x)^(2n-1) (x^n + A(x))^(2n-1).

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaThe escape witness is the general theorem hannaconjecturegeneral, covering every natural c ≥ 2: the normalized series defined by 1/2 = Σ{n∈ℤ} xⁿ·Aⁿ·(Aⁿ + x)^{cn−1}·(xⁿ + A)^{cn−1} has a(n) ≡ 2 (mod 4) for every n > 0.

Our contributionHanna's bilateral product family has every positive coefficient equal to two modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A381362 bilateral product coefficient congruence

OEIS A381362, Paul D. Hanna, Feb 21 2025:

NAME:

G.f. A(x) satisfies 1/2 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + x)^(2*n-1) * (x^n + A(x))^(2*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4).

OEIS A381363, Paul D. Hanna, Feb 21 2025:

NAME:

G.f. A(x) satisfies 1/2 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + x)^(3*n-1) * (x^n + A(x))^(3*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4).

This dossier resolves A381362, the c = 2 instance, and cites its own Library note Library/ArithSums/hanna2025a381362.md. The sibling quote records the shared family. The generating series is normalized by A(0) = 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/BilateralProductModFour.hanna_conjecture_a381362

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A381363 bilateral product coefficient congruenceHanna's bilateral product family has every positive coefficient equal to two modulo four.Proved

The questionA(x) satisfies 1/2 = Sum{n=-oo..+oo} x^nA(x)^n (A(x)^n + x)^(2n-1) (x^n + A(x))^(2n-1).

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaThe escape witness is the general theorem hannaconjecturegeneral, covering every natural c ≥ 2: the normalized series defined by 1/2 = Σ{n∈ℤ} xⁿ·Aⁿ·(Aⁿ + x)^{cn−1}·(xⁿ + A)^{cn−1} has a(n) ≡ 2 (mod 4) for every n > 0.

Our contributionHanna's bilateral product family has every positive coefficient equal to two modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A381363 bilateral product coefficient congruence

OEIS A381362, Paul D. Hanna, Feb 21 2025:

NAME:

G.f. A(x) satisfies 1/2 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + x)^(2*n-1) * (x^n + A(x))^(2*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4).

OEIS A381363, Paul D. Hanna, Feb 21 2025:

NAME:

G.f. A(x) satisfies 1/2 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + x)^(3*n-1) * (x^n + A(x))^(3*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4).

This dossier resolves A381363, the c = 3 instance, and cites its own Library note Library/ArithSums/hanna2025a381363.md. The sibling quote records the shared family. The generating series is normalized by A(0) = 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/BilateralProductModFour.hanna_conjecture_a381363

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOddness of all A397241 termsEvery coefficient of the normalized series in OEIS A397241 is odd.Proved

The questionA(x) satisfies n [x^n] A(x)^n = (n-1) [x^n] A(x)^(n+1) for n > 1.

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaFor positive n and m, the coefficient multiplying a(n) in [x^n] A^m is ma(0)^(m-1).

Our contributionEvery coefficient of the normalized series in OEIS A397241 is odd.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Oddness of all A397241 terms

OEIS A397241, Paul D. Hanna, Jun 19 2026:

G.f. A(x) satisfies n * [x^n] A(x)^n = (n-1) * [x^n] A(x)^(n+1) for n > 1.

The entry seeds are a(0) = a(1) = 1. Its first conjecture is:

Conjecture: all terms are odd.

These NAME and COMMENT sentences are reproduced verbatim from the claim in Library/Recurrence/hanna2026a397241b.md. The two further entry comments, "Conjecture: a(n) == 2 (mod 3) for n > 2." and "Conjecture: a(n) == 3 (mod 4) for n > 2.", are not resolved by this dossier. This dossier records one claim, on hanna_conjecture only.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalPowerRatioAllOdd.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Faal's Open Problem 1: non-chordal graphs with unstable restricted clique polynomialsFor every r ≥ 1, the join of a complete graph on max(r−2,0) vertices with a four-cycle is r-connected, K{r+3}-free and non-chordal.Proved

The questionTeimoori Faal, A Bivariate B-Restricted Clique Polynomial: From Local Neighborhoods to Global Expansion, arXiv:2602.24151v1, Section 7, Open Problem 1, asks:

Why it mattersIt separates the existence question from the source's proposed chordal stability theorem.

The ideaCliques of a join split into cliques of the two factors, and

Our contributionFor every r ≥ 1, the join of a complete graph on max(r−2,0) vertices with a four-cycle is r-connected, K{r+3}-free and non-chordal.

Verification details
Release source recordProved in Lean

arXiv

Faal's Open Problem 1: non-chordal graphs with unstable restricted clique polynomials

H. Teimoori Faal, A Bivariate B-Restricted Clique Polynomial: From Local Neighborhoods to Global Expansion, arXiv:2602.24151v1, Section 7, Open Problem 1, asks:

Necessity of Conditions: Are the conditions of r-connectivity and chordality also necessary? Is there an r-connected K_{r+3}-free non-chordal graph for which C_B(G;x,y) fails to be real-stable?

The second question has a positive answer for every integer $r\geq1$. The polynomial is the literal clique sum

$$C_B(G;x,y)=\sum_{K\subseteq V(G),\ K\text{ clique}}x^{|K|}y^{|K\cap B|},$$

including the empty clique. Real stability means that the polynomial is not identically zero and has no zero with $\operatorname{Im}(x)>0$ and $\operatorname{Im}(y)>0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/CliquePolynomial/FaalNonStable.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Operational comparison of qubit incompatibility without equal lengthsRestricted-state compatibility determines incompatible unbiased qubit pairs.Proved

The questionImamura, Comparing quantum incompatibility of device sets from an operational perspective, Phys.

Why it mattersThe operational ordering compares incompatibility by the state sets on which it can be detected.

The ideaEquality of all restricted compatibility relations forces the four vectors into one plane.

Our contributionRestricted-state compatibility determines incompatible unbiased qubit pairs.

Verification details
Release source recordProved in Lean

Open problem

Operational comparison of qubit incompatibility without equal lengths

K. Torii, R. Takakura and R. Imamura, Comparing quantum incompatibility of device sets from an operational perspective, Phys. Rev. A 112, 062228 (2025), arXiv:2509.08284v2, §V:

Future work will also address the generalization of Proposition 7, as it remains an open problem whether the result holds without the constraints of |a₁|=|a₂| and |b₁|=|b₂|.

Two pairs of unbiased qubit observables are operationally equivalent when, for every set $S$ of states, one pair is $S$-compatible exactly when the other is; $S$-compatibility means that some four-outcome POVM has both margins with the same Born probabilities as the pair on every state of $S$. Proposition 7 proves that, for $|a_1|=|a_2|$ and $|b_1|=|b_2|$ in $(1/\sqrt2,1]$, equivalence forces equality up to signs and exchange. The verbatim texts are in the literature note. Issue #15522 preregisters the general statement and the route.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/QubitIncompatibilityRigidity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Comparability of conical-design concurrence boundsConical two-design concurrence bounds have a uniform ratio ordering.Proved

The questionFei, Estimating the concurrence for quantum states via symmetric measurements, arXiv:2606.31010v2, Section IV, after Theorem 2:

Why it mattersIt proves one ordering for all states for every pair of finite designs.

Our contributionConical two-design concurrence bounds have a uniform ratio ordering.

Verification details
Release source recordProved in Lean

arXiv

Comparability of conical-design concurrence bounds

H.-F. Wang, W. Zhou, L. Chen and S.-M. Fei, Estimating the concurrence for quantum states via symmetric measurements, arXiv:2606.31010v2, Section IV, after Theorem 2:

The lower bounds of concurrence induced by arbitrary two distinct conical 2-designs are comparable.

A finite positive semidefinite family satisfies $\sum_i E_i\otimes E_i=\alpha_E I+\beta_E F$, where $\alpha_E\geq\beta_E>0$ and $F$ is the tensor swap. For a positive semidefinite trace-one bipartite state $\rho$, put $P_E(\rho)_{ij}=\operatorname{tr}[\rho(E_i\otimes E_j)]$ and $B_E(\rho)=c_d(|P_E(\rho)|1-\alpha_E-\beta_E)/\beta_E$, where $c_d=\sqrt{2/[d(d-1)]}$ and $d\geq2$. The bound is the untruncated two-constant expression in Siudzińska, J. Phys. A 58 (2025) 375302, Section 7, Theorem 6, Eqs. (73)–(74), with $S=\beta_E$ and $\mathcal C{\max}=\alpha_E+\beta_E$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ConicalDesignConcurrenceComparability.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

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Discrete structuresQuestion resolvedFirst preprint 2026The Arrow Patterns (12; 3 → 3) and (23; 1 → 1) Are Arrow-Wilf-EquivalentThe avoidance classes of the two three-letter arrow patterns have equal cardinality in every positive degree.Proved

The questionZhou and Xinyang Yu, Arrow-Wilf equivalences and enumerative results for short arrow patterns, arXiv:2609.29392v1, Section 7:

The idea1. f is a fixed point of π̂ exactly when f is a left-to-right maximum of π followed by a larger entry or by the end.

Our contributionThe avoidance classes of the two three-letter arrow patterns have equal cardinality in every positive degree.

Verification details
Release source recordProved in Lean

Open problem

The Arrow Patterns (12; 3 → 3) and (23; 1 → 1) Are Arrow-Wilf-Equivalent

Robin D.P. Zhou and Xinyang Yu, Arrow-Wilf equivalences and enumerative results for short arrow patterns, arXiv:2609.29392v1, Section 7:

We have computed the avoidance counts for the arrow patterns (12; 3 → 3) and (23; 1 → 1) and found that the resulting sequences agree for all values of n tested. This suggests that these two patterns may be arrow-Wilf-equivalent. However, the explicit formulas we obtained for the two sequences are quite different in form, and we have not been able to establish the equivalence by a direct bijection or by any other method. We leave this as an open problem.

Arrow-Wilf equivalence means |S_n(12; 3 → 3)| = |S_n(23; 1 → 1)| for every n ≥ 1, with containment of an arrow pattern as defined in Section 1 of the paper through π̂ = θ⁻¹(π).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ArrowWilfEquivalence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresCounterexample foundFirst preprint 2018Alternating binary words need not minimize positional autocorrelationA period-six binary word has smaller positional autocorrelation than the alternating word at length 348.Refuted

The questionAtashpendar, Mestel, Roscoe and Ryan, A Proof of Entropy Minimization for Outputs in Deletion Channels via Hidden Word Statistics, arXiv:1807.11609v1, printed page 7, state Conjecture IV.4:

Why it mattersThe separate finite entropy-ordering question in Section V has a different statement and a different settlement.

The ideaThe two words' 348 row sums are verified in 29 groups of 12 rows.

Our contributionA period-six binary word has smaller positional autocorrelation than the alternating word at length 348.

Verification details
Release source recordRefuted in Lean

arXiv

Alternating binary words need not minimize positional autocorrelation

Atashpendar, Mestel, Roscoe and Ryan, A Proof of Entropy Minimization for Outputs in Deletion Channels via Hidden Word Statistics, arXiv:1807.11609v1, printed page 7, state Conjecture IV.4:

The alternating subsequence of length m, i.e., x = 1010..., minimizes the autocorrelation coefficient κ².

Equation (11) on the same page defines

$$\kappa^2(x)=\sum_{1\le r,s\le m}\binom{r+s-2}{r-1}\binom{2m-r-s}{m-r}[[x_r=x_s]].$$

The universal minimization statement is false. At length $m=348$, the word $x=(011001)^{58}$ has strictly smaller coefficient than the alternating word $a=1010\ldots$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Binary word of length $m$ | x : Fin m → Bool in claim | false is 0 and true is 1; each position is counted once. | | Every positive length | The natural binder $m$ and hypothesis $1\le m$ | No restriction on positive lengths. | | Equation (11) | Both literal double sums in claim | Positions are shifted down by one; the binomials become $\binom{r+s}{r}\binom{2m-2-r-s}{m-1-r}$. | | $1010\ldots$ | decide (i.val % 2 = 0) inside the left sum | Even zero-based positions carry 1. | | Minimizes | The weak inequality against every word $x$ | All words of the same length are included. | | Refutation | result : ¬ claim | The strict instance at length 348 contradicts the universal inequality. |

Natural subtraction in the binomial arguments is exact on these indices. The binders represent source lengths, positions and words; there is no additional mathematical premise. The claim keeps the source expression literal, while the proof uses the frozen D5.S3.Entropy.DeletionChannelEntropyOrdering.kappa2, whose body is that expression.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hamming/AtashpendarKappaAlternatingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2022An exclusive Grassmannian dual need not be canonicalA frame can have a unique dual minimizing the off-diagonal cross-Gramian magnitude without that dual being canonical.Refuted

The questionKaczanowski, Cross-Frame Potential, arXiv:2205.05613v3, Section 4.2, Conjecture 42, p. 17:

Why it mattersThis is an external named problem and uses deposit-uncovered without an atom or coverage edge.

The ideaLet n = 1, k = 3, F = (3, 2, 1), and G = (1/5, 2/15, 2/15).

Our contributionA frame can have a unique dual minimizing the off-diagonal cross-Gramian magnitude without that dual being canonical.

Scope & next questionConjecture 43 remains open in this delivery: co-equidistribution and its equivalence with exclusivity are not formalized or separately settled.

Verification details
Release source recordRefuted in Lean

Open problem

An exclusive Grassmannian dual need not be canonical

R. Aceska and M. Kaczanowski, Cross-Frame Potential, arXiv:2205.05613v3, Section 4.2, Conjecture 42, p. 17:

If a frame F for F^n forms an exclusive Grassmannian pair with one of its duals, then that dual must be the canonical dual frame of F.

Definition 1 requires k ≥ n vectors in an n-dimensional Hilbert space and positive finite frame bounds. Definition 3 requires both dual reconstruction identities. Definition 34 minimizes the maximum off-diagonal cross-Gramian magnitude over all dual frames. The explanation in Section 4.2 says that an exclusive pair has exactly one dual attaining this minimum. The canonical dual is S⁻¹F for the frame operator S.

The exact real finite-family reading is:

∀ n k (F G : Fin k → EuclideanSpace ℝ (Fin n)), n ≤ k →
  IsFrame F → IsDualFrame F G →
  (∀ H, IsDualFrame F H → mu F G ≤ mu F H) →
  (∀ H, IsDualFrame F H → mu F H ≤ mu F G → H = G) →
  G = canonicalDual F

IsDualFrame records the first reconstruction identity. Over the reals, its adjoint gives the second; reconstruction forces both finite families to span the space, hence to be frames. A real counterexample refutes the source's assertion for real or complex Hilbert spaces. Zero-based indexing only relabels the finite families. The empty off-diagonal maximum is zero; the counterexample has three vectors and a nonempty off-diagonal set.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/Designs/CrossFrameExclusiveDualRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Discrete structuresQuestion resolvedBala's A258899 and A258903 totient-period conjecturesDivisor-factorial tails prove Bala's A258899 and A258903 totient-period conjectures.Proved

The questionPeter Bala's COMMENTS conjecture for OEIS A258899 and A258903 states:

The ideaThe proof uses factorization.

Our contributionDivisor-factorial tails prove Bala's A258899 and A258903 totient-period conjectures.

Scope & next questionThe least possible onset and least eventual period remain open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A258899 and A258903 totient-period conjectures

Peter Bala's COMMENTS conjecture for OEIS A258899 and A258903 states:

Conjecture: the sequence a(n) taken modulo a positive integer k is eventually periodic with the period dividing phi(k).

The two sequences use the FORMULA fields

$$a_{899}(n)=\sum_{d\mid n}2^d,\frac{n!}{d!(n/d)!},\qquad a_{903}(n)=\sum_{d\mid n}2^d3^{n/d},\frac{n!}{d!(n/d)!}$$

for $n\ge1$, with value $1$ at index zero. The settling declaration states the two eventual-period assertions literally, with natural-number division and Nat.ModEq.

Statement echo

For each positive natural modulus $m$, the result supplies an onset $N$, a positive natural period $T$ dividing Nat.totient m, and the congruence for every $n\ge N$, separately for A258899 and A258903.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/BalaDivisorFactorialTotientPeriod.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    Source 1
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Number theoryConjecture provedBala's A295462 binary-spacing parity conjectureConjecture: a(n) is odd iff the 1 bits in the binary expansion of n are separated by at least three 0's, that is, iff n belongs to A048718.Proved

The questionConjecture: a(n) is odd iff the 1 bits in the binary expansion of n are separated by at least three 0's, that is, iff n belongs to A048718.

Why it mattersThe retained assertions are unbounded symbolic statements; the finite floor reduction is internal arithmetic, with no retained standalone finite positive-instance certificate.

The ideaThese are descriptions of the named compiled helpers, not additional public theorems.

Our contributionConjecture: a(n) is odd iff the 1 bits in the binary expansion of n are separated by at least three 0's, that is, iff n belongs to A048718.

Scope & next questionBroader ratios and spacing lengths remain open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A295462 binary-spacing parity conjecture

Peter Bala's OEIS A295462, native record #19, defines $a(n)=(15n)!(2n)!/((10n)!(6n)!n!)$ and states in its REFERENCES (%D) field:

Conjecture: a(n) is odd iff the 1 bits in the binary expansion of n are separated by at least three 0's, that is, iff n belongs to A048718.

The offset includes $n=0$. The rational quotient is interpreted by exact division; its natural realization is explicit before oddness is tested. Preregistration #22428 fixes the statement and conventions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The displayed factorial quotient | a | All five natural factorials are cast to $\mathbb Q$; division is exact. | | Every natural index, including zero | claim | Universal natural quantifier, with no added hypotheses. | | The quotient is a natural integer | claim | A natural witness realizes the rational value. | | Odd iff at least three zero bits separate ones | spacedBits, claim | Every occupied pair $i<j$ satisfies $i+4\le j$, in both directions of the iff. | | Conjecture holds | result | The entire claim, without additional premises. |

The names belong to D5.S3.Arith.FactorialRatio.BalaA295462Parity. The dropped or weakened clause set is empty. All-pairs spacing is equivalent to successive spacing: the forward implication restricts to successive positions; the converse follows by ordering the occupied positions and adding successive gaps.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/BalaA295462Parity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025Even-depth permuted brickwork has a negative moment directionFor four sites, every local dimension q at least two and every even depth at least two give a permuted-brickwork Haar moment with a strictly negative quadratic form.Refuted

The questionBelkin, Allen and Clark, arXiv:2510.23726v2, Section 5.2, ask:

Why it mattersThe four-site restriction suffices to refute a general even-depth PSD assertion.

The ideaThe Haar module derives the two-copy commutant from quarter phases, permutation invariance and a Hadamard constraint, and integrates the literal four-factor moment.

Our contributionFor four sites, every local dimension q at least two and every even depth at least two give a permuted-brickwork Haar moment with a strictly negative quadratic form.

Verification details
Release source recordRefuted in Lean

arXiv

Even-depth permuted brickwork has a negative moment direction

Belkin, Allen and Clark, arXiv:2510.23726v2, Section 5.2, ask:

Like the brickwork, this architecture can be shown to have a PSD vectorization if the depth is odd. It is unclear if the vectorization is PSD at even depths.

The source defines a complete layer by a uniformly random perfect matching, with independent Haar-random two-site gates on its pairs. Permuted brickwork requires every adjacent pair of matchings to have connected union. Its second-moment vectorization is $\mathbb E[\overline U\otimes\overline U\otimes U\otimes U]$. The bars are entrywise conjugation. Matrix vectorization is column-first.

Issue #13744 preregisters the negative answer on four sites: for every integer $q\ge2$ and every even $d\ge2$, the depth-$d$ vectorization is not positive semidefinite. The public Lean claim states this negative answer and result : claim proves it; Refuted refers to even-depth PSD, not to claim. The three matchings are $A={01,23}$, $B={02,13}$ and $C={03,12}$. Equal matchings have disconnected union and distinct matchings form a four-cycle, so the connected-block condition is exactly no adjacent repeats.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/RandomCircuits/PermutedBrickworkEvenRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The exhaustive-pairs question for caterpillar principal specializationsThe eighteen-vertex caterpillars (2,2,2,2,2,1,4,3) and (2,2,3,2,3,4,2) refute the second, exhaustive-pairs question in Remark 3.11.Refuted

The questionBingham, Celano, Commins, Orellana, Pan and Simone, Principal specializations of the t-chromatic symmetric function, arXiv:2610.31785v1, Remark 3.11, second question:

Why it mattersThe source found only its listed pairs through total size sixteen.

The ideaUse the original Vertex, edgeRel, caterpillar, ps, alpha and beta definitions from BinghamCaterpillarPs3Family.

Our contributionThe eighteen-vertex caterpillars (2,2,2,2,2,1,4,3) and (2,2,3,2,3,4,2) refute the second, exhaustive-pairs question in Remark 3.11.

Verification details
Release source recordRefuted in Lean

Open problem

The exhaustive-pairs question for caterpillar principal specializations

Bingham, Celano, Commins, Orellana, Pan and Simone, Principal specializations of the t-chromatic symmetric function, arXiv:2610.31785v1, Remark 3.11, second question:

And if so, do they continue to be the only pairs of compositions which do?

The assertion is that for every n at least one and every two source-admissible positive compositions of n, unequal ordinary second and identical ordinary third principal specializations imply that their unordered graph-isomorphism pair is the displayed base pair {(5,4,3),(4,5,3)} or {(5,j,j,4,3),(4,j,j,5,3)} for some positive integer j. All graph relabelings, reversals and pair exchanges are included. There is no size cutoff. The first all-j question is a separate result.

Definition 2.1 admits (1), and every nonempty positive composition whose two endpoint parts are at least two, including length-one stars. The actual graph has one spine vertex per part and a_i−1 pendant leaves at the ith spine vertex. Definitions 2.4, 2.7 and 2.9 and equation (2.1) give the ordinary specialization as the sum over all proper colorings into {0,...,k−1} of X raised to the sum of vertex colors. Surjectivity, weights, the t refinement and numerical evaluation of X are absent.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PrincipalSpecialization/BinghamCaterpillarUniquenessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

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Geometry & algebraQuestion resolvedFirst preprint 2025Carnero Bravo's hereditary lexicographic product questionUnsuspended hereditary sphere-wedge coproduct types for actual lexicographic graph products.Proved

The questionThe hereditary classes require the respective property on induced subgraphs.

The ideaUse the actual heterogeneous graph-substitution induction and its occupied fiber transport.

Our contributionUnsuspended hereditary sphere-wedge coproduct types for actual lexicographic graph products.

Verification details
Release source recordProved in Lean

arXiv

Carnero Bravo's hereditary lexicographic product question

Andrés Carnero Bravo, On independence complexes of graph products, arXiv:2505.06457v2, Section 4, Question 22 asks whether, under the hypotheses of Theorem 21, $G\circ H$ belongs to $\mathcal W$. Both graphs are finite, have at least two vertices, and belong to $\mathcal{SW}$. Section 2 defines $\mathcal{SP}$ as contractibility or the homotopy type of an ordinary wedge of spheres, and $\mathcal P$ additionally permits a disjoint union of such wedges. The hereditary classes require the respective property on induced subgraphs. The requested conclusion is unsuspended.

The paper does not separately specify the vertexless convention. This settlement states it explicitly: sourceSW requires SP on every nonempty induced vertex set; the output sourceW includes every induced vertex set, including the empty one. The native realization of a vertexless graph is empty. Its output P witness is an empty outer coproduct, rather than a contractible point or a nonempty ordinary wedge. The initial preregistration sketch represented the vertexless input by an augmented empty-sphere model. The retained statement makes that boundary explicit through the nonempty-induced input convention and the independent empty-output proof; it does not assert ordinary SP for an empty realization.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/HereditaryLexicographicProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Further frontiersConjecture refutedFirst preprint 2026CayleyPy-4 Conjecture 11: consecutive four-cycle diameterConjecture 11 of CayleyPy-4 fails at n = 6: reversal cannot be reached in four consecutive-four-cycle moves, while the proposed diameter is four.Refuted

The questionTowards analogs of holographic string dualities for AI tasks, arXiv:2603.22195v1, section 8.5, printed page 57, states:

Why it mattersA counterexample at one permitted rank negates this entire universal clause.

The ideaUse native Equiv.Perm (Fin n) vertices. cycle n i h is List.formPerm on the four distinct consecutive indices with i+4 <= n. generators n contains these permutations and their inverses, and graph n is SimpleGraph.mulCayley (generators n) with right edges.

Our contributionConjecture 11 of CayleyPy-4 fails at n = 6: reversal cannot be reached in four consecutive-four-cycle moves, while the proposed diameter is four.

Verification details
Release source recordRefuted in Lean

Open problem

CayleyPy-4 Conjecture 11: consecutive four-cycle diameter

Chervov et al., CayleyPy-4: AI-Holography. Towards analogs of holographic string dualities for AI tasks, arXiv:2603.22195v1, section 8.5, printed page 57, states:

Conjecture 11. Case k = 4 from n = 6:

$n \equiv 0,1 \pmod{3}: D_4(n)=\frac{n(n-1)}{6}-1,$

$n \equiv 2 \pmod{3}: D_4(n)=\frac{n(n-1)}{6}+\frac{2}{3}.$

Section 8.1 specifies the cycles (i,i+1,i+2,i+3), for i = 0,...,n-4, with no wrapping. Section 8.5 specifies Cayley graphs, not Schreier coset graphs, and inverse-closed generators. The vertices are all permutations of {0,...,n-1} and distance is inverse-closed word distance. This dossier concerns the complete clause for every n >= 6.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/CayleyGrowth/ConsecutiveFourCycleDiameterRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

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Number theoryQuestion resolvedPairwise Coprime Ordered Tuples of DivisorsPairwise coprime ordered tuples of divisors of a number are counted by a product over its primes, and at tuple length equal to the number itself that product counts the divisors of the number raised to itself.Proved

The questionConjecture: The number of divisors of n^n equals the number of pairwise coprime ordered n-tuples of divisors of n.

The ideaBoth sides equal ∏ p ∈ primeFactors n, (1 + n · mp), where mp is the exponent of p in n.

Our contributionPairwise coprime ordered tuples of divisors of a number are counted by a product over its primes, and at tuple length equal to the number itself that product counts the divisors of the number raised to itself.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Pairwise Coprime Ordered Tuples of Divisors

OEIS A062319, "Number of divisors of n^n, or of A000312(n)", carries a comment by Gus Wiseman dated May 02 2021 and still standing at revision #60 of Aug 30 2026:

Conjecture: The number of divisors of n^n equals the number of pairwise coprime ordered n-tuples of divisors of n. Confirmed up to n = 30.

The comment goes on to print the tuples counted for n = 1 through n = 5. Those lists fix two readings the words leave open: the value one may occur and may repeat, so the constant tuple of ones counts, and the order of the coordinates matters, so (1,2) and (2,1) count separately.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/CoprimeDivisorTuples.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    Source 1
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    The exact recorded statement is proved.

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Discrete structuresConjecture provedFirst preprint 2026Cui–Sun Conjecture 2.1: integral Legendre telescopersIntegral coefficient polynomials provide telescopers for every positive odd-power Legendre sum.Proved

The questionLi-Li Cui and Zhi-Hong Sun, Curious identities involving Legendre polynomials and Apéry-like numbers, arXiv:2607.12330v1, page 10, state:

Why it mattersEquality as polynomials also includes x = 1 and entails every rational specialization without division by 1 − x.

The ideaThe operator lowers positive degree by one and annihilates constants.

Our contributionIntegral coefficient polynomials provide telescopers for every positive odd-power Legendre sum.

Scope & next questionSuch consequences still require the source's specialization formulas, denominator controls and congruence estimates; their all-m analogues remain open here.

Verification details
Release source recordProved in Lean

Open problem

Cui–Sun Conjecture 2.1: integral Legendre telescopers

Li-Li Cui and Zhi-Hong Sun, Curious identities involving Legendre polynomials and Apéry-like numbers, arXiv:2607.12330v1, page 10, state:

Suppose that m, p ∈ Z^+. Then there are integral polynomials f_i(t) with degree i (i = 0, 1, …, m − 1) such that (1 − x)^{m+1} Σ_{n=0}^{p−1} (2n + 1)^{2m+1} P_n(x) = pL_m(p, 1 − x)P_{p−1}(x) − pL_m(−p, 1 − x)P_p(x), where L_m(p, t) = (2p + 1)^{2m} t^m + f_{m−1}((2p + 1)^2) t^{m−1} + · · · + f_1((2p + 1)^2) t + f_0.

Equation (1.1), page 1, specifies:

The famous Legendre polynomials {P_n(x)} are given by P_0(x) = 1, P_1(x) = x and (n + 1)P_{n+1}(x) = (2n + 1)xP_n(x) − nP_{n−1}(x) (n ≥ 1).

The coefficient family is chosen once for each positive m and works for every positive p. The model is an identity in ℚ[X], with integral coefficient polynomials in ℤ[s]; the endpoint parameter of L is an integer.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Sun/LegendreOddPowerTelescoper.result

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Quantum informationQuestion resolvedFirst preprint 2026Every gate ordering in an open-boundary integrable quantum circuitEvery ordering that uses each boundary gate and each nearest-neighbour gate once is a nonzero scalar multiple of a double-row transfer matrix with signed inhomogeneities.Proved

The questionSection 5 restates the conjecture and says that an analytic proof remains open.

Why it mattersIt identifies every once-per-bond ordering with a transfer-matrix circuit of the source, including both boundary gates.

The ideaEvery inversion that this sort exchanges has disjoint physical support, so the operator product is preserved.

Our contributionEvery ordering that uses each boundary gate and each nearest-neighbour gate once is a nonzero scalar multiple of a double-row transfer matrix with signed inhomogeneities.

Scope & next questionSection 5 restates the conjecture and says that an analytic proof remains open.

Verification details
Release source recordProved in Lean

Open problem

Every gate ordering in an open-boundary integrable quantum circuit

García Fernández, Paletta and Retore, arXiv:2607.02093v1, §3.1.2, “From the circuit to $\vec{n}$”, state:

For the open-boundary case, similarly to the periodic setting \cite{Paletta:2025sap}, we conjecture that any circuit in which each gate $U_{i,i+1}$ (constructed from an $\check{R}$-matrix) appears exactly once per period to every nearest-neighbor pair of spins, and where each boundary gate is constructed from a $K$-matrix, is integrable.

Section 5 restates the conjecture and says that an analytic proof remains open. The bulk gate is $\check R(\kappa,-\kappa)$, where $\check R(u,v)=P R(u,v)$; the boundary gates are $K^R_1(\kappa)$ and $\widetilde K^L_L(\kappa)=\operatorname{tr}a(K^L_a(\kappa) \check R{La}(\kappa,-\kappa))$.

The formulation in issue #13232 uses the explicit double-row product and states mutual commutation of its transfer matrices as the hypothesis of the commutation conclusion. For every length $L\ge2$ and permutation of the $L+1$ gate labels, there are signed inhomogeneities $\theta_i\in{\kappa,-\kappa}$ and $c\ne0$ such that $M_\pi=c,t(\kappa;\theta)$. Whenever the transfer matrices $t(u;\theta)$ commute pairwise, $M_\pi$ commutes with every $t(u;\theta)$. The hypotheses of the identity are $R(u,u)=g(u)P$ and $g(\kappa)\ne0$, $g(-\kappa)\ne0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/OpenIntegrableCircuitIntegrability.result

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Discrete structuresCounterexample foundFirst preprint 2026Giacomelli's cone of infinitely log-concave initial dataTwo positive geometric solutions have infinitely log-concave initial data, but their midpoint has a negative first log-concavity determinant.Refuted

The questionGiacomelli, arXiv:2604.14391v1, Section 7 (1), printed page 13, asks:

Why it mattersGeometric solutions sit on the zero locus of the log-concavity determinant.

The ideaNo separate supporting or general module is required.

Our contributionTwo positive geometric solutions have infinitely log-concave initial data, but their midpoint has a negative first log-concavity determinant.

Scope & next questionThe uniform formal theorem and its recurrence construction remain open above.

Verification details
Release source recordRefuted in Lean

Open problem

Giacomelli's cone of infinitely log-concave initial data

P. Giacomelli, arXiv:2604.14391v1, Section 7 (1), printed page 13, asks:

(Cone of initial data for variable coefficients.) Proposition 3.4 gives a matrix condition on v₀ for log-concavity of a general P-recursive sequence. Characterize the geometry of the cone C∞ = {v₀ : Lⁱ(aₙ) ≥ 0 ∀ n, i} for d ≥ 3. Is it convex? Simply connected?

The convexity question is GIA-1 in preregistration #18698, a tier-1 published question. The Library note contains the source's recurrence and operator definitions.

Statement echo

The Lean namespace is D5.S0.Certificates.GiacomelliLogConcaveConeNonconvex.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (3): $a_{n+1}=\sum_{k=0}^{d-1}(p_kn+q_k)a_{n-k}$ for $n\ge d-1$ | seq | Literal recurrence, with argument $m=n+1$, increasing-order initial coordinates, and the natural subtraction performed before the real cast. | | Definition 2.1: $L(a)k=a_k^2-a{k+1}a_{k-1}$ | logConcaveOp | Literal expression; natural subtraction is truncated. | | $C_\infty={v_0:L^i(a_n)\ge0\ \forall n,i}$ | coneInf | Every $i,n\in\mathbb N$ with $i\le n$; these are the interior indices where the complete dependence window exists. | | Infinitely log-concave initial data | Iteration zero in coneInf | Requires original nonnegative terms. Both witnesses also satisfy the convention starting at $i=1$; the midpoint fails at $i=1$. | | “for d ≥ 3. Is it convex?” | claim | Every natural $d\ge3$ and every pair of real coefficient vectors, using Mathlib's Convex. | | Negative answer to convexity | result | Negates the complete affirmative assertion; $d=3$ supplies the counterexample. | | Characterization and simple connectedness | Separate open items in Triage | Neither is included in the settled assertion. |

There is no additional coefficient hypothesis. The binding definitions are literal source expressions with the index convention above.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/GiacomelliLogConcaveConeNonconvex.result

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026The fixed-support Davenport limit of integer boxesFixed-support box Davenport ratios converge to the largest augmented sign determinant in every positive dimension.Proved

The questionGirard and Plagne, The Davenport constant of low-dimensional boxes, arXiv:2610.16562v1, Section 2, printed page 4, ask:

Why it mattersThe identified limit connects an additive minimality problem to the maximal determinant of a cube simplex in every positive dimension.

Our contributionFixed-support box Davenport ratios converge to the largest augmented sign determinant in every positive dimension.

Scope & next questionHowever, this problem remains open when d ≥ 4.

Verification details
Release source recordProved in Lean

Open problem

The fixed-support Davenport limit of integer boxes

Girard and Plagne, The Davenport constant of low-dimensional boxes, arXiv:2610.16562v1, Section 2, printed page 4, ask:

It appears tempting to conjecture that for every integer d ≥ 1, the ratio D^(d+1)(⟦−m, m⟧^d)/m^d has a limit when m tends to infinity. This is clear when d = 1, and it follows from Theorems 1 and 5 when d ∈ {2, 3}. However, this problem remains open when d ≥ 4.

Here $D^{(d+1)}$ is the maximal length of a minimal zero-sum sequence in $\llbracket-m,m\rrbracket^d$ with exactly $d+1$ distinct support vectors. The result is Proved under the literal conventions of preregistration #17348.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Integer box | The coordinatewise absolute-value condition in CofactorKernel.lengths | Every coordinate is bounded by the natural radius | | Minimal zero-sum sequence | CofactorKernel.IsMinimalZeroSum | Nonzero sequence, zero total, no nonempty proper zero-sum subsequence | | Support of size $d+1$ | Support cardinality in CofactorKernel.lengths | Exactly $d+1$, not at most | | Maximal length | GirardPlagneBoxLimit.D | Natural supremum of the literal sum-of-multiplicities length set | | Every positive dimension | GirardPlagneBoxLimit.claim and result | Every natural $d\ge1$ | | A limit as the radius tends to infinity | GirardPlagneBoxLimit.girard_plagne_box_limit | Natural radii, real ratios, limit $\Delta_d$ | | Existence of the maximum | The nonemptiness and boundedness conjuncts of claim | Every $m\ge1$; makes the supremum the intended maximum |

The box and total are literal expressions; proofs use Finsupp.degree for length. Their correspondence to the preregistered formulas is definitional. The auxiliary constructions augment, cofactor, fromWeights, signMatrix and delta are repository constructions. In particular, $\Delta_d$ is the finite maximum of $|\det T(V)|$ over Boolean sign matrices, where $T$ adds a top row of ones.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Lattices/GirardPlagneBoxLimit.result

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Number theoryQuestion resolvedHanna's A369673 square classification and five-adic ordersHanna A369673 square classification and five-adic orders.Proved

The questionThe valuation agrees with A112765 on positive arguments.

Why it mattersThe conclusion is an unbounded symbolic classification and valuation formula.

The ideaAt odd indices the same recurrence preserves squareness and excludes divisibility by five.

Our contributionHanna A369673 square classification and five-adic orders.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A369673 square classification and five-adic orders

Paul D. Hanna's OEIS A369673, revision #21, states four conjectures for $a(N)=\prod_{k=0}^{N}(2^k+2^{N-k})$:

  • (C.1) $a(N)$ is a square if and only if $4\nmid N$.
  • (C.2) $5\nmid a(2n+1)$.
  • (C.3) $v_5(a(4n))=2b(n)$.
  • (C.4) $v_5(a(4n+2))=2b(n+1)$.

Here $b(n)=v_5((10n)!)-v_5((5n)!)$ is the signed integer difference in OEIS A127428. All indices are natural, including zero. The valuation agrees with A112765 on positive arguments. The source, conventions and literature screen are in preregistration #21915.

Statement echo

| Source item | Lean counterpart | Fidelity | | --- | --- | --- | | Product from $k=0$ through $k=N$ | HannaA369673.a | Literal finite product; natural subtraction has nonnegative arguments on the range. | | Signed factorial valuation difference | HannaA369673.b | Natural valuations cast into integers before subtraction. | | C.1 | First conjunct of HannaA369673.claim | Natural square root and both directions of the equivalence. | | C.2 | Second conjunct of HannaA369673.claim | Every natural index. | | C.3 | Third conjunct of HannaA369673.claim | Exact integer-valued equality, zero included. | | C.4 | Fourth conjunct of HannaA369673.claim | Exact integer-valued equality with the source shift. | | Entire conjecture package | HannaA369673.result | All four conjuncts, with no additional hypothesis. |

No source clause is dropped. The private products and sums are proof auxiliaries; they impose no hypothesis on the public claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/HannaA369673.result

Frozen module record

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Discrete structuresQuestion resolvedHanna's OEIS A338632 congruenceHanna's continued-fraction coefficients are two modulo four above degree one.Proved

The questionA(x) satisfies: 1 = A(x) - x/(A(x) - 3x/(A(x) - 5x/(A(x) - 7x/(A(x) - 9x/(A(x) - ...))))), a continued fraction relation.

The ideaThe proof constructs finite tails over a commutative ring with unit constant coefficient.

Our contributionHanna's continued-fraction coefficients are two modulo four above degree one.

Scope & next questionThe source's A191107 and A186776 classifications, and uniform algebraic reductions at other odd moduli, remain open in this delivery.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's OEIS A338632 congruence

Paul D. Hanna's OEIS A338632, revision 13, states:

G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 3*x/(A(x) - 5*x/(A(x) - 7*x/(A(x) - 9*x/(A(x) - ...))))), a continued fraction relation.

Its formula line is:

a(n) = 2 (mod 4) for n > 1 (conjecture).

The offset is zero. The source note is Library/RecurrenceComposition/hanna2026a338632.md.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | The displayed continued fraction | tail uses the literal rational-series finite truncations, ending at $A$ | The numerator at level $j$ is $(2j+1)x$. | | The equation equals one | IsGF | For each coefficient index, every sufficiently deep truncation has that coefficient equal to the constant series one. | | The generating series exists | The first conjunct of claim, proved by result | This makes the universal coefficient assertion nonvacuous. | | $a(n)$, offset zero | The degree-$n$ coefficient of any rational series satisfying IsGF | The source convention is unchanged. | | $n>1$ | $2\le n$ for natural $n$ | The two ranges coincide. | | $a(n)=2\pmod4$ | An integer representative $m$ of that rational coefficient with $m\bmod4=2$ | Integrality at the stated indices is part of the conclusion. |

The quantification over all series satisfying the relation and explicit existence are stronger than selecting one unnamed generating series; they do not weaken the source. The private gf_unique identifies every such series with the constructed solution.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaA338632.result

Frozen module record

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Discrete structuresQuestion resolvedHanna's Series A(x - A(x)) = x^2 + x A(x) Is Rational Modulo EightModulo eight the series fixed by substituting the variable minus itself is the rational function with coefficients one at even and three at odd degrees, because that function satisfies the equation modulo eight and the equation has only one solution.Proved

The questionIt appears that a(2n) == 1 (mod 8) and a(2n+1) == 3 (mod 8) for n >= 1.

The ideaOne, contraction. For a series f of order at least two put T(f) = x^2 + x f - (f(x - f) - f); the equation is T(f) = f.

Our contributionModulo eight the series fixed by substituting the variable minus itself is the rational function with coefficients one at even and three at odd degrees, because that function satisfies the equation modulo eight and the equation has only one solution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's Series A(x - A(x)) = x^2 + x A(x) Is Rational Modulo Eight

OEIS A392203, by Paul D. Hanna, Jan 03 2026:

G.f. A(x) satisfies A(x - A(x)) = x^2 + x*A(x).

The comment carrying the question:

It appears that a(2*n) == 1 (mod 8) and a(2*n+1) == 3 (mod 8) for n >= 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaShiftSquareModEight.result

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    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Huang–Ma Remark 2.7: the binary delta invariantA fixed positive skew-symmetrizer permits at most two delta residues.Proved

The questionMin Huang and Qiling Ma, A mutation invariant for skew-symmetrizable matrices, arXiv:2602.03194v1, Remark 2.7; Journal of Algebra 712 (2027), 546–558.

Why it mattersPreregistration 18784 identifies the published named conjecture, its literal conventions and the literature screen.

The ideaChoosing one attained nonzero residue therefore leaves at most that residue and zero.

Our contributionA fixed positive skew-symmetrizer permits at most two delta residues.

Verification details
Release source recordProved in Lean

Open problem

Huang–Ma Remark 2.7: the binary delta invariant

Min Huang and Qiling Ma, A mutation invariant for skew-symmetrizable matrices, arXiv:2602.03194v1, Remark 2.7; Journal of Algebra 712 (2027), 546–558. The source note gives the verified locator and verbatim definitions.

We also conjecture that for any fixed n and any choice of skew-symmetrizer d₁, …, dₙ, the invariant δ is binary. This is supported by several explicit computations.

For an integer matrix $B$, fix positive integer weights $d_i$ with $d_i b_{ij}=-d_j b_{ji}$. The real symmetric matrix $S(B)$ has diagonal $2$, upper entries $\operatorname{sgn}(b_{ij})\sqrt{|b_{ij}b_{ji}|}$ and lower entries $\operatorname{sgn}(b_{ji})\sqrt{|b_{ij}b_{ji}|}$. The invariant is $\delta(B)=\det S(B)\pmod4$. Binary means at most two residues for fixed size and weights, without asserting that both are attained.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Fixed matrix size and fixed positive integer skew-symmetrizer | claim quantifies the natural size and positive vector before choosing the two residues | Same quantifier order; the empty matrix is included as an extension | | The diagonal skew-symmetrizer makes the matrix skew-symmetric | IsSkewSymmetrizer is positivity and the literal entrywise skew relation | Diagonal zero follows from the relation and positivity | | The three cases defining the real symmetrization | symmetrization uses diagonal 2, integer sign, absolute value and real square root | Same ordered-index expression | | The determinant is an integer and its residue is binary | claim chooses an integer determinant and one of two residues for every admissible matrix; result proves it | Integrality is included explicitly; the two residues may coincide |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ClusterMutation/HuangMaBinaryDeltaInvariant.result

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Discrete structuresConjecture provedFirst preprint 2026Hughes iteration-depth no-gap conjectureLiteral fixed-degree insertion has a gap-free spectrum of minimum iteration depths.Proved

The questionFor a finite alphabet and arbitrary languages A and B, fix a positive insertion degree k throughout each iteration history.

Why it mattersLiteral insertion is the next operation considered here: it alternates source and base factors and then repeats that operation at a fixed degree.

The ideaPad any degree-j insertion factorization to a larger degree K by appending empty (x,y) factor pairs.

Our contributionLiteral fixed-degree insertion has a gap-free spectrum of minimum iteration depths.

Scope & next questionThe result supplies the universal proof for arbitrary alphabets and arbitrary languages, including empty and epsilon-only cases.

Verification details
Release source recordProved in Lean

arXiv

Hughes iteration-depth no-gap conjecture

For a finite alphabet and arbitrary languages A and B, fix a positive insertion degree k throughout each iteration history. A word has iteration depth m when m is the least stage at which the word occurs for some such degree. Hughes's Section 10 conjecture states that if depth r occurs, every depth from zero through r occurs. The formalization also proves this statement over arbitrary alphabets, with no finiteness or countability assumption.

This is the arbitrary-language iteration-depth conjecture on page 8. It is not the finite-language Theorem 8 and not the insertion-degree conjecture in Section 11.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/IterationDepthNoGap.result

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Discrete structuresConjecture provedIrwin's hook-sum conjecture for OEIS A138099The row-flattened triangle equals each hook-sum increment of the square spiral.Proved

The questionPaul Curtz's OEIS A138099, revision 26, NAME and FORMULA blocks, contains Benedict W.

Why it mattersIt proves the full identity, including both existence conclusions.

The ideaRow addresses construct a source arithmetic sequence, while insertion constructs and uniquely determines the spiral sequence.

Our contributionThe row-flattened triangle equals each hook-sum increment of the square spiral.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Irwin's hook-sum conjecture for OEIS A138099

Paul Curtz's OEIS A138099, revision 26, NAME and FORMULA blocks, contains Benedict W. J. Irwin's unnumbered conjecture:

Conjecture: a(n) = HookSum(T(n))-HookSum(T(n-1)), where HookSum(A) is the sum of all hooks across a tableau A, and T(n) are the spiraling sequence of tableaux:

The complete displayed diagrams and arithmetic rows are reproduced in Library/FiniteGeometry/irwin2016a138099.md. The arithmetic row is $R(k,j)=j+\lfloor k/2\rfloor$, for $k\geq1$ and $1\leq j\leq\lceil k/2\rceil$. Put $S(k)=\sum_{i=1}^{k}\lceil i/2\rceil$. A source sequence satisfies $a(S(k-1)+j)=R(k,j)$ on that domain, with no restriction on $a(0)$.

The literal square spiral begins at $(0,0)$, travels west, south, east, north, and repeats these directions with leg lengths $1,1,2,2,3,3,\ldots$. $T_0=\varnothing$ and $T_{n+1}$ inserts walk cell $n$ into $T_n$. For occupied integer cells, an ordinary hook contains its own cell and the occupied cells strictly right in its row and strictly below in its column. $H(D)$ sums these hook lengths; $H(\varnothing)=0$.

Statement echo

| Source clause or binding interpretation | Lean counterpart | Scope | | --- | --- | --- | | Irregular triangle read by rows | triangleEntry, trianglePrefix, IsSourceSequence | Every positive row and admissible column; index zero unused | | Spiraling sequence of the displayed tableaux | legDirection, WalkState, advance, walk, IsSourceSpiral | Actual occupied positions; no Ferrers replacement | | Sum of all ordinary hooks | hookLength, hookSum | Every finite integer-cell diagram, including holes | | Both implicitly specified sequences exist | First two conjuncts of claim | Conclusions, not hypotheses | | Every source sequence and source spiral | Universal binders and two source predicates in claim | No extra shape assumptions | | Increment at every positive index | Final conjunct of claim, proved by result | Both hook sums cast to integers before subtraction |

The finite cell carrier is the existing type $\mathbb Z\times\mathbb Z$. The counters implement the literal walk; they add no hypothesis to the source identity. Natural division has positive denominator two, remainder has positive modulus four, and the predecessor expressions are guarded by positive indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Polyomino/IrwinA138099HookSum.result

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    The exact recorded statement is proved.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026Infinite natural-basis quotient with finite full automorphismsFinite full automorphisms coexist with infinitely many natural bases modulo arbitrary permutation and independent nonzero scaling.Proved

The questionKaygorodov, Martin Gonzalez and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.2, Question 14 asks:

Why it mattersA rank-one square can admit many orthogonal coordinate decompositions while its distinguished nonzero idempotent restricts multiplication automorphisms.

The ideaThe square is the first coordinate line, so this algebra is non-perfect as in the question's focus.

Our contributionFinite full automorphisms coexist with infinitely many natural bases modulo arbitrary permutation and independent nonzero scaling.

Verification details
Release source recordProved in Lean

Open problem

Infinite natural-basis quotient with finite full automorphisms

Kaygorodov, Martin Gonzalez and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.2, Question 14 asks:

Describe the set of natural bases (up to permutation and scaling) of a non-degenerate evolution algebra. When is it finite? Can it be infinite for an algebra whose automorphism group is finite?

The endpoint is the entire final interrogative only. Its positive answer requires a permitted field and a finite-dimensional commutative bilinear, possibly nonassociative evolution algebra with zero annihilator, finite full linear multiplication-automorphism group, and infinitely many natural bases modulo all permutations and independent nonzero scalings. It does not ask for every field, a group-scheme classification, or algebra-automorphism orbits of bases.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/FiniteAutomorphismsInfiniteNaturalBases.result

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  • Problem significance · 1/4 · 30% weight

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedKurkov's three-parameter Stirling-minor Vandermonde sumKurkov's three-parameter Stirling-minor Vandermonde sum.Proved

The questionThe four public definitions are the literal source determinant, summand, normalized sum and quantified assertion.

The ideaFactorial cancellation and Cauchy–Binet turn the literal sum into a signed moment determinant.

Our contributionKurkov's three-parameter Stirling-minor Vandermonde sum.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's three-parameter Stirling-minor Vandermonde sum

Mikhail Kurkov's sole FORMULA conjecture in OEIS A396563, approved revision 15, asks whether the unsigned first-kind Stirling determinant $B(n,k,m)=\det[c(n+m+i,n+j)]_{1\le i,j\le k}$ equals

$$ \frac{\displaystyle\sum_{n\ge j_1\ge\cdots\ge j_m\ge0} (-1)^{j_1+\cdots+j_m} \prod_{1\le q<p\le m}(j_q-j_p+p-q)^2 \prod_{t=1}^m\frac{(n+t-1){j_t}(n-j_t+t)^{n+k}}{(j_t+m-t)!}} {\displaystyle\prod{i=0}^{m-1}(n+i)!}. $$

Here $n,k\ge0$, $m\ge1$, $c$ denotes unsigned first-kind Stirling numbers, and $(x)_r$ is the falling factorial. The empty determinant is one. The primary source is https://oeis.org/A396563/internal; the preregistration is https://github.com/the-omega-institute/trureturing/issues/22587.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The $k\times k$ first-kind minor with one-based indices | B | Fin k values plus one preserve both source indices. | | Nested ranges $n\ge j_1\ge\cdots\ge j_m\ge0$ | Reused KurkovNestedStirlingIdentity.Chain n m | Antitone maps into Fin (n+1) enumerate each source tuple once. | | Sign, squared pair differences, falling factorial, numerator power and denominator factorial | sourceTerm | Rational differences and division; natural factorial indices and exponents. | | Division by the product of $(n+i)!$ | sourceRHS | The entire chain sum retains the full normalization. | | Equality for all natural $n,k$ and $m\ge1$ | claim, result | The source boundaries $n=0$ and $k=0$ are included. |

The four public definitions are the literal source determinant, summand, normalized sum and quantified assertion. There is no extra hypothesis on $n$ or $k$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BinomialBases/KurkovStirlingMinorVandermondeSum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedMeeussen's formula for two-layer solid partitions A381265The two-layer solid partitions whose first layer is a plane partition of n and whose second layer is a plane partition of 3 number 3(2 A000219(n) - A000990(n) - 2 A000041(n) + 1), with A000219 the plane partitions, A000990 the plane partitions with at most two rows and A000041 the partitions of n, as conjectured by W.Proved

The questionConjecture: equal to 3(2A000219 -A000990 -2A000041 +1) tested up to n=20.

The ideaThe plane partitions of 3 are the three lines {0, ek, 2ek} and the three corners {0, ei, ej}.

Our contributionThe two-layer solid partitions whose first layer is a plane partition of n and whose second layer is a plane partition of 3 number 3(2 A000219(n) - A000990(n) - 2 A000041(n) + 1), with A000219 the plane partitions, A000990 the plane partitions with at most two rows and A000041 the partitions of n, as conjectured by W.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Meeussen's formula for two-layer solid partitions A381265

OEIS A381265 counts the solid partitions with two layers whose second layer is a plane partition of 3. Its COMMENTS line records:

Conjecture: equal to 3*(2A000219 -A000990 -2A000041 +1) tested up to n=20.

Issue #10460 fixes the reading. The index n is the size of the first layer, as the data and the example a(3) = 6 show. Plane partitions are lower sets of n cells of ℕ³, and a two-layer solid partition is a pair P₂ ⊆ P₁ of them. A000990 counts the plane partitions all of whose cells have row index at most 1, and A000041 counts Nat.Partition n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TwoLayerSolidPartitions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedStrict decrease in Mendonça's min-plus configuration triangleStrict decrease of the min-plus configuration counts from depth three onward.Proved

The questionMendonça's OEIS A400584, revision 28 states: “Conjecture: T(n,k) > T(n,k+1) for n >= 3 and 1 <= k < 2^n.” The associated technical note gives the counting recurrence as Proposition 2.

The ideaThe ordered root decomposition gives the min-plus counting recurrence.

Our contributionStrict decrease of the min-plus configuration counts from depth three onward.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Strict decrease in Mendonça's min-plus configuration triangle

J. R. G. Mendonça's OEIS A400584, revision 28 states: “Conjecture: T(n,k) > T(n,k+1) for n >= 3 and 1 <= k < 2^n.” The associated technical note gives the counting recurrence as Proposition 2. The complete source quotations and conventions are in preregistration #19612 and the Library note.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Complete plane binary tree of depth $n$ | LabeledTree n: one leaf at zero, one labeled root with two ordered equal-depth subtrees at successor depth | Left and right positions remain distinct; no quotient by swaps. | | Each internal vertex is labeled by addition or minimum | The two constructors of Op and the operator parameter of LabeledTree.node | Exactly the source's two labels. | | Every leaf equals one; each vertex applies its operator | All three recursion cases of value | Literal source evaluation. | | Number of labelings with root value $k$ | T n k, the Nat.card of the root-value subtype | Counts labelings rather than possible values; finiteness is constructed privately. | | $n\ge3$ and $1\le k<2^n$ | All natural-number binders and all three guards of claim | The same range, including every adjacent pair through the last column. | | $T(n,k)>T(n,k+1)$ | result : claim | The strict comparison has only its sides reversed syntactically. | | Depth-zero convention $T(0,1)=1$ | The leaf constructor and the private T_zero helper | The source recurrence's initial condition; no additional hypothesis on the conjecture. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/MendoncaMinPlusTriangle.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Proof methods · 3/4 · 15% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe A115805 to A115803 carryless containment questionThe carryless-multiplier equation for 13 and 29 implies the equation for 11 and 31.Proved

The questionA115803 revision 13 defines the natural numbers k satisfying 11k = A048720(31,k).

Why it mattersThe question compares solution sets for two different pairs of ordinary and carryless multipliers.

The ideaThe shared owner CarrylessMultiplierEquivalence defines genuine binary coefficient convolution and proves its equality to binaryProduct, the XOR recursion.

Our contributionThe carryless-multiplier equation for 13 and 29 implies the equation for 11 and 31.

Scope & next questionQuestions beyond this implication remain outside the result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A115805 to A115803 carryless containment question

OEIS A115805 revision 14 defines the natural numbers k satisfying 13k = A048720(29,k), and its cross-reference conjectures containment in A115803. A115803 revision 13 defines the natural numbers k satisfying 11k = A048720(31,k). Both sequences include zero. The exact question is

$$ \forall k\in\mathbb N,\qquad 13k=C(29,k)\ \Longrightarrow\ 11k=C(31,k). $$

Here C is the binary carryless product of A048720 revision 62. If a_i and b_i are the binary digits of a and b, the t-th binary digit of C(a,b) is the sum of a_i*b_(t-i) over 0 <= i <= t, reduced modulo two. Thus C is coefficient convolution over GF(2). Multiplication outside C is ordinary natural-number multiplication. Every occurrence of k denotes the same natural number, with no positivity or size restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/CarrylessThirteenElevenContainment.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBala's Lucas recurrence and its growing five-adic congruenceBala's quintic recurrence satisfies the conjectured power-of-five congruence.Proved

The questionPeter Bala's Nov 14 2022 recurrence (verbatim):

Why it mattersThe expression a(n)^2+4 gains two powers of 5 at each step, explaining all the moduli indexed by r <= n through one invariant.

The ideaInduction proves 5^(2n+1) ∣ a(n)^2+4, beginning with a(0)^2+4=5.

Our contributionBala's quintic recurrence satisfies the conjectured power-of-five congruence.

Scope & next questionOnly the Conjecture line is settled for the recurrence interpretation, covering every n >= 1 and also n=0.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's Lucas recurrence and its growing five-adic congruence

OEIS A144837, NAME (%N, verbatim):

a(n) = Lucas(5^n).

Peter Bala's Nov 14 2022 recurrence (verbatim):

a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.

The settled FORMULA line (%F, verbatim):

Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.

AUTHOR (%A, verbatim):

Artur Jasinski, Sep 22 2008

The offset is %O 1,1. Both n and r range over the natural numbers including zero, with the sole hypothesis r <= n. The sequence is integer-valued and defined by Bala's recurrence, extended to zero with a(0)=1=Lucas(5^0)=Lucas(1). This gives a(1)=11=A144837(1) and agrees with the OEIS domain n >= 1. Congruence means that the integer 5^(n+r+1) divides the integer difference a(n+1)-a(n).

The exact Lean statement is:

theorem result (n r : ℕ) (hr : r ≤ n) :
    (5 : ℤ) ^ (n + r + 1) ∣ a (n + 1) - a n

Only the Conjecture line is settled for the recurrence interpretation, covering every n >= 1 and also n=0. The Lucas representation, the 5-adic limit (A269591), and A268922 are not claimed. The formal definition does not use Lucas numbers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/BalaLucasFiveAdicCongruence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedGranvik's Möbius diagonal and Mertens rowThe diagonal and first row of the Granvik Mertens array.Proved

The questionMats Granvik's OEIS A192763 COMMENTS field states:

Why it mattersThe single result settles both all-index conjectures in the entry.

The ideaFor fidelity, a symmetric array with Somos's positive reduction and zero row-sum diagonal has periodic positive columns of index at least two; each full period sums to zero.

Our contributionThe diagonal and first row of the Granvik Mertens array.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Granvik's Möbius diagonal and Mertens row

Preregistration: #20559. Mats Granvik's OEIS A192763 COMMENTS field states:

The main diagonal is the Mobius function times the natural numbers A055615 (conjecture). For k>1 the first row is the Mertens function + 2 = A002321 + 2 (conjecture).

The positive-index array is defined by Michael Somos's Mathematica and PARI programs in the entry. It is symmetric, $T(1,1)=1$, $T(2,1)=2$, and [ T(n,1)=\frac{-T(n-1,1)-\sum_{i=2}^{n-1}T(n,i)}{n+1}+T(n-1,1) \quad(n\ge3). ] For $n>k\ge2$ it satisfies $T(n,k)=T(k,((n-1)\bmod k)+1)$; for $n\ge2$ its diagonal is $T(n,n)=-\sum_{i=1}^{n-1}T(n,i)$. The formal definition takes values in $\mathbb Q$ and extends zero-index cells by zero. It uses well-founded recursion with measure $4\max(n,k)$ plus phases three for the upper triangle, two for the diagonal, one for the lower border, and zero for the strict lower interior.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Somos's positive rational recurrence | GranvikMertensArray.T | The displayed branch order and inclusive sums; no integer rounding. | | Corner and first border cell | T 1 1 = 1, T 2 1 = 2 by definition | OEIS program normalization. | | Main diagonal for every positive index | First conjunct of result | ∀ n : ℕ, 1 ≤ n → T n n = (n : ℚ) * (ArithmeticFunction.moebius n : ℚ). | | First row for every index greater than one | Second conjunct of result | ∀ k : ℕ, 2 ≤ k → T 1 k = 2 + ∑ j ∈ Finset.Icc 1 k, (ArithmeticFunction.moebius j : ℚ). | | Mats Granvik and Olivier Gérard's first Mathematica program | Private granvik, T_eq_granvik and T_published_unique | Its literal border and reverse-window equations define exactly T. This equality is consumed by result. | | Math.SE 50719 | Source note | Its printed corner is three; this settlement uses the OEIS program's corner one. Its index slips are resolved by the executable Mathematica and PARI programs in OEIS. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/GranvikMertensArray.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedKrizek's phitorial divisor-sum parity characterizationThe divisor sum of phitorials is odd exactly away from twice a square.Proved

The questionConjecture: a(n) is odd for numbers in A183300; a(n) is even for numbers in A001105 (2n^2).

Why it mattersThe independent question is the parity characterization printed as Krizek's named A280258 conjecture.

The ideaThe phitorial is odd exactly at one and at even inputs.

Our contributionThe divisor sum of phitorials is odd exactly away from twice a square.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Krizek's phitorial divisor-sum parity characterization

OEIS A280258, NAME (%N, verbatim):

a(n) = Sum_{d|n} pxi(d), where pxi(m) is the product of totatives of m (A001783).

Jaroslav Krizek's COMMENT (%C, verbatim):

Conjecture: a(n) is odd for numbers in A183300; a(n) is even for numbers in A001105 (2*n^2).

The dependent definitions are A001783's product of the integers from one through m that are relatively prime to m, A183300's positive integers not of the form 2*k^2, and A001105's integers of the form 2*k^2. The complete quantified reading is

forall n : Nat, 0 < n -> (Odd (a n) <-> not exists k : Nat, n = 2*k^2),

where a(n) sums the phitorials over the positive divisors of n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/KrizekPhitorialDivisorSumParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresRecurrence provedA301420: positive monomial coefficients for five dataStrict positive monomial coefficients give the rational shifted generating function and the positive-index recurrence for A301420.Proved

The questionGregory Gerard Wojnar's OEIS A301420, NAME and COMMENTS, defines the sequence by:

Why it mattersThe conjecture relates an unbounded monomial positivity filter to a rational formal series.

The ideaMultinomial expansion groups multiplicity vectors by their factor count.

Our contributionStrict positive monomial coefficients give the rational shifted generating function and the positive-index recurrence for A301420.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A301420: positive monomial coefficients for five data

Gregory Gerard Wojnar's OEIS A301420, NAME and COMMENTS, defines the sequence by:

for a family of 5 data, the sum of the positive coefficients in these polynomials is a(n).

The FORMULA field states:

G.f.: (-x*(x+1)^4+1)/(x^6+4x^5+5x^4-5x^2-6x+1); this denominator equals (1-x)*(2-(x+1)^5) (conjectured).

It also states:

a(n+14) = 7*a(n+13) - a(n+12) - 6*a(n+11) + 2*a(n+10) - a(n+9) + 4*a(n+8) + a(n+7) + 4*a(n+5) + 2*a(n+4) - a(n+3) - 5*a(n+2) - 4*a(n+1) - a(n) (conjectured).

The source snapshot is A301420 revision #28; the full generating-function conjecture occurs in revision #2 and the recurrence is explicitly conjectural in revision #16. The published offset is 1,2. The shift and the universal recurrence domain $n\ge1$ are explicit interpretations accepted in #23723, rather than printed source quantifiers. The Library note gives the source and coefficient normalization.

For $k\ge1$ and $m=(m_1,\ldots,m_5)\in\mathbb N_0^5$ of weighted degree $\sum_{i=1}^5 i m_i=k$, put $r=\sum_i m_i$ and define

$$ c_m=\frac{k}{5r}(-1)^{k+r}\frac{r!}{\prod_i m_i!}\prod_{i=1}^5\binom5i^{m_i},\qquad a(k)=\sum_{\sum i m_i=k,\ c_m>0}c_m. $$

These are definitions, using equation (2) of Wojnar, Wojnar and Brin, arXiv:1706.08381v2, pp. 3–4. Each distinct monomial contributes separately, before substituting data values. No $a(0)$ is defined.

The exact target is the conjunction of the rational identity

$$ F(X)=\sum_{k\ge1}a(k)X^{k-1} =\frac{1-X(1+X)^4}{(1-X)(2-(1+X)^5)}\quad\text{in }\mathbb Q[[X]] $$

and the displayed recurrence for every integer $n\ge1$.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Five distinct variables and weights 1 through 5 | Multiplicity, count, degree, multiplicities | All nonnegative multiplicity vectors of a fixed weighted degree, each once. | | Normalized Girard–Waring coefficient | literalCoefficient | The displayed factorial, binomial and sign factors. | | Sum of positive coefficients | positiveCoefficient | Strict per-monomial positivity, before valuation. | | Offset-one generating function | positiveSeries, first conjunct of result | Coefficient at degree $j$ is $a(j+1)$. | | Printed recurrence | recurrence, second conjunct of result | Every positive index, with all signs and coefficients as printed. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PowerSeries/PositiveNewtonSums.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Strict positive monomial coefficients give the rational shifted generating function and the positive-index recurrence for A301420.

    Read source ↗
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    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresQuestion resolvedDyadic divisibility at Lang's Fibonacci indicesFibonacci numbers at Lang's dyadic indices have the conjectured power-of-two divisor.Proved

The questionAll entries from a(3) to a(n) appear in addition to 2^n as factors in the conjectured factorization of Fibonacci(2^(n-2)3m) for n >= 3 and all m >= 0.

Why it mattersThe stated power of two is a uniform divisor for every natural multiplier and every dyadic level starting at n=3.

The ideaApply Nat.leinduction to the level n, starting at n=3 with arbitrary natural m.

Our contributionFibonacci numbers at Lang's dyadic indices have the conjectured power-of-two divisor.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dyadic divisibility at Lang's Fibonacci indices

OEIS A319197, NAME (%N, verbatim):

All entries from a(3) to a(n) appear in addition to 2^n as factors in the conjectured factorization of Fibonacci(2^(n-2)*3*m) for n >= 3 and all m >= 0.

The first COMMENT sentence (%C, verbatim), which is the settled clause:

It appears that Fibonacci(2^(n-2)3m)/(2^n) is a nonnegative integer for n >= 3 and all m >= 0.

AUTHOR (%A, verbatim):

Wolfdieter Lang, Oct 09 2018

The exact Lean statement is theorem result (n m : ℕ) (hn : 3 ≤ n) : 2 ^ n ∣ Nat.fib (2 ^ (n - 2) * 3 * m). The natural-number domain includes m=0; n-2 is truncated subtraction, which agrees with ordinary subtraction under 3 <= n. The positive divisor 2^n makes natural divisibility equivalent to the quoted nonnegative-integral quotient assertion, not a claim about truncated division.

Only that first COMMENT sentence is settled. The product A(n) = Product_{j=3..n} a(j), the I(n; m) factorization conjecture, the specific factors from A049660 and A253368, and the rest of the entry are not claimed. The theorem does not assert an exact 2-adic valuation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/LangFibDyadicIndexDivisibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedKagey's eventual equality of power-sequence minimaPrime-power sequence minima eventually equal the rational-relation minimum.Proved

The questionConjecture: T(p,k) = A071537(k) for fixed k and sufficiently large prime p.

Why it mattersThis is a Tier 1 externally stated OEIS conjecture.

The ideaInteger-exponent relations give rational linear relations among these vectors.

Our contributionPrime-power sequence minima eventually equal the rational-relation minimum.

Scope & next questionThe three neighbouring questions in the table remain open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kagey's eventual equality of power-sequence minima

Conjecture: T(p,k) = A071537(k) for fixed k and sufficiently large prime p.

The source is P. Kagey's OEIS A343825, COMMENT, revision 22. The defining NAME and the linked A071537 NAME are quoted in the source note. The preregistration is #18687.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | A nondecreasing sequence beginning at $k$ and ending at $m$ | Valid n k m b: option-valued first and last elements and pairwise natural order | Literal source convention | | Every multiplicity is below $n$; the product is an integer $n$th power | The count bound and integer-power conjuncts of Valid | All conjuncts retained | | $T(n,k)$ is the least such $m$ | T uses Nat.find | Zero extends the function outside the existence domain | | Integer exponents on $[k,m]$, with nonzero first exponent and product one | Relation k m | Integer powers are interpreted in the rationals | | A071537 is the least relation endpoint $m\ge k$ | A071537 uses Nat.find | Zero extends the function outside the existence domain | | Fixed $k$ and sufficiently large prime $p$ | claim: $\forall k\ge1,\ \exists N,\ \forall p,\ p\text{ prime}\land N\le p\Rightarrow T(p,k)=A071537(k)$ | $k\ge1$ is A071537's offset; the threshold depends only on $k$ |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Factorization/OeisA343825KageyPowerSequences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedBala's A347855 auxiliary b integralitya(3n+1) = 3^(3n+1) b(n), where b(n) = (12n+4)!/( (6n+2)!(3n+1)!Product{k = 0..3n} (3n+3k+4) ) appears to be integral.Proved

The questiona(3n+1) = 3^(3n+1) b(n), where b(n) = (12n+4)!/( (6n+2)!(3n+1)!Product{k = 0..3n} (3n+3k+4) ) appears to be integral.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Our contributiona(3n+1) = 3^(3n+1) b(n), where b(n) = (12n+4)!/( (6n+2)!(3n+1)!Product{k = 0..3n} (3n+3k+4) ) appears to be integral.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347855 auxiliary b integrality

Peter Bala, OEIS A347855, states:

a(3*n+1) = 3^(3*n+1) * b(n), where b(n) = (12*n+4)!/( (6*n+2)!*(3*n+1)!*Product_{k = 0..3*n} (3*n+3*k+4) ) appears to be integral.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxBIntegral.b | Literal expression; natural factorials are cast to the rationals. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxBIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxBIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347855BalaAuxBIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedBala's A347855 auxiliary c integralitya(3n+2) = 3^(3n+2) c(n), where c(n) = (12n+8)!/( (6n+4)!(3n+1)!Product{k = 0..3n+2} (3n+3k+2) ) appears to be integral. (End)Proved

The questiona(3n+2) = 3^(3n+2) c(n), where c(n) = (12n+8)!/( (6n+4)!(3n+1)!Product{k = 0..3n+2} (3n+3k+2) ) appears to be integral. (End)

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Our contributiona(3n+2) = 3^(3n+2) c(n), where c(n) = (12n+8)!/( (6n+4)!(3n+1)!Product{k = 0..3n+2} (3n+3k+2) ) appears to be integral. (End)

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347855 auxiliary c integrality

Peter Bala, OEIS A347855, states:

a(3*n+2) = 3^(3*n+2) * c(n), where c(n) = (12*n+8)!/( (6*n+4)!*(3*n+1)!*Product_{k = 0..3*n+2} (3*n+3*k+2) ) appears to be integral. (End)

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxCIntegral.c | Literal expression; natural factorials are cast to the rationals. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxCIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347855BalaAuxCIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347855BalaAuxCIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedBala's A347856 fractional-factorial integralityProved

The questionThe integrality conjecture quantifies over every natural index, including zero.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347856 fractional-factorial integrality

Peter Bala, OEIS A347856, states:

a(n) = (6*n)!/((4*n)!*n!) * (n/2)!/(3*n/2)!.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347856BalaHalfIndexIntegral.a | Literal expression; natural factorials are cast to the reals. | | Fractional factorials use Gamma | OeisA347854BalaHalfIndexIntegral.ffact | $x! = \Gamma(1+x)$; the natural-index factorials are Nat.factorial. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347856BalaHalfIndexIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347856BalaHalfIndexIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347856BalaHalfIndexIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedBala's A347857 fractional-factorial integralityProved

The questionThe integrality conjecture quantifies over every natural index, including zero.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347857 fractional-factorial integrality

Peter Bala, OEIS A347857, states:

a(n) = (6*n)!/((3*n)!*(2*n)!) * (3*n/2)!/(5*n/2)!.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347857BalaHalfIndexIntegral.a | Literal expression; natural factorials are cast to the reals. | | Fractional factorials use Gamma | OeisA347854BalaHalfIndexIntegral.ffact | $x! = \Gamma(1+x)$; the natural-index factorials are Nat.factorial. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347857BalaHalfIndexIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347857BalaHalfIndexIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347857BalaHalfIndexIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedBala's A347858 fractional-factorial integralityProved

The questionThe integrality conjecture quantifies over every natural index, including zero.

Why it mattersAll declarations are symbolic, unbounded identities, inequalities or divisibility statements, rather than fixed numerical certificates or bounded enumerations.

The ideaLegendre's formula converts factorial divisibility to inequalities between sums of natural floor divisions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A347858 fractional-factorial integrality

Peter Bala, OEIS A347858, states:

a(n) = (9*n)!/((3*n)!*(2*n)!) * (n/2)!/(9*n/2)!.

Fractional factorials mean $x! = \Gamma(1+x)$. The integrality conjecture quantifies over every natural index, including zero.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The source's displayed ratio and product endpoints | D5.S3.Arith.FactorialRatio.OeisA347858BalaHalfIndexIntegral.a | Literal expression; natural factorials are cast to the reals. | | Fractional factorials use Gamma | OeisA347854BalaHalfIndexIntegral.ffact | $x! = \Gamma(1+x)$; the natural-index factorials are Nat.factorial. | | Integral for all natural indices | D5.S3.Arith.FactorialRatio.OeisA347858BalaHalfIndexIntegral.claim | For every $n\in\mathbb N$ there exists $z\in\mathbb Z$ whose cast equals the ratio. | | The conjecture is true | D5.S3.Arith.FactorialRatio.OeisA347858BalaHalfIndexIntegral.result | Unconditional proof of the complete claim. | | Proof-only arithmetic parameters | Consumed floor, divisibility and Gamma helpers | Internal parameters support the proof; they add no assumption to the conjecture. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/OeisA347858BalaHalfIndexIntegral.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedParity of the cubic Catalan composition A363308The coefficients of C(xC(x)^3) are odd exactly at zero and powers of two.Proved

The questionHanna, May 28, 2023), quoted verbatim from Library/Arith/hanna2023a363308.md:

Why it mattersThis is the first-tier recent OEIS conjecture selected in the lane brief.

The ideaThus C is the Catalan generating function specified by the NAME, not merely an unspecified solution of a functional equation.

Our contributionThe coefficients of C(xC(x)^3) are odd exactly at zero and powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the cubic Catalan composition A363308

OEIS A363308 (Paul D. Hanna, May 28, 2023), quoted verbatim from Library/Arith/hanna2023a363308.md:

NAME:

Expansion of g.f. C(x*C(x)^3), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

COMMENT:

Conjecture: a(n) is odd iff n is a power of 2 or n = 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CatalanCubicCompositionParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedModulo-four classification of the A372577 coefficientsThe normalized series of OEIS A372577 has a complete modulo-four classification.Proved

The questionA(x) satisfying A(x)^2 = A(A( xA(x) + xA(x)^2 )).

Why it mattersThis first-tier OEIS conjecture was recorded in 2024.

The ideaNormalize A = X·U and cancel X²·U to turn the double composition into a contracting operator.

Our contributionThe normalized series of OEIS A372577 has a complete modulo-four classification.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Modulo-four classification of the A372577 coefficients

OEIS A372577, Paul D. Hanna, May 30, 2024:

NAME:

Expansion of g.f. A(x) satisfying A(x)^2 = A(A( x*A(x) + x*A(x)^2 )).

COMMENT (target):

Conjecture: a(6*n - 3) == 3 (mod 4) and a(6*n - k) == 1 (mod 4) when k = {0,1,2,4,5} for n >= 1.

Separate COMMENT (a derived corollary, not a second resolution):

All terms appear to be odd.

The normalized generating series has constant coefficient zero and linear coefficient one; a(n) = [x^n]A(x) for positive indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DoubleCompositionModFourClassification.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedIntegrality and dyadic parity of OEIS A377106Every normalized rational solution of Hanna's cubic equation has integer coefficients odd exactly at powers of two.Proved

The questionConjecture: a(n) is odd iff n = 2^k for some k >= 0.

The ideaReduction modulo two and CubicFifteenSubstitutionParity.thueseriesequation identify its reduction with CubicFifteenSubstitutionParity.powerTwoSeries.

Our contributionEvery normalized rational solution of Hanna's cubic equation has integer coefficients odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Integrality and dyadic parity of OEIS A377106

Conjecture: a(n) is odd iff n = 2^k for some k >= 0.

The entry defines $A^3=A(x^3+9xA^3)$, $A(0)=0$, $[x]A=1$. A normalized rational solution exists, and every such solution has integer coefficients at each $n\geq1$, with odd coefficient exactly when $n=2^k$ for some natural $k$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | the displayed cubic equation and normalization | IsGF contains all three source conditions over rational power series | exact registered convention | | offset one | claim quantifies over every n at least one | exact registered convention | | odd iff a power of two | claim supplies an integer coefficient and its Odd equivalence | exact registered convention | | the entry describes a series | claim also asserts existence; uniqueness transfers the conclusion to every normalized solution | exact registered convention |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaA377106ParityReversion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedHanna's Modulo-Six ConjectureEvery normalized solution of Hanna's system has coefficients avoiding one and five modulo six beyond degree one.Proved

The questionHanna's OEIS A378258 Comments state: "Conjecture: a(n) is not congruent to 1 nor 5 (mod 6) for n > 1." The Name states: "G.f. satisfies A(x) = B(x) + 5A(x)^2 where B(A(x) - A(x)^2) = x." The separate preregistration is 25601.

Why it mattersThe missing information is divisibility by three at those degrees.

The ideaThe resulting integer series F maps exactly to A.

Our contributionEvery normalized solution of Hanna's system has coefficients avoiding one and five modulo six beyond degree one.

Scope & next questionThe separate first-negative-term assertion remains outside the result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a378258 hanna mod six

Paul D. Hanna's OEIS A378258 Comments state: "Conjecture: a(n) is not congruent to 1 nor 5 (mod 6) for n > 1." The Name states: "G.f. satisfies A(x) = B(x) + 5*A(x)^2 where B(A(x) - A(x)^2) = x." The separate preregistration is #25601.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Rational formal series with offset one and a(1)=1 | The original HannaA378258SquareRootParity.IsGF A: constant coefficient zero, linear coefficient one. | | Both equations, including existential B | The same original predicate, without replacing either equation by a weaker consequence. | | Existence | The first conjunct of claim, reused from the frozen parity result. | | Every normalized solution | Universal rational A satisfying that original predicate. | | Every natural n>1 | Universal n : ℕ with 1 < n; no cutoff. | | Integer coefficient excluding both residues | An integer z with coeff n A = (z : ℚ), z % 6 ≠ 1, and z % 6 ≠ 5. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaA378258ModSix.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedHanna's prime bilateral integrality conjectureThe normalized bilateral equation has an integral solution exactly at prime parameters.Proved

The questionHanna's OEIS A380681, Comments section, states:

Why it mattersThe equation relates an infinite bilateral Laurent sum to an integral coefficient recursion.

The ideaTogether with the positive summands this makes each coefficient a finite sum.

Our contributionThe normalized bilateral equation has an integral solution exactly at prime parameters.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's prime bilateral integrality conjecture

P. D. Hanna's OEIS A380681, Comments section, states:

Conjecture: if F(x) satisfies p*x = Sum_{n=-oo..+oo} (-1)^n * x^(p*n) * (F(x) + x^n)^(p*n) with F(0) = 1 for some fixed integer p, then F(x) is an integer series in x iff p is prime. The g.f. A(x) of this sequence is the case when p = 2; other cases include A380682 (p=3), A380683 (p=5), A380684 (p=7), A380685 (p=11), and A380686 (p=13).

Preregistration: #18831. The Library note records the source. The formal parameter is $p\in\mathbb N$ with $p\ge2$, and $F\in\mathbb Q[[X]]$. Integer powers are taken in $\mathbb Q((X))$. At degree $N$, the bilateral sum means every symmetric window $[-K,K]$ with $K\ge N+1$.

Statement echo

The Lean namespace is D5.S1.Recurrence.Bilateral.HannaPrimeBilateralIntegrality.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $(-1)^n x^{pn}(F+x^n)^{pn}$, $n\in\mathbb Z$ | term | Literal integer Laurent powers, with $x=X$ and the embedded $F$. | | $F(0)=1$ | First conjunct of IsSolution | The rational constant coefficient is one. | | $px$ equals the bilateral sum | Remaining conjunct of IsSolution | Every coefficient in every sufficiently large symmetric window has the required value. | | Integer series | Integer witness for every coefficient in claim | Every rational coefficient is the cast of an integer. | | If and only if $p$ is prime | Equivalence in claim | Both directions hold for every normalized rational solution. | | Fixed integer parameter | Natural $p\ge2$ in claim | The preregistered range; nonpositive parameters are outside the statement. | | A solution exists | Existential conjunct of claim | Additional conclusion for every $p\ge2$, with no additional hypothesis. |

No source clause within the preregistered interpretation is dropped. Both the claim and result Blueprint nodes display the complete proposition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/HannaPrimeBilateralIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThe valuation-step greedy sequence is a permutationThe valuation-step greedy sequence is a permutation of the positive integers.Proved

The questionThe supplied source states: “Conjecture: this sequence is a permutation of the positive integers.”

Why it mattersIts natural index zero corresponds to OEIS index one.

The ideaThe values at valuation level k form the increasing queue 2^k(2r+1).

Our contributionThe valuation-step greedy sequence is a permutation of the positive integers.

Scope & next question[open] The analogous permutation statements for A382374, using Omega, and A382376, using omega, remain open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The valuation-step greedy sequence is a permutation

OEIS A382357 is the lexicographically earliest sequence of distinct positive integers whose adjacent 2-adic valuations differ by exactly one. It starts at 1, with offset 1. The supplied source states: “Conjecture: this sequence is a permutation of the positive integers.”

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/ValuationStepPermutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe A388986 Euler-form inclusion questionEvery Euler-form number in A228058 satisfies the strict divisor-sum inequality of A388986.Proved

The questionDoes every number in A228058 belong to A388986?

Why it mattersA388986 revision 36, read on September 9, 2026, still asks whether A228058 is a subsequence.

The ideaParametrize a unitary divisor by its prime support: coprimality forces its exponents to be the full exponents in N.

Our contributionEvery Euler-form number in A228058 satisfies the strict divisor-sum inequality of A388986.

Scope & next questionThe separate 2026 prime-factorization-subset conjecture on the same OEIS page is outside this result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A388986 Euler-form inclusion question

Does every number in A228058 belong to A388986? Explicitly, let p be prime, p congruent to 1 modulo 4, a a natural number, and r an odd integer greater than one with gcd(p,r)=1. Put N=p^(4a+1)r^2. The question asks whether the sum of the unitary divisors of N plus the sum of its squarefree divisors is strictly less than 2N.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/EulerFormDivisorSum.euler_form_lt

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe full alternating Laurent-row formula in A391810For every positive row, the signed alternating sum of the original bilateral Laurent coefficients is twice round(2^n/3), and hence twice the offset-zero Jacobsthal number.Proved

The questionHere T(n,k) is the signed coefficient of x to the power -k in the formal expansion at zero of the literal bilateral sum Sum{j=-oo..oo} x^j(1-x^j)^(n+j).

Why it mattersThe zero row sum, first moment and squared-coefficient questions are separate events and do not supply this alternating-row conclusion.

The ideaThe existing original-Laurent supplier proves that every nonpositive coefficient is the corresponding coefficient of the integer polynomial Pn(y)=-1+Sum{j=1..n} y^j(1-y^j)^(n-j), and bounds its degree by the published endpoint.

Our contributionFor every positive row, the signed alternating sum of the original bilateral Laurent coefficients is twice round(2^n/3), and hence twice the offset-zero Jacobsthal number.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The full alternating Laurent-row formula in A391810

OEIS A391810 formula (5), in its original positive-index range, is

$$ -\sum_{k=0}^{\lfloor(n+1)/2\rfloor^2}(-1)^k T(n,k) =2\operatorname{round}(2^n/3)=2,\mathrm{A001045}(n),\qquad n\ge1. $$

Here T(n,k) is the signed coefficient of x to the power -k in the formal expansion at zero of the literal bilateral sum Sum_{j=-oo..oo} x^j*(1-x^j)^(n+j). Integer powers with negative exponent use the Laurent-field inverse. The range includes k=0 and its upper endpoint. Rounding acts on the exact rational number, with nearest-integer ties toward positive infinity. A001045 has offset zero and recurrence J(0)=0, J(1)=1, J(n+2)=J(n+1)+2*J(n).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/HannaLaurentAlternatingJacobsthal.full_formula_five

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedA393867: every logarithmic-derivative coefficient is oddEvery coefficient of the logarithmic derivative defining OEIS A393867 is odd.Proved

The questionThe second OEIS A393867 comment is exactly: "Conjecture: all terms are odd." The target is Odd (a393867 n) for every n >= 1, using the definition in PrimePowerShiftLogDerivative.

Why it mattersSkill context: lean4; Codex implementation worker, single source of reasoning, zero independent review seats in this worker.

The ideaProposed escape witness: the paired-coefficient theorem F(2m) = F(2m+1) modulo two, derived from the integral source equations.

Our contributionEvery coefficient of the logarithmic derivative defining OEIS A393867 is odd.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A393867: every logarithmic-derivative coefficient is odd

The second OEIS A393867 comment is exactly: "Conjecture: all terms are odd." The target is Odd (a393867 n) for every n >= 1, using the definition in PrimePowerShiftLogDerivative. This is the first-tier 2026 conjecture; the prime-divisibility comment and A393866's shifted printed formula are distinct.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/PrimePowerShiftLogDerivativeOdd.a393867_odd

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedDivisibility of the A396793 coefficientsEvery coefficient above degree one in OEIS A396793 is divisible by three.Proved

The questionThese verbatim quotations are recorded in Library/Recurrence/hanna2026a396793.md.

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaThe escape content is liftmodnine: if f(0) = 0 and f is coefficientwise congruent to X modulo 3, then f iterate f 2 is coefficientwise congruent to X^2 modulo 9.

Our contributionEvery coefficient above degree one in OEIS A396793 is divisible by three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A396793 coefficients

OEIS A396793, Paul D. Hanna, Jun 07 2026, states:

NAME:

G.f. A(x) satisfies A(x) * A(A(x)) = x^2 + 9*x^3.

COMMENT:

Conjecture: a(n) == 0 (mod 3) for n > 1.

These verbatim quotations are recorded in Library/Recurrence/hanna2026a396793.md. Here A is an integer formal power series normalized by A(0) = 0 and a(1) = 1, and a(n) is its coefficient of x^n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateProductNineModThree.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedOEIS A396843: all coefficients are oddEvery positive-index coefficient in the series of OEIS A396843 is odd.Proved

The questionHanna's comment of July 1, 2026 states, verbatim:

Why it mattersThe listed 23 coefficients are odd, but a finite computation does not prove the assertion for every positive index.

The ideaWrite F for the reduced series and H=x(F+F^2).

Our contributionEvery positive-index coefficient in the series of OEIS A396843 is odd.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A396843: all coefficients are odd

OEIS A396843 defines A(x) = Sum_{n>=1} a(n)*x^n by

G.f. A(x) satisfies A( x*A(x) - 3*x*A(x)^2 ) = x^2.

Paul D. Hanna's comment of July 1, 2026 states, verbatim:

Conjecture: all terms are odd.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/CompositionalSquareAllOddCoefficients.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedA397346: square-weight congruenceOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.Proved

The questionHanna's OEIS A397346 NAME is the formal equation A(x)=exp(x+Sum{n>=2}(2n^2-1)a(n)x^n/(2n^2)).

Why it mattersThis is one endpoint of the shared integer square-weight family; A397242's parity endpoint already exists and is reused.

The ideaDefine b(q,0)=0, b(q,1)=1 and, for n>=2, b(q,n)=q(n-1)b(q,n-1)+Sum{2<=j<n}(qj^2-1)(n-j)b(q,j)b(q,n-j).

Our contributionOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A397346: square-weight congruence

Paul D. Hanna's OEIS A397346 NAME is the formal equation A(x)=exp(x+Sum_{n>=2}(2*n^2-1)*a(n)x^n/(2n^2)). Its first COMMENT is the single target of this dossier:

Conjecture: a(n) == [4,2,0,2] repeating (mod 8) for n >= 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ParametricExponentialSquareCongruence.residues_q2

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedEventual periodicity of every A398589 rowEvery initialized infinite self-banning row of OEIS A398589 is eventually periodic.Proved

The questionConjecture: All of the rows in this sequence are eventually periodic.

Why it mattersThe Library note quotes the definition and targeted conjecture and gives the index correspondence.

The ideaThe recursion is total because among k,...,k+t one label is absent from the t earlier positions.

Our contributionEvery initialized infinite self-banning row of OEIS A398589 is eventually periodic.

Scope & next questionThe source's nonempty-preperiod conjecture remains open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Eventual periodicity of every A398589 row

Conjecture: All of the rows in this sequence are eventually periodic.

For every k in Nat, let a_k(0)=k. At time t>0 choose the least natural x>=k such that every prior occurrence a_k(s)=x, s<t, satisfies s+x<t. The original claim is that for every k there exist N in Nat and p>0 such that for every t>=N, a_k(t+p)=a_k(t). The row is its infinite continuation; OEIS display truncation is not a premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/OeisA398589EventualPeriodicity.eventual_periodicity

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedTebni's prime and dyadic row-divisibility conjectures for A341724Both original A341724 row-divisibility conjectures hold for the integral coefficients of the exact column EGF.Proved

The questionFor 0 < k < p and p prime, T(p,k) == 0 (mod p). For 0 <= k < n and n = 2^m (m natural number), T(n,k) == 0 (mod n).

Why it mattersThis is a tier-one external open-problem settlement under preregistration 23158.

The ideaStrong induction proves all coefficients integral.

Our contributionBoth original A341724 row-divisibility conjectures hold for the integral coefficients of the exact column EGF.

Scope & next questionThe diagonal value one explains the exclusion for row modulus greater than one. - Unresolved boundary: the public result does not settle arbitrary composite rows, exact valuations, or any stronger modulus.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Tebni's prime and dyadic row-divisibility conjectures for A341724

The Comments block of OEIS A341724, revision 26, attributes these conjectures to Mélika Tebni in 2023:

For 0 < k < p and p prime, T(p,k) == 0 (mod p).

For 0 <= k < n and n = 2^m (m natural number), T(n,k) == 0 (mod n).

The Formula block specifies the column EGF as $x^k/(k!(1-2\sinh(-x)))$. The conjectures concern $T(n,k)=n![x^n]x^k/(k!(1-2\sinh(-x)))$ over the rationals. The 2021 Shannon–Ollerton paper describes the array, rather than stating these 2023 conjectures.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/TebniA341724RowDivisibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The sharp six-level bound for extreme two-qutrit absolutely PPT statesThe achievable upper bound on the number of distinct eigenvalues of extreme two-qutrit absolutely PPT states is six.Proved

The questionSong, Extreme points of absolutely PPT states with exactly three distinct eigenvalues, arXiv:2603.20717v1, Physica Scripta 101 (2026), 315210, DOI 10.1088/1402-4896/ae9148, Section VII, Conclusions, page 29, asks:

Why it mattersPreregistration 15732 identifies the published question and its literature scope.

The ideaA block-preserving trace-zero vector survives the constraints, and FeasiblePerturbation.spectralradius gives a positive radius preserving order, positivity, trace and both PSD pencils.

Our contributionThe achievable upper bound on the number of distinct eigenvalues of extreme two-qutrit absolutely PPT states is six.

Scope & next questionThe computed pencil and determinant identities are separate evidence. - Open.

Verification details
Release source recordProved in Lean

Open problem

The sharp six-level bound for extreme two-qutrit absolutely PPT states

N. Wang, L. Chen and Z. Song, Extreme points of absolutely PPT states with exactly three distinct eigenvalues, arXiv:2603.20717v1, Physica Scripta 101 (2026), 315210, DOI 10.1088/1402-4896/ae9148, Section VII, Conclusions, page 29, asks:

Fourth, what is the achievable upper bound of distinct eigenvalues of extreme two-qutrit AP states?

The answer is six, including attainment.

Statement echo

A state is a positive semidefinite trace-one matrix on $\mathbb C^3\otimes\mathbb C^3$. The public definition apStates is exactly the set of such matrices satisfying the frozen QutritPerturbationAttainment.APPT: every global unitary conjugate has positive semidefinite partial transpose. Extremality is Set.extremePoints ℝ apStates; the number of distinct eigenvalues is (spectrum ℂ rho).ncard, without multiplicity. claim is the conjunction of the universal bound $\le 6$ and existence of an extreme state with exactly six eigenvalues. result : claim has no extra hypotheses. The real scalar field is the field of convex coefficients; the complex spectrum belongs to the original complex matrix, rather than a real surrogate. Finite-dimensional Hermitian spectra are finite.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsolutePPT/QutritExtremeSixLevels.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

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Number theoryConjecture provedPosed in 2026Wu’s adjacent-index Fibonacci divisibility conjectureAn index whose Fibonacci number is divisible by itself and both neighbours ends in zero, two or eight.Proved

The questionNumber k such that Fibonacci(k) is divisible by k, k + 1 and k - 1.

Why it mattersThe proof shape is bind-only: the settlement combines existing Fibonacci index divisibility with arithmetic normalization.

The ideaEvenness and these two exclusions leave exactly the three asserted decimal alternatives.

Our contributionAn index whose Fibonacci number is divisible by itself and both neighbours ends in zero, two or eight.

Scope & next questionInfinitude, density and frequencies remain open, as do infinitely many occurrences of each digit.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Wu’s adjacent-index Fibonacci divisibility conjecture

OEIS A221018, revision 35, NAME:

Number k such that Fibonacci(k) is divisible by k, k + 1 and k - 1.

Its first FORMULA states:

a(n) % 10 == 0, 2, or 8 (conjectured). - Chai Wah Wu, Aug 06 2026

The source generator starts at two. Write $F_k$ for the standard Fibonacci sequence, with $F_0=0$, $F_1=1$ and $F_{k+2}=F_k+F_{k+1}$. For every natural $k\ge2$, the three assumptions $(k-1)\mid F_k$, $k\mid F_k$ and $(k+1)\mid F_k$ imply $k\bmod10\in{0,2,8}$. No converse is asserted.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Generator starts at 2 | Natural index $k$ and $2\le k$ | Covers the source integers and makes natural subtraction ordinary subtraction. | | Fibonacci(k) | Existing Nat.fib k | Standard initial values and recurrence; no replacement sequence. | | Divisible by k, k+1 and k-1 | The three separate divisibility hypotheses of claim | Their implication chain is equivalent to the source conjunction. | | Final digit 0, 2 or 8 | The three remainder alternatives in claim and result | Exact necessary condition; no added hypotheses or converse. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/WuAdjacentFibonacciResidues.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Zero forcing the ordinary powers of equal-legged spidersThe ordinary kth power of the equal-legged spider has zero forcing number (m-1)k.Proved

The questionAida Abiad, Mary Flagg, Sina Ghasemi Nezhad and Sjanne Zeijlemaker, The Zero Forcing Number of Graph Powers, arXiv:2610.03892v1, section 3, Conjecture 10: “Let m ≥ 2, k ≥ 1 and ℓ ≥ k+1.

Why it mattersThe lower bound quantifies over every ordinary zero forcing initial set, and the published Proposition 9 gives a forcing set of the claimed cardinality.

The ideaUse the integer tridiagonal leg matrix with diagonal ell-1 and opposite directed weights ell-1-i and i+1.

Our contributionThe ordinary kth power of the equal-legged spider has zero forcing number (m-1)k.

Verification details
Release source recordProved in Lean

Open problem

Zero forcing the ordinary powers of equal-legged spiders

Aida Abiad, Mary Flagg, Sina Ghasemi Nezhad and Sjanne Zeijlemaker, The Zero Forcing Number of Graph Powers, arXiv:2610.03892v1, section 3, Conjecture 10: “Let m ≥ 2, k ≥ 1 and ℓ ≥ k+1. Then Z(S_m(ℓ)^k) = (m−1)k.” The spider consists of m paths on ell vertices, joined at their first endpoints to a new center. It has m*ell+1 vertices. The kth ordinary power joins distinct vertices at distance at most k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/SpiderPowerZeroForcing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2025The source-digraph polytopeProved

Why it mattersThe source's Section 5 asks whether kefalopoda inequalities are its only nontrivial facet inequalities.

The ideaAll seven theorem proofs are classified bind-only; there is no claimed escape witness.

Verification details
Release source recordProved in Lean

Open problem

The source-digraph polytope

Baumeler, Tselentis and Wolf, Graphical Tests of Causality, arXiv:2511.22552v1, after Definition 13, state: “We know that $\widetilde Q_n^{\rm src}=Q_n^{\rm src}$ for $n\leq4$. In general, this identity is unsolved.” The settlement is $\widetilde Q_n^{\rm src}=Q_n^{\rm src}$ for every natural $n\geq2$. The external question and literal conventions are preregistered in https://github.com/the-omega-institute/trureturing/issues/20672.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Section 2: vertices $[n]$, arcs $(s,r)$ with $s\ne r$ | Fin n and the off-diagonal pair subtype, inlined in the definitions | All loopless directed arcs, with sender first and receiver second. | | Section 2: adjacency is one exactly on the arc set | The inlined conditional in sourcePolytope | Every arc set is a finite set; the empty digraph is included. | | Theorem 3: some vertex has in-degree zero | HasSource | A vertex is never the receiver of an arc in the set. | | Definition 8: convex hull of source-digraph adjacency vectors | sourcePolytope | Real convex hull of precisely those vectors. | | Definition 13 (58): nonnegative coordinates | First conjunct of kefalopodaPolytope | Every arc coordinate. | | Definition 13 (59): coordinate upper bounds for $n\geq3$ | Second conjunct | No upper-bound hypothesis is added for $n=2$. | | Definitions 9, 13 (60): every fixed-point-free function gives a sum at most $n-1$ | Third conjunct | One incoming arc per receiver; the bound is real. | | Identity in every dimension | claim, proved by result | Every natural $n\geq2$; no bounded enumeration hypothesis. |

The arc subtype and adjacency expression remain inline, while the proof uses Mathlib's Set.indicator for that same adjacency function.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/FiniteGeometry/Polytopes/BaumelerTselentisWolfSourceDigraphPolytope.result

Frozen module record

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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The two identities of Remark 4.3Complete homogeneous symmetric functions of finite path spectra.Proved

The ideaPartial fractions at the distinct Chebyshev nodes express complete homogeneous polynomials through nodal derivative weights.

Our contributionComplete homogeneous symmetric functions of finite path spectra.

Verification details
Release source recordProved in Lean

Open problem

The two identities of Remark 4.3

Bouthat, Chávez and Sheng, A general framework for inequalities on simple graphs, arXiv:2606.17356v1, Remark 4.3, state the two path CHS identities as unproved and then assert a one-term simplification under $n+1\ge d/2$.

For every $n\ge1$ and even $d\ge2$, the symmetric value $h_d(\lambda(P_n))$ equals both the trigonometric sum and the bilateral zero-extended binomial sum printed in Remark 4.3.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $|P_n|_{X,d}^d=h_d(\lambda(P_n))$ | PathCHSNormClosedForm.chs | Same complete homogeneous symmetric value. | | All $n\ge1$ and even $d\ge2$ | Arithmetic and parity hypotheses in claim | No degenerate path witness is used. | | Trigonometric sum, $1\le k\le n$ | trigForm, with Finset.Icc 1 n | Every factor and exponent is literal. | | Sum over all integers | bilateralForm, using finsum | Zero-extended binomials have finite support. | | Both equalities | The two conjuncts in claim | Both are proved by result. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PathCHS/PathCHSNormBCSRemark43.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Six broadcast-dominating packing classes on the two-way infinite pathThe six-class packing-domatic palette of the two-way infinite path is exactly five.Proved

The questionBrešar, Ferme and Hu, Partitioning an S-packing coloring into broadcast dominating sets, arXiv:2610.03477v1, Section 3.2, immediately after Theorem 3.6, ask:

Why it mattersThe source gives a five-colour upper construction but leaves the exact six-class value to the reader.

The ideaFor the upper bound, consume the source's repeating colour pattern [5,1,2,4,1,3] and its period-30 partition.

Our contributionThe six-class packing-domatic palette of the two-way infinite path is exactly five.

Verification details
Release source recordProved in Lean

Open problem

Six broadcast-dominating packing classes on the two-way infinite path

Brešar, Ferme and Hu, Partitioning an S-packing coloring into broadcast dominating sets, arXiv:2610.03477v1, Section 3.2, immediately after Theorem 3.6, ask:

For k=6, the problem is reduced to determining which of the two possible values in {4,5} is attained by χ_{ρ,6}(P∞), a question we leave for the reader.

The graph P∞ has vertices Z and distance |u-v|. For t in Nat, Q(t) asserts the existence of f:Z→Nat and one common c:Z→Fin 6 such that 1≤f(x)≤t for every x, distinct vertices of the same colour j are at distance greater than j, and for every x and every class i there is a vertex a with c(a)=i and |x-a|≤f(a). Neither map is assumed periodic or regular. The exact answer is 5: Q(5) holds, and Q(t) implies 5≤t for every natural t.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PackingDomatic/PackingDomaticInfiniteSix.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The large-squeezing law of the genuine multi-entropy of fully symmetric Gaussian statesTwisting the replica coordinates of a real Gaussian state produces a positive-definite quadratic form.Proved

Our contributionTwisting the replica coordinates of a real Gaussian state produces a positive-definite quadratic form.

Verification details
Release source recordProved in Lean

Open problem

The large-squeezing law of the genuine multi-entropy of fully symmetric Gaussian states

H. A. Camargo and M. Nishida, Genuine Multi-Entropy of Fully Symmetric Gaussian States, arXiv:2609.30754v1, compute the tripartite genuine Rényi multi-entropy $\mathrm{GM}^{(3)}_n(A:B:C)$ of the fully symmetric pure Gaussian state of $N$ bosonic modes (diagonal $a\ge1$, off-diagonal $e^-$ of their Eq. (3.1)) for $N\le8$ and $n\le4$, observe $\mathrm{GM}^{(3)}_n\sim\frac{2-n}{2n}\log a$ as $a\to\infty$ (Eq. (3.8)), and write "we \emph{conjecture} that the asymptotic behavior \eqref{AsymB} is valid for the fully symmetric Gaussian states with any $n$ and $N$." The verbatim definitions are in the literature note.

Issue #13990 reads the conjecture for every replica order $n\ge2$ and every tripartition with positive party sizes $N_A,N_B,N_C$, with the replica partition functions as the literal Lebesgue integrals of products of the real Gaussian density kernels under the twists $g_A,g_B,g_C$, and asymptotic equivalence in the sense of Asymptotics.IsEquivalent (for $n=2$ both sides are the zero function).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GaussianMultiEntropyAsymptotic.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026The tribonacci and four-Bonacci local Dold observationsThe printed Tribonacci and four-Bonacci sequences have empty local-Dold good-prime sets, with prime-index violations above every natural bound.Proved

The questionChuysurichay, Jaidee, Panraksa and Ward, The local Dold congruence for Fibonacci and Lucas sequences, arXiv:2609.24544v1, section 4, PDF page 11:

Why it mattersThe local Dold condition tests whether arithmetic support of a recurrence can satisfy orbit-count congruences.

The ideaReduce the integral invertible matrix A modulo p to obtain a positive period T.

Our contributionThe printed Tribonacci and four-Bonacci sequences have empty local-Dold good-prime sets, with prime-index violations above every natural bound.

Verification details
Release source recordProved in Lean

arXiv

The tribonacci and four-Bonacci local Dold observations

Chuysurichay, Jaidee, Panraksa and Ward, The local Dold congruence for Fibonacci and Lucas sequences, arXiv:2609.24544v1, section 4, PDF page 11:

Another natural problem from the results here is to try and characterize these sets of primes for linear recurrence sequences in general, starting with the intriguing observation that the classical tribonacci sequence (1,1,2,4,...) and 4-bonacci sequence (1,1,1,2,5,...) seem (computationally) to have G_D = ∅.

The quotation normalizes mathematical notation and whitespace. It is an empirical open observation, without a numbered conjecture. Its positive sequences are interpreted by their preceding-three and preceding-four recurrences: T1=T2=1, T3=2, and B1=B2=B3=1, B4=2. Thus B=(1,1,1,2,5,9,17,33,...).

For a positive natural sequence a and prime p, put b(n)=p^valuation_p(a(n)). The assertion is that neither sequence has any prime p for which every positive n divides the integer sum of mu(n/d)b(d) over positive divisors d of n. Both sequences and every prime are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/LocalDold/TribonacciFourBonacci.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026A uniform mass gap for the π/2-flux Hofstadter modelThe third and fourth Hofstadter bands have a uniform gap at Fermi energy two for every nonzero mass in the interval from minus two to two.Proved

The questionGiuliani, The conductivity matrix at the topological phase transition, arXiv:2609.03832v1 (2026), Section 3.1, leave an analytic gap assertion supported by numerical evidence.

Why it mattersThe gap is a premise for applying the source's main conductivity theorem to its four-band Hofstadter example.

Our contributionThe third and fourth Hofstadter bands have a uniform gap at Fermi energy two for every nonzero mass in the interval from minus two to two.

Verification details
Release source recordProved in Lean

Open problem

A uniform mass gap for the π/2-flux Hofstadter model

S. Fabbri, G. Marcelli and A. Giuliani, The conductivity matrix at the topological phase transition, arXiv:2609.03832v1 (2026), Section 3.1, leave an analytic gap assertion supported by numerical evidence. In the source member FMG_v1.tex, lines 689–690, they write (mathematical macros expanded and the assumption cross-reference rendered):

In the MathematicaNotebook1.nb sheet we provide numerical evidence of the validity of Assumption 2.1, and specifically of the fact that for all $m\in[-2,2]\setminus{0}$, $\min_{\vec{k}\in\mathbb{B}}(\lambda_4(\vec{k};m)-\mu)>0$ and $\max_{\vec{k}\in\mathbb{B}}(\lambda_3(\vec{k};m)-\mu)<0$. We expect that, with some additional technical effort, this can be fully proved analytically, but we prefer not to belabor the details of such a proof here.

Their Bloch Hamiltonian is Eq. eqn:Hofst_Hamiltonian, lines 663–671:

$$ H(\vec k;m)=-\begin{pmatrix} d&1-i e^{2ik_1}&1-i e^{2ik_2}&0\ 1+i e^{-2ik_1}&0&0&-i+e^{2ik_2}\ 1+i e^{-2ik_2}&0&0&1+i e^{2ik_1}\ 0&i+e^{-2ik_2}&1-i e^{-2ik_1}&-d \end{pmatrix},\qquad d=2+m. $$

The substitution $d=2+m$ and the Fermi energy are specified at lines 686–687. Write $\lambda_1\le\lambda_2\le\lambda_3\le\lambda_4$ for its real eigenvalues, counted with multiplicity, $\vec k_F=(\pi/4,\pi/4)$, and $\mu:=\lambda_3(\vec k_F;0)$.

Issue #22177 fixes the reading: $\lambda_4(\vec k_F;0)=\mu$, and for every real $m\in[-2,2]\setminus{0}$ there is $\delta>0$ such that

$$ \mu+\delta\le\lambda_4(\vec k;m),\qquad \lambda_3(\vec k;m)\le\mu-\delta \quad\text{for every }\vec k\in\mathbb R^2. $$

The same $\delta$ works for all momenta at a fixed mass; it may depend on the mass. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/SpectralTopology/HalfFluxHofstadterMassGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Maximal connectivity of full mutual-visibility reconfiguration graphsMinimum rooted pairs and the full mixed mutual-visibility separator equality.Proved

The questionGoldfeder and Aldo Lozano-Piña, Reconfiguration graphs of mutual-visibility sets, arXiv:2610.24324v1, Section 3.3 immediately after Theorem 3.6, asks:

Why it mattersThe Library note records the primary definitions and literature attribution.

The idea1. SingletonCore.minDegreelehostorder uses the empty configuration's degree.

Our contributionMinimum rooted pairs and the full mixed mutual-visibility separator equality.

Verification details
Release source recordProved in Lean

Open problem

Maximal connectivity of full mutual-visibility reconfiguration graphs

Julián A. Fresán-Figueroa, Ilán A. Goldfeder and Aldo Lozano-Piña, Reconfiguration graphs of mutual-visibility sets, arXiv:2610.24324v1, Section 3.3 immediately after Theorem 3.6, asks:

A natural question is whether $\mathcal{R}{\mathcal{M}}(G)$ is maximally connected for every connected $G$, that is, whether $\kappa(\mathcal{R}{\mathcal{M}}(G))=\delta(\mathcal{R}_{\mathcal{M}}(G))$ holds, where $\delta$ denotes minimum degree?

This dossier anchors exactly this question, repeated in Section 7. The source's standing convention is finite, simple, undirected, connected graphs. Connected hosts are nonempty; orders one and two are included. The preceding theorem's order-at-least-three hypothesis does not restrict the question.

For every distinct selected pair, some shortest path in the original host must avoid the other selected vertices. Different pairs may use different paths. The configuration graph contains every feasible subset, including empty and all feasible cardinalities. Distinct configurations $S,T$ are adjacent when $|S\mathbin{\triangle}T|=1$, or when $S\setminus T={x}$, $T\setminus S={y}$ and $xy$ is an actual host edge. Thus a slide replaces an occupied vertex by an unoccupied neighbor.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Every finite nonempty simple undirected host | V, [Fintype V], [DecidableEq V], [Nonempty V], H : SimpleGraph V in FullMaximalConnectivityClaim | | Connected, with no order or rank restriction | H.Connected is the only graph hypothesis | | Some original-host geodesic avoids the other selected vertices | Visible uses an existential H.Walk of length H.dist; shortest walks are paths | | All feasible subsets, including empty | Configuration H is the subtype of all visible Finset V | | Addition/removal and actual host-edge slides | The two disjuncts of MixedMove, used by reconfiguration | | Standard vertex connectivity | Separates deletes to at most one vertex or to an induced graph that is not preconnected; vertexConnectivity minimizes cut size | | Equality to minimum degree | result : FullMaximalConnectivityClaim |

The separator convention gives $\kappa(K_q)=q-1$ for every positive $q$; Connectivity.connectivity_complete verifies this convention. No geodesic, separator, sparse-slide, girth or diameter premise is added to the conclusion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/MutualVisibility/MinimumPair.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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Discrete structuresConjecture provedFirst preprint 2026Han–Pedon Conjecture C: all higher-shift metallic Hankel residue polynomialsEvery higher shifted metallic Hankel sequence has the stated residue polynomial laws from the first term.Proved

The questionGuo-Niu Han and Emmanuel Pedon, Hankel determinants for the (n+2)-shifted q-metallic numbers, arXiv:2610.36132v1, introduction, Conjecture C.

Why it mattersConjecture C supplies that structure from the first term while retaining all metallic parameters and higher shifts.

The ideaIts ordered indices, sign, common-degree reversals and zero-support guards retain the original moments.

Our contributionEvery higher shifted metallic Hankel sequence has the stated residue polynomial laws from the first term.

Verification details
Release source recordProved in Lean

Open problem

Han–Pedon Conjecture C: all higher-shift metallic Hankel residue polynomials

Guo-Niu Han and Emmanuel Pedon, Hankel determinants for the (n+2)-shifted q-metallic numbers, arXiv:2610.36132v1, introduction, Conjecture C. The defining quadratic is equation (2.13); equations (1.4) and (1.5) define the shifted determinant and its empty value. The original preregistration is issue #23253. The follow-up is distinct from arXiv:2502.05993v2 and its Conjecture E on bounded-shift periodicity and higher-shift unboundedness.

For every natural $n\ge1$, there exists $\Phi\in\mathbb Z[[q]]$ with constant coefficient one satisfying

$$q\Phi^2+\bigl((1+q^n)(1-q)-q[n]_q\bigr)\Phi=1, \qquad [n]q=\sum{i=0}^{n-1}q^i.$$

For every series satisfying these conditions and every natural shift $\ell\ge n+3$, define

$$\Delta_j^{(\ell)}=\det\bigl([q^{\ell+a+b}]\Phi\bigr)_{0\le a,b<j}, \qquad \Delta_0^{(\ell)}=1,$$

and put $\rho=2n(n+1)$ for even $n$ and $\rho=4n(n+1)$ for odd $n$. There is a family $Q_r\in\mathbb Q[X]$, indexed by every $0\le r<\rho$, such that

$$\Delta_{r+m\rho}^{(\ell)}=Q_r(m)\quad\text{for every }m\in\mathbb N,$$

and $Q_r(z)\in\mathbb Z$ for every $z\in\mathbb Z$. Here $\mathbb N$ includes zero. There is no eventual threshold, excluded initial residue, target or neighboring-minor invertibility condition, degree bound, or assertion that $\rho$ is minimal. Existence is a conclusion, and every satisfying metallic series is covered.

Formal correspondence

Trureturing.OpenProblems.HanPedonShiftedQMetallicHankel.claim, defined in MetallicHankelQuasipolynomialDefs, retains these clauses literally: 1 ≤ n, an existential integral metallic series, universal quantification over every satisfying series, n + 3 ≤ ℓ, the displayed parity-dependent period, a family indexed by Fin ρ, an integer witness for each integer input, and equality at every natural m. IsMetallic and shiftedHankel are the original definitions in MetallicHankelDefs; the latter is the determinant on Fin j, so size zero is included.

The settling declaration is D5.S3.Combinatorics.MetallicHankel.MetallicHankelResiduePolynomials.result : Trureturing.OpenProblems.HanPedonShiftedQMetallicHankel.claim. Its Scribe carries this dossier's OpenProblemResolutionClaim with resolution Proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MetallicHankel/MetallicHankelResiduePolynomials.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedHanna's Thue/Roth Exponential IntegralityHanna's Thue/Roth weighted formal exponential has integer coefficients.Proved

The questionHanna's OEIS A174470 states: “G.F.: exp( Sum{n>=1} A014578(n)(3x)^n/n ), where A014578 is the binary expansion of Thue constant.” Its COMMENT field says: “Conjectured to consist entirely of integers.” The source is revision 2; A014578 is revision 68.

Why it mattersThe existing Dold exponential-integrality criterion supplies the final implication once the logarithmic weights satisfy its prime-power congruences.

The ideaThe proof uses padicValNat.mul, padicValNat.eqzeroofnotdvd, Nat.padicValNatltself, Nat.ModEq.powtotient, and Nat.totientprimepowsucc.

Our contributionHanna's Thue/Roth weighted formal exponential has integer coefficients.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

hanna a174470

Paul D. Hanna's OEIS A174470 states: “G.F.: exp( Sum_{n>=1} A014578(n)*(3x)^n/n ), where A014578 is the binary expansion of Thue constant.” Its COMMENT field says: “Conjectured to consist entirely of integers.” The source is revision 2; A014578 is revision 68. The source has no DOI or primary arXiv version.

Write $b(0)=0$ and, for $n>0$, let $b(n)=1$ when $v_3(n)$ is even and $b(n)=0$ otherwise. Set $c(n)=b(n)3^n$ and

$$ F(x)=\exp\left(\sum_{n\ge1}\frac{c(n)}n x^n\right)\in\mathbb Q[[x]]. $$

The assertion includes existence of this rational formal series and integrality of each ordinary coefficient at every $N\in\mathbb N$, including $N=0$.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | A014578's zero case and valuation-parity implementation | b | Integer-valued bits with explicit zero case. | | Weight A014578(n) times $3^n$ | c | Literal weight, with natural exponent. | | Formal exponential of the logarithmic sum | IsSourceSeries | Equality to PowerSeries.subst (logInput c) (PowerSeries.exp ℚ); logInput is reused from DoldExponentialIntegrality. | | “Conjectured to consist entirely of integers” | claim, proved by result | Existence and all-series/all-degree coefficient integrality; no additional hypotheses. |

The declarations are in D5.S1.Recurrence.Algebraic.HannaThueRothExponentialIntegrality. The claim spells out coefficient integrality and defines no second IntegralSeries predicate.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaThueRothExponentialIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Discrete structuresConjecture provedPosed in 2026Hanna’s OEIS A271960 dyadic vanishing conjectureReversion and dyadic coefficient decimation prove Hanna's A271960 conjecture.Proved

The questionHanna’s OEIS A271960, COMMENTS (unpaginated), states:

Why it mattersThe coefficient pattern is a support restriction propagated by compositional inversion and repeated halving of the degree and exponent.

The ideaThe coefficient-agreement steps reuse the existing public dev owners SignedCatalanCubicSubstitutionModThree.squarelinear and substtwoagree; its existing public inverseexists supplies the two-sided inverse.

Our contributionReversion and dyadic coefficient decimation prove Hanna's A271960 conjecture.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna’s OEIS A271960 dyadic vanishing conjecture

Paul D. Hanna’s OEIS A271960, COMMENTS (unpaginated), states:

Conjecture: [x^(2^n-2^k)] (x/A(x))^(2^n) = 0 for k = 0..n-2 and n >= 2. - Paul D. Hanna, Jan 16 2026

Its NAME is:

G.f. A(x) satisfies: A(x)^2 = A( (x + 2*A(x)^2)^2 ).

The OFFSET is 1 and the EXAMPLE begins $A(x)=x+2x^2+9x^3+50x^4+312x^5+\cdots$. The Library note records the source. Preregistration: #19138.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The defining equation | IsGF A | Literal rational power-series equation, with PowerSeries.subst u f meaning $f(u)$. | | Offset and leading coefficient | First two conjuncts of IsGF | Constant coefficient zero and linear coefficient one. | | $x/A(x)$ | xOverA A | Inverse of the series with coefficient $[x^{n+1}]A$ at index $n$. | | $n\ge2$, $0\le k\le n-2$ | Natural binders and the two bounds in claim | Natural subtraction is exact under the bounds. | | Coefficient and power | Universal conjunct of claim | $x^{2^n-2^k}^{2^n}=0$. | | Existence | Existential conjunct of claim | Additional conclusion; it does not restrict the universally quantified solutions. | | Every normalized solution | Universal $A$ with IsGF A | No selected solution replaces the source’s equation. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PowerSeries/HannaA271960DyadicVanishing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 3/4 · 15% weight

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Discrete structuresQuestion resolvedHanna's A377100 coefficient congruenceThe normalized generating equation has an integral solution, and every positive coefficient of every normalized solution is one modulo three.Proved

The questionHanna, OEIS A377100, Name and Comments (unpaginated), as quoted in the source note.

Why it mattersThe published entry explicitly states this coefficient congruence as a conjecture.

The ideaUniqueness identifies the reduced integral solution with that series.

Our contributionThe normalized generating equation has an integral solution, and every positive coefficient of every normalized solution is one modulo three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A377100 coefficient congruence

Paul D. Hanna, OEIS A377100, Name and Comments (unpaginated), as quoted in the source note.

G.f. A(x) satisfies A(x) = A(x^3)/A(x^2) + A(x)^2.

Conjecture: a(n) == 1 (mod 3) for n >= 1, which has been verified for the initial 2305 terms.

For $A(x)=x+\sum_{n\ge2}a(n)x^n$, the equation is $A(x)=\frac{A(x^3)}{A(x^2)}+A(x)^2$. The resolution is Proved: a normalized rational power-series solution exists, and every normalized solution has integral coefficients $a(n)\equiv1\pmod3$ for all $n\ge1$. Preregistration: #18851. The other Hanna problem is separately preregistered and has its own resolution claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The displayed generating equation | HannaCubeSquareQuotientModThree.IsGF | Literal preregistered equation over rational power series; the A377100 quotient is in Laurent series. | | The example begins $x+x^2+\cdots$ | Constant coefficient zero and coefficient one at $X$ in IsGF | The normalization is explicit. | | $a(n)$ for $n\ge1$ | coeff n A, with the hypothesis 1 ≤ n | Every positive natural index. | | Integral coefficients congruent to one modulo three | Integer witness equal to coeff n A and Int.ModEq 3 m 1 in claim | Coercion into the rationals is explicit. | | A generating function is defined by the equation | Existence conjunct in claim | Excludes a vacuous universal statement. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaCubeSquareQuotientModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedHanna's bilateral reversion parity conjecturePairing bilateral Laurent summands proves the evenness conjecture for OEIS A379202.Proved

The questionA(x) satisfies 1/x = Sum{n=-oo..+oo} A(x)^n (A(x)^n + 2)^(n+1).

Why it mattersBilateral Laurent summation connects the source equation to an integral power-series reversion problem.

The ideaLaurent normalization identifies every rational solution of the source equation with this inverse.

Our contributionPairing bilateral Laurent summands proves the evenness conjecture for OEIS A379202.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's bilateral reversion parity conjecture

P. D. Hanna's OEIS A379202, Name section, states:

G.f. A(x) satisfies 1/x = Sum_{n=-oo..+oo} A(x)^n * (A(x)^n + 2)^(n+1).

Its Comments section states:

Conjecture: a(n) is even for n > 1.

The offset is one and the first coefficient is one. The formal solution is $A\in\mathbb Q[[X]]$ with $A(0)=0$ and $[X]A=1$. Integer powers and inversion are taken in the field $\mathbb Q((X))$. For each natural $j$, the coefficient at degree $j-1$ is the sum over $-j-2\le n\le j+2$; all omitted summands have zero coefficient at that degree. The source is recorded in the Library note. Preregistration: #18876.

Statement echo

All Lean names below belong to D5.S1.Recurrence.Invariants.HannaBilateralReversionParity.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A^n(A^n+2)^{n+1}$ for integer $n$ | term | Literal Laurent-series integer powers. | | Sum over all integer indices | bilateralSum | Coefficientwise finite window; outside_window and literal_finite_window prove that omitted coefficients vanish. | | $1/X$ equals the bilateral sum | IsGF | Literal Laurent inverse of $X$. | | Offset one, first entry one | Two normalization conjuncts of IsGF | Zero constant coefficient and linear coefficient one. | | Coefficient $a(n)$ for every $n>1$ | Natural binder and strict inequality in claim | Index one is excluded. | | Integer, even coefficient | Integer witness and Even in claim | Witness is cast to the rational coefficient. | | Generating series exists | Existential conjunct of claim | An additional conclusion, with no added hypothesis. | | Uniqueness in the preregistered settlement statement | Private solution_unique | Every normalized solution equals the same rational image of the integral inverse. | | Any generating series satisfying the equation | Universal series binder and IsGF hypothesis | No selected solution replaces the source's solution class. |

No source clause is dropped. The full proposition appears at both the claim and result Blueprint nodes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaBilateralReversionParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The proved statement applies to the stated general family, not just a checked finite range.

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Quantum informationExact optimum establishedFirst preprint 2025The always-failing purification policy attains the sharp availability lower boundThe always-failing purification policy at q equal to one attains the minimum buffer availability.Proved

The questionWehner, Entanglement buffering with multiple quantum memories, arXiv:2502.20240v1 (npj Quantum Information 12, 64, 2026), pose the conjecture immediately after Corollary 1.

Why it mattersAvailability measures whether buffered entanglement is present when requested.

Our contributionThe always-failing purification policy at q equal to one attains the minimum buffer availability.

Verification details
Release source recordProved in Lean

Open problem

The always-failing purification policy attains the sharp availability lower bound

Á. G. Iñesta, B. Davies, S. Kar and S. Wehner, Entanglement buffering with multiple quantum memories, arXiv:2502.20240v1 (npj Quantum Information 12, 64, 2026), pose the conjecture immediately after Corollary 1. The v1 source archive's main.tex (MD5 c1b5c2a56b45349a94ac22e6fde7bd87), lines 500–502, reads verbatim:

Unlike the upper bound, we note that the lower bound from (\ref{eq.A_bounds}) has not yet shown to be tight. We believe that the availability at $q=1$ of a policy that always fails purification ($c_k=d_k=0$, $\forall k$) constitutes a tight lower bound for any other purification policy. We leave this proof as future work.

The system has one long-lived storage memory and $n$ short-lived generation memories per node. Write $p_g$ for the generation probability per short-lived memory, $p_c$ for the consumption probability, $q$ for the probability of attempting purification, $\Gamma$ for the depolarising decay parameter and $F_{\mathrm{new}}$ for the newly generated fidelity.

The module's admissible requires, for each $k\in\mathrm{Fin},n$ and every $f\in[1/4,1]$, $$ 0\le c_k(f-1/4)+d_k\le1,\qquad 0\le a_k(f-1/4)+b_k\le\frac34\bigl(c_k(f-1/4)+d_k\bigr). $$ Here $k$ represents the source's $k.\mathrm{val}+1$ generated links. These are the probability and unnormalised output-fidelity conditions of Appendix C, main.tex lines 1734–1767.

Write $A((a,b,c,d),q)$ for availability n pg pc q gam fnew a b c d, the real-valued closed formula in the source's Theorem 1 (main.tex lines 378–415). Issue #21747 fixes the reading: for every natural $n\ge1$, real $p_g,p_c\in(0,1]$, $\Gamma\ge0$, $F_{\mathrm{new}}\in(1/4,1]$, admissible real coefficient families $a,b,c,d:\mathrm{Fin},n\to\mathbb R$, and $q\in[0,1]$, $$ A((0,0,0,0),1)\le A((a,b,c,d),q). $$ This is claim, and result : claim proves it. The identification of the closed formula with the limiting fraction of consumption requests finding a link is the source's Theorem 1; that stochastic identification is not formalized in this module. The definitions and source locators are also in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Decoherence/EntanglementBufferAvailabilityLowerBound.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    The always-failing purification policy at q equal to one attains the minimum buffer availability.

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  • Historical context

    First preprint year: 2025.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026Jenzer's permuted rotor-loop formula without the star conditionPermuted rotor loops obey the subset formula without the star condition.Proved

The questionNik Jenzer, Star-Ordered Rotor Loops in Clifford Algebras, arXiv:2610.21593v1, Theorem 5.5 and the following scope remark, TeX Remark 5.7, rendered Remark 5.5, page 19, asks whether the permuted loop formula remains true without the star condition:

Why it mattersThe scalar-cyclicity and exterior-word modules supply consumed parts of this proof.

The ideaJenzerPermutedRotorLoopAlgebra.NativeBridge.scalarcyclic proves cyclicity of the native scalar coefficient.

Our contributionPermuted rotor loops obey the subset formula without the star condition.

Verification details
Release source recordProved in Lean

Open problem

Jenzer's permuted rotor-loop formula without the star condition

Nik Jenzer, Star-Ordered Rotor Loops in Clifford Algebras, arXiv:2610.21593v1, Theorem 5.5 and the following scope remark, TeX Remark 5.7, rendered Remark 5.5, page 19, asks whether the permuted loop formula remains true without the star condition:

Nevertheless, numerical evidence suggests that the formula also holds without the star condition in that case: tested on the triangle family from Remark 4.4, which violates the star condition, under all six permutations of {1, 2, 3} with random parameter sets; the largest deviation encountered was in the range of double machine precision. A general proof would have to capture the cancellation of the odd off-diagonal terms in coarser groups than pairs; we have not determined such a grouping.

For an anticommuting unit bivector family $B_1,\ldots,B_k$ in a finite-dimensional nondegenerate real Clifford algebra, $k\geq1$, put $c_i=\cos(\theta_i/2)$, $s_i=\sin(\theta_i/2)$, $w_i=s_i^2$, $U_i=c_i+s_iB_i$, and $\Theta_k(\tau)=(U_1\cdots U_k)(\widetilde U_{\tau(1)}\cdots\widetilde U_{\tau(k)})$. Then the proved statement is

$$ \langle\Theta_k(\tau)\rangle_0 =\sum_{S\subseteq{1,\ldots,k}} \eta(|S|)(-1)^{\iota(\tau,S)} \prod_{i\in S}w_i\prod_{i\notin S}(1-w_i), \qquad \eta(m)=(-1)^m(-1)^{m(m+1)/2}. $$

Here $\iota(\tau,S)$ counts pairs $i<j$ in $S$ with $\tau^{-1}(i)>\tau^{-1}(j)$. No star condition is assumed.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Real finite-dimensional space and nondegenerate quadratic form | claim binders V, real module and finite-dimensional instances, q, q.Nondegenerate | All source signatures | | Positive family size | k, 1 ≤ k | Every positive natural size | | Homogeneous bivectors, square $-1$, pairwise anticommutation | IsBivector, the three conjuncts of AnticommutingUnitFamily | No independence assumption | | Native scalar coefficient and reversion | scalarProjection, CliffordAlgebra.reverse in returnProduct | The exterior equivalence reads grades only; multiplication remains Clifford multiplication | | All real angles and all return permutations | θ, τ in claim | Zero sine and cosine values included | | Ordered products, subset weights and restricted inverse inversions | subsetProduct, rotor, chain, returnProduct, loop, a, c, s, w, inversionCount, eta | Fin k is the increasing relabelling of the one-based source indices | | Removal of the star condition | StarCondition is defined, but is absent from claim | No replacement hypothesis |

The canonical names in the table belong to D5.S3.QuadraticForms.JenzerPermutedRotorLoopAlgebra. The settling theorem is D5.S3.QuadraticForms.JenzerPermutedRotorLoop.result. The public definitions retain the source expressions and the binding conventions in #22266.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/JenzerPermutedRotorLoop.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Kasel's displacement ladderKasel's displacement ladder equals m minus two at every horizon four to the m, for m at least two.Proved

The questionKasel, arXiv:2609.02939, Appendix B.4, states: “The conjectured growth L(m) = m − 2”.

Why it mattersThe displacement ladder measures the finite-horizon structural obstruction behind Erdős Problem 197.

The ideaThe bounded numeric rank is injective and puts each target first.

Our contributionKasel's displacement ladder equals m minus two at every horizon four to the m, for m at least two.

Verification details
Release source recordProved in Lean

Open problem

kasel 2026 displacement ladder

Kasel, arXiv:2609.02939, Appendix B.4, states: “The conjectured growth L(m) = m − 2”. The exact target was preregistered in #14885.

Let $b(v)=\lceil\log_2 v\rceil$ for $v\ge2$, implemented by the total natural-number function Nat.clog 2 v. Set

$$S_A(m)={v\in[1,4^m]:b(v)\ge2,\ b(v)\text{ even}},\qquad D={3,4,9,10,11,12,13,14,15,16}.$$

A scheme consists of natural-valued stages $s$ and fibre positions $r$. It orders values by stage, then by fibre position. Validity requires injectivity within each stage and excludes both increasing and decreasing three-term arithmetic progressions in that concatenation. Normalization requires $s(v)\ge\lfloor b(v)/2\rfloor$ on $S_A(m)$. Write $\delta(v)=s(v)-\lfloor b(v)/2\rfloor$ and let $L(m)$ be the minimum, over valid normalized schemes, of $\max_{v\in D}\delta(v)$. The proved statement is $L(m)=m-2$ for every natural $m\ge2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/KaselDisplacementLadder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Kemp–Ramgoolam arbitrary-depth determinant and monomial-basis conjecturesActual source monomials form a basis in every commutative complex algebra with complete orthogonal leaf projectors.Proved

The questionKemp and Ramgoolam's arXiv:2603.05259v2 defines a layered source tree with recursive threshold sets in (3.1)–(3.4).

Why it mattersThe source is a named recent open problem whose exact arbitrary-depth clauses are retained as one cohesive target.

The ideaThe proof uses factorization.

Our contributionActual source monomials form a basis in every commutative complex algebra with complete orthogonal leaf projectors.

Verification details
Release source recordProved in Lean

arXiv

Kemp–Ramgoolam arbitrary-depth determinant and monomial-basis conjectures

Kemp and Ramgoolam's arXiv:2603.05259v2 defines a layered source tree with recursive threshold sets in (3.1)–(3.4). Its ordinary source columns are the disjoint union in (3.5)–(3.6). The leaf-path matrix is (6.21). The determinant conjecture is the signed sibling-only product (6.22): at the last layer (i=L), the exponent is one, while at every earlier layer (i<L), the exponent is the number of actual source-bounded positive threshold vectors whose literal two-root truncation retains both endpoints, as in (6.23)–(6.27). Labels need only separate siblings; parent-dependent fibers and equal labels under unrelated parents are permitted. The monomial-basis conjecture is the assertion in (3.6) that these ordinary source monomials span the required column space at arbitrary depth.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/DegeneracyGraphMonomialBasis.source_monomial_basis_and_expansions

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Pure-target quantum Černý complexityPure-target quantum Cerny complexity grows linearly on constant binary words.Refuted

The questionPui Hang Lee, Pui Yee Lee and Bjørn Kjos-Hanssen, arXiv:2609.40154v1, §6 item 5:

Why it mattersA pure common output imposes a support constraint beyond ordinary synchronization.

The ideaA word with z zeros sends the basis projector at j to the basis projector at max(j−z,0).

Our contributionPure-target quantum Cerny complexity grows linearly on constant binary words.

Verification details
Release source recordRefuted in Lean

Open problem

Pure-target quantum Černý complexity

Pui Hang Lee, Pui Yee Lee and Bjørn Kjos-Hanssen, arXiv:2609.40154v1, §6 item 5:

Does a general quadratic-type saving qc_pure(w) ≤ O(√|w|) hold, i.e. a pure-target analogue of Theorem 2.4?

An instance consists of two quantum channels on d by d density matrices and a start state. Letters act from left to right. The reachable set contains the images of the start state under all binary words. A word synchronizes when its channel is constant on this set. The complexity qc_pure(w) is the least dimension of an instance in which w is the unique shortest synchronizing word and its common output is pure.

The precise proposed uniform bound is the existence of a real constant C and a natural threshold m₀ such that qc_pure(w) ≤ C√|w| for every binary word with |w| ≥ m₀.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/QuantumCernyPureTargetRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
Quantum informationQuestion resolvedSource 2026Malkoun's two-matrix elementary symmetric inequalityElementary symmetric eigenvalue inequality for a Hermitian Jordan product.Proved

The questionThis dossier anchors only the explicit first question of Malkoun's Promoting polynomial equalities for nonnegative variables to elementary symmetric polynomial inequalities, MathOverflow 512579, revision 1, GUID DFD20136-D37A-462C-8A59-FD828AEEFF9A.

Why it mattersEigenvalue-by-eigenvalue monotonicity cannot simply be assumed for elementary symmetric functions when the Jordan product can have negative eigenvalues.

The ideaCauchy–Binet uses the same row selections for all three determinants.

Our contributionElementary symmetric eigenvalue inequality for a Hermitian Jordan product.

Verification details
Release source recordProved in Lean

Open problem

Malkoun's two-matrix elementary symmetric inequality

This dossier anchors only the explicit first question of Malkoun's Promoting polynomial equalities for nonnegative variables to elementary symmetric polynomial inequalities, MathOverflow 512579, revision 1, GUID DFD20136-D37A-462C-8A59-FD828AEEFF9A. The source first promotes real nonnegative variables to Hermitian positive semidefinite matrices and sets $L=XY+YX$ and $R=X^2+Y^2$. Its question reads (HTML formatting adapted):

Given an $n$ by $n$ matrix $A$, denote by $\lambda(A)$ its "set" of eigenvalues, taking multiplicity into account. Moreover, denote by $e_k$ the $k$-th elementary symmetric polynomial. Note that both $L$ and $R$ are hermitian, so that their eigenvalues are all real. Is it true that $$e_k(\lambda(L)) \leq e_k(\lambda(R)),$$ for any $n$ by $n$ hermitian positive semidefinite matrices $X$ and $Y$ and for any $k \in {1, \dots, n}$?

The precise assertion is: for every positive integer $n$, every pair of complex Hermitian positive semidefinite $n\times n$ matrices $X,Y$, and every integer $1\leq k\leq n$, the displayed inequality holds. Repeated eigenvalues retain their algebraic multiplicity. No positivity of $XY+YX$, commutation, invertibility, rank, simple-spectrum or dimension restriction is imposed. The source's later unspecified question about promoting arbitrary symmetric polynomial inequalities is not this dossier's assertion.

The source text is by Malkoun (user 81645) and is licensed under CC BY-SA 4.0. See the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/MalkounJordanElementary.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Sublinear edge completion for consecutive cycle sumsFor every positive natural a, every sufficiently large consecutively labelled cycle can be completed using at most n/a added edges; a hub and tail construction uses fewer than the square root of 2n + 8 edges for n at least eight.Proved

The questionMarotti and Needleman, Consecutive Cycle Sums, arXiv:2610.08889, Section 5, final paragraph on printed page 6, ask whether the cycle with labels 1, …, n can eventually be completed using at most n/a added edges for every positive integer a.

Why it mattersPreregistration 14954 identifies the eventual sublinear edge-completion question as an unsettled named problem from Section 5.

The ideaFor n ≥ 8, let k ≥ 4 be least with Tk ≥ n + 4.

Our contributionFor every positive natural a, every sufficiently large consecutively labelled cycle can be completed using at most n/a added edges; a hub and tail construction uses fewer than the square root of 2n + 8 edges for n at least eight.

Verification details
Release source recordProved in Lean

Open problem

Sublinear edge completion for consecutive cycle sums

Marotti and Needleman, Consecutive Cycle Sums, arXiv:2610.08889, Section 5, final paragraph on printed page 6, ask whether the cycle with labels 1, …, n can eventually be completed using at most n/a added edges for every positive integer a. A graph is complete when every integer from 1 through T_n = n(n + 1)/2 is the label sum of a nonempty vertex set inducing a connected subgraph. The label order remains fixed, and mEC(n) is the minimum number of edges added to this cycle to make it complete.

The precise target is: for every integer a ≥ 1, there exists N such that, for every integer n ≥ N, a complete graph containing the original n-cycle has an added-edge count e satisfying ea ≤ n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CycleSums/ConsecutiveCycleSums.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedFirst preprint 2026Nathanson's Problem 12: affine invariance of sum intersection setsInjective affine maps preserve the sum intersection set, answering Nathanson's Problem 12 affirmatively.Proved

The questionNathanson, Problems and results on intersections of product sets and sumsets in semigroups, arXiv:2604.04781v2, Problem 12, p. 21, asks:

Why it mattersThe source is recorded in Library/ArithUnits/nathanson2026sumintersections.md.

The ideaThe local hinj uses mulleftcancel₀ and addrightcancel.

Our contributionInjective affine maps preserve the sum intersection set, answering Nathanson's Problem 12 affirmatively.

Verification details
Release source recordProved in Lean

Open problem

Nathanson's Problem 12: affine invariance of sum intersection sets

Melvyn B. Nathanson, Problems and results on intersections of product sets and sumsets in semigroups, arXiv:2604.04781v2, Problem 12, p. 21, asks:

Let $(A_q)$ and $(A'_q)$ be asymptotically strictly decreasing sequences of sets of integers such that $A'_q=sA_q+t$ for all $q\in\mathbb N$ and some $s,t\in\mathbb Z$ with $s\notin{0,\pm1}$. Does this imply $H(A_q)=H(A'_q)$?

The paper takes $\mathbb N={1,2,3,\ldots}$, defines $A=\bigcap_q A_q$ and $H(A_q)={h\in\mathbb N:hA=\bigcap_q hA_q}$, and uses $hB$ for the $h$-fold sumset. Asymptotically strict decrease means $A_{q+1}\subseteq A_q$ at every index, with infinitely many strict changes. The answer to Problem 12 is affirmative.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Two sequences of integer sets indexed by positive integers | A A' : ℕ+ → Set ℤ in claim | The paper's positive-natural indexing is literal. | | Both sequences asymptotically strictly decreasing | AsymptoticallyStrictlyDecreasing A and its primed counterpart | The definition retains adjacent inclusion and infinitely many unequal neighbours. | | The $h$-fold sumset | Pointwise h • B | This is repeated set addition, with independent summands. | | Positive sum lengths and the sum-intersection equality | H | Its natural-number carrier is restricted by 0 < h. | | $A'_q=sA_q+t$ at every index | The affine Set.image hypothesis in claim | It is the literal set ${sa+t:a\in A_q}$. | | Integer $s,t$ and $s\notin{0,\pm1}$ | Integer parameters and the exclusion hypothesis in claim | All three excluded values are retained. | | $H(A_q)=H(A'_q)$ | Conclusion of claim; result : claim | No source premise is dropped from the delivered statement. |

Every explicit binder and conjunct comes from these clauses. The Boolean-free formulas in the Scribe display the three definition bodies and the complete proposition of the settling theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Additive/NathansonProblem12AffineInvariance.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe complete first growth assertion of OEIS A222759The prime threshold places the cubes of the first k primes below n.Proved

The questionNoe, OEIS A222759, revision 12, November 2, 2024.

Why it mattersThe 2013 first-comment growth assertion is a Tier 1 published OEIS conjectural assertion.

The ideaThis is the ordinary remainder-range mechanism; it uses no special binomial congruence.

Our contributionThe prime threshold places the cubes of the first k primes below n.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The complete first growth assertion of OEIS A222759

T. D. Noe, OEIS A222759, revision 12, November 2, 2024. The NAME is:

Conjectured number of primes p for which binomial(n*p,p) (mod p^3) does not equal n.

The entire first COMMENT is:

It appears that, for k > 2 and n >= prime(prime(k)^3), then a(n) >= k.

Let $P(j)$ be the one-based $j$-th prime for positive $j$, with $P(1)=2$. For every integer $k\ge3$ and every integer $n\ge1$ with $n\ge P(P(k)^3)$, the event is the existence of $k$ pairwise distinct positive primes $q_1,\ldots,q_k$ such that $\binom{nq_i}{q_i}\bmod q_i^3\ne n$ for every $i$. The remainder is the least nonnegative integer remainder, compared with $n$ itself. The threshold includes equality and cubes $P(k)$ inside the outer prime index. The event asserts at least $k$ witnesses without assuming finiteness of the entire set satisfying the predicate.

The formal domain is exactly the positive integer domain represented by naturals with $3\le k$ and $1\le n$. An injection $q:\mathrm{Fin}(k)\to \mathbb N$ supplies distinct witnesses, each explicitly positive and prime. The total extension $P(0)=2$ is unused on the event's indices; the natural remainder convention at modulus zero is unused on prime witnesses. The source identity and operator correspondence are in Library/Factorization/noe2013a222759.md.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/PrimeRemainders/OeisA222759Growth.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2025Schulte's A000680 half-factorial primality criterionSchulte's A000680 half-factorial sequence satisfies his primality criterion.Proved

The questionConjecture: For n > 0 (a(n) + 2^n) is a multiple of (2n + 1) if and only if (2n + 1) is a prime number. - Werner Schulte, Oct 05 2025

Why it mattersThe independent question is Schulte's named primality characterization of OEIS A000680 for every positive natural index.

The ideaUse 2^n a(n) = (2n)! from the exact natural-number quotient.

Our contributionSchulte's A000680 half-factorial sequence satisfies his primality criterion.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's A000680 half-factorial primality criterion

OEIS A000680, NAME (%N, verbatim):

a(n) = (2n)!/2^n.

The settled COMMENT (%C, verbatim) is:

Conjecture: For n > 0 (a(n) + 2^n) is a multiple of (2n + 1) if and only if (2n + 1) is a prime number. - Werner Schulte, Oct 05 2025

The offset is %O 0,3, and the AUTHOR line (%A, verbatim) is:

N. J. A. Sloane

The full-quantifier reading is: for every n : Nat, if 0 < n, then (2 * n + 1) ∣ a n + 2 ^ n if and only if (2 * n + 1).Prime. The scope wall is exactly this biconditional for every positive natural n; no additional sequence values, implications, or historical claims are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/SchulteHalfFactorialPrimeCriterion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedCorneth's second recurrence for OEIS A068012The complete residue table proves Corneth's second recurrence for OEIS A068012.Proved

The questionLet a(n) count subsets of {1,...,n} whose element sum is zero modulo six.

Why it mattersThis is the second recurrence of a first-tier recent OEIS conjecture selected in the orchestrator's v3 bind-first brief.

The ideaIn phase m mod 3=1 the residues 2 and 5 satisfy 6C(m,r)+22^q=2^m, and the other residues satisfy 6C(m,r)=2^m+2^q.

Our contributionThe complete residue table proves Corneth's second recurrence for OEIS A068012.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Corneth's second recurrence for OEIS A068012

Let a(n) count subsets of {1,...,n} whose element sum is zero modulo six. David A. Corneth's September 13, 2025 FORMULA contribution states:

a(3*k+1) = 2*a(3*k) - 2^(k-1), k >= 1, 3*k + 1 = n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/SubsetSums/ModSixResidueCounts.corneth_step

Frozen module record

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About this highlight
  • Contribution

    The complete residue table proves Corneth's second recurrence for OEIS A068012.

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe Somos prime integrality family in OEIS A091190/A091200The normalized rational series of the Somos family is integral exactly for prime exponents.Proved

The questionMore generally, given A(x) satisfies xA(x)^p = B(xA(x^p)) where B(x) = x/(1-px), then it appears that A(x) is an integer series only when p is prime.

The ideaPrime Frobenius congruence makes integral corrections possible through the shared lifting construction.

Our contributionThe normalized rational series of the Somos family is integral exactly for prime exponents.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Somos prime integrality family in OEIS A091190/A091200

More generally, given A(x) satisfies x*A(x)^p = B(x*A(x^p)) where B(x) = x/(1-p*x), then it appears that A(x) is an integer series only when p is prime.

For every natural $p\geq2$, a rational solution with $A(0)=1$ exists for $xA(x)^p=B(xA(x^p))$, where $B(x)=x/(1-px)$. Every such solution has integer coefficients if and only if $p$ is prime. The coefficientwise expression for $B$ is inlined in IsGF.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | xA(x)^p = B(xA(x^p)) | IsGF preserves the outer and inner substitutions | exact registered convention | | B(x)=x/(1-p*x) | IsGF inlines coefficient zero at n=0 and p to the power n-1 otherwise | exact registered convention | | offset-zero first term one | IsGF requires constant coefficient one | exact registered convention | | integer only when prime | claim proves equivalence with primality for every normalized solution | exact registered convention | | exponents in the registered domain | claim requires p at least two and asserts rational existence | exact registered convention |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SomosPrimeIntegrality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2026A165719 integrality and the sole primeThe positive integral values of k(k+9)/8 have 17 as their only prime.Proved

The question%N A165719 Integers of the form k(k+9)/8. %C A165719 Only one term is a prime number (17).

Why it mattersOrlovsky submitted the sequence in September 2009.

The ideaReduction modulo eight shows that only residues zero and seven make the product divisible by eight.

Our contributionThe positive integral values of k(k+9)/8 have 17 as their only prime.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A165719 integrality and the sole prime

The OEIS entry states, verbatim:

%N A165719 Integers of the form k*(k+9)/8. %C A165719 Only one term is a prime number (17). Are all others composite? %F A165719 Conjecture: Valid k values are pairs of the form (8m-1,8m) for m=1,2,3.... If true, 17 is the only prime term. - Bill McEachen, May 06 2026 %A A165719 Vladimir Joseph Stephan Orlovsky, Sep 24 2009 %O A165719 1,1

For every positive natural k, 8 | k*(k+9) is equivalent to the existence of a positive natural m with k+1=8*m or k=8*m. A natural a is an integral term exactly when some positive k satisfies 8*a=k*(k+9). The value 17 is a term and a prime; every other term is greater than one and not prime. The example's positive-parameter restriction is retained. The equations avoid natural-number floor division and truncated subtraction.

Not claimed: a result for nonpositive parameters, a new method of prime classification, any assertion about arbitrary quadratic forms, an exhaustive literature search, or priority beyond the settlement of these named assertions in the searched scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/PrimeForms/OrlovskyEighthQuadraticSolePrime.result

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Number theoryQuestion resolvedOEIS A167857: three distinct prime divisorsAn integral polynomial represents all increasing divisors of an integer with three distinct prime factors.Proved

The questionThe Online Encyclopedia of Integer Sequences, A167857, T.

Why it mattersThis is the original published existence question, classified as Tier 1 in the prospective registration.

The ideaThe same product n=pqr has eight positive divisors, all 1 modulo 5040.

Our contributionAn integral polynomial represents all increasing divisors of an integer with three distinct prime factors.

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Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A167857: three distinct prime divisors

The Online Encyclopedia of Integer Sequences, A167857, T. D. Noe, defines numbers whose divisors are represented by an integer polynomial:

That is, these numbers n have the property that there is a polynomial f(x) with integer coefficients whose values at x=0..tau(n)-1 are the divisors of n, where tau(n) is the number of divisors of n.

The published question is:

Are there terms with three distinct prime divisors?

The exact target is existence of a positive natural n with n.primeFactors.card = 3 and f : Polynomial Int such that, for every j < n.divisors.card, evaluation at the integer cast of j equals entry j of the actual increasingly sorted finset of all positive divisors of n. Both factor count and divisor list belong to this same n. The list has no duplicates; the inputs start at zero. No minimum, degree bound, numerical candidate, divisor subset, alternate order or rational-only conclusion is imposed. Preregistration: https://github.com/the-omega-institute/trureturing/issues/16601.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SortedDivisorPolynomialThreePrimes.result

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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

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Number theoryQuestion resolvedPosed in 2025Prime alternating divisor sums in OEIS A193351A prime alternating sum of decreasing divisors above three forces a square or twice a square.Proved

The questionNumbers k such that the alternating sum of all divisors of k (divisors nonincreasing, starting with k) is prime.

Why it mattersThe conjecture was recorded by Michel Lagneau in 2012.

The idea(a) Pair consecutive terms of the decreasing divisor list to obtain 2T(n) >= n.

Our contributionA prime alternating sum of decreasing divisors above three forces a square or twice a square.

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Integer sequences (OEIS)

Prime alternating divisor sums in OEIS A193351

OEIS A193351, NAME (%N, verbatim):

Numbers k such that A071324(k) is prime.

COMMENT (%C, verbatim):

Numbers k such that the alternating sum of all divisors of k (divisors nonincreasing, starting with k) is prime.

CONJECTURE (%C, verbatim):

For k > 3, if the alternating sum of all divisors of k (divisors nonincreasing, starting with k) is a prime number, then k is either a square or twice a square.

The entry also records:

This conjecture is verified up to k <= 10^8. - Shreyansh Jaiswal, Apr 25 2025

With T : ℕ → ℤ denoting that literal alternating sum, the formal claim is ∀ k : ℕ, 3 < k → Nat.Prime (T k).toNat → (∃ t : ℕ, k = t ^ 2) ∨ (∃ t : ℕ, k = 2 * t ^ 2). The use of .toNat makes the primality predicate explicit; the proof establishes the required nonnegativity before using it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/LagneauAlternatingDivisorSumPrimeSquare.lagneau_a193351

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Discrete structuresQuestion resolvedHanna's A204061 coefficients modulo fiveHanna's A204061 residues mod 5 are one exactly when no base-5 digit is two.Proved

The questionG.f.: exp( Sum{n>=1} A001333(n)^2 x^n/n ) where A001333(n) = A002203(n)/2, one-half the companion Pell numbers.

Why it mattersThe conjecture links a recurrence-defined exponential to a base-five digit condition at every natural index.

The ideaFormal differentiation proves constant coefficient one and f^4 ((1+X)^2 (1-6X+X^2)) = 1.

Our contributionHanna's A204061 residues mod 5 are one exactly when no base-5 digit is two.

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Integer sequences (OEIS)

Hanna's A204061 coefficients modulo five

OEIS A204061, NAME (%N, verbatim):

G.f.: exp( Sum_{n>=1} A001333(n)^2 * x^n/n ) where A001333(n) = A002203(n)/2, one-half the companion Pell numbers.

COMMENT conjecture (%C, verbatim):

a(n) == 1 (mod 5) iff n has no 2's in its base 5 expansion (A023729), otherwise a(n) == 0 (mod 5); this is a conjecture needing proof.

FORMULA (%F, verbatim):

G.f.: 1 / ( sqrt(1+x) * (1-6*x+x^2)^(1/4) ).

AUTHOR (%A, verbatim):

_Paul D. Hanna_, Jan 10 2012

The exact formal target is, for every n : Nat, exists z : Z_5, (z : Q_5) = a(n) and PadicInt.toZMod z = (if 2 in Nat.digits 5 n then 0 else 1). Here a(n) is coefficient n of the defining exponential in Q_5. Its exponent uses the frozen sequence Q : Nat → Nat from D5/S1/Recurrence/PellCompanionGcd: A001333, one-half the companion Pell numbers, canonically embedded into Q_5. This proves the residue conjecture in the 5-adic reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CompanionPellExpBaseFiveResidue.result

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    Source 1
Discrete structuresQuestion resolvedHanna's overpartition logarithmic derivative congruencesThe logarithmic derivative of square-indexed overpartitions has residue twice its index modulo four.Proved

The questionHanna's OEIS A219431 has Name “Logarithmic derivative of the g.f. of the overpartitions of n^2 (A219430).” Its Comment states: “Conjecture: a(2n+1) == 2 (mod 4), a(2n) == 0 (mod 4).” The offset is one.

The ideaThe finite-product identity productmodfour eliminates every product containing two factors of two.

Our contributionThe logarithmic derivative of square-indexed overpartitions has residue twice its index modulo four.

Verification details
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Integer sequences (OEIS)

Hanna's overpartition logarithmic derivative congruences

Paul D. Hanna's OEIS A219431 has Name “Logarithmic derivative of the g.f. of the overpartitions of n^2 (A219430).” Its Comment states: “Conjecture: a(2n+1) == 2 (mod 4), a(2n) == 0 (mod 4).” The offset is one. OEIS A219430 states: “Number of overpartitions of n^2; a(n) = A015128(n^2).”

Let $\bar p(m)$ be the coefficient of $q^m$ in $\prod_{k\geq1}(1+q^k)/(1-q^k)$, and put $G(x)=\sum_{n\geq0}\bar p(n^2)x^n$. The settlement proves that every positive-degree coefficient $a(m)=[x^m]xG'(x)/G(x)$ is an integer, with $a(m)\equiv 2$ for odd $m$ and $a(m)\equiv0$ for even $m$, modulo four. Preregistration: #18874.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Overpartition generating product | pbar | At degree $m$, factors with $k>m$ have no effect; the finite product uses $1\leq k\leq m$. | | Square-indexed coefficients, including index zero | G | PowerSeries.mk with coefficient pbar at $n^2$. | | Logarithmic derivative, offset one | logDerivative | $xG'(x)G(x)^{-1}$ over the rationals. | | Every index at least one | claim | All natural $m$ with $1\leq m$. | | Integer coefficients and two congruences | claim, result | One integer witness equals the rational coefficient and satisfies both parity implications. Integrality is an explicit strengthening, with no additional hypothesis. |

The four public definitions retain the expressions of the preregistration. Their parameters, coefficient rings and modulus are unchanged. The positive geometric tail and the divisibility-indicator geometric series are reused from their existing owners rather than declared again.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Partitions/OverpartitionSquareLogDerivativeModFour.result

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's triangular-exponent EGF congruenceEvery factorial-normalized coefficient of Hanna's original triangular-exponent EGF is an integer congruent to one modulo three.Proved

The questionHanna's OEIS A226839, returned revision 11, gives the NAME:

Why it mattersAll other declarations are definitions or live consumed helpers.

The ideaThus the actual integral inner coefficients reduce modulo three to the identity sequence, with zero constant term.

Our contributionEvery factorial-normalized coefficient of Hanna's original triangular-exponent EGF is an integer congruent to one modulo three.

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Integer sequences (OEIS)

Hanna's triangular-exponent EGF congruence

Paul D. Hanna's OEIS A226839, returned revision 11, gives the NAME:

E.g.f.: exp( Sum_{n>=1} x^(n*(n+1)/2) / n! ).

Its FORMULA is:

a(n) == 1 (mod 3) (conjecture - valid up to n=1024; if true for n>=0, why?).

The offset is 0,4. The immutable preregistration is #23823. The source note is Library/RecurrenceComposition/hanna2013a226839.md. The source question and source editorial approval do not supply a proof.

Statement echo

Over $\mathbb Q[[X]]$, set $T_m=\lfloor m(m+1)/2\rfloor$, $H(X)=\sum_{m\ge1}X^{T_m}/m!$ and $A(X)=\exp(H(X))$. For every $n\in\mathbb N$, including zero, there is a $k\in\mathbb Z$ with $(n!: \mathbb Q)[X^n]A=1+3(k:\mathbb Q)$.

The public definitions triangular, H and A retain the literal exponent, coefficientwise positive-index rational sum and formal exponential substitution. The denominator is $m!$, not $T_m!$. The private coeff_H proves local finiteness using $m\le T_m$. The public innerIntegral is the natural finite sum over $0\le m\le n$ of $(m+1)^{\overline{n-m}}$ when $0<m$ and $T_m=n$, zero otherwise. The private factorial_ratio and eCoeff_H derive its equality to $n![X^n]H$. There is no assumed integer representation or finite cutoff. The sole public theorem result is exactly the displayed all-index event.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA226839Congruence.result

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    The delivered argument documents a mathematical mechanism.

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedPrice's A228558 prime-exponent conjectureA prime quotient (x^k+y^k)/(x+y) for coprime positive bases forces the exponent k to be prime.Proved

The questionNumbers k such that (17^k + 4^k)/21 is prime.

Why it mattersThe entry defines its sequence by prime values of an exponential quotient and asks whether every exponent in that sequence is prime.

The ideaFirst, under positivity and coprimality, x+y divides x^k+y^k exactly when k is odd: the forward direction excludes even k, and Odd.natadddvdpowaddpow supplies the reverse direction.

Our contributionA prime quotient (x^k+y^k)/(x+y) for coprime positive bases forces the exponent k to be prime.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Price's A228558 prime-exponent conjecture

OEIS A228558, NAME (%N, verbatim):

Numbers k such that (17^k + 4^k)/21 is prime.

Robert Price's COMMENT (%C, verbatim):

Conjecture: All terms are prime.

The literal proved statement is

forall k : Nat, 21 | 17^k + 4^k -> Prime((17^k + 4^k)/21) -> Prime(k).

The divisibility hypothesis is explicit because the OEIS expression denotes an exact integer quotient, while natural-number division would otherwise truncate.

Not claimed here are the A228558 comment a(9) > 10^5, the A231329 comment a(5) > 10^5, infinitude of either sequence, primality of any particular term, or the classical general factorization fact as a new result. That fact is the same background argument as 2^n-1 prime -> n prime.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/PricePrimeQuotientExponent.result_a228558

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedPrice's A231329 prime-exponent conjectureA prime quotient (x^k+y^k)/(x+y) for coprime positive bases forces the exponent k to be prime.Proved

The questionNumbers n such that (19^n + 4^n)/23 is prime.

Why it mattersThe entry defines its sequence by prime values of an exponential quotient and asks whether every exponent in that sequence is prime.

The ideaFirst, under positivity and coprimality, x+y divides x^k+y^k exactly when k is odd: the forward direction excludes even k, and Odd.natadddvdpowaddpow supplies the reverse direction.

Our contributionA prime quotient (x^k+y^k)/(x+y) for coprime positive bases forces the exponent k to be prime.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Price's A231329 prime-exponent conjecture

OEIS A231329, NAME (%N, verbatim):

Numbers n such that (19^n + 4^n)/23 is prime.

Robert Price's COMMENT (%C, verbatim):

Conjecture: All terms are prime.

The literal proved statement is

forall k : Nat, 23 | 19^k + 4^k -> Prime((19^k + 4^k)/23) -> Prime(k).

The divisibility hypothesis is explicit because the OEIS expression denotes an exact integer quotient, while natural-number division would otherwise truncate.

Not claimed here are the A228558 comment a(9) > 10^5, the A231329 comment a(5) > 10^5, infinitude of either sequence, primality of any particular term, or the classical general factorization fact as a new result. That fact is the same background argument as 2^n-1 prime -> n prime.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/PricePrimeQuotientExponent.result_a231329

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The proved statement applies to the stated general family, not just a checked finite range.

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Discrete structuresQuestion resolvedThe antepenultimate-label count in OEIS A270236Restricted-growth words give Mathar's two closed label-occurrence counts.Proved

The questionTriangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.

Why it mattersThis is one of two independent formulas recorded as conjectures in the same OEIS entry.

The ideaIsRestrictedGrowth states the chronological source condition, including the positive-label bound, and memwordsiff identifies words n with its length-n solutions after reversing the generator's storage order.

Our contributionRestricted-growth words give Mathar's two closed label-occurrence counts.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The antepenultimate-label count in OEIS A270236

OEIS A270236, NAME (verbatim):

Triangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.

COMMENT defining the restricted growth functions (verbatim):

The RG functions used here are defined by f(1)=1, f(j) <= 1+max_{i<j} f(i).

FORMULA (verbatim; R. J. Mathar, Mar 13 2016):

Conjecture: T(n+1,n-1) = 2+n*(n+1)*(3*n^2-5*n+26)/24 for n>1.

Revision #22 records that Alois P. Heinz corrected the left index from T(n,n-2) to the current T(n+1,n-1) on Mar 18 2016.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/RestrictedGrowthLabelOccurrences.mathar_f2

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    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedSchulte's exponent-product multiplicativity conjectureProved

The questionConjecture: Let k be some fixed integer and ak(n) = A005361(n) k^A001221(n) for n > 0 with 0^0 = 1.

Why it matterscomputationalcontent.kind: none: neither the definition nor the theorem is a bounded enumeration, checker, numerical reduction, or certified finite instance.

The ideaFor coprime m,n, split the factorization-exponent product across the disjoint prime-factor sets, split omega(mn) additively, and rearrange the integer products to prove ak(mn)=ak(m)ak(n).

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's exponent-product multiplicativity conjecture

OEIS A322327, NAME (%N, verbatim):

a(n) = A005361(n) * A034444(n).

OFFSET (%O, verbatim):

1,2

AUTHOR (%A, verbatim):

Werner Schulte, Dec 03 2018

COMMENT (%C, verbatim):

Conjecture: Let k be some fixed integer and a_k(n) = A005361(n) * k^A001221(n) for n > 0 with 0^0 = 1. Then a_k(n) is multiplicative with a_k(p^e) = k*e for prime p and e > 0. For k = 0 see A000007 (offset 1), for k = 1 see A005361, for k = 2 see this sequence, for k = 3 see A226602 (offset 1), and for k = 4 see A322328.

A005361, NAME (%N, verbatim):

Product of exponents of prime factorization of n.

A001221, NAME (%N, verbatim):

Number of distinct primes dividing n (also called omega(n)).

For every integer k, define a_k(n) = (product over prime divisors p of n of v_p(n)) * k^omega(n), where v_p(n) is the exponent of p in the prime factorization and the empty product is one. The full-quantifier reading asks that for all positive naturals m,n with gcd(m,n)=1, a_k(mn)=a_k(m)*a_k(n), and that for all natural primes p and positive exponents e, a_k(p^e)=k*e. Lean's convention gives 0^0=1. The formal theorem establishes the same equations for all natural m,n satisfying coprimality, which includes the stated positive domain.

The scope wall contains only those two assertions in the conjecture sentence for every integer k. The trailing correspondences with A000007, A005361, A322327, A226602, and A322328 are not claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SchulteExponentProductOmegaPowerMultiplicative.result

Frozen module record

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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe divisor-count prime-factor bound in A328959The divisor-count expression in OEIS A328959 is nonnegative from n=2 onward.Proved

The questiona(n) = sigma0(n) - 2 - (omega(n) - 1) nu(n), where sigma0 = A000005, nu = A001221, omega = A001222.

Why it mattersThis 2019 OEIS conjecture asks for a uniform relation between the divisor count and the prime-factor counts.

The ideaWrite n = producti pi^(ei), put r = omega(n) >= 1, and set bi = ei - 1.

Our contributionThe divisor-count expression in OEIS A328959 is nonnegative from n=2 onward.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The divisor-count prime-factor bound in A328959

OEIS A328959, NAME (verbatim):

a(n) = sigma_0(n) - 2 - (omega(n) - 1) * nu(n), where sigma_0 = A000005, nu = A001221, omega = A001222.

Here the entry's omega is A001222, the number of prime factors counted with multiplicity, and its nu is A001221, the number of distinct prime factors. In conventional notation these are respectively Omega and omega.

COMMENT (verbatim; Gus Wiseman, Nov 02 2019; idea from Mats Granvik):

Conjecture: All terms are nonnegative except for a(1) = -1.

The exception is exact: a(1) = -1. The target is therefore the nonnegativity assertion for every n >= 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/DivisorCountPrimeFactorBound.wiseman_a328959

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe triangular-residue descent count in A329278Kagey's descent count for the triangular-number permutation modulo a positive power of two.Proved

The questionThe n-th row is the permutation of {0, 1, 2, ..., 2^n-1} given by T(n,k) = k(k+1)/2 (mod 2^n).

Why it mattersThis 2019 OEIS conjecture asks for the exact number of adjacent strict descents in every positive row of the triangular-number permutation modulo a power of two.

The ideagives s(k+1) = (sk + (k+1)) mod M; a wrap is equivalent to a descent.

Our contributionKagey's descent count for the triangular-number permutation modulo a positive power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The triangular-residue descent count in A329278

OEIS A329278, NAME (verbatim):

Irregular table read by rows. The n-th row is the permutation of {0, 1, 2, ..., 2^n-1} given by T(n,k) = k(k+1)/2 (mod 2^n).

COMMENT (verbatim; Peter Kagey, Nov 11 2019):

Conjecture: for n > 0, the n-th row has 2^(n-1)-1 descents.

A descent is an index k with 0 <= k < 2^n - 1 and T(n,k) > T(n,k+1). The target concerns every positive natural n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/TriangularResidueDescentCount.kagey_a329278

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedFirst occurrences in the A335925 self-referential quotient recurrenceThe first occurrences in Alkan's self-referential quotient recurrence are A000522 thresholds.Proved

The questiona(n) = a(floor((n-1)/a(n-1))) + 1 with a(1) = 1.

Why it mattersThis OEIS conjecture was recorded in 2020 and remained unchanged through the bounded source and literature checks recorded in preregistration issue 7464.

The ideaThe corrected block invariant is a(T(r)) = r+1; for T(r) <= n < T(r+1), the value a(n) belongs to {r,r+1}; and a(n)=r implies n < rT(r).

Our contributionThe first occurrences in Alkan's self-referential quotient recurrence are A000522 thresholds.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

First occurrences in the A335925 self-referential quotient recurrence

OEIS A335925, NAME (%N, verbatim):

a(n) = a(floor((n-1)/a(n-1))) + 1 with a(1) = 1.

COMMENT (%C, verbatim; Altug Alkan, Jun 30 2020):

Least k such that a(k) = n are 1, 2, 5, 16, 65, 326, 1957, ... (Conjecture: This sequence is A000522).

Here A000522(0) = 1 and A000522(m) = m*A000522(m-1) + 1. The phrase "least k such that a(k) = m" is formalized as a hit at that index together with minimality among every positive index carrying m.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SelfReferentialQuotientFirstOccurrence.alkan_a335925

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The first occurrences in Alkan's self-referential quotient recurrence are A000522 thresholds.

    Read source ↗
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  • Problem significance · 1/4 · 30% weight

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    Source 1 · Source 2
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  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

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Number theoryQuestion resolvedDe Vlieger's descending prime-power exponentsDescending prime-power exponents end at one, so their absolute first differences sum to n minus one.Proved

The questionDe Vlieger's conjecture in OEIS A347285, COMMENTS (unpaginated), states: "Conjecture: let S be the sum of the absolute values of the first differences of terms in row n.

Why it mattersA universal endpoint identity determines the total variation of every row without enumerating its prime powers.

The ideaA row that never reached zero would give an infinite strictly descending sequence of natural numbers, excluded by well-foundedness.

Our contributionDescending prime-power exponents end at one, so their absolute first differences sum to n minus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a347285 de vlieger prime power descent

M. De Vlieger's conjecture in OEIS A347285, COMMENTS (unpaginated), states: "Conjecture: let S be the sum of the absolute values of the first differences of terms in row n. For all n > 0, n - S = 1." The NAME defines the rows as "Irregular triangle T(n,k) starting with n followed by e_k = floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0." The FORMULA is "T(n,1) = n; T(n,k) = floor(log_p_k(p_(k-1)^T(n,k-1)))." The source text is pinned at e7d0073e. Preregistration: #18689. Source note: devlieger2021a347285.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The primes are indexed from one | Existing D5.S1.Recurrence.Parity.PrimePowerShiftLogDerivative.prime, with value Nat.nth Nat.Prime (k - 1) | The literal preregistered enumeration is reused. | | Initial row entry is $n$ | e n 0 | Equals $n$. | | Real logarithm and downward natural floor at the next prime | e n (j + 1) | The current exponent is raised on the current prime, and the logarithm base is the next prime. | | Keep entries greater than zero | len n, the first zero index, with zero fallback if none exists | e_zero_exists and len_spec establish existence and positivity before that index. | | Sum absolute first differences within the row | S n | Natural absolute value of each integer difference, over indices below len n - 1. | | Every $n>0$, with $n-S=1$ | claim, proved by result | Natural indices, strict positivity, and integer subtraction; no additional hypothesis. |

The source convention "T(0,1) = 0 by convention; 0 is not allowed for n > 0" leaves the separately listed row zero outside the conjecture. The recurrence extends through zero for proof purposes; only the positive prefix enters $S$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Primes/OeisA347285PrimePowerDescent.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's OEIS A352700 Catalan power-series and modular-coefficient conjecturesLagrange substitution and Frobenius pairing prove Hanna's six A352700 congruences.Proved

The questionHanna's OEIS A352700 (revision 7, March 31, 2022; unpaginated), NAME, defines the sequence by:

Our contributionLagrange substitution and Frobenius pairing prove Hanna's six A352700 congruences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's OEIS A352700 Catalan power-series and modular-coefficient conjectures

Paul D. Hanna's OEIS A352700 (revision #7, March 31, 2022; unpaginated), NAME, defines the sequence by:

G.f.: Sum_{n>=0} binomial((n+1)*(2*n+1),n)/(2*n+1) * x^n / C(x)^(n*(2*n+1)+1), where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).

Its COMMENTS state the six numbered conjectures:

Conjectures: given g.f. A(x), let C(x) = (1 - sqrt(1-4*x))/(2*x) be the Catalan power series (A000108), then (1) A(x)^3 = A(x^3) (mod 3), (2) A(x) = C(x) + x^2*C(x)^3 (mod 3) = (2 - x)*C(x) - 1 (mod 3), (3) A(x) = C(x) (mod 2), (4) a(n) = binomial(2*n+1,n)/(2*n+1) + 3*binomial(2*n-1,n-2)/(2*n-1) (mod 3) for n >= 0, (5) a(n) = 2*A000108(n) - A000108(n-1) (mod 3) for n >= 1, (6) a(n) = A000108(n) (mod 2) for n >= 0.

The published OFFSET is 0,3. Its EXAMPLE field records the coefficient expansion of $C(x) + x^2 C(x)^3$, under which the second term $3\binom{2n-1}{n-2}/(2n-1)$ vanishes for $n < 2$ by the negative-lower-index binomial convention $\binom{2n-1}{n-2} = 0$, and its PARI PROG defines $a(n)$ by the finite degree-$n$ truncation over $0 \le m \le n$. The Library note records the source and exact integer divisibility of each weight and Raney quotient. Preregistration: #23834.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Weight $\binom{(m+1)(2m+1)}{m}/(2m+1)$ | weight m | Exact natural quotient; $(2m+1) \mid \binom{(m+1)(2m+1)}{m}$ proved for every $m \in \mathbb N$. | | Catalan generating series $C(x) = 1 + x C(x)^2$ and $C(x)^{-1}$ | CatalanCompositionSquareParity.catalanUnit, catalanInvZ | Integral formal series in $\mathbb Z[[X]]$ with $[X^n]C(X) = \operatorname{catalan}(n)$, constant term CatalanCompositionSquareParity.catalan_unit_zero, and multiplicative unit inverse. | | Defining series $A(x)$ and $a(n) = [x^n]A(x)$ | coeffA, seriesA | Finite $(X)$-adically exact sum over $0 \le m \le n$ in $\mathbb Z$ and $\mathbb Z[[X]]$. | | Reductions modulo $2$ and $3$ | seriesMod p, catalanMod p | Coefficientwise ring-homomorphism reductions in $(\mathbb Z/p\mathbb Z)[[X]]$. | | Second term of (4) under $\binom{2n-1}{n-2}=0$ for $n < 2$ | raneyTerm n | Zero for $n < 2$ and exact natural quotient $(3\binom{2n-1}{n-2})/(2n-1)$ for $n \ge 2$. | | Conjectures (1)–(6) | Six conjuncts of claim, proved by result | Exact conjunction of (1) (seriesMod 3)^3 = (seriesMod 3)(X^3), (2) both mod-$3$ series identities, (3) mod-$2$ series identity, (4) mod-$3$ Raney congruence for all $n \ge 0$, (5) mod-$3$ Catalan difference congruence for all $n \ge 1, and (6) mod-$2$ Catalan congruence for all $n \ge 0. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PowerSeries/HannaA352700CatalanCongruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363558 prime-index coefficient formulaHanna's A363558 prime-index coefficient formula.Proved

The questiona(p) = p2^(p-1) + 4^p for primes p > 3 (conjecture).

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363558 prime-index coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363558 prime-index coefficient formula

Paul D. Hanna's OEIS A363558 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (2 + x^n)^(2*n).

a(p) = p*2^(p-1) + 4^p for primes p > 3 (conjecture).

The binding coefficient convention and the exact claim are in #18660. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (2 + x^n)^(2n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(p) = p2^(p-1) + 4^p for primes p > 3 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363558Prime.claim and D5.S1.Recurrence.HannaBilateral.OeisA363558Prime.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363558Prime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363559 prime-index coefficient formulaHanna's A363559 prime-index coefficient formula.Proved

The questiona(p) = p3^(p-1) + 9^p for primes p > 3 (conjecture).

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363559 prime-index coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363559 prime-index coefficient formula

Paul D. Hanna's OEIS A363559 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (3 + x^n)^(2*n).

a(p) = p*3^(p-1) + 9^p for primes p > 3 (conjecture).

The binding coefficient convention and the exact claim are in #18662. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (3 + x^n)^(2n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(p) = p3^(p-1) + 9^p for primes p > 3 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363559Prime.claim and D5.S1.Recurrence.HannaBilateral.OeisA363559Prime.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363559Prime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363561 prime-index coefficient formulaHanna's A363561 prime-index coefficient formula.Proved

The questionG.f.: Sum{n=-oo..+oo} x^n (sqrt(2) + x^n)^(2n).

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363561 prime-index coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363561 prime-index coefficient formula

Paul D. Hanna's OEIS A363561 states:

G.f.: Sum_{n=-oo..+oo} x^n * (sqrt(2) + x^n)^(2*n).

a(p) = p*2^((p-1)/2) + 2^p for primes p > 3 (conjecture).

The binding coefficient convention and the exact claim are in #18664. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | G.f.: Sum_{n=-oo..+oo} x^n * (sqrt(2) + x^n)^(2n). | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(p) = p2^((p-1)/2) + 2^p for primes p > 3 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363561Prime.claim and D5.S1.Recurrence.HannaBilateral.OeisA363561Prime.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363561Prime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A363569 prime-index coefficient formulaHanna's A363569 prime-index coefficient formula.Proved

The questionA(x) = Sum{n=-oo..+oo} x^n (i + x^n)^(2n), where i^2 = -1.

Why it mattersThe form is deposit-uncovered, with no atom or coverage edge.

The ideaPublic coeffApowtwo states the resulting identity.

Our contributionHanna's A363569 prime-index coefficient formula.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A363569 prime-index coefficient formula

Paul D. Hanna's OEIS A363569 states:

Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (i + x^n)^(2*n), where i^2 = -1.

a(p) = p*(-1)^((p-1)/2) - 1 for primes p > 3 (conjecture).

The binding coefficient convention and the exact claim are in #18666. Values are complex, and the bilateral expression is interpreted coefficientwise as finite sums of formal power-series coefficients.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Expansion of g.f. A(x) = Sum_{n=-oo..+oo} x^n * (i + x^n)^(2n), where i^2 = -1. | coeffA in D5.S1.Recurrence.HannaBilateral.OeisA363558PowerOfTwo; the source sequence a | Literal preregistered finite coefficient expression; complex formal power series, natural coefficient degree. | | a(p) = p(-1)^((p-1)/2) - 1 for primes p > 3 (conjecture). | D5.S1.Recurrence.HannaBilateral.OeisA363569Prime.claim and D5.S1.Recurrence.HannaBilateral.OeisA363569Prime.result | Every source quantifier and range is retained; result proves the entire claim. | | Sequence parameter | a | $c=2,3,\sqrt{2},i$ for A363558, A363559, A363561, A363569, respectively. | | Complex codomain and natural arithmetic | The types in claim | Integer expressions are cast to complex numbers; the half exponent in the two relevant prime formulas uses natural division. This is the binding convention, not an extra hypothesis. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/HannaBilateral/OeisA363569Prime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe one-common-Zeckendorf-term greedy sequence is a permutationThe least-unused one-common-Zeckendorf-term sequence is total and enumerates every positive integer exactly once.Proved

The questionThis dossier anchors exactly the OEIS A370630 conjecture: “Conjecture: this sequence is a permutation of the positive integers.” The record title is “Lexicographically earliest sequence of distinct positive integers such that the Zeckendorf expansions of two consecutive terms have exactly one common term.” Rémy Sigrist is the record author; the conjecture comment has no separate signed author.

Why it mattersThe registration body SHA-256 is be4c48b37d89250e2d18945944f59181886adf667620d4fa4e5ddfea8a660c15.

The ideaA fresh sufficiently distant Fibonacci index pairs with an occupied current index to give an eligible value larger than the sum of the used list.

Our contributionThe least-unused one-common-Zeckendorf-term sequence is total and enumerates every positive integer exactly once.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The one-common-Zeckendorf-term greedy sequence is a permutation

This dossier anchors exactly the OEIS A370630 conjecture: “Conjecture: this sequence is a permutation of the positive integers.” The record title is “Lexicographically earliest sequence of distinct positive integers such that the Zeckendorf expansions of two consecutive terms have exactly one common term.” Rémy Sigrist is the record author; the conjecture comment has no separate signed author. The source version is revision 16, revision time 2024-05-04T14:57:01-04:00, with offset 1,2.

Let F(0)=0, F(1)=1 and F(k+2)=F(k+1)+F(k). For every integer m>=1, let Z(m) be its unique finite set of nonconsecutive indices k>=2 whose Fibonacci weights sum to m. Set a(1)=1. At every integer n>=1, a(n+1) is the least integer m>=1 outside the actual prefix {a(1),...,a(n)} with |Z(m) intersect Z(a(n))|=1. The whole assertion includes existence of an eligible unused successor at every finite positive stage, distinctness at positive indices, and, for every m>=1, a finite n>=1 with a(n)=m, all for that same sequence.

The linked PARI definition is https://oeis.org/A370630/a370630.gp.txt. Its bit j encodes fibonacci(2+j); its candidate test is !seen(v) && hammingweight(bitand(zp, tozeck(v)))==1. Ordinary binary overlap, at-least-one overlap, zero overlap and finite checked coverage are different assertions. The posted 1..10000 driver does not prove the whole event.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/ZeckendorfGreedyPermutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPowers-of-four parity of the A374568 coefficientsThe coefficients of OEIS A374568 are odd exactly at powers of four.Proved

The questionHanna, Aug 14 2024) records the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2024a374568.md:

Why it mattersThe KPI is open problems resolved, rather than module counts or finite checks.

The ideaInteger coefficient stabilization constructs generatingSeries. generatingequation bridges its fixed-point recursion to the exact OEIS equation and proves its constant and linear coefficients are 0 and 1. generatingunique uses degree induction to identify every zero-constant integer solution, without a separate linear-coefficient hypothesis.

Our contributionThe coefficients of OEIS A374568 are odd exactly at powers of four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Powers-of-four parity of the A374568 coefficients

OEIS A374568 (Paul D. Hanna, Aug 14 2024) records the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2024a374568.md:

NAME:

Expansion of g.f. A(x) satisfying x = A(x - x^2) - A(x + x^2)^2.

COMMENT:

Conjecture: a(n) is odd iff n = 4^k for k >= 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CatalanDoubleCompositionPowersOfFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedKurkov's periodic boundary recurrences and the large Schroder diagonalTwo periodic shifts of Kurkov's original array have the large Schroder diagonal.Proved

The questionMikhail Kurkov's OEIS A376317, revision 25, states:

Why it mattersThe two periodic source shifts have the same diagonal sequence.

The ideaWriting Cf(n,d)=Pf(n,n+d) yields the boundary and step equations used by the proof.

Our contributionTwo periodic shifts of Kurkov's original array have the large Schroder diagonal.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's periodic boundary recurrences and the large Schroder diagonal

Mikhail Kurkov's OEIS A376317, revision 25, states:

Conjecture: cases f(n) = n mod 2 and f(n) = [(n mod 3) > 0] both gives A006318.

The source array is the natural recursion with P(0,m)=1, the positive-row branch P(n,m)=P(n-1,m) for m<=n, and P(n,m)=P(n,m-1)+P(n-1,m+f(m-n)) otherwise. The zeroth-row clause has priority. The settled statement keeps both shifts in one proposition:

∀ n : ℕ,
  P f2 (n + 1) (n + 1) = Nat.largeSchroder n ∧
  P f3 (n + 1) (n + 1) = Nat.largeSchroder n

Here f2 d = d % 2 and f3 d = 0 exactly when d % 3 = 0, and Nat.largeSchroder is A006318. The statement includes n=0; no alternate array, truncation, integrality premise, or second count is substituted. A376388 repeats this same conjecture and is not a second resolution.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/PeriodicBoundarySchroder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedPosed in 2025The A380056 divisibility conjecture at multiples of fourA roots-of-unity filter collapses the surviving binomial sum, which is the step the two published facts leave open.Proved

The questionA sequence is read off an exponential generating function whose numerator is the exponential minus one and whose denominator is the cosine of twice the argument.

Why it mattersDirect inspection on September 9, 2026 found it still labelled Conjecture.

The ideaState the two published facts as hypotheses on an abstract pair of sequences, and derive the conclusion.

Our contributionA roots-of-unity filter collapses the surviving binomial sum, which is the step the two published facts leave open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A380056 divisibility conjecture at multiples of four

A sequence is read off an exponential generating function whose numerator is the exponential minus one and whose denominator is the cosine of twice the argument. Paul D. Hanna conjectured on January 28, 2025 that five divides every term whose index is a multiple of four.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/A380056.a380056_div_five

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A380558 square-reversion seriesThe normalized A380558 series has odd coefficients exactly at twice the stated binary-prefix indices.Proved

The questionHanna (February 13, 2025), defines the generating series A(x) = Sum{n>=2} a(n)x^n through A(x - A(x)) = x^2/(1 - x^2).

Why it mattersThis was a named, literature-open parity conjecture in a 2025 OEIS entry.

The ideaThe new characteristic-two calculation sets T=L/(1-L) and derives T=X+X^2+(1+X)T^2.

Our contributionThe normalized A380558 series has odd coefficients exactly at twice the stated binary-prefix indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A380558 square-reversion series

OEIS A380558, by Paul D. Hanna (February 13, 2025), defines the generating series A(x) = Sum_{n>=2} a(n)*x^n through A(x - A(x)) = x^2/(1 - x^2). Its comment states:

Conjecture: a(n) is odd iff n = 2*A004760(k) for some k > 1, where A004760 lists numbers whose binary expansion does not begin 10.

The formal target constructs an integral series with zero coefficients at degrees zero and one, proves that source equation and uniqueness among all similarly normalized integral solutions, and proves for every natural n:

Odd ([x^n]A) <-> n = 2 or there exist m,j in N with n = 2*m and 3*2^j <= m < 4*2^j.

This is the source conjecture with its natural zero extension at degrees zero and one. A004760 has offset one: its first values are 0,1,3,6,7,.... The condition k>1 retains 1, giving n=2; all later values lie in the half-open dyadic intervals displayed above. Neither m nor j is the OEIS index k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/A380558.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A380711 absolute cubic congruenceCoefficient contraction and reduction modulo two prove Hanna's cubic congruence.Proved

The questionA(x) satisfies A(x) = 1 + xA(x)abs( 1/A(x)^3 ).

The ideaFor positive indices, quotientinteger uses AbsoluteReciprocalSquareShiftParity.ternarydvd, and result directly applies that owner's hannaconjecture.

Our contributionCoefficient contraction and reduction modulo two prove Hanna's cubic congruence.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A380711 absolute cubic congruence

Paul D. Hanna's OEIS A380711 states:

G.f. A(x) satisfies A(x) = 1 + x*A(x)*abs( 1/A(x)^3 ).

Conjecture: a(n) == binomial(3*n-1,n)/(3*n-1) (mod 2) for n >= 0.

The literal binding conventions and source screen are in #18828. All series are formal; no analytic convergence is assumed.

Statement echo

All unqualified names below belong to D5.S1.Recurrence.Algebraic.HannaAbsoluteCubicCongruence.

| Source clause or binding convention | Lean counterpart | Fidelity | | --- | --- | --- | | Coefficientwise absolute value | absSeries | The coefficient of degree $n$ is the absolute value of the original rational coefficient. | | The generating equation | IsGF | The displayed source equation is retained literally over rational formal power series. | | Generalized binomial quotient | quotient | Ring.choose uses the integer top argument $3n-1$ and rational division; at zero the quotient is $-1$. | | Congruence for every $n\ge0$ | claim, result | Integer witnesses represent both rational operands and satisfy Int.ModEq 2. | | Existence and integrality | claim, result | These strengthen the literal congruence's interpretation without adding hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/HannaAbsoluteCubicCongruence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A381364 scaled bilateral product congruenceHanna's scaled bilateral product family has every positive coefficient equal to six modulo nine.Proved

The questionThis dossier cites its own Library note, Library/ArithSums/hanna2025a381364.md, for the A381364 conjecture.

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaThe escape witness is the general theorem hannaconjecturegeneral, covering every c ≥ 1, rather than the two named instances: the unique normalized integer solution of 1/3 = Σ{n∈ℤ} xⁿ·Aⁿ·(Aⁿ + 2x)^{cn−1}·(xⁿ + 2A)^{cn−1} has a(c,n) ≡ 6 (mod 9) for every n > 0.

Our contributionHanna's scaled bilateral product family has every positive coefficient equal to six modulo nine.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A381364 scaled bilateral product congruence

This dossier cites its own Library note, Library/ArithSums/hanna2025a381364.md, for the A381364 conjecture. The two entry statements describe instances of one family:

OEIS A381364, Paul D. Hanna, Feb 21 2025, NAME:

G.f. A(x) satisfies 1/3 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + 2*x)^(n-1) * (x^n + 2*A(x))^(n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 6 (mod 9).

OEIS A381365, Paul D. Hanna, Feb 21 2025, NAME:

G.f. A(x) satisfies 1/3 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + 2*x)^(2*n-1) * (x^n + 2*A(x))^(2*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 6 (mod 9).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/ScaledBilateralProductModNine.hanna_conjecture_a381364

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A381365 scaled bilateral product congruenceHanna's scaled bilateral product family has every positive coefficient equal to six modulo nine.Proved

The questionThis dossier cites its own Library note, Library/ArithSums/hanna2025a381365.md, for the A381365 conjecture.

Why it mattersThis is a first-tier OEIS conjecture from 2025.

The ideaThe escape witness is the general theorem hannaconjecturegeneral, covering every c ≥ 1, rather than the two named instances: the unique normalized integer solution of 1/3 = Σ{n∈ℤ} xⁿ·Aⁿ·(Aⁿ + 2x)^{cn−1}·(xⁿ + 2A)^{cn−1} has a(c,n) ≡ 6 (mod 9) for every n > 0.

Our contributionHanna's scaled bilateral product family has every positive coefficient equal to six modulo nine.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A381365 scaled bilateral product congruence

This dossier cites its own Library note, Library/ArithSums/hanna2025a381365.md, for the A381365 conjecture. The two entry statements describe instances of one family:

OEIS A381364, Paul D. Hanna, Feb 21 2025, NAME:

G.f. A(x) satisfies 1/3 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + 2*x)^(n-1) * (x^n + 2*A(x))^(n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 6 (mod 9).

OEIS A381365, Paul D. Hanna, Feb 21 2025, NAME:

G.f. A(x) satisfies 1/3 = Sum_{n=-oo..+oo} x^n*A(x)^n * (A(x)^n + 2*x)^(2*n-1) * (x^n + 2*A(x))^(2*n-1).

COMMENT:

Conjecture: for n > 0, a(n) == 6 (mod 9).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/ScaledBilateralProductModNine.hanna_conjecture_a381365

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedHanna's A382125 residue patternThe cube has the characteristic-three coefficient pattern at every index.Proved

Why it mattersThe exact rational formal series carry integer witnesses for the source's congruence statements.

The ideaThe other primes use the Dold congruences of the divisor sum and their product and index-scaling closure.

Our contributionThe cube has the characteristic-three coefficient pattern at every index.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A382125 residue pattern

OEIS A382125 states: "Conjectures: a(3n) == A382124(n) (mod 3) for n >= 0; a(3n+1) == 0 (mod 3) and a(3*n+2) == 0 (mod 3) for n >= 0."

Write $A(x)=\exp(\sum_{n\ge1}\sigma(n)\sigma(2n)x^n/(3n))$ and $B(x)=\exp(\sum_{n\ge1}\sigma(n)\sigma(2n)x^n/n)$, with $\sigma(n)=\sum_{d\mid n}d$. The public rational-series definitions are exactly the preregistered expressions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every index $n\ge0$ | claim quantifies over every natural index | Exact | | A382124 generating function | HannaSigmaProductExpModThree.A | Exact rational exponential substitution | | A382125 generating function | HannaSigmaProductExpModThree.B | Exact rational exponential substitution | | $b(3n)\equiv a(n)\pmod3$ | First conjunct: coeff (3*n) B = u, coeff n A = v, u ≡ v [ZMOD 3] | Exact | | $b(3n+1)\equiv0\pmod3$ | Second conjunct: coeff (3*n+1) B = u and u ≡ 0 [ZMOD 3] | Exact | | $b(3n+2)\equiv0\pmod3$ | Second conjunct: coeff (3*n+2) B = v and v ≡ 0 [ZMOD 3] | Exact | | Integer coefficients | Both conjuncts supply integer witnesses and rational coefficient equalities | Witnesses make congruence well typed; no extra hypothesis |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Robin/HannaSigmaProductExpModThreeCube.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedFibonachos ties and first-player majorityFibonachos ties and large-heap majority follow Fibonacci blocks.Proved

The questionConjecture: a(n) = n/2 if and only if n is in {2, 8, 10, 32}.

Why it mattersIssue 12461 fixes the two literal comments, complete quantifiers, proposed resolution and bounded prior-work checks before the probe.

The ideaFor m>=3, a heap in the interval F(m)-1<=n<=F(m+1)-2 begins with exactly m-2 uninterrupted moves.

Our contributionFibonachos ties and large-heap majority follow Fibonacci blocks.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Fibonachos ties and first-player majority

Peter Kagey's OEIS A382814 comment states:

Conjecture: a(n) = n/2 if and only if n is in {2, 8, 10, 32}.

Conjecture: For n > 32, a(n) > n/2 if and only if F(m)-1 <= n <= F(m+1)-2 for some odd integer m, where F(n) = A000045(n).

The two players alternate taking successive Fibonacci numbers, beginning with F(1)=F(2)=1. When the next number exceeds the remaining heap, the index resets to one; the player's turn still alternates. The game stops at the empty heap, and a(n) is the first player's total. For every positive integer n, the tie condition is 2a(n)=n. For every n>32, first-player majority is n<2a(n) and the block condition has an existential natural index m with Odd m.

The formal interval is F(m)<=n+1 and n+2<=F(m+1). In this range the upper condition forces a positive, nontruncated upper endpoint, and these are equivalent to the displayed source inequalities. Multiplying the score comparison by two avoids natural-division rounding.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FibonacciAtomic/FibonachosScore.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedParity of the A384270 coefficientsThe coefficients of OEIS A384270 are odd exactly at powers of two.Proved

The questionThe following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2025a384270.md.

Why it mattersThis is a first-tier recent OEIS conjecture, selected as a 2025 small conjecture.

The ideaAn integral contracting update uses only addition, subtraction, multiplication, and substitution, without division.

Our contributionThe coefficients of OEIS A384270 are odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A384270 coefficients

OEIS A384270, Paul D. Hanna, Jul 10 2025. The following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2025a384270.md.

NAME:

G.f. satisfies A(x) = A(x^4 + 4*x*A(x)^4) / A(x^3 + 3*x*A(x)^3).

COMMENT:

Conjecture: a(n) is odd iff n = 2^k for k >= 0.

范围与诚实边界: The module formalizes the cross-multiplied identity A·A(x³+3xA³) = A(x⁴+4xA⁴), with A(0)=0 and a(1)=1. The displayed denominator has zero constant term, so it is not a unit in the integer formal power-series ring. The public parity theorem covers every positive natural index; the constant coefficient is zero by generating_equation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/QuarticCubicCatalanQuotientParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe Alternating Binomial Sum of OEIS A385590The Fibonacci triangle satisfies the alternating binomial sum in OEIS A385590.Proved

The questionThe single statement considered here is the formula-field conjecture in Werner Schulte's OEIS A385590, dated 2025-07-03.

Why it mattersThis question concerns a different alternating operation on a triangle whose rows are selected by Fibonacci intervals.

The ideaFix the row, write T(n,k)=A+(k-1)B, and cancel the constant and linear moments of the alternating binomial coefficients for n>=3.

Our contributionThe Fibonacci triangle satisfies the alternating binomial sum in OEIS A385590.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a385590 alternating binomial

The single statement considered here is the formula-field conjecture in Werner Schulte's OEIS A385590, dated 2025-07-03. For every integer n >= 1, let i > 1 be the unique index with F(i) <= n < F(i+1), with F(0)=0 and F(1)=1. Put T(n,k) = F(i-1)^2 + 1 - ((i-1) mod 2) + (n-F(i))*F(i-2)

  • (k-1)*F(i-1). Is the sum from k=1 to n of (-1)^k * binomial(n-1,k-1) * T(n,k) equal to (-1)^n when n<3, and zero otherwise? All triangle values and the sum are integers. The entry's separate permutation conjecture is outside this question.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/A385590.alternating_binomial_sum

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2025Multiple-of-Three Residue CoincidencesThe modulo-three coincidence occurs exactly for composites other than 4, 8, 10, and 25.Proved

The questionThe exact predicate is ∃ m : ℕ, 6 ≤ m ∧ m ≤ k ∧ 3 ∣ m ∧ 3((2k)%m) = m((2k)%3).

Why it mattersThis is a first-tier recent OEIS comment conjecture.

The ideaThe equality holds exactly when t divides 2k and (2k/t)%3=(2k)%3, with 2≤t and 3t≤k.

Our contributionThe modulo-three coincidence occurs exactly for composites other than 4, 8, 10, and 25.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Multiple-of-Three Residue Coincidences

OEIS A387319, revision 69 (November 10, 2025), conjectures that every composite k except 4, 8, 10, and 25 has a coincidence between m=3 and another multiple of 3 among the rational fractions (2k mod m)/m for 1≤m≤k, whereas no prime does. The exact predicate is ∃ m : ℕ, 6 ≤ m ∧ m ≤ k ∧ 3 ∣ m ∧ 3*((2*k)%m) = m*((2*k)%3). The theorem states that this is equivalent to 1 < k ∧ ¬Nat.Prime k ∧ k ∉ ({4,8,10,25} : Finset ℕ) for every natural k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/MultipleOfThreeCoincidence.classify

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe least simultaneous digit-sum witness in A389000Wu's conjecture for the least simultaneous digit-sum witness in OEIS A389000.Proved

The questionConjecture: a(9m+5) = 210^(2m+1)-1 for m >= 0.

Why it mattersThis is a first-tier recent OEIS conjecture selected by the search seat.

The ideaBoth K and K^2 have digit sum 2n, so K satisfies the two divisibility conditions.

Our contributionWu's conjecture for the least simultaneous digit-sum witness in OEIS A389000.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The least simultaneous digit-sum witness in A389000

Conjecture: a(9m+5) = 210^(2*m+1)-1 for m >= 0.

— Chai Wah Wu, Oct 01 2025. The OEIS rendering means 2·10^(2m+1) − 1. Here a(n) is the least positive integer k such that n divides both the base-10 digit sum of k and that of k^2. The entry was created Sep 22 2025.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/DigitSumSquareLeastWitness.wu_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresConjecture provedThe A389536 square-cube substitution conjectureThe normalized coefficients of OEIS A389536 are two modulo four exactly at powers of two plus one.Proved

The questionThe following verbatim NAME and COMMENT quotations are recorded in Library/Arith/hanna2025a389536.md:

Why it mattersThe KPI is open problems resolved: the target is the complete assertion for every index greater than one.

The ideaThe integer sequence is defined by a well-founded coefficient recursion, with a(0) = 0 and a(1) = 1.

Our contributionThe normalized coefficients of OEIS A389536 are two modulo four exactly at powers of two plus one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A389536 square-cube substitution conjecture

OEIS A389536, Paul D. Hanna, Oct 28 2025. The following verbatim NAME and COMMENT quotations are recorded in Library/Arith/hanna2025a389536.md:

NAME:

G.f. A(x) satisfies: A(x) = A(x^2 + 2*x^3) / x.

COMMENT:

Conjecture: for n > 1, a(n) == 2 (mod 4) when n = 2^k + 1 (for some k >= 0) otherwise a(n) is divisible by 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SubstitutionSquareCubeModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe mod-four coefficients of the Catalan-shift square seriesThe coefficients of OEIS A393172 above degree one are two modulo four exactly at powers of two, and zero otherwise.Proved

The questionThe following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2026a393172.md.

Why it mattersThe KPI is open problems resolved: the target is the full assertion at every index greater than one, not a finite coefficient check or a previously known special case.

The ideaInteger coefficient stabilization constructs generatingSeries and a. generatingequation supplies the equation bridge: A(0) = 0, a(1) = 1, and A(X - X^2) = X + A^2.

Our contributionThe coefficients of OEIS A393172 above degree one are two modulo four exactly at powers of two, and zero otherwise.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The mod-four coefficients of the Catalan-shift square series

OEIS A393172, Paul D. Hanna, Feb 04 2026. The following NAME and COMMENT are copied verbatim from the claim in Library/Arith/hanna2026a393172.md.

NAME:

G.f. A(x) satisfies A(x-x^2) = x + A(x)^2.

COMMENT:

Conjecture: for n > 1, a(n) == 2 (mod 4) iff n is a power of 2, otherwise a(n) is divisible by 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CatalanShiftSquareModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedEvery integer occurs in the factorial Heraclitus transformEvery integer occurs in the factorial Heraclitus transform.Proved

The questionEach subsequent term is the unused integer with smallest absolute value at factorial distance k!, k >= 1, from the previous term.

The ideaThe initial 25 terms cover [-12,12] without repetition and end at -11.

Our contributionEvery integer occurs in the factorial Heraclitus transform.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Every integer occurs in the factorial Heraclitus transform

OEIS A393434 starts at a(0) = 0. Each subsequent term is the unused integer with smallest absolute value at factorial distance k!, k >= 1, from the previous term. Positive values are preferred when absolute values tie. The source states: “It is conjectured that every integer appears in this sequence.”

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/FactorialHeraclitusTransform.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe A396846 logarithmic coefficients modulo eightThe positive-index coefficients of OEIS A396846 repeat 1, 7, 5, 7 modulo eight.Proved

The questionThe following NAME and COMMENT are copied verbatim from the quotation in Library/Recurrence/hanna2026a396846.md:

Why it mattersThis is a first-tier OEIS conjecture from 2026.

The ideaThe coefficients are constructed over the integers by well-founded recursion for the integral normalization cn = a(n)/(4n^2-1), for n >= 2. crecurrence gives cn = a(n-1) + 4 Sum{k=2}^{n-1} k ck a(n-k), and aeq gives a(n) = (4n^2-1)cn.

Our contributionThe positive-index coefficients of OEIS A396846 repeat 1, 7, 5, 7 modulo eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A396846 logarithmic coefficients modulo eight

Paul D. Hanna, OEIS A396846, Jun 24 2026. The following NAME and COMMENT are copied verbatim from the quotation in Library/Recurrence/hanna2026a396846.md:

L.g.f. Sum_{n>=1} a(n)*x^n/n = log(1+x + Sum_{n>=2} 4*n/(4*n^2-1) * a(n)*x^n ).

Conjecture: a(n) == [1,7,5,7] repeating (mod 8) for n >= 1.

This dossier resolves only the mod-8 conjecture. The entry's first conjecture, "a(n) == 1 (mod 3) iff n is a power of 3, otherwise a(n) is divisible by 3", is not resolved here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/LogarithmicQuarterNormalizationModEight.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A397241 coefficients modulo 4The coefficients of OEIS A397241 are congruent to 11 modulo 12 above degree two.Proved

The questionHanna, Jun 19 2026) records the following NAME:

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideamodthreeidentity and modfouridentity are the escape witnesses: ternary coefficient extraction and binary square-zero lifting turn the diagonal relation n·[xⁿ]Aⁿ = (n−1)·[xⁿ]A^{n+1} into contracting residual recurrences, with coefficient-comparison multiplier exactly one.

Our contributionThe coefficients of OEIS A397241 are congruent to 11 modulo 12 above degree two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A397241 coefficients modulo 4

OEIS A397241 (Paul D. Hanna, Jun 19 2026) records the following NAME:

G.f. A(x) satisfies n * [x^n] A(x)^n = (n-1) * [x^n] A(x)^(n+1) for n > 1.

The COMMENT targeted by this dossier is:

Conjecture: a(n) == 3 (mod 4) for n > 2.

The entry's third COMMENT, "Conjecture: all terms are odd", was already proved and frozen in this repository as D5/S1/Recurrence/Residue/DiagonalPowerRatioAllOdd.hanna_conjecture. It is NOT re-claimed here. This dossier claims only the modulo-4 conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalPowerRatioModTwelve.hanna_conjecture_mod_four

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A397241 coefficients modulo 3The coefficients of OEIS A397241 are congruent to 11 modulo 12 above degree two.Proved

The questionHanna, Jun 19 2026) records the following NAME:

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideamodthreeidentity and modfouridentity are the escape witnesses: ternary coefficient extraction and binary square-zero lifting turn the diagonal relation n·[xⁿ]Aⁿ = (n−1)·[xⁿ]A^{n+1} into contracting residual recurrences, with coefficient-comparison multiplier exactly one.

Our contributionThe coefficients of OEIS A397241 are congruent to 11 modulo 12 above degree two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A397241 coefficients modulo 3

OEIS A397241 (Paul D. Hanna, Jun 19 2026) records the following NAME:

G.f. A(x) satisfies n * [x^n] A(x)^n = (n-1) * [x^n] A(x)^(n+1) for n > 1.

The COMMENT targeted by this dossier is:

Conjecture: a(n) == 2 (mod 3) for n > 2.

The entry's third COMMENT, "Conjecture: all terms are odd", was already proved and frozen in this repository as D5/S1/Recurrence/Residue/DiagonalPowerRatioAllOdd.hanna_conjecture. It is NOT re-claimed here. This dossier claims only the modulo-3 conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/DiagonalPowerRatioModTwelve.hanna_conjecture_mod_three

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Geometry & algebraConjecture provedSource 2026OEIS A397315: the Three-Row blackout conjectureA three-row lattice grid of every width at least three admits exactly the conjectured maximum number of blacked-out points.Proved

The questionArjun Pemmasani's Three-Row conjecture states that the maximum number of points that can be blacked out in a three-row grid with m columns, while every lattice rectangle remains identifiable, is m+2 for every m>=3.

Why it mattersThe registration discloses the earlier unverified row-pair counting direction; it does not claim to precede every mathematical thought.

The ideaOnly the two necessary source definitions and the single complete result are delivered.

Our contributionA three-row lattice grid of every width at least three admits exactly the conjectured maximum number of blacked-out points.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A397315: the Three-Row blackout conjecture

Arjun Pemmasani's Three-Row conjecture states that the maximum number of points that can be blacked out in a three-row grid with m columns, while every lattice rectangle remains identifiable, is m+2 for every m>=3. The exact source definitions and complete target are:

def IsRectangle {m : ℕ} (C : Finset (Fin m × Fin 3)) : Prop :=
  C.card = 4 ∧
  ∃ a b c d : Fin m × Fin 3,
    C = {a, b, c, d} ∧
    (((b.1.val : ℤ) - (a.1.val : ℤ)) *
        ((c.1.val : ℤ) - (a.1.val : ℤ)) +
      ((b.2.val : ℤ) - (a.2.val : ℤ)) *
        ((c.2.val : ℤ) - (a.2.val : ℤ)) = 0) ∧
    (a.1.val : ℤ) + (d.1.val : ℤ) =
      (b.1.val : ℤ) + (c.1.val : ℤ) ∧
    (a.2.val : ℤ) + (d.2.val : ℤ) =
      (b.2.val : ℤ) + (c.2.val : ℤ)

def ValidBlackout {m : ℕ} (S : Finset (Fin m × Fin 3)) : Prop :=
  ∀ C D : Finset (Fin m × Fin 3),
    IsRectangle C → IsRectangle D → C \ S = D \ S → C = D

theorem result :
  ∀ m : ℕ, 3 ≤ m →
    (∃ S : Finset (Fin m × Fin 3),
      ValidBlackout S ∧ S.card = m + 2) ∧
    (∀ S : Finset (Fin m × Fin 3),
      ValidBlackout S → S.card ≤ m + 2)

The first coordinate is column and the second row. Translating both coordinates by one identifies this grid with the source convention. The parallelogram equations and perpendicular adjacent vectors describe Euclidean rectangles; four distinct corners exclude degeneracy. Squares and tilted rectangles are included. Rectangle identity is equality of unordered corner sets. Empty presentations are allowed. The result gives both an attaining valid blackout and the matching bound on every valid blackout, for every natural width at least three.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/FiniteGeometry/RectangleIdentifiability/ThreeRowBlackoutMaximum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA397345: square-weight congruenceOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.Proved

The questionHanna's OEIS A397345 NAME is the formal equation A(x)=exp(x+Sum{n>=2}(5n^2-1)a(n)x^n/(5n^2)).

Why it mattersThis is one endpoint of the shared integer square-weight family; A397242's parity endpoint already exists and is reused.

The ideaDefine b(q,0)=0, b(q,1)=1 and, for n>=2, b(q,n)=q(n-1)b(q,n-1)+Sum{2<=j<n}(qj^2-1)(n-j)b(q,j)b(q,n-j).

Our contributionOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A397345: square-weight congruence

Paul D. Hanna's OEIS A397345 NAME is the formal equation A(x)=exp(x+Sum_{n>=2}(5*n^2-1)*a(n)x^n/(5n^2)). Its first COMMENT is the single target of this dossier:

Conjecture: a(n) is odd iff n+1 is a power of 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ParametricExponentialSquareCongruence.parity_q5

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedA397348: square-weight congruenceOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.Proved

The questionHanna's OEIS A397348 NAME is the formal equation A(x)=exp(x+Sum{n>=2}(3n^2-1)a(n)x^n/(3n^2)).

Why it mattersThis is one endpoint of the shared integer square-weight family; A397242's parity endpoint already exists and is reused.

The ideaDefine b(q,0)=0, b(q,1)=1 and, for n>=2, b(q,n)=q(n-1)b(q,n-1)+Sum{2<=j<n}(qj^2-1)(n-j)b(q,j)b(q,n-j).

Our contributionOdd square-weight parameters have binary parity support; parameter two has period four modulo eight.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A397348: square-weight congruence

Paul D. Hanna's OEIS A397348 NAME is the formal equation A(x)=exp(x+Sum_{n>=2}(3*n^2-1)*a(n)x^n/(3n^2)). Its first COMMENT is the single target of this dossier:

Conjecture: a(n) is odd iff n+1 is a power of 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ParametricExponentialSquareCongruence.parity_q3

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Further frontiersQuestion resolvedOEIS A398726: the distinct cycle size limitThe total weight of distinct cycle lengths over all permutations is asymptotic to n times n factorial.Proved

The questionVaclav Kotesovec's August 14, 2026 FORMULA line in OEIS A398726 states:

The ideaThe delivery contains only the two necessary source definitions and one result.

Our contributionThe total weight of distinct cycle lengths over all permutations is asymptotic to n times n factorial.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A398726: the distinct cycle size limit

Vaclav Kotesovec's August 14, 2026 FORMULA line in OEIS A398726 states:

Conjecture: a(n) ~ n * n!.

The sequence, entered by Alois P. Heinz on August 8, 2026, is the sum of the distinct cycle sizes over all permutations of [n]. The exact source and full statement, preregistered in issue #8059, are:

noncomputable def DistinctCycleSizes {n : ℕ}
    (σ : Equiv.Perm (Fin n)) : Finset ℕ :=
  σ.partition.parts.toFinset

noncomputable def CycleSizeSum (n : ℕ) : ℕ :=
  ∑ σ : Equiv.Perm (Fin n), ∑ k ∈ DistinctCycleSizes σ, k

theorem result :
  Filter.Tendsto
    (fun n : ℕ => (CycleSizeSum n : ℝ) /
      ((n : ℝ) * (Nat.factorial n : ℝ)))
    Filter.atTop (nhds (1 : ℝ))

Partition parts include fixed points as lengths one; the finite set removes multiplicities. Each distinct length contributes its size. Both sums range over their entire finite domains. The empty permutation contributes zero. Fin n is a relabeling of [n], and the real normalization is positive for n>0. No sequence, counting identity or asymptotic estimate is an assumed premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Asymptotics/WeightedProbability/DistinctCycleSizeLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPositive values of A399155 have finite fibersEvery positive integer value of the prime-factor subtraction difference occurs at only finitely many indices.Proved

The questionIts empirical observation is that each positive value is attained only finitely often.

Why it mattersNonnegativity does not exclude infinitely many repetitions of a fixed positive value.

The ideaThis argument needs no classification of integers whose largest prime factor is two.

Our contributionEvery positive integer value of the prime-factor subtraction difference occurs at only finitely many indices.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a399155 value finiteness

Dilkhush's OEIS A399155 defines $a(n)=f(n)-g(n)$ using the smallest- and largest-prime-factor subtraction step counts. Its empirical observation is that each positive value is attained only finitely often. The exact assertion is preregistered in #14993: for every integer $v>0$, the set ${n\in\mathbb N: n\ge2\text{ and }a(n)=v}$ is finite.

The two walk definitions and their integer-valued difference are defined together in the finiteness module, reusing CenteredReducedResidueProgressions.GreatestPrimeFactor.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Primes/PrimeFactorSubtractionValueFiniteness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedSource 2026OEIS A399232 radical identityAn odd-power residue sum determines the radical of its odd modulus.Proved

The questionRui Ferreira's first COMMENT in OEIS A399232 is:

Why it mattersThe present question asks whether a literal sum of odd-power remainders determines that same prime-support quantity through its positive deficit.

The ideaFix n >= 1, and set m=2n+1, e=4n+1, r=rad(m) and q=m/r.

Our contributionAn odd-power residue sum determines the radical of its odd modulus.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A399232 radical identity

Rui Ferreira's first COMMENT in OEIS A399232 is:

a(n) appears to satisfy rad(2*n+1) = (2*n+1)^2 / ((2*n+1)^2 - a(n)).

Its NAME is:

a(n) = Sum_{k=1..4*n+2} (k^(4*n+1) mod (2*n+1)).

OFFSET 1,1 gives the domain n >= 1. The literal finite sum uses least nonnegative natural remainders. The radical is the product of distinct prime divisors. The full intended statement proves, for every such n, strict positivity of the signed rational deficit D=(2*n+1)^2-a(n) and the rational equality rad(2*n+1)=(2*n+1)^2/D. Natural truncated subtraction and integer division are not the meaning of the quotient. There is no squarefree or coprime restriction.

The immutable source is revision 23, Sep 01 2026, of the official export at 2ce625d68f85e35e26e15c0d9681f69280c095e7. Its author line is _Rui Ferreira_, Aug 23 2026. Source quotations and adapted description are attributed to Rui Ferreira and the OEIS Foundation Inc., under CC BY-SA 4.0. Source provenance is distinct from the repository proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/FerreiraOddPowerResidueRadical.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedOEIS A400429 second semi-meander diagonalConnected semi-meanders with n crossings and winding n minus four have the second-diagonal count for every n at least four.Proved

The questionThe source's equivalent arch model fixes a lower rainbow pairing of 2n vertices and counts noncrossing upper perfect matchings whose union with it is one loop.

Why it mattersThe entry still labels this all-order formula a conjecture.

The ideaIts noncrossing condition excludes a<b<M(a)<M(b).

Our contributionConnected semi-meanders with n crossings and winding n minus four have the second-diagonal count for every n at least four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A400429 second semi-meander diagonal

OEIS A400429 revision 16 states verbatim: "Conjecture: D_2(n) = (n^2 + 2*n + n mod 2 - 20)/2, for n >= 4." Its diagonal is D_j(n)=T(n,floor(n/2)-j+1), where T(n,k) counts one-loop semi-meanders with n crossings and winding 2*(k-1)+(n mod 2). The source's equivalent arch model fixes a lower rainbow pairing of 2*n vertices and counts noncrossing upper perfect matchings whose union with it is one loop. The second diagonal has exactly n-4 upper arches crossing the midpoint.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SemiMeanderSecondDiagonal.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Pan–Skandera–Wang's Permanental Inequality at Every Initial SplitAll-split permanental inequality for totally nonnegative real matrices.Proved

The questionSihong Pan, Mark Skandera and Jiayuan Wang, Permanental Inequalities and Unit Interval Orders, EPTCS 445 (2026), pp. 139–147, DOI 10.4204/EPTCS.445.17, arXiv:2606.13162v1, abstract:

Why it mattersThe supporting modules connect it to a pairing of permanent monomials and transport the resulting balanced inequality to every initial split.

The ideaThe rank-to-chain bridge constructs a finite chain of increasing transpositions.

Our contributionAll-split permanental inequality for totally nonnegative real matrices.

Verification details
Release source recordProved in Lean

Open problem

Pan–Skandera–Wang's Permanental Inequality at Every Initial Split

Sihong Pan, Mark Skandera and Jiayuan Wang, Permanental Inequalities and Unit Interval Orders, EPTCS 445 (2026), pp. 139–147, DOI 10.4204/EPTCS.445.17, arXiv:2606.13162v1, abstract:

We also conjecture the inequalities (∗) to hold for all TNN matrices and all h = 1, …, n−1.

Section 1 states:

We conjecture the inequalities to hold for all totally nonnegative matrices and I = [h].

For every real totally nonnegative matrix A of order n ≥ 2 and every 1 ≤ h ≤ n−1, the target is

$$ \operatorname{per}(A_{E,E})\operatorname{per}(A_{O,O}) \leq \operatorname{per}(A_{[h],[h]}) \operatorname{per}(A_{[n]\setminus[h],[n]\setminus[h]}). $$

Here [h] = {1,…,h}, and E and O are the even and odd one-based indices in [1,n]. Total nonnegativity means that every square minor selected in increasing row and column order has nonnegative determinant. The empty-block permanent is one. Lean's Fin n indices start at zero; evenIndices n tests (i.val + 1) % 2 = 0, and prefixIndices n h tests i.val < h. The source's introductory word “negative” conflicts with its displayed det(A_{I,J}) ≥ 0; the predicate follows that displayed inequality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/PanSkanderaWangAllSplits.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Comparing the Wu–Wu unified imaginarity monotones in every finite dimensionUnified imaginarity minimized over real states is bounded by conjugate divergence.Proved

The questionPhys. 77, 095101 (2025), DOI 10.1088/1572-9494/adbdbd; arXiv:2506.09799v1), Section 4, Example 1, state:

Why it mattersThe paper constructs two ways to quantify imaginarity: comparison with the conjugate state and optimization over real states.

The ideaA density matrix means IsDensity: positive semidefinite with complex trace equal to one.

Our contributionUnified imaginarity minimized over real states is bounded by conjugate divergence.

Verification details
Release source recordProved in Lean

Open problem

Comparing the Wu–Wu unified imaginarity monotones in every finite dimension

Chuanfa Wu and Zhaoqi Wu, Two imaginarity monotones induced by unified $(\alpha,\beta)$-relative entropy (Commun. Theor. Phys. 77, 095101 (2025), DOI 10.1088/1572-9494/adbdbd; arXiv:2506.09799v1), Section 4, Example 1, state:

We have not found a proof of $M_{\alpha,\beta}^H(\rho)\geq M_{\alpha,\beta}^E(\rho)$ for any $\alpha\in (0,1)$, $\beta\in (0,1]$ and any qubit state $\rho$, however, numerical experiments shows that it might be true, and it seems difficult to find a counterexample. It is also conjectured that $M_{\alpha,\beta}^H(\rho)\geq M_{\alpha,\beta}^E(\rho)$ for any $\alpha\in (0,1)$, $\beta\in (0,1]$ and any quantum state $\rho$.

This quotation is from lines 573–579 of the downloaded v1 TeX, 20250301WuCF-imaginarity_monotone_R2.tex (md5 eb104c03d3638b12044677758c200bae).

Fix the coordinate basis and let $\bar\rho$ mean entrywise complex conjugation, denoted $\rho^*$ in the paper. For density matrices, write

$D_{\alpha,\beta}(\rho\Vert\sigma)=\big[(\operatorname{Re}\operatorname{tr}(\rho^\alpha\sigma^{1-\alpha}))^\beta-1\big]/\big[(\alpha-1)\beta\big]$,

$M^H_{\alpha,\beta}(\rho)=D_{\alpha,\beta}(\rho\Vert\bar\rho)$, and

$M^E_{\alpha,\beta}(\rho)=\inf_{\sigma\in\mathcal F}D_{\alpha,\beta}(\rho\Vert\sigma)$,

where $\mathcal F$ consists of positive semidefinite matrices of trace one with real entries. Matrix powers are spectral powers from continuous functional calculus. These are the module's definitions; the source's Eqs. (7), (13) and (20) use the trace without an explicit real part, and Eq. (20) uses a minimum. The corresponding source definitions and locators are in the literature note.

Issue #20971 fixes the formal reading: for every natural $d\ge1$, every density matrix $\rho\in M_d(\mathbb C)$, every $0<\alpha<1$ and every $0<\beta\le1$, $M^E_{\alpha,\beta}(\rho)\le M^H_{\alpha,\beta}(\rho)$. Singular states are included. This infimum comparison is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Divergence/UnifiedImaginarityMonotoneComparison.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Exact Hilbert–Schmidt distances for noisy four-qubit Smolin statesThe noisy Smolin state has an attained Hilbert-Schmidt distance for each of four partitions.Proved

The questionLu, Universal entanglement detection with geometrically optimal witnesses, arXiv:2610.17592v1 [quant-ph]; Phys.

Why it mattersThe source constructs geometrically optimal entanglement witnesses from exact nearest separable states (lines 101–112).

The ideaThe same fully separable candidate and the same lower bound give all three least-distance assertions.

Our contributionThe noisy Smolin state has an attained Hilbert-Schmidt distance for each of four partitions.

Verification details
Release source recordProved in Lean

Open problem

Exact Hilbert–Schmidt distances for noisy four-qubit Smolin states

X. Zhu, K. Huang, C. Zhang, X. Nie, Q. Zhao and D. Lu, Universal entanglement detection with geometrically optimal witnesses, arXiv:2610.17592v1 [quant-ph]; Phys. Rev. A 114, 042409 (2026), Appendix, “Four-qubit Smolin state”. The source TeX is manuscript.tex (1053 lines, MD5 db9c376c24127cc118b3e270cf74af42). Lines 672–687 state:

A natural conjecture is that

$D(\mathcal{N}_p(\rho_S)) = \begin{cases} 0, & p \in [0, \frac{1}{3}], \ |\mathcal{N}p(\rho_S) - \mathcal{N}{\frac{1}{3}}(\rho_S)|, & p \in (\frac{1}{3}, 1], \end{cases}$

which simplifies to

$D(\mathcal{N}_p(\rho_S)) = \begin{cases} 0, & p \in [0, \frac{1}{3}], \ \frac{\sqrt{3}}{4} \left( p - \frac{1}{3} \right), & p \in (\frac{1}{3}, 1]. \end{cases}$

Here the norm is Hilbert–Schmidt, $|X|{\mathrm{HS}}=\sqrt{\operatorname{Re}\operatorname{tr}(X^\dagger X)}$. The appendix defines $\rho_S=(I+\sigma_x^{\otimes4}+\sigma_y^{\otimes4}+\sigma_z^{\otimes4})/16$ at lines 659–661, $D(\rho)=\min{\sigma\in\mathrm{SEP}(\mathcal P)}|\rho-\sigma|$ at lines 662–665, and $\mathcal N_p(\rho_S)=p\rho_S+(1-p)I/16$ at lines 669–671. Its four representative partitions are ${1}:{2,3,4}$, ${1}:{2}:{3,4}$, ${1}:{2}:{3}:{4}$ and ${1,2}:{3,4}$ (line 666).

Issue #20844 fixes the reading: for every $p\in[0,1]$, the distance is the displayed function $f(p)$ for the first three partitions, and zero for the two-pair partition, which the source identifies as separable at line 668. Every minimum is attained. This is proved. The source and definitions are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GeometricMeasure/SmolinNoisyHilbertSchmidtDistance.result

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Quantum informationCounterexample foundFirst preprint 2007Monotonicity of the residual-correlation sum of four qubitsBai, Yang and Wang (arXiv:quant-ph/0703098, Phys.Refuted

The questionWang, Multipartite quantum correlation and entanglement in four-qubit pure states, Phys.

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionBai, Yang and Wang (arXiv:quant-ph/0703098, Phys.

Verification details
Release source recordRefuted in Lean

Open problem

Monotonicity of the residual-correlation sum of four qubits

Y.-K. Bai, D. Yang, Z. D. Wang, Multipartite quantum correlation and entanglement in four-qubit pure states, Phys. Rev. A 76, 022336 (2007), arXiv:quant-ph/0703098v2:

Nevertheless, we conjecture that the correlation $M$ is an entanglement monotone

Here $M=\sum_k\tau_k-2\sum_{p>q}C_{pq}^2$ (Eq. (7)) for a four-qubit pure state, with the linear entropies $\tau_k=2(1-\mathrm{tr},\rho_k^2)$ of the one-qubit reduced states and Wootters' concurrences $C_{pq}=\max(\sqrt{\lambda_1}-\sqrt{\lambda_2}-\sqrt{\lambda_3}-\sqrt{\lambda_4},0)$ of the six two-qubit reduced states, $\lambda_1\ge\dots\ge\lambda_4$ the eigenvalues of $\rho_{pq}(\sigma_y\otimes\sigma_y)\rho_{pq}^\ast(\sigma_y \otimes\sigma_y)$. An entanglement monotone does not increase on average under LOCC; in particular, for every local instrument ${K_j}$ on one qubit, $\sum_jp_jM(\phi_j)\le M(\psi)$, where $p_j=|K_j\psi|^2$ and $\phi_j=K_j\psi/\sqrt{p_j}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/FourQubitResidualSumMonotoneRefutation.result

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Analysis & dynamicsCounterexample foundFirst preprint 2010Refutation of order log-convexity for the modified Bessel productHalf-integer modified Bessel values contradict log-convexity in the order.Refuted

The questionPonnusamy, On Turán type inequalities for modified Bessel functions, Proc.

Why it mattersThe paper distinguishes order ν from its square-root parametrization.

The ideaThe connection formula and the half-integer sine values then give

Our contributionHalf-integer modified Bessel values contradict log-convexity in the order.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of order log-convexity for the modified Bessel product

Á. Baricz and S. Ponnusamy, On Turán type inequalities for modified Bessel functions, Proc. Amer. Math. Soc. 141(2) (2013), 523–532, arXiv:1010.3346v1, final Conjecture in §3, printed p. 7:

Motivated by Theorems 1 and 2 we conjecture the following.

Conjecture. The function ν ↦ I_ν(u)K_ν(u) is log-convex on (−1, ∞) for all u > 0 fixed.

Formal reading

Lean names below have prefix D5.S3.Analytic.BesselProductOrderLogConvexity.Result, except the reused first-kind function.

| Source object | Lean name | Correspondence | | --- | --- | --- | | First-kind modified Bessel function | D5.S3.Analytic.BesselRatioLogConcavity.Result.besselI | Existing standard Gamma-denominator power series. | | Second-kind function at noninteger orders | besselK | Standard connection formula DLMF 10.27.4 using the reused first-kind function. | | Product at fixed positive argument | P | The first-kind function times the second-kind function. | | Necessary three-point inequality | claim | Every u>0, ν>−1 and h>0 with ν, ν+h and ν+2h noninteger satisfies P(u,ν)P(u,ν+2h)≥P(u,ν+h)². | | Refutation | result | Exactly ¬ claim. |

For a positive function, convexity of its logarithm implies the midpoint inequality after exponentiation. Restricting the three orders to nonintegers gives a weaker assertion than the conjecture, where the connection formula defines the standard second-kind function. Refuting this weaker assertion therefore refutes the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BesselProductOrderLogConvexity/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2009Positive couplings do not force a real spectrum for U_q(sl_2)-invariant spin chainsA positive-coupling quantum-group spin chain has a nonreal eigenvalue.Refuted

The questionBytsko, Non-Hermitian spin chains with inhomogeneous coupling, arXiv:0911.4476v3 (St.

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionA positive-coupling quantum-group spin chain has a nonreal eigenvalue.

Verification details
Release source recordRefuted in Lean

Open problem

Positive couplings do not force a real spectrum for U_q(sl_2)-invariant spin chains

A. G. Bytsko, Non-Hermitian spin chains with inhomogeneous coupling, arXiv:0911.4476v3 (St. Petersburg Math. J. 22, 393–410, 2011), Conjecture 3:

For $|\gamma| < \frac{\pi}{2S}$, the Hamiltonian (22) of a spin chain with inhomogeneous coupling has a real spectrum if all $a_n >0$.

Here:

  • The chain: $N$ sites, each carrying the spin-$S$ representation of $U_q(sl_2)$ with $q=e^{i\gamma}$, in the source's normalization.
  • The Hamiltonian (22): $H=\sum_{n=1}^{N-1}a_nP^{S,s}_{n,n+1}$, where $P^{S,s}$ is the non-orthogonal projector onto the spin-$s$ summand of $V^S\otimes V^S$, given by the source's Casimir polynomial.

The definitions are quoted in the literature note.

Issue #20040 fixes the reading: for every $S$, $s\in{0,\dots,2S}$, $N\ge2$, $0<|\gamma|<\pi/(2S)$ away from projector poles, and positive couplings, every eigenvalue of $H$ is real. This is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/QuantumGroupChainRealSpectrumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    First preprint year: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Further frontiersCounterexample foundFirst preprint 2010Khavinson–Pereira–Putinar–Saff–Shimorin: nonunivalence radiusThe polynomial z−z⁴ is injective on the closed disk of radius 5/8, although its order-one root moment times sin(π/4) is strictly smaller than 5/8.Refuted

The questionShimorin, Borcea's variance conjectures on the critical points of polynomials, arXiv:1010.5167v2, §5, Theorem 5.1 and Conjecture 4.

Why it mattersThe conjecture proposes an analytic route from a root-spread statistic to a critical-point bound.

The ideaNonnegativity of every root moment supplies the lower bound needed for this infimum comparison.

Our contributionThe polynomial z−z⁴ is injective on the closed disk of radius 5/8, although its order-one root moment times sin(π/4) is strictly smaller than 5/8.

Verification details
Release source recordRefuted in Lean

Open problem

Khavinson–Pereira–Putinar–Saff–Shimorin: nonunivalence radius

D. Khavinson, R. Pereira, M. Putinar, E. B. Saff and S. Shimorin, Borcea's variance conjectures on the critical points of polynomials, arXiv:1010.5167v2, §5, Theorem 5.1 and Conjecture 4. See khavinson2011borcea.

Theorem 5.1 (Alexander-Kakeya). If a polynomial F of degree n has no critical points in a closed disk of radius R, then it is univalent in the concentric disk of radius R sin(π/n).

If F is an nth degree polynomial with F(0) = 0, then F is not univalent in any closed disk of radius larger than σ_p(F) sin(π/n), p ≥ 1, centered at zero.

Of course, if the conjecture is true for p = 1, then it follows for all p ≥ 1. By the Alexander-Kakeya Theorem, the truth of this conjecture implies the truth of Conjecture 1 for the same value of p. It can be easily seen that the strongest case p = 1 of this conjecture is true for n = 2.

For every $n\ge1$, every complex polynomial $F$ of degree $n$ vanishing at zero, every real $p\ge1$ and every real $r>\sigma_p(F)\sin(\pi/n)$, the assertion forbids injectivity on the closed disk centered at zero with radius $r$. Zeros are counted with algebraic multiplicity, and $\sigma_p$ is the infimum over all complex centers of the normalized root moment. The formal definition includes every stated quantifier and imposes no simple-root or geometric restrictions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/CriticalPoints/BorceaNonunivalenceRadius.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2026A design with the maximally mixed state is not the most discriminable qubit ensemble at N = N′(2,3) + 1Seven qubit states with three copies outperform every minimal weighted design with the maximally mixed state appended.Refuted

The questionQuintino, The most discriminable quantum states in the multicopy regime, arXiv:2604.26927v3, study N equiprobable states received in k copies. - The discriminability Discr is the largest average success (1/N) Σi Tr(Mi ρi^{⊗k}) over POVMs. - A weighted state k-design (Definition 6) is a probability-weighted family whose weighted k-th tensor moment is Πsym/Tr Πsym, with every state of positive weight pure. - N′(d,k) is the least size of such a design (§3.3).

Why it mattersDesigns are the optimal pure-state ensembles once N is large enough to contain one (the paper's Corollary 5.1).

The idea- The six octahedron states with equal weights form a weighted 3-design.

Our contributionSeven qubit states with three copies outperform every minimal weighted design with the maximally mixed state appended.

Verification details
Release source recordRefuted in Lean

arXiv

A design with the maximally mixed state is not the most discriminable qubit ensemble at N = N′(2,3) + 1

M. Kvashchuk, P. Chernyshova, L. E. A. Porto, T.-A. Ohst, L. B. Vieira and M. T. Quintino, The most discriminable quantum states in the multicopy regime, arXiv:2604.26927v3, study N equiprobable states received in k copies.

  • The discriminability Discr is the largest average success (1/N) Σ_i Tr(M_i ρ_i^{⊗k}) over POVMs.
  • A weighted state k-design (Definition 6) is a probability-weighted family whose weighted k-th tensor moment is Π_sym/Tr Π_sym, with every state of positive weight pure.
  • N′(d,k) is the least size of such a design (§3.3).

For mixed states with N = N′(d,k) + 1, §11.2 proposes a design on N′(d,k) states together with the maximally mixed state I/d. §12, question 4, asks whether this construction is optimal in every dimension, citing numerical evidence for qubits. The verbatim statements are in the literature note.

Issue #15935 fixes the reading: for all d, k > 0, every weighted k-design on N′(d,k) states, appended with I/d, has Discr at least that of every ensemble of N′(d,k) + 1 states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/KvashchukMulticopyDesignRefutation.result

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    Source 1
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2026A four-state qubit ensemble where an adaptive post-measurement query beats a pre-measurement queryFour qubit states admit an adaptive one-bit query with success 99/100, while every pre-measurement query has success at most 483/500.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersFor three states the source finds ensembles, such as the trine, where the adaptive query wins, even with perfect discrimination of non-orthogonal states.

The idea1. Witness. Priors (33, 33, 33, 1)/100 and the pure states

Our contributionFour qubit states admit an adaptive one-bit query with success 99/100, while every pre-measurement query has success at most 483/500.

Scope & next questionThe uniform-prior form of the suggestion remains open.

Verification details
Release source recordRefuted in Lean

arXiv

A four-state qubit ensemble where an adaptive post-measurement query beats a pre-measurement query

H. Lee and T. Heinosaari, Perfect Discrimination of Non-Orthogonal Quantum States via Adaptive Post-Measurement Queries, arXiv:2608.01167v1, study state discrimination with a one-bit query.

  • Pre-measurement query. P_pre maximizes, over the binary partitions of the symbols and the measurements of the two subensembles, the success probability after the answer is known (Eq. (3)).
  • Adaptive post-measurement query. P^adap_post maximizes over a measurement whose outcome selects the partition and the guesses (Eq. (5)).
  • The suggestion. "A relevant question is whether there exists an ensemble of four qubit states with this property. … These examples suggest that P^adap_post > P_pre may never occur for qubit ensembles with n ≥ 4, though we leave a proof as an open question" (§VI, p. 6).

The verbatim statements are in the literature note.

Issue #17780 fixes the reading: for every n ≥ 4 and every ensemble of n qubit states (priors q_x ≥ 0 with Σ q_x = 1 and 2 × 2 density matrices), P^adap_post ≤ P_pre. This suggestion is refuted, and the §VI question is answered affirmatively, by an ensemble of four qubit states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/AdaptiveQueryQubitFourStateRefutation.result

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Quantum informationCounterexample foundFirst preprint 2026Local-unitary inequivalence of the Question 5.4 stabilizer pairsNeither the two [[5,2]] codespaces of SA and SB nor the two [[6,3]] codespaces of S+ and S- of Theorem 5.3 of arXiv:2607.26214 are related by a product unitary composed with a qubit permutation, which answers the first part of Question 5.4 of A.Refuted

The questionMahmoud (arXiv:2607.26214v1, 2026) constructs weight-isometric pairs of stabilizer codes that are not monomially equivalent.

Why it mattersFor the basic family of the paper, the codespaces are separated by the number of qubits on which the code projector acts as the identity (Lemma 3.2 and Theorem 3.3 (iii)).

The ideaFor a set T of qubits and a subspace C, let VT(C) be the span of M ψ with ψ ∈ C and M a product operator that is the identity on T; it is the whole space exactly when the reduced state of C on T has full rank.

Our contributionNeither the two [[5,2]] codespaces of SA and SB nor the two [[6,3]] codespaces of S+ and S- of Theorem 5.3 of arXiv:2607.26214 are related by a product unitary composed with a qubit permutation, which answers the first part of Question 5.4 of A.

Verification details
Release source recordRefuted in Lean

Open problem

Local-unitary inequivalence of the Question 5.4 stabilizer pairs

A. A. Mahmoud (arXiv:2607.26214v1, 2026) constructs weight-isometric pairs of stabilizer codes that are not monomially equivalent. Theorem 5.3 gives the full-support pairs

  • (ii) S_A = ⟨ZZZZI, XXIIX, IIXXX⟩ and S_B = ⟨ZZIZX, XIZXI, IXXYI⟩, two [[5,2]] codes;
  • (iii) S+ = ⟨XXIIXX, IIXXXX, ZZZZIX⟩ and S− = ⟨XXZXII, ZZXYII, IZZZXX⟩, two [[6,3]] codes.

The paper then asks:

Question 5.4 (Extension analogue of LU–LC). Are the codespaces of SA and SB of Theorem 5.3(ii)—or those of S+ and S− of Theorem 5.3(iii)—mapped to one another by some product unitary U1⊗···⊗Un composed with a qubit permutation?

and §6.2 lists it as open problem 1, with the [[5,2]] pair as "the minimal open instance". Issue #11210 fixes the reading: qubits are numbered from the left; the Pauli matrices are the frozen qubitX, qubitZ with Y = i X Z; a word and a product unitary act by the tensor product; the codespace is the joint +1 eigenspace of the three generators; the permutation acts by (P_σ ψ)(x) = ψ(x ∘ σ); and "mapped to one another" means that the image of the first codespace under U·P_σ is the second. The formal claim is the positive answer for either pair, and the formal result refutes it. The "More generally" part of the question is not addressed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/StabilizerPairLocalUnitaryInequivalence.result

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Quantum informationCounterexample foundFirst preprint 2026A qutrit obstruction to positive/CP factorization of the channel transposeRefuted

The questionMárquez González, Feasibility Ordering of Entanglement-Source Placement for Qubit Channels, arXiv:2609.18803v1, Section VI, Eq. (48), states the weak question:

Why it mattersA failure of the weak condition locates where the qubit proof strategy stops.

The ideaThe proof uses factorization.

Verification details
Release source recordRefuted in Lean

Open problem

A qutrit obstruction to positive/CP factorization of the channel transpose

Samuel A. Márquez González, Feasibility Ordering of Entanglement-Source Placement for Qubit Channels, arXiv:2609.18803v1, Section VI, Eq. (48), states the weak question:

The weaker question, which is sufficient for the source-placement theorem, asks only whether there exist a positive map $\mathcal{F}$ and a completely positive map $\mathcal{E}$ such that $\Upsilon^T=\mathcal{F}\circ\Upsilon\circ\mathcal{E}$.

and adds that "failure of the weak condition would identify the precise obstruction to this proof strategy". The channel transpose is Eq. (16), $\Lambda^T(X)=\sum_iK_i^TX\overline{K_i}$. The verbatim statements are in the literature note.

Issue #13926 reads Eq. (48) as the assertion that every unital qutrit channel admits the weak factorization: for every completely positive, trace-preserving, unital $\Upsilon$ on $M_3$ and every finite Kraus family of $\Upsilon$, there exist a positive $\mathcal F$ and a completely positive $\mathcal E$ on $M_3$ with $\Upsilon^T=\mathcal F\circ\Upsilon\circ\mathcal E$. Neither filter must preserve trace, be unital or be invertible.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/PositiveFilterTransposeRefutation.result

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Quantum informationConjecture refutedFirst preprint 2026Pathak's generic-state negativity conjecturePathak's generic-state negativity identity fails on a four-site kicked Ising chain.Refuted

The questionPathak, Mixed-State Entanglement in a Minimal Model of Quantum Chaos, arXiv:2603.14292v1, Conjecture 1, p. 3, states:

Why it mattersIts definitions retain the literal Hamiltonians, matrix exponentials, product state, partial trace, partial transpose and trace norm.

The ideaSubsystem local unitaries preserve both entanglement measures.

Our contributionPathak's generic-state negativity identity fails on a four-site kicked Ising chain.

Scope & next questionThese are the two tested product-of-pairs instances; a uniform statement for untested states or block sizes remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Pathak's generic-state negativity conjecture

T. Pathak, Mixed-State Entanglement in a Minimal Model of Quantum Chaos, arXiv:2603.14292v1, Conjecture 1, p. 3, states:

2𝓔(t) = I_{A:B}^{(α)}(t), hold for generic states at all times t.

The periodic kicked Ising chain has $J=π/4$, $b=-π/4$ and arbitrary real longitudinal fields, with $U=\exp(-iH_K)\exp(-iH_I)$. Its initial state is the source's product of Bloch-angle qubit states. “Generic” excludes both the all-transverse class and the all-longitudinal class. The target in #13296 is the exact finite-chain equality at $α=1/2$ for every chain length, field, generic product state, contiguous nonempty tripartition and integer time. Refuting this specialization refutes the equality asserted for all Rényi orders; it does not refute each order separately.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/KickedIsingNegativityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2009Four physical levels admit a $U(1)$-symmetric injective matrix-product familyFour physical levels with nonzero charge values admit a normalized, symmetric matrix-product family of bond dimension three that is injective at block length two.Refuted

The questionCirac, Matrix product states: Symmetries and two-body Hamiltonians, Phys.

Why it mattersThe source studies which restrictions a symmetry places on a single translationally invariant matrix-product state.

The idea4. Refutation. result applies the universal assertion claim to bond dimension three, these matrices, the identity unitary and these charges.

Our contributionFour physical levels with nonzero charge values admit a normalized, symmetric matrix-product family of bond dimension three that is injective at block length two.

Verification details
Release source recordRefuted in Lean

Open problem

Four physical levels admit a $U(1)$-symmetric injective matrix-product family

M. Sanz, M. M. Wolf, D. Pérez-García and J. I. Cirac, Matrix product states: Symmetries and two-body Hamiltonians, Phys. Rev. A 79, 042308 (2009), arXiv:0901.2223v1, Section IV, Symmetries191208.tex, lines 748–751:

Let us remark that this counter-example is applicable to spin $\ge \frac{5}{2}$. Indeed, one can prove Theorem \ref{} for $U(1)$ and spin $\frac{1}{2}$, which is the content of the following proposition. The case of spin $3/2$ remains an open question.

The preceding example, lines 726–746, uses general generator eigenvalues, all nonzero, at physical dimension $d=D^2-D$ with bond dimension $D\ge3$. The question concerns the remaining four-level case. Issue #23698 fixes its general-generator reading as follows.

For $D\in\mathbb N$, let $A:\mathrm{Fin},4\to\mathcal M_D(\mathbb C)$, let $W$ be a unitary $4\times4$ matrix, and let $\lambda:\mathrm{Fin},4\to\mathbb R$. Define

$$u(\varphi)=W\operatorname{diag}(e^{i\varphi\lambda_k})W^\dagger,\qquad \Psi_N(w)=\operatorname{tr}(A_{w_0}\cdots A_{w_{N-1}}).$$

IsSymmetric means that for every real $\varphi$ and every positive $N$ there is a real $\theta$, allowed to depend on both, with $u(\varphi)^{\otimes N}\Psi_N=e^{i\theta}\Psi_N$. IsInjective means that for some natural block length $n$ the map

$$\Gamma_n:X\longmapsto\bigl(w\longmapsto\operatorname{tr}(XA_{w_0}\cdots A_{w_{n-1}})\bigr)$$

is injective on $\mathcal M_D(\mathbb C)$. These are the trace-amplitude, symmetry and injectivity conventions of the source at lines 222–237, 134–140 and 267–276, respectively; the literature note records the locators.

claim asserts universally in $D,A,W,\lambda$ that if every $A_k\ne0$, $\sum_kA_kA_k^\dagger=I_D$, every $\lambda_k\ne0$, at least two charges differ, and IsSymmetric holds, then IsInjective does not hold. This assertion is refuted. The specified reading concerns four physical levels with general charges; it does not identify their generator with the usual spin-$3/2$ matrix $J_z$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/FourLevelU1SymmetricInjectiveMPS.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2026Selective LOCC can increase Shirokov's Moreau-Yosida entanglement of formationShirokov's Moreau-Yosida approximation of the entanglement of formation can increase on average under a local projective measurement.Refuted

The questionShirokov, The Moreau-Yosida approximation of the EoF: basic properties and accuracy estimates, arXiv:2609.30246v1, Section 7, p. 20, asks:

Why it mattersEquation (1), (EF-d), takes the infimum of average marginal entropy over all finite pure-state ensembles, with no bound on ensemble size.

The ideaThe proof uses convexity.

Our contributionShirokov's Moreau-Yosida approximation of the entanglement of formation can increase on average under a local projective measurement.

Verification details
Release source recordRefuted in Lean

arXiv

Selective LOCC can increase Shirokov's Moreau-Yosida entanglement of formation

M. E. Shirokov, The Moreau-Yosida approximation of the EoF: basic properties and accuracy estimates, arXiv:2609.30246v1, Section 7, p. 20, asks:

Open question: can the function $E^{\lambda}_F$ increase under selective LOCC-operations?

The conjectural sentence is:

So, we may conjecture, at the moment, that the function $E^{\lambda}_F$ does not increase under selective LOCC-operations as well.

Issue #11557 preregisters the source, quantified statement, Tier 1 classification and literature check. The selective inequality is the probability-weighted average of normalized terminal values being at most the input value, for every positive parameter and bipartite state. A finite-dimensional one-round local projective measurement is already sufficient to refute that universal claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/MoreauYosidaFormationSelectiveLoccRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationConjecture refutedFirst preprint 2026The finite-length Pareto conjectureAn actual iid stopping-time bridge and a universal five-column obstruction refute the finite-length Pareto conjecture at file dimensions (1,2).Refuted

The questionBar-Lev, arXiv:2603.17154v2, Section VII-A, Conjecture 2, asks whether every rank-k length-n code over Fq admits a rank-k length-n+1 code with both actual expected file-retrieval times no larger, whenever n>=k, s1,s2>0, s1+s2=k, and max(s1,s2)>=2.

Why it mattersThe arbitrary-successor Pareto question asks whether increasing code length can preserve both files' retrieval performance simultaneously.

The ideaTake k=3, n=4, file dimensions (1,2), and generator columns e0,e0,e1,e2 over F2.

Our contributionAn actual iid stopping-time bridge and a universal five-column obstruction refute the finite-length Pareto conjecture at file dimensions (1,2).

Verification details
Release source recordRefuted in Lean

Open problem

The finite-length Pareto conjecture

Bar-Lev, arXiv:2603.17154v2, Section VII-A, Conjecture 2, asks whether every rank-k length-n code over F_q admits a rank-k length-n+1 code with both actual expected file-retrieval times no larger, whenever n>=k, s1,s2>0, s1+s2=k, and max(s1,s2)>=2. The expected time is the minimum number of independent uniform-with-replacement draws required for the sampled columns to span the whole coordinate file. Issue #11618 fixes this target before the proof probe.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Resource/MinimumRetrievalTime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2025Fibonacci-word MAP maximumBound the longest monochromatic arithmetic progression in the infinite Fibonacci word.Proved

The questionJoshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, arXiv:2501.05830v2, Theoretical Computer Science 1050 (2025), 115391, Conjecture 4.24:

Why it mattersThe source gives exact expressions for A(d) in the small- and large-step regimes (Theorems 4.6 and 4.15), but leaves this strict uniform comparison with the golden ratio as Conjecture 4.24.

The ideaThe repository's infinite goldenWord is read through the golden mechanical letter at the i+1 fractional coordinate.

Our contributionBound the longest monochromatic arithmetic progression in the infinite Fibonacci word.

Scope & next questionThe result addresses the literal global Conjecture 4.24 and no claim about other words or asymptotic sharpness.

Verification details
Release source recordProved in Lean

Open problem

Fibonacci-word MAP maximum

Joshi and Rust, Monochromatic arithmetic progressions in the Fibonacci, Thue-Morse, and Rudin-Shapiro words, arXiv:2501.05830v2, Theoretical Computer Science 1050 (2025), 115391, Conjecture 4.24:

For all d >= 1, (A(d)-1)/d < sqrt(5)/tau.

Here tau=(1+sqrt(5))/2 and A(d) is the global maximum positive length of a monochromatic arithmetic progression of difference d in the zero-indexed infinite Fibonacci fixed point 0 -> 01, 1 -> 0. Both symbols and every nonnegative starting position are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/FibonacciMapBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2025Kok's maximum degree and literal A319433 deletionThe ordinary maximum degree of every positive-order recursive linear Jaco graph is its same-index literal Zeckendorf deletion.Proved

The questionJohan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500v1, Section 2.2, Conjecture 2.2, states:

Why it mattersThe theorem supplies a complete all-order resolution of the two clauses in Conjecture 2.2 for the source's actual recursive graph and sequence indexing.

The ideaJacoEndpointBeatty.endpointrecurrence identifies the actual recursive endpoint array, and endpointeqbeatty proves its all-order golden Beatty law. JacoMaximumDegree.degreeformula counts the actual incoming and outgoing neighbor intervals.

Our contributionThe ordinary maximum degree of every positive-order recursive linear Jaco graph is its same-index literal Zeckendorf deletion.

Scope & next questionUnresolved extensions: The result does not settle other graph parameters or other Jaco defining functions.

Verification details
Release source recordProved in Lean

Open problem

Kok's maximum degree and literal A319433 deletion

Johan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500v1, Section 2.2, Conjecture 2.2, states:

For the finite linear Jaco graphs J_{n}(x), n=2,3,4,\dots the respective maximum degrees \Delta(J_{n}(x)) are yielded by the sequence A319433 with closed formula \Delta(J_{n}(x))=\lfloor\frac{2(n+2)}{1+\sqrt{5}}\rfloor-1, n=2,3,4,\dots

Furthermore, it implies that sequence A319433 is p-graphical where p(G)=\Delta(G) and \mathcal{F}={G:G=J_{n}(x),n=1,2,3,\dots}.

The formal graph is the underlying simple undirected graph of the finite directed linear Jaco graph with f(x)=x. For vertices v_i, 1 ≤ i ≤ n, write d_i for the number of earlier vertices v_k satisfying 2*k-d_k ≥ i; for i < j ≤ n, the directed arc (v_i,v_j) exists exactly when 2*i-d_i ≥ j. J_n(x) forgets orientation and Delta is its ordinary maximum degree.

The sequence term uses its actual offset-zero index. For n > 0, write the canonical Zeckendorf representation with weights Fib_2=1, Fib_3=2, and no consecutive occupied positions. Delete the digit position of weight Fib_2, whether its digit is one or zero: remove occupied index 2 if present, lower every surviving occupied index by one, and decode in the same Fibonacci system. Call the result a(n)=deleteLeast(n), with a(0)=a(1)=0.

The complete assertion is

$$ \left(\forall n\in\mathbb N,;2\le n\Rightarrow \left(\Delta(J_n(x))=\left\lfloor\frac{2(n+2)}{1+\sqrt5}\right\rfloor-1 \land\Delta(J_n(x))=a(n)\right)\right) \land \left(\forall n\in\mathbb N,;1\le n\Rightarrow \Delta(J_n(x))=a(n)\right). $$

The second conjunct retains the singleton J_1(x), whose degree and deletion value are both zero. The first conjunct starts at n=2, exactly as in the source. The index-zero sequence term is not reassigned to J_1; no J_0, offset shift, order cutoff, subsequence, changed graph, or extra premise is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/JacoMaximumDegree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedFirst preprint 2025Lidar's QPU-restricted IQP anticoncentration conjectureRefuted

The questionThe Library note quotes Definition 1 and Conjecture 2 verbatim and identifies Eqs. (17)–(20).

Why it mattersA valid hardware family and one fixed field choice violating the uniform positive lower bound refute its universal scope.

The ideaThe hardware adjacency is literal; hardwaredegreefive checks its degree and factornonempty proves that the four-factor ensemble exists.

Verification details
Release source recordRefuted in Lean

Open problem

Lidar's QPU-restricted IQP anticoncentration conjecture

D. A. Lidar, “Digital-Analog-Digital Quantum Supremacy”, arXiv:2512.07127v1, p. 4, Conjecture 2, asks for positive constants $a,b$, depending only on $d$, such that every output string $s$ satisfies $\Pr_{G,\theta}[P_{U_{\mathrm{IQP}}^{(\theta)}}(s)\ge a2^{-n}]\ge b$. The hardware graph is fixed, simple and $D$-regular, $3\le d\le D$ is fixed, $G$ is uniform on its $d$-factors, the angles are independently uniform on $[0,2\pi)$, and the one-qubit fields are arbitrary but fixed. The Library note quotes Definition 1 and Conjecture 2 verbatim and identifies Eqs. (17)–(20).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/IQP/LidarRestrictedAnticoncentrationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryQuestion resolvedPalindromic composition of the A384222 rowsThe maximal two-dense divisor blocks form a palindromic composition.Proved

The questionConjecture 1: row n is a palindromic composition of A000005(n).

Why it mattersThis is the first-tier recent OEIS conjecture selected in the implementation brief.

The ideaFor positive n, the involution d ↦ n/d reverses the ordered positive-divisor list.

Our contributionThe maximal two-dense divisor blocks form a palindromic composition.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Palindromic composition of the A384222 rows

Conjecture 1: row n is a palindromic composition of A000005(n).

— Omar E. Pol (entry Jun 03 2025), OEIS A384222.

The defining comments supplied with the conjecture are:

The 2-dense sublists of divisors of n are the maximal sublists whose terms increase by a factor of at most 2.

In a sublist of divisors of n the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of n.

For positive n, the row lists the lengths of these maximal consecutive blocks; A000005(n) is the number of positive divisors, τ(n).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/TwoDenseDivisorBlocksPalindrome.row_palindrome

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2010Yoshida's rapid-decrease closure questionA positive sequence satisfies Yoshida's rapid-decrease inequalities but its first Turan transform does not.Refuted

The questionRintaro Yoshida, On some questions of Fisk and Brändén, arXiv:1005.4218v1, Open Problem 5, printed page 11, asks:

Why it mattersThe threshold depends on the first coefficient.

The ideaThe theorem result applies that contradiction to the closure conjunct.

Our contributionA positive sequence satisfies Yoshida's rapid-decrease inequalities but its first Turan transform does not.

Scope & next questionThe finite computation in item 3 does not settle them.

Verification details
Release source recordRefuted in Lean

Open problem

Yoshida's rapid-decrease closure question

Rintaro Yoshida, On some questions of Fisk and Brändén, arXiv:1005.4218v1, Open Problem 5, printed page 11, asks:

Let $f(x)=\sum_{k=0}^{\infty}a_kx^k$ where ${a_k}{k=0}^{\infty}$ is a rapidly decreasing sequence. If $S_r[f(x)]=\sum{k=0}^{\infty}(a_k^2-a_{k-r}a_{k+r})x^k:=\sum_{k=0}^{\infty}b_kx^k$, is it true that ${b_k}_{k=0}^{\infty}$ is a rapidly decreasing sequence?

Definition 29, printed page 10, requires positive terms and one real $\alpha$ with $\alpha\ge\max{2,\frac{\sqrt2}{2}(1+\sqrt{1+a_1})}$ such that $a_k^2\ge\alpha^2a_{k-1}a_{k+1}$ for every $k\ge1$. Negative indices are zero-extended, following Definition 8, printed page 3. The Library note carries these locators. The tier-1 preregistration is #23414.

Statement echo

All declaration names below belong to D5.S0.Certificates.YoshidaRapidDecreaseRefutation.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive real sequence indexed by all nonnegative integers | RapidlyDecreasing positivity conjunct | Every natural index, including zero; no finite truncation. | | A single threshold parameter and all positive-index inequalities | RapidlyDecreasing existential and universal clauses | The source threshold is literal, with real square roots and real division. | | $a_j=0$ outside the index domain | Integer conditionals in RapidlyDecreasing and sCoeff | Zero for negative indices; otherwise Int.toNat selects the coefficient. No separate forwarding definition. | | $b_k=a_k^2-a_{k-r}a_{k+r}$ | sCoeff | Subtraction occurs after casting indices to integers. | | The source's coefficient object and closure question | claim | An explicit existence conjunct plus closure for every real coefficient sequence and every natural $r\ge1$, as preregistered. The coefficient representation imposes no convergence condition. | | Negative answer | result | The literal negation of the complete claim; $r=1$ supplies the failure. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/YoshidaRapidDecreaseRefutation.result

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Quantum informationCounterexample foundFirst preprint 2017A unit-gain circulant on three vertices with universal perfect state transferThe circulant Circ(0, a, conj(a)) with a = (-4 sqrt(3) + i)/7 is a Hermitian circulant on three vertices whose nonzero entries have modulus 1 and whose continuous-time quantum walk has perfect state transfer between every pair of vertices, yet it is switching equivalent neither to K2 nor to Circ(0, -i, i).Refuted

The questionTamon (arXiv:1701.04145, quant-ph and math.CO; Linear Algebra Appl. 531 (2017) 516–532) study the continuous-time quantum walk U(t) = exp(−i t A) of a graph with Hermitian adjacency matrix A.

Why it mattersUniversal perfect state transfer is rare: real symmetric matrices on more than two vertices never have it, and a later paper (arXiv:2301.01473) proved that among oriented graphs, with weights ±i, only K₂ and the oriented triangle have it.

The ideaTake A = Circ(0, α, ᾱ) with α = (−4√3 + i)/7.

Our contributionThe circulant Circ(0, a, conj(a)) with a = (-4 sqrt(3) + i)/7 is a Hermitian circulant on three vertices whose nonzero entries have modulus 1 and whose continuous-time quantum walk has perfect state transfer between every pair of vertices, yet it is switching equivalent neither to K2 nor to Circ(0, -i, i).

Verification details
Release source recordRefuted in Lean

Open problem

A unit-gain circulant on three vertices with universal perfect state transfer

E. Connelly, N. Grammel, M. Kraut, L. Serazo and C. Tamon (arXiv:1701.04145, quant-ph and math.CO; Linear Algebra Appl. 531 (2017) 516–532) study the continuous-time quantum walk U(t) = exp(−i t A) of a graph with Hermitian adjacency matrix A. The walk has universal perfect state transfer if for all vertices u, v some time gives |U(t)_{v,u}| = 1. A graph has property 𝕋 if every nonzero entry of A has modulus 1. The paper states:

\begin{conjecture} $\Circ(0,-\ii,\ii)$ is the only circulant with property
$\TT$ which has universal perfect state transfer. \end{conjecture}

and, in its introduction,

The only known examples of complex unit gain graphs
with the universal property are the circulants $K_{2}$ and $\Circ(0,-\ii,\ii)$.
We conjecture that this set is unique.

Issue #11528 fixes the reading. Circ(a) has entries C_{jk} = a_{k−j} over ℤ/nℤ, "the only" is read up to the paper's switching equivalence (MA = BM with M a permutation matrix times an invertible diagonal matrix) and together with K₂, the order is n ≥ 2, and the circulant is Hermitian without loops. The formal claim is: every such circulant with property 𝕋 and universal perfect state transfer is switching equivalent to K₂ or to Circ(0, −i, i). The result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PropertyTCirculantUniversalTransfer.result

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Quantum informationCounterexample foundFirst preprint 2020The correlation-minor bound fails without filter normal formA separable qubit-ququart state violates the conjectured correlation-minor operator-norm bound without filter normal form.Refuted

The questionThe source then asks whether the filter-normal-form hypothesis can be removed.

Why it mattersThe paper presents correlation-minor norms as basis-invariant quantities for entanglement detection: exceeding a separable-state upper bound is intended to certify entanglement.

The ideaUsing this lemma with c = compound 3 coefficients, r = selectedRow, and s = selectedCols, witnesslower proves the squared lower bound for cmnInf 3 (corr rho alice bob).

Our contributionA separable qubit-ququart state violates the conjectured correlation-minor operator-norm bound without filter normal form.

Verification details
Release source recordRefuted in Lean

Open problem

The correlation-minor bound fails without filter normal form

B. Y. Peled, A. Te'eni, A. Carmi and E. Cohen, Correlation minor norms, entanglement detection and discord, arXiv:2005.12079v2, Scientific Reports 11 (2021) 2849, states in Theorem 3 that a separable state in filter normal form satisfies the $p=\infty$ correlation-minor bound $\mathcal{M}_{h,p=\infty} \leq \frac{1}{\sqrt{Dd}}\left[\frac{D-1}{D(h-1)}\frac{d-1}{d(h-1)}\right]^{(h-1)/2}$. The source then asks whether the filter-normal-form hypothesis can be removed. In the source TeX file CMN_16_12_20_arXiv.tex, line 281, it says:

As in Theorem \ref{thm_bound_p_1}, we conjecture this theorem still holds without the assumption that $\rho$ is in FNF.

The compiled Lean claim is the universal assertion over natural numbers $da$, $db$, and $h$ that, when $2 \leq da$, $2 \leq db$, $\max(da,db)\min(da,db) \leq h^2$, and $h \leq \min(da,db)^2$, every separable matrix $\rho$ on the product index space and every pair of IsHermBasis families satisfies $\operatorname{cmnInf}(h,\operatorname{corr}(\rho,a,b))\leq\operatorname{bound}(da,db,h)$. Here IsHermBasis, corr, compound, cmnInf, and bound are the definitions in the module, so the formal assertion has no filter-normal-form assumption.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/CorrelationMinorNormFNFRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2015Periodicity of generalized pseudostandard wordsGeneralized pseudostandard words and involutory reflection structure.Proved

The questionDvořáková, On periodicity of generalized pseudostandard words, Electronic Journal of Combinatorics 23(1), P1.2, Conjecture 29, asks whether a generalized pseudostandard word over any finite alphabet is periodic exactly when its normalized directive sequence satisfies the two tail conditions.

Why it mattersThe source proves periodicity criteria for restricted alphabets and states the arbitrary finite-alphabet characterization as Conjecture 29.

The ideaWrite each involutory antimorphism as reversal after its involutive letter permutation.

Our contributionGeneralized pseudostandard words and involutory reflection structure.

Scope & next questionAn exact least-period characterization is a neighboring question and remains open.

Verification details
Release source recordProved in Lean

Open problem

Periodicity of generalized pseudostandard words

J. Florian and Ľ. Dvořáková, On periodicity of generalized pseudostandard words, Electronic Journal of Combinatorics 23(1), P1.2, Conjecture 29, asks whether a generalized pseudostandard word over any finite alphabet is periodic exactly when its normalized directive sequence satisfies the two tail conditions. The source and literal definitions are in the Library note. The named assertion is proved by D5.S1.Words.Palindromes.GeneralizedPseudostandardPeriodicity.result. Preregistration: #15626.

For some common tail offset $k$, the conditions are:

  1. $\vartheta_{k+i}(\delta_{k+i+1})=\vartheta_{k+j}(\delta_{k+j+1})$ for all $i,j\ge1$.
  2. For all $i\ge1$, the image under $\vartheta_{k+i+1}$ of every $\vartheta_{k+i}$-palindrome is a $\vartheta_{k+i+2}$-palindrome.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every finite alphabet and every involutory antimorphism | claim quantifies over α : Type, Fintype α, DecidableEq α; InvAntimorphism has the literal concatenation-reversal and involution laws | No alphabet-cardinality restriction | | Successive shortest pseudopalindromic closures, starting with the empty word | Generates, with IsClosure expressing prefix, fixed word, and shortest length | Directive indices begin at one; the entries at zero are unused | | Normalized directive generating the same infinite word | The first conjunct of claim; Generates.normalization_exists | Every prefix fixed by any involutory antimorphism occurs as a stage | | The word is $d$-ary | Function.Surjective u | Every alphabet letter occurs | | Periodicity | A positive natural period in Function.Periodic | Pure periodicity; the sufficiency proof constructs a pure period directly | | A common finite directive prefix | One existential natural offset in TailConditions | Both directive sequences have the same offset | | Condition 1, all tail indices | The first conjunct of TailConditions, with images of singleton words and both indices at least one | Literal letter-image equality | | Condition 2, every fixed finite word | The second conjunct of TailConditions | Literal universal palindrome transport | | Both directions | The second conjunct of claim is an equivalence; result proves claim | No source clause is dropped |

Palindromes use Mathlib's Function.IsFixedPt and infinite prefixes use the literal List.ofFn expression. Finite decidable equality supports the finite-alphabet formalization and does not restrict finite alphabets. The normalization conjunct makes the normalized sequence in the source assertion available for every generated word.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Palindromes/GeneralizedPseudostandardPeriodicity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Geometry & algebraConjecture refutedFirst preprint 2020The Huang–Stoica–Xia–Zhong conjecture on $T_2$ is refutedA cubic tree admits positive nonconstant edge lengths satisfying the tree equations of motion.Refuted

The questionSection 5.2's second conjecture (ms.tex lines 1931–1934) asks:

Why it mattersThe conjecture would make the positive solutions of the tree equations on the 3-regular tree rigid: satisfying the edge equations would force one global edge length.

The idea3. ratiospine (n) proves that ratio graph lengths (Sum.inl n) = 2^n. ratiobranch (n) (w) proves that ratio graph lengths (Sum.inr (n,w)) = 2^n.

Our contributionA cubic tree admits positive nonconstant edge lengths satisfying the tree equations of motion.

Verification details
Release source recordRefuted in Lean

arXiv

The Huang–Stoica–Xia–Zhong conjecture on $T_2$ is refuted

A. Huang, B. Stoica, X. Xia and X. Zhong, Bounds on the Ricci curvature and solutions to the Einstein equations for weighted graphs, arXiv:2006.06716v1 (2020), define the tree equation of motion using the shortest-path lengths $P_{ij}$ and the vertex sums $c_i$ and $d_i$. For an edge $ij$ they write $$ \mathrm{tEoM}{ij} =\frac{1}{P{ij}}\left(\frac{c_i^2}{d_i^2}+\frac{c_j^2}{d_j^2}\right) -\frac{c_i}{d_i}-\frac{c_j}{d_j}=0. $$ For a tree, they identify $P_{ij}$ on an edge with its edge length, and use $d_i=\sum_{i'\sim i}P_{ii'}^{-2}$ and $c_i=\sum_{i'\sim i}P_{ii'}^{-1}$. They define $T_q$ as the infinite tree whose vertices have degree $q+1$, and call the setting with all edge lengths equal the constant solution. Section 5.2's second conjecture (ms.tex lines 1931–1934) asks:

There exists no non-constant solution to the tEoM for $T_2$ with all edge lengths positive.

The Lean definition claim quantifies this assertion over every locally finite simple graph with decidable adjacency that is a tree, has degree $3$ at every vertex, and carries a symmetric positive edge-length function whose tEoM vanishes on every edge. It concludes that one real constant equals every edge length.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/DiscreteGravity/TreeEquationsNonconstantRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2018The maximum number of pairwise strongly complementary permutation matricesThe maximum size of a strongly complementary permutation family is the largest power of two dividing its positive dimension.Proved

The questionThe v2 source unisto9.tex, lines 686–690, states:

Why it mattersThe ensuing discussion considers larger pairwise strongly complementary families.

The ideaThus that set is bounded above, and result applies csSuple and lecsSup with these hypotheses to identify its natural-number supremum with the attained value.

Our contributionThe maximum size of a strongly complementary permutation family is the largest power of two dividing its positive dimension.

Verification details
Release source recordProved in Lean

Open problem

The maximum number of pairwise strongly complementary permutation matrices

G. Rajchel, A. Gąsiorowski and K. Życzkowski, Robust Hadamard matrices, unistochastic rays in Birkhoff polytope and equi-entangled bases in composite spaces, Mathematics in Computer Science 12, 473–490 (2018), arXiv:1804.10715v2, Section 4, ask about sets of $n\times n$ permutation matrices whose distinct members are pairwise strongly complementary. The v2 source unisto9.tex, lines 686–690, states:

Due to the definition of strong complementarity each pair of matrices has non-zero entries in different places, so that $k \leq n$. We are not able to determine how the maximal number $k$ of such matrices depends on the dimension $n$, nor whether the bistochastic matrices belonging to the interior of this polytope are unistochastic.

Definitions 4.1 and 4.3 (unisto9.tex, lines 606–612 and 623–627) specify the relations:

  • Permutation matrices $P,Q$ are complementary if, for every $i,j,h,k$, the equalities $P_{ij}=P_{hk}=Q_{ik}=1$ imply $Q_{hj}=1$.
  • They are strongly complementary if they are complementary and, for every $i,j$, $P_{ij}=1$ implies $Q_{ij}=0$.

The definitions and source locators are recorded in the literature note.

Issue #23851 fixes the cardinality clause: for every natural $n\ge1$, the maximum is $k(n)=2^{v_2(n)}$, where $v_2(n)$ is the exponent of two in $n$. This identity answers the cardinality question. The question about unistochasticity is separate.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/StronglyComplementaryPermutationMaximum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2014No generalized Choi map detects absolutely PPT entanglementGeneralized Choi maps preserve positivity on absolutely PPT qutrit density matrices.Proved

The questionRusso, Is absolute separability determined by the partial transpose?, Quantum Inf.

Why it mattersThe absolute separability problem asks for a spectral characterization of states that stay separable under every global unitary.

Our contributionGeneralized Choi maps preserve positivity on absolutely PPT qutrit density matrices.

Verification details
Release source recordProved in Lean

arXiv

No generalized Choi map detects absolutely PPT entanglement

S. Arunachalam, N. Johnston and V. Russo, Is absolute separability determined by the partial transpose?, Quantum Inf. Comput. 15(7–8), 694–720 (2015); arXiv:1405.5853v3. Section 7 (p. 21):

Other open problems in this work include proving that all of the generalized Choi maps $\Phi_{b,c}$ in Section 5.3 are incapable of detecting absolutely PPT entanglement (including those in the light gray region of Figure 3) …

Here $\Phi_{b,c}(X)=\tfrac12\begin{pmatrix}ax_{11}+bx_{22}+cx_{33}&-x_{12}&-x_{13}\-x_{21}&cx_{11}+ax_{22}+bx_{33}&-x_{23}\-x_{31}&-x_{32}&bx_{11}+cx_{22}+ax_{33}\end{pmatrix}$ with $a=2-b-c$, positive but not completely positive exactly when $(b,c)\ne(0,0)$, $b,c\ge0$ and $b+c\le1$ or $bc\ge(b+c-1)^2$ (Section 5.3, p. 14). The verbatim texts are in the literature note. Issue #14738 preregisters the reading and the proof plan.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsolutePPT/GeneralizedChoiNonDetection.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023The minimal translation-invariant MPS bond dimension of the W-stateTwo 3 x 3 matrices give a translation-invariant matrix product state representation with periodic boundary conditions of the normalized W-state of order 7.Refuted

Why it mattersThe conjecture says that this upper bound is exact.

Our contributionTwo 3 x 3 matrices give a translation-invariant matrix product state representation with periodic boundary conditions of the normalized W-state of order 7.

Verification details
Release source recordRefuted in Lean

Open problem

The minimal translation-invariant MPS bond dimension of the W-state

P. Klimov, R. Sengupta and J. Biamonte, On Translation-Invariant Matrix Product States and advances in MPS representations of the $W$-state, arXiv:2306.16456v2, Section 3, conjecture after their Theorem 1:

These experiments lead us to conjecture that the representation obtained in Theorem \ref{W state representation} could be optimal both in terms of asymptotics and the constant factor or in other words it is impossible to come up with a TI MPS representation with PBC with bond dimension smaller than $\floor*{\frac{n}{2}}+1$ for the $W$-state of order $n$.

A TI MPS representation with PBC of bond dimension $d$ of the normalized $W$-state of order $n$ is a pair of complex $d\times d$ matrices $A_0$, $A_1$ with $\operatorname{Tr}(A_{i_1}\cdots A_{i_n})=1/\sqrt n$ when exactly one $i_j$ is $1$ and $0$ otherwise (Section 2). Problem 4 of Section 4 asks whether $d(n)=\lfloor n/2\rfloor+1$, and the table of Section 4 lists $d(7)=4$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/WStateTIMPSBondDimension.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2014Local-channel monotonicity of Rényi conditional quantum mutual informationComplete dephasing of A increases the Rényi conditional quantum mutual information of equation (3.18) for a faithful three-qubit state at alpha = 2.Refuted

The questionWilde, Rényi squashed entanglement, discord, and relative entropy differences, §3, Conjecture 1:

Why it mattersThe paper uses local-channel monotonicity to obtain LOCC monotonicity for its Rényi squashed entanglement.

Our contributionComplete dephasing of A increases the Rényi conditional quantum mutual information of equation (3.18) for a faithful three-qubit state at alpha = 2.

Verification details
Release source recordRefuted in Lean

Open problem

Local-channel monotonicity of Rényi conditional quantum mutual information

K. P. Seshadreesan, M. Berta, M. M. Wilde, Rényi squashed entanglement, discord, and relative entropy differences, §3, Conjecture 1:

Conjecture 1 The Rényi conditional quantum mutual information obeys monotonicity with respect to CPTP maps on system A, i.e., for ρABE ∈ S(HABE) and α ∈ (0, 1) ∪ (1, 2], Iα(A; B|E)ρ ≥ Iα(A′; B|E)τ, where τA′BE ≡ MA→A′(ρABE) and MA→A′ is a CPTP map acting on system A alone.

Equation (3.18) and the source's hypothesis statements appear in the literature note. Issue #16068 preregisters the witness.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/RenyiCMILocalChannelRefutation.result

D5/S3/Quantum/Information/RenyiCMILowAlphaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2018Divisibility does not force entanglement breaking for qutritsA divisible qutrit channel lies outside the closure of P-divisible dynamical endpoints and is not entanglement breaking.Refuted

The questionPineda, Divisibility of qubit channels and dynamical maps, Quantum 3, 144 (2019), arXiv:1812.11437v4, ask in the abstract:

Why it mattersThe qubit theorem makes entanglement breaking a consequence of being outside the P-divisible set.

The ideaComparing it with the explicit outputs proves wnotunitary and dnotunitary.

Our contributionA divisible qutrit channel lies outside the closure of P-divisible dynamical endpoints and is not entanglement breaking.

Verification details
Release source recordRefuted in Lean

Open problem

Divisibility does not force entanglement breaking for qutrits

D. Davalos, M. Ziman and C. Pineda, Divisibility of qubit channels and dynamical maps, Quantum 3, 144 (2019), arXiv:1812.11437v4, ask in the abstract:

Finally, we show that every divisible but not infinitesimal divisible qubit channel (in positive maps) is entanglement breaking, and open the question if something similar occurs for higher dimensions.

This sentence is in the source TeX paper.tex, lines 162–164. In Section 2.1 the authors define a divisible channel by the negation of indivisibility, where indivisibility means that a concatenation into two channels forces one factor to be unitary (lines 256–261). In Section 2.2 they write:

we say $\mathcal E\in\mathsf C^P$ if there exists a P-divisible dynamical map $\mathcal E_t$ such that $\mathcal E=\mathcal E_t$ (with arbitrary precision) for some value of $t$ (including $t=\infty$).

This is lines 337–339. Theorem 6 states:

Consider a qubit channel $\mathcal E$. If $\det\hat\mathcal E<0$, then $\mathcal E$ is entanglement breaking, i.e. all qubit channels outside $\mathsf C^P$ are entanglement breaking.

This is lines 1237–1241. The settled qutrit reading is the assertion that every completely positive trace-preserving qutrit map that is divisible and outside the formal pDivisibleSet is entanglement breaking. The module names this assertion claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/QutritDivisibilityEntanglementBreakingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2017Binegativity monotonicity for two qubitsBinegativity increases under a finite one-way LOCC channel on two qubits.Refuted

The questionGour, The binegativity of two qubits, arXiv:1701.02724v3, introduction, p. 1:

Why it mattersThus binegativity does not satisfy the LOCC monotonicity requirement of an entanglement measure.

The ideaPositive rank-one decompositions with vanishing positive-negative products identify both nested negative parts through CFC.posPartnegPartunique.

Our contributionBinegativity increases under a finite one-way LOCC channel on two qubits.

Scope & next questionThe separate upper bound in Eq. (9) is already refuted by the existing upper-bound module; the lower bound remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Binegativity monotonicity for two qubits

M. W. Girard and G. Gour, The binegativity of two qubits, arXiv:1701.02724v3, introduction, p. 1:

That is, we conjecture that for any two-qubit state $\sigma$ it holds that $N_2(\mathcal{E}(\sigma))\leq N_2(\sigma)$ for any LOCC (or PPT) channel $\mathcal{E}$ that outputs states of two qubits.

Here $N_2(\sigma)=\operatorname{Tr}[(\sigma^\Gamma)-]+ 2\operatorname{Tr}[(((\sigma^\Gamma)-)^\Gamma)_-]$, with partial transposition on Bob and the positive and negative components of self-adjoint matrices. S. Sazim and N. Awasthi, Binegativity of two qubits under noise, arXiv:1711.03717v2, introduction, p. 1, restate:

On the basis of numerical evidence, it is conjectured that the binegativity behaves monotonically under both LOCC and PPT channels [15].

Their abstract also states: "Our study supports the conjecture that the binegativity is a monotone."

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TwoQubitBinegativityMonotonicityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundSource 2021Quasi-Injective Composition RefutationSquaring and collapsing the squares are quasi-injective while their composite is not.Refuted

The questionThe same item goes on to ask for a classification of the pairs whose composite is quasi-injective, and records that "An obvious sufficient condition for f ∘ g to be quasi-injective is that g is both surjective and completely multiplicative, but there may be a weaker condition." Only the displayed question is settled here.

Why it mattersQuasi-injectivity is therefore not preserved by composition, so the classification the article asks for is not vacuous: the sufficient condition it names cannot be dropped entirely.

The ideag(n) = n^2 , f(n) = 1 if n is a perfect square, and f(n) = n otherwise.

Our contributionSquaring and collapsing the squares are quasi-injective while their composite is not.

Verification details
Release source recordRefuted in Lean

Open problem

Quasi-Injective Composition Refutation

Pongsriiam, Journal of Integer Sequences 24 (2021), Article 21.10.1, Definition 1, reads: "We call a function f : N → C a quasi-injective function if for all a, b ∈ N, the condition f(an) = f(bn) for all n ∈ N implies a = b." Section 5 asks:

Question 19. Suppose f and g are quasi-injective. Is the composition f ∘ g quasi-injective?

The same item goes on to ask for a classification of the pairs whose composite is quasi-injective, and records that "An obvious sufficient condition for f ∘ g to be quasi-injective is that g is both surjective and completely multiplicative, but there may be a weaker condition." Only the displayed question is settled here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/QuasiInjectiveCompositionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2021.

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Quantum informationCounterexample foundFirst preprint 2020Infinitely many extreme classes survive the commuting-unitary quotient in matrix D-majorizationA pure qubit has infinitely many extreme D-majorization classes under unitaries commuting with a positive diagonal D.Refuted

The questionFrederik vom Ende, Strict positivity and D-majorization, Linear and Multilinear Algebra 70(19) (2022), 4023–4048, arXiv:2004.05613v4, asks in the Conclusion and Outlook, TeX file strictlypositiveqcv2.tex, line 1064:

Why it mattersA finite list of extreme classes would reduce their extreme-point analysis to finitely many representatives, despite the continuous unitary symmetry.

The ideaPositivity then gives the annihilation hypothesis of purerigidity.

Our contributionA pure qubit has infinitely many extreme D-majorization classes under unitaries commuting with a positive diagonal D.

Scope & next questionWhether one ends up with finitely many extreme points after factoring out the unitary equivalents for now remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Infinitely many extreme classes survive the commuting-unitary quotient in matrix D-majorization

Frederik vom Ende, Strict positivity and D-majorization, Linear and Multilinear Algebra 70(19) (2022), 4023–4048, arXiv:2004.05613v4, asks in the Conclusion and Outlook, TeX file strictly_positive_qc_v2.tex, line 1064:

Whether one ends up with finitely many extreme points after factoring out the unitary equivalents for now remains open.

The equivalence relation immediately preceding the question, lines 1060–1064, is $X\sim UXU^*$ for unitary $U$ satisfying $[U,D]=0$, with $D=\operatorname{diag}(d)$, arbitrary $d\in\mathbb R_{++}^n$ and arbitrary $A\in\mathbb C^{n\times n}$. Here $M_D(A)={X:X\prec_D A}$, and $X\prec_D A$ means that a complex linear, completely positive, trace-preserving map sends $A$ to $X$ and fixes $D$ (Definition defi_matrix_D_maj, lines 466–469; equation eq:def_MD, lines 926–932). See the literature note.

Issue #22739 fixes the affirmative reading: for every positive dimension, strictly positive $d$ and complex matrix $A$, the real extreme points of $M_D(A)$ have finitely many classes under this relation. The module's claim quantifies over all natural dimensions, including zero, and asks for a finite set of ambient matrices representing every extreme point. Its negation is established using dimension two, so the positive-dimension reading is also refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/DMajorizationExtremeClassInfinitude.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2017Unimodality of the X-state conditional entropyProcess. 16, 249) writes the conditional entropy of a two-qubit X state through the function f1 of Eq. (A1) on [0, 1] and supposes that it is weakly unimodal for every choice of the parameters p1, ..., p5 with nonnegative Shannon arguments.Refuted

The questionYurischev, Extremal properties of conditional entropy and quantum discord for XXZ, symmetric quantum states, Quantum Inf.

Why it mattersThe hypothesis underlies the paper's bifurcation picture: interior extrema of the conditional entropy appear and disappear only at the end points, so the boundaries of the regions with an interior extremum are given by end-point conditions.

The ideaThe proof uses convexity.

Our contributionProcess. 16, 249) writes the conditional entropy of a two-qubit X state through the function f1 of Eq. (A1) on [0, 1] and supposes that it is weakly unimodal for every choice of the parameters p1, ..., p5 with nonnegative Shannon arguments.

Verification details
Release source recordRefuted in Lean

Open problem

Unimodality of the X-state conditional entropy

M. A. Yurischev, Extremal properties of conditional entropy and quantum discord for XXZ, symmetric quantum states, Quantum Inf. Process. 16, 249 (2017), arXiv:1702.03728v3.

Introduction:

we suppose the unimodal property for the function $S_{cond}(\theta)$ (see Appendix). So, if the unimodality hypothesis is valid the only possibility (except the trivial case $S_{cond}(\theta)=const$) for a single local extremum (minimum or maximum) to appear or disappear inside the open interval by continuous varying the parameters defining the X state is the doubling the extremun at the ends of interval $[0,\pi/2]$.

The Appendix writes the conditional entropy of an X state as $f_1(x)=-h_2\bigl(\frac{1+p_2x}2,\frac{1-p_2x}2\bigr)+h_4\bigl(\frac{1+p_2x\pm\sqrt{r_1}}4,\frac{1-p_2x\pm\sqrt{r_2}}4\bigr)$ on $x\in[0,1]$ (Eq. (A1)). Here $r_{1,2}=(p_1\pm p_5x)^2+4w^2(1-x^2)$, $w=(|p_3+p_4|+|p_3-p_4|)/4$, and $h_2,h_4$ are Shannon entropies in bits. The Appendix defines weak unimodality as follows:

A function $f(x)$ is a weakly unimodal function in the interval $[a,b]$ if there exists a value $x_m\in[a,b]$ for which it is weakly monotonically increasing for $x\le x_m$ and weakly monotonically decreasing for $x\ge x_m$. […] Analogous definitions are given for the minimum.

It then states the conjecture:

The functions $f_1(x)$ and $f_2(x)$ for every choice of parameters $p_1,\ldots,p_5$ for which all arguments of Shannon functions are non-negative can have at most only one local extremum (minimum or maximum) in the open interval $x\in(0,1)$. […] It is required to prove or refute this proposition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/XStateWeakUnimodalityRefutation.result

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Discrete structuresCounterexample foundFirst preprint 2012Bergman's Question 19: the field condition in Question 16An associative commutative nilpotent rational algebra of dimension three has a two-dimensional subspace on which every nonzero vector has nonzero square.Refuted

The questionBergman, Thoughts on Eggert's Conjecture, arXiv:1206.0326v2, Contemporary Mathematics 609 (2014), 1–17, Question 16, asks:

Why it mattersOver an infinite field a quadratic form can have no nonzero zero while taking values in a one-dimensional space.

The ideaSet F=Q and R=(Q × Q) × Q with the ordinary coordinatewise vector-space structure and the bilinear multiplication

Our contributionAn associative commutative nilpotent rational algebra of dimension three has a two-dimensional subspace on which every nonzero vector has nonzero square.

Verification details
Release source recordRefuted in Lean

Open problem

Bergman's Question 19: the field condition in Question 16

Bergman, Thoughts on Eggert's Conjecture, arXiv:1206.0326v2, Contemporary Mathematics 609 (2014), 1–17, Question 16, asks:

Let R be a finite-dimensional nilpotent commutative algebra over an algebraically closed field F, let V be a subspace of R, and let n be a positive integer such that every nonzero element of V has nonzero n-th power. Must dim R ≥ n dim V?

Question 19 explicitly asks whether, for Question 16, algebraic closedness can be dropped or weakened to “F is infinite”. The target is only this field-condition branch, retaining associativity and commutativity. Convention 2 assumes associativity unless explicitly stated otherwise; an identity is not required. All dimensions and all linear structures use the same F.

The affirmative infinite-field assertion is quantified over every infinite field F, every finite-dimensional F-vector space R with an actual F-bilinear multiplication satisfying associativity and commutativity, every linear subspace V, and every positive integer n. Common global nilpotence and nonvanishing n-th powers of every nonzero vector in V are its antecedents. The conclusion is n dim_F V ≤ dim_F R.

| Source clause | Lean counterpart | | --- | --- | | every infinite field | F : Type, [Field F], [Infinite F] | | F-vector space | R : Type, [AddCommGroup R], [Module F R] | | finite dimension | [FiniteDimensional F R] | | F-bilinear multiplication | μ : R →ₗ[F] R →ₗ[F] R | | associativity | ∀ x y z, μ (μ x y) z = μ x (μ y z) | | commutativity | ∀ x y, μ x y = μ y x | | common nilpotence | CommonNilpotent μ | | actual F-linear subspace | V : Submodule F R | | positive exponent | n : ℕ, 0 < n | | all nonzero vectors | ∀ v ∈ V, v ≠ 0 → … ≠ 0 | | n-th power without identity | (List.replicate (n-1) v).foldr (fun x y => μ x y) v | | dimension bound | n * Module.finrank F V ≤ Module.finrank F R | | exact negation | closed claim : Prop, designated result : ¬ claim |

CommonNilpotent μ means that there is k in N such that every list xs of length k and every terminal z have xs.foldr μ z = 0. The terminal is an additional factor: the common vanishing length is k+1, always positive. Under associativity this is global algebra nilpotence, equivalently the corresponding product-span being zero. It is stronger than pointwise nilpotence. For positive n the displayed power has exactly n factors; for n=1 it is v. The closed assertion uses ordinary small carriers, including arbitrary infinite carriers. The counterexample lies in these universes and also refutes the unrestricted mathematical assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Algebra/BergmanInfiniteFieldRefutation.result

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Quantum informationConjecture refutedFirst preprint 2020Refutation of the symmetric binary-outcome detection-threshold conjectureThe Section 4.3 conjecture eta = 4/(m+4) is refuted at m = 13 by a three-input leader model and a universal local extension at efficiency 6/25.Refuted

The questionCope, The Binary-Outcome Detection Loophole, New Journal of Physics 23 (2021) 073032, DOI 10.1088/1367-2630/ac10a5, arXiv:2005.03344v3, Section 4.3, printed p. 6, states:

Why it mattersThe quantity relevant to equal independent loss is the ratio of joint to single conclusive probability.

The ideaThe module first expresses the leader construction as a finite mixture of deterministic local strategies.

Our contributionThe Section 4.3 conjecture eta = 4/(m+4) is refuted at m = 13 by a three-input leader model and a universal local extension at efficiency 6/25.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the symmetric binary-outcome detection-threshold conjecture

T. Cope, The Binary-Outcome Detection Loophole, New Journal of Physics 23 (2021) 073032, DOI 10.1088/1367-2630/ac10a5, arXiv:2005.03344v3, Section 4.3, printed p. 6, states:

In order to prove that our conjecture of $\eta^=4/(m+4)$ for $m_{\mathrm{A}}=m_{\mathrm{B}}=m$ is correct, one would need to provide a extremal $\mathcal{NS}$ distribution $p^{\mathrm{NS}}$, and corresponding Bell inequality $s_{a'b'}^{xy}$, such that $\sum_{a',b',x,y}s_{a'b'}^{xy}p^{\mathrm{NS}}_{\eta}(a'b'|xy) \not \leq k,; \forall \eta > \eta^$.

The named conjecture asserts $\eta^*(m)=4/(m+4)$ for every $m\ge2$. Its definitions and the refutation criterion are fixed in #15572. The source definitions and locator are in the Library note.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Two parties, $m$ inputs each, binary outputs | Two Fin m inputs and two Bool outputs in IsNSBox | Equal input counts; failure is an additional output only after loss. | | Conditional probabilities and no signalling (Section 2) | Four conjuncts of IsNSBox: nonnegativity, normalization, and both input-independent marginals | Every input and output is quantified. | | Independent equal-efficiency loss, equation (3) | All four branches of lossy | some denotes a detected bit and none denotes $F$; the two single-failure terms use the original marginals. | | Convex hull of deterministic local responses | IsLocal | A finite weight for every pair of response functions, nonnegative weights, sum one, and pointwise reconstruction. | | No-signalling threshold, minimum of extremal-box thresholds | etaStar m, the supremum of $\eta\in[0,1]$ making every no-signalling box local after loss | The all-box formulation below is the convention of #15572 and expresses the same threshold. | | Symmetric conjecture $\eta^=4/(m+4)$ | claim | Quantifies every natural $m\ge2$; division is in $\mathbb R$. | | Proposed extremal-box Bell-inequality certificate above the conjectured threshold | No additional Lean hypothesis | This is the source's proposed way to establish tightness, not an assumption of its equality conjecture. The local construction above that conjectured value disproves the equality instead. | | A counterexample to the universal equality | result : ¬ claim | The private lower_bound proves $6/25\le\eta^(13)$, whereas the equality gives $4/17$. |

The threshold formulation follows from convexity. Every no-signalling box is a convex combination of extremal boxes. At a fixed efficiency, loss preserves normalized convex combinations and the local polytope is convex. Thus every box is local after loss if and only if every extremal box is. Further independent erasure preserves locality, so the admissible efficiencies of each extremal box form a downward interval. There are finitely many extremal boxes in this finite polytope; the supremum of efficiencies shared by all of them is the minimum of their thresholds. This is the mathematical interpretation of etaStar, not an additional Lean theorem in this module.

The efficiency interval $[0,1]$ is physical, and $m\ge2$ restricts the conjecture to its nontrivial input domain. Neither restricts the refuting instance $m=13$. Nonnegativity and normalization express a conditional probability distribution and are not extra assumptions about the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/BinaryDetectionThresholdRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2018Three product observables with useless pairs and a useful tripleThree Hermitian product observables have useless pairs and a useful triple on two qutrits.Proved

The questionGühne, The structure of ultrafine entanglement witnesses, J.

Why it mattersThe question asks whether entanglement information can require three product-observable expectations even when every pair of expectations admits a separable explanation.

The ideaThese matrices specify the settlement's construction, rather than a formula attributed to the paper.

Our contributionThree Hermitian product observables have useless pairs and a useful triple on two qutrits.

Scope & next questionParameter families, the smallest possible local dimensions, classification of useful triples, and optimal robustness bounds remain open in this settlement. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

Three product observables with useless pairs and a useful triple

M. Gachechiladze, N. Wyderka and O. Gühne, The structure of ultrafine entanglement witnesses, J. Phys. A: Math. Theor. 51, 365307 (2018), arXiv:1805.06404v2, Summary, TeX source uew-iop-revision.tex, lines 750–752:

For instance, it would be interesting to identify three product observables, for which each pair is useless for entanglement detection, but the whole triple is useful.

The same source defines separable states as the convex hull of pure product states (lines 113–114) and describes a useless pair by the fact that every possible pair of expectation values can originate from a separable state (lines 504–507).

Issue #25940 fixes the existential reading: find three Hermitian product observables $M_i=A_i\otimes B_i$ such that, for each distinct pair $i,j$ and every density matrix $\rho$, one separable density matrix $\sigma$ reproduces both expectations, whereas some density matrix has no separable density matrix reproducing all three expectations. The pairwise matching state may depend on the pair and on $\rho$. The module fixes both local dimensions to three. This existence statement is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/UltrafineThreeProductObservables.result

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Discrete structuresConjecture provedFirst preprint 2019Garg's Fibonacci prefix antipower conjectureThe first Fn−1 blocks of length Fn/2+F{n−1} in the Fibonacci word are distinct whenever Fn is even.Proved

The questionSwapnil Garg, Antipowers in Uniform Morphic Words and the Fibonacci Word, arXiv:1907.10816v4, DMTCS 23(3), article 14, Conjecture 18, printed page 8:

The ideaThis is the new cut construction on the active proof path.

Our contributionThe first Fn−1 blocks of length Fn/2+F{n−1} in the Fibonacci word are distinct whenever Fn is even.

Scope & next questionOther Sturmian slopes and optimal prefix block lengths remain open follow-up questions.

Verification details
Release source recordProved in Lean

Open problem

Garg's Fibonacci prefix antipower conjecture

Swapnil Garg, Antipowers in Uniform Morphic Words and the Fibonacci Word, arXiv:1907.10816v4, DMTCS 23(3), article 14, Conjecture 18, printed page 8:

Let $F_n$ be an even Fibonacci number. Then, there is an $(F_n-1)$-antipower with block length $\frac{F_n}{2}+F_{n-1}$ that is a prefix of $\mathbf f$.

The Fibonacci word is the fixed point starting with $0$ of $0\mapsto01$, $1\mapsto0$; $F_1=F_2=1$. A $k$-antipower consists of $k$ consecutive, pairwise distinct blocks of equal length. The source note quotes the definitions from §3 and the conjecture. Issue #14479 fixes the universal statement GARG-1: every $n\ge1$ with $F_n$ even, $K=F_n-1$, $L=F_n/2+F_{n-1}$, and the first $K$ blocks starting at $jL$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Antipowers/GargFibonacciPrefixAntipower.result

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Analysis & dynamicsCounterexample foundFirst preprint 2014Refutation of the complete original Jaco addition lawThe distinct orders three and eleven refute the complete original two-branch Jaco addition assertion.Refuted

The questionKok, Fisher, Wilkens, Mabula and Mukungunugwa, Characteristics of Finite Jaco Graphs, Jn(1), n in N, arXiv:1404.0484v1, printed page 7, Theorem 2.10 (Conjectured), states:

Why it mattersThis is the explicit author-posed addition question.

The ideaReuse JacoDomPathDiameterRefutation.Vertex and jaco, the positive-label finite truncation of JacoExponentialDominationRefutation's recursive adjacency.

Our contributionThe distinct orders three and eleven refute the complete original two-branch Jaco addition assertion.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the complete original Jaco addition law

Kok, Fisher, Wilkens, Mabula and Mukungunugwa, Characteristics of Finite Jaco Graphs, J_n(1), n in N, arXiv:1404.0484v1, printed page 7, Theorem 2.10 (Conjectured), states:

For the Jaco Graphs J_n(1), J_m(1) with n>=3, m>=3, n!=m we have Delta(J_(n+m)(1))=Delta(J_n(1))+Delta(J_m(1)), if J_n(1) or J_m(1) has a unique Jaconian vertex; Delta(J_n(1))+Delta(J_m(1))+1 otherwise.

This quotation normalizes only mathematical typography. Printed page 11 explicitly asks: “Prove Theorem (conjectured) 2.10.” Definitions 1.1--1.4 and 1.7, printed page 2, give positive vertices, the recursion i<j<=2i-d^-(v_i), finite truncation, ordinary total degree, and Jaconian vertices as all maximizing vertices.

Write Delta_n=(jaco n).maxDegree and U(n) iff exactly one v in Vertex n has degree Delta_n. The complete claim is

$$ \forall n,m\in\mathbb N,\quad 3\le n\Rightarrow3\le m\Rightarrow n\ne m\Rightarrow \left[ ((U(n)\lor U(m))\Rightarrow\Delta_{n+m}=\Delta_n+\Delta_m) \land ((\neg U(n)\land\neg U(m))\Rightarrow\Delta_{n+m}=\Delta_n+\Delta_m+1) \right]. $$

Both branches and every admissible ordered pair are one assertion. OR is inclusive. Unique means a unique vertex within that graph, not a unique degree value across orders. The source's prime Jaconian designation alone does not imply this uniqueness.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/JacoAdditionLawRefutation.result

Frozen module record

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Number theoryQuestion resolvedPosed in 2016Yanev's closed form for binary run compressionBinary run compression satisfies Yanev's formula at every natural index, including zero.Proved

The questionIn binary expansion of n: reduce contiguous blocks of 0's to 0 and contiguous blocks of 1's to 1.

Why it mattersThe value of the compressed binary word depends only on its number of runs and its least significant digit.

The ideaEstablish the local have alternatingvalue identity inside result by induction on an arbitrary binary list with unequal adjacent entries and last digit one: 3 Nat.ofDigits 2 l + 2 = 2 ^ (l.length + 1) + l.head!.

Our contributionBinary run compression satisfies Yanev's formula at every natural index, including zero.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Yanev's closed form for binary run compression

OEIS A090079, NAME (%N, verbatim):

In binary expansion of n: reduce contiguous blocks of 0's to 0 and contiguous blocks of 1's to 1.

Yanev's FORMULA conjecture (%F, verbatim):

Conjecture: a(n) = (2^(A005811(n)+1) + (1-(-1)^n)/2 - 2)/3. - Velin Yanev, Dec 12 2016

The AUTHOR line (%A, verbatim):

Reinhard Zumkeller, Nov 20 2003

The proved statement is forall n : Nat, 3 * a n = 2 ^ (runs n + 1) + n % 2 - 2, where runs n = ((Nat.digits 2 n).splitBy (· == ·)).length and a n = Nat.ofDigits 2 (((Nat.digits 2 n).splitBy (· == ·)).map List.head!). The list is little-endian; reversing a word preserves its run count and reverses the order of its compressed digits. Thus these definitions implement the literal operation. The number of binary runs is the quantity A005811(n) used in the conjecture. The integer identity (1-(-1)^n)/2 = n mod 2 identifies the parity term. The statement is multiplied by 3 to stay in ℕ. For positive n the numerator is nonnegative; at zero both sides are zero. Lean subtraction is truncated natural subtraction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/YanevRunCompressionClosedForm.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedPosed in 2013The prime-period conjecture for OEIS A224899A224899 has period p-1 modulo every odd prime, and the assertion fails at p=2.Refuted

The questionThe sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1.

Why it mattersThis is a first-tier OEIS conjecture: entry year 2013, conjecture year 2022.

The ideaThe escape content is the coefficient bridge H(n,k) = kⁿ·k!·Σ{r=k..n} C(n,r)(−k)^{n−r}2^{r−k}S(r,k) for the coefficient expression H(n,k) = n![xⁿ] sinh(kx)^k.

Our contributionA224899 has period p-1 modulo every odd prime, and the assertion fails at p=2.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The prime-period conjecture for OEIS A224899

OEIS A224899, NAME:

E.g.f.: Sum_{n>=0} sinh(n*x)^n.

COMMENT (Peter Bala, May 29 2022):

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [1, 1, 2, 0, 3, 0, 1, 1, 2, 0, 3, 0, ...], with an apparent period of 6. Cf. A245322.

The Library note attributes the entry to Paul D. Hanna, April 20, 2013, and the conjecture to Peter Bala, May 29, 2022. The conjecture quantifies over all primes. It is false at p = 2; its substantive content survives for every odd prime, so the failure is confined exactly to p = 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/SinhPowerPrimePeriod.bala_conjecture_two_false

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Discrete structuresRecurrence provedPosed in 2019Kurkov's full lower-bit recurrence for A284005Binary prefix products give the complete lower-bit recurrence for A284005.Proved

The questionThis dossier anchors exactly the November 10, 2019 FORMULA in OEIS A284005, attributed to Mikhail Kurkov:

Why it mattersThe defining recurrence follows successive binary prefixes.

The ideaThe result reuses KurkovWeightProductBinomial.a and DyadicRowPolynomialRecurrence.wt, together with the retained binary-weight zero, even and odd lemmas.

Our contributionBinary prefix products give the complete lower-bit recurrence for A284005.

Scope & next questionThe divisor-count representation in the entry's NAME is outside this result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's full lower-bit recurrence for A284005

This dossier anchors exactly the November 10, 2019 FORMULA in OEIS A284005, attributed to Mikhail Kurkov:

Conjecture: a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) + 2^k*(1 - T(n,k))) for n > 1 with a(0) = 1, a(1) = 2, f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2. - Mikhail Kurkov, Nov 10 2019

The NAME defines a(n)=(1+A000120(n))*a(floor(n/2)) for n>1. The initial data begins 1,2,4,6,8,12,18,24, including a(1)=2. Here wt(n)=A000120(n) is binary digit sum, L=floor(log_2(n)), and removing the leading binary one gives f(n)=n-2^L. All operations in the formal statement are natural-number operations. Every index 0 <= k < L occurs: an existing one bit contributes a(f(n)), and these repeated contributions are retained.

The exact native result is:

theorem result :
    a 0 = 1 ∧ a 1 = 2 ∧
    (∀ n : ℕ, 1 < n → a n = (1 + wt n) * a (n / 2)) ∧
    (∀ n : ℕ, 1 < n →
      a n = 2 * a (n - 2 ^ Nat.log 2 n) +
        ∑ k ∈ Finset.range (Nat.log 2 n),
          a (n - 2 ^ Nat.log 2 n + 2 ^ k * (1 - n / 2 ^ k % 2)))

This includes n=2 and every power of two. The lower-bit identity asserts nothing at zero or one. The April 24, 2023 two-parameter binomial conjecture and the A329369 modulo-two binomial transform are distinct questions. The divisor-count representation in the entry's NAME is outside this result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/KurkovLowerBitRecurrence.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Contribution

    Binary prefix products give the complete lower-bit recurrence for A284005.

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  • Historical context

    Dated conjecture statement: 2019.

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Discrete structuresQuestion resolvedFirst preprint 2020Okada's pre-merged Gaussian determinants of width at least three are not almost palindromicAlmost strictly pre-fitting tuples of width at least three give pre-merged Gaussian determinants that are not almost palindromic.Proved

The questionSo Okada, Merged-log-concavity of rational functions, almost strictly unimodal sequences, and phase transitions of ideal boson-fermion gases, arXiv:2003.02112v2 (2020 preprint, revised 2024), states the following in the subsection “On the almost palindromicity”, conjecture labelled conj:palindromic-width-three, phases.tex, lines 8770–8779:

Why it mattersOkada's paper studies merged-log-concavity, unimodality and palindromicity of expressions constructed from Gaussian binomial coefficients, with applications to ideal boson-fermion gases.

The ideaThese are the hypotheses of indexbounds, lowend, reverseddegreegap and highend below.

Our contributionAlmost strictly pre-fitting tuples of width at least three give pre-merged Gaussian determinants that are not almost palindromic.

Verification details
Release source recordProved in Lean

Open problem

Okada's pre-merged Gaussian determinants of width at least three are not almost palindromic

So Okada, Merged-log-concavity of rational functions, almost strictly unimodal sequences, and phase transitions of ideal boson-fermion gases, arXiv:2003.02112v2 (2020 preprint, revised 2024), states the following in the subsection “On the almost palindromicity”, conjecture labelled conj:palindromic-width-three, phases.tex, lines 8770–8779:

Let $\mu\in\Z_{\geq 3}$. Then, $\dt(U)_{a}^{b}(\mu,\iota^{\mu}(1),\iota^{\mu}(1), 0, \iota^{\mu}(q))$ is not an almost palindromic $q$-polynomial, provided $\tp{\mu,a,b}$ is almost strictly pre-fitting.

The quotation retains the source macros: \Z denotes $\mathbb Z$, \dt denotes $\tilde d$, and \tp prints a tuple. The source defines almost strictly pre-fitting by $0\le a_1<\cdots<a_\mu\le b_1<\cdots<b_\mu$ (lines 8337–8346), and almost palindromic by equality of the coefficient list with its reverse after zero coefficients are omitted (Definition def:almost, lines 8503–8512).

At these parameters, the pre-merged determinant is the integer polynomial $P=A-B$, where $A=\prod_{i=1}^{\mu}{b_i\brack a_i}q$ and $B=\prod{i=1}^{\mu}{b_i\brack a_{\mu+1-i}}_q$. This specialization follows from the source's tuple Gaussian product (lines 3505–3519), unit base-shift factors (lines 5942–5955), and pre-parcel and pre-merged determinant definitions (lines 6005–6048). The scalar degree shift is zero.

Issue #25498 fixes the reading: for every natural $\mu\ge3$ and every such pair of tuples, the nonzero coefficients of $P$, ordered by increasing degree, do not form a palindrome. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GaussianQuotient/OkadaPreMergedNonPalindromic.result

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Analysis & dynamicsCounterexample foundSource 2014Refutation and exact classification of exponential derivative ratiosExact complete monotonicity classification of consecutive signed derivative ratios.Refuted

The questionFor given i ∈ {0} ∪ ℕ, the functions ℱi(t) and 𝒢i(t) defined in (10) or (27) are completely monotonic on (0, ∞).

Why it mattersDecreasingness controls the first derivative; complete monotonicity requires simultaneous sign constraints at every order.

The ideaFor the remaining indices the closed forms are

Our contributionExact complete monotonicity classification of consecutive signed derivative ratios.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation and exact classification of exponential derivative ratios

Section 5 of the source identified by D5/L/Asymptotics/weiguo2014completemonotonicity states:

Conjecture 12. For given i ∈ {0} ∪ ℕ, the functions ℱ_i(t) and 𝒢_i(t) defined in (10) or (27) are completely monotonic on (0, ∞).

Equation (6) defines

$$F_i(t)=(-1)^i\left(\frac1{e^t-1}\right)^{(i)},\qquad G_i(t)=(-1)^i\left(\frac1{1-e^{-t}}\right)^{(i)},$$

and equation (10) defines

$$\mathcal F_i(t)=\frac{F_{i+1}(t)}{F_i(t)},\qquad \mathcal G_i(t)=\frac{G_{i+1}(t)}{G_i(t)}.$$

The source verifies $i=0,1$ in (30). Complete monotonicity means smoothness on $(0,\infty)$ and $(-1)^kh^{(k)}(t)\ge0$ for every nonnegative integer $k$ and every $t>0$.

Formal reading

All Lean names below have prefix D5.S3.Analytic.ExpRatioCM.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $F_i(t)$ in (6) | Defs.F | Exactly the globally defined signed iterated derivative of $1/(e^t-1)$. | | $G_i(t)$ in (6) | Defs.G | Exactly the globally defined signed iterated derivative of $1/(1-e^{-t})$. | | $\mathcal F_i$ in (10) | Defs.ratioF | Exactly $F_{i+1}/F_i$. The denominator is positive for $t>0$. | | $\mathcal G_i$ in (10) | Defs.ratioG | Exactly $G_{i+1}/G_i$. The denominator is positive for $t>0$. | | Complete monotonicity | Defs.CompletelyMonotoneOn | ContDiffOn with smoothness index ⊤ on Set.Ioi 0, and the sign condition for every natural derivative order at every positive argument. | | Conjecture 12 | Defs.claim | Both complete-monotonicity assertions for every natural index, including zero. | | Exact index sets | Positive.cm_of_le_two, Result.failure_on_every_tail | The positive theorem covers $i\le2$, while the failure theorem excludes every $i\ge3$. | | Failure on every right half-line | Result.failure_on_every_tail | For every $i\ge3$ and real $T$, there are a natural $k$ and $t>\max{T,0}$ at which both alternating derivatives are negative. | | Universal refutation | Result.result | Exactly $\neg,\texttt{Defs.claim}$. |

The globally defined derivatives agree with derivatives of the smooth restrictions on the open positive half-line by neighborhood equality. No derivative value at zero is used.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/ExpRatioCM/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2014.

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Quantum informationQuestion resolvedFirst preprint 2021Classical permutations admit a strict two-query sequential advantageTwo adaptive queries distinguish four permutations with success one, while parallel success is at most three quarters.Proved

The questionQuintino, Unitary channel discrimination beyond group structures: Advantages of sequential and indefinite-causal-order strategies (Journal of Mathematical Physics 63, 042203, 2022; arXiv:2105.13369v3), ask in the Conclusion, main.tex, line 514:

Why it mattersThe question separates the benefit of adapting a later query to an earlier answer from the quantum features of the source's discrimination tasks.

Our contributionTwo adaptive queries distinguish four permutations with success one, while parallel success is at most three quarters.

Verification details
Release source recordProved in Lean

Open problem

Classical permutations admit a strict two-query sequential advantage

J. Bavaresco, M. Murao and M. T. Quintino, Unitary channel discrimination beyond group structures: Advantages of sequential and indefinite-causal-order strategies (Journal of Mathematical Physics 63, 042203, 2022; arXiv:2105.13369v3), ask in the Conclusion, main.tex, line 514:

Assuming the analog of a unitary channel in classical information theory to be a classical channel that maps deterministic probability distributions into deterministic probability distributions, i.e., channels that can be described by a permutation matrix, it would be interesting to investigate whether their discrimination could also be enhanced by sequential strategies.

The quotation is from the v3 TeX source (main.tex, md5 7b7f57add94cbe20dcb088bf7c16f071); see also the literature note.

Issue #22253 fixes the existential reading: there are a finite alphabet, a finite permutation ensemble with a probability vector, and a two-use sequential strategy whose average success exceeds a real number $c$ bounding every two-use parallel strategy. Both strategies query the same unknown permutation twice. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ClassicalPermutationSequentialAdvantage.result

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Quantum informationQuestion resolvedFirst preprint 2016The relabelled chained inequality holds for every odd-input binary COD modelThe relabelled chained inequality holds for every odd-input binary COD model.Proved

The questionChaves, Bell scenarios with communication, J.

Why it mattersThe inequality constrains classical explanations that allow Alice's outcome to influence Bob's outcome.

The ideaThe last diagonal condition then contradicts the vanishing closing indicator.

Our contributionThe relabelled chained inequality holds for every odd-input binary COD model.

Verification details
Release source recordProved in Lean

Open problem

The relabelled chained inequality holds for every odd-input binary COD model

J. B. Brask and R. Chaves, Bell scenarios with communication, J. Phys. A: Math. Theor. 50(9), 094001 (2017), arXiv:1607.08182v1, subsection “Generalization to more inputs”, conjecture following Eq. (25). The source member Bell_comm_arXiv_v1.tex, line 325, states verbatim:

For the inequality to hold for the COD model, Alice's outputs must be relabelled for every second input. That is $a=0 \leftrightarrow a=1$ for even $x$. With this relabelling, we conjecture that \eqref{eq.validchained} holds for any odd number of inputs.

The source label eq.validchained denotes Eq. (25), at TeX lines 318–324:

$$[a_1-b_1]+[b_1-a_2]+[a_2-b_2]+\cdots+[a_n-b_n]+[b_n-a_1-1]\geq1.$$

At line 325, the bracket is defined by $[a_x-b_y]=\sum_{j=0}^1j,p(a-b=j\bmod2\mid x,y)$. For binary outputs, the ordinary brackets are unequal-output probabilities and the closing bracket is an equal-output probability.

Issue #24183 fixes the finite hidden-variable reading of the source's causal outcome-dependent (COD) model, Eq. (7), TeX lines 150–155:

$$p(a,b\mid x,y)=\sum_{\lambda\in\Lambda}w_\lambda A_x(a\mid\lambda)B_y(b\mid a,\lambda),$$

where $\Lambda$ is finite, $w_\lambda\geq0$, $\sum_\lambda w_\lambda=1$, and both response functions are nonnegative and normalized over their binary output, including Bob's normalization for each raw Alice output. Let $r_x=1$ for even one-based inputs $x$, and $r_x=0$ otherwise. The inequality reads Alice's relabelled output $c=a\oplus r_x$, while Bob receives the raw $a$. Writing $q(c,b\mid x,y)=p(c\oplus r_x,b\mid x,y)$, the statement is

$$\sum_{i=1}^nq(c\ne b\mid i,i)+\sum_{i=1}^{n-1}q(c\ne b\mid i+1,i)+q(c=b\mid1,n)\geq1$$

for every odd natural $n$ and every such model. This is proved by result : claim. The definitions and source locator are also in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/OutcomeCommunicationChainedInequality.result

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Number theoryQuestion resolvedPosed in 2018Schulte's Ω/ω binomial convolutionSchulte's binomial convolution relates the total and distinct prime-factor counts.Proved

The questionNumber of prime divisors of n counted with multiplicity (also called big omega of n, bigomega(n) or Omega(n)).

Why it mattersThe independent question is Schulte's published 2018 binomial convolution, a first-tier external named small conjecture.

The ideaExtend F(z,n) = z^Ω(n) and H(n) = (x+y)^(Ω(n)−ω(n)) y^ω(n) by zero at n = 0.

Our contributionSchulte's binomial convolution relates the total and distinct prime-factor counts.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's Ω/ω binomial convolution

OEIS A001222, NAME (%N, verbatim):

Number of prime divisors of n counted with multiplicity (also called big omega of n, bigomega(n) or Omega(n)).

COMMENT (%C, verbatim):

Conjecture: Let f(n) = (x+y)^a(n), and g(n) = x^a(n), and h(n) = (x+y)^A046660(n) * y^A001221(n) with x, y complex numbers and 0^0 = 1. Then f(n) = Sum_{d|n} g(d)*h(n/d). This is proved for x = 1-y (see Dressler and van de Lune link). - Werner Schulte, Feb 10 2018

The A001222 offset (%O) is 1,3. A046660, NAME (%N, verbatim):

Number of prime factors of n counted with multiplicity minus number of distinct prime factors. Omega(n) - omega(n).

A001221, NAME (%N, verbatim):

Number of distinct primes dividing n (also called omega(n)).

Write Ω for ArithmeticFunction.cardFactors and ω for ArithmeticFunction.cardDistinctFactors. The full-quantifier reading is

theorem result (x y : ℂ) (n : ℕ) (hn : 0 < n) :
    (x + y) ^ Ω n = ∑ d ∈ n.divisors,
      x ^ Ω d * ((x + y) ^ (Ω (n / d) - ω (n / d)) * y ^ ω (n / d))

Every complex pair x, y and every positive natural n are quantified. The sum uses all positive divisors, including 1 and n; n / d is natural division and is exact for these divisors. Exponent subtraction is natural subtraction; ω(m) ≤ Ω(m) follows from deduplication of the prime-factor list, so it equals the ordinary nonnegative difference. Natural powers implement 0^0 = 1. The statement includes n = 1 and allows x, y, and x+y to vanish, without any nonzero-base assumption.

Only the general x, y assertion in the quoted Conjecture sentence is settled. The historical x = 1 − y case is Dressler and van de Lune, Proc. AMS 41 (1973), 403–406, doi 10.1090/S0002-9939-1973-0340191-8. Other OEIS assertions and historical priority are outside the formal claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SchulteOmegaBinomialConvolution.result

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Number theoryQuestion resolvedPosed in 2020Rotondo's A006972 Lucas-Carmichael sufficient conditionRotondo's factorization conditions imply the Lucas-Carmichael divisibility criterion.Proved

The questionConjecture: if k = pqr, p = ad - 1, q = bd - 1, r = cd - 1 are distinct odd primes, with d = gcd(p + 1, q + 1, r + 1) and abcd divides k + 1, then k is a Lucas-Carmichael number. - Davide Rotondo, Dec 23 2020

Why it mattersThe independent question is Rotondo's named sufficient condition for a three-prime product to be Lucas-Carmichael.

The ideaDistinct primality makes pqr squarefree and makes the product composite and greater than one.

Our contributionRotondo's factorization conditions imply the Lucas-Carmichael divisibility criterion.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Rotondo's A006972 Lucas-Carmichael sufficient condition

OEIS A006972, NAME (%N, verbatim):

Lucas-Carmichael numbers: squarefree composite numbers k such that p | k => p+1 | k+1.

Davide Rotondo's COMMENT (%C, verbatim) is:

Conjecture: if k = pqr, p = ad - 1, q = bd - 1, r = cd - 1 are distinct odd primes, with d = gcd(p + 1, q + 1, r + 1) and abcd divides k + 1, then k is a Lucas-Carmichael number. - Davide Rotondo, Dec 23 2020

The full-quantifier reading takes a,b,c,d,p,q,r,k in the natural numbers, keeps the stated odd-prime, distinctness, product, gcd, and divisibility hypotheses, adds the derivable positivity hypotheses, and concludes that k is squarefree, composite, greater than one, and satisfies s+1 | k+1 for every prime divisor s. The equations p+1=a*d, q+1=b*d, and r+1=c*d are equivalent to the source's subtraction equations under the corresponding prime hypotheses: primality gives positivity, which removes the natural-subtraction truncation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/RotondoLucasCarmichaelCriterion.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedPosed in 2015Schulte's A008287 alternating-binomial expansionSchulte's conjectured A008287 alternating-binomial expansion is proved.Proved

The questionTriangle of quadrinomial coefficients, row n is the sequence of coefficients of (1 + x + x^2 + x^3)^n.

Why it mattersThe independent question is Schulte's named expansion of all A008287 quadrinomial coefficients at indices 0 ≤ k ≤ 3n.

The ideaApply addpow and regroup powers, with 3(n−j)+j = 3n−2j for j≤n.

Our contributionSchulte's conjectured A008287 alternating-binomial expansion is proved.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's A008287 alternating-binomial expansion

OEIS A008287, NAME (%N, verbatim):

Triangle of quadrinomial coefficients, row n is the sequence of coefficients of (1 + x + x^2 + x^3)^n.

The settled FORMULA (%F, verbatim) is:

T(n,k) = Sum_{j=0..k} (-2)^j*binomial(n,j)*binomial(3*n-2*j,k-j) for n >= 0 and 0 <= k <= 3*n (conjectured). - Werner Schulte, Sep 09 2015

The full-quantifier reading is: for all n k : ℕ, if k ≤ 3 * n, the integer coefficient T n k of X^k in (1 + X + X^2 + X^3 : ℤ[X]) ^ n equals ∑ j ∈ Finset.range (k + 1), (-2 : ℤ)^j * (n.choose j : ℤ) * ((3 * n - 2 * j).choose (k - j) : ℤ). Subtraction in binomial indices is truncated natural subtraction, and n.choose j = 0 for j > n. The scope wall is only this Schulte 2015 formula. The Shevelev 2010 and Bala 2013 formulas in the same entry are different and are not settled here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/SchulteQuadrinomialAlternatingBinomialExpansion.result

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Number theoryQuestion resolvedPosed in 2017The A008590 gcd-filtered sums characterize the positive multiples of eightBoth of Ratajczak's gcd-filtered sums are even exactly at multiples of eight above one.Proved

The questionLechoslaw Ratajczak's COMMENT conjecture (%C, verbatim):

Why it mattersThe conjecture gives an arithmetic characterization of the positive terms of A008590 through two independently defined gcd-filtered sums.

The ideaIn ZMod 2 all nonfixed pairs cancel, and the unique possible fixed point is m/2.

Our contributionBoth of Ratajczak's gcd-filtered sums are even exactly at multiples of eight above one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A008590 gcd-filtered sums characterize the positive multiples of eight

OEIS A008590, NAME (%N, verbatim):

Multiples of 8.

Lechoslaw Ratajczak's COMMENT conjecture (%C, verbatim):

From Lechoslaw Ratajczak, Sep 03 2017: (Start)

Conjecture: let gcd_2(b,c) be the second greatest common divisor and lcd_2(b,c) be the second least common divisor of not coprime integers b and c. Consecutive elements of this sequence (for a(n) > 0) are consecutive integers m for which both Sum_{k=1..m, gcd(k,m)<>1} gcd_2(k,m) and Sum_{k=1..m, gcd(k,m) <>1} lcd_2(k,m) are even numbers.

a(1) = 8 because 1+2+1+4 = 8 (8 is even) and 2+2+2+2 = 8 (8 is even).

a(2) = 16 because 1+2+1+4+1+2+1+8 = 20 (20 is even) and 2+2+2+2+2+2+2+2 = 16 (16 is even).

a(3) = 24 because 1+1+2+3+4+1+1+6+1+1+4+3+2+1+1+12 = 44 (44 is even) and 2+3+2+2+2+3+2+2+2+3+2+2+2+3+2+2 = 36 (36 is even).

The conjecture was checked for 5*10^4 consecutive integers. (End)

For d = gcd(k,m), the example at m=24 fixes gcd2(k,m) = d / minFac(d) and lcd2(k,m) = minFac(d). Thus gcd2 is the greatest proper divisor of a nontrivial gcd and lcd2 is its least divisor greater than one. Define G(m) = Sum_{1<=k<=m, gcd(k,m) != 1} gcd2(k,m) and L(m) = Sum_{1<=k<=m, gcd(k,m) != 1} lcd2(k,m).

The literal proved statement is forall m : Nat, 1 < m -> ((Even (G m) and Even (L m)) iff 8 divides m). At m=1 both filtered sums are empty and equal zero, hence both are even, while one is not a positive multiple of eight; this is the unique mismatch in the independent scan on [1,6000], and the theorem explicitly assumes 1 < m. The knight-move, odd-square-difference, lattice-pair, square-frame, and Catalan/Berdellé eight-odd-squares comments in A008590 are NOT claimed. Values of gcd2 and lcd2 on coprime pairs are also NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/RatajczakGcdSumParityCharacterization.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2018The immediate successor is the least weak pseudoprime exactly when it is an odd compositeOrdowski's immediate-successor weak-pseudoprime condition holds exactly at odd composite successors.Proved

The questionSmallest composite c > n such that n^c == n (mod c).

Why it mattersOrdowski's 2018 comment asks for an exact characterization of the bases at which the first possible modulus, n+1, is already the sequence value.

The ideaIf n+1 is odd, this power is -1, so the required congruence holds.

Our contributionOrdowski's immediate-successor weak-pseudoprime condition holds exactly at odd composite successors.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The immediate successor is the least weak pseudoprime exactly when it is an odd composite

OEIS A239293, NAME (%N, verbatim):

Smallest composite c > n such that n^c == n (mod c).

The neighbouring terminology COMMENT (%C, verbatim):

a(n) is the smallest weak pseudoprime to base n that is > n.

The neighbouring sufficient-condition COMMENT (%C, verbatim):

If n is even and n+1 is composite, then a(n) = n+1. [Corrected by Thomas Ordowski, Aug 03 2018]

Thomas Ordowski's target COMMENT (%C, verbatim):

Conjecture: a(n) = n+1 if and only if n+1 is an odd composite number. - Thomas Ordowski, Aug 03 2018

The literal proved statement is

forall n : Nat, 1 <= n -> ((1 < n+1 and not Prime(n+1) and n^(n+1) mod (n+1) = n mod (n+1)) iff (not Prime(n+1) and Odd(n+1) and 1 < n+1)).

By the NAME, a(n) is the least composite c>n satisfying the displayed congruence. Since n+1 is the immediate successor of n, no natural number lies strictly between them. Thus a(n)=n+1 holds exactly when n+1 itself qualifies; minimality at the immediate successor is automatic. This is the reduction used by the theorem, and it does not require a total definition of a at bases for which no qualifying composite has been supplied.

The neighbouring sufficient-condition COMMENT is not claimed as a new result. The weak-pseudoprime terminology COMMENT is not claimed as a new result. Totality of a is not claimed; proving it would require a different existence theorem asserting a qualifying composite above every base.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/OrdowskiImmediateSuccessorWeakPseudoprime.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    Source 1
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    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryQuestion resolvedPosed in 2017No A254074 term is a multiple of fiveWu's A254074 conjecture excludes multiples of five from the specified decimal digit range.Proved

The questionNumbers k such that the decimal expansions of both k and k^2 have 5 as the digit with the smallest value and 9 as the digit with the largest value.

Why it mattersWu's comment isolates a simple divisibility restriction seen in the known terms.

The ideaIf 5 divides k, while every decimal digit of k is at least 5, its units digit is forced to be 5.

Our contributionWu's A254074 conjecture excludes multiples of five from the specified decimal digit range.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

No A254074 term is a multiple of five

OEIS A254074, NAME (%N, verbatim):

Numbers k such that the decimal expansions of both k and k^2 have 5 as the digit with the smallest value and 9 as the digit with the largest value.

Chai Wah Wu's COMMENT (%C, verbatim):

The first digit of a term is either 7, 8 or 9 and the last digit is either 5, 6 or 7. Conjecture: no term is a multiple of 5. - Chai Wah Wu, Sep 10 2017

The literal proved sentence is "Conjecture: no term is a multiple of 5." Formally, every positive natural k whose decimal digits and square's decimal digits both have minimum 5 and maximum 9 is not divisible by 5. The first sentence of Wu's comment, concerning the first and last digits, is NOT claimed. The sparsity, finiteness or infinitude of A254074, and every other property of the sequence are also NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/WuDigitRangeSquareNotMultipleOfFive.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Dated conjecture statement: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryQuestion resolvedPosed in 2020Oudra's divisor-count formula for the Stechkin functionThe Stechkin function counts a divisibility condition that splits into divisors of two adjacent integers.Proved

The questionBoris Stechkin function: a(n) is the number of m with 2 <= m <= n and floor(n(m-1)/m) divisible by m-1.

Why it mattersIt gives a closed arithmetic description of every term of the sequence rather than a finite-prefix verification.

The ideaFor 1 <= k < n, Euclidean division shows that k divides nk/(k+1) exactly when k+1 divides n or n-1.

Our contributionThe Stechkin function counts a divisibility condition that splits into divisors of two adjacent integers.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Oudra's divisor-count formula for the Stechkin function

OEIS A274010, NAME (%N, verbatim):

Boris Stechkin function: a(n) is the number of m with 2 <= m <= n and floor(n(m-1)/m) divisible by m-1.

Ridouane Oudra's FORMULA conjecture (%F, verbatim):

Conjecture: a(n) = tau(n) + tau(n-1) - 2, for n>=2. - Ridouane Oudra, Feb 28 2020

AUTHOR (%A, verbatim):

Robert Israel, Jun 06 2016

The formal definition counts the natural numbers m in the closed interval from two through n for which m-1 divides the natural quotient n*(m-1)/m. Natural division is the floor of the corresponding nonnegative rational quotient. For every n >= 2, the formal theorem states stechkinFunction n + 2 = n.divisors.card + (n-1).divisors.card. Both divisor finsets contain one, so their cardinalities sum to at least two; the subtraction-free equality is therefore equivalent to the quoted formula.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/StechkinFunctionDivisorCount.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Dated conjecture statement: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2021A circulant unistochastic matrix without a circulant unitary liftA circulant unistochastic matrix of order ten has no circulant unitary lift.Refuted

The questionPuchała and K. Życzkowski, Algebraic and geometric structures inside the Birkhoff polytope, arXiv:2101.11288, J.

Why it mattersThe source's Lemma 18 supplies circulant unitary lifts at dimensions at most four.

The ideaThe consumed uunitary transfers it to the complex unitary group.

Our contributionA circulant unistochastic matrix of order ten has no circulant unitary lift.

Verification details
Release source recordRefuted in Lean

Open problem

A circulant unistochastic matrix without a circulant unitary lift

G. Rajchel-Mieldzioć, K. Korzekwa, Z. Puchała and K. Życzkowski, Algebraic and geometric structures inside the Birkhoff polytope, arXiv:2101.11288, J. Math. Phys. 63, 012202, DOI 10.1063/5.0046581, Section IV, states:

Conjecture 19. The sets Cd ∩ Ud and Cd² coincide for all d, i.e., for every circulant unistochastic matrix it is possible to find a corresponding circulant unitary matrix.

The universal assertion is Refuted at dimension ten. Preregistration #16445 fixes the source, quantifiers and Lean conventions. Definition 6 defines doubly circulant unistochastic matrices as those having a circulant unitary realization. The source quotations and locator are in the existing Library note.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every dimension $d$ | claim quantifies over natural $d$ | No restriction to even or positive dimensions in the assertion | | Every circulant matrix $B$ | Every real Matrix (Fin d) (Fin d) with the reused IsCirculant | Matrix.circulant uses $v(i-j)$ with modular indices; the source's reversed convention describes the same class | | Unistochastic $B$ | Reused IsUnistochastic, using Matrix.unitaryGroup and RHLinalg.normSqMatrix | Allows every complex unitary realization, including a real orthogonal one | | A corresponding circulant unitary matrix | HasCirculantLift quantifies over a complex vector $z$, requires its circulant to be unitary, and requires entrywise squared norms to equal $B$ | The full source condition, with no certificate restriction | | Equality of the two sets | claim gives the inclusion from circulant unistochastic matrices to matrices with a circulant unitary lift | The reverse inclusion follows from the source definitions | | Refutation of the all-dimensional claim | result proves the literal negation of claim | One instance at $d=10$ suffices |

All binders and conjuncts implement these clauses. There is no additional hypothesis on $B$ or on a realizing unitary. HasCirculantLift mirrors Definition 6 verbatim: “The set of d × d doubly circulant unistochastic matrices Cd² consists of these circulant unistochastic matrices for which there exists the corresponding unitary matrix that is circulant.” claim mirrors the quoted Conjecture 19. The two other source predicates are reused from CirculantBraceletNonUnistochastic, without aliases.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/CirculantUnitaryLiftRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2015The zero higher-order EPR bound is tight at every positive orderThe zero higher-order EPR bound is tight for both signs at every positive order.Proved

The questionShchukin and P. van Loock, Higher-order EPR correlations and inseparability conditions for continuous variables, arXiv:1511.08444v1 (Phys.

Why it mattersTightness asks whether arbitrarily small values are available at each order, rather than merely values below the separable threshold.

Our contributionThe zero higher-order EPR bound is tight for both signs at every positive order.

Verification details
Release source recordProved in Lean

Open problem

The zero higher-order EPR bound is tight at every positive order

E. Shchukin and P. van Loock, Higher-order EPR correlations and inseparability conditions for continuous variables, arXiv:1511.08444v1 (Phys. Rev. A 93, 032114, 2016), subsection “Approach two”, state the following conjecture in bip-ent2.tex, lines 961–963:

Our conjecture is that the inequality \eqref{eq:nS:1} cannot be violated by Gaussian states (including all inseparable ones) and the inequality \eqref{eq:nS:2} is tight. We verified this by numerical computations, but we could not find rigorous proofs.

Here the source's inequalities (nS:1) and (nS:2), at lines 914–924, concern two bosonic modes and every positive integer $n$:

$$ \left\langle(\hat a^{\dagger n}\pm\hat b^n)(\hat a^n\pm\hat b^{\dagger n})\right\rangle \ge \begin{cases} n! & \text{for separable states},\ 0 & \text{for all states}. \end{cases} $$

Issue #24246 fixes the reading to the tightness clause: for every $n\ge1$, either sign $\sigma\in{1,-1}$ and every $\varepsilon>0$, there is a normalized finitely supported occupation-basis vector $\psi$ with $|(\hat a^n+\sigma\hat b^{\dagger n})\psi|^2<\varepsilon$. This is proved. The Gaussian clause is a separate question.

The source and its bibliographic locator are recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/FockSpace/HigherOrderEPRTightness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2020Three volume ratios for generalized Pauli eigenvalue regionsThree volume ratios for generalized Pauli eigenvalue regions hold in every natural dimension at least two.Proved

The questionKatarzyna Siudzińska, Geometry of generalized Pauli channels, Phys.

Why it mattersThe ratios measure how much of the source's eigenvalue parameter space belongs to each nested class.

The ideaThe following helper names are local to GeneralizedPauliVolumeRatios unless otherwise qualified.

Our contributionThree volume ratios for generalized Pauli eigenvalue regions hold in every natural dimension at least two.

Verification details
Release source recordProved in Lean

Open problem

Three volume ratios for generalized Pauli eigenvalue regions

Katarzyna Siudzińska, Geometry of generalized Pauli channels, Phys. Rev. A 101, 062323 (2020), arXiv:2002.04657v2, states three conjectures in Section 4.2, “$N=d+1$ and $N=d$ mutually unbiased bases”. The v2 TeX source GGPC_volume5.tex, lines 243–246, reads:

and conjecture that the general formula reads $\frac{V_{CP}(d)}{V_P(d)}=\frac{d}{(d+1)!}.$

Lines 276–279 read:

and conjecture that the general formula reads $\frac{V_{G}(d)}{V_{CP}(d)}=\frac{d^2-1}{d^2}\left(\frac{d-1}{d}\right)^{d}.$

Lines 302–305 read:

from which we conjecture that $\frac{V_{EB}(d)}{V_{G}(d)}=\frac{1}{d+1}.$

For a natural number $d\ge2$, put $m=d+1$ and $s(x)=\sum_{i=1}^{m}x_i$ for $x\in\mathbb R^m$. The module defines precisely these four sets:

  • $P={x:\ -1/(d-1)\le x_i\le1\text{ for every }i}$ (boxRegion).
  • $C={x:\ -1/(d-1)\le s(x)\text{ and }s(x)\le1+dx_j\text{ for every }j}$ (cpRegion).
  • $G=C\cap{x:\ x_i\ge0\text{ for every }i}$ (generatorRegion).
  • $E=G\cap{x:\ s(x)\le1}$ (ebRegion).

For these sets write $V_A=\operatorname{channelVolume}(d,A)=(\sqrt{d-1}/d)^m\operatorname{Leb}m(A)$, for $A=P,C,G,E$. In Lean, channelVolume uses (volume A).toReal; the proof obtains finite positive Lebesgue measures for all four sets when $d\ge2$. Thus $V_P,V_C,V_G,V_E$ match the source's $V_P(d),V{CP}(d),V_G(d),V_{EB}(d)$ as coordinate-region volumes. The source locates the box and its weight at lines 210–228, complete positivity at lines 359–364, nonnegative eigenvalues at lines 256–267, and the nonnegative entanglement-breaking region at lines 282–293; see also the literature note.

Issue #22754 fixes the statement as the conjunction of these three identities for every natural $d\ge2$. This is proved for the specified coordinate regions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/GeneralizedPauliVolumeRatios.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024One generating measurement does not make pseudo-broadcasting equivalent to non-contextualityOne four-outcome qubit POVM generates a contextual subtheory that admits one-copy pseudo-broadcasting.Refuted

The questionUola, No-Broadcasting Characterizes Operational Contextuality, Phys.

Why it mattersIt would make feasibility at that single level a certificate of non-contextuality.

The ideaThe post-processing set equals its convex hull: postprocessingsconvex applies to every finite matrix family.

Our contributionOne four-outcome qubit POVM generates a contextual subtheory that admits one-copy pseudo-broadcasting.

Verification details
Release source recordRefuted in Lean

Open problem

One generating measurement does not make pseudo-broadcasting equivalent to non-contextuality

P. Jokinen, M. Weilenmann, M. Plávala, J.-P. Pellonpää, J. Kiukas and R. Uola, No-Broadcasting Characterizes Operational Contextuality, Phys. Rev. Lett. 133, 240201 (2024), arXiv:2406.07305v1, Appendix D, state:

One can also see that $\mathcal S(\mathcal A)$ is a Cartesian product of $k$ simplexes if $\mathcal A$ is generated by $k$ independent measurements. This leads to a conjecture that if $\mathcal A$ is generated by $k$ measurements, then $1\mapsto k$-pseudobroadcasting is equivalent to non-contextuality. This is only a conjecture since even if $\mathcal A$ is generated by $k$ measurements, $\mathcal S(\mathcal A)$ in general is only a subset of a Cartesian product of $k$ simplexes. We leave resolving this issue and proving or disproving the conjecture for future work.

This quotation is from Arxiv.tex, line 482, in the arXiv v1 source archive (534 lines; md5 cc2fa845c0b0f4155ed18dcb76e4c92e). The source macros \hS, \hA and \btn{k} are expanded to $\mathcal S$, $\mathcal A$ and $1\mapsto k$; the wording is unchanged. The main-text conjecture is on line 348. The definitions and further quotations are in the literature note.

Issue #21862 fixes the reading of generation: $\mathcal A$ is the real convex hull of the binary classical post-processings $\sum_o c_oM_j(o)$, with $0\le c_o\le1$, of $k$ finite POVMs. A subtheory has a convex set $\mathcal T$ of density matrices and a convex set $\mathcal A$ of effects, contains $I$ among its effects, and is closed under $E\mapsto I-E$. The conjectured equivalence is refuted by a qubit subtheory generated by one four-outcome POVM.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumContext/SingleMeasurementPseudobroadcastingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2022Positive Kraus operators do not exclude a side-entanglement advantageTwo qubit depolarizing channels have strictly positive-definite tetrahedral Kraus operators, yet a Bell-assisted discrimination strategy succeeds with probability 33/56 while every unassisted strategy succeeds with probability at most 47/84.Refuted

The questionMancini, Discrimination of dephasing channels, J.

Why it mattersThe success probabilities depend on the channels represented.

Our contributionTwo qubit depolarizing channels have strictly positive-definite tetrahedral Kraus operators, yet a Bell-assisted discrimination strategy succeeds with probability 33/56 while every unassisted strategy succeeds with probability at most 47/84.

Verification details
Release source recordRefuted in Lean

Open problem

Positive Kraus operators do not exclude a side-entanglement advantage

M. Rexiti, L. Memarzadeh and S. Mancini, Discrimination of dephasing channels, J. Phys. A 55, 245301 (2022), arXiv:2201.00388v2, Section 7, conjecture:

In contrast, if all Kraus operators have simply the identity as unitary part in the polar decomposition, there is no benefit from side entanglement.

The complete source passage is in the Library note. Issue #16535 preregisters the negative clause, its representation-dependent reading and the exact finite-dimensional conventions.

Statement echo

| Source clause or convention | Lean counterpart and scope | | --- | --- | | Kraus operators have identity polar unitary | claim assumes every Kraus operator is strictly positive definite, which makes that unitary uniquely the identity; this stronger hypothesis suffices to refute the source statement. | | Channels and side entanglement | Both channels act on complex qubit matrices; the ancilla is a qubit and the first tensor factor. These are instances of the source's unrestricted systems. | | Trace-preserving Kraus representations | Both right Gram sums equal the identity, with the same finite number of operators. | | No benefit | For every assisted density and binary effect, some unassisted density and binary effect achieve at least the same success probability, with equal priors (successAs and successUn). | | Density and effect | Frozen GHZMeasureBiseparableBound.IsDensity is the positive-semidefinite, trace-one predicate. The preregistered effect condition and the two Kraus sums are inlined literally; proofs use the frozen Kraus and Bell owners. | | Negative clause fails | result : ¬ claim, with four operators per channel. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/PositiveKrausSideEntanglementRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundSource 2023Malkoun: the unrestricted subgroup permanent lower boundTwo cyclic subgroups of the permutation group on seven points give a negative cross-sum on a complex Hermitian positive semidefinite Gram matrix.Refuted

The questionMalkoun, Do these cousins of permanents satisfy the following inequality?, MathOverflow question 456248, revision 8, 382F6B28-9E4F-456B-9254-E0A041A674B5.

Why it mattersThe source relates the expression to two subgroup symmetrizations of a pure tensor and to graph-associated functions.

The ideaTheir seven powers therefore enumerate the actual subgroups bijectively, using the pinned finEquivZPowers equivalence.

Our contributionTwo cyclic subgroups of the permutation group on seven points give a negative cross-sum on a complex Hermitian positive semidefinite Gram matrix.

Verification details
Release source recordRefuted in Lean

Open problem

Malkoun: the unrestricted subgroup permanent lower bound

Malkoun, Do these cousins of permanents satisfy the following inequality?, MathOverflow question 456248, revision 8, 382F6B28-9E4F-456B-9254-E0A041A674B5. The source note gives the identity, attribution, license and bounded literature comparison.

Let $H$ denote an $n$ by $n$ hermitian positive semidefinite matrix. Let $G$ and $K$ be two subgroups of the symmetric group $\Sigma_n$. Define $f_{G,K}(H)=\sum_{(\sigma,\tau)\in G\times K}\prod_{i=1}^n h_{\sigma(i),\tau(i)}$. Does there exist a positive constant $c(n,G,K)$ depending on $n$, $G$ and $K$ such that $\Re(f_{G,K}(H))\geq c(n,G,K)\prod_{i=1}^n h_{ii}$?

The complete quantifier order is

$$ \forall n\in\mathbb N\ \forall G,K\leq\operatorname{Sym}(n)\quad \exists c\in\mathbb R\quad c>0\ \land
\forall H\in M_n(\mathbb C),\quad (H=H^*\land H\succeq0)\Longrightarrow c\prod_i\Re(h_{ii})\leq\Re(f_{G,K}(H)). $$

claim has this proposition as its definition and result has type ¬ claim. Native index $i:\mathrm{Fin}(n)$ corresponds to source label $i+1$; every permutation acts on that same index set. crossSum sums over the actual subgroup types, without normalization. No Young-subgroup, rank, entry-sign or strictly positive diagonal condition is added to the universal assertion. diagonal_product_eq_real supplies the Hermitian real-diagonal bridge.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/MalkounSubgroupLowerBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2023.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year recorded in the source bibliography: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundRefutation of the A049591 divisor-count prime-gap characterizationThe value 529 refutes Cloitre's divisor-count prime-gap characterization.Refuted

The questionBenoit Cloitre's COMMENT of April 13, 2002 (verbatim):

Why it mattersCloitre's comment identifies the sequence with all integers satisfying a divisor-count prime-gap condition.

The ideaIts divisors are 1, 23, and 529, so the divisor count is three and the upper endpoint is 529+3^2=538.

Our contributionThe value 529 refutes Cloitre's divisor-count prime-gap characterization.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A049591 divisor-count prime-gap characterization

OEIS A049591, NAME (verbatim):

Odd primes p such that p+2 is composite.

Benoit Cloitre's COMMENT of April 13, 2002 (verbatim):

Sequence appears also to give all n > 1 such that there is no prime p satisfying the inequality n < p < n+tau(n)^2 where tau(n)=A000005(n).

For natural numbers, define Term(n) := Odd(n) and Prime(n) and not Prime(n+2), and define NoPrimeInGap(n) := for every natural p, Prime(p) implies not (n < p and p < n + card(divisors(n))^2). The literal refuted statement is for every natural n, 1 < n implies (Term(n) iff NoPrimeInGap(n)).

The NAME definition is not refuted. The Bernoulli, f(2p)=p, Ianakiev 2019 quotient, and Ordowski 2020 congruence comments are not claimed or assessed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/CloitrePrimeGapDivisorCharacterizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Conceptual contribution · 2/4 · 20% weight

    The delivered argument gives a structural characterization or normal form.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2015Skewness at most two does not force randomness at most one in multicyclic enzyme networksAn eight-state enzyme network has waiting-time skewness at most two and randomness greater than one.Refuted

The questionSeifert, Skewness and Kurtosis in Statistical Kinetics, Phys.

Why it mattersA completion-time statistic above its unicyclic upper bound provides a signature that the network is more complex than a single cycle.

The ideaThe theorem result applies the proposed universal implication to this network with its positive variance and skewness bound, contradicting randomnessgt.

Our contributionAn eight-state enzyme network has waiting-time skewness at most two and randomness greater than one.

Scope & next questionWhether the implication holds throughout the source's four-state two-route topology or its off-pathway specialization remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Skewness at most two does not force randomness at most one in multicyclic enzyme networks

A. C. Barato and U. Seifert, Skewness and Kurtosis in Statistical Kinetics, Phys. Rev. Lett. 115, 188103 (2015), arXiv:1510.03279v1, ask in skewkurtstatkinarxiv.tex, line 358, at the end of the multicyclic discussion:

It remains as an open question whether this property is true for arbitrary networks that do not consist of a single cycle.

The preceding numerical investigation concerns a four-state network with two product-forming routes and its off-pathway specialization. At lines 350–352 the authors report that constraining the skewness to be at most two prevents the randomness parameter from exceeding one. The question asks whether the superior detection of departures from a single cycle by higher moments extends to arbitrary networks.

The source defines cycle completion at an irreversible product-release transition (Eq. fullreaction, lines 110–117). For completion-time moments $t_n$, its definitions at lines 133–149 are $$ \sigma^2=t_2-t_1^2,\qquad \mathcal R=\frac{\sigma^2}{t_1^2},\qquad \mathcal S=\frac{t_3-3t_1\sigma^2-t_1^3}{\sigma^3}. $$

The source's kurtosis is $\mathcal K=(t_4-4t_3t_1+6\sigma^2t_1^2+3t_1^4)/\sigma^4$ (line 149); it is not used by the settlement.

Issue #26046 fixes the implication to be refuted: for every finite network with nonnegative off-diagonal rates, a positive marked rate $p\to r$, every state able to reach $p$, and positive variance, $\mathcal S\le2$ implies $\mathcal R\le1$. Moments use the first-passage matrix definition below. This implication is refuted.

The literature note records the source and its scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/WaitingTimeRandomnessSkewnessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2016The Fuchs–Caves fidelity measurement need not give optimal ROC pointsA three-dimensional fidelity measurement point lies below the optimal quantum ROC curve.Refuted

The questionKoniorczyk, Receiver operation characteristics of quantum state discrimination (J.

Why it mattersThe question asks whether one measurement selected to attain fidelity also supplies optimal binary discriminators at all its cumulative thresholds.

The ideaThe proof uses spectral estimates.

Our contributionA three-dimensional fidelity measurement point lies below the optimal quantum ROC curve.

Verification details
Release source recordRefuted in Lean

Open problem

The Fuchs–Caves fidelity measurement need not give optimal ROC points

A. Bodor and M. Koniorczyk, Receiver operation characteristics of quantum state discrimination (J. Russ. Laser Res. 38, 150–163, 2017; arXiv:1603.06801v1), ask whether the cumulative points of the fidelity measurement lie on the optimal receiver operating characteristic (ROC) curve. The v1 TeX member Bodor_Koniorczyk_QuantumROC_submit_1.tex, lines 940–944, in the subsection “The quantum Bhattacharyya coefficient”, states:

This obviously defines $d+1$ points in the feasible ROC domain, where $d$ stands for the dimension of the joint support of the two density operators. (It is an interesting open question whether these points are actually on the optimal ROC curve.)

The observable in the preceding equation, labelled eq:fidelitymeasurable, is $$M=(\sqrt{\rho})^{-1}\sqrt{\sqrt{\rho},\sigma\sqrt{\rho}},(\sqrt{\rho})^{-1},$$ where $\rho=\varrho_N$ and $\sigma=\varrho_P$. For an ordered orthonormal eigenbasis $(a_i){i=0}^{d-1}$, put $q_i=\operatorname{Re}(a_i^\dagger\rho a_i)$ and $p_i=\operatorname{Re}(a_i^\dagger\sigma a_i)$, ordered by nonincreasing $p_i/q_i$, and define $Q_k=\sum{i<k}q_i$, $P_k=\sum_{i<k}p_i$.

Issue #24220 fixes the universal reading: for every finite dimension $d$, every pair of positive-definite trace-one complex matrices $\rho,\sigma$, every such ordered eigenbasis, and every $0\le k\le d$, every binary effect $0\le F\le I$ satisfying $\operatorname{Re}\operatorname{tr}(F\rho)=Q_k$ must satisfy $\operatorname{Re}\operatorname{tr}(F\sigma)\le P_k$. This is claim; result refutes it.

The definitions and source locators are collected in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/FidelityMeasurementROCRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2015The dimer bounds for k at most 4 on regular biconnected graphsButera, Federbush and Pernici (arXiv:1502.06734) ask whether the bounds Delta^k ln(i!Refuted

The questionPernici, Positivity of the virial coefficients in lattice dimer models and upper bounds on the number of matchings on graphs, Physica A 437 (2015) 278–294, arXiv:1502.06734v2, Section IV B:

Why it mattersThe paper derives Eq. (1) on finite regular graphs as the counterpart of the positivity of the virial coefficients of the monomer-dimer model on infinite regular lattices, and tests it on lattice graphs and on regular biconnected graphs.

The ideaFor two edge sets without a common vertex, the matching sequence of the union is the Cauchy product of the two sequences.

Our contributionButera, Federbush and Pernici (arXiv:1502.06734) ask whether the bounds Delta^k ln(i!

Verification details
Release source recordRefuted in Lean

Open problem

The dimer bounds for k at most 4 on regular biconnected graphs

P. Butera, P. Federbush, M. Pernici, Positivity of the virial coefficients in lattice dimer models and upper bounds on the number of matchings on graphs, Physica A 437 (2015) 278–294, arXiv:1502.06734v2, Section IV B:

For all the graphs examined in this section, Eq. (\ref{Delta0}) is satisfied for $k \le 4$. It would be interesting to know whether these bounds, Eq. (\ref{Delta0}) for $k \le 4$, are always satisfied for regular biconnected graphs.

Eq. (1) is $\Delta^k\ln(i!,N(i))\le0$ for $k=2,\dots,\nu$ and $i=0,\dots,\nu-k$, where $N(i)$ is the number of configurations of $i$ dimers on the graph ($i$-edge matchings), $\Delta$ is the forward difference in $i$, and $\nu$ is the matching number.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/DimerVirialFourthDifferenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2020The naive orthostochastic equations are insufficient at order fiveThe naive orthostochastic equations are insufficient at order five.Refuted

The questionDey, The 4 × 4 orthostochastic variety, arXiv:2001.10691v1, Experimental Mathematics 32(4) (2023), 692–700, DOI 10.1080/10586458.2021.1982427, Question 5 (cf.

Why it mattersPairwise orthogonality can choose different signs for different pairs.

The ideaThe row and column sign choices need not be mutually compatible.

Our contributionThe naive orthostochastic equations are insufficient at order five.

Scope & next questionNext, for n = 4, 5 it remains open whether the naive equations cut out the orthostochastic variety inside the Birkhoff polytope set-theoretically.

Verification details
Release source recordRefuted in Lean

Open problem

The naive orthostochastic equations are insufficient at order five

J. Chen and P. Dey, The 4 × 4 orthostochastic variety, arXiv:2001.10691v1, Experimental Mathematics 32(4) (2023), 692–700, DOI 10.1080/10586458.2021.1982427, Question 5 (cf. Chterental–Đoković, Remark 3.4), asks:

Do the naive equations define the orthostochastic variety inside B_n? In other words, if a doubly stochastic matrix satisfies C_{i,j}, R_{i,j} = 0 for all i < j, must it be orthostochastic?

Section 5 states:

Next, for n = 4, 5 it remains open whether the naive equations cut out the orthostochastic variety inside the Birkhoff polytope set-theoretically.

The order-five clause is Refuted by $B=(4J_5-3I_5)/17$. Preregistration #17606 fixes the source, quantifiers and conventions. The Library note contains the source locators and quotations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A real orthostochastic matrix is the entrywise square of an orthogonal matrix | IsOrthostochastic quantifies over every real orthogonal matrix and uses RHLinalg.normSqMatrix | For real entries, the squared norm is the square | | The equivalent sign condition after Definition 4 for $C_{i,j}=0$ | ColumnNaive quantifies over all real signs equal to $1$ or $-1$ and requires the signed radical sum to vanish | A sign on every term is equivalent to fixing the first sign to $1$, by reversing all signs together | | Row equations are obtained “by transposing indices” | RowNaive exchanges the row and column positions in that sum | No symmetry assumption on the quantified matrix | | Every pair $i<j$, both $C_{i,j}=0$ and $R_{i,j}=0$ | SatisfiesNaive requires both conditions for every such pair | Zero-based Fin n replaces the source's one-based indices | | Every doubly stochastic matrix at $n=5$ | claim quantifies over every real five by five matrix in Mathlib's doublyStochastic set | The only restriction is the order-five clause being settled | | The naive equations imply orthostochasticity | The final implication in claim | No restriction on an orthogonal lift or its signs | | A negative answer | result proves the literal negation of claim | The witness B has diagonal $1/17$ and off-diagonal $4/17$ |

These are all binders and conjuncts in the public predicates. The concrete witness is additional data used to refute the universal assertion, not an additional hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/ChenDeyNaiveEquationsFive.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2021Ghanbari's integer Sombor energy conjectureThe graph formed by three four-cycles sharing one vertex has Sombor spectrum 16, -16, 4, -4, 4, -4, 0, 0, 0, 0.Refuted

The questionGhanbari, On the Sombor characteristic polynomial and Sombor energy of a graph, Computational and Applied Mathematics 41 (2022), article 242, Conjecture 3.8:

The ideaThe proof uses spectral estimates and polynomial identities.

Our contributionThe graph formed by three four-cycles sharing one vertex has Sombor spectrum 16, -16, 4, -4, 4, -4, 0, 0, 0, 0.

Verification details
Release source recordRefuted in Lean

Open problem

Ghanbari's integer Sombor energy conjecture

N. Ghanbari, On the Sombor characteristic polynomial and Sombor energy of a graph, Computational and Applied Mathematics 41 (2022), article 242, Conjecture 3.8:

There is no graph with integer-valued Sombor energy.

The paper studies finite simple graphs. For a graph $G$, the Sombor matrix has entry $\sqrt{d_i^2+d_j^2}$ on an adjacent pair and zero otherwise, and the Sombor energy is the sum of the absolute values of its eigenvalues.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SomborEnergyIntegerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2017Refutation of reciprocal q-gamma function positivityA convergent reciprocal q-gamma series whose function Turanian is negative.Refuted

The questionKarp, Inequalities for series in q-shifted factorials and q-gamma functions, J.

Why it mattersA negative power-series coefficient alone need not produce a negative function value.

The ideaThe first ten normalized series terms use these bounds with six product factors each.

Our contributionA convergent reciprocal q-gamma series whose function Turanian is negative.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of reciprocal q-gamma function positivity

S. I. Kalmykov and D. B. Karp, Inequalities for series in q-shifted factorials and q-gamma functions, J. Math. Anal. Appl. 460 (2018), 332–351, DOI 10.1016/j.jmaa.2017.11.041, arXiv:1702.03788v1, Conjecture 1, printed p. 12:

All conclusions of Theorem 3 are valid for all α > 0, while conditions on other parameters remain intact. In particular, µ → g(µ; x) is log-concave for each fixed 0 < x < Rg.

Theorem 3, printed p. 11:

For a real sequence {gₙ}ₙ≥₀ and fixed 0 < q < 1 define

$g(\mu;x)=\displaystyle\sum_{n=0}^{\infty}\frac{g_nx^n}{\Gamma_q(\mu+n)}.$ (16)

Suppose {gₙ}∞ₙ₌₀ is doubly positive. Then (16) has a positive radius of convergence Rg and the generalized Turánian

$\Delta_g(\alpha,\beta;x)=g(\mu+\alpha;x)g(\mu+\beta;x)-g(\mu;x)g(\mu+\alpha+\beta;x)$

is positive for 0 < x < Rg if α ∈ N, β > 0 and µ ≥ max(−β, −1). If, in addition, α ≤ β + 1, then Δg(α, β; x) has positive power series coefficients at all powers of x.

Equation (6), printed p. 4:

The q-gamma function is defined by

$\Gamma_q(z)=(1-q)^{1-z}\displaystyle\frac{(q;q)\infty}{(q^z;q)\infty}$ (6)

for |q| < 1 and all complex z such that qᶻ⁺ᵏ ≠ 1 for k ∈ N₀.

The reciprocal is extended by zero at nonpositive integers. Doubly positive means nontrivial, nonnegative, log-concave and without internal zeros, as on printed p. 3. The function assertion has no $\alpha\le\beta+1$ restriction.

Formal reading

The names below belong to D5.S3.Analytic.QGammaTuran.FunctionResult, except for the reused reciprocal and sequence condition.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Reciprocal of (6) | Result.rqgamma | The same real infinite-product quotient as in the coefficient settlement. | | Doubly positive sequence | D5.S3.Analytic.ReciprocalGammaTuran.Defs.AdmissibleSeq | Nontriviality, nonnegativity, log-concavity and no internal zeros. | | Series (16) | qseries | The real infinite sum, in the printed order of factors. | | Radius of (16) at the given $\mu$ | radius | Mathlib scalar-series radius in $[0,\infty]$, for coefficients $g_n,\texttt{rqgamma}(q,\mu+n)$. | | Function clause | claim | All permitted parameters and every $0<x<R_g$, with $R_g=\texttt{radius}\ q\ \mu\ g$. | | Refutation | result | Exactly $\neg,\texttt{claim}$. |

The radius is taken for series (16) at the theorem's given $\mu$. As the source states on printed p. 12, “Writing Rg > 0 for the radius of convergence in (16), we claim that the next conjecture is true.” Mathlib's FormalMultilinearSeries.ofScalars represents its coefficient sequence, and FormalMultilinearSeries.radius supplies the standard extended nonnegative radius. For this scalar series the coefficient-map norm is exactly the absolute value of the coefficient.

At the example, absolute summability at $x=3/2$ gives $3/2\le R_g$. Thus $0<93/100<3/2\le R_g$, so the refutation holds strictly inside the disk of convergence of (16). The same geometric comparison proves summability of all four shifted series at $x=93/100$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/QGammaTuran/FunctionResult.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2018Circular hypergraphs do not maximise orientation-count entropyA four-edge three-uniform multihypergraph has greater entropy than the circle.Refuted

The questionThe competitor class consists of multi-hypergraphs and permits repeated edges and isolated vertices.

Why it mattersUniform placement of edges around a circle is proposed to maximise entropy.

The ideaThe claim applied to this uniform witness gives the opposite weak inequality.

Our contributionA four-edge three-uniform multihypergraph has greater entropy than the circle.

Verification details
Release source recordRefuted in Lean

Open problem

Circular hypergraphs do not maximise orientation-count entropy

Pelekis, Le Matematiche 74(1), pp. 128–129, Section 4: “We conjecture that circular hypergraphs are such that H(X_{C_{n,r}}) ≥ H(X_H), for all H ∈ D_{n,n,r}.” The preceding definition assumes $r\ge3$ and $n>r$; $C_{n,r}$ has all $n$ cyclic paths of $r$ vertices as edges. The competitor class consists of multi-hypergraphs and permits repeated edges and isolated vertices. Every edge independently selects one uniformly random vertex, and $X$ counts the selected vertices. The title's complementary zero-in-degree count has the same entropy.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $r\ge3$ and $n>r$ | claim quantifies over r n, with 3 ≤ r and r < n | No additional hypothesis. | | All cyclic paths of size $r$ | circular n r maps each starting vertex to the image of natural residues $(i+j)\bmod n$, $0\le j<r$ | Its bound derives $n>0$ from the actual index i : Fin n; for $n=0$ there is no index. | | $D_{n,n,r}$ | E : Fin n → Finset (Fin n) with all cardinalities equal to r | Exactly $n$ labelled edges; multiplicities retained. | | Uniform independent orientations and Shannon entropy | Shared X and entropyX r E from OrientationEntropyBalance | Literal filter, denominator $(r:\mathbb R)^n$, base-two logarithm and zero-probability convention. | | Circular entropy at least every competitor | entropyX r E ≤ entropyX r (circular n r) for every uniform E | No restriction of the competitor class. | | Refutation | result : ¬ claim | Unconditional negation; the numbered asymptotic Conjecture 4.1 is a separate statement. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Hypergraphs/CircularOrientationEntropy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundFirst preprint 2018Sun's weighted distinct-sum boundSun's weighted distinct-sum bound is refuted over the rationals at n = 2 by the set {0, 1, 2}.Refuted

The questionZhi-Wei Sun, On permutations of {1,...,n} and related topics, J.

Our contributionSun's weighted distinct-sum bound is refuted over the rationals at n = 2 by the set {0, 1, 2}.

Scope & next questionThe closest surviving statement in this delivery is the exact rational cardinality; the unrestricted amended bound remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Sun's weighted distinct-sum bound

Zhi-Wei Sun, On permutations of {1,...,n} and related topics, J. Algebraic Combin. 54 (2021), 893–912, DOI 10.1007/s10801-021-01028-8, arXiv:1811.10503v6, Conjecture 4.9, printed page 18:

Conjecture 4.9. Let $n\in\mathbb Z^+$ and let $F$ be a field with $p(F)>n+1$, where $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if the characteristic of $F$ is zero. Let $A$ be any finite subset of $F$ with $|A|\ge n+\delta_{n,3}$, where $\delta_{n,3}$ is 1 or 0 according as $n=3$ or not. Then, for the set $S(A):={\sum_{k=1}^n k a_k:a_1,\ldots,a_n\text{ are distinct elements of }A}$, we have $|S(A)|\ge\min{p(F),(|A|-n)n(n+1)/2+n(n^2-1)/6+1}$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A field $F$ | F : Type, [Field F] in claim | The preregistered carrier is literal. | | Characteristic zero gives infinity; prime characteristic gives that prime | pF | ringChar, zero test and ENat top/cast. | | Positive $n$ | n : Nat, 1 ≤ n | Positive integers represented by positive naturals. | | $p(F)>n+1$ | Characteristic hypothesis in claim | Strict comparison in extended naturals. | | Finite $A$ | A : Finset F, [DecidableEq F] | Decidable equality is a classical representation requirement; it excludes no classical field. | | $|A|\ge n+\delta_{n,3}$ | Size hypothesis in claim | The conditional is exactly 1 at $n=3$ and 0 otherwise. | | Distinct selected elements; weights $1,\ldots,n$ | weightedDistinctSums | Embeddings from Fin n into $A$, weighted by the natural value plus 1. | | Numerical expression and minimum with $p(F)$ | sunBound and the conclusion of claim | Natural subtraction is exact under the size hypothesis; both divided products are integral, so natural division preserves the source expression. | | Refutation | result : ¬ claim | The witness is $F=\mathbb Q$, $n=2$, $A={0,1,2}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Additive/SunWeightedDistinctSums.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021A mixed four-dimensional state has equal coherence roofsA rank-two state in four dimensions has equal coherence roofs, refuting the universal strict-gap claim.Refuted

The questionFei, Coherence of assistance and assisted maximally coherent states, Scientific Reports 11, 5935 (2021), arXiv:2103.08818v1, ask in the paragraph “Relation between the convex roof extension and the least concave majorant extension” (ZhaoMJ-AMC4.tex, line 303):

Why it mattersThe question asks whether changing the pure-state ensemble with assistance always creates a strict increase over the convex roof, for every mixed state and every function in the source's class.

The idea1. Admissible function and state. For this four-dimensional function, quadraticcoherence proves IsCoherenceFunction f, and quadraticnonzero gives a nonzero simplex value.

Our contributionA rank-two state in four dimensions has equal coherence roofs, refuting the universal strict-gap claim.

Scope & next questionA characterization of all states or all functions with equal roofs remains open here. - Consequences for dependent assertions.

Verification details
Release source recordRefuted in Lean

Open problem

A mixed four-dimensional state has equal coherence roofs

M.-J. Zhao, R. Pereira, T. Ma and S.-M. Fei, Coherence of assistance and assisted maximally coherent states, Scientific Reports 11, 5935 (2021), arXiv:2103.08818v1, ask in the paragraph “Relation between the convex roof extension and the least concave majorant extension” (ZhaoMJ-AMC4.tex, line 303):

The strict relation between the coherence of convex roof and the coherence of assistance, that is, whether $C_c(\rho)< C_a(\rho)$ holds for all mixed quantum states is an interesting topic.

The source's standing assumption is $f\in\mathfrak F\setminus{0}$ (lines 189–190): $f$ is real, symmetric and concave on the probability simplex and vanishes at its vertices. The pure-state coherence is $C_f(|\psi\rangle)=f(\mu(|\psi\rangle))$, where $\mu(|\psi\rangle)_i=|\psi_i|^2$ (lines 207–211).

In the formal reading, for every natural dimension $n$, IsCoherenceFunction f requires concavity on $\Omega_n={p\in\mathbb R^n:p_i\ge0,\ \sum_i p_i=1}$, invariance under every coordinate permutation on $\Omega_n$, and $f(e_i)=0$ for every vertex. Nonzeroness means $\exists p\in\Omega_n,\ f(p)\ne0$.

A finite pure-state decomposition has nonnegative real weights $p_j$ summing to one, unit vectors $\psi_j\in\mathbb C^n$ for every index, and $\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|$. Let $D_f(\rho)$ be the set of all averages $\sum_jp_jf(\mu(\psi_j))$ over these decompositions. The module defines $C_c(\rho)=\inf D_f(\rho)$ and $C_a(\rho)=\sup D_f(\rho)$, using real sInf and sSup; the source uses minimum and maximum (lines 212–216 and 232–236). IsMixed ρ means $\rho\succeq0$, $\operatorname{tr}\rho=1$ and $\operatorname{tr}(\rho^2)\ne1$.

Issue #22770 fixes the claim: for every $n$, every such nonzero $f$, and every mixed $\rho$, $C_c(\rho)<C_a(\rho)$. This universal statement is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CoherenceAssistanceRoofEquality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2020A max multiflow need not reproduce the Ingleton entropy combinationA capacitated star separates the Ingleton combination of minimum cuts from that of a max multiflow.Refuted

The questionNing Bao and Jonathan Harper, Bit threads on hypergraphs (2020), arXiv:2012.07872v1, Conjecture 1, Section 3.2, BTHyper.tex lines 419–424:

Why it mattersBao and Harper seek a bit-thread interpretation of entropy inequalities on hypergraphs.

The idea4. Contradiction. starsymmetric, starloopless and starcapacitynonnegative verify all capacity hypotheses for this exact star.

Our contributionA capacitated star separates the Ingleton combination of minimum cuts from that of a max multiflow.

Verification details
Release source recordRefuted in Lean

Open problem

A max multiflow need not reproduce the Ingleton entropy combination

Ning Bao and Jonathan Harper, Bit threads on hypergraphs (2020), arXiv:2012.07872v1, Conjecture 1, Section 3.2, BT_Hyper.tex lines 419–424:

For a given geometry consider a linear combination of entropies $R(S)$. If $R(V)\geq0$ for all multiflows and there exists a multiflow such that $R(V)\neq0$, then for all max multiflows

$R(S)=R(V).$

The source is the TeX member BT_Hyper.tex from the v1 source archive, md5 e001e89a911f72e1614de54e82230780. Lines 409–416 define the same linear combination on entropies and subflows; the footnote at line 417 expressly includes ordinary 2-graphs among the geometries. The definitions and source locators are collected in the literature note.

Issue #22654 fixes the finite-graph reading. Let $X$ be any finite vertex set, $k$ any natural number, $b:\mathrm{Fin},k\hookrightarrow X$ an injective boundary labelling, and $c:X\times X\to\mathbb R$ a symmetric capacity function with $c(u,u)=0$ and $c(u,v)\ge0$. For each boundary region $I$:

  • $S_I$ is the minimum of $\sum_{u\in W,,v\notin W}c(u,v)$ over cuts $W\subseteq X$ containing precisely the boundary vertices labelled by $I$.
  • A multiflow consists of real pair flows $f_{ij}$ and thread numbers $N_{ij}$, using $i<j$ once for each unordered pair. Each such flow is antisymmetric on vertex pairs, has zero divergence away from $b(i),b(j)$, and has divergences $N_{ij}\ge0$ and $-N_{ij}$ at these two endpoints. On every ordered vertex pair, $\sum_{i<j}|f_{ij}(u,v)|\le c(u,v)$.
  • $V_I$ sums $N_{ij}$ over unordered pairs with exactly one endpoint in $I$; a max multiflow is a multiflow satisfying $V_{{i}}=S_{{i}}$ for every boundary index $i$.
  • For arbitrary real coefficients $\alpha_I$, write $R(x)=\sum_I\alpha_Ix_I$.

The assertion claim is that for every such choice of $X,k,c,b,\alpha$, if $R(V)\ge0$ for every multiflow and $R(V)\ne0$ for some multiflow, then $R(S)=R(V)$ for every max multiflow. The result is its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/BitThreadIngletonMultiflowRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2017Refutation of the normalized Bessel Redheffer lower boundCounterexamples to the normalized Bessel Redheffer lower bound.Refuted

The questionMehrez, Redheffer type bounds for Bessel and modified Bessel functions of the first kind, arXiv:1701.08446v1, §3, Conjecture 1, inequality (3.2):

Why it mattersThe proposed exponent makes the quadratic terms at the origin equal.

The ideaThe normalized series is entire, equals one at zero, and satisfies

Our contributionCounterexamples to the normalized Bessel Redheffer lower bound.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the normalized Bessel Redheffer lower bound

Á. Baricz and K. Mehrez, Redheffer type bounds for Bessel and modified Bessel functions of the first kind, arXiv:1701.08446v1, §3, Conjecture 1, inequality (3.2):

If ν > −1 and |x| < j_{ν,1}, then the following inequalities are valid (j_{ν,1}²/(j_{ν,1}² − x²))^{ϑ_ν} ≤ 𝒥_{ν+1}(x)/𝒥_ν(x) ≤ (j_{ν,1}²/(j_{ν,1}² − x²))^{η_ν}, where ϑ_ν = j_{ν,1}²/(4(ν+1)(ν+2)) and η_ν = 1 are the best possible constants.

The source states the left-hand inequality as the conjecture; the right-hand inequality is a known result. The source proves the lower bound in the case $\nu=-1/2$. Only the lower bound is refuted here.

Formal reading

All Lean names below have prefix D5.S3.Analytic.BesselRedheffer.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $\mathcal J_\nu(x)$ | Defs.nbessel | The real sum $\sum_{k\ge0}(-x^2/4)^k/[k!(\nu+1)k]$ with ascending Pochhammer evaluations. | | $j{\nu,1}$ | Defs.firstZero | The infimum of ${x>0:\mathcal J_\nu(x)=0}$; existence, positivity and attainment are proved for $\nu\ge30$. | | $\vartheta_\nu$ | Exponent in Defs.claim | Exactly $j_{\nu,1}^2/[4(\nu+1)(\nu+2)]$, using the real power. | | Left-hand inequality of (3.2) | Defs.claim | Every real $\nu>-1$ and every real $x$ with $|x|<j_{\nu,1}$. | | Strict violations | Result.failure_family | For every real $\nu\ge30$, some $0<x<j_{\nu,1}$ has the ratio strictly below the proposed lower bound. | | Universal refutation | Result.result | Exactly $\neg,\texttt{Defs.claim}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BesselRedheffer/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2020The Caro–Graswald maximal index is $2d-5$ for $d\ge5$, and $2$ for $d=3$The greatest admissible index is 2d minus 5 for dimensions at least five, and two in dimension three.Proved

The questionGraswald, Necessary criteria for Markovian divisibility of linear maps, Journal of Mathematical Physics 62, 042203 (2021), arXiv:2009.06666v3, Conjecture 4.19 (CnjAMGMMatrices), main.tex, lines 729–737.

Why it mattersThe remark at lines 719–740 reduces a possible increase in the number of factors to the probability-vector conjecture.

The ideaAll dimensions and indices below are natural numbers.

Our contributionThe greatest admissible index is 2d minus 5 for dimensions at least five, and two in dimension three.

Verification details
Release source recordProved in Lean

Open problem

The Caro–Graswald maximal index is $2d-5$ for $d\ge5$, and $2$ for $d=3$

M. C. Caro and B. R. Graswald, Necessary criteria for Markovian divisibility of linear maps, Journal of Mathematical Physics 62, 042203 (2021), arXiv:2009.06666v3, Conjecture 4.19 (CnjAMGMMatrices), main.tex, lines 729–737. The following quotation is checked against the v3 TeX; its macro \IR is expanded as defined in commands.tex:

Let $p\in\mathbb{R}^d_{\geq 0}$ with $\sum\limits_{i=1}^d p_i = 1$. Define the matrices $a,g\in\mathbb{R}^{d\times d}$ via

$a_{ij} = \frac{p_i + p_j}{2},\quad g_{ij}=\sqrt{p_i p_j}.$

Denote by $a_k^\downarrow$ and $g_k^\downarrow$ the $k^{th}$ largest entry of $a$ and $g$, respectively. Define

$A = \sum\limits_{k=1}^{h(d)} a_k^\downarrow,\quad G = \sum\limits_{k=1}^{h(d)} g_k^\downarrow.$

We conjecture that the maximal integer $h(d)$ such that $A+G\leq d$ holds for any probability vector $p$ is given by $h(d) = 2d-5$.

For a natural index $m$, write $A_m(p)$ and $G_m(p)$ for the sums of the largest $\min(m,d^2)$ entries of the two matrices, sorted independently, counting both ordered positions and retaining equal values at distinct positions. Let $S_d={m\in\mathbb{N}:\forall p\in\mathbb{R}^d_{\ge0},\ \sum_i p_i=1\Rightarrow A_m(p)+G_m(p)\le d}$. The module's admissibleIndices d is $S_d$.

Issue #22748 fixes the settlement as two greatest-element assertions: $\max S_d=2d-5$ for every $d\ge5$, and $\max S_3=2$. Thus the conjectured formula is proved in the specified range, while the three-dimensional value differs from the printed expression $2\cdot3-5=1$. The source states no dimension restriction; the settlement does not assert its formula in every dimension.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/MarkovianDivisibilityMaximalIndex.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2015Weak Abelian Periodicity Does Not Force Finitely Many Weakly Abelian Unbordered FactorsAn explicit ternary word gives a negative answer to Question 2 of Charlier, Harju, Puzynina, and Zamboni.Refuted

The questionÉmilie Charlier, Tero Harju, Svetlana Puzynina and Luca Q.

The ideaWith A = 110022222000111 and B = 012, the word w = B A^{e1} B A^{e2} … with unbounded exponents is a concatenation of blocks of length at most fifteen, all with letter frequencies (1/3, 1/3, 1/3), so it is bounded weakly abelian periodic.

Our contributionAn explicit ternary word gives a negative answer to Question 2 of Charlier, Harju, Puzynina, and Zamboni.

Verification details
Release source recordRefuted in Lean

Open problem

Weak Abelian Periodicity Does Not Force Finitely Many Weakly Abelian Unbordered Factors

Émilie Charlier, Tero Harju, Svetlana Puzynina and Luca Q. Zamboni, Abelian bordered factors and periodicity, arXiv:1501.07464v1, Section 7, Question 2: let w be an infinite bounded weakly abelian periodic word over a k-letter alphabet such that its graph G_w belongs to a k-dimensional cylinder with axis with rational coefficients, and each tangential line to G_w has points of G_w on it with bounded gaps. Does it follow that w has only finitely many weakly abelian unbordered factors?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/AbelianBorders/AbelianBorderQuestion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsQuestion resolvedFirst preprint 2016Effective resistance is the optimal finite-graph moving particle constantLinear occupation observables force every moving-particle constant k to satisfy one at most k times the effective conductance.Proved

The questionChen, The moving particle lemma for the exclusion process on a weighted graph, Electron.

Why it mattersChen's Theorem 1 bounds the exclusion exchange cost by effective resistance times the exclusion Dirichlet energy.

Our contributionLinear occupation observables force every moving-particle constant k to satisfy one at most k times the effective conductance.

Scope & next questionOptimality on resistance spaces, a classification of all equality observables, and general equality conditions for the octopus inequality remain open in this settlement.

Verification details
Release source recordProved in Lean

Open problem

Effective resistance is the optimal finite-graph moving particle constant

J. P. Chen, The moving particle lemma for the exclusion process on a weighted graph, Electron. Commun. Probab. 22 (2017), paper 47, arXiv:1606.01577v4, §1, OctopusMoving_v3.tex, lines 177–179, states:

However, it is unclear in general when \eqref{Reffinf} becomes an equality.

That being said, we conjecture that Theorem \ref{thm:dirichletex} is optimal on any finite weighted graph, and more generally, on resistance spaces \cite{KigamiRF}, in light of the connection between the moving particle lemma and the spectral gap of an interacting particle system \cite{Morris, Jara, Quastel}.

These are verbatim TeX excerpts; the labels refer to equation (1.9) and Theorem 1. The definitions and source locators are also recorded in the literature note.

For a finite vertex set $V$, let $c:V\times V\to\mathbb R$ be nonnegative and symmetric, with zero weights representing absent edges. A configuration is $\zeta\in{0,1}^V$, $\zeta^{xy}$ exchanges its occupations at $x,y$, and $\nabla_{xy}f(\zeta)=f(\zeta^{xy})-f(\zeta)$. Write $\nu_\alpha$ for the product Bernoulli measure of constant density $\alpha$, and put

$\mathcal E^{\rm el}(h)=\tfrac12\sum_{u\in V}\sum_{w\in V}c_{uw}(h(u)-h(w))^2$,

$\mathcal E^{\rm EX}{\nu\alpha}(f)=\tfrac14\sum_{u\in V}\sum_{w\in V}c_{uw}\nu_\alpha[(\nabla_{uw}f)^2]$,

$C(x,y)=\inf{\mathcal E^{\rm el}(h):h:V\to\mathbb R,\ h(x)=1,\ h(y)=0}$.

The factors remove ordered-pair double counting and retain the source's additional Dirichlet-energy factor $\tfrac12$ for exclusion. For positive $C(x,y)$, the source's effective resistance is $R_{\rm eff}(x,y)=1/C(x,y)$.

Issue #24189 fixes the finite-graph reading: for every such $V,c$, every distinct $x,y\in V$, every $0<\alpha<1$ and every real $k$, if

$\tfrac12\nu_\alpha[(\nabla_{xy}f)^2]\le k\mathcal E^{\rm EX}{\nu\alpha}(f)$ for every $f:{0,1}^V\to\mathbb R$,

then $1\le kC(x,y)$. This is the proposition claim proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/ParticleProcesses/ExclusionMovingParticleOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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    Year in the primary source attribution: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2017Christoffel words need not attain the Lyndon-factor minimumA binary Lyndon word of length 506 has sixteen distinct Lyndon factors, fewer than every Christoffel word of that length.Refuted

The questionGlen, Simpson and Smyth, Counting Lyndon factors, Section 6, p. 9, Open Problem 1 (arXiv:1701.00928v1; DOI 10.37236/6915), ask:

Why it mattersThus optimizing within the Christoffel family does not solve the minimum problem over all Lyndon words.

The ideaThe proof uses the literal word and certificates, without assuming general morphism identities or the paper's continued-fraction count.

Our contributionA binary Lyndon word of length 506 has sixteen distinct Lyndon factors, fewer than every Christoffel word of that length.

Scope & next questionThe alphabet-reduction and first-counterexample questions remain open.

Verification details
Release source recordRefuted in Lean

Open problem

Christoffel words need not attain the Lyndon-factor minimum

Glen, Simpson and Smyth, Counting Lyndon factors, Section 6, p. 9, Open Problem 1 (arXiv:1701.00928v1; DOI 10.37236/6915), ask:

Is it true that the minimum number of distinct Lyndon factors over all Lyndon words of the same length is attained by at least one Christoffel word of that length?

The resolution is Refuted, under the literal finite-word conventions of #18581.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Lyndon words over an ordered alphabet | Frozen LyndonOrder.IsLyndon, applied to List.ofFn w | Nonempty and strictly least among nontrivial rotations; the imported suffix characterization is used in the proof. | | A factor is a contiguous subword | factor w i l | Drops the first $i$ letters and takes at most $l$; valid nonempty occurrences satisfy $1\le l$ and $i+l\le n$. | | Number of distinct Lyndon factors | distinctLyndon w | Cardinality after taking the image of valid occurrences; equal factor lists count once, including letters and the whole word. | | At every common length | claim, quantified over $n\ge2$ | All lengths remain in the claim; $n=506$ refutes it. | | At least one Christoffel word | A numerator $0<p<n$ with $\gcd(p,n)=1$ and christoffel p n | Lower mechanical word, letter $1$ exactly when $\lfloor(i+1)p/n\rfloor-\lfloor ip/n\rfloor=1$. | | Minimum over all Lyndon competitors | Every $k$ and every $w:\operatorname{Fin}(n)\to\operatorname{Fin}(k)$ satisfying IsLyndon | Every finite ordered alphabet; the counterexample already has $k=2$. | | Negative resolution | result proves the negation of claim | No restricted replacement of the affirmative claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Complexity/LyndonFactorChristoffelMinimumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2016Coefficients of the shifted Gamma product TuranianA family of negative first Turanian coefficients.Refuted

The questionKarp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, Conjecture 1:

Why it mattersThe coefficient conclusion is one of the conclusions asserted by the conjecture.

The ideaThe half-integer Gamma identities supply the displayed values.

Our contributionA family of negative first Turanian coefficients.

Verification details
Release source recordRefuted in Lean

Open problem

Coefficients of the shifted Gamma product Turanian

S. I. Kalmykov and D. B. Karp, Log-concavity and Turán-type inequalities for the generalized hypergeometric function, arXiv:1611.06724v1, Section 3, Conjecture 1:

All conclusions of Theorem 3 hold for all α, β ≥ 0 if µ ≥ 1.

For positive parameter vectors of lengths $p$ and $q$, put

$$ c(t)=\frac{\prod_{i=1}^p\Gamma(a_i+t)}{\prod_{j=1}^q\Gamma(b_j+t)}, \qquad f(\mu;x)=\sum_{n=0}^{\infty}c(\mu+n)\frac{x^n}{n!}. $$

Equation (9) defines

$$ \Delta_f(\alpha,\beta;x)= f(\mu+\alpha;x)f(\mu+\beta;x) -f(\mu;x)f(\mu+\alpha+\beta;x), $$

and the coefficient of degree $m$ is

$$ \delta_m=\sum_{k=0}^m \frac{c(\mu+\alpha+k)c(\mu+\beta+m-k) -c(\mu+k)c(\mu+\alpha+\beta+m-k)}{k!(m-k)!}. $$

For $p\le q$, condition (13) consists of

$$ \frac{e_{q-j}(\mathbf b)}{e_{p-j}(\mathbf a)} \le\frac{e_{q-j-1}(\mathbf b)}{e_{p-j-1}(\mathbf a)} \qquad(0\le j<p), $$

where $e_0=1$. Empty parameter vectors are allowed; in particular, the condition is vacuous at $p=0$. Defs.claim retains all parameter vectors and all coefficient degrees with $\mu\ge1$, $\alpha,\beta\ge0$ and $\alpha\le\beta+1$.

Formal correspondence. All Lean names below have prefix D5.S3.Analytic.HypergeometricShiftTuran.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Equation (10) coefficient ratio | Defs.coeff | Product of Gamma values for numerator parameters divided by the product for denominator parameters. | | Equation (9) coefficient | Defs.turanCoeff | Exact finite Cauchy sum for the degree-$m$ coefficient of the shifted Turanian. | | Condition (13) | Defs.esymm, Defs.cond13 | Elementary symmetric functions and their consecutive ratio chain. | | Conjecture 1 coefficient clause | Defs.claim | Universal nonnegativity for the printed parameter and shift range. | | Negative coefficient family | Result.half_parameter_family_neg | For $p=0$, repeated $b_j=1/2$, $\mu=1$, $\alpha=\beta=1/2$, every $q\ge6$ gives $\delta_1<0$. | | Refutation | Result.result | Exactly $\neg,\texttt{Defs.claim}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/HypergeometricShiftTuran/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

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    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2016The two Karlsson families are the only affine suborbits of BC_9^(2)Every nonconstant affine dephasing of the order-nine BCCB sum-one locus lies in the cancelling-pair or equilateral-triple family.Proved

The questionKarlsson, BCCB complex Hadamard matrices of order 9, and MUBs, arXiv:1604.03770v1, Linear Algebra Appl. 504 (2016), Conjecture 3:

Why it mattersThe paper constructs BC9^{(2)}, a two-parameter non-affine orbit of order-9 complex Hadamard matrices that does not lie in the Fourier orbit F9^{(4)}.

The idea1. Bridge. The dephased entries at (1,2), (3,6), (4,8) and (5,7) (zero-based) are 1/x, 1/y, 1/u and 1/w.

Our contributionEvery nonconstant affine dephasing of the order-nine BCCB sum-one locus lies in the cancelling-pair or equilateral-triple family.

Verification details
Release source recordProved in Lean

Open problem

The two Karlsson families are the only affine suborbits of BC_9^(2)

B. R. Karlsson, BCCB complex Hadamard matrices of order 9, and MUBs, arXiv:1604.03770v1, Linear Algebra Appl. 504 (2016), Conjecture 3:

$BC_{9A}^{(1)}$ and $BC{}{9B}^{(1)}$ are the only affine suborbits of $BC{9}^{(2)}$.

The definition of an affine orbit, the matrix, the unitarity locus x + y + u + w = 1 and the two families are quoted in the literature note.

Issue #19395 fixes the reading. Consider:

  • a nonconstant family of four unit parameters on a nonempty open subset of ℝ^p with sum 1;
  • whose canonically dephased matrix has entries of the form c e^{iL(t)} with L linear.

Up to a fixed permutation of the parameters, such a family is (μ, −μ, −ω, −ω²) or (1, ξ, ωξ, ω²ξ). The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Tomography/MUB/BCCBOrderNineAffineSuborbits.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is proved.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2021Biased post-processed interferometric improvements are not faithfulEvery biased prior admits a coherence-creating qubit channel with zero post-processed improvement.Proved

The questionPlenio, Coherence of operations and interferometry, Phys.

Why it mattersThe discrimination game tests whether a channel's ability to create coherence improves the success probability beyond guessing from the prior alone.

The ideaA coordinate projector attains this value, and the trace bound bounds every value on the unit ball; the proof uses IsGreatest.csSupeq.

Our contributionEvery biased prior admits a coherence-creating qubit channel with zero post-processed improvement.

Scope & next questionThe other half, a general method to compute post-processed improvements for arbitrary channels, remains open in this settlement.

Verification details
Release source recordProved in Lean

Open problem

Biased post-processed interferometric improvements are not faithful

M. Masini, T. Theurer and M. B. Plenio, Coherence of operations and interferometry, Phys. Rev. A 103, 042426 (2021), arXiv:2102.04863v2, Section IV (Conclusions), PaperMasiniTheurerPlenio.tex, line 300:

Whilst we provided a method to compute the pre-processed improvements and showed that they are not faithful for $\lambda\ne\mu$, it is an open question whether there exist analogous results for the post-processed cases.

The source defines faithfulness in Section II A, line 153:

$M(\Theta)=0$ exactly if $\Theta$ is a free channel.

Here $\mu=1-\lambda$, $\Delta$ removes off-diagonal entries in the chosen incoherent basis, and the free channels for coherence creation are the maximally incoherent operations, characterized by $\Psi\Delta=\Delta\Psi\Delta$ (Section II A, lines 132–134). The phase channel is conjugation by $U=\operatorname{diag}(e^{i\phi_k})$ (Section II B, lines 171–175). The post-processed improvement is defined in Section III B, Eq. (14), line 273:

$N_{\lambda,\vec{\phi}}(\Theta):=\max_{\Psi\in\mathcal{MIO}}\left|\left|\left(\lambda-\mu \Lambda_{\vec{\phi}}\right) \Psi \Theta\Delta\right|\right|_1-|\lambda-\mu |$.

The source calls the operation norm the induced trace norm (Section III A, line 202). The formal reading uses $|\Phi|{1\to1}=\sup{|A|_1\le1}|\Phi(A)|_1$ over all complex matrices, with no positivity or Hermiticity restriction on $A$.

Issue #22607 fixes the faithfulness half of the question: for every $\lambda\in[0,1]$ with $\lambda\ne\tfrac12$, every natural $M\ge1$, and every $\vec\phi\in\mathbb R^M$, there exists a qubit-to-qubit channel $\Theta$ that is not maximally incoherent and has $N_{\lambda,\vec\phi}(\Theta)=0$, optimizing over all MIO channels from the qubit to dimension $M$. This is proved.

The definitions and source locators are collected in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/PostprocessedImprovementUnfaithful.result

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2014A cooling-map transition has no finite ground-space thermal realizationA four-level cooling-map transition cannot be realized by any finite ancilla in its normalized ground-space state and any energy-conserving unitary.Refuted

The questionGour, Low-temperature thermodynamics with quantum coherence, Nature Communications 6, 7689 (2015), arXiv:1409.7740v3, Supplementary subsection “Cooling maps and thermal operations”, Conjecture 1 (LTTarXiv.tex, lines 924–927):

Why it mattersNarasimhachar and Gour (2015), arXiv:1409.7740v3, give a state-transition criterion for cooling maps, while thermal operations impose an operational requirement involving one energy-conserving unitary and a thermal ancilla.

The ideaThe input is the existing matrix NiceErrorBasisNonNormalStabilizer.Pm.

Our contributionA four-level cooling-map transition cannot be realized by any finite ancilla in its normalized ground-space state and any energy-conserving unitary.

Verification details
Release source recordRefuted in Lean

Open problem

A cooling-map transition has no finite ground-space thermal realization

V. Narasimhachar and G. Gour, Low-temperature thermodynamics with quantum coherence, Nature Communications 6, 7689 (2015), arXiv:1409.7740v3, Supplementary subsection “Cooling maps and thermal operations”, Conjecture 1 (LTT_arXiv.tex, lines 924–927):

Cooling maps are equivalent to low-temperature thermal operations, with regard to the feasibility of state transitions.

The next sentence distinguishes state-transition feasibility from equality of channel sets:

Note that this could be true even if the set of cooling maps is strictly larger than that of thermal operations---there could still be a thermal operation achieving every state transition that is possible through cooling maps.

Issue #26067 fixes the following reading of the cooling-to-thermal direction. For every finite dimension $d$, every strictly increasing real energy spectrum $E$ with distinct positive gaps, and every pair of density matrices $\rho,\sigma$, a cooling-map transition $\rho\mapsto\sigma$ has a low-temperature thermal realization. Here:

  • Density matrices are positive semidefinite complex matrices of trace one (IsDensity).
  • A cooling map has a finite Kraus family, each member diagonal or of the form $c|j\rangle\langle k|$ with $j<k$, satisfying $\sum_a K_a^\dagger K_a=I$ and $\sum_a K_a\rho K_a^\dagger=\sigma$ (IsCoolingKraus, IsCoolingMap). This is the source's definition in LTT_arXiv.tex, lines 693–700.
  • A thermal realization uses an arbitrary positive finite ancilla dimension $m$, arbitrary real ancilla energies $e$, and a unitary $U$ commuting with the diagonal total Hamiltonian $H_S\otimes I+I\otimes\operatorname{diag}(e)$ (IsEnergyConserving). If $G={a:e_a\le e_b\text{ for every }b}$ and $g=|G|$, its output is $\operatorname{Tr}_A[U(\rho\otimes P_G/g)U^\dagger]$ (groundSet, groundState, thermalOutput). The full ground eigenspace, including every degeneracy, is retained.

The source writes the bath state as a low-temperature approximation to the normalized ground-space projector (lines 137–141 and 651–654). The formal assertion uses that projector exactly, as stipulated in #26067; it makes no assertion about an error tolerance or a positive-temperature Gibbs state. The resulting universal assertion is claim, and the module proves result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/CoolingMapThermalOperationRefutation.result

Frozen module record

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2024The normalized nonlocality–contextuality trade-off holds for every binary n-cycleNormalized Bell and contextuality inequalities satisfy the trade-off for every binary n-cycle.Proved

The questionKurzyński, “Trade-off relations between Bell nonlocality and local Kochen–Specker contextuality in generalized Bell scenarios”, New Journal of Physics 26, 083028 (2024), arXiv:2405.02964v2, states the following conjecture in finalversion.tex, lines 373–375:

Why it mattersAn all-cycle result identifies the geometric reason that a contextuality violation consumes the same unit budget that otherwise permits a generalized Bell violation.

The ideaThe proof uses convexity.

Our contributionNormalized Bell and contextuality inequalities satisfy the trade-off for every binary n-cycle.

Scope & next questionTheir instantiations for the formal binary cycle model remain open here. - Whether an analogous trade-off holds with more than two Alice measurements or for other compatibility structures is open.

Verification details
Release source recordProved in Lean

Open problem

The normalized nonlocality–contextuality trade-off holds for every binary n-cycle

L. E. A. Porto, G. Ruffolo, R. Rabelo, M. Terra Cunha and P. Kurzyński, “Trade-off relations between Bell nonlocality and local Kochen–Specker contextuality in generalized Bell scenarios”, New Journal of Physics 26, 083028 (2024), arXiv:2405.02964v2, states the following conjecture in final_version.tex, lines 373–375:

Consider a bipartite generalized Bell scenario where Bob has an $n$-cycle contextuality setup. For any normalized generalized-Bell inequality $S_{GBell}$ and any normalized noncontextuality inequality $S_{NC}$ of Bob's local experiment, every $\mathcal{NSND}$ behavior satisfies expression (12).

The compiled formalization fixes Behavior n to Fin 2 → Fin n → Bool → Bool → Bool → ℝ, BobBehavior n to Fin n → Bool → Bool → ℝ, and AliceResponse to Fin 2 → Bool → ℝ. IsNSND p imposes nonnegativity, normalization in every Alice setting and cyclic Bob context, equality of the shared Alice–Bob marginal in consecutive contexts, independence of Bob's marginal from Alice's setting, and independence of Alice's marginal from Bob's context. IsGeneralizedLocal is the real convex hull of products of a row-stochastic Alice response with an IsND Bob behavior, and IsNoncontextual is the real convex hull of deterministic global Boolean assignments. The formal claim is: for every $n$ with $3 \le n$, every $f$ of type ((Behavior n →ₗ[ℝ] ℝ) × ℝ) and every $t$ of type ((BobBehavior n →ₗ[ℝ] ℝ) × ℝ) satisfying IsNormalizedGBell f and IsNormalizedNC t, and every $p$ satisfying IsNSND p, $$ f.1(p)+f.2+\bigl(t.1(\operatorname{bobMarginal}(p))+t.2\bigr)\le 1. $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumContext/NCycleNonlocalityContextualityTradeoff.result

Frozen module record

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Quantum informationQuestion resolvedFirst preprint 2023First-level SDP inexactness on the weighted-crown intervalThe first-level SDP is inexact throughout the open weighted-crown interval.Proved

Why it mattersThe formal carrier uses hubs none and some none, leaves some (some i), two-element pair subsets, the real moment matrix, and the literal qubit singlet Hamiltonian.

The ideaThe proof uses spectral estimates.

Our contributionThe first-level SDP is inexact throughout the open weighted-crown interval.

Verification details
Release source recordProved in Lean

Open problem

First-level SDP inexactness on the weighted-crown interval

J. Takahashi, C. Rayudu, C. Zhou, R. King, K. Thompson and O. Parekh, An SU(2)-symmetric Semidefinite Programming Hierarchy for Quantum Max Cut, arXiv:2307.15688v3, Appendix C.1, leaves the interval $(n+2)^2/(4(n+1))<x<(n+2)/3$ open for $H=xh_{ab}+\sum_{j=1}^{n}(h_{aj}+h_{bj})$, with $n\geq3$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $n\geq3$ | claim binder 3 ≤ n | Exact | | $(n+2)^2/(4(n+1))<x$ | claim binder L n < x | Exact | | $x<(n+2)/3$ | claim binder x < U n | Exact | | Weighted crown Hamiltonian | crownH n x | Exact literal singlet sum | | First-level SDP value | npa1Value n x | Exact real moment supremum | | Strict inexactness | result : claim proves lambdaMax (crownH n x) < npa1Value n x | Exact |

The definitions of $\mathcal{Proj}(V,w)$ and $NPA_1$ and the real level-one equality are recorded in the cited Library note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Optimization/WeightedCrownFirstLevelGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024The $\mathrm F_3$ steering inequality has the sharp ellipsoid-centre threshold two thirdsThe three-setting steering value is at most max(sqrt(3(1 - c)), 1), where c is Bob's steering-ellipsoid centre length.Proved

The questionSkrzypczyk, Locally inaccessible hidden quantum correlations, Phys.

Why it mattersThe source uses a centre-only impossibility region to detect hidden correlations that cannot be accessed by filtering on just one side.

The ideaThe lemma names below belong to SteeringEllipsoidFThreeThreshold unless an owner is specified.

Our contributionThe three-setting steering value is at most max(sqrt(3(1 - c)), 1), where c is Bob's steering-ellipsoid centre length.

Verification details
Release source recordProved in Lean

Open problem

The $\mathrm F_3$ steering inequality has the sharp ellipsoid-centre threshold two thirds

A. F. Ducuara, C. E. Susa and P. Skrzypczyk, Locally inaccessible hidden quantum correlations, Phys. Rev. A 110, 022435 (2024), arXiv:2406.03945v1, state the QSE conjecture in QC_LI.tex, lines 282–284:

For any two-qubit state $\rho \in D(\mathds{C}^2 \otimes \mathds{C}^2)$, its maximal violation of the $\rm CHSH$-inequality \eqref{eq:HC} and the $\rm F_3$-inequality \eqref{eq:CAC} are upper bounded by functions that depend on the magnitude of the QSE's centre as ${\rm B}(\rho) \leq f_{\rm CHSH}(c_B)$ and ${\rm F_3}(\rho)\leq f_{\rm F_3}(c_B)$, respectively. With $c_B=|\bm{c}_B|$.

Line 288 asks:

Exploring the form of function $f_{\rm F_3}(c_B)$ is an interesting task which we leave for future research.

The same line describes the numerical thresholds:

In particular, we highlight that there seem to exist values $c_{\rm , CHSH}=0.5$ and $c_{\rm , F_3} = 0.66$, such that states with QSE centres' magnitudes greater than these values, can no longer lead to a violation of the $\rm CHSH$-inequality and the $\rm F_3$-inequality, respectively.

For a positive semidefinite trace-one complex two-qubit matrix $\rho$, write $\boldsymbol a$, $\boldsymbol b$ for its Bloch vectors and $T_{ij}=\operatorname{Re}\operatorname{tr}[\rho(\sigma_i\otimes\sigma_j)]$. The module defines Bob's centre by

$\boldsymbol c_B=(\boldsymbol b-T^{\mathsf T}\boldsymbol a)/(1-|\boldsymbol a|^2)$ when $|\boldsymbol a|\ne1$, and $\boldsymbol c_B=\boldsymbol b$ when $|\boldsymbol a|=1$; put $c_B=|\boldsymbol c_B|$.

Its $\mathrm F_3(\rho)$ is the real supremum of

$\displaystyle \frac1{\sqrt3}\left|\sum_{i=1}^3\operatorname{tr}[\rho,((\boldsymbol\alpha^i\cdot\boldsymbol\sigma)\otimes(\boldsymbol\beta^i\cdot\boldsymbol\sigma))]\right|$

over real unit Alice directions $\boldsymbol\alpha^i$ and real orthonormal Bob frames $\boldsymbol\beta^i$. The absolute value here is the complex norm. The source's Eq. (CAC), lines 209–214, gives its optimized value as $\sqrt{s_1^2+s_2^2+s_3^2}$, where the $s_i$ are the singular values of $T$; the relationship to the frame formulation is discussed below. The source definitions and locator are in the literature note.

Issue #26164 fixes the settlement as the conjunction of

$\displaystyle \mathrm F_3(\rho)\le\max{\sqrt{3(1-c_B)},1}$

for every positive semidefinite trace-one two-qubit matrix, and the existence, for every real $0\le t<2/3$, of such a matrix with $c_B=t$ and $\mathrm F_3(\rho)>1$. This conjunction is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/SteeringEllipsoidFThreeThreshold.result

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Quantum informationQuestion resolvedFirst preprint 2023Non-negativity of the photon-addition coefficientsFor every k at least 2, the coefficients cn^(kk) that compare k-photon addition with single-photon addition on a two-mode squeezed vacuum are non-negative.Proved

The questionA 110 (2024) 042430) compare a two-mode squeezed vacuum with k photons added to each mode against the single-photon-added state.

Why it mattersNon-negative cn^{(kk)} make D column stochastic, so q^{(kk)} is majorized by q^{(11)} and σ^{k,k} is at least as entangled as σ^{1,1} for every entanglement monotone.

The ideaThe expansion for all n is (1 + Σn cn x^{n+1}) N1 = Nk for the series Nk = Σ C(n+k,k)² xⁿ.

Our contributionFor every k at least 2, the coefficients cn^(kk) that compare k-photon addition with single-photon addition on a two-mode squeezed vacuum are non-negative.

Verification details
Release source recordProved in Lean

Open problem

Non-negativity of the photon-addition coefficients

Z. Van Herstraeten, N. J. Cerf, S. Guha and C. N. Gagatsos (arXiv:2312.02066, quant-ph; Phys. Rev. A 110 (2024) 042430) compare a two-mode squeezed vacuum with k photons added to each mode against the single-photon-added state. The Schmidt coefficients are q_n^{(kk)} = λⁿ C(n+k,k)² / N_kk, and the paper writes q^{(kk)} = D q^{(11)} with a lower-triangular Toeplitz matrix D whose entries are λN_11/N_kk · c_m^{(kk)} λ^m, where

\binom{n+k+1}{k}^2 = \sum_{i=0}^{n+1} c_{n-i}^{(kk)} (i+1)^2

and c_{-1}^{(kk)} = 1. The columns of D sum to 1. The appendix ends:

At this point, we conjecture the non-negativity of Eq. \eqref{eqapp:Coeffcients}, \textit{i.e.}, we conjecture that the matrix $\mathbf{D}$ of Eq. \eqref{eqapp:Dvector} is column stochastic for all $k\geq 2$, implying the majorization relation $\hat{\sigma}^{k,k} \prec \hat{\sigma}^{1,1}$.

Issue #11700 fixes the reading: c_n^{(kk)} (n ≥ 0) are the numbers with c_{-1}^{(kk)} = 1 that satisfy the expansion for every n ≥ 0, and the conjecture is c_n^{(kk)} ≥ 0 for all k ≥ 2 and n ≥ 0. The formal claim states that a solution exists and that every solution is non-negative; the result proves it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PhotonAddedMajorizationCoefficients.result

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Quantum informationCounterexample foundFirst preprint 2022Tight purification circuits need not have monotone output errorA tight purification circuit has strictly decreasing output error between two preparation error rates.Refuted

The questionTaylor, Post-selection-free preparation of high-quality physical qubits, Quantum 7, 994 (2023), arXiv:2209.05391v3, Section 3.2, PDF page 20 (purification.tex, lines 859–862), ask:

Why it mattersThe source uses purification circuits to improve preparation quality without post-selection.

The ideaIts eight input cases use weightgetwo: for any bitstring and two distinct indices whose bits are both true, its Hamming distance from the all-false string is at least two.

Our contributionA tight purification circuit has strictly decreasing output error between two preparation error rates.

Scope & next questionWhich noisy purification circuits satisfy this response condition remains open here. - A surviving special case.

Verification details
Release source recordRefuted in Lean

Open problem

Tight purification circuits need not have monotone output error

B. Barber, N. I. Gillespie and J. M. Taylor, Post-selection-free preparation of high-quality physical qubits, Quantum 7, 994 (2023), arXiv:2209.05391v3, Section 3.2, PDF page 20 (purification.tex, lines 859–862), ask:

We conjecture that it should be possible to prove monotonicity under some reasonable set of conditions. For instance, suppose that $\big|\binom n 0 + \cdots + \binom n e\big| = 2^{n-k}$ and $p_0 < 0.5$. Is $p_{\text{out}}(p_0,p_I,p_C,p_T)$ an increasing function of $p_0$, $p_I$, $p_C$ and $p_T$? We leave this as an open question.

Their $(n,k,e)$ purification circuit consists of CNOT and Toffoli gates and sends every input bitstring of Hamming weight at most $e$ to a bitstring zero in $k$ nominated output positions (Section 2, line 203). Each input bit independently starts at one with probability $p_0$. Each idle qubit depolarises independently with probability $p_I$, and each CNOT or Toffoli independently depolarises its entire support with probability $p_C$ or $p_T$. Depolarisation replaces that support by uniformly random bits (Section 2.1, lines 219–243). The quantity $p_{\text{out}}$ is the expected number of erroneous output bits divided by $k$ (lines 711–713).

Issue #22506 fixes the nondecreasing reading: for all natural $n,k,e$ with $1\le k\le n$ and $1\le e\le n$, every gate list executed one gate per round, and every set of exactly $k$ outputs, purification through weight $e$ together with $\sum_{j=0}^{e}\binom nj=2^{n-k}$ implies separate nondecreasing dependence on each noise coordinate. Both parameter vectors must lie in $0\le p_0<1/2$, $0\le p_I,p_C,p_T\le1$, and the other three coordinates are held fixed in each comparison. This is claim. Its refutation already occurs in the preparation coordinate. One gate per round is an allowed subclass of the source's scheduled circuits.

The definitions and source locators are also in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/PurificationCircuitMonotonicityRefutation.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2024At dimension five the full and non-symmetric aggregates of a two-qudit state fall below their conjectured lower boundsIn dimension five, the stochastic Lagrangian subspaces are exactly the stochastic orthogonal graphs.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionIn dimension five, the stochastic Lagrangian subspaces are exactly the stochastic orthogonal graphs.

Verification details
Release source recordRefuted in Lean

Open problem

At dimension five the full and non-symmetric aggregates of a two-qudit state fall below their conjectured lower bounds

H. Zhu, C. Mao and C. Yi, Third moments of qudit Clifford orbits and 3-designs based on magic orbits, arXiv:2410.13575v1, Section VII.1, Conjecture 2, asserts for an odd prime $d$ and $|\Psi\rangle\in\mathcal H_d^{\otimes n}$ the pointwise bounds $0\le\kappa(\Psi,\mathcal T)\le1$ on $\Sigma(d)$ together with the aggregate bounds $6\le\kappa(\Psi,\Sigma(d))\le2d+2$, $0\le\kappa(\Psi,\mathscr T_{\rm ns})\le2d-4$ and $\kappa(\Psi,\mathscr T_{\rm iso})\ge6$. Here $\Sigma(d)$ is the set of stochastic Lagrangian subspaces of $\mathbb F_d^6$, $\mathscr T_{\rm sym}={\mathcal T_O\mid O\in S_3}$, $\mathscr T_{\rm ns}=\Sigma(d)\setminus\mathscr T_{\rm sym}$ and $\kappa(\Psi,\mathscr T)=\sum_{\mathcal T\in\mathscr T}\kappa(\Psi,\mathcal T)$. The verbatim statements are in the literature note. The pointwise clause and the clause $\kappa(\Psi,\mathscr T_{\rm iso})\ge6$ are refuted (pointwise, isotropic aggregate). Issue #14620 targets the two remaining lower bounds, $6\le\kappa(\Psi,\Sigma(d))$ and $0\le\kappa(\Psi,\mathscr T_{\rm ns})$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/CliffordThirdMomentSigmaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024A CSS-T doubling map with better distance than the identityA binary cyclic map gives an admissible CSS-T extension with distance at least five, while identity doubling has distance at most four.Proved

The questionSanz Larrarte, Asymptotically good CSS-T codes and a new construction of triorthogonal codes, arXiv:2412.08586v2, Remark 1:

Why it mattersThe source proves that the identity map turns every [[n, k, d]] CSS code into a [[2n, k, ≥ d]] CSS-T code, which supports a transversal T gate, and uses this to obtain asymptotically good CSS-T codes.

The ideaLet A be the 13 × 13 circulant with (Ax)i = xi + x{i+1} + x{i+2} (indices mod 13).

Our contributionA binary cyclic map gives an admissible CSS-T extension with distance at least five, while identity doubling has distance at most four.

Scope & next questionDoubled inputs make the parity condition automatic; whether non-doubled admissible maps give further gains remains open.

Verification details
Release source recordProved in Lean

Open problem

A CSS-T doubling map with better distance than the identity

E. Berardini, R. Dastbasteh, J. Etxezarreta Martinez, S. Jain and O. Sanz Larrarte, Asymptotically good CSS-T codes and a new construction of triorthogonal codes, arXiv:2412.08586v2, Remark 1:

Finding maps $\phi$ giving CSS-T codes with better parameters than the choice $\phi=I$ or a permutation is an interesting open question.

For a binary CSS code (C₁, C₂) and a linear map φ on C₁ satisfying the triple-product parity condition of the source's Proposition 1, the CSS-T code (C₁^{Nφ}, C₂^{Nφ}) has C^{Nφ} = {(x, φ(x)) : x ∈ C}. Every such φ gives length 2n and dimension k = dim C₁ − dim C₂. The definitions are quoted in the literature note.

Issue #19814 fixes the reading. The question asks for some CSS code and some admissible φ for which the φ-extension has strictly larger CSS distance than the identity extension. Such a pair is constructed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Codes/CssTDoublingMapImprovement.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Analysis & dynamicsQuestion resolvedFirst preprint 2023Polynomial content has no maximizer on suitable unit-area polygonsPolynomial content has no maximizer on suitable unit-area polygons.Proved

The questionSimanek, New Perspectives on Torsional Rigidity and Polynomial Approximations of z-bar, arXiv:2309.16450v1, Conjecture 2.1, p. 3:

Why it mattersAn all-degree result requires a polygon family whose symmetry cancels every polynomial correction of the specified degree while its second moment grows without bound.

The ideaPlanarFanPolynomialApproximation.polarfanconiccoverage supplies directional coverage; adjacenttrianglesspokeinterior and finitecyclicfanfrontier identify the interior and frontier.

Our contributionPolynomial content has no maximizer on suitable unit-area polygons.

Verification details
Release source recordProved in Lean

Open problem

Polynomial content has no maximizer on suitable unit-area polygons

A. Kraus and B. Simanek, New Perspectives on Torsional Rigidity and Polynomial Approximations of z-bar, arXiv:2309.16450v1, Conjecture 2.1, p. 3:

For each N ∈ ℕ, there is an n ∈ ℕ so that among all n-gons with area 1, ρ_N has no maximizer.

The source characterizes $\rho_N(\Omega)$ as the squared distance from $\bar z$ to complex polynomials of degree at most $N$ in $L^2(\Omega,dA)$. The settlement is Proved by D5/S3/Analytic/Torsion/PolynomialContentNoMaximizer.result, with preregistration #16916.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | For each $N\in\mathbb N$ | claim quantifies over all natural degrees, including zero | Covers the source domain and zero. | | There is a polygon side count | The existential index in claim represents index plus three sides | Every polygon has at least three sides; the constructed count is $2\max(3,N+2)$. | | A simple polygonal Jordan region | IsSimplePolygon requires nonadjacent edge disjointness, adjacent-edge intersection exactly at the shared vertex, and consecutive noncollinearity; IsPolygonRegion requires a bounded connected open region with frontier equal to Mathlib’s Polygon.boundary ℝ of those vertices | The closed path $v_0\to v_1\to\cdots\to v_{n+2}\to v_0$ is the union of Polygon.edgeSet ℝ, identified with vector-space segments by affineSegment_eq_segment and finRotate_apply. | | Area one | volume of the region equals one | Complex planar Lebesgue measure. | | Polynomial content | rhoN is the infimum of the squared integral errors over Polynomial.degreeLE | The source's squared-distance expression. | | Has no maximizer | claim negates the existence of an admissible polygon dominating all other admissible polygons | No additional assumption on the competing polygon. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Torsion/PolynomialContentNoMaximizer.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2022The sharp lower bound on the kurtosis of a unidirectional unicyclic currentThe kurtosis of every finite positive unidirectional cycle exceeds −(1 + √5)/10, and this constant is the greatest lower bound over all cycle sizes.Proved

The questionPtaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys.

Why it mattersThe source uses higher current cumulants to constrain transport models and diagnose departures from unicyclic dynamics.

The ideaThis identity has no positivity hypothesis and uses total real division.

Our contributionThe kurtosis of every finite positive unidirectional cycle exceeds −(1 + √5)/10, and this constant is the greatest lower bound over all cycle sizes.

Verification details
Release source recordProved in Lean

Open problem

The sharp lower bound on the kurtosis of a unidirectional unicyclic current

K. Ptaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys. Rev. E 106, 024119 (2022), arXiv:2204.08421v4, Appendix, main.tex line 792:

As discussed in Sec.~\ref{subsec:unmark}, it was numerically conjectured that the minimum and maximum values of $\mathcal{S}$, $\mathcal{K}$, $\mathcal{S}+\mathcal{K}$, $\mathcal{S}-\mathcal{K}$ and $\mathcal{S} \times \mathcal{K}$ are always obtained for a specific type of network topology, with all rates $k_{ij}$ equal to $k$ apart from one (here $k_{21}$) equal to $ak$, with the parameter $a$ depending on the optimized quantity and the number of states $N$.

Section III E 1, main.tex lines 424–428, gives the kurtosis lower endpoint $C=-(1+\sqrt5)/10$ in the limit $N\to\infty$ along rates $(A/N,1,\ldots,1)$, where $A=(3-\sqrt5)/2$. The asymptotic family supplies a candidate endpoint; the conjectured extremal topology leaves a universal bound over arbitrary positive rates to establish.

Issue #25526 fixes the reading. For each natural $n\ge2$ and strictly positive real rate vector $k:\mathrm{Fin},n\to\mathbb R$, put

$R_j=\sum_{i=0}^{n-1}k_i^{-j},\qquad \mathcal K(k)=\frac{15R_2^2}{R_1^4}-\frac{20R_3}{R_1^3}+\frac{6R_4}{R_1^2R_2},\qquad C=-\frac{1+\sqrt5}{10}.$

The claim is that $\mathcal K(k)>C$ for every such $n$ and $k$, and that $C$ is the greatest lower bound of the set of all these values, allowing $n$ to vary. This is proved. The endpoint is an infimum over state counts, with no attainment by a finite positive cycle of at least two states.

The source defines $\mathcal K$ as the fourth scaled current cumulant divided by the second (main.tex line 54). Substituting its current-cumulant formulas at lines 773–775 and waiting-time cumulants at lines 787–790 gives the displayed expression. This is the quantity encoded by kurtosis, rather than the standardized fourth central moment. See the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/UnicyclicKurtosisLowerBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2022The sharp lower bound on skewness plus kurtosis of a unidirectional unicyclic currentThe sum of skewness and kurtosis of every finite positive unidirectional cycle exceeds −8/27, the greatest lower bound over all cycle sizes.Proved

The questionPtaszyński, Bounds on skewness and kurtosis of steady state currents, Phys.

Why it mattersA sharp bound on the sum constrains the joint values of two current-cumulant ratios for the same rate vector.

The ideaFinally result : claim joins the strict bound and the greatest-lower-bound statement.

Our contributionThe sum of skewness and kurtosis of every finite positive unidirectional cycle exceeds −8/27, the greatest lower bound over all cycle sizes.

Verification details
Release source recordProved in Lean

Open problem

The sharp lower bound on skewness plus kurtosis of a unidirectional unicyclic current

K. Ptaszyński, Bounds on skewness and kurtosis of steady state currents, Phys. Rev. E 106, 024119 (2022), arXiv:2204.08421v4, Appendix, main.tex line 792:

As discussed in Sec.~\ref{subsec:unmark}, it was numerically conjectured that the minimum and maximum values of $\mathcal{S}$, $\mathcal{K}$, $\mathcal{S}+\mathcal{K}$, $\mathcal{S}-\mathcal{K}$ and $\mathcal{S} \times \mathcal{K}$ are always obtained for a specific type of network topology, with all rates $k_{ij}$ equal to $k$ apart from one (here $k_{21}$) equal to $ak$, with the parameter $a$ depending on the optimized quantity and the number of states $N$.

Section III E 1, main.tex line 424, specifies the limiting interpretation:

The other bounds have been derived in the assymptotic limit $N \rightarrow \infty$ in which the saturating value of $a$ scales as $a=A/N$.

The lower endpoint displayed at lines 434–437 is $\min(\mathcal S+\mathcal K)=-8/27$ for $A=1/2$.

The reading fixed in issue #25636 concerns this lower endpoint for positive unidirectional cycles. For $n\ge2$ and $k:\mathrm{Fin},n\to\mathbb R$ with every $k_i>0$, put $R_j=\sum_i k_i^{-j}$ and

$$ \mathcal S(k)=\frac{3R_2^2}{R_1^4}-\frac{2R_3}{R_1^3},\qquad \mathcal K(k)=\frac{15R_2^2}{R_1^4}-\frac{20R_3}{R_1^3}+\frac{6R_4}{R_1^2R_2}. $$

The claim is that $\mathcal S(k)+\mathcal K(k)>-8/27$ for every such $n,k$, and that $-8/27$ is the greatest lower bound of the set $V$ of all these values as $n$ and $k$ vary. This is proved.

The source defines $\mathcal S$ and $\mathcal K$ as the third-to-first and fourth-to-second scaled current-cumulant ratios (main.tex lines 53–54). Substitution of its winding-number cumulants at lines 772–775 and waiting-time cumulants at lines 787–790 gives the displayed expressions. These are the reused Lean definitions UnicyclicSkewnessMinimum.skewness and UnicyclicKurtosisLowerBound.kurtosis; they are not standardized central moments. See the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/UnicyclicSkewnessPlusKurtosisLowerBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Geometry & algebraConjecture provedSource 2024Tărnăuceanu's reciprocal subgroup-order conjectureThe subgroup-order reciprocal sum has no universal lower bound above one on finite non-cyclic groups.Proved

The questionMarius Tărnăuceanu, On the sum of inverses of subgroup orders in finite groups, Scientific Studies and Research, Series Mathematics and Informatics 34 (2024), no. 1, 115–120, printed p. 119, Conjecture 3.1:

Why it mattersThe source proves a conditional construction using infinitely many Sophie Germain primes.

The ideaD5.S3.FiniteGroups.TarnauceanuReciprocalSubgroupOrders.result proves claim.

Our contributionThe subgroup-order reciprocal sum has no universal lower bound above one on finite non-cyclic groups.

Verification details
Release source recordProved in Lean

Open problem

Tărnăuceanu's reciprocal subgroup-order conjecture

Marius Tărnăuceanu, On the sum of inverses of subgroup orders in finite groups, Scientific Studies and Research, Series Mathematics and Informatics 34 (2024), no. 1, 115–120, printed p. 119, Conjecture 3.1:

There is no constant c > 1 such that f(G) > c, for all finite non-cyclic groups G.

On printed p. 116 the source defines:

In the current note, we consider the function f(G) = Σ_{H∈L(G)} 1/|H|.

Here $L(G)$ is the whole subgroup lattice, including the trivial and full subgroups.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Sum of reciprocal subgroup orders | f | finsum over Subgroup G, with real reciprocal of Nat.card H; for finite groups this is a finite sum. | | No constant $c>1$ | claim | Negated existential over real $c$, retaining the strict inequality. | | All finite non-cyclic groups | claim | Every G : Type, its group and finite instances, and the hypothesis ¬ IsCyclic G. Every finite group has such a carrier up to isomorphism. | | $f(G)>c$ | claim | Exactly the strict inequality c < f G. | | Conjecture holds | result | The complete proposition claim, with no extra hypothesis. |

The definitions are the literal conventions in preregistration #18701.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FiniteGroups/TarnauceanuReciprocalSubgroupOrders.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2024.

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Analysis & dynamicsCounterexample foundFirst preprint 2017Refutation of reciprocal q-gamma coefficient positivityA negative Cauchy coefficient for a reciprocal q-gamma series.Refuted

The questionKarp, Inequalities for series in q-shifted factorials and q-gamma functions, J.

Why it mattersThe unit sequence meets the sequence hypotheses.

Our contributionA negative Cauchy coefficient for a reciprocal q-gamma series.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of reciprocal q-gamma coefficient positivity

S. I. Kalmykov and D. B. Karp, Inequalities for series in q-shifted factorials and q-gamma functions, J. Math. Anal. Appl. 460 (2018), 332–351, DOI 10.1016/j.jmaa.2017.11.041, arXiv:1702.03788v1, Conjecture 1, printed p. 12:

All conclusions of Theorem 3 are valid for all α > 0, while conditions on other parameters remain intact. In particular, µ → g(µ; x) is log-concave for each fixed 0 < x < Rg.

Theorem 3, printed p. 11:

For a real sequence {gₙ}ₙ≥₀ and fixed 0 < q < 1 define

$g(\mu;x)=\displaystyle\sum_{n=0}^{\infty}\frac{g_nx^n}{\Gamma_q(\mu+n)}.$ (16)

Suppose {gₙ}∞ₙ₌₀ is doubly positive. Then (16) has a positive radius of convergence Rg and the generalized Turánian

$\Delta_g(\alpha,\beta;x)=g(\mu+\alpha;x)g(\mu+\beta;x)-g(\mu;x)g(\mu+\alpha+\beta;x)$

is positive for 0 < x < Rg if α ∈ N, β > 0 and µ ≥ max(−β, −1). If, in addition, α ≤ β + 1, then Δg(α, β; x) has positive power series coefficients at all powers of x.

The sequence is nontrivial, nonnegative, log-concave and has no internal zeros, the definition of doubly positive on printed p. 3. Equation (6), printed p. 4, defines

$$\Gamma_q(z)=(1-q)^{1-z}\frac{(q;q)\infty}{(q^z;q)\infty}.$$

Its reciprocal is extended by zero at the poles $z\in-\mathbb N_0$. Only the coefficient-positivity conclusion is refuted here.

Formal reading

The new Lean names have prefix D5.S3.Analytic.QGammaTuran.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Reciprocal of (6) | rqgamma | Real infinite-product quotient, extended by zero at nonpositive integers. | | Doubly positive sequence | D5.S3.Analytic.ReciprocalGammaTuran.Defs.AdmissibleSeq | Existing nontriviality, nonnegativity, log-concavity and no-internal-zero hypotheses. | | Degree $m$ coefficient | qturanCoeff | Exact finite Cauchy sum over $0\le k\le m$. | | Coefficient clause of Conjecture 1 | claim | Every $0<q<1$, admissible sequence, permitted real shifts and natural degree. | | Refutation | result | Exactly $\neg,\texttt{claim}$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/QGammaTuran/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2021The output maximal coherence of channels is not tensorization-stableA qubit Pauli channel has output maximal coherence 3/4, while two copies on a Bell input give maximal coherence 225/272, contradicting Conjecture 1 of Lee, Baek, Park, Kim and Nha.Refuted

The questionNha, Fundamental limits on concentrating and preserving tensorized quantum resources, Phys.

Why it mattersConjecture 1 would imply that correlated, in particular entangled, inputs never preserve more maximal coherence through parallel channels than the best single channel.

The idea1. Witness. The qubit Pauli channel Φ(ρ) = Σσ pσ σρσ with (pI, pX, pY, pZ) = (31/40, 1/10, 1/10, 1/40).

Our contributionA qubit Pauli channel has output maximal coherence 3/4, while two copies on a Bell input give maximal coherence 225/272, contradicting Conjecture 1 of Lee, Baek, Park, Kim and Nha.

Verification details
Release source recordRefuted in Lean

Open problem

The output maximal coherence of channels is not tensorization-stable

J. Lee, K. Baek, J. Park, J. Kim and H. Nha, Fundamental limits on concentrating and preserving tensorized quantum resources, Phys. Rev. Research 4, 043070 (2022), arXiv:2104.12307, ask whether correlated inputs preserve more of a resource through parallel noisy channels than product inputs.

  • For coherence they use the maximal coherence η(ρ) = max_{i≠j} |ρ_ij|/√(ρ_ii ρ_jj) of Lami, Regula and Adesso (Eq. (15)). It governs the fidelity of asymptotic coherence distillation under strictly incoherent operations.
  • They define the output maximal coherence η̃(Φ) = max_ρ η(Φ(ρ)) (Eq. (28)).
  • Conjecture 1 (Eq. (29)): η̃(Φ₁ ⊗ Φ₂) = max{η̃(Φ₁), η̃(Φ₂)}. The authors support it with numerics on random channels in dimensions (2, 2) and (2, 3).

The verbatim statements are in the literature note.

Issue #18019 fixes the reading: for all finite-dimensional channels, given by Kraus families, Eq. (29) holds, with the supremum over all density matrices. This conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/OutputMaximalCoherenceTensorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2016A negative third difference of exponential remainder ratiosForward differences of consecutive exponential remainder ratios.Refuted

The question11. Определим функцию при x > 0. gₙ(x) = Rₙ₋₁(x)/Rₙ(x). Исследовать, при каких k разности этой функции при любом фиксированном x неотрицательны ∆ᵏgₙ(x) > 0? Гипотеза: при любых k. При k = 1 гипотеза сводится к (42), при k = 2 к неравенству задачи 1 на стр. 20.

Why it mattersThe conjecture asks whether the positivity known for the first two orders extends to every order.

Our contributionForward differences of consecutive exponential remainder ratios.

Verification details
Release source recordRefuted in Lean

arXiv

A negative third difference of exponential remainder ratios

S. M. Sitnik, Inequalities for the exponential remainder of the Taylor series (Неравенства для остатка ряда Тейлора экспоненты), Vladivostok preprint (1993), https://arxiv.org/abs/1608.08784v2, appendix (Дополнение), additional problem list, Problem 11, PDF pages 27–28:

  1. Определим функцию при x > 0. gₙ(x) = Rₙ₋₁(x)/Rₙ(x). Исследовать, при каких k разности этой функции при любом фиксированном x неотрицательны ∆ᵏgₙ(x) > 0? Гипотеза: при любых k. При k = 1 гипотеза сводится к (42), при k = 2 к неравенству задачи 1 на стр. 20.

Translation: For $x>0$, define $g_n(x)=R_{n-1}(x)/R_n(x)$. Investigate for which $k$ the differences of this function, at every fixed $x$, are nonnegative, $\Delta^k g_n(x)>0$. Conjecture: for all $k$. For $k=1$ the conjecture reduces to (42), and for $k=2$ to the inequality of Problem 1 on page 20.

Equation (1) defines

$$R_n(x)=e^x-\sum_{j=0}^{n}\frac{x^j}{j!},\qquad x\ge0.$$

The difference is forward in the natural index:

$$\Delta^k g_n(x)=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}g_{n+j}(x).$$

The Lean assertion quantifies over every natural $k\ge1$, every natural $n\ge1$ and every positive real $x$, and asserts $\Delta^k g_n(x)\ge0$. This weakens the source's displayed strict inequality. A negative instance refutes both the nonnegative and the strict readings.

| Source object | Lean declaration | Correspondence | | --- | --- | --- | | $R_n(x)$ | R | Exponential minus the finite Taylor sum through degree $n$. | | $g_n(x)$ | g | $R_{n-1}(x)/R_n(x)$, used only at $n\ge1$. | | $\Delta^k g_n(x)$ | (fwdDiff 1)^[k] (fun m => g m x) n | Iteration of Mathlib's positive unit-step difference. | | Universal assertion | claim | All $k\ge1$, $n\ge1$, $x>0$, with a nonnegative conclusion. | | Negative instance | witness_neg (private) | $\Delta^3 g_1(1)<0$. | | Refutation | result | Exactly $\neg,\texttt{claim}$. |

All declaration names have prefix D5.S3.Analytic.ExpRemainderDifferences.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/ExpRemainderDifferences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2023Refutation of the symmetric Bell cross-polytope conjectureAt three parties and seventeen inputs, single-frequency strategies remain exposed while a mixed strategy lies outside their convex hull; at least nineteen extreme points exclude a nine-dimensional cross-polytope.Refuted

The questionThe source defines deterministic strategies and their local polytope in Eqs. (1)--(2), and the symmetry generators in Eq. (25).

Why it mattersThe conjecture predicts that the projected deterministic strategy polytope has exactly the cross-polytope extreme-point structure in every allowed dimension.

The ideaThe Reynolds average is the source's Eqs. (26)--(27).

Our contributionAt three parties and seventeen inputs, single-frequency strategies remain exposed while a mixed strategy lies outside their convex hull; at least nineteen extreme points exclude a nine-dimensional cross-polytope.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the symmetric Bell cross-polytope conjecture

S. Designolle, T. Vértesi and S. Pokutta, Symmetric multipartite Bell inequalities via Frank-Wolfe algorithms, arXiv:2310.20677v3 and Physical Review A 109, 022205 (2024), DOI 10.1103/PhysRevA.109.022205, conjecture in Section V that the symmetrised local polytope is affinely equivalent to the cross-polytope in dimension $\lceil m/2\rceil$ for the stated multipartite range. The source defines deterministic strategies and their local polytope in Eqs. (1)--(2), and the symmetry generators in Eq. (25). The literal claim is encoded by D5/S3/Quantum/Entanglement/SymmetricBellPolytopeNotCrossPolytope.claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SymmetricBellPolytopeNotCrossPolytope.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2022The upper local degree comparison fails on P6A six-vertex path disproves the upper local comparison between adjacency and normalized Laplacian energies.Refuted

The questionGuerrero, Laplacian Energies of Vertices, arXiv:2201.01252v1, Remark 6 (Conjecture), page 13, states:

Why it mattersThe source compares the global trace energies by minimum and maximum degree, then asks for the analogous comparison at each diagonal entry.

The ideaThe public result applies the conjecture to this connected graph and vertex and contradicts its upper inequality.

Our contributionA six-vertex path disproves the upper local comparison between adjacency and normalized Laplacian energies.

Verification details
Release source recordRefuted in Lean

Open problem

The upper local degree comparison fails on P6

J. Guerrero, Laplacian Energies of Vertices, arXiv:2201.01252v1, Remark 6 (Conjecture), page 13, states:

According to numerical experiments the analog result is true for the energy of vertices, that is: $d_{\min}LE_G(v)\le E_G(v)\le d_{\max}LE_G(v)$ for all $v\in V$.

The standing assumptions are that $G$ is simple and connected. The definitions are $E_G(v)=|A|{vv}$, $L=I-D^{-1/2}AD^{-1/2}$ and $LE_G(v)=|L-I|{vv}$. The Library note contains the source quotations and locators. The preregistration is #19702. The v1 conjecture is refuted by the path on six vertices at zero-based index $2$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite simple graph | V, Fintype V, SimpleGraph V in claim | Undirected loopless graph; arbitrary finite vertex type | | Connected | G.Connected | Hypothesis on the entire conjunction | | Every vertex | ∀ v : V | Universal quantifier | | $\lvert A\rvert$ and $E_G(v)=\lvert A\rvert_{vv}$ | matAbs, vertexEnergy, SimpleGraph.adjMatrix | Real symmetric spectral absolute value via cfc | | $D^{-1/2}=\operatorname{diag}(1/\sqrt{d_v})$ | degInvSqrt | Natural degrees coerced to real numbers | | $L=I-D^{-1/2}AD^{-1/2}$ | normLap | Literal real matrix expression | | $LE_G(v)=\varphi_v(\lvert L-I\rvert)$ | normLapVertexEnergy | $\varphi_v(M)=M_{vv}$ | | Both degree inequalities | Conjunction in claim | minDegree and maxDegree coerced to real numbers | | Negative answer | result : ¬ claim | One graph and one vertex refute the universal conjunction | | Implementation instances | DecidableEq V, DecidableRel G.Adj | Finite graph operations; no additional mathematical restriction |

Real division fixes reciprocal zero to zero. The counterexample has positive degrees, so this totalization does not affect the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/GuerreroVertexEnergyDegreeBounds.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2022The reduced function of the partial negativity is not concaveThe function ĥ(ρ) = √(δ₁δ₂) of the two largest eigenvalues δ₁ ≥ δ₂ of a density matrix, the reduced function of the partial negativity, is not concave: the commuting qutrit states diag(1/2, 1/3, 1/6) and diag(1/2, 1/6, 1/3) have ĥ = √(1/6), and their midpoint diag(1/2, 1/4, 1/4) has ĥ = √(1/8).Refuted

The questionGuo, Partial-Norm of Entanglement: Entanglement Monotones That are not Monogamous, New J.

Why it mattersThe source's Proposition 1 states that an entanglement monotone's reduced function is concave.

The ideaBoth matrices are positive semidefinite and have trace one.

Our contributionThe function ĥ(ρ) = √(δ₁δ₂) of the two largest eigenvalues δ₁ ≥ δ₂ of a density matrix, the reduced function of the partial negativity, is not concave: the commuting qutrit states diag(1/2, 1/3, 1/6) and diag(1/2, 1/6, 1/3) have ĥ = √(1/6), and their midpoint diag(1/2, 1/4, 1/4) has ĥ = √(1/8).

Verification details
Release source recordRefuted in Lean

Open problem

The reduced function of the partial negativity is not concave

Y. Guo, Partial-Norm of Entanglement: Entanglement Monotones That are not Monogamous, New J. Phys. 25 (2023) 083047, arXiv:2212.06521v6, states:

We conjecture that $\hat{h}$ is concave [Suppl].

For decreasing Schmidt coefficients $\lambda_1\ge\lambda_2\ge\cdots$, the paper defines $\hat N(|\psi\rangle)=\lambda_1\lambda_2$ and the reduced function $\hat h(\rho)=\sqrt{\delta_1\delta_2}$, where $\delta_j=\lambda_j^2$. Its definition of concavity is

$$ h(t\rho+(1-t)\sigma)\ge t h(\rho)+(1-t)h(\sigma),\qquad 0\le t\le1. $$

Issue #14495 preregisters the reading over every finite dimension $d\ge2$ and all positive semidefinite complex $d\times d$ matrices of trace one. The later paper Complete Genuine Multipartite Entanglement Monotone, arXiv:2301.00334v3, explicitly identifies $\delta_1,\delta_2$ as the two largest eigenvalues of $\rho$ and repeats the conjecture in its table footnote. Eigenvalues are counted with multiplicity; their positions are chosen separately for each state, rather than tied to fixed eigenvectors.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PartialNegativityConcavityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2024Conjecture 29 on the path Steiner coefficient peak is falseThe path on sixteen vertices refutes the proposed Steiner coefficient peak.Refuted

The questionSivasubramanian, Unimodality and peak location of the characteristic polynomials of two distance matrices of trees, Computational and Applied Mathematics 44, 223 (2025), arXiv:2407.03309v1, Conjecture 29, page 18:

Why it mattersThe coefficient peak changes location beyond the small sizes displayed in the source.

The ideaThe existing owner theorem D5.S3.Constants.Moments.CoefficientDrivenJacobiCharacteristicPolynomial.multiplicationmatrixcharpoly identifies the companion polynomial after the injective integer-to-real cast and finite reindexing in companioncharpoly.

Our contributionThe path on sixteen vertices refutes the proposed Steiner coefficient peak.

Verification details
Release source recordRefuted in Lean

Open problem

Conjecture 29 on the path Steiner coefficient peak is false

S. Jana, I. Mahato and S. Sivasubramanian, Unimodality and peak location of the characteristic polynomials of two distance matrices of trees, Computational and Applied Mathematics 44, 223 (2025), arXiv:2407.03309v1, Conjecture 29, page 18:

For a path P_n on n > 5 vertices, if CharPoly_{𝔇_{P_n}}(x) = Σ_{i=0}^{2n−3} a_i x^i and |a_ℓ| = max{|a_0|, |a_1|, …, |a_{2n−4}|}, then ℓ = n − 1.

The Library note quotes the distance, basis and principal-submatrix definitions. Preregistration: #19152.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Minimum edges of a connected subtree containing the four vertices | steiner n S | A connected subtree of a path is an interval with $b-a$ edges; the natural infimum ranges over exactly the containing intervals | | Ordered basis $(e_1,f_1,…,e_{n-2},f_{n-2},e_{n-1})$ | basis n t | Even zero-based indices give $e_i={i,i+1}$ and odd indices give $f_j={j,j+2}$ | | Principal matrix $\mathfrak D_2(P_n)[B,B]$ | D n | Each integer entry is the Steiner distance of the union of its row and column pairs | | Every path with $n>5$ | claim quantifies over natural $n$ with $5<n$ | No extra graph or numerical hypothesis | | Largest magnitude among coefficients $0,…,2n-4$ | The quantified coefficient inequalities in claim | The leading coefficient is excluded exactly as in the source | | Every attaining index equals $n-1$ | The conclusion of claim | Quantifying over all attaining indices does not assume uniqueness | | Refutation | result : ¬ claim | The witness has $n=16$ and $\ell=16\ne15$ |

All natural subtractions and finite-index types retain the literal source conventions. The interval bounds are the definition of a subtree, not an additional hypothesis on the conjecture. The characteristic polynomial is $\det(xI-D)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/JanaMahatoSivasubramanianPeakRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2023Mbirika-Schrader-Spilker Associated Pell GCD Entry-Point RefutationThe value k = 12 refutes the printed associated Pell gcd-entry-point biconditional.Refuted

The questionMbirika, Schrader, and Spilker define the Pell and associated Pell sequences in Definition 3 on printed page 4:

The ideaOn the right side, p = 3 is prime and divides 12.

Our contributionThe value k = 12 refutes the printed associated Pell gcd-entry-point biconditional.

Verification details
Release source recordRefuted in Lean

Open problem

Mbirika-Schrader-Spilker Associated Pell GCD Entry-Point Refutation

Mbirika, Schrader, and Spilker define the Pell and associated Pell sequences in Definition 3 on printed page 4:

Definition 3. The Pell sequence (P_n){n≥0} and the associated Pell sequence (Q_n){n≥0} are defined by the recurrence relations P_n = 2P_{n−1} + P_{n−2} and Q_n = 2Q_{n−1} + Q_{n−2}, respectively, with initial conditions P_0 = 0, P_1 = 1, Q_0 = 1, and Q_1 = 1.

Section 6.2 on printed page 22 gives the entry-point convention:

When (S_n)_{n≥0} is the Pell or associated Pell sequence, we have partial results towards closed forms for gcd(S_k, k) that involve the entry point (or rank of apparition), e_S(p), which is the smallest index r > 0 such that p divides S_r where p is a prime.

The same printed page states:

Conjecture 32. We claim that gcd(Q_k, k) > 1 if and only if there exists a prime p such that p divides k and the rank of apparition (or entry point), e_Q(p) divides k. For example, gcd(Q_21, 21) = 7 and for the prime p = 7, we have p divides 21 and e_Q(p) = 3 divides 21.

The formal claim follows the printed universal biconditional and makes the entry point explicit as the least positive index at which the prime divides the associated Pell sequence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/MbirikaAssociatedPellGcdEntryPointRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2024The steady-state block of a Markov generator with a commuting symmetry need not admit a Markov formAn irreducible steady-state block of a four-state Markov generator is not similar to any real Markov generator, even by a complex change of basis.Refuted

The questionBernard, Symmetry Classes of Classical Stochastic Processes, J.

Why it mattersThe classification of the source assumes either no commuting symmetry or a restriction to the steady-state block, read as a Markov generator in its own right.

The idea1. Witness. The four-state generator L = (1/256)[[−295, 25, 25, 281], [273, −303, 17, 17], [7, 263, −313, 7], [15, 15, 271, −305]].

Our contributionAn irreducible steady-state block of a four-state Markov generator is not similar to any real Markov generator, even by a complex change of basis.

Verification details
Release source recordRefuted in Lean

Open problem

The steady-state block of a Markov generator with a commuting symmetry need not admit a Markov form

L. Sá, P. Ribeiro, T. Prosen and D. Bernard, Symmetry Classes of Classical Stochastic Processes, J. Stat. Phys. 192, 41 (2025), arXiv:2406.17955, classify the symmetries of Markov generators L, real matrices with nonnegative off-diagonal rates and zero column sums (Eqs. (2)–(4)).

A commuting symmetry U splits L into blocks, and the classification applies to each irreducible block that is itself a Markov generator. The authors write after Eq. (5): "It is, however, natural to conjecture that the block containing the invariant measure of the Markov process (henceforth referred to as the steady-state block) can still be recast into Markov form." Appendix A proves this under the extra assumption (A3).

The verbatim statements are in the literature note.

Issue #17999 fixes the reading. Take a Markov generator L, a commuting real orthogonal U, and a stationary probability p. Let the steady-state block be a whole eigenspace of U that contains p, with no further unitary symmetry. Then the restriction of L to it is similar to a Markov generator. This conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/SteadyStateBlockMarkovFormRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023Uhlmann's unrestricted diagonal-majorization questionA positive definite similarity in dimension seven defeats diagonal majorization.Refuted

The questionJeffrey Uhlmann, Partial Proof of a Conjecture with Implications for Spectral Majorization, arXiv:2309.01302v1, Section 4, asks about removing the dimensional restriction from the diagonal-spectrum majorization relationship:

The ideaThe proof uses spectral estimates and convexity.

Our contributionA positive definite similarity in dimension seven defeats diagonal majorization.

Verification details
Release source recordRefuted in Lean

Open problem

Uhlmann's unrestricted diagonal-majorization question

Jeffrey Uhlmann, Partial Proof of a Conjecture with Implications for Spectral Majorization, arXiv:2309.01302v1, Section 4, asks about removing the dimensional restriction from the diagonal-spectrum majorization relationship:

However, the fact that such a construction does not generally exist for n > 6 does not necessarily imply that the diagonal-spectrum majorization relationship does not hold in general.

Empirical investigation may be able to provide sufficient evidence to justify a conjecture or, alternatively, empirical and/or analytical efforts may identify counterexamples that constrain such generalizations

The assertion is that for every natural dimension $n$, every real symmetric positive definite $P$, and every real diagonal $D$, the diagonal of $PDP^{-1}$ majorizes its spectrum. The spectrum is the vector of diagonal entries of $D$. Section 3 defines $x\prec y$ by $x=Sy$ for a doubly stochastic matrix $S$. Preregistration: #18572.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $x\prec y$ iff $x=Sy$, with $S$ doubly stochastic | Majorizes y x | Literal existential definition using Mathlib's doublyStochastic. | | Every dimension | n : ℕ in claim | No upper bound is imposed. | | Real symmetric positive definite $P$ | P : Matrix (Fin n) (Fin n) ℝ and P.PosDef | Over the reals, Mathlib's Hermitian condition is symmetry. | | Every real diagonal $D$ | d : Fin n → ℝ and Matrix.diagonal d | All real entries are allowed. | | Spectrum of $PDP^{-1}$ is majorized by its diagonal | Final Majorizes expression in claim | The diagonal is the majorizing vector and d the majorized vector. | | The unrestricted assertion is false | result : ¬ claim | One seven-dimensional instance refutes the universal statement. |

No positivity requirement on $D$ is added; the chosen witness also has positive diagonal entries. The source's small-dimension theorem retains its dimensional hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/UhlmannDiagonalMajorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024The Eq. (10) set does not generate the 3 × 3 stochastic matricesA column-stochastic matrix with six-cycle support cannot be obtained from the generators of Eq. (10).Refuted

The questionShahbeigi, Generating Sets of Stochastic Matrices, arXiv:2411.18946, study sets G of column-stochastic matrices whose finite products give every stochastic matrix.

Why it mattersThe paper proposes Eq. (10) as an explicit, physically interpretable generating set for three-state classical channels, with a uniform bound on the number of elementary operations.

The ideaWitness: A = [[1/4, 0, 1/4], [3/4, 1/2, 0], [0, 1/2, 3/4]], column-stochastic with a six-cycle support.

Our contributionA column-stochastic matrix with six-cycle support cannot be obtained from the generators of Eq. (10).

Verification details
Release source recordRefuted in Lean

arXiv

The Eq. (10) set does not generate the 3 × 3 stochastic matrices

F. vom Ende and F. Shahbeigi, Generating Sets of Stochastic Matrices, arXiv:2411.18946, study sets G of column-stochastic matrices whose finite products give every stochastic matrix. N_G(s(n)) is the number of factors needed in the worst case (Definition 3). These are the classical counterparts of generating sets of quantum channels: a generating set models the operations an experiment can implement, and N_G bounds the circuit length needed to implement every classical channel.

Theorem 3 states that G = {P} ∪ conv{1, V1, V2, V3} of Eq. (10) generates s(3) with G^20 = s(3). The Outlook asks "is the upper bound N_G(S) ≤ 20 in Theorem 3 optimal? We conjecture that this is true".

The verbatim statements are in the literature note.

Issue #17954 fixes the reading: N_G(s(3)) = 20, that is, G^20 = s(3) and G^19 ≠ s(3). This conjecture is refuted, together with the generation claim of Theorem 3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/StochasticGeneratorSetRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2023The entropic tetrahedron is not a genuine four-qudit entanglement measureThe eight-level four-party pure state has an admissible zero-volume fill but is not biseparable, refuting the qudit genuine-entanglement claim.Refuted

The questionEberly, Multipartite Entanglement: A Journey Through Geometry, arXiv:2304.03281v2; Entropy 26(3), 217 (2024), define the genuine requirement in tetrahedron.tex, line 57:

Why it mattersThe source proposes using four groups of a larger system as the four parties of an entropic tetrahedron.

The ideaThe compiled proof checks all membership choices for the four parties and excludes every nontrivial cut.

Our contributionThe eight-level four-party pure state has an admissible zero-volume fill but is not biseparable, refuting the qudit genuine-entanglement claim.

Scope & next questionThe witness does not settle the source's four-qubit claims, additivity argument or numerical LOCC study.

Verification details
Release source recordRefuted in Lean

Open problem

The entropic tetrahedron is not a genuine four-qudit entanglement measure

S. Xie, D. Younis, Y. Mei and J. H. Eberly, Multipartite Entanglement: A Journey Through Geometry, arXiv:2304.03281v2; Entropy 26(3), 217 (2024), define the genuine requirement in tetrahedron.tex, line 57:

The ``genuine'' requirement requires that a GME measure should only, and always, vanish for biseparable states \cite{ma2011,xie2021}.

Their Scalability paragraph, line 220, states:

Nonetheless, our tetrahedron measure presented in this work maintains its scalability. Specifically, we can consistently partition a large system into four groups and explore genuine quadripartite entanglement among these groups. Essentially, the tetrahedron measure is a genuine four-qudit entanglement measure, where each party possesses more than two dimensions. The extension of entropic fill to qudit systems has been made possible by a recent study \cite{yang2022}. Additionally, we have conducted numerical verifications, demonstrating the existence of unique nonnegative solutions for Eq.~\eqref{eqn1} and \eqref{eqn2} in qudit systems up to $d=50$. Consequently, we conjecture that our tetrahedron measure is also applicable to qudit systems.

These quotations were checked against the complete v2 TeX source, tetrahedron.tex (233 lines, md5 da0b24bf21c330f90979f63e7679749c), fetched from https://arxiv.org/src/2304.03281v2. The definitions and further quotations are collected in the literature note.

Issue #20861 fixes the following reading. Each of four parties has Hilbert space $\mathbb C^d$, with $d\ge2$, and $\psi$ is a coefficient function with $\sum_x|\psi(x)|^2=1$. Write $S_X$ for the base-two von Neumann entropy of its reduced density matrix on $X$. Biseparability means factorization across some nonempty proper subset of the four parties.

An admissible fill consists of six nonnegative real coordinates $\sigma=(\sigma_{AB},\sigma_{AC},\sigma_{AD},\sigma_{BC},\sigma_{BD},\sigma_{CD})$ and a real parameter $\lambda$. The face equations are $\sigma_{ij}+\sigma_{ik}+\sigma_{il}=S_i$ for each party $i$. Put $p=\sqrt{\sigma_{AB}\sigma_{CD}}$, $q=\sqrt{\sigma_{AC}\sigma_{BD}}$ and $r=\sqrt{\sigma_{AD}\sigma_{BC}}$. The radical equations are $-p+q+r=\lambda S_{AB}$, $p-q+r=\lambda S_{AC}$ and $p+q-r=\lambda S_{AD}$. These are the source equations labelled eqn1 and eqn2, TeX lines 103–121.

The source volume, TeX lines 92–101, is $V=(\sqrt2/3)\sqrt{2\sum_{ij}\sigma_{ij}}\big((p+q+r)(-p+q+r)(p-q+r)(p+q-r)\big)^{1/4}$. The formal claim is: for every $d\ge2$ and every unit $\psi$, if an admissible fill exists and every admissible fill has $V=0$, then $\psi$ is biseparable. This implication is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/EntropicFill/QuditTetrahedronGenuineRefutation.result

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Discrete structuresQuestion resolvedFirst preprint 2024Cyclic Permutations Avoiding 4123 with 1324-Avoiding Cycle FormsThe cyclic avoidance numbers for 4123 and 1324 follow the Tetranacci sequence.Proved

The questionKassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves and Millie Jeske, Pattern-restricted cyclic permutations with a pattern-restricted cycle form, arXiv:2408.15000v1, Section 4:

Why it mattersLet T0 = T1 = T2 = 0, T3 = 1 and T{k+4} = T{k+3} + T{k+2} + T{k+1} + Tk (OEIS A000078).

The ideaEncode a cyclic permutation by its cycle word rooted at 1.

Our contributionThe cyclic avoidance numbers for 4123 and 1324 follow the Tetranacci sequence.

Verification details
Release source recordProved in Lean

Open problem

Cyclic Permutations Avoiding 4123 with 1324-Avoiding Cycle Forms

Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves and Millie Jeske, Pattern-restricted cyclic permutations with a pattern-restricted cycle form, arXiv:2408.15000v1, Section 4:

we conjecture that |A°_n(4123;1324)| is the (n−2)nd Tetranacci number

Here A°_n(4123; 1324) is the set of cyclic permutations of 1, …, n whose one-line notation avoids 4123 and all of whose cycle forms, the rotations of the cycle word, avoid 1324.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ArcherCyclicTetranacci.result

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Discrete structuresQuestion resolvedFirst preprint 2024Cyclic Permutations Avoiding 4132 with 1324-Avoiding Cycle FormsThe cyclic avoidance numbers for 4132 and 1324 follow a subsequence of the Padovan numbers.Proved

The questionKassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves and Millie Jeske, Pattern-restricted cyclic permutations with a pattern-restricted cycle form, arXiv:2408.15000v1, Section 4:

Why it mattersLet P0 = 1, P1 = P2 = 0 and P{k+3} = P{k+1} + Pk (OEIS A000931).

The ideaEncode a cyclic permutation by its cycle word rooted at 1.

Our contributionThe cyclic avoidance numbers for 4132 and 1324 follow a subsequence of the Padovan numbers.

Verification details
Release source recordProved in Lean

Open problem

Cyclic Permutations Avoiding 4132 with 1324-Avoiding Cycle Forms

Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves and Millie Jeske, Pattern-restricted cyclic permutations with a pattern-restricted cycle form, arXiv:2408.15000v1, Section 4:

we conjecture that … |A°_n(4132;1324)| is the (3n)th Padovan number

Here A°_n(4132; 1324) is the set of cyclic permutations of 1, …, n whose one-line notation avoids 4132 and all of whose cycle forms, the rotations of the cycle word, avoid 1324.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ArcherCyclicPadovan.result

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Quantum informationQuestion resolvedFirst preprint 2022The classical value of the odd polygon game is its Fibonacci boundFor every odd P at least three, the polygon game's optimal classical winning probability is one minus F(P minus one) divided by two to the power P.Proved

The questionSondhi, Playing nonlocal games with phases of quantum matter, Phys.

Why it mattersThe paper compares classical performance with a cluster-state quantum strategy and with strategies using states in a symmetry-protected topological phase.

The ideaThe proof uses a bijective argument.

Our contributionFor every odd P at least three, the polygon game's optimal classical winning probability is one minus F(P minus one) divided by two to the power P.

Scope & next questionThe argument gives a bound, not an exact optimum for arbitrary nonuniform input laws; that optimum remains open here. - The large-player limit.

Verification details
Release source recordProved in Lean

Open problem

The classical value of the odd polygon game is its Fibonacci bound

V. B. Bulchandani, F. J. Burnell and S. L. Sondhi, Playing nonlocal games with phases of quantum matter, Phys. Rev. B 107, 045412 (2023), arXiv:2206.11252v2, Section 4.1, "Classical winning probabilities", state in ArxivResub.tex, lines 714–717, the equation labelled eq:classwinbound and its conjecture:

$p_{\mathrm{cl}}^*(P) \geq \frac{2^P - F_{P-1}}{2^P} = 1 - \frac{F_{P-1}}{2^P}, \quad P ,\mathrm{odd}.$

We conjecture that this bound is tight.

Here $P\ge3$ is the number of players and $F_1=F_2=1$. The game's rules are in the same section, ArxivResub.tex, lines 629–655; see also the literature note.

Index players and oriented edges by $\mathbb Z/P$, with edge $i$ joining $i$ to $i+1$. A deterministic strategy assigns triples $(a_i,b_i^0,c_i)$ on input $0$ and $(d_i,b_i^1,e_i)$ on input $1$. The module allows the middle bit to depend on the input. For a matching $M$, let $x_M$ be the input whose ones are exactly its covered vertices. On input $x$, the strategy wins when every returned triple has even parity, the returned middle bits have even total parity, and, for every matching with $x_M=x$,

$$\sum_{i\in M}(d_i+e_{i+1})+\sum_{j\text{ uncovered by }M}a_j\equiv |M|\pmod2.$$

The empty matching gives Condition 3; matchings with one, two and more edges give Conditions 4–6. Inputs are uniform over all $2^P$ Boolean strings, including those arising from no matching. Write $w_P(s)$ for the fraction of these inputs won by a deterministic strategy $s$, and $W_P(\mu)=\sum_s\mu(s)w_P(s)$ for a nonnegative weight function of total mass one on all deterministic strategies.

Issue #26289 fixes the reading: for every odd $P\ge3$, every such $\mu$ satisfies $W_P(\mu)\le1-F_{P-1}/2^P$, and some deterministic $s$ attains equality. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/PolygonGameClassicalValue.result

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Quantum informationQuestion resolvedFirst preprint 2022No reverse perfect state transfer in the seven-state bipartite walkNeither seven-state matrix transfers the sixth coordinate state to the first at any natural step.Proved

The questionZhan, Hamiltonians of bipartite walks, arXiv:2207.01673v1; The Electronic Journal of Combinatorics 31(4) (2024), P4.10, Section “Open questions”, Hamiltonian7.tex, line 1288:

Why it mattersThe suspicion concerns an asymmetry between the two directions of exact state transfer, rather than the absence of transfer altogether.

The ideaThe proof uses spectral estimates.

Our contributionNeither seven-state matrix transfers the sixth coordinate state to the first at any natural step.

Scope & next questionThe requested graph condition for symmetry remains open here.

Verification details
Release source recordProved in Lean

Open problem

No reverse perfect state transfer in the seven-state bipartite walk

Q. Chen, C. Godsil, M. Sobchuk and H. Zhan, Hamiltonians of bipartite walks, arXiv:2207.01673v1; The Electronic Journal of Combinatorics 31(4) (2024), P4.10, Section “Open questions”, Hamiltonian_7.tex, line 1288:

State $e_i$ is the characteristic vector of $i$. It is easy to see that there is perfect state transfer from state $e_1$ to $e_6$ at step $k=1$. But up to $k=300000$ steps, there is no perfect state transfer observed from $e_6$ to $e_1$. We suspect that there is no perfect state transfer from $e_6$ to $e_1$. We would like to find a condition on graph $G$ that determines whether or not perfect state transfer is symmetric.

The checked v1 TeX has md5 2e3f9a1a91cb695794df49b0123dfc37. Its evolution convention is $U^kz$ (lines 135–142), and its bipartite-walk definition is $U=(2P-I)(2Q-I)$ (lines 154–163). The projections for the seven-edge example are printed at lines 320–338, and its displayed transition matrix at lines 340–351 and 1278–1286. These definitions are documented in the literature note.

Use the source's one-based coordinate labels throughout this dossier. Let $U_{\rm def}=(2P-I)(2Q-I)$ for those explicit projections and let $U_{\rm disp}$ be the printed matrix. Exact rational comparison gives one discrepancy: their entries at row 3, column 5 are respectively $-1/3$ and $+1/3$; all other entries agree.

Issue #21532 fixes a reading covering both matrices. In Lean, PerfectStateTransfer U a b means that there exist $k\in\mathbb N$ and $\gamma\in\mathbb C$ with $|\gamma|=1$ and $U^ke_a=\gamma e_b$. The predicate imposes no unitarity assumption and includes step zero. The declaration claim says that neither $U_{\rm def}$ nor $U_{\rm disp}$ has such transfer from $e_6$ to $e_1$. This claim is proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Walks/BipartiteWalkReverseStateTransfer.result

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Analysis & dynamicsCounterexample foundSource 2024Dmitrović: constant-one mixed-norm point evaluationThe constant-one point evaluation estimate fails for every finite positive p, every larger q including infinity, and every positive alpha.Refuted

The questionDmitrović, Norm of composition operator on mixed norm spaces, Operators and Matrices 18(3) (2024), 697–710, doi:10.7153/oam-2024-18-42, end of section 2, page 705, states:

Why it mattersThe conjecture asks whether the constant can nevertheless be one throughout the remaining parameter range.

The ideaThe full-period angular integral gives the exact identity

Our contributionThe constant-one point evaluation estimate fails for every finite positive p, every larger q including infinity, and every positive alpha.

Verification details
Release source recordRefuted in Lean

Open problem

Dmitrović: constant-one mixed-norm point evaluation

D. Dmitrović, Norm of composition operator on mixed norm spaces, Operators and Matrices 18(3) (2024), 697–710, doi:10.7153/oam-2024-18-42, end of section 2, page 705, states:

CONJECTURE. From the previous discussion, we conjecture that for all 0 < p, q ⩽ ∞, 0 < α < ∞ and f ∈ H^{p,q,α}: |f(z)| ⩽ ‖f‖_{p,q,α} / (1 − |z|²)^{α + 1/p}.

The definitions on page 697 are:

$$ |f|_{p,q,\alpha} =\left(2\alpha q\int_0^1 r(1-r^2)^{\alpha q-1}M_p^q(r,f),dr\right)^{1/q} \qquad(0<q<\infty), $$

$$ |f|{p,\infty,\alpha} =\sup{0<r<1}(1-r^2)^\alpha M_p(r,f), \qquad M_p(r,f)=\left(\frac1{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^p,d\theta\right)^{1/p}. $$

Here $M_\infty$ is the supremum over $\theta$. The space $H^{p,q,\alpha}$ consists of holomorphic functions on the unit disc with finite mixed norm. The convention is $1/\infty=0$. The Lean claim multiplies the inequality by the strictly positive weight $(1-|z|^2)^{\alpha+1/p}$; this is equivalent to the displayed quotient form.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MixedNorm/DmitrovicPointEvaluation/Result.result

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Discrete structuresQuestion resolvedFirst preprint 2024Nonnesting Permutations Avoiding 1231, 1312, 2231 and 3221The four-pattern avoidance class has the asserted rational generating function.Proved

The questionSergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1231, 1312, 2231, 3221}: the conjectured ordinary generating function of cn({1231, 1312, 2231, 3221}) is (1 - 3x + 2x^2)/((1 - 3x)(1 - x - x^2)), with the remark

The ideaA doubled word avoids 1221 and 2112 exactly when its first and second occurrences appear in the same letter order.

Our contributionThe four-pattern avoidance class has the asserted rational generating function.

Verification details
Release source recordProved in Lean

Open problem

Nonnesting Permutations Avoiding 1231, 1312, 2231 and 3221

Sergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1231, 1312, 2231, 3221}: the conjectured ordinary generating function of c_n({1231, 1312, 2231, 3221}) is (1 - 3x + 2x^2)/((1 - 3x)(1 - x - x^2)), with the remark

All the conjectures have been checked for n up to 8.

Here a nonnesting permutation of size n is a permutation of {1, 1, 2, 2, …, n, n} avoiding 1221 and 2112, and pattern containment keeps equal letters equal.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Nonnesting/NonnestingFour.result

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Discrete structuresQuestion resolvedFirst preprint 2024Nonnesting Permutations Avoiding 1233 and 1322The generating function for nonnesting permutations avoiding 1233 and 1322 satisfies the stated quadratic equation.Proved

The questionSergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1233, 1322}: the conjectured ordinary generating function is ((1-x)^2 - √((1-x)^4 - 4x(1-x)^2))/(2x) (OEIS A006319), with the remark

Why it mattersThis equation has a unique power-series solution, the branch of the displayed square root with constant term one, so the counts are 1, 1, 4, 16, 68, 304, 1412, 6752, ….

The ideaA nonnesting word is a permutation (its first-occurrence order) together with a Dyck word.

Our contributionThe generating function for nonnesting permutations avoiding 1233 and 1322 satisfies the stated quadratic equation.

Verification details
Release source recordProved in Lean

Open problem

Nonnesting Permutations Avoiding 1233 and 1322

Sergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1233, 1322}: the conjectured ordinary generating function is ((1-x)^2 - √((1-x)^4 - 4x(1-x)^2))/(2x) (OEIS A006319), with the remark

All the conjectures have been checked for n up to 8.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Nonnesting/NonnestingRoyalHigh.result

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Discrete structuresQuestion resolvedFirst preprint 2019Splythoff Nim: all Grundy values in every row and columnEvery Splythoff Nim row and column contains each Sprague–Grundy value exactly once.Proved

The questionRobbert Fokkink and Dan Rust, Queen reflections: a modification of Wythoff Nim, arXiv:1904.08339v2, Section 9, printed page 16; International Journal of Game Theory 53 (2024), 1099–1117, DOI 10.1007/s00182-022-00824-1:

Why it mattersIt establishes that each infinite row and each infinite column is a permutation of the natural Grundy values.

The ideaA single move connects any two distinct positions in one row, so movevaluene gives rowinjective; symmetry gives columninjective.

Our contributionEvery Splythoff Nim row and column contains each Sprague–Grundy value exactly once.

Scope & next questionThe source's losing-position and Tribonacci results remain separate, and the reflected-diagonal question remains open; no other source result is claimed to follow from this proof.

Verification details
Release source recordProved in Lean

Open problem

fokkink rust 2024 splythoff grundy rows

Robbert Fokkink and Dan Rust, Queen reflections: a modification of Wythoff Nim, arXiv:1904.08339v2, Section 9, printed page 16; International Journal of Game Theory 53 (2024), 1099–1117, DOI 10.1007/s00182-022-00824-1:

For Splythoff Nim, it appears that every row or column contains each Sprague-Grundy value once and no more. We do not have a proof for that.

Preregistration: #15902. The statement is proved by D5/S0/Certificates/Games/SplythoffGrundyRows.result. Its bound theorem row_exists_bound supplies an occurrence at $y\le 5x+2g+3$ for every row $x$ and value $g$.

Statement echo

| Source clause or convention | Lean counterpart | Fidelity | | --- | --- | --- | | Two piles, including an empty pile | Natural coordinates in splythoffMove and grundy | No positivity restriction on positions | | Reduce one pile | First two disjuncts of splythoffMove | The other coordinate is unchanged | | Remove an equal positive number from both piles | Third disjunct of splythoffMove | The same $k$ satisfies $1\le k\le\min(x,y)$ | | Split only after removing counters from both piles | Fourth disjunct of splythoffMove | The smaller pile is positive and the target sum equals $\max(x,y)-\min(x,y)$ | | Mex of all legal follower values | followers, grundy, private followers_iff and grundyPair_eq | Complete legal-follower set and its exact mex equation | | Every value in each row once and no more | First conjunct of claim; result | Universal natural row and value, unique natural column | | Every value in each column once and no more | Second conjunct of claim; result | Universal natural column and value, unique natural row | | The preregistered finite bound | row_exists_bound | Additional quantitative conclusion, with no extra hypothesis |

The bounded-follower convention repair leaves the literal move relation unchanged, including the unbounded existential for double moves. The finite rectangle with both coordinates at most $\max(x,y)$, filtered by that relation, is exactly its follower set: follower_bound proves the coordinate bound and followers_iff proves the equivalence. Every move strictly decreases the total pile size, and grundyPair_eq supplies the mex equation used by the proof. No legal move is discarded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/SplythoffGrundyRows.result

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Discrete structuresCounterexample foundFirst preprint 2024The Number of Indecomposable 321- and 1342-Avoiders with k InversionsThe number of indecomposable 321- and 1342-avoiders with k inversions is k(k - 1)/2 + 1, refuting the conjectured count k(k + 1)/2 + 1.Refuted

The questionAtli Fannar Franklín, Pattern avoiding permutations enumerated by inversions, arXiv:2410.07467v4, Section 1, printed pages 2–3:

The ideaEvery member of Ik(321, 1342) is either a star of length k + 1 or a five-block permutation B(r, t, d, h) with explicit parameter ranges; the two families are disjoint and the parametrization is injective.

Our contributionThe number of indecomposable 321- and 1342-avoiders with k inversions is k(k - 1)/2 + 1, refuting the conjectured count k(k + 1)/2 + 1.

Verification details
Release source recordRefuted in Lean

Open problem

The Number of Indecomposable 321- and 1342-Avoiders with k Inversions

Atli Fannar Franklín, Pattern avoiding permutations enumerated by inversions, arXiv:2410.07467v4, Section 1, printed pages 2–3:

Another case is I_k(321, 1342), which we conjecture to have k(k + 1)/2 + 1 elements.

Here I_k is the set of direct-sum indecomposable permutations, of any length, with exactly k inversions, and I_k(321, 1342) is its subset avoiding the classical patterns 321 and 1342.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/IndecomposableInversion/FranklinInversion.result

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Analysis & dynamicsCounterexample foundFirst preprint 2024Refutation of Fucci–Stanfill coefficient nonvanishingReflection of the generalized Bernoulli coefficient polynomials forces g(2,3,beta,3,1) to vanish.Refuted

The questionStanfill, The exotic structure of the spectral ζ-function for the Schrödinger operator with Pöschl–Teller potential, arXiv:2411.17860v1, Appendix B, Remark B.3, state:

Why it mattersThe source definitions and locators are recorded in the Library note.

The ideaThe square has no degree-five term, since its nonzero input degrees below five can only be two.

Our contributionReflection of the generalized Bernoulli coefficient polynomials forces g(2,3,beta,3,1) to vanish.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Fucci–Stanfill coefficient nonvanishing

G. Fucci and J. Stanfill, The exotic structure of the spectral ζ-function for the Schrödinger operator with Pöschl–Teller potential, arXiv:2411.17860v1, Appendix B, Remark B.3, state:

From calculations performed for particular choices of $p$, $q$, and $\beta$, we in fact conjecture that all of the $g_{m,j}(p/q,\beta)$ are nonzero for the choices of $p,\ q,$ and $\beta$ considered here.

The source range is relatively prime integers $0<p<q$ and $\beta\in(0,\pi)\setminus{\pi/2}$. The admissible indices satisfy $lp+kq=m$ for $l,k\in\mathbb N_0$, and $0\le j\le k_m$, where $k_m$ is the largest feasible $k$. The journal locator is Annales Henri Poincaré 27 (2026), 2073–2116, DOI 10.1007/s00023-025-01587-7.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/SpectralZeta/PoschlTellerCoefficientVanishing.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2023An unsharp qubit Lüders image violates ambient Sob sum closureA two-outcome qubit Luders instrument maps an effect to a sub-observable whose double is still a sub-observable but is outside the image of effects.Proved

The questionStanley Gudder, Dual Quantum Instruments and Sub-Observables, arXiv:2302.12243v1, Section 3, after Theorem 3.1:

Why it mattersThe image of effects has an induced notion of orthogonality, but ambient sub-observable addition can be weaker than simultaneous addition of effect preimages.

Our contributionA two-outcome qubit Luders instrument maps an effect to a sub-observable whose double is still a sub-observable but is outside the image of effects.

Verification details
Release source recordProved in Lean

arXiv

An unsharp qubit Lüders image violates ambient Sob sum closure

Stanley Gudder, Dual Quantum Instruments and Sub-Observables, arXiv:2302.12243v1, Section 3, after Theorem 3.1:

If $\mathcal I\in\operatorname{In}(H)$ and $U_{\mathcal I}={\mathcal I_a^*:a\in\mathcal E(H)}$, then we conjecture that $U$ need not be a Sob effect algebra.

Example 3 asks:

If $\mathcal L_x(\rho)=a_x\circ\rho$ is an arbitrary Lüders instrument, we do not know whether $U_{\mathcal L}$ is a Sob effect algebra. However, in the case where $a_x$ are projections (we then call $\mathcal L$ sharp) we can show that it is.

This is a Tier 1, explicitly published conjecture and question, preregistered as #16759. The conjecture is Proved by a finite counterexample to universal closure; the arbitrary Lüders question is answered negatively. The universal closure assertion is claim, and the settling theorem is result : ¬ claim.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Common outcome space, positive additive outcome values, total at most $I$ | Fin n, finite event sums, IsSubObservable | Each outcome is PSD and the complement of the total is PSD. Finite all-events space suffices for an existential counterexample. | | Effect $0\le a\le I$ | PSD body in ludersImage; TypedEscapeNeedsPositivity.IsEffect in the proofs | Both $a$ and $I-a$ are PSD, equivalently $0 \le a \le I$ under MatrixOrder. | | $U\subseteq\operatorname{Sob}(H)$ | First conjunct of IsSobEffectAlgebra | Every member is a sub-observable. | | (S1) an observable $Z\in U$ | Existential $Z$, with sub-observability and total $I$ | The observable body is inlined literally; no duplicate IsObservable definition is added. | | (S2) $Z-A\in U$ for $A\in U$ | Complement conjunct with the same $Z$ | No separate choice of $Z$ is allowed. | | (S3) $A,B\in U$ and $A+B\in\operatorname{Sob}(H)$ imply $A+B\in U$ | Final conjunct | Pointwise addition with ambient sub-observability as antecedent. | | Lüders action $K_xaK_x$, $K_x=a_x^{1/2}$ | ludersImage; PSD $K_x$ and $\sum_x K_xK_x=I$ in claim | A single effect supplies every outcome; the image equality has the preregistered orientation. | | Image need not be a Sob effect algebra | result | Refutes universal closure at $d=n=2$; it proves the existential conjecture. |

All binders describe the source's finite matrix representation. There is no additional mathematical hypothesis on the settling theorem and no dropped clause.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/LudersSubObservableClosure.result

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    Source 1
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    The exact recorded statement is proved.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2022Absolute two-incoherence beyond the purity boundAbsolute two-incoherence beyond the purity bound.Refuted

The questionJohnston, Moein, Pereira and Plosker, Absolutely k-incoherent quantum states and spectral inequalities for factor width of a matrix, Phys.

Why it mattersThe purity ball in Theorem 7 is sufficient for absolute two-incoherence.

The ideaThe proof uses spectral estimates.

Our contributionAbsolute two-incoherence beyond the purity bound.

Verification details
Release source recordRefuted in Lean

Open problem

Absolute two-incoherence beyond the purity bound

Johnston, Moein, Pereira and Plosker, Absolutely k-incoherent quantum states and spectral inequalities for factor width of a matrix, Phys. Rev. A 106, 052417, arXiv:2205.05110v2, Section 5, printed page 17:

Theorem 7 gives a sufficient condition for absolute 2-incoherence that is also necessary in dimension n = 3. It is unclear whether or not it is necessary in dimension n = 4 or higher.

The condition is $\operatorname{Tr}(\rho^2)\le1/(n-1)$. The universal necessity assertion for all $n\ge4$ is refuted by one density matrix in dimension five. The source and literal definitions are in the Library note; the external question is preregistered in #17863.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | A vector has at most $k$ nonzero entries | FactorWidthTwoCone.IsKSparse | The finite support-cardinality expression. | | A finite sum $\sum_jv_jv_j^*$ of such vectors | FactorWidthTwoCone.IsKIncoherentSum | Every summand has the required support; individual vectors need not be normalized. | | A mixed quantum state | GHZMeasureBiseparableBound.IsDensity | Positive semidefinite matrix with complex trace one. | | Every unitary conjugate is $k$-incoherent | AbsoluteTwoIncoherencePurityRefutation.IsAbsolutelyKIncoherent | All complex unitary matrices are quantified. | | Necessity for dimension four or higher | AbsoluteTwoIncoherencePurityRefutation.claim | Every natural $n\ge4$, every density matrix satisfying absolute two-incoherence. | | The eigenvalue purity bound | Real part of the trace of the square | For Hermitian matrices, the spectral theorem identifies this with the sum of squared eigenvalues. | | Refutation | AbsoluteTwoIncoherencePurityRefutation.result | Exact negation of claim, witnessed at $n=5$. |

FactorWidthTwoCone.IsLocally2 is the Hermitian support-two quadratic-form condition in Definition 1(b). It is an auxiliary dual-cone predicate, not an extra hypothesis in claim or result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/AbsoluteTwoIncoherencePurityRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2022Zero belongs to the spectrum of a finite-range translation-invariant lattice LindbladianFinite-range translation-invariant lattice Lindblad matrices define a bounded operator with spectral zero.Proved

The questionFrederik Ravn Klausen, Spectra of generators of Markovian evolution in the thermodynamic limit: From non-Hermitian to full evolution via tridiagonal Laurent matrices, arXiv:2206.09879v3 (2025), asks in “Further questions” (section04.tex, line 574; verbatim TeX):

Why it mattersThe missing spectral-membership premise is exactly the premise in Klausen's thm:nogap that is additional to the three structural assumptions.

The ideaThe commutator terms cancel and the jump gain cancels the two half losses, with all relevant sums finite.

Our contributionFinite-range translation-invariant lattice Lindblad matrices define a bounded operator with spectral zero.

Scope & next questionRemoving rank one from that alternative remains open in this settlement, as do the source's separate finite-volume gap-closing and spectral-convergence questions.

Verification details
Release source recordProved in Lean

Open problem

Zero belongs to the spectrum of a finite-range translation-invariant lattice Lindbladian

Frederik Ravn Klausen, Spectra of generators of Markovian evolution in the thermodynamic limit: From non-Hermitian to full evolution via tridiagonal Laurent matrices, arXiv:2206.09879v3 (2025), asks in “Further questions” (section04.tex, line 574; verbatim TeX):

Prove that if $\Li$ is \eqref{eq:finite-range}, \eqref{eq:translation-invariance}, and \eqref{eq:Rank1} then $0 \in \sigma(\Li)$, thereby strengthening \Cref{thm:no_gap}.

Here the source macro \Li denotes $\mathcal L$, the three assumption labels mean (Finite Range), (TransInv), and (Rank 1), and the theorem label is thm:no_gap. The source assumes bounded self-adjoint $H$ and $G>0$; (Rank 1) means $\operatorname{rank}(L_k)=1$ for every $k\in\mathbb Z$ (section02.tex, lines 12 and 50). It takes $\mathcal L$ on $\mathrm{HS}(\ell^2(\mathbb Z))$ (section02.tex, lines 6–13 and 54–57). Its theorem (section03.tex, lines 506–509) reads:

If $\Li$ is \eqref{eq:finite-range}, \eqref{eq:translation-invariance}, \eqref{eq:Rank1} and $0 \in \sigma(\Li)$. Then $\Li$ is gapless or $\dim(\ker(\Li))=\infty.$

Issue #23482 fixes the matrix-kernel reading. Put $E=\ell^2(\mathbb Z\times\mathbb Z;\mathbb C)$, the module's ObserverHilbertSpace (ℤ × ℤ). For every $r\in\mathbb N$, every $g\in\mathbb R$, every $h:\mathbb Z^2\to\mathbb C$, and every $\ell:\mathbb Z^3\to\mathbb C$, assume:

  • IsFiniteRange r h ℓ: $h(x,y)=0$ whenever $|x-y|>r$, and $\ell(k,x,y)=0$ whenever $\max{|x-k|,|y-k|}>r$;
  • IsTranslationInvariant h ℓ: $h(x+1,y+1)=h(x,y)$ and $\ell(k+1,x+1,y+1)=\ell(k,x,y)$ for all integer indices.

The conclusion of result is that there exists a continuous complex linear operator $T:E\to E$ such that, for every $\rho\in E$ and every $x,y\in\mathbb Z$, $(T\rho)(x,y)=\operatorname{lindbladEntry}(g,h,\ell,\rho,x,y)$, and $0\in\sigma_{\mathbb C}(T)$. Explicitly, the entry is

$$ \begin{aligned} &-i\left(\sum_z h(x,z)\rho(z,y)-\sum_z\rho(x,z)h(z,y)\right)\ &\quad+g\sum_k\left[ \sum_u\sum_v\ell(k,x,u)\rho(u,v)\overline{\ell(k,y,v)} -\frac12\sum_z\left(\sum_w\overline{\ell(k,w,x)}\ell(k,w,z)\right)\rho(z,y) -\frac12\sum_z\rho(x,z)\left(\sum_w\overline{\ell(k,w,z)}\ell(k,w,y)\right) \right]. \end{aligned} $$

All indices in these sums range over $\mathbb Z$; the Lean definition uses unconditional sums, whose nonzero terms are finite under IsFiniteRange. The theorem imposes no rank condition on the jumps, no self-adjointness condition on $h$, and no sign condition on $g$. Its specialization to $g>0$ and self-adjoint $h$ answers the registered question, including its rank-one case.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/LatticeLindbladianZeroSpectrum.result

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    Source 1
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Analysis & dynamicsQuestion resolvedFirst preprint 2022Strength three of every Mariot–Manzoni MOCA arrayEvery three distinct columns in the binary array of mutually orthogonal bipermutive cellular automata contain each binary triple equally often.Proved

The questionManzoni, Building Correlation Immune Functions from Sets of Mutually Orthogonal Cellular Automata, arXiv:2207.08280v1, Section 5, asks to “prove that the binary OA obtained from a set of k-MOCA has always at least strength 3”.

Why it mattersSection 5 asks whether strength three always holds; this is the tier-1 external named question settled here.

The ideaA coordinate flip reverses a column's endpoint bit.

Our contributionEvery three distinct columns in the binary array of mutually orthogonal bipermutive cellular automata contain each binary triple equally often.

Verification details
Release source recordProved in Lean

Open problem

Strength three of every Mariot–Manzoni MOCA array

L. Mariot and L. Manzoni, Building Correlation Immune Functions from Sets of Mutually Orthogonal Cellular Automata, arXiv:2207.08280v1, Section 5, asks to “prove that the binary OA obtained from a set of k-MOCA has always at least strength 3”. The precise target and source quotations are preregistered in #19624. The source is described in the Library note.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Definition 1: sliding windows of length $d$ | ca | $n=2b$, $d=b+1$, zero-based finite indices | | Bipermutive: $f(y)=y_0+\phi(y_1,\ldots,y_{b-1})+y_b$ over $\mathbf F_2$ | IsBipermutive | existential arbitrary central Boolean function | | Eq. (6): unequal superposed pairs at distinct cells | AreOrthogonal | input halves joined by Fin.append | | A family of $k$ pairwise orthogonal bipermutive CA | IsMOCA | every rule bipermutive, distinct rule indices orthogonal | | Eq. (7): juxtaposition of all $k$ output vectors | mocaArray | all $2^{2b}$ input rows, columns in Fin k × Fin b | | Each pattern occurs $N/2^t$ times in any $t$ distinct columns | IsOrthogonalArray | injective column selections, subtype cardinality, natural division | | Always strength at least three | claim, result | all $b,k$ and all such families; strength $t=3$ |

For $b=0$ the output is empty; this degenerate extension exceeds Definition 1's $d\le n$ parameter range and introduces no three-column selection. The same vacuity applies whenever $kb<3$. The theorem retains the complete source range $b\ge1$ and does not assert that three columns exist.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/MariotManzoniMOCAStrength.result

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    Source 1
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    The exact recorded statement is proved.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2023Monkman–Sirker configurational entropy is Schur concavePoisson-binomial configurational entropy is nondecreasing under majorization throughout the closed unit cube.Proved

The questionSirker, Symmetry-resolved entanglement: general considerations, calculation from correlation functions, and bounds for symmetry-protected topological phases, J.

Why it mattersThe entropy of N is Schur concave, as is the sum of binary entropies, but subtracting these two functions does not determine the majorization order of their difference.

The ideaExpand the Bernoulli-sum law as a finite subset sum.

Our contributionPoisson-binomial configurational entropy is nondecreasing under majorization throughout the closed unit cube.

Scope & next questionStrict Schur concavity and equality classification also remain open here.

Verification details
Release source recordProved in Lean

Open problem

Monkman–Sirker configurational entropy is Schur concave

D. Monkman and J. Sirker, Symmetry-resolved entanglement: general considerations, calculation from correlation functions, and bounds for symmetry-protected topological phases, J. Phys. A 56 (2023), 495001; arXiv:2307.05820, Section 3.3, “A conjecture on the concavity of Sc”:

We note that the configurational entropy Sc = S + (−SN) is a sum of a concave and a convex function. Therefore, one cannot directly apply majorization to Sc based on the already obtained results for S and SN. However, this does not rule our that Sc is also a concave function of the parameters {Ck}. Similar to the original Shepp-Olkin paper [22] we will here just investigate the cases M = 2 and M = 3. Based on these results, we conjecture that Sc({Ck}) is indeed Schur concave and majorization thus applies.

For independent Bernoulli variables X₁,…,X_M with probabilities c₁,…,c_M, put N=Σk Xk, S(c)=Σk h(ck) and SN(c)=H(N). Natural logarithms and 0 log 0=0 are used. The claim is that F(c)=Sc(c)=S(c)−SN(c) is Schur concave on [0,1]^M for every natural M: if d has the same coordinate sum as c and every descending partial sum of d is no larger than the corresponding one of c, then F(c)≤F(d).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/PoissonBinomial/MonkmanSirkerConfigurationalEntropy.result

Frozen module record

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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2022The separable CHSH optimum under imprecise qubit measurements is attainedThe sharp separable CHSH value for two qubits with imprecisely calibrated binary observables is attained and optimal.Proved

The questionTavakoli, Entanglement detection with imprecise measurements, Phys.

Why it mattersThe source studies entanglement detection with operationally bounded measurement inaccuracies.

The ideaThis supplies the upper bound on the full state set in claim.

Our contributionThe sharp separable CHSH value for two qubits with imprecisely calibrated binary observables is attained and optimal.

Scope & next questionThe source's other witnesses, higher-dimensional tightness question involving SIC-POVMs, entangled values and SDP methods are outside this result, as are its general-correction, auxiliary-setting and multipartite questions at lines 182–183.

Verification details
Release source recordProved in Lean

Open problem

The separable CHSH optimum under imprecise qubit measurements is attained

S. Morelli, H. Yamasaki, M. Huber and A. Tavakoli, Entanglement detection with imprecise measurements, Phys. Rev. Lett. 128, 250501 (2022), arXiv:2202.13131v2, main text, Manuscript_epsilonEW.tex, line 111:

However, as we show in Supplementary Material through an explicit separable model that we conjecture to be optimal, this fact does not qualitatively improve the robustness of idealised ($\varepsilon=0$) entanglement detection to small inaccuracies.

The witness and candidate value are given at lines 111–115; the supplement, Section “Entanglement detection based on the CHSH quantity” (AppCHSH, lines 232–256), specifies two qubits, targets $X=\sigma_X$ and $Z=\sigma_Z$ on both sites, and a common inaccuracy bound $\varepsilon$. See the literature note.

In the formal statement, each of $A_1,A_2,B_1,B_2$ is a complex $2\times2$ matrix satisfying IsBinaryObservable: it is Hermitian and $I-A$ and $I+A$ are positive semidefinite. Admissible additionally requires

$\operatorname{Re}\operatorname{Tr}(A_1X),\operatorname{Re}\operatorname{Tr}(A_2Z),\operatorname{Re}\operatorname{Tr}(B_1X),\operatorname{Re}\operatorname{Tr}(B_2Z)\ge 2-4\varepsilon$.

Separable means membership in the real convex hull of $\sigma\otimes\tau$, where $\sigma,\tau$ are positive semidefinite complex $2\times2$ matrices with trace one. The witness is

$W(\rho;A_1,A_2,B_1,B_2)=\operatorname{Re}\operatorname{Tr}[\rho(A_1\otimes B_1+A_1\otimes B_2+A_2\otimes B_1-A_2\otimes B_2)]$.

Put $\varepsilon_0=\frac12-\frac1{2\sqrt2}$. The definition separableValue is the source's candidate:

  • for $\varepsilon\le\varepsilon_0$, $V(\varepsilon)=4(1-2\varepsilon)\sqrt{\varepsilon(1-\varepsilon)}+\sqrt{2-16\varepsilon(1-\varepsilon)(1-2\varepsilon)^2}$;
  • for $\varepsilon>\varepsilon_0$, $V(\varepsilon)=2$.

Issue #22442 fixes the reading: for every $\varepsilon\in[0,1]$, $V(\varepsilon)$ is a greatest element of the set of witness values attained by separable states and admissible observables. This is claim, proved by result : claim. Greatestness includes both attainment and an upper bound on every value in that set.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/ImpreciseCHSHSeparableOptimum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2022The largest root of Dukes's permaspin cubic dominates the two explicit eigenvaluesFor positive real c and d, the supremum of the real roots of Dukes's cubic dominates c(1-d)^2 and c(1-d^2).Proved

The questionMark Dukes, An Ising model having permutation spin motivated by a permutation complexity measure, Physica A 626 (2023) 129090, DOI 10.1016/j.physa.2023.129090, arXiv:2211.08908v3, Section 4, permaspin.tex, lines 510–513:

Why it mattersThe source identifies the cubic's three roots and the two explicit expressions as its eigenvalue candidates, with one explicit eigenvalue repeated.

The ideaIt applies the intermediate value theorem to the continuous polynomial evaluation, its tendency to positive infinity at positive infinity, and cubicatcomparisonneg; degree three and leading coefficient one supply the degree and leading-coefficient hypotheses.

Our contributionFor positive real c and d, the supremum of the real roots of Dukes's cubic dominates c(1-d)^2 and c(1-d^2).

Verification details
Release source recordProved in Lean

Open problem

The largest root of Dukes's permaspin cubic dominates the two explicit eigenvalues

Mark Dukes, An Ising model having permutation spin motivated by a permutation complexity measure, Physica A 626 (2023) 129090, DOI 10.1016/j.physa.2023.129090, arXiv:2211.08908v3, Section 4, permaspin.tex, lines 510–513:

Let $\lambda^{*}$ be the largest root of $t(\lambda)=0$. It remains to see which of the eigenvalues $\lambda^{*}$, $\lambda_4$, and $\lambda_5$ is largest. Numerical investigations indicate that $\lambda^{*}$ appears to be the largest and this is supported by the evidence presented by in Figures~\ref{figblue1} and \ref{figblue2}.

The source sets $c=a^4$ and $d=b^2$, where $a=e^{H'/2}$ and $b=e^{J'}$ (lines 458 and 490). These parameters are positive for real $H',J'$. Its two explicit expressions are $\lambda_4=c(1-d)^2$ and $\lambda_5=c(1-d^2)$ (lines 500–503), and its cubic is (lines 504–509)

$t_{c,d}(x)=x^3-(cd^2+c^2+2cd+c+1)x^2-c(cd^3+3c^2d+cd^2+c^2+cd+c+3d+1)(d-1)x+(d^2+4d+1)c^3(d+1)(d-1)^3$.

The definitions and source context are in the literature note. The module's dukesCubic c d is this polynomial, defined for all real $c,d$, and largestRoot c d is $\lambda^*(c,d)=\sup S_{c,d}$, where $S_{c,d}={x\in\mathbb R:t_{c,d}(x)=0}$.

Issue #22514 fixes the reading: for every real $c>0$ and $d>0$, $S_{c,d}$ is nonempty and bounded above, and $\max(\max(\lambda^(c,d),c(1-d)^2),c(1-d^2))=\lambda^(c,d)$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/PermaspinTransferEigenvalueOrder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding 2143, 1423 and 3124Fishburn permutations and classical permutations avoiding specified patterns are classified by their forms and permitted insertion positions.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaInserting a new maximum into a Fishburn permutation keeps it Fishburn exactly at the far left or right after an entry x with x = 1 or x − 1 to its left; deleting the maximum is the parent operation for all three classes.

Our contributionFishburn permutations and classical permutations avoiding specified patterns are classified by their forms and permitted insertion positions.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding 2143, 1423 and 3124

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.10. For all n ≥ 0, |Fn(2143, 1423, 3124)| = |Sn(321, 2143, 3124)| = |Sn(231, 4132, 2134)|.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenTen.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding 1243 and 3124Mathematical definitions and results for Fishburn permutations and classical pattern avoidance.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaPaths with nonzero sites in the Fishburn tree correspond to the blocks of this decomposition, so both sides have the same counts.

Our contributionMathematical definitions and results for Fishburn permutations and classical pattern avoidance.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding 1243 and 3124

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.12. For all n ≥ 0, |Fn(1243, 3124)| = |Sn(231, 4123)|.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenTwelve.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresQuestion resolvedFirst preprint 2022Fishburn Permutations Avoiding Pairs Counted by (n − 1)2^(n−2) + 1Mathematical definitions and results for Fishburn permutations and classical pattern avoidance.Proved

The questionEgge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

The ideaInserting a new maximum right after an entry x keeps a Fishburn permutation Fishburn exactly when x = 1 or x − 1 lies to the left of x; insertion at the far left is always allowed, and deleting the maximum keeps the Fishburn condition.

Our contributionMathematical definitions and results for Fishburn permutations and classical pattern avoidance.

Verification details
Release source recordProved in Lean

Open problem

Fishburn Permutations Avoiding Pairs Counted by (n − 1)2^(n−2) + 1

Eric S. Egge, Pattern-Avoiding Fishburn Permutations and Ascent Sequences, arXiv:2208.01484v1, Section 10:

Conjecture 10.4. For all n ≥ 1, |Fn(1324, 2143)| = |Fn(1423, 2143)| = |Fn(1423, 3124)| = (n − 1)2^{n−2} + 1.

A permutation p is Fishburn when there are no indices i and j > i + 1 with p_i = p_j + 1 < p_{i+1}; F_n(B) and S_n(B) are the Fishburn permutations and all permutations of length n avoiding the classical patterns in B.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Fishburn/FishburnTenFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2022Furuichi–Eghbali Amlashi: the hyperbolic ratio below oneThe logarithmic mean hyperbolic ratio below parameter one.Proved

The questionEghbali Amlashi, On bounds of logarithmic mean and mean inequality chain, arXiv:2203.01134v2, Remark 3.2(i), proposes the positive definite function conjecture.

Why it mattersNumerical counterexamples at individual parameters do not decide the full real interval below one.

The ideaRestricting the complex power series to the real axis shows that its coefficients are real.

Our contributionThe logarithmic mean hyperbolic ratio below parameter one.

Verification details
Release source recordProved in Lean

Open problem

Furuichi–Eghbali Amlashi: the hyperbolic ratio below one

S. Furuichi and M. Eghbali Amlashi, On bounds of logarithmic mean and mean inequality chain, arXiv:2203.01134v2, Remark 3.2(i), proposes the positive definite function conjecture. The supplied source identifies this with Remark 2(i) in Mathematical Inequalities & Applications 27(2) (2024), 379–399.

The ratio is

$$ \frac{K_r(e^{2t},1)}{L(e^{2t},1)} =\frac{t(\cosh t)^{1-r}}{\sinh t}=: \phi_r(t),\qquad \phi_r(0)=1, $$

where $K_r=G^r A^{1-r}$ and $L$ is the logarithmic mean. The source states:

Thus these numerical computations support the conjecture that the function ϕr (t) for r < 1, is not positive definite.

The exact target is: for every real $r<1$, the complex cast of $\phi_r$ is not positive definite. Positive definiteness means that for every natural $m$ and every $x:\operatorname{Fin}(m)\to\mathbb R$, the matrix with entries $f(x_j-x_i)$ is positive semidefinite. The existing FiniteInversePoissonCriterion.PositiveDefinite supplies this definition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/FuruichiAmlashiHyperbolicRatio.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Analysis & dynamicsQuestion resolvedFirst preprint 2023Optimal hyperbolic cotangent constantThe optimal real coefficient in the Hirai–Nieuwboer–Walter hyperbolic cotangent inequality is one over the square root of two.Proved

The questionHirai–Nieuwboer–Walter, arXiv:2303.04771v2, §5, Lemma 5.13 and Remark 5.14:

Why it mattersThe scalar inequality controls third derivatives of squared Riemannian distance on positive definite matrices.

Our contributionThe optimal real coefficient in the Hirai–Nieuwboer–Walter hyperbolic cotangent inequality is one over the square root of two.

Verification details
Release source recordProved in Lean

Open problem

Optimal hyperbolic cotangent constant

Hirai–Nieuwboer–Walter, arXiv:2303.04771v2, §5, Lemma 5.13 and Remark 5.14:

"Lemma 5.13. The constant C = √2 is such that for all a, b ∈ ℝ, one has |coth((a − b)/2) (a coth(a/2) − b coth(b/2))| ≤ C √(ab(a − b) coth(a/2) coth(b/2) coth((a − b)/2))."

"Remark 5.14. We conjecture, based on numerical evidence, that the optimal constant in the above inequality is C = 1/√2."

The source proves the coefficient $\sqrt2$ and conjectures the optimal coefficient $1/\sqrt2$. The repository resolves that conjecture: the optimal constant in

$$ \left|\coth\frac{a-b}{2} \left(a\coth\frac a2-b\coth\frac b2\right)\right| \le C\sqrt{ab(a-b)\coth\frac a2\coth\frac b2\coth\frac{a-b}{2}} $$

is $C=1/\sqrt2$. The quantifiers range over all real $a,b$ with $a\ne0$, $b\ne0$, and $a\ne b$. Optimality means that for every real $C<1/\sqrt2$, some such pair violates the estimate strictly. Here $\coth x=\cosh x/\sinh x$ and square roots are nonnegative real roots.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Hyperbolic/HiraiNieuwboerWalterCothConstant.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2023The Meer–Weeks fifth-moment bounds for distributions bounded on one sideFifth central moment inequalities for probability laws bounded on one side.Proved

The questionWeeks, Bounded distributions place limits on skewness and larger moments, PLOS ONE 19, e0297862 (2024), arXiv:2308.05006v2, subsection “Higher order generalized moments”, ploslatextemplate.tex, lines 506–510:

Why it mattersThe paper relates support bounds, the coefficient of variation and standardized moments.

The ideaThis gives the upper-support implication. result supplies both implications for every probability measure satisfying the finite fifth absolute moment and positive variance hypotheses.

Our contributionFifth central moment inequalities for probability laws bounded on one side.

Verification details
Release source recordProved in Lean

Open problem

The Meer–Weeks fifth-moment bounds for distributions bounded on one side

D. J. Meer and E. R. Weeks, Bounded distributions place limits on skewness and larger moments, PLOS ONE 19, e0297862 (2024), arXiv:2308.05006v2, subsection “Higher order generalized moments”, plos_latex_template.tex, lines 506–510:

One can immediately put in a value for $a$ of interest and get a potential limit of $D_n$. For example, for distributions bounded from below by $x_{\rm min}$ we conjecture

$D_5 \geq D_{5,{\rm min}} = z^3 - z^1 + z^{-1} - z^{-3}$

with $z = \delta / (1 - x_{\rm min}/\mu)$ as above. As with $D_3$, our conjectured $D_{5,{\rm max}}$ is a similar equation using $z = \delta / (x_{\rm max}/\mu - 1)$.

The quotation preserves the source's text and formula, with its displayed equation rendered as a separate mathematical paragraph. The v2 TeX was read directly; its MD5 is 4452c28a26c7a76c87642af46fe0c8a9. The source defines central moments and standardized moments at lines 228–236, the original coefficient of variation at lines 290–295, and the shifted one-sided ratios at lines 397–409. See the literature note.

Write $\nu$ for the probability law, distinguishing it from the paper's mean $\mu$, take $X(x)=x$, and put

$m=\int x,d\nu(x)$, $v=\int(x-m)^2,d\nu(x)$, $\sigma=\sqrt v$, $D_5=\int(x-m)^5,d\nu(x)/\sigma^5$, and $F(z)=z^3-z+z^{-1}-z^{-3}$.

In Lean, these are Defs.mean, centralMoment id 2, Real.sqrt, the ratio involving centralMoment id 5, and fifthBound, respectively. Issue #21300 fixes the statement: for every Borel probability law on $\mathbb R$ with $\int|x|^5,d\nu(x)<\infty$ and $v>0$, both of the following implications hold:

  • for every real $L$, if $X\ge L$ almost surely, then $F(\sigma/(m-L))\le D_5$;
  • for every real $U$, if $X\le U$ almost surely, then $D_5\le-F(\sigma/(U-m))$.

This statement is proved by result : claim. Attainment by two-point laws, also discussed in #21300, is not part of claim or result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MomentMinors/BoundedFifthMomentLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2020The attained coherence optimum for arbitrary finite qubit Lüders measurementsEvery qubit density matrix has an attained maximum of output coherence under finite nonselective measurements, with exact zero and maximal gain criteria.Proved

The questionPati, Creation of quantum coherence with general measurement processes, Quantum Information Processing 21, 48 (2022), arXiv:2004.10875v2, Section 3.2, coh-measure.tex, line 298:

Why it mattersThe paper asks how much coherence a measurement can create when its outcome is discarded.

The ideaThe lemma names below belong to LudersCoherenceCreationOptimum unless qualified otherwise.

Our contributionEvery qubit density matrix has an attained maximum of output coherence under finite nonselective measurements, with exact zero and maximal gain criteria.

Verification details
Release source recordProved in Lean

Open problem

The attained coherence optimum for arbitrary finite qubit Lüders measurements

S. D. Mohanty, G. Sharma, Sk Sazim, B. Pradhan and A. K. Pati, Creation of quantum coherence with general measurement processes, Quantum Information Processing 21, 48 (2022), arXiv:2004.10875v2, Section 3.2, coh-measure.tex, line 298:

We couldn't get the expression for $\max_{{\vec{a},a_{\pm}}}C_{l_1}(\rho'{\pm})$ analytically, so we have plotted it against the initial state coherence $C{l_1}(\rho_{\pm})$.

However in this case, we don't know yet for which states there can not be any coherence gain and for which states the coherence gain will be maximum.

The Conclusion, line 505, asks:

As a future line of work, it would be interesting to obtain the analytical results for two outcome and n-outcome POVMs.

The source defines $C_{l_1}(\rho)=\sum_{i\ne j}|\rho_{ij}|$ at line 109 and the nonselective Lüders action $\Phi_E(\rho)=\sum_i\sqrt{E_i}\rho\sqrt{E_i}$ at lines 127–130. A qubit density matrix is a positive semidefinite complex $2\times2$ matrix of trace one; an $n$-outcome POVM is a family of positive semidefinite effects indexed by $\mathrm{Fin},n$ whose sum is $I$.

Put $s=C_{l_1}(\rho)$, $R=\sqrt{(\operatorname{Re}(\rho_{00}-\rho_{11}))^2+4|\rho_{01}|^2}$ and $\lambda=(s+R)/2$. Issue #22649 fixes the statement: for every qubit density matrix and every integer $n\ge2$, $\lambda$ is the attained greatest output coherence; the optimal gain $\lambda-s$ vanishes exactly when $\rho_{00}=\rho_{11}$, is at most $1/2$, and equals $1/2$ exactly for $\operatorname{diag}(1,0)$ and $\operatorname{diag}(0,1)$. This is proved.

The definitions and source locators are also in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/LudersCoherenceCreationOptimum.result

Frozen module record

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Number theoryQuestion resolvedPosed in 2024The A091259 congruenceThe reduction can consume whatever it likes: if every divisor on the denominator side is one modulo three, the reduced numerator is determined anyway.Proved

The questionTake the ratio of the divisor sum of cubes to the ordinary divisor sum, reduce it to lowest terms, and read the numerator modulo three.

Why it mattersDirect inspection on September 9, 2026 found it still labelled Conjecture, and the entry is marked easy.

The ideaAt a prime power the ratio of the two divisor sums equals a ratio of values of the quadratic whose roots are the primitive cube roots of unity, with the argument raised one exponent higher in the numerator.

Our contributionThe reduction can consume whatever it likes: if every divisor on the denominator side is one modulo three, the reduced numerator is determined anyway.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A091259 congruence

Take the ratio of the divisor sum of cubes to the ordinary divisor sum, reduce it to lowest terms, and read the numerator modulo three. Michel Marcus conjectured on August 11, 2024 that the result is the indicator of a prime-exponent condition: one exactly when every prime congruent to two modulo three occurs to an even exponent, zero otherwise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/A091259.a091259_mod_three

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    Source 1
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Number theoryQuestion resolvedPosed in 2024Bala's alternating gcd-sum formula for A344598Bala's alternating gcd sum equals the floor-square totient sequence A344598.Proved

The questiona(n) = Sum{k=1..n} phi(k) (floor(n/k)^2 - floor((n-1)/k)^2).

Why it mattersKPI = open problems resolved, not module counts or a finite prefix checked.

The ideaTwo gcd bookkeeping identities finish the proof. pillaifourmul proves pillai(4n) + 4pillai(n) = 4pillai(2n) by an even/odd range decomposition and equality of shifted odd-index gcd blocks. altGcdSumeq proves 2altGcdSum(n) + 2n = 4pillai(2n) - pillai(4n) by a reflected-index construction with endpoint accounting.

Our contributionBala's alternating gcd sum equals the floor-square totient sequence A344598.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's alternating gcd-sum formula for A344598

OEIS A344598 NAME (Seiichi Manyama, May 24 2021):

a(n) = Sum_{k=1..n} phi(k) * (floor(n/k)^2 - floor((n-1)/k)^2).

FORMULA (Peter Bala, Jan 01 2024):

Conjecture: a(n) = Sum_{k = 1..2*n} (-1)^k * gcd(k, 4*n). Cf. A344372. - Peter Bala, Jan 01 2024

Both quotations are copied verbatim from the claim in Library/Arith/bala2024a344598.md. The target is the equality for every positive n.

范围与诚实边界: This is a modest identity (tier 1). The two divisor-sum formulas a(n) = Sum_{d|n} phi(d)*(2*n/d - 1) and a(n) = Sum_{d|n} (2*d*tau(d) - sigma(d))*mu(n/d) were already proved elsewhere by Daniel Weber, as credited in Mikhail Kurkov's Mar 31 2026 comment. They are not the target. The target is the alternating-gcd formula, which the search seat's 2026-09-08 reading recorded as still marked Conjecture. Natural division and truncated subtraction define a; the alternating sum and theorem subtraction are in the integers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/AlternatingGcdSumPillai.bala_conjecture

Frozen module record

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Number theoryQuestion resolvedPosed in 2024The A373561 divisor-bucketed quadruple sumA divisor condition sorts the summands into classes rather than restricting them, leaving a square-sum identity.Proved

The questionThree indices run over the interval from one to a modulus and contribute the quantity x^2 + y^2 - z^2.

Why it mattersDirect inspection on September 9, 2026 found the statement still labelled Conjecture, added at revision 17 with no later revision converting it to a theorem.

The ideaThe divisor condition sorts rather than restricts.

Our contributionA divisor condition sorts the summands into classes rather than restricting them, leaving a square-sum identity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A373561 divisor-bucketed quadruple sum

Three indices run over the interval from one to a modulus and contribute the quantity x^2 + y^2 - z^2. A fourth index selects, for each triple, only those whose greatest common divisor with the modulus equals it. Mats Granvik conjectured on June 10, 2024 that the resulting fourfold sum equals the cube of the modulus times the sum of the first squares.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/A373561.a373561

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsQuestion resolvedFirst preprint 2022The sharp lower bound on skewness times kurtosis for unidirectional unicyclic currentsThe product of skewness and kurtosis of every finite positive unidirectional cycle strictly exceeds its greatest lower bound over all cycle sizes.Proved

The questionPtaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys.

Why it mattersThe product bound constrains two current-cumulant ratios jointly.

The ideaAll lemma names below refer to UnicyclicSkewnessTimesKurtosisLowerBound unless another module is named.

Our contributionThe product of skewness and kurtosis of every finite positive unidirectional cycle strictly exceeds its greatest lower bound over all cycle sizes.

Verification details
Release source recordProved in Lean

Open problem

The sharp lower bound on skewness times kurtosis for unidirectional unicyclic currents

K. Ptaszyński, Bounds on skewness and kurtosis of steady-state currents, Phys. Rev. E 106, 024119 (2022), arXiv:2204.08421v4, gives the following numerical conjecture in the Appendix (main.tex, line 792):

it was numerically conjectured that the minimum and maximum values of $\mathcal{S}$, $\mathcal{K}$, $\mathcal{S}+\mathcal{K}$, $\mathcal{S}-\mathcal{K}$ and $\mathcal{S} \times \mathcal{K}$ are always obtained for a specific type of network topology, with all rates $k_{ij}$ equal to $k$ apart from one (here $k_{21}$) equal to $ak$, with the parameter $a$ depending on the optimized quantity and the number of states $N$.

For the product's lower endpoint, the subsection “Unidirectional networks” gives (main.tex, line 445):

$\min(\mathcal{S} \times \mathcal{K})=\min_A \frac{1-10A+22A^2-12A^3}{(1+A)^8}$.

Line 424 specifies the limit $N\to\infty$ with $a=A/N$; line 439 gives the approximate minimum $-0.054$ at $A\approx0.23$. The summary's Eq. produni states $\mathcal{K}\times\mathcal{S}\in[-0.054,1]$ (lines 129–137), explicitly treating decimals as approximate. These quotations were checked against the v4 source member main.tex, MD5 5b72a7d5de1e5c34c76db21badb56492.

Issue #25992 fixes the reading to the sharp lower endpoint for unidirectional cycles. Write $R_j=\sum_{i=1}^N k_i^{-j}$ for a strictly positive real rate vector $k:\mathrm{Fin},N\to\mathbb R$, and use the existing definitions

$\mathcal S(k)=3R_2^2/R_1^4-2R_3/R_1^3$,

$\mathcal K(k)=15R_2^2/R_1^4-20R_3/R_1^3+6R_4/(R_1^2R_2)$.

These are UnicyclicSkewnessMinimum.skewness and UnicyclicKurtosisLowerBound.kurtosis, respectively. They match the source's current-cumulant ratios (lines 53–54) upon substitution of its waiting-time formulas (Appendix, lines 772–790); they are not the usual standardized third and fourth moments of a probability distribution.

Define $E(A)=(1-10A+22A^2-12A^3)/(1+A)^8$ and $C=\inf{E(A):A>0}$, the module's lowerBound. The statement is that $\mathcal S(k)\mathcal K(k)>C$ for every $N\ge2$ and every strictly positive real rate vector, and that $C$ is the greatest lower bound of all these products, allowing $N$ to vary. Both factors use the same vector; repeated rates are allowed. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/UnicyclicSkewnessTimesKurtosisLowerBound.result

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2022Two unbounded classicalization gaps with a qubitBoth classicalization gaps for tripartite negativity are unbounded with a qubit measured party.Proved

The questionZhen-Peng Xu, Satoya Imai and Otfried Gühne, Fate of multiparticle entanglement when one particle becomes classical, arXiv:2206.12834v2, Phys.

Why it mattersA single qubit can control two distinct losses: the unavoidable loss when its quantum information is replaced by any measurement-and-prepare register, and the further loss caused by discarding that register.

The ideaThe output-register cut is positive under partial transpose and has unit trace, hence contributes zero negativity.

Our contributionBoth classicalization gaps for tripartite negativity are unbounded with a qubit measured party.

Verification details
Release source recordProved in Lean

Open problem

Two unbounded classicalization gaps with a qubit

Zhen-Peng Xu, Satoya Imai and Otfried Gühne, Fate of multiparticle entanglement when one particle becomes classical, arXiv:2206.12834v2, Phys. Rev. A 107, L040401, Section V.B, page 5:

Inspired by those two examples, an interesting question arises whether there exist entanglement measures E and states ρ_ABC such that both Δ_E(ρ_ABC) and Δ̃_E(ρ_ABC) − Δ_E(ρ_ABC) can be arbitrarily large in the sense that they are not limited by the size of C, even if C is only a qubit. We leave this question for further research.

The answer is Proved for the fixed tripartite negativity of Equation (4), with no dimension-dependent rescaling. The source and conventions are in the Library note; #19638 is the preregistration.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Equation (4), sum of three bipartite negativities | tripartiteNegativity | Partial transposes on C, A and B; positive negativity $(|\rho^{T_Y}|_1-1)/2$. | | Equation (1), $\sum_i p_i\sigma_i\otimes\tau_i$ | branch, classicalize | Unnormalized branches $p_i\sigma_i=\operatorname{tr}_C(\rho m_i)$, including zero-probability outcomes. | | Equation (2), minimum entanglement loss | delta | Infimum over every finite POVM and every finite register dimension; attainment is proved for the constructed family. | | Equation (9), discard C and prepare its zero flag | deltaTilde | The prepared flag is the zero basis state of a qubit. | | Both gaps arbitrarily large on the same state | claim, result | Every real threshold admits one positive semidefinite trace-one state on $\operatorname{Fin}(a)\times\operatorname{Fin}(b)\times\operatorname{Fin}(2)$ exceeding it in both gaps. |

The density and POVM conditions appear literally in the public proposition and the infimum. The proofs use the frozen IsDensity and IsPOVM predicates with those same defining conditions. The source prints a sum of negative eigenvalues without an absolute value; its examples use the standard positive negativity adopted here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Classicalization/XuImaiGuhneSimultaneousUnboundedness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2026El Zein-Mortada Saturated Packing RefutationA connected seven-vertex graph refutes the printed saturated-subcubic packing conjecture.Refuted

The questionEl Zein and Mortada, arXiv:2603.25113v1, Conjecture 3, printed page 27, state: "Every 2-saturated subcubic graph is (1, 1, 2)-packing colorable." A subcubic graph has maximum degree at most three.

Why it mattersIt separates the statement as printed from the preceding discussion of local girth and makes no assertion about a strengthened local-girth version.

The ideaUse the connected graph with graph6 encoding FhcYG, vertices 0 through 6, and edges 01, 04, 12, 16, 23, 34, 35, 45, and 56.

Our contributionA connected seven-vertex graph refutes the printed saturated-subcubic packing conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

El Zein-Mortada Saturated Packing Refutation

El Zein and Mortada, arXiv:2603.25113v1, Conjecture 3, printed page 27, state: "Every 2-saturated subcubic graph is (1, 1, 2)-packing colorable." A subcubic graph has maximum degree at most three. It is 2-saturated when each degree-three vertex has at most two degree-three neighbors. A (1,1,2)-packing coloring partitions the vertices into two independent classes and one class whose distinct vertices have distance greater than two. The printed conjecture contains no local-girth hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ElZeinMortadaSaturatedPackingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025The original unnormalized achievement-set residue converseIndexed infinite achievement sets refute the unnormalized residue converse.Refuted

The questionSzymon Głąb and Franciszek Prus-Wiśniowski, Achievement sets – current results and open problems, arXiv:2512.17285v1, Section 2, Problem 9, PDF p.6.

Why it mattersThe question connects the Lebesgue measure of an actual infinite subsum set to the arithmetic residues of one finite indexed block.

The ideaThe two positions of coefficient three remain distinct.

Our contributionIndexed infinite achievement sets refute the unnormalized residue converse.

Verification details
Release source recordRefuted in Lean

arXiv

The original unnormalized achievement-set residue converse

Szymon Głąb and Franciszek Prus-Wiśniowski, Achievement sets – current results and open problems, arXiv:2512.17285v1, Section 2, Problem 9, PDF p.6. The complete source question, transcribed in preregistration #23055, is:

Let (k₁,…,kₘ;1/n) be a multigeometric series such that kᵢ∈ℕ, Σ={σ₀<⋯<σₙ₋₁} has n elements, none of k₁,…,kₘ is divisible by n, and E(k₁,…,kₘ;1/n) has positive Lebesgue measure. Is it true that the set of remainders tᵢ<n with σᵢ≡tᵢ (mod n) equals {0,1,…,n−1}?

The same question appears in Szymon Głąb and Mateusz Kula, Kenyon theorem revisited, arXiv:2508.19454v1, Section 5.1, Problem 1, PDF p.18. These source aliases identify one target.

Statement echo

For every $m\in\mathbb N$ with $m>0$, indexed coefficients $k:\operatorname{Fin}(m)\to\mathbb N$ and $n\in\mathbb N$, define

$$ \Sigma(k)=\left{\sum_{i\in\operatorname{Fin}(m)}\varepsilon_i k_i:\varepsilon:\operatorname{Fin}(m)\to{0,1}\right},\qquad a_t=\frac{k_{t\bmod m}}{n^{\lfloor t/m\rfloor+1}}, $$

$$ E(k,n)=\left{x\in\mathbb R:\exists A\subseteq\mathbb N,\ \operatorname{HasSum}\bigl((a_t)_{t\in A},x\bigr)\right}. $$

Here the coefficient subscript includes the proof that $t\bmod m<m$. Equal coefficients at distinct indices remain independently selectable; $\Sigma$ identifies only equal sum values. The first denominator exponent is one. blockSums, sourceTerm and achievement are these three definitions.

The closed proposition claim is the complete universal assertion

$$ \forall m>0\ \forall k:\operatorname{Fin}(m)\to\mathbb N\ \forall n\in\mathbb N,\quad \Bigl[(\forall i,\ k_i>0)\land \operatorname{Antitone}(k)\land n>1\land |\Sigma(k)|=n\land(\forall i,\ n\nmid k_i)\land\lambda(E(k,n))>0\Bigr] \Longrightarrow {s\bmod n:s\in\Sigma(k)}={0,\ldots,n-1}. $$

Antitone means weakly decreasing, so repeated entries are allowed. The $n>1$ condition expresses the source ratio $1/n\in(0,1)$. There is no primitive, gcd-one, normalization or distinct-coefficient hypothesis.

The closed theorem result : ¬ claim proves exactly

$$ \neg!\left(\forall m>0\ \forall k:\operatorname{Fin}(m)\to\mathbb N\ \forall n\in\mathbb N,\quad \Bigl[(\forall i,\ k_i>0)\land \operatorname{Antitone}(k)\land n>1\land |\Sigma(k)|=n\land(\forall i,\ n\nmid k_i)\land\lambda(E(k,n))>0\Bigr] \Longrightarrow {s\bmod n:s\in\Sigma(k)}={0,\ldots,n-1}\right). $$

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Topology/Measures/AchievementResidueConverseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryConjecture refutedPosed in 2004Refutation of the A100952 log-bounded semiprime-partition conjectureA residue-class obstruction refutes the log-bounded partition conjecture in A100952.Refuted

The questionNumbers that cannot be written as pq+r with three distinct primes p, q and r.

Why it mattersA symbolic obstruction for the entire residue class five modulo six disproves every possible threshold, including thresholds larger than sixty.

The ideaFor m = 6t+5, suppose a representation exists.

Our contributionA residue-class obstruction refutes the log-bounded partition conjecture in A100952.

Scope & next questionIt does not resolve the completeness conjecture for A100952 or Seidov's pq-r conjecture.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A100952 log-bounded semiprime-partition conjecture

OEIS A100952, NAME (verbatim):

Numbers that cannot be written as p*q+r with three distinct primes p, q and r.

COMMENT (verbatim; Jonathan Vos Post, Nov 25 2004):

A weaker conjecture: every integer greater than 60 (or some larger value based on further search) may be partitioned into a prime p and a semiprime qr, where the prime p is bounded by log(min(q,r)). Chen (1978) showed that all sufficiently large even numbers are the sum of a prime and the product of at most two primes. Zumkeller's conjecture effectively extends this from "even" to both even and odd integers. - Jonathan Vos Post, Nov 25 2004

AUTHOR (verbatim):

Reinhard Zumkeller, Nov 23 2004

The literal refuted statement is exists B : Nat, 60 <= B and forall m : Nat, B < m -> Rep m, where Rep m means that there exist natural primes p, q, and r satisfying m = p + q*r and (p : Real) <= Real.log ((min q r : Nat) : Real). Here Real.log is the natural logarithm, equal factors q = r are allowed, and no distinctness condition is imposed. The refutation therefore covers every stricter reading that requires the three primes to be distinct. It does not resolve the completeness conjecture for A100952 or Seidov's p*q-r conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/VosPostLogBoundedSemiprimePartitionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2004. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationConjecture refutedFirst preprint 2025Malkoun's Conjecture D fails for Young subgroupsFifteen copies of a three-column Gram matrix violate the proposed lower bound for two class-preserving permutation groups.Refuted

The questionJoseph Malkoun, Finite graphs and configurations of points (2025 preprint; arXiv:2508.13472v3; Journal of Geometry and Physics, DOI 10.1016/j.geomphys.2026.105993), Section "A conjecture in matrix analysis", defines the setting as follows (TeX line 405):

Why it mattersThe source connects this matrix inequality to graph amplitudes generalizing the Atiyah problem on configurations of points.

The ideaThese are blockRelation m (seedRelation 2) and blockRelation m (seedRelation 0).

Our contributionFifteen copies of a three-column Gram matrix violate the proposed lower bound for two class-preserving permutation groups.

Scope & next questionWhether Conjecture D survives a rank-at-most-two restriction or a restriction to real positive semidefinite matrices remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Malkoun's Conjecture D fails for Young subgroups

Joseph Malkoun, Finite graphs and configurations of points (2025 preprint; arXiv:2508.13472v3; Journal of Geometry and Physics, DOI 10.1016/j.geomphys.2026.105993), Section "A conjecture in matrix analysis", defines the setting as follows (TeX line 405):

Given an equivalence relation, say $\sim$, on $[n]$, we let $P_\sim$ be the group of all permutations of $[n]$ which preserve the equivalence classes of $[n]$ under $\sim$. Let $\sim_i$, for $i = 1, 2$, be two equivalence relations on $[n] = {1, \dots, n}$, with associated permutation groups $P_1$, $P_2$, respectively (we write $P_i$ rather than $P_{\sim_i}$ to alleviate the notation).

At TeX lines 407–408:

Let $H$ be a Hermitian positive semidefinite $n \times n$ matrix. We define $f_{\sim_1, \sim_2}(H) = \sum_{\sigma_1 \in P_1} \sum_{\sigma_2 \in P_2} \prod_{i = 1}^n H_{\sigma_1(i), \sigma_2(i)}.$

Conjecture D (TeX lines 410–412) reads, with the source's macro \Real expanded as $\operatorname{Re}$:

Under the above hypotheses, we have $\operatorname{Re}( f_{\sim_1, \sim_2}(H) ) \geq |P_1 \cap P_2| , \prod_{i = 1}^n H_{i i}.$

These quotations were checked in finite_graphs_and_configurations_of_points_v5.tex from the v3 source archive, md5 3c520a352e066a90cdcffb75a2ee568c. See also the literature note.

Issue #22163 fixes the reading: for every natural number $n$, every pair of equivalence relations $r_1,r_2$ on $\mathrm{Fin},n$, and every complex Hermitian positive semidefinite matrix $H$, the proposed inequality holds, with $P_j={\sigma:\forall i,\ r_j(\sigma(i),i)}$. The real-valued formulation uses $\prod_i\operatorname{Re}H_{ii}$. The universal assertion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/MalkounYoungSubgroupConjectureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresQuestion resolvedHanna's mixed-parity block partition product in A124418Parity restriction and gluing prove Hanna's mixed-block partition product.Proved

The questionTriangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having exactly k blocks that contain both odd and even entries (0<=k<=floor(n/2)).

Why it mattersThis is a first-tier OEIS conjecture entered in 2006 and still presented as a conjecture in the checked 2026 revision.

The ideaSplit the carrier into its odd and even entries.

Our contributionParity restriction and gluing prove Hanna's mixed-block partition product.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's mixed-parity block partition product in A124418

OEIS A124418, %N (verbatim):

Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having exactly k blocks that contain both odd and even entries (0<=k<=floor(n/2)).

%C (verbatim):

Row n has 1+floor(n/2) terms. Row sums are the Bell numbers (A000110). T(n,0)=A124419(n).

%F (verbatim; Paul D. Hanna, Nov 08 2006):

Conjecture: T(n,k) = k!*A049020([n/2],k)*A049020([(n+1)/2],k) where A049020(n,k)=Sum_{i=0..n} S2(n,i)*C(i,k) and S2(n,k)=(1/k!)*Sum_{j=0..k} (-1)^(k-j)*C(k,j)*j^n (the Stirling numbers of 2nd kind).

The Lean definition stirling2 r i counts partitions of an r-element set into i blocks, which is the object the source names as the Stirling numbers of the second kind. The alternating-sum closed form for S2 printed in the FORMULA line is a standard identity for this object and is not formalized here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Partitions/MixedParityBlockPartitionProduct.hanna_a124418

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    Source 1
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    Source 1
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    Source 1
Quantum informationCounterexample foundFirst preprint 2023Both proposed spectral-radius bounds for asynchronous iteration are falseA real matrix of spectral radius one half has a unit-delay asynchronous iteration of spectral radius at least five quarters.Refuted

The questionAksoy et al., Seven open problems in applied combinatorics, Journal of Combinatorics 14(4), 559–601 (2023), arXiv:2303.11464v1, includes the section “Price of Asynchrony” by Carlos Ortiz Marrero and Stephen J.

Why it mattersThe section asks how communication delays change the convergence of distributed linear iterations.

The ideaThe reverse inclusion uses noninvertibility of the nilpotent shift. baseradius therefore gives radius M = ENNReal.ofReal (1 / 2).

Our contributionA real matrix of spectral radius one half has a unit-delay asynchronous iteration of spectral radius at least five quarters.

Scope & next questionThe formal module does not settle the proposed bound for all matrices with zero diagonal blocks. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

Both proposed spectral-radius bounds for asynchronous iteration are false

S. G. Aksoy et al., Seven open problems in applied combinatorics, Journal of Combinatorics 14(4), 559–601 (2023), arXiv:2303.11464v1, includes the section “Price of Asynchrony” by Carlos Ortiz Marrero and Stephen J. Young. Its Open Problem reads:

Let $M$ be a $n \times n$ real matrix with $p(M)= \lambda < 1$ partitioned into an $\ell \times \ell$ array of blocks. Is it the case that for all elements $\delta$ of $[k]^{\ell \times \ell}$ with zero diagonals, $\rho(M^{(\delta)}) \leq \lambda^{\nicefrac{1}{k+1}}?$ If not, is there some function $f \colon (0,1) \times \N \rightarrow (0,1)$ such that $\rho(M^{(\delta)}) \leq f(\lambda,k)$?

This quotation is from the downloaded v1 TeX, Sections/ratfish.tex, lines 75–77, md5 08f1f56ce6f899219ee74094fe3c0e53. The surrounding model, lines 48–49, specifies:

$\tilde{x}i^{(k+1)} = \sum_j M{ij}\tilde{x}j^{(k - \delta{ji})}$

and says “we assume that $\delta_{ii}=0$”. Here $p(M)$ in the question is read as the spectral radius $\rho(M)$ of that model. The operator $M^{(\delta,k)}$ acts on the newest iterate and its $k$ preceding iterates, updating the newest slot by this recurrence and shifting the remaining slots. The source uses $M^{(\delta)}$ for a history operator with enough slots (lines 51–62).

Issue #21351 fixes the formal reading: claim is Q1 ∨ Q2, where Q1 is the universal power bound and Q2 is the existence of the uniform bound strictly below one. Both alternatives are refuted, even with two scalar blocks, positive remote delays and $0<\lambda<1$. The source and its locator are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/AsynchronousIterationSpectralRadiusRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2024Two four-cycles sharing a vertex refute elliptic Sombor Conjecture 3.9Two four-cycles sharing a vertex have elliptic Sombor energy 144.Refuted

The questionAlikhani, Ghanbari and Dehghanizadeh, Elliptic Sombor energy of a graph, arXiv:2404.18622v1, Conjecture 3.9, states:

Why it mattersThe conjecture is a universal nonintegrality assertion.

The ideaThe proof uses spectral estimates.

Our contributionTwo four-cycles sharing a vertex have elliptic Sombor energy 144.

Verification details
Release source recordRefuted in Lean

Open problem

Two four-cycles sharing a vertex refute elliptic Sombor Conjecture 3.9

Alikhani, Ghanbari and Dehghanizadeh, Elliptic Sombor energy of a graph, arXiv:2404.18622v1, Conjecture 3.9, states:

There is no graph with integer-valued elliptic Sombor energy.

For a simple graph, the elliptic Sombor matrix has entry (d_i + d_j) sqrt(d_i^2 + d_j^2) on an adjacent pair and zero otherwise; the elliptic Sombor energy is the sum of the absolute values of its eigenvalues.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/EllipticSomborEnergyIntegerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2022A closest fully separable state of an X-MEMS need not have the conjectured shapeA fully separable three-qubit state is closer to an X-MEMS than every fully separable state of the proposed shape.Refuted

The questionWieśniak, A Variational Approach to the Quantum Separability Problem, arXiv:2209.01430v2 (Phys.

Why it mattersConsiglio, Apollaro and Wieśniak (2022, arXiv:2209.01430v2) use this reduction for their three-qubit calculation and their proposed general X-MEMS entanglement curves.

The ideaFor every pair of three-bit words, phaseaverageentry proves that this character average removes every off-diagonal entry except the opposite corners.

Our contributionA fully separable three-qubit state is closer to an X-MEMS than every fully separable state of the proposed shape.

Verification details
Release source recordRefuted in Lean

Open problem

A closest fully separable state of an X-MEMS need not have the conjectured shape

M. Consiglio, T. J. G. Apollaro and M. Wieśniak, A Variational Approach to the Quantum Separability Problem, arXiv:2209.01430v2 (Phys. Rev. A 106, 062413, 2022), Appendix A, subsection “$n$-qubit $X$-MEMS”, state in PRA_v2.tex, line 459 (acronym macros expanded):

While we have not explicitly derived the CSS of $n$-qubit X-MEMS in general, we conjecture that it is of the form

The following matrix is Eq. (A11), lines 460–473. Put $N=2^{n-1}$ and order computational basis words with qubit 1 most significant. Its diagonal entries are $a/2$ at the all-zero and all-one words, $b/(N-1)$ at every other first-bit-zero word, and $(1-a-b)/(N-1)$ at every other first-bit-one word. Its opposite corners are $d,d^*$ and all other entries vanish. Line 474 restricts $a,b\in[0,1]$, $a+b\le1$, and $|d|\le a/2$.

The source defines the CSS as a minimizer over fully separable states of $D_{HS}(\rho,\sigma)=\operatorname{Tr}((\rho-\sigma)^2)$ (Section II, lines 70–93). The definitions and source locator are in the literature note.

Issue #25134 fixes the low-branch reading: for every integer $n\ge2$ and every $\gamma\in\mathbb C$ with $|\gamma|\le1/(N+1)$, there exists a fully separable $\sigma$ of Eq. (A11) such that

$\operatorname{hsSq}(\rho,\sigma)\le\operatorname{hsSq}(\rho,\tau)$ for every fully separable $\tau$.

Here $\rho=\operatorname{xmems}(n,\gamma)$ has diagonal $1/(N+1)$ at all first-bit-zero words and the all-one word, zero at the remaining words, and opposite corners $\gamma,\gamma^*$; this is Eq. (A1) on the branch where $f=g=1/(N+1)$ (lines 341–373). The module defines $\operatorname{hsSq}(\rho,\sigma)=\operatorname{Re}\operatorname{Tr}((\rho-\sigma)^\dagger(\rho-\sigma))$. The statement asserts attainment as well as comparison with every fully separable competitor. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GeometricMeasure/XMemsClosestSeparableShapeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2019Two-site uniform matrix product states need not have expected dimensionTwo-site uniform matrix product states with bond dimension two and five symbols miss the expected dimension.Refuted

The questionSeynnaeve, Uniform matrix product states from an algebraic geometer's point of view, arXiv:1904.07563v1 (2019; Adv.

Why it mattersUniform matrix product states represent tensors by repeating one matrix tuple at every site.

The idea5. Contradiction. The positive triple satisfies every hypothesis of claim.

Our contributionTwo-site uniform matrix product states with bond dimension two and five symbols miss the expected dimension.

Scope & next questionConjecture 2.14 cannot justify a universal expected-dimension equality or an unrestricted inference of ambient filling from the two numerical bounds.

Verification details
Release source recordRefuted in Lean

Open problem

Two-site uniform matrix product states need not have expected dimension

A. Czapliński, M. Michałek and T. Seynnaeve, Uniform matrix product states from an algebraic geometer's point of view, arXiv:1904.07563v1 (2019; Adv. Appl. Math. 142 (2023), 102417), Section 2, Conjecture 2.14:

For every choice of $(D,d,N)$, $\IMPS(D,d,N)$ has expected dimension.

This is the verbatim text of uMPS-final.tex, line 312, inside the conjecture at lines 311–313; the source's macro \IMPS denotes uMPS. Definition 2.1 fixes $D,d,N\ge1$. For $d$ complex $D\times D$ matrices, the coefficient indexed by a word $(i_1,\ldots,i_N)$ is $\operatorname{tr}(M_{i_1}\cdots M_{i_N})$, and $\mathrm{uMPS}(D,d,N)$ is the image of this parametrization (lines 180–193).

The expected-dimension equality in lines 299–305 is

$\dim\overline{\mathrm{uMPS}(D,d,N)}=\min{(d-1)D^2+1,\dim\mathrm{Cyc}^N(\mathbb C^d)}$,

where $\mathrm{Cyc}^N(\mathbb C^d)$ is the space of cyclically invariant tensors (line 209). Dimension is interpreted through the coordinate ring of the Zariski closure, as in the preregistration and the module's varietyDim. The source introduces this closure at line 211 and its dimension bound at lines 272–274.

Issue #23659 fixes this universal reading. It is refuted at $(D,d,N)=(2,5,2)$. The definitions and source locators are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/UniformMPSExpectedDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2023Convexity of the spatial numerical range in non-unital normed algebrasA three-dimensional complex normed algebra has two spatial numerical range points whose midpoint is absent.Refuted

The questionPatel, Spatial numerical range in non-unital, normed algebras and their unitizations, arXiv:2306.16172v1, Sarajevo Journal of Mathematics 20(2) (2024), §1, p. 2, write:

Why it mattersConvexity would allow norming functionals at different unit vectors to contribute a common convex range.

The ideaThe algebra instances use the associativity, distributivity, scalar compatibility and submultiplicative norm lemmas.

Our contributionA three-dimensional complex normed algebra has two spatial numerical range points whose midpoint is absent.

Scope & next questionThe stronger proposed conclusion that every nonconvex spatial range of an idempotent transfers to a nonconvex algebra spatial range remains open: inclusion alone is insufficient, because other norming pairs can add points.

Verification details
Release source recordRefuted in Lean

arXiv

Convexity of the spatial numerical range in non-unital normed algebras

H. V. Dedania and A. B. Patel, Spatial numerical range in non-unital, normed algebras and their unitizations, arXiv:2306.16172v1, Sarajevo Journal of Mathematics 20(2) (2024), §1, p. 2, write:

Let $S(A)={x\in A:|x|=1}$ be the unit sphere in $A$. Let $A^*$ be the Banach space dual of $A$. Let $D_A(x)={\varphi\in A^*:|\varphi|=1=\varphi(x)}$ for each $x\in S(A)$. Further, let $V_A(a;x)={\varphi(ax):\varphi\in D_A(x)}$ for $a\in A$ and $x\in S(A)$. Then $V_A(a)=\cup{V_A(a;x):x\in S(A)}$ is the spatial numerical range (SNR) … It is still open problem wether $V_A(a)$ is always convex? This problem is discussed in [5].

The question is whether $V_A(a)$ is convex as a subset of the real vector space underlying $\mathbb C$, for every element of every complex non-unital normed algebra.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/NumericalRange/SpatialNumericalRangeNonconvex.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2022Erdos Deep Triple Classification RefutationThree progressions in the integers modulo twenty-seven fall outside the conjectured list.Refuted

The questionFor a family F = {A1, …, As}, ΔF is the multiset of distances dist(x, y) over pairs {x, y} ⊆ Ai, and F is Erdős-deep when "the multiplicities of distances that occur in ΔF are precisely 1, 2, …, k − 1 for some integer k".

Why it mattersThe classification of Erdős-deep triples therefore remains open, and the list published for it is incomplete rather than merely unproved.

The ideaA₁ = AP₂₇(1, 12) = {0,1,2,3,4,5,6,7,8,9,10,11} A₂ = AP₂₇(12, 6) = {0, 12, 24, 9, 21, 6} A₃ = AP₂₇(7, 5) = {0, 7, 14, 21, 1}.

Our contributionThree progressions in the integers modulo twenty-seven fall outside the conjectured list.

Verification details
Release source recordRefuted in Lean

Open problem

Erdos Deep Triple Classification Refutation

Dukes and Gaede, INTEGERS 23 (2023), Article #A25, work in Z_n with |x|_n := min(x, −x) and dist(x, y) = |x − y|_n. For a family F = {A_1, …, A_s}, ΔF is the multiset of distances dist(x, y) over pairs {x, y} ⊆ A_i, and F is Erdős-deep when "the multiplicities of distances that occur in ΔF are precisely 1, 2, …, k − 1 for some integer k". They write AP_n(g, k) := {0, g, 2g, …, (k−1)g}, and Theorem 2 fixes the conventions for families of APs: lengths at least three and gcd(n, g_1, …, g_s) = 1. Section 5 states:

Conjecture 1. An Erdős-deep family of three APs of lengths k1 ≥ k2 ≥ k3 in Z_n exists if and only if (k1, k2, k3) ∈ {(4, 4, 3), (6, 3, 3)}, each for infinitely many n, or (k1, k2, k3) ∈ {(6, 5, 3), (6, 6, 4), (6, 6, 6), (7, 7, 3), (9, 4, 3), (8, 7, 4), (8, 8, 5), (10, 6, 4), (12, 4, 4), (13, 5, 3), (13, 7, 4), (16, 5, 4), (21, 6, 4)}, each for a finite number of values of n.

Conjecture 4.5 of Gaede's University of Victoria thesis states the same with one further hypothesis, 3 ≤ k3 ≤ k2 ≤ k1 ≤ ⌊n / (2·gcd(n, g1))⌋ + 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ErdosDeepTripleClassificationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023Positivity of the multi-entropy replica invariantThe fourth replica invariant of the antisymmetric three-qutrit state is -35/58773123072, so the replica invariant need not be positive on normalized pure states.Refuted

The questionSharma, Monotonicity conjecture for multi-party entanglement.

Why it mattersA negative normalized pure-state instance tests that requirement without assuming a holographic, Gaussian or stabilizer description.

The ideaTracing the unshifted third party gives the two-qutrit marginal

Our contributionThe fourth replica invariant of the antisymmetric three-qutrit state is -35/58773123072, so the replica invariant need not be positive on normalized pure states.

Scope & next questionWhether reflection-positive invariants are exactly the positive ones remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Positivity of the multi-entropy replica invariant

A. Gadde, S. Jain, V. Krishna, H. Kulkarni and T. Sharma, Monotonicity conjecture for multi-party entanglement. Part I, JHEP 02 (2024) 025, arXiv:2308.16247v1, Section 7:

“Although we found that E_n is positive in the examples we studied, it would be nice to prove this in general and to see if there is any relation between the positivity of E_n and the coarse-graining monotonicity.”

Footnote 2 makes positivity the assumption needed for the logarithm in Eq. (1.11). The literature note gives the full quotation and the replica convention in Eqs. (1.10)–(1.11). Issue #17540 fixes the finite-dimensional statement with all quantifiers before formalization.

The question is refuted. For the normalized totally antisymmetric three-qutrit state $\psi_{ijk}=\varepsilon_{ijk}/\sqrt{6}$,

$$E_4^{(3)}(\psi)=\frac{-1680}{6^{16}}=\frac{-35}{58773123072}<0.$$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $n^{q-1}$ replicas on $(\mathbb Z/n)^{q-1}$ | E, replica index Fin (q - 1) → ZMod n | Literal contraction in Eq. (1.10). | | Independent translations for the first $q-1$ parties; last party fixed | StabilizerMultiEntropyCollapse.shift | The source labels parties from one; Lean labels from zero. | | Arbitrary finite local bases and complex pure-state amplitudes | E, dependent family L, its Fintype instances and amplitude ψ | No dimension restriction in claim. | | Normalization | IsUnitState | Sum of Complex.normSq is one, exactly the convention in #17540. The proof uses the existing unitVector predicate through the private consumed correspondence unit_norm_iff. | | Every $n,q\ge2$ has positive real invariant | claim | Positive real part and zero imaginary part; NeZero n is implied by $n\ge2$ and supports ZMod n. | | Negative answer at $n=4,q=3$ | result : ¬ claim | One normalized instance disproves the universal assertion. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/MultiEntropyPositivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedFirst preprint 2022A connected cubic counterexample to Jean–Seo Conjecture 1A connected cubic graph on fifty vertices has an error-detecting identifying code with thirty-six detectors.Refuted

The questionJean and Seo, Optimal Error-detection system for Identifying Codes, arXiv:2208.06052v1, Section 4.2.2, Conjecture 1:

The ideaAn explicit symmetric loopless neighbor table defines the graph.

Our contributionA connected cubic graph on fifty vertices has an error-detecting identifying code with thirty-six detectors.

Verification details
Release source recordRefuted in Lean

Open problem

A connected cubic counterexample to Jean–Seo Conjecture 1

Jean and Seo, Optimal Error-detection system for Identifying Codes, arXiv:2208.06052v1, Section 4.2.2, Conjecture 1:

For a cubic graph G, DET:IC%(G) ≥ 3/4.

The source explicitly defines density for finite graphs. Its Theorem 2.1 characterizes a DET:IC by two-domination and asymmetric two-separation. The source statements and locators are in the Library note. Issue #19618 preregisters the finite connected reading.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | $N[u]=N(u)\cup{u}$ and $N_S[u]=N[u]\cap S$ | closedDominators | Closed neighborhood, intersected with the finite detector set. | | Every vertex is at least 2-dominated | First conjunct of IsDETIC | At least two closed detectors at every vertex. | | Distinct vertices are 2#-distinguished | Second conjunct of IsDETIC | Either directed neighborhood difference has cardinality at least two. | | Cubic graph | SimpleGraph and IsRegularOfDegree 3 in claim | Finite simple graph; degree three everywhere. | | Lowest density is at least $3/4$ | Every code has rational density at least $3/4$ in claim | Equivalent universal reading of a lower bound on the minimum. | | Connectedness | Connected in claim | Additional restriction; a connected counterexample also refutes the unrestricted conjecture. | | Finite enumerability and decidability | Fintype, DecidableEq, DecidableRel | Computational presentation of finite graphs, not an extra mathematical restriction. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/JeanSeoCubicErrorDetectingCode.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2022Negativity and structured negativity differ for a pure two-qutrit stateIt refutes the coincidence conjecture of Kumari and Adhikari.Refuted

The questionAdhikari, “Structured negativity: A physically realizable measure of entanglement based on structural physical approximation”, arXiv:2209.03909v1 (2022-09-08, quant-ph), state in the abstract:

Why it mattersEquation (7) defines ρ̃ = (d/(d³+1)) I + (1/(d³+1)) ρ^(TB).

The ideaThe vector has squared norm (1+4+4)/9=1, so ρ is a positive semidefinite, trace-one matrix.

Our contributionIt refutes the coincidence conjecture of Kumari and Adhikari.

Verification details
Release source recordRefuted in Lean

Open problem

Negativity and structured negativity differ for a pure two-qutrit state

A. Kumari and S. Adhikari, “Structured negativity: A physically realizable measure of entanglement based on structural physical approximation”, arXiv:2209.03909v1 (2022-09-08, quant-ph), state in the abstract:

For d ⊗ d dimensional state, we conjecture from the result obtained in this work that negativity coincide with the structured negativity when the number of negative eigenvalues of the partially transposed matrix is equal to d(d−1)/2.

The conclusion repeats: “Thus, we conjecture that the negativity and structured negativity coincides when q=d(d−1)/2.” Issue #11475 preregisters this quantified reading and the counterexample before any Lean. The claim ranges over natural dimensions d ≥ 2 and positive semidefinite, trace-one matrices on ℂ^d ⊗ ℂ^d. The partial transpose on B sends the entry ρ_(i,j),(k,l) to ρ_(i,l),(k,j); q counts its strictly negative eigenvalues with multiplicity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/StructuredNegativityCoincidenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedFirst preprint 2023An order-19 counterexample to Conjecture 5.4At order nineteen the broom with seven star leaves has larger Laplacian ratio than the broom with eight.Refuted

The questionCai, The Laplacian spectral ratio of connected graphs, arXiv:2302.10491v1, Conjecture 5.4, page 16, state:

Why it mattersThe conjecture specifies one broom as a maximizer among all trees of a fixed order.

The ideaRayleigh bounds from two explicit rational vectors give

Our contributionAt order nineteen the broom with seven star leaves has larger Laplacian ratio than the broom with eight.

Scope & next questionExact attainment for all orders remains open.

Verification details
Release source recordRefuted in Lean

Open problem

An order-19 counterexample to Conjecture 5.4

Z. Lin, J. Wang and M. Cai, The Laplacian spectral ratio of connected graphs, arXiv:2302.10491v1, Conjecture 5.4, page 16, state:

Let T be a tree with n ≥ 8 vertices. Then R_L(K_{1,n−1}) ≤ R_L(T) ≤ R_L(B_n^{(n−3)/2}) if n is odd; R_L(B_n^{(n−4)/2}) if n is even. The left (right) equality holds if and only if T = K_{1,n−1} (T = B_n^{(n−3)/2} or T = B_n^{(n−4)/2}).

The source note records the definitions and version; the preregistration is #19460. The upper inequality is false: $R_L(B_{19}^{8})<R_L(B_{19}^{7})$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Laplacian $L=D-A$, eigenvalues in nonincreasing order | lapEig uses SimpleGraph.lapMatrix and Matrix.IsHermitian.eigenvalues₀ | Index $i$ denotes $\mu_{i+1}$ | | $R_L=\mu_1/\mu_{n-1}$ | spectralRatio uses indices $0$ and $n-2$ | Exact for $n\ge2$; the total extension below two is irrelevant | | Figure 2 broom, center $v_0$, leaves $v_1,\ldots,v_t$, path $v_{t+1},\ldots,v_{n-1}$ | broom symmetrizes precisely those edges with SimpleGraph.fromRel | Literal labelled construction | | Odd and even broom parameters | conjParam | Natural subtraction and division; at nineteen it is eight | | Every tree with $n\ge8$ | claim quantifies over n, hn, T and adjacency decidability, then assumes T.IsTree | No additional mathematical hypothesis | | Both bounds | First and second conjuncts of claim | Both retained | | Both equality characterizations | Final two equivalences in claim | Graph equality means existence of an isomorphism | | Refutation | result proves Not claim | The upper-bound conjunct fails at broom 19 7 |

Fin n fixes vertex labels without restricting finite graph isomorphism types. Adjacency decidability supplies finite matrix entries, rather than a spectral hypothesis. No source clause is dropped or weakened.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/LinWangCaiBroomSpectralRatio.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2022Mitra's instrument-independent unsharpness measure is not monotone under white noise for qubit observablesA four-outcome diagonal qubit observable contradicts the first inequality in Mitra's white-noise conjecture.Refuted

The questionMitra, Quantifying unsharpness of observables in an outcome-independent way, Int.

Why it mattersE would be an unsharpness measure consistent with a resource-theoretic framework only if adding white noise never decreased it; the source states that this follows if Conjecture 1 is proved (§IV.B, §VI).

The ideaTake n = 4 and the diagonal effects A1 = diag(12, 1)/40, A2 = diag(13, 1)/40, A3 = diag(14, 1)/40, A4 = diag(1, 37)/40.

Our contributionA four-outcome diagonal qubit observable contradicts the first inequality in Mitra's white-noise conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Mitra's instrument-independent unsharpness measure is not monotone under white noise for qubit observables

A. Mitra, Quantifying unsharpness of observables in an outcome-independent way, Int. J. Theor. Phys. 61, 236 (2022), arXiv:2201.02578v3, defines instrument-independent unsharpness measures of an observable A = {A_i}.

  • The measures. With the X-matrix X^A = Σ_i ‖A_i‖A_i, they are E(A) = ‖I − X^A‖ and E′(A) = Tr[(I/d)(I − X^A)] (Eqs. (44) and (48)).
  • Monotonicity criteria. Theorems 3 and 4 show that E and E′ are non-decreasing under the white noise A^λ_i = λA_i + (1 − λ)I/n exactly when Σ^A_1 ≥ Σ^A_2 (Eq. (50)) and Σ′^A_1 ≥ Σ′^A_2 (Eq. (60)) hold. Here Σ^A_1 = x^A_min − 1/n and Σ^A_2 = Σ_i ‖A_i‖/n − x^A_min, with x^A_min the lowest eigenvalue of X^A.
  • The conjecture. Propositions 1–2 prove both inequalities for dichotomic qubit observables. Then: "Conjecture 1. For any qubit observable A, inequality Σ^A_1 ≥ Σ^A_2 and inequality Σ′^A_1 ≥ Σ′^A_2 hold and therefore, E(A^λ) ≥ E(A) and E′(A^λ) ≥ E′(A) for all 1 ≥ λ ≥ 0."

The verbatim statements are in the literature note.

Issue #17118 fixes the reading: every qubit POVM with any number n ≥ 1 of outcomes, the operator norm, and all four clauses. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/MitraUnsharpnessWhiteNoiseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023The proposed pruning ranges do not prepare every qutrit Dicke stateThe pruned qutrit circuit has zero amplitude at a word required by the Dicke state with occupation (1,3,1).Refuted

The questionNepomechie and David Raveh, Qudit Dicke state preparation, Quantum Information and Computation 24 (2024), 37–56, arXiv:2301.04989v4, §4.5, “Simplifying the circuit”, propose occupation-dependent ranges for deleting gates from their qutrit circuit.

Why it mattersDeleting gates for a fixed occupation can reduce the resources needed to prepare a Dicke state.

Our contributionThe pruned qutrit circuit has zero amplitude at a word required by the Dicke state with occupation (1,3,1).

Scope & next questionA general branch-aware range satisfying it remains open here; the single-instance repair above supplies no such classification. - A surviving special case.

Verification details
Release source recordRefuted in Lean

Open problem

The proposed pruning ranges do not prepare every qutrit Dicke state

Rafael I. Nepomechie and David Raveh, Qudit Dicke state preparation, Quantum Information and Computation 24 (2024), 37–56, arXiv:2301.04989v4, §4.5, “Simplifying the circuit”, propose occupation-dependent ranges for deleting gates from their qutrit circuit. The v4 source qdit_v3.tex (md5 fcb86a69ac9cdb2a7353251021ea3983), line 1006, introduces the formulas labelled W3genericsimple, W3specialsimple and kkdefs:

we conjecture that, similarly to the $d=2$ case \eqref{W2explicit2},

Line 1031 states:

We have not yet succeeded to prove the result \eqref{W3genericsimple}-\eqref{kkdefs}, which we found through experimentation.

The reading fixed in issue #21375 is the following universal preparation identity: for every natural $n\ge2$ and all natural occupations $k_0,k_1,k_2$ with $k_0+k_1+k_2=n$, the specified pruned circuit maps $|\mathrm e(\vec k)\rangle=|0\rangle^{\otimes k_0}|1\rangle^{\otimes k_1}|2\rangle^{\otimes k_2}$ to $|D^n(\vec k)\rangle$.

More precisely, put $l_1=k_1+k_2$, $l_2=k_2$, and $\tilde k=k_2$ if $k_0=0$, otherwise $\tilde k=\max(k_1,k_2)$. The ranges in that reading are inclusive:

  • The single-index product retains $\mathrm I_{m,l}$ for $\max(\tilde k+m-n,1)\le l\le\min(\tilde k,m-1)$, applied in increasing $l$.
  • The double-index product retains $\mathrm{II}_{m,a,b}$ for $\max(l_2+m-n,1)\le b\le\min(l_2,m-2)$ and $\max(l_1+m-n,b+1)\le a\le\min(l_1,m-1)$, applied with increasing $b$ and increasing $a$ within each $b$.
  • Each $\mathcal W_m$ applies the single-index product before the double-index product; $\mathcal U_n$ applies the blocks in time order $n,n-1,\ldots,2$, each on the leftmost $m$ qutrits. Empty ranges contribute the identity.

These are prunedTilde, prunedTildeTilde, prunedW and prunedU. The universal equality is claim; it is refuted. The definitions and source locators are also documented in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/PureState/QutritDickePruningRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundFirst preprint 2024The Order of a Pisano Period Is Not Confined to TwoA two-term recurrence modulo thirteen has three zeros in one period, so the conjectured range for the number of such zeros is wrong.Refuted

The questionBenfield and Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048v2, 30 January 2025, close with Conjecture 5.3 on the range of the order as the parameters of a two-term recurrence vary.

The ideaThe repository already carries the machinery, so the settlement reuses it rather than rebuilding it.

Our contributionA two-term recurrence modulo thirteen has three zeros in one period, so the conjectured range for the number of such zeros is wrong.

Verification details
Release source recordRefuted in Lean

arXiv

The Order of a Pisano Period Is Not Confined to Two

Benfield and Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048v2, 30 January 2025, close with Conjecture 5.3 on the range of the order as the parameters of a two-term recurrence vary. Clause (v) reads, from the LaTeX source fiborder.tex:

\item $\{0,1,2\}$ if $b \neq \pm1$ and $|a|-|b| = 1$,

Section 4 defines the objects: F_0 = 0, F_1 = 1, F_n = a F_{n-1} + b F_{n-2}. Section 1 defines the order: "The number of zeros in a Pisano period is the order of m, denoted omega(m)." Section 5 fixes the degenerate convention: "define the order to be zero if the sequence is eventually periodic modulo m and this period contains no multiples of m."

Written out, the assertion is that for all integers a, b and all m > 1, if b ≠ ±1 and |a| - |b| = 1, then omega_(a,b)(m) ∈ {0,1,2}.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/PisanoOrderRangeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023A commuting obstruction to the nonlinear measurement monotoneA three-outcome measurement on a two-dimensional complex space refutes the nonlinear 2-norm prefix inequality under commuting blockwise bistochastic dynamics.Refuted

The questionRico and K. Życzkowski, Discrete dynamics in the set of quantum measurements, J.

Why it mattersA three-outcome, two-dimensional diagonal measurement is sufficient to refute the universally quantified assertion.

The ideaUsing zero-based indices, the formal witness is

Our contributionA three-outcome measurement on a two-dimensional complex space refutes the nonlinear 2-norm prefix inequality under commuting blockwise bistochastic dynamics.

Verification details
Release source recordRefuted in Lean

Open problem

A commuting obstruction to the nonlinear measurement monotone

A. Rico and K. Życzkowski, Discrete dynamics in the set of quantum measurements, J. Phys. A 57 (2024) 435302, arXiv:2308.05835v2, Appendix C.1, Conjecture 1, asks whether blockwise bistochastic dynamics preserve the centered-prefix Hilbert–Schmidt norm in the following sense:

Let $P\in\Delta_{n,d}$ and $Q\in\Delta_{n,d}$ be blockwise probability vectors. If there exists a blockwise bistochastsic matrix $B\in B_{n,d}$ such that $Q=B\ast P$, then for any ordering of ${Q_i}$ there exists an ordering of ${P_i}$ such that $\left|\sum_{i=1}^{k}(P_i-1/n)\right|2\geq \left|\sum{i=1}^{k}(Q_i-1/n)\right|_2$ for all $1\leq k\leq n$, where $|A|_2=\sqrt{\operatorname{tr}(A^\dagger A)}$ is the 2-norm.

Definitions 1, 3 (Eq. (19)) and 5 give positive semidefinite effects summing to $I_d$, positive semidefinite blocks with $d\geq2$ and both block row and column sums $I_d$, and the principal-square-root product $Q_i=\sum_j\sqrt{P_j}B_{ij}\sqrt{P_j}$. Here $1/n$ in the centered prefix denotes $(1/n)I_d$. Issue #14670 preregisters this tier-1 question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/RicoZyczkowskiMonotoneRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023The extended MUB mixed-norm equality is falseThe doubly stochastic matrix I/2 + J/6 in dimension three satisfies the spectral condition of the extended MUB conjecture at mu = lambda = 2/3, but a nonzero real vector gives a mixed norm ratio strictly above the predicted norm.Refuted

The questionRotundo and René Schwonnek, arXiv:2303.11382v1, Conjecture 1 (Extended MUB regime), states that a doubly stochastic matrix has mixed operator norm d^(1−lambda−mu) when ((1−mu)/mu)((1−lambda)/lambda) is at least the square of its second largest singular value.

Why it mattersThe witness is a three-dimensional real doubly stochastic matrix, so it also lies in the complex vector domain used by the source's norm definition.

The ideaUse d = 3, C = I/2 + J/6, mu = lambda = 2/3, and x = (4, 1, 1).

Our contributionThe doubly stochastic matrix I/2 + J/6 in dimension three satisfies the spectral condition of the extended MUB conjecture at mu = lambda = 2/3, but a nonzero real vector gives a mixed norm ratio strictly above the predicted norm.

Verification details
Release source recordRefuted in Lean

Open problem

The extended MUB mixed-norm equality is false

Antonio F. Rotundo and René Schwonnek, arXiv:2303.11382v1, Conjecture 1 (Extended MUB regime), states that a doubly stochastic matrix has mixed operator norm d^(1−lambda−mu) when ((1−mu)/mu)((1−lambda)/lambda) is at least the square of its second largest singular value. Equation (7) gives the proposed value for the 1/mu to 1/(1−lambda) norm. Issue #11781 preregistered this statement, its quantifiers, and the refutation route.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ExtendedMubNormRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2023The 2-swap greedy shortest-path conjecture is falseBinary words at distance one force arbitrarily many greedy swaps.Refuted

The questionAdamson, Flaherty, Potapov and Spirakis, Structural and Combinatorial Properties of 2-swap Word Permutation Graphs, arXiv:2307.01648v3, Section 5, page 23:

Why it mattersA pair of adjacent words for which every admissible run generates more than one nontrivial pair suffices.

Our contributionBinary words at distance one force arbitrarily many greedy swaps.

Verification details
Release source recordRefuted in Lean

Open problem

adamson flaherty potapov spirakis 2023 two swap greedy

Adamson, Flaherty, Potapov and Spirakis, Structural and Combinatorial Properties of 2-swap Word Permutation Graphs, arXiv:2307.01648v3, Section 5, page 23:

For the 2-swap graph specifically, we leave open the problem of determining the shortest path between two given words w and v. We conjecture that the simple greedy algorithm used to derive the upper bound in Lemma 4 can be used to find the shortest path between any pair of vertices.

The target is the greedy optimality conjecture, preregistered in issue #19317. For words with the same Parikh vector, the formal claim asks whether some admissible order makes the number of generated pairs with distinct indices equal the configuration-graph distance. An admissible order omits any maximal-count reference symbol and lists every other symbol once in non-increasing count. All tied choices are allowed. The count does not require that the generated swaps be legal at application time or reach the target, so refuting this necessary condition also refutes the stronger shortest-path assertion.

Statement echo

| Source clause | Formal counterpart and fidelity | | --- | --- | | Words in $\Sigma^n$ | Functions Fin n → Fin s; zero-based relabelling of the source's positions and alphabet. | | Definition 1 exchanges positions $i<j$ holding different symbols | twoSwap precomposes with Equiv.swap; IsTwoSwapOf imposes both restrictions. | | $P(w)[a]$ counts occurrences of $a$ | parikh counts the filtered finite set of positions. | | $G(P)$ has exactly the words of Parikh vector $P$ | configGraph induces swapGraph on that subtype; the equality binder puts both endpoints in the same graph. | | Algorithm 1 pairs corresponding occurrences | occurrences lists positions increasingly; stageSwaps zips the lists before the stage is applied. | | Apply a stage and continue over the remaining symbols | applySwaps folds the transpositions; greedySwaps recursively concatenates stages. | | $P[1]\ge\cdots\ge P[\sigma]$, processing $\Sigma\setminus{1}$ | AdmissibleOrder permits every maximal reference and every count-compatible order. | | A shortest path for any pair of vertices | claim universally quantifies alphabet size, length and both words; its conclusion requires only equality of count and distance. | | Refutation | result proves the negation of the full universal claim. |

swapGraph on all words, list application order and greedyLength are explicit encoding conventions. They introduce no extra hypothesis on the source's vertices. List application order cannot alter the count in the binary family, which has only one stage. No positivity restriction is added on alphabet size or length.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/AdamsonFlahertyPotapovSpirakisTwoSwapGreedy.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresBijective proofFirst preprint 2024Avoiders of 231 and 312 Fixed by the Third Iterate of the Fundamental BijectionThe generating functions for 231- and 312-avoiders fixed by the third iterate of the fundamental bijection are rational.Proved

The questionLaudone, Pattern avoidance and the fundamental bijection, arXiv:2407.06338v1, Section 5, Conjecture 5.6:

The idea1. θ commutes with direct sums, and every 231- or 312-avoider is uniquely a direct sum of sum-indecomposable ones, so Fσ^3 = 1/(1 − Iσ^3) where Iσ^3 counts the indecomposable avoiders fixed by θ^3.

Our contributionThe generating functions for 231- and 312-avoiders fixed by the third iterate of the fundamental bijection are rational.

Verification details
Release source recordProved in Lean

Open problem

Avoiders of 231 and 312 Fixed by the Third Iterate of the Fundamental Bijection

Kassie Archer and Robert P. Laudone, Pattern avoidance and the fundamental bijection, arXiv:2407.06338v1, Section 5, Conjecture 5.6:

For σ ∈ {231, 312}, the generating functions F_σ^k(x) are rational. A few examples of conjectured generating functions for these patterns are found below. F_σ^3(x) = 1/(1 − x − x^2 − 2x^3)

Here F_σ^k(x) = Σ f_n^k(σ) x^n and f_n^k(σ) counts the permutations of [n] avoiding σ and fixed by the k-th iterate of the fundamental bijection θ.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/FundamentalBijection/ThetaCube.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The generating functions for 231- and 312-avoiders fixed by the third iterate of the fundamental bijection are rational.

    Read source ↗
  • Historical context

    First preprint year: 2024.

    Read source ↗
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2024Permutations Whose First Iterates under the Fundamental Bijection Avoid 132Avoidance of 132 through successive iterates has a cubic quasipolynomial count followed by linear and constant counts.Proved

The questionLaudone, Pattern avoidance and the fundamental bijection, arXiv:2407.06338v1, Section 4, Conjecture 4.5:

Why it mattersThe table does not state where its rows start; they hold from n = 3, and t1^k = 1, t2^k = 2.

The ideaA 132-avoider splits at its maximum into a larger left part and a smaller right part; cutting a word before its left-to-right maxima gives a block criterion for 132-avoidance and the inverse of θ.

Our contributionAvoidance of 132 through successive iterates has a cubic quasipolynomial count followed by linear and constant counts.

Verification details
Release source recordProved in Lean

Open problem

Permutations Whose First Iterates under the Fundamental Bijection Avoid 132

Kassie Archer and Robert P. Laudone, Pattern avoidance and the fundamental bijection, arXiv:2407.06338v1, Section 4, Conjecture 4.5:

For n ≥ 2, t_n^2(132) = k^3 + 3k^2 + 2k − 1 (n = 3k), k^3 + 4k^2 + 4k (n = 3k + 1), k^3 + 5k^2 + 7k + 2 (n = 3k + 2). The conjectured values for k ≥ 3 are found in the table below: 3n − 4, 2n − 1, n + 2 and 5 for k = 3, 4, 5 and k ≥ 6.

Here t_n^k(σ) counts the permutations π of [n] such that π, θ(π), …, θ^k(π) all avoid σ.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/FundamentalBijection/ThetaIterate.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023The gcd-three equator formula for optimized positive maps is falseAn exact positive-input certificate refutes the gcd-three equator formula.Refuted

The questionThe Library note gives the source statements and locators. Preregistration 19282 specifies the external named problem and the source conventions.

Why it mattersSubtracting a rank-one Hadamard map from a positive map asks how much positivity remains.

The ideaPositivity would require this output to be positive semidefinite, hence its quadratic form to be nonnegative.

Our contributionAn exact positive-input certificate refutes the gcd-three equator formula.

Verification details
Release source recordRefuted in Lean

Open problem

The gcd-three equator formula for optimized positive maps is false

A. Bera, G. Sarbicki and D. Chruściński, Optimizing positive maps in the matrix algebra M_n, arXiv:2309.09621v1, Conjecture 2, Eq. (13), asserts that when $1\le k<n$, $\gcd(n,k)=3$, and $v=\alpha v_1+\beta v_2$ with $|\alpha|^2+|\beta|^2=1$ and $|\alpha|=|\beta|$, the greatest nonnegative parameter for which $\tau_{n,k}-\lambda H_v$ is positive is

$$n-k+\frac{n-3}{n-2k/3}.$$

The assertion is Refuted. The Library note gives the source statements and locators. Preregistration #19282 specifies the external named problem and the source conventions.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Eqs. (3)–(5), diagonal projection and cyclic shift | tau | The diagonal projection composes Mathlib's diagonalLinearMap and diagLinearMap; the shift has entry one exactly at $i=j+1$ in Fin n; conjugation uses mulLeft and mulRight. | | Eq. (6), $H_v(X)=(vv^*)\circ X$ | hadamardMap | The rank-one matrix is Matrix.vecMulVec v (star v); multiplication is entrywise. | | Eqs. (10)–(11), the two Fourier vectors | vr n 1, vr n 2 | Coordinate $j$ is $\exp(2\pi i rj/3)/\sqrt n$. | | Greatest positivity parameter | IsLambdaMax | IsGreatest requires attainment and maximality among nonnegative real parameters, using the frozen FiniteKrausChannel.PhyslibLeaf.MatrixMap.IsPositive. | | $1\le k<n$, $\gcd(n,k)=3$ | The first three hypotheses of claim | Natural-number parameters with NeZero n. The instance excludes dimension zero; the source hypotheses already force positive dimension. | | $|\alpha|^2+|\beta|^2=1$, $|\alpha|=|\beta|$ | The coefficient hypotheses of claim | Complex coefficients and real norms. | | Eq. (13) | The conclusion of claim | The displayed real parameter is used literally. | | Negation of the universal assertion | result | The witness has $n=6$, $k=3$, $\alpha=\beta=1/\sqrt2$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/BeraSarbickiChruscinskiOptimizedMaps.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023Removing intermediate memory resets can increase permutation approximation errorRemoving intermediate memory resets increases the permutation approximation error in a finite reduced protocol.Refuted

The questionKorzekwa, Thermal Recall: Memory-Assisted Markovian Thermal Processes, PRX Quantum 4, 040304 (2023), arXiv:2303.12840v2, introduce a memory-assisted protocol for approximating permutations by sequential two-level thermalisations.

Why it mattersThe conjecture would make the protocol without intermediate memory resets, which lets memory correlations pass between successive transpositions, uniformly at least as accurate as the protocol that resets the memory after each transposition.

The ideaThe module defines the infinite-temperature elementary operation average, the nested-round operation block, the uniform-memory coupling joint, the memory reset, and the two folded outputs.

Our contributionRemoving intermediate memory resets increases the permutation approximation error in a finite reduced protocol.

Verification details
Release source recordRefuted in Lean

Open problem

Removing intermediate memory resets can increase permutation approximation error

J. Czartowski, A. de Oliveira Junior and K. Korzekwa, Thermal Recall: Memory-Assisted Markovian Thermal Processes, PRX Quantum 4, 040304 (2023), arXiv:2303.12840v2, introduce a memory-assisted protocol for approximating permutations by sequential two-level thermalisations. Conjecture 1, “Improved convergence,” is stated in the source TeX TR.tex, lines 417–427:

In the infinite temperature limit, $\beta=0$, and for an $N$-dimensional memory, $\widetilde{\mathcal{P}}^{{\Pi}}$ gives a better approximation of a permutation $\Pi$ than $\mathcal{P}^{\Pi}$: $$\delta\left( \Pi \mathbf p,\tilde{\mathbf q}\right) \leq \delta\left( \Pi \mathbf p, \mathbf{q}\right),$$ where $\tilde{\mathbf{q}}$ and $\mathbf{q}$ are defined via $\widetilde{\mathcal{P}}^{{\Pi}}(\mathbf{p}\otimes \boldsymbol\eta_M) = \tilde{\mathbf{q}}\otimes \boldsymbol\eta_M$ and $\mathcal{P}^{{\Pi}}(\mathbf{p}\otimes \boldsymbol\eta_M) = \mathbf{q}\otimes \boldsymbol\eta_M$.

The repository proposition claim is the following exact finite reading. For every $d,n\in\mathbb N$, $p:\mathrm{Fin},d\to\mathbb R$, and word $w:\mathrm{List}(\mathrm{Fin},d\times\mathrm{Fin},d)$, if $0<n$, $(\forall i,\ 0\le p(i))$, $\sum_i p(i)=1$, StrictAnti p, and IsReducedNeighbourWord p w, then [ \operatorname{totalVariation}(\operatorname{applySwaps}(p,w),\operatorname{truncatedOutput}(n,p,w)) \le \operatorname{totalVariation}(\operatorname{applySwaps}(p,w),\operatorname{resetOutput}(n,p,w)). ] Here applySwaps folds the word by precomposing with each Equiv.swap; totalVariation is one half of the finite sum of absolute coordinate differences. average replaces the values at two selected joint states by their arithmetic mean. block applies these averages for every first memory index and, inside each round, every second memory index, both in increasing List.finRange order. joint couples $p$ to a uniform memory, reset replaces a joint state by the product of its system marginal and that uniform memory, truncatedOutput folds the blocks without intermediate resets, and resetOutput folds a reset after each block. IsReducedNeighbourWord requires, at every position, an increasing index pair whose two current population values are ordered with no current value strictly between them.

The Lean declaration is result : \lnot claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/ThermalRecallTruncationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresQuestion resolvedFirst preprint 2022The Thue–Morse two-block fixed point has frequency one halfThe Thue-Morse two-block fixed point has one-letter frequency one half.Proved

The questionArXiv v2 and the journal version change the substitution; this conclusion concerns the displayed v1 table.

Why it mattersThe assertion asks for the limit over every natural prefix length, rather than a numerical approximation at selected lengths.

Our contributionThe Thue-Morse two-block fixed point has one-letter frequency one half.

Verification details
Release source recordProved in Lean

Open problem

The Thue–Morse two-block fixed point has frequency one half

Dekking and Keane, Two-block substitutions and morphic words, arXiv:2202.13548v1, Section 4, define $\kappa_{\rm TM}(00)=001$, $\kappa_{\rm TM}(01)=010$, $\kappa_{\rm TM}(10)=101$, and $\kappa_{\rm TM}(11)=110$. Their fixed point $x^{(00)}$ starts with $00$ and satisfies $x_{3n}=x_{2n}$, $x_{3n+1}=x_{2n+1}$, and $x_{3n+2}=1-x_{2n+1}$.

Conjecture 4. The frequency of 1 in $x^{(00)}$ exists and equals $\frac12$.

The source is v1, Section 4, pp. 5–6. ArXiv v2 and the journal version change the substitution; this conclusion concerns the displayed v1 table.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/TwoBlockSubstitution/ThueMorseFrequency.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresQuestion resolvedFirst preprint 2024Nonnesting Permutations Avoiding 1132 and 2213The generating function for nonnesting permutations avoiding 1132 and 2213 satisfies the stated quadratic equation.Proved

The questionSergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1132, 2213}: the conjectured ordinary generating function is ((1-x)^2 - √((1-x)^4 - 4x(1-x)^2))/(2x) (OEIS A006319), with the remark

Why it mattersThis equation has a unique power-series solution, the branch of the displayed square root with constant term one, so the counts are 1, 1, 4, 16, 68, 304, 1412, 6752, ….

The ideaA nonnesting word is a permutation (its first-occurrence order) together with a Dyck word.

Our contributionThe generating function for nonnesting permutations avoiding 1132 and 2213 satisfies the stated quadratic equation.

Verification details
Release source recordProved in Lean

Open problem

Nonnesting Permutations Avoiding 1132 and 2213

Sergi Elizalde and Amya Luo, Pattern avoidance in nonnesting permutations, arXiv:2412.00336v6, Section 4, Table 4, row {1132, 2213}: the conjectured ordinary generating function is ((1-x)^2 - √((1-x)^4 - 4x(1-x)^2))/(2x) (OEIS A006319), with the remark

All the conjectures have been checked for n up to 8.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Nonnesting/NonnestingRoyalLow.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryCounterexample foundFirst preprint 2024Stationary characterization of Gini mean comparisonsNormalized stationary tests do not characterize Gini comparisons on the positive half-line.Refuted

The questionPáles, Comparison of Gini means with fixed number of variables, arXiv:2408.07658v1, §4, state:

Why it mattersThe proposed criterion would certify global comparison on intervals with arbitrarily large coordinate ratios using only interior stationary tuples after normalization.

The ideaThe constant normalized tuple has every coordinate one and gives equality.

Our contributionNormalized stationary tests do not characterize Gini comparisons on the positive half-line.

Verification details
Release source recordRefuted in Lean

arXiv

Stationary characterization of Gini mean comparisons

R. Grünwald and Zs. Páles, Comparison of Gini means with fixed number of variables, arXiv:2408.07658v1, §4, state:

Conjecture 2. Assume that 0 = inf I · I⁻¹ and let n ≥ 3. Then ((r, s), (p, q)) ∈ Γ_n(I) if and only if, for all u₁, …, u_n ∈ ℝ₊ such that G^{[n]}{r,s}(u₁, …, u_n) = 1, ∃ρ ∈ ℝ such that χ′{p,q}(u_k) + ρχ′{r,s}(u_k) = 0 (k ∈ {1, …, n}) the inequality 1 ≤ G^{[n]}{p,q}(u₁, …, u_n) holds.

Here $I$ is a nonempty interval in $(0,\infty)$ and $\Gamma_n(I)$ consists of the parameter pairs satisfying $G_{r,s}(x)\le G_{p,q}(x)$ for every $x\in I^n$. The stationary test universally quantifies over positive normalized tuples satisfying the common-multiplier equations. It does not assert that every normalized tuple admits such a multiplier.

The formal reading is:

| Source object or condition | Lean encoding and meaning | |---|---| | $n\ge3$, $x\in I^n$ | n : ℕ, 3 ≤ n, x : Fin n → ℝ, and ∀ i, x i ∈ I | | Nonempty positive interval $I$ | I ⊆ Set.Ioi 0, I.OrdConnected, and I.Nonempty | | $\inf(I\cdot I^{-1})=0$ | sInf ((fun z : ℝ × ℝ => z.1 / z.2) '' (I ×ˢ I)) = 0; the set is precisely ${a/b:a,b\in I}$, rather than a product of two scalar infima | | $G_{p,q}$, $p\ne q$ | Defs.giniMean delegates to GiniMeanEndpointComparison.giniMean, namely $(\sum x_i^p/\sum x_i^q)^{1/(p-q)}$ | | $G_{p,p}$ | $\exp(\sum x_i^p\log x_i/\sum x_i^p)$, the equal-parameter branch of Defs.giniMean | | $\chi_{p,q}$ | Defs.chi: $(t^p-t^q)/(p-q)$ for $p\ne q$, and $t^p\log t$ for $p=q$ | | $\chi'{p,q}$ | Defs.chiDeriv p q t := deriv (Defs.chi p q) t | | $\Gamma_n(I)$ | Defs.Gamma n I: comparison for every tuple from $I$ | | Stationary criterion | Defs.stationaryTest n r s p q: positivity, $G{r,s}(u)=1$, and existence of a common real multiplier imply $1\le G_{p,q}(u)$ | | Claimed equivalence | Defs.claim: all the displayed interval conditions, all $n\ge3$, and all real $r,s,p,q$ |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/GiniStationaryTest/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2023Order-three Rényi information-combining equalityBoth branches of the binary quantum information-combining expression are equalities at Renyi order three, in arbitrary finite dimensions including singular marginals.Proved

The questionTomamichel, “Chain Rules for Rényi Information Combining”, arXiv:2305.02589v1, Section V, Conjecture V.5, state:

Why it mattersIt provides the quantum, noncommuting counterpart of the classical order-three equality and the explicit expression anticipated in the remark after Proposition V.4 of the pinned PDF.

The ideaThe proof reuses GHZMeasureBiseparableBound.IsDensity, PartialTraceMutualInformation.partialTraceLeft, PartialTraceMutualInformation.kroneckereqconjdiagonaleigenvalues, ProjectionProbabilityFlow.tracehermitianproductreal, and RHLinalg.tracemulnonnegofposSemidef.

Our contributionBoth branches of the binary quantum information-combining expression are equalities at Renyi order three, in arbitrary finite dimensions including singular marginals.

Verification details
Release source recordProved in Lean

Open problem

Order-three Rényi information-combining equality

C. Hirche, X. Guan and M. Tomamichel, “Chain Rules for Rényi Information Combining”, arXiv:2305.02589v1, Section V, Conjecture V.5, state:

For $\alpha\in[2,3]$ the same holds with $\geq$ exchanged by $\leq$ and for $\alpha\in{2,3}$ the above holds with equality.

The settlement is the $\alpha=3$ equality clause for two independent, equiprobable binary classical-quantum inputs. Each input has two positive semidefinite, trace-one complex matrices in an arbitrary finite dimension. Put $H_i=\widetilde H_3^\downarrow(X_i|B_i){\rho_i}$ and $H{\rm out}=\widetilde H_3^\downarrow(X_1+X_2|B_1B_2)_\tau$, with $X_2$ traced out of the CNOT state. Then

$$ \begin{aligned} H_1+H_2\leq\log2&\Longrightarrow H_{\rm out}=h_3(h_3^{-1}(H_1)\ast h_3^{-1}(H_2)),\ \log2\leq H_1+H_2&\Longrightarrow H_{\rm out}=H_1+H_2-\log2+ h_3(h_3^{-1}(\log2-H_1)\ast h_3^{-1}(\log2-H_2)). \end{aligned} $$

Here $p\ast q=p(1-q)+(1-p)q$, and $h_3^{-1}$ is the inverse on probabilities in $[0,1/2]$. Both guards hold at the common boundary. The binding matrix, inverse and natural-logarithm conventions are in #15115.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/RenyiInformationCombiningOrderThree.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2022A quantum behavior with rank of contextuality threeA quantum behavior has rank of contextuality three.Proved

The questionHorodecki et al., The rank of contextuality, arXiv:2205.10307v1, New Journal of Physics 25 (2023), 073003, Section IX (Section 9 in the journal):

Why it mattersThe Library attestation is Library/QuantumContext/horodecki2022rank.md.

The ideaEach square has five positive parity contexts and one negative parity context.

Our contributionA quantum behavior has rank of contextuality three.

Verification details
Release source recordProved in Lean

Open problem

A quantum behavior with rank of contextuality three

K. Horodecki et al., The rank of contextuality, arXiv:2205.10307v1, New Journal of Physics 25 (2023), 073003, Section IX (Section 9 in the journal):

Finally, it would be interesting to construct a quantum behavior having the rank of contextuality different from the power of 2 (the power of 2 is achieved by a tensor product of the Peres-Mermin behaviors).

The question asks for an example. The quantum behavior $Q=P\otimes P$ formed from two Peres–Mermin squares has rank $3$.

Statement echo

| Source clause or representation choice | Lean counterpart | Scope | | --- | --- | --- | | An ordered list of observable outcomes | restrict C x g | Boolean outcomes on finite ordered contexts | | A family of conditional distributions | Behavior C, constrained by IsQuantum C P | Behavior is the repository's unrestricted real-table carrier; the quantum predicate imposes the Born rule | | Mixtures of deterministic assignments, equation (2) | NonContextual C N | Nonnegative real weights with total mass one and every restriction-fiber sum | | Minimum number of distinct noncontextual behaviors covering every context, equation (3) | coverSizes and IsRankOfContextuality | One member agrees on all outcomes at each context; the member can depend on that context | | Quantum observables and state | IsQuantum | Positive semidefinite trace-one state; Hermitian involutions; within-context commutation; ordered spectral-projection products | | Construct a quantum behavior | claim, proved by result | Existential observable count, context count, context lists, table and rank | | Rank different from every power of two | Final conjunct of claim | All natural exponents, including zero |

The witness has $18$ observable labels, $36$ contexts, state $I_{16}/16$ and rank $3$. Finite labels, ordered contexts and binary outcomes restrict the construction to a subclass of quantum behaviors; existence in that subclass answers the source's existence question. No hypothesis is added to the question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumContext/RankOfContextualityNonPowerTwo.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2020Complex testers do not detect every entangled Werner stateEntangled Werner states with dimension at least three and parameter between one over the dimension and one half escape every pair of complex testers.Refuted

The questionNechita, Multipartite Entanglement Detection Via Projective Tensor Norms, arXiv:2010.06365v1 (Annales Henri Poincaré 23 (2022), 3791–3838), Section sec:conclusions, posed the following question in main.tex, line 1365:

Why it mattersThe paper presents testers as a way to turn mixed-state entanglement detection into a projective norm calculation on Euclidean output spaces.

The ideaThe module follows a finite fourth-root design and a swap witness.

Our contributionEntangled Werner states with dimension at least three and parameter between one over the dimension and one half escape every pair of complex testers.

Scope & next questionIt does not settle completeness after an index permutation, the source's separate statements about realignment or SIC POVM testers, or restricted variants of the tester class. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

Complex testers do not detect every entangled Werner state

M. A. Jivulescu, C. Lancien and I. Nechita, Multipartite Entanglement Detection Via Projective Tensor Norms, arXiv:2010.06365v1 (Annales Henri Poincaré 23 (2022), 3791–3838), Section sec:conclusions, posed the following question in main.tex, line 1365:

The main question that remains unanswered at this point is whether our family of entanglement criteria is complete, without allowing for permutation of indices before applying the testers. Concretely, we would like to know if, for any entangled state $\rho$ on $\mathbb C^{d_1}\otimes\cdots\otimes\mathbb C^{d_m}$, there exist testers $\mathcal E_i:S_1^{d_i}\to\ell_2^{n_i}$, $1\leq i\leq m$, such that $|\mathcal E_1\otimes\cdots\otimes\mathcal E_m(\rho)|{\ell_2^{n_1}\otimes\pi\cdots\otimes_\pi\ell_2^{n_m}}>1$. Even for bipartite states we are only able to show a weaker version of this statement (see Section \ref{sec:completeness}).

The source defines the projective tensor norm by the infimum of the sums of products of factor norms (Definition, main.tex, lines 234–238), the trace norm by $|X|1=\operatorname{Tr}\sqrt{X^*X}$ (main.tex, lines 288–290), and a complex tester as a complex-linear map of norm exactly one from the Schatten-1 space to the Euclidean space (main.tex, lines 340–352). Its Werner state is $\sigma\mu=\mu\frac{I+F}{d(d+1)}+(1-\mu)\frac{I-F}{d(d-1)},\qquad 0\leq\mu\leq1$ (main.tex, line 946).

The Lean declaration claim : Prop is the bipartite instance. For every $d_1,d_2:\mathbb N$ and every $\rho:\operatorname{Matrix}(\operatorname{Fin}d_1\times\operatorname{Fin}d_2)(\operatorname{Fin}d_1\times\operatorname{Fin}d_2),\mathbb C$, its hypotheses are ρ.PosSemidef, ρ.trace = 1, and ¬ Separable ρ; its conclusion requires $n_1,n_2:\mathbb N$ and complex-linear maps $e_i:\operatorname{Matrix}(\operatorname{Fin}d_i)(\operatorname{Fin}d_i),\mathbb C\to\operatorname{EuclideanSpace},\mathbb C,(\operatorname{Fin}n_i)$ satisfying IsTester e₁ ∧ IsTester e₂ ∧ 1 < ‖testerImage e₁ e₂ ρ‖.

Here IsTester e is exactly the conjunction $\forall x,\ |e x|\leq\operatorname{traceNorm}(x)$ and $\exists x,\ x\neq0\land|e x|=\operatorname{traceNorm}(x)$. The formal Separable predicate is membership in the real convex hull of matrices Matrix.kroneckerMap (· * ·) ρ₁ ρ₂, where each factor is positive semidefinite with trace $1$.

The output of testerImage has type ⨂[ℂ] s : Bool, EuclideanSpace ℂ (Fin (cond s n₂ n₁)): the false factor is the first Euclidean output space and the true factor is the second. Its norm $|\cdot|$ is Mathlib's bundled PiTensorProduct.projectiveSeminorm on this tensor product. This represents the source's $\ell_2^{n_1}\otimes_\pi\ell_2^{n_2}$ norm from main.tex, lines 234–238: the two factors carry their Euclidean norms, and each finite pure-tensor expansion has cost equal to the sum of the products of those factor norms. Mathlib permits explicit complex scalar coefficients in these expansions; absorbing each coefficient into either factor gives the source's convention, with the same cost. The tensor norm notation and application of PiTensorProduct.projectiveSeminorm are definitionally equal.

The settlement is the negation of this bipartite claim. The source's universal multipartite question is therefore already false at two parties.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/WernerTesterIncompleteness.result

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Quantum informationQuestion resolvedFirst preprint 2022The coefficient system of Quesne's Kepler–Coulomb extensions on the sphere has the conjectured closed forms for every mFor every natural m at least two, the Kepler-Coulomb coefficient system has a unique solution on indices zero through m, with positive constants in the conjectured closed forms.Proved

The questionQuesne, Quasi-exactly solvable extensions of the Kepler-Coulomb potential on the sphere, Ann.

Why it mattersBy (4.28)–(4.29) the energies of the two known states depend on a0 alone, and the source states that the wavefunctions (4.30)–(4.31) are normalizable provided am > 0, which it had checked only for the values of m computed explicitly.

The ideaWrite m = 2µ + ε with ε ∈ {0, 1}, x = Q²/κ, tk = m − 2k − 1 and F(m, r) = m!/(m − r)!.

Our contributionFor every natural m at least two, the Kepler-Coulomb coefficient system has a unique solution on indices zero through m, with positive constants in the conjectured closed forms.

Scope & next questionThey share the denominator Q^{2µ} + b1κQ^{2µ−2} + ⋯ + bµκ^µ, and the odd-index numerators contain brackets with constants cj^{(k)}, "where b1, b2, …, bµ, and c^{(k)}1, c^{(k)}2, …, c^{(k)}{µ−k} or c^{(k)}{µ−k−1} are some positive constants." - The open problem. §5: "The problem that remains open is to prove our conjecture by finding the explicit values of the constants appearing …"

Verification details
Release source recordProved in Lean

Open problem

The coefficient system of Quesne's Kepler–Coulomb extensions on the sphere has the conjectured closed forms for every m

C. Quesne, Quasi-exactly solvable extensions of the Kepler-Coulomb potential on the sphere, Ann. Phys. 451, 169265 (2023), arXiv:2206.01033v3, builds extensions of the Kepler–Coulomb potential on a sphere of curvature κ with 2m additional terms and known ground and first excited states.

  • The system. The generating-function method (§4.1, Eq. (4.10)) expresses the first two superpotentials through m + 1 constants a_0, …, a_m, which must solve the linear system (4.13)–(4.17) in the rescaled coupling Q and κ (§4.2).
  • The conjecture. From the solutions for m ≤ 7: "Conjecture For any m > 1, the solution of the system of equations (4.13)–(4.17) can be written as" the closed forms (4.19)–(4.21) for m = 2µ + 1 and (4.22)–(4.24) for m = 2µ. They share the denominator Q^{2µ} + b_1κQ^{2µ−2} + ⋯ + b_µκ^µ, and the odd-index numerators contain brackets with constants c_j^{(k)}, "where b_1, b_2, …, b_µ, and c^{(k)}1, c^{(k)}2, …, c^{(k)}{µ−k} or c^{(k)}{µ−k−1} are some positive constants."
  • The open problem. §5: "The problem that remains open is to prove our conjecture by finding the explicit values of the constants appearing …"

The verbatim statements are in the literature note.

Issue #17076 fixes the reading: the constants are independent of Q and κ, and for all Q, κ > 0 the system has a solution, every solution has the closed forms, and solutions agree on 0, …, m. The conjecture is proved for every m ≥ 2, with explicit constants.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Analysis/KeplerCoulombSphereQESCoefficients.result

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Discrete structuresConjecture provedFirst preprint 2024Ulfarsson's alternating mesh-pattern basis conjectureEvery mesh-pattern basis of the even permutations consists exactly of fully shaded odd permutations.Proved

The questionUlfarsson, BiSC: An algorithm for discovering generalized permutation patterns, arXiv:2411.17778v1, Conjecture 2.4.2, page 8; DOI 10.48550/arXiv.2411.17778.

Why it mattersBiSC outputs fully shaded patterns for the alternating subgroups.

The ideaInsert two points there, using either order for their values.

Our contributionEvery mesh-pattern basis of the even permutations consists exactly of fully shaded odd permutations.

Verification details
Release source recordProved in Lean

Open problem

Ulfarsson's alternating mesh-pattern basis conjecture

H. Ulfarsson, BiSC: An algorithm for discovering generalized permutation patterns, arXiv:2411.17778v1, Conjecture 2.4.2, page 8; DOI 10.48550/arXiv.2411.17778. The source is recorded in ulfarsson2024bisc. The external named problem is preregistered in #19120.

Let A be the union of the alternating subgroups, and let P be a basis of patterns such that A = Av(P). Then P = {(p, {0, …, n} × {0, …, n}) : p ∈ S_n ∖ A for some n}.

The settlement is Proved under the literal mesh-pattern conventions in #19120. Every mesh-pattern basis of the even permutations consists exactly of fully shaded odd permutations.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A mesh pattern is a permutation together with shaded cells | MeshPattern, with length k, permutation p, and shading R | Positions are zero-based; cells range from $0$ through $k$ on each axis | | An occurrence has order-isomorphic selected points and no extra point in a shaded region | Occurs | Increasing selected positions, relative value order, and both cell cardinalities are retained | | $S$ contains permutations of every length | The dependent sum in Av and Alt | Length zero is included | | $\operatorname{Av}(P)$ avoids all members of $P$ | Av | Universal quantification over the forbidden set | | $P$ is a basis of $A$ | IsBasis | $\operatorname{Av}(P)=A$ and every subset with that avoidance set equals $P$; equality orientation is immaterial | | $A$ is the union of the alternating subgroups | Alt | The sign is $1$ | | The forbidden patterns are precisely fully shaded odd permutations | claim and result | $R=\mathrm{univ}$ and sign different from $1$, quantified over every basis $P$ |

Every public definition has the literal body of its binding convention in #19120. The result adds no binder or hypothesis beyond claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/MeshPattern/UlfarssonAlternatingMeshBasis.result

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Quantum informationConjecture refutedFirst preprint 2023Van Herstraeten–Guha–Cerf Conjecture 2: coherent inputs for phase-covariant attenuatorsThe full bosonic attenuator and the coherent-state output-entropy question.Refuted

The questionConjecture 2 (Phase-covariant attenuator), section IV.B, arXiv:2312.15623v2, PDF p. 12, states: “The minimum output entropy of a phase-covariant attenuator channel ℳη,𝐩 is achieved by coherent states, ∀η,𝐩.” The domain is every transmissivity η in [0,1] and every probability vector p on the natural occupations.

Why it mattersThe model uses lp (fun : ℕ => ℂ) 2, with unrestricted states, amplitudes and environment support, a global two-mode beam-splitter unitary, and its square-root Kraus realization of the partial trace.

The ideaTotalness of the exponential family constructs the global displacements and beam splitter on the Hilbert completion.

Our contributionThe full bosonic attenuator and the coherent-state output-entropy question.

Scope & next questionThis refutation does not settle Conjecture 3 because global minimality of the competitor is not asserted.

Verification details
Release source recordRefuted in Lean

Open problem

Van Herstraeten–Guha–Cerf Conjecture 2: coherent inputs for phase-covariant attenuators

Conjecture 2 (Phase-covariant attenuator), section IV.B, arXiv:2312.15623v2, PDF p. 12, states: “The minimum output entropy of a phase-covariant attenuator channel ℳη,𝐩 is achieved by coherent states, ∀η,𝐩.” The domain is every transmissivity η in [0,1] and every probability vector p on the natural occupations. The assertion is that some complex coherent amplitude α minimizes the output entropy over all density operators on the full one-mode Fock space. Preregistration #11681 gives this quantifier order and the counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/PhaseCovariantAttenuatorCoherentRefutation.result

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Discrete structuresQuestion resolvedFirst preprint 2024Schröder Enumeration of the Sorting Class of the Stack Avoiding 23-1The sorting class of the stack avoiding 23-1 is enumerated by the large Schroeder numbers.Proved

The questionZhao, arXiv:2410.17057v1, Conjecture 3.30 (Section 3.2, printed page 11):

The ideaThe sortable permutations are characterized directly from the stack rule by avoidance of an explicit mesh pattern.

Our contributionThe sorting class of the stack avoiding 23-1 is enumerated by the large Schroeder numbers.

Verification details
Release source recordProved in Lean

Open problem

Schröder Enumeration of the Sorting Class of the Stack Avoiding 23-1

Zhao, arXiv:2410.17057v1, Conjecture 3.30 (Section 3.2, printed page 11):

The sorting class of SC_{23̲1} is enumerated by |Sort_n(SC_{23̲1})| = S_{n−1}.

SC_{23̲1} is the right-greedy stack map whose stack, read from top to bottom, must avoid the vincular pattern 23̲-1 (the entries playing 2 and 3 adjacent); Sort_n(SC_{23̲1}) is the set of permutations of length n whose image avoids 231, and S_m is the large Schröder number.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/VincularStack/VincularStackSort.result

Frozen module record

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2024The isotropic aggregate of a two-qudit state in dimension five is below sixThe isotropic aggregate in Zhu, Mao and Yi's Conjecture 2 sums the third-moment expectations over stochastic orthogonal graph subspaces.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersThe paper derives the aggregate bounds from the pointwise clause and uses the aggregate sums for the shadow norms of stabilizer projectors and for the third frame potential of Clifford orbits.

Our contributionThe isotropic aggregate in Zhu, Mao and Yi's Conjecture 2 sums the third-moment expectations over stochastic orthogonal graph subspaces.

Verification details
Release source recordRefuted in Lean

Open problem

The isotropic aggregate of a two-qudit state in dimension five is below six

H. Zhu, C. Mao and C. Yi, Third moments of qudit Clifford orbits and 3-designs based on magic orbits, arXiv:2410.13575v1, Section VII.1, Conjecture 2, asserts for an odd prime $d$ and $|\Psi\rangle\in\mathcal H_d^{\otimes n}$ the pointwise bounds $0\le\kappa(\Psi,\mathcal T)\le1$ on $\Sigma(d)$ together with the aggregate bounds $6\le\kappa(\Psi,\Sigma(d))\le2d+2$, $0\le\kappa(\Psi,\mathscr T_{\rm ns})\le2d-4$ and $\kappa(\Psi,\mathscr T_{\rm iso})\ge6$. Here $\kappa(\Psi,\mathscr T)=\sum_{\mathcal T\in\mathscr T}\kappa(\Psi,\mathcal T)$ and $\mathscr T_{\rm iso}={\mathcal T_O : O\in O_3(d)}$. The verbatim statements are in the literature note.

The pointwise clause was refuted earlier with one qudit at $d=11$ (dossier); for that state the aggregate bounds hold. Issue #14561 targets the remaining aggregate lower bound in its printed form $\kappa(\Psi,\mathscr T_{\rm iso})\ge6$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/CliffordThirdMomentAggregateRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2022The controlled-rotation channel assemblages do not obey the proposed LOSR angle orderingThe controlled-rotation channel assemblage at pi/3 converts by local operations and shared randomness to the assemblage at pi/2, contradicting the proposed angle ordering.Refuted

The questionSainz, The resource theory of nonclassicality of channel assemblages, Quantum 7, 1134 (2023), arXiv:2209.10177v3, Conjecture 3:

Why it mattersThe paper uses LOSR convertibility to compare the nonclassicality of channel assemblages.

The ideaHere IsCPTP means complete positivity at every finite amplification and preservation of the trace of every matrix.

Our contributionThe controlled-rotation channel assemblage at pi/3 converts by local operations and shared randomness to the assemblage at pi/2, contradicting the proposed angle ordering.

Scope & next questionThe module does not settle the correctness of the general SDP formulation or the later comparisons of different post-quantum assemblages.

Verification details
Release source recordRefuted in Lean

Open problem

The controlled-rotation channel assemblages do not obey the proposed LOSR angle ordering

B. Zjawin, D. Schmid, M. J. Hoban and A. B. Sainz, The resource theory of nonclassicality of channel assemblages, Quantum 7, 1134 (2023), arXiv:2209.10177v3, Conjecture 3:

Let the two resources $\In^{,\theta_1}{\A|\X}$ and $\In^{,\theta_2}{\A|\X}$ belong to the infinite family of assemblages $\FA$. Then, the conversion $\In^{,\theta_1}{\A|\X} \overset{\text{LOSR}}{\longrightarrow} \In^{,\theta_2}{\A|\X}$ is possible if and only if $\theta_1 \geq \theta_2$.

This is the verbatim text of channel-quantum.tex, line 354, in the v3 source, md5 1478712c3cb6058c040039d9ae9978e6. The source macros \FA, \In, \A and \X denote $\mathsf R$, $\mathbf I$, $\mathbb A$ and $\mathbb X$. The definitions and locators are also in the literature note.

The family has binary outcomes and settings, $a,x\in{0,1}$, and angles $\theta\in(0,\pi/2]$. Alice and Bob share $|\phi\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Bob applies the controlled gate $|0\rangle\langle0|B\otimes I+|1\rangle\langle1|B\otimes U\theta$, where $U\theta$ has rows $(\cos(\theta/2),-\sin(\theta/2))$ and $(\sin(\theta/2),\cos(\theta/2))$. The construction traces out $B$, applies Alice's effect $M_{a|0}=(I+(-1)^a\sigma_z)/2$ or $M_{a|1}=(I+(-1)^a\sigma_x)/2$, and traces out $A$ (Eqs. eq:CR and eq:CRfam, TeX lines 316–333).

Issue #21490 fixes the reading: for every pair of angles in $(0,\pi/2]$, LOSR conversion between these literal assemblages is possible exactly when the source angle is at least the target angle. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ChannelAssemblageLOSRRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2025Chauve--Zhang condensed-neighborhood unary minimalityBinary words refute unary minimality for condensed and super condensed neighborhoods.Refuted

The questionChauve and Zhang, On the size of the neighborhoods of a word, arXiv:2505.13796v2, Section 6, printed page 7, ask:

Why it mattersThe paper proves exact cardinality formulas for condensed neighborhoods of unary words.

The ideaAt length three and radius two, CN(000,2) = {0,10,110} while CN(001,2) = {0,1}.

Our contributionBinary words refute unary minimality for condensed and super condensed neighborhoods.

Verification details
Release source recordRefuted in Lean

Open problem

Chauve--Zhang condensed-neighborhood unary minimality

Chauve and Zhang, On the size of the neighborhoods of a word, arXiv:2505.13796v2, Section 6, printed page 7, ask:

It was also shown in [6] that unary words have the smallest neighborhoods among all words of the same length over a given alphabet, thus leading to lower bounds for the size of neighborhoods. It is thus natural to ask if a similar property holds for condensed and super condensed neighborhoods, namely that unary words have the smallest condensed or super condensed neighborhoods.

This dossier tracks the condensed assertion. Equations (2.1)--(2.2) define N(w,d) as the words at Levenshtein distance at most d and CN(w,d) as the members of N(w,d) with no proper prefix in N(w,d). The assertion asks whether, for every finite alphabet, length n, radius d, letter a, and word w of length n,

|CN(a^n,d)| <= |CN(w,d)|.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/ChauveZhangUnaryNeighborhoodRefutation.resultCondensed

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2025Chauve--Zhang super-condensed-neighborhood unary minimalityBinary words refute unary minimality for condensed and super condensed neighborhoods.Refuted

The questionChauve and Zhang, On the size of the neighborhoods of a word, arXiv:2505.13796v2, Section 6, printed page 7, ask:

Why it mattersThe paper proves exact cardinality formulas for super-condensed neighborhoods of unary words.

The ideaAt length four and radius one, SCN(0000,1) = {000,0010,0100} while SCN(0011,1) = {001,011}.

Our contributionBinary words refute unary minimality for condensed and super condensed neighborhoods.

Verification details
Release source recordRefuted in Lean

Open problem

Chauve--Zhang super-condensed-neighborhood unary minimality

Chauve and Zhang, On the size of the neighborhoods of a word, arXiv:2505.13796v2, Section 6, printed page 7, ask:

It was also shown in [6] that unary words have the smallest neighborhoods among all words of the same length over a given alphabet, thus leading to lower bounds for the size of neighborhoods. It is thus natural to ask if a similar property holds for condensed and super condensed neighborhoods, namely that unary words have the smallest condensed or super condensed neighborhoods.

This dossier tracks the super-condensed assertion. Equations (2.1) and (2.3) define N(w,d) as the words at Levenshtein distance at most d and SCN(w,d) as the members of N(w,d) with no proper contiguous subword in N(w,d). The assertion asks whether, for every finite alphabet, length n, radius d, letter a, and word w of length n,

|SCN(a^n,d)| <= |SCN(w,d)|.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/ChauveZhangUnaryNeighborhoodRefutation.resultSuperCondensed

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture provedFirst preprint 2026Božović Conjecture 12: k-limited domination of Sierpiński graphsThe k-limited domination number of the Sierpiński graph S(n,m) is proved exactly for n ≥ 1, 2 ≤ m, and 1 ≤ k ≤ m − 1.Proved

The questionBožović, On k-limited domination: complexity and Sierpiński graphs, arXiv:2610.01584v1, Section 6, Conjecture 12, states:

Why it mattersThe exact formula determines the parameter for all graph sizes and all 1 ≤ k ≤ m−1.

The ideaFor the upper bound, a recursive ZMod m coloring is constant on linking edges and bijective within every base clique.

Our contributionThe k-limited domination number of the Sierpiński graph S(n,m) is proved exactly for n ≥ 1, 2 ≤ m, and 1 ≤ k ≤ m − 1.

Verification details
Release source recordProved in Lean

Open problem

Božović Conjecture 12: k-limited domination of Sierpiński graphs

D. Božović, On k-limited domination: complexity and Sierpiński graphs, arXiv:2610.01584v1, Section 6, Conjecture 12, states:

Let n, m, and k be integers such that n ≥ 1 and 1 ≤ k ≤ m − 1. Then γ_k^L(S(n, m)) = (m − k) m^(n−1).

Section 2 defines S(n,m) for n ≥ 1 and m ≥ 2. Its vertices are words of length n over {0,…,m−1}. Adjacency means that for some coordinate h the preceding coordinates agree, the h-coordinates differ, and each later coordinate equals the other word's h-coordinate. A dominating set D is k-limited when every selected vertex has at most k neighbors outside D; γ_k^L is its minimum cardinality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SierpinskiLimitedDomination.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedSource 2025Pronko's commutation of the Dyck-class operator Ξ with the periodic Fredkin HamiltonianSpin words of the periodic Fredkin chain are read as lattice paths and sorted into Dyck classes.Proved

The questionPronko reads spin words as lattice paths in section 2.2, printed p. 4:

Why it mattersThe paper uses this relation, with C Ξ C⁻¹ = −Ξ, to explain the double degeneracy it observes numerically in the S^z = 0 sector of the spectrum.

The ideaA word is balanced when it has as many up letters as down letters; a balanced word lies in exactly one class C{a,a}(N), with a = −min{0 ≤ i ≤ N} si for the prefix heights si, and Ξ takes the value (−1)^a there; Ξ vanishes on unbalanced words.

Our contributionSpin words of the periodic Fredkin chain are read as lattice paths and sorted into Dyck classes.

Verification details
Release source recordProved in Lean

Open problem

Pronko's commutation of the Dyck-class operator Ξ with the periodic Fredkin Hamiltonian

Pronko reads spin words as lattice paths in section 2.2, printed p. 4:

A Dyck path of length N starts at (x,y)=(a,0) and ends at (x,y)=(b,N), at each step Δx = 1 and Δy = ±1, with the condition that at least once the path hits the x-axis, i.e., min y = 0 along the path. … we will call all such paths as belonging to the equivalence class C_{a,b}(N).

A letter ↑ is a step up and a letter ↓ a step down. Section 4.1, printed pp. 9-10, defines for even N

Ξ = Σ_{k=0}^{N/2} (−1)^k Σ_{ℓ₁ℓ₂…ℓ_N ∈ C_{k,k}(N)} n_1^{ℓ₁} n_2^{ℓ₂} ⋯ n_N^{ℓ_N}, (4.1)

with n^{ℓ} the spin-up and spin-down projections, and states on printed p. 10:

It commutes with the Hamiltonian and anti-commutes with the cyclic shift operator: [Ξ, H] = 0, C Ξ C⁻¹ = −Ξ. The first relation may require a proof though we find it satisfied in all checks; the second one is obvious from (4.1).

H is the periodic Fredkin Hamiltonian (2.3) with density (2.2), F_{j,j+1,j+2} = n↑_j Π_{j+1,j+2} + Π_{j,j+1} n↓_{j+2}, sites modulo N.

Issue #9996 fixes the readings: (i) ℓ ∈ C_{a,b}(N) when the path started at height a stays at height ≥ 0, touches 0, and ends at b; (ii) the projections and H are those of the frozen module D5/S3/Quantum/Dynamics/PronkoFredkinNonCyclicAnnihilation; (iii) N is even and N ≥ 3; (iv) only the first relation [Ξ, H] = 0 is settled.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PronkoFredkinXiCommutes.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2025.

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Quantum informationQuestion resolvedFirst preprint 2026Global dynamic cooling strictly beats equal parallel groups except at (N, M) = (6, 2)For every initial excited population strictly between zero and one half, global unitary cooling gives a strictly smaller common target population than equal parallel single-target groups whenever M is at least two and N is at least M plus two, except for the pair (6, 2).Proved

The questionGuarnieri, Optimal Dynamic Cooling of Multiple Qubits, arXiv:2607.23641v1, cool the first M of N identical thermal qubits, in the state ρ = τ^{⊗N} with τ = diag(1 − P₁, P₁), by one global unitary.

Why it mattersThe comparison decides when joint control of the whole register is worth its cost over independent single-target refrigerators.

The ideaNo sorting or Birkhoff decomposition is needed.

Our contributionFor every initial excited population strictly between zero and one half, global unitary cooling gives a strictly smaller common target population than equal parallel single-target groups whenever M is at least two and N is at least M plus two, except for the pair (6, 2).

Scope & next questionComputed, not formalized (orchestrator, exact rational arithmetic, 14 October 2026): - the source's passive formula, Eqs. (12)–(15), gives p★(N, M) = p★(⌊N/M⌋, 1) only at (6, 2) among the 275 cases with M ≥ 2, N ≥ M + 2, N ≤ 13 and P₁ ∈ {1/10, 1/4, 1/3, 2/5, 49/100}; - the margin above is respected in every case. - the coherent step is needed: at (N, M) = (4, 2) and P₁ = 1/10, every computational-basis permutation with equal target populations leaves the common population at least 1/10 = ppar.

Verification details
Release source recordProved in Lean

Open problem

Global dynamic cooling strictly beats equal parallel groups except at (N, M) = (6, 2)

M. Reda, M. Sacchi, C. Macchiavello and G. Guarnieri, Optimal Dynamic Cooling of Multiple Qubits, arXiv:2607.23641v1, cool the first M of N identical thermal qubits, in the state ρ = τ^{⊗N} with τ = diag(1 − P₁, P₁), by one global unitary. The aim is the lowest common local temperature: every target must end with the same diagonal one-qubit marginal. The optimum is p_glob(N, M; P₁) = p★(N, M; P₁).

§IX compares it with M disjoint groups of n = ⌊N/M⌋ qubits, each running the optimal n-qubit single-target protocol: p_par(N, M; P₁) = p★(n, 1; P₁) (Eqs. (88)–(89)). The source proves p_glob ≤ p_par, strictness for M + 2 ≤ N ≤ 3M − 1 (Eq. (91)), and the tie p_glob(6, 2) = p_par(6, 2) = 3P₁² − 2P₁³ (Eq. (92)). It then writes: "Exact symbolic checks over the finite range investigated revealed no other equality, although the uniqueness of this tie remains a conjecture." The verbatim statements are in the literature note.

Issue #16483 fixes the reading: for all M ≥ 2, N ≥ M + 2 with (N, M) ≠ (6, 2), and every 0 < P₁ < 1/2, p_glob(N, M; P₁) < p_par(N, M; P₁). It is proved. The tie (92) is the source's and is not re-proved here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Thermal/DynamicCoolingParallelTieUniqueness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2025The geometric measure of the flip-conjugate symmetric subspaceThe geometric measure is constant on each flip-conjugate symmetric subspace in every local dimension at least two.Proved

The questionGühne, Chiral Symmetries and Multiparticle Entanglement, arXiv:2506.15609v1, Appendix H.2:

Why it mattersThe source builds highly entangled three-qudit subspaces from chiral and flip-conjugate symmetries; every pure state in such a subspace is genuinely multipartite entangled, and a large uniform geometric measure makes the subspace useful for entanglement certification.

The idea4. The conjugate subspace follows by complex conjugation.

Our contributionThe geometric measure is constant on each flip-conjugate symmetric subspace in every local dimension at least two.

Verification details
Release source recordProved in Lean

arXiv

The geometric measure of the flip-conjugate symmetric subspace

S. Denker, S. Imai and O. Gühne, Chiral Symmetries and Multiparticle Entanglement, arXiv:2506.15609v1, Appendix H.2:

Conjecture: The geometric measure of entanglement for any state in the flip-conjugate symmetric subspace $\mathcal{H_I}$ (and respectively $\mathcal{H}{\bar{\mathcal{I}}}$) is given by $G(\ket{\psi}) = 1-\frac{d^2}{(d^2-1)(d+1)}$, where $d$ denotes the local dimension, which coincides with the dimension $\dim(\mathcal{H}\mathcal{I})$ of the subspace.

Here:

  • The subspace: $\mathcal H_{\mathcal I}\subset(\mathbb C^d)^{\otimes3}$ is spanned by $\phi_n=(\bar z_d\sum_i|ini\rangle+z_d\sum_i|iin\rangle)/\sqrt{d+1}$ with $z_d=\tfrac12(1+i\sqrt{(d+1)/(d-1)})$.
  • The geometric measure: $G(\psi)=1-\max|\langle abc|\psi\rangle|^2$ over unit product vectors.

The definitions are quoted in the literature note.

Issue #20037 fixes the reading: for every $d\ge2$ and every unit vector of either subspace, $G=1-d^2/((d^2-1)(d+1))$. This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GeometricMeasure/FlipConjugateGeometricMeasure.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Gupta's rectangular-grid dispersion inequalityInteger grid Lipschitz functions satisfy Gupta's dispersion bound.Proved

The questionGupta, Height functions on the m × n Miura-ori flip graph: degree sequence and diameter, arXiv:2606.22614v2, Section 4, inequality (4), p. 14, asks for

Why it mattersTheir combination with inequality (4) identifies the diameter for positive rectangles.

The ideaThese permutations preserve every sum of a function of the values.

Our contributionInteger grid Lipschitz functions satisfy Gupta's dispersion bound.

Verification details
Release source recordProved in Lean

Open problem

Gupta's rectangular-grid dispersion inequality

Gupta, Height functions on the m × n Miura-ori flip graph: degree sequence and diameter, arXiv:2606.22614v2, Section 4, inequality (4), p. 14, asks for

$\operatorname{disp}(\phi)\le D(m,n)$

for every integer $1$-Lipschitz function on the rectangular grid, realisable or not. Here

$\operatorname{disp}(\phi)=\min_{K\in\mathbb Z}\sum_v|\phi(v)-K|$,

and

$D(m,n)=\min_{K\in\mathbb Z}\sum_{i=1}^m\sum_{j=1}^n|i+j-K|$.

The named external problem and literal conventions are preregistered in #19839.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Rectangular grid with horizontal and vertical neighbours | Fin m × Fin n, adjacency in SimpleGraph.pathGraph m □ SimpleGraph.pathGraph n | Source cell $(i,j)$ is represented by $(i-1,j-1)$. | | Every integer function, realisable or not | Universal function binder in claim | No origami realisability assumption. | | Adjacent differences at most one | IsLipschitz | Literal edge predicate. | | Minimum over integer centres, each vertex counted once | disp | Natural-valued infimum of a nonempty integer-cost range, hence its minimum. | | One-indexed coordinate sum | D | Both zero-based coordinates are incremented by one. | | Inequality (4) for all positive rectangles | result : claim | All natural dimensions are covered; empty rectangles add the zero-cost case. |

The public definitions IsLipschitz, disp, D and claim display the source expressions. The existing rectangular graph owner LeeLiuThreeRowIntegerDomination.grid is used through the private correspondence lipschitz_owner; the public predicate retains its literal graph expression.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/GuptaMiuraGridDispersion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Every Interior Multiple of n Is a Latin-Square Ascent TotalEvery permitted interior multiple of the order occurs as the total column-ascent number of a Latin square.Proved

The questionMadjid Mirzavaziri and Daniel Yaqubi, Latin Eulerian Numbers, arXiv:2609.25100v1, Remark 4.16:

Why it mattersTogether with the paper's results it shows that the set of attained values of Σ is exactly [n − 1, (n − 1)²] with the two values n and (n − 1)² − 1 removed.

The ideaFor a permutation p of 0, …, n − 1 let Lp(i, c) = τ((pi + c) mod n), where τ exchanges the symbols 0 and 1.

Our contributionEvery permitted interior multiple of the order occurs as the total column-ascent number of a Latin square.

Verification details
Release source recordProved in Lean

Open problem

Every Interior Multiple of n Is a Latin-Square Ascent Total

Madjid Mirzavaziri and Daniel Yaqubi, Latin Eulerian Numbers, arXiv:2609.25100v1, Remark 4.16:

Precisely two things remain: (i) that every multiple of n strictly between n and (n − 1)^2 − 1 is attained by some Latin square which Proposition 4.14 reduces to an explicit, checkable question about the function F, verified for n ≤ 11 but open in general; (ii) unimodality of (T_n(m)) …

Here Σ(L) is the sum over the columns of an order-n Latin square of the number of ascents read from top to bottom. Part (i) asks, for every n ≥ 5 and 2 ≤ k ≤ n − 3, for an order-n Latin square with Σ(L) = kn.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/LatinEulerianMultiples.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026The Wagner graph and the rank-5 eight-vertex graph have no qutrit violation, so d* = 4The Wagner graph and the rank-five graph of Tamer et al. have greatest real qutrit score three, independence number three, and least dimension of a real contextuality violation four.Proved

The questionThey mark d = 4 for both with "†Numerically indicated; a rigorous three-dimensional upper bound is not established here." - §IV: "A formal demonstration that d(W) = 4 would require an independent upper bound on η₃(W)."

Why it mattersThe minimum dimension separates contextuality witnesses that a single qutrit can show from those that need a four-level system or two qubits.

The idea1. Square lemma. In ℝ³, unit vectors with a ⊥ b, b ⊥ c, c ⊥ d and d ⊥ a satisfy a = ±c or b = ±d.

Our contributionThe Wagner graph and the rank-five graph of Tamer et al. have greatest real qutrit score three, independence number three, and least dimension of a real contextuality violation four.

Verification details
Release source recordProved in Lean

arXiv

The Wagner graph and the rank-5 eight-vertex graph have no qutrit violation, so d* = 4

U. Tamer, B. Ünal, A. Dizdar, Z. Gedik and Ö. E. Müstecaplıoğlu, The Quad-C₅ Graph: Maximum Contextuality Gap on Eight Vertices, arXiv:2605.12828v3, rank eight-vertex exclusivity graphs by their contextuality gap.

  • For a graph G they define the dimension-restricted value η_d(G), the largest λ_max(Σ_i |v_i⟩⟨v_i|) over unit vectors v_i ∈ ℝ^d orthogonal on every edge (Eq. (13)).
  • They define the minimum dimension d*(G) = min{d : η_d(G) > α(G)} (Eq. (25)).
  • For the Wagner graph W and the graph ranked fifth in Table V(a), their non-convex search finds no three-dimensional value above α = 3. They mark d* = 4 for both with "†Numerically indicated; a rigorous three-dimensional upper bound is not established here."
  • §IV: "A formal demonstration that d*(W) = 4 would require an independent upper bound on η₃(W)."

The verbatim statements are in the literature note.

Issue #16223 fixes the reading. For G ∈ {W, G₅}, with the edge lists of Table VI:

  • every realization in ℝ^d with d ≤ 3 satisfies Σ_i ⟨ψ, v_i⟩² ≤ 3‖ψ‖², that is λ_max ≤ 3;
  • the value 3 is attained in ℝ³;
  • some realization in ℝ⁴ exceeds 3;
  • α(G) = 3.

Hence d*(G) = 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/EightVertexQutritBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2026A four-level system whose triplets all violate the three-level conditions, yet shows the Markovian Mpemba effectFour reversible energy levels exhibit the Mpemba effect although every triplet fails the proposed three-level condition.Refuted

The questionThe follow-up arXiv:2609.09988v1 (§II) restates the conjecture as shown.

Why it mattersThe conjecture would reduce the search for the Mpemba effect in N-level systems to checking three-level conditions.

The idea1. Witness. Four levels with energies (0, 1, 2, 64)·log 16, βb = 1, and r = 16⁻⁶⁴.

Our contributionFour reversible energy levels exhibit the Mpemba effect although every triplet fails the proposed three-level condition.

Scope & next questionReductions that keep the eigenvalue dependence of the eliminated block remain open. - Whether the conjecture holds under additional hypotheses on the energy spectrum, for example bounded level spacing.

Verification details
Release source recordRefuted in Lean

arXiv

A four-level system whose triplets all violate the three-level conditions, yet shows the Markovian Mpemba effect

I. Avitan, R. Factor and D. Gelbwaser-Klimovsky, Necessary conditions for the Markovian Mpemba effect, arXiv:2603.04567v1, study relaxation under a rate matrix M with detailed balance.

  • Slow-mode criterion. The Mpemba effect occurs iff the slow-mode coefficient |b_2| of the initial Gibbs state Π(β_in) has a maximum at some β_in ≠ β_b (p. 2).
  • Three-level conditions. For three levels they derive the necessary condition (3) on the rates.
  • Triplets. For N levels they apply (3) to every triplet of levels, keeping the original rates.
  • The conjecture (p. 4): "we conject that if for an N−level system, all its triplets violate conditions (3), then the N−level system does not experience the Mpemba effect."

The follow-up arXiv:2609.09988v1 (§II) restates the conjecture as shown.

The verbatim statements are in the literature note.

Issue #17871 fixes the reading:

  • the slow-mode coefficient is ℓ·Π(β) for a left eigenvector ℓ of the slowest nonzero eigenvalue, assumed simple;
  • the effect is two initial temperatures on the same side of β_b, the farther one with the smaller |ℓ·Π(β)|. This implies the source's criterion.

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/MarkovMpembaTripletRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2025The purity of a state under loss is not always log-convex in the transmissionThe purity of a three-level mixed Fock state under vacuum-environment optical loss fails the midpoint inequality for log-convexity.Refuted

The questionGoldberg, Equalities and inequalities from entanglement, loss, and beam splitters, arXiv:2501.02047v1, Quantum Stud.

Why it mattersThe source uses Conjecture 7 to extend its loss inequalities beyond 50% loss.

The ideaThe partial trace over the environment, slice by slice, gives the output diag(q₀, q₁, q₂) with

Our contributionThe purity of a three-level mixed Fock state under vacuum-environment optical loss fails the midpoint inequality for log-convexity.

Verification details
Release source recordRefuted in Lean

Open problem

The purity of a state under loss is not always log-convex in the transmission

A. Hertz, N. Lupu-Gladstein, K. Heshami and A. Z. Goldberg, Equalities and inequalities from entanglement, loss, and beam splitters, arXiv:2501.02047v1, Quantum Stud. Math. Found. 13, 34 (2026), study the vacuum-environment loss channel ρ_T = Tr₂[B(T)(ρ₁ ⊗ |0⟩⟨0|)B(T)†], with B(T) = exp(arccos(√T)(a₁a₂† − a₁†a₂)).

  • They prove that the purity Tr[ρ_T²] is convex in the transmission T.
  • Conjecture 7 asserts the stronger "The purity of a state ρ₁ subject to loss is log-convex in T ∈ [0, 1]; ∂² log Tr[ρ_T²]/∂T² ≥ 0", equivalently (26) (P′)² ≤ P·P″ with P(T) = Tr[ρ_T²].
  • Appendix B extends it to mixtures of identical two-copy states (Conjecture 19, (B4)).

The verbatim statements are in the literature note.

Issue #16345 fixes the reading: for every single-mode state ρ₁, the function T ↦ log Tr[ρ_T²] is convex on [0, 1]. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/FockAttenuator/LossPurityLogConvexityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2026One ancilla qubit and bounded rounds cannot simulate every channel with constant successA family of sign-frame channels has exact postselection success at most 4 times 4 to the R divided by n for every adaptive protocol with one ancillary qubit and at most R measurement rounds.Refuted

The questionKotowski, Randomized simulation of quantum channels using small ancilla, arXiv:2606.08784v1, Question 6.1:

Why it mattersThe paper shows that one ancilla qubit suffices for success probability Ω(1/log d) on unital channels and Ω(1) on broad classes of channels.

The ideaIndex the coordinates by m basis labels and by all sign strings, and let A have rows ej and s/n.

Our contributionA family of sign-frame channels has exact postselection success at most 4 times 4 to the R divided by n for every adaptive protocol with one ancillary qubit and at most R measurement rounds.

Verification details
Release source recordRefuted in Lean

arXiv

One ancilla qubit and bounded rounds cannot simulate every channel with constant success

M. Kotowski and M. Kotowski, Randomized simulation of quantum channels using small ancilla, arXiv:2606.08784v1, Question 6.1:

Consider the model of simulation with postselection that allows: 1) unitary operations with ancilla; 2) classical randomization; 3) projective measurements; 4) adaptive unitary operations based on measurement results. Can \textit{every} channel $\Phi$ be simulated in this way with success probability $\Omega(1)$ using one additional qubit and $O(1)$ rounds?

The verbatim passages are in the literature note.

Issue #19196 fixes the reading: universal constants R₀ and q₀ > 0 such that every channel on M_d(ℂ), for every d, has a randomized adaptive protocol on ℂ^d ⊗ ℂ² with at most R₀ projective-measurement stages and success map qΦ with q ≥ q₀. The answer is no.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/OneQubitAdaptiveSimulationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2026The polar rank-one encoder of Li–Jiang is not exactly optimal at fixed noiseA feasible qubit rank-one encoder strictly improves the source polar pair.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersThe polar construction is how the source shows that rank-one encoders achieve the optimal leading-order behaviour near perfect recovery.

The ideaBoth encoders are isometries, so they are completely positive and trace-preserving with a single Kraus operator.

Our contributionA feasible qubit rank-one encoder strictly improves the source polar pair.

Verification details
Release source recordRefuted in Lean

arXiv

The polar rank-one encoder of Li–Jiang is not exactly optimal at fixed noise

B. Li and L. Jiang, High-Rank Encoding Can Improve Approximate Quantum Error Correction, arXiv:2609.00778v1, compare encoders of a logical qudit into two qudits under a family of noise channels N_p (Eqs. (18)–(23)).

  • The pair. For a unit vector χ, the encoder C_{V_p(χ)} uses the polar isometry V_p(χ) = C_p(χ)[C_p(χ)†C_p(χ)]^{−1/2} with C_p(χ) = Q_p(I ⊗ |χ⟩) (Eq. (S60)), and the decoder is the partial trace tr_S. The supplement shows that this rank-one pair attains the optimal quadratic coefficient of the rank-one gap as p → 0 (Eqs. (S63)–(S64)).
  • The question. "Exact optimality of this pair at fixed p > 0 remains open" (after Eq. (S64)).

The verbatim statements are in the literature note.

Issue #13324 fixes the reading. For every d ≥ 2, 0 < p < 1 and unit χ, the pair is claimed to be optimal against every pair of completely positive trace-nonincreasing encoder and decoder whose encoder has input-first Choi rank at most one (the source's rank-one class, Eq. (13)), for the entanglement fidelity at τ = I/d. The question is answered negatively.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Recovery/LiJiangPolarPairOptimalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2025The kernel criterion for witness optimality does not imply the trace criterionRefuted

The questionCichy, Simple Sufficient Criteria for Optimality of Entanglement Witnesses, arXiv:2505.15615v2, Section IV (p. 12):

Verification details
Release source recordRefuted in Lean

arXiv

The kernel criterion for witness optimality does not imply the trace criterion

F. vom Ende and S. Cichy, Simple Sufficient Criteria for Optimality of Entanglement Witnesses, arXiv:2505.15615v2, Section IV (p. 12):

Next: is our trace-based criterion (Coro. 2) actually equivalent to our kernel criterion (Thm. 2)? This is the question mark in Fig. 2, and while the kernel criterion seems stronger we were not able to find a witness for which only the kernel criterion holds.

The kernel criterion asks for a vector of Schmidt rank $\min{m,n}$ in the kernel of $W+{\rm tr}_2(W)\otimes{\bf1}$ (if $m\le n$) or of $W+{\bf1}\otimes{\rm tr}_1(W)$ (if $m\ge n$); the trace criterion asks for a maximally entangled $\Omega$ with $\langle\Omega|W|\Omega\rangle=-{\rm tr}(W)/\min{m,n}$. The verbatim texts are in the literature note. Issue #14763 preregisters the reading and the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/WitnessKernelTraceCriteriaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2025The plus-count statistic is insufficient for exact-weight discriminationTwo queries to a fixed four-address memory contain discrimination information beyond the plus count.Refuted

The questionBohac, Bias-Class Discrimination of Universal QRAM Boolean Memories, arXiv:2512.17503v1, asks in Remark 5.3, “Open question: do collective measurements help?” The source TeX biasclassdiscrimination.tex, lines 454–457, states:

Why it mattersThe source's separable strategy measures the all-plus projector on each address register and retains only the number of plus outcomes.

The ideaThe exact weight-one and weight-two classes are given by classone and classtwo.

Our contributionTwo queries to a fixed four-address memory contain discrimination information beyond the plus count.

Scope & next questionThe source's optimal error exponent, the collective Helstrom optimum, and any claim for a different prior model remain open.

Verification details
Release source recordRefuted in Lean

Open problem

The plus-count statistic is insufficient for exact-weight discrimination

L. Bohac, Bias-Class Discrimination of Universal QRAM Boolean Memories, arXiv:2512.17503v1, asks in Remark 5.3, “Open question: do collective measurements help?” The source TeX bias_class_discrimination.tex, lines 454–457, states:

The $t$-copy induced state $\rho_p^{(t)}$ generally does not factor as $\rho_p^{\otimes t}$, so in principle a collective Helstrom measurement on $\rho_{p_0}^{(t)}$ vs.\ $\rho_{p_1}^{(t)}$ could outperform any separable strategy.

An interesting open problem is to characterize the optimal error exponent for distinguishing these \emph{exchangeable} but potentially correlated mixtures, and to determine whether the ``$\ket{+}$-count'' statistic is sufficient for optimal discrimination in the exact-weight prior model.

The module formalizes the second question through claim. Its weightClass n m is the finite set of Boolean memories $f : \mathrm{Fin}(2^n) \to \mathrm{Bool}$ having exactly $m$ true addresses. Its copyState n t m averages the $t$-fold product of one addressState n f over that class, with the same $f$ in every copy. For every $n \ge 1$, $t \ge 1$, $m_0,m_1 \le 2^n$, every $e : \operatorname{Matrix}(\mathrm{Fin}(t) \to \mathrm{Fin}(2^n))(\mathrm{Fin}(t) \to \mathrm{Fin}(2^n)),\mathbb{C}$ with e.PosSemidef and (1 - e).PosSemidef, and successProb as defined in the module, claim requires a Boolean rule $g : \mathrm{Fin}(t+1) \to \mathrm{Bool}$ such that $\operatorname{successProb}(n,t,m_0,m_1,e) \le \operatorname{successProb}(n,t,m_0,m_1,\operatorname{countTest}(n,t,g))$. This is the equal-prior special case: if the count were sufficient for every prior, it would in particular satisfy this equal-prior assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Tomography/QramCountStatisticInsufficiency.result

Frozen module record

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Geometry & algebraQuestion resolvedFirst preprint 2026Feldman–Wilce Question 1: finite, unbounded cliquesThe actual principal extension of a section-dense selector carries a closed fibre graph with finite cliques of every size.Proved

The questionFeldman and Alexander Wilce, A Note on Compactness and Clique Size, arXiv:2608.13320v1, Question 1, printed page 5:

Why it mattersThe paper proves that compact spaces have the bounded clique property and that, for T1 spaces, countable compactness is equivalent to every closed orthogonality relation having only finite cliques.

The ideaEach fibre over b is the finite range of Sigma.mk b; injectivity gives n+1 points in the n-th fibre.

Our contributionThe actual principal extension of a section-dense selector carries a closed fibre graph with finite cliques of every size.

Verification details
Release source recordProved in Lean

Open problem

Feldman–Wilce Question 1: finite, unbounded cliques

David V. Feldman and Alexander Wilce, A Note on Compactness and Clique Size, arXiv:2608.13320v1, Question 1, printed page 5:

Is there a countably compact T1 (or Hausdorff) space with a closed orthogonality relation having finite but unbounded cliques?

An orthogonality relation is symmetric, irreflexive and closed in the entire product, including the diagonal. A clique is a subset with every distinct pair related. The requested relation must have every clique finite and no natural number bounding all clique sizes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Topology/StoneCech/SectionClosedGraph.exists_countablyCompact_closed_fibreGraph

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedFirst preprint 2025Labelle's Toda numerator unimodality conjectureLabelle's recursively defined Toda specialization has a non-unimodal numerator in type C2 at the positive-root coordinates (2,2).Refuted

The questionLabelle, "On a specialization of Toda eigenfunctions", arXiv:2502.10655v3, Section 7, Conjecture 7.3:

Why it mattersThe paper proves polynomiality and positive coefficients in type A and proposes unimodality in its general setting.

The ideaCartanDatum r encodes finite-type symmetrizable generalized Cartan matrices at every finite rank r: integer matrices on Fin r, a symmetric positive-definite Gram form, positive integer d, G(i,j)=d(i)C(i,j), and G(i,i)=2d(i). shortnormalization requires a vertex with d=1 in every connected Dynkin component: every index reaches an index with d=1 in the graph of nonzero off-diagonal Cartan entries.

Our contributionLabelle's recursively defined Toda specialization has a non-unimodal numerator in type C2 at the positive-root coordinates (2,2).

Scope & next questionThe C2 witness satisfies positivity, so the refutation does not settle Conjecture 7.2.

Verification details
Release source recordRefuted in Lean

Open problem

Labelle's Toda numerator unimodality conjecture

A. Labelle, "On a specialization of Toda eigenfunctions", arXiv:2502.10655v3, Section 7, Conjecture 7.3:

The polynomial $(q)\alpha^2\mathfrak{J}\alpha$ is unimodal.

Section 1 fixes a split semisimple Lie algebra over the rationals, its simple roots, and its invariant form normalized so short roots have squared length 2. Definition 1.1 defines $\mathfrak{J}_\alpha$ in $\mathbb{Q}(q)$ for every nonnegative root-lattice coordinate vector. Unimodality means weak increase up to a peak and weak decrease thereafter.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/TodaSpecializationUnimodalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2025Every q-transversal matroid is representable over a finite extensionEvery q-transversal rank function has a representation over a finite field extension.Proved

The questionSaaltink, A theory of q-transversals, arXiv:2503.12201v1, Section 4.2, page 14, states:

Why it mattersThe source proves the coordinate-family case in Theorem 24.

The ideaThere are finitely many subspaces over the finite base field.

Our contributionEvery q-transversal rank function has a representation over a finite field extension.

Scope & next questionThe four-slot escape audit is unfinished under 22087; the sharp uniform extension-degree bound remains open.

Verification details
Release source recordProved in Lean

Open problem

Every q-transversal matroid is representable over a finite extension

M. Saaltink, A theory of q-transversals, arXiv:2503.12201v1, Section 4.2, page 14, states:

Conjecture 3. All q-transversal matroids are representable.

The source note records Definition 14, Theorem 15 and the representation convention of Section 3.2. The preregistration is #19965. The settling declaration is D5/S3/Combinatorics/Matroids/QTransversalRepresentability.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite field $\mathbb F_q$ | F, Field F, Fintype F | Arbitrary finite field | | Finite family of subspaces of a finite-dimensional vector space | m, k, X : Fin k → Submodule F (Fin m → F) | Coordinates identify the ambient space with $F^m$ | | Every basis of a partial q-transversal admits a partial avoiding transversal | IsPartialQTransversal | Every independent finite spanning set admits an injective assignment to indices, with each vector outside its assigned subspace | | Rank of the q-transversal matroid | qTransversalRank | Supremum of dimensions of partial q-transversals contained in the argument subspace | | Section 3.2 matrix rank | representedRank | Rank of the representing matrix times the transpose of a basis-coordinate matrix | | One representing matrix for the whole q-matroid | claim, result | One extension and one matrix give equality of the two ranks for every subspace | | A field extending $F$ | Algebra F K, FiniteDimensional F K | A finite-dimensional extension strengthens the source requirement |

The source's phrase “a dimension n subspace” in the partial definition is read with Theorems 15 and 18, which treat partial q-transversals in every dimension. The public definitions and claim retain the literal conventions of #19965. No assumption that the family is coordinate or nested is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Matroids/QTransversalRepresentability.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedFirst preprint 2022The Higher Order C*-algebraic Smale Mean Value ConjectureThe higher-order C-algebraic Smale mean value conjecture fails at degree three in Complex x Complex with its supremum norm.Refuted

The questionMahesh Krishna, C-algebraic Smale Mean Value Conjecture and Dubinin-Sugawa Dual Mean Value Conjecture, arXiv:2206.08154v1, Section 2, Conjecture HIGHERMEAN; journal DOI 10.26117/2079-6641-2026-54-2-38-47, Conjecture 2.3, p. 44:

Our contributionThe higher-order C-algebraic Smale mean value conjecture fails at degree three in Complex x Complex with its supremum norm.

Verification details
Release source recordRefuted in Lean

Open problem

The Higher Order C*-algebraic Smale Mean Value Conjecture

K. Mahesh Krishna, C-algebraic Smale Mean Value Conjecture and Dubinin-Sugawa Dual Mean Value Conjecture*, arXiv:2206.08154v1, Section 2, Conjecture HIGHERMEAN; journal DOI 10.26117/2079-6641-2026-54-2-38-47, Conjecture 2.3, p. 44:

Let $\mathcal A$ be a commutative C*-algebra. Let $P(z)\coloneqq(z-a_1)\cdots(z-a_n)$ be a polynomial of degree $n\ge2$ over $\mathcal A$, $a_1,\ldots,a_n\in\mathcal A$. If $z\in\mathcal A$ is not a critical point of $P$, then there exists a critical point $w\in\mathcal A$ of $P$ such that

$$\frac{|P^{(k)}(z)|}{k!}\frac{|P(z)-P(w)|^{k-1}}{|P'(z)|^k}\le4^{k-1},\qquad\forall,2\le k\le n.$$

The binding convention of #14982 uses Polynomial A, iterates of Polynomial.derivative, and CommCStarAlgebra A. The factor index Fin n retains every root with multiplicity. Noncritical means $P'(z)\ne0$; the same $w$ must satisfy every inequality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarSmaleHigherOrder.result

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Quantum informationCounterexample foundFirst preprint 2024A negative Clifford third-moment expectation at dimension elevenZhu, Mao and Yi define the third-moment expectation kappa of a normalized qudit state against a stochastic Lagrangian subspace.Refuted

The questionYi, Third moments of qudit Clifford orbits and 3-designs based on magic orbits, arXiv:2410.13575v1 (Commun.

Our contributionZhu, Mao and Yi define the third-moment expectation kappa of a normalized qudit state against a stochastic Lagrangian subspace.

Verification details
Release source recordRefuted in Lean

Open problem

A negative Clifford third-moment expectation at dimension eleven

H. Zhu, C. Mao and C. Yi, Third moments of qudit Clifford orbits and 3-designs based on magic orbits, arXiv:2410.13575v1 (Commun. Math. Phys. 407, 242 (2026)), Conjecture 2: for an odd prime $d$ and $|\Psi\rangle\in\mathcal H_d^{\otimes n}$, $0\le\kappa(\Psi,\mathcal T)\le1$ for every stochastic Lagrangian subspace $\mathcal T\in\Sigma(d)$, together with three aggregate inequalities that the authors derive from it. Here $\kappa(\Psi,\mathcal T)=\operatorname{tr}[R(\mathcal T)(|\Psi\rangle\langle\Psi|)^{\otimes3}]$, $R(\mathcal T)=r(\mathcal T)^{\otimes n}$ and $r(\mathcal T)=\sum_{(x;y)\in\mathcal T}|x\rangle\langle y|$. The verbatim statements are in the literature note.

Issue #14367 targets the pointwise clause for every odd prime $d$, every $n$, every normalized $\Psi$ and every $\mathcal T\in\Sigma(d)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/CliffordThirdMomentNegativity.result

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Quantum informationCounterexample foundFirst preprint 2018The three-distinguishability region in dimension four is not of product formThree orthogonal pure states exclude every labelling of the product ansatz.Refuted

The questionPuchała and K. Życzkowski, Distinguishing classically indistinguishable states and channels, J.

Our contributionThree orthogonal pure states exclude every labelling of the product ansatz.

Verification details
Release source recordRefuted in Lean

Open problem

The three-distinguishability region in dimension four is not of product form

K. Korzekwa, S. Czachórski, Z. Puchała and K. Życzkowski, Distinguishing classically indistinguishable states and channels, J. Phys. A: Math. Theor. 52, 475303 (2019), arXiv:1812.09083v3, conjecture in Section III.B, Eq. (30), Distinguishability.tex lines 462–467 (notation macros expanded):

Moreover, based on numerical evidence, we conjecture that points belonging to $\mathcal A^3_4$ have the following product structure:

$a b \boldsymbol{f}^1+a (1-b) \boldsymbol{f}^2+(1-a) b \boldsymbol{f}^3+(1-a) (1-b) \boldsymbol{f}^4,$

with $a,b\in[0,1]$.

The source defines $\mathcal A_d^M$ by the existence of $M$ perfectly distinguishable quantum states with the same classical version (Distinguishability.tex, lines 147–149), defines that version as the diagonal in the preferred basis (lines 236–241), and identifies perfect distinguishability with pairwise orthogonal supports (lines 297–300). The $\boldsymbol f^i$ are the four face centres of the probability simplex, each a permutation of $(1,1,1,0)/3$ (line 462). The definitions and quotation are also recorded in the literature note.

For the formal statement, index the four coordinates and face centres by $0,1,2,3$, and put $f^j_k=(1-\delta_{jk})/3$. InRegion p means that $p$ is nonnegative with coordinate sum one and that there exist three complex $4\times4$ positive semidefinite matrices $\rho_i$ of trace one satisfying $\rho_i\rho_j=0$ for $i\ne j$ and $\operatorname{Re}(\rho_i)_{kk}=p_k$ for every $i,k$. ProductForm f p is precisely the displayed combination with $a,b\in[0,1]$.

Issue #25867 fixes claim: there exists a permutation $\sigma$ of the four face centres such that every $p$ satisfying InRegion p has ProductForm with centres $f^{\sigma(0)},\ldots,f^{\sigma(3)}$. The existential labelling accommodates the different conventions in the source's text and figure. The module proves result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/ThreeDistinguishabilityProductRefutation.result

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Analysis & dynamicsCounterexample foundFirst preprint 2025Lazy and invertible cellular automata do not generate all cellular automataOver the Klein four-group with the binary alphabet, the cellular automaton that complements the configurations with exactly one cell equal to 1 and fixes all others is not a finite composition of invertible and lazy cellular automata.Refuted

The questionCastillo-Ramirez (arXiv:2510.14841, cs.FL with nlin.CG; Theoret.

Why it mattersA positive answer would reduce every cellular automaton to invertible ones and automata that act at a single local pattern, the analogue for A^G of the generation of Tran(A) by Sym(A) and the idempotents of defect one; the paper notes that the lazy automata with neighborhood {e} are exactly those idempotents.

The ideaLet G be the Klein four-group, A = {0, 1}, w(x) the number of cells of x equal to 1, and F(x) the complement of x when w(x) = 1, and x otherwise.

Our contributionOver the Klein four-group with the binary alphabet, the cellular automaton that complements the configurations with exactly one cell equal to 1 and fixes all others is not a finite composition of invertible and lazy cellular automata.

Verification details
Release source recordRefuted in Lean

Open problem

Lazy and invertible cellular automata do not generate all cellular automata

E. Alcalá-Arroyo and A. Castillo-Ramirez (arXiv:2510.14841, cs.FL with nlin.CG; Theoret. Comput. Sci. 1076 (2026) 115965) study the lazy cellular automata over A^G: a local map on a finite neighborhood S ∋ e returns the central value z(e) on every pattern z except one, the unique active transition p, where it writes a symbol different from p(e). Motivated by the generation of the full transformation monoid of a finite set by its permutations and its idempotents of defect one, they ask:

If $\mathcal{L}(G,A)$ is the set of all lazy cellular automata over $A^G$, prove or disprove the following: $\text{CA}(G,A) = \langle \text{ICA}(G,A) \cup \mathcal{L}(G,A) \rangle$.

Issue #11773 fixes the reading: every group G and every finite alphabet A with at least two symbols, cellular automata and lazy automata as in the paper, ICA(G, A) the units of the monoid CA(G, A) under composition, and ⟨C⟩ the submonoid generated by C. The result disproves the equality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/LazyInvertibleGeneration.result

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Discrete structuresCounterexample foundFirst preprint 2026The one-term formula at the boundaryComplete homogeneous symmetric functions of finite path spectra.Refuted

The ideaExact evaluation gives one, while the numerator of the proposed expression vanishes.

Our contributionComplete homogeneous symmetric functions of finite path spectra.

Verification details
Release source recordRefuted in Lean

Open problem

The one-term formula at the boundary

Bouthat, Chávez and Sheng, A general framework for inequalities on simple graphs, arXiv:2606.17356v1, Remark 4.3, state the two path CHS identities as unproved and then assert a one-term simplification under $n+1\ge d/2$.

The closing assertion is false. At $n=2,d=6$, $h_6(\lambda(P_2))=1$, whereas $((n^2+n-d)/((n+d)(n+d+1)))\binom{{n+d+1}}{{d/2}}=0$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $|P_n|_{X,d}^d=h_d(\lambda(P_n))$ | PathCHSNormClosedForm.chs | Same complete homogeneous symmetric value. | | All $n\ge1$ and even $d\ge2$ | Arithmetic and parity hypotheses in claim | No degenerate path witness is used. | | $d/2\le n+1$ | Natural quotient guard in claim | Exact for even $d$; equality is included. | | One-term rational-binomial expression | Literal right side of claim | Real arithmetic and Nat.choose. | | Failure of the universal formula | result : ¬ claim | Witness $n=2,d=6$ satisfies every guard. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PathCHS/PathCHSNormBCSRemark43ClosedRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedSource 2025Bradshaw's Conjecture 20At a = 17, b = 7, the nonsquarefree input 125 satisfies Bradshaw's commutation equation.Refuted

The questionBradshaw, On a Family of Solutions to Arithmetic Differential Equations Involving the Collatz Map, Journal of Integer Sequences 28 (2025), Article 25.1.8, Conjecture 20, printed page 16:

Why it mattersAt (a,b,n)=(17,7,125), both sides equal 641, while 125=5^3 is not squarefree.

The ideaConstruct an arithmetic derivative internally as D0(n) = sum (p,k) in n.factorization, k NatDiv(n,p).

Our contributionAt a = 17, b = 7, the nonsquarefree input 125 satisfies Bradshaw's commutation equation.

Verification details
Release source recordRefuted in Lean

Open problem

Bradshaw's Conjecture 20

Z. P. Bradshaw, On a Family of Solutions to Arithmetic Differential Equations Involving the Collatz Map, Journal of Integer Sequences 28 (2025), Article 25.1.8, Conjecture 20, printed page 16:

The solutions to the commutation problem D(C_{a,b}(n)) = C_{a,b}(D(n)) are squarefree for every a ≡ b (mod 2).

The same page defines C(a,b,n) = (a*n+b)/2 for odd n and n/2 for even n. Printed page 3 defines D : Nat -> Nat by D(0)=D(1)=0, D(p)=1 at each prime, and D(m*n)=D(m)*n+m*D(n) for all m,n. The closed formal claim quantifies over such D and all natural a,b,n, with equal parity of a,b and 1 <= n, and concludes Squarefree n from the displayed commutation equality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/BradshawConjectureTwentyRefutation.result

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    Year in the primary source attribution: 2025.

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Discrete structuresConjecture refutedFirst preprint 2026CayleyPy-4 Conjecture 16 fails at k = 4, L = 2, N = 5Conjecture 16 of CayleyPy-4 is false under both readings of the diameter: with rotations of four consecutive letters on words with two zeros and three ones, the word 11010 is three moves from 00111, one more than the conjectured formula.Refuted

The questionCayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph SN / (SL × S{N−L}) on binary words with L zeros and N − L ones, generated by the consecutive k-cycles (i, …, i + k − 1) for N > k.

The ideaAt k = 4, L = 2, N = 5 the ten vertices lie at distances 0, 1, 1, 1, 1, 2, 2, 2, 2, 3 from 00111, and 11010 is the only vertex at distance 3.

Our contributionConjecture 16 of CayleyPy-4 is false under both readings of the diameter: with rotations of four consecutive letters on words with two zeros and three ones, the word 11010 is three moves from 00111, one more than the conjectured formula.

Verification details
Release source recordRefuted in Lean

Open problem

CayleyPy-4 Conjecture 16 fails at k = 4, L = 2, N = 5

CayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph S_N / (S_L × S_{N−L}) on binary words with L zeros and N − L ones, generated by the consecutive k-cycles (i, …, i + k − 1) for N > k. Conjecture 16 treats the inverse-closed case, in which a move rotates k consecutive letters one place in either direction:

For $k=3$ from $L\ge 2$(?): $D(L, N) = \left\lceil \frac{L (N - L)}{2} \right\rceil$ For $k=4$ from $L\ge 2$: $D(L, N) = \frac{L}{3}(N - L)$ if $3 \mid L$, $\left\lfloor \frac{L(N - L) + 2}{3} \right\rfloor$ if $3 \nmid L$. For $k=5$ for $N,L,N-L$ large enough: $D(L, N) = \left\lfloor \frac{LN + 2}{4} \right\rfloor - 2\left\lfloor \frac{L^2}{8} \right\rfloor - \mathbf{1}{L \equiv 2 \pmod{4}} , \mathbf{1}{N \equiv 2 \pmod{4}}$

Issue #10252 fixes the readings as in #10230: words are List Bool, and the diameter is read both as the largest distance from the central state [0]^L + [1]^{N−L} and as the largest distance between two vertices; the k = 5 range reads "for some constant C, all L ≥ C and N − L ≥ C".

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ShrunkenGrassmannianConjectureSixteenRefutation.result

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Analysis & dynamicsCounterexample foundFirst preprint 2025The box partition function of the six-vertex modelThe determinant formula conjectured by Kade for the partition function of the six-vertex model in a box with arrow-reflecting walls is false already for the smallest box: with one pair of horizontal and one pair of vertical spectral lines, crossing parameter 2, spectral parameters 2 and 3 and all boundary parameters 1, the partition function is -400400/81 while the formula gives -7150.Refuted

The questionKade (arXiv:2509.03416v1, section 2.3) puts the six-vertex model in a square box with arrow-reflecting walls: 2M horizontal spectral lines, in pairs xi, 1/xi, run from the right wall to the left wall, and 2M vertical lines, in pairs yj, 1/yj, run from the top wall to the bottom wall.

Why it mattersThe six-vertex model is the square-ice model of statistical mechanics, and the box boundary conditions are an integrable way to close the lattice on all four sides; a closed determinant for the box partition function would play the role of the Izergin–Korepin determinant for domain-wall boundaries.

The ideaAt M = 1 both products in the prefactor are over a single pair or empty, and the determinant is the single entry, so the formula reads (x/y − y/x) W(x, y) · c² a(xy) a(1/(xy)) F^LU(x) F^DR(y) / ((x/y − y/x) W(x, y)).

Our contributionThe determinant formula conjectured by Kade for the partition function of the six-vertex model in a box with arrow-reflecting walls is false already for the smallest box: with one pair of horizontal and one pair of vertical spectral lines, crossing parameter 2, spectral parameters 2 and 3 and all boundary parameters 1, the partition function is -400400/81 while the formula gives -7150.

Verification details
Release source recordRefuted in Lean

Open problem

The box partition function of the six-vertex model

Kade (arXiv:2509.03416v1, section 2.3) puts the six-vertex model in a square box with arrow-reflecting walls: 2M horizontal spectral lines, in pairs x_i, 1/x_i, run from the right wall to the left wall, and 2M vertical lines, in pairs y_j, 1/y_j, run from the top wall to the bottom wall. The bulk weights are a = p t − 1/(p t), b = t − 1/t, c = p − 1/p with t = x/y, and the walls carry the diagonal K-matrices K_L(x) = diag(b(xξ_L), b(x/(qξ_L))), K_R(x) = diag(b(xξ_R), b(xq/ξ_R)), K_U(y) = diag(b(yξ_U), b(yq/ξ_U)), K_D(y) = diag(b(yξ_D), b(y/(qξ_D))). The thesis proposes

Z_M({x_i}|{y_i}) = ∏_{i,j} (x_i/y_j − y_j/x_i) W(x_i, y_j) / ∏_{i<j} (x_j/x_i − x_i/x_j)(y_i/y_j − y_j/y_i) · det[ c² a(x_i y_j) a(1/(x_i y_j)) F^LU(x_i) F^DR(y_j) / ((x_j/y_i − y_i/x_j) W(x_i, y_j)) ]

with W(x, y) = a(xy) a(1/(xy)) a(x/y) a(y/x), F^LU(x) = b(xξ_L) b(xq/ξ_U) + b(x/(qξ_L)) b(xξ_U) and F^DR(y) = b(yξ_D) b(yq/ξ_R) + b(y/(qξ_D)) b(yξ_R), and its concluding chapter says that the conjecture "has to be proven or discarded".

Issue #10444 fixes the readings: the absolute arrow conventions of the six vertices and of the diagonal wall components are those of the thesis's figures (6Vvertices, KLeft_mat, KRight_mat, KUp_mat, KDown_mat); rows carry x_1, 1/x_1, x_2, 1/x_2, … from the top and columns y_1, 1/y_1, … from the left; q is the crossing parameter p, the only value among those tried at which the 2 × 1 box partition function is symmetric in its two horizontal spectral parameters, as integrability requires; the parameters are complex and the formula is asserted wherever none of its printed denominators vanishes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/VertexModels/KadeBoxBoundaryRefutation.result

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Discrete structuresCounterexample foundFirst preprint 2026Sublattice dismantling of the priority latticeThe priority lattice is not always sublattice dismantlable.Refuted

The questionAdrián Lillo and Mercedes Rosas, The Priority Lattice, arXiv:2603.28905v1, Section 6, item 4, ask:

Why it mattersThus one positive parameter refutes the whole universal question.

The ideaThe parents of vertices 3 and 6 distinguish all sixteen forests; top is distinct from every forest.

Our contributionThe priority lattice is not always sublattice dismantlable.

Verification details
Release source recordRefuted in Lean

Open problem

Sublattice dismantling of the priority lattice

Adrián Lillo and Mercedes Rosas, The Priority Lattice, arXiv:2603.28905v1, Section 6, item 4, ask:

A sublattice dismantling is a subdivision of a lattice into two non-empty sublattices. A lattice is sublattice dismantlable if it can be decomposed into 1-element lattices by consecutive sublattice dismantlings. We have checked with the help of SageMath that $\Pi(n)$ is sublattice dismantlable for $n=$ 1 to 5. Is this always the case?

The question quantifies over every positive integer $n$. The source's priority forests are increasing rooted forests on $0,\ldots,n$ with interval components. Their order is edge containment, with an additional greatest element. The existing PrincipalFilters.Pi is this literal carrier, and coreOrderIso identifies it with WithTop (IntervalForest n).

A subdivision is read as a disjoint partition into nonempty sublattices, closed under the ambient binary meet and join, without requiring the ambient bounds. The identical definition and the interval/complement split in Sage's immutable implementation support this standard meaning. The source does not separately say “disjoint”. The authors' historical Sage version and exact call are unknown. Evidence of an intended overlapping-cover meaning would invalidate this semantic bridge.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PriorityLattice/SublatticeDismantling.result

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Discrete structuresCounterexample foundFirst preprint 2025A bad Anderson potential without nontrivial shared symmetryA potential on the eight-cycle has an eigenvector with a zero coordinate at every real coupling, while every matrix commuting with its Laplacian and potential is scalar.Refuted

The questionGuerrero, Simple Eigenvalues and Non-vanishing Eigenvectors of the Anderson Model, arXiv:2512.00278v1, ask after Theorem 1.3:

Why it mattersThe converse would identify all bad potentials through shared symmetries and support symmetry-based counting.

The ideaThe proof uses spectral estimates.

Our contributionA potential on the eight-cycle has an eigenvector with a zero coordinate at every real coupling, while every matrix commuting with its Laplacian and potential is scalar.

Verification details
Release source recordRefuted in Lean

arXiv

A bad Anderson potential without nontrivial shared symmetry

O. Lindblad and E. Guerrero, Simple Eigenvalues and Non-vanishing Eigenvectors of the Anderson Model, arXiv:2512.00278v1, ask after Theorem 1.3:

Conversely, we also ask whether every bad potential shares a nontrivial symmetry with the laplacian.

Section 4 also states:

In general, we do not know whether the converse of lemma 3.2 is true.

On the cycle with $L>2$, the source's operator is $$(H_t\psi)(j)=\sum_{k\sim j}(\psi(j)-\psi(k))+t v_j\psi(j).$$ Definition 1.1 calls $V=\operatorname{diag}(v)$ bad when for every real $t$ at least one of simple eigenvalues and non-vanishing eigenvectors fails. The target LG-CONV quantifies over every $L>2$ and every $v:\operatorname{Fin}L\to\mathbb R$ taking values in ${-1,1}$. Issue #13591 fixes nontrivial shared symmetry as a real orthogonal matrix $O\ne I$, $O\ne-I$, commuting with both $\Delta$ and $V$. Both sufficient branches of Lemma 3.2 exclude $-I$, so refuting this weak reading also refutes their converse. A one-dimensional counterexample is a counterexample to the general-grid question. An affine change $V\mapsto\alpha V+\beta I$, $\alpha\ne0$, reparametrizes the coupling and shifts eigenvalues without changing eigenvectors or shared commutants; normalization to ${-1,1}$ therefore represents any two distinct values.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Simple eigenvalues | bad: ∀ μ : ℂ, H.charpoly.IsRoot μ → H.charpoly.rootMultiplicity μ = 1, with H = (SimpleGraph.cycleGraph L).lapMatrix ℂ + (t : ℂ) • Matrix.diagonal (fun j => (v j : ℂ)) | Each characteristic-polynomial root has algebraic multiplicity one. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/AndersonSymmetryConverseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 3/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundRefutation of Barker's A181278 recurrenceThe n = 4 term refutes Barker's published recurrence for OEIS A181278.Refuted

The questionNumber of 2 X n binary matrices M with rows in strictly increasing order and rows of MMtranspose (mod 2) in strictly decreasing order.

The ideaRepresent each length-n binary row by a natural number below 2^n.

Our contributionThe n = 4 term refutes Barker's published recurrence for OEIS A181278.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Barker's A181278 recurrence

OEIS A181278 defines the sequence as follows:

Number of 2 X n binary matrices M with rows in strictly increasing order and rows of M*Mtranspose (mod 2) in strictly decreasing order.

Colin Barker's FORMULA line of March 27, 2018 states verbatim:

a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) for n>3.

The offset is 1, so a(1) = 0 is a sequence term. For natural-valued terms, the subtraction-free formal claim is the equivalent assertion a(n) + 16*a(n-3) = 4*a(n-1) + 4*a(n-2) for every n >= 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/BarkerBinaryGramRecurrenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsCounterexample foundThe ratio-limit assertion for A394578A uniform enclosure for the native moving-point interpolant excludes the asserted consecutive-ratio limit of one quarter.Refuted

The questionIts complete native definition is quoted in the corresponding Library note.

Why it mattersMoving the evaluation point with the interpolation order changes the sequence's asymptotic behavior.

The ideaThe first factor contributes 1/2 and all subsequent factors are bounded by the corresponding positive denominator factors.

Our contributionA uniform enclosure for the native moving-point interpolant excludes the asserted consecutive-ratio limit of one quarter.

Scope & next questionThe independent D-finiteness conjecture, exact limiting value, recurrence and integrality remain outside this result.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The ratio-limit assertion for A394578

The selected source COMMENT is:

Conjecture: a(n+1)/a(n) -> 1/4.

Its complete native definition is quoted in the corresponding Library note. For every natural n, p_n is the rational Lagrange polynomial through (1,0) and (k,1) for all integers k from 2 through n+1. Define $a_n=(4^n/2)p_n(n+1/2)$, including zero. The exact settling declaration is OEISA394578RatioRefutation.result : Not claim, where claim : Prop is convergence of the real consecutive ratio at natural atTop to 1/4. The degree and node conditions have no extra hypothesis; no shifted sequence or recurrence replaces the interpolation definition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Interpolation/OEISA394578RatioRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryCounterexample foundA396596 divisor-power union: a counterexampleA divisor-power record at exponent minus four refutes the highly-composite or deeply-composite union in OEIS A396596.Refuted

The questionSwitkay, OEIS A396596, contribution August 31, 2026, revision 65:

Why it mattersquestionanswered: whether every positive natural that is a strict record for some real exponent at most zero is highly composite or deeply composite, with the converse included in the registered iff.

The ideaWrite S(n,x) = DivisorPowerSum n x, R(n,x) = StrictDivisorPowerRecord n x, N = 32125373280 and L = 232792560.

Our contributionA divisor-power record at exponent minus four refutes the highly-composite or deeply-composite union in OEIS A396596.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

A396596 divisor-power union: a counterexample

Hal M. Switkay, OEIS A396596, contribution August 31, 2026, revision 65:

Natural numbers k such that the sum of the x-th powers of the divisors of k achieves a record, for some x <= 0.

The selected COMMENTS assertion is:

We conjecture that the present sequence can be constructed simply as a union of highly composite numbers and deeply composite numbers.

The primary URL in this dossier is byte-for-byte the URL in the source note's Verified locator. The source note is Library/notes/switkay2026divisorpowerunion.md, referenced as D5/L/switkay2026divisorpowerunion.

The complete statement mirror, including the two needed definitions and the closed proposition that the single result negates, is:

noncomputable def DivisorPowerSum (n : ℕ) (x : ℝ) : ℝ :=
  ∑ d ∈ n.divisors, (d : ℝ) ^ x

def StrictDivisorPowerRecord (n : ℕ) (x : ℝ) : Prop :=
  0 < n ∧ ∀ m : ℕ, 0 < m → m < n →
    DivisorPowerSum m x < DivisorPowerSum n x

def claim : Prop :=
  ∀ n : ℕ, 0 < n →
    ((∃ x : ℝ, x ≤ 0 ∧ StrictDivisorPowerRecord n x) ↔
      (StrictDivisorPowerRecord n 0 ∨
        ∃ B : ℝ, B < 0 ∧ ∀ x : ℝ, x ≤ B →
          StrictDivisorPowerRecord n x))

theorem result : ¬ claim

The last line is a type display; the proof remains solely in the Lean module. Every positive divisor is summed, and every smaller positive natural is compared strictly at one common real exponent. The empty predecessor case at 1 is allowed; zero candidates are excluded. The right-hand eventual quantifier ranges over all real exponents below one negative real threshold, following A095848's NAME, “all sufficiently low (i.e., negative) values of k,” read in the real-parameter setting of A396596. It is not a test of finitely many exponents or an assumed lexicographic characterization.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/DivisorRecords/DivisorPowerUnionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 3/4 · 15% weight

    The delivered argument combines documented mathematical mechanisms.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Four colours in small online majority edge-colouring gamesFour colours cannot guarantee online majority edge-colouring for every order from five to seven.Refuted

The questionProblem 11 of Pękała (2026) asks whether Algorithm can guarantee an online majority edge-colouring with at most four colours for each of n=5,6,7 vertices when Presenter must finish with minimum degree exactly two.

Why it mattersThis is a named external finite-game question.

The ideaThe Presenter has a finite adaptive strategy on five vertices.

Our contributionFour colours cannot guarantee online majority edge-colouring for every order from five to seven.

Verification details
Release source recordRefuted in Lean

arXiv

Four colours in small online majority edge-colouring games

Problem 11 of Pękała (2026) asks whether Algorithm can guarantee an online majority edge-colouring with at most four colours for each of n=5,6,7 vertices when Presenter must finish with minimum degree exactly two. Presenter chooses each next edge after seeing previous colours. Algorithm must colour it immediately and cannot recolour old edges.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/PekalaOnlineMajorityFourColourRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Density-independent Alhasan–Fawwaz boundednessThe Alhasan–Fawwaz quotient is bounded by dimension independently of every positive definite density.Proved

The questionAbed, Nikolaeva and Novikov, How characterisations of traces and central elements fail, arXiv:2610.34388v1, Section 11.2(5), ask:

Why it mattersA dimension-dependent bound controls every positive definite density on the same matrix algebra, independently of its spectrum and scale.

The ideaThe square identity gives norm(X ej)=sj; Hermitian B and Cauchy–Schwarz give inner(ej,A ej)<=sj.

Our contributionThe Alhasan–Fawwaz quotient is bounded by dimension independently of every positive definite density.

Scope & next questionSharpness, a rank improvement, the entire function a(rho), and the II1-factor question remain outside this result.

Verification details
Release source recordProved in Lean

Open problem

Density-independent Alhasan–Fawwaz boundedness

Abed, Nikolaeva and Novikov, How characterisations of traces and central elements fail, arXiv:2610.34388v1, Section 11.2(5), ask:

Is a bounded on M_n(C) independently of rho?

The selected question fixes a positive dimension and varies every strictly positive density. Section 5.12 defines the quotient set using positive semidefinite A, Hermitian B and nonzero X=A+iB. Section 3.1 uses the ordinary trace and modulus |X|=(X*X)^(1/2), without a trace-one requirement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/AbedNikolaevaNovikovAFDensityBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryCounterexample foundSource 2026Agoh's alternating-numerator characterizationA repeated real root is compatible with every alternating numerator factor.Refuted

The questionThe sole target is the unnumbered Conjecture on final printed page 9 of INTEGERS 26 (2026), A9.

Why it mattersThe printed discussion asks whether simple roots are necessary, not only sufficient, for the all-order factor property.

The ideaSet ti=f(X^i), D=producti ti and N=sumi (-1)^i binom(n,i) product{j != i} tj, indexed over 0 through n.

Our contributionA repeated real root is compatible with every alternating numerator factor.

Verification details
Release source recordRefuted in Lean

Open problem

Agoh's alternating-numerator characterization

The sole target is the unnumbered Conjecture on final printed page 9 of INTEGERS 26 (2026), A9. Its exact source and provenance are in the linked Library note. For every nonzero real polynomial f splitting over the reals, the asserted equivalence is: (for every natural n, (X-1)^n divides num(Q_n(f))) iff (f is a monomial or every real root is simple and different from 1). Q_n is independently the literal rational sum in Equation (3.1), and num is its actual reduced RatFunc.num. The all-n condition is inside the equivalence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/AgohAlternatingNumeratorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2026.

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2024The exact five-dimensional bad science optimumThe displayed five-dimensional bad science matrix maximizes the unit-row functional.Proved

The questionThe journal reference is Involve 19 (2026), 443–460, DOI 10.2140/involve.2026.19.443.

Our contributionThe displayed five-dimensional bad science matrix maximizes the unit-row functional.

Verification details
Release source recordProved in Lean

Open problem

The exact five-dimensional bad science optimum

Albors, Bhatti, Ganjoo, Guo, Kunisky, Mukherjee, Stepin and Zeng, On the Structure of Bad Science Matrices, arXiv:2408.00933v2, Section 1.4, ask: “We do not know whether this matrix is indeed optimal.” The matrix is their displayed real $5\times5$ matrix $A_*$, and the admissible matrices have rows of Euclidean length one. The journal reference is Involve 19 (2026), 443–460, DOI 10.2140/involve.2026.19.443.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Real square matrices and the uniform sign-cube average in Section 1 | beta, with Matrix (Fin n) (Fin n) ℝ, Boolean sign vectors and weight $2^{-n}$ | The function-space norm is the sup norm. The sign expression is literal; the proofs reuse IsingUniquenessSets.sgn through definitional equality. | | “rows are normalized to have ℓ²-length 1” | UnitRows | Every row's sum of squared real entries equals one. | | The displayed matrix of Section 1.4 | Astar | All 25 entries and the factor $1/(2\sqrt3)$ agree with the source. | | “this matrix is indeed optimal” | claim and result | Every real $5\times5$ matrix satisfying UnitRows has $\beta(A)\le\beta(A_*)$; there is no extra hypothesis. |

The declarations belong to D5.S3.Combinatorics.Hamming.BadScienceFiveOptimality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hamming/BadScienceFiveOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedSource 2026Archer-Bourne cube-avoidance counting equalityThe permutations avoiding 312 and 321 are exactly the direct sums of cyclic rotations indexed by compositions, and the cube criterion gives the corresponding counting identity.Proved

The questionThis dossier deliberately anchors only the unnumbered counting conjecture in Section 5, "Further directions and open questions", page 13 of arXiv:2505.05218v3.

Why it mattersIts cube specialization provides the composition-side condition in the quoted conjecture.

The ideaThe completed formal half is the decomposition of every 312/321-avoiding permutation as a direct sum of cyclic rotations, its bijectivity with compositions of n, and the restriction of that bijection to transport cardinalities using the cube criterion.

Our contributionThe permutations avoiding 312 and 321 are exactly the direct sums of cyclic rotations indexed by compositions, and the cube criterion gives the corresponding counting identity.

Scope & next questionThe paper's broader questions about other patterns and higher powers are deliberately out of scope.

Verification details
Release source recordProved in Lean

Open problem

Archer-Bourne cube-avoidance counting equality

This dossier deliberately anchors only the unnumbered counting conjecture in Section 5, "Further directions and open questions", page 13 of arXiv:2505.05218v3. The worker extracted this sentence from the PDF; line wrapping is removed and mathematical glyphs are transcribed into inline LaTeX:

For example, based on the ideas similar to the ones in this paper, we conjecture that the number of permutations that avoid the chain $(312,321 : \varnothing : 2143)$ (i.e, those with the property that $\pi$ avoids ${312,321}$ and $\pi^3$ avoids $2143$) is equal to the number of compositions $d = (d_1,d_2,\ldots,d_m)$ of $n$ so that all $d_i$ are 1 or 3, except for at most one.

Thus, for each positive n, count permutations of n avoiding 312 and 321 whose cubes avoid 2143, and compare with ordered compositions of n having at most one part outside {1,3}. The paper's broader questions about other patterns and higher powers are deliberately out of scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/PatternAvoidance/ArcherBourneDecomposition.card_avoids_312_321_cube_2143_eq_compositions

Frozen module record

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    Year recorded in the source bibliography: 2026.

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    Source 1
Discrete structuresCounterexample foundFirst preprint 2025The Archer–Bourne Square Recurrence Fails at n = 11There are nine indecomposable permutations of length eleven avoiding 312 and 54321 whose squares avoid 132, which refutes the Archer-Bourne recurrence conjecture.Refuted

The questionKassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218v3, Section 5, printed page 14:

The ideaA nonempty 312-avoider is ρ ⊕ τ with ρ its first direct-sum component.

Our contributionThere are nine indecomposable permutations of length eleven avoiding 312 and 54321 whose squares avoid 132, which refutes the Archer-Bourne recurrence conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

The Archer–Bourne Square Recurrence Fails at n = 11

Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218v3, Section 5, printed page 14:

For example, we conjecture that if we take a_n := a_n(312, 54321 : 132) then a_n = a_{n−1} + a_{n−2} + a_{n−3} + a_{n−4} + n − 1 for n ≥ 6.

Here a_n(312, 54321 : 132) is the number of permutations π of [n] that avoid 312 and 54321 and whose square π ∘ π avoids 132.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PermutationSquare/PermutationSquareRefutation.result

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryStructural characterizationBala's A361035 parity conjectureConjecture: a(n) is odd iff n = 2^k - 3 for some k >= 2.Proved

The questionConjecture: a(n) is odd iff n = 2^k - 3 for some k >= 2.

Why it mattersThe retained declarations are unbounded symbolic statements, with no retained finite certificate or enumeration.

The ideaThe imported OeisA347854BalaHalfIndexIntegral.muldivreduce reduces floor terms to quotients and residues, and its factorialproddvdoffloor turns their sum inequality into factorial divisibility.

Our contributionConjecture: a(n) is odd iff n = 2^k - 3 for some k >= 2.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Bala's A361035 parity conjecture

Peter Bala's OEIS A361035 defines $a(n)=9979200(4n)!/(n!(n+3)!^3)$ and states:

Conjecture: a(n) is odd iff n = 2^k - 3 for some k >= 2.

The index $n$ is any natural number, including zero. The prerequisite integrality and the exact exponent bound are part of the source's statement in preregistration #18607.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Factorial quotient with coefficient 9979200 and shift 3 | BalaA361035Parity.a | Literal source expression, using natural-number division. | | Every natural index, including zero | BalaA361035Parity.claim | Universal quantifier over the naturals. | | Odd exactly when the shifted index is a power of two | BalaA361035Parity.claim | Both directions of the iff; the exponent is natural and at least 2. | | Quotient is an integer | Private BalaA361034Parity.integrality | Denominator divides numerator; its hypotheses are discharged in the settling proof. | | Conjecture holds | BalaA361035Parity.result | Complete claim without extra assumptions. | | Parameters $C,j$ in the shared proof | BalaA361034Parity.quotient_odd | Internal generalization, instantiated with $C=9979200$ and $j=3$; no added premise in result. |

The dropped or weakened source-clause set is empty.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FactorialRatio/BalaA361035Parity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedPaired three-adic valuationsInteger quadratic products and Bala's valuation conjectures.Proved

The questionPreregistration 18636 maps to D5.S3.Arith.Congruence.BalaA363412Valuation.result (Proved).

Why it mattersThis is an explicit OEIS conjecture in the source family documented in bala3634-valuations.md.

The ideaThe primary dossier gives the four issue-to-result mappings.

Our contributionInteger quadratic products and Bala's valuation conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Paired three-adic valuations

$v_3(b(3n))=v_3(b(3n-1))$ for every $n\in\mathbb N$ with $n\ge1$. Here $a(n)$ and $b(n)$ are the integer coordinates of $\prod_{k=0}^n(1+k\sqrt{-3})$. Preregistration #18636 maps to D5.S3.Arith.Congruence.BalaA363412Valuation.result (Proved).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BalaA363412Valuation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedType-one primes through 101Integer quadratic products and Bala's valuation conjectures.Proved

The questionPreregistration 18634 maps to D5.S3.Arith.Congruence.BalaA363411TypeOne.result (Proved).

Why it mattersThis is an explicit OEIS conjecture in the source family documented in bala3634-valuations.md.

The ideaThe primary dossier gives the four issue-to-result mappings.

Our contributionInteger quadratic products and Bala's valuation conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Type-one primes through 101

$\forall p\le101$, the primes dividing no $a(n)$ are exactly ${3,11,17,23,29,41,47,53,59,83,89,101}$, with the term index $n$ unbounded. Here $a(n)$ and $b(n)$ are the integer coordinates of $\prod_{k=0}^n(1+k\sqrt{-3})$. Preregistration #18634 maps to D5.S3.Arith.Congruence.BalaA363411TypeOne.result (Proved).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BalaA363411TypeOne.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedBalanced Compositions and Nonnegative Walks from Height TwoProved

The questionand carries a comment by Gus Wiseman dated Mar 17 2018, still standing at revision 55 of Oct 13 2025:

The ideaIt is even in z and is Pascal's array in displacement coordinates.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Balanced Compositions and Nonnegative Walks from Height Two

OEIS A026010 is defined by

a(n) = number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n and s(0) = 2

and carries a comment by Gus Wiseman dated Mar 17 2018, still standing at revision #55 of Oct 13 2025:

Conjecture: a(n) is the number of integer compositions of n + 2 in which the even parts appear as often at even positions as at odd positions (confirmed up to n = 19).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BalancedCompositionPaths.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026The normal-modulus triangle constant in dimension twoA Hermitian pair in dimension two rules out every normal-modulus triangle constant below one quarter.Refuted

The questionLee, Triangle inequalities for the operator symmetric modulus, arXiv:2602.19607v1, §4, Corollary 4.2, state verbatim:

Why it mattersCorollary 4.2 provides a uniform upper bound.

The ideaThe explicit positive matrices satisfying the modulus square identities are

Our contributionA Hermitian pair in dimension two rules out every normal-modulus triangle constant below one quarter.

Verification details
Release source recordRefuted in Lean

Open problem

The normal-modulus triangle constant in dimension two

J.-C. Bourin and E.-Y. Lee, Triangle inequalities for the operator symmetric modulus, arXiv:2602.19607v1, §4, Corollary 4.2, state verbatim:

Corollary 4.2. Let A, B ∈ Mn be normal. Then, for some unitary V ∈ Mn, |A + B| ≤ |A| + |B| + (1/4) V(|A| + |B|)V*. If n ≥ 3, the constant 1/4 is the smallest possible one.

The following paragraph asks verbatim:

However, contrarily to the Schur version (4.1), it seems necessary to request n ≥ 3 in Corollary 4.2 to get sharpness of 1/4 (what is the best constant for n = 2 ?).

Here $|X|=(X^*X)^{1/2}$ is the positive modulus and the inequality uses positive-semidefinite order. The article is in press in Proceedings of the American Mathematical Society, doi:10.1090/proc/17861; the quotations and locators refer to the arXiv version.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/NormalModulusTriangleConstant.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedSource 2025Weak Ascent Sequences Avoiding {100, 101, 110, 201} and {021, 101, 201, 210}The two weak ascent sequence classes in Class 215 have equal cardinalities in every length.Proved

The questionDavid Callan and Toufik Mansour, Ascent Sequences and Weak Ascent Sequences Avoiding a Quadruple of Length-3 Patterns, Integers 25 (2025), A80, Section 1 and Table 2:

Why it mattersWith the rest of the classification this gives WAS4 = 228.

The ideaThe append criteria of both classes are determined exactly; for the left class a stack of closed intervals spanned by inversions, together with the largest repeated value, decides which letters can be appended.

Our contributionThe two weak ascent sequence classes in Class 215 have equal cardinalities in every length.

Verification details
Release source recordProved in Lean

Open problem

Weak Ascent Sequences Avoiding {100, 101, 110, 201} and {021, 101, 201, 210}

David Callan and Toufik Mansour, Ascent Sequences and Weak Ascent Sequences Avoiding a Quadruple of Length-3 Patterns, Integers 25 (2025), #A80, Section 1 and Table 2:

Theorem 2. The number WAS4 of WA-Wilf-equivalence classes among quadruples of length-3 patterns is either 228 or 229. The slight ambiguity in Theorem 2 is because the status of Class 215 is left open in Table 2 below.

Table 2 lists Class 215 as the pair {100, 101, 110, 201} and {021, 101, 201, 210}, marked "Open".

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/WeakAscent/WeakAscentClass215.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

    Read source ↗
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Simple Permutations Sortable by Two Parallel Pop Stacks with BypassThe numbers of simple permutations in C of sizes zero, one and two are respectively one, one and two.Proved

The questionLapo Cioni, Luca Ferrari and Rebecca Smith, Sorting permutations using a pop stack with a bypass, arXiv:2503.08285v1, Section 9.2:

The ideaThe simple permutations of C, together with one extra non-simple permutation for each odd length, are split by the position of their minimum: position 2, position 3, or later.

Our contributionThe numbers of simple permutations in C of sizes zero, one and two are respectively one, one and two.

Verification details
Release source recordProved in Lean

Open problem

Simple Permutations Sortable by Two Parallel Pop Stacks with Bypass

Lapo Cioni, Luca Ferrari and Rebecca Smith, Sorting permutations using a pop stack with a bypass, arXiv:2503.08285v1, Section 9.2:

Let a_n be the number of simple permutations of size n that can be sorted by a machine consisting of two pop stacks in parallel where entries are allowed to bypass the pop stacks. Then a_0 = a_1 = 1, a_2 = 2, a_n = F_{2n−5} − 1 if n ≥ 3 is odd, and a_n = F_{2n−5} if n > 3 is even.

The sortable permutations are the class C = Av(2341, 25314, 42513, 42531, 45213, 45231, 52314, 642135, 642153); a permutation is simple when it has no interval of length strictly between 1 and n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PopStack/PopStackSimple.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedA Multiple with Fewer than Twice as Many Divisors Divides the SquareA multiple whose divisor count stays below twice that of the original divides its square, because one prime exponent exceeding twice its original value, or one new prime, already doubles the divisor count on its own.Proved

The questiona(n) is the largest multiple of n with fewer than twice as many divisors as n.

The ideaWrite f for the exponent in the prime factorisation, so that τ(k) = ∏ (fk(p) + 1) over the primes of k.

Our contributionA multiple whose divisor count stays below twice that of the original divides its square, because one prime exponent exceeding twice its original value, or one new prime, already doubles the divisor count on its own.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A Multiple with Fewer than Twice as Many Divisors Divides the Square

OEIS A225004, at revision #24 of Mar 17 2022, by J. Lowell, Apr 23 2013:

a(n) is the largest multiple of n with fewer than twice as many divisors as n.

Conjecture: a(n) is always a divisor of n^2.

The entry's worked example: "a(6) = 18 because 6 has 4 divisors and term must have fewer than 8 divisors. Only 6, 12, and 18 are multiples of 6 with fewer than 8 divisors."

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/DivisorDoublingMultipleDividesSquare.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Minimal bipartite Erdős–Gyárfás counterexamples are 2-edge-connectedA minimal bipartite graph of minimum degree at least three without a power-of-two cycle is connected and has no bridge.Proved

The questionDucoffe–Dumitru, Towards a more structured search for Erdős–Gyárfás counter-examples, arXiv:2609.28594v1, §1:

Why it mattersA structural restriction on minimal counterexamples reduces the graph families that must be considered in the Erdős–Gyárfás problem.

The ideaIf G has more than one connected component, induce G on the vertex support of one component.

Our contributionA minimal bipartite graph of minimum degree at least three without a power-of-two cycle is connected and has no bridge.

Verification details
Release source recordProved in Lean

Open problem

Minimal bipartite Erdős–Gyárfás counterexamples are 2-edge-connected

Ducoffe–Dumitru, Towards a more structured search for Erdős–Gyárfás counter-examples, arXiv:2609.28594v1, §1:

for any class of graphs F, a counter-example within F refers to any graph G ∈ F such that G has minimum degree at least three but no cycle of length some power of two. It is called a minimal counter-example within F if it minimizes (|V(G)|, |E(G)|) with respect to the lexicographic order.

The bipartite clause in §6 reads:

we also conjecture … that any minimal bipartite counter-example is 2-edge connected.

For a nonempty finite simple graph G, let HasPowTwoCycle(G) mean that G has an IsCycle walk of length 2^k for a natural k ≥ 2. A bipartite counterexample is a graph admitting a proper two-colouring, with degree at least three at every vertex and no such cycle. Minimality compares (|V(G)|, |E(G)|) lexicographically against every finite nonempty bipartite counterexample, independently of its vertex type. The target conclusion is that G is connected and deleting any edge of G leaves a connected graph.

This is a structural question about potential counterexamples to Erdős Problem #64, the Erdős–Gyárfás conjecture. It asserts neither that a counterexample exists nor that every graph of minimum degree three has a power-of-two cycle.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ErdosGyarfasBipartiteBridgeless.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Filimonova–Puzynina's rank-one iteration-bound questionA primitive binary rank-one morphism has an effective bounded criterion for eventual abelian periodicity.Proved

The questionFilimonova and Puzynina ask in Section 4 of On abelian periodicity of purely morphic words (arXiv:2605.30306v1) whether the iteration variable in their primitive binary rank-one criterion has an upper bound making the criterion algorithmic.

Why it mattersThe question asks for an effective version of the source criterion.

The ideaThe proof extracts the coprime rank-one parameters, presents the actual iterates through a finite uniform expansion, and encodes cyclic witnesses as constant-output subset paths.

Our contributionA primitive binary rank-one morphism has an effective bounded criterion for eventual abelian periodicity.

Verification details
Release source recordProved in Lean

arXiv

Filimonova–Puzynina's rank-one iteration-bound question

Filimonova and Puzynina ask in Section 4 of On abelian periodicity of purely morphic words (arXiv:2605.30306v1) whether the iteration variable in their primitive binary rank-one criterion has an upper bound making the criterion algorithmic.

The source's fixed word is allowed an arbitrary finite preperiod. The target is the complete nonerasing, prolongable, primitive, rank-one binary domain, with exact Parikh equality in the cyclic block witness.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/RankOneMorphismIterationBound.effective_iteration_bound

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Gaussian atemporality implies entanglementPositive partial transpose implies temporal compatibility in both directions for physical two-mode Gaussian covariance matrices; Gaussian atemporality therefore implies entanglement in the covariance PPT formulation.Proved

The questionSong et al., “Gaussian Atemporality: When Gaussian Quantum Correlations Imply Common Cause”, arXiv:2508.11804v2, Section III.D (the fourth subsection of Section III), p. 6, state:

Why it mattersThe result replaces the source's numerical implication by a covariance-matrix proof.

The ideaAfter identifying the single-mode symplectic coordinates, D5.S3.Quantum.Measurements.GeneralDyneSteadyStateOptimality.physicalforward gives positive definiteness and determinant at least one for each real block.

Our contributionPositive partial transpose implies temporal compatibility in both directions for physical two-mode Gaussian covariance matrices; Gaussian atemporality therefore implies entanglement in the covariance PPT formulation.

Verification details
Release source recordProved in Lean

Open problem

Gaussian atemporality implies entanglement

Song et al., “Gaussian Atemporality: When Gaussian Quantum Correlations Imply Common Cause”, arXiv:2508.11804v2, Section III.D (the fourth subsection of Section III), p. 6, state:

From Fig. 3, Gaussian-atemporality appears to imply entanglement. The converse, on the other hand is not generally true.

Preregistration #19029 supplies the source quotations, Tier 1 classification, binding covariance conventions and literature screen. The assertion concerns every physical two-mode Gaussian covariance matrix in standard form, in quadrature order $(q_A,p_A,q_B,p_B)$ and units $\hbar=2$.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Eq. (2), $V_{AB}=\begin{pmatrix}V_A&C\C^T&V_B\end{pmatrix}$ | covariance | Arbitrary real two-by-two blocks, indexed by Fin 2; the combined index is Fin 2 ⊕ Fin 2. | | Definition 1, “with locally diagonal covariance matrices” | IsStandardForm | Both local blocks satisfy Matrix.IsDiag. | | Spatial uncertainty inequality | IsSpatiallyPhysical | $V_{AB}+i(\Omega\oplus\Omega)\succeq0$, including Hermiticity. | | Eqs. (7)–(8) | forwardX | $T=C^TV_A^{-1}$, $N=V_B-C^TV_A^{-1}C$, $X_f=N+i\Omega-iT\Omega T^T$. | | Reverse temporal direction | forwardX VB VA Cᵀ | Exchange the local blocks and transpose $C$. | | Section III.B, “Gaussian-atemporal in both temporal direction” | IsGaussianAtemporal | Both directional matrices fail to be positive semidefinite. | | Section III.D, the two-mode Gaussian PPT criterion | partialTransposeA, IsEntangled | $\Lambda=\operatorname{diag}(1,-1,1,1)$ and failure of $\Lambda V_{AB}\Lambda+i(\Omega\oplus\Omega)\succeq0$. | | Section III.D, the quoted implication | claim, result | All three hypotheses imply the covariance NPT predicate. |

The symplectic form is the existing D5.S3.Quantum.Thermal.PositiveOneModeWilliamson.physicalJ2, namely $\Omega=\begin{pmatrix}0&1\-1&0\end{pmatrix}$. Real-to-complex conversion is Matrix.map Complex.ofReal; no additional conversion or reverse-direction alias is introduced. The inverse is Mathlib's total nonsingular matrix inverse. Physicality supplies the invertibility used by the proof. No assumption beyond the displayed source hypotheses is added to claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/GaussianAtemporalityEntanglement.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2026Ghosh–Ramachandran's capacity-one area decay questionThe closed unit disc together with the isolated point 4 has capacity one and minimal lemniscate area of neither order 1/n nor order 1/log n.Refuted

The questionRamachandran, A note on the Erdős minimal area problem, arXiv:2604.03036v3, beginning of Section 4:

Our contributionThe closed unit disc together with the isolated point 4 has capacity one and minimal lemniscate area of neither order 1/n nor order 1/log n.

Verification details
Release source recordRefuted in Lean

Open problem

Ghosh–Ramachandran's capacity-one area decay question

S. Ghosh and K. Ramachandran, A note on the Erdős minimal area problem, arXiv:2604.03036v3, beginning of Section 4:

This shows that while $\vartheta(K)=0$ may hold for all compacts $K$ with $\mathrm{cap}(K)=1$, the decay rate of $A_n(K)$ is not universal. It will be interesting to determine whether $A_n(K)$ can only decay like $\frac1n$ or $\frac1{\log n}$, or whether other rates are possible.

For compact $K\subseteq\mathbb C$, let $P_n(K)$ consist of the monic degree-$n$ polynomials whose zeros all lie in $K$. Write $\Lambda_p={z:|p(z)|<1}$ and $A_n(K)=\inf_{p\in P_n(K)}m(\Lambda_p)$, where $m$ is planar Lebesgue measure. The paper defines capacity as logarithmic capacity, equivalently the transfinite diameter.

The Lean claim says that for every nonempty compact $K$ of transfinite diameter one, the real-valued minimal-area sequence is either $\Theta(1/n)$ or $\Theta(1/\log n)$ as $n\to\infty$. Thus result : ¬ claim refutes the proposed two-rate restriction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Lemniscates/CapacityOneAreaDecayRate.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresConjecture provedHanna’s ternary-digit residue conjectureHanna's binomial-square generating function has the asserted residues at every coefficient.Proved

The questionHanna, OEIS A246056, Name and Comments; preregistration 18887.

Why it mattersThe two generating functions differ only in the linear denominator.

The ideaEvery source summand has order at least its index, so every coefficient uses a finite sum.

Our contributionHanna's binomial-square generating function has the asserted residues at every coefficient.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna’s ternary-digit residue conjecture

P. D. Hanna, OEIS A246056, Name and Comments; preregistration #18887.

Conjecture: a(n) == 1 (mod 3) when n = 2*A005836(k) for k >= 0, and a(n) == 0 (mod 3) otherwise, where A005836 gives numbers whose base-3 representation contains no 2.

Its generating function is $\sum_{n\ge0}x^n(1-2x)^{-2n-1}Q_n(2x)Q_n(3x)$, with offset zero. The conjecture is proved.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The Name generating function | term and coefficientwise A | Literal preregistered rational expressions; the coefficient at $m$ uses $n\le m$. | | Every natural index | Outer universal quantifier in claim | Includes zero. | | Integer coefficient | Existential integer $z$ equal to the rational coefficient | Explicit cast to $\mathbb Q$; integrality is retained. | | $n=2,A005836(k)$ | An $m$ with $n=2m$ and $2\notin\operatorname{Nat.digits}(3,m)$ | Zero has the empty digit list and qualifies. | | Residue one on that support, zero otherwise | Both implications in claim | Neither support nor complement is dropped. |

The public result proves exactly claim, without additional hypotheses. The definitions are mirrored in the Library note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaLegendreCentralBinomialModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedHanna’s power-of-two residue conjectureHanna's binomial-square generating function has the asserted residues at every coefficient.Proved

The questionHanna, OEIS A246423, Name and Comments; preregistration 18888.

Why it mattersThe two generating functions differ only in the linear denominator.

The ideaEvery source summand has order at least its index, so every coefficient uses a finite sum.

Our contributionHanna's binomial-square generating function has the asserted residues at every coefficient.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna’s power-of-two residue conjecture

P. D. Hanna, OEIS A246423, Name and Comments; preregistration #18888.

a(n) == 2 (mod 4) iff n = 2^k for k>=2, and a(n) == 0 (mod 4) elsewhere except at a(0)=1 (conjecture).

Its generating function is $\sum_{n\ge0}x^n(1-3x)^{-2n-1}Q_n(2x)Q_n(3x)$, with offset zero. The conjecture is proved.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The Name generating function | term and coefficientwise A | Literal preregistered rational expressions; the coefficient at $m$ uses $n\le m$. | | $a(0)=1$ | First conjunct of claim | The constant term is explicit. | | Every positive index, integral coefficients | Universal $n>0$ and integer witness equal to its coefficient | Integrality at zero follows from the first conjunct. | | Residue two iff $n=2^k$, $k\ge2$ | The biconditional in claim | The exponent bound is retained. | | Zero modulo four elsewhere | Complement implication in claim | No complementary case is omitted. |

The public result proves exactly claim, without additional hypotheses. The definitions are mirrored in the Library note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaLegendreCentralBinomialModFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThe Nested-Square Generating Function Has Even Coefficients at Odd IndicesA series equal to x plus a square has even coefficients at every odd index above one, because the square pairs each term of an odd coefficient with its mirror image.Proved

The questionG.f.: x + sq(x/(1-x) + sq(x/(1-2x) + sq(x/(1-3x) + sq(x/(1-4x) + ...)))) where sq(x) = x^2, an infinitely nested square.

The ideaThe coefficient of x^(2m+1) in F^2 is Σ{i=0}^{2m+1} fi f(2m+1-i).

Our contributionA series equal to x plus a square has even coefficients at every odd index above one, because the square pairs each term of an odd coefficient with its mirror image.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Nested-Square Generating Function Has Even Coefficients at Odd Indices

OEIS A392210, by Paul D. Hanna, Feb 02 2026:

G.f.: x + sq(x/(1-x) + sq(x/(1-2x) + sq(x/(1-3x) + sq(x/(1-4*x) + ...)))) where sq(x) = x^2, an infinitely nested square.

The comment carrying the question:

It appears that a(2*n-1) is even for n > 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/NestedSquareOddEven.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedHanna's derivative-substitution congruences for A232686Both nominated congruences hold for every factorial-normalized coefficient of OEIS A232686.Proved

The questionThe source equation and its nominated congruences are:

Why it mattersThe preregistration 21488 supplies the source conventions and literature screen for A232686.

The ideaThe shared machinery constructs the unique integral coefficient recurrence, transports coefficient congruences from an integral approximate solution, and cancels integral unit powers.

Our contributionBoth nominated congruences hold for every factorial-normalized coefficient of OEIS A232686.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's derivative-substitution congruences for A232686

Paul D. Hanna's OEIS A232686 COMMENTS conjectures concern integer exponential coefficients $a_n=n![x^n]A$ of rational formal power series with $A(0)=1$. The source equation and its nominated congruences are:

%N A232686 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^2. %C A232686 CONJECTURES. %C A232686 a(n) == 0 (mod 2) for n>=2. %C A232686 a(n) == 0 (mod 2^2) for n>=4.

| Record | Formal equation | Nominated congruences | Settling declaration | | --- | --- | --- | --- | | A232686 | $A'(x)=A(xA'(x))^2$ | $2\mid a_n$ for $n\ge2$; $4\mid a_n$ for $n\ge4$ | HannaA232686Congruences.result |

The settling declaration belongs to D5.S1.Recurrence.Residue. It universally quantifies over rational formal solutions and concludes all-index integrality as well as the two displayed congruences. The source note records the OEIS record and its literal source text.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $A'(x)=A(xA'(x))^2$ and $A(0)=1$ | IsSolution in HannaA232686Congruences | The formal derivative, substitution and constant coefficient are literal. | | $a_n=n![x^n]A$ is integral | The existential a : ℕ → ℤ with eCoeff A n = (a n : ℚ) | Integrality is a conclusion at every index. | | $2\mid a_n$ for $n\ge2$ | The first divisibility conjunct in claim | The threshold and modulus are literal. | | $4\mid a_n$ for $n\ge4$ | The second divisibility conjunct in claim | The threshold and modulus are literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA232686Congruences.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
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Number theoryCounterexample foundFirst preprint 2026Hou–Zhao: minimum-weight independence of sum-free codesTwo first-order sum-free functions on the ternary plane have original codes with different attained minimum nonzero weights.Refuted

The questionXiang-dong Hou and Shujun Zhao, Further Results on Sum-Freedom of Binary and q-ary Functions, arXiv:2609.31489v1 (2026-09-25), printed p.14, immediately after Theorem 4.6:

Why it mattersFunction-independent parameters in the binary APN case need not persist under the weaker general sum-free hypothesis.

The ideaOn a line a+tv, with v nonzero, the sum of either homogeneous quadratic map Q is −Q(v).

Our contributionTwo first-order sum-free functions on the ternary plane have original codes with different attained minimum nonzero weights.

Verification details
Release source recordRefuted in Lean

Open problem

Hou–Zhao: minimum-weight independence of sum-free codes

Xiang-dong Hou and Shujun Zhao, Further Results on Sum-Freedom of Binary and q-ary Functions, arXiv:2609.31489v1 (2026-09-25), printed p.14, immediately after Theorem 4.6:

There are some open questions. In Theorem 4.1, the dimension and minimum weight of the code C(f) are determined and they are independent of f as long as f is APN, i.e., 2nd order sum-free. Do we have the same conclusion for the code C_s(f) in Theorems 4.5 and 4.6, where f is an sth order sum-free function? If the dimension k and the minimum weight d of C_s(f) depend on the sum-free function f, what are the ranges for k and d? Likely, these questions will lead the investigation of sum-free functions to new directions.

The component considered here is minimum-weight independence alone. For every finite field K of cardinality q, every n ≥ 2, every 1 ≤ s ≤ n−1 and every f,g:K^n→K^n whose vector sums on every affine s-plane are nonzero, assert d(C_s(f)) = d(C_s(g)). There is no additional planarity, permutation, non-degeneracy, rank or equal-dimension hypothesis.

The original definition (1.1) uses affine subspaces. A basis of the direction subspace parametrizes such a subspace bijectively by K^s; independence of the directions prevents repetitions. SumFree uses exactly these sums. Definitions (4.2) and (4.7) make C_s(h) the kernel of the evaluations of all reduced monomials of total degree at most s(q−1)−1, together with all n coordinate rows of h. Each exponent is less than q. These monomials span the source Reed–Muller space, so redundant rows leave the kernel unchanged. All q^n points, including zero, are present.

claim : Prop retains the full universal assertion over small finite-field carriers. Every finite field has a small isomorphic carrier. Hamming weight counts nonzero entries; minimumWeight takes the infimum over nonzero kernel words in the extended naturals. The zero code has minimum infinity. This convention adds no premise and is irrelevant to the displayed nonzero codes. The conclusion is literally result : ¬ claim.

The separate dimension result, issue14810 and PR14846, neither proves nor refutes this minimum-weight component. The present result does not classify the complete ranges of k and d and does not contradict Theorem 4.6's bounds.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SumFreeCodeMinimumWeightRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2025Cartesian-control angle matrices are invertible for mixed qudit dimensionsThe real angle matrix of a finite Cartesian control set is invertible for every pair of enumerations and every choice of coordinate dimensions.Proved

The questionDi Matteo, Non-Abelian dynamics on a cube: Improving quantum compilation through qudit-based simulations, arXiv:2506.10945v1 (Phys.

Why it mattersThe matrix converts desired controlled-rotation angles into the angles of the decomposed circuit.

The ideaThe finite product-sum identity Fintype.prodsum factors the sum over all control words into coordinate sums.

Our contributionThe real angle matrix of a finite Cartesian control set is invertible for every pair of enumerations and every choice of coordinate dimensions.

Scope & next questionFinding the source's broader hypothesized decomposition remains open here, as do numerical conditioning and circuit optimality. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

Cartesian-control angle matrices are invertible for mixed qudit dimensions

J. Jiang, N. Klco and O. Di Matteo, Non-Abelian dynamics on a cube: Improving quantum compilation through qudit-based simulations, arXiv:2506.10945v1 (Phys. Rev. D 112, 074512, 2025), state the following conjecture in subsec_3c_restricted-rotations.tex, lines 38–46 (the display labelled eq:conjecture). The quotation expands the source macro \ctrlseq as $\boldsymbol C$ and renders its displayed equation inline:

By keeping $\boldsymbol C$ general, we sacrifice the guarantee that the matrix $M$ is invertible, although we conjecture our particular implementation yields an invertible $M$ when ${\boldsymbol C} = \mathbb{Z}{d_1} \times \mathbb{Z}{d_2} \times \cdots \times \mathbb{Z}_{d_k}$, i.e., uniform control but non-uniform qudit dimension.

Equation (5), subsec_3a_uniform-rotations.tex, lines 38–49, defines $\boldsymbol\theta=M\boldsymbol\beta$ and $M_{ij}=\prod_l(-1)^{f(b_l^{(i)},g_l^{(j)})}$, where $f(b,g)=1$ when $0<b\le g$ and $f(b,g)=0$ otherwise. Here $b$ and $g$ are control-word sequences. The definitions and source locators are collected in the literature note.

Issue #23818 fixes the reading as invertibility for every pair of enumerations of the full Cartesian control set. Precisely, for every $k\in\mathbb N$, every $d:\mathrm{Fin},k\to\mathbb N$, and every pair of bijections $b,g:\mathrm{Fin},N\simeq C$, where $N=\prod_l d_l$ and $C=\prod_{l:\mathrm{Fin},k}\mathrm{Fin}(d_l)$, angleMatrix d b g is the real matrix with entries $M_{ij}=\prod_l\bigl(\text{if }0<b(i)_l\le g(j)_l\text{ then }-1\text{ else }1\bigr)$. The proposition claim says that this matrix is a unit in the real square-matrix ring, so it has a two-sided inverse; result : claim proves it. There is no lower-bound hypothesis on the dimensions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/CartesianControlAngleMatrixInvertible.result

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Geometry & algebraCounterexample foundFirst preprint 2026Natural orthonormal bases of a dense-square Hilbert evolution algebraDense algebraic square does not force uniqueness of a complete natural orthonormal basis of a Hilbert evolution algebra.Refuted

The questionKaygorodov, Martin Gonzalez and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.8, Question 46 (page 26), ask:

Why it mattersThe condition that all products span densely need not distinguish individual basis squares.

The ideaEquality of the first two squares cancels the only potentially nonzero mixed product in the changed block.

Our contributionDense algebraic square does not force uniqueness of a complete natural orthonormal basis of a Hilbert evolution algebra.

Scope & next questionThat strengthened direction remains open here. - Effect: a claim relying on uniqueness cannot obtain that premise solely from density of the algebraic square.

Verification details
Release source recordRefuted in Lean

Open problem

Natural orthonormal bases of a dense-square Hilbert evolution algebra

Kaygorodov, Martin Gonzalez and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.2.8, Question 46 (page 26), ask:

If the square of a Hilbert evolution algebra is dense, is its natural orthonormal basis unique up to permutation and unimodular scalars?

The assertion quantifies over real or complex complete separable Hilbert spaces, bilinear products with bounded left multiplication for every fixed vector, and two complete natural orthonormal bases indexed by the positive integers. The algebraic square is the linear span of all pair products; density means norm density. Completeness of a basis means closed linear span equal to the Hilbert space. The following sentence about derivations, automorphisms and spectral theory is excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/HilbertNaturalBasisNonuniqueness.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2026Local distinguishability at every boundary projection stateAdjacent trace-one blockpositive sets are locally distinguishable at every boundary projection state.Proved

The questionYoung-Hoon Kiem and Seung-Hyeok Kye, Local geometry for Schmidt number witnesses, arXiv:2608.14199v1, Section 5, states:

Why it mattersThis is an external named problem with no digestion atom.

The ideaCompactness of unit low-rank vectors makes the projection of the nonpositive quadratic-form tests closed.

Our contributionAdjacent trace-one blockpositive sets are locally distinguishable at every boundary projection state.

Verification details
Release source recordProved in Lean

Open problem

Local distinguishability at every boundary projection state

Young-Hoon Kiem and Seung-Hyeok Kye, Local geometry for Schmidt number witnesses, arXiv:2608.14199v1, Section 5, states:

Nevertheless, we conjecture that the inclusion BP_ℓ ⊋ BP_{ℓ+1} is locally distinguishable around ϱ_{E⊥} for ℓ ≥ k, whenever ϱ_{E⊥} is on the boundary of the convex set BP_k.

For a nonzero proper complex subspace $E\subset\mathbb C^m\otimes\mathbb C^n$, write $\rho=P_{E^\perp}/\dim E^\perp$. The result says that if $\rho\in\partial_H BP_k$, then every neighbourhood of $\rho$ meets $BP_\ell\setminus BP_{\ell+1}$ whenever $1\le k\le\ell<\min(m,n)$. Here $H$ is the trace-one Hermitian affine space, and $BP_j$ tests nonnegativity against all vectors of Schmidt rank at most $j$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Trace-one Hermitian affine space $H$ | ambient m n | Both Hermiticity and complex trace one. | | $j$-blockpositivity | BP m n j; imported schmidtRank | The real quadratic form is nonnegative on every vector whose coefficient matrix has rank at most $j$. | | Normalized projection onto $E^\perp$ | rhoPerp E | Submodule.starProjection converted by Matrix.toEuclideanLin.symm and divided by the complex dimension. | | Nonzero proper subspace | E ≠ ⊥, E ≠ ⊤ in claim | The entire source range is retained. | | Boundary in $H$ | OnBoundary | Membership in the closed set $BP_k$ and in the closure of $H\setminus BP_k$. | | Adjacent inclusions in the finite chain | 1 ≤ k, k ≤ ℓ, ℓ < min m n | The range in Sections 3 and 5; it implies $m,n\ge2$. | | Every neighbourhood meets the difference | LocallyDistinguishable | Ambient and relative neighbourhoods give the same test because the difference lies in $H$. | | The conjecture for every subspace | result : claim | No genericity hypothesis. |

All public definition bodies are the literal conventions of #17787, with the matrix abbreviation expanded. The imported Schmidt rank is the rank of the coefficient matrix.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/LocalDistinguishability/KiemKyeLocalDistinguishability.result

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Further frontiersQuestion resolvedKotesovec's weighted distinct-partition logarithmic limitsThe logarithm of the weighted distinct-partition coefficient has Kotesovec's ratio asymptotic.Proved

The questionConjecture: log(a(n)) ~ sqrt(n/2) (log(2n) - 2).

Why it mattersThe source definitions and attributions are in the Library note.

The ideaBounding each partition product and combining these two estimates gives the finite sandwich used by the ratio-limit proof.

Our contributionThe logarithm of the weighted distinct-partition coefficient has Kotesovec's ratio asymptotic.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kotesovec's weighted distinct-partition logarithmic limits

OEIS A022629 defines $a_1(n)=[x^n]\prod_{k\ge1}(1+kx^k)$ and states:

Conjecture: log(a(n)) ~ sqrt(n/2) * (log(2*n) - 2).

The coefficient index ranges over every natural number, including zero. The claim means that the quotient tends to one along all natural indices. For a positive natural exponent $r$, write $a_r(n)=[x^n]\prod_{k\ge1}(1+k^r x^k)$. The related entries A092484, A265840, A265841 and A265842 take $r=2,3,4,5$, respectively. Their denominators are $\sqrt{2n}(\log(2n)-2)$ for $r=2$, and $r\sqrt{n/2}(\log(2n)-2)$ for $r=3,4,5$.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | A022629: “Expansion of Product_{m>=1} (1 + mq^m).” | KotesovecA022629.a | The coefficient of degree $n$ in factors $1$ through $n$; later factors cannot affect that coefficient. | | A022629: the quoted conjecture | KotesovecA022629.claim, KotesovecA022629.result | Exact source denominator; no assumed positivity or asymptotic hypothesis. | | A092484: “Expansion of Product_{m>=1} (1 + m^2q^m).” and its logarithmic conjecture | KotesovecA092484.a, .claim, .result | The literal $r=2$ normalization. | | A265840, A265841, A265842: products with exponents $3,4,5$ and their logarithmic conjectures | The corresponding modules' a, claim, result | Each source denominator is retained. | | Natural indices, including zero | Each a has domain $\mathbb N$ | Total real logarithm and division affect only initial exceptional indices. | | Every fixed positive natural exponent | KotesovecA022629.all_exponents | A consumed family theorem; fixing $r$ introduces no additional hypothesis into the five source claims. |

All Lean names above have prefix D5.S0.Asymptotics.GeneratingFunctions. The coefficient expression and the triangular sum are inline where needed; no higher-stratum triangular-number definition is imported.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Asymptotics/GeneratingFunctions/KotesovecA022629.result

D5/S0/Asymptotics/GeneratingFunctions/KotesovecA092484.result

D5/S0/Asymptotics/GeneratingFunctions/KotesovecA265840.result

D5/S0/Asymptotics/GeneratingFunctions/KotesovecA265841.result

D5/S0/Asymptotics/GeneratingFunctions/KotesovecA265842.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture provedFirst preprint 2026Kulhandjian–Hanzo Conjecture 25: the odd-prime distance boundProved

The questionThe centralizer consists of pairs with zero signed symplectic pairing against every stabilizer, and Pauli weight counts positions where either component is nonzero.

Why it mattersThe bound supplies a nontrivial logical operator of controlled weight in every odd-prime flagship code.

Verification details
Release source recordProved in Lean

Open problem

Kulhandjian–Hanzo Conjecture 25: the odd-prime distance bound

Kulhandjian and Hanzo, arXiv:2610.02392v1, Section VII, Conjecture 25 (“Flagship distance, original”), state: “For prime $p$ odd, $d_{\min}(\mathcal Q(p,2))\le p+1$.” Issue #13799 records the complete statement and its literature checks.

For every odd prime $p$, take a primitive $\alpha\in\mathbb F_{p^3}$ and $n=p^2+p+1$. The source's literal trace definitions are $\tau_H(i)=[\operatorname{Tr}(\alpha^i)=0]$ and $\tau_Q(i)=[\operatorname{Tr}(\alpha^{2i})=0]$. Set $A=\operatorname{circ}(\tau_H)$, $M_Q=\operatorname{circ}(\tau_Q)$, and $H=(A\mid M_QA)$ over $\mathbb F_p$, using entries $v_{i-j}$ for circulants. Stabilizers are $\operatorname{rowspan}(H)$. The centralizer consists of pairs with zero signed symplectic pairing against every stabilizer, and Pauli weight counts positions where either component is nonzero. The result constructs a centralizer vector outside the stabilizers of weight at most $p+1$. It quantifies over every primitive $\alpha$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SingerQuadricQuditDistanceBound.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedZero Slope Fits a Sample Best Exactly When Its Weighted Index Sum Is BalancedA finite sample is fitted best by a line of zero slope exactly when its position-weighted total is balanced against its total, because completing the square in the slope and the intercept leaves the centred cross term as the only obstruction.Proved

The questionNumber of n X 1 0..1 arrays with every row and column least squares fitting to a zero slope straight line, with a single point array taken as having zero slope.

The ideaAn n X 1 array is a binary word written as a column.

Our contributionA finite sample is fitted best by a line of zero slope exactly when its position-weighted total is balanced against its total, because completing the square in the slope and the intercept leaves the centred cross term as the only obstruction.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Zero Slope Fits a Sample Best Exactly When Its Weighted Index Sum Is Balanced

OEIS A222955, by R. H. Hardin, Mar 10 2013:

Number of n X 1 0..1 arrays with every row and column least squares fitting to a zero slope straight line, with a single point array taken as having zero slope.

The comment carrying the question, by Gus Wiseman, Jan 07 2023:

Conjecture: A binary word is counted iff it has the same sum of positions of 1's as its reverse, or, equivalently, the same sum of partial sums as its reverse.

The companion entry OEIS A359402, which lists the numbers whose binary expansion satisfies the position condition, repeats the same bridge in its own comments:

Conjecture: Also numbers whose binary expansion has as least squares fit a line of zero slope, counted by A222955.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/LeastSquaresBinaryFit.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedFirst preprint 2026Principal ideals of the priority latticePriority-forest interval structure and counting.Proved

The questionThe Library note quotes their interval-component, increasing-tree and edge-inclusion definitions.

Why it mattersThese literal cardinalities use the public parent-function priority forest, edge-inclusion order and adjoined greatest element.

The ideaidealCodeOrderIso compresses the ideal to the poset of closed subsets of its ordered support. forestidealshapeiff classifies the qualifying forests: one edge, or two or three edges with the OneLong property.

Our contributionPriority-forest interval structure and counting.

Scope & next questionThese formulas add the interval enumeration; the other final-section questions remain open.

Verification details
Release source recordProved in Lean

Open problem

Principal ideals of the priority lattice

Adrián Lillo and Mercedes Rosas, The Priority Lattice, arXiv:2603.28905v1, Section 6, item 3, p. 24, ask how many principal ideals are isomorphic to $\Pi(m)$, initially with $1\le m\le n$ and then with $m\le n$. They give $2,4,8,14,22,32,44,\ldots$ and conjecture OEIS A014206, displaying $n^2+n+2$.

The Library note quotes their interval-component, increasing-tree and edge-inclusion definitions. The binding claim is #14857, comment 6090966050: natural subtraction is ordered as n ^ 2 + 2 - 2 * n, and both counts include the principal ideal of the greatest element.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PriorityLattice/PrincipalIdeals.result

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Geometry & algebraQuestion resolvedSource 2026Inertia of a singular diagonal matrix after a structured rank-one downdateThe original finite-puncture simplex domain supplies nonemptiness, constant pivot sign and one globally unique negative diagonal coordinate.Proved

The questionThe source is Lucid (MathOverflow user 596445), revision 2 of MathOverflow question 513954, revision GUID 1084D752-1BBD-405E-B335-28FBDD1F4D6D.

Why it mattersThe question asks whether the structured downdate removes exactly the pivot sign and adds one null direction, even when the original diagonal form is singular, and whether the source domain forces one global sign and one global negative coordinate in the negative case.

The ideaCharacteristic-polynomial roots give positive and negative counts with multiplicity, and the real diagonal count handles singular entries.

Our contributionThe original finite-puncture simplex domain supplies nonemptiness, constant pivot sign and one globally unique negative diagonal coordinate.

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Release source recordProved in Lean

Open problem

Inertia of a singular diagonal matrix after a structured rank-one downdate

The source is Lucid (MathOverflow user 596445), revision 2 of MathOverflow question 513954, revision GUID 1084D752-1BBD-405E-B335-28FBDD1F4D6D. The Library note is the sole source acknowledgement and records the CC BY-SA 4.0 license.

The source asks:

I ran into this while working through a rank-one perturbation problem and got stuck on the case where $D$ is singular. Let $D = \operatorname{diag}(d_1, \dots, d_n)$ be real diagonal, possibly singular, and let $x \in \mathbb{R}^n$ satisfy $$\Delta := x^\top D x \neq 0.$$ Set $w = Dx$ and $$M = D - \frac{ww^\top}{\Delta}.$$ Writing $\operatorname{inertia}(A) = (n_+(A), n_-(A), n_0(A))$, is the following formula always valid? $$\operatorname{inertia}(M) = \begin{cases} (n_+(D) - 1,\ n_-(D),\ n_0(D) + 1), & \Delta > 0, \ (n_+(D),\ n_-(D) - 1,\ n_0(D) + 1), & \Delta < 0. \end{cases}$$ I specifically want a proof that doesn't assume $D^{-1}$ exists. Note: this isn't the general rank-one update $D + \rho vv^\top$ for arbitrary $v$ (the classical secular-equation setting) , here $w = Dx$ is tied to $D$ itself, with coefficient exactly $1/\Delta$, chosen so $x$ becomes an exact null vector of $M$. I'm after the inertia count specifically, including the singular-$D$ case and the index-tracking behavior in the corollary below. Here's what I've tried. Define $H = {z : w^\top z = 0}$. Since $w^\top x = \Delta \neq 0$, $$\mathbb{R}^n = H \oplus \operatorname{span}{x}.$$ If $Z$ contains a basis of $H$ and $P = [Z \ \ x]$, then $$P^\top D P = \begin{pmatrix} Z^\top D Z & 0 \ 0 & \Delta \end{pmatrix}, \qquad P^\top M P = \begin{pmatrix} Z^\top D Z & 0 \ 0 & 0 \end{pmatrix}.$$ Does Sylvester's law of inertia immediately establish the displayed formula, including all additional null directions when $D$ is singular? A related corollary: suppose $D(y)$, $x(y)$, and hence $M(y)$, vary continuously on $X \setminus A$, where $X$ is homeomorphic to an open $(n-1)$-simplex, $n \geq 3$, and $A$ is finite. If $\Delta(y) \neq 0$ and $M(y) \succeq 0$, does path-connectedness imply that $\Delta$ has constant sign and, when negative, exactly one fixed diagonal index of $D(y)$ remains negative throughout? I've checked the inertia formula numerically for random $D$, $x$ up to $n = 9$, including singular $D$, and haven't found a counterexample.

This is one full unit: the complete two-sign inertia triple and the continuous-family sign/fixed-negative-index corollary. The matrix clause ranges over every finite dimension and retains singular diagonal matrices; the family clause keeps the open-simplex-minus-finite-points domain, continuous families, nonzero pivot and pointwise positive semidefiniteness. No invertibility, connected-family replacement, dimension cutoff, tolerance or spectral-gap premise is added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/StructuredRankOneFamily.mo513954_revision2

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  • Historical context

    Year recorded in the source bibliography: 2026.

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Number theoryConjecture refutedFirst preprint 2026Narayan's expanded-generator delta-set conjectureExpanded generators refute Narayan's delta-set formula by unit-length normalization.Refuted

The questionAshwin Narayan, The Delta Set of a Numerical Monoid Generated by a Generalized Arithmetic Sequence, arXiv:2610.02495v1, Conjecture 18.3, p. 22, states:

The ideaEach rule preserves value and raises length by one (rewritevalue, rewritesize).

Our contributionExpanded generators refute Narayan's delta-set formula by unit-length normalization.

Verification details
Release source recordRefuted in Lean

Open problem

Narayan's expanded-generator delta-set conjecture

Ashwin Narayan, The Delta Set of a Numerical Monoid Generated by a Generalized Arithmetic Sequence, arXiv:2610.02495v1, Conjecture 18.3, p. 22, states:

For $S_h=\langle a,ah,ah+d,ah+2d,\ldots,ah+kd\rangle$, $\Delta(S_h)=E(e,d)$.

Section 3 assumes $\gcd(a,d)=1$, $1<k<a$, $h>1$, and puts $e=h-1$. All four parameters are positive integers. Every displayed generator, including the redundant $ah$, is retained as a separate nonnegative factorization coordinate. The delta set is the union over represented natural numbers of positive differences between consecutive distinct lengths. The subtractive Euclidean algorithm starts at $(e,d)$; its final equal pair may be repeated without adding visited positive coordinates. These interpretations resolve the unbound $s$, omitted endpoint order and malformed initialization in Definitions 3.1–3.2, as fixed by #23185.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | All positive $a,h,d,k$ | The four natural-number binders and four positivity premises of claim | No parameter restriction is omitted. | | $\gcd(a,d)=1$, $1<k<a$, $h>1$ | Nat.Coprime a d and the three strict inequalities of claim | Literal restrictions. | | $(a,ah,ah+d,\ldots,ah+kd)$ | generators, indexed by Fin (k + 2) | The $ah$ coordinate is retained. | | Nonnegative coordinate factorizations with weighted sum $n$ | Z | The complete coordinate equation is retained. | | Length is the coordinate sum; $L(n)$ contains all such lengths | The image in L | No minimal-generator substitution. | | Positive successive length differences, over represented $n$ | delta | Strict endpoint order and no intermediate member are explicit. | | Subtractive Euclidean states and visited positive coordinates | euclidStep, euclidState, E | The initial and terminal coordinates are included. | | $\Delta(S_h)=E(h-1,d)$ | The equality in claim; result proves its universal assertion false | The witness is $(3,2,2,2)$. |

The displayed finite sums use Mathlib directly; no separate coordinate-sum alias is introduced. The implementation carrier is a product of a natural number and a finite list. Every public definition displays its complete body in the Blueprint. There are no additional source-independent hypotheses in the public claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Additive/NarayanExpandedGeneratorDeltaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

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  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryQuestion resolvedContainment of the A115809 carryless multiplier conditionThe seventeen-times binary carryless condition implies the nine-times condition.Proved

The questionIn the notation of the source, the exact question is

Why it mattersThe source asks for containment of two sets of nonnegative integers defined by different ordinary and carryless multiplier pairs.

The ideaThe proof uses the imported binary convolution definition and its equality with the recursive binary product.

Our contributionThe seventeen-times binary carryless condition implies the nine-times condition.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Containment of the A115809 carryless multiplier condition

OEIS A115809 asks whether the integers satisfying its defining condition are a subset of A115801. In the notation of the source, the exact question is

$$ \forall k\in\mathbb{N},\qquad 17k=C(49,k)\ \Longrightarrow\ 9k=C(25,k). $$

Here $\mathbb{N}$ includes zero and the same $k$ occurs in both sides. The products $17k$ and $9k$ are ordinary natural-number products. The operation $C$ is binary coefficient convolution over $\mathrm{GF}(2)$:

$$ \operatorname{bit}(C(u,k),t)

\left(\sum_{i=0}^{t} \operatorname{bit}(u,i)\operatorname{bit}(k,t-i)\right)\bmod 2. $$

This is the carryless polynomial product described by OEIS A048720, not ordinary multiplication. The source question and operation are recorded at https://oeis.org/A115809.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/CarrylessSeventeenNineContainment.result

Frozen module record

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresRecurrence provedPosed in 2025Kurkov's reversed q-Bell row polynomialKurkov's triangular recurrence produces the reversed q-Bell row polynomials.Proved

The question1,1,1,1,1,2,1,1,1,3,3,4,2,1,1,1,4,6,10,9,7,7,4,2,1,1,1,5,10,20,25,26,

Why it mattersKurkov's recurrence builds each row from the preceding diagonal and its prefix sums.

The ideaProve by strong induction that qBell n is monic of triangular degree.

Our contributionKurkov's triangular recurrence produces the reversed q-Bell row polynomials.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kurkov's reversed q-Bell row polynomial

OEIS A126347 has offset %O 0,6. Its coefficient data begins (%S, verbatim):

1,1,1,1,1,2,1,1,1,3,3,4,2,1,1,1,4,6,10,9,7,7,4,2,1,1,1,5,10,20,25,26,

Its NAME line (%N, verbatim) is:

Triangle, read by rows, where row n lists coefficients of q in B(n,q) that satisfies: B(n,q) = Sum_{k=0..n-1} C(n-1,k)*B(k,q)*q^k for n>0, with B(0,q) = 1; row sums equal the Bell numbers: B(n,1) = A000110(n).

Mikhail Kurkov's conjecture (%F, verbatim) is:

Conjecture: R(n,n) is the (n+1)-th reversed row polynomial where R(0,0) = 1, R(n,k) = R(n-1,n-1) + x^n * Sum_{j=0..k-1} R(n-1,j) for 0 <= k <= n. - Mikhail Kurkov, Jul 06 2025

The AUTHOR line (%A, verbatim) is:

Paul D. Hanna, Dec 31 2006, May 28 2007

The referenced Bell-number entry A000110 has NAME line (%N, verbatim):

Bell or exponential numbers: number of ways to partition a set of n labeled elements.

The exact formal statement is:

forall n : Nat, kurkovR n n = (qBell (n + 1)).reverse

Here qBell uses Wagner's q-Bell recurrence. The formal reversal is reflection about the natural degree. The locally proved facts that qBell n is monic of degree Nat.choose n 2 ensure that this is the fixed-row coefficient reversal meant by the source.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/KurkovReversedQBellRowPolynomial.result

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About this highlight
  • Contribution

    Kurkov's triangular recurrence produces the reversed q-Bell row polynomials.

    Read source ↗
  • Historical context

    Dated conjecture statement: 2025.

    Read source ↗
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedZeros of the A131853 binary Gaussian evaluationA vanishing binary digit polynomial at the Gaussian unit has an argument divisible by five.Proved

The questionNumbers m such that z(m)=(0,0) with z as defined in A131851.

Why it mattersThe formal result proves the full unbounded divisibility statement for the binary Gaussian evaluation, rather than only extending the displayed sequence of examples.

The ideaModulo 5 the Gaussian unit behaves like 2 because 2² ≡ −1.

Our contributionA vanishing binary digit polynomial at the Gaussian unit has an argument divisible by five.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Zeros of the A131853 binary Gaussian evaluation

OEIS A131853, NAME (%N, verbatim):

Numbers m such that z(m)=(0,0) with z as defined in A131851.

COMMENTS (%C, verbatim):

Intersection of A131854 and A131855: A131851(a(n))=0, A131852(a(n))=0;

conjecture: a(n) mod 5 = 0.

OEIS A131851, NAME (%N, verbatim):

Real part of the function z(n)=Sum(d(k)*i^k: d as in n=Sum(d(k)*2^k), i=sqrt(-1)).

FORMULA (%F, verbatim):

z(n) = if n=0 then (0, 0) else z(floor(n/2))*(0, 1) + (n mod 2, 0), complex multiplication.

The literal claim is ∀ m : ℕ, z m = 0 → 5 ∣ m, where z : ℕ → GaussianInt is defined by Nat.binaryRec 0 (fun b _ w => (if b then 1 else 0) + (⟨0, 1⟩ : GaussianInt) * w). The Lean statement quantifies over every natural m, including zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/ZumkellerBinaryGaussianEvaluationZeroMultipleOfFive.zumkeller_a131853

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedCurtz's six-residue denominator conjectureThe reduced denominators of Curtz's rational triangle satisfy all six residue formulas.Proved

The questionConjecture: a(6n)=1, a(1+6n)=1+3n, a(2+6n)=1+2n, a(3+6n)=2+3n, a(4+6n)=1, a(5+6n)=6+6n. a(n) = 2a(n-6) - a(n-12).

Why it mattersThis is Paul Curtz's externally named OEIS conjecture.

The ideaThe finite coefficient moment J(m,P) sums coeff(P,i)/(m+i+1) over the support.

Our contributionThe reduced denominators of Curtz's rational triangle satisfy all six residue formulas.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Curtz's six-residue denominator conjecture

OEIS A189733, revision 36, states:

a(n) = denominator(B(0,n)). Conjecture: a(6*n)=1, a(1+6*n)=1+3*n, a(2+6*n)=1+2*n, a(3+6*n)=2+3*n, a(4+6*n)=1, a(5+6*n)=6+6*n. a(n) = 2*a(n-6) - a(n-12).

All indices are natural numbers. Define the actual rational triangle by T(m,0)=0, T(m,1)=(-1)^m/(m+1), and T(m,k+2)=T(m,k+1)+T(m+1,k), with boundary cases taking precedence. The public triangle implements these three branches, and the public a is exactly its zeroth-row reduced denominator. In particular, a zero rational has denominator one. The theorem result is the equality of the six-tuples for every natural k, including zero, with no extra premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/OEISA189733DenominatorConjecture.result

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedKrizek's Fermat-Mersenne classification for A249759Prime divisor sums force a prime-power input and the Fermat-Mersenne forms conjectured for OEIS A249759.Proved

The question%N A249759 Primes p such that sigma(p-1) is a prime q. %S A249759 3,5,17,65537 %C A249759 Conjectures: 1) sequence is finite; 2) sequence is a subsequence of A019434 (Fermat primes); 3) sequence consists of Fermat primes p such that sigma(p-1) is a Mersenne prime; 4) a(n) = (A249761(n)+3)/2. %A A249759 Jaroslav Krizek, Nov 13 2014

Why it mattersThe result places every member of A249759 in the Fermat-prime sequence A019434 and shows that its divisor sum is a Mersenne prime.

The ideaExclude m = 0 and m = 1, then write σ 1 m as the product, over all prime factors q of m, of the geometric sums ∑ i < factorization q + 1, q^i.

Our contributionPrime divisor sums force a prime-power input and the Fermat-Mersenne forms conjectured for OEIS A249759.

Scope & next questionNot claimed: item 1 is a finiteness statement about the resulting class of Fermat primes and remains open.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Krizek's Fermat-Mersenne classification for A249759

OEIS A249759, revision #40 Sep 10 2025 16:53:46, records these lines verbatim:

%N A249759 Primes p such that sigma(p-1) is a prime q.

%S A249759 3,5,17,65537

%C A249759 Conjectures: 1) sequence is finite; 2) sequence is a subsequence of A019434 (Fermat primes); 3) sequence consists of Fermat primes p such that sigma(p-1) is a Mersenne prime; 4) a(n) = (A249761(n)+3)/2.

%A A249759 Jaroslav Krizek, Nov 13 2014

Items 2 and 3 have the following formal joint statement, where σ 1 is Mathlib's sum-of-divisors function:

open scoped ArithmeticFunction.sigma in
theorem result {p : ℕ} (hp : p.Prime) (hq : (σ 1 (p - 1)).Prime) :
    (∃ m : ℕ, p = 2 ^ 2 ^ m + 1) ∧
      (∃ r : ℕ, r.Prime ∧ σ 1 (p - 1) = 2 ^ r - 1)

Natural subtraction does not weaken this statement. Nat.Prime.two_le gives 2 ≤ p, and the proof separately rules out p = 2 from σ 1 (2 - 1) = σ 1 1 = 1, which is not prime.

Not claimed: item 1 is a finiteness statement about the resulting class of Fermat primes and remains open. Item 4 concerns the definition of the separate sequence A249761. Neither item appears on the Lean delivery surface.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Mersenne/KrizekSigmaPrimeFermatMersenne.result

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedKagey's modulo-ten congruence for no-leaf grid subgraphsBarker's recurrence and Kagey's modulo-ten congruence for no-leaf subgraphs of the three-by-n grid.Proved

The questionNumber of no-leaf subgraphs of the 3 X n grid.

Why it mattersThe congruence was recorded as a conjecture in 2018.

The ideaThe companion theorem proves Barker's order-four recurrence for the same literal counting function.

Our contributionBarker's recurrence and Kagey's modulo-ten congruence for no-leaf subgraphs of the three-by-n grid.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Kagey's modulo-ten congruence for no-leaf grid subgraphs

OEIS A301976, NAME (verbatim from Library/Words/kagey2018a301976.md):

Number of no-leaf subgraphs of the 3 X n grid.

COMMENT (verbatim; attributed there to Peter Kagey):

Conjecture: a(n) mod 10 = 3 for n > 2.

The literal object is the set of spanning edge-subgraphs of the 3 X n grid graph, with all 3n vertices retained, in which every vertex has degree different from one. Degree zero is allowed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/GridNoLeafSubgraphRecurrence.kagey_a301976_mod10

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresRecurrence provedFirst Occurrences in OEIS A335901Alkan's self-referential doubling recurrence has first-occurrence positions A117261.Proved

The questiona(n) = 2a(floor((n-1)/a(n-1))) with a(1) = 1.

Why it mattersThe conjecture identifies the first occurrence of every power of two in a self-referential recurrence with a named OEIS sequence.

The ideaThe proof establishes the block invariant: a(T(r)) = 2^r; for r >= 1 and T(r) <= n < T(r+1), a(n) is either 2^(r-1) or 2^r; and if a(n) = 2^(r-1), then n < 2^(r-1) T(r).

Our contributionAlkan's self-referential doubling recurrence has first-occurrence positions A117261.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

First Occurrences in OEIS A335901

OEIS A335901, %N (verbatim):

a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

%C (verbatim):

Least k such that a(k) = 2^n are 1, 2, 5, 21, 169, 2705, ... (Conjecture: This sequence is A117261).

The comparison sequence A117261 is given by A117261(0) = 1 and A117261(r+1) = 2^r * A117261(r) + 1. The phrase "least k such that a(k) = 2^r" is formalized as a hit together with minimality among positive indices.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SelfReferentialDoublingFirstOccurrence.alkan_a335901

Frozen module record

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About this highlight
  • Contribution

    Alkan's self-referential doubling recurrence has first-occurrence positions A117261.

    Read source ↗
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThe A363184 bilateral quarter-double-shift theta congruenceThe integer solution of the A363184 bilateral equation is theta3(x^2) modulo four.Proved

The questionHanna, May 20, 2023), quoted verbatim from Library/Recurrence/hanna2023a363184.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2023 entry.

The ideaNo A363104 definitions are imported or aliased.

Our contributionThe integer solution of the A363184 bilateral equation is theta3(x^2) modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A363184 bilateral quarter-double-shift theta congruence

OEIS A363184 (Paul D. Hanna, May 20, 2023), quoted verbatim from Library/Recurrence/hanna2023a363184.md:

NAME:

Expansion of g.f. A(x) satisfying 4 = Sum_{n=-oo..+oo} (-1)^n * x^n * (4*A(x) + x^(2*n-1))^(n+1).

COMMENT:

Conjecture: g.f. A(x) == theta_3(x^2) (mod 4); a(n) == 2 (mod 4) if n = 2*k^2 for integer k > 0, and a(n) == 0 (mod 4) if floor(n/2) is nonsquare.

The NAME is formalised coefficientwise with exact finite windows after cancelling the two Laurent terms. The g.f. clause is mod_four_identity, the full coefficient classification is hanna_conjecture, and the COMMENT's weaker zero clause is hanna_conjecture_floor. This dossier has one resolution claim, attached only to hanna_conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Bilateral/BilateralQuarterDoubleShiftThetaModFour.hanna_conjecture

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedParity of q-Catalan row square sumsThe square sum of a Carlitz-Riordan q-Catalan row is odd exactly one below a power of two.Proved

The questionHanna's comment of October 11, 2024 states, verbatim:

Why it mattersThe assertion covers every natural row index, including zero.

The ideaCoefficient extraction from the defining bivariate equation gives R0=1 and R(n+1)=Sum{i+j=n} q^i Ri Rj.

Our contributionThe square sum of a Carlitz-Riordan q-Catalan row is odd exactly one below a power of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of q-Catalan row square sums

OEIS A376527 defines

a(n) = Sum_{k=0..n*(n-1)/2} A227543(n,k)^2.

Paul D. Hanna's comment of October 11, 2024 states, verbatim:

Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

Here A227543 is the Carlitz-Riordan q-Catalan triangle defined by A(x,q) = 1 + x*A(q*x,q)*A(x,q).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/QCatalanSquareSumParity.hanna_conjecture

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedOdd-index parity of the A389472 coefficientsEvery odd-index coefficient beyond the first in OEIS A389472 is even.Proved

The questionHanna, Oct 15 2025, gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2025a389472.md:

Why it mattersThis is a first-tier recent OEIS conjecture, selected in the lane brief.

The ideaFor n >= 3, coeffrecurrence gives a(n) = Σ{k=0}^{floor(n/2)+1} a(k)·C(k, n+2−2k).

Our contributionEvery odd-index coefficient beyond the first in OEIS A389472 is even.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Odd-index parity of the A389472 coefficients

OEIS A389472, Paul D. Hanna, Oct 15 2025, gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2025a389472.md:

G.f. satisfies A(x) = A(x^2 + x^3)/x^2 - 1.

Conjecture: a(2*n-1) = 0 (mod 2) for n > 1.

范围与诚实边界: The separate comment "Conjecture: a(3*n-1) = 0 (mod 3) for n > 1." is not resolved here. The generating equation alone leaves a(2) free. The module fixes a(0)=0, a(1)=a(2)=1, matching the OEIS PARI program and DATA, and formalizes the cross-multiplied identity x²(A+1) = A(x²+x³) over integer formal power series. Only the first, odd-index parity conjecture for that normalized sequence is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/SquareCubeSubstitutionOddIndexParity.hanna_conjecture

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedG.f. A(x) satisfies A(x)^2 = A(2*x - 2*A(x)) / 2.The odd coefficients of OEIS A389540 occur exactly at powers of two.Proved

The questionA(x) satisfies A(x)^2 = A(2x - 2A(x)) / 2." COMMENT "Conjecture: a(n) is odd iff n = 2^k for k >= 0."

The ideaThe compositional inverse R is defined first by the integer recursion r (inverseSeries), and A is its inverse (generatingSeries). generatingequation gives the normalized integral solution of the cleared equation 2A² = A(2X − 2A), and generatingunique proves uniqueness by inverse-coefficient induction.

Our contributionThe odd coefficients of OEIS A389540 occur exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

G.f. A(x) satisfies A(x)^2 = A(2*x - 2*A(x)) / 2.

NAME "G.f. A(x) satisfies A(x)^2 = A(2*x - 2*A(x)) / 2."

COMMENT "Conjecture: a(n) is odd iff n = 2^k for k >= 0."

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HalfScaledReflectionParity.hanna_conjecture

Frozen module record

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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe full signed Laurent second moment in A391810The full signed second moment of Hanna's bilateral Laurent rows.Proved

The questionHanna's OEIS A391810 revision 39 formula (4) states:

Why it mattersThis is the explicitly posed formula (4), distinct from formula (2)'s zero row sum and formula (3)'s first moment.

The ideaThe inherited HannaLaurentRowSums.series is the full formally summable bilateral Laurent series.

Our contributionThe full signed second moment of Hanna's bilateral Laurent rows.

Scope & next questionThese prerequisites are reused rather than counted as a new resolution. - Open: the support bound's sharpness, other moment identities and other source conclusions are outside this result.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The full signed Laurent second moment in A391810

Paul D. Hanna's OEIS A391810 revision 39 formula (4) states:

(4) Sum_{k=0..floor((n+1)/2)^2} k^2 * T(n,k) = -(2*n-5) for n >= 2 (conjecture).

The source defines T(n,k) as the coefficient at zero of x^k times the bilateral series Sum_{j=-oo..oo} x^j*(1-x^j)^(n+j), expanded formally at zero. Equivalently it is the series coefficient at -k. All integer outer indices, the constant coefficient, and both finite-sum endpoints are included. Negative integer exponents are Laurent-field inverses. The bound is natural division (n+1)/2 followed by squaring. The right side uses signed integer subtraction and negation, then a rational cast.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/HannaLaurentSecondMoment.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

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Number theoryQuestion resolvedPrime twin-center sigma-gcd values are two or threeFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k)) is prime, then gcd(k, sigma(k))=2 or gcd(k, sigma(k))=3.Proved

The questionFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k)) is prime, then gcd(k, sigma(k))=2 or gcd(k, sigma(k))=3.

Why it mattersThis is the second A394757 conjecture supplied in preregistration 15004.

The ideaAn even sigma(k) gives two dividing the prime gcd, so that gcd equals two.

Our contributionFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k)) is prime, then gcd(k, sigma(k))=2 or gcd(k, sigma(k))=3.

Scope & next questionThis involves unresolved twin-prime infinitude and the additional sigma-gcd restrictions; twin-prime infinitude alone does not settle those restrictions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime twin-center sigma-gcd values are two or three

For every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k)) is prime, then gcd(k, sigma(k))=2 or gcd(k, sigma(k))=3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Robin/TwinPrimeSigmaGcdTwoOrThree.result

Frozen module record

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    Source 1
Discrete structuresQuestion resolvedThe A396099 period-four coefficient patternThe unique integer series of OEIS A396099 satisfies all four congruence conjectures.Proved

The questionHere A^k denotes the k-fold compositional iterate, as the entry's formulas and PARI program make explicit: A^2(x) = A(A(x)) and A^3(x) = A(A(A(x))).

Why it mattersThis is a first-tier OEIS conjecture from the 2026 entry.

The ideaOver ZMod 4, modfouridentity identifies Ā with F := mobius 1 + 2X⁴·((1−X)(1+X²))⁻¹.

Our contributionThe unique integer series of OEIS A396099 satisfies all four congruence conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A396099 period-four coefficient pattern

OEIS A396099 (Paul D. Hanna, Jun 02 2026) gives the NAME:

G.f. satisfies A(x) = x + A^2(x)*A^3(x).

Here A^k denotes the k-fold compositional iterate, as the entry's formulas and PARI program make explicit: A^2(x) = A(A(x)) and A^3(x) = A(A(A(x))). The single resolution anchor of this dossier is the COMMENT:

Conjecture: a(n) = [1,3,3,1] repeating (mod 4) for n > 2.

The entry also states three companion conjectures:

Conjecture: all terms are odd.

Conjecture: [x^n] A(A(x)) == 0 (mod 4) for n > 2.

Conjecture: [x^n] ( A(x) - x*A(A(A(x))) ) == 2 (mod 4) for n > 2.

These are settled by all_odd, iterate_two_mod_four, and shift_mod_four, respectively, without adding a second resolution claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/IterateProductTwoThreeModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedDivisibility of the A396794 coefficientsEvery coefficient above degree one in OEIS A396794 is divisible by four.Proved

The questionThe following exact NAME and COMMENT quotations are recorded in Library/Arith/hanna2026a396794.md.

Why it mattersThis is the first-tier recent OEIS conjecture selected in the implementation brief.

The ideaCoefficient lifting constructs a normalized integer solution.

Our contributionEvery coefficient above degree one in OEIS A396794 is divisible by four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A396794 coefficients

OEIS A396794, Paul D. Hanna, Jun 07 2026. The following exact NAME and COMMENT quotations are recorded in Library/Arith/hanna2026a396794.md.

NAME:

G.f. A(x) satisfies A(x) * A(A(A(x))) = x^2 + 16*x^3.

COMMENT:

Conjecture: a(n) == 0 (mod 4) for n > 1.

The formal target uses the normalized integer series with constant coefficient zero and degree-one coefficient one; a(n) is its degree-n coefficient.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/TripleIterateProductModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    Source 1
Discrete structuresQuestion resolvedA396803 coefficients modulo threeThe positive-index coefficients of OEIS A396803 agree with their indices modulo three.Proved

The questionHanna's OEIS A396803 entry, dated June 8, 2026, states:

Why it mattersThis first-tier external conjecture is the remaining A396803 clause after the parity result in IterateExponentialParity.

The ideaLet l(n)=n in ZMod 3 and let g be its third compositional iterate in integral EGF coordinates.

Our contributionThe positive-index coefficients of OEIS A396803 agree with their indices modulo three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

A396803 coefficients modulo three

Paul D. Hanna's OEIS A396803 entry, dated June 8, 2026, states:

%N A396803 E.g.f. satisfies A(x) = x*exp( A^3(x) ).
%C A396803 Conjecture: a(n) == [1,2,0] repeating (mod 3) for n >= 1.

The formula field defines A^3 as the third compositional iterate. For the unique zero-constant rational formal solution of A=X*exp(A composed with A composed with A), put a(n)=n![X^n]A. The assertion is forall n >= 1, a(n) = n (mod 3).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/IterateExponentialModThree.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedParity of the A397349 logarithmic-weight coefficientsThe logarithmic-weight coefficients are odd exactly at powers of two.Proved

The questionHanna's NAME and both conjecture COMMENT lines, quoted verbatim:

The ideaForm the series A, C and S over ZMod 2 from a, from b shifted to have zero constant coefficient, and from s.

Our contributionThe logarithmic-weight coefficients are odd exactly at powers of two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A397349 logarithmic-weight coefficients

Paul D. Hanna's NAME and both conjecture COMMENT lines, quoted verbatim:

%N A397349 L.g.f. Sum_{n>=1} a(n)*x^n/n = log(1+x + Sum_{n>=2} 3*n/(3*n^2 - 1) * a(n)*x^n ).
%C A397349 Conjecture: a(n) is odd iff n is a power of 2.
%C A397349 Conjecture: a(n) = [1,2] repeating (mod 3) for n >= 1.

A397349 carries two conjectures. The modulo-three alternation is already settled in this repository by LogarithmicWeightCatalanParity.hanna_conjecture_a397349_mod_three. This lane settles the other one, the parity clause, which that module deliberately left as the bare proposition parity_conjecture carrying no proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/LogarithmicWeightBinaryParity.parity_conjecture_holds

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedParity of the A397356 reciprocal square-exponent diagonalsThe coefficients of OEIS A397356 are odd exactly when their index plus one is a power of two.Proved

The questionHanna's NAME and BOTH conjecture COMMENT lines, quoted verbatim from the entry snapshot oeis-A397356.src:

Why it mattersThe candidate is taken from the repository's own triage note Library/Words/oeis2026triage0911b.md, which scored it as a first-tier target and recorded that the entry supplies only conjectures, programs, a data table and an asymptotic estimate.

The ideaWork with the reciprocal R = 1/A throughout, so that the entry's relation reads coeff n (R ^ (n²)) = coeff n (R ^ (n² − 1)) for n > 1 with natural-number exponents and no negative powers anywhere.

Our contributionThe coefficients of OEIS A397356 are odd exactly when their index plus one is a power of two.

Scope & next questionThe divisibility-by-three conjecture remains open; its finite numerical support is not a second resolution.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Parity of the A397356 reciprocal square-exponent diagonals

Paul D. Hanna's NAME and BOTH conjecture COMMENT lines, quoted verbatim from the entry snapshot oeis-A397356.src:

%N A397356 G.f. A(x) satisfies: [x^n] 1/A(x)^(n^2) = [x^n] 1/A(x)^(n^2-1) for n > 1 with A(0) = A'(0) = 1.
%C A397356 Conjecture: for n >= 0, a(n) is odd iff n+1 is a power of 2.
%C A397356 Conjecture: for n >= 0, a(n) is not divisible by 3 iff 2*(n+1) is a sum of two powers of 3 (A055235).
%A A397356 Paul D. Hanna, Jul 03 2026

A397356 carries two conjectures. This lane settles ONLY the parity classification; the assertion about divisibility by three is not claimed here and appears nowhere in this module as a theorem or as a definition. That second conjecture is settled separately in D5/S1/Recurrence/Residue/ReciprocalSquareExponentDiagonalModThree.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ReciprocalSquareExponentDiagonalParity.hanna_conjecture_a397356

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
Number theoryQuestion resolvedPosed in 2026OEIS A400585: the odd-cofactor summatory asymptoticThe odd-cofactor binomial divisor sum has the cubic asymptotic proposed in A400585.Proved

The questionConjecture: Sum{k=1..n} a(k) ~ 7zeta(3)n^3/48. - Vaclav Kotesovec, Oct 01 2026

Why it mattersIt records the quoted OEIS statement, the revision-22 source identity, the full quantifiers, and the intended ratio-limit target before implementation.

The ideaReindexing by the odd cofactor gives the finite sum of binomial(N / q + 1, 3) over odd q.

Our contributionThe odd-cofactor binomial divisor sum has the cubic asymptotic proposed in A400585.

Scope & next questionIt does not settle the entry's other comments or imply a literature-novelty or first-publication claim; those questions remain outside this theorem.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A400585: the odd-cofactor summatory asymptotic

OEIS A400585, revision 22, defines

a(n) = Sum_{d|n, n/d odd} binomial(d, 2).

Its FORMULA field states:

Conjecture: Sum_{k=1..n} a(k) ~ 7*zeta(3)*n^3/48. - Vaclav Kotesovec, Oct 01 2026

The delivered statement keeps the source's full natural range. It defines the natural divisor sum a, the real positive-prefix sum S, and the real series Z3 = sum_{q >= 0} 1/(q+1)^3, then proves

Tendsto (fun N : ℕ => S N / ((7 * Z3 / 48) * (N : ℝ)^3))
  atTop (𝓝 1)

There is no parity, primality, or finite-range restriction. The natural value at zero is harmless because the limit is at atTop.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/A400585OddCofactorAsymptotic.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedOEIS A400644 rounded three-term norm ratioThe exact rounded three-term norm sequence has the Tribonacci square-root ratio limit.Proved

The questiona(n+1) = round(sqrt(a(n)^2 + a(n-1)^2 + a(n-2)^2)).

Why it mattersThe fixed initial data do not make this infinite asymptotic theorem a certified finite instance.

The ideaUse the exact zero-based sequence a(0)=a(1)=0, a(2)=1, and set sourceA(n)=a(n-1) on positive indices.

Our contributionThe exact rounded three-term norm sequence has the Tribonacci square-root ratio limit.

Scope & next questionExtensions to other initial values or rounded higher-order norm recurrences remain open here.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A400644 rounded three-term norm ratio

OEIS A400644 revision 28 has the name:

a(n+1) = round(sqrt(a(n)^2 + a(n-1)^2 + a(n-2)^2)).

Its conjectural comment is:

Numerical evidence suggests that the ratio of consecutive terms (a(n+1)/a(n)) tends to 1.356203065626295... which is the largest positive real root of x^6-x^4-x^2-1=0. Setting x^2=t gives t^3-t^2-t-1 which yields the tribonacci constant (A058265) as the largest positive root.

The Mathematica source initializes a(1)=a(2)=0, a(3)=1 and uses Round[Sqrt[a[n]^2+a[n-1]^2+a[n-2]^2]]. The Python source starts [0,0,1], computes s as the sum of the last three squares, puts r=isqrt(s), and appends r+(s>r*r+r).

The target quantifies over the unique sequence with values in the nonnegative integers on every positive index and over every recurrence index n>=3. It asks for the full infinite consecutive-ratio limit.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/OEISA400644.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresQuestion resolvedSource 2026Perrier's period-one generating functionsPerrier's period-one series satisfy both numerator identities over commutative rings.Proved

The questionWe conjecture that solving these recurrences yields

Why it mattersThis settles the two conjectured generating-function formulas for every dimension in the period-one case.

Our contributionPerrier's period-one series satisfy both numerator identities over commutative rings.

Scope & next questionThe subsequent conjecture concerning the generalized recurrence in equation (7) is outside this result, as are periods greater than one and analytic convergence of the series.

Verification details
Release source recordProved in Lean

Open problem

Perrier's period-one generating functions

Rachel Perrier, Multidimensional Continued Fractions and Riordan Arrays, Enumerative Combinatorics and Applications 6:4 (2026), Section 5, printed pp. 14–15, gives the period-one recurrence with initial values $r_1^{(-1)}=1$ and $r_{j+1}^{(-1)}=a_j$ for $1\le j\le k-1$:

$$ r_1^{(n+1)}=r_k^{(n)},\qquad r_{j+1}^{(n+1)}=r_j^{(n)}+m_jr_k^{(n)}. $$

The source states on printed p. 15:

We conjecture that solving these recurrences yields

$$ R_1(t)=\frac{1+\sum_{j=1}^{k-1}(a_j-m_j)t^{k-j}} {1-\sum_{j=1}^{k-1}m_jt^{k-j}-t^k},\qquad R_k(t)=\frac{t^{k-1}+\sum_{j=1}^{k-1}a_jt^{k-1-j}} {1-\sum_{j=1}^{k-1}m_jt^{k-j}-t^k}. $$

The finite sums reproduce the two printed numerators and their common denominator. The shifted convention is $R_i(t)=\sum_{n\ge0}r_i^{(n-1)}t^n$, explicitly defined on printed pp. 8 and 12. Section 5 introduces higher dimensions with the sentence ‘The same method extends to higher dimensions.’ It does not repeat the definition of the generating functions; retaining that shift is the contextual reading used here.

Set $k=d+1$. Lean time $n$ represents source time $n-1$, coordinate $j:\operatorname{Fin}(d+1)$ represents source coordinate $j+1$, and $m,a:\operatorname{Fin}(d)\to K$ likewise use zero-based indices. For every commutative ring $K$, every natural $d$, every parameter pair, and every sequence satisfying the initial values and both recurrence equations, the formal result is:

theorem result {K : Type*} [CommRing K] {d : ℕ}
    (m a : Fin d → K) (r : ℕ → Fin (d + 1) → K)
    (h : Recurrence m a r) :
    R r 0 * D m = P m a ∧ R r (Fin.last d) * D m = Q a

Here R is the power series with coefficients r; D, P, and Q are exactly the denominator and two numerators above, with constants embedded in $K[[X]]$. All finite sums are empty when $d=0$; the result includes this case. The initial vector and next-vector rule recursively determine a unique sequence for every parameter pair.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Algebraic/PerrierPeriodicGeneratingFunctions.result

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    Year in the primary source attribution: 2026.

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Quantum informationQuestion resolvedFirst preprint 2025The Haar mean of two-qubit non-local nonstabilizerness is 10/3 − πThe Appendix C density has integrable first moment 10/3 minus pi.Proved

The questionWang, Quantum Non-Local Nonstabilizerness, arXiv:2502.06393v4, Phys.

Why it mattersThe mean is the typical amount of magic that no local basis change can remove from a random two-qubit state.

The idea1. The primitive. Put q(t) = t⁴ − t² + 1 and F(t) = −t³ ln q(t) + (4/3)t³ + 2t − 2(arctan(2t + √3) + arctan(2t − √3)).

Our contributionThe Appendix C density has integrable first moment 10/3 minus pi.

Verification details
Release source recordProved in Lean

Open problem

The Haar mean of two-qubit non-local nonstabilizerness is 10/3 − π

D. Qian and J. Wang, Quantum Non-Local Nonstabilizerness, arXiv:2502.06393v4, Phys. Rev. A 111, 052443 (2025), define the non-local nonstabilizerness (NN) of a state as the stabilizer Rényi entropy minimized over local unitaries. For two-qubit pure states it depends only on the Schmidt weight x = cos²θ (Eq. (C1)): M(x) = ln(1/(1 + 4x(x − 1)(1 − 2x)²)). For Haar-random states x has density 3(1 − 2x)², and Appendix C derives the density (C5) of y = M(x) on 0 < y < ln(4/3).

§III states that "a closed-form expression for the average value of M … remains unclear" and reports approximately 0.192. Appendix C writes: "we have not been able to find a simple closed-form expression for the average amount of NN, ȳ ≡ ∫ yP(y)dy, for the Haar random state", and gives ȳ ≈ 0.1917 by numerical integration. The verbatim statements are in the literature note.

Issue #16544 fixes the reading. With P the literal (C5) on (0, ln(4/3)) and 0 elsewhere, y·P(y) is integrable and ȳ = ∫ y P(y) dy = 10/3 − π. It is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/NonlocalNonstabilizernessHaarMean.result

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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedSource 2026Second adjacency eigenvalue of the inner-face Apollonian stagesThe exact all-natural-stage second ordinary-adjacency eigenvalue limit.Proved

The questionRad Sarar Anando's MathOverflow question 516405, revision 1, GUID 8C9D6CD5-E3A3-4B25-BDB6-FF5A9A3D3A2B, asks whether the second largest ordinary adjacency eigenvalue, counted with multiplicity, satisfies

Why it mattersThe dominant old vertices have exponentially growing degree, while their common late neighbours grow only linearly.

The ideaThe proof uses spectral estimates.

Our contributionThe exact all-natural-stage second ordinary-adjacency eigenvalue limit.

Scope & next questionThe source's broader question about other leading eigenvalues remains outside this result.

Verification details
Release source recordProved in Lean

Open problem

Second adjacency eigenvalue of the inner-face Apollonian stages

Rad Sarar Anando's MathOverflow question 516405, revision 1, GUID 8C9D6CD5-E3A3-4B25-BDB6-FF5A9A3D3A2B, asks whether the second largest ordinary adjacency eigenvalue, counted with multiplicity, satisfies

$$\lim_{m\to\infty}\theta_2(G_m)/2^{m/2}=\sqrt2.$$

Here $G_0$ is plane $K_4$ with outer triangle $abc$ and centre $o$. Every step inserts one vertex in each inner triangular face and joins it exactly to the three corners; the outer triangle is never subdivided. The source note attributes the quotation and adapted exposition to the author under CC BY-SA 4.0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Apollonian/SecondEigenvalueLimit.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2026.

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Quantum informationQuestion resolvedFirst preprint 2025Equal multipartite powers of diagonal phase gatesDiagonal phase gates have equal multipartite entangling and disentangling powers.Proved

The questionSen (De), Disparity between multipartite entangling and disentangling powers of unitaries: Even vs Odd, arXiv:2505.18539v1, Appendix B.3, page 8, states:

The ideaIt preserves the squared-modulus sum and transforms every product factor into its conjugate.

Our contributionDiagonal phase gates have equal multipartite entangling and disentangling powers.

Verification details
Release source recordProved in Lean

Open problem

Equal multipartite powers of diagonal phase gates

M. Samanta, S. Mondal and A. Sen (De), Disparity between multipartite entangling and disentangling powers of unitaries: Even vs Odd, arXiv:2505.18539v1, Appendix B.3, page 8, states:

It also indicates that such an equivalence may hold for arbitrary diagonal unitary operators acted on arbitrary number of qubits.

The equivalence is $E_N(U)=D_N(U)$, where Eq. (1) maximizes the generalized geometric measure over fully separable inputs, Eq. (2) defines $D_N(U)=E_N(U^\dagger)$, and Appendix A.1 defines the generalized geometric measure by overlap with non-genuinely multipartite entangled pure states.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Any number of qubits | claim quantifies over every natural N | The physical multipartite interpretation concerns $N\ge2$; the same formal definitions include $N=0,1$. | | Every diagonal unitary | Real phase function θ on Fin N → Fin 2, with diagonal entries $e^{i\theta_x}$ | Every computational-basis phase is independent. | | Fully separable normalized inputs | IsFullySeparable | Squared-modulus sum one and a product of local amplitudes. | | Non-genuinely multipartite entangled pure states | IsBiseparable | Normalization and factorization across a bipartition with both parts nonempty. | | GGM | G | One minus the supremum of squared overlap over the biseparable subtype. | | Entangling and disentangling powers | E, D | Suprema over fully separable inputs, with the adjoint in D. | | Equality | result : claim | No additional hypotheses. |

The normalization expression is inlined; the overlap is Mathlib's dotProduct applied to the conjugated first argument. Maxima are represented by suprema. The argument identifies the entire value families, so it needs no attainment assertion or restriction on input local phases.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/DiagonalUnitary/SamantaMondalSenDiagonalPowers.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Shin's four-point sumset compression questionThe eleven-fold sumset of {0,7,17,80} has 347 elements, a cardinality absent from every four-element integer set in [0,79].Refuted

The questionFor every natural h>=2, put Dh=choose(h+2,2)+1 when h is odd, and Dh=choose(h+2,2) otherwise.

Why it mattersFiniteExhaustion provides the existing setting for complete finite certificate arguments.

The ideaUse h=11 and A={0,7,17,80}, whose repeated sumset has 347 elements.

Our contributionThe eleven-fold sumset of {0,7,17,80} has 347 elements, a cardinality absent from every four-element integer set in [0,79].

Verification details
Release source recordRefuted in Lean

Open problem

Shin's four-point sumset compression question

For every natural h>=2, put D_h=choose(h+2,2)+1 when h is odd, and D_h=choose(h+2,2) otherwise. Must every four-element integer set A have a four-element integer set B in [0,D_h] with |hB|=|hA|? Here hA is the set of sums of exactly h elements of A with repetitions allowed. This is the question after Corollary 10.5 and equation (193) in arXiv:2609.01690v1, equivalently nu(h,4)=D_h for all h>=2. It is restated as open in arXiv:2609.08915v1, Question 12.5.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/ShinFourPointCompression.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedThe higher exterior traces of Spezia’s serpentine matrixAll exterior traces above order two of the serpentine matrix vanish.Proved

The questionStefano Spezia, OEIS A322844, COMMENTS revision 68, states:

Why it mattersThe two conjectures ask for uniform identities across every matrix order.

The ideaThe first result therefore proves the absolute-value formula.

Our contributionAll exterior traces above order two of the serpentine matrix vanish.

Scope & next questionThe result changes their status in the characteristic-coefficient formulation; it makes no claim about unrelated Spezia matrix families or a new general rectangular-product identity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The higher exterior traces of Spezia’s serpentine matrix

Stefano Spezia, OEIS A322844, COMMENTS revision 68, states:

For k > 2, the trace of the k-th exterior power of the matrix M(n) is equal to zero.

The matrix definition is shared with A317614 revision 194. With one-based indices $1\leq i,j\leq n$, its integer entries are $M_{ij}=j+n(i-1)$ for odd $i$ and $M_{ij}=ni-j+1$ for even $i$. Put $\varepsilon=n\bmod 2$ and

$$ a(n)=\frac{n^2}{12}\bigl(3(1+n^2)-2(2+n^2)\varepsilon\bigr). $$

Statement echo

| Source clause or convention | Lean counterpart | Fidelity | | --- | --- | --- | | One-based integer serpentine entries | SpeziaSerpentineSecondExteriorTrace.M | Fin n indices are increased by one; entry arithmetic is integer arithmetic. | | Source rational expression for $a(n)$ | SpeziaSerpentineSecondExteriorTrace.a | Rational division and natural remainder; no truncated rational quotient. | | Second exterior trace, characteristic coefficient, principal-minor sum | SpeziaSerpentineSecondExteriorTrace.traceExt | The binding convention uses $(-1)^k[x^{n-k}]\chi_M$ for $k\leq n$ and zero for $k>n$. It does not construct an exterior-power linear map. | | First conjecture, every $n>1$ | SpeziaSerpentineSecondExteriorTrace.claim, result | Integer absolute value, cast to the rationals, equals $a(n)$. | | Second conjecture, every $n$ and $k>2$ | SpeziaSerpentineHigherExteriorTrace.claim, result | Includes $n<k$, including the empty matrix, using the zero exterior-space convention. | | Centered coordinates and two rectangular factors | w, u, U, V in the second-trace module | Auxiliary constructions, not additional hypotheses on either claim. |

All declaration names in this table have prefix D5.S3.Quantum.Matrix.. The source calls the two coefficient/minor descriptions equivalent views of the first conjecture. No additional source conjecture is counted from those equivalent descriptions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/SpeziaSerpentineHigherExteriorTrace.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Sra's symmetric alpha-permanent questionSymmetry of an inverse M matrix does not force log-supermodularity of every positive-order alpha-permanent.Refuted

The questionSuvrit Sra, A Koteljanskii inequality for permanents, arXiv:2609.39979v1, Section 4(b), asks:

Why it mattersThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

The ideaThe proof uses factorization.

Our contributionSymmetry of an inverse M matrix does not force log-supermodularity of every positive-order alpha-permanent.

Verification details
Release source recordRefuted in Lean

Open problem

Sra's symmetric alpha-permanent question

Suvrit Sra, A Koteljanskii inequality for permanents, arXiv:2609.39979v1, Section 4(b), asks:

For symmetrizable inverse $M$-matrices and $\alpha\notin{\tfrac12,1}$, decide whether $S\mapsto\operatorname{per}_\alpha(A_S)$ is log-supermodular.

The order is positive as in equation (3.1). The universal affirmative assertion quantifies over $n\ge1$, every real symmetrizable inverse $M$-matrix, every permitted positive order, and every pair of index subsets. The source definitions are quoted in the Library note.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Every $n\ge1$, real inverse $M$-matrix | claim, imported InverseMMatrix | Exactly the source's nonnegativity, invertibility and inverse off-diagonal sign conditions. | | Positive diagonal similarity to symmetry | Symmetrizable | The diagonal function is strictly positive; its pointwise reciprocal gives the inverse diagonal. | | Equation (3.1), all cycles including fixed points | alphaPermanent | cycleType.card counts nontrivial cycles, and the fixed-point cardinality adds one-cycles. | | Principal restriction, including the empty case | principalAlphaPermanent | Matrix.submatrix on the index subtype; empty product and permutation sum give one. | | $\alpha>0$, $\alpha\ne1/2,1$ and every $S,T$ | claim | All three order conditions and both subset binders are retained. | | Log-supermodular inequality | claim | The $S,T$ product is at most the union-intersection product. | | Universal affirmative assertion is false | result | Literal negation of claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/SraSymmetricAlphaPermanentRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Bound-coherent states exist in every dimension at least threeEvery dimension at least three admits a coherent, nonmaximal density matrix whose positive modes cannot increase under an allowed interaction of two identical copies.Proved

The questionSkrzypczyk, Unspeakable coherence concentration, arXiv:2512.04255v1 (2025), published as Phys.

Why it mattersThe question distinguishes having coherence from being able to concentrate it through a number-conserving interaction of two copies.

The ideaThese facts supply the existential state in result : claim.

Our contributionEvery dimension at least three admits a coherent, nonmaximal density matrix whose positive modes cannot increase under an allowed interaction of two identical copies.

Verification details
Release source recordProved in Lean

Open problem

Bound-coherent states exist in every dimension at least three

B. Stratton, C.-Y. Hsieh and P. Skrzypczyk, Unspeakable coherence concentration, arXiv:2512.04255v1 (2025), published as Phys. Rev. A 113, 062429 (2026), Section IX, Discussion, main.tex line 387:

Finally, a family of states with bound-coherence was shown to exist, i.e., states that are not maximally coherent, nor incoherent, but for which no allowed unitary exists to perform coherence concentration. We leave it open to determine if such states exist in higher dimensions.

Issue #22599 fixes the reading for two identical copies with the truncated number observable $L=\sum_{n=0}^{d-1}n|n\rangle\langle n|$. For every integer $d\ge3$, there is a positive semidefinite matrix $\rho$ of trace one with $[\rho,L]\ne0$, which is not an equal-amplitude pure state, such that for every unitary $V$ commuting with $L\otimes I+I\otimes L$,

$M^{(j)}\bigl(\operatorname{tr}_B[V(\rho\otimes\rho)V^\dagger]\bigr)\le M^{(j)}(\rho)\quad(1\le j\le d-1).$

Here $\rho^{(j)}$ retains the entries with row number minus column number equal to $j$, and $M^{(j)}(\rho)=|\rho^{(j)}|_1$. An equal-amplitude pure state means $|\psi\rangle\langle\psi|$ with $\psi_n=e^{i\theta_n}/\sqrt d$ for real phases $\theta_n$. This reading is proved. The source definitions and their locators are in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/BoundCoherenceHigherDimensions.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2026Boundedness of the Damascus violation setsThe positive product-one Damascus violation sets have a complete boundedness classification.Proved

The questionChanatip Sujsuntinukul and Christophe Chesneau, Variants of the Damascus inequality, arXiv:2601.00916v4, Section 5, page 19, ask:

Why it mattersThe delivery has one general module, four public definitions and one public theorem.

The ideaThe total is negative, contradicting violation.

Our contributionThe positive product-one Damascus violation sets have a complete boundedness classification.

Scope & next questionThe source concludes: “Thus this question, together with question (iii), remains open.” This dossier settles question (ii).

Verification details
Release source recordProved in Lean

Open problem

Boundedness of the Damascus violation sets

Chanatip Sujsuntinukul and Christophe Chesneau, Variants of the Damascus inequality, arXiv:2601.00916v4, Section 5, page 19, ask:

In addition to the above, it is natural to ask whether analogous structural properties hold for I_n^m when m is arbitrary. Can one prove or disprove that I_n^m (i) stays uniformly away from 1, (ii) is bounded, or (iii) forms an increasing family in n?

The source concludes: “Thus this question, together with question (iii), remains open.” This dossier settles question (ii).

Write $f_n(x)=(x^n-1)/(x^{n+1}+1)$, $H_m={x\in(0,\infty)^m:\prod_jx_j=1}$ and $I_n^m={x\in H_m:\sum_j f_n(x_j)>0}$. For all natural numbers $m,n\ge1$, the result is

$$I_n^m\text{ is bounded}\quad\Longleftrightarrow\quad m\le3\ \lor\ (m=4\land n\le4)\ \lor\ (m\in{5,6}\land n=1).$$

The classification is the repository's answer, rather than a proposition quoted from the paper.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Summand of inequality (1), page 2 | term | Literal real quotient $(x^n-1)/(x^{n+1}+1)$; no transformed predicate. | | $H_m$ consists of positive coordinates with product one | H | Functions Fin m → ℝ; every coordinate is positive and occurs once in the product. | | $I_n^m\subset H_m$ has strictly positive summand sum | I | Membership in H m and a strict inequality $0<\sum_j f_n(x_j)$. | | Arbitrary dimension and positive integer exponent | claim, binders $m,n$ and guards $1\le m$, $1\le n$ | Both parameters vary without upper bounds. Dimensions one and two are included; dimension zero and exponent zero are outside the question. | | Boundedness | Bornology.IsBounded (I n m) | The usual sup norm on a finite real coordinate space. | | Answer to question (ii) | result : claim | Exactly the displayed three-branch classification, in both directions. The classification branches express the answer; they are not additional hypotheses. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/DamascusViolationSetBoundedness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedComposite prime divisors of the gcd determinantPrime divisors of composite gcd determinants are smaller than the index.Proved

Why it mattersThe result settles the composite-index prime-support assertion for the source's gcd determinant while retaining the source's separate positivity and divisibility assertions as distinct questions.

The ideaThe gcd Fourier transform gives an integer eigenvalue product on each prime-power component.

Our contributionPrime divisors of composite gcd determinants are smaller than the index.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Composite prime divisors of the gcd determinant

Zhi-Wei Sun's OEIS A228884, COMMENTS clause (ii), states: “For any composite number $n$, all prime divisors of $a(n)$ are smaller than $n$.” The source defines $a(n)$ as the determinant of the $n\times n$ matrix whose $(i,j)$ entry is $gcd(i-j,n)$ for $1\leq i,j\leq n$.

The formal claim uses the literal integer-entry matrix C, its determinant D, the hypotheses $4\leq n$ and $\neg\operatorname{Nat.Prime}(n)$, and integer divisibility by a natural prime $q$.

Statement echo

The settling declaration is D5/S3/Arith/Divisors/SunA228884.result : D5/S3/Arith/Divisors/SunA228884.claim. Its Scribe mirror expands the complete proposition, including every binder and the integer cast in the divisibility clause.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Divisors/SunA228884.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Further frontiersQuestion resolvedUlgenes's truncated-binomial Hadamard limitThe truncated binomial limit equals the entire Hadamard factorial function at every complex parameter.Proved

The questionConjecture: H(x+1) = lim{n->oo} n^{x+1} Sum{k=0..n} (-1)^k/(n+k) binomial(x, k) such that H(1/4) = 0.79564036... .

Why it mattersThe settlement concerns a moving finite cutoff, distinct from the infinite Euler-beta sum.

The ideaEuler's Gamma-sequence limit gives convergence of the scaled generalized-binomial coefficient.

Our contributionThe truncated binomial limit equals the entire Hadamard factorial function at every complex parameter.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Ulgenes's truncated-binomial Hadamard limit

OEIS A398064, revision 39, FORMULA:

Conjecture: H(x+1) = lim_{n->oo} n^{x+1} * Sum_{k=0..n} (-1)^k/(n+k) * binomial(x, k) such that H(1/4) = 0.79564036... . Changing the upper bound of the sum to oo yields Euler's product representation of the Gamma function.

The binding interpretation in #22706 is over every $x\in\mathbb C$. Put

$$S_n(x)=\exp((x+1)\log n)\sum_{k=0}^n\frac{(-1)^k}{n+k}\binom{x}{k}.$$

The logarithm is real for positive $n$. The entire source function satisfies $H(m+1)=m!$ and, away from positive integers,

$$H(y)=\frac{\psi(1-y/2)-\psi((1-y)/2)}{2\Gamma(1-y)}.$$

The assertion includes existence of such an entire function and convergence $S_n(x)\to H(x+1)$ for every such function and every complex parameter. No existence assumption is substituted for this conclusion.

Statement echo

| Source or binding clause | Lean counterpart | | --- | --- | | The inclusive finite sum, both endpoints retained | truncation | | Generalized binomial coefficients | Ring.choose in truncation | | Positive-integer power with real logarithm | Complex.exp and Real.log in truncation | | Entire factorial interpolation and the specified quotient | IsHadamard | | Existence and convergence for every admissible function, every complex parameter | claim, proved by result | | Quoted decimal example | Not an accuracy assertion of claim; the exact quarter-value identity is computed below | | Separate infinite-sum sentence | Outside claim; its omitted-tail limit remains open as a separate Lean result |

All local names below belong to D5.S3.Weil.ZetaGamma.HadamardTruncatedBinomialLimit; intermediate lemmas are private. The definitions retain the preregistered expressions. The all-complex domain and existence conjunct specify the source's otherwise unstated conventions without restricting the parameter domain.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Weil/ZetaGamma/HadamardTruncatedBinomialLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The quantum value of balanced hypercube rendezvous is strictly below its no-signalling valueFor t ≥ 2 and d > 2t, the quantum supremum is strictly below the no-signalling supremum in balanced hypercube rendezvous.Proved

The questionLemma 5.1 gives the no-signalling value ωNS(d,t) = C(2t,t)/C(d,t).

Why it mattersNon-attainment by every finite-dimensional strategy does not separate the supremum from the no-signalling value: quantum correlations could approach the no-signalling optimum as the dimension grows.

The idea1. Vectors. On operators of the joint system the form ⟨X, Y⟩ = Re Tr(X Y ρ) is positive semidefinite.

Our contributionFor t ≥ 2 and d > 2t, the quantum supremum is strictly below the no-signalling supremum in balanced hypercube rendezvous.

Scope & next questionTheorems 5.5 and 6.4 show that for t ≥ 2 and d > 2t no finite-dimensional quantum strategy attains it. §7 asks: "It remains open whether ωQ(d,t) is then strictly below ωNS(d,t)", where ωQ is the supremum over finite-dimensional quantum strategies (Eq. (2.8)).

Verification details
Release source recordProved in Lean

arXiv

The quantum value of balanced hypercube rendezvous is strictly below its no-signalling value

C. Wang, Quantum-assisted rendezvous on hypercubes: a quantum Bernoulli noise approach, arXiv:2610.12949v1 (9 October 2026), studies one-step rendezvous on the hypercube {0,1}^d.

  • Inputs: Alice and Bob receive vertices σ, τ with |σ △ τ| = 2t, uniformly.
  • Outputs: each moves t coordinates, Alice A and Bob B, both t-subsets.
  • Winning: they meet, σ △ A = τ △ B, that is A ∩ B = ∅ and A ∪ B = σ △ τ.

Lemma 5.1 gives the no-signalling value ω_NS(d,t) = C(2t,t)/C(d,t). Theorems 5.5 and 6.4 show that for t ≥ 2 and d > 2t no finite-dimensional quantum strategy attains it. §7 asks: "It remains open whether ω_Q(d,t) is then strictly below ω_NS(d,t)", where ω_Q is the supremum over finite-dimensional quantum strategies (Eq. (2.8)). The verbatim statements are in the literature note.

Issue #15814 fixes the reading:

  • arbitrary finite-dimensional systems, density matrices and POVMs;
  • no-signalling correlations with input-independent marginals;
  • both values as suprema;
  • the claim ω_Q(d,t) < ω_NS(d,t) for all t ≥ 2 and d > 2t.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/HypercubeRendezvousQuantumGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    Existing proof account records a substantive content argument.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedFirst preprint 2025Wu-Lou permutation-invariant Niven digit-sum boundWu and Lou's permutation-invariant decimal Niven digit-sum bound.Proved

The questionWu and Lou, Permutation-Invariant Niven Numbers, arXiv:2508.01611v3, section 8.4, state that the relevant nontrivial distinct-digit permutation-invariant Niven numbers have digit sums between 3 and 81 and conjecture that no exceptions exist beyond the cases excluded there.

Why it mattersThe formal target is the universal bound for the same decimal permutation-invariant object, with the hypotheses excluding the two elementary families named in the source sentence.

The ideaThe divisibility-by-three assertion is literature-attested; the lower bound is its positive immediate consequence; the upper bound is the repository-derived settlement of the conjectural part.

Our contributionWu and Lou's permutation-invariant decimal Niven digit-sum bound.

Verification details
Release source recordProved in Lean

Open problem

Wu-Lou permutation-invariant Niven digit-sum bound

Wu and Lou, Permutation-Invariant Niven Numbers, arXiv:2508.01611v3, section 8.4, state that the relevant nontrivial distinct-digit permutation-invariant Niven numbers have digit sums between 3 and 81 and conjecture that no exceptions exist beyond the cases excluded there.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/WuLouPermutationInvariantNivenDigitSum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationCounterexample foundSource 2025Zhang–Jing Theorem 2.3: the upper power intervalThe power block characterizes normality for 1/2 ≤ a < 1, including singular matrices; at a = 1 it is positive semidefinite for every matrix.Refuted

The questionFeng Zhang and Hefang Jing, More inequalities for positive semidefinite matrices, Electronic Journal of Linear Algebra 41 (2025), 595–601, state after Remark 2.13:

The ideaIt transports the block by unitary coordinates to diagonal powers and a unitary conjugate.

Our contributionThe power block characterizes normality for 1/2 ≤ a < 1, including singular matrices; at a = 1 it is positive semidefinite for every matrix.

Verification details
Release source recordRefuted in Lean

Open problem

Zhang–Jing Theorem 2.3: the upper power interval

Feng Zhang and Hefang Jing, More inequalities for positive semidefinite matrices, Electronic Journal of Linear Algebra 41 (2025), 595–601, state after Remark 2.13:

“Due to the limitations of our research method, we cannot solve the problem that Theorem 2.3 holds for $1/2 \le a \le 1$.”

Their Theorem 2.3 states:

“Let $A \in M_n$. Then $[[|A|^{2a}, A^*], [A, |A|^{2(1-a)}]] \ge 0$, for $0 \le a \le 1/2$ if and only if $A$ is normal.”

The notation is “$|A| = (A^A)^{1/2}$ and $A^$ is the conjugate transpose of $A$” (p. 595). The delivered claim extends the displayed equivalence to every real $a$ with $1/2 \le a \le 1$, every natural dimension, and every complex square matrix, using the existing block construction with $|A|=\operatorname{CFC.sqrt}(A^*A)$ and real powers.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $|A|=(A^A)^{1/2}$ and $A^$ is conjugate transpose | The reused ZhangJingBlockNormalityRefutation.block owner | The source expression is preserved through the frozen owner; no duplicate absolute-value definition is introduced. | | $1/2\le a\le1$ | claim quantifies a, with both inequalities | Both endpoints are included. | | Every n and $A\in M_n$ | claim quantifies n and A : Matrix (Fin n) (Fin n) Complex | The full finite-dimensional family is retained, including n=0. | | The block is positive semidefinite iff A is normal | claim uses Matrix.PosSemidef and $A^A=A A^$ | Every block and the normality equality are retained. | | $1/2\le a<1$ survives | The owning Scribe displays the full strict-interval quantified application | Both directions, singular matrices, the half endpoint and dimension zero are retained as an application of canonical suppliers. | | The closed interval fails | result : \lnot claim | The conclusion is the literal refutation. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/ZhangJingPowerCaseSettlement.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Further frontiersQuestion resolvedPosed in 2026The exact halting time of Wolfram machine 3626The signed Moebius and totient divisor sum counts every transition of the one-sided machine on every positive binary input.Proved

The questionPeter Bala, OEIS A393210, conjecture of September 11, 2026, revision 18:

Why it mattersOperational execution is defined independently of the arithmetic formula.

The ideaA block of k zero bits followed by one and arbitrary higher cells admits a restoring trace.

Our contributionThe signed Moebius and totient divisor sum counts every transition of the one-sided machine on every positive binary input.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The exact halting time of Wolfram machine 3626

Peter Bala, OEIS A393210, conjecture of September 11, 2026, revision 18:

Conjecture: a(n) = Sum_{d | n} m(d)^2 * phi(d), where m(n) = Sum_{d | n} (-1)^(d+1) * moebius(d) = A069517(n) and phi(n) denotes Euler's totient function A000010(n). Cf. A296306.

The sequence has offset 1. The equivalent conjectural relation in A069517, revision 31, is the same question, counted once. Here a(n) is the actual number of transitions of the one-sided two-state two-symbol machine 3626 on every positive natural input n in binary. Its half-tape extends infinitely leftward, blank zero beyond the most significant bit. The head starts in state A at the right edge on the least significant bit. Its rules are A0 → A0L, A1 → B1R, B0 → A1L, B1 → B0R. A simultaneous read, write, state change and move counts as one transition. Moving right off the edge terminates the run, and that final transition is counted. Input zero is excluded. There is no time cap, tape bound or counter width. Both divisor sums range over positive divisors; the signed Moebius arithmetic is in the integers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Computability/BinaryMachine/Machine3626.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026A SUNTF_R(2, 5, 3) existsAn explicit real SUNTF with parameters (D,N,R) = (2,5,3).Proved

The questionMixon, Incoherent filter banks, arXiv:2610.05302v1, Section 4, Example 4.4, asks whether a SUNTFR(2, 5, 3) exists.

Why it mattersThe present result settles the concrete real instance by an explicit exact construction.

The ideaThe three layers have columns (1,0), (u, s{r,1}v), (u, s{r,2}v), (u, s{r,3}v), (u, s{r,4}v), with sign words (+,−,−,+), (+,−,+,−), and (+,+,−,−).

Our contributionAn explicit real SUNTF with parameters (D,N,R) = (2,5,3).

Verification details
Release source recordProved in Lean

Open problem

A SUNTF_R(2, 5, 3) exists

W. J. Brinkley, M. Fickus, J. Jasper and D. G. Mixon, Incoherent filter banks, arXiv:2610.05302v1, Section 4, Example 4.4, asks whether a SUNTF_R(2, 5, 3) exists. The last paragraph states:

We do not yet know whether a SUNTF_R(2, 5, 3) exists, and we leave this for future research.

Definition 3.2 requires each of the R layers to be a unit norm tight frame in dimension D and requires the averaged squared inner products between every two distinct columns to be constant.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/Designs/StratifiedTightFrameTwoFiveThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresExact optimum establishedFirst preprint 2026Optimal Cycle–Clique Unions Without Regular Induced Subgraphs of Prime OrderThe cycle-clique unions of Dyson and McKay attain the largest possible order without a regular induced subgraph of prime order p for every prime p at least thirteen.Proved

The questionMcKay, Ramsey numbers for regular induced subgraphs, arXiv:2604.08215v3, Section 4: after Theorem 4.2 the authors write that they believe the graphs Gp of Theorem 4.2 are optimal for p ≥ 13 within the class of disjoint unions of lexicographic products of cycles and cliques.

The ideaThe connected induced regular subgraphs of Cr[Ks] for r ≥ 4 are cliques of order at most 2s and subgraphs meeting every bag whose bag counts are periodic of period three; this gives the exact order spectrum of each component at each degree.

Our contributionThe cycle-clique unions of Dyson and McKay attain the largest possible order without a regular induced subgraph of prime order p for every prime p at least thirteen.

Verification details
Release source recordProved in Lean

Open problem

Optimal Cycle–Clique Unions Without Regular Induced Subgraphs of Prime Order

Paul W. Dyson and Brendan D. McKay, Ramsey numbers for regular induced subgraphs, arXiv:2604.08215v3, Section 4: after Theorem 4.2 the authors write that they believe the graphs G_p of Theorem 4.2 are optimal for p ≥ 13 within the class of disjoint unions of lexicographic products of cycles and cliques. Precisely: for every prime p ≥ 13, every disjoint union of products C_r[K_s] (r ≥ 3, s ≥ 1) with no induced regular subgraph on exactly p vertices has at most 9(p−1)²/8 vertices when p ≡ 1, 5 (mod 12), (p−1)(9p−7)/8 when p ≡ 7 (mod 12) and (p−1)(9p−11)/8 when p ≡ 11 (mod 12), and G_p attains the bound.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/RegularInduced/DysonMcKay.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The cycle-clique unions of Dyson and McKay attain the largest possible order without a regular induced subgraph of prime order p for every prime p at least thirteen.

    Read source ↗
  • Historical context

    First preprint year: 2026.

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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The Order Gadget of the Erdős–Graham Two-Set Problem Is Infeasible at Every ScaleEvery order gadget on the integers from M plus one through twice M fails once M is at least sixteen: arithmetic-progression constraints propagate two neighbouring centres around ladders whose guard endpoints force both possible rank orders.Proved

The questionWilliam Kasel, Structural rigidity in the Erdős–Graham two-set permutation problem, arXiv:2609.02939v1, Remark 30:

Why it mattersThe paper needed only the residue class M ≡ 0 (mod 8), which contains the dyadic scales, to eliminate the canonical partition for Erdős Problem 197.

The ideaAP-freeness gives the four midpoint rules of the paper, from which follow its zigzag lemma (on any ladder of a fixed difference, one parity class of rungs precedes all its neighbours), phase dichotomy, and the phase-independent flood: for a center q and a class C of step g ∈ {2, 4} with q ≡ C + g/2 (mod g), one relation q ≺ v with v in the admissible mirror range forces q ≺ w for every class member w in that range.

Our contributionEvery order gadget on the integers from M plus one through twice M fails once M is at least sixteen: arithmetic-progression constraints propagate two neighbouring centres around ladders whose guard endpoints force both possible rank orders.

Verification details
Release source recordProved in Lean

arXiv

The Order Gadget of the Erdős–Graham Two-Set Problem Is Infeasible at Every Scale

William Kasel, Structural rigidity in the Erdős–Graham two-set permutation problem, arXiv:2609.02939v1, Remark 30:

The full order gadget is nevertheless conjectured infeasible at every M ≥ 16 (machine-verified for 16 ≤ M ≤ 200 and M = 512, Proposition 18); at every swept scale with M ≢ 0 (mod 8) the C3 core was satisfiable, and other attack subsets were observed computationally to take over (Proposition 19); a proof beyond the swept scales would need per-residue analogues of Theorems 27–28 built from the same flood toolkit.

The order gadget OG(M) of Definition 15 asks for a linear order of (M, 2M] in which no in-block three-term arithmetic progression is monotone and, for x ∈ {15, 16} and 1 ≤ j ≤ x/2, the guard 2M + 2j − x precedes the bottom M + j.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ErdosGrahamOrderGadget.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Once-subdivided complete graphs have infinite edge multiset dimensionOnce-subdivided complete graphs have no edge multiset resolving setProved

The questionProblem 24, section 7.4, printed page 27 of the versioned source asks whether the edge multiset dimension of Kn^1 is infinity for every integer n at least three.

The ideaExact shortest-distance histograms force distinct augmented selected degrees for any hypothetical resolving set.

Our contributionOnce-subdivided complete graphs have no edge multiset resolving set

Verification details
Release source recordProved in Lean

arXiv

Once-subdivided complete graphs have infinite edge multiset dimension

Problem 24, section 7.4, printed page 27 of the versioned source asks whether the edge multiset dimension of K_n^1 is infinity for every integer n at least three. The definition on printed page 21 inserts one midpoint into every complete-graph edge. Section 6.1 on printed page 20 uses minimum actual shortest endpoint distances and arbitrary landmark subsets of the full vertex set, including originals and midpoints. Profiles are multisets with multiplicity; infinity means that no resolving subset exists.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Walks/CompleteSubdivisionResolution.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2026The diploid cellular automaton 10-245 has exact density 1/4 + λ/2For every mixing parameter in [0,1], diploid rule 10-245 has density 1/4 + λ/2 at every time from one onwards, starting from the fair Bernoulli cylinder law.Proved

The questionStarting from the symmetric Bernoulli measure, dt = Pr(si(t) = 1) is the density and d∞ = lim dt (§2).

Why it mattersThe paper sorts the 168 endogamous diploid rules into six classes by the shape of d∞(λ) and tests how far local structure theory predicts them.

The idea2. Two moments. Let dt be the density and qt the probability of a 0 followed by a 1 at distance two.

Our contributionFor every mixing parameter in [0,1], diploid rule 10-245 has density 1/4 + λ/2 at every time from one onwards, starting from the fair Bernoulli cylinder law.

Verification details
Release source recordProved in Lean

Open problem

The diploid cellular automaton 10-245 has exact density 1/4 + λ/2

H. Fukś and N. Fatès, Analysis of the asymptotic density of endogamous diploid cellular automata, arXiv:2607.01917v1 (EPJ ST 2026), study probabilistic cellular automata on ℤ in which every cell independently applies elementary rule f with probability 1 − λ and rule g with probability λ: w(1|x₁x₂x₃) = (1−λ) f(x₁,x₂,x₃) + λ g(x₁,x₂,x₃) (Eq. (2)). Starting from the symmetric Bernoulli measure, d_t = Pr(s_i(t) = 1) is the density and d∞ = lim d_t (§2). For rule 10-245, third-order local structure approximation predicts d∞ = 1/4 + λ/2 (§3.2), while mean-field and second-order approximation give a nonlinear curve. §5 asks: "for rule 10-245, the expression for the asymptotic density, d = 1/4 + λ/2, appears to be exact. Could this be formally demonstrated?" The verbatim statements are in the literature note.

Issue #16027 fixes the reading: for every λ ∈ [0, 1], with the lattice process encoded by its cylinder probabilities (independent simultaneous updates of each interval from its n + 2 parents, fair Bernoulli start), d_t → 1/4 + λ/2. The settlement proves the stronger d_t = 1/4 + λ/2 for every t ≥ 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/Diploid10245Density.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsConjecture provedFirst preprint 2025Gürses and Pekcan's four-soliton conjecture for higher order Hirota formsFor the Hirota bilinear equation Dx(Dx^3 + a1 Dt + a2 Dy)^(2k+1){f.f} = 0, the four-soliton condition holds identically on the dispersion relation when k = 0 and fails at the wave numbers 1, 3, 4, 5 for every k at least one.Proved

The questionGürses and Pekcan study the Hirota bilinear equations

Why it mattersThe k = 0 equation is a Kadomtsev–Petviashvili-type model of shallow-water waves; the result says that raising its dispersive operator to an odd power 2k + 1 ≥ 3 keeps three-soliton solutions but destroys Hietarinta's four-soliton test.

The ideaOn the dispersion relation α₁ωi + α₂li = −ki³, and the map (k, ω, l) ↦ (k, α₁ω + α₂l) is additive, so every value of P at a signed sum of parameters equals x(x³ + e)^m with x the signed sum of the wave numbers and e minus the signed sum of their cubes.

Our contributionFor the Hirota bilinear equation Dx(Dx^3 + a1 Dt + a2 Dy)^(2k+1){f.f} = 0, the four-soliton condition holds identically on the dispersion relation when k = 0 and fails at the wave numbers 1, 3, 4, 5 for every k at least one.

Verification details
Release source recordProved in Lean

Open problem

Gürses and Pekcan's four-soliton conjecture for higher order Hirota forms

Gürses and Pekcan study the Hirota bilinear equations

D_x(D_x³+α₁D_t+α₂D_y)^{2k+1}{f·f}=0, k=0,1,2,…

For soliton parameters p_i = (k_i, ω_i, l_i) the operator has the polynomial P(p) = k(k³+α₁ω+α₂l)^{2k+1}, and the dispersion relation is ω_i = −(k_i³+α₂l_i)/α₁. The paper proves three-soliton solutions for every k and (α₁,α₂) ≠ (0,0), records the four-soliton condition for k = 1, and states, after "we conjecture that the following lemma is valid for all k":

The equation (bilinearProb2) has four-soliton solution only for k=0. For k≥1, it does not satisfy the four-soliton solution condition (4SC) directly.

Issue #10059 fixes the readings: the dispersion relation is k³ + α₁ω + α₂l = 0, which also covers α₁ = 0; the four-soliton condition is the eight-term display of the source's introduction with the last factor of the fifth term read as P(p₁+p₂+p₃−p₄) (the source prints P(p₁−p₂+p₃−p₄), the last factor of the seventh term, and with that factor the condition fails already for k = 0); "has four-soliton solution for k = 0" is read as the condition vanishing at all parameters satisfying the dispersion relation, and "does not satisfy (4SC) directly" as the existence, for each k ≥ 1, of such parameters at which it does not vanish.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/Solitons/GursesPekcanFourSoliton.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026The Number of Distinct Terms of the q-Derivatives of y₀ in the Grammar G_SecThe number of distinct terms in every positive iterate of the Sec q-derivative is the formula of Conjecture III.7.Proved

The questionJi and Huan Xiong, q-Derivative Grammar, arXiv:2604.23959v2, Appendix III, Conjecture III.7.

The ideaEvery coefficient of Dⁿ(y₀) is a nonzero polynomial in q with nonnegative coefficients, so no cancellation occurs and the support of the next derivative is the set of normalized successors of the support.

Our contributionThe number of distinct terms in every positive iterate of the Sec q-derivative is the formula of Conjecture III.7.

Verification details
Release source recordProved in Lean

Open problem

The Number of Distinct Terms of the q-Derivatives of y₀ in the Grammar G_Sec

Guo-Niu Han, Kathy Q. Ji and Huan Xiong, q-Derivative Grammar, arXiv:2604.23959v2, Appendix III, Conjecture III.7. The grammar G_Sec has rules x_j ↦ q^j(1 + x_j x_{j+1}) and y_j ↦ q^j y_j x_{j+1}, and its order sorts the variables of a word by descending index, x before y at equal index. The q-derivative is D(w₁⋯wₙ) = Σ_j ρ(w₁⋯w_{j−1} R(w_j) ↑(w_{j+1}⋯wₙ)), where ↑ raises every index by one, extended linearly. With Ω(E) the number of distinct words of E with nonzero coefficient, the conjecture states Ω(Dⁿ(y₀)) = 1 for n = 1, 3 for n = 2, (20k³ + 33k² + k − 6)/6 for n = 2k + 1 with k ≥ 1, and (20k − 17)(k + 1)k/6 for n = 2k with k ≥ 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/QGrammar/SecGrammar.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's second congruence for A365771Conjectures: given A033042 is the sums of distinct powers of 5, then (2) a(5A033042(n) + 1) == 2 (mod 5) for n > 0,Proved

The questionConjectures: given A033042 is the sums of distinct powers of 5, then (2) a(5A033042(n) + 1) == 2 (mod 5) for n > 0,

Why it mattersThis is one of the three residue assertions in A365771; the family dossier is hanna365771-mod5.md, and the source note is Library/Digit/hanna2023a365771.md.

The ideaThe same-delivery HannaA365771.digitsstep and HannaA365771.residueclassification supply the digit decomposition and the residue.

Our contributionConjectures: given A033042 is the sums of distinct powers of 5, then (2) a(5A033042(n) + 1) == 2 (mod 5) for n > 0,

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's second congruence for A365771

OEIS A365771 revision 23, Comments, states:

Conjectures: given A033042 is the sums of distinct powers of 5, then (2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,

Preregistration: #18657.

Statement echo

Every positive $m$ in A033042 satisfies $a(5m+1)\bmod5=2$. The sequence and digit predicate are exactly D5.S1.Recurrence.Residue.HannaA365771.a and D5.S1.Recurrence.Residue.HannaA365771.InA033042. The claim uses natural-number arithmetic.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA365771Two.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Linear spin measurements never violate CHSH for any spin $s\ge1$Linear spin components obey the CHSH bound for every spin at least one.Proved

The questionLoubenets, Nonviolation of the CHSH inequality under local spin-1 measurements on two spin qutrits, Phys.

Why it mattersThe question separates entanglement from the correlations accessible through linear spin projections.

The ideaThe reused LandauCommutingCollapse.chshsquareeqfouroflocalpaircommutes requires finite index types with decidable equality, four Hermitian involutions, and commutation of at least one local pair; the slot lemmas supply these hypotheses on the finite subset index types.

Our contributionLinear spin components obey the CHSH bound for every spin at least one.

Verification details
Release source recordProved in Lean

Open problem

Linear spin measurements never violate CHSH for any spin $s\ge1$

L. Hanotel and E. R. Loubenets, Nonviolation of the CHSH inequality under local spin-1 measurements on two spin qutrits, Phys. Scr. 100, 055115 (2025), arXiv:2503.05172v2, Introduction, Final_Revision_Hanotel_Loubenets.tex, line 201:

However, up to the moment the problem whether or not the CHSH inequality is violated under Alice and Bob high spin $s\geq1$ measurements is still open -- though specifically this Bell inequality has been used for proving \cite{1, Lou.Kuz.Han:24} nonlocality of every pure nonseparable two-qudit state.

The same source's Conjecture section, lines 913–914, states:

Under local Alice and Bob spin-$1$ measurements in an arbitrary nonseparable pure two-qutrit state, the CHSH inequality is not violated.

For spin $s$, the measurements are the linear projections $S_u=u\cdot S$ for unit $u\in\mathbb R^3$. The companion paper, E. R. Loubenets and L. Hanotel, High-spin measurements in an arbitrary two-qudit state, J. Phys. A: Math. Theor. 58, 265304 (2025), arXiv:2412.03470v2, Revision_Loubenets_Hanotel.tex, lines 210–237 and 253–261, specifies the spin matrices and the CHSH operator $\mathcal B=S_a\otimes S_b+S_a\otimes S_d+S_c\otimes S_b-S_c\otimes S_d$. Its lines 348–363 give the local-hidden-variable bound $2s^2$.

Issue #24190 fixes the reading: for every integer $n=2s\ge2$, every positive semidefinite trace-one complex matrix $\rho$ on $\mathbb C^{n+1}\otimes\mathbb C^{n+1}$, and every four unit real directions $a,c,b,d$, $|\operatorname{tr}(\rho\mathcal B)|\le n^2/2=2s^2$. This is proved, including integer and half-integer spins $s\ge1$ and both pure and mixed states.

The source definitions and bibliographic locators are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/HighSpinCHSHNonviolation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Identical TW measurements attain $2(d-1)$ for every $d\ge2$Identical TW measurements attain the symmetric SATWAP value 2(d−1) on a swap-invariant maximally entangled vector in every dimension d at least two.Proved

The questionLiang, Trading symmetry for Hilbert-space dimension in Bell-inequality violation, arXiv:2601.02893v4; Quantum Science and Technology 11(3), 035003 (2026), conjecture following Eq. Eq:symSATWAP:

Why it mattersThe source compares the Hilbert-space dimension needed for maximal Bell violation with and without the requirement of a symmetric quantum strategy.

Our contributionIdentical TW measurements attain the symmetric SATWAP value 2(d−1) on a swap-invariant maximally entangled vector in every dimension d at least two.

Verification details
Release source recordProved in Lean

Open problem

Identical TW measurements attain $2(d-1)$ for every $d\ge2$

H.-Y. Hsu, G. N. M. Tabia, K.-S. Chen, M.-E. Liu, T. Vértesi, N. Brunner and Y.-C. Liang, Trading symmetry for Hilbert-space dimension in Bell-inequality violation, arXiv:2601.02893v4; Quantum Science and Technology 11(3), 035003 (2026), conjecture following Eq. Eq:symSATWAP:

Our computation results show that for the symmetric inequalities of~\cref{Eq:symSATWAP}, the measurement of \cref{Eq:CGLMP-Opt}, in conjunction with a symmetric maximally entangled two-qudit state, indeed reproduces the maximal quantum violation~\cite{SAT+17} of $2(d-1)$ for $d\le 100$. Together with the fact~\cite{Sarkar:2021aa} that this maximal quantum violation self-tests a maximally entangled two-qudit state, we know that the symmetric family of SATWAP inequalities, \cref{Eq:symSATWAP}, also cannot exhibit a trade-off for $d\le 100$. Moreover, we conjecture that this holds true for any higher $d$ as well.

This quotation is from line 532 of the downloaded v4 TeX member SymmetricBIV-arXiv_v2.tex (2736 lines, md5 6116dc142244f252c16a835d291626cf). The symmetric-strategy definition is at lines 285–297, the measurements at lines 488–497, and the correlators and symmetric SATWAP expression at lines 513–530. See also the literature note.

The module uses computational indices $0,\ldots,d-1$ and real-exponent phases $\omega^t=e^{2\pi i t/d}$. Its measurement matrices are

$M_{a|0}=|a\rangle\langle a|$, $M_{a|1}=U|a\rangle\langle a|U^\dagger$, $U=TW$,

where $T=\operatorname{diag}(-1,1,\ldots,1)$ and $W_{ij}=1/[d\sin((i-j-1/2)\pi/d)]$. Both parties use this same measurement function. Module settings $0,1$ correspond respectively to settings $1,2$ in the source's symmetric SATWAP expression.

For a vector $\psi$, put $P_\psi(a,b|x,y)=\langle\psi|(M_{a|x}\otimes M_{b|y})|\psi\rangle$ and $C_{xy}^{(k)}=\sum_{a,b=0}^{d-1}\omega^{-k(a+b)}P_\psi(a,b|x,y)$. With $a_k=(1-i)\omega^{k/4}/2$, satwapValue is the complex-valued finite sum

$\tilde I_d(\psi)=\sum_{k=1}^{d-1}\left[\omega^{-2k}\left(a_k C_{00}^{(k)}+a_k^*C_{01}^{(k)}+a_k^*C_{10}^{(k)}\right)+\omega^{-3k}a_k C_{11}^{(k)}\right]$.

Issue #22009 fixes the attainment statement: for every natural $d\ge2$, there exists $\psi\in\mathbb C^d\otimes\mathbb C^d$ with $|\psi|=1$, $S\psi=\psi$, $\operatorname{Tr}_B|\psi\rangle\langle\psi|=I/d$, and $\tilde I_d(\psi)=2(d-1)$. This is proved. The universal quantum upper bound and the self-testing statement used by the source to infer minimal dimension are separate literature inputs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/SymmetricSatwapMaximalViolation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
Number theoryQuestion resolvedColliding blocks: the collision count leaves floor(π√n) only at recordsFor the number a(n) of elastic collisions between a block of mass n, a block of mass 1 and a wall, every n at which a(n) differs from floor(pi sqrt(n)) is a position of a record of a (OEIS A331859, conjecture of Kagey).Proved

The questionSuppose there is a block A of mass n sliding left toward a stationary block B of mass 1, to the left of which is a wall.

Why it mattersFor large n the collision count is close to π√n; the conjecture says that it can exceed ⌊π√n⌋ only at the masses where the count reaches a new maximum.

The ideaWrite t(m) = arctan(√(1/m)); a(m) is the largest integer strictly below π/t(m).

Our contributionFor the number a(n) of elastic collisions between a block of mass n, a block of mass 1 and a wall, every n at which a(n) differs from floor(pi sqrt(n)) is a position of a record of a (OEIS A331859, conjecture of Kagey).

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Colliding blocks: the collision count leaves floor(π√n) only at records

OEIS A331859 (Peter Kagey, 2020) counts collisions in the colliding-blocks system of the 3Blue1Brown videos:

Suppose there is a block A of mass n sliding left toward a stationary block B of mass 1, to the left of which is a wall. Assuming the sliding is frictionless and the collisions are elastic, a(n) is the number of collisions between A and B plus the number of collisions between B and the wall.

Its FORMULA section gives a(n) = ceiling(Pi/arctan(sqrt(1/n))) - 1, and its COMMENTS state

Conjecture: The values of n for which a(n) != A121854(n) is a subset of A331903.

where A121854 is floor(Pi*(sqrt(n))) and A331903 is the positions of records in A331859.

Issue #10098 fixes the readings: a(n) is the FORMULA closed form for n ≥ 1, whose identification with the collision count is the published count of Galperin (2003) and is not formalized here; A121854 is ⌊π√n⌋; A331903 is the set of n ≥ 1 with a(k) < a(n) for all 1 ≤ k < n (strict records, as the entry's data 1, 2, 4, 5, 6, 8, … shows).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Constants/Billiards/CollidingBlocksRecords.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2026Optimality of the coefficient in the entanglement-breaking norm inequalityProved
Verification details
Release source recordProved in Lean

Open problem

Optimality of the coefficient in the entanglement-breaking norm inequality

E. Kopel, A sharp norm inequality for entanglement-breaking channels, arXiv:2609.27906v1, fixes a Hermitian traceless basis $\lambda_1,\dots,\lambda_{d^2-1}$ of $M_d$ with $\mathrm{Tr}[\lambda_i\lambda_j]=2\delta_{ij}$, writes a channel in Bloch form $r\mapsto Ar+c$ and proves (Theorem 1) that every entanglement-breaking channel satisfies $|A|_*^2+\tfrac{d(d-1)}{2}|c|^2\le(d-1)^2$. Item 3 of its open questions asks: "Is the coefficient $d(d-1)/2$ optimal?" The verbatim statements are in the literature note.

Issue #14492 reads the question as: for every $d\ge2$, every admissible basis and every real $B>d(d-1)/2$, some entanglement-breaking channel has $|A|_*^2+B|c|^2>(d-1)^2$. The inequality with a coefficient $\kappa$ becomes stronger as $\kappa$ grows, so this says that $d(d-1)/2$ is the largest admissible coefficient.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/EntanglementBreakingNormCoefficient.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
Number theoryQuestion resolvedFirst preprint 2026Rational collision density of bracket expressionsEvery finite rational-modulus bracket expression has an existing rational collision density.Proved

The questionUrban Larsson and Hikaru Manabe, Additive Subtraction Games, arXiv:2603.10414v2, Section 7, Question 20(iii), define the collision density of an arbitrary bracket expression and separately ask: “Is it always rational?” Section 1 specifies integer coefficients and positive rational moduli.

Why it mattersThe source definition and question are literature-attested in Library/ArithUnits/larssonmanabe2026additive.md; the proof is repository-derived.

The ideabracketaffine clears the finitely many rational denominators to produce a common positive index period and integer drift. affineimagetail decomposes the image into finitely many residue rays.

Our contributionEvery finite rational-modulus bracket expression has an existing rational collision density.

Verification details
Release source recordProved in Lean

Open problem

Rational collision density of bracket expressions

Urban Larsson and Hikaru Manabe, Additive Subtraction Games, arXiv:2603.10414v2, Section 7, Question 20(iii), define the collision density of an arbitrary bracket expression and separately ask: “Is it always rational?” Section 1 specifies integer coefficients and positive rational moduli.

For every finite $t$, coefficients $c_i\in\mathbb Z$ and positive $\alpha_i\in\mathbb Q$, set $w_n=\sum_{i=1}^t c_i\lfloor n\alpha_i\rfloor$ for $n\geq0$ and $W={w_n:n\geq0}$. For every positive integer $k$, set $C(k)={x\in W:x+k\in W}$. The result proves that

$$\lim_{N\to\infty}\frac{|C(k)\cap{0,\ldots,N}|}{N}$$

exists and is rational. This is the complete rationality subquestion, including existence of the limit, rather than the other requests in Question 20.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Every finite number of terms, including zero | t : ℕ and Fin t | | Integer coefficients of either sign | c : Fin t → ℤ | | Positive rational moduli | α : Fin t → ℚ and ∀ i, 0 < α i | | Natural sequence indices including zero | bracket c α n, n : ℕ | | Actual distinct image values at $x$ and $x+k$ | collision has two independent existential indices; the count filters values once | | Every positive integer shift | k : ℕ with 0 < k | | Inclusive nonnegative interval | Finset.range (N + 1) | | Denominator $N$ | Real cardinality divided by (N : ℝ) | | Existence and rationality of the whole limit | ∃ a : ℚ, Tendsto … atTop (𝓝 (a : ℝ)) in claim |

No monotonicity, injectivity, positive drift or nonnegative-coefficient premise is imposed. The empty expression, zero coefficients and all drift signs remain in scope.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Density/RationalBracketCollisionDensity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025The coefficient matrix of a 2p-term separable qubit state has 2^(p-1) formsThe corresponding coefficient matrices of an ordered complementary two-row support are indexed by compositions of the number of columns.Proved

The questionLi, A necessary and sufficient condition for genuinely entangled n-qubit states with six non-zero coefficients, arXiv:2510.16561v1, Section 6, closes with the question: "What to do next is how to find the forms of the corresponding coefficient matrices.

Why it mattersLi's criterion for genuine entanglement checks the support against the separable shapes and the coefficient matrix for proportional rows.

The ideaWithin a group every first-row entry precedes every second-row entry, and groups are consecutive intervals.

Our contributionThe corresponding coefficient matrices of an ordered complementary two-row support are indexed by compositions of the number of columns.

Verification details
Release source recordProved in Lean

Open problem

The coefficient matrix of a 2p-term separable qubit state has 2^(p-1) forms

D. Li, A necessary and sufficient condition for genuinely entangled n-qubit states with six non-zero coefficients, arXiv:2510.16561v1, Section 6, closes with the question: "What to do next is how to find the forms of the corresponding coefficient matrices. When $m=4$, there are two forms [Li-24]. When $m=6$, there are four forms. How many forms of the corresponding coefficient matrices are there when $m=2p$?" Here an $n$-qubit state with $m=2p$ non-zero coefficients $b_i$ on basis states $B_1<\cdots<B_m$, $p$ prime, is separable and not trivially separable. It factors as a two-term state with complementary strings on qubits $p_1<\cdots<p_k$ times a $p$-term state on the remaining qubits, and its corresponding coefficient matrix is the $2\times p$ array $(\alpha_r\beta_j)$, whose entries are the $b_i$. The verbatim statements are in the literature note.

Issue #14459 reads a form as the $2\times p$ array of indices $i$, with the source's conventions: the first row is the one whose row-factor string has bit 0 at $p_1$, and the columns follow the increasing column-factor strings.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/QubitSupportCoefficientForms.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Tightness of the lossy-qubit partial joint-measurability bound η ≤ 2/(2 + sin θ)Partial joint measurability bounds the efficiency of two lossy projective qubit measurements.Proved

The questionThe verbatim statements are in the literature note.

Why it mattersPartial joint measurability bounds the detection efficiency below which lossy measurements are useless for device-independent and semi-device-independent protocols: an eavesdropper who learns the guessed outcomes can simulate the device.

Our contributionPartial joint measurability bounds the efficiency of two lossy projective qubit measurements.

Verification details
Release source recordProved in Lean

Open problem

Tightness of the lossy-qubit partial joint-measurability bound η ≤ 2/(2 + sin θ)

E. P. Lobo, M. Balanzó-Juandó and S. Pironio, Generalized measurement incompatibility, arXiv:2605.16151v1, call measurements $G$-jointly measurable when a common parent reproduces them while deterministically fixing only the outcomes in the subsets $G_y$ (Definition 2). For two lossy projective qubit measurements (Eqs. (13), (18)–(19)) with detection efficiency $\eta$, whose axes are separated by an angle $\theta\in[0,\pi/2]$, and $G=({+,-},\emptyset)$ (case (d) of §IV.B), the paper proves $G$-joint measurability for $\eta\le 2/(2+\sin\theta)$ (Eq. (36)) and reports that "up to machine precision, this bound is tight", from numerical solutions of the $G$-JM semidefinite program. The verbatim statements are in the literature note.

Issue #15638 fixes the reading: for all unit Bloch vectors $r_1,r_2$ and every $\eta\in[0,1]$, $G$-joint measurability implies $\eta,(2+\lVert r_1\times r_2\rVert)\le2$, where $\lVert r_1\times r_2\rVert=\sin\theta$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/LossyQubitPartialJMTightness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2026The Lotka–Volterra integrals are birational coordinates in every dimensionIn every dimension at least two over a characteristic-zero field, the specified Lotka–Volterra integrals and the last coordinate generate the rational function field.Proved

The questionQuispel, Subalgebras of integrals, commutants, and superintegrable deformations of Lotka–Volterra systems, arXiv:2602.17945v1, Section 2:

Why it mattersBirational coordinates make every rational observable a rational function of the known integrals and one evolving coordinate.

The idea- Odd n. Two families of product identities hold: Fk G{m−k}/F1 = S{2k−1}(H − S{2k})/x{2k} and Fk G{m−k+1}/F1 = S{2k−1}(H − S{2k−2})/x{2k−1}.

Our contributionIn every dimension at least two over a characteristic-zero field, the specified Lotka–Volterra integrals and the last coordinate generate the rational function field.

Verification details
Release source recordProved in Lean

arXiv

The Lotka–Volterra integrals are birational coordinates in every dimension

I. Marquette, P. H. van der Kamp and G. R. W. Quispel, Subalgebras of integrals, commutants, and superintegrable deformations of Lotka–Volterra systems, arXiv:2602.17945v1, Section 2:

In arbitrary dimensions, we have not been able to prove the birationality of the change of variables, but we conjecture this to be the case, and we provide explicit formulas in Appendix \ref{secA}.

Appendix A restates it: "We conjecture that the transformation from the $n$ variables $x_i$ to the $n-1$ integrals and distinguished variable $x_n$ is a birational map." The integrals F_k and their coordinate reversals G_k are quoted in the literature note.

Issue #19118 fixes the reading: for every n ≥ 2 and every field of characteristic zero, the following n elements generate the rational function field F(x_1, …, x_n):

  • odd n = 2m − 1: F_1, …, F_m, G_2, …, G_{m−1}, x_n;
  • even n = 2m: F_1, …, F_m, G_1, …, G_{m−1}, x_n.

This is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/Solitons/LotkaVolterraIntegralBirationality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2025Identity and transposition classes have the two largest tensor high-temperature degreesFor every n at least two and every positive rank, the all-ones and transposition classes have the two largest tensor high-temperature degrees.Proved

The questionRamgoolam, Gauged permutation invariant tensor quantum mechanics, least common multiples and the inclusion-exclusion principle, arXiv:2506.18813v2 (2025), Section 7.2, state:

Why it mattersIdentifying the two largest degrees supplies the discrete ordering used in Section 7.2's analysis of the breakdown of the high-temperature expansion.

Our contributionFor every n at least two and every positive rank, the all-ones and transposition classes have the two largest tensor high-temperature degrees.

Verification details
Release source recordProved in Lean

Open problem

Identity and transposition classes have the two largest tensor high-temperature degrees

D. O'Connor and S. Ramgoolam, Gauged permutation invariant tensor quantum mechanics, least common multiples and the inclusion-exclusion principle, arXiv:2506.18813v2 (2025), Section 7.2, state:

Computations with Mathematica and the analysis of special families give strong evidence that the leading contributions in the high temperature limit $ x \rightarrow 1 $ come from $ p = [ 1^N ] $ followed by $ p = [ 2,1^{ N-2} ] $.

Section 9 asks:

It would be very interesting to find the proof of the general $s$ conjecture for the leading two terms.

These quotations are from lines 1068 and 1322 of the v2 e-print member Perm-Invt-Tensor-Counting-Functions-Journal.tex, md5 202b33a8e55dec1592ee90b7e5fbd8d9.

For a partition $p$ of $N$, let $A(p)$ be its set of distinct part values and $m_a$ the multiplicity of $a$. The module's rational-valued degree is the first sum in the source's equation (7.21), Section 7.2, TeX lines 1079–1087:

$$D_s(p)=\sum_{t:\mathrm{Fin},s\to A(p)} \frac{\prod_{j\in\mathrm{Fin},s}t_jm_{t_j}} {\operatorname{lcm}{t_j:j\in\mathrm{Fin},s}}.$$

Tuple entries may repeat. The part values, multiplicities and natural-number least common multiple are cast to the rationals for this expression.

Write $I_N=[1^N]$ and $T_N=[2,1^{N-2}]$. Issue #24869 fixes the reading: for every natural $N\ge2$ and $s\ge1$,

$$D_s(I_N)=N^s,\qquad D_s(T_N)=\frac{N^s+(N-2)^s}{2}<D_s(I_N),$$

and every partition $p$ of $N$ different from both $I_N$ and $T_N$ satisfies $D_s(p)<D_s(T_N)$. This assertion is proved. The subtraction $N-2$ is natural subtraction; it agrees with integer subtraction in the stated range.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/TensorHighTemperatureLeadingClasses.result

Frozen module record

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Analysis & dynamicsQuestion resolvedForberg's exponential formula for the quarter-squaresForberg's exponential quotient has the shifted quarter-square floor at every positive integer.Proved

The questionEmpirical: a(n-1) = floor(n/(e^(4/n)-1)). - Richard R.

Why it mattersThe result explains the agreement through uniform analytic bounds and the spacing of integer floor intervals.

Our contributionForberg's exponential quotient has the shifted quarter-square floor at every positive integer.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Forberg's exponential formula for the quarter-squares

OEIS A002620, revision 880, contains the formula attributed to Richard R. Forberg:

Empirical: a(n-1) = floor(n/(e^(4/n)-1)). - Richard R. Forberg, Jul 24 2013

The entry defines the quarter-squares by $a(m)=\lfloor m/2\rfloor\lceil m/2\rceil=\lfloor m^2/4\rfloor$, with offset zero. The complete assertion is

$$ \forall n\in\mathbb N,\quad 1\le n\ \Longrightarrow \left\lfloor\frac{n}{\exp(4/n)-1}\right\rfloor =\left\lfloor\frac{(n-1)^2}{4}\right\rfloor. $$

All operations inside the floors are in $\mathbb R$, with $n$ coerced before division and subtraction. The exponential is Real.exp; both floors are Int.floor and the equality is in $\mathbb Z$. The index shift is $m=n-1$. The assertion includes $n=1$, both parities, and arbitrarily large indices, with no additional hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/SeriesInequalities/ForbergExponentialFloor.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedEvery row sum of the A105594 triangle is oddEvery row sum of the A105594 triangle is odd.Proved

The questiona(n)=sum{k=0..n, mod(sum{j=0..n, abs(mu(binomial(n, j)))mod(binomial(j, k), 2)}, 2)}

Why it mattersThe conjecture is present from revision 1 in 2005 and remains labelled Conjecture in 2026, a span of 21 years.

The ideaCast each already reduced inner sum to ZMod 2.

Our contributionEvery row sum of the A105594 triangle is odd.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Every row sum of the A105594 triangle is odd

OEIS A105595, NAME (%N, verbatim):

Row sums of number triangle A105594.

Its COMMENT conjecture (%C, verbatim) is:

Conjecture : all terms are odd.

Its FORMULA (%F, verbatim) is:

a(n)=sum{k=0..n, mod(sum{j=0..n, abs(mu(binomial(n, j)))*mod(binomial(j, k), 2)}, 2)}

OEIS A105594, FORMULA (%F, verbatim):

T(n, k) = mod(Sum_{j=0..n}(abs(mu(binomial(n,j)))*mod(binomial(j,k),2)), 2).

The literal proved statement defines rowEntry n k = (sum j in range (n+1), abs(mu(choose n j)) * (choose j k mod 2)) mod 2, defines rowSum n = sum k in range (n+1), rowEntry n k, and proves forall n : Nat, Odd (rowSum n).

The following are not claimed:

  • any other property of A105594;
  • equivalence between the %N matrix-product phrasing abs(A103447)*A047999 mod 2 and the displayed %F sum;
  • exhaustive literature coverage or a priority claim.

The delivery formalizes the two displayed %F sums and discloses that source choice rather than identifying the matrix-product description with them.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/BarryTriangleRowSumsOdd.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
Number theoryQuestion resolvedThe zero indices of A110454The only positive indices without an eligible composite concatenation are one, two, and four.Proved

The questionLargest composite number obtained by concatenation of parts of a distinct partition of n, or 0 if no such number exist.

Why it mattersThe conjecture identifies every positive index where the admissible set is empty.

The ideaAt n = 3, the ordered distinct partition [2,1] gives the concatenation 21, which is greater than four and composite.

Our contributionThe only positive indices without an eligible composite concatenation are one, two, and four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The zero indices of A110454

OEIS A110454, NAME (verbatim):

Largest composite number obtained by concatenation of parts of a distinct partition of n, or 0 if no such number exist.

COMMENT (verbatim; Amarnath Murthy, Aug 04 2005):

Conjecture a(n) = 0 only for n = 1, 2 and 4.

The Maple program by R. J. Mathar, dated Feb 08 2008, examines every permutation of every distinct partition and accepts a concatenation exactly when m > 4 and m is not prime. Its loop includes single-part partitions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/DistinctPartitionConcatenationComposite.murthy_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedPosed in 2026Morken's first A112399 divisor-sum conjectureThe coprime Moebius sum equals a double sum over squarefree divisors.Proved

The questiona(n) = Sum{k=1..n, gcd(k,n)=1} mu(k), where mu(k) = A008683(k) (the Moebius function).

Why it mattersThe identity connects a coprime Moebius partial sum to divisor-indexed partial sums.

The ideaFor each divisor d and positive m, express the filtered summand as mu(dm)mu(d).

Our contributionThe coprime Moebius sum equals a double sum over squarefree divisors.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Morken's first A112399 divisor-sum conjecture

OEIS A112399, revision 41, source timestamp 2026-09-19T09:22:12-04:00, NAME defines:

a(n) = Sum_{k=1..n, gcd(k,n)=1} mu(k), where mu(k) = A008683(k) (the Moebius function).

Its FORMULA section attributes the first of two conjectures to Jamie Morken, Sep 18 2026:

a(n) = Sum_{d|n, mu(d)!=0} (Sum_{m=1..n/d, gcd(m,d)=1} mu(m)).

The question is this equality for every positive integer n. Both intervals include their endpoints. The function mu is integer-valued and signed, with mu(1)=1. The outer sum ranges over positive divisors of n for which mu(d) is nonzero, and n/d is exact integer division. The right inner coprimality is with d; the left coprimality is with n. There is no squarefree assumption on n and no finite upper bound. The conjecture has AI-assisted provenance, which supplies no mathematical proof. The second A112399 conjecture, also indexed by A400203, and the A372706 product-parameter identity are excluded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/MoebiusLogarithmic/MorkenFirstDivisorSum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPrime square perfect-power representationEvery prime can be added to a square to obtain a perfect power with exponent at least two.Proved

The questionConjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p.

Why it mattersThe assertion asks for one nontrivial perfect-power representation for every prime, including the even prime.

The ideaThese choices have positive bases and exponents at least two.

Our contributionEvery prime can be added to a square to obtain a perfect power with exponent at least two.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime square perfect-power representation

OEIS A115039 and A115038 state the same conjecture:

Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p.

The quantified reading is: for every prime natural number p, there are natural numbers x, y, and n with 2 <= n and x^2 + p = y^n. The lower bound is the standard perfect-power condition; without it, exponent one makes the equation immediate.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/PrimeSquarePerfectPowerRepresentation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedPrime values of the divisor-pair product sum have exactly two preimagesPrime values of the second elementary symmetric divisor function have exactly two preimages when attained at composites.Proved

The questionComposite numbers for which the second elementary symmetric function of divisors (s2) is prime.

Why it mattersThis is a first-tier OEIS conjecture explicitly recorded in 2006.

The ideaStarting from the verbatim definition S2(n)=e2(divisors(n)), the module internally proves the equivalent divisor identity 2S2(n) = sigma1(n)^2 - sigma2(n) from A119616 %F.

Our contributionPrime values of the second elementary symmetric divisor function have exactly two preimages when attained at composites.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Prime values of the divisor-pair product sum have exactly two preimages

OEIS A119623, NAME (Zak Seidov, Jun 08 2006):

Composite numbers for which the second elementary symmetric function of divisors (s2) is prime.

OEIS A119623, COMMENTS (Zak Seidov, Jun 08 2006):

Terms in A119616 are always prime if n is prime p and s2(p)=p, hence it is interesting to find composite numbers for which s2 is also prime. Relative values of s2 are: s2=47,97,163,457,733,2203,3733,7993,10723,11317,21313,22147,26557,33403,57283,61417,67153,79393,101467,149323,160453,162727,174337,272683,296827,318793,358273,432907,440383,486583,551767,639007,832687,843043,911917,961183,1152913,1202017,1277593,1322743,1375303,1462897,1567327,1824997,1878883. Otherwise the sequence s2 gives numbers which appear in A119616 at least twice (and conjecture is that exactly twice).

OEIS A119616, NAME:

Second elementary symmetric function of divisors of n.

Accordingly, S2(n) is defined here verbatim as e_2(divisors(n)), the second elementary symmetric function of the finite divisor multiset. Lean's Nat.divisors 0 is empty, so this definition gives S2(0)=0. For positive n, it is equivalently the sum of d*e over unordered pairs of distinct divisors, or (sigma_1(n)^2 - sigma_2(n))/2 as recorded in A119616 %F; these are equivalent descriptions, not the adopted definition.

The adopted reading is: for every composite n for which p = S2(n) is prime, {m >= 1 : S2(m) = p} = {n, p}. This is the only reading consistent with both "at least twice" and "exactly twice": the identity S2(p) = p provides the second occurrence, while "exactly" excludes every other positive preimage.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/DivisorPairProductPrimePreimages.seidov_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture refutedRefutation of Stephan's A229140 least-coordinate distinctness conjectureThe values 628 and 673 refute the distinctness conjecture for OEIS A229140.Refuted

The question%N A229140 Smallest k such that k^2 + l^2 = n-th number expressible as sum of two squares (A001481). %C A229140 Conjecture: the values between two zeros are always distinct from each other. %A A229140 Ralf Stephan, Sep 15 2013 %O A229140 1,6

Why it mattersThe conjecture first appeared in OEIS revision 8 on September 17, 2013 and has stood since then; revision 39 of September 23, 2025 still carries it with no proof or refutation comment.

The ideaFour finite obligations refute the universal claim:

Our contributionThe values 628 and 673 refute the distinctness conjecture for OEIS A229140.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Stephan's A229140 least-coordinate distinctness conjecture

OEIS A229140 states (verbatim):

%N A229140 Smallest k such that k^2 + l^2 = n-th number expressible as sum of two squares (A001481). %C A229140 Conjecture: the values between two zeros are always distinct from each other. %A A229140 Ralf Stephan, Sep 15 2013 %O A229140 1,6

The literal refuted statement is: for every pair L<R of representable natural values with least first coordinate zero and with no intervening value having least first coordinate zero, distinct representable values strictly between L and R have distinct least first coordinates.

The relation IsLeastCoord m x states that m=x^2+y^2 for some natural y and that no natural x'<x is the first coordinate of any such representation. Consequently, IsLeastCoord m 0 is exactly the assertion that m is a square: the representation clause becomes m=y^2, and the exclusion clause is vacuous because no natural number is below zero. In claim, the two endpoint hypotheses say that L and R are zero values, L<R orders them, and the universal no-zero hypothesis says they are consecutive. The final quantifiers assert the distinctness of least coordinates for every two distinct intermediate values.

This result does not claim an enumeration of A001481; the identities added by Zhuorui He in 2025, a(n) = sqrt(A001481(n)-A385236(n)^2), a(n) = A328803(n) - A385236(n), or a(n) = A064875(A001481(n)); any property of A064875, A385236, A328803, A283303, or A283304; whether the conjecture holds on any other interval; or exhaustive literature coverage.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/PrimeForms/StephanLeastCoordinateDistinctnessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The argument establishes an obstruction to a proposed characterization or class of methods.

    Source 1
Number theoryQuestion resolvedThe A254748 twin-prime mean has no superdivisorTwin-prime pair averages have no superdivisor.Proved

The questionNumbers without superdivisors: numbers n such that n/k + n fails to divide at least one of (n/k)^(n/k) + n, (n/k)^n + n/k or n^(n/k) + n/k for any divisor k of n.

Why it mattersThe conjecture asserts that the arithmetic mean of every twin-prime pair is a zerosuperdivisor number.

The ideaProve the stronger statement for every even n at least 4 with n - 1 prime.

Our contributionTwin-prime pair averages have no superdivisor.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A254748 twin-prime mean has no superdivisor

OEIS A254748, NAME (%N, verbatim):

Numbers without superdivisors: numbers n such that n/k + n fails to divide at least one of (n/k)^(n/k) + n, (n/k)^n + n/k or n^(n/k) + n/k for any divisor k of n.

COMMENT (%C, verbatim):

Zerosuperdivisor numbers. Numbers n such that A247477(n) = 0.

COMMENT (%C, verbatim):

Conjecture: Average of twin prime pairs (A014574) are zerosuperdivisor numbers.

AUTHOR (%A, verbatim):

Juri-Stepan Gerasimov, Feb 07 2015

The A247477 superdivisor definition is: a divisor k is a superdivisor of n when n/k + n divides each of (n/k)^(n/k) + n, (n/k)^n + n/k, and n^(n/k) + n/k, with all division interpreted in ℕ (exact when k ∣ n).

The literal claim is ∀ p : ℕ, Nat.Prime p → Nat.Prime (p + 2) → ∀ k : ℕ, 1 ≤ k → k ∣ p + 1 → ¬ IsSuperdivisor (p + 1) k, where IsSuperdivisor is the three-clause predicate above and the quotient is natural-number division. The perfect-number sentence in A254748 is NOT claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/GerasimovTwinPrimeMeanNoSuperdivisor.gerasimov_a254748

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe divisor-difference gcd and Heinz decoding identity in A258409For every n at least two, the divisor-minus-one gcd agrees with the gcd of successive divisor gaps and with the prime-index gcd of their Heinz number.Proved

The questionGreatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.

Why it mattersThis is a first-tier OEIS conjecture recorded in 2019.

The ideaWrite the sorted divisors as 1 = d0 < ... < dk = n, and put Deltai = di - d(i-1).

Our contributionFor every n at least two, the divisor-minus-one gcd agrees with the gcd of successive divisor gaps and with the prime-index gcd of their Heinz number.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The divisor-difference gcd and Heinz decoding identity in A258409

OEIS A258409, NAME:

Greatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.

COMMENT (Gus Wiseman, Oct 16 2019):

Conjecture: a(n) = A289508(A328023(n)) = GCD of the differences between consecutive divisors of n. See A328163 and A328164.

OEIS A328023, NAME:

Heinz number of the multiset of differences between consecutive divisors of n.

Its Heinz-number convention is:

The Heinz number of an integer partition or multiset {y_1,...,y_k} is prime(y_1)*...*prime(y_k).

OEIS A289508, NAME:

a(n) is the GCD of the indices j for which the j-th prime p_j divides n.

These OEIS prime indices are one-based, with prime(1)=2. Mathlib's Nat.nth Nat.Prime is zero-based, so the Lean encoding of prime(delta) is Nat.nth Nat.Prime (delta - 1).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/DivisorDifferenceGcdHeinz.wiseman_a258409

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedThe penultimate-label count in OEIS A270236Restricted-growth words give Mathar's two closed label-occurrence counts.Proved

The questionTriangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.

Why it mattersThis is one of two independent formulas recorded as conjectures in the same OEIS entry.

The ideaIsRestrictedGrowth states the chronological source condition, including the positive-label bound, and memwordsiff identifies words n with its length-n solutions after reversing the generator's storage order.

Our contributionRestricted-growth words give Mathar's two closed label-occurrence counts.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The penultimate-label count in OEIS A270236

OEIS A270236, NAME (verbatim):

Triangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.

COMMENT defining the restricted growth functions (verbatim):

The RG functions used here are defined by f(1)=1, f(j) <= 1+max_{i<j} f(i).

FORMULA (verbatim; R. J. Mathar, Mar 13 2016):

Conjecture: T(n,n-1) = 2+n*(n-1)/2 for n>1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/RestrictedGrowthLabelOccurrences.mathar_f1

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedKrizek's A300657 noncomposite characterizationKrizek's divisor sum of sigma residues equals the final residue exactly at one and the primes.Proved

The questionConjecture: a(n) = A054024(n) only for the noncomposite numbers A008578.

Why it mattersThe conjecture asks for an exact characterization, rather than a one-sided bound or a verification over a finite range.

The ideaFor n = 1 and for prime n, the divisor set evaluates directly and the equality follows.

Our contributionKrizek's divisor sum of sigma residues equals the final residue exactly at one and the primes.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Krizek's A300657 noncomposite characterization

OEIS A300657 defines

a(n) = Sum_{d|n} sigma(d) mod d.

Its Comment line states verbatim:

a(n) >= A054024(n). Conjecture: a(n) = A054024(n) only for the noncomposite numbers A008578.

The entry's Author line states verbatim:

Jaroslav Krizek, Mar 10 2018

Here A054024(n) = sigma(n) mod n, and A008578 consists of one and the primes. With

a300657(n) = Sum_{d|n} (sigma(d) mod d),

the proved equivalence, with all quantifiers explicit, is: for every natural number n, if 1 <= n, then

a300657(n) = sigma(n) mod n if and only if n = 1 or n is prime.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/KrizekDivisorSigmaModNoncomposite.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedCaceres' A308090 gcd-factorial-power primality criterionCaceres' A308090 gcd-factorial-power condition implies primality.Proved

The questionConjecture: Conversely, if gcd(2^n + n!, 3^n + n!, n+1) = n+1, then n+1 is prime.

Why it mattersThe independent question is Caceres' named converse primality criterion for OEIS A308090 at every positive natural index.

The ideaThe gcd equality makes n+1 divide both 2^n+n! and 3^n+n!.

Our contributionCaceres' A308090 gcd-factorial-power condition implies primality.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Caceres' A308090 gcd-factorial-power primality criterion

OEIS A308090, NAME (%N, verbatim):

a(n) = gcd(2^n + n!, 3^n + n!, n+1).

The preceding COMMENT (%C, verbatim) is:

From observation: For n > 3, if n+1 is prime, then a(n) = n+1.

The settled COMMENT (%C, verbatim) is:

Conjecture: Conversely, if gcd(2^n + n!, 3^n + n!, n+1) = n+1, then n+1 is prime.

The offset is %O 1,4, and the AUTHOR line (%A, verbatim) is:

Pedro Caceres, May 11 2019

The full-quantifier reading is: for every n : Nat, if 0 < n and a n = n + 1, then (n + 1).Prime, where a n = gcd(gcd(2^n + n!, 3^n + n!), n+1). The scope wall is exactly the unsigned Conjecture sentence for every natural n ≥ 1. The positivity hypothesis is part of the statement: at n = 0, the gcd equality holds but 1 is not prime. The preceding one-way observation is context, not an additional theorem in this resolution.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/CaceresGcdFactorialPowerPrimeCriterion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedThe Karttunen–Munn prime-values conjecture for OEIS A344695The prime values of gcd of the divisor sum and Dedekind psi are the odd primes.Proved

The questionThe public definitions are the binding expressions of 18688.

Why it mattersThe classification explains why 2 is absent and why every odd prime occurs, including primes absent from a finite prefix.

Our contributionThe prime values of gcd of the divisor sum and Dedekind psi are the odd primes.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The Karttunen–Munn prime-values conjecture for OEIS A344695

OEIS A344695, revision 39, defines “a(n) = gcd(sigma(n), psi(n)), where sigma is the sum of divisors function, A000203, and psi is the Dedekind psi function, A001615.” Its comment states: “Conjecture: the set of primes that appear in the sequence is A065091 (the odd primes).” The assertion is preregistered in #18688.

For each natural prime $q$, the formal claim says that some natural $n\ge1$ has $a(n)=q$ if and only if $q$ is odd.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A001615: “Dedekind psi function: n * Product_{p|n, p prime} (1 + 1/p).” | psi, using natural exact division by the product of distinct prime factors | The binding integer expression of #18688; positive indices make the division exact | | A344695 defines the gcd of sigma and psi | a | ArithmeticFunction.sigma 1 and Nat.gcd retain the source operations | | “the set of primes that appear” | claim quantifies over every natural prime and an existential positive index | The guard excludes composite candidate values, and positivity matches the sequence's domain | | “is A065091 (the odd primes)” | The existential occurrence is equivalent to Odd q | Both directions, with no upper bound |

The public definitions are the binding expressions of #18688. The value at zero is outside the source sequence. The settling declaration result proves the complete claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Divisors/OeisA344695GcdSigmaPsiPrimes.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedOEIS A347293 shifted-product gcd row-sum convolutionThe row sum of the OEIS A347293 shifted-product gcd triangle equals a divisor convolution of squares and squared totients.Proved

The questionTriangle read by rows: T(n, k) = Sum{i=1..n} gcd(1 + (i-1) (k-1),n) for 1 <= k <= n.

Why it mattersThe problem is a named 2022 OEIS conjecture giving a closed divisor formula for a two-dimensional gcd sum.

The ideaThe first coordinate must be a unit, and the second is its unique negative inverse.

Our contributionThe row sum of the OEIS A347293 shifted-product gcd triangle equals a divisor convolution of squares and squared totients.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A347293 shifted-product gcd row-sum convolution

OEIS A347293, NAME (%N, verbatim):

Triangle read by rows: T(n, k) = Sum_{i=1..n} gcd(1 + (i-1) * (k-1),n) for 1 <= k <= n.

FORMULA (%F, verbatim):

Conjecture: Row sums equal Dirichlet convolution of A000290 and A127473.

A000290 is the square function and A127473 is the square of Euler's totient. Thus the full-quantifier statement is that for every positive natural n,

Sum_{x=0}^{n-1} Sum_{y=0}^{n-1} gcd(1+x*y,n) = Sum_{d|n} d^2 * phi(n/d)^2.

The left side is the triangle row sum after the index changes x=i-1 and y=k-1. The divisor sum is the pointwise definition of the stated Dirichlet convolution.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/SchulteShiftedProductGcdRowSumConvolution.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedSchulte's gcd-quotient row-sum convolutionSchulte's A350900 row sum is a Dirichlet convolution of two totient sums.Proved

The questionTriangle read by rows: T(n, k) = Sum{i=1..n} gcd(i,n) / gcd(gcd(i,k),n) for 1 <= k <= n.

Why it mattersThe two-level gcd quotient reduces, for fixed i, to a periodic sum in k whose divisor fibres have totient cardinalities.

The ideaRewrite gcd(gcd(i,k),n) as gcd(gcd(i,n),k) and rotate the 1..n sums to complete residue ranges; endpoint equality follows from gcd periodicity.

Our contributionSchulte's A350900 row sum is a Dirichlet convolution of two totient sums.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's gcd-quotient row-sum convolution

OEIS A350900, NAME (%N, verbatim):

Triangle read by rows: T(n, k) = Sum_{i=1..n} gcd(i,n) / gcd(gcd(i,k),n) for 1 <= k <= n.

The settled FORMULA (%F, verbatim) is:

Conjecture: Row sums equal Dirichlet convolution of A002618 and A057660.

The full-quantifier reading is: for every positive natural n, define rowSum(n) = Sum_{i=1..n} Sum_{k=1..n} gcd(i,n)/gcd(gcd(i,k),n); then

rowSum(n) = Sum_{d|n} d*phi(d) * (Sum_{e|(n/d)} e*phi(e)).

Here phi is Euler's totient, A002618(n) is n*phi(n), and A057660(n) is Sum_{d|n} d*phi(d). The identity gcd(gcd(i,k),n) = gcd(gcd(i,n),k) shows the denominator of each row summand divides its numerator; these are exact natural quotients. For each d|n the complementary quotient n/d is also exact. Only the quoted row-sum formula is settled. The %C conjecture for an arbitrary arithmetic function f is outside this claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/SchulteGcdQuotientRowSumConvolution.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedCoefficient congruences for the A355872 bilateral theta reversionThe bilateral theta equation of OEIS A355872 implies Hanna's three congruences.Proved

The questionHanna, Aug 09 2022), as quoted in Library/Recurrence/hanna2022a355872.md, gives the NAME:

Why it mattersThis is a first-tier OEIS conjecture from the 2022 entry.

The ideaThe NAME's bilateral sum is formalised coefficientwise.

Our contributionThe bilateral theta equation of OEIS A355872 implies Hanna's three congruences.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Coefficient congruences for the A355872 bilateral theta reversion

OEIS A355872 (Paul D. Hanna, Aug 09 2022), as quoted in Library/Recurrence/hanna2022a355872.md, gives the NAME:

G.f. A(x) satisfies: x = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x)^((n-1)^2).

The single conjecture anchor of this dossier is the COMMENT:

Conjecture: a(n) == 2 (mod 4) for n >= 1.

The entry also states two further conjectures:

Conjecture: a(2*n-1) == 2 (mod 8) for n >= 1.

Conjecture: a(2*n) == 6 (mod 8) for n >= 1.

The latter two are settled by the companion hanna_conjecture_mod_eight; they do not create additional resolution anchors in this dossier.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/BilateralThetaReversionModSixteen.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's OEIS A356508 product-partition congruenceHanna's OEIS A356508 product equation and its coefficient residues modulo four.Proved

The questionA(x) satisfies: 2 = Product{n>=1} (1 + x^nA(x)) (1 + x^(n-1)/A(x)).

Why it mattersThe conjecture gives a residue description for the coefficients of the generating function specified by the entry.

The ideaCoefficientwise agreement gives a contraction: the coefficient of degree m in the product depends only on coefficients below degree m.

Our contributionHanna's OEIS A356508 product equation and its coefficient residues modulo four.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's OEIS A356508 product-partition congruence

P. D. Hanna, OEIS A356508, created 11 August 2022. The NAME field states:

G.f. A(x) satisfies: 2 = Product_{n>=1} (1 + x^n*A(x)) * (1 + x^(n-1)/A(x)).

The COMMENTS field states:

Conjecture: a(n) == 2 (mod 4) at n = 2*k for all k > 0 that have an odd number of partitions (A052002), otherwise a(n) == 0 (mod 4) when n > 0.

The formal statement uses rational formal power series, takes the product coefficientwise through the finite window needed for each coefficient, and interprets the partition number as Fintype.card (Nat.Partition k). It asserts existence of a solution and the quoted integer and residue properties for every positive coefficient.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | G.f. A(x) satisfies: 2 = Product_{n>=1} (1 + x^n*A(x)) * (1 + x^(n-1)/A(x)) | IsGF A over PowerSeries ℚ; at degree m the product is Finset.Icc 1 (m + 1), and A⁻¹ is invOfUnit A 1. | | “a(n) == 2 (mod 4) at n = 2*k for all k > 0 that have an odd number of partitions (A052002)” | For 0 < n, an integer z equal to coeff n A satisfies z ≡ 2 [ZMOD 4] whenever some k > 0 has n = 2 * k and Odd (Fintype.card (Nat.Partition k)). | | “otherwise a(n) == 0 (mod 4) when n > 0” | For the same positive n, if no such k exists then z ≡ 0 [ZMOD 4]. | | “The terms are integers.” | The claim supplies an integer z equal to every positive coefficient. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaProductPartitionModFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSquare divisibility of the A365095 perturbed diagonalHanna's vanishing diagonal defines a unique integer series and implies square divisibility for every integer perturbation.Proved

The questionA(x) satisfying [x^(n-1)] (1 + (n-1)xA(x)^2)^n / A(x)^n = 0 for n > 1.

Why it mattersThis first-tier OEIS conjecture is from the 2023 entry.

The ideaThe escape content is the integral triangular construction together with the perturbation congruence (B + n·C)^n ≡ B^n (mod n²).

Our contributionHanna's vanishing diagonal defines a unique integer series and implies square divisibility for every integer perturbation.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Square divisibility of the A365095 perturbed diagonal

OEIS A365095, Paul D. Hanna, Sep 03 2023. NAME (the defining formula (1)):

Expansion of g.f. A(x) satisfying [x^(n-1)] (1 + (n-1)*x*A(x)^2)^n / A(x)^n = 0 for n > 1.

FORMULA (2), the target:

[x^(n-1)] (1 + (k*n-1)xA(x)^2)^n / A(x)^n is divisible by n^2 for n > 0 and all integer k (conjecture).

The normalization is A(0) = 1; the parameter ranges over all integers.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/PerturbedDiagonalSquareDivisibility.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedBinomial parity of the A369083 coefficientsThe integer coefficients of OEIS A369083 have the conjectured binomial parity.Proved

The questionHanna, Jan 12 2024, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2024a369083.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2024 entry.

The ideaConstruct A over the integers by coefficient stabilisation with exact division.

Our contributionThe integer coefficients of OEIS A369083 have the conjectured binomial parity.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Binomial parity of the A369083 coefficients

OEIS A369083, Paul D. Hanna, Jan 12 2024, gives the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2024a369083.md:

Expansion of g.f. A(x) satisfying A(x) = 1 + x*(5*A(x)^2 - A(-x)^2)/4.

Conjecture: a(n) == binomial(4*n+3,n) (mod 2) for n >= 0 (cf. A263133).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/ReflectedQuadraticQuarterParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryConjecture provedHuber's minimum-volume cuboid conjectureThe smallest first edge, then the smallest second edge, gives minimum cuboid volume.Proved

The questionThe least-pair hypothesis is the lexicographic condition on (p.1, p.2.1), and the conclusion is volume p ≤ volume q for every member q of the same finset.

Why it mattersThe conjecture identifies the minimum volume directly from the two smallest ordered edges.

Our contributionThe smallest first edge, then the smallest second edge, gives minimum cuboid volume.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Huber's minimum-volume cuboid conjecture

F. Huber's OEIS A375786 asks whether, among integer-sided cuboids with the same surface $6n^2$ as a cube of edge $n$, the cuboid with the smallest first edge, breaking ties by the second edge, always has minimum volume. The formal claim is the non-strict comparison for every $n \ge 1$, using ordered positive edges $x \le y \le z$ and the surface equation $xy+yz+xz=3n^2$.

Statement echo

The source's positive integer cuboids are the cuboids finset, whose three inclusive intervals are bounded by $3n^2$ and whose filter records ordering and the surface equation. Its volume is volume (x,y,z) = xyz. The least-pair hypothesis is the lexicographic condition on (p.1, p.2.1), and the conclusion is volume p ≤ volume q for every member q of the same finset. These are the bodies of cuboids, volume, and claim in the settling module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/OeisA375786CuboidMinimumVolume.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's Lambert W coefficient congruencesIntegrality and a period-six congruence for Hanna's Lambert W exponential coefficients.Proved

The questionThe formal series interpretation makes no assertion about analytic convergence or values outside that series.

Why it mattersThe four result declarations settle the four explicitly published congruence conjectures.

The ideaThe shared module uses NonnestingOneThreeTwoTwo.lagrangecoefficient to compute the coefficients of an exponential after substitution into the inverse of QuarticEGFModFour.F.

Our contributionIntegrality and a period-six congruence for Hanna's Lambert W exponential coefficients.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's Lambert W coefficient congruences

Preregistration: #18668. Paul D. Hanna's OEIS A386655 NAME reads: "E.g.f.: Sum_{n>=0} (2^n*x + LambertW(x))^n / n!." Its COMMENTS sentence is: "Conjecture: for n >= 6, a(n) (mod 6) equals [1, 0, 3, 0, 3, 2] repeating." The corresponding entries A386656, A386657 and A386658 replace $2$ by $3$, $4$ and $5$, respectively. Their COMMENTS sentences specify $[4,3,2,1,0,1]$, $[1,2,3,2,3,4]$ and $[4,3,0,3,0,5]$. All four statements include integrality and the index $(n-6)\bmod6$ for every natural $n\ge6$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Principal formal Lambert W | OeisA386655LambertWEgfModSix.W | Constant coefficient zero and positive coefficient $(-m)^{m-1}/m!$. | | Exponential generating coefficient | egf q m | $m!$ times the coefficient of $x^m$; summation through $m$ is exact because each later summand has order greater than $m$. | | Bases $q=2,3,4,5$ | Each module's a | The respective egf specialization. | | Every $n\ge6$ | Each module's claim and result | Universal natural index and the literal lower bound. | | Integral remainder with the published pattern | Each module's claim and result | An integer representative with rational cast equal to a n, and integer remainder equal to the zero-based List.getD entry. | | Finite-sum endpoint | closedZ q n | The source's last summand is $q^{n^2}$; all remaining exponents are natural numbers. |

The public definitions retain the preregistered expressions. The formal series interpretation makes no assertion about analytic convergence or values outside that series.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/HannaLambertW/OeisA386655LambertWEgfModSix.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedQuadratic square reversion modulo four for A389542The coefficients of OEIS A389542 satisfy Hanna's mod-four classification.Proved

The questionHanna, Nov 01 2025, supplies the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a389542.md:

Why it mattersThis is a first-tier OEIS conjecture from the 2025 entry.

The ideaDegree contraction constructs the integral fixed point H = -X - H^2 - 2XH(4X^3).

Our contributionThe coefficients of OEIS A389542 satisfy Hanna's mod-four classification.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Quadratic square reversion modulo four for A389542

OEIS A389542, Paul D. Hanna, Nov 01 2025, supplies the following NAME and COMMENT, quoted verbatim from Library/Recurrence/hanna2025a389542.md:

G.f. A(x) satisfies A( A(x)^2 - x^2 ) = 4*A(x)^3.

Conjecture: for n > 1, a(n) == 2 (mod 4) if n = 2^k + 1 (for some k >= 0) otherwise a(n) is divisible by 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/QuadraticSquareReversionModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedThe first weighted moment of A391810The first weighted moment of every positive A391810 Laurent row is one.Proved

The questionHanna's OEIS A391810, revision 39, formula (3), states:

Why it mattersThis is a Tier 1 explicitly published universal coefficient identity.

The ideaUse the existing polynomial Pn(y)=-1+Sum{j=1..n} y^j(1-y^j)^(n-j).

Our contributionThe first weighted moment of every positive A391810 Laurent row is one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The first weighted moment of A391810

Paul D. Hanna's OEIS A391810, revision 39, formula (3), states:

(3) Sum_{k=0..floor((n+1)/2)^2} k * T(n,k) = 1 for n >= 1 (conjecture).

The original definition is:

Irregular triangle, read by rows, where row n lists the coefficients of nonpositive powers of x in the doubly infinite series Sum_{k=-oo..oo} x^k * (1 - x^k)^(n+k) for n >= 1.

Thus, for every positive natural n, expand S_n(x)=Sum_{j in Z} x^j*(1-x^j)^(n+j) formally at zero and let T(n,k)=[x^(-k)]S_n(x). The outer sum includes every integer j; negative integer powers use Laurent-field inverses. The row includes its constant coefficient at k=0 and its exact inclusive final index floor((n+1)/2)^2. Coefficients and the weighted row sum are signed integers. The literal Laurent realization uses rational coefficients and the exact integer-to-rational coefficient bridge. The shared library note's claim quotation concerns formula (2); the quoted formula (3) is the distinct target here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/HannaLaurentFirstMoment.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedArbitrarily high simultaneous sigma peaksProved

The questionExistence of a witness entails existence of a least witness by well-ordering.

Why it mattersThe assertion asks whether divisor-sum peaks can simultaneously exceed both neighbours by an arbitrarily large factor.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

oeis a397578 sigma neighbour peaks

Ratushnyak's OEIS A397578 defines a(n) as the least k satisfying sigma(k) > n sigma(k-1) and sigma(k) > n sigma(k+1), and states: “Conjecture: a(n) exists for all n.” The exact assertion is preregistered in #14990.

Writing $\sigma(k)=\sum_{d\mid k}d$, the closed claim is

$$\forall n\in\mathbb N,\ \exists k\in\mathbb N,\quad k\ge2\ \land\ n\sigma(k-1)<\sigma(k)\ \land\ n\sigma(k+1)<\sigma(k).$$

The natural domain includes zero. Existence of a witness entails existence of a least witness by well-ordering.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Robin/SigmaNeighbourPeak.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Rather's real-rootedness question for general position polynomialsA finite nonempty connected graph has a real-rooted general position polynomial exactly when it is complete.Proved

The questionRather, Explicit Formulas and Unimodality Phenomena for General Position Polynomials, arXiv:2603.06930v3, Section 8 item 1, page 29, asks:

Why it mattersThe classification answers the real-rootedness question simultaneously for all nonempty finite connected simple graphs.

The ideaIts sole public theorem is RatherGPRealRooted.result; its other public declarations are the four source definitions.

Our contributionA finite nonempty connected graph has a real-rooted general position polynomial exactly when it is complete.

Verification details
Release source recordProved in Lean

Open problem

Rather's real-rootedness question for general position polynomials

B. A. Rather, Explicit Formulas and Unimodality Phenomena for General Position Polynomials, arXiv:2603.06930v3, Section 8 item 1, page 29, asks:

For no nontrivial family is it currently known whether ψ(G) is always real-rooted. Even partial real-rootedness results for chordal graphs, bipartite graphs, or claw-free graphs (in analogy with [8, 13]) would be significant.

Section 1 says: “All graphs considered are finite, simple and undirected.” Section 2, page 4, defines a general position set by the absence of three distinct selected vertices $x,y,z$ with $d(x,z)=d(x,y)+d(y,z)$, and says:

For each k ≥ 0, let α_k(G) denote the number of general position sets of size k.

The general position polynomial of G is ψ(G) = ∑_{k≥0} α_k(G)x^k.

The settlement is Proved: for every nonempty finite connected simple graph, $\psi(G)$ is real-rooted if and only if $G$ is complete.

Statement echo

| Source clause or scope | Formal counterpart | Fidelity | | --- | --- | --- | | Finite, simple, undirected graphs | V : Type, Fintype V, SimpleGraph V | Exact finite simple-graph carrier. | | Selected vertices avoid the distance equality | Frozen CylinderSevenTen.IsGeneralPosition and OnGeodesic | On connected graphs SimpleGraph.dist is shortest-path distance. | | Number of general position sets of size $k$ | alpha | Cardinality of the complete finite-subset filter; each subset occurs once. | | $\psi(G)=\sum_{k\geq0}\alpha_k(G)x^k$ | psi | Integer coefficients; the finite range ends at $|V|$, since larger subsets do not exist. | | Real-rootedness | IsRealRooted | Every complex zero has imaginary part zero. | | Classify connected graphs | claim, result | The nonempty connected scope is explicit; the conclusion is equivalence with G = ⊤. | | Decidable equality | DecidableEq V | Finite-set infrastructure; it places no additional mathematical restriction on finite graphs. |

The four public definitions are the literal preregistered conventions. IsGeneralPosition and OnGeodesic are reused from their frozen owner. No statement about disconnected graphs under raw SimpleGraph.dist is part of the formal classification.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/GeneralPosition/RatherRealRootedComplete.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationSharp bound provedFirst preprint 2025The termwise local-perception CHSH witness is at most 2 for every measurement settingThe termwise local-perception CHSH sum has sharp upper bound two for all two-qubit states and dichotomic settings.Proved

The questionRay, Local perception operators and classicality: new tools for old tests, arXiv:2511.00314v1, replaces each correlator Cxy = Tr(ρ Ax ⊗ By) of a Bell functional by its "locally perceived" version ax by Cxy.

Why it mattersThe witness is meant to certify that locally perceived statistics admit a local hidden-variable description.

The idea1. Probabilities. For ε, η ∈ {1, −1} the effects (1 + εAx)/2 and (1 + ηBy)/2 are positive, so Pxy(ε, η) = Tr(ρ (1 + εAx)/2 ⊗ (1 + ηBy)/2) = (1 + εax + ηby + εηCxy)/4 ≥ 0.

Our contributionThe termwise local-perception CHSH sum has sharp upper bound two for all two-qubit states and dichotomic settings.

Verification details
Release source recordProved in Lean

Open problem

The termwise local-perception CHSH witness is at most 2 for every measurement setting

R. K. Ray, Local perception operators and classicality: new tools for old tests, arXiv:2511.00314v1, replaces each correlator C_xy = Tr(ρ A_x ⊗ B_y) of a Bell functional by its "locally perceived" version a_x b_y C_xy. Here a_x = Tr(ρ_A A_x) and b_y = Tr(ρ_B B_y) are the local means of dichotomic observables (Eqs. (10), (11) and (35)). For CHSH (Eq. (19)) this gives the symmetric witness W^s_LPO(ρ) = sup over settings of a₀b₀C₀₀ + a₀b₁C₀₁ + a₁b₀C₁₀ − a₁b₁C₁₁ (Eq. (21)).

  • Theorem 4 proves W^s_LPO ≤ 2 when each party's settings are mutually orthogonal Bloch directions (Eq. (31)).
  • The author then writes: "we conjecture that for all other conceivable measurement settings, the upper-bound of W^s_LPO(ρ) saturates at 2."

The verbatim statements are in the literature note.

Issue #16628 fixes the reading.

  • The observables are the source's dichotomic observables "(POVM, or projective)": Hermitian A with 1 ± A ⪰ 0.
  • The witness is the termwise sum. The first line of Eq. (21), Tr(ρ (B)^A_ρ ⊗ (B)^B_ρ), mixes different settings and is not equal to that sum.

The conjecture is proved: the sum is at most 2 for every two-qubit state and all dichotomic settings, and 2 is attained.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/LocalPerceptionCHSH.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The termwise local-perception CHSH sum has sharp upper bound two for all two-qubit states and dichotomic settings.

    Read source ↗
  • Historical context

    First preprint year: 2025.

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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025No finite measure-and-prepare process perfectly clones the amplitude-damping familyNo finite measure-and-prepare ensemble perfectly clones every amplitude-damping channel into a positive number of outputs.Proved

The questionSkotiniotis, Cloning Quantum Channels, arXiv:2509.08059v1 (2025), Section V, amplitude-damping channels, performance comparison, state:

Why it mattersThe conjecture asks whether passing the unknown channels coherently can achieve an exact task that extracting a classical outcome and preparing a channel cannot.

The idea4. Contradiction. Each outcome polynomial vanishes on the interval except possibly at one point, hence has infinitely many roots and is zero.

Our contributionNo finite measure-and-prepare ensemble perfectly clones every amplitude-damping channel into a positive number of outputs.

Verification details
Release source recordProved in Lean

Open problem

No finite measure-and-prepare process perfectly clones the amplitude-damping family

P. Sekatski, Y. Guryanova, N. B. T. Kothakonda and M. Skotiniotis, Cloning Quantum Channels, arXiv:2509.08059v1 (2025), Section V, amplitude-damping channels, performance comparison, state:

In particular, it achieves ${\rm F}_{\rm CJ}^{\ADset|\rm coh}=1$ at $N=M$, which, for amplitude-damping channels, we conjecture to be impossible for any measure-and-prepare process.

This quotation is from the downloaded v1 source arxiv_v1.tex, lines 1684–1685, md5 abcdd29ea99d1c5adf0206c6b9a9d211. The source macro \ADset denotes the amplitude-damping family; the coherent process is the subject of “it”. The arXiv abstract page supplies the arXiv-issued DOI in the front matter.

The source defines measure-and-prepare processes as a measurement consuming $N$ channel queries, followed by an outcome-dependent global $M$-qubit channel preparation (TeX lines 1094–1103). Its preparations need not be tensor powers of an estimated channel; that narrower class is introduced separately at lines 1105–1115.

Issue #20683 fixes the target as impossibility of one such process reproducing $\mathcal E_\gamma^{\otimes N}$ for every $\gamma\in[0,1]$, at $N=M\geq1$. This is proved for finite outcome sets with the trace-pairing interpretation specified below. The compiled statement also allows every $N\geq0$ and every $M\geq1$ independently.

Precisely, the module defines the following objects:

  • adChoi is the unnormalized, reference-first Choi matrix $J_\gamma$ of the source's Kraus operators $A_0=\operatorname{diag}(1,\sqrt{1-\gamma})$ and $A_1=\begin{pmatrix}0&\sqrt\gamma\0&0\end{pmatrix}$ (source TeX lines 1538–1553). Its entry at $((r,o),(r',o'))$ is $\sum_{k=0}^1 A_k(o,r)\overline{A_k(o',r')}$.
  • D5.S3.Quantum.Recovery.ProductPrefixRigidity.productMap (H := fun _ => Fin 2 × Fin 2) (fun _ : Fin n => J_γ) supplies the coordinate tensor power: for assignments $x,y:\operatorname{Fin}n\to\operatorname{Fin}2\times\operatorname{Fin}2$, its entry is $\prod_i J_\gamma(x_i,y_i)$. At $n=0$ this is the one-dimensional matrix with entry one.
  • IsChannelChoi requires a preparation matrix to be positive semidefinite and to have reference marginal equal to the identity: $\sum_o K((r,o),(r',o))=\delta_{r,r'}$, where $r,r',o:\operatorname{Fin}M\to\operatorname{Fin}2$ and each pair denotes its sitewise assignment.
  • claim says that, for every pair of natural numbers $N,M$ with $M\geq1$, there are no finite $L$, fixed complex matrices $T_\ell$ on the $N$-fold Choi space and fixed preparation matrices $K_\ell$ on the $M$-fold Choi space satisfying IsChannelChoi, such that for every $\gamma\in[0,1]$ the numbers $p_\ell(\gamma)=\operatorname{tr}(T_\ell J_\gamma^{\otimes N})$ are real and nonnegative, sum to one, and satisfy $\sum_{\ell=0}^{L-1}p_\ell(\gamma)K_\ell=J_\gamma^{\otimes M}$. The real nonnegative requirement is expressed using the complex order. No further positivity or tester constraints are imposed on $T_\ell$.

The definitions and the source quotations are also described in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/AmplitudeDampingMeasurePrepareCloning.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSource 2026Zhan–Bie consecutive cyclic-stack fibresThe consecutive cyclic [123] stack has the exact even and odd fibres conjectured by Zhan and Bie.Proved

The questionAlex Zhan and Stella Bie, Cyclic-Pattern-Avoiding Stacks, INTEGERS 26 (2026), A15, Conjectures 3 and 4 on page 12 ask for the complete fibre of the consecutive cyclic [123] stack map.

Why it mattersExact enumeration of pattern-constrained objects requires both a construction and a converse classification.

The ideaSuccessful inputs start high and have at most one low after each high.

Our contributionThe consecutive cyclic [123] stack has the exact even and odd fibres conjectured by Zhan and Bie.

Verification details
Release source recordProved in Lean

Open problem

Zhan–Bie consecutive cyclic-stack fibres

Alex Zhan and Stella Bie, Cyclic-Pattern-Avoiding Stacks, INTEGERS 26 (2026), A15, Conjectures 3 and 4 on page 12 ask for the complete fibre of the consecutive cyclic [123] stack map. Write the stack top-first. For an incoming value x and top entries a,b, pop a exactly when x<a<b, b<x<a, or a<b<x; repeat until pushing x is allowed. Read the input left-to-right and flush the remaining stack top-first. The definition and Figure 3 on page 4 fix this orientation by 3124 -> 4213.

For every integer n >= 4, let m=floor(n/2), let y_n=(1,...,m,n,...,m+1), and let S_n contain every permutation of 1,...,n. The two claims are

|{p in S_n : cyclicStackSort(p)=y_n}| = 1 for every even n >= 4,

|{p in S_n : cyclicStackSort(p)=y_n}| = ceil(n/2) for every odd n >= 4.

Equivalently, for every natural m >= 2, the cardinalities at 2*m and 2*m+1 are 1 and m+1. There is no additional condition on the input beyond being a permutation, and no finite upper bound on n.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/CyclicStackPreimages.zhan_bie_conjectures_3_4

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2026.

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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 3/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025The spectral degree exponent is not submodularFive-vertex graphs refute edge-set submodularity of the spectral degree exponent.Refuted

The questionVan Mieghem, The spectral degree exponent of a graph, arXiv:2502.01815v1, Section 7, p. 15, asks:

Why it mattersThere is no anchor atom or coverage operation: this is an external named open-problem resolution through deposit-uncovered.

The ideaThese are the statements of degrees0, degreese, degreesf and degreesef; nonregular excludes regularity.

Our contributionFive-vertex graphs refute edge-set submodularity of the spectral degree exponent.

Scope & next questionPositive classes with a nontrivial edge-set lattice remain open here.

Verification details
Release source recordRefuted in Lean

Open problem

The spectral degree exponent is not submodular

M. A. Achterberg and P. Van Mieghem, The spectral degree exponent of a graph, arXiv:2502.01815v1, Section 7, p. 15, asks:

Several other properties of the SDE have not yet been established, e.g. whether the SDE is submodular or if the SDE can be used successfully in applications to enhance or reduce assortativity.

This is the tier-2 external question preregistered in https://github.com/the-omega-institute/trureturing/issues/19958. The settlement is Refuted for edge sets on a fixed vertex set.

Definition 2.1 and equation (2.1), p. 1, define the exponent by $\lambda_1=((1/N)\sum_i d_i^q)^{1/q}$ for a simple undirected nonregular graph. Theorem 2.2 places it in $[2,\infty]$. Here every exponent used is a finite real number and $\lambda_1$ is the supremum of the real adjacency spectrum.

Statement echo

| Source clause or explicit interpretation | Lean counterpart | Fidelity | | --- | --- | --- | | Simple undirected graph on $N$ vertices | SimpleGraph (Fin n) | Unweighted finite graphs form a subclass of the source's graphs. | | Degree mean $((1/N)\sum_i d_i^q)^{1/q}$ | degreePowerMean | Natural degrees and $n$ are cast to real numbers; both powers are real powers. | | Nonregular graph | IsRegular, negated in IsSpectralDegreeExponent | Regularity is exactly existence of a common natural degree, using SimpleGraph.IsRegularOfDegree. | | Largest adjacency eigenvalue | Real-spectrum supremum in IsSpectralDegreeExponent | The literal preregistered expression is inlined. The private correspondence spectral_sup_eq_edgeMax connects it to the reused SpikeEdgeEstimate.edgeMax in the proofs. | | Finite solution with $q\ge2$ | IsSpectralDegreeExponent G q | Includes nonregularity, the lower bound and the defining equation. Infinite exponents are outside this restricted claim. | | Submodular on the possible edges of one vertex set | claim | All $n,G,H,q_G,q_H,q_U,q_I$ are quantified; all four defining equations precede $q_U+q_I\le q_G+q_H$. | | Answer to the question | result | Negation of that whole universal proposition. |

The source does not specify the ground set. The fixed-vertex edge-set interpretation is fixed by #19958. Refuting its finite-exponent restriction also refutes a submodularity claim whose domain includes these instances. The induced-vertex interpretation has separate numerical evidence below. Uniqueness of the exponent and monotonicity of all degree means are not hypotheses of the Lean refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/PerronFrobenius/AchterbergVanMieghemSubmodularityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026A 23-edge coupling realizes the ten-mode Fourier transformBarriga et al. (arXiv:2609.05644) realize the discrete Fourier transform FN with one waveguide array, Phiout exp(-i H) Phiin = FN, and conjecture edge ranges for the coupling graphs; for N = 10 their third conjecture, with l = 5, requires at least 25 edges.Refuted

The ideaThe proof uses spectral estimates.

Our contributionBarriga et al. (arXiv:2609.05644) realize the discrete Fourier transform FN with one waveguide array, Phiout exp(-i H) Phiin = FN, and conjecture edge ranges for the coupling graphs; for N = 10 their third conjecture, with l = 5, requires at least 25 edges.

Verification details
Release source recordRefuted in Lean

arXiv

A 23-edge coupling realizes the ten-mode Fourier transform

Barriga et al., N-dimensional discrete Fourier transform via bosonic Hamiltonian, arXiv:2609.05644v1, realize the discrete Fourier transform $F_N$ with a single waveguide array, $\Phi^{\rm out}e^{-i\mathcal C}\Phi^{\rm in}=F_N$. The third conjecture of §III.D states that the remaining solutions for non-prime $N$ have between $N\ell/2$ and $|E(K_{N-1})|-\ell$ edges, and the same section fixes $\ell=5$ for $N=10$:

Third, all the other possible solutions for non-prime $N$ are in the following range of edges $\lvert E(R_{\ell})\rvert \leq \lvert E\rvert < \lvert E(K_{N-1})\rvert-\ell$

Issue #13297 fixes the reading, taking the most restrictive notion of solution. claim: every real symmetric $10\times10$ coupling matrix with nonnegative off-diagonal entries, connected support, and diagonal unimodular phases $\Phi^{\rm out},\Phi^{\rm in}$ with $\Phi^{\rm out}e^{-iH}\Phi^{\rm in}=F_{10}$ has at least 25 edges. The propagator is the frozen ProjectionProbabilityFlow.hamiltonianPropagator.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/BarrigaTenModeFourierEdgeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Bartoli--Stănică reduced critical-point conjectureA fixed reduced polynomial over F32 refutes the reduced critical-point conjecture.Refuted

The questionDaniele Bartoli and Pantelimon Stănică, Reduced polynomial lifts of APN permutations over Galois rings and effective non-APN bounds, arXiv:2608.30808v1 (31 August 2026), Conjecture 2 in Section 2 states:

Why it mattersConjecture 2 is a named open problem in a current arXiv paper.

The ideaUse the field with 32 elements represented by five-bit vectors for F2[t]/(t^5+t^2+1).

Our contributionA fixed reduced polynomial over F32 refutes the reduced critical-point conjecture.

Verification details
Release source recordRefuted in Lean

arXiv

Bartoli--Stănică reduced critical-point conjecture

Daniele Bartoli and Pantelimon Stănică, Reduced polynomial lifts of APN permutations over Galois rings and effective non-APN bounds, arXiv:2608.30808v1 (31 August 2026), Conjecture 2 in Section 2 states:

For every q = 2^m, the reduced representative f ∈ F_(q)[x] of every APN permutation of F_(q) has a critical point in F_(q).

Here APN means differential uniformity exactly two. The reduced representative has degree less than q and the critical point is a finite rational a with f'(a) = 0 in the same field. The dossier addresses this exact universal statement. It does not assert a result about unnormalized polynomial lifts, affine-invariance, Galois-ring permutations, or non-rational ramification.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/BartoliStanicaReducedCriticalPointRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The complex matrix exponential does not determine the halved commutatorThe exponential of a complex matrix does not determine its halved commutator operator.Refuted

The questionBathory, Commutator calculus and symbolic differentiation of matrix functions (2025 preprint; SIAM J.

Why it mattersBathory's calculus expresses derivatives of matrix functions through multiplication, anticommutator and commutator operators.

The ideaConsequently their halved commutator operators would agree.

Our contributionThe exponential of a complex matrix does not determine its halved commutator operator.

Scope & next questionIt does not settle other branch conventions or the other identities.

Verification details
Release source recordRefuted in Lean

Open problem

The complex matrix exponential does not determine the halved commutator

M. Bathory, Commutator calculus and symbolic differentiation of matrix functions (2025 preprint; SIAM J. Matrix Anal. Appl., 2026), arXiv:2505.07987v3, Conjecture 3.7. The following is the verbatim TeX text of cc_arxiv_1_.tex, line 506 (conjecture environment at lines 505–507), MD5 3f5e6f983ebd4bc8e0036ae4a4b880db:

With a~suitable generalization of Definition~\ref{Def}, all identities of Theorem~\ref{Tform}, Lemma~\ref{Had}, and Corollaries~\ref{corpow},~\ref{Cor} (except \eqref{eq1}, \eqref{eq2}) continue to hold for every $\mathbf{G}\in\mathbb{C}^{d\times d}$ and for any $\mathbf{A}\in\mathbb{C}^{d\times d}$ such that $\log\mathbf{A}$ exists.

The references in this quotation denote Definition 2.1, Theorem 3.2, Lemma 3.4, Corollaries 3.5 and 3.6, and the exceptions (3.55) and (3.56). The source and operator definitions are also documented in the literature note.

For a matrix $G$, write $L_G(X)=GX$, $R_G(X)=XG$, and $\operatorname{ad}G=(L_G-R_G)/2$, following Eq. (1.1), TeX line 38. Lemma 3.4 assumes $G\in\mathbb R^{d\times d}{\rm sym}$ and $A=e^G$; its Eq. (3.42), at TeX line 385, expresses $\operatorname{ad}_G$ as $(\log L_A-\log R_A)/2$.

Issue #20790 fixes the complex extension to be settled. Put $M=\mathbb C^{2\times2}$. The claim is that there exist two functions $\Lambda_L,\Lambda_R:M\to\operatorname{End}_{\mathbb C}(M)$ such that, for every $G\in M$,

$\displaystyle \frac12(L_G-R_G)=\frac12\bigl(\Lambda_L(e^G)-\Lambda_R(e^G)\bigr).$

Both operator-valued functions depend only on $A=e^G$. No continuity, analyticity, or logarithm identities are imposed on them. This reading of the conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/CommutatorLogarithmBranchRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2025Refutation of Bhagat, Kulkarni, Larsson and Murali Conjecture 4For {4,22,35,38} at heap 161, Bob gains from 77 to 78 under AvA instead of FvF.Refuted

The questionConsider any subtraction set S, and suppose both players act friendly in case of indifference.

Why it mattersThis is a named external conjecture in a published arXiv paper.

The ideaUse the repository's finite subtraction-game outcome recurrence with the four-step list [4, 22, 35, 38].

Our contributionFor {4,22,35,38} at heap 161, Bob gains from 77 to 78 under AvA instead of FvF.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of Bhagat, Kulkarni, Larsson and Murali Conjecture 4

Conjecture 4 of arXiv:2510.24280v2 states:

Consider any subtraction set S, and suppose both players act friendly in case of indifference. Then each player's PSPE utility is never worse than if both players have antagonistic tie-breaking rules.

The paper orders outcome pairs coordinatewise. In the formal reading, a finite nonempty duplicate-free list of positive natural numbers represents S, and both coordinates of the AvA outcome are required to be at most the corresponding FvF coordinates for every natural heap.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/BhagatKulkarniLarssonMuraliConjectureFourRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Three-central edge extremality against all metricsA four-point ultrametric has more central edges than a three-central metric.Refuted

The questionThe comparator is unrestricted: it need not be ultrametric.

Why it mattersIt refutes the published assertion under its unrestricted comparator class.

Our contributionA four-point ultrametric has more central edges than a three-central metric.

Verification details
Release source recordRefuted in Lean

Open problem

Three-central edge extremality against all metrics

Dovgoshey and Rovenska, On the Graph Corresponding to Center of Distances of Metric Space, arXiv:2610.11596v1, Conjecture 4.2, p. 13, assert an equivalence for every finite ultrametric space $X$ with $|X|\ge4$ and three central distances. The center is $C(X)={t\ge0:\forall p\in X;\exists y\in X,\ d(y,p)=t}$. The central graph joins distinct vertices whose distance belongs to $C(X)$; the diametrical graph joins vertices at the diameter.

Clause (i) says that the diametrical graph is complete bipartite on a partition $X=A\sqcup B$ with $\min(|A|,|B|)=2$, and that the central graph is its union with one star on each part. Clause (ii) says that its central graph has at most as many edges as every metric space $Y$ with $|Y|=|X|$ and $|C(Y)|=3$. The comparator is unrestricted: it need not be ultrametric.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Definition 1.1, nonnegative center including zero | center | The realizing point may depend on the base point. | | Definition 1.2, simple central graph | centralGraph | SimpleGraph.fromRel excludes loops; metric symmetry gives the stated unordered adjacency. | | Definition 2.14, diameter graph | diametricalGraph | Metric.diam on the whole carrier. | | Finite ultrametric, order at least four, three central values | claim quantifiers and hypotheses on $X$ | No restriction omitted. | | Complete bipartite partition, smaller part of size two, two stars | conditionI | Disjoint sets cover the carrier; subtype star centers ensure nonempty parts. Graph maps preserve the original carrier. | | Every equal-order three-central metric comparator | claim quantifiers on $Y$ and edge inequality | No ultrametric assumption on $Y$. | | Equivalence of (i) and (ii) | claim biconditional | result proves its literal negation. |

Fintype supplies the source finiteness. Nat.card counts actual center values and unordered edges. No binder adds a condition absent from the source on the nonempty finite carriers in question. The five public definitions retain the preregistered expressions; the module uses the existing metric and graph constructions rather than redefining those constructions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Geometry/DovgosheyRovenskaCentralEdges.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2025Coulter–Do Conjecture 5.4(a)Coulter and Do's b-deformed Jucys--Murphy operators fail to commute on the cyclic orbit X(6).Refuted

The questionXavier Coulter and Norman Do, arXiv:2506.04002v1, Conjecture 5.4(a), page 34:

Why it mattersThe source verifies the conjecture numerically for k ≤ 5.

The ideaPair partitions are fixed-point-free involutive partner maps on Fin (2k), with the paper's labels shifted down by one.

Our contributionCoulter and Do's b-deformed Jucys--Murphy operators fail to commute on the cyclic orbit X(6).

Verification details
Release source recordRefuted in Lean

Open problem

Coulter–Do Conjecture 5.4(a)

Xavier Coulter and Norman Do, arXiv:2506.04002v1, Conjecture 5.4(a), page 34:

The 𝒥-operators commute when restricted to 𝒳(k) — that is, 𝒥ₘ 𝒥ₙ (v) = 𝒥ₙ 𝒥ₘ (v) for 1 ⩽ m ⩽ n ⩽ k and for all v ∈ 𝒳(k).

Here k is a positive integer, the scalars are ℂ(b) with b an indeterminate, and 𝒳(k) is the orbit of the consecutive identity pair partition under the unital algebra generated by the operators in Definition 5.1. The theorem result proves ¬ claim, with all these quantifiers retained.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/DeformedJucysMurphyNoncommutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Further frontiersCounterexample foundFirst preprint 2006Cuntz's integral Fourier zero-free column questionEvery zero-free column of an integral Fourier matrix consists of integer units.Refuted

The questionMichael Cuntz, Integral modular data and congruences, Section 7, Question (5), arXiv:math/0611233v2 p. 21, journal p. 380, asks:

Why it mattersA zero-free column is necessarily a sign vector.

The ideaThe second identity follows from the first for square matrices.

Our contributionEvery zero-free column of an integral Fourier matrix consists of integer units.

Verification details
Release source recordRefuted in Lean

Open problem

Cuntz's integral Fourier zero-free column question

Michael Cuntz, Integral modular data and congruences, Section 7, Question (5), arXiv:math/0611233v2 p. 21, journal p. 380, asks:

Does an integral Fourier matrix $s$ exist with a column $k$ such that $\forall i$ $s_{ik}\ne0$ and $\exists i$ $s_{ik}\ne\pm1$?

The answer is negative for the exact existential assertion. Preregistration: #22614. The source note is Cuntz.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive finite size and an integer square matrix | claim quantifies n, hn, s | All sizes; source indices $1,\ldots,n$ correspond to Fin n, with source index one represented by zero. | | Definition 4.1(1): first column one and nonzero determinant | First two conjuncts of FourierAxioms | Integer determinant, unchanged. | | Definition 3.1: positive square roots | Third conjunct of FourierAxioms | Every square root is positive. | | Definition 4.1(2): $ss^T=\operatorname{diag}(d)$ | Fourth conjunct of FourierAxioms | Exact integer-to-real cast; $d$ is existentially quantified by IsIntegralFourierMatrix. | | Definition 4.1(3): $\sqrt{d_j}s_{ij}=\sqrt{d_i}s_{ji}$ | Fifth conjunct of FourierAxioms | Every pair of indices. | | Definition 4.1(4): integral structure constants | Sixth conjunct of FourierAxioms | Every triple; each real sum equals an integer witness. | | A zero-free column containing a nonunit | Final witnesses of claim | Every entry nonzero and some entry different from both $1$ and $-1$. | | Negative answer | result | The exact negation of claim. |

Definition 4.1 names integral Fourier matrices before adding congruence conditions (5)–(6); neither these conditions nor a $T$-matrix is required. The matrix equation fixes row norms. The reversed indices in Definition 3.1's explanatory norm sum are not an additional hypothesis. No extra binder or conjunct weakens the source question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Fourier/CuntzZeroFreeColumnRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    First preprint year: 2006. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryCounterexample foundFirst preprint 2026Refutation of the printed barrycade omitted-element relationThe printed i >= 3 relation in Conjecture 2 (2) is false at i = 3.Refuted

The questionMichał Dębski, Jarosław Grytczuk, Paweł Naroski, Bartłomiej Pawlik, Jakub Przybyło, and Małgorzata Śleszyńska-Nowak, Finite and infinite barrycades, arXiv:2609.18476v1, Section 3.1, Conjecture 2 (2), print:

Why it mattersIssue 8675 records the source quotation, its full universal scope, the literature-status search, and the proposed i = 3 counterexample.

The ideaStrong induction gives the first two rows in closed form.

Our contributionThe printed i >= 3 relation in Conjecture 2 (2) is false at i = 3.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the printed barrycade omitted-element relation

Michał Dębski, Jarosław Grytczuk, Paweł Naroski, Bartłomiej Pawlik, Jakub Przybyło, and Małgorzata Śleszyńska-Nowak, Finite and infinite barrycades, arXiv:2609.18476v1, Section 3.1, Conjecture 2 (2), print:

A2(i) = A1(i) + 1 for every i >= 3.

The paper defines A1(i) as the least positive integer omitted from the quasi-permutation rho_i and A2(i) as its first entry rho_i(1).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/DebskiBarrycadeOmittedElementRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Quantum informationCounterexample foundFirst preprint 2026The positive-branch Wigner-distance deficit does not factor through the equatorial inputThe Wigner tensor deficit cannot factor as the equatorial state's Wigner distance times a function of the other state.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionThe Wigner tensor deficit cannot factor as the equatorial state's Wigner distance times a function of the other state.

Verification details
Release source recordRefuted in Lean

Open problem

The positive-branch Wigner-distance deficit does not factor through the equatorial input

S. Dutta and Tushar, A Phase-Space Geometric Measure of Magic in Qubit Systems, arXiv:2603.20792v4, Section 5.4, Conjecture 5.5 (Factored deficit): "For equatorial magic states $\rho$ and qubit states $\sigma$ with $s(\sigma)>0$: $\mathrm{deficit}(\rho,\sigma)=C(\rho)\cdot f(\sigma)$ for a universal function $f\geq0$ of the Pauli invariants $(|r_x|,|r_y|,|r_z|)$ of $\sigma$, vanishing iff $s(\sigma)\leq0$." Here $C$ is the $\ell_1$ distance of the Wootters Wigner function to the stabilizer Wigner polytope, $s(\sigma)=\operatorname{sgn}(r_xr_yr_z)$ and $\mathrm{deficit}(\rho,\sigma)=(1+C(\rho))(1+C(\sigma))-1-C(\rho\otimes\sigma)$. The verbatim statements are in the literature note.

Issue #14651 reads the factorization clause: there is a function $f$ of $\sigma$ with $\mathrm{deficit}(\rho,\sigma)=C(\rho)f(\sigma)$ for every density matrix $\rho$ with $\langle Z\rangle_\rho=0$ and $C(\rho)>0$ and every density matrix $\sigma$ with $r_xr_yr_z>0$. Any $f$ satisfying the conjecture satisfies this clause.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/QubitWignerDeficitFactorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2025A nice-error-basis code whose stabilizer group is not normalIn the projective error model of Proposition 8.1 with n = 2, a group of order 16 acting on C^4, the code spanned by (1, 1, 1, 1) has four logical operators and four stabilizers, so their orders multiply to the order of the group, yet the stabilizers do not form a normal subgroup.Refuted

The questionEidesen models quantum errors by a projective error model (G, π) on a finite-dimensional Hilbert space V: a projectively faithful irreducible projective representation of a finite group.

Why it mattersThe follow-up paper arXiv:2606.02531 assumes that the logical group is normal, citing that it was normal in all examples studied so far; in this example the logical group is not normal either.

The ideaAll sixteen matrices have entries in {0, ±1, ±i}.

Our contributionIn the projective error model of Proposition 8.1 with n = 2, a group of order 16 acting on C^4, the code spanned by (1, 1, 1, 1) has four logical operators and four stabilizers, so their orders multiply to the order of the group, yet the stabilizers do not form a normal subgroup.

Verification details
Release source recordRefuted in Lean

Open problem

A nice-error-basis code whose stabilizer group is not normal

Eidesen models quantum errors by a projective error model (G, π) on a finite-dimensional Hilbert space V: a projectively faithful irreducible projective representation of a finite group. For a subspace W with orthogonal projection P_W, the logical operators and the stabilizers are

L_{(G,π)}(W) := { x ∈ G : P_W π(x) = π(x) P_W }, S_{(G,π)}(W) := { x ∈ G : P_W π(x) P_W ∈ 𝕋 P_W }.

After recording that no code has been found that is a Clifford code and a weak stabilizer code but not a stabilizer code, §11 of arXiv:2506.01843v3 asks Question 11.3:

Does there exist a Hilbert space V, a subspace W ⊂ V, and a projective error model (G, π) ∈ PEM_V with |G| = (dim V)^2, such that the equation |G| = |L_{(G,π)}(W)| · |S_{(G,π)}(W)| holds, but S_{(G,π)}(W) is not normal in G?

Issue #10156 fixes the readings: V = EuclideanSpace ℂ (Fin d); π takes values in d × d complex matrices acting on V; a projective representation has unitary values with π(x)π(y) = c π(xy) for some |c| = 1; projective faithfulness is the injectivity of q ∘ π; irreducibility means that the only invariant subspaces are 0 and V; P_W is the orthogonal projection; the orders are Set.ncard; and "not normal" means that some conjugate g s g⁻¹ of an element of S_{(G,π)}(W) lies outside it. The formal claim is the negative answer, and the formal result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/NiceErrorBasisNonNormalStabilizer.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Geometry & algebraCounterexample foundFirst preprint 2026Centred convex bodies and Gaussian dilationA centred 16-dimensional convex pyramid has non-log-concave Gaussian dilations.Refuted

The questionPolterovich, The Faber–Krahn position of convex bodies and Gaussian measure inequalities, arXiv:2607.21539v2, §5.2, p. 38, state verbatim:

Why it mattersCentral symmetry implies zero Lebesgue barycenter, but zero barycenter does not imply central symmetry.

The ideaThis is a compact convex pyramid with nonempty interior.

Our contributionA centred 16-dimensional convex pyramid has non-log-concave Gaussian dilations.

Verification details
Release source recordRefuted in Lean

Open problem

Centred convex bodies and Gaussian dilation

D. Faifman and I. Polterovich, The Faber–Krahn position of convex bodies and Gaussian measure inequalities, arXiv:2607.21539v2, §5.2, p. 38, state verbatim:

The B-theorem requires central symmetry. To prove Theorem 1.1, we circumvent this by proving a weaker concavity property of the Gaussian measure in a special case. However, we do not know a convex counterexample to the B-conjecture with its linear barycenter at the origin. Question 5.3. Let K ⊂ R^n be a convex body whose Lebesgue barycenter is at the origin. Is t ↦ γ_n(e^t K) log-concave? More generally, is t ↦ γ_n(e^{tS} K) log-concave for every symmetric matrix S?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/CentredGaussianDilationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Quantum informationCounterexample foundSource 2025Subadditivity of the entropy of quantum states over timeThe Werner–Holevo channel on the maximally mixed five-level state gives a two-time pseudo-density matrix whose signed entropy exceeds the sum of its marginal entropies.Refuted

The questionParzygnat, On Dynamical Measures of Quantum Information, Entropy 27(4), 331 (2025), Remark 1:

Why it mattersSubadditivity fails for general unit-trace Hermitian matrices with positive marginals; the conjecture asserts that the matrices arising from processes are an exception.

The ideaThe proof uses spectral estimates.

Our contributionThe Werner–Holevo channel on the maximally mixed five-level state gives a two-time pseudo-density matrix whose signed entropy exceeds the sum of its marginal entropies.

Verification details
Release source recordRefuted in Lean

Open problem

Subadditivity of the entropy of quantum states over time

J. Fullwood and A. J. Parzygnat, On Dynamical Measures of Quantum Information, Entropy 27(4), 331 (2025), Remark 1:

At present, we do not know of any examples of quantum states over time that violate subadditivity, which leads us to conjecture that the entropy functional S restricted to quantum states over time does in fact satisfy subadditivity.

A quantum state over time is $\varrho=\tfrac12{\rho\otimes I,\mathscr J[\mathcal E]}$ for a density matrix $\rho$ and a CPTP map $\mathcal E$, with the Jamiołkowski matrix $\mathscr J[\mathcal E]=\sum_{i,j}|i\rangle\langle j|\otimes\mathcal E(|j\rangle\langle i|)$, and $S(X)=-\mathrm{tr}(X\log|X|)$. The verbatim texts, including the restatement in J. Fullwood and B. Yang, arXiv:2608.28946 §6, are in the literature note. Issue #14994 preregisters the reading over finite Kraus channels and the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/FullwoodParzygnatSubadditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2025.

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Quantum informationCounterexample foundFirst preprint 2026The six-cycle graph state is not strongly 2-resistantThe C5 clause of the question is not answered here.Refuted

The questionHan, Zhang and Zhang, arXiv:2606.08561v1, page 8, Discussion, ask:

The ideaWrite its bits t,u and the retained bits r,x,y,z on qubits 1,3,4,5.

Our contributionThe C5 clause of the question is not answered here.

Verification details
Release source recordRefuted in Lean

arXiv

The six-cycle graph state is not strongly 2-resistant

Han, Zhang and Zhang, arXiv:2606.08561v1, page 8, Discussion, ask:

Several open problems remain. First, do C₅ and C₆ give strongly m-resistant graph states for m = 1 and m = 2, respectively? Here, “strong” means genuine multipartite entanglement rather than mere entanglement.

This dossier concerns the C₆, m = 2 clause. On qubits 0,...,5 its graph-state amplitudes are (-1) raised to the sum of the six cyclic edge products, divided by 8. Definition 2 of arXiv:2505.06567v1 requires initial genuine multipartite entanglement (GME), GME after every two-qubit loss, and full separability after every three-qubit loss. GME excludes a finite convex mixture of products across nontrivial bipartitions. The formal claim is IsStrongResistant 2 (cycleGraphState 6); result proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/CycleSixStrongTwoResistanceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2026Refutation of the Okada regular-trace discriminant identityHivert--Scott Conjecture 7.9 fails at N = 3 over the rationals with X = (1,2) and Y = (1): the regular-trace determinant is -16 and the cell-determinant product is 1.Refuted

The questionFlorent Hivert and Jeanne Scott, Diagrammatic Okada monoid and cellularity of the Okada algebra, arXiv:2609.01440v1, Conjecture 7.9, page 60:

Why it mattersThe conjecture connects the discriminant in the permutation-word basis to the Gram determinants of the half-diagram cell forms.

The ideaUse the quotient of the free algebra by the paper's three relations.

Our contributionHivert--Scott Conjecture 7.9 fails at N = 3 over the rationals with X = (1,2) and Y = (1): the regular-trace determinant is -16 and the cell-determinant product is 1.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the Okada regular-trace discriminant identity

Florent Hivert and Jeanne Scott, Diagrammatic Okada monoid and cellularity of the Okada algebra, arXiv:2609.01440v1, Conjecture 7.9, page 60:

det G_N = ∏_{S ∈ 𝕐𝔽𝕊_N} [det G^S_N]^{2 dim(Fib(S))}.

The preamble defines G_N(σ,τ) as tr[ϱ^reg(E_σ)ϱ^reg(E_τ)], where ϱ^reg is the left-regular representation of the Okada algebra. The parameters are a positive integer N, an arbitrary field K, X = (x₁,…,x_{N−1}) and Y = (y₁,…,y_{N−2}). The algebra has relations E_i² = x_i E_i, distant commutation, and E_{i+1}E_iE_{i+1} = y_i E_{i+1}. The conjecture concerns exact equality with this regular trace.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/OkadaDiscriminantRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedSource 2025Li–Guo Conjecture 2.22The triangle has signless Laplacian eigenvalues 4, 1, 1.Refuted

The questionWen-Jun Li and Ji-Ming Guo, A new conjecture on Laplacian eigenvalues and degree sequence of a graph, Filomat 39:13 (2025), 4551–4564, Conjecture 2.22, asserts

The ideaThe proof uses polynomial identities.

Our contributionThe triangle has signless Laplacian eigenvalues 4, 1, 1.

Verification details
Release source recordRefuted in Lean

Open problem

Li–Guo Conjecture 2.22

Wen-Jun Li and Ji-Ming Guo, A new conjecture on Laplacian eigenvalues and degree sequence of a graph, Filomat 39:13 (2025), 4551–4564, Conjecture 2.22, asserts

$$ \sum_{q_i\geq2}(q_i-2)^2\leq \left(1-\frac1{d_1}\right)\sum_{i=1}^n d_i(d_i-1). $$

The graph is finite, simple, undirected and loopless; $Q=D+A$ is its signless Laplacian, and all eigenvalues occur with algebraic multiplicity. Degrees are ordered decreasingly in the source. The denominator restricts the assertion to $d_1=\Delta(G)>0$. There is no connectedness or regularity hypothesis. Preregistration: #23307.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A graph with $n$ vertices | n : ℕ, G : SimpleGraph (Fin n) in claim | All finite labelled simple graphs; no additional structural restriction. | | $Q=D+A$ and the eigenvalues $q_i$ | Existing sourceMatrix G .signless and graphSpectrum G .signless | Actual real eigenvalues, retaining multiplicity. | | Sum over $q_i\geq2$ | TQ | Sum of $\max(q-2,0)^2$ over the entire spectral multiset; the summand at $q=2$ is zero. | | $d_1$ | SimpleGraph.maxDegree | The maximum vertex degree. | | Degree-sequence sum | R | Sum over vertices; permutation of the degree sequence preserves it. | | $1/d_1$ is defined | 0 < G.maxDegree | Positive denominator domain; an edgeless graph is excluded. | | The displayed upper bound | TQ G ≤ R G | Every degree is cast to $\mathbb R$ before subtraction and division. | | Universal assertion is false | result : ¬ claim | A triangle contradicts the unchanged assertion. |

TQ, R and claim retain the literal binding definitions in #23307. Graph matrices, degrees and multiplicity-preserving spectra reuse their existing owners.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SignlessLaplacian/LiGuoDegreeSequenceRefutation.result

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Quantum informationCounterexample foundFirst preprint 2026Data processing fails for the cumulant-based Rényi functional at every order above oneMeunson and Deesuwan (arXiv:2606.31205) define a cumulant-based quantum relative Renyi functional for alpha > 1 and state in the abstract and conclusion that its quantum data-processing inequality under arbitrary CPTP maps remains open.Refuted

The questionThe abstract and the conclusion state that the quantum data-processing inequality (QDPI) for arbitrary CPTP maps remains open.

Our contributionMeunson and Deesuwan (arXiv:2606.31205) define a cumulant-based quantum relative Renyi functional for alpha > 1 and state in the abstract and conclusion that its quantum data-processing inequality under arbitrary CPTP maps remains open.

Scope & next questionThe abstract and the conclusion state that the quantum data-processing inequality (QDPI) for arbitrary CPTP maps remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Data processing fails for the cumulant-based Rényi functional at every order above one

A. Meunson and T. Deesuwan, Cumulant-based quantum relative Rényi functional, arXiv:2606.31205v1, define for $\alpha>1$ (Definition 3, Section IV)

$$S_\alpha^{Q}(\rho\Vert\sigma)=\frac{1}{\alpha-1}\ln\operatorname{Tr} \left(\rho,e^{(\alpha-1)(\ln\rho-\ln\sigma)}\right)$$

on pairs with $\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma$. The abstract and the conclusion state that the quantum data-processing inequality (QDPI) for arbitrary CPTP maps remains open. The verbatim sentences and Definition 3 are in the literature note. This question differs from Conjecture 15 of the same paper ($\alpha=0$, commutativity-preserving channels), settled in the earlier dossier.

Issue #13439 fixes the reading. QDPI α asserts $S_\alpha(\mathcal N\rho\Vert\mathcal N\sigma)\le S_\alpha(\rho\Vert\sigma)$ for every dimension $d$, every pair of positive definite density matrices and every CPTP map $\mathcal N$ with positive definite outputs. Matrix logarithms are continuous functional calculus, as in the frozen $\alpha=0$ module. claim asserts that some $\alpha>1$ satisfies QDPI α. Faithful inputs meet the support condition, so refuting this restricted claim refutes QDPI for the paper's functional at every $\alpha>1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/CumulantRenyiDataProcessingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Geometry & algebraCounterexample foundFirst preprint 2025Misawa--Nishimura harmonic-strength pair refutationThe pair {2,4} refutes the proposed universal value five for two-element harmonic strength.Refuted

The questionMisawa and Nishimura, Spherical Designs on S¹ of Finite Harmonic Strength, arXiv:2505.06893v2, Conjecture 3.3, printed page 6:

Why it mattersThe assertion fixes the same minimum for every pair of distinct indices greater than one.

The ideaThe natural-number infimum then belongs to its nonempty set of attainable sizes, yielding a five-point unit-circle set with exactly this harmonic strength.

Our contributionThe pair {2,4} refutes the proposed universal value five for two-element harmonic strength.

Verification details
Release source recordRefuted in Lean

Open problem

Misawa--Nishimura harmonic-strength pair refutation

Misawa and Nishimura, Spherical Designs on S¹ of Finite Harmonic Strength, arXiv:2505.06893v2, Conjecture 3.3, printed page 6:

Let p ≠ q be integers with p, q > 1. Then N({p, q}, 2) = 5.

The quantity N(T, 2) is defined on the same printed page 6:

For a nonempty finite set T ⊂ N, define N(T, 2) := min{|X| | X ⊂ S¹, Hst(X) = T}.

and the harmonic strength is taken in the paper's working form (printed pages 2--3):

Finally, we identify S¹ with {z ∈ C | |z| = 1} and write P_k(X) := Σ_{x∈X} x^k for the k-th complex moment of a finite set X ⊂ S¹. By [2, Lemma 1], for X ⊂ S¹ one has Hst(X) = {k ∈ N | P_k(X) = 0}.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/MisawaHarmonicStrengthPairRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2025Integer diminished Sombor energy of edgeless graphsThe edgeless graph has zero diminished Sombor matrix, zero eigenvalues and zero energy.Refuted

The questionMovahedi, Diminished Sombor matrix, spectral radius, and energy of the graphs, arXiv:2508.06531v1, Section 5, page 19, states:

Why it mattersThe one-vertex edgeless graph has energy zero, an integer.

The ideaFor every edgeless graph, adjacency is false at every ordered vertex pair, so the non-edge branch gives the zero matrix without evaluating any degree quotient.

Our contributionThe edgeless graph has zero diminished Sombor matrix, zero eigenvalues and zero energy.

Verification details
Release source recordRefuted in Lean

Open problem

Integer diminished Sombor energy of edgeless graphs

F. Movahedi, Diminished Sombor matrix, spectral radius, and energy of the graphs, arXiv:2508.06531v1, Section 5, page 19, states:

Conjecture 5.1 There does not exist a graph whose diminished Sombor energy is an integer value.

For vertex degrees $d_i$, the real symmetric matrix has entry $\sqrt{d_i^2+d_j^2}/(d_i+d_j)$ on an edge and zero otherwise. Its energy is the sum of the absolute values of all eigenvalues, counted with multiplicity. The conjecture places no edge or connectedness restriction on the graph. Corollary 4.3, page 17, explicitly includes the edgeless graph as an equality case. Issue #13380 preregisters the positive order, finite simple graph statement and its edgeless counterexample.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/DiminishedSomborEnergyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
Number theoryConjecture refutedPosed in 2002Refutation of the A072872 prime-index upper-bound conjectureThe n = 6298 certificate refutes Cloitre's A072872 prime-index upper bound.Refuted

The questiona(n) is the smallest positive number k such that n divides 2^k - k

Why it mattersThis first-tier OEIS conjecture was printed in 2002 and is a universal bound on the least positive exponent satisfying the power-minus-index divisibility condition.

The ideaAt n = 6298, the endpoint certificate proves 6298 ∣ 2^77742 - 77742, and the incremental residue certificate proves that no positive k < 77742 satisfies the same divisibility.

Our contributionThe n = 6298 certificate refutes Cloitre's A072872 prime-index upper bound.

Scope & next questionIt asserts no corrected bound and no claim about other values of n.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A072872 prime-index upper-bound conjecture

OEIS A072872, NAME (verbatim):

a(n) is the smallest positive number k such that n divides 2^k - k

COMMENTS (verbatim; Benoit Cloitre, Jul 28 2002):

If n is a power of 2, a(n) = n. Conjecture : if n > 47, a(n) < prime(n). - Benoit Cloitre, Jul 28 2002

The literal claim is ∀ n > 47, a(n) < prime(n), where prime(n) is the n-th prime. The formal definition uses a(n) = sInf {k ∈ ℕ | 0 < k ∧ n ∣ 2^k - k} and discloses the convention sInf ∅ = 0; no general existence claim for this set is made.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/CloitrePowerMinusIndexPrimeBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2002. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedPosed in 2002Refutation of the A076502 nested-recurrence floor-offset conjectureCloitre's floor-offset conjecture for A076502 fails at n = 1167.Refuted

The questionConjecture : a(n) - cn is bounded and a(n)=floor(cn) + 0 + 1 or +2 - Benoit Cloitre, Nov 08 2002

Why it mattersCloitre's 2002 comment gives a precise universal floor-offset conjecture for a concrete nested recurrence and algebraic constant.

The ideaThe index invariant 1 <= a(k) <= k makes the nested recursion total and shows that both clamps in its definition are identities, yielding the literal source recurrence.

Our contributionCloitre's floor-offset conjecture for A076502 fails at n = 1167.

Further research on a Padovan-automatic description of a nested recurrence — see more details here: arXiv.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A076502 nested-recurrence floor-offset conjecture

OEIS A076502 %N (verbatim):

a(1)=1, a(n)=n-a(n-a(n-a(n-1)))

%C (verbatim; Benoit Cloitre, Nov 08 2002):

Conjecture : a(n) - c*n is bounded and a(n)=floor(c*n) + 0 + 1 or +2 - Benoit Cloitre, Nov 08 2002

%F (verbatim; Benoit Cloitre, Nov 08 2002):

a(n) is asymptotic to c*n with c=0.5698..is the positive root of x^3-x^2+2*x-1 - Benoit Cloitre, Nov 08 2002

The literal objects are the nested recurrence, c as the unique positive root of x^3-x^2+2*x-1, and the conjunction that a(n)-c*n is bounded and a(n) = floor(c*n) + 0, 1 or 2 for every positive n. The formal theorem refutes only the second conjunct, which refutes the conjunction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/CloitreNestedRecurrenceFloorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Dated conjecture statement: 2002. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedPosed in 2010Refutation of the A104863 floor-square-root recurrence conjectureThe term at n = 17 refutes the conjectured recurrence for OEIS A104863.Refuted

The questiona(n) = floor(sqrt(a(n-1)^2 + a(n-2)^2)), a(1)=10, a(2)=30.

Why it mattersThis OEIS conjecture was printed in 2010 as a universal recurrence for a sequence introduced in 2005.

The ideaThe Nat.eqsqrt' certificates evaluate the defining recurrence through index 17.

Our contributionThe term at n = 17 refutes the conjectured recurrence for OEIS A104863.

Scope & next questionIt asserts no corrected recurrence, no closed form, and no claim about any other value of n.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A104863 floor-square-root recurrence conjecture

OEIS A104863, NAME (verbatim):

a(n) = floor(sqrt(a(n-1)^2 + a(n-2)^2)), a(1)=10, a(2)=30.

FORMULA conjecture line (verbatim; Ralf Stephan, Nov 15 2010):

For n>=17, a(n) = a(n-2) + a(n-4) + 1 (conjectured). If true then for m>5, a(2*m+1) = 4*F(m) + 25*F(m+1) + 1 and a(2*m+2) = 8*F(m) + 30*F(m+1) + 1 with F(n) = A000045(n). - Ralf Stephan, Nov 15 2010

The sequence begins with a(1) = 10 and a(2) = 30, and each later term is the natural floor square root of the sum of the squares of the two preceding terms. The literal conjecture is ∀ n ≥ 17, a(n) = a(n−2) + a(n−4) + 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/StephanFloorSqrtRecurrenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Dated conjecture statement: 2010. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundA397436: the unrestricted product supercongruenceA one-factor positive instance refutes the unrestricted signed product congruence.Refuted

The questionThe selected external assertion is OEIS A397436 revision 44, Conjecture 3, quoted in Library/Certificates/oeis2026a397436.md.

Why it mattersquestionanswered: does the published unrestricted universal congruence hold?

The ideaInstantiate the same r everywhere by 1, take m(0)=1, M=1 and p=5.

Our contributionA one-factor positive instance refutes the unrestricted signed product congruence.

Scope & next questionIt does not settle Conjectures 1 or 2, or prove or refute the special negative-parameter two-factor sequence's supercongruence.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

A397436: the unrestricted product supercongruence

The selected external assertion is OEIS A397436 revision 44, Conjecture 3, quoted in Library/Certificates/oeis2026a397436.md. It asserts the following complete same-r event, with arbitrary signed m and inclusive endpoints:

noncomputable def U {r : ℕ} (m : Fin r → ℤ) (M n : ℕ) : ℤ :=
  ∑ k ∈ Finset.range (M * n + 1), ∏ i : Fin r, Ring.choose (m i * (n : ℤ)) k

def claim : Prop :=
  ∀ (r : ℕ), 0 < r → ∀ (m : Fin r → ℤ) (M : ℕ), 0 < M →
    ∀ (p : ℕ), Nat.Prime p → 5 ≤ p →
      (p : ℤ) ^ (3 * r) ∣ U m M (p ^ r) - U m M (p ^ (r - 1))

theorem result : ¬ claim

The last line displays the theorem type; its proof lives in D5. Natural subtraction r-1 agrees with the source exponent because r is positive.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/A397436GeneralSupercongruenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsConjecture refutedFirst preprint 2025Refutation of Ostrovskii–Shcherbakov Conjecture 6.2A self-conjugate pair in the closed unit disk violates the shifted Schur-ratio bound in Conjecture 6.2 of Ostrovskii and Shcherbakov.Refuted

The questionOstrovskii and Shcherbakov, Amplitude maximization in stable systems, Schur positivity, and some conjectures on polynomial interpolation, arXiv:2508.13554v2 [math.CV], state on page 15:

Why it mattersThe conjecture is a 2025 externally named open problem.

The ideaSorting the arm and leg gives the weak row and strict column, counting each tableau once.

Our contributionA self-conjugate pair in the closed unit disk violates the shifted Schur-ratio bound in Conjecture 6.2 of Ostrovskii and Shcherbakov.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Ostrovskii–Shcherbakov Conjecture 6.2

Ostrovskii and Shcherbakov, Amplitude maximization in stable systems, Schur positivity, and some conjectures on polynomial interpolation, arXiv:2508.13554v2 [math.CV], state on page 15:

For all $0 \le k < n \le t$ and self-conjugate $z_{1:n} \in \mathbb{D}^n$, $|Q_{t,n,k}(z_{1:n} + 1_n)| \le 1$. Moreover, if the grid additionally satisfies $\Re(z_{1:n}) \in \mathbb{R}^n_+$, then $|s_{(t-n|n-k-1)}(z_{1:n})| \le \binom{t}{n} |e_{n-k}(z_{1:n})|$.

The same paper labels this Conjecture 6.2 equivalent to Conjecture 6.1. The formal claim retains both clauses, the self-conjugacy conditions, the closed unit disk, and the natural inequalities k<n<=t.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Interpolation/SelfConjugateGridSchurRatioRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026A 1225-vertex graph refutes the star bound on the minimal-energy uniform overlap of quantum batteriesA triangulated complete graph refutes the proposed star bound for the uniform overlap of the least adjacency eigenspace.Refuted

The questionTkatchenko, Star Topology Optimizes the Charging Power of Quantum Batteries, arXiv:2603.10853v1, study a battery of N fermionic cells coupled along a graph G. - Overlaps and mode energies. The adjacency matrix A has orthonormal eigenvectors uk with eigenvalues εk and uniform overlaps wk = (1ᵀuk)² (Eq. (11)).

Why it mattersThe conjecture was the authors' primary evidence that the star stays optimal for the charging power in a non-zero local field (§III.C).

The ideaLet Gm be the q-triangulation of Km with q = 3(m − 1).

Our contributionA triangulated complete graph refutes the proposed star bound for the uniform overlap of the least adjacency eigenspace.

Verification details
Release source recordRefuted in Lean

Open problem

A 1225-vertex graph refutes the star bound on the minimal-energy uniform overlap of quantum batteries

M. Sarkis, O. A. Prośniak, S. Nigro and A. Tkatchenko, Star Topology Optimizes the Charging Power of Quantum Batteries, arXiv:2603.10853v1, study a battery of N fermionic cells coupled along a graph G.

  • Overlaps and mode energies. The adjacency matrix A has orthonormal eigenvectors u_k with eigenvalues ε_k and uniform overlaps w_k = (1ᵀu_k)² (Eq. (11)). With a local field h, the mode energies are E_k(h) = (N/M)ε_k + h (Eq. (14)).
  • The charging power. It is a weighted mean of the w_k, and the authors single out w_min = Σ_{k′} w_{k′} over the modes of least energy.
  • The star. It gives w_min(S) = N/2 − √(N − 1) (Eq. (23)).
  • The conjecture. "Conjecture 1 (Star optimality for the uniform overlap). Let G be a non-empty graph on N vertices. Its minimal energy uniform overlap w_min(G) is upper bounded by w_min(G) ≤ w_min(S) = N/2 − √(N − 1)." The authors base it on an exhaustive check for N ≤ 7 and random samples.

The verbatim statements are in the literature note.

Issue #16578 fixes the reading.

  • Because E_k(h) is increasing in ε_k, w_min(G) is the squared length of the projection of 1 onto the lowest adjacency eigenspace.
  • "Non-empty" means at least one edge.
  • The conjecture is refuted.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/QuantumBatteryUniformOverlapRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025A Clifford unitary that changes the operator norm of the MUB autocorrelation matrixIn dimension five, a Fourier Clifford changes the Euclidean operator norm of the matrix of cyclic MUB probability autocorrelations.Refuted

The questionStacey, SICs, the Stabilizer Polytope and Majorization, arXiv:2509.16819v1, Section III, defines A(ψ){mk} = Σj p{m,j} p{m,j+k}, where p{m,j} are the outcome probabilities in the Weyl–Heisenberg mutually unbiased bases.

Why it mattersA magic monotone must be invariant under the free Clifford unitaries.

The idea1. Clifford. F is unitary and conjugates the shift and the clock to displacement operators.

Our contributionIn dimension five, a Fourier Clifford changes the Euclidean operator norm of the matrix of cyclic MUB probability autocorrelations.

Verification details
Release source recordRefuted in Lean

arXiv

A Clifford unitary that changes the operator norm of the MUB autocorrelation matrix

B. C. Stacey, SICs, the Stabilizer Polytope and Majorization, arXiv:2509.16819v1, Section III, defines A(ψ){mk} = Σ_j p{m,j} p_{m,j+k}, where p_{m,j} are the outcome probabilities in the Weyl–Heisenberg mutually unbiased bases. It then states:

Acting upon $\ket{\psi}$ with a Clifford unitary can at most permute the rows and columns of the matrix $A$. … Indeed, let $||\cdot||$ be any unitarily invariant norm; because permutation matrices are a very special subset of the unitary matrices, left- and right-multiplying $A$ by any pair of permutation matrices will leave $||A||$ unchanged. It seems, then, that we have a large family of new candidate monotones that are all extremized by Weyl--Heisenberg SIC fiducials.

The verbatim passage is in the literature note.

Issue #18928 fixes the reading: for d = 5, the Euclidean operator norm of A is invariant under every Clifford unitary, for every pure state. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/StabilizerAutocorrelationNormRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2026The traceless BPS cohomology at (p, N) = (5, 4) is not free over the diagonal creatorsThe traceless cohomology at p = 5 and N = 4 is not free over the exterior algebra of the diagonal creators.Refuted

The questionTierz, BPS spectra of Tr[Ψ^p] matrix models for odd p, arXiv:2604.27164 (v1 April 2026, v2 August 2026), studies the cohomology of wedge multiplication by Qp = Tr(Ψ^p) on the exterior algebra of the N² fermionic creators Ψij. - The trace fermion χ splits off (v2 Theorem 2), leaving the cohomology B•0 of Qp = Tr(A^p) on the traceless creators A = Ψ − (χ/√N)·1. - The traceless diagonal creators anticommute with Qp, so B•0 is a graded module over their exterior algebra Λ•(C^{N−1}).

Why it mattersFreeness would explain the factors of 1 + x in the BPS generating functions beyond the structural ones (v2 Corollary 3) and would reduce the BPS spectrum to a primitive space M^{(p,N)} with Hilbert series p^{b{p,N}} x^{qmin} T^{(p)}N(x).

The ideaCoordinates: the fifteen traceless creators uij = Eij − δij I/4 are ordered so that A = [[a,d,e,f],[g,b,h,i],[j,k,c,l],[m,n,o,−a−b−c]], and Q = Tr(A⁵) is computed from the matrix product.

Our contributionThe traceless cohomology at p = 5 and N = 4 is not free over the exterior algebra of the diagonal creators.

Verification details
Release source recordRefuted in Lean

arXiv

The traceless BPS cohomology at (p, N) = (5, 4) is not free over the diagonal creators

M. Tierz, BPS spectra of Tr[Ψ^p] matrix models for odd p, arXiv:2604.27164 (v1 April 2026, v2 August 2026), studies the cohomology of wedge multiplication by Q_p = Tr(Ψ^p) on the exterior algebra of the N² fermionic creators Ψ_ij.

  • The trace fermion χ splits off (v2 Theorem 2), leaving the cohomology B•_0 of Q_p = Tr(A^p) on the traceless creators A = Ψ − (χ/√N)·1.
  • The traceless diagonal creators anticommute with Q_p, so B•_0 is a graded module over their exterior algebra Λ•(C^{N−1}).

The source states freeness in two places, verbatim in the literature note:

  • v1 §5 marks "(conjectural) H•_red is free over Λ•(C^{N−1})" as "the content of the open problem";
  • v2 §6.3 calls an isomorphism of graded modules B•_0 ≅ Λ•(C^{N−1}) ⊗ M^{(p,N)} "a sufficient explanation for the remaining factors of 1+x" and says that "such freeness … is not proved here".

Issue #16298 fixes the reading: for odd p ≥ 5 with p ≤ 2N − 1, B•_0 is a free Λ•(C^{N−1})-module. The admissible case (p, N) = (5, 4) is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Fermionic/TierzTracelessFreenessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Analysis & dynamicsConjecture refutedFirst preprint 2026Visomirski–Griffin frozen-wave conjectureAssuming periodic boundary conditions and initial conditions of the form,Refuted

The questionMatthew Visomirski and Christopher Griffin, Coherent Structures and Travelling Waves in Spatial Replicators from a Biased Volterra Lattice, arXiv:2601.06314v1, Section 4.2, Conjecture 2 (page 19), asserts:

Why it mattersThe proof also establishes the family homogenization statement VG-R on its live path.

The ideaUniqueness follows from finite-strip boundedness, local Lipschitz continuity of the finite-dimensional reaction, and a vector energy estimate.

Our contributionAssuming periodic boundary conditions and initial conditions of the form,

Verification details
Release source recordRefuted in Lean

Open problem

Visomirski–Griffin frozen-wave conjecture

Matthew Visomirski and Christopher Griffin, Coherent Structures and Travelling Waves in Spatial Replicators from a Biased Volterra Lattice, arXiv:2601.06314v1, Section 4.2, Conjecture 2 (page 19), asserts:

Consider the one-dimensional spatial replicator, i.e., Eq. (6) with reaction term generated by a $2n$-cycle with $n\geq2$. Assuming periodic boundary conditions and initial conditions of the form,

$$ u_i(x,0)=\begin{cases}(1+\sin x)/(2n),&i\text{ odd},\ (1-\sin x)/(2n),&i\text{ even},\end{cases} $$

the dynamics will evolve to a frozen wave stationary solution. That is, ecological niches will be maintained.

The simulated parameters are $a=-1/2$, $D=1/100$, on $[-\pi,\pi]$ with periodic boundary conditions. Preregistration: #14205.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/FluidDynamics/ReactionDiffusion/VisomirskiGriffinFrozenWaveRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2025Critical values of parametrized quantum circuitsA product of fifteen two-qubit Pauli rotations has full derivative rank almost everywhere, while its critical values have Hausdorff dimension at least fourteen.Refuted

The questionKilloran, Unitary synthesis with optimal brick wall circuits, arXiv:2511.16736v1, §V (page 9):

Why it mattersWith it, they "know [their] circuit families to be universal for n≤5".

Our contributionA product of fifteen two-qubit Pauli rotations has full derivative rank almost everywhere, while its critical values have Hausdorff dimension at least fourteen.

Verification details
Release source recordRefuted in Lean

Open problem

Critical values of parametrized quantum circuits

D. Wierichs, K. Kottmann and N. Killoran, Unitary synthesis with optimal brick wall circuits, arXiv:2511.16736v1, §V (page 9):

Conjecture 3. The critical values F(C) of a PQC that has full rank almost everywhere in T^d form an at most (d−2)-dimensional subset of G.

A PQC (Appendix F, Definition 1) consists of Clifford gates and $d$ rotations from ${R_X,R_Y,R_Z}$, each input feeding one rotation without preprocessing, viewed as $F:T^d\to G\subset\mathbb C^{2^n\times 2^n}$ with $d=\dim G$; $C$ is the set of inputs at which the Jacobian has rank below $d$. The verbatim texts are in the literature note. Issue #15435 preregisters the two-qubit, $G=SU(4)$, $d=15$ specialization with Hausdorff dimension and the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/PauliRotationCriticalValues.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresConjecture refutedFirst preprint 2026The phase-shifted extension of Conjecture 2A pi-flux four-cycle refutes the phase-shifted parity conjecture.Refuted

The questionWong, Bravyi, Gosset and Liu, Lee-Yang tensors and Hamiltonian complexity, arXiv:2602.03605v1, Section 5.5, Conjecture 2:

Why it mattersThe original zero-phase Conjecture 2 is outside this negative conclusion.

The ideaIncreasing order with multiplicity then gives the strict inequality needed to negate claim.

Our contributionA pi-flux four-cycle refutes the phase-shifted parity conjecture.

Scope & next questionThe original zero-phase Conjecture 2 and the phase-shifted gap conjecture remain open in this delivery.

Verification details
Release source recordRefuted in Lean

Open problem

The phase-shifted extension of Conjecture 2

Wong, Bravyi, Gosset and Liu, Lee-Yang tensors and Hamiltonian complexity, arXiv:2602.03605v1, Section 5.5, Conjecture 2:

Suppose n ≥ 2, s ∈ [0,1), and the interaction graph is connected. Then the second smallest eigenvalue of H_s coincides with the smallest eigenvalue of H_s restricted to the odd-weight subspace.

Section 5.6 asks:

We leave as an open question whether Conjectures 1, 2, 3 hold for phase-shifted EPR-like Hamiltonians.

The question settled here is the extension of Conjecture 2. Preregistration: #18759. This is an external published question, tier 1.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | $n\ge2$ | claim quantifies over natural n with 2 ≤ n. | | Connected interaction graph | SimpleGraph (Fin n), its adjacency decision procedure, and G.Connected. The decision procedure is computational infrastructure. | | $0\le s<1$ | The two inequalities in claim. | | $h(z)=\lvert00\rangle\langle00\rvert+z\lvert11\rangle\langle00\rvert+\overline z\lvert00\rangle\langle11\rvert+\lvert z\rvert^2\lvert11\rangle\langle11\rvert$ | hpq has exactly these four entries, with the identity on all other bits. | | $H_s(\theta)=-\sum_{{p,q}\in E}h_{p,q}(se^{i\theta_{p,q}})$ | H counts each undirected edge once using p < q. | | Arbitrary edge angles in $[0,2\pi)$ | Real representatives in the exponential; the counterexample uses only $0$ and $\pi$, both in the source interval. | | Second smallest eigenvalue | eigAsc, sorted with multiplicity, and List.getD at index $1$. | | Smallest eigenvalue on the odd-weight subspace | Restriction by Matrix.submatrix to the odd computational basis and index $0$. The odd weight count is inlined literally in claim; the private certificate relates it to Mathlib's hammingNorm. | | Negative settlement | result : ¬ claim. |

The arbitrary default in List.getD and the empty non-Hermitian branch of eigAsc totalize the definitions. The witness matrices are Hermitian and the relevant lists have the required entries. The computational basis is Fin n → Fin 2 directly.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/LeeYangPhaseShiftedParityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025The Negative Coefficients of the Two-Color Partition Series D'(2,3,n)The two-color partition series D-prime at parameters 2 and 3 has negative coefficients exactly at degrees 10 and 22.Proved

The questionAndrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 3, Conjecture 4.

The ideaTheorem 4 and Theorem 1 of the paper for k = 2, m = 3, together with finite partial fractions, give an exact formula for D'(2,3,n) in terms of the triangular-pair count of Gauss's identity and divisor sums twisted by a character modulo small moduli.

Our contributionThe two-color partition series D-prime at parameters 2 and 3 has negative coefficients exactly at degrees 10 and 22.

Verification details
Release source recordProved in Lean

Open problem

The Negative Coefficients of the Two-Color Partition Series D'(2,3,n)

George E. Andrews and Mohamed El Bachraoui, Certain positive q-series and inequalities for two-color partitions, arXiv:2507.09276v1, Section 3, Conjecture 4. With Σ_n D'(k,m,n) qⁿ = Σ_{j≥0} q^{m(2j+2)} (q^{2j+4}, q^{2j+2+2k}; q²)_∞ / (q^{2j+3}; q²)_∞² (equation (1.3)), the conjecture states that the only negative coefficients of Σ_n D'(2,3,n) qⁿ occur at n = 10 and n = 22.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/TwoColorPartition/AndrewsElBachraouiFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThree-adic nondivisibilityInteger quadratic products and Bala's valuation conjectures.Proved

The questionPreregistration 18635 maps to D5.S3.Arith.Congruence.BalaA363412.result (Proved).

Why it mattersThis is an explicit OEIS conjecture in the source family documented in bala3634-valuations.md.

The ideaThe primary dossier gives the four issue-to-result mappings.

Our contributionInteger quadratic products and Bala's valuation conjectures.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Three-adic nondivisibility

$3\nmid b(3n+1)$ for every $n\in\mathbb N$, including zero. Here $a(n)$ and $b(n)$ are the integer coordinates of $\prod_{k=0}^n(1+k\sqrt{-3})$. Preregistration #18635 maps to D5.S3.Arith.Congruence.BalaA363412.result (Proved).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BalaA363412.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedTwo or three prime occurrences in Bouras A363102Every occurring prime has two or three indices in Bouras A363102.Proved

The questionBouras, OEIS A363102, COMMENTS, Conjecture 2:

Why it mattersThe result determines the occurrence count for every prime value, uniformly in the prime and in the index.

Our contributionEvery occurring prime has two or three indices in Bouras A363102.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Two or three prime occurrences in Bouras A363102

M. Bouras, OEIS A363102, COMMENTS, Conjecture 2:

Conjecture 2: All prime numbers appear either twice (same as A356247 and A357127) or three times.

For $n\ge3$, $a(n)$ is the positive reduced denominator of the rational continued fraction $1/(2-3/(3-4/(\cdots-n/(-2))))$. The claim quantifies over primes that occur, as specified in preregistration #18627. It does not assert that every prime occurs. This dossier anchors only Conjecture 2; the $-4$ sequence has its own dossier.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Rational continued fraction with tail $-2$ | BourasA363102.a uses BourasA356360.K n (-2) and Rat.den | Literal reciprocal right fold | | Prime values | Nat.Prime and the existential occurrence hypothesis in claim | Excludes primes absent from the sequence | | OEIS offset $n=3$ | $3\le n$ in the occurrence guard and counted set | All larger indices are included | | Twice or three times | The disjunction of Set.ncard = 2 and Set.ncard = 3 | The proof exhibits a pair or a triple |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/BourasA363102PrimeMultiplicity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Packing k-Domatic Colourings of Paths Need More Than k + 1 ColoursThe path on 56 vertices refutes the proposed bound for packing 28-domatic colourings.Refuted

The questionBoštjan Brešar, Jasmina Ferme and Wenjie Hu, Partitioning an S-packing coloring into broadcast dominating sets, arXiv:2610.03477v1, Section 5: "Problem 2.

Why it mattersIts counting lemma proves that a packing k-domatic colouring of Pn with colours at most t and n ≥ t + 1 satisfies 15t ≥ 16k − 12; hence χ{ρ,k}(Pn) ≥ ⌈(16k − 12)/15⌉ for k ≥ 3 and n ≥ 2k, which exceeds k + 1 for k ≥ 28.

The ideaWith d = t − k and m = 8d + 8, the vertices of low colours in E cannot reach vertex m, vertices outside E inside [1, t + 1] are few, and every broadcaster beyond t + 1 reaching m has a large colour and is constrained by the packing distance against the vertex of the same colour in E.

Our contributionThe path on 56 vertices refutes the proposed bound for packing 28-domatic colourings.

Verification details
Release source recordRefuted in Lean

Open problem

Packing k-Domatic Colourings of Paths Need More Than k + 1 Colours

Boštjan Brešar, Jasmina Ferme and Wenjie Hu, Partitioning an S-packing coloring into broadcast dominating sets, arXiv:2610.03477v1, Section 5: "Problem 2. Is it true that χ_{ρ,k}(P_n) ≤ k + 1 for any k ≥ 3 and any n ≥ 2k?" A packing k-domatic colouring f with colours in [t] keeps two vertices of colour j at distance greater than j and admits a partition of the vertices into k classes, each broadcast-dominating every vertex x through some a in the class with d(x, a) ≤ f(a); χ_{ρ,k}(G) is the least such t.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/PackingDomatic/PackingDomaticPath.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationCounterexample foundFirst preprint 2008The trace exponential is not exponentially convex on density matricesThe trace exponential fails exponential convexity on density matrices, already on four qubit states.Refuted

The questionJop Briët and Peter Harremoës, Properties of classical and quantum Jensen-Shannon divergence, Physical Review A 79, 052311 (2009), arXiv:0806.4472v4, Definition 4 (Exponential convexity):

Why it mattersThe conjecture proposes a route from exponential convexity to negative definiteness of the quantum Jensen–Shannon divergence.

The ideaUse the maximally mixed state and three real rank-one projectors:

Our contributionThe trace exponential fails exponential convexity on density matrices, already on four qubit states.

Verification details
Release source recordRefuted in Lean

Open problem

The trace exponential is not exponentially convex on density matrices

Jop Briët and Peter Harremoës, Properties of classical and quantum Jensen-Shannon divergence, Physical Review A 79, 052311 (2009), arXiv:0806.4472v4, Definition 4 (Exponential convexity):

Let X be a convex set and φ : X → R a mapping. Then φ is said to be exponentially convex if the function X × X → R given by (x, y) → φ((x+y)/2) is positive definite.

§VI (Conclusions and open problems):

Whether (QJD1)^{1/2} is a metric for all mixed states remains unknown. However, based on a large amount of numerical evidence, we conjecture the function A → Tr(e^A) to be exponentially convex for density matrices A. Proving this would imply that QJDα is negative definite for α ∈ (0, 2], and hence the square of a metric that can be embedded in a real Hilbert space.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/TraceExpExponentialConvexityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
  • Historical context · 3/4 · 10% weight

    First preprint year: 2008. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Further frontiersQuestion resolvedFirst preprint 2025Brown–George's argument-dependent degree-root boundEvery nonzero degree root satisfies the Brown–George argument-dependent modulus bound.Proved

The questionGeorge, On the roots of degree polynomials, Australasian Journal of Combinatorics 92(2), 164–183, Conjecture 4.1, p. 180, states:

Why it mattersPreregistration 19987 selects Conjecture 4.1.

The ideaNonnegativity excludes argument zero; conjugation handles negative argument.

Our contributionEvery nonzero degree root satisfies the Brown–George argument-dependent modulus bound.

Verification details
Release source recordProved in Lean

Open problem

Brown–George's argument-dependent degree-root bound

J. I. Brown and I. C. George, On the roots of degree polynomials, Australasian Journal of Combinatorics 92(2), 164–183, Conjecture 4.1, p. 180, states:

If z is a nonzero degree root of a graph of order n with argument arg(z) ∈ (−π, π], then |z| ≤ (n − 1)^{|arg(z)/π|}.

Section 1 defines $D(G;x)=\sum_{v\in V(G)}x^{\deg(v)}$. The arXiv version is The Degree Polynomial, arXiv:2505.04882v1. The settled proposition is

$$ \forall n\ge1;\forall G\text{ simple on }\operatorname{Fin}(n);\forall z\in\mathbb C, \quad z\ne0\land D(G;z)=0\Longrightarrow |z|\le(n-1)^{|\arg z|/\pi}. $$

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Graph of positive order $n$ | Natural n, 1 ≤ n, SimpleGraph (Fin n) in claim | The preregistered simple-graph convention; empty graphs are excluded | | Degree polynomial | degreePolynomial, the sum of Polynomial.X ^ G.degree v over every vertex | Includes degree-zero vertices | | Nonzero degree root | Complex z, z ≠ 0, polynomial evaluation equal to zero | Both hypotheses retained | | $\arg z\in(-\pi,\pi]$ | Complex.arg z | Pinned Mathlib branch convention | | Modulus and power | Complex norm and real power | $\pi>0$ gives $|\arg(z)/\pi|=|\arg z|/\pi$ | | All nonzero roots | result : claim | Universal inequality; no extra root hypothesis |

The public definitions degreePolynomial and claim reproduce the two preregistered expressions literally. Positive order is explicit: at order zero the vertex sum is the zero polynomial, so a universal root bound would have a different meaning.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Zeros/DegreeRoots/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Burri's SW–Petz decoder comparisonOpposite strict comparisons for the actual Schumacher-Westmoreland and ordinary Petz decoders.Proved

The questionLaura Burri, Entanglement fidelity of Petz decoder for one-shot entanglement transmission, arXiv:2502.17411v1, Section 3.3, Question 1 asks:

Why it mattersThe question compares two concrete recovery constructions on the full source domain, while the numerical examples in the source use flat code states.

The ideaUse two source-faithful Appendix A instances with opposite strict comparisons.

Our contributionOpposite strict comparisons for the actual Schumacher-Westmoreland and ordinary Petz decoders.

Scope & next questionThe result does not settle Question 2, classify all channels, assert uniform strict advantage, or claim that every input admits both directions.

Verification details
Release source recordProved in Lean

arXiv

Burri's SW–Petz decoder comparison

Laura Burri, Entanglement fidelity of Petz decoder for one-shot entanglement transmission, arXiv:2502.17411v1, Section 3.3, Question 1 asks:

Which performs better: the SW decoder or the Petz decoder?

The source domain is every density state and CPTP noise channel (Section 3.1, Theorem 1). The question concerns the actual Appendix A SW decoder and ordinary Petz decoder, measured by canonical entanglement fidelity. It is separate from Question 2's comparison with a lower bound.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Recovery/BurriDecoderComparison.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Cullinan–Young Remark 8 leading coefficientThe even Krawtchouk sign interpolant has the leading coefficient stated by Cullinan and Young.Proved

Why it mattersThe observation gives an exact degree-leading-term formula for the interpolant used in the paper's Fourier–Krawtchouk approximation.

The ideaSince the centered nodes sum to zero, the next coefficient is the signed sum of omitted nodes divided by those denominators.

Our contributionThe even Krawtchouk sign interpolant has the leading coefficient stated by Cullinan and Young.

Verification details
Release source recordProved in Lean

Open problem

Cullinan–Young Remark 8 leading coefficient

J. Cullinan and E. Young, The Gibbs phenomenon for the Krawtchouk polynomials, arXiv:2603.05408v1, Remark 8, state that numerical experimentation suggests that for even $N$ the leading coefficient of the sign interpolant $I_N$ is

$$(-1)^{(N-2)/2} C((N-2)/2)/(N-1)!,$$

where $C(n)$ is the $n$th Catalan number. Section 2 defines

$$I_N(x)=\sum_{i=0}^{N}\operatorname{sgn}(i-N/2)\ell_i(x),qquad \ell_i(x)=\prod_{0\le j\le N,,j\ne i}\frac{x-(j-N/2)}{i-j}.$$

The settling declaration is D5.S1.Recurrence.Interpolation.CullinanYoungKrawtchouk.result.

Statement echo

The source nodes $j-N/2$, signs, products, sums, and the even case are literal in ell, I, and claim. The Lean proposition quantifies $N\in\mathbb N$, assumes $2\le N$ and Even N, and equates Polynomial.leadingCoeff (I N) with the rational Catalan expression using Nat.div and Nat.factorial. The source's Proposition 6 identification $F_N=I_N$ is outside this claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Interpolation/CullinanYoungKrawtchouk.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedFirst preprint 2026Easlick's complete minimal-advantage dice cycle conjectureThe complete dice conjecture follows from an exhaustive six-face upper bound and uniform longer cycles.Proved

The questionTerry Easlick, Computer-assisted results for n-sided non-transitive dice taking face values in [n], arXiv:2610.40321v1, Conjecture 1, states:

Why it mattersA local winning margin does not determine a global cycle bound.

The ideaThe four mathematical modules establish the following live chain.

Our contributionThe complete dice conjecture follows from an exhaustive six-face upper bound and uniform longer cycles.

Scope & next questionThis candidate remains open here and has no new prediction or research terminal.

Verification details
Release source recordProved in Lean

arXiv

Easlick's complete minimal-advantage dice cycle conjecture

Terry Easlick, Computer-assisted results for n-sided non-transitive dice taking face values in [n], arXiv:2610.40321v1, Conjecture 1, states:

n=6 is the largest n≥3 such that the set of non-standard n-sided dice admits a non-transitive cycle of maximum length n with uniform and minimal winning advantage.

The exact target is the conjunction $S(6)$ and $\forall n\in\mathbb N,\ n\ge7\Rightarrow\neg S(n)$. For $n\ge3$, a die is a nondecreasing list of exactly $n$ labels in $[1,n]$, with repetitions permitted. The standard tuple is included, as in the source's Example 1 and its full graph enumeration. Each of the $n^2$ independent face-index pairs has equal weight. A strict win excludes ties from the numerator and retains them in the denominator. Let $m_n$ be the least attainable strict-win probability greater than one half. An edge $d\to e$ means that the actual unconditional probability that $d$ beats $e$ equals $m_n$. A cycle is a nonempty list of pairwise distinct dice, with every successive edge and its closing edge. The proposition $S(n)$ says that a cycle of length $n$ exists and that no cycle of any length $k>n$ exists. The theorem quantifies over all such $k$ and all $n\ge7$; it introduces no finite cutoff.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Dice/EaslickSixBound.conjecture_one

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The Three-Colour Restricted Schur Number S_3(k; 2) Exceeds k^3 + 3k^2 + k − 1A seven-block three-colouring gives a negative answer to Open Question 6.2 of Gaiser.Refuted

The questionCollier Gaiser, Restricted generalized Schur numbers, arXiv:2608.08789v1, Section 6: Sr(k; l) is the least n such that every r-colouring of {1, …, n} has a monochromatic solution of x1 + ⋯ + xk = x{k+1} with exactly l + 1 distinct integers.

The ideaA solution with exactly three distinct integers uses two distinct summand values a < b, j and k − j times with 1 ≤ j < k, and the sum ja + (k − j)b.

Our contributionA seven-block three-colouring gives a negative answer to Open Question 6.2 of Gaiser.

Verification details
Release source recordRefuted in Lean

Open problem

The Three-Colour Restricted Schur Number S_3(k; 2) Exceeds k^3 + 3k^2 + k − 1

Collier Gaiser, Restricted generalized Schur numbers, arXiv:2608.08789v1, Section 6: S_r(k; l) is the least n such that every r-colouring of {1, …, n} has a monochromatic solution of x_1 + ⋯ + x_k = x_{k+1} with exactly l + 1 distinct integers. Proposition 6.1 proves S_3(k; 2) ≥ k^3 + 3k^2 + k − 1, and "Open Question 6.2. Is it true that S_3(k; 2) = k^3 + 3k^2 + k − 1 for all large enough k?"

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/RestrictedSchur/RestrictedSchur.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2026Ghosh, Birbonshi and Ojha: the unique polynomial rootFor positive real s and t and zero less than q less than one, the Ghosh-Birbonshi-Ojha polynomial has a unique root between one third and one.Proved

The questionSection 2, Open question 1 of arXiv:2608.17486v1 asks:

Why it mattersThis is a recent explicit external question in the first problem tier.

The ideaThe endpoint evaluations satisfy g(1/3)>0 and g(1)<0.

Our contributionFor positive real s and t and zero less than q less than one, the Ghosh-Birbonshi-Ojha polynomial has a unique root between one third and one.

Scope & next questionThe checked result does not settle these extensions or prove endpoint optimality.

Verification details
Release source recordProved in Lean

arXiv

Ghosh, Birbonshi and Ojha: the unique polynomial root

Section 2, Open question 1 of arXiv:2608.17486v1 asks:

For $s,t>0$ and $0<q<1$ does the polynomial $g(z)$ defined in \eqref{g(z)} have a unique root in $(\frac{1}{3},1)$?

The exact public definitions and result are:

def g (s t q z : ℝ) : ℝ :=
  t*q^7*z^8 + q^6*(t*q-1)*z^7 + 3*q^5*(q-s)*z^6 + 5*q^4*(s*q-1)*z^5 +
  q^3*(7*q-9*t)*z^4 + 7*q^2*(t*q-1)*z^3 + 5*q*(q-s)*z^2 + 3*(s*q-1)*z + 1

theorem result (s t q : ℝ) (hs : 0 < s) (ht : 0 < t)
    (hq : 0 < q) (hq1 : q < 1) :
    ∃! z : ℝ, (1 / 3 : ℝ) < z ∧ z < 1 ∧ g s t q z = 0

All four variables are real. Positivity of s and t and the strict bounds on q are the only parameter hypotheses. The conclusion includes both existence and uniqueness with strict interval bounds. There is no relation between s and t and no compact restriction on their values.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/GhoshWeightedShiftRootUniqueness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A382318 congruence modulo threeThe Lambert transform connects Hanna's two cubic equations; cubic Frobenius gives their coefficient congruences modulo three.Proved

The questionHanna's OEIS A382318, Comments (unpaginated), states:

Why it mattersThe congruence links two implicit generating functions rather than two independently chosen coefficient lists.

The ideaThe contraction of agreement below a degree gives existence and uniqueness.

Our contributionThe Lambert transform connects Hanna's two cubic equations; cubic Frobenius gives their coefficient congruences modulo three.

Scope & next questionThe entry's asymptotic formula remains open here, since coefficient residues do not supply an asymptotic estimate.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A382318 congruence modulo three

Paul D. Hanna's OEIS A382318, Comments (unpaginated), states:

Conjectures: a(3*n) == A382319(n) (mod 3) for n >= 0; a(3*n+1) == 0 (mod 3) and a(3*n+2) == 0 (mod 3) for n >= 1.

The source equations are $A=X+L(A)^3$ and $B=X(1-X)^{-1}+L(B^3)$, where $L(F)=\sum_{k\ge1}F(X^k)$ and both series have zero constant term. At degree $m$, $L(F)$ has coefficient $\sum_{1\le k\le m,\ k\mid m}[X^{m/k}]F$. The empty degree-zero sum is zero. The cube in the second equation is taken before Lambert summation. The source is recorded in the Library note. Preregistration: #18861.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Two generating equations, with offset one | IsA382318 A, IsA382319 B in HannaLambertCubeModThree | Literal rational equations and zero constant coefficients. | | Infinite substitution sum | lambertSum | Positive-divisor coefficient sum; natural-number division. | | Sequence values | PowerSeries.coeff | Integer witnesses are cast into rational coefficient equalities. | | Nonnegative index | Natural-number binder | Includes zero. | | First congruence | First universal conjunct of claim | $a(3n)\equiv b(n)\pmod3$. | | Two other residue classes, $n\ge1$ | Second universal conjunct of claim | Both integer residues are zero. | | Existence | Existential conjunct of claim | Additional conclusion; no restriction of the source's solution class. | | Every solution pair | Universal quantifiers and the two predicate hypotheses | No selected solution replaces the source's generating functions. |

The first congruence is $a(3n)\equiv b(n)\pmod3$ for every natural $n$. The other two are $a(3n+1)\equiv a(3n+2)\equiv0\pmod3$ for $n\ge1$. No source clause is dropped. The complete proposition is displayed at both claim and result in the Blueprint.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaLambertCubeModThree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Source 1
Number theoryCounterexample foundFirst preprint 2026Hou–Zhao: the dimension of a sum-free function's codeFirst-order sum-freedom on the ternary plane does not determine the dimension of the code obtained by adjoining function coordinates to Reed–Muller rows.Refuted

The questionXiang-Dong Hou and Shujun Zhao, “Further Results on Sum-Freedom of Binary and q-ary Functions”, arXiv:2609.31489v1, §4.2, printed p. 14, immediately after Theorem 4.6:

Why it mattersThe binary APN case has function-independent code parameters in Theorem 4.1.

The ideaFor every a and every nonzero v, summing over a + tv gives

Our contributionFirst-order sum-freedom on the ternary plane does not determine the dimension of the code obtained by adjoining function coordinates to Reed–Muller rows.

Verification details
Release source recordRefuted in Lean

Open problem

Hou–Zhao: the dimension of a sum-free function's code

Xiang-Dong Hou and Shujun Zhao, “Further Results on Sum-Freedom of Binary and q-ary Functions”, arXiv:2609.31489v1, §4.2, printed p. 14, immediately after Theorem 4.6:

There are some open questions. In Theorem 4.1, the dimension and minimum weight of the code C(f ) are determined and they are independent of f as long as f is APN, i.e., 2nd order sum-free. Do we have the same conclusion for the code Cs (f ) in Theorems 4.5 and 4.6, where f is an sth order sum-free function? If the dimension k and the minimum weight d of Cs (f ) depend on the sum-free function f , what are the ranges for k and d? Likely, these questions will lead the investigation of sum-free functions to new directions.

Definition (1.1) requires a nonzero function sum on every s-dimensional affine subspace. Equation (4.7) defines C_s(f) as the kernel of the parity-check matrix consisting of generator rows of R_q(s(q−1)−1,n), followed by the n coordinate rows of f.

The dimension-only universal assertion, specialized to q = 3, is

$$ \forall n,s\in\mathbb N,\quad n\ge2,\ 1\le s\le n-1\ \Longrightarrow
\forall f,g:\mathbb F_3^n\to\mathbb F_3^n,\quad \operatorname{SumFree}_s(f)\land\operatorname{SumFree}_s(g) \ \Longrightarrow\ \dim C_s(f)=\dim C_s(g). $$

The closed Lean definitions claim : Prop and result : ¬ claim retain exactly these quantifiers. They settle dimension independence negatively, without settling minimum-distance independence or either full range.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SumFreeCodeDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2025Infinite odd fibers of Thue-Morse global maximaInfinitely many exact global Thue-Morse maxima have infinite positive odd fibers.Proved

The questionJoshi and Rust, arXiv:2501.05830v2, Section 3.2.1, Question 3.7, second clause asks: “Are there infinitely many values of n for which Omax(n)=∞?” The first and third clauses are outside this target.

Why it mattersThis question concerns the distribution of actual progression maxima, rather than one explicit family of long progressions.

The ideaFix odd m>=3 and put M=2^m, c=M^2+M+1, B=2^(4m+1), a=c(M^2-1), U=2B+2.

Our contributionInfinitely many exact global Thue-Morse maxima have infinite positive odd fibers.

Verification details
Release source recordProved in Lean

Open problem

Infinite odd fibers of Thue-Morse global maxima

Joshi and Rust, arXiv:2501.05830v2, Section 3.2.1, Question 3.7, second clause asks: “Are there infinitely many values of n for which O_max(n)=∞?” The first and third clauses are outside this target.

Exact question locator: https://arxiv.org/html/2501.05830v2#S3.Thmtheorem7.

The staged claim is that the set of natural lengths n for which the set of positive odd differences d satisfying ExactMax(d,n) is infinite is itself infinite. ExactMax requires a start attaining length n and a universal bound on every monochromatic AP length at every natural start. MAP and the actual zero-indexed thueMorse are reused from the existing Words sources. This covers both colors and the entire infinite word.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/ThueMorseMapInfiniteFibers.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Geometry & algebraQuestion resolvedFirst preprint 2026The largest automorphism groups of perfect evolution algebrasThe partition weight is the attained maximum of the full automorphism orders of perfect complex evolution algebras in every positive finite dimension.Proved

The questionKaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.3, Conjecture 3, pages 27–28, state:

Why it mattersBounding these two parts separately by their independent maxima loses their shared support structure.

The ideaLet P be the least successor-compatible quotient of the support relation, let beta be its successor permutation, and let k(B) be the size of an actual original fiber.

Our contributionThe partition weight is the attained maximum of the full automorphism orders of perfect complex evolution algebras in every positive finite dimension.

Scope & next questionNo claim about them follows from this settlement.

Verification details
Release source recordProved in Lean

Open problem

The largest automorphism groups of perfect evolution algebras

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, section 2.3, Conjecture 3, pages 27–28, state:

Over C, the maximal order of the automorphism group of a perfect n-dimensional evolution algebra is max { product_l (l(2^l−1))^(m_l) m_l! : sum_l l m_l=n }, that is 1, 6, 21, 72, 155, 1296 for n=1,...,6 and n! for n>=7.

The dimensions are positive. Multiplication is bilinear and need not be associative. A natural basis has zero mixed products. Perfectness means that the span of all pair products is the whole algebra. Automorphisms are all multiplication-preserving complex linear equivalences.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/EvolutionAutomorphismMaximum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026A Permutation Grid with 155 Rook PlacementsThe permutation 1,6,3,7,0,5,2,4 on zero-based indices gives a grid with 155 complete rook placements.Refuted

The questionJoel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Section 3.3:

Why it mattersThe module also proves, for any grid with a certified list of its words, that complete rook placements are exactly the perfect matchings of the word-intersection graph and that a recursive matching count equals the rook count.

The ideaFor every grid given with a list of across words partitioning its white cells and a labelling of its down words, placements are the perfect matchings of the word graph.

Our contributionThe permutation 1,6,3,7,0,5,2,4 on zero-based indices gives a grid with 155 complete rook placements.

Verification details
Release source recordRefuted in Lean

Open problem

A Permutation Grid with 155 Rook Placements

Joel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Section 3.3:

Conjecture 3.10. A positive integer r occurs as the rook placement count of a permutation grid if and only if r ≠ 4, r ≠ 12, and r ≢ 3 (mod 4).

The paper supports the conjecture by exhaustive computation over permutations with n ≤ 8, recorded for the counts r ≤ 108.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CrosswordPermutationGridRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationConjecture provedFirst preprint 2025McNulty's path-robustness conjectureThe operator certificate gives the upper bound for P₂ₙ.Proved

The questionDaniel McNulty, “A Graph-Theoretic Approach to Quantum Measurement Incompatibility”, arXiv:2511.15954v1, Conjecture 1, p. 9:

Why it mattersThe source's Corollary 7 supplies lower bounds for the two path sizes and numerical semidefinite-programming evidence motivates their sharpness.

The ideaThe upper bound uses sine-product weights, a Majorana extension, a shifted Fourier positive quadratic decomposition, and the signed-marginal trace identity.

Our contributionThe operator certificate gives the upper bound for P₂ₙ.

Verification details
Release source recordProved in Lean

arXiv

McNulty's path-robustness conjecture

Daniel McNulty, “A Graph-Theoretic Approach to Quantum Measurement Incompatibility”, arXiv:2511.15954v1, Conjecture 1, p. 9:

For all $n\geq 1$, paths satisfy $\eta(P_{2n})=\eta(P_{2n+1})=\eta(C_{2n+2})= \frac{2}{2n+2}\csc\bigl(\frac{\pi}{2n+2}\bigr)$.

The source's Definition 1 and Eqs. (8)–(10) fix the observables and the visibility convention. MCN-1 in #13934 fixes the strong statement for every finite-dimensional realization: Hermitian involutions on $\mathbb C^d$, $d\geq1$, anticommutation on distinct adjacent vertices, and commutation on distinct nonadjacent vertices. No irreducibility is assumed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/CliffordJointMeasurability/McNultyPathRobustness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Further frontiersQuestion resolvedFirst preprint 2025Neighborhood two-shadowing without quasi-finite typeAn explicit binary two-sided subshift with neighborhood two-shadowing and without original exact-length quasi-finite type.Proved

The questionJonathan Meddaugh and Elyssa Stephens, Shadowing with small diameter sets of bounded cardinality, arXiv:2509.04312v1, Question 5.3:

Why it mattersThe explicit example separates the finite tracing-set property with an initial diameter bound from the exact-length language gluing property.

The ideaThe space includes finite supports, singleton supports and the configuration with no ones.

Our contributionAn explicit binary two-sided subshift with neighborhood two-shadowing and without original exact-length quasi-finite type.

Scope & next questionIt does not settle the full Question 5.2 hierarchy or assert nonsoficity.

Verification details
Release source recordProved in Lean

arXiv

Neighborhood two-shadowing without quasi-finite type

Jonathan Meddaugh and Elyssa Stephens, Shadowing with small diameter sets of bounded cardinality, arXiv:2509.04312v1, Question 5.3:

Are there subshifts with neighborhood 2-shadowing which are not a subshift of quasi-finite type?

Use neighborhood shadowing from Definition 2.1 and exact-length quasi-finite type from Definition 3.5, with their full quantifiers as retained in the source note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Dynamics/Shadowing/PowersTwoReturnDistances.powers_two_return_distance_example

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2025Miña-Díaz–Rubin's Chebyshev–Faber equipotential questionA degree-one monic Faber polynomial is Chebyshev on a compact set but fails on an exterior equipotential.Refuted

The questionRubin, Chebyshev polynomials on equipotential curves, arXiv:2505.03967v1, Section 2, Problem 2.2:

Why it mattersEquality at every finite radius is stronger than this limit.

The ideaDifferentiability and the complex open mapping theorem make the open exterior image open; the two-sided bounds make the closed exterior image closed.

Our contributionA degree-one monic Faber polynomial is Chebyshev on a compact set but fails on an exterior equipotential.

Verification details
Release source recordRefuted in Lean

arXiv

Miña-Díaz–Rubin's Chebyshev–Faber equipotential question

E. Miña-Díaz and O. Rubin, Chebyshev polynomials on equipotential curves, arXiv:2505.03967v1, Section 2, Problem 2.2:

Suppose the compact set $K$ is such that for some $n\in\mathbb N$, $T_{n,K}=\hat F_n$, where $\hat F_n$ is the $n$th monic Faber polynomial associated to the map $\phi$ that satisfies (1.2). Is it true then that $T_{n,L_r}=\hat F_n$ for all $r>1$?

Here $\Omega$ is the unbounded component of $\mathbb C\setminus K$, $g_\Omega(z,\infty)=\log|\phi(z)|$, and $\phi(z)=cz+c_0+c_{-1}/z+\cdots$, with $c=\operatorname{cap}(K)^{-1}$. When $\Omega$ is simply connected, $\psi=\phi^{-1}$. The equipotential curve is $L_r={z\in\Omega:|\phi(z)|=r}=\psi({|w|=r})$. The polynomial $F_n$ is the polynomial part of $\phi^n$ at infinity, $\hat F_n=c^{-n}F_n$, and $T_{n,E}$ is the unique monic degree-$n$ sup-norm minimizer on $E$.

The Lean claim quantifies over exterior maps $\psi$ differentiable and injective on $|w|>1$, a positive capacity $\kappa=1/c$, compact complement $K$, and a bounded correction $\psi(w)-\kappa w$ for $|w|>2$. Its IsMonicFaber requires a monic degree-$n$ polynomial $F$ with $F(\psi(w))-(\kappa w)^n\to0$ at infinity. Its IsChebyshev compares every uniform real bound for every monic degree-$n$ competitor. For compact nonempty sets this expresses sup-norm minimization. The implication asks whether such an $F$ Chebyshev on $K$ remains Chebyshev on every $L_r$, $r>1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Chebyshev/FaberEquipotentialRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedDivisibility of the A338636 coefficients by eightThe coefficients of the odd-square continued fraction in OEIS A338636 are divisible by eight above degree one.Proved

The questionHanna, Nov 04 2020) records the following NAME and FORMULA lines, quoted verbatim from Library/Recurrence/hanna2020a338636.md:

Why it mattersThis first-tier OEIS conjecture comes from Hanna's 2020 entry.

The ideastabilization is the escape content: finite-tail comparison gains one coefficient of agreement per level, so each coefficient converges with depth in the sense of eventual exact agreement. hannaconjecture uses this result through generatingequation, placing stabilization on the live proof path.

Our contributionThe coefficients of the odd-square continued fraction in OEIS A338636 are divisible by eight above degree one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A338636 coefficients by eight

OEIS A338636 (Paul D. Hanna, Nov 04 2020) records the following NAME and FORMULA lines, quoted verbatim from Library/Recurrence/hanna2020a338636.md:

NAME:

G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 3^2*x/(A(x) - 5^2*x/(A(x) - 7^2*x/(A(x) - 9^2*x/(A(x) - ...))))), a continued fraction relation.

FORMULA:

a(n) = 0 (mod 8) for n > 1 (conjecture).

Only this FORMULA conjecture is claimed proved. The entry's two further conjectures are outside this resolution:

For n > 0, a(n) = 1 (mod 3) iff n = A191107(k) for some k >= 1

For n > 0, a(n) = 2 (mod 3) iff n = A186776(k) for some k >= 2 where A186776 is the Stanley sequence S(0,2)

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/ContinuedFractionOddSquare.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedHanna's OEIS A375449 diagonal derivative parity conjectureThe diagonal derivative equation has integral coefficients, with odd coefficients at exactly two Mersenne supports above degree one.Proved

The questionHanna's OEIS A375449 asks about the normalized rational power series whose coefficients satisfy

Why it mattersThe settlement answers the explicitly stated A375449 conjecture and supplies the shared diagonal parity theorem used by the sibling A375452 settlement.

Our contributionThe diagonal derivative equation has integral coefficients, with odd coefficients at exactly two Mersenne supports above degree one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's OEIS A375449 diagonal derivative parity conjecture

P. D. Hanna's OEIS A375449 asks about the normalized rational power series whose coefficients satisfy

Expansion of g.f. A(x) satisfying [x^n] x*A'(x) / (1 - n*A(x)) = n^(n+1) for n >= 1.

Its Comments field states:

Conjecture: for n > 1, a(n) is odd iff n is of the form 2^k-2 or 2^k-1 for some k > 1.

The formal statement uses zero constant coefficient and the literal formal derivative and unit-denominator inverse. It asserts existence of a rational solution and integral coefficients with the stated parity support.

Statement echo

| source clause | Lean counterpart | fidelity | | --- | --- | --- | | $[x^n]xA'(x)/(1-nA(x))=n^{n+1}$ for $n\ge1$ | D5.S1.Recurrence.Invariants.HannaDiagonalDerivativeParity.IsGF | The formal derivative, coefficient operator, rational scalar, and inverse are literal. | | offset one and normalized solution | constantCoeff A = 0 in IsGF | The constant coefficient condition is explicit. | | odd exactly at $2^k-2$ or $2^k-1$ for $n>1$ | claim and result | The quantifiers and natural subtraction are literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaDiagonalDerivativeParity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedThe theta self-composition coefficients modulo fourThe positive coefficients of OEIS A378581 are two modulo four exactly at square degrees.Proved

The questionHanna, Jan 08 2025), as quoted in Library/Recurrence/hanna2025a378581.md:

Why it mattersThis is a first-tier OEIS conjecture from a 2025 entry.

The ideathetaSeries is 1 + 2·Σ{j≥1} X^{j²}, defined coefficientwise: coeffthetaSeries gives 1 at n = 0, 2 at positive squares, and 0 otherwise.

Our contributionThe positive coefficients of OEIS A378581 are two modulo four exactly at square degrees.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The theta self-composition coefficients modulo four

OEIS A378581 (Paul D. Hanna, Jan 08 2025), as quoted in Library/Recurrence/hanna2025a378581.md:

NAME:

G.f. A(x) satisfies: A(x*A(x)) = theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2).

COMMENT:

Conjecture: for n > 0, a(n) == 2 (mod 4) iff n is square, else a(n) is divisible by 4 if n is nonsquare.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/ThetaSelfCompositionModFour.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is proved.

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Discrete structuresQuestion resolvedDyadic parity support of the A389476 quadratic reversionHanna's A389476 series has odd coefficients exactly at the indices of A027383.Proved

The questionHanna, Oct 05 2025) gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2025a389476.md.

Why it mattersThis is a first-tier recent OEIS conjecture, selected in the lane's dispatch brief.

The ideaThe module constructs A = generatingSeries over the integers by coefficient stabilization. generatingequation proves A(0) = 0, a(1) = 1, A.subst innerSeries = X, and the pinning identity (1 - A)^2 (X - innerSeries) = A^2.

Our contributionHanna's A389476 series has odd coefficients exactly at the indices of A027383.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Dyadic parity support of the A389476 quadratic reversion

OEIS A389476 (Paul D. Hanna, Oct 05 2025) gives the following NAME and COMMENT, quoted verbatim from Library/Arith/hanna2025a389476.md.

NAME:

G.f. A(x) satisfies: A( x - A(x)^2/(1 - A(x))^2 ) = x.

COMMENT:

Conjecture: a(n) is odd iff n is a term of A027383, where A027383(2*m) = 3*2^m - 2 and A027383(2*m+1) = 4*2^m - 2 for m >= 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/QuadraticReversionDyadicSupportParity.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedTwin-prime sigma-gcd three forces twice a squareFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k))=3, then there is a natural number m with k=2m².Proved

The questionFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k))=3, then there is a natural number m with k=2m².

Why it mattersThis is the A394399 conjecture supplied in preregistration 15003.

The ideaThe existing divisor-sum parity criterion gives a square or twice a square.

Our contributionFor every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k))=3, then there is a natural number m with k=2m².

Scope & next questionThis involves unresolved twin-prime infinitude and the additional sigma-gcd restrictions; twin-prime infinitude alone does not settle those restrictions.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Twin-prime sigma-gcd three forces twice a square

For every natural number k, if k-1 and k+1 are prime and gcd(k, sigma(k))=3, then there is a natural number m with k=2m². Here sigma(k) is the sum of positive divisors.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Robin/TwinPrimeSigmaGcdThreeTwiceSquare.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedDivisibility of the A394757 termsEvery twin-prime center with prime gcd(k, sigma(k)) is divisible by 18.Proved

The questionThe entry states that every such term is divisible by 18.

Why it mattersThis is the first-tier recent OEIS conjecture selected in the implementation brief.

The ideaEither case contradicts primality of that gcd.

Our contributionEvery twin-prime center with prime gcd(k, sigma(k)) is divisible by 18.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Divisibility of the A394757 terms

OEIS A394757 lists the centers k of twin-prime pairs for which the greatest common divisor of k and the sum-of-divisors function sigma(k) is prime. The entry states that every such term is divisible by 18.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.sigma_gcd_divisibility

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresQuestion resolvedG.f. A(x) satisfies A(A(A(x))) = (1+x) * A(A(x))Every positive-index coefficient of OEIS A396102 is congruent to one modulo three.Proved

The questionThese are the verbatim NAME and COMMENT lines supplied for OEIS A396102 by Paul D.

The ideaUniqueness over ℤ and over ZMod 3 uses the live first-difference lemma: the first differing coefficient of two agreeing series enters the j-fold compositional iterate with multiplicity j, so f + (1 + X)·iterate f 2 − iterate f 3 gains one degree of agreement because 1 + 2 − 3 = 0 (generatingunique).

Our contributionEvery positive-index coefficient of OEIS A396102 is congruent to one modulo three.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

G.f. A(x) satisfies A(A(A(x))) = (1+x) * A(A(x))

NAME "G.f. A(x) satisfies A(A(A(x))) = (1+x) * A(A(x))."

COMMENT "Conjecture: a(n) == 1 (mod 3) for n >= 1."

These are the verbatim NAME and COMMENT lines supplied for OEIS A396102 by Paul D. Hanna, Jun 06 2026.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/TripleIterateShiftModThree.hanna_conjecture

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedFirst preprint 2025Skalski and Stroiński's conjecture on the smallest set of uniqueness for the Ising coneOn the cube of sign vectors, the Walsh functions of degree at most two span the Ising space.Proved

The questionSkalski and Stroiński work on X = {-1,+1}^k with the Rademacher functions rj(x) = xj and the Walsh functions wL(x) = ∏{j∈L} rj(x); for 1 ≤ q ≤ k they set

Why it mattersSets of uniqueness control the existence of maximum likelihood estimators in the Ising exponential family, and u(k,2) is the sample size below which no sample can guarantee it.

The ideaSince every coordinate squares to one, the square φ(x) = (v0 + Σi vi xi)^2 is a combination of w∅, the w{i} and the w{i,j}, so φ ∈ B^k2; it is nonnegative and vanishes on U, and it is not identically zero, because comparing the all-plus point with the point where coordinate i is flipped forces vi = 0 and then v0 = 0.

Our contributionOn the cube of sign vectors, the Walsh functions of degree at most two span the Ising space.

Verification details
Release source recordProved in Lean

Open problem

Skalski and Stroiński's conjecture on the smallest set of uniqueness for the Ising cone

Skalski and Stroiński work on X = {-1,+1}^k with the Rademacher functions r_j(x) = x_j and the Walsh functions w_L(x) = ∏_{j∈L} r_j(x); for 1 ≤ q ≤ k they set

B^k_q = Lin{w_L : L ⊂ {1,…,k} and |L| ≤ q}.

Following Bogdan, Bosy and Skalski, a subset U ⊂ X is a set of uniqueness for (B^k_q)_+, the nonnegative functions of B^k_q, if φ = 0 is the only such function vanishing on U, and u(k,q) is the size of the smallest one. The section on extremal sizes closes with the conjecture

u(k,2) = k+1, i.e., for every k there are no sets of uniqueness having at most k elements.

Issue #10022 fixes the readings: points of X are functions Fin k → Bool with true read as +1; B^k_2 is the real span of the w_L with |L| ≤ 2; the equality is read for k ≥ 3, and the edge k = 2 is recorded separately.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/IsingUniquenessSets.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025A steady prior does not imply tabletop time reversibilityA full-rank steady qutrit prior need not admit fixed-dilation tabletop reversal.Refuted

The questionScarani, Exact and approximate conditions of tabletop reversibility: when is Petz recovery cost-free?, arXiv:2510.26895v1, Section V, question 2, page 9:

Why it mattersStationarity is a constraint on the forward channel and prior.

The ideaTheir inequality excludes every environmental matrix, hence every environmental density matrix.

Our contributionA full-rank steady qutrit prior need not admit fixed-dilation tabletop reversal.

Verification details
Release source recordRefuted in Lean

Open problem

A steady prior does not imply tabletop time reversibility

M. Song, H. Kwon and V. Scarani, Exact and approximate conditions of tabletop reversibility: when is Petz recovery cost-free?, arXiv:2510.26895v1, Section V, question 2, page 9:

In all of the examples we considered, a dynamics appears to be tabletop time-reversible whenever a prior state is a steady state. We do not have a proof that this is true in general, and leave it for a future investigation.

The source note gives the definitions and locators. Preregistration #20652 specifies the finite-matrix statement. The universal assertion, including positive definite steady priors, is Refuted by D5.S3.Quantum.Information.SongKwonScaraniSteadyPriorReversibility.result.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (1), forward dilation | channel | The environmental partial trace of $U(X\otimes\xi)U^\dagger$, with system-first product indices | | Hilbert–Schmidt adjoint used in Definition 1 | channelAdjoint | The environmental partial trace of $U^\dagger(Y\otimes I_E)U(I_S\otimes\xi)$ | | Definition 1, equation (2), Petz recovery | petz | $\sqrt\gamma,N^\dagger(N(\gamma)^{-1/2}X N(\gamma)^{-1/2})\sqrt\gamma$; matrix inverse and CFC.sqrt | | Definition 2, equation (3), same interaction | tabletopReverse | $\operatorname{Tr}_E[U^\dagger(X\otimes\eta)U]$; $U$ is fixed | | Definition 2, equation (4), allowed environment | IsTTR | A positive semidefinite trace-one $\eta$ whose reverse equals the Petz map on every complex matrix $X$ | | Finite system and environment | The types S, E and their Fintype, DecidableEq instances in claim | Matrix indices, with no dimension restriction on the universal claim | | Physical forward dynamics | Unitarity of U, positivity and trace one of xi in claim | The source's unitary dilation and environmental density state | | Full-rank prior | Positive definiteness and trace one of gamma in claim | A restriction of the source's priors; a refutation in this subclass refutes the general assertion | | Steady state | channel U xi gamma = gamma in claim | Exact stationarity | | TTR whenever steady | IsTTR U xi gamma conclusion of claim | Same interaction, arbitrary reverse environmental density matrix | | Negative answer | result | Negation of the complete universal claim |

The public definitions keep the preregistered expressions. The proof reuses PartialTraceMutualInformation.partialTraceRight, NMLWMeasuredFidelitySeparation.sqrt_diagonal, and the public Fuchs qutrit-state positivity and trace lemmas. The environmental matrix is literal witness data.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SongKwonScaraniSteadyPriorReversibility.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    An explicitly identified research question in the original source.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Sra's reciprocal-permanent Lorentzian questionThe reciprocal-permanent polynomial of a positive inverse M matrix need not be Lorentzian.Refuted

The questionSuvrit Sra, A Koteljanskii inequality for permanents, arXiv:2609.39979v1, Section 4(d), asks:

Why it mattersThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionThe reciprocal-permanent polynomial of a positive inverse M matrix need not be Lorentzian.

Verification details
Release source recordRefuted in Lean

Open problem

Sra's reciprocal-permanent Lorentzian question

Suvrit Sra, A Koteljanskii inequality for permanents, arXiv:2609.39979v1, Section 4(d), asks:

Is $f_A$ Lorentzian for every inverse $M$-matrix $A$?

Here, for $n\ge1$,

$$f_A(x_0,x)=\sum_{S\subseteq[n]}\operatorname{per}(A_S)^{-1} x_0^{n-|S|}\prod_{i\in S}x_i.$$

An inverse $M$-matrix is entrywise nonnegative, invertible, and has inverse off-diagonal entries at most zero. The empty principal permanent is one. Lorentzian means the coefficient-topology closure of the recursively defined strict Lorentzian class of Brändén–Huh, Definition 2.1; it is not defined here by only a necessary Hessian test. The complete source expressions are in the source note and the Lorentzian note.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Every positive dimension, every real inverse $M$-matrix | claim: $n>0$, real Matrix (Fin n) (Fin n), InverseMMatrix | No symmetry or strict-entry-positivity assumption is added. | | Principal-submatrix permutation permanent | PSW.principalPermanent, Matrix.permanent | The subtype supplies exactly the selected indices, including the empty case. | | Displayed reciprocal polynomial | reciprocalPolynomial | none is $x_0$ and some i is source variable $x_{i+1}$. | | Brändén–Huh positive homogeneous class, strict recursion, closure | positiveHomogeneous, quadraticHessian, strictlyLorentzian, Lorentzian | All degree-$d$ coefficients are quantified; eigenvalue indices count multiplicity; finite-dimensional coefficientwise epsilon approximation expresses the closure. | | Affirmative universal assertion is false | result | The theorem proves the literal negation of claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permanental/SraReciprocalPermanentLorentzianRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationCounterexample foundFirst preprint 2024A three-qubit counterexample to the Shannon branch-and-bound lower boundA three-qubit state violates the proposed Shannon entropy lower bound for stabilizer bases.Refuted

The questionPiroli, A nonstabilizerness monotone from stabilizerness asymmetry, arXiv:2411.05766v1 (2024; Quantum Sci.

Why it mattersThe basis-minimized stabilizerness asymmetry (BMSA) minimizes measurement entropy over stabilizer bases.

The ideaThe zero matrix meets IsStrictUpper because every entry is zero.

Our contributionA three-qubit state violates the proposed Shannon entropy lower bound for stabilizer bases.

Scope & next questionEstablishing it for any proposed restricted family remains open. - Dimension.

Verification details
Release source recordRefuted in Lean

Open problem

A three-qubit counterexample to the Shannon branch-and-bound lower bound

P. S. Tarabunga, M. Frau, T. Haug, E. Tirrito and L. Piroli, A nonstabilizerness monotone from stabilizerness asymmetry, arXiv:2411.05766v1 (2024; Quantum Sci. Technol. 10 (2025) 045026), Section V.C, conjecture following Eq. (69). The source member draft.tex, line 782, states verbatim:

Note that, while we are unable to prove Eq. \eqref{eq:bound_r} for the case $\alpha=1$, we conjecture that the lower bound remains valid, as we have not found any counter-examples.

The label eq:bound_r is the inequality at draft.tex lines 776–779: for fixed rank-$k$ binary $R$, $\min_{Q',c}S_\alpha(p_{Q',c,R})\ge S_\alpha(q_R)$, stated there for integer $\alpha>1$. The states indexed by the diagonal part of $Q$ and the translation $t$ give the squared-overlap law $p$; $q_R$ uses zero off-diagonal matrix and zero phase vector after replacing the amplitudes of $\psi$ by their absolute values. The definitions and source locators are in the literature note.

Issue #24993 fixes the reading at $k=N=n\ge1$ and $R=I_n$, where the translation quotient has one element. Write $B_n=\mathbb F_2^n$. For every unit vector $\psi:B_n\to\mathbb C$, strictly upper-triangular $Q'\in\mathbb F_2^{n\times n}$ and $c\in B_n$, define

$\phi_d(x)=2^{-n/2}(-1)^{x^\top(\operatorname{diag}d+Q')x}i^{\sum_j\widetilde c_j\widetilde x_j}$,

$p(d)=\left|\sum_{x\in B_n}\overline{\phi_d(x)}\psi(x)\right|^2$, and $q(d)=\left|\sum_{x\in B_n}2^{-n/2}(-1)^{d\cdot x}|\psi(x)|\right|^2$.

Here the quadratic and dot-product exponents of $-1$ are computed in $\mathbb F_2$ and represented by zero or one, whereas each tilde is the natural representative zero or one and the exponent of $i$ is summed in $\mathbb N$. Put $S_1(r)=-\sum_d r(d)\ln r(d)$, with $0\ln0=0$. The module's claim asserts $S_1(q)\le S_1(p)$ for every such $n,\psi,Q',c$, with the unit condition $\sum_x|\psi(x)|^2=1$. Its result proves the negation. This refutes a subcase of the source's Shannon conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/StabilizerAsymmetryShannonBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Three-weight Pauli channels have Bayesian inverses only at the maximally mixed stateTing, Fullwood and Wu (arXiv:2605.10375) characterize when a unital qubit channel has a Bayesian inverse with respect to a state, and for Pauli channels with exactly three non-zero weights leave open whether the maximally mixed state is the only such state.Proved

The questionThe verbatim statements are in the literature note.

Why it mattersA Bayesian inverse is the operational time reversal of a sequential measurement through the channel.

The ideaThe three active weights lie strictly between zero and one.

Our contributionTing, Fullwood and Wu (arXiv:2605.10375) characterize when a unital qubit channel has a Bayesian inverse with respect to a state, and for Pauli channels with exactly three non-zero weights leave open whether the maximally mixed state is the only such state.

Verification details
Release source recordProved in Lean

Open problem

Three-weight Pauli channels have Bayesian inverses only at the maximally mixed state

O. Ting, J. Fullwood and Z. Wu, Operational time-reversal symmetry for unital qubit channels, arXiv:2605.10375v1, call a channel $\mathcal F$ a Bayesian inverse of $\mathcal E$ with respect to $\rho$ when ${\mathcal E(\rho)\otimes\mathbb 1,\mathscr J[\mathcal F]}={\mathbb 1\otimes\rho,\mathscr J[\mathcal E^\dagger]}$, $\mathscr J[\mathcal N]=(\mathrm{id}\otimes\mathcal N)(\mathtt{SWAP})$. For Pauli channels $\mathcal P(A)=\sum_\mu p_\mu\sigma_\mu A\sigma_\mu$ they settle the case of two non-zero weights and characterize the general case by three inequalities, and in Section V state that for exactly three non-zero weights they found no Bayesian inverse away from the maximally mixed state, which is "probably due to the fact that either the maximally mixed state is the only state for which a Bayesian inverse exists in such a case, or that such states are restricted to a measure-zero subset of the Bloch ball". The verbatim statements are in the literature note.

Issue #13585 fixes the reading: for every probability vector with exactly three non-zero entries and every qubit density matrix $\rho$ (pure states included), a CPTP Bayesian inverse of $\mathcal P$ with respect to $\rho$ exists if and only if $\rho=\mathbb 1/2$; $\mathcal P^\dagger=\mathcal P$ is part of the statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/PauliThreeWeightBayesianInverseUniqueness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025The token-graph cut bound fails on a five-cycle blow-upA phase product on the 100-token graph of the five-cycle blow-up with block size 40 has Rayleigh quotient below the cut and matching bound of Apte, Parekh and Sud.Refuted

The questionApte, Parekh and Sud, Conjectured Bounds for 2-Local Hamiltonians via Token Graphs, arXiv:2506.03441v2, Appendix D.2, Conjecture 7, page 39, states:

Why it mattersThe source uses the proposed cut bound to obtain an XY-Hamiltonian upper bound and conditional approximation guarantees.

Our contributionA phase product on the 100-token graph of the five-cycle blow-up with block size 40 has Rayleigh quotient below the cut and matching bound of Apte, Parekh and Sud.

Verification details
Release source recordRefuted in Lean

arXiv

apte parekh sud 2025 token graph cut bound

Apte, Parekh and Sud, Conjectured Bounds for 2-Local Hamiltonians via Token Graphs, arXiv:2506.03441v2, Appendix D.2, Conjecture 7, page 39, states:

In contrast, the following bound was shown to hold on our dataset

For all $G$ and for all $1\le k\le\lfloor n/2\rfloor$, $\lambda_{\min}(A(F_k(G)))\ge-(C(G)+M_k(G))/2$.

Here $F_k(G)$ is the token graph, $C(G)$ is the maximum cut weight, and $M_k(G)$ is the maximum matching weight among matchings with at most $k$ edges. The settlement is Refuted by a unit-weight simple graph with $n=200$ and $k=100$.

Statement echo

| source clause | Lean counterpart | fidelity | | --- | --- | --- | | Finite graph $G$ on $n$ vertices | claim quantifies SimpleGraph (Fin n). | A labelled finite simple graph with unit edge weights is within the source's universal range. | | Token graph $F_k(G)$ | Frozen TokenGraphLaplacianRadiusRefutation.tokenGraph | Vertices are $k$-subsets; adjacency moves one token along an original edge. | | Maximum cut $C(G)$ | crosses and maxCut | Exactly one endpoint is in the cut; each unordered unit-weight edge is counted once; the supremum ranges over all cuts. | | Matching weight with at most $k$ edges | maxMatchingLe | Edge subsets have pairwise disjoint endpoint sets and cardinality at most $k$; weight equals cardinality. | | Minimum adjacency eigenvalue | lambdaMin applied to the real adjacency matrix | Minimum of the finite real Hermitian eigenvalue family; the zero extensions for empty or non-Hermitian matrices are unused. | | $1\le k\le\lfloor n/2\rfloor$ | The two hypotheses of claim | Natural division by two gives the floor. | | The proposed lower bound | The conclusion of claim | Cut and matching cardinalities are cast to real numbers before division by two. | | Failure of the universal assertion | result : ¬ claim | The instance uses the five-cycle blow-up with block size 40. |

Decidable adjacency is an implementation instance, not a restriction of the finite graphs. Connectedness and bipartiteness are not extra hypotheses. Public definitions display the literal source notions; the phase vector, block occupancies and other proof constructions are private.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/TokenGraph/CutBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025Unitary entanglement-assisted encoding beats every unassisted strategy in an energy-restricted four-message taskFour unitary encodings of a partially entangled state outperform every unassisted strategy at the same energy.Refuted

The questionTavakoli, The role of entanglement in energy-restricted communication and randomness generation, arXiv:2510.27473v1, Quantum Sci.

Why it mattersThe source shows that entanglement helps energy-restricted communication only through non-unitary encodings for one bit.

The idea- Alice encodes x = (u, v) by U{uv} = Z^u R Z^v, with R = (1/5)[[4, −3], [3, 4]] and Z = diag(1, −1).

Our contributionFour unitary encodings of a partially entangled state outperform every unassisted strategy at the same energy.

Verification details
Release source recordRefuted in Lean

Open problem

Unitary entanglement-assisted encoding beats every unassisted strategy in an energy-restricted four-message task

C. Roch i Carceller and A. Tavakoli, The role of entanglement in energy-restricted communication and randomness generation, arXiv:2510.27473v1, Quantum Sci. Technol. 11, 025020 (2026), study prepare-and-measure communication whose only restriction is the vacuum component of the message, ⟨0|τ_x^C|0⟩ ≥ 1 − ω (Eq. (3)).

  • Result 4. For the probabilistic transmission of a bit, unitary encodings of a shared pure entangled state give no advantage over unassisted quantum communication with shared randomness.
  • The question (§VI, p. 9): "Whether unitary encoding remains resourceless for general tasks is an interesting open problem."

The verbatim statements are in the literature note.

Issue #17895 fixes the reading:

  • resourceless for general tasks means that, for every linear task and every energy 0 ≤ ω < 1/2, each unitary-encoded entanglement-assisted value is matched by an unassisted value;
  • at ω ≥ 1/2 the question is settled by dense coding with a maximally entangled pair.

This statement is refuted: unitary encoding is a resource for a four-message task at ω = 288/625.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/EnergyRestrictedUnitaryEncodingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Source 1
Quantum informationConjecture refutedFirst preprint 2014A resonant memory parameter refutes Krawec's recycled-coin uniformity conjectureA noninteger memory parameter gives a nonuniform recycled-coin return average.Refuted

The questionKrawec, History dependent quantum walk on the cycle with an unbalanced coin, Physica A 428 (2015), arXiv:1411.6298v1, studies a two-coin walk on the d-cycle. - The coin flip applies C(π/4) or C(π(1+φ)/4) to the second coin according to the first coin. - The second coin moves the walker one step left or right. - The coins are then swapped.

Why it mattersThe paper uses the time-averaged distribution to classify the walk's behaviour by memory parameter, initial state and cycle size.

The ideaFix d = 12, a = √2/2, κ = (√5 − 1)/2, c = aκ, θ = arccos c and φ★ = 4θ/π − 1 ≈ 0.4241.

Our contributionA noninteger memory parameter gives a nonuniform recycled-coin return average.

Verification details
Release source recordRefuted in Lean

Open problem

A resonant memory parameter refutes Krawec's recycled-coin uniformity conjecture

W. O. Krawec, History dependent quantum walk on the cycle with an unbalanced coin, Physica A 428 (2015), arXiv:1411.6298v1, studies a two-coin walk on the d-cycle.

  • The coin flip applies C(π/4) or C(π(1+φ)/4) to the second coin according to the first coin.
  • The second coin moves the walker one step left or right.
  • The coins are then swapped.

The memory parameter is φ ∈ [0, 8). §III.B and §IV conjecture, from numerics for d ≤ 300 on the grid φ = m/10, that the time-averaged position distribution is uniform for every φ outside E_d. Here E_d = {0, 2, 4, 6} if 12 ∤ d and E_d = {0, 1, 2, 4, 5, 6} if 12 ∣ d. The source states it for every initial coin state at position 0. The verbatim statements are in the literature note.

Issue #15747 fixes the reading: d ≥ 2, every unit coin α ∈ ℂ⁴ at position 0, and the Cesàro averages over times 1 … T converging to 1/d.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/KrawecRecycledCoinNonUniformity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2013The overlap entropy bound is not c-optimal for qutritsThe Tsallis-two overlap bound is not attained by qutrit observables at interior overlaps.Refuted

The questionPortesi, General entropy-like uncertainty relations in finite dimensions, arXiv:1311.5602v3, J.

Why it mattersThe bound depends only on the overlap c, so its c-optimality decides whether the overlap alone can determine the best entropic uncertainty relation.

The ideaFix 1/√2 < c < 1 and γ = arccos c, so 0 < γ < π/4.

Our contributionThe Tsallis-two overlap bound is not attained by qutrit observables at interior overlaps.

Scope & next questionThe exact c-optimal bound for qutrits remains open; at c = 3/4 the smallest value found, 0.443040, lies above B(3/4) = 7/16.

Verification details
Release source recordRefuted in Lean

Open problem

The overlap entropy bound is not c-optimal for qutrits

S. Zozor, G. M. Bosyk and M. Portesi, General entropy-like uncertainty relations in finite dimensions, arXiv:1311.5602v3, J. Phys. A 47 (2014) 495302, Section 3, after Corollary 2:

We suspect that this corollary is also valid when $N=3$, but we have not been able to prove it yet.

Corollary 2 states that for c > 1/√2 and N = 2 or N ≥ 4 the bound (22), B(c) = min over θ ∈ [0, arccos c] of D(θ) + D(arccos c − θ), is c-optimal. That is, it equals the minimum of the entropy sum over all states and all unitary T ∈ U(N) with c(T) = max |T_ij| = c. The entropies, the overlap and the function D are quoted in the literature note.

Issue #19610 fixes the reading for the Tsallis pair of order 2, S₂(p) = 1 − Σ p_i², an admissible pair of the corollary. The suspicion then says that, for every c with 1/√2 < c ≤ 1, some T ∈ U(3) with c(T) = c and some state attain B(c). The suspicion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/QutritOverlapBoundNonOptimality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

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Quantum informationCounterexample foundFirst preprint 2017A mixed density tensor attains the nuclear-norm maximumA mixed flagged Bell density tensor attains the conjectured nuclear-norm maximum.Refuted

The questionWang, Theoretical and computational aspects of entanglement (2017), arXiv:1705.07160v1, §6, immediately following Lemma 6.2, conjecture:

Why it mattersLemma 6.2 gives an upper bound attained by pure densities of maximally entangled vectors.

The ideaIt uses PiTensorProduct.projectiveSeminormtprodle on the finite family of complex Euclidean factor spaces.

Our contributionA mixed flagged Bell density tensor attains the conjectured nuclear-norm maximum.

Scope & next questionThose classifications remain open here. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

A mixed density tensor attains the nuclear-norm maximum

H. Derksen, S. Friedland, L.-H. Lim and L. Wang, Theoretical and computational aspects of entanglement (2017), arXiv:1705.07160v1, §6, immediately following Lemma 6.2, conjecture:

We conjecture that equality in (6.4) implies that $\mathcal A$ is a pure density state corresponding to maximum entangled state.

This is the wording on PDF page 13. The corresponding TeX is dflw19May17.tex, line 878, with the equation label maxinsepin. The source TeX has md5 1a948baafea109a303b169459eff9371. See also the literature note.

For $\mathbf n=(n_1,\ldots,n_d)$, the source defines $|\mathcal T|1$ as the complex tensor nuclear norm, the minimum of $\sum_a\prod_j|x{a,j}|2$ over finite product decompositions $\mathcal T=\sum_a\bigotimes_j x{a,j}$ (TeX lines 494–497), and defines $\alpha(\mathbf n,\mathbb C)=\max_{|\mathcal T|_2=1}|\mathcal T|_1$ (lines 525–526). A density tensor is the coefficient tensor of a positive semidefinite trace-one matrix, with its row and column modes kept separate (lines 680–695). Equation (6.4) is $\log_2|\mathcal A|_1\le 2\log_2\alpha(\mathbf n,\mathbb C)$.

Issue #23962 fixes the reading: for every mode count and mode-size function, every positive semidefinite trace-one complex matrix $A$ satisfying $|\operatorname{densityTensor}(A)|_1=\alpha(\mathbf n,\mathbb C)^2$ is $tt^*$ for some $t$ with $|t|_2=1$ and $|t|_1=\alpha(\mathbf n,\mathbb C)$. This universal implication is refuted at $d=2$, $\mathbf n=(2,4)$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/DensityTensorNuclearNormEqualityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    Source 1 · Source 2
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    Source 1
Quantum informationCounterexample foundFirst preprint 2008Bourin's matrix subadditivity Question 1Bourin's Question 1 has a negative answer for the Frobenius norm and the square-root function.Refuted

The questionJean-Christophe Bourin, Matrix subadditivity inequalities and block-matrices, arXiv:0805.1954v1, p. 3, Question 1; published as Matrix subadditivity inequalities and block-matrices, International Journal of Mathematics 20 (2009), 679–691, DOI 10.1142/S0129167X09005509.

Why it mattersThe question asks whether reflecting positive contractions suffices to extend an expansive-congruence inequality to positive linear maps.

The ideaThe proof applies CommutatorGap.frobSqunitaryleft, CommutatorGap.frobSqunitaryright, GNSMatrix.frobeniusnormsqeqtrace, and EntropyProductionCoherenceDeletionIdentity.cfcdiagonal through their imported owners.

Our contributionBourin's Question 1 has a negative answer for the Frobenius norm and the square-root function.

Verification details
Release source recordRefuted in Lean

Open problem

Bourin's matrix subadditivity Question 1

Jean-Christophe Bourin, Matrix subadditivity inequalities and block-matrices, arXiv:0805.1954v1, p. 3, Question 1; published as Matrix subadditivity inequalities and block-matrices, International Journal of Mathematics 20 (2009), 679–691, DOI 10.1142/S0129167X09005509. The source note is bourin2008matrixsubadditivity.

Let Φ : M_n → M_k be a positive linear map between the algebras of matrices of size n and k and let F_n and F_k be the sets of non-expansive, positive operators in M_n and M_k respectively. Does Φ⁻¹(F_k) ⊂ F_n imply ‖f ∘ Φ(M)‖ ≤ ‖Φ ∘ f(M)‖ for all M ≥ 0, all symmetric norms and all non-negative concave functions on the positive half line?

The matrix algebras are over $\mathbb C$. Write $F_j={X:0\preceq X\preceq I_j}$. The inclusion means non-strict set inclusion, and the inverse image is taken in the entire domain algebra. A symmetric norm is invariant under left and right multiplication by unitaries. The preregistration is #18573.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Matrix sizes $n,k$ and positive linear map $\Phi:M_n\to M_k$ | claim quantifies over natural sizes and complex LinearMap; its first premise is positivity | All sizes and all positive maps | | $\Phi^{-1}(F_k)\subseteq F_n$ | Full-domain preimage of the literal sets ${X\mid 0\le X\land X\le1}$ | No extra predicate or domain restriction | | Every $M\ge0$ | Universal matrix binder and nonnegativity premise | Loewner order | | Every symmetric norm | Frozen D5.S3.Quantum.Schatten.UINorm k and its field N | The six preregistered norm and two-sided unitary-invariance laws | | Every nonnegative concave function on the positive half line | Real function, ConcaveOn on Set.Ici 0, pointwise nonnegativity there | Both function premises retained | | $|f(\Phi(M))|\le|\Phi(f(M))|$ | Norm evaluations of the two cfc expressions in claim | Same inequality direction |

There is no additional public hypothesis. The positive-contraction condition is inlined literally; the existing UINorm supplies the preregistered norm structure.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Schatten/BourinQuestionOneRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    Year in the primary source attribution: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundSource 1979Paul ErdősRefutation of the one-divisor interval-density unimodality suggestionExact periodic densities refute unimodality of the one-divisor interval density.Refuted

The questionErdős, Some unconventional problems in number theory, Astérisque 61 (1979), printed page 78:

Why it mattersThe 1979 paper poses an elementary distribution-of-divisors question, and the 2013 chapter records it as still open.

The ideaFor fixed n,m, membership in the set of integers having exactly one divisor in (n,m) is periodic with period the least common multiple of the integers in that interval.

Our contributionExact periodic densities refute unimodality of the one-divisor interval density.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the one-divisor interval-density unimodality suggestion

Erdős, Some unconventional problems in number theory, Astérisque 61 (1979), printed page 78:

Perhaps ε₁(n,m) is unimodular for m > n+1, but I know nothing about this.

Here ε₁(n,m) is the natural density of integers N having precisely one divisor d with n < d < m. Tenenbaum, Some of Erdős' Unconventional Problems in Number Theory, Thirty-four Years Later (2013), printed page 18, states:

To my knowledge, the question of unimodality of ε₁(y,z) as a function of z is still open.

Tenenbaum writes the upper endpoint as d <= z; the formal refutation follows the strict upper endpoint in Erdős's original sentence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Density/ErdosExactlyOneDivisorUnimodalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 1979.

    Read source ↗
  • Scholarly context

    Paul Erdős made foundational contributions to number theory and combinatorics and posed many influential problems. This background does not enlarge the particular claim resolved here.

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  • Not headlined

    Already disproved by S. Cambie (arXiv:2501.10333, 2025); erdosproblems.com #692. The release re-verifies a known answer and is not headlined.

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Quantum informationCounterexample foundFirst preprint 2026The minimum depth of open integrable circuitsConjectures 1 and 2 of Garcia Fernandez, Paletta and Retore on the minimum depth of open-boundary integrable quantum circuits are false: with eight sites and two sites carrying -kappa the configuration (6, 3) runs in three layers, one fewer than conjectured, and with eleven sites the configuration (8, 4) runs in four, one fewer than conjectured.Refuted

The questionGarcía Fernández, Paletta and Retore (arXiv:2607.02093v1) build open-boundary integrable quantum circuits for every set of sites carrying the inhomogeneity -κ and group them by the number κ₋ of such sites.

Why it mattersThese circuits are the Floquet evolutions of Yang–Baxter integrable open spin chains, and the minimum depth selects the shallowest circuit in each spectral-equivalence class.

The ideaFor N = 8 and the -κ sites {3, 6} (n⃗ = (6, 3)), Theorem 1 gives U₆₇U₃₄K₁U₁₂U₂₃U₄₅U₅₆U₇₈K̃₈.

Our contributionConjectures 1 and 2 of Garcia Fernandez, Paletta and Retore on the minimum depth of open-boundary integrable quantum circuits are false: with eight sites and two sites carrying -kappa the configuration (6, 3) runs in three layers, one fewer than conjectured, and with eleven sites the configuration (8, 4) runs in four, one fewer than conjectured.

Verification details
Release source recordRefuted in Lean

Open problem

The minimum depth of open integrable circuits

García Fernández, Paletta and Retore (arXiv:2607.02093v1) build open-boundary integrable quantum circuits for every set of sites carrying the inhomogeneity -κ and group them by the number κ₋ of such sites. They conjecture:

Conjecture 1 For odd $N$ and $0<\kappa_-\le \frac{N-1}{2}$, the minimum depth is given by $d=\frac{1}{2}(N+3)-\kappa_-.$

Conjecture 2 For even $N$ and $0<\kappa_-\le \frac{N}{2}$, the minimum depth is given by $d=\frac{1}{2}(N+4)-\kappa_-.$

Issue #10379 fixes the readings: the circuits are those of the paper's Theorems 1 and 2, with U_{j,j+1} on the sites j, j + 1, K₁^R on site 1 and K̃^L_N on site N, and the rightmost factor of the operator product acting first; the depth is the least number of layers into which the gates can be placed so that any two gates sharing a site keep their time order; the minimum is taken over all sets of κ₋ sites among 1, …, N.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/OpenIntegrableCircuitDepthRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Pick functions do not preserve strong infinite divisibilityA positive Pick function sends a 4×4 strongly infinitely divisible matrix to a matrix with a negative entry.Refuted

The questionTsatsomeros, Powers, Roots, and Positivity, arXiv:2609.06298v1, §1, ask verbatim:

Why it mattersFractional powers preserve SIDM according to row 12.

Our contributionA positive Pick function sends a 4×4 strongly infinitely divisible matrix to a matrix with a negative entry.

Verification details
Release source recordRefuted in Lean

Open problem

Pick functions do not preserve strong infinite divisibility

C. R. Johnson, S. Mondal, and M. J. Tsatsomeros, Powers, Roots, and Positivity, arXiv:2609.06298v1, §1, ask verbatim:

Given A ∈ C and a Pick function f, is f[A] ∈ C?

Table 2, “Behavior of Matrix Classes under Pick Functions and Fractional Powers”, p. 22, row 12 reads verbatim:

Strongly Infinitely Divisible Matrix (SIDM) | Open | Yes

The note below the table states verbatim:

The entries marked Open indicate cases for which preservation under Pick functions remains an open question.

Definition 2.7 states verbatim:

Let D := C(−∞, 0]. A complex function f analytic on D is called a Pick function if ℑf(z) ℑz ≥ 0, ∀z ∈ D

Definition 3.25 states verbatim:

A nonnegative matrix A ∈ Mn(R) is said to be an infinitely divisible matrix (IDM) if, for every positive integer m, there exists a nonnegative matrix Km such that (Km)^m = A. If, in addition, A is invertible, then A is called a strongly infinitely divisible matrix (SIDM).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/InfinitelyDivisiblePickRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2024Log-concavity of modified Bessel ratiosA positive rational series disproves log-concavity of a modified Bessel ratio.Refuted

The questionZhang, Unimodality preservation by ratios of functional series and integral transforms, Results Math. (2025), DOI 10.1007/s00025-025-02431-4, arXiv:2408.01755v2, Conjecture 1, printed page 12, states verbatim (line breaks joined):

Why it mattersLog-concavity imposes a three-point midpoint inequality on actual function values.

The ideaThe reversed strict inequality is equivalent to

Our contributionA positive rational series disproves log-concavity of a modified Bessel ratio.

Verification details
Release source recordRefuted in Lean

Open problem

Log-concavity of modified Bessel ratios

D. Karp, A. Vishnyakova and Y. Zhang, Unimodality preservation by ratios of functional series and integral transforms, Results Math. (2025), DOI 10.1007/s00025-025-02431-4, arXiv:2408.01755v2, Conjecture 1, printed page 12, states verbatim (line breaks joined):

Conjecture 1. Suppose b ≥ 0, ν1 ≥ ν2 > −1 and 0 < a1 ≤ a2. Then the ratio x → Iν1(a1 x)/Iν2(a2 x) is unimodal on (0, ∞). Moreover, if ν1 ≥ ν2 > 0 it is log-concave.

The modified Bessel function of the first kind has the standard series

$$ I_\nu(z)=\sum_{k=0}^{\infty}\frac{(z/2)^{2k+\nu}}{k!\Gamma(k+\nu+1)}. $$

The encoded claim is

$$ \forall\nu_1\ge\nu_2>0\ \forall,0<a_1\le a_2,\qquad x\longmapsto\log\frac{I_{\nu_1}(a_1x)}{I_{\nu_2}(a_2x)} \quad\hbox{is concave on }(0,\infty). $$

The source's $b\ge0$ is immaterial because $b$ does not occur in the ratio. The unimodality assertion is outside this claim.

| Source object | Lean name in D5.S3.Analytic.BesselRatioLogConcavity.Result | | --- | --- | | Standard real modified Bessel series | besselI | | Universal positive-order log-concavity clause | claim | | Negation of that clause | result |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BesselRatioLogConcavity/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsConjecture refutedFirst preprint 2025Refutation of Kok's dom-path diameter conjectureThe finite linear Jaco graph on 33 vertices refutes Kok's dom-path diameter conjecture.Refuted

The questionKok, arXiv:2507.16500v1, defines the object in Observation 2.7 on printed pages 6--7, verbatim:

Why it mattersThe paper presents Conjecture 2.9 as an open bound and leaves its conjectures for future work.

The ideaEvery dominating set of this graph has at least four vertices.

Our contributionThe finite linear Jaco graph on 33 vertices refutes Kok's dom-path diameter conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Kok's dom-path diameter conjecture

Kok, arXiv:2507.16500v1, defines the object in Observation 2.7 on printed pages 6--7, verbatim:

For any finite linear Jaco graph Jn (x), n ≥ 2 there exists a pair of vertices i.e. v1 , vn for which a minimal (v1 , vn )-path (not necessarily a diam-path) i.e. Pd (Jn (x)) exists such that a γ-set of Pd (Jn (x)) is a γ-set of Jn (x). We call the path Pd (Jn (x)) the primary minimal dom-path.

and states Conjecture 2.9 on printed page 7, verbatim:

For any linear Jaco graph Jn (x), n ≥ 1 the length of a diam-path and a primary minimal dom-path Pd satisfy |Pd | − |diam(Jn (x))| ≤ 1.

The paper measures path length in edges. The formal target is the weakest selection-independent consequence: for every positive n, some dom-path has edge length at most the Mathlib diameter of J_n(x) plus one. Refuting this existential consequence refutes every possible choice of primary minimal dom-path satisfying the source's defining property.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/JacoDomPathDiameterRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 3/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Refutation of Kok's A000149 domination clauseThe A000149 vertices fail the domination clause of Kok's Conjecture 2.12.Refuted

The questionKok, arXiv:2507.16500v1, section 2.3, Conjecture 2.12, printed page 9, states these two clauses verbatim:

Why it mattersThe paper presents the two-clause statement as a conjecture and leaves its conjectures for future work.

The ideaFor each positive vertex n, let jacoRight n = 2n - dn, where dn counts earlier positive vertices whose right endpoints reach n.

Our contributionThe A000149 vertices fail the domination clause of Kok's Conjecture 2.12.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Kok's A000149 domination clause

Kok, arXiv:2507.16500v1, section 2.3, Conjecture 2.12, printed page 9, states these two clauses verbatim:

The vertex subscripts of a γ-set X = ”{v1 }” ∪ {v2 , v7 , v20 , v54 , . . . } of the infinite linear Jaco graph J∞ (x) is given by the sequence A000149: a(t) = ⌊e^t⌋, t = 0, 1, 2, . . . where e ≈ 2.71828 is the Euler number (or Napier’s constant).

Furthermore, it implies that sequence A000149 is p-graphical where p(G) = {i : j the subscript of vj ∈ X with X some γ-set of G} and F = {G : G = Jn(x), n = 1, 2, 3, . . . }.

The formal target is only the domination clause implied by the first quotation. A gamma-set is a minimum dominating set, so a failure of domination refutes the gamma-set assertion. Gamma-set minimality and the separate p-graphical clause are not encoded.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/JacoExponentialDominationRefutation.result

Frozen module record

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Discrete structuresConjecture refutedSource 2011Merca's even-order residue-sum conjecture refutationMerca's printed even-order residue-sum conjecture fails for a = 2 and m = 15.Refuted

The questionMerca, Journal of Integer Sequences 14 (2011), Article 11.9.1, defines the remainder operation on printed page 2:

Why it mattersBoth coprimality conditions hold, ord15(2) = 4, and the residues are 2, 4, 8, 1, whose sum is 15.

The ideaAt (a, m) = (2, 15), the equality orderOf (2 : ZMod 15) = 4 follows from the fourth power being one and the smaller positive exponents being non-one.

Our contributionMerca's printed even-order residue-sum conjecture fails for a = 2 and m = 15.

Verification details
Release source recordRefuted in Lean

Open problem

Merca's even-order residue-sum conjecture refutation

Merca, Journal of Integer Sequences 14 (2011), Article 11.9.1, defines the remainder operation on printed page 2:

When m is integer and n is a positive integer the quotient of m divided by n is ⌊m/n⌋ and the value m mod n = m − n⌊m/n⌋ is the remainder (or residue) of the division.

Section 3.2 on printed page 17 defines the multiplicative order:

For every positive integer m and every integer a relatively prime to m, we denote by ord_m(a) the multiplicative order of a modulo m, i.e., the smallest positive integer n such that a^n ≡ 1 (mod m), namely ord_m(a) = min {n ∈ N∗ | a^n ≡ 1 (mod m)}

Section 4 on printed page 23 states the conjecture:

Conjecture 1. Let a and m be relatively prime positive integers. If a−1 and m are relatively prime and ord_m(a) is even then Σ_{i=1}^{ord_m(a)} (a^i mod m) = m · ord_m(a) / 2.

The formal claim retains both positivity conditions, both coprimality conditions, even multiplicative order, the inclusive index range from 1 through the order, and the least non-negative remainder operation. Its equality doubles both printed sides to express the exact integer statement in natural numbers.

| Printed expression | Formal expression | | --- | --- | | positive integers a, m | 0 < a, 0 < m with a m : ℕ | | a and m relatively prime | Nat.Coprime a m | | a−1 and m relatively prime | Nat.Coprime (a - 1) m | | ord_m(a) | Mathlib's orderOf (a : ZMod m) | | ord_m(a) is even | Even (orderOf (a : ZMod m)) | | Σ_{i=1}^{ord_m(a)} (a^i mod m) | residueSum m a over inclusive Finset.Icc 1 (orderOf (a : ZMod m)) | | sum = m · ord_m(a) / 2 | 2 * residueSum m a = m * orderOf (a : ZMod m) |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/MercaResidueSumEvenOrderRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2011.

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Discrete structuresCounterexample foundFirst preprint 2026A negative answer to Vatter's Question 4.3Av(12) has one permutation per length and a nonconvergent derangement ratio.Refuted

The questionVincent Vatter, "An Assortment of Problems in Permutation Patterns: Unimodality, Equivalence, Derangements, and Sorting", arXiv:2602.16355v2, Section 4:

Why it mattersThis first-tier numbered question from a 2026 paper was selected in the orchestrator's brief.

The ideaTake Av(12), the class of decreasing permutations.

Our contributionAv(12) has one permutation per length and a nonconvergent derangement ratio.

Verification details
Release source recordRefuted in Lean

arXiv

A negative answer to Vatter's Question 4.3

Vincent Vatter, "An Assortment of Problems in Permutation Patterns: Unimodality, Equivalence, Derangements, and Sorting", arXiv:2602.16355v2, Section 4:

A permutation class is a downset (or order ideal) in this poset: a set 𝒞 of permutations such that if π∈𝒞 and σ≤π, then σ∈𝒞.

For a permutation class 𝒞, let 𝒞° denote the set of derangements in 𝒞.

Question 4.3. Does the ratio |𝒞°_n|/|𝒞_n| converge for every permutation class 𝒞?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/DerangementRatioNonconvergence.vatter_question_4_3_answer_no

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2025Fan–Wu–Fei channel coherence domination is falseA qubit measure-and-prepare channel has Choi coherence below its commutator coherence at alpha = 2.Refuted

The questionFei, Quantifying coherence of quantum channels based on the generalized α-z-relative Rényi entropy, Quantum Information Processing 23, 100 (2024), DOI 10.1007/s11128-024-04309-w; arXiv:2506.10796v1, Section 3, page 8:

Why it mattersThe two quantities observe different aspects of a channel.

The ideaThe bound inputdivle establishes boundedness above, allowing commutativitylower to conclude

Our contributionA qubit measure-and-prepare channel has Choi coherence below its commutator coherence at alpha = 2.

Scope & next questionA classification of channels satisfying domination remains open; the source's depolarizing and amplitude-damping examples are not universal class results supplied by this module.

Verification details
Release source recordRefuted in Lean

Open problem

Fan–Wu–Fei channel coherence domination is false

J. Fan, Z. Wu and S.-M. Fei, Quantifying coherence of quantum channels based on the generalized α-z-relative Rényi entropy, Quantum Information Processing 23, 100 (2024), DOI 10.1007/s11128-024-04309-w; arXiv:2506.10796v1, Section 3, page 8:

From the examples in the next section and numerical results, it is conjectured that C_{α,1}(φ) ≥ C̃_{α,1}(φ) holds for all quantum channels φ, but we have not yet found a proof.

Preregistration #20020 fixes the source conventions and the settlement as Refuted. The Library note gives the source locators. The parameter range is $\alpha\in(0,1)\cup(1,2]$ from Eq. (8).

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Eqs. (1)–(2), $z=1$ | renyiDiv | The real part of the trace of the three-factor power sandwich, outer power $1/\alpha$, subtraction of one and division by $\alpha-1$ are literal. | | Eq. (7), incoherent states on $A\times B$ | coherence | Infimum over positive semidefinite trace-one diagonal matrices; normalized Choi entries retain the factor $1/|A|$. | | Eq. (9), computational-basis dephasing | Diagonal expressions inside coherence and commutativityCoherence | The public formulas inline the source expression; proofs use ProjectedUnistochasticDynamics.projectiveReadout. | | Eq. (10), supremum over every input state | commutativityCoherence | The full positive semidefinite trace-one domain is retained. | | Every CPTP channel between finite systems | claim | Finite index types with decidable equality, MatrixMap, complete positivity and trace preservation on every matrix. The counterexample uses two nonempty two-dimensional systems. | | Every allowed $\alpha$ and $\widetilde C_{\alpha,1}\le C_{\alpha,1}$ | claim | Both intervals and the non-strict comparison are retained. | | Refutation | result | The theorem proves the negation of that universal proposition, with no hypotheses. |

Finite enumerations and decidable equality implement the matrix coordinates. Matrix powers use the pinned continuous functional calculus. Density and Choi expressions are inlined literally, with private consumed correspondences to existing owners. No additional channel hypothesis restricts the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/FanWuFeiChannelCoherenceDomination.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2015The Allen–Meyer family does not give the entire three-qutrit negativity regionA normalized asymmetric three-qutrit vector has a negativity triple outside the family in equation (13) of Allen and Meyer.Refuted

The questionMeyer, Polynomial Monogamy Relations for Entanglement Negativity (Phys.

Why it mattersThe achievable region of negativity triples encodes the polynomial monogamy relations of the paper.

The idea1. Separator over the family. Write r² = a² + b² + c², u = b² + c² and k = d/r.

Our contributionA normalized asymmetric three-qutrit vector has a negativity triple outside the family in equation (13) of Allen and Meyer.

Verification details
Release source recordRefuted in Lean

Open problem

The Allen–Meyer family does not give the entire three-qutrit negativity region

G. W. Allen and D. A. Meyer, Polynomial Monogamy Relations for Entanglement Negativity (Phys. Rev. Lett. 118, 080402 (2017); arXiv:1502.04807), study the triples of squared negativities (N²_A|C, N²_A|B, N²_A|BC) of pure three-party states, with N(ρ) = ‖ρ^{T_A}‖₁ − 1.

  • Three qubits. They describe the achievable region by the states d|000⟩ + a|101⟩ + b|110⟩ + c|111⟩.
  • Qudits. They generalize these states to the family (13), d|000⟩ + Σ_{j=1}^{D−1} (a|j0j⟩ + b|jj0⟩ + c|jjj⟩).
  • The conjecture. "we conjecture that it is the entire achievable set of negativities, (N²_A|C, N²_A|B, N²_A|BC), for D > 2 as well".

The verbatim statements are in the literature note.

Issue #18420 fixes the reading for D = 3: every unit three-qutrit vector has the negativity triple of some normalized family state. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AllenMeyerNegativityRegionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2021Antipin's four-qubit subspaces are not genuinely entangled for every parameterThe four-qubit subspace obtained from Antipin's vectors (35) and substitutions (39)–(40) contains a nonzero product vector at a parameter tuple in (0,1)⁷.Refuted

The questionAntipin, Construction of genuinely multipartite entangled subspaces and the associated bounds on entanglement measures for mixed states, arXiv:2104.09664v1 (J.

Why it matters§III.B.2 bounds the dimension of a four-qubit GES by seven, using the bipartitions with a single qubit on one side, and the construction is the source's example reaching that bound. §IV turns GESs into lower bounds on the concurrence and the convex-roof extended negativity of mixed states, so the conjecture would give a seven-parameter family of maximal GESs for these bounds.

The ideaTake λ = (1/2, 16/25, 1/2, 1/2, 1/17, 1/2, 1/2) and let v1, …, v7 be the vectors (35).

Our contributionThe four-qubit subspace obtained from Antipin's vectors (35) and substitutions (39)–(40) contains a nonzero product vector at a parameter tuple in (0,1)⁷.

Verification details
Release source recordRefuted in Lean

Open problem

Antipin's four-qubit subspaces are not genuinely entangled for every parameter

K. V. Antipin, Construction of genuinely multipartite entangled subspaces and the associated bounds on entanglement measures for mixed states, arXiv:2104.09664v1 (J. Phys. A: Math. Theor. 54, 505303 (2021)), constructs genuinely entangled subspaces (GESs): subspaces in which every nonzero vector is entangled across every bipartition of the parties.

  • The construction. §III.B.2 starts from the seven orthonormal vectors (35a)–(35g) of ℂ⁴ ⊗ ℂ⁴, each of the form √λ_i |p_i⟩⊗|q_i⟩ + √(1 − λ_i) |p′_i⟩⊗|q′_i⟩ with 0 < λ_i < 1. It splits each ququart into two qubits by the substitutions (39) for the first factor and (40) for the second. This gives a seven-dimensional subspace S_λ of four qubits A₁A₂B₁B₂.
  • The conjecture. With every λ_i = 1/2 the source obtains a GES. Then: "Actually, we've done calculations for several different sets of values of λ_i and each time obtained a GES. We can conjecture that eqs. (35) together with eqs. (39) and (40) determine a GES for any set of values of λ_i."

The verbatim statements are in the literature note.

Issue #17193 fixes the reading: S_λ is a GES for every λ ∈ (0, 1)⁷. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumStates/FourQubitGenuineSubspaceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2015The no-signalling optimum of the relaxed multipartite Hardy testFor four parties with two outcomes each, every no-signalling box that satisfies the relaxed Hardy conditions of S.Refuted

The questionRahaman, Witnessing Genuine Mutipartite Non-locality, arXiv:1507.07327v1, Section 3, conjecture:

Our contributionFor four parties with two outcomes each, every no-signalling box that satisfies the relaxed Hardy conditions of S.

Verification details
Release source recordRefuted in Lean

Open problem

The no-signalling optimum of the relaxed multipartite Hardy test

S. S. Bhattacharya, A. Roy, A. Mukherjee and R. Rahaman, Witnessing Genuine Mutipartite Non-locality, arXiv:1507.07327v1, Section 3, conjecture:

The optimal value of the probability of success, $q$ in(\ref{relH}) is $\frac{1}{3}$ for any arbitrary dimension for any number of parties in GNST.

Here (relH) are the relaxed Hardy conditions for $N$ parties with inputs $u$, $v$ and outcomes $1,\dots,d$: $P(1\dots1|u\dots u)=q>0$; for every $r$, with $v$ at $r$ only, every outcome string with $1$ elsewhere and a value other than $d$ at $r$ has probability $0$; and for a fixed $j$ and every $i\ne j$, with $v$ at $i$ and $j$, the string with $d$ at $i$ and $j$ and $1$ elsewhere has probability $0$. GNST is the theory of no-signalling boxes. The paper's Observation reports the value $1/3$ for $n=3,4$ and $d=2,\dots,5$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.result

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2016The unified-entropy correlation triangle inequality is not universally validA two-qubit density matrix violates the unified-entropy correlation triangle inequality at q = 2 and s = -1.Refuted

The questionLamberti, Unified entropic measures of quantum correlations induced by local measurements, Physica A 462 (2016) 930–939, DOI 10.1016/j.physa.2016.06.131, arXiv:1604.00329v1, Concluding remarks, Unifiedarxiv.tex, lines 944–951:

Why it mattersA universal triangle-like inequality would control the disturbance from measuring both factors by the sum of the separately minimized disturbances, throughout the unified-entropy family.

The ideaThe underlying identity genPuritytwo requires a Hermitian matrix and identifies its spectral-square trace with its matrix-square trace.

Our contributionA two-qubit density matrix violates the unified-entropy correlation triangle inequality at q = 2 and s = -1.

Scope & next questionWe provide numerical counterexamples where the local contractivity property of unilocal disturbances fails in a wide range of the entropic index of R\'enyi and Tsallis entropies, but it remains open if the triangle-like inequality is fulfilled for other entropic measures.

Verification details
Release source recordRefuted in Lean

Open problem

The unified-entropy correlation triangle inequality is not universally valid

G. M. Bosyk, G. Bellomo, S. Zozor, M. Portesi and P. W. Lamberti, Unified entropic measures of quantum correlations induced by local measurements, Physica A 462 (2016) 930–939, DOI 10.1016/j.physa.2016.06.131, arXiv:1604.00329v1, Concluding remarks, Unifiedarxiv.tex, lines 944–951:

In the general case, we only proof this for the von Neumann and Tsallis with entropic index of order $2$, which follows from the contractivity property under a projective measurement of quantum relative entropy and Hilbert Schmidt distance, respectively. We provide numerical counterexamples where the local contractivity property of unilocal disturbances fails in a wide range of the entropic index of R'enyi and Tsallis entropies, but it remains open if the triangle-like inequality is fulfilled for other entropic measures.

For $q>0$, $q\ne1$ and $s\ne0$, the source defines $S_{(q,s)}(\rho)=((\operatorname{Tr}\rho^q)^s-1)/((1-q)s)$ and $D^{\Pi^K}{(q,s)}(\rho)=[S{(q,s)}(\Pi^K(\rho))-S_{(q,s)}(\rho)]/(\operatorname{Tr}\rho^q)^s$. The correlations $D^K_{(q,s)}$ are the minima over complete local rank-one projective measurements without postselection, for $K\in{A,B,AB}$. These definitions occur at lines 187–194, 371–394 and 418–433; the triangle-like inequality (38) is at lines 697–701. See also the literature note.

Issue #24210 fixes the universal reading: for every admissible pair $(q,s)$ and every density matrix on a product of finite-dimensional complex spaces, $D^{AB}{(q,s)}(\rho)\le D^A{(q,s)}(\rho)+D^B_{(q,s)}(\rho)$. This reading is refuted at $(q,s)=(2,-1)$ by the two-qubit state

$\rho_0=\frac1{16}\begin{pmatrix}6&1&1&0\1&2&0&1\1&0&2&1\0&1&1&6\end{pmatrix}$,

in coordinate order $00,01,10,11$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/UnifiedEntropyTriangleRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2020Campbell’s classification of irreducible reciprocal pairs is falseA sign-coupled group on three edges and two four-cliques refutes the reciprocal-pair classification.Refuted

The questionKirsty Campbell, Group-Graph Reciprocal Pairs, arXiv:2005.13566v1, Conjecture on page 3:

Why it mattersThe full internal symmetric groups are permitted, but the two component layers share the parity of their top action.

The ideaFixed maps are constant on SameCycle classes, so their number is a power of the number of colours.

Our contributionA sign-coupled group on three edges and two four-cliques refutes the reciprocal-pair classification.

Verification details
Release source recordRefuted in Lean

Open problem

Campbell’s classification of irreducible reciprocal pairs is false

Kirsty Campbell, Group-Graph Reciprocal Pairs, arXiv:2005.13566v1, Conjecture on page 3:

Conjecture: Let (Γ, G) be an irreducible reciprocal pair. Then one of the following hold:

  1. (Γ, G) is one of the two trivial cases: (a) Γ is a null graph and G is any group with no odd permutations, (b) Γ is the complete graph with n vertices and G = Sn.
  2. Γ is a 4-cycle and G ≅ Aut(Γ).
  3. Γ is a k-star and G is defined in Theorem 1.

The source definitions and locators are in Campbell’s Library note. Cameron–Semeraro, The cycle polynomial of a permutation group, DOI 10.37236/7299, supplies the literature background. The quantified conventions and external Tier-1 settlement are preregistered in #16260.

Statement echo

| Source clause | Formal counterpart | Scope and fidelity | | --- | --- | --- | | Finite simple graph and automorphism group | claim quantifies a finite vertex type, a simple graph and a finite permutation subgroup; IsAut preserves adjacency in both directions | The decidability instances support finite counting and do not restrict classical finite graphs. | | Cycles include fixed points | cyclePoly uses the cardinality of the SameCycle quotient | The literal cycle-count convention is inlined. Its proof correspondence cycleConvention uses the existing NikolovSegal.Equation47TypeII.actualCycleCount. | | Proper colourings fixed by a permutation | fixedColourings filters all maps to the finite colour set by properness and fixedness | Both conditions and every vertex are quantified. | | Reciprocal polynomial identity | IsReciprocal equates the colouring total with the signed evaluation of the cycle polynomial for every positive integer | These infinitely many evaluations determine the polynomial identity. | | No nontrivial direct-product decomposition into reciprocal factors | First branch of Reducible: at least two nonempty parts, no cross-part edges, generation by subgroups supported on individual parts | Omitting the reciprocity requirements enlarges reducibility. The corresponding irreducibility premise is stronger. | | No nontrivial wreath-product decomposition with isomorphic part graphs | Second branch of Reducible: all induced part graphs are pairwise isomorphic | Omitting all group conditions enlarges reducibility. | | Null, complete, four-cycle, or k-star with the specified groups | Four alternatives of Listed, retaining the graph conditions | Omitting group restrictions enlarges the conclusion. The k-star centre is nonempty and proper, has the stated cardinality, and gives exactly the source adjacency rule. | | Every irreducible reciprocal pair belongs to the list | claim | A stronger premise and a weaker conclusion make the formal claim no stronger than the source conjecture. | | Refutation | result negates claim | A counterexample to this conservative consequence refutes the source conjecture. No source clause is dropped without the stated conservative direction. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ReciprocalPairs/GroupGraphReciprocalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2016The maximal three-qubit violation of Cabello's nonlocality argumentA genuinely entangled three-qubit state with rational amplitudes, measured with sigmaz and with an observable whose eigenvectors are (3, 4)/5 and (-4, 3)/5 on every qubit, satisfies the three vanishing conditions of Cabello's argument with Q > 0 and gives C = P - Q = 3209679/22562500, which exceeds 9/64.Refuted

The questionCereceda, Cabello's nonlocality for generalized three-qubit GHZ states, arXiv:1609.04763v2, Section 2, and Quantum Stud.

The ideaThe proof uses spectral estimates.

Our contributionA genuinely entangled three-qubit state with rational amplitudes, measured with sigmaz and with an observable whose eigenvectors are (3, 4)/5 and (-4, 3)/5 on every qubit, satisfies the three vanishing conditions of Cabello's argument with Q > 0 and gives C = P - Q = 3209679/22562500, which exceeds 9/64.

Verification details
Release source recordRefuted in Lean

Open problem

The maximal three-qubit violation of Cabello's nonlocality argument

J. L. Cereceda, Cabello's nonlocality for generalized three-qubit GHZ states, arXiv:1609.04763v2, Section 2, and Quantum Stud. Math. Found. 4(3) (2017) 205–215, conjectures:

we conjecture that, for the case in which $Q >0$, the maximum quantum violation of inequality (ineq2) (subject to the fulfillment of conditions (hc2)-(hc4)) is achieved, over all possible states and choice of observables, for the GHZ state $|\psi \rangle =\frac{1}{\sqrt{2}}(|v_1, v_2,v_3 \rangle + |w_1, w_2, w_3 \rangle)$, with the maximum violation being equal to $C_{\text{max}} =\frac{9}{64}$.

Here $P = P(U_1,U_2,U_3|+++)$, $Q = P(D_1,D_2,D_3|---)$, the conditions (hc2)–(hc4) are $P(D_1,U_2,U_3|+++) = P(U_1,D_2,U_3|+++) = P(U_1,U_2,D_3|+++) = 0$, and (ineq2) is $C = P - Q \le 0$; each $U_k$, $D_k$ is a $\pm1$-valued projective qubit observable. The conclusions restate the conjecture for all entangled states of three qubits.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ThreeQubitCabelloRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2018Completely symmetric three-qutrit states need not be fully separableA completely symmetric density matrix on three qutrits has no fully separable decomposition.Refuted

The questionShen, Separability of completely symmetric states in a multipartite system, Phys.

Why it mattersThe question asks whether the linear constraints of complete index symmetry force a density matrix to admit a convex decomposition into product projectors at local dimension three.

The idea1. Density and complete symmetry of the fixed instance. statecs proves complete symmetry of state, since a permutation of its six index positions preserves their combined counts. gramfactor expresses this same matrix as seven rank-one positive semidefinite terms with the nonnegative weights gramWeight; statepsd proves positivity and statetrace proves trace one.

Our contributionA completely symmetric density matrix on three qutrits has no fully separable decomposition.

Scope & next questionThus complete symmetry permits separable states as well as the entangled instance; a classification of the separable part remains open here. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

Completely symmetric three-qutrit states need not be fully separable

L. Chen, D. Chu, L. Qian and Y. Shen, Separability of completely symmetric states in a multipartite system, Phys. Rev. A 99, 032312 (2019), arXiv:1812.04182v4, Conclusion, item 2 (v4 TeX, line 1857):

Are multi-qutrit CS states separable?

The source's Definition 1 (TeX lines 360–391) requires invariance of matrix entries under every permutation of all $2d$ ket and bra index positions, including permutations mixing ket and bra. Its multipartite convention is explicit (lines 449–452):

For simplicity, a separable state will indicate the fully separable one in the multipartite case.

Issue #23788 fixes the assertion as follows. For every natural $d\ge3$, every positive semidefinite complex matrix $\rho$ of trace one on $(\mathbb C^3)^{\otimes d}$ that is completely symmetric is fully separable. Here full separability means a finite convex combination, with strictly positive real weights summing to one, of projectors onto products of unit local complex vectors. This assertion is claim; it is refuted.

The source definitions and locator are also recorded in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/ThreeQutritCompletelySymmetricEntangled.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2018A non-exhaustive periodic orbit in Fernandez's three-cell degrade-and-fire chainA saturated three-cell degrade-and-fire system has a nine-firing periodic trajectory with repeated firing labels.Refuted

The questionBastien Fernandez, Revisiting asymptotic periodicity in networks of degrade-and-fire oscillators (Eur.

Why it mattersExhaustive firing is an explicit hypothesis of the source's periodic-orbit and asymptotic-periodicity results.

Our contributionA saturated three-cell degrade-and-fire system has a nine-firing periodic trajectory with repeated firing labels.

Verification details
Release source recordRefuted in Lean

Open problem

A non-exhaustive periodic orbit in Fernandez's three-cell degrade-and-fire chain

Bastien Fernandez, Revisiting asymptotic periodicity in networks of degrade-and-fire oscillators (Eur. Phys. J. Spec. Top. 227, 1267–1279, 2018; arXiv:1809.03228v1; DOI 10.1140/epjst/e2018-800068-8), states in the Concluding Remarks, DynamicsDFNetworksRevised.tex, line 373:

We believe that similar impossibility prevails for any other periodic trajectory with non-exhaustive firing in this graph.

The graph is the right transition graph in Figure 1, for the Section 4 system with $N=3$, $\epsilon>1/2$ and coupling matrix

$$ W=\begin{pmatrix}1-\epsilon&0&\epsilon\ \epsilon&1-\epsilon&0\ 0&\epsilon&1-\epsilon\end{pmatrix}. $$

The source assumes $0<\eta<1$, $\epsilon<1-\eta$, initially $(Wx)_j>\eta$ for every cell, and restricts to trajectories with distinct coordinates and no simultaneous firings (Section 4, lines 239–255). Exhaustiveness means that every three consecutive firing labels form a permutation of the three cells (line 68). The quoted belief extends the source's exclusion of a particular six-firing itinerary to every non-exhaustive periodic trajectory in this graph.

Issue #25922 fixes the formal reading. Write $\theta$ for the source's $\eta$. For every real $\theta,\epsilon$ with $0<\theta<1$ and $1/2<\epsilon<1-\theta$, and every state sequence $s:\mathbb N\to\mathbb R^3$, label sequence $\mathrm{lab}:\mathbb N\to\mathrm{Fin},3$ and waiting-time sequence $\tau:\mathbb N\to\mathbb R$, claim says that initial repressors strictly above $\theta$, IsRun ε θ s lab τ, and IsPeriodic s imply IsExhaustive lab. Lean labels $0,1,2$ correspond to the source's $1,2,3$. This assertion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/DegradeFireNonExhaustiveOrbitRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundSource 2021Fu–Luo decoherence convexity is falseAn explicit four-dimensional system with four qubit sub-environments violates the conjectured convexity of the Fu-Luo decoherence quantifier.Refuted

The questionLuo, Quantifying Decoherence via Increases in Classicality, Entropy 23 (2021), 1594, Section 3, after Proposition 3:

Why it mattersThe conjecture asks whether the increase in the square-root classicality functional behaves convexly under mixing.

The ideaEach theorem is bind-only: finite cases, applications of existing results and normalization.

Our contributionAn explicit four-dimensional system with four qubit sub-environments violates the conjectured convexity of the Fu-Luo decoherence quantifier.

Verification details
Release source recordRefuted in Lean

Open problem

Fu–Luo decoherence convexity is false

S. Fu and S. Luo, Quantifying Decoherence via Increases in Classicality, Entropy 23 (2021), 1594, Section 3, after Proposition 3:

We conjecture that D(ρ_S|Π, U) is convex in ρ. It seems that a proof may require some deep mathematics.

The source note identifies DOI 10.3390/e23121594. Issue #19026 fixes the universally quantified reading and the literal conventions.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Eq. (17), $J(\rho,\Pi)=1+\sum_i\langle i|\sqrt\rho|i\rangle^2$ | J | Standard pointer basis of Fin d; real part of the squared diagonal entry. Positive matrix square roots have real diagonal entries. | | Eq. (18), $H_E=\bigotimes_i H_{E_i}$, each $U_{E_i}$ acts only on its own part | EnvIdx, liftE | Arbitrary local dimensions k; equality on every other coordinate. | | Product initial state $\rho_E=\bigotimes_i\rho_{E_i}$ | rhoE | Each factor is the reused DensityState. | | Eq. (21), $\omega_{ij}=\operatorname{tr}(U_{E_i}\rho_EU_{E_j}^\dagger)$ | omega | Exact lifted-operator trace. | | Eq. (22), $\rho'_S=\rho_S\circ\Omega$ | finalState | Entrywise multiplication. | | Eq. (27), $D=J(\rho'_S,\Pi)-J(\rho_S,\Pi)$ | D | Same environment and detector family for each input. | | Convexity in $\rho$ | claim | Every dimension, product environment, local-unitary family, two density states and $0\le t\le1$; Jensen inequality. | | Refutation | result | Negation of that complete claim. |

There is no extra physical hypothesis. CStarMatrix.ofMatrix.symm is the coordinate equivalence from the reused density-state representation to matrices. The conventions in claim are the literal preregistered expressions; DensityState and ProbeComponentRigidity.probe are reused in the proof construction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Decoherence/FuLuoDecoherenceConvexityRefutation.result

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    Year in the primary source attribution: 2021.

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Number theoryConjecture refutedSource 2015Refutation of the divergence sentence in Kreh's Conjecture 18An infinite decimal-subsequence set has minimal-layer sizes 1, 2, and then 1 forever.Refuted

The questionDefinition 16 (Unicode transcription, preserving the printed superscripts):

Why it mattersThe divergence sentence asserts a universal asymptotic consequence from the first two layer sizes.

The ideaLet u j = 11010^j and M = {1, 10, 11} ∪ {u j : j ∈ ℕ}.

Our contributionAn infinite decimal-subsequence set has minimal-layer sizes 1, 2, and then 1 forever.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the divergence sentence in Kreh's Conjecture 18

Definition 16 (Unicode transcription, preserving the printed superscripts):

Definition 16. For a given set M ⊂ N define a sequence δⁿ(M) of sets recursively by δ⁰(M) := M, δ(M) := δ¹(M) := M \ S(M), δⁿ⁺¹(M) := δ(δⁿ(M)) and let ηⁿ(M) := |S(δⁿ(M))|, η(M) := η¹(M).

Conjecture 18 (Unicode transcription, preserving the printed superscripts):

Conjecture 18. There are only countably many infinite sets M ⊂ N with η(M) ≤ η⁰(M). For all other sets we have ηⁿ(M) → ∞.

Kreh writes N for the positive integers, writes x ⊳ y when the decimal string of x is obtained from that of y by deleting zero or more digits, and writes S(M) for the elements of M minimal under that order. The Lean definition digitSubseq a b is a ⊳ b; minimal, peel, and eta are Kreh's S, δ, and η constructions.

The literal refuted statement is the second sentence: for every infinite set M of positive natural numbers with eta M 0 < eta M 1, every natural bound B eventually satisfies B ≤ eta M k at every later layer. This is exactly claim, including the explicit positivity premise. Lean's Set.ncard is zero on an infinite set, but that convention is never used in the refutation: every displayed minimal layer is proved to be a singleton or a pair. That theorem does not address the first sentence, concerning countably many infinite sets with η(M) ≤ η⁰(M). A companion result for that sentence is described below; this dossier's resolution binding remains the divergence theorem.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/KrehMinimalSetLayerGrowthRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year recorded in the source bibliography: 2015.

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Discrete structuresConjecture refutedPosed in 2012Refutation of the A192023 comb-Wiener determinant-count conjectureAt n = 3, the printed comb Wiener-index and determinant-count conjecture fails.Refuted

The questionThe Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph.

Why it mattersThis first-tier OEIS conjecture was printed in 2012 and remains unchanged in the live revision recorded by preregistration issue 7441.

The ideaThe comb indexed by n-2 = 1 is a single edge on two vertices, so its only unordered pair has distance one and A192023(1) = 1.

Our contributionAt n = 3, the printed comb Wiener-index and determinant-count conjecture fails.

Scope & next questionIt makes no claim about equality or inequality at unverified inputs and asserts no corrected indexing.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A192023 comb-Wiener determinant-count conjecture

OEIS A192023, NAME (verbatim):

The Wiener index of the comb-shaped graph |||...|_| with 2n (n>=1) nodes. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph.

FORMULA (verbatim):

a(n) = n*(2*n^2 + 6*n - 5)/3.

COMMENT (verbatim; Clark Kimberling, Mar 31 2012):

Conjecture: for n>2, A192023(n-2) is the number of 2 X 2 matrices with all terms in {1,2,...,n} and determinant 2n. - Clark Kimberling, Mar 31 2012

Here A192023(n-2) means the Wiener index of the comb on 2(n-2) vertices, as specified by the NAME, rather than an assumed use of the closed formula.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KimberlingCombWienerDeterminantRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresConjecture refutedPosed in 2013Refutation of the A231548 antisigma gap-three conjectureAt n = 332640, antisigma decreases across a gap of three.Refuted

The questionNumbers n such that 2n - 1 < sigma(n) - sigma(n-2).

Why it mattersThis first-tier OEIS conjecture was printed in 2013 and remains a universal claim about the behavior of the antisigma function.

The ideaFor every natural n, splitting the interval from one through n into divisors and non-divisors gives antisigma(n) = n(n+1)/2 - sigma(n).

Our contributionAt n = 332640, antisigma decreases across a gap of three.

Scope & next questionIt asserts no corrected monotonicity statement and no claim about other values of n.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A231548 antisigma gap-three conjecture

OEIS A231548, NAME (verbatim):

Numbers n such that 2*n - 1 < sigma(n) - sigma(n-2).

COMMENT defining antisigma (verbatim):

Also numbers n such that antisigma(n) < antisigma(n-2), where antisigma(n) = A024816(n) = the sum of the non-divisors of n that are between 1 and n.

COMMENT conjecture line (verbatim; Jaroslav Krizek, Nov 12 2013):

Conjecture: there are no numbers n such that antisigma(n) < antisigma(n-3). - Jaroslav Krizek, Nov 12 2013

OEIS A024816, NAME (verbatim):

Antisigma(n): Sum of the numbers less than n that do not divide n.

The literal object is antisigma(n) = Σ of d ∈ [1,n] with d ∤ n. Since n always divides itself, this agrees with the A024816 interval ending before n. The conjecture is the claim ∀ n ≥ 3, antisigma(n-3) ≤ antisigma(n).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KrizekAntisigmaDecreaseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Number theoryConjecture refutedPosed in 2015Refutation of the A260310 balanced-pair mod-six conjectureThe balanced pair (140140, 141601) refutes Wilson's mod-six conjecture for OEIS A260310.Refuted

The questionPairs with balanced sums of prime divisors (A008472) and inverse prime divisors (A069359), ordered by larger members.

Why it mattersWilson's 2015 comment makes a universal divisibility assertion about every composite smaller member of a balanced pair.

The ideaThe certificate factors 140140 = 2^2 5 7^2 11 13 and derives Nat.primeFactors 140140 = {2, 5, 7, 11, 13} using Nat.primeFactorsmul, Nat.primeFactorsprimepow, and Nat.Prime.primeFactors.

Our contributionThe balanced pair (140140, 141601) refutes Wilson's mod-six conjecture for OEIS A260310.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A260310 balanced-pair mod-six conjecture

OEIS A260310, NAME (verbatim):

Pairs with balanced sums of prime divisors (A008472) and inverse prime divisors (A069359), ordered by larger members.

COMMENTS definition line (verbatim):

Consider pairs (x,y) of numbers where sum(p|x) p + sum(q|y) q = xsum(p|x) 1/p + ysum(q|y) 1/q where p, q are primes and sum(p|x) p > sum(q|y) q.

COMMENTS conjecture line (verbatim; Robert G. Wilson v, Jul 22 2015):

Conjecture: if a(2n-1) is prime then a(2n) is composite and vice versa. And when a(2n-1) is composite, it is congruent to 0 (mod 6). - Robert G. Wilson v, Jul 22 2015

AUTHOR line (verbatim):

Juri-Stepan Gerasimov, Jul 22 2015

The first three pairs in the SEQUENCE lines are (3, 8), (7, 16), and (11, 18). The formal definitions use S(n) = ∑ p ∈ n.primeFactors, p and the frozen repository definition squarefreeDeriv(n) = ∑ p ∈ n.primeFactors, n / p, where the division is natural-number division and is exact because each indexed prime divides n. The predicate IsPair(x, y) means x < y ∧ S(x) + S(y) = squarefreeDeriv(x) + squarefreeDeriv(y) ∧ S(x) ≠ S(y) ∧ S(y) ≠ squarefreeDeriv(y). The literal refuted clause is ∀ x y : ℕ, IsPair x y → 1 < x → ¬ Nat.Prime x → 6 ∣ x.

The prime/composite alternation sentence is not claimed. The separate remark a(2n) < a(2n+2) is also not claimed; the 2015 b-file already refutes it at a(6552) = a(6554) = 35407.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/WilsonBalancedPrimeDivisorPairModSixRefutation.result

Frozen module record

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Discrete structuresConjecture refutedFirst preprint 2018Oh and Scrimshaw's two-shifted q-Motzkin Hankel conjectureAt n = 3 the two-shifted Hankel determinant of Cigler's q-Motzkin numbers takes the value 3 at q = 1, while the printed factor f3 takes the value 2, so no power of q times f3 equals the determinant.Refuted

The questionOh and Scrimshaw consider Cigler's q-Motzkin numbers

Why it mattersThe formal statement makes no claim about a corrected summation range.

The ideaAt q = 1 the recursion is the first-return recursion of the Motzkin numbers, giving 1, 1, 2, 4, 9, 21, 51, and the determinant becomes

Our contributionAt n = 3 the two-shifted Hankel determinant of Cigler's q-Motzkin numbers takes the value 3 at q = 1, while the printed factor f3 takes the value 2, so no power of q times f3 equals the determinant.

Verification details
Release source recordRefuted in Lean

Open problem

Oh and Scrimshaw's two-shifted q-Motzkin Hankel conjecture

Oh and Scrimshaw consider Cigler's q-Motzkin numbers

M̃†_{n+1}(q) = M̃†n(q) + Σ{k=0}^{n−1} q^{k+1} M̃†k(q) M̃†{n−k−1}(q), M̃†_0(q) = 1,

and state, based on numerical computations, the first conjecture labelled conj:factored_motzkin_2shifted in their appendix on other q-determinants:

Define f_n(q) := Σ_{1 ≤ k ≤ n, k ≢ 1 mod 3} q^k if n ≡ 0 mod 3, and (q+1)(Σ_{k=0}^{⌊n/3⌋} q^{3k}) otherwise. Then we have (det[M̃†{i+j+2}(q)]{i,j=0}^{n−1}){n=1}^∞ = (q^{c_n} f_n(q)){n=1}^∞ for some c_n ∈ ℤ_{≥0}.

Issue #10046 fixes the readings: the polynomials and the determinant live in ℤ[q]; "for some c_n" means that for each n ≥ 1 some natural number c_n makes the polynomial identity hold; the second conjecture carrying the same label, on the three-shifted determinants, is not addressed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/OhScrimshawQMotzkinHankelRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2020Level-one QAOA reaches a genuine four-state superposition on three qubitsLevel-one QAOA on three qubits reaches a genuine superposition of exactly four computational basis states.Refuted

The questionLeib, Forbidden subspaces for level-1 QAOA and IQP circuits, arXiv:2007.13563v1 (Phys.

Why it mattersThe source's statement would exclude every level-one QAOA output whose nonzero computational-basis support has an intermediate cardinality.

The ideaThe module defines mixer as the sum of the single-qubit bit-flip permutation matrices, qaoaState as the mixer propagator applied to the phase propagator of plusState, IsGenuineSuperposition by exact nonvanishing support, and claim as the universal exclusion above.

Our contributionLevel-one QAOA on three qubits reaches a genuine superposition of exactly four computational basis states.

Verification details
Release source recordRefuted in Lean

Open problem

Level-one QAOA reaches a genuine four-state superposition on three qubits

M. Streif and M. Leib, Forbidden subspaces for level-1 QAOA and IQP circuits, arXiv:2007.13563v1 (Phys. Rev. A 102, 042416 (2020)), study when a level-one QAOA state can have support on a target subspace. In the paper's Section “More than two target states,” main.tex, lines 335–341, they write:

Based on the assumption that the only solution can be found for $\beta = \frac{1}{4} \pi, \frac{3}{4} \pi, \frac{5}{4} \pi, \frac{7}{4} \pi$ we show in the following that level-1 QAOA can not map to a genuine superposition of more than $2$ but less than $2^n$ target states. We consider a genuine superposition of target states a superposition where no amplitude vanishes.

They then state:

Based on this observation as well as the calculation for the 2 target state case and numerical investigations we conducted we conjecture that the only possible solutions are $\beta = \frac{1}{4} \pi, \frac{3}{4} \pi, \frac{5}{4} \pi, \frac{7}{4} \pi$.

The module formalizes the unrestricted exclusion as $\mathit{claim} := \forall n,e,\beta,\gamma,t,\quad 2 < t.card \mathbin{\to} t.card < 2^n \mathbin{\to} \neg\mathit{IsGenuineSuperposition}(\mathit{qaoaState}\ n\ e\ \beta\ \gamma)\ t$, where qaoaState is the one phase-layer and one mixer-layer state and IsGenuineSuperposition means nonzero amplitudes exactly on $t$. The settlement is result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/LevelOneQAOAGenuineSuperposition.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2016Volkoff's exact-ground-state pair-tunnelling ansatzVolkoff's exact-ground-state conjecture for the two-mode pair-tunnelling ansatz fails in the eight-boson sector.Refuted

The questionVolkoff, arXiv:1610.05807v1, Section V.A, Eq. (20), pp. 6–7, fixes even particle number and writes:

Why it mattersIt separates exact eigenstate representation from the paper's numerical high fidelity and metrological usefulness.

The ideaThus the sum state is not an eigenvector for any real parameter.

Our contributionVolkoff's exact-ground-state conjecture for the two-mode pair-tunnelling ansatz fails in the eight-boson sector.

Verification details
Release source recordRefuted in Lean

arXiv

Volkoff's exact-ground-state pair-tunnelling ansatz

T. J. Volkoff, arXiv:1610.05807v1, Section V.A, Eq. (20), pp. 6–7, fixes even particle number and writes:

We conjecture that for each N ≥ 4 there exists a value c_N for which |ω_+(c_N)⟩ or |ω_−(c_N)⟩ is the exact ground state.

The sector has basis $|N-k,k\rangle$, indexed by $k\in\operatorname{Fin}(N+1)$, and Hamiltonian $H=a_0^{\dagger 2}a_1^2+a_1^{\dagger 2}a_0^2$. With $M=N/2$ and real $c$, the unnormalized ansatz is the coefficient vector of $(x^2+2icxy-y^2)^M\pm(x^2-2icxy-y^2)^M$, multiplied coordinatewise by $\sqrt{(N-k)!k!}$. Normalization is irrelevant for a nonzero eigenvector. An exact ground state has the least real eigenvalue among all nonzero eigenvectors in the full sector. Issue #11667 preregisters this quantified reading and its refutation at $N=8$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PairTunnelingAnsatzGroundStateRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2016. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2015Xu's oblique-discord mutual-information conjectureA normalized oblique operation can increase two-qubit mutual information.Refuted

The questionJianwei Xu, Oblique discord, arXiv:1506.00404v1, p. 3, Conjecture (22):

Why it mattersThe entropy and mutual-information objects are the existing vonNeumannEntropy and quantumMutualInformation, with actual partial-trace marginals.

The ideaIn order 00,01,10,11 the input matrix is (1/65)[[64,0,0,8],[0,0,0,0],[0,0,0,0],[8,0,0,1]].

Our contributionA normalized oblique operation can increase two-qubit mutual information.

Verification details
Release source recordRefuted in Lean

Open problem

Xu's oblique-discord mutual-information conjecture

Jianwei Xu, Oblique discord, arXiv:1506.00404v1, p. 3, Conjecture (22):

Conjecture: I(ρ^{AB}) ≥ I(Φ_A ρ^{AB}) for any Φ_A and any ρ^{AB}, where I(ρ^{AB}) is the mutual information, Φ_A is defined in Eq.(13).

The normalized basis immediately before Eq. (13) consists of unit vectors v_i and their biorthogonal duals w_i, with ⟨v_i,w_j⟩ = δ_ij. The operation is T_b(ρ) = ∑_i (|v_i⟩⟨w_i| ⊗ 1) ρ (|w_i⟩⟨v_i| ⊗ 1), N_b(ρ) = tr_B(∑_i ⟨w_i|ρ|w_i⟩), Φ_b(ρ) = T_b(ρ)/N_b(ρ). The Lean normalizer uses partialTraceRight over B on the sum of rectangular bra-vector sandwiches, extracting the scalar on the Unit factor. The unit norms prove equality with Re(tr(T_b(ρ))) inside phi, and a local have in result checks the same equality at the witness. Mutual information is S(ρ_A) + S(ρ_B) − S(ρ). The conjecture quantifies over all finite dimensions, all normalized bases and their duals, and all density states on the domain of the normalized operation. Its Lean domain explicitly requires a positive normalizing trace.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ObliqueDiscordMutualInformationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2015. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2019The incompatibility probability of random qubit measurementsThe qubit incompatibility probability is strictly less than one quarter.Refuted

The questionFei, Incompatibility probability of random quantum measurements, Phys.

Why it mattersThe exact reduction and certificate in its proof give

The ideaOrderedRootCoordinates.rootcoordinates identifies the squared Yu factors in ordered-root coordinates, and angularvolumeordered computes the incompatible angular interval. coordinatechangelintegral uses the explicit inverse coordinateMapsurjective to transfer volume over the strict interiors. volumeNJMfoursectors, bodyboundarynull and vectorzeronull restore the complete original body.

Our contributionThe qubit incompatibility probability is strictly less than one quarter.

Scope & next questionThe source's independent unbiased-qubit calculation survives; other source results are not re-established by this refutation, and no claim about consequences outside the conjecture's dependency scope is made.

Verification details
Release source recordRefuted in Lean

Open problem

The incompatibility probability of random qubit measurements

L. Zhang, H. Xiang, X. Li-Jost and S.-M. Fei, Incompatibility probability of random quantum measurements, Phys. Rev. E 100, 062139, arXiv:1912.12321, Section III.C, PDF p. 5, states:

We conjecture that the incompatibility probability of a pair of random qubit measurements $\mathbf A$ and $\mathbf B$, generated by $(a_0,\boldsymbol a)$ and $(b_0,\boldsymbol b)$ in Eq. (4), is given by $\mathbf{Pr}[\Theta^{\mathrm{NJM}}_{2,2}]=\frac14.$

Issue #15842 fixes the literal statement, the squared-form criterion and the source and literature boundaries.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Uniform random qubit observables in $V={(x_0,\boldsymbol x): \lvert x_0\rvert+\lVert\boldsymbol x\rVert\le1}$ | OrderedRootCoordinates.V on $\mathbb R\times\mathbb R^3$ | The entire closed body is retained. | | Yu–Liu–Li–Oh square-root function | OrderedRootCoordinates.hYu | Both $(1+x_0)^2$ and $(1-x_0)^2$ occur under the square roots. | | The set of pairs violating Eq. (4) | OrderedRootCoordinates.NJM | Both body memberships and the strict violation of the squared criterion are retained. | | Product Lebesgue volume divided by $\operatorname{vol}(V)^2$ | OrderedRootCoordinates.incompProb | The measure is actual product Lebesgue measure. | | Probability equals $1/4$ | OrderedRootCoordinates.claim | Literal equality, without extra hypotheses. | | Negation of the conjecture | QubitIncompatibilityProbability.result | Unconditional negation of that claim. |

Definition conventions

The main-text display of $h$ omits squares on $1\pm x_0$. The squared form is the expression in Appendix C and in Yu–Liu–Li–Oh, Eq. (5), arXiv:0805.1538. The formal definitions use that preregistered squared form. The strict-interior coordinate maps are smooth bijections; the closed body is not replaced by an interior in the claim. Body boundaries, zero vectors and cosine endpoints are removed only within the integration proof, where their nullity is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/QubitIncompatibilityProbability.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2012Zhang–Wu–Fei: relative-entropy superadditivity after unitary conjugationA maximally mixed two-qubit input and a full-rank correlated diagonal reference contradict relative-entropy superadditivity.Refuted

The questionZhang, Wu and Fei, arXiv:1210.4720v4, p. 6, §4, conjecture (4.2), state:

Why it mattersUnitary optimization does not repair relative-entropy superadditivity when the input is maximally mixed: every unitary fixes that input and both of its marginals.

The ideaAll three unitary orbits of the input states are constant.

Our contributionA maximally mixed two-qubit input and a full-rank correlated diagonal reference contradict relative-entropy superadditivity.

Verification details
Release source recordRefuted in Lean

Open problem

Zhang–Wu–Fei: relative-entropy superadditivity after unitary conjugation

Zhang, Wu and Fei, arXiv:1210.4720v4, p. 6, §4, conjecture (4.2), state:

Based on the above result, we propose the following conjecture: There exist unitary operators U_A ∈ U(H_A), U_B ∈ U(H_B) and U_AB ∈ U(H_A ⊗ H_B) such that S(U_AB ρ_AB U_AB† ‖ σ_AB) ⩾ S(U_A ρ_A U_A† ‖ σ_A) + S(U_B ρ_B U_B† ‖ σ_B), (4.2) where the reference state σ_AB is required to be full-ranked state. Indeed, our numerical calculations show that inequality (4.2) is true.

The source is recorded in the Library note. The assertion ranges over every pair of finite subsystem dimensions and every joint density $\rho_{AB}$ and full-rank joint reference $\sigma_{AB}$; all four marginals are actual partial traces. The relative entropy is $S(\rho\Vert\sigma)=\operatorname{Re}\operatorname{Tr}(\rho(\log\rho-\log\sigma))$ with natural logarithms. The independent existential joint and local unitaries are unrestricted.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite $H_A,H_B$ | Natural dimensions a, b, with coordinates Fin a, Fin b | All finite dimensions; a density state supplies trace one | | Joint density matrices $\rho_{AB},\sigma_{AB}$ | DensityState (Fin a × Fin b) | Positive semidefinite complex matrices of trace one | | Full-ranked $\sigma_{AB}$ | (CStarMatrix.ofMatrix.symm σ.1).PosDef | Positive definiteness of the reference matrix | | $\rho_A,\sigma_A$ | marginalRight ρ, marginalRight σ | Keeps the first product factor | | $\rho_B,\sigma_B$ | marginalLeft ρ, marginalLeft σ | Keeps the second product factor | | Three existential unitaries | UAB, UA, UB and Matrix.unitaryGroup membership proofs | The membership proofs express unitarity, with no extra hypotheses | | Inequality (4.2) | claim | Sum of the two marginal divergences is at most the joint divergence | | Refutation | result | The negation of the complete universally quantified claim |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ZhangWuFeiSuperadditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 2/4 · 15% weight

    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2018Andrade–Severini star-graph evolution maps are unitarily similarThe two evolution maps of every quantum star graph are unitarily similar.Proved

The questionSeverini, Unitary equivalence between the Green's function and Schrödinger approaches for quantum graphs, arXiv:1808.01154v2, state Claim 1 after defining the star-graph maps.

Why it mattersThe Schrödinger and Green's-function constructions put the same scattering and propagation matrices in opposite orders.

Our contributionThe two evolution maps of every quantum star graph are unitarily similar.

Verification details
Release source recordProved in Lean

Open problem

Andrade–Severini star-graph evolution maps are unitarily similar

F. M. Andrade and S. Severini, Unitary equivalence between the Green's function and Schrödinger approaches for quantum graphs, arXiv:1808.01154v2, state Claim 1 after defining the star-graph maps. In the source TeX qgl.tex, lines 522–530 read:

The quantum evolution maps $U^{S}{S_n}(k)$ and $U^{G}{S_n}(k)$ are unitarily similar.

Given the properties of these maps, there are strong reasons to believe that this claim works for every $n$. Although a proof for a general $n$ is not known, we checked this for $n=2,3,4,5$ by using Specht's theorem \cite{Book.2012.Horn}.

The formal problem defines scatteringMatrix and the diagonal propagationMatrix on the index type Fin (n - 1) × Fin 2. It asks whether, for every $n : \mathbb N$ with $2 \le n$, every $r : \mathrm{Fin}(n-1) \to \mathbb C$, every $r1,t1 : \mathbb C$, every $k : \mathbb R$, and every $\ell : \mathrm{Fin}(n-1) \to \mathbb R$, there is a $w$ in Matrix.unitaryGroup (Fin (n - 1) × Fin 2) ℂ such that

$w, (\operatorname{scatteringMatrix}(r,r1,t1),\operatorname{propagationMatrix}(k,\ell)),\operatorname{star}(w) = \operatorname{propagationMatrix}(k,\ell),\operatorname{scatteringMatrix}(r,r1,t1)$.

The source identifies these products as $U^S_{S_n}(k)=S_{S_n}(k)D_{S_n}(k)$ and $U^G_{S_n}(k)=D_{S_n}(k)S_{S_n}(k)$ at lines 383–405 and 472–474.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Transport/StarGraphEvolutionMapsUnitarilySimilar.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationEquivalence provedFirst preprint 2013Bob cannot perfectly hedge more than n/2 of n parallel gamesCertain success in a strict majority of repetitions of the qubit game is equivalent to certain single-game success.Proved

The questionRusso, Quantum hedging in two-round prover-verifier interactions, arXiv:1310.7954v5, TQC 2017, LIPIcs 73 (2018) 5:1–5:30, Section 5:

Why it mattersThe source characterizes when Bob can win at least one of n repetitions with certainty, which is the perfect-hedging phenomenon behind the failure of strong error reduction by parallel repetition for quantum interactive proofs.

The ideaWrite c = cos θ, s = sin θ, β = √(1 − α²) and R = diag(α², β²), the reduced state of one game on Alice's retained qubit.

Our contributionCertain success in a strict majority of repetitions of the qubit game is equivalent to certain single-game success.

Verification details
Release source recordProved in Lean

Open problem

Bob cannot perfectly hedge more than n/2 of n parallel games

S. Arunachalam, A. Molina and V. Russo, Quantum hedging in two-round prover-verifier interactions, arXiv:1310.7954v5, TQC 2017, LIPIcs 73 (2018) 5:1–5:30, Section 5:

However, it still remains open to determine the conditions under which Bob can always win at least $k$ out of $n$ repetitions for some $k > 1$.

Based on our current numerical evidence, it is possible that Bob cannot perfectly hedge more than $k = n/2$ games.

The game, the shared state u_α = α|00⟩ + √(1 − α²)|11⟩, Alice's measurement {P₀,θ, P₁,θ} with P₁,θ = v_θ v_θ* and v_θ = cos θ|00⟩ + sin θ|11⟩, and the parallel repetition with one joint channel Φₙ are quoted in the literature note.

Issue #19590 fixes the reading. Bob perfectly hedges k games if some channel on his n registers makes Alice record at least k wins with certainty. For all n and k with n < 2k ≤ 2n, all α ∈ (0, 1] and all θ, this is possible exactly when α² = cos²θ, that is, when a single game can already be won with certainty. The conjecture is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/HedgingMajorityThreshold.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    Certain success in a strict majority of repetitions of the qubit game is equivalent to certain single-game success.

    Read source ↗
  • Historical context

    First preprint year: 2013.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2017Sharp return-probability bounds for conference graphsConference walks have sharp square-order floors and nonsquare vanishing infima.Proved

The questionChris Godsil, Sedentary quantum walks, arXiv:1710.11192v1 (2017; Linear Algebra and its Applications 614 (2021), 356–375), Section “Questions?”, TeX source conemix.tex, lines 1037–1038:

Why it mattersGodsil's Section “Other Strongly Regular Graphs”, conemix.tex, lines 1009–1023, discusses a lower bound for the diagonal of the average mixing matrix and explains that it provides no useful bound for the instantaneous diagonal entries in the conference-graph case.

The ideaThe lemma requires no square-order hypothesis; in the square case it attains the lower bound.

Our contributionConference walks have sharp square-order floors and nonsquare vanishing infima.

Verification details
Release source recordProved in Lean

Open problem

Sharp return-probability bounds for conference graphs

Chris Godsil, Sedentary quantum walks, arXiv:1710.11192v1 (2017; Linear Algebra and its Applications 614 (2021), 356–375), Section “Questions?”, TeX source conemix.tex, lines 1037–1038:

We do not know whether, for conference graphs, the diagonal entries of $M(t)$ can be bounded away from zero.

In the Introduction, conemix.tex, lines 190–207, Godsil defines $U(t)=\exp(itA)$ for the adjacency matrix $A$ and $M(t)=U(t)\circ\overline{U(t)}$, where $\circ$ is the Schur product. Thus $M(t){u,u}=|U(t){u,u}|^2$.

Issue #26302 fixes the reading as a sharp classification, distinguishing a positive bound for each fixed graph from a bound uniform across orders. Write $n=4m+1$, with $m\in\mathbb N$ and $m\ge1$. The module takes a finite vertex type $V$ with decidable equality and a simple graph $G$ with decidable adjacency satisfying G.IsSRGWith (4 * m + 1) (2 * m) (m - 1) m. This means that $|V|=n$, every vertex has degree $2m$, adjacent vertices have $m-1$ common neighbours, and distinct nonadjacent vertices have $m$ common neighbours. For $u\in V$, put $p_u(t)=|\exp(itA)_{u,u}|^2$.

The classification gives a sharp positive bound $1/n^2$ exactly at square orders; at nonsquare orders, return probabilities become arbitrarily small at positive times. Its precise quantified statement is given under “Proved by result”.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/ConferenceGraphSedentariness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedFirst preprint 2021Conditional randomness implies uniform attenuation without sharpnessConditional randomness implies uniform attenuation for all finite quantum observables.Proved

The questionStan Gudder, Mutually Unbiased Quantum Observables (2021), arXiv:2110.03099v1, Section 2, asks:

Why it mattersThe source compares sequential measurement, conditional randomness and value complementarity in its study of mutually unbiased observables.

Our contributionConditional randomness implies uniform attenuation for all finite quantum observables.

Scope & next questionSuch stability bounds, and extensions to infinitely many outcomes or infinite-dimensional spaces, remain open here.

Verification details
Release source recordProved in Lean

Open problem

Conditional randomness implies uniform attenuation without sharpness

Stan Gudder, Mutually Unbiased Quantum Observables (2021), arXiv:2110.03099v1, Section 2, asks:

We do not know whether the converse of Theorem~\ref{thm21}(i) or (ii) holds.

This is the first sentence of line 223 of the downloaded v1 TeX, MutualUnbiasedQO.tex (576 lines, MD5 ea23a0dd97764fa2b7444408b910866f); the reference resolves to Theorem 2.1. Its first two parts, lines 186–187, read, with typographical commands expanded:

(i) Condition (1) implies Condition (2).

(ii) If $A$ and $B$ are sharp, then Condition (1) and Condition (2) are equivalent.

For finite observables $A=(A_x){x\in\Omega_A}$ and $B=(B_y){y\in\Omega_B}$, with $|\Omega_A|=m\ge1$ and $|\Omega_B|=n\ge1$, the source specifies at line 182:

Condition~(1) is shorthand for $A_x\circ B_y=\tfrac{1}{n},A_x$ and $B_y\circ A_x=\tfrac{1}{m}B_y$ for all $x\in\Omega _A$, $y\in\Omega _B$ and Condition~(2) is shorthand for $(B\mid A)_y=\tfrac{1}{n},I$ and $(A\mid B)_x=\tfrac{1}{m},I$ for all $x\in\Omega _A$, $y\in\Omega _B$.

Here $E\circ F=E^{1/2}FE^{1/2}$, using the positive square root, and $(B\mid A)_y=\sum_x A_x\circ B_y$ (Section 1, lines 97 and 117–120). Each observable consists of positive semidefinite effects summing to $I$. The source calls Condition (1) uniform attenuation and Condition (2) conditional randomness. The definitions and source locators are also in the literature note.

Issue #21653 fixes the statement: for every finite-dimensional complex Hilbert space and all finite observables with positive outcome counts, Condition (2) implies Condition (1), without assuming sharpness. This converse of Theorem 2.1(i) is proved. The separately worded, ambiguous question about a converse of part (ii) is outside the registered statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/ConditionallyRandomUniformAttenuation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2018The rank vector $(3_2,2_6)_7$ is realized by an extreme POVMAn explicit extreme eight-outcome POVM on complex seven-space refutes the proposed rank-vector exclusion.Refuted

The questionPellonpää, Effective methods for constructing extreme quantum observables, arXiv:1809.09935v1 (J.

Why it mattersThe source presents ranks as a resource measure for implementing an extreme observable and asks which rank combinations can occur.

The ideaThe hypothesis-free theorem frameposDef proves frame.PosDef by an exact coordinate calculation.

Our contributionAn explicit extreme eight-outcome POVM on complex seven-space refutes the proposed rank-vector exclusion.

Verification details
Release source recordRefuted in Lean

Open problem

The rank vector $(3_2,2_6)_7$ is realized by an extreme POVM

E. Haapasalo and J.-P. Pellonpää, Effective methods for constructing extreme quantum observables, arXiv:1809.09935v1 (J. Phys. A: Math. Theor. 53, 245301 (2020)), define finite-outcome POVMs as tuples of positive operators summing to the identity and define extremality by the midpoint condition (source TeX laatikkopaperi_v4.tex, lines 139 and 148–152). They order rank vectors and omit rank-one outcomes (lines 219–222). Their low-dimensional discussion says:

In dimensions 1-4, we are able to completely characterize the possible rank vectors of extreme POVMs but in dimensions 5-7 some open questions remain. (line 307)

and, for dimension seven,

The new problematic case now is $(3_2,2_6)_7$. (line 314)

The formal question is whether there is an extreme POVM on $\mathbb C^7$ whose rank-at-least-two multiset is two 3s and six 2s. The module's claim states that every n and every family $M:\operatorname{Fin}(n)\to\operatorname{Matrix}(\operatorname{Fin}(7),\operatorname{Fin}(7),\mathbb C)$ satisfying IsPOVM M and IsExtreme M has bigRanks M different from ${3,3,2,2,2,2,2,2}$. The declaration result has type $\lnot\claim$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/ExtremePOVMRankThreeThreeTwoSix.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2020Hedemann's proposed diagonal-correlance normalization is exceeded by seven qubitsA seven-qubit diagonal state exceeds the proposed diagonal-correlance normalization.Refuted

The questionHedemann, Correlance and Discordance: Computable Measures of Nonlocal Correlation, Quantum Information Processing 19, 189 (2020), arXiv:2001.03453v2, Section II.C, defines the normalization as the maximum of raw correlance over all diagonal states in Eq. (16), and states:

Why it mattersDiagonal correlance is intended to normalize correlation in classical probability distributions and diagonal quantum states against the largest value available within that class.

The ideaThe definitions reductionProduct and correlanceRaw use real diagonal matrices and their matrix trace; FunOnFinite.map supplies the coordinate-fiber sums.

Our contributionA seven-qubit diagonal state exceeds the proposed diagonal-correlance normalization.

Scope & next questionEquations (15) and (18) together permit diagonal correlance above one at the witness.

Verification details
Release source recordRefuted in Lean

Open problem

Hedemann's proposed diagonal-correlance normalization is exceeded by seven qubits

Samuel R. Hedemann, Correlance and Discordance: Computable Measures of Nonlocal Correlation, Quantum Information Processing 19, 189 (2020), arXiv:2001.03453v2, Section II.C, defines the normalization as the maximum of raw correlance over all diagonal states in Eq. (16), and states:

Instead, one simple example that maximizes $\mathcal{X}D$ in all systems is $\rho{D_{\max}}=\frac{1}{2}(|1\rangle\langle1|+|n\rangle\langle n|),$ yielding the normalization factor, valid for all systems, $\mathcal{N}{\mathcal{X}D}=\widetilde{\mathcal{X}}(\rho{D{\text{max}}})=\frac{1}{2}-\frac{1}{2^N}.$

This quotation is from CorrelanceAndDiscordance.tex, lines 456–465, Eqs. (17)–(18); mathematical macros are expanded and layout commands omitted. Here $N$ is the number of modes and $n$ is their total Hilbert-space dimension. Appendix L, Step 9, line 2385, acknowledges:

However, a rigorous proof of this is still lacking.

For $N$ qubits, write a diagonal state as $\rho=\operatorname{diag}(p)$ with $p:{0,1}^N\to\mathbb{R}$, $p(x)\ge0$ and $\sum_xp(x)=1$. Put $m_i(b)=\sum_{x:x_i=b}p(x)$ and $q(x)=\prod_i m_i(x_i)$. The source's raw correlance and reduction product, Eqs. (10)–(11), become $R_N(p)=\operatorname{tr}[(\operatorname{diag}(p)-\operatorname{diag}(q))^2]=\sum_x(p(x)-q(x))^2$.

Issue #22752 fixes the qubit consequence of the source's assertion: for every natural $N\ge2$ and every such probability table, $R_N(p)\le\frac12-2^{-N}$. This universal bound is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/DiagonalCorrelanceNormalizationRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2020The flag partition maximizes the squared sizes of affine parts outside the originThe flag partition maximizes the squared sizes of affine parts outside the origin.Proved

The questionGross, The axiomatic and the operational approaches to resource theories of magic do not coincide, arXiv:2011.11651v3 (J.

Why it mattersThe source compares stabiliser operations (SO), defined by Clifford gates, Pauli measurements and classical control, with completely stabiliser-preserving channels (CSP).

The ideaTogether with partitionbound, these are the three conjuncts of claim.

Our contributionThe flag partition maximizes the squared sizes of affine parts outside the origin.

Scope & next questionA classification of all maximizing partitions, and whether every maximizer arises from a complete flag, remain open here. - Corrected arithmetic.

Verification details
Release source recordProved in Lean

Open problem

The flag partition maximizes the squared sizes of affine parts outside the origin

A. Heimendahl, M. Heinrich and D. Gross, The axiomatic and the operational approaches to resource theories of magic do not coincide, arXiv:2011.11651v3 (J. Math. Phys. 63, 112201, 2022), conjecture in the remark following Corollary cor:SO-cap-AD. The checked TeX member paper_new.tex, lines 1219–1227, reads:

We conjecture that such a projective measurement which maximizes \eqref{eq:estimation-of-overlapII} has the following form:

Measure the first qudit in the computational basis.

If $ x \neq 0 $ is measured, do nothing. If $ 0 $ is measured, measure the second qudit in the computational basis. Continue in this fashion until all qubits are measured.

If $ 0 $ has been measured on every qudit, apply a Hadamard gate to every qudit.

Write $p$ for the source's prime dimension $d$, and $V=\mathbb F_p^n$, represented by Fin n → ZMod p. An affine partition (isAffinePartition) is a finite set of nonempty affine subspaces, with distinct parts disjoint, such that a vector belongs to a part if and only if it is nonzero. Its objective is sqSum, namely $\sum_{K\in\mathcal K}|K|^2$.

The cells $K_{i,x}={v\in V:(\forall j<i,\ v_j=0),\ v_i=x}$, for $0\le i<n$ and $x\in\mathbb F_p\setminus{0}$, form flagPartition p n. Put $S_p(n)=\sum_{k=0}^{n-1}(p-1)p^{2k}$, the definition flagValue p n.

Issue #25329 fixes the reading: for every prime $p$ and every $n\in\mathbb N$, the flag family is an affine partition, its objective equals $S_p(n)$, and every affine partition of $V\setminus{0}$ has objective at most $S_p(n)$. This is the proposition claim, proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/AffinePartitionFlagMaximum.result

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Quantum informationQuestion resolvedFirst preprint 2016Eighteen-mode bosonic occupation assignments exist exactly for even particle numbersThe eighteen-mode bosonic occupation model admits an assignment exactly for even particle numbers.Proved

The questionKurzyński, Contextuality of identical particles, Phys.

Why it mattersThe source distinguishes an obstruction to assigning occupations from contextuality witnessed only after imposing constraints from a particular quantum state.

Our contributionThe eighteen-mode bosonic occupation model admits an assignment exactly for even particle numbers.

Scope & next questionThe separate possibility of other state-independent contextuality inequalities for two particles (line 113), and the questions for arbitrary dimensions or other mode sets (line 242), remain open in this settlement.

Verification details
Release source recordProved in Lean

Open problem

Eighteen-mode bosonic occupation assignments exist exactly for even particle numbers

P. Kurzyński, Contextuality of identical particles, Phys. Rev. A 95(1), 012133 (2017), arXiv:1609.01250v1, asks in “Contextuality of $N>2$ bosons” (contextuality_fermions.tex, line 233):

The Hardy-like proof shows that there is the state-dependent contextuality in the 18-mode scenario for any $N>0$. However, the lack of the state-independent contextuality was only shown for $N=2$. It is possible that for some $N>2$ the bosonic occupation numbers of 18 modes cannot be assigned in a non-contextual way. This would require further research, since for large numbers of particles there are many ways in which the particles can be distributed between the modes.

The source's bosonic model assigns a nonnegative integer to each mode, independently of the context, with the occupations in every context summing to the particle number $N$ (lines 70–76, Eq. (1)). A mode $v_{ij}$ belongs to contexts $i$ and $j$ (line 83); the nine contexts are displayed in lines 132–221. The definitions and source locator are also recorded in the literature note.

The formal definition modes is the set

$$ E={(1,2),(1,6),(1,7),(1,8),(2,3),(2,8),(2,9),(3,4),(3,7), (3,9),(4,5),(4,7),(4,8),(5,6),(5,8),(5,9),(6,7),(6,9)}. $$

For $i\in{1,\ldots,9}$, put $C_i={(a,b)\in E:a=i\text{ or }b=i}$. The definition IsNoncontextualAssignment requires one function $v:\mathbb N\times\mathbb N\to\mathbb N$ satisfying $\sum_{p\in C_i}v(p)=N$ for every such $i$. Values outside $E$ are unrestricted and do not enter any context sum.

Issue #24135 fixes the statement claim: for every $N\in\mathbb N$, such an assignment exists if and only if $N$ is even. This classification is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/EighteenModeBosonicAssignmentParity.result

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Geometry & algebraQuestion resolvedFirst preprint 2020Three continuous invariant parity-odd variables are necessary for two interchangeable pairsAny two continuous invariant parity-odd real variables vanish together on a chiral four-particle event in each of the two event classes.Proved

The questionGripaios, Lorentz and permutation invariants of particles III: constraining non-standard sources of parity violation, Int.

Why it mattersThe source seeks finite sets of observables that detect every chiral event while respecting the Lorentz and permutation symmetries.

Our contributionAny two continuous invariant parity-odd real variables vanish together on a chiral four-particle event in each of the two event classes.

Scope & next questionRestricting to a fixed-mass, equal-mass, massless or momentum-conserving subdomain can remove the displayed sphere; the minimum number on such a subdomain remains open here.

Verification details
Release source recordProved in Lean

Open problem

Three continuous invariant parity-odd variables are necessary for two interchangeable pairs

C. G. Lester, W. Haddadin and B. Gripaios, Lorentz and permutation invariants of particles III: constraining non-standard sources of parity violation, Int. J. Mod. Phys. A 37(16) (2022) 2250093, arXiv:2008.05206v6, Introduction, introduction.tex, lines 103–104:

No claim is made that either set $S$ or $\hat S$ is globally \textbf{minimal} for the class of events with which it is concerned. That is to say: it has not been proved that there does not exist some other set $S_?$ containing only one or two event variables but which retains all the stated properties. Nonetheless, we \textit{conjecture} that there is indeed no smaller set $S_?$ and that $S$ and $\hat S$ are therefore examples of \textbf{minimal} sets.

The quotation is from the downloaded v6 TeX member introduction.tex (md5 2b21a1456fcb7bb5df295f300f8b61ae). The definitions and source locators are collected in the literature note.

An event is an ordered quadruple $(p,q,a,b)$ of real four-vectors. Write $u\cdot v=u_0v_0-u_1v_1-u_2v_2-u_3v_3$ and $u^2=u\cdot u$. The forward cone consists of vectors with $u_0\ge0$ and $u_1^2+u_2^2+u_3^2\le u_0^2$, including zero. The two domains are:

  • $\Omega_{(pq)(ab)X}$: all quadruples of forward-cone vectors (allEvents);
  • $\Omega_{(pq)\to(ab)X}$: those quadruples additionally satisfying $p^2q^2<(p\cdot q)^2$ (collisionEvents).

The symmetry uses a common matrix $\Lambda$ satisfying $\Lambda^{\mathsf T}\eta\Lambda=\eta$, $\det\Lambda=1$ and $\Lambda_{00}>0$, where $\eta=\operatorname{diag}(1,-1,-1,-1)$, together with independent permutations of $(p,q)$ and $(a,b)$. Parity $\mathscr P$ fixes each energy and negates each spatial momentum. An event is chiral when no such symmetry carries it to its parity image. These match prelims.tex, lines 25–27, 190–193, 228–243, 265–268, 433–443 and 656–666; neither domain adds momentum conservation or fixed masses.

Issue #25338 fixes the assertion: for either domain and every pair $f_1,f_2:\mathrm{event}\to\mathbb R$ continuous on that domain, invariant under these symmetries there, and parity odd there, some chiral event in the domain satisfies $f_1=f_2=0$. This assertion is proved.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Geometry/Lorentz/ChiralEventParityMinimality.result

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Quantum informationConjecture provedFirst preprint 2016Lovas and Andai's conjecture on the least trace-distance contraction over a classical channelThe trace-distance contraction coefficient isProved

The questionThe trace-distance contraction coefficient is

Why it mattersThe trace-distance contraction coefficient is the quantum analogue of the Dobrushin coefficient of ergodicity and bounds mixing times of quantum Markov processes; the theorem says that fixing the classical channel of a qubit channel forces no contraction beyond that of the classical channel itself.

The ideaFor states ρ, σ with t = ρ00 − σ00, it sends ρ − σ to diag((a−f)t, (f−a)t).

Our contributionThe trace-distance contraction coefficient is

Verification details
Release source recordProved in Lean

Open problem

Lovas and Andai's conjecture on the least trace-distance contraction over a classical channel

Lovas and Andai represent a qubit channel Q : M_2 → M_2, a completely positive trace-preserving map, by its Choi block matrix, whose blocks Q_11 = Q(|0⟩⟨0|) and Q_22 = Q(|1⟩⟨1|) have diagonals (a, 1−a) and (f, 1−f) in the parametrization (eq:matQ). The trace-distance contraction coefficient is

η^Tr(Q) = sup { Tr|Q(ρ) − Q(σ)| / Tr|ρ − σ| : ρ, σ ∈ M_2 },

and Q_C(a,f) denotes the set of qubit channels over the classical channel ((1−a, a), (1−f, f)) with respect to that parametrization. The subsection on the distribution of η^Tr over classical channels states

We conjecture that inf{η(Q) : Q ∈ Q_C(a,f)} = |a−f| which is equal to the trace-distance contraction coefficient of the underlying classical channel.

Issue #10033 fixes the readings: M_2 is the set of qubit states and a qubit channel is a completely positive trace-preserving map; Q_C(a,f) consists of the channels for which Q(|0⟩⟨0|) and Q(|1⟩⟨1|) have diagonals (a, 1−a) and (f, 1−f); the source writes positivity of the Choi matrix as Q > 0, and complete positivity is read as a positive semidefinite Choi matrix; the supremum defining η^Tr runs over all pairs of states, the ratio of trace norms Tr|·| being the ratio of trace distances (the factors ½ cancel), and a pair ρ = σ counts as 0/0 = 0, which does not change the supremum; a, f ∈ [0,1].

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/LovasAndaiDobrushinInfimum.result

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Discrete structuresQuestion resolvedPosed in 2015The trinomial odd diagonal of OEIS A077864Schulte's conjectured odd diagonal equals the coefficients of a rational power series.Proved

The questionTriangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

Why it mattersThe formal target keeps the entry's generating-function definition and expresses the proposed identity for every natural index.

The ideaFor the trinomial triangle, write G for the inverse of 1 - X^2(1+X+X^2).

Our contributionSchulte's conjectured odd diagonal equals the coefficients of a rational power series.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The trinomial odd diagonal of OEIS A077864

Expansion of (1-x)^(-1)/(1-x-2*x^2-x^3).

Triangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2*n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

a(0)=1, a(1)=2, a(2)=5, a(3)=11, a(n)=2*a(n-1)+a(n-2)-a(n-3)-a(n-4) for n>3. - Philippe Deléham, Oct 25 2006

Conjecture: a(n) = Sum_{j=0..n/2} A027907(n+1-j,2*j+1), n >= 0. - Werner Schulte, Sep 29 2015

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/TrinomialOddDiagonalRationalSeries.schulte_a077864

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Number theoryQuestion resolvedPosed in 2017The A177680 twin-prime averages modulo fiveTwin-prime pairs centered at both 6n and 12n force n to be one or divisible by five.Proved

The questionNumbers n such that 6n and 12n are both the average of twin prime pairs.

Why it mattersDale's 2017 comment asserts that every sequence member after the first is a multiple of five.

The ideaIf n ≡ 1 (mod 5), then 5 ∣ 6n−1; primality forces 6n−1 = 5, hence n = 1.

Our contributionTwin-prime pairs centered at both 6n and 12n force n to be one or divisible by five.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A177680 twin-prime averages modulo five

OEIS A177680, NAME (%N, verbatim):

Numbers n such that 6n and 12n are both the average of twin prime pairs.

COMMENT (%C, verbatim; Harvey P. Dale, Sep 26 2017):

Conjecture: all terms but the first term are multiples of 5. - Harvey P. Dale, Sep 26 2017

FORMULA (%F, verbatim):

a(n)=A066388(n)/6.

The literal claim is ∀ n : ℕ, IsMember(n) → n = 1 ∨ 5 ∣ n, where IsMember(n) means 1 ≤ n ∧ Prime(6n−1) ∧ Prime(6n+1) ∧ Prime(12n−1) ∧ Prime(12n+1). Lean natural-number subtraction truncates in general, but the hypothesis n ≥ 1 keeps each displayed subtraction equal to ordinary subtraction here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/DaleTwinPrimeAverageMultipleOfFive.dale_a177680

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Quantum informationQuestion resolvedFirst preprint 2021Four clues are necessary for unique four by four SudoQ solvabilityNo uniquely solvable four by four SudoQ has at most three clues.Proved

The questionBurchardt and K. Życzkowski, Genuinely quantum SudoQ and its cardinality, arXiv:2106.02967v2, Section III.A, page 4, state:

Why it mattersThe source's Theorem 13 supplies uniquely solvable four-clue grids.

The ideaThe complement is a line, and its unit ray occurs once in every row, column and block.

Our contributionNo uniquely solvable four by four SudoQ has at most three clues.

Verification details
Release source recordProved in Lean

Open problem

Four clues are necessary for unique four by four SudoQ solvability

J. Paczos, M. Wierzbiński, G. Rajchel-Mieldzioć, A. Burchardt and K. Życzkowski, Genuinely quantum SudoQ and its cardinality, arXiv:2106.02967v2, Section III.A, page 4, state:

Conjecture. Similarly to the classical Sudoku there are no $4\times4$ SudoQ with 3 clues that are uniquely solvable.

G. Rajchel-Mieldzioć, Quantum mappings and designs, arXiv:2204.13008, Chapter 7, restates it:

Uniquely solvable grids of $4 \times 4$ SudoQ admit at least 4 clues.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | A quantum Latin square has orthonormal rows and columns. | SudoQClueGeometry.IsSudoQ | Four vectors in $\mathbb C^4$ in each row and column. | | Every disjoint block is an orthonormal basis. | IsSudoQ and blockCell | Four $2\times2$ blocks; zero-based coordinates $(2b_1+k_1,2b_2+k_2)$. | | A quantum grid has blank entries; its clues are the nonblank entries. | C and v in IsSolution | A finite set of cells with literal prescribed vectors; values outside C impose no constraint. | | Solvability includes existence of a completion. | UniquelySolvable | Its existential quantifier supplies an IsSolution. | | Solutions agree up to phase. | SameRays | A separate complex scalar of norm one for each entry. Clues remain literal vector equalities. | | No uniquely solvable grid with three clues. | SudoQMinimumClues.claim and result | At most three clues, hence also exactly three; arbitrary clue values, including unsolvable grids. |

RayEq and HasRepeatedRay are auxiliary predicates for a pair of vectors and for one ray occurring once in every unit. They add no hypothesis to the settled claim. The four-dimensional specialization is the source's $N=2$ SudoQ, not the $N=4$ Sudoku parameter.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/LatinSquares/SudoQMinimumClues.result

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Analysis & dynamicsCounterexample foundFirst preprint 2018The expected two-bath Bose-selection bound is refuted for cubic spectraTwo cubic product-coupled baths admit an open set with five Bose-selected modes.Refuted

The questionEckardt, On the number of Bose-selected modes in driven-dissipative ideal Bose gases, Phys.

Why it mattersThe source distinguishes a few macroscopically occupied modes from an extensive selected set.

The ideaFor each pair of indices, asymcontinuousOn proves continuity of the cubic matrix entry on this domain; the positive sums of the coupling coordinates supply the nonzero denominators.

Our contributionTwo cubic product-coupled baths admit an open set with five Bose-selected modes.

Scope & next questionThe settlement makes no claim about the source's continuum-rate or driven-rotor results.

Verification details
Release source recordRefuted in Lean

Open problem

The expected two-bath Bose-selection bound is refuted for cubic spectra

A. Schnell, R. Ketzmerick and A. Eckardt, On the number of Bose-selected modes in driven-dissipative ideal Bose gases, Phys. Rev. E 97, 032136 (2018), DOI 10.1103/PhysRevE.97.032136, arXiv:1802.04088v2, section “Rates with product structure”, main.tex lines 1173–1178 (TeX macros expanded and mathematical delimiters normalized):

We expect that there is a generalization of the above arguement also to non-ohmic systems described by arbitrary spectral densities. We have checked functions of the form (remember that $J$ must be odd) $J(\varepsilon)\propto\vert\varepsilon\vert^d\mathrm{sgn}(\varepsilon)$ with some power $d$ (not necessarily integer) and similarly observe $M_S\leq2M_B-1$.

Here $M_B$ is the number of baths and $M_S=|S|$ the number of selected modes. The two-bath matrix in main.tex lines 1062–1083 uses opposite bath spectra and projector couplings, with weights $f_i=|\langle i|f\rangle|^2$ and $g_i=|\langle i|g\rangle|^2$. The selection condition is Eq. MS-problem, lines 374–382. These definitions are also recorded in the literature note.

Issue #25984 fixes a generic reading for $M_B=2$. For $m\in\mathbb N$, take unrestricted real energies $\varepsilon$ and coordinatewise positive real vectors $F,G$, and put $f_i=F_i/\sum_jF_j$, $g_i=G_i/\sum_jG_j$. The module's asym is $A_{ij}=J_d(\varepsilon_j-\varepsilon_i)(f_if_j-g_ig_j)$, where spow defines $J_d(0)=0$ and $J_d(x)=x|x|^{d-1}$ for $x\ne0$.

IsSelected A S w means that, for every index $i$, membership in $S$ implies $w_i>0$ and $(Aw)i=0$, while nonmembership implies $w_i=0$ and $(Aw)i<0$. Write $V{d,m}$ for violating d m: the parameter triples $(\varepsilon,F,G)$ with $F,G>0$ admitting such $S,w$ with $|S|\ge4$. The proposition claim says that for every real $d>0$ and every natural $m$, $\operatorname{int}(V{d,m})=\varnothing$ in the full real parameter space $\mathbb R^{3m}$. This generic reading is weaker than an unconditional three-mode ceiling; it is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/BoseSelectionCubicTwoBathRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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Quantum informationQuestion resolvedFirst preprint 2020Pure preparations cannot violate the HTMB inequality in finite dimensionPure preparations satisfying both trit parity equivalences have HTMB success probability at most two thirds in every finite complex dimension.Proved

The questionAbbott, Bounding and simulating contextual correlations in quantum theory, PRX Quantum 2, 020334 (2021), arXiv:2010.04751v3, subsection “Mixed states as resources for quantum contextuality”, state:

Why it mattersAn exact pure-state HTMB bound determines whether a violation of this particular inequality can have a pure-preparation realization, even when finite dimension and the two POVMs are unrestricted.

The ideaPositivity and purity are not hypotheses of this lemma.

Our contributionPure preparations satisfying both trit parity equivalences have HTMB success probability at most two thirds in every finite complex dimension.

Scope & next questionThe tight mixed-state maximum, explicitly left open at line 509, remains unresolved by this theorem. - Neighbouring questions.

Verification details
Release source recordProved in Lean

Open problem

Pure preparations cannot violate the HTMB inequality in finite dimension

A. Tavakoli, E. Zambrini Cruzeiro, R. Uola and A. A. Abbott, Bounding and simulating contextual correlations in quantum theory, PRX Quantum 2, 020334 (2021), arXiv:2010.04751v3, subsection “Mixed states as resources for quantum contextuality”, state:

This gives us strong evidence (i.e., up to numerical precision) that pure states cannot violate the HTMB inequality~\eqref{eq:HTMB}, and we conjecture this to indeed be the case exactly.

The quotation is from line 506 of the e-print member Manuscript_infocontext_reresubmission.tex, md5 f767464510150f9e02bc504393cf40e8. The scenario and Eq. (33) are at lines 492–500; the numerical pure-state bound is at lines 502–507. See also the literature note.

Alice receives two trits $x=(x_1,x_2)\in\mathbb Z_3^2$, and Bob guesses either coordinate using one of two three-outcome measurements. Write $P_x=\operatorname{projector}(\psi_x)=|\psi_x\rangle\langle\psi_x|$. Issue #25326 fixes the finite-dimensional reading: for every $d\in\mathbb N$, every family of unit vectors $\psi_x\in\mathbb C^d$ and every pair of POVMs $(M_b){b\in\mathbb Z_3}$ and $(N_b){b\in\mathbb Z_3}$, assume

$$ \sum_{x:,x_1+a x_2=k}P_x=\sum_{x:,x_1+a x_2=0}P_x \qquad(a\in{1,2},\ k\in\mathbb Z_3). $$

Each POVM has positive semidefinite effects summing to the identity. Then the module's htmbValue satisfies

$$ \operatorname{htmbValue}(\psi,M,N) =\frac1{18}\sum_{x\in\mathbb Z_3^2} \left(\operatorname{Re}\operatorname{tr}(P_xM_{x_1})+ \operatorname{Re}\operatorname{tr}(P_xN_{x_2})\right) \le\frac23. $$

This is claim, and result : claim proves it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/PureStateHTMBBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationQuestion resolvedSource 2018Every localized quantum Bernoulli noise walk has the standard Gaussian limitEvery unit localized quantum Bernoulli noise state has the standard Gaussian scaling limit.Proved

The questionRen, Quantum Walk in Terms of Quantum Bernoulli Noise and Quantum Central Limit Theorem for Quantum Bernoulli Noise, Advances in Mathematical Physics 2018, Article 2507265, DOI 10.1155/2018/2507265, Section 6, “Conclusion Remarks,” published PDF p. 8:

Why it mattersThe question asks whether the classical diffusive Gaussian limit persists throughout the infinite-dimensional coin space.

The ideaThe definitions annihilation, creation, R, L and walk implement the operators and recursion above.

Our contributionEvery unit localized quantum Bernoulli noise state has the standard Gaussian scaling limit.

Scope & next questionThe spectral-measure identification itself is not a formalized statement of result. - Neighbouring questions.

Verification details
Release source recordProved in Lean

Open problem

Every localized quantum Bernoulli noise walk has the standard Gaussian limit

C. Wang, C. Wang, Y. Tang and S. Ren, Quantum Walk in Terms of Quantum Bernoulli Noise and Quantum Central Limit Theorem for Quantum Bernoulli Noise, Advances in Mathematical Physics 2018, Article 2507265, DOI 10.1155/2018/2507265, Section 6, “Conclusion Remarks,” published PDF p. 8:

It is still unclear whether or not the walk has a limit probability distribution for a general localized initial state.

Thus we conjecture that, for a general localized initial state, the QBN-based walk might still have the same limit probability distribution as the classical random walk.

The source's coin space $h$ has orthonormal basis ${Z_\sigma:\sigma\in\Gamma}$, where $\Gamma$ consists of the finite subsets of $\mathbb N$ (Section 2, Eq. (7), p. 2). Its annihilation and creation operators remove or add a coordinate when permitted, and otherwise give zero (Lemmas 1–2, Eqs. (8)–(9)). Lemma 5 and Definition 6, p. 3, specify

$$ R_k=\tfrac12(\partial_k^+\partial_k+I),\qquad L_k=\tfrac12(\partial_k^+\partial_k-I),\qquad \Phi_{n+1}(x)=R_n\Phi_n(x-1)+L_n\Phi_n(x+1). $$

A localized initial state is $\Phi_0(0)=\varphi$, $\Phi_0(x)=0$ for $x\ne0$, with $|\varphi|=1$ (Eq. (53), p. 6); the position probabilities are $\Pr(X_n=x)=|\Phi_n(x)|^2$ (Section 3, p. 3).

Issue #22700 fixes the reading: for every unit vector $\varphi\in h$, the law of $X_n/\sqrt n$ converges weakly to $N(0,1)$. This is proved in the coefficient-space model below. The source definitions and their locators are also described in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/BernoulliNoiseWalkGaussianLimit.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Historical context

    Year recorded in the source bibliography: 2018.

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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year recorded in the source bibliography: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Negativity and the partial transpose distance to the PPT statesThe equality fails: a rank-five two-qutrit state has negativity 1/34, while its distance to every PPT state is at least 2/51.Refuted

The questionPaterek, M. Żukowski, Hierarchy of correlation quantifiers comparable to negativity, Quantum 6, 654 (2022), arXiv:2111.11887v2, Conjecture 1:

Why it mattersThe conjecture would make the negativity a member of a family of distance-based correlation measures built from one distance, so that entanglement measured by negativity could be compared with discord and total correlations measured by the same distance.

Our contributionThe equality fails: a rank-five two-qutrit state has negativity 1/34, while its distance to every PPT state is at least 2/51.

Verification details
Release source recordRefuted in Lean

Open problem

Negativity and the partial transpose distance to the PPT states

R. Ganardi, M. Miller, T. Paterek, M. Żukowski, Hierarchy of correlation quantifiers comparable to negativity, Quantum 6, 654 (2022), arXiv:2111.11887v2, Conjecture 1:

Let $\rho$ be a density matrix. Then $\inf_{\sigma \in \mathrm{PPT}} d_T(\rho, \sigma) = N(\rho)$.

Here $d_T(\rho,\sigma)=\frac12|\rho^{T_B}-\sigma^{T_B}|_1$ is the partial transpose distance, $N(\rho)=\frac12(|\rho^{T_B}|1-1)$ is the negativity, $T_B$ is the partial transpose on the second system, and PPT is the set of states with $\sigma^{T_B}\ge0$. The triangle inequality gives $N(\rho)\le\inf\sigma d_T(\rho,\sigma)$, so the conjecture asserts the reverse inequality.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/NegativityPartialTransposeDistanceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2020A SIC-compound in dimension 8 from Hoggar SICsEight orthonormal fiducials generate a SIC-compound in complex dimension eight under the three-qubit Pauli action.Refuted

The questionRenes, Compounds of symmetric informationally complete measurements and their application in quantum key distribution (Phys.

Why it mattersA SIC-compound gives d orthogonal SICs, so that each SIC measurement can be paired with a projective measurement in the bases of the compound.

The ideaFor x, z ∈ {0, …, 7}, let (D(x, z)u)(b) = (−1)^{z·(b ⊕ x)} u(b ⊕ x) on ℂ⁸, with z·y the parity of the bitwise AND.

Our contributionEight orthonormal fiducials generate a SIC-compound in complex dimension eight under the three-qubit Pauli action.

Verification details
Release source recordRefuted in Lean

Open problem

A SIC-compound in dimension 8 from Hoggar SICs

A. Tavakoli, I. Bengtsson, N. Gisin and J. M. Renes, Compounds of symmetric informationally complete measurements and their application in quantum key distribution (Phys. Rev. Research 2, 043122 (2020); arXiv:2007.01007), define a SIC-compound: d³ unit vectors ψ_{jk} in ℂ^d, j ∈ [d²], k ∈ [d], such that

  • for every k the vectors ψ_{jk} form a SIC, |⟨ψ_{jk}, ψ_{lk}⟩|² = (dδ_{jl} + 1)/(d + 1);
  • for every j the vectors ψ_{jk} form an orthonormal basis.

They construct compounds for d = 2 and d = 4 and prove that none exists for d = 3.

  • For d = 5, 6, 7, 8 they enumerate the Weyl–Heisenberg fiducials and find no compound. The introduction states: "for d = 5, 6, 7, 8 we give evidence in support of the same conclusion."
  • For d = 8 they add: "dimension 8 also houses SICs that are not based on the WH group … Whether a SIC-compound can be formed from these exceptional SICs is left as an open question."

The verbatim statements are in the literature note.

Issue #18207 reads the exceptional SICs as the Hoggar SICs: the orbits of a fiducial vector under the three-qubit Pauli group. Its claim is the negative answer to the question. The claim is refuted, which answers the question in the positive: a SIC-compound in dimension 8 is formed from eight Hoggar SICs.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/HoggarSicCompoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2020The l1-norm correlated coherence is not monogamous for tripartite pure statesA pure state of three qubits has a negative l1 correlated-coherence monogamy difference.Refuted

The questionMaziero, Monogamy and trade-off relations for correlated quantum coherence (Phys.

Why it mattersA monogamy relation would make correlated coherence a quantity that cannot be freely shared, like entanglement.

The idea2. The seven coherences. Each reduced state is a finite sum of products wu wv/49922.

Our contributionA pure state of three qubits has a negative l1 correlated-coherence monogamy difference.

Verification details
Release source recordRefuted in Lean

Open problem

The l1-norm correlated coherence is not monogamous for tripartite pure states

M. L. W. Basso and J. Maziero, Monogamy and trade-off relations for correlated quantum coherence (Phys. Scr. 95, 095105 (2020); arXiv:2006.14037), study the coherence contained in correlations.

  • The quantities. For the l1-norm coherence C in a fixed product basis, the correlated coherence of a bipartite marginal is C^c(ρ_XY) = C(ρ_XY) − C(ρ_X) − C(ρ_Y), and C^c(ρ_A|BC) = C(ρ_ABC) − C(ρ_A) − C(ρ_BC).
  • The known case. They prove the monogamy relation C^c(ρ_A|BC) ≥ C^c(ρ_AB) + C^c(ρ_AC) under the conditions (22)–(24) and in several examples.
  • The conjecture. They state it for all pure states: "seems reasonably to conjecture that the monogamy relation C^c_l1(ρ_A|BC) ≥ C^c_l1(ρ_AB) + C^c_l1(ρ_AC) holds for any tripartite pure quantum state".

The verbatim statements are in the literature note.

Issue #18402 fixes the reading. By Eq. (33) the relation says 0 ≤ M, where M = C(ρ_ABC) + C(ρ_A) + C(ρ_B) + C(ρ_C) − C(ρ_AB) − C(ρ_AC) − C(ρ_BC). The conjecture is refuted for three qubits.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/CorrelatedCoherenceMonogamyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2012Chaffin-Linderman-Sloane-Wilks Gijswijt Merge RefutationThe starting sequence 1 2 refutes the merge identity once a 1 is allowed inside S.Refuted

The questionChaffin, Linderman, Sloane and Wilks, Journal of Integer Sequences 16 (2013), Article 13.4.3 = arXiv:1212.6102v3, Section 5, Theorem 23, state: "Assume the curling number conjecture is true.

Why it mattersThe consequent of the weakened statement is false outright, so the weakened Theorem 23, read as an implication whose hypothesis is the curling number conjecture, is equivalent to the negation of that conjecture: a proof of the weakened theorem would refute it.

The ideaSince cn(1 2) = 1 the tail length is zero and the extension is 1 2 itself, so the extension exists without appeal to the curling number conjecture.

Our contributionThe starting sequence 1 2 refutes the merge identity once a 1 is allowed inside S.

Verification details
Release source recordRefuted in Lean

Open problem

Chaffin-Linderman-Sloane-Wilks Gijswijt Merge Refutation

Chaffin, Linderman, Sloane and Wilks, Journal of Integer Sequences 16 (2013), Article 13.4.3 = arXiv:1212.6102v3, Section 5, Theorem 23, state: "Assume the curling number conjecture is true. Let S be an initial sequence not containing a 1, let S^(e) be its 'extension' (defined in §1), and let S^(inf) be its infinite continuation. Then S^(inf) = S^(e) G." Here G is Gijswijt's sequence A090822, cn(S) is the largest k with S = X Y^k for nonempty Y, the continuation appends cn repeatedly, and the extension is the first prefix whose curling number equals one.

The section closes: "We do not know if the theorem is still true if S is allowed to contain a 1 but does not end with 1." Section 6 lists this as open question 12: "The question implicit in the last sentence of §5."

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ChaffinSloaneGijswijtMergeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    Year in the primary source attribution: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresCounterexample foundFirst preprint 2021Refutation of the printed partition-polynomial derivative questionThe partitions (1,1) and (2) refute the printed derivative-separation question.Refuted

The questionDawsey, Russell, and Urban define the partition polynomial on printed page 2:

Why it mattersIts literal universal statement has a two-partition counterexample.

The ideaThey are unequal partitions of 2, with largest parts 1 and 2.

Our contributionThe partitions (1,1) and (2) refute the printed derivative-separation question.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the printed partition-polynomial derivative question

Dawsey, Russell, and Urban define the partition polynomial on printed page 2:

To define this new polynomial, let λ = ⟨1^{m_1}, 2^{m_2}, . . . , k^{m_k}⟩ be a partition written in frequency notation. We define the partition polynomial f_λ by f_λ(x) = Σ_{i=1}^{k} m_i x^i. (1)

Printed page 9 gives the range convention and asks Question 9:

For the following question, let lg(λ) denote the largest part of the partition λ, and recall that f_λ^{(d)}(x) = 0 for all d > lg(λ).

Question 9. If λ, λ′ are any two unequal partitions, is it true that f_λ^{(d)}(1) ≠ f_{λ′}^{(d)}(1) for some positive integer d ≤ min{lg(λ), lg(λ′)}?

The formal claim follows these printed quantifiers: the two partitions may have different sizes and lengths.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/DawseyPartitionPolynomialDerivativeQuestionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresCounterexample foundSource 2018Flórez-Junes-Ramírez xz-plane path-count refutationBoth printed plane-count conjectures fail at path length three.Refuted

The questionFlórez, Junes, and Ramírez, Journal of Integer Sequences 21 (2018), Article 18.1.2, Section 2 (Background) on printed page 3, define the path family as follows:

Why it mattersThe printed definition requires the third coordinate to be nonnegative along every nonempty prefix and to return to zero after the complete path.

The ideaEnumerate all 6^3 = 216 signed three-step paths.

Our contributionBoth printed plane-count conjectures fail at path length three.

Verification details
Release source recordRefuted in Lean

Open problem

Flórez-Junes-Ramírez xz-plane path-count refutation

Flórez, Junes, and Ramírez, Journal of Integer Sequences 21 (2018), Article 18.1.2, Section 2 (Background) on printed page 3, define the path family as follows:

We use C_n^±(k) to mean the set of all paths of length k in the n-dimensional cubic lattice. We divide C_n^±(k) into subfamilies depending on the behavior of the path. We now give definitions and notation for those families. If P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}), then we define V_r := (±e_{j_1}) + (±e_{j_2}) + · · · + (±e_{j_r}), the algebraic combination of the first r components of P for 0 < r ≤ k, i.e., V_r is the sum of the components of any initial subpath of P with r steps. We denote C_n(k) the subset of C_n^±(k) formed by all paths P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}) that satisfy that the nth coordinate of V_k is zero. We use C_n^≥(k) to denote all paths in C_n^±(k) with P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}) and that nth coordinate of V_r is non-negative for all 0 < r ≤ k. We now let C_n^+(k) be C_n^≥(k) ∩ C_n(k). […] For example, Figure 1 depicts the 14 paths in C_2^+(3). Figure 2 depicts the 17 paths in C_3^+(2).

The paper also specifies that paths start at p_0 = (0, ..., 0) and each step is a positive or negative coordinate direction.

Section 6 on printed page 23 states:

Conjecture 1: For k ≥ 1, the number of paths in C_3^+(k) that are completely contained in the xz−plane is (see Table 4 first line) Σ_{i=1}^{k+1} \binom{2i}{i}\binom{k}{i−1}/(i+1).

The formal claim1 retains the universal quantifier, lower bound, printed path predicate, plane-containment condition, and binomial sum. The theorem result1 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/FlorezCubicLatticePlanePathCountRefutation.result1

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2018.

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Discrete structuresCounterexample foundSource 2018Flórez-Junes-Ramírez yz-plane path-count refutationBoth printed plane-count conjectures fail at path length three.Refuted

The questionFlórez, Junes, and Ramírez, Journal of Integer Sequences 21 (2018), Article 18.1.2, Section 2 (Background) on printed page 3, define the path family as follows:

Why it mattersThe printed definition requires the third coordinate to be nonnegative along every nonempty prefix and to return to zero after the complete path.

The ideaEnumerate all 6^3 = 216 signed three-step paths.

Our contributionBoth printed plane-count conjectures fail at path length three.

Verification details
Release source recordRefuted in Lean

Open problem

Flórez-Junes-Ramírez yz-plane path-count refutation

Flórez, Junes, and Ramírez, Journal of Integer Sequences 21 (2018), Article 18.1.2, Section 2 (Background) on printed page 3, define the path family as follows:

We use C_n^±(k) to mean the set of all paths of length k in the n-dimensional cubic lattice. We divide C_n^±(k) into subfamilies depending on the behavior of the path. We now give definitions and notation for those families. If P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}), then we define V_r := (±e_{j_1}) + (±e_{j_2}) + · · · + (±e_{j_r}), the algebraic combination of the first r components of P for 0 < r ≤ k, i.e., V_r is the sum of the components of any initial subpath of P with r steps. We denote C_n(k) the subset of C_n^±(k) formed by all paths P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}) that satisfy that the nth coordinate of V_k is zero. We use C_n^≥(k) to denote all paths in C_n^±(k) with P = (±e_{j_1})(±e_{j_2}) · · · (±e_{j_k}) and that nth coordinate of V_r is non-negative for all 0 < r ≤ k. We now let C_n^+(k) be C_n^≥(k) ∩ C_n(k). […] For example, Figure 1 depicts the 14 paths in C_2^+(3). Figure 2 depicts the 17 paths in C_3^+(2).

The paper also specifies that paths start at p_0 = (0, ..., 0) and each step is a positive or negative coordinate direction.

Section 6 on printed page 23 states:

Conjecture 2: For k ≥ 1, the number of paths in C_3^+(k) that are completely contained in the yz−plane is Σ_{i=1}^{k+1} \binom{2i}{i}\binom{k}{i−1}/(i+1).

The formal claim2 retains the universal quantifier, lower bound, printed path predicate, plane-containment condition, and binomial sum. The theorem result2 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/FlorezCubicLatticePlanePathCountRefutation.result2

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2018.

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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2020A misclassified configuration for a ternary two-rule density classifierThe ternary cellular automaton with Wolfram number 6478767664173 sends the configuration 210 of length 3 to 111, which it and the rule 7580606234490 both fix, although 210 contains a zero and has density 1/2.Refuted

The questionHenryk Fukś and Roman Procyk, Explorations of Ternary Cellular Automata and Ternary Density Classification Problems, arXiv:2002.08924v1 (Conjecture 1 of Acta Phys.

Why it mattersDensity classification asks a cellular automaton to decide, from a random initial configuration, which density interval it belongs to; two-rule solutions are known in the binary case (rules 184 and 232).

Our contributionThe ternary cellular automaton with Wolfram number 6478767664173 sends the configuration 210 of length 3 to 111, which it and the rule 7580606234490 both fix, although 210 contains a zero and has density 1/2.

Verification details
Release source recordRefuted in Lean

Open problem

A misclassified configuration for a ternary two-rule density classifier

Henryk Fukś and Roman Procyk, Explorations of Ternary Cellular Automata and Ternary Density Classification Problems, arXiv:2002.08924v1 (Conjecture 1 of Acta Phys. Pol. B Proc. Suppl. 12(1) (2019) 75–89), write:

Let $F$ be the ternary nearest-neighbour rule with Wolfram number 6478767664173, and $G$ be the rule with Wolfram number 7580606234490. For any finite ternary string $\mathbf{x}$ of length $L$, containing at least one zero, let $\displaystyle \rho=\frac{1}{2L}\sum_{i=0}^{L-1} x_i$. Then $G^LF^L(\mathbf{x})=0^L$ if $\rho(\mathbf{x}) \in [0, 2/3)$, $G^LF^L(\mathbf{x})=1^L$ if $\rho(\mathbf{x}) \in (2/3, 3/4)$, $G^LF^L(\mathbf{x})=2^L$ if $\rho(\mathbf{x}) \in (3/4, 1)$.

Configurations are periodic, $(F(\mathbf{x}))i=f(x{i-1},x_i,x_{i+1})$ with indices in $\mathbb{Z}/L$, and the rule with Wolfram number $N$ has $f(a,b,c)=\lfloor N/3^{9a+3b+c}\rfloor \bmod 3$. The precise question, preregistered in issue #11845, is whether the three implications hold for every $L$ and every configuration containing a zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CellularAutomata/TernaryDensityClassificationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2019Three-state overlap interval propagation is not always optimalFive overlaps of four pure states force the sixth to be one quarter, although intersection of the three-state lower bounds gives zero.Refuted

The questionBrod, Quantum and classical bounds for two-state overlaps, Phys.

Why it mattersInferring unmeasured overlaps from measured ones is the paper's organizing problem.

The ideaThe lemma witnessrealizes verifies all four unit norms and every required ordered overlap for this family.

Our contributionFive overlaps of four pure states force the sixth to be one quarter, although intersection of the three-state lower bounds gives zero.

Scope & next questionA general sufficient compatibility criterion, or a characterization of the larger graphs and data for which interval propagation is exact, remains open here. - Effect on the paper.

Verification details
Release source recordRefuted in Lean

Open problem

Three-state overlap interval propagation is not always optimal

E. F. Galvão and D. J. Brod, Quantum and classical bounds for two-state overlaps, Phys. Rev. A 101, 062110 (2020), arXiv:1902.11039v3, Section II.B, overlap_pra_v3.tex, line 165:

We leave it as an open question whether this procedure is optimal, but emphasize that even the initial iteration that produces a complete graph already provides bounds for every pair of vertices, which may be useful even if not tight.

The procedure is specified at lines 151–156: use the three-state bounds through each common neighbour of an unmeasured pair, intersect the inferred intervals, add the edge, and continue. Line 165 also allows revisiting inferred edges. The source's definitions and bounds are at lines 97–131; see the literature note.

Issue #23699 fixes the reading to graphs with exactly one unknown pair. Write $U(x,y)=(\sqrt{xy}+\sqrt{(1-x)(1-y)})^2$, and let $L(x,y)=(\sqrt{xy}-\sqrt{(1-x)(1-y)})^2$ when $x+y>1$, and $L(x,y)=0$ otherwise. These are the module's upper and lower.

For $m$ labelled states, distinct indices $i_0,j_0$, and an array $r$, a realization is a family of unit vectors $\psi_i\in\mathbb C^n$ satisfying $|\langle\psi_i,\psi_j\rangle|^2=r_{ij}$ for every ordered pair of distinct indices except the unordered pair ${i_0,j_0}$. Diagonal entries impose no conditions. The set $F$ of feasible unknown overlaps ranges over every finite dimension $n$. The procedure's endpoints are $\ell=\sup{L(r_{ki_0},r_{kj_0}):k\ne i_0,j_0}$ and $u=\inf{U(r_{ki_0},r_{kj_0}):k\ne i_0,j_0}$, encoded by procLower and procUpper.

The proposition claim asserts that, for every $m\ge3$, every distinct $i_0,j_0$, and every $r$ with $F\ne\varnothing$, both $\ell=\inf F$ and $u=\sup F$. This universal tightness statement is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/PureState/OverlapIntervalPropagationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier explains a connection to a broader mathematical question.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021The Geiger–Kedem spin-1 spin-entropy depends on the relative phasesThe spin-one entropy in equations (15)–(18) depends on a relative phase.Refuted

The questionKedem, Spin Entropy, Entropy 24, 1292 (2022), arXiv:2111.11605v3, define the spin-entropy of a pure spin state as the Shannon entropy of the Sz probabilities plus the differential entropy of the polarization-angle density (Eq. (4)). - Spin 1. For the state (15), with amplitudes sin θα cos θβ e^{iφx}, cos θα and sin θα sin θβ e^{iφz}, Proposition 2 gives the spin-entropy (17).

Why it mattersThe conjecture underlies the source's Figure 2, which plots Eq. (18) against (θα, θβ), and its further conjecture (§II) that the ordering of states by spin-entropy is determined by the z-axis component alone.

The ideaTake (a, b, c) = (3, 1, 1)/√11, that is θα = arctan √10 and θβ = arctan(1/3), with φx = 0 and φz = 0 or π.

Our contributionThe spin-one entropy in equations (15)–(18) depends on a relative phase.

Verification details
Release source recordRefuted in Lean

Open problem

The Geiger–Kedem spin-1 spin-entropy depends on the relative phases

D. Geiger and Z. M. Kedem, Spin Entropy, Entropy 24, 1292 (2022), arXiv:2111.11605v3, define the spin-entropy of a pure spin state as the Shannon entropy of the S_z probabilities plus the differential entropy of the polarization-angle density (Eq. (4)).

  • Spin 1. For the state (15), with amplitudes sin θ_α cos θ_β e^{iφ_x}, cos θ_α and sin θ_α sin θ_β e^{iφ_z}, Proposition 2 gives the spin-entropy (17). Its density is ρ_1(φ) = (1/2π)|sin θ_α cos θ_β e^{i(φ_x+2φ)} + cos θ_α e^{iφ} + sin θ_α sin θ_β e^{i(φ_z−2φ)}|².
  • The conjecture. "Conjecture 1. The spin-entropy depends only on the variables that define the spin-entropy component along the z axis. In particular, the spin-entropy (17) of a state given by (15) can be simplified to" its zero-phase form (18).

The verbatim statements are in the literature note.

Issue #17162 fixes the reading: for all θ_α, θ_β ∈ [0, π/2] and φ_x, φ_z ∈ [0, 2π), Eq. (17) equals its value at φ_x = φ_z = 0. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/GeigerKedemSpinEntropyPhaseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundFirst preprint 2021Prefix closure for squares with three digitsA fifteen-block decimal word has a square with digits in {0,2,5}, but deleting its last block gives a square containing digit one.Refuted

The questionThe source identified by geisser2021squares, Section 3.3, Conjecture 1, printed pages 25–26, states:

Why it mattersIt motivates this problem's encoding discipline; its theorem is not a premise or a claimed solution of this different conjecture.

The ideaUse the tuple [0,1,3,7,15,31,63,127,2,6,14,62,14,6,2].

Our contributionA fifteen-block decimal word has a square with digits in {0,2,5}, but deleting its last block gives a square containing digit one.

Verification details
Release source recordRefuted in Lean

arXiv

Prefix closure for squares with three digits

The source identified by geisser2021squares, Section 3.3, Conjecture 1, printed pages 25–26, states:

If ⟨a₁′,...,aₗ′⟩ is a solution with l ≥ 2, then ⟨a₁′,...,aₗ₋₁′⟩ is also a solution.

For a finite list of natural numbers, form each block as 4 followed by exactly that entry's number of nines, concatenate the blocks, and append 5. Let encode be the resulting ordinary decimal natural number, and let GoodSquare n mean that every actual decimal digit of n² belongs to {0,2,5}. The closed claim is exactly:

∀ runs : List ℕ, 2 ≤ runs.length →
  GoodSquare (encode runs) → GoodSquare (encode runs.dropLast)

Zero runs are included. Neither list length nor entries have an upper bound. Deletion removes the entire final 4/nines block and preserves the final 5. All three allowed digits need not occur.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/SquaresThreeDigitsPrefixRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021The unital-block overlap lower bound fails for qutrit channelsTwo completely positive trace-preserving qutrit maps have unital-block overlap minus four thirds, contradicting the proposed lower bound of minus one.Refuted

The questionJennings, Estimation of correlations and non-separability in quantum channels via unitarity benchmarking, arXiv:2104.04352v3, Physical Review Research 4, 023041 (2022), Appendix C.5 after Lemma C.3, state:

Why it mattersThey are completely positive and trace preserving.

The ideaComplete positivity follows by applying FiniteKrausChannel.PhyslibLeaf.MatrixMap.ofkrausisCompletelyPositive.

Our contributionTwo completely positive trace-preserving qutrit maps have unital-block overlap minus four thirds, contradicting the proposed lower bound of minus one.

Verification details
Release source recordRefuted in Lean

Open problem

The unital-block overlap lower bound fails for qutrit channels

M. Girling, C. Cîrstoiu and D. Jennings, Estimation of correlations and non-separability in quantum channels via unitarity benchmarking, arXiv:2104.04352v3, Physical Review Research 4, 023041 (2022), Appendix C.5 after Lemma C.3, state:

However, we conjecture a stronger result that for any two quantum channels E₁, E₂ that ⟨T₁, T₂⟩ ≥ −1.

They also state:

Since it is not essential to our work we leave it as an open problem.

Section II.C chooses a Hilbert-Schmidt orthonormal operator basis with $X_0=I/\sqrt d$. Its positive-index Liouville entries are $T_{ij}=\operatorname{tr}(X_i^\dagger\Phi(X_j))$. The overlap is $\operatorname{tr}(T_\Phi^\dagger T_\Psi)$. The Library note gives the literal source and locators.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Any two quantum channels | claim quantifies over d, Phi, Psi; both satisfy the frozen CoPRelativeQuantumnessRefutation.IsCPTP | Complete positivity and trace preservation on all complex matrices | | Orthonormal operator basis | IsLiouvilleBasis | All $d^2$ inner products, and normalized identity at zero | | Positive-index unital block | unitalBlock, restricted in overlap | Both index values are nonzero | | Hilbert-Schmidt overlap | overlap | Sum of conjugate products; claim compares its real part | | Lower bound $-1$ | claim | Universal implication, with all channel and basis hypotheses | | Refutation | result | Negation of claim, using one Hermitian qutrit basis |

The natural dimension and proof h0 encode a nonempty finite-dimensional system. Defining entries at every index does not add zero-index terms to the overlap. The real-part comparison agrees with the paper on Hermitian operator bases. Quantifying over complex orthonormal bases adds no restriction on the explicit Hermitian witness.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/UnitalBlockOverlapRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2013The d-orthostochastic dimension is not eventually bounded by 3Star matrices contradict an eventual upper bound of three on Gutkin's internal dimension.Refuted

The questionGutkin, On a multi-dimensional generalization of the notions of orthostochastic and unistochastic matrices, arXiv:1301.2537v1, Conjecture 1:

Why it mattersThe source reads P ∈ OS(F, n, d) as a quantization of the classical system P with d internal degrees of freedom. dmin(F, n) is then the number of internal degrees of freedom that suffice for every bistochastic n × n matrix.

The idea1. The star matrix. Sn is symmetric and bistochastic for n ≥ 2.

Our contributionStar matrices contradict an eventual upper bound of three on Gutkin's internal dimension.

Verification details
Release source recordRefuted in Lean

Open problem

The d-orthostochastic dimension is not eventually bounded by 3

E. Gutkin, On a multi-dimensional generalization of the notions of orthostochastic and unistochastic matrices, arXiv:1301.2537v1, Conjecture 1:

Denote by $d_{\min}(n)$ the smallest value of $d$ such that the set of $d$-orthostochastic matrices coincides with $\B_n$. Then there exists $n_0$ such that for $n\ge n_0$ we have $2 \le d_{\min}(n) \le 3.$

A bistochastic real n × n matrix P is d-orthostochastic when P_rc = ‖V_rc‖² for vectors V_rc ∈ ℝ^d satisfying the source's equations (1) and (2): column orthogonality, Σ_r ⟨V_rj, V_rk⟩ = δ_jk, and row orthogonality, Σ_c ⟨V_jc, V_kc⟩ = δ_jk. The definitions are quoted in the literature note.

Issue #19782 fixes the reading:

  • d_min(n) is the natural-number infimum of the set of d for which every bistochastic n × n matrix is d-orthostochastic;
  • the claim is that some n₀ has 2 ≤ d_min(n) ≤ 3 for every n ≥ n₀.

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/GutkinOrthostochasticDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2013. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2020Square-root Jensen–Shannon balls need not admit majorization-minimizersA three-point Jensen-Shannon ball has no majorization-minimizer.Refuted

The questionHanson, Entropic Continuity Bounds & Eventually Entanglement-Breaking Channels (PhD thesis, Cambridge, 2020; arXiv:2010.02408), derives continuity bounds for Schur-concave functions from the majorization structure of total-variation balls. - Every total-variation ball in the probability simplex contains a majorization-minimizer r∗, with r∗ ≺ q for every q in the ball (Eq. (3.1)).

Why it mattersA majorization-minimizer of a ball maximizes every Schur-concave function over the ball at once, which gives universal local continuity bounds for entropies.

The ideaLet J(p, q) = (1/2) Σᵢ [pᵢ ln(2pᵢ/(pᵢ + qᵢ)) + qᵢ ln(2qᵢ/(pᵢ + qᵢ))].

Our contributionA three-point Jensen-Shannon ball has no majorization-minimizer.

Verification details
Release source recordRefuted in Lean

arXiv

Square-root Jensen–Shannon balls need not admit majorization-minimizers

E. P. Hanson, Entropic Continuity Bounds & Eventually Entanglement-Breaking Channels (PhD thesis, Cambridge, 2020; arXiv:2010.02408), derives continuity bounds for Schur-concave functions from the majorization structure of total-variation balls.

  • Every total-variation ball in the probability simplex contains a majorization-minimizer r∗, with r∗ ≺ q for every q in the ball (Eq. (3.1)). Every Schur-concave function, such as an entropy, attains its maximum over the ball at r∗.
  • For p-norm balls with 1 < p < ∞ no such element exists (Section 3.7).
  • Chapter 8, Question 1 asks: "Do balls in Hellinger distance, Bures distance, the (square-root) Shannon-Jensen divergence, Wasserstein distances, or other metrics on P admit majorization-minimizers?"

The verbatim statements are in the literature note.

Issue #18189 fixes the reading of the Jensen–Shannon clause: for every probability vector p and every positive radius, the square-root Jensen–Shannon ball around p has a majorization-minimizer. This clause is answered in the negative.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/JensenShannonBallMajorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2021A matrix refutation of Krishna's C*-algebraic Novak conjectureTheir differences have scalar squares, and the exponential series gives cosine values 1, 1 and -1.Refuted

The questionMahesh Krishna, C-algebraic Schur product theorem, Pólya-Szegő-Rudin question and Novak's conjecture, arXiv:2108.06662v1, Conjecture 4.3, p. 10; journal DOI 10.4134/JKMS.j210627, states:

Why it mattersThe conjecture extends the commutative theorem to all unital C-algebras.

Our contributionTheir differences have scalar squares, and the exponential series gives cosine values 1, 1 and -1.

Scope & next questionThe experiment entry contains check.py; command python3 check.py, exit 0, SHA-256 fbd6153f6b485f05f2c1d7809a623bc49ebabf8b22a6fcfe99388595074d28e4.

Verification details
Release source recordRefuted in Lean

Open problem

A matrix refutation of Krishna's C*-algebraic Novak conjecture

K. Mahesh Krishna, C-algebraic Schur product theorem, Pólya-Szegő-Rudin question and Novak's conjecture*, arXiv:2108.06662v1, Conjecture 4.3, p. 10; journal DOI 10.4134/JKMS.j210627, states:

Let $\mathcal A$ be a unital C*-algebra. Then the matrix $[\prod_{l=1}^d(1+\cos(x_{j,l}-x_{k,l}))/2-1/n]{1\le j,k\le n}$ is positive for all $n,d\ge2$ and all choices of $x_j=(x{j,1},\ldots,x_{j,d})\in\mathcal A_{\rm sa}^d$.

The source defines $\cos X=(e^{iX}+e^{-iX})/2$ using the exponential series, and positivity means self-adjointness together with $\langle Mv,v\rangle\ge0$ for every $v\in\mathcal A^n$, where the source's left-linear form is $\sum_j\sum_k M_{jk}v_kv_j^*$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarNovak.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2016A signed sum of three-qubit Pauli words of norm 21For three qubits, one choice of signs makes the signed sum of all 63 non-identity Pauli words an operator with eigenvalue 21, hence of norm at least 21, above (sqrt(3) + 1)^3 - 1 = 9 + 6 sqrt(3).Refuted

The questionA 95, 052315 (2017)) studies maximal (4^m − 1, m, p) quantum random access codes.

Why it mattersIf the conjecture held, the matrix I − (I + σx + σy + σz)^{⊗m} would be extremal, and the worst-case success probability of these codes would equal (1 + 1/((1+√3)^m − 1))/2.

The ideaThe base-4 digits of k = 1, …, 63 run once over the 63 non-identity words.

Our contributionFor three qubits, one choice of signs makes the signed sum of all 63 non-identity Pauli words an operator with eigenvalue 21, hence of norm at least 21, above (sqrt(3) + 1)^3 - 1 = 9 + 6 sqrt(3).

Verification details
Release source recordRefuted in Lean

Open problem

A signed sum of three-qubit Pauli words of norm 21

O. Liabøtrø (arXiv:1607.02667, quant-ph; Phys. Rev. A 95, 052315 (2017)) studies maximal (4^m − 1, m, p) quantum random access codes. Their worst-case success probability is governed by the signed sums Σ(β) = Σ_{k=1}^{4^m−1} (−1)^{β(k)} σ_{c_1(k)} ⊗ ⋯ ⊗ σ_{c_m(k)} of all non-identity Pauli words, for arbitrary signs β. The paper states:

We have checked numerically that no $\Sigma(\beta)$ has an eigenvalue less than $1-(1+\sqrt(3))^m$ for $m=1,2$. … We conjecture that this is the case for all $m$, i.e.: $||\sum_{k=1}^{4^m-1}(-1)^{\beta(k)}\bigotimes_{i=1}^m\sigma_{c_i(k)}||\leq (\sqrt{3}+1)^m-1$

Issue #11468 fixes the reading. k ↦ (c_1(k), …, c_m(k)) is a bijection onto the non-identity words, ‖·‖ is the operator norm, and the conjecture is the displayed inequality for every m and every β. The formal claim is that inequality; the result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SignedPauliSumNormRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2012Lin's operator Wielandt norm conjecture failsA unital 2-positive complex-linear map violates Lin's operator Wielandt product norm bound.Refuted

The questionMinghua Lin, On an operator Kantorovich inequality for positive linear maps, J.

Why it mattersThe order inequality in Theorem 3.1 suggests a corresponding product norm bound.

Our contributionA unital 2-positive complex-linear map violates Lin's operator Wielandt product norm bound.

Verification details
Release source recordRefuted in Lean

Open problem

Lin's operator Wielandt norm conjecture fails

Minghua Lin, On an operator Kantorovich inequality for positive linear maps, J. Math. Anal. Appl. 402 (2013), 127–132; arXiv:1212.5690v1, §3, pp. 6–7.

§3, p. 6, Theorem 3.1 states verbatim:

Theorem 3.1. [3] Let 0 < m ≤ A ≤ M and let X, Y be two partial isometries on H whose final spaces are orthogonal to each other. Then for every 2-positive linear map Φ, Φ(XAY)Φ(YAY)⁻¹Φ(YAX) ≤ ((M−m)/(M+m))² Φ(XAX).

§3, pp. 6–7, immediately before Conjecture 3.4:

Despite the invalidity of (3.6), there is some evidence that the following assertion could be true. We have been unable to prove (or disprove) it.

§3, p. 7, Conjecture 3.4 states verbatim:

Conjecture 3.4. Under the same condition as in Theorem 3.1. Then for every 2-positive linear map Φ, ‖Φ(XAY)Φ(YAY)⁻¹Φ(YAX)Φ(XAX)⁻¹‖ ≤ ((M−m)/(M+m))².

§3, p. 6, defines 2-positivity verbatim:

Let Φ be a unital positive linear map between C*-algebras. We say that Φ is 2-positive if whenever the 2 × 2 operator matrix [[A, B],[B*, C]] is positive, then so is [[Φ(A), Φ(B)],[Φ(B*), Φ(C)]].

P. Zhang, Remarks on an operator Wielandt inequality, arXiv:1506.00737, describes Conjecture 3.4 as still open.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/OperatorWielandtNormRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021The Liu–DeBrota fewer-effects construction does not minimize orthogonalityFour effects on a ten-dimensional complex space disprove the proposed orthogonality minimum.Refuted

The questionDeBrota, Relating measurement disturbance, information and orthogonality, arXiv:2105.02074v2 (Phys.

Why it mattersThe source studies the relation between measurement disturbance, measurement strength and orthogonality.

Our contributionFour effects on a ten-dimensional complex space disprove the proposed orthogonality minimum.

Scope & next questionFor nonzero remainder, optimality of the balanced replacement among all POVMs remains open in this dossier. - Effect on the source.

Verification details
Release source recordRefuted in Lean

Open problem

The Liu–DeBrota fewer-effects construction does not minimize orthogonality

Y. Liu and J. B. DeBrota, Relating measurement disturbance, information and orthogonality, arXiv:2105.02074v2 (Phys. Rev. A 104, 052216, 2021), Appendix C, subsection “Fewer effects”, Conjecture 2 (apssamp.tex, lines 795–797):

The above constructed POVM has minimal orthogonality $O$ among POVMs having $n(n<d)$ nonvanishing effects.

For a POVM $F=(F_i){i=1}^n$ on $\mathbb C^d$, Section IV (apssamp.tex, lines 403–420) defines $S(F){ij}=\operatorname{tr}(\sqrt{F_i}\sqrt{F_j})$ and $O(F)=|S(F)-I_n|F^2$. The square roots are the positive semidefinite square roots. Equivalently, $O(F)=\sum{i,j}|S(F){ij}-\delta{ij}|^2$.

For natural numbers $0<n<d$, write $d=an+b$, where $a=\lfloor d/n\rfloor$ and $0\le b<n$. The proposed construction $E^{(n,d)}$ consists of diagonal coordinate projectors on consecutive disjoint blocks: the first $n-1$ blocks have size $a$, and the last has size $a+b$. Its Gram matrix is $\operatorname{diag}(a,\ldots,a,a+b)$, as displayed in Appendix C (apssamp.tex, lines 765–778).

Issue #25254 fixes the formal reading: for every $n,d\in\mathbb N$ with $0<n<d$ and every family of nonzero positive semidefinite complex $d\times d$ matrices $F_i$ satisfying $\sum_iF_i=I_d$, one has $O(E^{(n,d)})\le O(F)$. This universal assertion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/FewerEffectsOrthogonalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Mitra broadcasting-channel extremaA three-dimensional observable with two outcomes has a nonempty set of broadcasting channels and has neither extremum under nondisturbing precomposition.Refuted

The questionMitra, Layers of classicality in the compatibility of measurements, Phys.

The ideaTwo distinct fixed channels exclude a lowest element.

Our contributionA three-dimensional observable with two outcomes has a nonempty set of broadcasting channels and has neither extremum under nondisturbing precomposition.

Verification details
Release source recordRefuted in Lean

Open problem

Mitra broadcasting-channel extrema

A. Mitra, Layers of classicality in the compatibility of measurements, Phys. Rev. A 104, 022206 (2021), arXiv:2101.05752v4, Section III.A asks whether the broadcasting channels of an observable have a highest element and whether they have a lowest element under precomposition by channels that are nondisturbing for that observable. The source repeats both questions in its Conclusions.

The formal convention uses an observable with outcomes Fin m, input matrices on Fin d, output matrices on Fin d × Fin d, finite Kraus channels, the broadcasting predicate, the nondisturbing predicate, and the precomposition relation. The settling declaration D5/S3/Quantum/Measurements/BroadcastingChannelExtremaRefutation.result has type ¬ claim and carries OpenProblemResolutionClaim(Refuted).

Statement echo

The source's broadcasting condition is the pair of marginal trace equalities in Broadcasts; its nondisturbance condition is equality of each observable effect's trace after channel application in Nondisturbing; ConstructedFrom is precomposition by such a channel. HasHighest and HasLowest quantify over broadcasting channels and this relation. claim combines observability, existence of a broadcasting channel, and the disjunction of the two extrema. The fixed witness uses d = 3, m = 2, and the observable A; the remaining binders and hypotheses are those displayed by the corresponding Scribe nodes.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/BroadcastingChannelExtremaRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Circulant bracelet sufficiency fails in dimension twelveA circulant bracelet matrix of order twelve is not unistochastic.Refuted

The questionPuchała and K. Życzkowski, Algebraic and geometric structures inside the Birkhoff polytope, arXiv:2101.11288, Section V, asks:

Why it mattersTheorem 17 identifies circulant bracelet and circulant unistochastic matrices in dimension four.

The ideaEvery row and column has six nonzero entries.

Our contributionA circulant bracelet matrix of order twelve is not unistochastic.

Verification details
Release source recordRefuted in Lean

Open problem

Circulant bracelet sufficiency fails in dimension twelve

G. Rajchel-Mieldzioć, K. Korzekwa, Z. Puchała and K. Życzkowski, Algebraic and geometric structures inside the Birkhoff polytope, arXiv:2101.11288, Section V, asks:

Another observation that holds for d ≤ 4, and should be further investigated for d > 4, is whether the bracelet condition is sufficient for unistochasticity if we restrict the analysis to the set of circulant matrices.

The universal sufficiency assertion is refuted. Preregistration: #16299. The source locator and quoted definitions are in the Library note.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Definition 5: circulant, with indices modulo $d$ | IsCirculant expresses existence of a vector whose Mathlib circulant is the matrix | Mathlib uses $v(i-j)$; reversing the first row gives the source's class | | Definition 4: entries are squared moduli of a complex unitary matrix | IsUnistochastic uses Matrix.unitaryGroup and the frozen RHLinalg.normSqMatrix | The unitary may be arbitrary; no circulant-unitary restriction is imposed | | Definition 7: bistochasticity and column and row bracelet inequalities | IsBracelet uses doublyStochastic and both quantified inequalities | Pairs are distinct: $k\ne l$ for columns, $j\ne k$ for rows. The source counts $d(d-1)$ conditions; ordered pairs repeat the same inequalities | | Section V: all $d>4$, all circulant bracelet matrices | claim quantifies over natural $d$, $4<d$, and all real square matrices | Finite real suprema represent the source's maxima; $d>4$ ensures nonempty index sets | | A negative answer to the universal assertion | result proves the literal negation of claim | A counterexample at $d=12$ suffices |

All parameters and conjuncts in these definitions implement the source clauses. There is no additional hypothesis on a unitary realizing the matrix.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/CirculantBraceletNonUnistochastic.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2021A ring of four complete blocks refutes the flat-graph clique-root conjectureFour complete four-vertex graphs joined in a ring give a twelve-vertex counterexample to Conjecture 1 of On Clique Roots of Flat Graphs.Refuted

The questionTeimoori Faal, On Clique Roots of Flat Graphs, arXiv:2112.09721v1, Conjecture 1, page 6, states:

Why it mattersFlatness controls the number of triangles through each edge.

The ideaPublic result applies this to claim without assuming the conjecture's multiplicity clause.

Our contributionFour complete four-vertex graphs joined in a ring give a twelve-vertex counterexample to Conjecture 1 of On Clique Roots of Flat Graphs.

Verification details
Release source recordRefuted in Lean

Open problem

A ring of four complete blocks refutes the flat-graph clique-root conjecture

H. Teimoori Faal, On Clique Roots of Flat Graphs, arXiv:2112.09721v1, Conjecture 1, page 6, states:

The class of 2-connected K₅-free flat graphs without isolated triangles has only clique roots. In particular, this class has always the clique root r = −1 of multiplicity 2.

The source note contains the definitions and locators. The preregistration is #19709. The quoted conjecture is Refuted by a twelve-vertex graph.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite simple graph | V, Fintype V, SimpleGraph V in claim | Arbitrary finite vertex type; undirected and loopless | | $C(G,x)=\sum_j c_j(G)x^j$, with $c_0=1$ | cliquePoly | Integer clique counts from SimpleGraph.cliqueFinset; sum through the vertex count, with higher terms zero | | Every edge is in at most two triangles (Definition 2.2) | IsFlat | Counts the common neighbours of each adjacent pair | | No isolated triangle (Definition 4.1) | NoIsolatedTriangle | Every triangle has a common neighbour of all three vertices | | 2-connected | Existing D5.S3.StatisticalMechanics.DimerVirialFourthDifferenceRefutation.Biconnected and 3 ≤ Fintype.card V | Connected before and after deletion of any vertex; more than two vertices | | $K_5$-free | SimpleGraph.CliqueFree 5 | No five-vertex clique | | Has only clique roots | First conjunct of claim | Every complex zero has imaginary part zero | | $-1$ has multiplicity exactly two | Second conjunct of claim | Polynomial.rootMultiplicity (-1) = 2 | | Implementation instances | DecidableEq V, DecidableRel G.Adj | Finite graph operations; no additional graph restriction | | Negative answer | result : ¬ claim | The first conjunct fails on one graph satisfying every hypothesis |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/CliquePolynomial/TeimooriFaalFlatCliqueRoots.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2020The coherent discrimination bias is not a metric on quantum channelsCoherent discrimination bias fails the triangle inequality for three qubit replacement channels.Refuted

The questionThe success probability p^cohs(N⁰, N¹) is the supremum of Eq. (17) over the prover's strategies (Eq. (18)), and 1/2 ≤ p^cohs ≤ 1 (Proposition 2).

Why it mattersA distinguishability measure that is a metric supports triangle-inequality arguments, such as hybrid arguments over sequences of channels.

The ideaThrough the replacement structure, the uncomputed amplitude equals (1/2)Σj (tj Q{0j} + uj Q{1j}) applied to the components of ψ.

Our contributionCoherent discrimination bias fails the triangle inequality for three qubit replacement channels.

Verification details
Release source recordRefuted in Lean

Open problem

The coherent discrimination bias is not a metric on quantum channels

M. M. Wilde, Coherent Quantum Channel Discrimination (IEEE ISIT 2020; arXiv:2001.02668), introduces a coherent version of channel discrimination.

  • The verifier controls the two channels by a qubit R₁ in |+⟩.
  • The prover supplies an input ψ_RA and processes the channel output with a unitary V_RBE′F, returning B and a flag qubit F.
  • The verifier uncomputes the controlled isometric extensions and projects R₁F onto a Bell state.

The success probability p^coh_s(N⁰, N¹) is the supremum of Eq. (17) over the prover's strategies (Eq. (18)), and 1/2 ≤ p^coh_s ≤ 1 (Proposition 2). Section VI asks: "An intriguing open question is to determine if p^coh_s − 1/2 is a metric on quantum channels." The incoherent analogue p^inc_s − 1/2 is a metric: by Eq. (8) it is half the normalized diamond distance (1/2)‖N⁰ − N¹‖_⋄.

The verbatim statements are in the literature note.

Issue #18289 fixes the reading: the triangle inequality for d = p^coh_s − 1/2 on qubit channels, which a metric would satisfy. The question is answered in the negative.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2020. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024A full-rank qutrit refutes the purification lower bound on Krylov operator complexityA positive definite qutrit state has greater time-dependent purification spread complexity than operator Krylov complexity.Refuted

The questionMori, Krylov complexity of purification, Phys.

Why it mattersThe left inequality would make the purified state's spread complexity a lower bound on the mixed state's operator complexity, which the authors use to interpret CK(ρ(t)) through a pure state.

The ideaTake ρ = diag(1, 4, 16)/21, H = [[2, 1, 0], [1, 2, 0], [0, 0, 4]] and t = π/2.

Our contributionA positive definite qutrit state has greater time-dependent purification spread complexity than operator Krylov complexity.

Verification details
Release source recordRefuted in Lean

Open problem

A full-rank qutrit refutes the purification lower bound on Krylov operator complexity

R. N. Das and T. Mori, Krylov complexity of purification, Phys. Rev. Lett. 136, 030201 (2026), arXiv:2408.00826v4, compare the Krylov operator complexity C_K(ρ(t)) of a mixed state ρ(t) = U_t ρ U_t† (U_t = e^{−iHt}) with complexities of its purifications.

  • The time-dependent purification. |Ψ^{U*}_ρ(t)⟩ = (U_t ⊗ U_t^)|Ψ_ρ⟩ (Eq. (2)), generated by H ⊗ I − I ⊗ H^.
  • The complexities. C_S is the spread complexity Σ_n n|ψ_n(t)|² of a state in its Lanczos basis (Eq. (S.4)), and C_K is its operator analogue for the Liouvillian L = [H, ·] with the Hilbert–Schmidt inner product (Eqs. (S.5)–(S.7)).
  • The conjecture. "Krylov operator complexity of the mixed state is upper bounded by the operator complexity of the time-independent purification and is lower bounded by the state complexity of the time-dependent purification, C_S(|Ψ^{U*}_ρ(t)⟩) ≤ C_K(ρ(t)) ≤ C_K(|Ψ^I_ρ(t)⟩) (4)". The authors confirm it analytically for one-qubit states and numerically for Werner states.

The verbatim statements are in the literature note.

Issue #16717 fixes the reading.

  • The claim is the left inequality of Eq. (4), for every full-rank density matrix, Hermitian H and t ≥ 0. Full rank is a special case of the source's statement, so its failure refutes the source's.
  • For full-rank ρ, |Ψ_ρ⟩ is the canonical purification Σ_{ij}(√ρ)_{ij}|i⟩|j⟩.
  • The left inequality is refuted.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PurificationLowerBoundQutritRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2024The twin end vertex of P'_9 is not sharply 1/9-sedentarySo vertex 1 is not sharply 1/9-sedentary, which refutes the conjecture of H.Refuted

The questionMonterde (arXiv:2401.00362, math.CO and quant-ph; Discrete Math. 349(4) (2026) 114959) studies the quantum walk U(t) = e^{itA} of a graph with adjacency matrix A.

Why it mattersA sedentary vertex keeps a bounded-below part of the walk's amplitude for all time.

The idea4. α = β + γ because cos(5π/18) + cos(7π/18) = 2cos(π/3)cos(π/18), and cos a + cos b + cos(a + b) ≥ −3/2 because (1 + cos a + cos b)² + (sin a − sin b)² = 3 + 2(cos a + cos b + cos(a + b)).

Our contributionSo vertex 1 is not sharply 1/9-sedentary, which refutes the conjecture of H.

Verification details
Release source recordRefuted in Lean

Open problem

The twin end vertex of P'_9 is not sharply 1/9-sedentary

H. Monterde (arXiv:2401.00362, math.CO and quant-ph; Discrete Math. 349(4) (2026) 114959) studies the quantum walk U(t) = e^{itA} of a graph with adjacency matrix A. A vertex u is sharply C-sedentary if inf_{t>0} |U(t)_{u,u}| = C with 0 < C ≤ 1. The graph P_n' is the path 1, …, n with an added vertex n + 1 joined to vertex 2, so that 1 and n + 1 are non-adjacent twins. Example 18 proves |U(t)_{1,1}| ≥ 1/n for odd n and states:

We conjecture that $u$ is sharply $(\frac{1}{n})$-sedentary in $P_n'$ for all odd $n\geq 5$.

Issue #11607 fixes the reading: U(t) = e^{itA} = hamiltonianPropagator A (−t), u = 1, and the claim is inf_{t>0} |U(t)_{1,1}| = 1/n for every odd n ≥ 5. The result refutes it at n = 9.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/PathTwinSharpSedentariness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Source 1 · Source 2
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Analysis & dynamicsConjecture refutedFirst preprint 2024Acosta Babb's weighted Bessel-kernel positivity conjectureRefuted

The questionAcosta Babb, A partial converse to the Riemann–Lebesgue lemma for Bessel–Fourier series of order zero, arXiv:2410.17681v1, Section 4, Conjecture 1, states:

Why it mattersThe conjecture supplies a proposed positivity premise for a stronger Bessel–Fourier coefficient construction.

The ideaThe existing normalized real Bessel series, coefficient, convergence and derivative facts supply the analytic inputs.

Verification details
Release source recordRefuted in Lean

Open problem

Acosta Babb's weighted Bessel-kernel positivity conjecture

Ryan L. Acosta Babb, A partial converse to the Riemann–Lebesgue lemma for Bessel–Fourier series of order zero, arXiv:2410.17681v1, Section 4, Conjecture 1, states:

For all, $0\leq\alpha\leq\frac12$, $M\in\mathbb N$ and $x\in[0,1]$, $K_M^\alpha(x)\geq0$.

The selected definition is the source's explicit weighted display

$$ K_M^\alpha(x)=\sum_{m=1}^M\left(1-\frac{m}{M+1}\right)j_m^\alpha J_0(j_mx)\quad(M>0), \qquad K_0^\alpha(x)=1. $$

Here $j_m$ is the $m$th positive zero of $J_0$. The printed partial sum is $S_m^\alpha(x)=\sum_{k=1}^m j_m^\alpha J_0(j_kx)$. Its factor has index $m$, so the printed Cesàro equality does not follow. The explicit weighted display, rather than that equality, is the target fixed by #22175.

Statement echo

| Source clause or convention | Lean counterpart | Fidelity | | --- | --- | --- | | Standard real $J_0$ | Existing BesselRedheffer.Defs.nbessel 0 | Reused directly; no new Bessel definition. | | Positive zeros indexed from one | j 1 = firstZero 0; later j are infima of positive zeros strictly above the preceding entry | Literal binding recursive convention; first two infima identified with actual zeros in the proof of root_brackets. Index zero is unused. | | Printed $S_m^\alpha$ | S | The printed index $m$ in the weight is preserved; this definition is separate from K. | | Weighted formula and exceptional $M=0$ | K | Natural indices $1$ through $M$, real weights and real powers; zero branch equals one. | | Every real $0\leq\alpha\leq1/2$ | First quantifier and two hypotheses of claim | Both endpoints included. | | Every natural $M$ and real $0\leq x\leq1$ | Remaining quantifiers and hypotheses of claim | Includes zero modes and both interval endpoints. | | $K_M^\alpha(x)\geq0$ | Conclusion of claim | No extra hypothesis. | | Refutation | result : Not claim | Exact negation; witness $(\alpha,M,x)=(1/2,2,999/1000)$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/BesselRedheffer/AcostaBabbKernelRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2023The cross-integral bound for continuous-time walks on two cyclesEarly coherent amplitudes contradict the uniform lattice-walk cross-integral estimate.Refuted

The questionDeng, Quantum walk mixing is faster than classical on periodic lattices, Physica A 630, 129252 (2023), arXiv:2309.16352v1, Conjecture 1 in its case d = 2 (Eq. (15)):

Why it mattersThe source supports it by simulations for odd coprime pairs in [10, 200].

Our contributionEarly coherent amplitudes contradict the uniform lattice-walk cross-integral estimate.

Verification details
Release source recordRefuted in Lean

Open problem

The cross-integral bound for continuous-time walks on two cycles

S. Dhamapurkar and X.-H. Deng, Quantum walk mixing is faster than classical on periodic lattices, Physica A 630, 129252 (2023), arXiv:2309.16352v1, Conjecture 1 in its case d = 2 (Eq. (15)):

$\left| \int_{0}^{T} n_1(t) n_2(t) dt \right|\leq 32 n_1(n_2 \log(n_2))^2+ 32 n_2(n_1 \log(n_1))^2$

for odd, relatively prime cycle orders $n_1 > n_2$. Here $n_i(t)$ is the time-dependent part of the transition amplitude product on $\mathbb{Z}{n_i}$, Eq. (B8): $n(t) = \sum{j \neq k,, j+k \neq n} e^{-it\sin(\pi(j+k)/n)\sin(\pi(j-k)/n)} w^{l(j-k)}$ with $w = e^{2\pi i/n}$ and displacement $l$. The verbatim statements are in the literature note.

Issue #18528 fixes the reading: the bound for all odd coprime $n_1 > n_2$, all displacements and every $T \ge 0$. This is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/LatticeWalkCrossIntegralRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Analysis & dynamicsCounterexample foundFirst preprint 2023A shorter relaxation time does not imply a faster energy relaxation in Glauber annealingA four-state Glauber annealing run has a larger spectral gap but a smaller energy relaxation rate than a partial swap process.Refuted

The questionDahlsten, Bound on annealing performance from stochastic thermodynamics, with application to simulated annealing, Phys.

Why it mattersEquation (50) is the step the paper uses to turn a spectral bound on the Glauber chain into the inequality µ(k) ≥ µ^PS(k), and from it Ep(k) ≤ E^PSp(k) and the performance bound I(κ) ≤ I^PS(κ) (§IV.C–D).

The ideaTake the cycle 0–1–2–3–0 (two neighbours per state), energies E = (0, 1, 1, 2), and e^{β(0)} = 2, e^{β(1)} = 32, e^{β(2)} = 33, e^{β(k)} = k + 31 for k ≥ 3.

Our contributionA four-state Glauber annealing run has a larger spectral gap but a smaller energy relaxation rate than a partial swap process.

Verification details
Release source recordRefuted in Lean

Open problem

A shorter relaxation time does not imply a faster energy relaxation in Glauber annealing

Y. Luo, Y.-Z. Zhen, X. Liu, D. Ebler and O. Dahlsten, Bound on annealing performance from stochastic thermodynamics, with application to simulated annealing, Phys. Rev. E 108, 054119 (2023), arXiv:2311.10424v1, bound the performance of simulated annealing (SA) by comparing it with a partial swap (PS) model.

  • The dynamics. SA with Glauber acceptance moves p(k + 1) = P(k + 1)p(k), with P_ij = γ_i/(|N(j)|(γ_i + γ_j)) on neighbours (Eqs. (30)–(32)). The PS model moves p′ = p + µ^PS(γ′ − p) (Eq. (43)), and its second largest eigenvalue is 1 − µ^PS (Proposition 2).
  • The rate. The energy relaxation rate of SA is µ = (E′_p − E_p)/(E′_γ − E_p) (Eq. (46)), under E_p ≥ E′_p and E_p ≥ E_γ ≥ E′_γ.
  • The conjecture. With τ_rel(P) = 1/(1 − λ_2(P)): "we conjecture that the order of relaxation times of the two processes corresponding to µ and µ^PS is the inverse order of µ and µ^PS, i. e. τ_rel ≤ τ^PS_rel ⇒ µ ≥ µ^PS. (50)"

The verbatim statements are in the literature note.

Issue #17238 fixes the reading: for every regular neighbourhood, energies, strictly cooling schedule, step k and rate 0 < m ≤ 1, m ≤ 1 − λ_2(P(β(k + 1))) implies µ(k) ≥ m. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/GlauberRelaxationOrderRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2024Subsystem size does not order coherent information in noisy graph statesA uniformly dephased graph state on five plus four qubits has greater coherent information on its smaller subsystem.Refuted

The questionGuha, Lower bounds on bipartite entanglement in noisy graph states, arXiv:2404.09014v2; Quantum 10, 2111 (2026), §5.4, “The subsystem with more qubits has higher coherent information?” (main.tex, line 951):

Why it mattersThe source uses coherent information as a lower bound on distillable entanglement.

The ideaHere the graph is exactly bipartiteGraph N, with no edges within either part.

Our contributionA uniformly dephased graph state on five plus four qubits has greater coherent information on its smaller subsystem.

Scope & next questionIt does not settle a universal rank-two ordering, the separate higher-rank robustness conjecture in §5.3, or the extensions to other noise models raised in the conclusion.

Verification details
Release source recordRefuted in Lean

Open problem

Subsystem size does not order coherent information in noisy graph states

A. Sajjad, E. Kaur, K. Goodenough, D. Towsley and S. Guha, Lower bounds on bipartite entanglement in noisy graph states, arXiv:2404.09014v2; Quantum 10, 2111 (2026), §5.4, “The subsystem with more qubits has higher coherent information?” (main.tex, line 951):

In all of the examples that we have considered, we have observed that the subsystem with the most qubits has the largest coherent information regardless of the types of the two subsystems and their distributions. Therefore, we conjecture that this is generally true but are unable to make a definitive statement.

The quotation is from the downloaded v2 TeX, main.tex, md5 230ee16b992b9567441d518d7471253e; the trailing TeX comment is omitted. The source's noise model is in §2.2, lines 186–194, and its entropy and coherent-information definitions are in §2.3, lines 201–216. See also the literature note.

Issue #21154 fixes the following non-strict reading. For natural numbers $n_A,n_B$ with $n_B<n_A$, let $G$ be any simple graph on $A\sqcup B$, where $|A|=n_A$ and $|B|=n_B$. Each qubit independently starts in

$\sigma_P=P\lvert+\rangle\langle+\rvert+(1-P)\lvert-\rangle\langle-\rvert$,

with the same $P\in[0,1]$. Put $\rho_G(P)=U_G\sigma_P^{\otimes(n_A+n_B)}U_G^\dagger$, where $U_G$ is the product of controlled-Z gates over the edges. Writing $\rho_A=\operatorname{Tr}_B\rho_G(P)$ and $\rho_B=\operatorname{Tr}_A\rho_G(P)$, define

$\mathcal I_A=H(\rho_A)-H(\rho_G(P))$, $\mathcal I_B=H(\rho_B)-H(\rho_G(P))$,

where $H$ is base-two von Neumann entropy. The module's claim asserts $\mathcal I_B\le\mathcal I_A$ for every such pair of sizes, every such graph with decidable adjacency, and every such $P$. Edges within either part are allowed. This non-strict assertion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/NoisyGraphStateCoherentInformationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Source 1
Quantum informationCounterexample foundFirst preprint 2023Strict averaging need not improve detector-sensor probesTwo optimal three-sensor probes refute strict improvement by coefficient-square averaging.Refuted

The questionHillery, Optimizing Initial State of Detector Sensors in Quantum Sensor Networks, ACM Transactions on Quantum Computing 5(2), Article 10 (2024), DOI 10.1145/3655028, arXiv:2306.17401v6, Section 5, Conjecture 2 (optimal.tex, lines 131–149, label conj:avg):

Why it mattersThe source uses Conjecture 2 as a route to Conjecture 1, its proposed optimal Dicke probe.

Our contributionTwo optimal three-sensor probes refute strict improvement by coefficient-square averaging.

Scope & next questionA strict statement restricted to nonoptimal probes, or requiring different fired-state Gram matrices, remains open here. - A weak replacement.

Verification details
Release source recordRefuted in Lean

Open problem

Strict averaging need not improve detector-sensor probes

C. Zhan, H. Gupta and M. Hillery, Optimizing Initial State of Detector Sensors in Quantum Sensor Networks, ACM Transactions on Quantum Computing 5(2), Article 10 (2024), DOI 10.1145/3655028, arXiv:2306.17401v6, Section 5, Conjecture 2 (optimal.tex, lines 131–149, label conj:avg):

For a given $U$, consider two initial states $\ket{\psi} = \sum\limits_j \psi_j \ket{j}$ and $\ket{\psi'} = \sum\limits_j \psi'_j \ket{j}$ such that (i) they are two unequal coefficient-squares, i.e., for some $j$, $|\psi_j|^2 \neq |\psi'j|^2$, and (ii) they have the same objective function value, i.e., $P(\ket{\psi}, U) = P(\ket{\psi'}, U)$. We claim that the ``average'' state given by $\ket{\psi{avg}} = \sum_j \sqrt{\frac{|\psi_j|^2 + |\psi'j|^2}{2}} \ket{j}$ has a lower objective function value, i.e., $P(\ket{\psi{avg}}, U) < P(\ket{\psi'}, U)$.

The quotation is from the v6 source TeX. The Library note records the paper and its encoding. In the source's setting (problem-related.tex, lines 17–47), exactly one of $n$ qubit sensors fires, with equal priors $1/n$; its unitary has eigenvalues $e^{\pm i\theta}$. The objective is the minimum error of a global POVM discriminating the $n$ possible fired states.

Write $X_n={0,1}^{\mathrm{Fin},n}$, with $1$ representing the plus eigenvector. A probe is an amplitude function $\psi:X_n\to\mathbb C$, normalized by $\sum_x|\psi(x)|^2=1$. For $n\ge1$, the reused fire and isoSuccess give $\phi_i(x)=e^{i\theta(2x_i-1)}\psi(x)$ and $S_\theta(\psi,M)=n^{-1}\sum_i\operatorname{Re}\langle\phi_i,M_i\phi_i\rangle$. The module defines optSuccess as the real indexed supremum of $S_\theta(\psi,M)$ over LocalPOVM (Fin n → Bool) (Fin n), errorProb as $P_\theta(\psi)=1-\operatorname{optSuccess}_\theta(\psi)$, and average as $a(\psi,\psi')(x)=\sqrt{(|\psi(x)|^2+|\psi'(x)|^2)/2}$, included in $\mathbb C$ with zero imaginary part.

Issue #26032 fixes claim: for every natural $n$, every $0<\theta<\pi$, and every pair of normalized complex probes, unequal coefficient-squares somewhere and equal $P_\theta$ imply $P_\theta(a(\psi,\psi'))<P_\theta(\psi')$. Radians replace the source's degrees. This universal strict statement is refuted at $n=3$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/DetectorSensorStrictAveragingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
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Quantum informationCounterexample foundFirst preprint 2023The Grassmann entwining Yang–Baxter map S over Γ(3) does not preserve the measure y₁₁⁻³The coefficient Jacobian of the Grassmann map S in three generators differs from the conjectured density ratio.Refuted

The questionPapamikos, Entwining Yang–Baxter maps over Grassmann algebras, Physica D 472, 134469 (2025), arXiv:2311.18673v1, construct birational entwining Yang–Baxter maps R{a,b}, S{a,b}, T{a,b} with variables in the Grassmann algebra Γ(n) (Eqs. (17)–(19)). - The map. S{a,b} (Eq. (18)) sends (x₁, x₂, χ₁, χ₂, y₁, y₂, ψ₁, ψ₂), with x, y even and χ, ψ odd, to u₁ = y₁²(1 + χ₁ψ₂)/(b x₁) + y₁(y₂ − a − ψ₁ψ₂)/b, u₂ = b/y₁, v₁ = x₁, v₂ = a + x₁x₂ − y₁/x₁ + χ₁χ₂, ξ₁ = ψ₁ − (y₁/x₁)χ₁, ξ₂ = ψ₂, η₁ = χ₁, η₂ = χ₂ + (y₁/x₁)ψ₂.

Why it mattersInvariant measures are a standard integrability property of discrete maps, and the conjecture predicts the measure for the whole hierarchy of commutative maps obtained by fixing n.

The ideaTake a = 1, b = 2 and the point z₀ whose scalar coefficients are x₁ = 1, x₂ = 3, y₁ = 2, y₂ = 4, with every other coefficient 0.

Our contributionThe coefficient Jacobian of the Grassmann map S in three generators differs from the conjectured density ratio.

Verification details
Release source recordRefuted in Lean

Open problem

The Grassmann entwining Yang–Baxter map S over Γ(3) does not preserve the measure y₁₁⁻³

P. Adamopoulou and G. Papamikos, Entwining Yang–Baxter maps over Grassmann algebras, Physica D 472, 134469 (2025), arXiv:2311.18673v1, construct birational entwining Yang–Baxter maps R_{a,b}, S_{a,b}, T_{a,b} with variables in the Grassmann algebra Γ(n) (Eqs. (17)–(19)).

  • The map. S_{a,b} (Eq. (18)) sends (x₁, x₂, χ₁, χ₂, y₁, y₂, ψ₁, ψ₂), with x, y even and χ, ψ odd, to u₁ = y₁²(1 + χ₁ψ₂)/(b x₁) + y₁(y₂ − a − ψ₁ψ₂)/b, u₂ = b/y₁, v₁ = x₁, v₂ = a + x₁x₂ − y₁/x₁ + χ₁χ₂, ξ₁ = ψ₁ − (y₁/x₁)χ₁, ξ₂ = ψ₂, η₁ = χ₁, η₂ = χ₂ + (y₁/x₁)ψ₂. Expanding every variable in a basis of Γ(n) gives a map on 2^{n+2} commutative coordinates.
  • Measure preservation. A map preserves the measure m when its Jacobian determinant is J = m(z)/m(S(z)) (§4). The authors verify m = 1/y₁ for n = 1 and m = 1/y₁₁² for n = 2.
  • The conjecture. "We conjecture that in Γ(n) the map R_{a,b} is volume preserving for every n, while the maps S_{a,b} and T_{a,b} preserve measures of the form y₁₁^{−n} and x₁₁^{−n}, respectively".

The verbatim statements are in the literature note.

Issue #16775 fixes the reading: the S clause at n = 3, with J the determinant of the derivative of the 32-coordinate map. Refuting one clause at one n refutes the conjecture, a conjunction over all n of clauses for R, S and T. The S clause is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/GrassmannEntwiningMeasureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2024A local counterexample to the triangle inequality ineq_l1The inequality s111(p) - 0.475 Delta1(p) <= 0.289, which E.Refuted

The questionA 111, 052453 (2025)) build Bell-type inequalities for the triangle network with four outcomes per party.

Why it mattersThe paper notes that on the symmetric subspace, where the penalty vanishes, ineql1 is the more constraining of its two inequalities, and the same construction with l = 2 is used to certify experimental data in arXiv:2401.15428.

The ideaEach source sends one of the 24 orderings of the four outcomes, uniformly; every party applies one rule to its two sources in cyclic order.

Our contributionThe inequality s111(p) - 0.475 Delta1(p) <= 0.289, which E.

Verification details
Release source recordRefuted in Lean

Open problem

A local counterexample to the triangle inequality ineq_l1

E. Bäumer, V. Gitton, T. Kriváchy, N. Gisin and R. Renner (arXiv:2405.08939, quant-ph; Phys. Rev. A 111, 052453 (2025)) build Bell-type inequalities for the triangle network with four outcomes per party. They penalise asymmetry by

Δ_l = Σ_{X ∈ {111,112,123}} Σ_{{a,b,c} ∈ I_X} |M_X − p(a,b,c)|^l, M_X = (1/|I_X|) Σ_{{a,b,c} ∈ I_X} p(a,b,c),

with I_X the outcome triples of type X (4, 36 and 24 of them), and from neural-network estimates they state:

… getting that approximately s₁₁₁(p) − 0.475 Δ_{l=1}(p) ≤ 0.289 (eq. ineq_l1), s₁₁₁(p) − 5.211 Δ_{l=2}(p) ≤ 0.316 (eq. ineq_l2), should both hold for all local models.

Issue #11263 fixes the reading. Locality is eq. trilocal with the frozen IsTriangleLocal, s₁₁₁ is the frozen s111, and deltaL1 is Δ_{l=1} summed over the three outcome types. The formal claim is eq. ineq_l1 with its printed constants for every local p; the result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TriangleInequalityL1Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2021Edge marginals at least one half do not imply GHS for general exponential random graphsAn edge-triangle Gibbs law has every edge marginal above one half and a positive third Ursell function.Refuted

The questionMagnanini, The GHS and other correlation inequalities for the two-star model, arXiv:2107.08889v2 (ALEA 19 (2022), 1679–1695), Section 4, closing paragraph, conjecture:

Why it mattersGHS controls third joint cumulants of edge indicators.

The ideaFor every real observable on these graphs, sumgraphs reindexes its sum by that bijection.

Our contributionAn edge-triangle Gibbs law has every edge marginal above one half and a positive third Ursell function.

Scope & next questionA replacement sufficient criterion for arbitrary motif families remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Edge marginals at least one half do not imply GHS for general exponential random graphs

A. Bianchi, F. Collet and E. Magnanini, The GHS and other correlation inequalities for the two-star model, arXiv:2107.08889v2 (ALEA 19 (2022), 1679–1695), Section 4, closing paragraph, conjecture:

Since $\text{Cov}{n;\boldsymbol{\beta}}(x_i,x_j) \geq 0$, due to the GKS inequality \eqref{GKS_g}, we can conjecture that the necessary and sufficient condition for the GHS inequality to hold in the present general setting is again only the requirement $\mathbb{E}{n;\boldsymbol{\beta}}(x_i) \geq 1/2$.

This is the verbatim sentence at line 1322 of the downloaded v2 source, GHS_arXiv-replacement.tex (md5 b353bfa014348d3c29f4b9c509b5665d). The general setting is specified at lines 1248–1287; the model and third Ursell function are defined at lines 286–312 and 577–589, respectively. The definitions and source locators are also in the literature note.

Issue #22390 fixes the following reading. For every $n,k\in\mathbb N$, take a nonempty family $h_j$ of finite simple graphs on $\mathrm{Fin}(m_j)$, indexed by $j\in\mathrm{Fin}(k+1)$, with $h_0=K_2$, and real coefficients $\beta_j$ with $\beta_j\geq0$ for every $j\ne0$. The coefficient $\beta_0$ is unrestricted. The source indexes this family from one instead of zero. On the labeled simple graphs $G$ with vertex set $\mathrm{Fin}(n)$, put

$t(h_j,G)=|\operatorname{hom}(h_j,G)|/n^{m_j}$, $H(G)=n^2\sum_{j=0}^k\beta_jt(h_j,G)$ and $\mu(G)=e^{H(G)}/\sum_{G'}e^{H(G')}$.

Write $\mathcal E_n=E(K_n)$, $X_e(G)=\mathbf1_{e\in E(G)}$ and

$u_3(e,f,g)=\mathbb E(X_eX_fX_g)-\mathbb E(X_e)\mathbb E(X_fX_g)-\mathbb E(X_f)\mathbb E(X_eX_g)-\mathbb E(X_g)\mathbb E(X_eX_f)+2\mathbb E(X_e)\mathbb E(X_f)\mathbb E(X_g)$.

The proposition claim asserts, for every such choice,

$(\forall e\in\mathcal E_n,\ \mathbb E(X_e)\geq1/2)\ \Longleftrightarrow\ (\forall e,f,g\in\mathcal E_n,\ u_3(e,f,g)\leq0)$.

The triple quantifier includes repeated edges. The equivalence is refuted by its sufficiency direction.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/ErgmGhsMarginalConjectureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2022A maximally entangled (3, 3, 7) state of length 9 refutes the length formula Σ d_i − 2Kephart, Decoherence and the classes of maximally entangled states, Int.Refuted

The questionKephart, Decoherence and the classes of maximally entangled states, Int.

Why it mattersThe refutation shows that the law, read off from dimensions up to 5, does not extend: unbalanced dimensions admit maximally entangled states with far fewer terms.

The ideaLet u at D = (3, 3, 7) be the indicator of the nine coordinates (0,1,0), (0,2,1), (1,0,2), (1,2,3), (2,0,4), (2,1,5), (0,0,6), (1,1,6), (2,2,6).

Our contributionKephart, Decoherence and the classes of maximally entangled states, Int.

Verification details
Release source recordRefuted in Lean

Open problem

A maximally entangled (3, 3, 7) state of length 9 refutes the length formula Σ d_i − 2

R. V. Buniy, R. P. Feger and T. W. Kephart, Decoherence and the classes of maximally entangled states, Int. J. Quantum Inf. 22, 2450035 (2024), arXiv:2210.07618v1, classify tripartite qudit states by algebraic entanglement invariants.

  • The invariants. For a state v of ℂ^{d₁} ⊗ ℂ^{d₂} ⊗ ℂ^{d₃} with coordinates v_{ijk}, they are (n₁, n₂, n₃, n₁,₂,₃) (Appendix A):
    • n_i, the kernel dimensions of the three flattenings;
    • n₁,₂,₃, the dimension of the space of arrays z with Σ_{j,k} v_{ajk}z_{bjk} = Σ_{i,k} v_{iak}z_{ibk} = Σ_{i,j} v_{ija}z_{ijb} = 0 for all a, b.
  • The class. A state is maximally entangled when every invariant takes its least value, and these states form the maximally entangled class C_D (Definition II.1).
  • The length. It is the number of nonzero coordinates, and L_D is the least length in C_D (Eq. (2)).
  • The conjecture. From exact values of L_D for dimensions up to 5 (Table XII), Conjecture III.4 states "L_(d₁,d₂,d₃) = Σ_{i=1}^{3} d_i − 2 for min(d₁, d₂, d₃) ≥ 3", together with two clauses for d₁ = 2.

The verbatim statements are in the literature note.

Issue #16992 fixes the reading: the lower-bound half of the third clause, i.e. every v ∈ C_D has L(v) ≥ d₁ + d₂ + d₃ − 2 when all d_i ≥ 3. Refuting it refutes the conjecture. It is refuted at D = (3, 3, 7).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/TripartiteMaximalClassLengthRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2023The literal numerator degree in Cigler's Conjecture 16The literal exact-degree requirement in Cigler's v2 Conjecture 16 fails at k=1 and m=0.Refuted

The questionJohann Cigler, Some experimental observations about Hankel determinants of convolution powers of Catalan numbers, arXiv:2308.07642v2, Section 3, Conjecture 16, equation (30), printed page 10, asserts the following whole conjunction.

Why it mattersThe named conjecture combines rationality of every full determinant generating series with exact numerator degree and reciprocity.

The ideaAn actual polynomial of exact degree one has a nonzero coefficient at one, giving the contradiction.

Our contributionThe literal exact-degree requirement in Cigler's v2 Conjecture 16 fails at k=1 and m=0.

Verification details
Release source recordRefuted in Lean

Open problem

The literal numerator degree in Cigler's Conjecture 16

Johann Cigler, Some experimental observations about Hankel determinants of convolution powers of Catalan numbers, arXiv:2308.07642v2, Section 3, Conjecture 16, equation (30), printed page 10, asserts the following whole conjunction. For every integer $k\ge1$ and every integer $m\ge1-k$, there is an actual polynomial $Q=Q_{2k-1,m}$ such that

$$ \left(1+(-1)^k x^{2k-1}\right)^{\binom{k+m}{2}+1} \sum_{N\ge0}D_{2k-1,m}(N)x^N=Q(x), \qquad \deg Q=\pi_{2k+1}(m+k-1)+k. $$

Here $C(x)=\sum_{n\ge0}C_nx^n$, where $C_n=\binom{2n}{n}/(n+1)$ is the Catalan number; $C_{r,t}=[x^t]C(x)^r$ for $t\ge0$, and $C_{r,t}=0$ for $t<0$. The actual determinant is

$$ D_{r,m}(N)=\det(C_{r,i+j+m}){0\le i,j<N},\qquad D{r,m}(0)=1, \qquad \pi_j(t)=\frac{t((j-2)t+(j-4))}{2}. $$

Every $Q$ is either palindromic or skew palindromic: for its actual degree $d$, $x^dQ(1/x)=Q(x)$ or $x^dQ(1/x)=-Q(x)$. Palindromicity is required if $k\equiv1\pmod4$ and $m$ is even, if $k\equiv3\pmod4$ and $m$ is odd, and if $k\equiv0,2\pmod4$ and $m\equiv0,1\pmod4$. These implications include no converses. The generating identity is coefficientwise.

The source does not specify a coefficient ring. The formal claim uses complex polynomials and series. Its exact-degree clause uses an actual natural degree $d$ with $Q.\mathrm{degree}=d$, then equates $d$ to the source's expression in exact rational arithmetic. It therefore excludes the zero polynomial and does not assign a nominal or padded degree. The reciprocity equations are stated directly at every nonzero complex argument with that actual degree.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CatalanPowerHankel/CiglerSixteenDegreeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2022A seventeen-vertex tree admits normalized-Laplacian perfect state transferA seventeen-vertex tree admits normalized-Laplacian perfect state transfer at time three pi.Refuted

The questionBaptista, Quantum Walks in the Normalized Laplacian, arXiv:2209.04668v1 (2022; journal publication online 2023, DOI 10.1080/03081087.2023.2236282), Conclusion, norm-laplacian-v2.tex, line 427:

Why it mattersThe source studies trees as quantum-network architectures requiring few edges.

The ideaThe proof uses spectral estimates.

Our contributionA seventeen-vertex tree admits normalized-Laplacian perfect state transfer at time three pi.

Verification details
Release source recordRefuted in Lean

Open problem

A seventeen-vertex tree admits normalized-Laplacian perfect state transfer

G. Coutinho and P. F. Baptista, Quantum Walks in the Normalized Laplacian, arXiv:2209.04668v1 (2022; journal publication online 2023, DOI 10.1080/03081087.2023.2236282), Conclusion, norm-laplacian-v2.tex, line 427:

For trees, we were not able to show a complete analogous result to the combinatorial Laplacian case --- that is, that perfect state transfer does not occur. We were however able to significantly restrict the possible cases, and therefore we conjecture that apart from the paths on $2$ and $3$ vertices, no tree admits perfect state transfer according to the normalized Laplacian.

This quotation was checked against the v1 TeX source, md5 b274bca76e7a9e881ea5b092556075a9. The definitions preceding equation (1), lines 89–98, and the transfer definition, lines 161–164, use $\mathcal L=D^{-1/2}(D-A)D^{-1/2}$, $U(t)=\exp(it\mathcal L)$ and $U(\tau)e_a=\lambda e_b$ for a complex unit $\lambda$. See also the literature note.

Issue #20154 fixes the statement: for every natural number $n\ge4$, every simple graph $G$ on $\operatorname{Fin}(n)$ with decidable adjacency, and every pair of distinct vertices $a,b$, if $G$ is a tree then there are no $\tau\in\mathbb R$ and $c\in\mathbb C$ with $|c|=1$ and $U(\tau)e_a=c e_b$. This is claim, and it is refuted. The two- and three-vertex trees are precisely the excluded paths.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/NormalizedLaplacianTreeStateTransfer.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Number theoryCounterexample foundPosed in 2023Fibonacci Numbers That Are Pythagorean PerimetersA Fibonacci number at an index that is not a multiple of six is the perimeter of a Pythagorean triangle, so the conjectured characterisation fails.Refuted

The questionIts COMMENTS section carries, at revision 62 of Sep 22 2025:

The ideaEvery Pythagorean triple is d times a primitive one with parameters m > n ≥ 1, gcd(m, n) = 1, m - n odd, so its perimeter is 2 d m (m + n).

Our contributionA Fibonacci number at an index that is not a multiple of six is the perimeter of a Pythagorean triangle, so the conjectured characterisation fails.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Fibonacci Numbers That Are Pythagorean Perimeters

OEIS A134492 is the sequence a(n) = Fibonacci(6n). Its COMMENTS section carries, at revision #62 of Sep 22 2025:

Conjecture: For n >= 2, the terms of this sequence are exactly those Fibonacci numbers which are the sum of the three numbers of a Pythagorean triple (checked up to F(80)). - Felix Huber, Nov 03 2023

With quantifiers written out, and P(N) for ∃ a b c > 0, a² + b² = c² ∧ a + b + c = N:

∀ N, (∃ k, F(k) = N) → (P(N) ↔ ∃ n ≥ 2, F(6n) = N).

The comment says "a Pythagorean triple" with no qualifier, so the triples are all triples of positive integers, primitive or not.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/FibonacciPythagoreanPerimeterRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationConjecture refutedFirst preprint 2024Refutation of the unconditional entropic Glibichuk–Konyagin conjectureA zero atom and a uniform component refute the unconditional entropic sum-product assertion for every fixed k.Refuted

The questionGoh, On an entropic analogue of additive energy, Essential Number Theory 5 (2026), no. 2, 243–269, DOI 10.2140/ent.2026.5.243, arXiv:2406.18798v4, §6, Conjecture 15, equation (89):

Why it mattersA large zero atom can suppress products without suppressing the entropy contributed by a small uniform component.

The ideaThe proof uses induction.

Our contributionA zero atom and a uniform component refute the unconditional entropic sum-product assertion for every fixed k.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the unconditional entropic Glibichuk–Konyagin conjecture

M. K. Goh, On an entropic analogue of additive energy, Essential Number Theory 5 (2026), no. 2, 243–269, DOI 10.2140/ent.2026.5.243, arXiv:2406.18798v4, §6, Conjecture 15, equation (89):

Conjecture 15 (Entropic Glibichuk–Konyagin lemma). There is an integer $k\geq 3$ such that the following holds. Let $X$ be a random variable taking values in $\mathbf F_p$ for a prime $p$. Then for independent copies $X_1,\ldots,X_k,X'_1,\ldots,X'_k$ of $X$, we have $$H{X_1X'_1+\cdots+X_kX'_k} \geq \min(2H{X},\log p)-1. \qquad (89)$$

In the original additive-combinatorial version of this statement, $k$ can be taken to be $3$, but any fixed integer is enough for our purposes.

The paper uses binary logarithms and discrete random variables. No additional hypothesis on the law of X is stated.

Formal reading

All local Lean names have prefix D5.S3.Entropy.GohGlibichukKonyagin.Result. Here $p$ is prime, $\mathbf F_p$ is ZMod p, $\delta_0$ is the point mass at zero, and $U_p$ assigns mass $1/p$ to each field element. Outside the source quotation, $H$ denotes Shannon entropy in nats and $H_{\mathrm{bits}}=H/\log 2$; shannonEntropy is reused from D5.S3.Entropy.MaxEntropy.

A law $\mu:\mathbf F_p\to\mathbb R$ satisfies $\mu(a)\ge0$ and $\sum_a\mu(a)=1$. sumLaw k μ is the pushforward of the product weights $\prod_{i\in\operatorname{Fin}k}\mu(x_i)\mu(x'_i)$, indexed by $(x,x'):(\operatorname{Fin}k\to\mathbf F_p)^2$, under $(x,x')\mapsto\sum_i x_i x'_i$. Here $\operatorname{Fin}k={0,\ldots,k-1}$. For the convolution computation, iter_product uses tuples of pairs $v:\operatorname{Fin}k\to\mathbf F_p\times\mathbf F_p$. The standard equivalence Equiv.arrowProdEquivProdArrow sends $v$ to its two projection functions; its inverse sends $(x,x')$ to $i\mapsto(x_i,x'_i)$. In sum_spike, Equiv.sum_comp proves that this change of indexing preserves the entire pushforward sum. Thus $Z=\sum_iX_iX'_i$ has exactly this law for the $2k$ independent copies. claim retains the existential $k\ge3$, all primes $p$, all such laws, and the constant $1$ in bits. result states exactly ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/GohGlibichukKonyagin/Result.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2024The Jeudy–Rouleux global-permutation Lie-algebra expectationThe spin-1 three-particle exchange Hamiltonian is outside the Lie algebra generated by global permutations of the spin labels.Refuted

The questionRouleux, Generalized exchange operators for a system of spin-1 particles, arXiv:2411.03952v1, Section 3.1, pp. 6–7; DOI 10.1088/1742-6596/2912/1/012043.

Why it mattersThe expectation would connect particle exchange Hamiltonians with the Lie algebra generated by global spin-label permutations.

The ideawhere each configuration denotes its basis vector.

Our contributionThe spin-1 three-particle exchange Hamiltonian is outside the Lie algebra generated by global permutations of the spin labels.

Verification details
Release source recordRefuted in Lean

Open problem

The Jeudy–Rouleux global-permutation Lie-algebra expectation

C. Jeudy and M. Rouleux, Generalized exchange operators for a system of spin-1 particles, arXiv:2411.03952v1, Section 3.1, pp. 6–7; DOI 10.1088/1742-6596/2912/1/012043. The source is recorded in jeudyrouleux2024exchange. The named expectation is preregistered in #19130.

Moreover since that the CSCO of the P- and Q-reps are the same in this case (see [Jin], p.70), we expect that Schrödinger Hamiltonian P_S, as well as other Hamiltonians, belongs to L.

The settlement refutes the membership assertion for the Schrödinger Hamiltonian at $(S,N)=(1,3)$. It uses all six global permutations of the three spin labels, hence also refutes membership in the smaller Lie algebra generated by the transpositions alone. It asserts nothing about other Hamiltonians.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Spin one, three particles, $\mathcal H=\mathbb C^{3^3}$ | Matrix indices Fin 3 → Fin 3, coefficients ℂ | Exactly three labels at each of three particle positions | | $Q(ij)=\widetilde Q(ij)^{\otimes3}$ | Global-action entry in L is one iff $x=\alpha\circ y$ | A single label permutation acts on every particle | | The $Q(\alpha)$ generate $L$ | L, using LieSubalgebra.lieSpan and all six label permutations | A larger generator set strengthens the exclusion | | $P_{ij}$ exchanges particles $i,j$ | Entries in schrodingerHamiltonian are one iff $x=y\circ\operatorname{swap}(i,j)$ | Zero-based indexing | | $H_S(1,3)=\sum_{\langle i,j\rangle}P_{ij}$ | schrodingerHamiltonian, with pairs $(0,1),(0,2),(1,2)$ | Each unordered pair occurs once | | The Hamiltonian belongs to $L$ | claim | Literal membership assertion | | Refutation | result : ¬ claim | No additional hypotheses |

The public definitions display the literal global-action and exchange entries; the generic matrix and Lie constructions come directly from Mathlib. The equality projector and all three auxiliary theorems are private.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/JeudyRouleuxGlobalPermutationLie.result

Frozen module record

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    Source 1
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Quantum informationCounterexample foundFirst preprint 2024Li–Tan circulant positivity is false at order fourThe Li–Tan circulant transform of a positive semidefinite correlation matrix can have a negative quadratic value at order four.Refuted

The questionMao-Sheng Li and Yi-Xi Tan, On the Bargmann invariants for quantum imaginarity, arXiv:2412.08022v1, Section III, page 5, state:

Why it mattersThe source proposes this particular transform as a route from correlation matrices to circulant matrices.

The ideaThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionThe Li–Tan circulant transform of a positive semidefinite correlation matrix can have a negative quadratic value at order four.

Scope & next questionThe positivity-preserving input classes beyond this counterexample family and a full classification of this map remain open here.

Verification details
Release source recordRefuted in Lean

Open problem

Li–Tan circulant positivity is false at order four

Mao-Sheng Li and Yi-Xi Tan, On the Bargmann invariants for quantum imaginarity, arXiv:2412.08022v1, Section III, page 5, state:

For the case when n = 3, we deduced that the positive semidefiniteness of C_H follows from the positive semidefiniteness of H. This leads us to the conjecture that the positive semidefiniteness of H implies the positive semidefiniteness of C_H (which may depend on n) for general n.

The universal implication is Refuted, even on real positive-entry correlation matrices at $n=4$. Preregistration #19081 specifies the source, conventions and literature screen. The Library note quotes the definitions of $\mathbf H_n$, $C_n$ and $z_k$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Every positive dimension $n$ | claim: every natural n with [NeZero n] | The source matrices have positive order | | Every complex matrix $H\in\mathbf H_n$ | Complex matrices indexed by Fin n, H.PosSemidef, and every diagonal entry equal to one | Hermitianity is included in Matrix.PosSemidef; no extra restriction | | $l\oplus k$ | l + Fin.ofNat n k in z | Modular addition with zero-based labels | | $r_{ij}\ge0$ and principal $\theta_{ij}\in(-\pi,\pi]$ | Complex norm and Complex.arg | The source's modulus and phase conventions | | Geometric modulus and arithmetic phase means | z, including the real $1/n$ power | All indices occur exactly once | | $C_H=\sum_{k=0}^{n-1}z_k C_n^k$ | CH, with the source shift written entrywise | The shift has a one exactly when $j=i+1$ | | Positive semidefiniteness of $C_H$ | (CH n H).PosSemidef | Full positivity conclusion | | Refutation | result : ¬ claim | One admissible order-four instance disproves the universal assertion |

All binders and conjuncts implement these clauses. The unit-diagonal restriction makes the claimed implication weaker than an unrestricted matrix implication, so its refutation also refutes that stronger assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/LiTanCirculantPositivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2024L∞-contraction for a weightA damped rotation on two points is L∞-bounded but admits no contractive positive weight.Refuted

The questionVafadar (eds.), Open problems in one-parameter operator semigroups theory, arXiv:2410.00416v2, §6 “L∞-bounds”, equation (1), state verbatim:

Why it mattersA bounded positive weight with bounded inverse defines an equivalent weighted sup norm.

The ideaThus the matrix satisfies IsSectorialMatrix with constant 2.

Our contributionA damped rotation on two points is L∞-bounded but admits no contractive positive weight.

Verification details
Release source recordRefuted in Lean

Open problem

L∞-contraction for a weight

F. Madou, I. Remizov, M. Vafadar (eds.), Open problems in one-parameter operator semigroups theory, arXiv:2410.00416v2, §6 “L∞-bounds”, equation (1), state verbatim:

The semigroup (T(t))t≥0 is called L∞-bounded, if there exists M > 0 such that ‖T(t)f‖∞ ≤ M‖f‖∞, for all f ∈ L²(Ω) ∩ L∞(Ω). (1)

Equation (3) states verbatim:

(3) ‖gT(t)f‖∞ ≤ ‖gf‖∞ for all f ∈ L∞(Ω) ∩ L²(Ω)

Problem 22 and its comment state verbatim:

Problem 22 (L∞-contraction for a weight problem). In the setting of Problem 21 assume that the semigroup (T(t))t≥0 is L∞-bounded, i.e., satisfies (1) for some M ≥ 1. Does there exist a function g ∈ L∞(Ω) satisfying g > 0 µ-a.e., 1/g ∈ L∞(Ω), and (3)? Comments. In general, this might be more than one can hope for.

The setting of Problem 21 is a σ-finite measure space $(\Omega,\mu)$ and a closed sectorial sesquilinear form $a$ on a dense $V\subseteq L^2(\Omega)$, with associated operator $A$ and bounded analytic semigroup $T(t)$ generated by $-A$. The locator binds to the edited problem list, without assigning individual authorship to this question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Semigroups/LinftyWeightContraction.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2021Two strongly cospectral vertices of $K_{2,3}$ exceed the multiplicity boundThe connected graph K{2,3} has two pairwise strongly cospectral vertices and an adjacency eigenspace of dimension at least three, contradicting the proposed multiplicity bound.Refuted

The questionMonterde, Strong cospectrality and twin vertices in weighted graphs, Electronic Journal of Linear Algebra 38 (2022), 494–518, DOI 10.13001/ela.2022.6721; arXiv:2111.01265v2, Section 3, immediately following Theorem 3.11:

Why it mattersThe proposed bound would control the size of a pairwise strongly cospectral set using only the vertex count and the largest adjacency-eigenspace dimension.

The idea1. Connected witness and a bounded supremum. The imported K23connected states that this fixed graph is connected.

Our contributionThe connected graph K{2,3} has two pairwise strongly cospectral vertices and an adjacency eigenspace of dimension at least three, contradicting the proposed multiplicity bound.

Verification details
Release source recordRefuted in Lean

Open problem

Two strongly cospectral vertices of $K_{2,3}$ exceed the multiplicity bound

H. Monterde, Strong cospectrality and twin vertices in weighted graphs, Electronic Journal of Linear Algebra 38 (2022), 494–518, DOI 10.13001/ela.2022.6721; arXiv:2111.01265v2, Section 3, immediately following Theorem 3.11:

We mention an interesting result of {'{A}}rnad{'{o}}ttir and Godsil which states that if $X$ is a normal Cayley graph and $n$ is the largest multiplicity of an eigenvalue of $A$, then the number of pairwise adjacency strongly cospectral vertices is bounded above by $|V (X)|/n$ \cite[Theorem 6.1]{Arn2021}. It would be interesting to check whether this bound holds for other classes of graphs, and in general whenever $M=\textbf{A}$.

This quotation is from line 727 of the downloaded v2 TeX, ELA-sample1_copy.tex (1444 lines, MD5 21a1ee53b9e64ae0b3a6773aa4fdbb45). The definitions and source locators are collected in the literature note.

Issue #21638 fixes the general-extension reading for ordinary adjacency: for every $n\in\mathbb N$, every connected simple graph $G$ on $\operatorname{Fin}n$ with decidable adjacency, and every finite set $S$ of pairwise adjacency strongly cospectral vertices, $|S|\le n/m(G)$. Here $V_\theta={x\in\mathbb R^n:A(G)x=\theta x}$ and $m(G)=\operatorname{sSup}{\dim_{\mathbb R}V_\theta:\theta\in\mathbb R}\in\mathbb N$, the module's maxMultiplicity. Cardinalities and $m(G)$ are cast to the reals before division. The universal extension is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/StrongCospectralMultiplicityBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedPosed in 2022Refutation of the all-prime A221077 period conjectureThe double Stirling formula has period p-1 at every odd prime and fails at p=2.Refuted

The questionThe Library note records Peter Bala's conjecture dated Jun 01 2022; that date and attribution belong to the conjecture, not to an asserted entry-creation date.

Why it mattersThis is a first-tier OEIS conjecture recorded in the entry in 2022.

The ideaThe exact positive-integer closed form a(n) = Σ{m=1..n} 2^{n−m}·m!·(m−1)!·S(n,m)·S(n+1,m) is the escape content.

Our contributionThe double Stirling formula has period p-1 at every odd prime and fails at p=2.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the all-prime A221077 period conjecture

OEIS A221077 (Paul D. Hanna's entry) specifies the following NAME. The Library note records Peter Bala's conjecture dated Jun 01 2022; that date and attribution belong to the conjecture, not to an asserted entry-creation date. The following NAME and COMMENT are quoted verbatim from the supplied Library note.

NAME:

E.g.f.: Sum_{n>=0} tanh(n*x)^n.

COMMENT (Peter Bala, Jun 01 2022):

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [1, 1, 6, 1, 0, 4, 1, 1, 6, 1, 0, 4, 1, 1, 6, 1, 0, 4 ...], with an apparent period of 6.

This is one conjecture quantified over all primes. It is false at p = 2; its substantive period assertion survives for every odd prime.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/TanhPowerPrimePeriod.bala_conjecture_two_false

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2024Natural valuations cannot preserve every abelian complexity functionA single ternary infinite word has fewer additive classes than original-symbol abelian classes at some positive length for every natural valuation.Refuted

The questionPopoli, Shallit and Stipulanti, Additive word complexity and Walnut, arXiv:2410.02409v1, Section 5, Question 31, PDF page 15, ask:

Why it mattersA scalar evaluation can identify two different original-symbol Parikh vectors.

The ideaConcatenate every decoded finite ternary list, appending a zero to each block.

Our contributionA single ternary infinite word has fewer additive classes than original-symbol abelian classes at some positive length for every natural valuation.

Verification details
Release source recordRefuted in Lean

Open problem

Natural valuations cannot preserve every abelian complexity function

Popoli, Shallit and Stipulanti, Additive word complexity and Walnut, arXiv:2410.02409v1, Section 5, Question 31, PDF page 15, ask:

Given an alphabet A, is there a valuation such that the additive complexity of a given sequence is equal to its abelian complexity?

The preceding definition is:

A map v:A→ℕ is called a valuation over an alphabet A.

The identical question is HTML Section 4, Question 30, element Thmtheorem30, and official FSTTCS 2024 Question 31, printed page 32:14, DOI 10.4230/LIPIcs.FSTTCS.2024.32. The full general assertion was preregistered in #23272.

Statement echo

| Source clause | Formal counterpart | | --- | --- | | Every finite alphabet and each given infinite sequence | Closed claim: every A : Type with [Fintype A] and every w : ℕ → A. | | One-sided infinite word, indexed from zero | The function w; the witness universalWord : ℕ → Fin 3 is defined at every natural index. | | Ordinary contiguous length-n factors | Reused MorseHedlund.wordFactor and wordFactorSet, with every natural starting index. | | Abelian equivalence of original symbols | parikh counts every original letter; parikhSet takes its finite image. List.perm_iff_count identifies equal counts with permutation. | | A valuation into the natural numbers | v : A → ℕ, including zero, equal and repeated weights; the alphabet is retained. | | Additive equivalence | sumSet is the finite image of actual factors under the sum of their letter values. | | Equality of complexities at all lengths | ∀ n : ℕ, additiveComplexity w v n = abelianComplexity w n; length zero is included. | | General existence order | ∀ A, ∀ w, ∃ v, ∀ n; the witness word is fixed before any valuation is extracted. | | Negative answer | The only public theorem is literally result : ¬ claim. |

No recurrence, automaticity, morphicity or balancedness restriction is present in the original question. Its neighboring positive results have their own hypotheses.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Complexity/PSSValuationRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundSource 2022Unrestricted two-state dequantumness power can exceed its maximal-quantumness restrictionA translated qubit contraction has strictly greater unrestricted two-state dequantumness power than its restriction to maximally quantum inputs.Refuted

The questionLi, Quantumness and Dequantumness Power of Quantum Channels, Entropy 24(8), 1146 (2022), DOI 10.3390/e24081146, Section 3, page 5 of the publisher PDF, in the paragraph following the amplitude-damping calculation:

Why it mattersA strict separation means that an input with slightly less quantumness can lose more quantumness under the same channel.

The ideaTheir Hermitian, idempotent and trace-one certificates satisfy all hypotheses of projectiondensity.

Our contributionA translated qubit contraction has strictly greater unrestricted two-state dequantumness power than its restriction to maximally quantum inputs.

Verification details
Release source recordRefuted in Lean

Open problem

Unrestricted two-state dequantumness power can exceed its maximal-quantumness restriction

H. Song and N. Li, Quantumness and Dequantumness Power of Quantum Channels, Entropy 24(8), 1146 (2022), DOI 10.3390/e24081146, Section 3, page 5 of the publisher PDF, in the paragraph following the amplitude-damping calculation:

Though $D_2(\Lambda)=D_2'(\Lambda)$ in this case, we conjecture this equality may fail in certain cases. But we have not found the counterexample satisfying $D_2(\Lambda)>D_2'(\Lambda)$ yet.

For a two-state qubit ensemble $\mathcal E={(p,\rho),(1-p,\sigma)}$, let $p\in[0,1]$ and let $\rho,\sigma$ be positive semidefinite complex two-by-two matrices of trace one. Equation (1), on page 2, specializes to

$$Q(\mathcal E)=-\sum_{i,j=1}^2p_ip_j\operatorname{tr}[\rho_i,\rho_j]^2.$$

In the module, quantumness takes the real part of each trace. With all suprema below over these two-state ensembles, its definitions are

$$Q_{\max,2}=\sup_{\mathcal E}Q(\mathcal E),\qquad D_2(\Lambda)=\sup_{\mathcal E}\bigl(Q(\mathcal E)-Q(\Lambda(\mathcal E))\bigr),$$

$$D_2'(\Lambda)=\sup_{\mathcal E:,Q(\mathcal E)=Q_{\max,2}}\bigl(Q(\mathcal E)-Q(\Lambda(\mathcal E))\bigr).$$

This is the two-ingredient reading used in the source's page-5 calculation, where the maximal initial quantumness is $1/4$; it does not optimize over ensembles of arbitrary size. The definitions and source locations are also given in the literature note.

Issue #22778 fixes claim as universal equality: every completely positive trace-preserving complex-linear qubit map satisfies $D_2=D_2'$. The resolution refuted negates that equality claim and supplies the strict separation sought by the authors; it confirms their quoted conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/DequantumnessPowerStrictGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Year in the primary source attribution: 2022.

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Quantum informationCounterexample foundFirst preprint 2024Condensation need not preserve LC-equivalenceTwo connected six-vertex graphs are related by local complementations and satisfy the bound of at most one outside neighbour for every vertex of the condensation set in both graphs, but their condensed graphs are not LC-equivalent.Refuted

The questionVandré, de Jong, Hahn, Burchardt, Gühne and Pappa, Distinguishing Graph States by the Properties of Their Marginals, arXiv:2406.09956v2, Section V, Conjecture 16, p. 13:

Why it mattersThe independent question is exactly Conjecture 16, not a positive finite instance of a known theorem.

The ideaThe first graph has edges {01, 02, 05, 14, 23}; the second has edges {05, 14, 23, 35, 45}.

Our contributionTwo connected six-vertex graphs are related by local complementations and satisfy the bound of at most one outside neighbour for every vertex of the condensation set in both graphs, but their condensed graphs are not LC-equivalent.

Verification details
Release source recordRefuted in Lean

Open problem

Condensation need not preserve LC-equivalence

Vandré, de Jong, Hahn, Burchardt, Gühne and Pappa, Distinguishing Graph States by the Properties of Their Marginals, arXiv:2406.09956v2, Section V, Conjecture 16, p. 13:

Given two graphs G and G′ and a condensation set C such that each node in C is connected to at most one node in the neighborhood in V \ C. If G and G′ are LC-equivalent, it follows that G_c and G′_c are LC-equivalent.

The paper considers simple connected graphs. Definition 13 (p. 12) replaces C by one fresh vertex c, retains edges between vertices outside C, and joins c to an outside vertex i exactly when some vertex of C is adjacent to i. Local complementation (Definition 4, pp. 4–5) complements adjacency among the selected vertex's neighbours. The correspondence with local Clifford operations of the associated graph states is stated on p. 5 and attributed to Van den Nest, Dehaene and De Moor (2004), DOI 10.1103/PhysRevA.69.022316. The formal statement uses finite labelled graphs and finite sequences of these local complementations.

The complete quantified claim is:

∀ (V : Type) [Fintype V] [DecidableEq V] (G H : SimpleGraph V) (C : Finset V),
  G.Connected → H.Connected →
  (∀ s ∈ C, (Finset.univ.filter fun i => i ∉ C ∧ G.Adj s i).card ≤ 1) →
  (∀ s ∈ C, (Finset.univ.filter fun i => i ∉ C ∧ H.Adj s i).card ≤ 1) →
  LCEquivalent G H → LCEquivalent (condense G C) (condense H C)

Both graphs must satisfy the outside-neighbour bound. This is the more restrictive reading of the premise, so a counterexample satisfying both bounds also refutes a reading requiring the bound on only one graph.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GraphCondensationLCRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2022The optimized GHZ witness bound need not equal the geometric GME measureThe optimized GHZ witness bound is strictly below the geometric convex roof for three qutrits at noise three eighths.Refuted

The questionChen, A generic framework for genuine multipartite entanglement detection, arXiv:2211.05563v1; published as General framework for genuine multipartite entanglement detection, Phys.

Why it mattersThe source distinguishes detecting entanglement from estimating its amount.

The ideaTrace expectations below mean the real part of the trace, as in witnessValue.

Our contributionThe optimized GHZ witness bound is strictly below the geometric convex roof for three qutrits at noise three eighths.

Verification details
Release source recordRefuted in Lean

Open problem

The optimized GHZ witness bound need not equal the geometric GME measure

X.-Y. Xu, Q. Zhou, S. Zhao, S.-M. Hu, L. Li, N.-L. Liu and K. Chen, A generic framework for genuine multipartite entanglement detection, arXiv:2211.05563v1; published as General framework for genuine multipartite entanglement detection, Phys. Rev. A 107 (2023) 052405, DOI 10.1103/PhysRevA.107.052405. In the arXiv TeX member GMEwitness.tex, subsection “Estimating entanglement measures”, line 325, they ask:

It remains open whether $\epsilon_o^{n,d}(p)$ equals $E_G(\rho_{n,d}(p))$.

The definitions are at lines 284–311:

  • $|GHZ_n^d\rangle=d^{-1/2}\sum_{i=0}^{d-1}|i\rangle^{\otimes n}$ and $\rho_{n,d}(p)=pI/d^n+(1-p)|GHZ_n^d\rangle\langle GHZ_n^d|$.
  • $E_G(\psi)=1-\sup_{\phi\text{ biseparable}}|\langle\phi,\psi\rangle|^2$, where both vectors are unit vectors and biseparability means a product across some nonempty proper bipartition; the mixed-state measure is its convex roof.
  • $\mathcal W=P/d-|GHZ_n^d\rangle\langle GHZ_n^d|$, where $P$ projects onto computational words using at most two distinct symbols. This is the orthogonal sum in lines 297–298, with the distinct arrangements specified by the orthogonal set $S$ at lines 690–692.
  • $w_{n,d}(p)=\operatorname{Tr}(\rho_{n,d}(p)\mathcal W)$, $\hat E_G(r\mathcal W)=\sup_{|\psi|=1,\ \phi\text{ unit biseparable}}{\langle\psi,(r\mathcal W+|\phi\rangle\langle\phi|)\psi\rangle-1}$ over unit vectors, and $\epsilon_o^{n,d}(p)=\sup_{r\in\mathbb R}{r w_{n,d}(p)-\hat E_G(r\mathcal W)}$.

The source locators are also collected in the literature note.

Issue #25044 fixes the universal reading: for every pair of integers $n,d\ge3$ and every real $p\in[0,1]$, $\epsilon_o^{n,d}(p)=E_G(\rho_{n,d}(p))$. The formal convex roof takes the real infimum over all finite ensembles with nonnegative weights summing to one, unit pure vectors and exact reconstruction of the density matrix. This universal equality is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/NoisyGhzWitnessBoundGap.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationConjecture refutedFirst preprint 2024Zhang's three-operator Clarkson–McCarthy conjecture is falseThree complex matrices of order two violate the three-operator Clarkson–McCarthy inequality at exponent nine quarters.Refuted

The questionThis quotation is from lines 574–580 of the downloaded TeX member v1ClarksonMcCarthytypeinequalitiespartI.tex, MD5 3ba34aa0913b51484593b3594a9f69d6; only mathematical display layout and whitespace are normalized.

Why it mattersThe conjecture proposes a sharp converse companion to the paper's Euler-type Clarkson–McCarthy estimate.

The ideaRaising nonnegative quantities to the fourth or eighth power certifies these bounds.

Our contributionThree complex matrices of order two violate the three-operator Clarkson–McCarthy inequality at exponent nine quarters.

Scope & next questionRequiring mutual commutativity or requiring all three matrices to be positive semidefinite excludes this witness; whether either restriction restores the proposed coefficient remains open here. - A neighbourhood of the transition.

Verification details
Release source recordRefuted in Lean

arXiv

Zhang's three-operator Clarkson–McCarthy conjecture is false

Teng Zhang, Clarkson–McCarthy type inequalities, part I: $\ell_p$–$\ell_p$ and $\ell_q$–$\ell_p$ Schatten $p$-estimates (2024 preprint, arXiv:2410.21961v4), Section 2, subsection “Euler's identity”, states Conjecture conj:sharp-inequality:

Let $A,B,C\in S_p$. Then for $2\le p<\infty$,

$|A+B+C|_p^p+|A|_p^p+|B|_p^p+|C|_p^p\le\frac{3^{p-1}+1}{2^p}\left(|A+B|_p^p+|B+C|_p^p+|C+A|_p^p\right).$

For $0<p\le2$, the inequality is reversed. Moreover, the constant $\frac{3^{p-1}+1}{2^p}$ is optimal: equality holds for all $p>0$ by taking $A=B=C\neq0$.

This quotation is from lines 574–580 of the downloaded TeX member v1_Clarkson_McCarthy_type_inequalities_part_I.tex, MD5 3ba34aa0913b51484593b3594a9f69d6; only mathematical display layout and whitespace are normalized. The conjecture is introduced by “numerical evidence suggests the following sharp inequality” (line 571). Zhang restates it in On five questions of Bourin and Lee (2026), arXiv:2602.01463v2, TeX member On_five_questions_of_Bourin_and_Lee.tex, lines 411–419, MD5 b3e91436b76494a53687b26bf5b2d06e.

For a finite complex matrix and $p>0$, write $|X|_p^p=\sum_j s_j(X)^p$. Abbreviate the four-term expression on the left by $L_p(A,B,C)$, the sum of the three pairwise terms by $P_p(A,B,C)$, and the proposed coefficient by $c(p)=(3^{p-1}+1)/2^p$.

Issue #22304 fixes the reading settled by claim: for every real $p\ge2$, every natural dimension $n$, and every $A,B,C\in M_n(\mathbb C)$, $L_p(A,B,C)\le c(p)P_p(A,B,C)$. This universal statement is refuted. The reverse clause for $0<p\le2$ is outside claim.

The source and norm definition are documented in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/ZhangThreeOperatorClarksonRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2023Independence polynomials do not determine T2-threshold hypergraphsTwo five-vertex T2 hypergraphs share an independence polynomial but are not isomorphic.Refuted

The questionZhang, On the Independence Polynomial and Threshold of an Antiregular k-Hypergraph, arXiv:2303.01024v1, Section 5, asks:

Why it mattersAn independence polynomial records the number of independent subsets of each size.

The ideaAll five helpers occur on the live proof path of result and are classified bind-only.

Our contributionTwo five-vertex T2 hypergraphs share an independence polynomial but are not isomorphic.

Scope & next questionThe source's separate antiregular uniqueness question remains open within this delivery.

Verification details
Release source recordRefuted in Lean

Open problem

Independence polynomials do not determine T2-threshold hypergraphs

E. Zhang, On the Independence Polynomial and Threshold of an Antiregular k-Hypergraph, arXiv:2303.01024v1, Section 5, asks:

Alternatively, a more general question is, given two T2-threshold k-hypergraphs with the same independence polynomial, are they isomorphic? These are left for future work.

The universal positive answer is false. The settling declaration is D5/S3/Combinatorics/Hypergraphs/ZhangThresholdIndependenceRefutation.result. The preregistration is #19406. The formal result uses five vertices and $k=3$; no arbitrary-$N$ theorem is delivered.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | A finite $k$-uniform hypergraph | KHypergraph k V, its edges and uniform fields | Every edge has cardinality $k$. | | No hyperedge is contained in an independent subset | IsIndependent | Every edge and subset are tested. | | Equation (8), the sum of $x^{|S|}$ over independent subsets | indepPoly | Integer coefficients; includes the empty set. | | Definition 4.1(T2), real labels and a strict threshold on every $k$-subset | IsT2Threshold | Literal strict inequality, with no positivity restriction on labels. | | Isomorphism | Isomorphic | A vertex equivalence preserves membership of every edge. | | Any two such hypergraphs with equal polynomials are isomorphic | claim | Quantifies over $k$, both finite vertex types and both hypergraphs. | | Negative answer | result | Negates the complete universal implication using $k=3$ and Fin 5. |

The encoding permits $k=0,1,2$, whereas the source question concerns $k\ge3$; the counterexample lies within the source range and all its edges are nonempty. The finite-type instances enumerate vertices and decide equality; they add no mathematical restriction to the finite source objects. The carrier Polynomial ℤ is the formal integer-coefficient expression of Equation (8).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hypergraphs/ZhangThresholdIndependenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2022A non-extreme minimum-rank superchannel extensionA two-qubit quantum superchannel has reduced Choi rank four for every extension and a minimizing extension that is a midpoint of distinct extensions.Refuted

The questionDaly, Axiomatic Approach to Quantum Superchannels, arXiv:2210.00370v1, Remark 4.1, asks:

The ideaextensionfixedcoherence identifies the shared block; allextensionse applies the obstruction to every legal extension. superextension provides the interval of legal extensions, and supernotextreme uses the distinct endpoints and their midpoint.

Our contributionA two-qubit quantum superchannel has reduced Choi rank four for every extension and a minimizing extension that is a midpoint of distinct extensions.

Scope & next questionThe numerical sample makes no claim about all boundary points or all dimensions.

Verification details
Release source recordRefuted in Lean

Open problem

A non-extreme minimum-rank superchannel extension

P. C. Daly, Axiomatic Approach to Quantum Superchannels, arXiv:2210.00370v1, Remark 4.1, asks:

The set of possible superchannel extensions of a QSC is a convex set. In the example given the extensions with minimal e are extreme points. A natural question to ask is whether it is generally true that the extensions which give minimal dimensions e are extreme points.

The Library note quotes the definitions and locators. Here $e$ is exactly the source's rank of the Choi matrix after tracing out both output factors. Convexity and extremality use real scalars. This question is the universal assertion about every minimizer; existence of some extreme minimizer is a weaker assertion.

Statement echo

All names in this table belong to D5.S3.QuantumChannels.DalySuperchannelMinimalExtension.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Definition 3.1: complex span of CPTP maps | SCPTP | Submodule.span of completely positive maps satisfying trace preservation on every matrix | | Remark 3.3: tensor product of matrix maps | kronBilin, kronLift | Bilinear MatrixMap.kron, lifted to all tensors | | Definition 3.4, conditions (1) and (2) | IsQSC | CPTP preservation and complete CP preservation on channel spans, for every auxiliary dimension and tensor | | Remark 3.7, all three bullets | IsQuantumSuperchannel | CP preservation, complete CP preservation and TP preservation on the full map spaces | | Full superchannels agreeing with the QSC | extensions | Equality of matrix maps on every element of the channel span | | Section 2: $C_L=(L(E_{ij}))$ | choi, choiEquiv | Input-first block entries and the double-sum inverse | | Induced map $\widetilde S=R_2SR_1^{-1}$ | induced | Composition with the two Choi equivalences | | Section 4: $e=\operatorname{rank}(\operatorname{Tr}{r_1}\operatorname{Tr}{r_2}C_{\widetilde S})$ | e | Reindex to input-pair/output-pair order and trace out the output pair | | All positive dimensions, all QSCs, every minimizing extension is extreme | claim | Four lower bounds, QSC hypothesis, membership, comparison with every extension, real extreme points | | Negative answer | result | Exact negation of claim |

Fin n uses zero-based indices. Complex-linear spaces and real convex combinations express the two different scalar roles in the source; neither adds a hypothesis to the question. The input-first Choi owner used in e has the same matrix-unit expression as choi.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/DalySuperchannelMinimalExtension.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2024A counterexample to the universal Kummer modulus lower boundAn exact nonreal zero contradicts the universal Kummer modulus lower bound.Refuted

The questionSimon, On some inequalities for the two-parameter Mittag-Leffler function in the complex plane, J.

Why it mattersA nonreal zero with a positive value at its real part prevents a global modulus lower bound, even though the function is real on the real axis.

The ideaCombining the exponential series and its first two factorial moments yields

Our contributionAn exact nonreal zero contradicts the universal Kummer modulus lower bound.

Verification details
Release source recordRefuted in Lean

Open problem

A counterexample to the universal Kummer modulus lower bound

R. Garrappa, S. Gerhold, M. Popolizio and T. Simon, On some inequalities for the two-parameter Mittag-Leffler function in the complex plane, J. Math. Anal. Appl. 551(1) (2025), 129588, DOI 10.1016/j.jmaa.2025.129588, arXiv:2410.11852v4, Remark 6:

In view of the preceding result and Remark 2, one may ask if for all α, β > 0 one has |M(α, α−β, z)| ≥ M(α, α−β, ℜz), z ∈ ℂ. (2.4) … In particular, (2.4) holds globally for α > 1 ≥ β > 0 … The question whether (2.4) holds globally for all α, β > 0 seems to be still open.

Remark 2(a) identifies M as "a Kummer confluent hypergeometric function": M(a, b, z) = Σ_{n≥0} (a)_n z^n / ((b)_n n!), b ∉ {0, −1, −2, …}.

The verified locators are Remark 6 and Remark 2(a) of v4; a page number is not asserted from the supplied extract.

Formal reading

Names without a prefix belong to D5.S3.Analytic.KummerModulusLowerBound.Result.

| Source object | Lean object | Correspondence | | --- | --- | --- | | $(a)_n/((b)_n n!)$ | D5.S3.Analytic.KummerRatio.Defs.kummerCoeff | The frozen real coefficient, cast into ℂ for complex arguments. | | $M(a,b,z)$ | kummerC | Infinite sum of those coefficients times $z^n$. | | $M(a,b,x)$ for real $x$ | D5.S3.Analytic.KummerRatio.Defs.kummer | The original real series. | | The universal question | claim | All positive real $\alpha,\beta$, nonpole second parameter, and all complex $z$. | | Negative answer | result | Exactly ¬ claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/KummerModulusLowerBound/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The refutation is supported by a documented parameterized argument or family.

    Source 1
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Analysis & dynamicsCounterexample foundFirst preprint 2024The reverse Mittag-Leffler modulus characterization is falseA closed beta interval of strict deficits exists for every alpha between one and two.Refuted

The questionGarrappa, Gerhold, Popolizio and Simon, On some inequalities for the two-parameter Mittag-Leffler function in the complex plane, arXiv:2410.11852v4, p. 18, Conjecture 25 (journal DOI 10.1016/j.jmaa.2025.129588):

Why it mattersA strict deficit at one nonreal argument disproves the claimed global lower bound even when the parameter interval holds.

The ideaThe literal source series is identified with the Barnes integral on the one-half line using actual residues, horizontal decay and the moving-left limit.

Our contributionA closed beta interval of strict deficits exists for every alpha between one and two.

Verification details
Release source recordRefuted in Lean

arXiv

The reverse Mittag-Leffler modulus characterization is false

Garrappa, Gerhold, Popolizio and Simon, On some inequalities for the two-parameter Mittag-Leffler function in the complex plane, arXiv:2410.11852v4, p. 18, Conjecture 25 (journal DOI 10.1016/j.jmaa.2025.129588):

For every α ∈ (1, 2), one has The inequality (1.4) holds globally ⇐⇒ β ∈ [α − 1, α] ⇐⇒ 1/Eα,β(x) is CM on (0, ∞).

Here $\beta\ge0$, and (1.4) is $|E_{\alpha,\beta}(z)|\ge E_{\alpha,\beta}(\operatorname{Re}z)$ for every $z\in\mathbb C$. Equation (1.1), p. 1, defines $E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}z^k/\Gamma(\alpha k+\beta)$ for $\alpha>0$ and every real $\beta$. The reciprocal Gamma value at a pole is zero. Complete monotonicity means a smooth positive-valued function on $(0,\infty)$ whose signed derivatives of all natural orders are nonnegative.

Statement echo

Unqualified Lean names below belong to D5.S3.Analytic.Asymptotics.MittagLefflerModulusRefutation.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Equation (1.1), all positive $\alpha$, real $\beta$, complex $z$ | IsSourceSeries | HasSum includes convergence; Gamma and ordinary complex division are Mathlib's. | | $1/E_{\alpha,\beta}(x)$ | reciprocal | The real argument is embedded in $\mathbb C$ and the real part reads the real series value. | | Global inequality | First equivalence in claim | Every complex argument is quantified; its real part is embedded without rescaling. | | $1<\alpha<2$, $\beta\ge0$, $\alpha-1\le\beta\le\alpha$ | Parameter binders and interval in claim | Both endpoints and all domain restrictions are literal. | | Positive smooth reciprocal with signed derivatives | Second equivalence in claim | Existing D5.S3.Analytic.ExpRatioCM.Defs.CompletelyMonotoneOn is used, with a separate positivity conjunct. | | Existence of the source family | Leading existential in claim | Explicit convergence/existence requirement; it does not weaken the universal conjecture. | | Negative answer | result | Exactly the negation of the complete claim. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/Asymptotics/MittagLefflerModulusRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Analysis & dynamicsCounterexample foundFirst preprint 2022ColorSym optimality is falseA colorable K4 list assignment defeats every ColorSym execution.Refuted

The questionJean-Christophe Godin, Rémi Grisot and Olivier Togni, On List Coloring with Separation of the Complete Graph and Set System Intersections, arXiv:2209.03436v1, Section 3.1; Electronic Journal of Graph Theory and Applications 13(1), 57–74, DOI 10.5614/ejgta.2025.13.1.5, Section 3.1, p. 63:

Why it mattersA colorable list assignment can have an ordering of scarce shared supply that defeats balancing by current load.

The ideaThere is only one active support of each size.

Our contributionA colorable K4 list assignment defeats every ColorSym execution.

Verification details
Release source recordRefuted in Lean

Open problem

ColorSym optimality is false

Jean-Christophe Godin, Rémi Grisot and Olivier Togni, On List Coloring with Separation of the Complete Graph and Set System Intersections, arXiv:2209.03436v1, Section 3.1; Electronic Journal of Graph Theory and Applications 13(1), 57–74, DOI 10.5614/ejgta.2025.13.1.5, Section 3.1, p. 63:

We conjecture that this algorithm is optimal (i.e., produces an (L, b)-coloring if K_n is (L, b)-colorable).

The algorithm is ColorSym. An $a$-list assignment associates to every vertex a set of $a$ integers. An $(L,b)$-coloring assigns a $b$-element subset of each vertex's list, with disjoint subsets at adjacent vertices. For $K_n$ every pair of distinct vertices is adjacent.

Statement echo

The binding semantics are recorded in preregistration #16679. The public definitions and the claim use those expressions. The paper's vertices $1,\ldots,n$ correspond to Lean's Fin n values $0,\ldots,n-1$.

| Source clause | Lean counterpart | Correspondence | | --- | --- | --- | | An $a$-list assignment assigns $a$ integers to each vertex. | The claim parameters $n,a,b$, the finite integer lists $L$, and the list-cardinality hypothesis. | All natural parameters are quantified, including degenerate cases. | | Proper intersections are the intersection of selected lists minus the union of the other lists. | supp and Ip. | A color is selected exactly when its support is the specified set; the empty support contributes no color from the union. | | A coloring assigns $b$ colors from each list, disjoint on every edge. | IsColoring and Colorable. | The exact cardinality, list inclusion and pairwise disjointness are all retained. | | “for each set S with |S| = i” in steps of increasing $i$. | AdmissibleOrder and runSupports. | Every support of size at least two occurs once; sizes are nondecreasing; equal-size order is arbitrary. | | “Remove a color from Ip(S) and assign it to vj.” | AdmissiblePick and assign. | Every choice from a nonempty integer pool is allowed; insertion and load increment are explicit. | | “let j ∈ S be the index for which w(j) is minimum (if more than one j with minimum w, then take the smallest index)” | best. | The minimum-load vertices are sorted by index and the first is selected. | | “If wi ≥ b for each i or i = n + 1, then Stop.” | stopped, runPool and the terminal case of runSupports. | Global stopping is rechecked at each color assignment; exhausting the finite support order implements the terminal step. | | “Goto Step i until Ip(S) is empty.” | runPool. | The pool loses one chosen color per assignment; initial fuel is its cardinality. | | “Color every proper intersections of sets of size 1: vi gets |Ip({i})| colors. Thus wi = |Ip({i})|.” | initial. | Singleton atoms determine both the initial outputs and loads. | | “produces an (L, b)-coloring if K_n is (L, b)-colorable” | colorSym, claim and result. | The claim asks for some admissible order and pick whose output contains a valid $b$-coloring; result negates it. |

The existential choice of order and pick and the allowance to trim an output to $b$ colors give the algorithm the weakest favorable reading. Every admissible execution fails on the witness, so this allowance does not rescue the conjecture. The fuel parameter, option for an empty support, product state and shifted indices implement the algorithm; they impose no additional hypothesis on the public claim. There is no separation bound in the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ConceptDynamics/GraphColoring/ColorSymOptimalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Number theoryQuestion resolvedPosed in 2024Granvik's A292273 real-part identityGranvik's real-part identity for the even-supported signed Mobius function.Proved

The questionMats Granvik, OEIS A292273, revision 24, Jul 06 2024:

Why it mattersThe necessary public definition a uses the two literal parity clauses.

The ideaReduce the arbitrary natural exponent modulo four.

Our contributionGranvik's real-part identity for the even-supported signed Mobius function.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Granvik's A292273 real-part identity

Mats Granvik, OEIS A292273, revision 24, Jul 06 2024:

For odd n: a(n) = 0, and for even n: a(n) = -mu(n), where mu is Moebius function (A008683).

Conjecture: a(n) = Re(A008683(n)*(i^n)). - Mats Granvik, Jul 06 2024

The offset is one. The assertion concerns every positive integer, with no upper bound. Here mu is the standard integer Mobius function and i is the ordinary complex imaginary unit.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Mobius/GranvikMobiusRealPart.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2023Every finite-dimensional signalling channel has a pure product input with positive PDM negativityEvery finite-dimensional signalling channel admits a pure product input with strictly positive PDM negativity.Proved

The questionWe prove this conjecture for a quite general case of 2-qubit unitary evolutions in the Supplementary Information.

Why it mattersPure product inputs would give a simple witness family, while the source had proved the assertion only for a broad two-qubit unitary case.

The idea1. signallingdichotomy assumes [Nonempty α₁], [Nonempty β₁], [Nonempty α₂], [Nonempty β₂], a channel p, and hp : IsSignallingBA p.

Our contributionEvery finite-dimensional signalling channel admits a pure product input with strictly positive PDM negativity.

Verification details
Release source recordProved in Lean

Open problem

Every finite-dimensional signalling channel has a pure product input with positive PDM negativity

X. Liu, Y. Qiu, O. Dahlsten and V. Vedral, Quantum causal inference with extremely light touch, arXiv:2303.10544v4 (npj Quantum Information 11, 54, 2025), define the causal monotone in Eq. (causmono) as $f(R) := \lVert R\rVert_{tr}-1 = \Tr \sqrt{ R R^\dag}-1$ (arxiv.tex lines 143–145). For a channel with Choi-Jamiolkowski matrix $M_{12}:= \sum_{i,j}\ket{i}\bra{j}^T \otimes \mathcal{M}{2|1}(\ket{i}\bra{j})$ (Eq. (CHOI), lines 161–164), their two-time PDM is $R{12}=\frac12(M_{12}\rho+\rho M_{12})$ with $\rho=\rho_1\otimes\mathbbm{1}_2$ (Theorem 2timesPDM, lines 166–173). After the null-negativity theorem for channels that do not signal from $B$ to $A$, the paper states:

We moreover conjecture that if the channel is signalling from B to A, we can always find a pure {\em product} input state such that $f(R_{B_1A_2} )> 0$. We prove this conjecture for a quite general case of 2-qubit unitary evolutions in the Supplementary Information.

This is the conjecture at arxiv.tex line 231. The formal statement uses finite nonempty index types $\alpha_1,\beta_1,\alpha_2,\beta_2$, a QuantumChannel $p$ from $\alpha_1\times\beta_1$ to $\alpha_2\times\beta_2$, and the module's IsSignallingBA predicate. It asks for vectors $a:\alpha_1\to\mathbb C$ and $b:\beta_1\to\mathbb C$ with $a^!\cdot a=1$, $b^!\cdot b=1$, and $0<\operatorname{traceNorm}(\operatorname{pdmBA}(p,a,b)^\mathsf{H})-1$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/SignallingPureProductNegativity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundFirst preprint 2024O'Bryant §5.7: the golden minimizerA seven-point positive-real set with at most seventeen products can have at most twenty-two sums, while the corresponding golden set has at least twenty-four.Refuted

The questionKevin O'Bryant, Visualizing the Sum-Product Conjecture, arXiv:2411.08139v2, §5.7 asks:

Why it mattersThe question proposes the consecutive golden powers as a universal minimizer under a product-cardinality constraint.

The ideaThe intermediate value theorem supplies r with 1 < r < 2 and r³ = r + 1.

Our contributionA seven-point positive-real set with at most seventeen products can have at most twenty-two sums, while the corresponding golden set has at least twenty-four.

Verification details
Release source recordRefuted in Lean

Open problem

O'Bryant §5.7: the golden minimizer

Kevin O'Bryant, Visualizing the Sum-Product Conjecture, arXiv:2411.08139v2, §5.7 asks:

What is the smallest possible value of |A + A|, where A is a set of n positive real numbers with |AA| ≤ 3n − 4? Is it achieved by {φ^i : 1 ≤ i ≤ n}, where φ = (1 + √5)/2?

The Lean claim retains the universal quantifiers over n and finite positive real sets and compares the pointwise sumsets. The theorem result : ¬ claim refutes this universal minimizer assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SumProductGoldenMinimizerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationQuestion resolvedFirst preprint 2024A strict Bayesian feedback advantage for the qutrit Werner–Holevo channelBayesian feedback exceeds every two-step Markovian fidelity for the qutrit Werner–Holevo channel.Proved

The questionMancini, On the effectiveness of Bayesian discrete feedback for quantum information reclaiming, arXiv:2402.18941v1, Quantum Information Processing 23, 258 (2024), Section 5, Conclusion, ask:

Why it mattersThe comparison concerns recovery of entanglement fidelity through measurements of the environment and outcome-dependent corrections.

The ideaAll lemma names below refer to WernerHolevoBayesianFeedbackAdvantage.

Our contributionBayesian feedback exceeds every two-step Markovian fidelity for the qutrit Werner–Holevo channel.

Scope & next questionThe exact optimal Bayesian value, higher-dimensional analogues, more than two feedback steps and history-dependent choices of environmental measurements remain open in this settlement.

Verification details
Release source recordProved in Lean

Open problem

A strict Bayesian feedback advantage for the qutrit Werner–Holevo channel

M. Rexiti, S. Khabbazi Oskouei and S. Mancini, On the effectiveness of Bayesian discrete feedback for quantum information reclaiming, arXiv:2402.18941v1, Quantum Information Processing 23, 258 (2024), Section 5, Conclusion, ask:

There are a number of problems that this study poses in future perspective. First, find other examples of channels for which Bayesian feedback is more effective than Marokovian one. Then, for these channels see if by increasing the dimension the advantage increases.

This quotation is from the downloaded v1 source, EffOfBayesianFB.tex, lines 910–912, md5 aeee6c3398ec2e49e99d0f461ed53290; the spelling “Marokovian” is the source's. The literature note gives the feedback definitions and source locators.

Issue #23263 specifies an answer to the first request using the qutrit Werner–Holevo channel $\Phi(X)=\tfrac12(\operatorname{tr}(X)I_3-X^{\mathsf T})$. For a finite family $K=(K_a)_{a\in\mathrm{Fin},m}$ of complex $3\times3$ matrices, IsKrausDecomposition K means $\sum_a K_a X K_a^\dagger=\Phi(X)$ for every complex $3\times3$ matrix $X$.

The module specializes the source's two-step formulas in Section 2, lines 245–253, to dimension three:

$F_{\mathrm M}(K,L)=\tfrac19\sum_{a,b}\bigl(\operatorname{Re}\operatorname{tr}(|L_b|,|K_a|)\bigr)^2$,

$F_{\mathrm B}(K,L)=\tfrac19\sum_{a,b}|L_b|K_a||_1^2$,

where $|T|=\sqrt{T^\dagger T}$ and $|M|1=\operatorname{Re}\operatorname{tr}\sqrt{M^\dagger M}$. The settled statement is that every pair of finite Kraus decompositions has $F{\mathrm M}\le\tfrac12$, while some pair of three-operator Kraus decompositions has $F_{\mathrm B}>\tfrac12$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Recovery/WernerHolevoBayesianFeedbackAdvantage.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    Source 1
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Quantum informationQuestion resolvedFirst preprint 2022The signalling game has equal classical, quantum and no-signalling valuesThe signalling game has classical, quantum and no-signalling value 1/max(m,n).Proved

Why it mattersThe signalling game is the example used by the source to separate unrestricted communication from classical, quantum-assisted and no-signalling-assisted sum rates of a multiple-access channel.

The ideaSumming each local POVM over its answers gives the identity; positivity of the state and effects gives nonnegative Born probabilities, normalization gives total mass one, and the resulting marginal traces are independent of the other question.

Our contributionThe signalling game has classical, quantum and no-signalling value 1/max(m,n).

Verification details
Release source recordProved in Lean

Open problem

The signalling game has equal classical, quantum and no-signalling values

A. Seshadri, F. Leditzky, V. Siddhu and G. Smith, On the separation of correlation-assisted sum capacities of multiple access channels, arXiv:2205.13538v3 (2023), Section 3.2.1, define the signalling game using question sets $\mathcal X_1$ and $\mathcal X_2$, answers $\mathcal Y_1=\mathcal X_2$ and $\mathcal Y_2=\mathcal X_1$, and the winning condition

Formally, we consider question sets $\mathcal{X}_1$, $\mathcal{X}_2$ and answer sets $\mathcal{Y}_1 = \mathcal{X}_2$ and $\mathcal{Y}_2 = \mathcal{X}_1$, and the winning condition is defined as $\mathcal{W} = {(x_1, x_2, y_1, y_2) \in (\mathcal{X}_1 \times \mathcal{X}_2) \times (\mathcal{Y}_1 \times \mathcal{Y}_2) \mid y_1 = x_2,\ y_2 = x_1}$.

The source proves the classical value $1/\max(|\mathcal X_1|,|\mathcal X_2|)$ and then asks whether the quantum and no-signalling values coincide with it. Its Section 3.2.1, line 880, states:

For $2 \leq |\mathcal{X}1|, |\mathcal{X}2| \leq 10$, we verify that the numerically computed value for $\omega^{\text{NS}}(G{\text{s}})$ matches $\omega^{\text{cl}}(\mathcal{N}{G_{\text{s}}}) = 1/\max(|\mathcal{X}1|, |\mathcal{X}2|)$. Thus, we expect $\omega^{\text{cl}}(\mathcal{N}{G{\text{s}}}) = \omega^{\text{Q}}(\mathcal{N}{G{\text{s}}}) = \omega^{\text{NS}}(\mathcal{N}{G{\text{s}}})$ for the signalling game.

The preregistration reads this as the finite-game value question for $m=|\mathcal X_1|\ge1$ and $n=|\mathcal X_2|\ge1$, with uniformly drawn questions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SignallingGameValueEquality.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2024Interaction-picture asymptoticsFinite past interaction integrals do not imply polynomial-generator asymptotics.Refuted

The questionVafadar (eds.), Open problems in one-parameter operator semigroups theory, arXiv:2410.00416v2, p. 21, Problem 29 (A.

Why it mattersThe problem concerns perturbative corrections to Markovian quantum master equations and generalization of E.

Our contributionFinite past interaction integrals do not imply polynomial-generator asymptotics.

Verification details
Release source recordRefuted in Lean

Open problem

Interaction-picture asymptotics

F. Madou, I. Remizov, M. Vafadar (eds.), Open problems in one-parameter operator semigroups theory, arXiv:2410.00416v2, p. 21, Problem 29 (A. E. Teretenkov), state verbatim:

Problem 29 (A.E. Teretenkov). Let B be a Banach space. Let P be a projection on a finite-dimensional Banach subspace of B. Let L0 + λL be a generator of C0-semigroup U^λ_t on B for all λ ∈ [0, λsup). Let U^0_t leave both PB and its complement (I − P)B invariant, let PLP = 0. Denote L_t ≡ (U^0_t)^{−1} L U^0_t. Let the integrals ∫_{−∞}^{t} dt_1 … ∫_{−∞}^{t_{k−1}} dt_k P L_{t_1} … L_{t_k} P, k = 1, …, n + 1 be finite for all t ⩾ 0. If is it possible to find such a λ-dependent operator r^{n,λ} on PB, which is polynomial in λ, and λ-dependent semigroup u^{n,λ}t on PB, whose generator is polynomial in λ, such that P(U^0{t/λ²})^{−1} U^λ_{t/λ²} P = u^{n,λ}_t r^{n,λ} + O(λ^{2n+2}), λ → +0 for all t > 0 ? Let us emphasize that we do not assume here asymptotic behaviour to be uniform in t, it is just assumed for each fixed t > 0. If there are counterexamples, what further restrictions should be assumed to obtain such an asymptotic estimate?

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Semigroups/InteractionPictureAsymptotics.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationQuestion resolvedFirst preprint 2024The temporal Bell endpoint contraction identity for every open XX chainThe endpoint vacuum contractions of the open XX chain equal a difference of one-particle propagator entries.Proved

The questionAndrea Tononi and Maciej Lewenstein, Temporal Bell inequalities in non-relativistic many-body physics, Quantum Science and Technology 10 (2025), arXiv:2409.17290v4, Methods, “Derivation of Eq. ICHexact”, conjecture in arxiv.tex, lines 343–353 (md5 225051e786fe4712ac07f5226a0a99ac):

Why it mattersThe paper evaluates temporal Clauser–Horne probabilities for a Bell pair at the endpoints of an open XX chain.

The ideaIts vacuumIndex N is the all-false occupation configuration, and FullOperator N is the complex matrix algebra on these configurations.

Our contributionThe endpoint vacuum contractions of the open XX chain equal a difference of one-particle propagator entries.

Scope & next questionPairing terms and interactions require different evolution formulas and remain open here. - The printed spectral sign.

Verification details
Release source recordProved in Lean

Open problem

The temporal Bell endpoint contraction identity for every open XX chain

Andrea Tononi and Maciej Lewenstein, Temporal Bell inequalities in non-relativistic many-body physics, Quantum Science and Technology 10 (2025), arXiv:2409.17290v4, Methods, “Derivation of Eq. ICHexact”, conjecture in arxiv.tex, lines 343–353 (md5 225051e786fe4712ac07f5226a0a99ac):

However, we conjecture that

$\text{Re}\big[ \bra{0} f_N f_1 \sigma^x_{N}(t) f_1^\dagger \ket{0} + \bra{0} f_N f_1 \sigma^x_{N}(t) f_N^\dagger \ket{0} \big] = \text{Re}\big[G_{NN}(t) - G_{1N}(t) \big]$,

and we verified \cite{sneg} that the exact analytical form of the left- and right-hand sides coincides for $2\leq N \leq 5${, and with exact numerics} up to $N=8$ \cite{mullerrigat2025}. In the absence of a general proof, we assume the relation to hold for any value of $N$.

The quotation preserves the TeX wording and citation keys, with display delimiters and the equation label omitted. The source's references identify sneg as R. Žitko's SNEG package (2011) and mullerrigat2025 as preliminary results of G. Müller-Rigat, M. Płodzień, D. Farina, M. Lewenstein and A. Tononi (2025).

Issue #21106 fixes the reading: for every integer $N\ge2$ and every real $J,\mu,t$, use the Jordan–Wigner annihilators $f_j$ on the full occupation space $(\mathbb C^2)^{\otimes N}$, its empty vacuum $|0\rangle$, and

$H=-J\sum_{j=1}^{N-1}(f_j^\dagger f_{j+1}+f_{j+1}^\dagger f_j)-\mu\sum_{j=1}^N f_j^\dagger f_j$,

$h_{jj}=-\mu$, $h_{j,j+1}=h_{j+1,j}=-J$, with all other entries zero, and $G(t)=e^{ith}$.

The observable is $\sigma_N^x=\big\prod_{j<N}(1-2f_j^\dagger f_j)\big$, with its prefix product in increasing site order, and $\sigma_N^x(t)=e^{itH}\sigma_N^xe^{-itH}$. With these definitions, the displayed real-part identity is proved.

The source introduces $G$ through the relation $f_j^\dagger(t)=\sum_iG_{ij}(t)f_i^\dagger$ (arxiv.tex, lines 153–154 and 291). The settlement follows this relation. Its matrix exponential is not identified with the source's printed spectral formula containing $(-1)^{j-1}$ in $u_{jk}$; the sign distinction is addressed below. The source quotations and bibliography are also in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/TemporalBellContractionIdentity.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2023Ruleset domination does not imply nimber dominationEqual outcomes under the enforce operator do not imply equal nimbers.Refuted

The questionAbuku, Kimura, Kiya, Larsson, Saha, Suetsugu and Yamashita, Enforce and selective operators of combinatorial games, arXiv:2311.01006v2, Section 8, Problem 42, printed page 23 asks:

Why it mattersOutcome equality retains a zero/nonzero distinction but does not specify positive nimber values.

The ideaEvery option strictly decreases its index, so every play in the union terminates.

Our contributionEqual outcomes under the enforce operator do not imply equal nimbers.

Verification details
Release source recordRefuted in Lean

Open problem

Ruleset domination does not imply nimber domination

Abuku, Kimura, Kiya, Larsson, Saha, Suetsugu and Yamashita, Enforce and selective operators of combinatorial games, arXiv:2311.01006v2, Section 8, Problem 42, printed page 23 asks:

If a ruleset dominates another, does it also ‘nimber dominate’ that ruleset? That is, does it follow that all nimbers of the combined ruleset are the same as those of the dominating ruleset?

The source is also published in International Journal of Game Theory 53, 1249–1273, DOI 10.1007/s00182-024-00910-6. The preregistration is issue #19251.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Positions and option maps, Definition 2 | Arbitrary type X and fA fB : X → Set X | No finite-board assumption in claim. | | Every play has a finite bound from its starting position, Definition 4 | Short quantifies over starts, bounds and sequences | The displayed source condition is used directly. | | Union ruleset is short, Definition 10 | JointlyShort fA fB | Both option maps share the same position type. | | Perfect-play outcomes, Definition 5 | Outcome, IsOutcome fA OA | P exactly when every option is N. | | Opponent enforces one ruleset before the move, Definition 13 | IsEnforceOutcome fA fB OE | P exactly when at least one ruleset has only N options. | | Domination, Definition 14 | Pointwise equality of OE and OA | Equality holds at every position. | | Least excluded natural, Definition 28 | mex s = sInf sᶜ | This totalizes the source's finite-set mex; empty complement gives 0 by the Nat sInf convention. The witness uses only finite sets. | | Ordinary nimbers, Definition 29 | IsNimValue fA GA | Mex of the ordinary option-value image. | | Enforce nimbers, Definition 33 | IsEnforceNimValue fA fB GE | Mex of the intersection of the two option-value images. | | All combined nimbers equal the dominating ruleset's nimbers, Problem 42 | claim concludes pointwise equality of GE and GA | result proves the negation of this universal implication. |

The four function parameters satisfy the source recursions; they introduce no extra restriction on the refuting board. The witness uses three positions and provides each of the four functions explicitly. No general uniqueness or existence theorem for these recursions is claimed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/AbukuEnforceNimberDomination.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2023A sign obstruction to Chirvasitu–Cusick Conjecture 6.1A zero trace of the twenty-fourth power contradicts the proposed order at (8,4).Refuted

The questionCusick, Quadratic rotation symmetric Boolean functions, arXiv:2304.12734v1, Section 6, Conjecture 6.1:

Why it mattersBinary transport of Pauli labels does not retain all signs.

The ideaThe endpoint permutation is the identity, while its phase reverses sign.

Our contributionA zero trace of the twenty-fourth power contradicts the proposed order at (8,4).

Verification details
Release source recordRefuted in Lean

Open problem

A sign obstruction to Chirvasitu–Cusick Conjecture 6.1

A. Chirvasitu and T. W. Cusick, Quadratic rotation symmetric Boolean functions, arXiv:2304.12734v1, Section 6, Conjecture 6.1:

For i > j ≥ 1 the eigenvalues of the matrix R(i, j) of (6-3) are all √2-scaled roots of unity. Moreover, the multiplicative order of R(i,j)/√2 is precisely 2 · lcm(4, i − j, i + j).

The source and its definitions are recorded in the Library note. The preregistration is #19397. The settlement refutes the conjunction through the order conjunct at $(i,j)=(8,4)$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $H_0=[[1,1],[1,-1]]$, $H_1=[[1,1],[-1,1]]$ | H0, H1 | Integer entries in source order. | | Append $k$ zero columns after each column | pad | Column $b$ reads column $b/(k+1)$ precisely when $b\bmod(k+1)=0$. | | Rotate every row rightward $k$ times | rot | Column $b$ reads column $b-k$ modulo the width; [NeZero n] expresses a nonempty cyclic index set. | | Alternate the blocks in batches of $2^{j-1}$ | blockH | Starts with $H_0$, then alternates with $H_1$. | | Stack the blocks of Eq. (6-3) | R | Row $2k+c$ reads row $c$ of block $k$. | | The same rules matrix over the real numbers | Rsq | Row-bound reindexing and the integer-to-real entry map; no altered entries. | | For all $i>j\ge1$ | claim, natural binders and the two inequalities | The source range is retained. The dimension is $2\cdot2^{i-1}=2^i$ throughout this range. | | Eigenvalues are $\sqrt2$-scaled roots of unity | First conjunct of claim | Complex spectrum; a positive exponent may depend on the eigenvalue. | | Order is precisely $2\operatorname{lcm}(4,i-j,i+j)$ | Second conjunct of claim | orderOf in the monoid of real square matrices. | | Refutation of the complete assertion | result | The proposition is exactly the negation of claim. |

The public definitions agree with the binding definitions in #19397. Integer entries, zero-based finite indices and real scalar multiplication specify the source matrix without extra hypotheses on the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/BooleanFunctions/ChirvasituCusickRulesMatrixOrder.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2022A rank-one partner breaks the negative-q bound for a traceless matrixA traceless complex matrix of size three and a rank-one projector violate the proposed q-deformed commutator inequality at q = −1.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionA traceless complex matrix of size three and a rank-one projector violate the proposed q-deformed commutator inequality at q = −1.

Verification details
Release source recordRefuted in Lean

Open problem

A rank-one partner breaks the negative-q bound for a traceless matrix

D. Chruściński, G. Kimura, H. Ohno and T. Singal, Bounding the Frobenius norm of a q-deformed commutator, arXiv:2202.11520v2, state Conjecture 2: "For any $q\le 0$, if $A$ or $B$ is traceless, the sharp bound is $|[A,B]_q|_F^2\le\max[g(n)(1-q)^2,1+q^2],|A|_F^2|B|_F^2$", with $[A,B]_q=AB-qBA$ and $g(n)=(n^2-3n+3)/(n(n-1))$. Their Proposition 3 proves the case $n=2$ with $\mathrm{tr}A=0$ for every real $q$. The verbatim statements are in the literature note.

Issue #14632 reads the inequality clause for every $n\ge2$, every $q\le0$ and all $A,B\in M_n(\mathbb C)$ with $\mathrm{tr}A=0$ or $\mathrm{tr}B=0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/QDeformedCommutatorNegativeQRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2022A traceless counterexample to the q-deformed Böttcher–Wenzel boundTwo traceless complex matrices of size five violate the proposed q-deformed commutator inequality at q = 2.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionTwo traceless complex matrices of size five violate the proposed q-deformed commutator inequality at q = 2.

Verification details
Release source recordRefuted in Lean

Open problem

A traceless counterexample to the q-deformed Böttcher–Wenzel bound

D. Chruściński, G. Kimura, H. Ohno and T. Singal, Bounding the Frobenius norm of a q-deformed commutator, arXiv:2202.11520v2, prove $|AB-qBA|_F^2\le(1+q^2)|A|_F^2|B|_F^2$ for $q>0$ when $A$ or $B$ is normal, note that it fails in general, and state Conjecture 1: "For any $q > 0$, if $A$ or $B$ is traceless, the inequality ... holds and is sharp." They prove it for $n=2$. The verbatim statements are in the literature note.

Issue #13930 reads the inequality clause for every $n$, every $q>0$ and all $A,B\in M_n(\mathbb C)$ with $\mathrm{tr}A=0$ or $\mathrm{tr}B=0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/QDeformedCommutatorTracelessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2022Jacobi and reducedness obstruct the second E6 and truck Poisson bracket conjecturesJacobi and reducedness rule out the proposed diagonal spinor Poisson brackets for the second E6 and truck families.Refuted

The questionKirsty Gledhill and Amihay Hanany, Poisson brackets for some Coulomb branches, JHEP 03 (2023) 154, arXiv:2210.02966v2, conjecture bracket tables for the second E6 family and the truck family.

Why it mattersThe paper seeks Poisson algebra presentations of Coulomb branch chiral rings, including magnetic quivers describing Higgs branches of five-dimensional theories at infinite coupling.

The ideaThe theorem result rules out either side of claim.

Our contributionJacobi and reducedness rule out the proposed diagonal spinor Poisson brackets for the second E6 and truck families.

Scope & next questionA correct replacement table, including whether additional representation components are needed, remains open. - Nearby surviving relations.

Verification details
Release source recordRefuted in Lean

Open problem

Jacobi and reducedness obstruct the second E6 and truck Poisson bracket conjectures

Kirsty Gledhill and Amihay Hanany, Poisson brackets for some Coulomb branches, JHEP 03 (2023) 154, arXiv:2210.02966v2, conjecture bracket tables for the second E6 family and the truck family. The v2 TeX source main.tex has md5 e6664eac5bf9635a3684d84de4e80a97.

For the $E_{6,n}^{II}$ family, main.tex, lines 1845–1856, display labelled final e62 pbs:

We conjecture that the non-zero Poisson brackets of (\ref{gens of e6 II fam}) are given by:

The spinor entries of that display, lines 1851–1853, are:

${C_1,s^{\alpha}}=+ , s^{\alpha}$

${C_1,s_\alpha}=- , s_\alpha$

${s^{\alpha},s_\beta} = {C_1}^{n-1} , {\delta^{\alpha}}_{\beta}$

The sentence following the display is:

where $C_1$ is normalised such that there are no additional numerical coefficients in (\ref{final e62 pbs}). All other Poisson brackets between (\ref{gens of e6 II fam}) that are not listed in (\ref{final e62 pbs}) vanish.

For the truck family, main.tex, lines 2185–2197, display labelled final truck pbs:

We conjecture that the non-zero Poisson brackets of (\ref{gens of truck fam}) are given by:

The two weight brackets are the same, and the mixed entry, lines 2193–2194, is:

${ s^{\alpha}, s_{\beta} } = {C_1}^{n+\sigma-1} , {\delta^{\alpha}}_{\beta}$

The following sentence specifies:

All other Poisson brackets between (\ref{gens of truck fam}) that are not listed in (\ref{final truck pbs}) vanish.

The second E6 table concerns $n\ge3$ (line 1807); the truck quiver has $n\ge2$ and hypermultiplet multiplicity $\sigma$ (line 2149). The spinor indices have $N=2^{2n}$ values (lines 1843 and 2185). Appendix A, line 2237, states:

A Poisson bracket ${ \cdot , \cdot }$ is simply a bilinear map which obeys anticommutativity, Leibniz and the Jacobi identity.

The source context is collected in the literature note.

Issue #21464 fixes the proposition claim as $\mathrm{ConjE6}\lor\mathrm{ConjTruck}$. Here ConjE6 requires, for every natural $n\ge3$, a reduced commutative complex algebra $R$, a Poisson bracket, a nonzero element $h$, and two spinor families indexed by $\mathrm{Fin}(2^{2n})$ satisfying the displayed spinor rules with exponent $e=n-1$. ConjTruck requires the same for every natural $n\ge2$ and every natural $\sigma\ge1$, with $e=n+\sigma-1$. The algebra and bracket may depend on these parameters. This disjunction of universal family assertions is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CoulombBranchPoissonBracketRefutation.result

Frozen module record

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    Source 1
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    Source 1
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Number theoryCounterexample foundFirst preprint 2024Endpoint comparison of Gini means on a bounded intervalA three-point rational witness refutes endpoint characterization of Gini mean comparisons.Refuted

The questionPáles, Comparison of Gini means with fixed number of variables, arXiv:2408.07658v1, §4, state verbatim:

Why it mattersThe conjecture would reduce a comparison on every tuple in a bounded positive interval to one parameter-sum inequality and two tests involving only the extreme endpoints.

The ideaThe means are nonnegative, so their order can be checked by taking fourth powers.

Our contributionA three-point rational witness refutes endpoint characterization of Gini mean comparisons.

Verification details
Release source recordRefuted in Lean

arXiv

Endpoint comparison of Gini means on a bounded interval

R. Grünwald and Zs. Páles, Comparison of Gini means with fixed number of variables, arXiv:2408.07658v1, §4, state verbatim:

We conclude this paper by formulating two conjectures that we have not been able to verify. The first conjecture seems to be a natural extension of the formula for Γ2(I), if 0 < inf I < sup I < ∞.

Conjecture 1 states verbatim:

Conjecture 1. Let n ≥ 3 and let I be a subinterval of R+ such that 0 < a := inf I < sup I =: b < ∞. Then Γn(I) = {((r, s), (p, q)) ∈ R² × R² | r + s ≤ p + q, G[n]r,s(a, …, a, b) ≤ G[n]p,q(a, …, a, b), and G[n]r,s(a, b, …, b) ≤ G[n]p,q(a, b, …, b)}.

Here $\Gamma_n(I)$ is the set of parameter pairs satisfying $G_{r,s}(x)\le G_{p,q}(x)$ for all $x\in I^n$. For $r\ne s$ and positive coordinates,

$$ G_{r,s}(x)=\left(\frac{\sum_{i=1}^n x_i^r}{\sum_{i=1}^n x_i^s}\right)^{1/(r-s)}. $$

The journal bibliographic reference is Results in Mathematics 80 (2025), article 54; the quotations and locators refer to the arXiv version.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/GiniMeanEndpointComparison.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Geometry & algebraCounterexample foundFirst preprint 2024Johnston–Martín Duro–Rumynin Question 2.5A rank-five commutative ring has no finite subset that computes all its Knutson indices.Refuted

Why it mattersIt supplies a finitely generated nonreduced rank-five ring in which every finite set fails for a suitable element.

The ideaEvery eligible coefficient is therefore even, so the finite-set subindex is even or infinite and cannot equal one.

Our contributionA rank-five commutative ring has no finite subset that computes all its Knutson indices.

Scope & next questionWhether finite subsets suffice for restricted classes of nonreduced rings remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Johnston–Martín Duro–Rumynin Question 2.5

Johnston, Martín Duro and Rumynin define the Knutson index and the finite-set subindex for a finitely generated commutative ring $R$, a homomorphism $\alpha : R \to \mathbb{Z}$ and a regular element $r$. Question 2.5 asks whether there is a finite $S \subset R$ such that $\mathcal{K}_r(x)=\mathcal{K}_r^S(x)$ for every $x\in R$.

Statement echo

The Lean definitions IsRegularElement, knutsonIndex, eligible, eligibleIndex and knutsonSubindex mirror the source conventions. The settling declaration result proves the negation of the universal claim. The ring hypotheses are [CommRing R] and [Algebra.FiniteType ℤ R], and the comparison is in $\mathbb{N}_\infty$ after casting the ordinary index.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/KnutsonFiniteSubindexRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationConjecture refutedFirst preprint 2022The C*-algebraic Dubinin–Sugawa norm mean value conjecture is falseA cubic in the product of two complex fields refutes the dual mean value bound.Refuted

The questionMahesh Krishna, C-algebraic Smale Mean Value Conjecture and Dubinin-Sugawa Dual Mean Value Conjecture, arXiv:2206.08154v1, Section 3, p. 7, Conjecture DUALSMALE (journal Conjecture 3.1), states:

Why it mattersA supremum norm can select different coordinates for the derivative and the displacement.

The ideaCStarSchoenberg.orderedDeriv supplies the imported ordered derivative. orderedPoly is the literal ordered product.

Our contributionA cubic in the product of two complex fields refutes the dual mean value bound.

Scope & next questionNo general settlement of CSMALE or STRONGERDUALSMALE follows from result. - Smaller constants — open.

Verification details
Release source recordRefuted in Lean

Open problem

The C*-algebraic Dubinin–Sugawa norm mean value conjecture is false

K. Mahesh Krishna, C-algebraic Smale Mean Value Conjecture and Dubinin-Sugawa Dual Mean Value Conjecture*, arXiv:2206.08154v1, Section 3, p. 7, Conjecture DUALSMALE (journal Conjecture 3.1), states:

Let $\mathcal A$ be a commutative C*-algebra. Let $P(z)=(z-a_1)\cdots(z-a_n)$ be a polynomial of degree $n\geq2$ over $\mathcal A$, with $a_1,\ldots,a_n\in\mathcal A$. If $z\in\mathcal A$ is not a critical point of $P$, then there exists a critical point $w\in\mathcal A$ such that $$\frac{|P'(z)|}{n}\leq\frac{|P(z)-P(w)|}{|z-w|}.$$

Section 2 defines $P'$ as the sum of ordered products obtained by omitting one factor. The source's gloss makes noncritical mean $P'(z)\ne0$; critical means $P'(w)=0$. The Lean definitions preserve this convention, use zero-based Fin n root indices, and cast $n$ to $\mathbb R$ in the displayed quotient.

The carrier A : Type with [CommCStarAlgebra A] restricts the universal claim to unital commutative C*-algebras in Type. This weakens the source assertion, so its negation refutes that assertion. The witness algebra $\mathbb C\times\mathbb C$ is in this subclass.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CStarDualMeanValue.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Quantum informationCounterexample foundFirst preprint 2024A three-level QSP segment that collapses without a common monomial factorA two-step three-level quantum signal protocol returns from effective dimension three to two without a common monomial action on any subspace of dimension at least two.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionA two-step three-level quantum signal protocol returns from effective dimension three to two without a common monomial action on any subspace of dimension at least two.

Verification details
Release source recordRefuted in Lean

Open problem

A three-level QSP segment that collapses without a common monomial factor

L. Laneve and S. Wolf, On multivariate polynomials achievable with quantum signal processing, arXiv:2407.20823v2 (Quantum 9, 1641 (2025)), Conjecture 8: if a Protocol C segment $A_m\tilde W\cdots A_1\tilde W$ ($\tilde W=\operatorname{diag}(1,a,b)$, $A_k\in SU(3)$) maps a polynomial state of effective dimension $\le2$ to one of effective dimension $\le2$ while every intermediate state has effective dimension $>2$, then the segment acts as $a^kb^h\cdot U$ on some pair of subspaces $H\to H'$ of dimension $d\ge2$. The verbatim statements are in the literature note.

Issue #14369 reads the conclusion permissively — subspaces $H,H'$ of $\mathbb C^3$ with its Euclidean (Hermitian) norm, any linear isometry $U:H\to H'$ and any $k,h\in\mathbb N$ — so that its refutation covers the printed form, where $U\in SU(d)$ is taken in orthonormal bases of $H$ and $H'$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/MultivariateQSPCollapseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024Subsystem monotonicity of generalized concentratable entanglementLiu, Knörzer, Wang and Tura (arXiv:2406.18517, Phys.Refuted

The questionTura, Generalized Concentratable Entanglement via Parallelized Permutation Tests, Phys.

Why it mattersMonotonicity under enlarging the subsystem is a basic property expected of a multipartite entanglement measure built from subsystem purities.

Our contributionLiu, Knörzer, Wang and Tura (arXiv:2406.18517, Phys.

Verification details
Release source recordRefuted in Lean

Open problem

Subsystem monotonicity of generalized concentratable entanglement

X. Liu, J. Knörzer, Z. J. Wang and J. Tura, Generalized Concentratable Entanglement via Parallelized Permutation Tests, Phys. Rev. Research 7, L032022 (2025), arXiv:2406.18517v1. For an $n$-qubit pure state $\psi$ and a set $s$ of qubits,

$\mathcal{C}^{(K)}{\ket{\psi}}(s):=\frac{1}{K-1}\left(1-\frac{1}{2^{|s|}}\sum{\alpha\in\mathcal{P}(s)}\Tr(\rho_{\alpha}^K)\right)$ for any $K>1$ […] We take $\Tr(\rho^K_{\varnothing})=1$ in the sum.

The conjecture (arXiv Conjecture 6, journal Conjecture 1), clause 1:

$\mathcal{C}^{(K)}{\ket{\psi}}(s')\leqslant \mathcal{C}^{(K)}{\ket{\psi}}(s)$ if $s'\subseteq s$.

The paper states that this clause and the subadditivity clause "have been proven to be true for $K=2$" and conjectures that "they also hold for $K>1$".

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ConcentratableEntanglementSubsystemRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2022A block measurement refutes mutual-unbiasedness necessity for two-classical dephasingA three-level block measurement has time-dependent two-classical dephasing statistics without mutual unbiasedness.Refuted

The questionChruściński, On the classicality of quantum dephasing processes, arXiv:2211.02014v2; Frontiers in Quantum Science and Technology 1, 1090022 (2022), Section 3.2, immediately following Proposition prop:2classicalitymubs.

Why it mattersTwo-classicality means that at most two sharp measurements do not reveal a violation of Kolmogorov consistency.

The ideaThe theorem result : ¬ claim derives the contradiction.

Our contributionA three-level block measurement has time-dependent two-classical dephasing statistics without mutual unbiasedness.

Verification details
Release source recordRefuted in Lean

Open problem

A block measurement refutes mutual-unbiasedness necessity for two-classical dephasing

D. Lonigro and D. Chruściński, On the classicality of quantum dephasing processes, arXiv:2211.02014v2; Frontiers in Quantum Science and Technology 1, 1090022 (2022), Section 3.2, immediately following Proposition prop:2classicality_mubs. The v2 TeX source (dephasing-v2.tex, md5 e2904085775dc9d0af3b2dba4d50c065), line 347, states:

It may be conjectured that this is the only measurement choice yielding a genuinely time-dependent $2$-classical process---excepting the particular case in which $\rho_{t_0}$ is maximally mixed, cf.~the discussion in the next subsection.

The antecedent is the phase-modified Fourier measurement basis of Eq. eq:mubs, whose squared overlaps with every dephasing basis vector are $1/d$ (TeX lines 284–289). Proposition prop:2classicality_mubs assumes an initial state diagonal in the dephasing basis and gives two-classicality for that measurement family, for arbitrary bath Hamiltonians and bath state (lines 326–347). The definitions and source locators are also in the literature note.

Issue #21441 fixes the following necessity statement, encoded by claim. For every $d,m\in\mathbb N$, every family of Hermitian bath matrices $H_j$ on $\mathbb C^m$, positive semidefinite unit-trace matrices $\sigma$ and $\rho$ on the bath and system respectively, unitary measurement matrix $V$, and initial time $t_0\in\mathbb R$, assume:

  • $\rho$ is diagonal in the dephasing basis and $\rho\ne I_d/d$;
  • some rank-one measurement projector $P_x=|m_x\rangle\langle m_x|$, with $m_x$ the $x$th column of $V$, does not commute with some dephasing projector $E_j=|e_j\rangle\langle e_j|$;
  • for every $t_0\le t_1\le t_2$, the one-time law is normalized and both marginals of the two-time law equal the corresponding one-time law;
  • for some fixed outcomes $x,y$, two ordered time pairs at or after $t_0$ give different two-time values.

The conclusion is $|V_{jx}|^2=1/d$ for every $j,x$. The laws here are the literal joint-system trace formula of the paper's Eq. (2), with Hamiltonian $\mathbf H=\sum_jE_j\otimes H_j$, propagator $e^{-i(t-s)\mathbf H}$, initial product $\rho\otimes\sigma$, and system Lüders operations $X\mapsto(P_x\otimes I_m)X(P_x\otimes I_m)$. There is no bath reset. The definitions prob1 and prob2 take values in $\mathbb C$; twoClassical imposes the stated normalization and marginal equalities. No positive lower bound on $d$ or $m$ is added to claim.

This is a consequence of the conjectured necessity: it restricts to finite baths and time-independent Hamiltonians, asks only for unbiasedness rather than membership in the named Fourier family, and additionally requires actual nonconstancy of the two-time law. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Decoherence/DephasingTwoClassicalityBlockRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    Year in the primary source attribution: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryConjecture refutedFirst preprint 2022Refutation of Miska–Ulas Conjecture 4.8A positive integer family refutes the degree-six symmetric-product bound.Refuted

The questionMiska and Ulas, On the Diophantine equation σ₂(X̄ₙ) = σₙ(X̄ₙ), arXiv:2203.03942v1, page 16, Conjecture 4.8 states:

The ideaMultiset expansion, binomial coefficient identities, and polynomial normalization give the family, its domain conditions and its factored gap.

Our contributionA positive integer family refutes the degree-six symmetric-product bound.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Miska–Ulas Conjecture 4.8

Miska and Ulas, On the Diophantine equation σ₂(X̄ₙ) = σₙ(X̄ₙ), arXiv:2203.03942v1, page 16, Conjecture 4.8 states:

Let n ∈ N≥4 and suppose that m ∈ N satisfies m = σ_3(X_n) = σ_n(X_n). Then m ≤ (1/12)(3021n⁶ − 77575n⁵ + 920361n⁴ − 6235705n³ + 24764202n² − 53664304n + 48986640) and x_n ≤ (1/12)(27n⁴ − 190n³ + 471n² − 500n + 216).

Write $Q(n)$ and $D(n)$ for the two bounds, respectively. The source defines $\sigma_j$ as the elementary symmetric sum of products over $j$ distinct increasing indices, and a solution as a nondecreasing tuple of positive integers. The target is the conjunction as printed in version 1. Preregistration: #19678.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | $n\ge4$ | claim quantifies n : ℕ with hn : 4 ≤ n | Exact | | Positive integer tuple $X_n$ | X : Fin n → ℕ and positive entries | Exact; zero-based indices | | $x_i\le x_{i+1}$ | Monotone X | Equivalent on the finite linear order | | $m=\sigma_3(X_n)=\sigma_n(X_n)$ | Both equalities to MvPolynomial.eval X (MvPolynomial.esymm (Fin n) ℕ j) | Literal subset-product sum; no new symmetric-sum alias | | $m\le Q(n)$ and $x_n\le D(n)$ | The two rational inequalities in claim; last entry indexed by n - 1 | Both conjuncts retained | | Refutation | result : ¬ claim | The first conjunct fails at $n=11$ |

The rational casts preserve the integer inequalities and interpret the source's factor $1/12$ exactly. The index proof follows from $n\ge4$ and adds no mathematical hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/MiskaUlasSigmaThreeBound.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Number theoryCounterexample foundFirst preprint 2022Refutation of Onnis's square convergence prohibitionA positive perfect square first reaches one after three iterations of the nonzero-decimal-digit-product square map.Refuted

The questionLuca Onnis, On a variant of the happy numbers and their generalizations, arXiv:2203.03381v1, Section 3, "Conjecture 1" (verbatim):

Why it mattersOne positive perfect-square member with first hitting time three contradicts the whole universal prohibition.

The idea100. Nat.digitsofDigits supplies these exact decimal representations, with digit bounds and nonzero final digits.

Our contributionA positive perfect square first reaches one after three iterations of the nonzero-decimal-digit-product square map.

Scope & next questionSection 4's at-most-three assertion, higher-exponent variants, image-restricted claims, a census and least-counterexample questions are outside this result.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of Onnis's square convergence prohibition

Luca Onnis, On a variant of the happy numbers and their generalizations, arXiv:2203.03381v1, Section 3, "Conjecture 1" (verbatim):

There are no perfect square that are term of the sequence with length of convergence equal to 3.

Section 1.2 defines F(n) to be the square of the product of the nonzero decimal digits of n. A positive integer is a sequence member when some finite iterate reaches one. With F^[0](n) = n, exact convergence length three means F^[3](n) = 1 and F^[k](n) != 1 for every natural k < 3.

The closed proposition claim is exactly:

∀ n : ℕ, 0 < n → (∃ r : ℕ, n = r ^ 2) →
  SequenceMember n → ¬ FirstHit n 3

SequenceMember n is ∃ t : ℕ, step^[t] n = 1, and FirstHit n t is step^[t] n = 1 ∧ ∀ k : ℕ, k < t → step^[k] n ≠ 1. The square root is an arbitrary natural number. There is no smoothness, image, digit-alphabet, root-bound or exponent-variation condition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/OnnisSquareThreeStepRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2023The boundary α-z Rényi relative entropies of magic are not additive on three qubitsBoundary alpha-z Rényi magic additivity fails for a pair of three-qubit states at alpha = 2 and z = 1.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersAdditivity would reduce the magic of a product of few-qubit states to single-system computations, which the source uses for bounds on magic state distillation (§6).

The ideaLet S be the sum of the seven pairwise anticommuting three-qubit Paulis XII, YII, ZXI, ZYI, ZZX, ZZY, ZZZ, so that S² = 7I.

Our contributionBoundary alpha-z Rényi magic additivity fails for a pair of three-qubit states at alpha = 2 and z = 1.

Verification details
Release source recordRefuted in Lean

Open problem

The boundary α-z Rényi relative entropies of magic are not additive on three qubits

R. Rubboli, R. Takagi and M. Tomamichel, Mixed-state additivity properties of magic monotones based on quantum relative entropies for single-qubit states and beyond, Quantum 8, 1492 (2024), arXiv:2307.08258v4, define the α-z Rényi relative entropies of magic D_{α,z}(ρ) = inf over σ ∈ S̄_n of D_{α,z}(ρ‖σ) (Eq. (10)), where:

  • S̄_n is the convex hull of the pure stabilizer states, the Clifford images of |0^n⟩ under the group generated by H, S and CNOT;
  • D_{α,z} is the α-z Rényi divergence with the support conventions of Eq. (2).

They prove additivity for all single-qubit states on the boundary |(1 − α)/z| = 1 of the data-processing region. §5 states: "It is still an open problem whether the α-z Rényi relative entropies of magic in the range |(1−α)/z| = 1 are additive for all two and three-qubit states". Table 1 and §10 conjecture that they are. The verbatim statements are in the literature note.

Issue #15943 fixes the reading: for every (α, z) in the region D with |(1 − α)/z| = 1, every pair of states on two or three qubits each satisfies D_{α,z}(ρ₁ ⊗ ρ₂) = D_{α,z}(ρ₁) + D_{α,z}(ρ₂).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/RenyiBoundaryMagicAdditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2024A passive incoherent memory realizes a non-classical dephasing superchannel in dimension 3In dimension 3, a passive memory implementation exists whose Gram matrix is not classical.Refuted

The questionShahbeigi, Dephasing noise simulation for coherence-generating devices, Phys.

Why it mattersThe question decides whether memory that stores no information about the input can still produce gate noise beyond classically correlated pre- and post-dephasing.

The idea1. Passive. The memory marginal of W(ρ ⊗ τ)W† is Σi ρii Ui τ Ui† = τ for every density matrix ρ.

Our contributionIn dimension 3, a passive memory implementation exists whose Gram matrix is not classical.

Verification details
Release source recordRefuted in Lean

Open problem

A passive incoherent memory realizes a non-classical dephasing superchannel in dimension 3

R. Salazar and F. Shahbeigi, Dephasing noise simulation for coherence-generating devices, Phys. Scr. 100 (2025) 085110, arXiv:2403.19247v1, study dephasing superchannels, which multiply the Choi matrix J(E) of every channel E entrywise by a Gram matrix C. Such a superchannel is implemented with a memory state τ and encoding and decoding channels on system ⊗ memory: Ξ(E)(ρ) = Tr_M[N_de((E ⊗ id)(N_en(ρ ⊗ τ)))] (Eq. (9)).

  • The memory is passive if its marginal after encoding does not depend on the input (Definition 2).
  • C is classical if C = Σ_i q_i C₁^{(i)} ⊗ C₂^{(i)}, the Gram matrix of dephasing channels applied before and after E under classical randomness.

Proposition 1 shows that for qubit channels a passive memory realizes exactly the classical Gram matrices. §V.C asks "whether a passive memory in any dimension is equivalent to a classical memory". §V.D argues that it is, for incoherent implementations. The verbatim statements are in the literature note.

Issue #15829 fixes the reading:

  • every system dimension d and finite memory dimension r;
  • every memory state and every Kraus encoder and decoder;
  • the source's output-first normalised Choi matrix and Eq. (13)'s equal diagonal blocks;
  • passivity for every pair of input density matrices;
  • classical Gram matrices as finite convex combinations of Kronecker products of unit-diagonal positive semidefinite matrices.
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/PassiveMemoryNonclassicalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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Discrete structuresConjecture refutedFirst preprint 2024Conjecture 1 on paired coalition numbers of perfect binary trees is falseThe perfect binary tree of height five has a paired coalition partition with five parts.Refuted

The questionSamadzadeh, Mojdeh and Nadimi, Paired coalition in graphs, arXiv:2402.10842v4, Section 4.2, p. 12, Conjecture 1:

Why it mattersThe Library note is Paired coalition in graphs.

The ideaThey are nonempty, disjoint and cover the tree.

Our contributionThe perfect binary tree of height five has a paired coalition partition with five parts.

Verification details
Release source recordRefuted in Lean

Open problem

Conjecture 1 on paired coalition numbers of perfect binary trees is false

Samadzadeh, Mojdeh and Nadimi, Paired coalition in graphs, arXiv:2402.10842v4, Section 4.2, p. 12, Conjecture 1:

If T(h) is a perfect binary tree with height h, then PC(T(h)) = 5 if h = 2; 3 if h is odd; 0 otherwise.

The domain is every positive natural height. The root has depth zero and all leaves have depth $h$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Perfect binary tree of height $h$ | PairedCoalitionBinaryTree.perfectBinaryTree h | Heap labels $1,\ldots,2^{h+1}-1$, children $2r,2r+1$. | | Dominating set whose induced subgraph contains a perfect matching | IsPairedDominating | Reused SimpleGraph.IsDominating and a matching subgraph with exactly the selected vertices. | | Disjoint sets, neither paired dominating, with paired dominating union | IsPairedCoalition | All four requirements are conjuncts. | | Vertex partition; each part has another coalition partner | IsPCPartition | Nonempty, pairwise disjoint, covering parts; distinct partners in the same finset. | | Maximum partition order; zero if none exists | pcNumber | Natural supremum of attained orders; finite vertex sets bound all orders. | | Every positive height | claim | $h:\mathbb N$ with $1\le h$. | | Height two first, then odd heights, then otherwise | claim | Values $5,3,0$ in exactly that order. | | Conjecture is false | result | Negation of the complete positive-height claim. |

All declaration names in this dossier belong to D5.S3.Combinatorics.Graph.Domination.SamadzadehMojdehNadimiRefutation.PairedCoalitionBinaryTree unless another owner is explicit. No additional hypothesis restricts the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/SamadzadehMojdehNadimiRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

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Quantum informationCounterexample foundFirst preprint 2023Formation is not constant on the Song–Chen–Đoković subspaceA nonorthogonal ensemble lowers formation on the Song–Chen–Djokovic subspace.Refuted

The questionSong, Chen and Đoković, The existence of distinguishable bases in three-dimensional subspaces of qutrit-qudit systems under one-way local operations and classical communication, arXiv:2307.01980v3, p. 3, asks:

Why it mattersThere is no digestion atom and no coverage claim.

The ideaThe finite-ensemble upper bound then contradicts the universal equality.

Our contributionA nonorthogonal ensemble lowers formation on the Song–Chen–Djokovic subspace.

Scope & next questionThe source's distinguishability conclusions do not depend on the proposed EOF equality; this refutation changes no claim about existence or uniqueness of those bases.

Verification details
Release source recordRefuted in Lean

Open problem

Formation is not constant on the Song–Chen–Đoković subspace

Song, Chen and Đoković, The existence of distinguishable bases in three-dimensional subspaces of qutrit-qudit systems under one-way local operations and classical communication, arXiv:2307.01980v3, p. 3, asks:

Further, by calculation one can see that every state in the span of the three states has entanglement more than 1.52 ebits. Hence, the entanglement of formation (EOF) of every mixed state whose range is contained in the span is also more than 1.52 ebits [19]. We don't know whether this EOF is exactly equal to that of the pure state |Ψ₁⟩.

The vectors are those of Eqs. (2)–(3), on $\mathbb C^3\otimes\mathbb C^9$. Write $P_j=|\Psi_j\rangle\langle\Psi_j|$, $S=\operatorname{span}_{\mathbb C}{\Psi_1,\Psi_2,\Psi_3}$ and $H_0=H(4/9,1/3,2/9)$, with entropy in bits.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Eq. (2), the twelve nonzero entries of $G$ | UniqueBasisFormationRefutation.G | Literal real square roots embedded in $\mathbb C$; indices decreased by one. | | Eq. (3), the columns $r_{3a+j}$ | Psi k (a,b) | Entry $G(b,3a+k)$, with zero-based Fin indices. | | The span of the three states | spanPsi | Complex linear span of Set.range Psi. | | Every density operator supported in $S$ | claim: positivity, trace one, range inclusion | All matrices on Fin 3 × Fin 9, with no added restriction. | | EOF equal to that of $\Psi_1$ | FormationBounds.E_F, PureStateHandshake.rankOneDensity (Psi 0) | Existing finite-ensemble formation in bits and rank-one density. | | A negative answer | result : ¬ claim | Refutes the complete universal equality. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GraphDensity/UniqueBasisFormationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2022Bell-diagonal entanglement distance is not twice the squared concurrenceThe entanglement distance of a Bell-diagonal state is not twice its squared concurrence.Refuted

The questionFranzosi, Entanglement and Quantum Correlation Measures from a Minimum Distance Principle, arXiv:2205.07143v2; published as Entanglement and quantum correlation measures for quantum multipartite mixed states, Sci.

Why it mattersThe source proposes the regularization of QCD as a measure of mixed-state entanglement.

The idea4. The contradiction. Each inner infimum has an actual zero-valued rotation candidate.

Our contributionThe entanglement distance of a Bell-diagonal state is not twice its squared concurrence.

Scope & next questionRestricting to the spectral decomposition would remove this particular decomposition, but would change the optimization; correctness of the corresponding restricted identity remains open here. - Lower bounds for the ordinary infima.

Verification details
Release source recordRefuted in Lean

Open problem

Bell-diagonal entanglement distance is not twice the squared concurrence

A. Vesperini, G. Bel-Hadj-Aissa and R. Franzosi, Entanglement and Quantum Correlation Measures from a Minimum Distance Principle, arXiv:2205.07143v2; published as Entanglement and quantum correlation measures for quantum multipartite mixed states, Sci. Rep. 13, 2852 (2023), DOI 10.1038/s41598-023-29438-7. In Applications, “Bell diagonal states”, the v2 source QCDED-V4.tex, line 368, states:

We were not able to find a simple analytic solution of the minimization procedure for the most general case of BD states. However, numerical minimization provided us with empirical evidence that the procedure (22) also leads for these states to the squared concurrence, as shown in figure \ref{figBD_ED}, which represent a face of the tetrahedral domain of BD states.

This is a reported numerical belief, rather than a proved identity. The source read is the arXiv v2 TeX, MD5 232776c9cef202cc624a80958cb4c818. Its figure caption at line 372 plots $E[\rho_{BD}]/2$; the Werner-state formula at lines 396–400 is described as “twice the square of the concurrence”. These fix the factor-two reading registered in issue #20089.

Let $B_0,B_1,B_2,B_3$ be the projectors onto $(|00\rangle+|11\rangle)/\sqrt2$, $(|00\rangle-|11\rangle)/\sqrt2$, $(|01\rangle+|10\rangle)/\sqrt2$ and $(|01\rangle-|10\rangle)/\sqrt2$, respectively. For $p_\alpha\ge0$ and $\sum_{\alpha=0}^3p_\alpha=1$, put $\rho_{BD}(p)=\sum_\alpha p_\alpha B_\alpha$ and $c(p)=\max(0,2\max_\alpha p_\alpha-1)$. Here $c$ denotes Bell-diagonal concurrence; $C_\mu$ below denotes site quantum correlation distance (QCD).

The registered claim is that $E(\rho_{BD}(p))=2c(p)^2$ for every such probability vector, where $E$ minimizes the site QCD of locally rotated mixtures over all finite density-matrix decompositions and component-dependent local rotations, summing both sites. This claim is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/EntanglementDistanceBellDiagonalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresConjecture refutedSource 2024Refutation of Zhao's maximum preimage conjectureZhao's maximum and second-largest vincular-stack fibre conjectures are false.Refuted

The questionZhao, arXiv:2410.17057v1, Conjecture 4.14 (Section 4.1.4, printed page 17), states:

Why it mattersThe claimed maximum for the 1-underline(23) map at n = 9 is at most 2⁷ = 128.

The ideaApplying the claimed upper bound at n = 9 yields 129 ≤ 128, and arithmetic closes the contradiction.

Our contributionZhao's maximum and second-largest vincular-stack fibre conjectures are false.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Zhao's maximum preimage conjecture

Zhao, arXiv:2410.17057v1, Conjecture 4.14 (Section 4.1.4, printed page 17), states:

For n ≥ 2, it holds that max over π in 𝔖ₙ of |SC₁₂̲₃̲⁻¹(π)| equals max over π in 𝔖ₙ of |SC₃₂̲₁̲⁻¹(π)| equals 2ⁿ⁻².

The formal claim uses the paper's one-based permutation entries and the right-greedy stack convention fixed by its four worked figures. The flag false denotes 1-underline(23), and true denotes 3-underline(21).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/ZhaoVincularPreimageRefutations.resultMaximum

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Discrete structuresConjecture refutedSource 2024Refutation of Zhao's second-largest preimage conjectureZhao's maximum and second-largest vincular-stack fibre conjectures are false.Refuted

The questionZhao, arXiv:2410.17057v1, Conjecture 5.2 (Section 5, printed page 20), states:

Why it mattersAt n = 5 the asserted second-largest value is 2² = 4.

The ideaUnder the conjecture's first clause, every fibre value strictly above 4 must equal one common larger value.

Our contributionZhao's maximum and second-largest vincular-stack fibre conjectures are false.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Zhao's second-largest preimage conjecture

Zhao, arXiv:2410.17057v1, Conjecture 5.2 (Section 5, printed page 20), states:

The second-largest number of preimages under SC₁₂̲₃̲ that a permutation in 𝔖ₙ can have is 2ⁿ⁻³, for n ≥ 3. Furthermore, the number of permutations π in 𝔖ₙ satisfying |SC₁₂̲₃̲⁻¹(π)| = 2ⁿ⁻³ is 2n − 2.

The formal claim retains both clauses under the same universal quantifier. The flag false denotes 1-underline(23), with one-based permutation entries and the right-greedy convention fixed by the paper's figures.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/ZhaoVincularPreimageRefutations.resultSecondLargest

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Quantum informationCounterexample foundFirst preprint 2024Logical Z operators with Z support on every stabilizer coset need not existA five-qubit stabilizer code has a Z-free word in every nontrivial logical coset.Refuted

The questionLiu, From Magic State Distillation to Dynamical Systems, Quantum 9, 1858 (2025), arXiv:2412.04402, model magic state distillation by a stabilizer code as a map of the Bloch ball.

Why it mattersThe fixed-plane property lets the authors reduce the three-dimensional distillation dynamics to a two-dimensional flow on the equator, the plane of magic states that can be injected directly.

The idea1. Witness. The five-qubit code with generators XZZXI, IXZZZ, XIXZY and ZXIXY.

Our contributionA five-qubit stabilizer code has a Z-free word in every nontrivial logical coset.

Verification details
Release source recordRefuted in Lean

Open problem

Logical Z operators with Z support on every stabilizer coset need not exist

Y. Zheng and D. E. Liu, From Magic State Distillation to Dynamical Systems, Quantum 9, 1858 (2025), arXiv:2412.04402, model magic state distillation by a stabilizer code as a map of the Bloch ball.

Proposition 3 gives a sufficient condition for z = 0 to be a fixed plane of the map: for every logical operator Z̄_i and every stabilizer element s_j, the product s_j Z̄_i contains a Pauli Z tensor factor. The authors write after Eq. (23): "This condition is satisfied by most known protocols that we have examined and we conjecture it to be generally true for any stabilizer codes by properly choosing the logical Z operators without modifying the stabilizer groups."

The verbatim statements are in the literature note.

Issue #17951 fixes the reading. For every stabilizer code with at least one logical qubit, some nontrivial logical Pauli word L has a literal Z factor in every product s·L with s in the stabilizer group. This conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/StabilizerLogicalZSupportRefutation.result

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Quantum informationCounterexample foundFirst preprint 2023In-between-state optimality for transmissivity sensing with a beta priorAt mean photon number one quarter and beta parameters alpha=3, beta=1, a vacuum–two-photon probe has strictly smaller Bayesian error over bounded self-adjoint Fock-space observables than every phased in-between state.Refuted

The questionGagatsos, “Bayesian minimum mean square error for transmissivity sensing”, arXiv:2304.05539v1, Phys.

Why it mattersThe paper gives the first Bayesian treatment of quantum-limited transmissivity estimation.

Our contributionAt mean photon number one quarter and beta parameters alpha=3, beta=1, a vacuum–two-photon probe has strictly smaller Bayesian error over bounded self-adjoint Fock-space observables than every phased in-between state.

Verification details
Release source recordRefuted in Lean

Open problem

In-between-state optimality for transmissivity sensing with a beta prior

B. Zhou, B. A. Bash, S. Guha and C. N. Gagatsos, “Bayesian minimum mean square error for transmissivity sensing”, arXiv:2304.05539v1, Phys. Rev. Research 5, 043033 (2023), Section VI (p. 6):

The numerical results indicate that the optimal input states with non-integer mean photon number, have the form of Eq. (27), which means that for $\bar{n} \in \mathbb{N}$ the optimal states are Fock states.

and Section VII (p. 7):

We note that for the beta prior PDF, we were not able to prove analytically that the Fock states or the in-between states of Eq. (27) states are optimal, even though our numerical computations support such conjecture.

The beta prior is eq. (16), $P(\tau)=\tau^{\alpha-1}(1-\tau)^{\beta-1}/\mathcal B(\alpha,\beta)$ with $\alpha,\beta>0$. The verbatim definitions are in the literature note. Issue #14698 preregisters the reading with the parameters free, the witness and the exact certificates.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Estimation/TransmissivityBetaPriorProbeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Discrete structuresCounterexample foundFirst preprint 2025Multiset rook-Eulerian interlacing on Ferrers boardsMultiset rook-Eulerian interlacing fails on a six-row Ferrers board.Refuted

The questionAlexandersson, Jal and Quemener, Real-rootedness of rook-Eulerian polynomials, arXiv:2502.05939v1, Section 3.3, Conjecture 28:

Our contributionMultiset rook-Eulerian interlacing fails on a six-row Ferrers board.

Verification details
Release source recordRefuted in Lean

Open problem

Multiset rook-Eulerian interlacing on Ferrers boards

Alexandersson, Jal and Quemener, Real-rootedness of rook-Eulerian polynomials, arXiv:2502.05939v1, Section 3.3, Conjecture 28:

For Ferrers boards, the polynomial $R(\lambda,\alpha;t)$ is real-rooted. Moreover, $R_{\lambda_1}(\lambda,\alpha;t),R_{\lambda_1-1}(\lambda,\alpha;t),\dotsc,R_2(\lambda,\alpha;t),R_1(\lambda,\alpha;t)$ forms an interlacing sequence.

Here $\lambda_1\le\cdots\le\lambda_n$ is the board, $\alpha$ is a vector of nonnegative integer multiplicities summing to $n$, and $W$ consists of words with those multiplicities satisfying $0<w_i\le\lambda_i$. The exponent $\operatorname{asc}(w)$ counts strict adjacent ascents. Equations (10) and (11) define $R=\sum_{w\in W}t^{\operatorname{asc}(w)}$ and $R_j=\sum_{w\in W,,w_1=j}t^{\operatorname{asc}(w)}$. Definitions 8–9 require positive leading coefficients, real nonpositive roots with multiplicity, equal degrees or one larger degree on the right, and the chain $\dotsm\le a_2\le b_2\le a_1\le b_1\le0$ for every ordered pair in the sequence. Issue #12894 specifies nonzero guards for the two refinements.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/MultisetRookEulerianInterlacingRefutation.result

D5/S3/Combinatorics/Permutation/MultisetRookEulerianRealRootednessRefutation.result

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Quantum informationQuestion resolvedFirst preprint 2025Deterministic-input violations exist at every positive energy budget below one halfQubit channels give a strict deterministic-input violation of the separable energy bound for every positive energy budget below one half.Proved

The questionAcín, Entanglement in the energy-constrained prepare-and-measure scenario: applications to randomness certification and channel discrimination, arXiv:2510.27559v3 (2025 preprint; v3, 2026), Section IV.A, state:

Why it mattersThe source uses energy constraints to certify randomness without specifying the devices' internal implementation.

The ideaAll parameters in the following lemmas are real.

Our contributionQubit channels give a strict deterministic-input violation of the separable energy bound for every positive energy budget below one half.

Scope & next questionThe exact minimum over this family or all allowed devices remains open here.

Verification details
Release source recordProved in Lean

Open problem

Deterministic-input violations exist at every positive energy budget below one half

R. D'Avino, G. Senno, M. Alimuddin and A. Acín, Entanglement in the energy-constrained prepare-and-measure scenario: applications to randomness certification and channel discrimination, arXiv:2510.27559v3 (2025 preprint; v3, 2026), Section IV.A, state:

We conjecture that this holds for all $\omega > 0$.

This is verbatim from paper_Quantum.tex, line 284 (MD5 012a72f4d8cfd30e14dc3813eda8f546). The preceding sentence's assertion is the existence of behaviors allowing entanglement that are deterministic for input $x=0$ but violate the separable deterministic-input inequality. The global restriction is $\omega\in[0,1/2)$ (line 154), so the conjecture concerns every $0<\omega<1/2$.

The paper considers a shared state $\sigma_{PM}$, local channels $\Lambda_x:P\to S$ for $x\in{0,1}$, and a binary POVM ${\Pi^0,\Pi^1}$ on $S\otimes M$. Its behavior formula is $p(b|x)=\operatorname{Tr}[\Pi^b(\Lambda_x\otimes\mathrm{id}M)(\sigma{PM})]$ (lines 137–148), and $E_x=p(0|x)-p(1|x)$ (lines 165–166). For separable devices with $p(0|0)=1$, its displayed inequality is:

$E_1\geq 2(1-2\omega)^2-1=:I_\mathrm{det}^{\omega,\mathrm{sep}}$

This is line 268, equation labelled eq:deterministic-ineq-for-fixed-input-0. The source defines $\mathcal Q_\omega^{\mathrm{ent}}$ by allowing entanglement, rather than by requiring an additional entanglement test (line 158).

Issue #22022 fixes the following finite-dimensional reading, which is proved. For every real $0<\omega<1/2$, there exist natural dimensions $d_p,d_s,d_m$, a complex matrix $\sigma$ on $P\otimes M$, two complex-linear matrix maps $\Lambda_x:P\to S$, a matrix $H$ and vector $g$ on $S$, and two effects $\Pi^b$ on $S\otimes M$, such that:

  • $\sigma\succeq0$ and $\operatorname{Tr}\sigma=1$;
  • each $\Lambda_x$ is completely positive and preserves the trace of every complex matrix;
  • $H=H^\dagger$, $g^\dagger g=1$, $Hg=0$, and $H-(I-|g\rangle\langle g|)\succeq0$;
  • $\Pi^b\succeq0$ for both outcomes and $\Pi^0+\Pi^1=I$;
  • $\operatorname{Re}\operatorname{Tr}[H\Lambda_x(\operatorname{Tr}_M\sigma)]\leq\omega$ for both inputs;
  • $p(0|0)=1$ and $p(0|1)-p(1|1)<2(1-2\omega)^2-1$.

Here the Lean definition probability takes the real part of the Born trace displayed above. The claim has no separate separability or entanglement predicate. The source permits arbitrary, including infinite, dimensions; finite witnesses are sufficient for its existential question. See also the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/EnergyConstrainedDeterministicViolation.result

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Quantum informationCounterexample foundFirst preprint 2001Gingrich's proposed three-qubit monotone built from the Kempe invariantA 65, 052302) proposes sigmaABC = 3 - (I1 + I2 + I3) I4, built from three purities and the Kempe invariant I4, as a fifth entanglement monotone of three-qubit pure states, supported by numerical tests.Refuted

The questionGingrich, Properties of entanglement monotones for three-qubit pure states, Phys.

Our contributionA 65, 052302) proposes sigmaABC = 3 - (I1 + I2 + I3) I4, built from three purities and the Kempe invariant I4, as a fifth entanglement monotone of three-qubit pure states, supported by numerical tests.

Verification details
Release source recordRefuted in Lean

Open problem

Gingrich's proposed three-qubit monotone built from the Kempe invariant

R. M. Gingrich, Properties of entanglement monotones for three-qubit pure states, Phys. Rev. A 65, 052302 (2002), arXiv:quant-ph/0106042v2, Eq. (31) and the text after it:

$\sigma_{ABC} = 3 - (I_1 + I_2 + I_3) I_4$ and numerical results suggest that it is an EM. ... For the rest of the paper I will assume that $\sigma_{ABC}$ is an EM.

For $\psi=\sum t_{ijk}|ijk\rangle$, the polynomial invariants are $P_{\sigma,\tau}=\sum t_{i_1j_1k_1}\cdots t_{i_nj_nk_n}, \bar t_{i_1j_{\sigma(1)}k_{\tau(1)}}\cdots\bar t_{i_nj_{\sigma(n)}k_{\tau(n)}}$ (Eq. (8)), with $I_1=P_{e,(12)}$, $I_2=P_{(12),e}$, $I_3=P_{(12),(12)}$ (the three one-qubit purities) and the Kempe invariant $I_4=P_{(123),(132)}$. An entanglement monotone (EM) in the paper's sense satisfies $E(\psi)\ge\sum_kp_kE(A_k\psi/\sqrt{p_k})$ for every complete instrument ${A_k}$ on one party, $p_k=|A_k\psi|^2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GingrichKempeMonotoneRefutation.result

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Quantum informationConjecture refutedFirst preprint 2026The affine Lie algebra refutes Heib–Bruschi Conjecture 83The two-dimensional affine Lie algebra refutes Heib–Bruschi Conjecture 83 over every field.Refuted

The questionA vertex subset has the ideal-graph-property when no edge leaves it.

Why it mattersThe paper uses the graphs to read off centers, derived and lower central series and ideals, for Lie algebras that arise in quantum dynamics.

Our contributionThe two-dimensional affine Lie algebra refutes Heib–Bruschi Conjecture 83 over every field.

Scope & next questionThis settlement makes no claim about Conjecture 82 in general.

Verification details
Release source recordRefuted in Lean

Open problem

The affine Lie algebra refutes Heib–Bruschi Conjecture 83

T. Heib and D. E. Bruschi, On the structural properties of Lie algebras via associated labeled directed graphs, arXiv:2601.16161v1, attach to a basis of a Lie algebra in which every bracket of two basis vectors is a multiple of a basis vector (a minimal-graph-admissible basis) the directed graph with an edge from $x_j$ to $x_\ell$, labelled $x_k$, whenever $[x_j,x_k]$ is a nonzero multiple of $x_\ell$. A vertex subset has the ideal-graph-property when no edge leaves it. Conjecture 83 states that a minimal-graph-admissible Lie algebra with trivial center either (i) has no proper non-empty vertex subset with the ideal-graph-property, or (ii) is a direct sum of components with trivial center whose every minimal graph has no such subset. The verbatim statements are in the literature note.

Issue #13676 fixes the reading over every field, as the paper's convention "We denote any field with the symbol $\mathbb F$" requires, with the direct sum read as an internal direct sum of ideals over an arbitrary index type.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/HeibBruschiCenterlessIdealGraphRefutation.result

Frozen module record

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Discrete structuresConjecture refutedFirst preprint 2026Mohan-Neetu Conjecture 6.2Two explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.Refuted

The questionThe formal claim62 retains every premise; result62 proves its negation.

Why it mattersConjecture 6.2 asks whether lowering that threshold by one restores commutativity.

The ideaIn the torsion-free Klein bottle group (r,n)(s,m) = (r + (-1)^n s,n+m), let S = {(1,1),(2,1),(3,1)} and take its three singleton subsets as A, B and C.

Our contributionTwo explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.

Verification details
Release source recordRefuted in Lean

Open problem

Mohan-Neetu Conjecture 6.2

Conjecture 6.2 in Mohan and Neetu, arXiv:2607.11194v1, section 6, asserts that a nonempty finite subset S of a torsion-free group, partitioned into three pairwise disjoint abelian sets A, B and C, generates an abelian subgroup when |S^2| <= 3k - 4. Here k = |S|, S^2 is the product set, and an abelian set has an abelian generated subgroup. The formal claim62 retains every premise; result62 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Groups/MohanNeetuSmallDoublingRefutation.result62

Frozen module record

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Quantum informationCounterexample foundFirst preprint 2025A local measure-and-reset refutes the LOCC monotonicity of the MUB correlation $I_2$A local measure-and-reset raises the mutually unbiased correlation I2 from 1 to 3/2, refuting its monotonicity under LOCC in the fixed and the optimized reading.Refuted

The questionThe verbatim statements are in the literature note.

Our contributionA local measure-and-reset raises the mutually unbiased correlation I2 from 1 to 3/2, refuting its monotonicity under LOCC in the fixed and the optimized reading.

Verification details
Release source recordRefuted in Lean

Open problem

A local measure-and-reset refutes the LOCC monotonicity of the MUB correlation $I_2$

S. Nandi, Genuine multipartite entanglement detection with mutually unbiased bases (MUBs), arXiv:2509.24045v1 (EPL 154 (2026)), §II, measures a bipartite state $\rho$ on $\mathbb C^d\otimes\mathbb C^d$ in a basis ${a_i}$ for Alice and ${b_i}$ for Bob, and in a second pair ${a'_i}$, ${b'_i}$ mutually unbiased to the first, and sets $I_2=\sum_i\langle a_ib_i|\rho|a_ib_i\rangle+\sum_i\langle a'_ib'_i|\rho|a'_ib'_i\rangle$. Its Conjecture states that "the quantity $I_2$ is monotonically non-increasing under LOCC operations", tested in Eq. (10) on binary local instruments $E_1,E_2$ with $E_1^\dagger E_1+E_2^\dagger E_2=I$: $I_2(\rho)-p_1I_2(\rho_1)-p_2I_2(\rho_2)\ge0$. The verbatim statements are in the literature note.

Issue #13804 fixes two readings of "the quantity $I_2$", both for $d\ge2$: for every fixed pair of local MUB settings, and for the supremum over settings (the paper evaluates $I_2$ of pure states in their Schmidt basis). The conjecture is read as the disjunction of the two. Because $I_2$ is linear in $\rho$, $p_kI_2(\rho_k)=I_2(\tau_k)$ with $\tau_k=(E_k\otimes I)\rho(E_k\otimes I)^\dagger$, which also covers $p_k=0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/NandiMutualPredictabilityLOCCRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Concealment-equivalent channels with different adjoint kernelsRefuted

The questionIssue 14233 reads the affirmative answer with the hypothesis in its strongest form.

Verification details
Release source recordRefuted in Lean

Open problem

Concealment-equivalent channels with different adjoint kernels

M. A. Siddiqui and Z. Wang, Operational Concealment of Measurement Incompatibility by Quantum Channels: Rank Loss versus Contraction, arXiv:2607.11762v2, Section III, prove that two channels with a common output space and equal adjoint kernels conceal the same POVM pairs, and state:

Whether equality of adjoint kernels is also necessary for concealment-equivalence remains open.

The definitions are quoted in the literature note.

  • A pair of POVMs is concealed by a channel $\mathcal E$ relative to an input set $\mathcal T$ if compatible POVMs reproduce its statistics on every output $\mathcal E(\rho)$, $\rho\in\mathcal T$.
  • The adjoint kernel is $\ker(\mathcal E^\dagger)$ on Hermitian operators.

Issue #14233 reads the affirmative answer with the hypothesis in its strongest form. For all finite dimensions, all trace-preserving Kraus channels $\mathcal E_1,\mathcal E_2$ with common input and output spaces, and every tomographically complete $\mathcal T$: if the two channels conceal the same pairs for every pair of finite outcome sets, then their adjoint kernels coincide.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ConcealmentKernelNecessityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The separable modified Wasserstein distance can exceed the De Palma–Trevisan oneFor two mixed states in dimension four, the modified separable transport cost is at least 15/16, whereas the modified De Palma–Trevisan cost is at most 9/16.Refuted

The questionPitrik, Quantum Wasserstein distance and its relation to several types of fidelities, arXiv:2506.14523v6, Conjecture 1:

Why it mattersThe source proves the inequality when one state is pure and tests 1000 random pairs.

The ideaThe instance is d = 4, N = 3, with ρ = diag(3,3,1,1)/8 and σ = diag(1,1,3,3)/8.

Our contributionFor two mixed states in dimension four, the modified separable transport cost is at least 15/16, whereas the modified De Palma–Trevisan cost is at most 9/16.

Verification details
Release source recordRefuted in Lean

arXiv

The separable modified Wasserstein distance can exceed the De Palma–Trevisan one

G. Tóth and J. Pitrik, Quantum Wasserstein distance and its relation to several types of fidelities, arXiv:2506.14523v6, Conjecture 1:

{\bf Conjecture 1.} The following inequality seems to hold $\tilde D_{\rm DPT, sep}(\varrho,\sigma)^2 \le \tilde D_{\rm DPT}(\varrho,\sigma)^2$ if $\varrho$ and $\sigma$ are arbitrary mixed states.

The two squared distances and their modified forms are quoted in the literature note.

Issue #19312 fixes the reading: the inequality for every d, every family of Hermitian observables and all density matrices ρ, σ. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SeparableWassersteinComparisonRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Refutation of order-33 minimality for domination root minus fourA connected 21-vertex ordinary-domination witness at minus four refutes the complete order-33 minimality claim.Refuted

The questionSaeid Alikhani and Max Griswold, On the Integer Domination Root Conjecture, arXiv:2608.00109v1, Section 4 item 2, printed page 5 asks:

Why it mattersThe source's Theorem 2.1 already supplies G33, a 33-vertex, 36-edge graph whose domination polynomial has a factor x+4.

The idea~~~text (0,5), (0,6), (0,7), (0,13), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (3,13), (4,13), (5,6), (7,8), (7,9), (7,10), (7,11), (7,12), (13,14), (13,15), (13,16), (16,17), (16,19), (17,18), (19,20) ~~~

Our contributionA connected 21-vertex ordinary-domination witness at minus four refutes the complete order-33 minimality claim.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of order-33 minimality for domination root minus four

Saeid Alikhani and Max Griswold, On the Integer Domination Root Conjecture, arXiv:2608.00109v1, Section 4 item 2, printed page 5 asks:

Is n=33 the minimal order of a graph admitting -4 as a domination root?

Section 1 uses finite nonempty connected simple undirected graphs. An ordinary dominating subset S contains or neighbors every vertex, and each subset is counted exactly once:

$$D(G,x)=\sum_{S\text{ dominates }G}x^{|S|}.$$

Connectedness applies to G; the dominating subsets need not induce connected graphs. There is no degree, planarity, cyclomatic-number or root-multiplicity restriction.

The complete positive claim includes both attainment at order 33 and exclusion at every smaller positive order. Its exact Lean definition is:

def claim : Prop :=
  (∃ G : SimpleGraph (Fin 33),
      G.Connected ∧ ∑ S ∈ domSets G, (-4 : ℤ) ^ S.card = 0) ∧
    (∀ n : ℕ, 1 ≤ n → n < 33 →
      ∀ G : SimpleGraph (Fin n), G.Connected →
        ∑ S ∈ domSets G, (-4 : ℤ) ^ S.card ≠ 0)

Here domSets is DominatingSetAverage.domSets. The designated result is literally "theorem result : ¬ claim", with no parameters or extra hypotheses. Its declaration GID is D5/S3/Combinatorics/Graph/Domination/AlikhaniGriswoldMinOrderRefutation.result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/AlikhaniGriswoldMinOrderRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The delivered argument documents a mathematical mechanism.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025The fixed-polynomial hypergraph Alon-Tarsi boundOne fully unbalanced F3 multihypergraph refutes the two printed fixed-polynomial AT bounds.Refuted

The questionConjectures 4 and 28 of Anholcer et al., Electronic Journal of Combinatorics 33(3), P3.74 (arXiv:2501.00157v2), assert AT(p) ≤ 2 ceil(ed(H)) + 1 for every hypergraph polynomial, respectively every fully unbalanced hypergraph polynomial.

Why it mattersThe proposed density bound applies to the supplied polynomial without coefficient permutations.

The ideaThree full edges on three vertices over F3, each with coefficients (1,1,-1), are jointly admissible.

Our contributionOne fully unbalanced F3 multihypergraph refutes the two printed fixed-polynomial AT bounds.

Verification details
Release source recordRefuted in Lean

Open problem

The fixed-polynomial hypergraph Alon-Tarsi bound

Conjectures 4 and 28 of Anholcer et al., Electronic Journal of Combinatorics 33(3), P3.74 (arXiv:2501.00157v2), assert AT(p) ≤ 2 ceil(ed(H)) + 1 for every hypergraph polynomial, respectively every fully unbalanced hypergraph polynomial. The field is arbitrary; multiple edges are allowed. Each edge has at least two vertices and every incident coefficient is nonzero. Full unbalance requires two unequal incident coefficients on each edge. Density counts induced occurrences over nonempty vertex subsets, and AT uses nonzero maximum-total-degree monomials.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Hypergraphs/AlonTarsiRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Further frontiersCounterexample foundFirst preprint 2025Blecher's reversible four-by-four algebra is not unique up to unitary equivalenceA weighted anticommuting algebra of four-by-four complex matrices disproves unitary uniqueness of reversible algebras.Refuted

Why it mattersBlecher's reversible operator algebras retain their matrix norms when multiplication is reversed.

The ideaApplying claim to witness, witnessalgebra, witnessnoncommutative, and witnessreversible yields result : ¬ claim.

Our contributionA weighted anticommuting algebra of four-by-four complex matrices disproves unitary uniqueness of reversible algebras.

Verification details
Release source recordRefuted in Lean

Open problem

Blecher's reversible four-by-four algebra is not unique up to unitary equivalence

D. P. Blecher, Commutativity of operator algebras, arXiv:2509.17673v2, Example 5.2, ends its construction of the CAR algebra $A={\rm Span}{U,V,UV}\subseteq M_4$ with the sentence in 3commfarx.tex, line 684:

We conjecture that up to unitary equivalence $A$ is the only noncommutative reversible $4 \times 4$ matrix algebra.

The preregistered formal reading takes a complex submodule $b$ of $M_4(\mathbb C)$ to be a matrix algebra when it is closed under multiplication, noncommutative when it contains a noncommuting pair, and reversible when some $m>0$ and complex-linear map $\rho:M_4(\mathbb C)\to M_m(\mathbb C)$ reverse multiplication on $b$ and preserve every block operator norm. UnitarilyEquivalent b carAlgebra means that some $q\in\operatorname{Matrix.unitaryGroup}(\operatorname{Fin} 4,\mathbb C)$ satisfies $x\in b$ exactly when $q^*xq\in carAlgebra$ for every $x$. Thus claim is the universal statement that every such $b$ is unitarily equivalent to the CAR submodule carAlgebra.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HilbertOperators/ReversibleFourByFourNonuniqueness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026The one-negative-eigenvalue bound for Hildebrand's APPT matrices fails in dimension fourAn admissible Hildebrand matrix for a four-by-four bipartite spectrum has at least two negative eigenvalues.Refuted

The questionThe verbatim statements are in the literature note.

Why it mattersThe source's analytic maximum purity of qutrit-qutrit APPT states (§II C) rests on Theorem II.2.

The idea- d = 4, λ = (13, 13, 13, 13, 1, …, 1)/64 (four entries 13/64, twelve entries 1/64).

Our contributionAn admissible Hildebrand matrix for a four-by-four bipartite spectrum has at least two negative eigenvalues.

Verification details
Release source recordRefuted in Lean

arXiv

The one-negative-eigenvalue bound for Hildebrand's APPT matrices fails in dimension four

M. Chen and N. Klco, Susceptibility of qudit-qudit entanglement to quantum noise: insights from the negativity rank, arXiv:2610.10820v1, derive the maximum purity of a qutrit-qutrit APPT state from Hildebrand's spectral criterion.

  • The criterion. For the ordered eigenvalues λ1 ≥ … ≥ λ9 ≥ 0, ρ is APPT iff the two matrices L1, L2 of Eq. (5) are positive semidefinite (Theorem II.1, citing Hildebrand).
  • The rank bound. "If λ⃗ is an ordered vector of qutrit-qutrit density matrix eigenvalues, then L1 and L2 of Eq. (5) have at most one negative eigenvalue" (Theorem II.2, p. 5).
  • The question. "it may be prudent to explore whether components of the proof, e.g., the limited rank available for inequality violation established in Theorem II.2, persists for the increasing number of corresponding inequalities in qudit-qudit systems with d > 3" (§V, p. 16).

The verbatim statements are in the literature note. The corresponding inequalities are Hildebrand's (note): for a pair of orderings (σ₊, σ₋) in Σ±, Λ(σ₊, σ₋) + Λ(σ₊, σ₋)ᵀ ⪰ 0.

Issue #17687 fixes the reading:

  • S₊ = {(k, l) : k ≤ l} and S₋ = {(k, l) : k < l}, with orderings as bijections onto ranks.
  • Σ± is the set of pairs compatible, in the sense of Hildebrand's Definition 1, with some x₁ > … > x_d > 0.
  • L = Λ + Λᵀ has diagonal entries 2λ_{d²+1−σ₊(k,k)} and off-diagonal entries λ_{d²+1−σ₊(k,l)} − λ_{σ₋(k,l)}. For d = 3 the two pairs of Σ± give L1 and L2.
  • The property: for every d > 3, every ordered spectrum (λ_k ≥ 0, Σλ_k = 1, no APPT condition, as in Theorem II.2) and every pair in Σ±, L has at most one negative eigenvalue counted with multiplicity.

The property is refuted in this reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/AbsolutePPT/HildebrandNegativeIndexRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Further frontiersConjecture refutedFirst preprint 2025CayleyPy Growth Conjecture 1 is falseA polynomial-size generator family with non-quasipolynomial Cayley diameter.Refuted

The questionChervov et al., CayleyPy Growth (arXiv:2509.19162v2), Section 3.2, conjecture that every generator family of Sn output by an algorithm of polynomial complexity in n has an eventually linear or quadratic quasipolynomial Cayley-graph diameter.

Why it mattersThe conjecture quantifies over every polynomial-time output family, so a single explicit family with non-quasipolynomial diameter settles its literal statement.

The ideaAt nonsquare sizes use Tn; at square sizes use every product of at most three elements of Tn.

Our contributionA polynomial-size generator family with non-quasipolynomial Cayley diameter.

Verification details
Release source recordRefuted in Lean

Open problem

CayleyPy Growth Conjecture 1 is false

Chervov et al., CayleyPy Growth (arXiv:2509.19162v2), Section 3.2, conjecture that every generator family of S_n output by an algorithm of polynomial complexity in n has an eventually linear or quadratic quasipolynomial Cayley-graph diameter.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/CayleyGrowth/CayleyPyConjectureOneRefutation.cayleyPy_conjecture1_refuted

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The delivered argument documents a mathematical mechanism.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026CayleyPy-4 Conjecture 15 fails at k = 5, L = 4, N = 13The k = 5 clause of Conjecture 15 of CayleyPy-4 is false: without inverses, rotations of five consecutive letters need at least 9 moves to turn 0000111111111 into its reversal, whichever way the rotation acts, while the clause gives 8.Refuted

The questionCayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph SN / (SL × S{N−L}) on binary words with L zeros and N − L ones, generated by the consecutive k-cycles (i, …, i + k − 1).

The ideaThe inversion count of a word is the number of positions p < q holding a one at p and a zero at q.

Our contributionThe k = 5 clause of Conjecture 15 of CayleyPy-4 is false: without inverses, rotations of five consecutive letters need at least 9 moves to turn 0000111111111 into its reversal, whichever way the rotation acts, while the clause gives 8.

Verification details
Release source recordRefuted in Lean

Open problem

CayleyPy-4 Conjecture 15 fails at k = 5, L = 4, N = 13

CayleyPy-4 (arXiv:2603.22195v1) studies the Schreier coset graph S_N / (S_L × S_{N−L}) on binary words with L zeros and N − L ones, generated by the consecutive k-cycles (i, …, i + k − 1). Conjecture 15 treats generators without their inverses and gives the diameters as quasipolynomials in n and t = n − (l + 1); its k = 5 clause reads:

For $k=5$ with period~$4$ (again $t = n - (l+1)$, for all $L \ge 4$, starting from $t+1 \ge 9$ that is, $N \ge L + 9$): $D(L,N) = L \left\lfloor \frac{t+1}{4} \right\rfloor + \mathbf{1}_{(t+1) \equiv 0 ; (\mathrm{mod}\ 4)}$.

Issue #10253 fixes the readings: vertices and the central state as in #10230; a directed move rotates a window of five consecutive letters one place, to the left in one convention for the action of a cycle on a word and to the right in the other; the diameter is read both as the largest directed distance from the central state and as the largest directed distance between two vertices. The formal claim is that the clause holds for some convention and some reading.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/ShrunkenGrassmannianConjectureFifteenRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Existing proof account records a substantive content argument.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Geometry & algebraCounterexample foundFirst preprint 2025Coşkun–Eden Remark 5.9: the octonion clauseThe exceptional seven-dimensional Vidinli product is not a linear push-forward of octonion multiplication.Refuted

The questionNeither map is assumed multiplicative or unital, and no retraction identity is assumed. result proves ¬ claim.

Our contributionThe exceptional seven-dimensional Vidinli product is not a linear push-forward of octonion multiplication.

Verification details
Release source recordRefuted in Lean

Open problem

Coşkun–Eden Remark 5.9: the octonion clause

Remark 5.9 of Coşkun–Eden, arXiv:2511.09395v1, asks whether the exceptional Vidinli algebra $V_7$ has a push-forward description using octonion multiplication. The affirmative assertion is the existence of real-linear maps $\nu:V_7\to\mathbb O$ and $\pi:\mathbb O\to V_7$ such that $a\mathbin{\veebar_7}b=\pi(\nu(a)\nu(b))$ for every $a,b$. The preregistration is #20661.

Statement echo

vidinliMul is the coordinate expression of Definition 5.1, with zero-based indices in Fin 7 → ℝ; index $i$ denotes the paper's coordinate $i+1$. claim quantifies over both arbitrary real-linear maps and every pair of vectors. Neither map is assumed multiplicative or unital, and no retraction identity is assumed. result proves ¬ claim. The octonion carrier, multiplication and multiplicative square norm are reused from the closed source module D5/S3/VertexAlgebra/MonsterTwistedOctonionNorm. That owner has no frozen state record on the searched dev revision. The source's higher Cayley–Dickson clause is a separate question.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/CoskunEdenVidinliOctonion.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Continued-fraction tails with u = 2 and v = 1 miss the Beatty sequencePositive continued-fraction tails need not round down to their asymptotic Beatty line.Refuted

The questionJoshi, On Cloitre's hiccup sequences, arXiv:2507.16956v4, Section 4 after Equation (14), state the two-parameter family

Why it mattersThe source's generalization would transfer an asymptotic estimate into an exact floor identity.

The ideaTherefore the universal rounding assertion is false.

Our contributionPositive continued-fraction tails need not round down to their asymptotic Beatty line.

Verification details
Release source recordRefuted in Lean

Open problem

Continued-fraction tails with u = 2 and v = 1 miss the Beatty sequence

R. Fokkink and G. Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956v4, Section 4 after Equation (14), state the two-parameter family

$$s_n=nu+v+\frac{n^2}{s_{n+1}}$$

and its asymptotic line

$$\alpha=\frac{u+\sqrt{u^2+4}}2,\qquad \beta=\frac{v\alpha^2-\alpha}{\alpha^2+1}.$$

They state that positive tails satisfying $s_n=\alpha n+\beta+O(1/n)$ round down to the Beatty sequence with slope $\alpha$ and offset $\beta$. The settling declaration is D5.S1.Recurrence.ContinuedFractionTailBeattyRefutation.result.

Statement echo

The Lean beta, IsTail and claim definitions retain the source's radical slope, offset, positive-tail recurrence, uniform $C/n$ error bound, and floor equality. The claim body is displayed in the Scribe and the result proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/ContinuedFractionTailBeattyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The work stored in a stabilizer battery is not a function of the degree differenceFourteen-qubit circulant stabilizer batteries with degree pairs (4,8) and (2,6) have the same degree difference but store different work at time π/4.Refuted

The questionSen(De), Quantum advantage unlocked: Charging quantum batteries with K-regular graph stabilizers, arXiv:2512.22908v2, Section III, Remark:

Why it mattersThe paper compares stabilizer batteries and chargers of different graph degrees and looks for a quantum advantage in the stored work.

The ideaThe proof uses spectral estimates.

Our contributionFourteen-qubit circulant stabilizer batteries with degree pairs (4,8) and (2,6) have the same degree difference but store different work at time π/4.

Scope & next questionThese overlaps depend on the two neighbourhoods themselves, not only on the difference of their radii.

Verification details
Release source recordRefuted in Lean

arXiv

The work stored in a stabilizer battery is not a function of the degree difference

D. Ghosh, T. K. Konar, G. Francica, A. K. Pal and A. Sen(De), Quantum advantage unlocked: Charging quantum batteries with K-regular graph stabilizers, arXiv:2512.22908v2, Section III, Remark:

While our initial investigation with $N\leq 10$ indicates that $W_{(N,K^\prime,K)}$ remains unchanged as long $\Delta = |K-K^{\prime}|$ is constant for $N\neq 4N^\prime$ with $N^\prime=1,2,\cdots$, no such trend is observed in the case of $N\neq 4N^\prime$. However, concrete conclusion requires extensive numerical analysis with large $N$, which is challenging.

Here:

  • Hamiltonians: $H_{(N,K)}=\sum_iH_{(i,K)}$ is the stabilizer Hamiltonian of the circulant $K$-regular graph on $N$ sites, $H_{(i,K)}=[\bigotimes_{j\le K/2}Z_{i-j}]X_i[\bigotimes_{j\le K/2}Z_{i+j}]$.
  • Battery and charger: the battery $H_B=H_{(N,K')}$ starts in a ground state $\rho(0)$, and the charger $H_C=H_{(N,K)}$ drives $\rho(t)=e^{-iH_Ct}\rho(0)e^{iH_Ct}$.
  • Work: $W_{(N,K',K)}(t)=\mathrm{Tr}[(\rho(t)-\rho(0))H_B]$.

The definitions are quoted in the literature note.

Issue #20091 fixes the reading: for every $N>2$ with $4\nmid N$ and all even interior degrees with $K'\ne K$, $L'\ne L$ and $|K-K'|=|L-L'|$, every pair of ground states gives $W_{(N,K',K)}(t)=W_{(N,L',L)}(t)$ for all $t\ge0$. Excluding the endpoint degrees $0$ and $K_{\max}$ makes the claim weaker than the literal one. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/StabilizerBatteryDegreeDifferenceRefutation.result

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Quantum informationConjecture refutedFirst preprint 2026A hidden central direction refutes Heib–Bruschi Conjecture 80A hidden central direction refutes Heib–Bruschi Conjecture 80.Refuted

The questionA vertex subset has the ideal-graph-property when no edge leaves it.

Why it mattersThe paper reads centers, derived series and ideals off these graphs for Lie algebras that arise in quantum dynamics.

Our contributionA hidden central direction refutes Heib–Bruschi Conjecture 80.

Verification details
Release source recordRefuted in Lean

Open problem

A hidden central direction refutes Heib–Bruschi Conjecture 80

T. Heib and D. E. Bruschi, On the structural properties of Lie algebras via associated labeled directed graphs, arXiv:2601.16161v1, attach to a minimal-graph-admissible basis of a Lie algebra (every bracket of two basis vectors is a multiple of a basis vector) the directed graph with an edge from $x_j$ to $x_\ell$, labelled $x_k$, whenever $[x_j,x_k]$ is a nonzero multiple of $x_\ell$. A vertex subset has the ideal-graph-property when no edge leaves it. Conjecture 80 (§IV.D, p. 46) states that if such an algebra has a nonzero center, then some proper non-empty vertex subset has the ideal-graph-property and spans a solvable ideal. The verbatim statements are in the literature note.

Issue #15636 fixes the reading over every field, as the paper's convention "We denote any field with the symbol $\mathbb F$" requires, for the given minimal graph $G(V,E)$ of the conjecture.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/HeibBruschiHiddenCenterRefutation.result

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Discrete structuresCounterexample foundKimberling's A341249 strict bound is falseThe term n = 2 refutes the upper bound in the OEIS A341249 conjecture.Refuted

The questionClark Kimberling's OEIS A341249, NAME and COMMENTS, revision 10, states:

Why it mattersThe delivery contains exactly the four definitions needed to state this conjecture and its single refutation theorem.

Our contributionThe term n = 2 refutes the upper bound in the OEIS A341249 conjecture.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Kimberling's A341249 strict bound is false

Clark Kimberling's OEIS A341249, NAME and COMMENTS, revision 10, states:

a(n) = floor(r*floor(s*n)), where r = 2 + sqrt(2) and s = sqrt(2).

Conjecture: 1 < r*s*n - a(n) < 2 for n >= 1.

This is the tier-1 external conjecture preregistered in https://github.com/the-omega-institute/trureturing/issues/18673.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | r = 2 + sqrt(2) | r | Real constant $2+\sqrt 2$. | | s = sqrt(2) | s | Real constant $\sqrt 2$. | | a(n) = floor(rfloor(sn)) | a | Natural index; both floors integer valued; products real. | | for n >= 1 | claim | All natural $n$ satisfying $1\le n$. The formula-defined value at zero imposes no extra condition. | | 1 < rsn - a(n) < 2 | claim | Conjunction of both strict inequalities, with integer values cast to real. | | Refutation | result | Negation of that complete universal conjunction. |

The namespace is D5.S0.Certificates.KimberlingA341249BoundRefutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KimberlingA341249BoundRefutation.result

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Quantum informationCounterexample foundFirst preprint 2026The Maity–Furuichi–Dutta noncommuting Tsallis bound ordering is falseA noncommuting positive definite qubit pair reverses the proposed Tsallis bound order.Refuted

The questionSrikrishna Maity, Shigeru Furuichi and Supriyo Dutta, r-deformed α-z-Rényi relative entropy, arXiv:2607.01805v1, equations (8)–(9), printed page 2:

Why it mattersThe two proposed upper bounds need not have the claimed relative order.

The ideaThese are steps within the proof of result, not separate public statements.

Our contributionA noncommuting positive definite qubit pair reverses the proposed Tsallis bound order.

Verification details
Release source recordRefuted in Lean

arXiv

The Maity–Furuichi–Dutta noncommuting Tsallis bound ordering is false

Srikrishna Maity, Shigeru Furuichi and Supriyo Dutta, r-deformed α-z-Rényi relative entropy, arXiv:2607.01805v1, equations (8)–(9), printed page 2:

For 0 ≤ α < 1 and q > (1 − α), we compare two upper bounds B₁(ρ||σ) = D^{(r)}{α,z}(ρ||σ) and B₂(ρ||σ) = −Tr[ρ ln{1−α}((ρ^{−q/2} σ^q ρ^{−q/2})^{1/q})] of the Tsallis relative entropy, numerically. We observe that for non-commutative density operators ρ and σ, D_α(ρ||σ) ≤ B₁(ρ||σ) ≤ B₂(ρ||σ).

Section 4, Case 5, printed page 12:

Let Tr(ρ) = 1 and Tr(σ) = 1, that is both ρ and σ are density operators. Here, we numerically observe that for any two operators ρ and σ, B₁(ρ||σ) is a tighter bound of Tsallis relative entropy than B₂(ρ||σ).

The target is the second inequality, $B_1\le B_2$. Preregistration #18947 supplies the source quotations, parameter conventions and literature screen.

Statement echo

| Source clause | Formal counterpart | Scope | | --- | --- | --- | | Definition 1: $\ln_r(x)=(x^r-1)/r$ | lnr | The literal real-power expression; $r\ne0$ is imposed by claim. | | Definition 2: $D^{(r)}{\alpha,z}$ is the deformed logarithm of the trace of the powered sandwich, divided by $\alpha-1$ | renyiRZ | The source's $B_1$, using cfc and the real trace; no separate forwarding definition. | | Equation (6): $D\alpha=\operatorname{Tr}[\rho-\rho^\alpha\sigma^{1-\alpha}]/(1-\alpha)$ | tsallis | Literal expression, with real trace. | | Equations (8), (47): the inverse-power sandwich and negative weighted deformed-log trace | B2 | Literal $B_2$, including the order of factors. | | Noncommuting density operators | claim quantifies $n$, two complex $n\times n$ matrices, positive definiteness, trace one and noncommutation | Positive definiteness is a sufficient subclass of density operators; it satisfies support and negative-power conditions. Refutation in this subclass refutes the stated assertion. | | $0\le\alpha<1$, $q>1-\alpha$; Definitions 1–2 give $r\ne0$, $z>0$ | All four real parameters and all five hypotheses in claim | The counterexample uses plotted parameters $\alpha=r=1/2$, $z=2$, $q=1$. | | $D_\alpha\le B_1\le B_2$ | Both conjuncts of claim | result negates the full conjunction by contradicting its second conjunct. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/MaityFuruichiDuttaOrderingRefutation.result

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Discrete structuresConjecture refutedRefutation of the A000364 pure-periodicity conjectureThe Euler secant numbers refute pure periodicity with period dividing phi(27).Refuted

The questionEuler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

Why it mattersThis first-tier OEIS conjecture was entered as a comment in 2023.

The ideaThe escape content is strong induction reducing the integer recurrence a 0 = 1, a n = Σ{j=1..n} (−1)^{j+1} C(2n,2j) a(n−j) into ZMod 27 at every index.

Our contributionThe Euler secant numbers refute pure periodicity with period dividing phi(27).

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A000364 pure-periodicity conjecture

OEIS A000364, NAME (verbatim from Library/Recurrence/bala2023a000364.md):

Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

COMMENT (verbatim; attributed in that Library note to Peter Bala, May 08 2023):

Conjecture: taking the sequence [a(n) : n >= 1] modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/SecantNumberPurePeriodRefutation.bala_conjecture_false

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Number theoryConjecture refutedRefutation of Cloitre's A090825 Bernoulli-integrality conjectureThe integer 833 refutes Cloitre's Bernoulli-integrality conjecture for A090825.Refuted

The questionNonprimes n such that (3/2)(1/n)(2n+1)(3^n+1)B(2n) is an integer, where B(k) denotes the k-th Bernoulli number.

Why it mattersCloitre's comment asserts a universal sufficient condition for membership in A090825.

The ideaIts prime divisors are 7 and 17; the corresponding values 27+1=15 and 217+1=35 are composite, so the prime-factor premise holds.

Our contributionThe integer 833 refutes Cloitre's Bernoulli-integrality conjecture for A090825.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Cloitre's A090825 Bernoulli-integrality conjecture

OEIS A090825, NAME (verbatim):

Nonprimes n such that (3/2)(1/n)(2n+1)(3^n+1)B(2n) is an integer, where B(k) denotes the k-th Bernoulli number.

Benoit Cloitre's COMMENT of February 11, 2004 (verbatim):

Conjecture: composite numbers with all prime factors in A053176 are in the sequence.

For natural numbers, define F(n) := (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) in the rationals and A053176(p) := Prime(p) and not Prime(2*p+1). The literal refuted statement is: for every composite natural n > 1, if every prime divisor p of n satisfies A053176(p), then F(n) is an integer.

The separate COMMENT clause asserting F(p) = 1 (mod p) for primes, and the question whether the subsequence beginning 91, 247 is finite, are not claimed or assessed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/CloitreBernoulliIntegralityRefutation.result

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Discrete structuresCounterexample foundThe printed index range of the sixth A329369 sumThe rows at powers of two are linear plus quadratic, which contradicts a printed sum.Refuted

The questionWrite a(n) for A329369 and T(n,k) for the A373183 row coefficients.

Why it mattersThe two entries define a(n) by a(2n+1)=a(n) and, for positive n, a(2n)=a(n)+a(n-2^f(n))+a(2n-2^f(n)) with f the two-adic valuation, and they define T through the row polynomials R(2n+1,x)=xR(n,x), R(2n,x)=x(R(n,x+1)-R(n,x)), R(0,x)=x.

The ideaThe rows at a positive power of two have the closed form R(2^k, x) = (2^k - 1)x + 2^kx^2, proved by induction: the difference operator sends the coefficient pair (a, c) to (a + c, 2c), and the row at index two is x + 2x^2.

Our contributionThe rows at powers of two are linear plus quadratic, which contradicts a printed sum.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The printed index range of the sixth A329369 sum

Write a(n) for A329369 and T(n,k) for the A373183 row coefficients. Mikhail Kurkov printed on June 05 2024 the assertion that

a(2^mn + q) = Sum over i from A001511(n+1) to A000120(n)+1 of T(n, i) * a(2^m(2^(i-1)-1) + q)

holds for n at least zero, m at least zero, and q at least zero. The same assertion appears in A373183 as its second conjecture. The question is whether the printed range for q is correct.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Parity/DyadicPowerRowClosedForm.not_kurkovRowRecurrence

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Discrete structuresConjecture refutedHanna's square-residue conjecture for OEIS A380680Hanna's square-residue conjecture for OEIS A380680 fails at n = 1.Refuted

The questionA(x) = Sum{n=-oo..+oo} x^(n^2) (2 - x^n)^n/(1 - 2x^n)^n.

Why it mattersThe conjecture would locate exactly the positive coefficients with residue two modulo four.

The ideaAt degree one, the finite sum contains only index one.

Our contributionHanna's square-residue conjecture for OEIS A380680 fails at n = 1.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Hanna's square-residue conjecture for OEIS A380680

Paul D. Hanna's OEIS A380680 states:

G.f. A(x) = Sum_{n=-oo..+oo} x^(n^2) * (2 - x^n)^n/(1 - 2*x^n)^n.

Conjecture: for n > 0, a(n) == 2 (mod 4) iff n is a square, and a(n) is divisible by 4 when n is nonsquare.

Formula (2) is:

A(x) = 1 + 2*Sum_{n>=1} x^(n^2) * (2 - x^n)^n/(1 - 2*x^n)^n.

The source's comment says each summand is invariant under sign reversal of its integer index. The zero summand is one. The citation is the Library note. Preregistration #18667 classifies this explicitly published conjecture as tier 1.

Statement echo

All Lean names below are in D5.S0.Certificates.OeisA380680BilateralModFourRefutation.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Bilateral generating function and formula (2) | a takes the rational power-series coefficient of the unilateral expression | The source's equal opposite summands and zero term supply the folding convention. | | Coefficient at degree $m$ | Finite sum over $1\le n\le m$ in a m | The factor $X^{n^2}$ makes larger indices irrelevant to that coefficient. | | Positive index | claim quantifies natural $n$ with $0<n$ | Degree zero is excluded. | | Congruence of an integer coefficient | Integer $z$ with $(z:\mathbb Q)=a(n)$ | This makes the source's integer congruence explicit; it adds a conclusion, not a hypothesis. | | Residue two iff square | z % 4 = 2 iff IsSquare n | Both directions use the same integer coefficient. | | Divisibility at nonsquares | Non-square $n$ implies $4\mid z$ | The second source clause is retained. | | Refutation | result negates all of claim | The positive square index one suffices. |

No source clause is dropped. The two public definitions retain the literal binding expressions from #18667; there is no additional source hypothesis.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/OeisA380680BilateralModFourRefutation.result

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Number theoryCounterexample foundA400168 has a term greater than oneA400168 has a term above one: at n = 2^49 the term is two.Refuted

The questionAntti Karttunen, OEIS A400168, revision 8, September 23, 2026:

Why it mattersThe factorization sum and maximum over all prime-power divisors retain the original component sequences, including their zero and one conventions.

The ideaThe proof uses factorization.

Our contributionA400168 has a term above one: at n = 2^49 the term is two.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

A400168 has a term greater than one

Antti Karttunen, OEIS A400168, revision 8, September 23, 2026:

Question: Are there terms greater than 1?

The NAME is a(n) = A235224(A003415(n)) - A400164(n), with offset one. Let $P_k$ be the product of the first $k$ primes and $P_0=1$. Set $L(0)=0$ and, for $m>0$, let $L(m)$ be the largest $k\ge1$ with $P_{k-1}\le m$, equivalently the least $k\ge0$ with $m<P_k$. Set $D(0)=D(1)=0$ and $D(n)=\sum_{p\mid n}v_p(n)(n/p)$. Let $M(1)=0$ and, for $n\ge2$, $M(n)=\max_{p^e\mid n,\ p\text{ prime},\ e\ge1} L(n/p^e)$. One is not a prime power. The question is exactly whether $\exists n\ge1,\ L(D(n))>M(n)+1$.

The closed claim is the universal negative answer $\forall n\ge1,\ L(D(n))\le M(n)+1$. Karttunen asked the existence question; he did not assert this bound. result : Not claim answers the original question positively.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/OeisA400168PrimorialGap.result

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Quantum informationCounterexample foundFirst preprint 2025The optimal W-state PPT measurement is outside SEPThe optimal three-qubit W-state PPT measurement is not fully separable.Refuted

The questionRossetti, Mancini, Winter and Schindler, Observational entropy of quantum correlations and entanglement, arXiv:2510.10058v1, Section X.0.4, ask:

Why it mattersSEP membership of this optimal PPT measurement would give the analytic SEP observational entropy claimed conditionally in the source.

The ideaFor a hypothetical finite positive-product sum, positivity makes every summand's four quadratic forms vanish separately.

Our contributionThe optimal three-qubit W-state PPT measurement is not fully separable.

Scope & next questionIts proposed route to a SEP entropy formula through this specific POVM is unavailable; the optimal SEP value, a strict gap, and other attaining measurements remain open here.

Verification details
Release source recordRefuted in Lean

Open problem

The optimal W-state PPT measurement is outside SEP

Rossetti, Mancini, Winter and Schindler, Observational entropy of quantum correlations and entanglement, arXiv:2510.10058v1, Section X.0.4, ask:

Nevertheless, none of the tests conducted so far have succeeded in detecting entanglement, leaving open the question of whether (M′, 𝟙 − M′) belongs to the SEP class.

The source and its equations (62), (79), and (83) are identified in the literature note. The positive membership assertion is refuted under the literal conventions of preregistration #19016.

Statement echo

| Source clause | Lean counterpart | Meaning | | --- | --- | --- | | Three ordered qubits | Fin 3 → Fin 2 | Eight computational basis configurations | | Normalized W and anti-W vectors | W3, W3bar | The source's three-term vectors with real coefficient $\sqrt{1/3}$ | | Weight-one and weight-two subspaces | P1, P2 | Sums of the three corresponding basis projectors | | Six invariant projectors | Q1 through Q6 | The six right-hand sides on printed page 22 | | Equation (83) | Mprime | Coefficients $1,1/2,0,0,2/3,1/12$ | | Equation (62): effects proportional to separable states | IsSEPEffect | A nonnegative real multiple of IsFullySeparableLocal, with three sites and two local basis states | | Both effects belong to SEP | claim | IsSEPEffect Mprime and IsSEPEffect (1 - Mprime) | | Negative answer | result | The negation of the entire conjunction |

Pi.single and Matrix.vecMulVec supply the basis vectors and rank-one operators. The existing PronkoFredkinNonCyclicAnnihilation.tensor supplies the tensor product in the proof. These choices preserve the source expressions. The real scalar and density-state representation follow the source's explicit gloss of equation (62); they impose no extra hypothesis on the asserted membership.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/RossettiW3PPTMeasurement.result

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Quantum informationCounterexample foundFirst preprint 2025Noisy entanglement curvature can exceed quantum Fisher informationA two-qubit amplitude-damped probe starting at diag(1,11,3,1)/16 has smooth concurrence curvature 1 and quantum Fisher information 8/15 at t=1, kappa=log 2 and g=pi/4.Refuted

The questionGray, Quantum Fisher Information and the Curvature of Entanglement, arXiv:2504.13729v4, Section IV.B, p. 9:

Why it mattersA bound on entanglement curvature by QFI would connect the sharpness of dynamically generated entanglement to metrological sensitivity.

Our contributionA two-qubit amplitude-damped probe starting at diag(1,11,3,1)/16 has smooth concurrence curvature 1 and quantum Fisher information 8/15 at t=1, kappa=log 2 and g=pi/4.

Verification details
Release source recordRefuted in Lean

Open problem

Noisy entanglement curvature can exceed quantum Fisher information

Z. H. Saleem, A. Shaji, A. M. Babu, D.-W. Luo, Q. Langfitt, T. Yu and S. K. Gray, Quantum Fisher Information and the Curvature of Entanglement, arXiv:2504.13729v4, Section IV.B, p. 9:

For the case with decay also there are more general examples we can consider but in all cases we conjecture that CoE ≤ QFI with equality at specific points.

The source note locates Section II, Eq. (7), the flip-flop Hamiltonian (21), the open evolution (41), the particular examples (42)–(43), and Appendix B. Preregistration #17681 fixes the literal conventions and the unrestricted-initial-density-matrix reading. Section III explicitly introduces losses and mixed probe states after treating pure lossless evolution. The resolution is Refuted.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Two qubits in the computational basis | Matrices indexed by $\mathrm{Fin},2\times\mathrm{Fin},2$ over $\mathbb C$ | The matrix type is written out, with no forwarding abbreviation. | | Appendix B: $\sigma_-=(\sigma_x-i\sigma_y)/2=|1\rangle\langle0|$ | Reused PronkoFredkinNonCyclicAnnihilation.sigmaMinus; jump1, jump2 | The two jumps are $\sigma_-\otimes I$ and $I\otimes\sigma_-$. | | Eq. (21), $\eta_{xy}=1$ | hamiltonian | Only the $01,10$ and $10,01$ entries equal $g$. | | Eq. (41), dissipator and commutator | dissipator, generator | Both jumps have the same positive rate $\kappa$. | | Parameter-independent initial density matrix | claim quantifies over rhoZero; reused StructuredNegativityCoincidenceRefutation.IsDensity | Positive semidefinite, trace one. | | Evolution with that initial state | IsTrajectory | Initial equality for every real $g$ and entrywise time derivatives for every real $g,t$. | | Section II, Wootters concurrence | sigmaYY, concurrence | Characteristic-polynomial roots with multiplicity, real square roots in decreasing order, and $\max(0,\lambda_1-\lambda_2-\lambda_3-\lambda_4)$. | | Eq. (7), $\mathrm{CoE}=-\partial_g^2C$ | derivG and the negative iterated scalar derivative in claim | Fixed observation time. | | Hermitian SLD, $\partial_g\rho=(\rho L+L\rho)/2$ | IsSLD; every such matrix $L$ is quantified in claim | QFI is $\operatorname{Re}\operatorname{Tr}(\rho L^2)$. | | Positive damping and observation time | claim hypotheses $0<\kappa$, $0<t$ | The witness uses $\kappa=\log2$, $t=1$. | | Comparison where curvature is smooth | ContDiffAt of order two in claim | A twice continuously differentiable witness refutes even this restricted comparison. | | In all cases, CoE $\le$ QFI | claim; result proves its negation | The witness coupling is $g=\pi/4$. | | Equality at specific points | No universal equality assertion is added | Failure of the universal inequality refutes the quoted conjunction; the source's particular equality examples are not negated. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SaleemNoisyCurvatureRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The unweighted fourth moment does not determine four-edge forest countsTwo trees have equal Laplacian characteristic polynomials, lower support counts and moments through degree four, but different four-edge path counts.Refuted

The questionBoris Shapiro, “The (n−2,2)-Spectrum of a Graph”, arXiv:2605.17501v2, Section 12, Outlook item 1:

The ideaThe supporting module SupportAndEdgeWordReflection proves twoPointscard by permutation cycle induction and evalMomentsound by sound reflection of the literal edge-word sum.

Our contributionTwo trees have equal Laplacian characteristic polynomials, lower support counts and moments through degree four, but different four-edge path counts.

Verification details
Release source recordRefuted in Lean

arXiv

shapiro 2026 unweighted fourth moment inversion

Boris Shapiro, “The (n−2,2)-Spectrum of a Graph”, arXiv:2605.17501v2, Section 12, Outlook item 1:

Compute an explicit closed formula for the unweighted fourth moment and invert it, modulo Laplacian and cubic data, on the finite list of four-edge support forests. The degree-three inversion is complete by the cubic inversion theorem above.

This settlement refutes the inversion implication SHAP-1 of #14353. It does not assert that an explicit fourth-moment formula is impossible. The source locator is Library/GraphInvariants/shapiro2026spectrum.md.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SupportForestMoments/ShapiroQuarticInversionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Complex hopping can increase parity-twisted ground-state coherencesComplex hopping can increase parity-twisted ground-state coherences.Refuted

The questionSokolovs, Parity superselection obstructs monogamy of mutual information in free fermions, arXiv:2603.14154v2, Section 7:

Why it mattersBy the source's Proposition 1, the D-parity-changing blocks of the fermionic reduced state are those of ρ^tw.

The ideaThe instance. Four ordered modes, with A = {1}, B = {2}, D = {3} and E = {4}.

Our contributionComplex hopping can increase parity-twisted ground-state coherences.

Verification details
Release source recordRefuted in Lean

arXiv

Complex hopping can increase parity-twisted ground-state coherences

A. Sokolovs, Parity superselection obstructs monogamy of mutual information in free fermions, arXiv:2603.14154v2, Section 7:

Theorem~\ref{thm:damping} settles the coherence damping inequality for ground states with real non-positive hopping; whether it extends to systems with complex hopping (e.g., in a magnetic field) or frustrated lattices remains open.

Theorem 1 states |ρ^tw_ij| ≤ ρ^spin_ij = |ρ^spin_ij| for D-parity-changing elements, where ρ^spin is the ordinary partial trace over B and ρ^tw = tr_B[(−1)^{N_B} ρ_ABD]. The verbatim passages are in the literature note.

Issue #19153 fixes the reading: for number-conserving free-fermion Hamiltonians with complex hopping, the ground state satisfies |ρ^tw_ij| ≤ |ρ^spin_ij| for every D-parity-changing pair and every tripartition into adjacent blocks. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/SpinChains/ComplexHoppingCoherenceDampingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Suvagiya Conjecture 28 refutationA signed square-cycle on 32 vertices has radius below the proposed universal optimum.Refuted

The questionConjecture 28 in section 12.3 of arXiv:2607.17343v2 states that for every integer m >= 4, the minimum spectral radius over all independent edge signings of C(8m)(1,2) equals the greatest real root of x^4−2x^3−6x^2+12x−4.

Why it mattersThe concrete counterexample has 32 vertices and does not belong to the periodic gauge slice supplying Theorem 26's upper bound.

The ideaThe step-two signs repeat (1,1,−1,1,−1,−1,1,−1), with b30 = −1 and b31 = +1.

Our contributionA signed square-cycle on 32 vertices has radius below the proposed universal optimum.

Verification details
Release source recordRefuted in Lean

Open problem

Suvagiya Conjecture 28 refutation

Conjecture 28 in section 12.3 of arXiv:2607.17343v2 states that for every integer m >= 4, the minimum spectral radius over all independent edge signings of C_(8m)(1,2) equals the greatest real root of x^4−2x^3−6x^2+12x−4. The radius is the attained maximum absolute eigenvalue of the real symmetric adjacency matrix. Both Hamilton-cycle sign products are included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/SuvagiyaSignedSquareCycleRefutation.result

Frozen module record

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryCounterexample foundA320043: an additional square valueA prime row and the Pell equation give an additional square value of OEIS A320043.Refuted

The questionStefano Spezia's sole COMMENTS conjecture in OEIS A320043, revision 43, says: “Conjecture: a(n) is not a perfect square except for n = 1, 6 and 96.” The source and its literal first FORMULA are cited in the source note.

Our contributionA prime row and the Pell equation give an additional square value of OEIS A320043.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

A320043: an additional square value

Stefano Spezia's sole COMMENTS conjecture in OEIS A320043, revision 43, says: “Conjecture: a(n) is not a perfect square except for n = 1, 6 and 96.” The source and its literal first FORMULA are cited in the source note. For $n\ge1$ the sequence is

$$ A(n)=\sum_{k=1}^{n}\frac{(n+1-k)^2k}{\gcd(n+1-k,k)^3}. $$

The assertion is $\forall n\in\mathbb N,\ 1\le n\Rightarrow \forall m\in\mathbb N,\ A(n)=m^2\Rightarrow(n=1\lor n=6\lor n=96)$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | First FORMULA | a, the sum over Finset.Icc 1 n with the literal numerator and gcd-cubed denominator | Exact natural quotients on the source domain | | OFFSET begins at 1 | 1 ≤ n in claim | The empty sum at zero is excluded | | Perfect square | Natural root m, equality a n = m ^ 2 | All nonnegative square roots | | Except for 1, 6, 96 | Three alternatives in claim | No added exclusion or hypothesis | | Refutation | result : ¬ claim | Unchanged source assertion negated |

For every selected summand, put $u=n+1-k$, $v=k$ and $d=\gcd(u,v)$. Both $u$ and $v$ are positive. Writing $u=du_0$ and $v=dv_0$ gives $u^2v/d^3=u_0^2v_0$, so natural division equals the source's exact quotient. This is a prose derivation of the encoding correspondence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/SpeziaA320043SquareValuesRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Discrete structuresQuestion resolvedFirst preprint 2026Bădică–Bădică: the Jaccard triangle inequality on arbitrary latticesReciprocal supermodularity gives the Jaccard triangle inequality on every lattice.Proved

The questionBădică, On the Triangle Inequality for the Jaccard Distance in Arbitrary Lattices, arXiv:2608.18194v1, Section 8, asks whether the generalized Jaccard distance

Why it mattersTheorem 2 assumes a relatively complemented distributive lattice.

The ideaMonotonicity and the two supermodularity assumptions give

Our contributionReciprocal supermodularity gives the Jaccard triangle inequality on every lattice.

Verification details
Release source recordProved in Lean

Open problem

Bădică–Bădică: the Jaccard triangle inequality on arbitrary lattices

C. Bădică and A. Bădică, On the Triangle Inequality for the Jaccard Distance in Arbitrary Lattices, arXiv:2608.18194v1, Section 8, asks whether the generalized Jaccard distance

$$d_{f,J}(A,B)=1-\frac{f(A\wedge B)}{f(A\vee B)}$$

satisfies the triangle inequality on arbitrary lattices under either of two sets of hypotheses: strict positivity, monotonicity, supermodularity and log-submodularity of $f$; or strict positivity, monotonicity and supermodularity of both $f$ and $1/f$. These are the second and fourth questions, preregistered in #20617.

Statement echo

| source clause | Lean counterpart | | --- | --- | | Arbitrary lattice, without distributivity or relative complementation | Universe-polymorphic $L$ with only a Lattice L instance in claim and claimSufficiency | | Strict positivity | IsStrictlyPositive: $\forall X,\ 0<f(X)$ | | Monotonicity | Mathlib Monotone f | | Supermodularity | IsSupermodular: $f(X)+f(Y)\le f(X\vee Y)+f(X\wedge Y)$ | | Submodularity and log-submodularity | IsSubmodular is the reversed sum inequality; IsLogSubmodular is strict positivity conjoined with submodularity of $\log\circ f$ | | Equation (17) | jaccardDist is the literal real-valued meet-over-join quotient above | | Second question | claim, proved by result | | Fourth question, with Theorem 4's positivity and monotonicity | claimSufficiency, proved by resultSufficiency |

The definitions preserve the preregistered source expressions literally. The repeated strict-positivity hypothesis inside log-submodularity is part of that convention. The generic Mathlib IsStrictlyPositive has a corresponding pointwise specialization; its use does not replace the binding source definition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Lattices/BadicaJaccardArbitraryLattices.result

Frozen module record

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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture refutedPosed in 2026The A126156 eventual totient-period conjecture is falseThe square-root secant sequence has no eventual period dividing the totient at modulus four.Refuted

The questionExpansion of e.g.f. sqrt(sec(sqrt(2)x)), showing coefficients of only the even powers of x.

Why it mattersPreregistration 22077 identifies this first-tier external named conjecture.

Our contributionThe square-root secant sequence has no eventual period dividing the totient at modulus four.

Scope & next questionThe finite four-cycle is compatible with allowing period four at this modulus; a universal twice-totient bound remains open.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The A126156 eventual totient-period conjecture is false

OEIS A126156, revision 110, NAME:

Expansion of e.g.f. sqrt(sec(sqrt(2)*x)), showing coefficients of only the even powers of x.

The third COMMENTS entry states:

Conjecture: the sequence {a(n)} reduced modulo a positive integer k is eventually periodic with the period dividing phi(k), where phi(n) = A000010(n). An example is given below. - Peter Bala, Mar 01 2026

The source has offset zero. For a rational formal power series $A$ with $A(0)=1$ and $A(x)^2\cos(\sqrt2x)=1$, put $a_A(n)=(2n)![x^{2n}]A$. The assertion includes existence of such a series and says that, for every such $A$ and every positive natural modulus $k$, there are natural numbers $N,p$ with $p>0$, $p\mid\varphi(k)$ and $a_A(n+p)-a_A(n)=kz$ for some integer $z$ at every $n\ge N$. The delivered claim is this positive universal assertion; result negates it.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Rational expansion of $\cos(\sqrt2x)$ | cosSqrtTwo | At even degree $j$, $(-2)^{j/2}/j!$; at odd degree, zero. | | Square root with constant term one | IsSourceEGF | Both the constant-coefficient condition and the square equation. | | Even-power EGF coefficients; offset zero | Factorial coefficient expressions in claim | The expressions are literal; proofs use the existing IntegralEGFComposition.eCoeff. | | Every positive integer modulus | Natural k and 0 < k | Positive integers have the same domain. | | Eventually | Natural N; every natural n with N ≤ n | No onset is fixed or bounded. | | Period dividing the totient | Natural p, 0 < p, p ∣ Nat.totient k | Every positive divisor is retained. | | Reduced modulo $k$ | Existential integer multiple of $k$ in the rational field | The coefficients' integrality is derived, not assumed. | | The sequence exists | Existential source-series conjunct in claim | A preregistered existence conjunct, not an extra hypothesis on result. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Periodic/SquareRootSecantTotientPeriodRefutation.result

Frozen module record

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Number theoryQuestion resolvedFirst preprint 2025Linear growth for every Gamma-label splitAs noted in Remark 2.7, the sequence grows exponentially fast (faster than the Fibonacci sequence).Proved

The questionMiller and Garrett Tresch, Problems Regarding a Pair of Diophantine Equations, arXiv:2512.12681v1, Section 3.4, p. 11, ask:

The ideaTheir ten edges have polynomial nonnegative-solution certificates and relatively prime endpoints.

Our contributionAs noted in Remark 2.7, the sequence grows exponentially fast (faster than the Fibonacci sequence).

Verification details
Release source recordProved in Lean

Open problem

Linear growth for every Gamma-label split

Hùng Việt Chu, Steven J. Miller and Garrett Tresch, Problems Regarding a Pair of Diophantine Equations, arXiv:2512.12681v1, Section 3.4, p. 11, ask:

As noted in Remark 2.7, the sequence grows exponentially fast (faster than the Fibonacci sequence). Is there a construction involving a strictly increasing sequence that grows slower than the Fibonacci sequence, or perhaps even like n^k for some k?

The question extends Corollary 2.6 to every prescribed split $p\in[0,1]$. The answer is affirmative. The construction has $-9\le a_n-8n\le15$, so $a_n=8n+O(1)$, and its zero-label count $C_N$ satisfies $0\le pN-C_N<5$.

Preregistration: #20691. Source note: Library/ArithUnits/chumillertresch2025diophantine.md.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Section 1: the gcd-reduced first equation has a nonnegative integral solution exactly when the label is zero | Gamma | Literal natural-number equation, integer division, truncated natural subtraction; complementary branch returns one. The source's exactly-one theorem supplies the interpretation for positive inputs. | | Corollary 2.6: count the first $N$ pairs of a sequence indexed from zero | zeroCount | Inclusive indices $1\le n\le N$ and pair $(a_{n-1},a_n)$. | | Every split $p\in[0,1]$ | claim, binders p : ℝ, 0 ≤ p, p ≤ 1 | Both endpoints included. | | A strictly increasing positive integer sequence | claim, existential a : ℕ → ℕ, strict monotonicity and positivity | Index zero and natural-valued entries. | | Growth like $n^k$ for some $k$ | Polynomial-bound conjunct of claim; result chooses exponent one and coefficient 23 | Upper bound for every $n\ge1$. The sharper two-sided linear estimate is private. | | Limiting zero proportion $p$ | Final conjunct of claim and result | Real division, natural index tending to infinity. |

The definitions Gamma, zeroCount and claim retain the literal preregistered bodies. The proof uses the existing D5.S1.Words.Mechanical.lowerMechanicalLetter p 0 schedule; no second definition of that schedule is introduced. The exponent, coefficient and positivity clauses specify the polynomial-growth question without adding a hypothesis on $p$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Density/ChuMillerTreschGammaSplitGrowth.result

Frozen module record

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    Source 1
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    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Analysis & dynamicsQuestion resolvedFirst preprint 2025Positivity of the JNR triple sumThe JNR triple sum is positive for every configuration of distinct poles with positive weights.Proved

The questionCork and Disney-Hogg, Locations of JNR skyrmions, Appendix A, Remark 13, conjecture:

Why it mattersThis is the tier-1 external named conjecture registered in 19039.

The ideaAll names are in D5.S3.Analytic.JNRRemarkThirteenPositivity.

Our contributionThe JNR triple sum is positive for every configuration of distinct poles with positive weights.

Verification details
Release source recordProved in Lean

Open problem

Positivity of the JNR triple sum

Cork and Disney-Hogg, Locations of JNR skyrmions, Appendix A, Remark 13, conjecture:

The above proof shows that this is positive so long as Σᵢ (λᵢ²/|yᵢ|²) yᵢ = 0, however we conjecture that the sum in (18) is positive for any choice of yᵢ ∈ ℝ³∖{aᵢ}.

The source and equation are recorded in the Library note. The exact statement is preregistered in #19039.

Statement echo

| Source clause | Lean counterpart | Interpretation | | --- | --- | --- | | $N+1$ poles, at least two | N : ℕ, 1 ≤ N, Fin (N + 1) | Every finite size in the stated range | | Pure imaginary quaternion positions in $\mathbb R^3$ | EuclideanSpace ℝ (Fin 3), a, x | The source's Euclidean norm and real inner product | | $\lambda_i>0$, $\mu>0$ | ∀ i, 0 < lam i, 0 < mu | Strict positivity retained | | Distinct poles | Function.Injective a | The source configuration assumption | | $y_i=a_i-x_*$ and the printed exclusion | fun i => a i - x, ∀ i, x ≠ a i | The off-pole interpretation fixed by #19039 | | Equation (3) at zero | rho0 | The full weighted reciprocal-square sum | | Equation (18) | S | All ordered triples, denominator factors and normalization retained | | Positive without condition (15) | result : claim | No balance premise |

No further binder or hypothesis restricts the claim. Division in the definitions is total real division; the theorem's hypotheses make every relevant denominator strictly positive.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/JNRRemarkThirteenPositivity.result

Frozen module record

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    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2024The NHCRB/HCRB ratio for full orthonormal qudit tomography is at most $d+1$Full orthonormal qudit tomography has NHCRB at most d+1 times HCRB.Proved

The questionAssad, Holevo Cramér-Rao bound: How close can we get without entangling measurements?, Quantum 9, 1867 (2025), DOI 10.22331/q-2025-09-30-1867; arXiv:2405.09622v2, Section 3.2.1, after Theorem 3, DraftQuantumSubmission.tex, lines 1147–1155:

Why it mattersThe two bounds compare precision limits in multiparameter quantum estimation.

The ideaThus holevoexact identifies the Holevo cost set with a singleton and evaluates its infimum.

Our contributionFull orthonormal qudit tomography has NHCRB at most d+1 times HCRB.

Scope & next questionThe corresponding extensions remain open here. - Effect on the paper.

Verification details
Release source recordProved in Lean

Open problem

The NHCRB/HCRB ratio for full orthonormal qudit tomography is at most $d+1$

A. Das, L. O. Conlon, J. Suzuki, S. K. Yung, P. K. Lam and S. M. Assad, Holevo Cramér-Rao bound: How close can we get without entangling measurements?, Quantum 9, 1867 (2025), DOI 10.22331/q-2025-09-30-1867; arXiv:2405.09622v2, Section 3.2.1, after Theorem 3, DraftQuantumSubmission.tex, lines 1147–1155:

Although we prove an upper bound of~$d+2$, based on numerical evidence shown in Fig.~\ref{fig:estimateallarbitstate}, we expect the attainable bound to be~$d+1$ and expect~$\rat[{\rho_\theta}]$ to be maximised at~$\theta=0$ over~$\Theta$. Thus for the linear GMM model, we propose~$\rat[{\rho_\theta}]\leq d+1$.

The quotation preserves the TeX macros: \rat denotes $\mathcal R^{\mathrm{NH}}$. The source defines $\mathcal R^{\mathrm{NH}}[{\rho_\theta}]=\max_{\theta\in\Theta}\mathrm C_{\mathrm{NHCRB}}[\rho_\theta]/\mathrm C_{\mathrm{HCRB}}[\rho_\theta]$ in Eq. ratiodef, lines 798–801.

For $d\ge2$, put $n=d^2-1$. Let $b_1,\ldots,b_n$ be traceless Hermitian $d\times d$ complex matrices with $\operatorname{tr}(b_jb_k)=\delta_{jk}$, and let $\theta\in\mathbb R^n$. The model is $\rho_\theta=I_d/d+\sum_j\theta_jb_j$, with $\Theta$ consisting of the parameters making $\rho_\theta$ positive semidefinite. Its trace is one. This includes the source's generalized Gell-Mann model, whose normalization is specified in Section 5.2, lines 2071–2074.

The bounds use identity weight and the source's uncentered locally unbiased operators. Precisely:

  • IsLUB requires every $X_k$ to be Hermitian, $\operatorname{tr}(\rho_\theta X_k)=\theta_k$, and $\operatorname{tr}(b_jX_k)=\delta_{jk}$.
  • zMatrix has entries $Z_{jk}=\operatorname{tr}(\rho_\theta X_jX_k)$; imPart takes the entrywise imaginary part and embeds it back into complex matrices. The trace norm is $\lVert A\rVert_1=\operatorname{Re}\operatorname{tr}\sqrt{A^\dagger A}$.
  • IsAdmissible requires Hermitian blocks $L_{jk}=L_{kj}$ and positive semidefiniteness of the matrix with blocks $L_{jk}-X_jX_k$, indexed by parameter–quantum pairs.
  • hcrb is $\inf_X{\operatorname{Re}\operatorname{tr}Z+\lVert\operatorname{Im}Z\rVert_1}-\sum_j\theta_j^2$, over IsLUB families; nhcrb is $\inf_{X,L}\operatorname{Re}\sum_j\operatorname{tr}(\rho_\theta L_{jj})-\sum_j\theta_j^2$, over IsLUB families and IsAdmissible blocks.

These are the infimum forms of the source's Eqs. HNCRBdefn, origunbiaseq and Holevodefn1, lines 588–602 and 658–666. The definitions are also described in the literature note.

Issue #25805 fixes the settlement as the following multiplicative inequality: for every such $d,b,\theta$ with $\rho_\theta$ positive semidefinite, $\mathrm C_{\mathrm{NHCRB}}[\rho_\theta]\le(d+1)\mathrm C_{\mathrm{HCRB}}[\rho_\theta]$. This is claim, proved by result.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurements/OnbTomographyHolevoRatio.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Trace-distance anticoherence can decrease under covariant spin noiseA covariant spin-four channel decreases order-seven trace-distance anticoherence.Refuted

The questionSerrano-Ensástiga, Total, quantum, and classical measures of anticoherence for mixed spin states (2026), arXiv:2605.29436v1, leave the trace-distance case of axiom (T4) open.

Why it mattersAxiom (T4) is required for a total anticoherence measure in the source's resource-theoretic framework: covariant noise must not lower the measure.

The ideaAll lemma names below refer to AnticoherenceTraceDistanceRefutation unless qualified otherwise.

Our contributionA covariant spin-four channel decreases order-seven trace-distance anticoherence.

Verification details
Release source recordRefuted in Lean

Open problem

Trace-distance anticoherence can decrease under covariant spin noise

J. Denis, T. Lacaille, J. Martin and E. Serrano-Ensástiga, Total, quantum, and classical measures of anticoherence for mixed spin states (2026), arXiv:2605.29436v1, leave the trace-distance case of axiom (T4) open. The downloaded TeX member arxiv.tex, line 363, immediately following Proposition 2, states verbatim:

We also find numerical evidence that axiom~\eqref{ax:T4} remains valid for the trace-distance case $p=1$; a proof is left open.

The TeX label ax:T4 denotes monotonicity under every completely positive trace-preserving channel covariant under the spin representation of $\mathrm{SU}(2)$ (Definition 2, lines 180–200). Identify a spin-$j$ state $\rho$ with a symmetric state $\rho_S$ of $N=2j$ qubits and put $\rho_t=\operatorname{Tr}{N-t}(\rho_S)$. In the normalized Dicke basis the reference state is $\rho_0^{(t)}=I{t+1}/(t+1)$. The source's candidate (lines 333–358) is

$\mathcal A_t^{T,d_1}(\rho)=1-|\rho_t-\rho_0^{(t)}|1/K_t$, with $K_t=\max{\sigma\in\mathcal D(\mathcal H_{t/2})}|\sigma-\rho_0^{(t)}|_1$.

The source uses the full Schatten one-norm. Its mixed-state domain includes every $t\in{1,\ldots,N}$; the footnote at line 178 says:

Whenever we consider pure states, we usually assume $t\le N/2$.

Issue #24441 fixes the universal reading: for every integer $N\ge1$, every integer $1\le t\le N$, every such covariant channel $\Phi$, and every positive semidefinite spin-$N/2$ matrix $\rho$ of trace one, $\mathcal A_t^{T,d_1}(\Phi(\rho))\ge\mathcal A_t^{T,d_1}(\rho)$. This assertion is refuted.

The source definitions and locators are collected in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/AnticoherenceTraceDistanceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The symmetric coin maximizes full-mode entanglement of the Hadamard walk on every finite circleThe symmetric localized coin maximizes full-mode geometric entanglement at every time on every finite circle.Proved

The questionSperling, Multipartite entanglement dynamics in quantum walks (2026), arXiv:2603.24679v1, Section III C 2, “Arbitrary localized initial conditions”, state in arxiv-v1.tex, lines 653–654:

Why it mattersFor each circle size and time, the question asks which localized initial coin produces the most geometric entanglement across the individual optical modes.

The idea1. The supremum and phases. For any finite mode type and any complex amplitude family, overlapbounded bounds the squared overlaps above, and overlapattained supplies unit local factors attaining gMax.

Our contributionThe symmetric localized coin maximizes full-mode geometric entanglement at every time on every finite circle.

Scope & next questionIts conjecture in lines 645–649 about the asymptotic local-maximum great circle on the infinite line remains open here: result supplies neither an infinite-line limit nor a characterization of a contour of local maxima.

Verification details
Release source recordProved in Lean

Open problem

The symmetric coin maximizes full-mode entanglement of the Hadamard walk on every finite circle

E. K. F. Donkersloot, R. Sondenheimer and J. Sperling, Multipartite entanglement dynamics in quantum walks (2026), arXiv:2603.24679v1, Section III C 2, “Arbitrary localized initial conditions”, state in arxiv-v1.tex, lines 653–654:

Hence, we expect this walk to always yield the maximum entanglement. This is indeed the observed behavior for all graph sizes and time steps we have tested.

The preceding sentence, line 652, identifies the initial coin by $\theta=\phi=\pi/2$. The walk, measure and initial states are specified in Sections II A–C and the localized initial-state equation at TeX lines 423–425, respectively; see the literature note.

For $P\ge1$, the modes are $(b,x)\in{0,1}\times\mathbb Z/P\mathbb Z$. The evolution is $T=S(H\otimes I)$, where $H=\frac1{\sqrt2}\begin{pmatrix}1&1\1&-1\end{pmatrix}$ and $S$ shifts coin component $0$ left and component $1$ right, periodically. A normalized coin $c=(\alpha,\beta)$ starts at position zero. Write $\lambda_{P,n,c}=T^n(c\otimes|0\rangle)$ and $c_*=(1,i)/\sqrt2$.

For a finite amplitude family $\lambda$, the encoded single-photon vector is $|\psi_\lambda\rangle=\sum_k\lambda_k|1_k\rangle$, with one tensor factor per mode. The module defines $g_{\max}(\lambda)=\sup_{|z_k|=1}|\langle\bigotimes_k z_k|\psi_\lambda\rangle|^2$, with $z_k\in\mathbb C^2$, and $E_g(\lambda)=1-g_{\max}(\lambda)$.

Issue #22527 fixes the reading: for every natural $P\ge1$, every $n\in\mathbb N$ and every $c\in\mathbb C^2$ with $|\alpha|^2+|\beta|^2=1$, $E_g(\lambda_{P,n,c})\le E_g(\lambda_{P,n,c_*})$. This is proved, including $P=1$ and $n=0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Walks/SymmetricCoinMaximalWalkEntanglement.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Exact preparation-noncontextuality threshold of the twenty-four-cellThe twenty-four-cell preparation subtheory admits a preparation-noncontextual model exactly up to visibility one half.Proved

The questionA two-level system with four mutually unbiased questions, arXiv:2609.10078v1, “Polytopes, quasiprobabilities and contextuality”, after Eq. (sbound), source main.tex line 182:

Why it mattersIt includes all twenty-four preparations, all twelve sharp questions, all mixture equivalences and the zero-visibility collapse.

The ideaIts ontic measure equals the coordinate midpoint measure.

Our contributionThe twenty-four-cell preparation subtheory admits a preparation-noncontextual model exactly up to visibility one half.

Scope & next questionThe separate question of measurement contextuality remains open.

Verification details
Release source recordProved in Lean

arXiv

Exact preparation-noncontextuality threshold of the twenty-four-cell

Jonte R. Hance, Why three? A two-level system with four mutually unbiased questions, arXiv:2609.10078v1, “Polytopes, quasiprobabilities and contextuality”, after Eq. (sbound), source main.tex line 182:

The 24-cell subtheory is therefore preparation contextual at least for $\lambda>2/3$; whether its further equivalences lower the threshold, and whether it is measurement contextual, we leave open, and both are linear programs.

Only the further-preparation-equivalences threshold clause is addressed. Let $V={\pm e_i:1\leq i\leq4}\cup{s/2:s\in{\pm1}^4}$. Take one direction from each of its twelve antipodal pairs as a sharp binary question, with both outcomes. For every real $\lambda\in[0,1]$ the statistics are $p(b\mid n,x)=(1+b\lambda n\cdot x)/2$. All classical convex mixtures are included.

A model has an arbitrary measurable ontic space, countably additive probability preparations and arbitrary measurable nonnegative normalized binary responses. Every pair of operationally equal convex preparation mixtures has equal ontic mixture measures. There is no measurement-noncontextuality or determinism premise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumContext/HanceTwentyFourCellThreshold.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's square-composition congruence for OEIS A372527An integral power-series solution has residue two modulo four exactly at shifted powers of two above degree one.Proved

The questionHanna's OEIS A372527 defines its generating function by:

Why it mattersThis is a first-tier externally stated OEIS conjecture.

The ideaThe compiled proof uses UnitReversionSquareParity.agreepow, agreemul, agreeX, substinneragree, substouteragree and substconstant, together with its Agree definition.

Our contributionAn integral power-series solution has residue two modulo four exactly at shifted powers of two above degree one.

Scope & next questionThe modulo-eight classification remains open here.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's square-composition congruence for OEIS A372527

Paul D. Hanna's OEIS A372527 defines its generating function by:

Expansion of g.f. A(x) satisfying A( A(x)^2 ) = x*A(x)/(1 - 2*A(x)).

Its conjecture reads:

Conjecture: a(2^k + 1) == 2 (mod 4) for k >= 0, and a(n) == 0 (mod 4) for n > 0 when n is not of the form 2^k + 1.

The offset is one and the data start with $a(1)=1$. Accordingly, the zero-residue clause is asserted for $n>1$, as specified in #18859. The literal clause for every $n>0$ is false at $n=1$.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | The displayed generating equation | IsGF | The literal rational formal-power-series equation, with the formal inverse of $1-2A$. | | Offset one, first coefficient one | IsGF requires constant coefficient zero and degree-one coefficient one | These are the source's normalization. | | Existence of the generating series | First conjunct of claim, proved by result | Explicit existence excludes vacuity. | | Every $k\ge0$, coefficient at $2^k+1$ congruent to two | First universal residue clause in claim | An integer witness equals the rational coefficient and is congruent to two modulo four. | | Other indices | Second residue clause in claim, with $n>1$ and no shifted-power representation | An integer witness equals that coefficient and is congruent to zero modulo four. | | A selected generating series | Universal quantification over all series satisfying IsGF | Private exists_integral_solution also proves uniqueness; universality strengthens the assertion. |

The two public definitions are the expressions registered in #18859. Coefficient agreement and the power-of-two series use existing owners rather than new forwarding definitions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaSquareCompositionModFour.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's A382319 congruence modulo threeThe A382319 coefficient at n and the A382318 coefficient at 3n have the same residue modulo three.Proved

The questionHanna's OEIS A382319, Comments (unpaginated), states:

Why it mattersThe congruence links two implicit generating functions rather than two independently chosen coefficient lists.

The ideaThe contraction of agreement below a degree gives existence and uniqueness.

Our contributionThe A382319 coefficient at n and the A382318 coefficient at 3n have the same residue modulo three.

Scope & next questionThe entry's asymptotic formula remains open here, since coefficient residues do not supply an asymptotic estimate.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's A382319 congruence modulo three

Paul D. Hanna's OEIS A382319, Comments (unpaginated), states:

Conjecture: a(n) == A382318(3*n) (mod 3) for n >= 0.

The source equations are $A=X+L(A)^3$ and $B=X(1-X)^{-1}+L(B^3)$, where $L(F)=\sum_{k\ge1}F(X^k)$ and both series have zero constant term. At degree $m$, $L(F)$ has coefficient $\sum_{1\le k\le m,\ k\mid m}[X^{m/k}]F$. The empty degree-zero sum is zero. The cube in the second equation is taken before Lambert summation. The source is recorded in the Library note. Preregistration: #18862.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Two generating equations, with offset one | IsA382318 A, IsA382319 B in HannaLambertCubeModThree | Literal rational equations and zero constant coefficients. | | Infinite substitution sum | lambertSum | Positive-divisor coefficient sum; natural-number division. | | Sequence values | PowerSeries.coeff | Integer witnesses are cast into rational coefficient equalities. | | Nonnegative index | Natural-number binder | Includes zero. | | Reciprocal congruence | Universal conjunct of claim | $b(n)\equiv a(3n)\pmod3$. | | Existence | Existential conjunct of claim | Additional conclusion; no restriction of the source's solution class. | | Every solution pair | Universal quantifiers and the two predicate hypotheses | No selected solution replaces the source's generating functions. |

The congruence is $b(n)\equiv a(3n)\pmod3$ for every natural $n$, including zero. No source clause is dropped. The complete proposition is displayed at both claim and result in the Blueprint.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaLambertCubeModThreeConverse.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedHanna's third congruence for A365771Conjectures: given A033042 is the sums of distinct powers of 5, then (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5A033042(k) for some k >= 0.Proved

The questionConjectures: given A033042 is the sums of distinct powers of 5, then (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5A033042(k) for some k >= 0.

Why it mattersThis is one of the three residue assertions in A365771; the family dossier is hanna365771-mod5.md, and the source note is Library/Digit/hanna2023a365771.md.

The ideaThe same-delivery HannaA365771.digitsstep and HannaA365771.residueclassification supply the digit decomposition and the residue.

Our contributionConjectures: given A033042 is the sums of distinct powers of 5, then (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5A033042(k) for some k >= 0.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's third congruence for A365771

OEIS A365771 revision 23, Comments, states:

Conjectures: given A033042 is the sums of distinct powers of 5, then (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.

Preregistration: #18658.

Statement echo

Every $n>0$ with no $m\geq0$ in A033042 satisfying $n=5m$ or $n-1=5m$ satisfies $a(n)\bmod5=0$. The sequence and digit predicate are exactly D5.S1.Recurrence.Residue.HannaA365771.a and D5.S1.Recurrence.Residue.HannaA365771.InA033042. The claim uses natural-number arithmetic.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Residue/HannaA365771Zero.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026The one-tenth synchronous doily valueThe classical value of Tony Lau's one-tenth synchronous doily game is exactly 22/25.Proved

The questionTony Lau, Beyond the Magic Square Game: Widening the Gap for Two Bell States, arXiv:2603.20748v2, §5, asks whether a value below 31/35 occurs for some p < 1/7 and identifies the one-tenth candidate: if the asymmetric strategy bound in Lemma 4.3 can be strengthened to twelve synchronous wins out of fifteen, then the 1/10-synchronous doily game has classical value 22/25.

Why it mattersThe source leaves the 1/10 value as an explicit exhaustive-check question and records 22/25 as the candidate value.

The ideaThe lower bound uses the explicit strategy returning 000 on equations 0 through 11 and 100 on equations 12 through 14 for both players.

Our contributionThe classical value of Tony Lau's one-tenth synchronous doily game is exactly 22/25.

Scope & next questionThe listed values of the full finite value function and the crossing of the (L,r)=(10,5) and (12,0) strategy lines at p=1/16 are computed by exhaustive search; a formal theorem about the minimum over p remains open.

Verification details
Release source recordProved in Lean

Open problem

The one-tenth synchronous doily value

Tony Lau, Beyond the Magic Square Game: Widening the Gap for Two Bell States, arXiv:2603.20748v2, §5, asks whether a value below 31/35 occurs for some p < 1/7 and identifies the one-tenth candidate: if the asymmetric strategy bound in Lemma 4.3 can be strengthened to twelve synchronous wins out of fifteen, then the 1/10-synchronous doily game has classical value 22/25. The settled statement is

classicalValue (1/10) = 22/25,

where classicalValue is the supremum over all deterministic pairs of functions from the fifteen equations to Boolean triples, the nonsynchronous distribution is uniform over the 90 ordered intersecting pairs, and the synchronous distribution is uniform over the fifteen diagonal pairs. The referee predicate uses the fifteen equations in Table 3, their stated right-hand sides, parity validity, and equality of every shared variable.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SynchronousDoilyClassicalValue.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Number theoryQuestion resolvedPosed in 2026The largest number neither dividing nor coprime to nMathar's A073759 conjecture holds for every natural index at least five.Proved

The questionMathar's conjecture in OEIS A073759, FORMULA (unpaginated), states: "Conjecture: a(n) = 2A046667(n) if n>=5. - R.

Why it mattersThe maximum excludes both divisors and coprime numbers.

Our contributionMathar's A073759 conjecture holds for every natural index at least five.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

mathar oeis a073759 least prime factor

R. J. Mathar's conjecture in OEIS A073759, FORMULA (unpaginated), states: "Conjecture: a(n) = 2*A046667(n) if n>=5. - R. J. Mathar, May 15 2026". The NAME is "Largest number < n that is neither a divisor of nor relatively prime to n, or 0 if no such number exists." The companion entries define A046666 by "a(n) = n - (smallest prime dividing n)." and A046667 by "a(n) = A046666(n)/2." The offsets are respectively 1, 2 and 2. Preregistration: #18937.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Positive number below $n$ | a: the finite interval from 1 inclusive to $n$ exclusive | Literal preregistered convention | | Neither divisor nor relatively prime | a: the conjunction of nondivisibility and failure of Nat.Coprime | Both exclusions retained | | Largest, or zero if empty | a: Finset.sup id of the filtered interval | Natural supremum defaults to zero | | Smallest prime dividing $n$ | A046666: natural subtraction of Nat.minFac n | Exact for the source domain $n\ge2$ | | Divide A046666 by two | A046667: Nat.div by 2 | Integer division; parity supplies exactness | | Every $n\ge5$ | claim, proved by result | No extra hypothesis or conjunct |

All names belong to D5.S3.Arith.Primes.MatharA073759. The definitions extend to natural indices outside the OEIS offsets; the settlement quantifies only over $n\ge5$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Primes/MatharA073759.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2026Sharpness of the high-degree fermionic Gaussian constantsThe fermionic high-degree energy and free-energy constants are sharp.Proved

The questionEisert, Approximation theorems for fermionic Gaussian states, arXiv:2610.01860v1 (2026), Section VI, printed page 26, state:

The ideaThe coordinate Hamiltonian is a normalized sum over all edges.

Our contributionThe fermionic high-degree energy and free-energy constants are sharp.

Scope & next questionIt remains open, however, whether the constants in the resulting high-degree energy and free-energy estimates are optimal, since their derivation also uses edge averaging, Cauchy–Schwarz, and only the operator-norm normalization of the interactions.

Verification details
Release source recordProved in Lean

Open problem

Sharpness of the high-degree fermionic Gaussian constants

A.-R. Negari, F. Salek, Z. Zimborás, A. Harrow, P. Hayden and J. Eisert, Approximation theorems for fermionic Gaussian states, arXiv:2610.01860v1 (2026), Section VI, printed page 26, state:

It remains open, however, whether the constants in the resulting high-degree energy and free-energy estimates are optimal, since their derivation also uses edge averaging, Cauchy–Schwarz, and only the operator-norm normalization of the interactions.

The source estimates are Theorem III.1, equation (44), and Theorem III.5, equation (79), both with prefactor $\sqrt{2m/D}$. The formal claim asks whether, for every $m\ge1$, every $c<1$ and every degree cutoff $D_0$, a finite regular graph and admissible quadratic fermionic edge interactions attain the energy value $\sqrt{2m/D}$ and have a finite positive inverse temperature whose free-energy gap is larger than $c\sqrt{2m/D}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Fermionic/GaussianHighDegreeSharpness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresRecurrence provedSquare values in the A097602 square-counting recurrenceThe square positions and square values in Zumkeller's recurrence follow a nine-term pattern.Proved

The questiona(n+1) = a(n) + number of squares so far; a(1) = 1.

Why it mattersThe square-value conjecture was recorded in 2004 and remained labelled as a conjecture through the bounded source and literature checks recorded in preregistration issue 7497.

The ideaWrite c(n) for the number of square values among a(1),...,a(n).

Our contributionThe square positions and square values in Zumkeller's recurrence follow a nine-term pattern.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Square values in the A097602 square-counting recurrence

OEIS A097602, NAME (%N, verbatim):

a(n+1) = a(n) + number of squares so far; a(1) = 1.

COMMENT (%C, verbatim):

Conjecture: a(n) = m^2 iff m mod 3 > 0.

COMMENT (%C, verbatim):

a(n) is a square iff n is congruent to {1, 4} mod 9. - Vladeta Jovovic, Aug 30 2004

FORMULA (%F, verbatim):

a(9*n+1) = (3*n+1)^2; a(9*n+4) = (3*n+2)^2.

The conjecture is read as a statement about attained values: for every positive m, the value m^2 occurs as a(n) at some positive index n if and only if 3 does not divide m. The literal reading with independent free n and m is false: at n=2 the sequence has a(2)=2, while for example m=1 makes the right side true and a(2)=m^2 false.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/SquareCountingRecurrenceSquarePositions.zumkeller_a097602

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Contribution

    The square positions and square values in Zumkeller's recurrence follow a nine-term pattern.

    Read source ↗
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedThe A140869 Sundaram-type triangle complementLibrandi's triangle complement maps every outside value to a prime 4h+5.Proved

The questionConjecture: If h does not belong to the sequence, then 4h+5 is prime.

Why it mattersThe 2012 OEIS conjecture gives a universal complement-primality statement.

The ideaIf 4h+5 is odd composite, choose odd factors d·e = 4h+5 with d,e ≥ 3, write d = 2a+1 and e = 2b+1, and order them so 1 ≤ n ≤ m after setting the smaller half-factor to n and the larger to m.

Our contributionLibrandi's triangle complement maps every outside value to a prime 4h+5.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The A140869 Sundaram-type triangle complement

OEIS A140869, NAME (%N, verbatim):

T(m,n) = floor((2mn+m+n-2)/2), m >= n >= 1.

COMMENT (%C, verbatim; Vincenzo Librandi, Nov 18 2012):

Conjecture: If h does not belong to the sequence, then 4*h+5 is prime.

The literal claim is ∀ h, ¬(∃ m n, 1 ≤ n ≤ m ∧ T(m,n) = h) → 4h+5 prime. Here T(m,n) and the membership predicate use natural-number truncated subtraction and division. On the stated domain m ≥ n ≥ 1, the numerator 2mn+m+n−2 is at least 2, so these operations agree with the intended expression.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Congruence/LibrandiSundaramTriangleComplementPrime.librandi_a140869

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedOEIS A249621 terminal digits for square concatenationsWu's A249621 conjecture for squares preserved by a decimal prefix and suffix.Proved

The questionSquares that remain squares when prepended with 1 and appended with 25 in base 10.

Why it mattersThis 2014 OEIS conjecture asks for the terminal two digits of every positive square whose decimal prefix and suffix preserve squareness.

The ideaFor d >= 2, the leading power of ten vanishes modulo 10000; cancelling 25 reduces the equation to u^2 = 4z^2 + 1 (mod 400).

Our contributionWu's A249621 conjecture for squares preserved by a decimal prefix and suffix.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

OEIS A249621 terminal digits for square concatenations

OEIS A249621 %N:

Squares that remain squares when prepended with 1 and appended with 25 in base 10.

OEIS A249621 %C:

Conjecture: all the numbers in the sequence end in either 00 or 56.

— Chai Wah Wu, Nov 02 2014.

For a natural number x, membership means that x is a square and, if d is the number of its base-ten digits, then 10^(d+2) + 100x + 25 is also a square. This is the literal base-ten concatenation 1||x||25.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Digit/Admissibility/SquarePrependOneAppendTwentyFive.wu_a249621

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryQuestion resolvedSchulte's Liouville and cube-Möbius product formulaSchulte's A299406 coefficients equal the Liouville function times A210826.Proved

The questionDirichlet g.f.: Sum{n>0} a(n)/n^s = (zeta(s)zeta(6s))/(zeta(2s)zeta(3s)).

Why it mattersThe independent question is Schulte's published product conjecture, relating three coefficient sequences for every positive natural index.

The ideaUse coprime power extraction and the exact-power guard to transfer multiplicativity to the power lifts.

Our contributionSchulte's A299406 coefficients equal the Liouville function times A210826.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Schulte's Liouville and cube-Möbius product formula

OEIS A299406, NAME (%N, verbatim):

Dirichlet g.f.: Sum_{n>0} a(n)/n^s = (zeta(s)*zeta(6*s))/(zeta(2*s)*zeta(3*s)).

FORMULA (%F, verbatim):

Conjecture: a(n) = A008836(n) * A210826(n).

AUTHOR (%A, verbatim):

Werner Schulte, Feb 20 2018

The offset (%O) is 1,1. A008836, NAME (%N, verbatim):

Liouville's function lambda(n) = (-1)^k, where k is number of primes dividing n (counted with multiplicity).

A210826, NAME (%N, verbatim):

G.f.: Sum_{n>=1} a(n)*x^n/(1 - x^n) = Sum_{n>=1} x^(n^3).

Let zeta be the arithmetic function equal to one on positive inputs, mu the Möbius function, and ⋆ Dirichlet convolution. For positive k, liftPow k f has value f(m) at m^k and zero off exact k-th powers. The total Lean definition uses the guard (Nat.floorRoot k n)^k = n. Here Nat.floorRoot divides prime-factor exponents by k using natural-number division; it returns zero if k = 0 or n = 0. This is the root for divisibility, not a real-root approximation.

The coefficient reading of zeta(6s) is liftPow 6 zeta, and that of 1/zeta(ks) is liftPow k mu for k = 2, 3. Hence set A = zeta ⋆ liftPow 6 zeta ⋆ liftPow 2 mu ⋆ liftPow 3 mu. The Lambert-series reading is sum_{d|n} B(d) = [n is a positive cube], where brackets are an indicator. Möbius inversion gives B = mu ⋆ liftPow 3 zeta, representing A210826. Both functions take integer values and vanish at zero.

The exact formal claim is forall n : Nat, 0 < n -> A n = ArithmeticFunction.liouville n * B n. Mathlib's liouville_apply identifies the named function with (-1 : Int) ^ Omega(n) for positive n, matching A008836's %N. Only the quoted A299406 %F line is settled under these coefficient interpretations. Analytic convergence, analytic uniqueness of the series, and other OEIS assertions are outside the claim. The hypothesis 0 < n excludes zero, as required by A299406's offset one.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SchulteLiouvilleCubeMobius.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryConjecture provedThe permutation conjecture for A363956The greedy sequence A363956 is a permutation of the positive integers.Proved

The questionThe sequence is conjectured to be a permutation of the positive integers.

Why it mattersMissing primes in a finite prefix do not decide it.

The ideaInfinitely activated queues exhaust all positive multiples.

Our contributionThe greedy sequence A363956 is a permutation of the positive integers.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

The permutation conjecture for A363956

The sequence is conjectured to be a permutation of the positive integers.

The sequence begins with 1 and 2. Thereafter take the smallest positive unused multiple of the omega-th prime, where omega counts distinct prime factors of the previous term.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/OmegaGreedyPermutation.a363956_surjective

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Discrete structuresConjecture provedHanna's OEIS A375452 diagonal derivative parity conjectureThe diagonal derivative equation has integral coefficients, with odd coefficients at exactly two Mersenne supports above degree one.Proved

The questionHanna's OEIS A375452 asks about the normalized rational power series whose coefficients satisfy

Why it mattersThe settlement answers the explicitly stated A375452 conjecture and reuses the general diagonal parity theorem proved for the sibling A375449 settlement.

Our contributionThe diagonal derivative equation has integral coefficients, with odd coefficients at exactly two Mersenne supports above degree one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Hanna's OEIS A375452 diagonal derivative parity conjecture

P. D. Hanna's OEIS A375452 asks about the normalized rational power series whose coefficients satisfy

Expansion of g.f. A(x) satisfying [x^n] x*A'(x) / (1 - n*A(x)) = n^n for n >= 1.

Its Comments field states:

Conjecture: for n > 1, a(n) is odd iff n is of the form 2^k-2 or 2^k-1 for some k > 1.

The formal statement uses zero constant coefficient and the literal formal derivative and unit-denominator inverse. It asserts existence of a rational solution and integral coefficients with the stated parity support.

Statement echo

| source clause | Lean counterpart | fidelity | | --- | --- | --- | | $[x^n]xA'(x)/(1-nA(x))=n^n$ for $n\ge1$ | D5.S1.Recurrence.Invariants.HannaDiagonalDerivativeParityA375452.IsGF | The formal derivative, coefficient operator, rational scalar, and inverse are literal. | | offset one and normalized solution | constantCoeff A = 0 in IsGF | The constant coefficient condition is explicit. | | odd exactly at $2^k-2$ or $2^k-1$ for $n>1$ | claim and result | The quantifiers and natural subtraction are literal. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Invariants/HannaDiagonalDerivativeParityA375452.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryConjecture provedHuber's prime conjecture for OEIS A394407Every prime other than three first attains its totient-to-divisor-count ratio.Proved

The questionPositive integers k such that there is no m < k with phi(m)/tau(m) = phi(k)/tau(k).

Why it mattersThe source note is Library/Factorization/huber2026a394407.md.

Our contributionEvery prime other than three first attains its totient-to-divisor-count ratio.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Huber's prime conjecture for OEIS A394407

Felix Huber's OEIS A394407 has the Name:

Positive integers k such that there is no m < k with phi(m)/tau(m) = phi(k)/tau(k).

Its Comment states:

Conjecture: All primes p != 3 are in the sequence; equivalently, for every prime p != 3 there is no m < p with phi(m)/tau(m) = phi(p)/tau(p) = (p - 1)/2; equivalently, the sequence is a supersequence of A045344.

The source is an explicitly stated conjecture in an approved OEIS entry. The settled proposition is the first form of that conjecture, preregistered in #18591. For positive integers, write $r(m)=\varphi(m)/\tau(m)$ with rational division.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Quotient phi(m)/tau(m) | ratio m | Nat.totient m divided by m.divisors.card, with both cast to rational numbers. | | Positive integers k | First conjunct of IsTerm k | Zero is excluded. | | No smaller m has the same ratio | Second conjunct of IsTerm k | Every positive natural m less than k is compared; no search bound is imposed. | | All primes p != 3 | claim | Universal natural p, Nat.Prime p, and p different from 3. | | Membership in the sequence | Conclusion IsTerm p of result | The first-occurrence predicate is unchanged. |

The positive domain for m follows the sequence's positive-integer convention. Lean totalizes the ratio at zero to zero; no zero input enters the claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/OeisA394407TotientTauPrimes.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025Direct four-copy universal entanglement concentration beats the bound of Eq. (8)A fixed local singlet-sector measurement on four copies exceeds the conjectured pointwise concentration bound.Refuted

The questionStreltsov, Optimality of universal conclusive entanglement concentration protocols, Phys.

Why it mattersUniversal protocols must work without knowing the Schmidt basis of the input, and the paper quantifies the cost of universality by comparing Eq. (8) with Vidal's state-dependent optimum.

The idea- u and w are the products of two singlets on positions (1, 2), (3, 4) and on positions (1, 3), (2, 4);

Our contributionA fixed local singlet-sector measurement on four copies exceeds the conjectured pointwise concentration bound.

Verification details
Release source recordRefuted in Lean

Open problem

Direct four-copy universal entanglement concentration beats the bound of Eq. (8)

A. C. Orthey, A. Philip, T. V. Kondra and A. Streltsov, Optimality of universal conclusive entanglement concentration protocols, Phys. Rev. A 113, 062404 (2026), arXiv:2509.09423v2, study LOCC protocols that turn copies of an unknown two-qubit pure state into the Bell state ϕ+.

  • Universality. Definition 1: an n-to-1 LOCC map E1 is a universal conclusive protocol if ⟨ϕ+|E1(ρ⊗n)|ϕ+⟩ = Tr E1(ρ⊗n) for every input in the set, here every pure state |ψ⟩ = c1|00⟩ + c2|01⟩ + c3|10⟩ + c4|11⟩ (Eq. (1)).
  • The bound. Theorem 3 bounds the four-copy success probability of concatenated protocols based on two-qubit operations by B(c) = 2|c2⁴c3⁴| + 2|c1⁴c4⁴| − 4Re[c1²c4²(c2*)²(c3*)²] (Eq. (8)).
  • The conjecture. §VIII: "protocols that process four copies directly without intermediate Schmidt-state conversion warrant investigation. We conjecture that such protocols cannot surpass the success probability bound in Eq. (8)".

The verbatim statements are in the literature note.

Issue #17404 fixes the reading: for every universal conclusive LOCC map on four copies and every normalized ψ, Tr E1(ψ⊗4) ≤ B(c).

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/PureState/FourCopyConcentrationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationQuestion resolvedFirst preprint 2025Distinct optimal entanglement witnesses can have an optimal sumDistinct optimal witnesses have an optimal sum and overlapping detection sets.Proved

The questionBae, Mirrored Entanglement Witnesses for Multipartite and High-Dimensional Quantum Systems, arXiv:2510.06863v2, Quantum Sci.

Why it mattersThe source’s generalized mirroring framework allows overlapping detection sets and asks about decomposing non-extremal witnesses into optimal ones (lines 1692–1696).

The ideaUnder the same dimension and index conditions, wrhermitian and wrblockPositive give Hermiticity and block positivity.

Our contributionDistinct optimal witnesses have an optimal sum and overlapping detection sets.

Verification details
Release source recordProved in Lean

Open problem

Distinct optimal entanglement witnesses can have an optimal sum

J. Seong, A. Bera, B. C. Hiesmayr, D. Chruściński and J. Bae, Mirrored Entanglement Witnesses for Multipartite and High-Dimensional Quantum Systems, arXiv:2510.06863v2, Quantum Sci. Technol. 11, 015037 (2026), ask in “Open problems”, item 2:

Optimal EWs can be mirrored via another EW, as shown in Eq. (\ref{eq:gmeew_form}). Since optimal EWs are generally not extremal, yet, it remains unanswered whether a pair of optimal EWs can be mirrored through another optimal EWs.

This quotation is from the downloaded v2 source mn_revision_clean.tex, line 1749, MD5 08daa608d307e41e61bcd9c0d9f06c31. The generalized mirroring relation is $W+M=K$ in lines 1688–1692; the witness definition, detection sets, optimality and spanning criterion are in lines 276–281, 388–396 and 485–490. See the literature note.

Issue #20970 fixes a nontrivial reading: for every natural number $d\ge3$, there are optimal witnesses $W,M$ on $\mathbb C^d\otimes\mathbb C^d$ such that $W+M$ is optimal, each detects a density matrix the other does not, and some density matrix is detected by both. This is proved.

Here a density matrix is positive semidefinite with trace one; a witness is Hermitian, block-positive on every product vector, and has negative real trace pairing with some density matrix. Optimality means that for every real $\epsilon>0$ and every nonzero positive semidefinite $P$, the operator $A-\epsilon P$ is not block-positive. Write $D_A={\rho:\rho\text{ is a density matrix},\ \operatorname{Re}\operatorname{tr}(A\rho)<0}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/Witness/MirroredOptimalWitnessSum.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The proved statement applies to the stated general family, not just a checked finite range.

    Source 1
Discrete structuresQuestion resolvedSquarefree Numbers Have No Partition Whose Parts and Multiplicities Agree in MeanA squarefree number above one has no partition whose parts and whose multiplicities share a mean, because squarefreeness turns the mean condition into a count of distinct parts that exceeds the count of parts unless every multiplicity is one.Proved

The questionNumbers for which there exists an integer partition such that the parts have the same mean as the multiplicities.

The ideaPresent a partition by its set D of distinct parts together with a multiplicity function m.

Our contributionA squarefree number above one has no partition whose parts and whose multiplicities share a mean, because squarefreeness turns the mean condition into a count of distinct parts that exceeds the count of parts unless every multiplicity is one.

Verification details
Release source recordProved in Lean

Integer sequences (OEIS)

Squarefree Numbers Have No Partition Whose Parts and Multiplicities Agree in Mean

OEIS A360070, at revision #18 of Jan 29 2023, by Gus Wiseman, Jan 27 2023:

Numbers for which there exists an integer partition such that the parts have the same mean as the multiplicities.

Conjecture: No term > 1 is squarefree.

The entry's worked example:

A partition of 20 with the same mean as its multiplicities is (5,4,3,2,1,1,1,1,1,1), so 20 is in the sequence.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SquarefreeMeanPartition.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The exact recorded statement is proved.

    Source 1
Number theoryCounterexample foundFirst preprint 2025Refutation of Zhang's character-sum constant restrictionTwo quadratic-polynomial Legendre sums differ by one for every odd prime.Refuted

The questionWenpeng Zhang, Some interesting number theory problems, arXiv:2506.17235v1, printed page 3, states problem (B) and question (D) verbatim as:

Why it mattersIssue 8626 records the source quotation, full quantifiers, literature-status scope, and the proposed quadratic counterexample.

The ideaFor every odd prime p, all terms in the first character sum equal 1, so its value is p-1.

Our contributionTwo quadratic-polynomial Legendre sums differ by one for every odd prime.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Zhang's character-sum constant restriction

Wenpeng Zhang, Some interesting number theory problems, arXiv:2506.17235v1, printed page 3, states problem (B) and question (D) verbatim as:

(B). Whether there are infinitely many pairs of fundamentally different integer coefficients polynomials f(x) and g(x) (That is, (f(x)/p) ≠ (g(x)/p)) such that Σ_{x=1}^{p−1} (f(x)/p) − Σ_{x=1}^{p−1} (g(x)/p) = c, (2) where c is a fixed constant.

(D). Whether the values of c can only be 0 or 2?

Identity (2) compares the Legendre character sums of two fundamentally different integer-coefficient polynomials. Following the preceding corollary, the formal statement uses one constant for every odd prime.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithSums/ZhangCharacterSumDifferenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 2/4 · 15% weight

    The refutation is supported by a documented parameterized argument or family.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Negativity at most (χ − 1)/2 does not imply positivity under the χ-reduction mapA pure two-qutrit state has the attained negativity bound but fails positive reduction.Refuted

The questionSchmidt number at most χ implies N(ρ) ≤ (χ − 1)/2 (Lemma 3). - Reduction. REDκ(ρ) = Tr(ρ)1 − ρ/κ (Eq. (35)).

Why it mattersBoth criteria are necessary conditions for Schmidt number at most χ, and the paper compares them through their spectral versions (Table 1, Eq. (44)).

The ideaSo (ρ^Γ)†ρ^Γ is diagonal with entries (bi bj/18)², and its square root is diagonal with entries bi bj/18.

Our contributionA pure two-qutrit state has the attained negativity bound but fails positive reduction.

Scope & next questionSchmidt number at most κ implies 1 ⊗ REDκ ≥ 0 (Theorem 7). - The question. §5.4: "the complete inclusions of the considered EFSγ sets is only pending for the unknown case NEGFS{(χ−1)/2} ?⊂ PREDFSχ, which is also unknown for the non-FS case."

Verification details
Release source recordRefuted in Lean

arXiv

Negativity at most (χ − 1)/2 does not imply positivity under the χ-reduction map

J. Abellanet-Vidal, G. Müller-Rigat, A. Rico and A. Sanpera, Bounding the entanglement of a state from its spectrum, arXiv:2604.02420v2, compare supersets of the states of Schmidt number at most χ.

  • Negativity. N(ρ) = (||ρ^Γ||_1 − 1)/2 (Definition 2, Eq. (7)). Schmidt number at most χ implies N(ρ) ≤ (χ − 1)/2 (Lemma 3).
  • Reduction. RED_κ(ρ) = Tr(ρ)1 − ρ/κ (Eq. (35)). Schmidt number at most κ implies 1 ⊗ RED_κ ≥ 0 (Theorem 7).
  • The question. §5.4: "the complete inclusions of the considered EFS_γ sets is only pending for the unknown case NEGFS_{(χ−1)/2} ?⊂ PREDFS_χ, which is also unknown for the non-FS case."

The verbatim statements are in the literature note.

Issue #17440 fixes the reading of the non-FS case: for every density matrix ρ on C^N ⊗ C^M and every integer 1 ≤ χ ≤ min(N, M), does N(ρ) = (χ − 1)/2 imply 1 ⊗ RED_χ = ρ_A ⊗ 1 − ρ/χ ≥ 0? The attained value is the stronger hypothesis, so a counterexample also refutes the sublevel reading N(ρ) ≤ (χ − 1)/2.

The non-FS inclusion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/NegativityReductionInclusionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundSource 2025The order-seven unicyclic CDSO minimizer has no universal vertexAmong connected simple graphs of order seven and cyclomatic number one, every CDSO minimizer lacks a vertex adjacent to all others.Refuted

Why it mattersThe source presents the universal-vertex assertion after numerical tests on fixed-order connected graphs.

Our contributionAmong connected simple graphs of order seven and cyclomatic number one, every CDSO minimizer lacks a vertex adjacent to all others.

Verification details
Release source recordRefuted in Lean

Open problem

The order-seven unicyclic CDSO minimizer has no universal vertex

Conjecture 4.1 states: “A graph minimizing (maximizing, respectively) the CDSO index (HSO index, respectively) among fixed-order connected graphs with cyclomatic number $\ell(\ge 1)$ has a vertex adjacent to all other vertices.” This dossier settles the CDSO half. The HSO half is outside the delivered claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/CdsoUniversalVertexRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2025.

    Read source ↗
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026The universal hyperbolic retrieval boundA nonsystematic twenty-column code over F11 refutes the universal hyperbolic tradeoff for actual whole-file iid retrieval.Refuted

The questionBar-Lev, arXiv:2603.17154v2, Section V-B, Conjecture 1, asserts s1/E1+s2/E2<=1 for every finite field, every positive two-file partition s1+s2=k with max(s1,s2)>=2, every length n>=k, and every rank-k generator.

Why it mattersThe assertion proposes a universal tradeoff between complementary files under arbitrary linear encoding.

The ideaParameters 0,1,2 give three independent Vandermonde columns, so the matrix has rank three.

Our contributionA nonsystematic twenty-column code over F11 refutes the universal hyperbolic tradeoff for actual whole-file iid retrieval.

Verification details
Release source recordRefuted in Lean

Open problem

The universal hyperbolic retrieval bound

Bar-Lev, arXiv:2603.17154v2, Section V-B, Conjecture 1, asserts s1/E1+s2/E2<=1 for every finite field, every positive two-file partition s1+s2=k with max(s1,s2)>=2, every length n>=k, and every rank-k generator. Each physical column index is drawn independently, uniformly and with replacement. Ei is the expectation of the minimum time at which the sampled-column span contains the entire coordinate file. Duplicate indices remain separate draws. Preregistration #11729 fixes this complete statement and a nonzero-column refutation criterion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Resource/VandermondeHyperbolicRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryCounterexample foundBouras's unique prime occurrence conjecture is falseThe prime 11 occurs at indices 10 and 21 in OEIS A372763.Refuted

The questionTotal rational division and truncated natural subtraction extend the definitions to all natural indices; claim restricts the assertion to the source range.

Why it mattersDistinct admissible indices disprove uniqueness without a global occurrence classification.

The ideaThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionThe prime 11 occurs at indices 10 and 21 in OEIS A372763.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

bouras372763 once

OEIS A372763, COMMENTS (unpaginated), states: "Conjecture 2: Except for 2 and 3, all primes appear in the sequence once." Its NAME is "Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+5)))))." The source index range is $n \ge 3$. The source and statement are preregistered in #18628.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Inner fraction from $2$ through $n-1$, terminal value $n+5$ | nested, K, a | Right fold with rational division and the literal terminal value. | | Denominator | a | Positive denominator of the reduced rational number. | | All primes except $2$ and $3$ | claim | Natural prime binder and the two explicit exclusions. | | Appear once in the sequence | claim | Unique natural index with $3 \le n$ and $a(n)=p$. | | Conjecture is false | result | Negation of the complete claim. |

The parameter $x$ in nested and K exposes the source's terminal argument; a specializes it to $n+5$. Total rational division and truncated natural subtraction extend the definitions to all natural indices; claim restricts the assertion to the source range. There are no additional hypotheses.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/Primes/BourasA372763OnceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationConjecture refutedFirst preprint 2026The damped angular-momentum kernel positivity conjectureThat is, if e^{−½γL(L+1)} ( C(2J,L)/C(2J+L+1,L) )^{−σ/2} ≤ ( C(2J,L)/C(2J+L+1,L) )^{1/2} (53) f…Refuted

The questionMelanathuru, Phase-space measurements and decoherence for angular momentum systems, arXiv:2605.02696v1, Section VI, after equation (52):

Our contributionThat is, if e^{−½γL(L+1)} ( C(2J,L)/C(2J+L+1,L) )^{−σ/2} ≤ ( C(2J,L)/C(2J+L+1,L) )^{1/2} (53) f…

Verification details
Release source recordRefuted in Lean

Open problem

The damped angular-momentum kernel positivity conjecture

D. C. Brody, E.-M. Graefe and R. Melanathuru, Phase-space measurements and decoherence for angular momentum systems, arXiv:2605.02696v1, Section VI, after equation (52):

For J > 1/2 we do not have an exact result, but it seems reasonable to conjecture that positivity is ensured provided that the damped σ kernel in (44), due to decoherence, becomes no sharper than the Husimi (σ = −1) kernel. That is, if e^{−½γL(L+1)} ( C(2J,L)/C(2J+L+1,L) )^{−σ/2} ≤ ( C(2J,L)/C(2J+L+1,L) )^{1/2} (53) for all L, then we restore positivity.

Equation (44) includes the complex conjugate of the harmonic:

$$ F^\sigma(\theta,\phi,t)=a_J\sum_{L,k}e^{-\gamma L(L+1)t/2} \left(\frac{\binom{2J}{L}}{\binom{2J+L+1}{L}}\right)^{-\sigma/2} \rho_{Lk}(0)\overline{Y_L^k(\theta,\phi)},\qquad a_J=(2J+1)^{-1/2}. $$

The tensor entries and multipoles follow equations (12) and (13); the harmonics have the Condon–Shortley phase and the paper's factor $\sqrt{4\pi}$ (footnote 1). Equation (53) omits $t$ in print. The refuting example has $t=1$, where the printed and time-dependent readings coincide.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Decoherence/DampedSpinKernelPositivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The equiprobable pretty-good-measurement comparison of Cha and LeeCha and Lee (arXiv:2507.05778, Quantum Information Processing 25, 310) compare two upper bounds on the optimal success probability of minimum-error state discrimination, one built from the pretty good measurement on all states and one built from it on the active set I+ of labels with nonzero optimal effects.Refuted

The questionLee, Structural perspectives from quantum states and measurements in optimal state discrimination, Quantum Information Processing 25, 310 (2026), arXiv:2507.05778v2.

Why it mattersEq. (18) is the equal-prior form of the statement that the upper bound on the optimal success probability obtained from the pretty good measurement on the active set is at least as tight as Renes's bound from the pretty good measurement on all states.

Our contributionCha and Lee (arXiv:2507.05778, Quantum Information Processing 25, 310) compare two upper bounds on the optimal success probability of minimum-error state discrimination, one built from the pretty good measurement on all states and one built from it on the active set I+ of labels with nonzero optimal effects.

Verification details
Release source recordRefuted in Lean

Open problem

The equiprobable pretty-good-measurement comparison of Cha and Lee

H. Cha and J. Lee, Structural perspectives from quantum states and measurements in optimal state discrimination, Quantum Information Processing 25, 310 (2026), arXiv:2507.05778v2. For an ensemble $\tilde\sigma_i=p_i\sigma_i$, $I_+$ is the set of labels whose optimal minimum-error effect is nonzero, and

$P_\mathcal{E}^{\text{PGM}+} = \sum_{i \in \text{I}+(\mathcal{E})} \text{tr}(\Tilde{\sigma}i E_i)$, where $S \equiv \sum{i \in \text{I}+(\mathcal{E})} \Tilde{\sigma}_i$ and $E_i \equiv S^{-1 / 2} \Tilde{\sigma}_i S^{-1 / 2}$.

For equal priors the paper conjectures (journal Eq. (18)):

$(|\text{I}+(\mathcal{E})| - 1) \left( P\mathcal{E}^{\text{PGM}+} - \frac{1}{N} \right) \leq (N - 1) \left( P_\mathcal{E}^{\text{PGM}} - \frac{1}{N} \right).$ […] Consequently, we conjecture that it holds universally for equiprobable states.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/EquiprobablePgmActiveSetRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Geometry & algebraCounterexample foundFirst preprint 2025The HOMFLYPT Yang–Baxter Künneth subcomplex need not splitThe HOMFLYPT Yang-Baxter Kunneth subcomplex has no chain retraction for the two-letter decomposition A = {2}, B = {1}.Refuted

The questionChristiana, Clingenpeel, Guo, Oh, Przytycki, Wang and Yun, Low Dimensional Homology of the Yang-Baxter Operators Yielding the HOMFLYPT Polynomial, arXiv:2502.20659v1, §5.1, Conjecture 5.6:

Why it mattersA degreewise module splitting alone does not supply the homology consequence asserted in the conjecture.

The ideaThe boundary uses the paper's one-based alternating signs.

Our contributionThe HOMFLYPT Yang-Baxter Kunneth subcomplex has no chain retraction for the two-letter decomposition A = {2}, B = {1}.

Verification details
Release source recordRefuted in Lean

arXiv

The HOMFLYPT Yang–Baxter Künneth subcomplex need not split

Christiana, Clingenpeel, Guo, Oh, Przytycki, Wang and Yun, Low Dimensional Homology of the Yang-Baxter Operators Yielding the HOMFLYPT Polynomial, arXiv:2502.20659v1, §5.1, Conjecture 5.6:

The short exact sequence of chain complexes $0 \to C_\bullet^{(m,A,B)} \to C_\bullet^m \to C_n^m/C_\bullet^{(m,A,B)} \to 0,$ splits. Thus homology $H_n(C_\bullet^m)$ splits.

Definition 5.4 includes every cutoff from zero to the word length. The hypotheses are $A\cup B=X_{(m)}$ and every letter of $A$ at least every letter of $B$. Splitting means a chain-map retraction of the inclusion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/HomflyptYangBaxterKunnethSplittingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Refutation of the CRIM square-rectair formula at its boundaryThe printed square-rectair formula gives 2 at r=1 and k=0, but the value is 1.Refuted

The questionBašić, Gottlieb, and Krnc, CRIM: A Natural Game on Integer Partitions, arXiv:2606.16828v1, printed page 18, Conjecture 2 states a three-branch Sprague-Grundy formula for every square rectair R^k{r,r} with 0 <= k < r.

Why it mattersThe boundary values r = 1 and k = 0 give the one-cell partition [1].

Our contributionThe printed square-rectair formula gives 2 at r=1 and k=0, but the value is 1.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the CRIM square-rectair formula at its boundary

Bašić, Gottlieb, and Krnc, CRIM: A Natural Game on Integer Partitions, arXiv:2606.16828v1, printed page 18, Conjecture 2 states a three-branch Sprague-Grundy formula for every square rectair R^k_{r,r} with 0 <= k < r.

The formal claim quantifies natural numbers r and k with k < r and uses the published three branches: value zero when r is even or k < r - 1, value one for r equal to 3 or 5, and value two otherwise.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/CrimRectairSquareBoundaryRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundSource 2001Refutation of Crux Problem 2623The cyclic ratio-sum monotonicity assertion in Crux Problem 2623 is false.Refuted

The questionZejnulahi and Arslanagic, Problem 2623, Crux Mathematicorum with Mathematical Mayhem 27, no. 2 (2001), displayed page 139, define positive cyclic variables by x(n+r) = xr and

Why it mattersThe published question asks for either a proof or a disproof of a universal monotonicity assertion.

The ideaTake n=4, k=1, and the positive cyclic tuple (1,2,1,2).

Our contributionThe cyclic ratio-sum monotonicity assertion in Crux Problem 2623 is false.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Crux Problem 2623

Zejnulahi and Arslanagic, Problem 2623, Crux Mathematicorum with Mathematical Mayhem 27, no. 2 (2001), displayed page 139, define positive cyclic variables by x_(n+r) = x_r and

S_k = sum_(j=1)^n (sum_(i=0)^k x_(j+i)) / (sum_(i=0)^k x_(j+i+1)).

They ask verbatim:

Prove or disprove that S_k >= S_(k+1).

The formal claim quantifies over every n >= 2, every positive real-valued cyclic function on ZMod n, and every interior index k with k+2 <= n. It asserts S_(k+1) <= S_k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/CruxProblem2623Refutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2001.

    Read source ↗
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    Year recorded in the source bibliography: 2001. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2026The within-DOF minimum clause of equipartition dualityAt two degrees of freedom, a nonuniform phase density has the same joint action marginal as its uniform-phase product comparator and strictly smaller uncertainty in both degrees of freedom.Refuted

The questionDinov, arXiv:2607.07851v1, Conjecture 3.22 (Equipartition duality):

Why it mattersCovariance inequalities concern the actual coordinate law; uniformity of a different marginal does not determine that covariance.

The ideaBoth strict comparisons occur in the contradiction with the asserted minimum inequalities.

Our contributionAt two degrees of freedom, a nonuniform phase density has the same joint action marginal as its uniform-phase product comparator and strictly smaller uncertainty in both degrees of freedom.

Verification details
Release source recordRefuted in Lean

Open problem

The within-DOF minimum clause of equipartition duality

Dinov, arXiv:2607.07851v1, Conjecture 3.22 (Equipartition duality):

Fix $n\geq2$, an entropy value $s$, and an action marginal $\rho_{\mathbf J}$ on $(0,\infty)^n$. Among all states on $\mathbb T^n\times(0,\infty)^n$ with entropy $\geq s$ and action marginal $\rho_{\mathbf J}$, the phase-equipartitioned product state (unique when it exists) simultaneously (i) maximizes the entropy, (ii) minimizes every within-DOF uncertainty $u_j$, and (iii) is the unique state at which the per-DOF conjectured bound of Problem 3.17(b) is saturated for all $j$; moreover it is the unique fixed point, with the given marginal, of the multi-DOF kime-deformed semigroup $\partial_t\tilde\rho=\sum_j(-\omega_j\partial_{\theta_j}+\varepsilon\partial^2_{\theta_j})\tilde\rho$.

This dossier anchors the necessary minimum clause (ii). The other clauses are quoted to identify the original simultaneous assertion and receive no separate settlement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/DinovOriginalMinimumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Refutation of the equality conclusion in Fazekas et al. Conjecture 4The binary word abab refutes the equality conclusion in Conjecture 4 on abelian squares.Refuted

The questionConjecture 4 of Fazekas, Mammoliti, Mercaş, and Simpson, Binary Words Containing Few Abelian Squares, arXiv:2604.23188v1, states that every binary word of length n contains at least floor(n/4) distinct abelian-square factors and that equality forces all of those factors to be trivial even powers of one letter.

Why it mattersThus its count equals floor(4/4), while the factor uses both letters and is not a positive even power of either one.

The ideaRepresent the binary alphabet by Bool and distinct factors by a filtered finite set of prefixes of suffixes.

Our contributionThe binary word abab refutes the equality conclusion in Conjecture 4 on abelian squares.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the equality conclusion in Fazekas et al. Conjecture 4

Conjecture 4 of Fazekas, Mammoliti, Mercaş, and Simpson, Binary Words Containing Few Abelian Squares, arXiv:2604.23188v1, states that every binary word of length n contains at least floor(n/4) distinct abelian-square factors and that equality forces all of those factors to be trivial even powers of one letter.

The formal claim quantifies over every finite binary word, counts distinct factors whose equal-length halves have the same Parikh vector, and expresses both the lower bound and the equality conclusion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Powers/FazekasAbelianSquareEqualityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryConjecture refutedFirst preprint 2026An odd midpoint refutes Fiore-Nasr-Stone Conjecture 6.8At the distinct odd primes three, five and eleven, both complete midpoint polynomials have nonnegative integer coefficients.Refuted

The questionThe complete Conjecture 6.8 of fiore2026squaresideddice states:

Why it mattersThe source expressly places this midpoint in Conjecture 6.8.

The ideaAll these sums have nonnegative coefficients.

Our contributionAt the distinct odd primes three, five and eleven, both complete midpoint polynomials have nonnegative integer coefficients.

Verification details
Release source recordRefuted in Lean

arXiv

An odd midpoint refutes Fiore-Nasr-Stone Conjecture 6.8

The complete Conjecture 6.8 of fiore2026squaresideddice states:

In the remaining 33 pairs of the polynomials A_abcd and B_(2-a,2-b,2-c,2-d) (including the case where p != 2 for when a=b=c=d=1), one of the two polynomials will have a negative coefficient.

Definition 6.1 uses distinct natural primes p,q,r and exponents a,b,c,d in {0,1,2}. At the midpoint a=b=c=d=1 both complementary tuples are (1,1,1,1). The two complete integer polynomials are

$$A(p,q,r)=X\Phi_p^2\Phi_q^2\Phi_{pq}\Phi_{pr}\Phi_{qr}\Phi_{pqr},$$

$$B(p,q,r)=X\Phi_r^2\Phi_{pq}\Phi_{pr}\Phi_{qr}\Phi_{pqr}.$$

The literal closed claim : Prop universally quantifies over natural p,q,r with Nat.Prime p, Nat.Prime q, Nat.Prime r, p != q, p != r, q != r and p != 2, and asserts that some natural-index coefficient of A or of B is strictly negative. There is no ordering premise, prefix bound or scalar proxy.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Cyclotomic/FioreNasrStoneMidpointRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Gobel-Misra Colored-Path Determinant RefutationA seven-vertex colored path pair refutes Conjecture 5.3 on determinant equality.Refuted

The questionGobel and Misra, arXiv:2506.23936v1, Conjecture 5.3, printed page 27, assert that two colored paths on m vertices with equal generic concentration-matrix determinants are identical, reflections, or instances of the color configuration in Theorem 3.6 when m is even and Theorem 3.8 when m is odd, allowing the listed configuration to be reflected.

Why it mattersIt compares the generic determinants in the same multivariable polynomial ring used to encode the source's vertex and edge color constraints.

The ideaUse one vertex color a, two edge colors u and v, and m = 7.

Our contributionA seven-vertex colored path pair refutes Conjecture 5.3 on determinant equality.

Verification details
Release source recordRefuted in Lean

Open problem

Gobel-Misra Colored-Path Determinant Refutation

Gobel and Misra, arXiv:2506.23936v1, Conjecture 5.3, printed page 27, assert that two colored paths on m vertices with equal generic concentration-matrix determinants are identical, reflections, or instances of the color configuration in Theorem 3.6 when m is even and Theorem 3.8 when m is odd, allowing the listed configuration to be reflected.

The formal claim quantifies over the vertex-color type, the edge-color type, the path length, and both labelled paths. It reads the final reflection clause in the widest form: neither path, the first path, the second path, or both paths may be reflected. The repository result is Not claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GobelMisraColoredPathDeterminantRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Refutation of Gonzalez D'Leon-Wachs Conjecture 4.13(2)Three disjoint triangles and three spanning paths refute Conjecture 4.13(2).Refuted

The questionConjecture 4.13(2) of arXiv:2608.08692v1 states that if G is a graph on n vertices with k connected components and H is a spanning subgraph with the same k components, then (-1)^(n-k) (muG - muH) is real-rooted.

Why it mattersThe conjecture gives a root-location prediction for differences of the weighted bond-poset polynomials introduced in the paper.

The ideaLet G be the union of three triangle fibers and H the union of the corresponding three path fibers.

Our contributionThree disjoint triangles and three spanning paths refute Conjecture 4.13(2).

Scope & next questionIt does not refute part (1), and it does not settle any connected-only variant obtained by adding a hypothesis absent from the source statement.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Gonzalez D'Leon-Wachs Conjecture 4.13(2)

Conjecture 4.13(2) of arXiv:2608.08692v1 states that if G is a graph on n vertices with k connected components and H is a spanning subgraph with the same k components, then (-1)^(n-k) (mu_G - mu_H) is real-rooted. Here mu_G is the Section 2 weighted bond-poset Mobius polynomial. The statement permits disconnected graphs. This dossier concerns only part (2) as written.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/GonzalezDLeonWachsWeightedBondDifferenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2025Refutation of the PNim heavy-interval conjectureThe PNim heavy-interval conjecture fails at a = 8, b = 7.Refuted

The questionGottlieb, Krnc, and Mursic, Nim on Integer Partitions and Hyperrectangles, arXiv:2506.04991v1, printed page 14, Conjecture 2 states that whenever the rectangle [a+1]^(b+1) is heavy, every partition in the Young interval from [a+1,a,...,a-b+1] to that rectangle is heavy.

Why it mattersThe upper rectangle [9,9,9,9,9,9,9,9] is heavy, while the intervening partition [9,9,8,8,8,5,5,5] has Grundy value 3 and longest-play length 16.

Our contributionThe PNim heavy-interval conjecture fails at a = 8, b = 7.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the PNim heavy-interval conjecture

Gottlieb, Krnc, and Mursic, Nim on Integer Partitions and Hyperrectangles, arXiv:2506.04991v1, printed page 14, Conjecture 2 states that whenever the rectangle [a+1]^(b+1) is heavy, every partition in the Young interval from [a+1,a,...,a-b+1] to that rectangle is heavy.

The formal claim is the fully quantified natural-number version with b <= a, the PNim row and column moves, the recursive Grundy value, and the definition of heaviness from Proposition 2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Games/PnimHeavyIntervalRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025Pure-state ECQC fails in dimension fiveA maximally entangled pure state in dimension five has log₂ 5 bits of mutual information in each of Iqbal's six measurements, contradicting pure-state ECQC.Refuted

The questionIqbal, On the CQC Conjecture: A sufficient condition and an extension, arXiv:2509.08286v2, Closing Remarks and Future Works, p. 24, asks:

Why it mattersThis convention reproduces the source's dimension-three isotropic calculation, which has zero mutual information in the two quadratic bases.

Our contributionA maximally entangled pure state in dimension five has log₂ 5 bits of mutual information in each of Iqbal's six measurements, contradicting pure-state ECQC.

Scope & next questionThe dimension-three universal pure-state question remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Pure-state ECQC fails in dimension five

H. Iqbal, On the CQC Conjecture: A sufficient condition and an extension, arXiv:2509.08286v2, Closing Remarks and Future Works, p. 24, asks:

So it would be interesting to develop a method to prove ECQC for pure states analytically, similar to the case of the original CQC conjecture.

The ECQC assertion is

$$ I(A:B)\geq\min_{S\subseteq{0,\ldots,d},\ |S|=d} \sum_{i\in S}I(M_i^A:M_i^B). $$

The source says: "The elements of the set $\mathcal{S}$ are collections of $d$ mutual information resulting from $d+1$ bipartite MUB measurements." Issue #12896 preregisters its pure-state specialization at prime dimension five, with Shannon and von Neumann entropy in bits and $0\log 0=0$. The arXiv PDF labels the conjecture 3.1, §3.1, p. 6, and the dimension-five bases §4.4.2, pp. 22–23; the preregistration identifies the same quoted content as Conjecture 2.1 and §4.5.2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/PureStateECQCRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Refutation of Javaheri's cyclic arrangement dimension conjectureThe cyclic arrangement conjecture fails at the nonnegative parameters a=6 and b=1.Refuted

The questionMohammad Javaheri, A new arrangement inequality (revised and extended version), arXiv:2610.33653v1, Section 1, Conjecture 1.3, states that for all real a,b ≥ 0 with a+b ≠ 1, the largest dimension N(a,b) for which

Why it mattersAt a=6 and b=1, the threshold condition is 36 ≤ 48 sin²(π/n).

The ideaThe threshold set has greatest element 3: the inequality holds at n=3, while for every n ≥ 4 the bound sin²(π/n) ≤ 1/2 makes the threshold inequality fail.

Our contributionThe cyclic arrangement conjecture fails at the nonnegative parameters a=6 and b=1.

Scope & next questionTheir independent formal verification remains open here; result does not prove them.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Javaheri's cyclic arrangement dimension conjecture

Mohammad Javaheri, A new arrangement inequality (revised and extended version), arXiv:2610.33653v1, Section 1, Conjecture 1.3, states that for all real a,b ≥ 0 with a+b ≠ 1, the largest dimension N(a,b) for which

(Σᵢ xᵢ)(Σᵢ xᵢᵃ xᵢ₊₁ᵇ) ≤ (Σᵢ xᵢ^((a+b+1)/2))²

holds for every nonnegative cyclic tuple equals

N*(a,b) = max { n ≥ 1 : n = 1 or
    (a+b−1)² ≤ 8ab sin²(π/n) }.

The cyclic successor is xₙ₊₁ = x₁. The source writes the largest valid dimension and the threshold maximum as named quantities. The Lean encoding makes those implicit extrema explicit: IsN is the extended-natural least-upper-bound predicate for the set of valid positive dimensions, and IsNStar is the greatest-element predicate for the threshold set. It then states the full universal claim together with existence of the two specified objects, so the refutation does not depend on an underspecified maximum.

The claim was preregistered in GitHub issue #22620 before the Lean probe and proof.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Polynomials/JavaheriArrangementDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationConjecture refutedFirst preprint 2009Kay's transfer-rate conjecture for perfect state transfer chainsKay's review of perfect state transfer (arXiv:0903.4274v3, subsection Transfer Rate) conjectures that no set of eigenvalues fulfilling the perfect state transfer condition can make all the sums Rk, k = 0, ..., M - 1, of its rate lemma equal for an integer M > 2, and states that it has a proof only for M > N/2.Refuted

The questionKay, A Review of Perfect State Transfer and its Application as a Constructive Tool, arXiv:0903.4274v3 (Int.

Why it mattersThe lemma turns the question of how fast a perfect state transfer chain can be reused as a quantum channel into a condition on its spectrum: the standard protocol has M = 2, and M > 2 would mean that a new state can be placed on the first site before the previous one has arrived.

The ideaThe gaps are 31, 15, 19, 23, 19, 15, 31, that is 2m + 1 with m = 15, 7, 9, 11, 9, 7, 15, all positive.

Our contributionKay's review of perfect state transfer (arXiv:0903.4274v3, subsection Transfer Rate) conjectures that no set of eigenvalues fulfilling the perfect state transfer condition can make all the sums Rk, k = 0, ..., M - 1, of its rate lemma equal for an integer M > 2, and states that it has a proof only for M > N/2.

Verification details
Release source recordRefuted in Lean

Open problem

Kay's transfer-rate conjecture for perfect state transfer chains

A. Kay, A Review of Perfect State Transfer and its Application as a Constructive Tool, arXiv:0903.4274v3 (Int. J. Quantum Inf. 8, 641 (2010)), subsection "Transfer Rate". For the ordered eigenvalues $\lambda_1 < \dots < \lambda_N$ of a mirror-symmetric chain the perfect state transfer condition is

$\lambda_n-\lambda_{n-1}=(2m_n+1)\pi/t_0$ where $t_0$ is the state transfer time, and $m_n$ is a positive integer (which can vary with $n$).

and $B'(\lambda_n)=\prod_{m=1\neq n}^N(\lambda_n-\lambda_m)$. The lemma of that subsection (the sixth lemma of the source) reads:

If a set of eigenvalues is chosen to fulfill the perfect state transfer condition of Eqn.~(\ref{eqn:st_cond}), then a necessary and sufficient condition to perfectly achieve the rate $M/2t_0$ for integer $M$ is that all the $R_k$ for $k=0\ldots M-1$ should be equal, where $R_k=\sum_{n=1}^N\frac{(-1)^n}{B'(\lambda_n)}$, and the sum is restricted to those terms satisfying the condition $\frac{t_0}{\pi}(\lambda_n-\lambda_1)\mod M=k.$

It is followed by the conjecture:

We conjecture that it is impossible to fulfill the condition of Lemma \ref{lemma:rate} for any $M>2$, although we only have a proof for $M>N/2$.

Issue #14475 fixes the reading. A spectrum is a function lam : Fin N → ℝ; the condition is 0 < t₀ together with positive integers m such that every gap between consecutive levels is (2 m + 1) π / t₀; the level n lies in the class k modulo M when the real number (t₀/π)(λ_n − λ_1) is an integer congruent to k; the sign (−1)^n uses the source's index, which starts at one. The conjecture is the statement that for every N ≥ 1, every such spectrum and transfer time, and every integer M > 2, the sums R_0, …, R_{M−1} are not all equal. The case N = 0 is excluded because all sums of the empty spectrum are empty and therefore equal.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/KayTransferRateRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    First preprint year: 2009. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2025The Koch-Narayan bipartite edge bound is falseA graph with fourteen vertices, twenty-eight edges and a unique minimum dominating set of size four exceeds the printed bound twenty-six.Refuted

The questionWith a=ceil(gamma/2), b=floor(gamma/2), the exact printed expression is

Why it mattersIts graph has 14 vertices, 28 edges and a unique minimum dominating set of cardinality four, whereas m(14,4)=26.

The ideaThe four centers 0,7,8,9 have disjoint closed neighborhoods partitioning all vertices.

Our contributionA graph with fourteen vertices, twenty-eight edges and a unique minimum dominating set of size four exceeds the printed bound twenty-six.

Scope & next questionThe true extremal function, other corrected conjectures, and the unrestricted predecessor question remain outside this result.

Verification details
Release source recordRefuted in Lean

Open problem

The Koch-Narayan bipartite edge bound is false

Koch and Narayan, Maximal bipartite graphs with a unique minimum dominating set, arXiv:2511.01719v1, Conjecture 1, assert |E(G)| <= m(n,gamma) for every finite simple bipartite graph without isolated vertices, with a unique minimum dominating set, gamma >= 2 and n >= 3gamma. With a=ceil(gamma/2), b=floor(gamma/2), the exact printed expression is

Phi = max(0,n-3gamma-2a-b+1),
m = 2gamma+2ab+min(n-3gamma,2a-b+1)(2a+1)
    + sum(i=1,...,Phi) ((2a+1)+ceil(i/2)).
Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/KochNarayanRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationConjecture refutedFirst preprint 2026Conjecture 15 fails for a commutativity-preserving qubit channelA completely positive, trace-preserving qubit channel that preserves commuting density matrices increases relative quantumness at alpha zero.Refuted

The questionDeesuwan, Cumulant-based quantum relative Rényi functional, arXiv:2606.31205v1, Section VIII.A, PDF p. 18, conjecture the following data-processing inequality at alpha zero:

Why it mattersThe regularized states and their channel outputs are

The ideaFor a=(1+u)(1+w), b=(1-u)(1+w), and delta=(1-u^2)(1-w^2)-v^2, the four Kraus matrices are

Our contributionA completely positive, trace-preserving qubit channel that preserves commuting density matrices increases relative quantumness at alpha zero.

Verification details
Release source recordRefuted in Lean

Open problem

Conjecture 15 fails for a commutativity-preserving qubit channel

A. Meunson and T. Deesuwan, Cumulant-based quantum relative Rényi functional, arXiv:2606.31205v1, Section VIII.A, PDF p. 18, conjecture the following data-processing inequality at alpha zero:

Let rho and sigma be noncommuting density matrices, with regularized states rho_epsilon and sigma_epsilon. If N is a commutativity-preserving CPTP channel and the support of N(rho_epsilon) is contained in that of N(sigma_epsilon), then Q(rho_epsilon || sigma_epsilon) is at least Q(N(rho_epsilon) || N(sigma_epsilon)), which is at least zero.

The literal source formulas and sentences are in the literature note. The source's malformed pair \rho,\sigma) is transcribed literally there; the Blueprint display uses the normalized pair (rho,sigma).

Equation state_regularized defines

$$A_\varepsilon=(1-\varepsilon)A+\frac{\varepsilon}{d}I, \qquad 0<\varepsilon<1.$$

Equation regularized_Cu-Q and the definition of relative quantumness give

$$Q(A\Vert B)=-S_0^Q(A\Vert B) =\ln\operatorname{Tr}\left[A\exp(\ln B-\ln A)\right].$$

Density means positive semidefinite with trace one. CPTP means complete positivity at every finite amplification and trace preservation on every matrix. CoP quantifies over every pair of commuting density matrices. Support is the operator range. The Lean claim includes both conclusion inequalities and all source hypotheses. It quantifies over d : ℕ and complex matrices on Fin d; one witness with d = 2 refutes that universal claim. Every matrix logarithm in the witness is of a positive definite matrix, so the refutation uses no convention for a singular output logarithm.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/CoPRelativeQuantumnessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedSource 2025Mi-Xu-Yang Conjecture 5.2 refutationMi, Xu and Yang's Conjecture 5.2 is false at n=3 for the permutation 132.Refuted

The questionMi, Xu and Yang, ECA 5:3 (2025), Article S2R22, introduce the stack machines used in the conjecture.

Why it mattersAt n = 3, every nonidentity permutation is sent to the identity by one application of s ∘ t.

The ideaDefine the paper's stack operation directly from finite lists, and define TakesSorts as reaching List.range' 1 pi.length at iterate j while every earlier iterate differs from it.

Our contributionMi, Xu and Yang's Conjecture 5.2 is false at n=3 for the permutation 132.

Verification details
Release source recordRefuted in Lean

Open problem

Mi-Xu-Yang Conjecture 5.2 refutation

Mi, Xu and Yang, ECA 5:3 (2025), Article #S2R22, introduce the stack machines used in the conjecture. Printed page 1 states:

The map sends permutations through a stack that always avoids the permutation 21 when read from top to bottom.

We further generalize s to s_{(T,k)}. We define the maps s_{(T,k)} that avoid containing more than k distinct permutations from T in the stack at once. More specifically, the map sorts a permutation π via the following stack sorting algorithm: If adding the leftmost element of the input to the stack keeps the stack (T, k)-avoiding, push that element into the stack. Otherwise, pop the top element off the stack and append it to the output.

Printed page 2 defines containment and avoidance:

Given two permutations π and σ, we say π contains σ if there exist a_1, a_2, . . . , a_k such that the sub-permutation π_{a_1} π_{a_2} . . . π_{a_k} is order-isomorphic to σ. Otherwise, we say π avoids σ.

Furthermore, for a set T of permutations and a nonnegative integer k, we say π is (T, k)-avoiding if π contains at most k elements of T.

The same page fixes the second map:

Throughout the paper, we use t to denote s_{({12,21},1)}.

Printed page 6 defines the sorting time:

Let m_{(s∘t)}(π) be the smallest nonnegative integer j such that (s ∘ t)^j(π) = id_{|π|}.

Printed page 7 gives the proved sufficient direction:

Lemma 5.5. For n ≥ 3 and π ∈ S_n, if π = 2σ1n where σ ∈ Av_{n−3}(213) then m_{(s∘t)}(π) = 2n − 5.

It then states the named open assertion:

Conjecture 5.2. The permutation π takes 2n − 5 sorts to be sorted to the identity by (s ∘ t) if and only if π = 2σ1n and σ ∈ Av_{n−3}(213).

The formal reading takes the quantifier to be n ≥ 3, reads m_{(s∘t)}(π) = 2n−5 as TakesSorts π (2n−5), and reads π = 2σ1n with σ a permutation of {3,…,n−1} avoiding 213. The printed conjecture carries no lower-bound clause; the n ≥ 3 reading is taken from Lemma 5.5 and Conjecture 5.1 and was preregistered in issue #9318. The reading uses s := s_{({21},0)}, the page-1 West machine. Figure 1 on printed page 1 shows a different restricted map and supplies no evidence about t.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/Patterns/MiXuYangPartiallyRestrictedStackMaximalSortRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year in the primary source attribution: 2025.

    Read source ↗
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    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The Cyclic Latin Eulerian Numbers Are Not Fully SymmetricThe column-ascent counts of cyclic Latin squares change only when an endpoint symbol wraps, which refutes full permutation symmetry of the ascent vector.Refuted

The questionMadjid Mirzavaziri and Daniel Yaqubi, Cyclic Latin Eulerian Numbers, arXiv:2609.28808v1, Remark 3.9:

The idea1. shiftstep: for every n ≥ 2, the ascent number of column c + 1 differs from that of column c by [the wrapping value is in the first row] − [it is in the last row], where the wrapping value is the entry equal to n − 1 in column c.

Our contributionThe column-ascent counts of cyclic Latin squares change only when an endpoint symbol wraps, which refutes full permutation symmetry of the ascent vector.

Verification details
Release source recordRefuted in Lean

Open problem

The Cyclic Latin Eulerian Numbers Are Not Fully Symmetric

Madjid Mirzavaziri and Daniel Yaqubi, Cyclic Latin Eulerian Numbers, arXiv:2609.28808v1, Remark 3.9:

Cyclically shifting π naturally explains an order-n cyclic symmetry for ⟨⟨n k⟩⟩_c. However, computations for n ≤ 5 (Table 1) reveal a surprising fact: ⟨⟨n k⟩⟩_c is actually invariant under the full symmetric group action on k … Identifying the hidden mechanism behind this full invariance remains an open problem.

Here ⟨⟨n k⟩⟩_c counts permutations π whose row-reordered cyclic square L_π has column-ascent vector k.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/CyclicLatinEulerianRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Mohan-Neetu Conjecture 6.1Two explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.Refuted

The questionThe formal claim61 quantifies over the group and all three finite sets with these premises; result61 proves its negation.

Why it mattersThe conjecture proposes a small-doubling threshold for the union of two abelian pieces.

The ideaUse the Klein bottle group of pairs (r,n) with multiplication (r,n)(s,m) = (r + (-1)^n s,n+m).

Our contributionTwo explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.

Verification details
Release source recordRefuted in Lean

Open problem

Mohan-Neetu Conjecture 6.1

Conjecture 6.1 in Mohan and Neetu, arXiv:2607.11194v1, section 6, asserts that a nonempty finite subset S of a torsion-free group, partitioned into two disjoint abelian sets A and B, generates an abelian subgroup whenever |S^2| <= 3k - 3. Here k = |S|, S^2 = {xy : x,y in S}, and an abelian set means one whose generated subgroup is abelian. The formal claim61 quantifies over the group and all three finite sets with these premises; result61 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Groups/MohanNeetuSmallDoublingRefutation.result61

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Mohan-Neetu Conjecture 6.3Two explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.Refuted

The questionThe formal claim63 states these quantified premises and conclusion; result63 proves not claim63.

Why it mattersThe paper exhibits a nonabelian small-doubling set without the identity and proposes identity membership as a sufficient additional hypothesis.

The ideaUse the torsion-free Klein bottle group of integer pairs with multiplication (r,n)(s,m) = (r + (-1)^n s,n+m).

Our contributionTwo explicit three-element subsets of the Klein bottle group contradict all three small-doubling conjectures printed in section 6.

Verification details
Release source recordRefuted in Lean

Open problem

Mohan-Neetu Conjecture 6.3

Conjecture 6.3 in Mohan and Neetu, arXiv:2607.11194v1, section 6, asserts that a finite subset S of a torsion-free group with |S| >= 3, identity in S, and |S^2| <= 3|S| - 3 generates an abelian subgroup. The formal claim63 states these quantified premises and conclusion; result63 proves not claim63.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Groups/MohanNeetuSmallDoublingRefutation.result63

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryConjecture refutedPosed in 2003Refutation of the A079063 square-root prime-gap lower-bound conjecturePrime counting refutes the eventual square-root lower bound proposed for A079063.Refuted

The questionLeast k such that sqrt(prime(n+k))-sqrt(prime(n))>1

Why it mattersThe conjecture was added to OEIS A079063 in 2003.

The ideaAssume constants c > 0 and N provide the claimed lower bound.

Our contributionPrime counting refutes the eventual square-root lower bound proposed for A079063.

Scope & next questionIt asserts no corrected asymptotic and no claim about unverified individual indices.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A079063 square-root prime-gap lower-bound conjecture

OEIS A079063, NAME (verbatim):

Least k such that sqrt(prime(n+k))-sqrt(prime(n))>1

COMMENT (verbatim):

Inspired by Andrica's conjecture. If it is true, a(n)>1 for all n.

FORMULA (verbatim; Benoit Cloitre, Feb 02 2003):

Conjecture: there is a constant c>0 such that for n large enough, a(n)>c*sqrt(n) and we can take c=0.4. More precisely, there are 2 constants A and B such that A=lim sup n ->infinity a(n)/sqrt(n) exists = 0.75....; B=lim inf n ->infinity a(n)/sqrt(n) exists =0.46.... - Benoit Cloitre, Feb 02 2003

The literal object is a(n), the least k >= 1 such that sqrt(p_(n+k)) - sqrt(p_n) > 1, where prime(n) = p_n is the n-th prime. The refuted claim is its weakest quantified eventual lower-bound form: there exist c > 0 and N such that every n >= N satisfies a(n) > c*sqrt(n). This refutes the proposed c = 0.4 and the claimed positive liminf = 0.46... a fortiori. The existence or value of either limsup or liminf is not formalized.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/PrimeGaps/CloitreSquareRootPrimeGapRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 3/4 · 10% weight

    Dated conjecture statement: 2003. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2026Minimal matrices of fixed RSK shape need not be HankelMinimal matrices of RSK shape (8,8,1,1) cannot be Hankel.Refuted

The questionThe shape is that of the insertion tableau of the lexicographic matrix biword, with the first strictly larger entry bumped during row insertion.

Why it mattersThe source and definitions are recorded in preregistration 14454.

The ideaNatural-number well-ordering supplies a minimizer.

Our contributionMinimal matrices of RSK shape (8,8,1,1) cannot be Hankel.

Verification details
Release source recordRefuted in Lean

arXiv

Minimal matrices of fixed RSK shape need not be Hankel

Nimisha Pahuja, Minimal Inversions in Integer Matrices of Fixed RSK Shape, arXiv:2602.14931v1, Conjecture 1.1, asserts that every minimal nonnegative integer $n\times n$ matrix whose shape is a partition with exactly $n$ positive parts is symmetric and Hankel. Minimality compares the inversion count $I(M)=\sum_{i<k,,j>l}m_{i,j}m_{k,l}$ against all matrices of that shape. The Hankel clause asks for integers $s_2,\ldots,s_{2n}$ with $m_{i,j}=s_{i+j}$. The shape is that of the insertion tableau of the lexicographic matrix biword, with the first strictly larger entry bumped during row insertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Permutation/RSKMinimalInversionHankelRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Pandey's Parity Conjecture 4.1For all integers n ≥ 2k+1, the independence polynomial I(GP(n,k),x) has only real roots if and only if k is even.Refuted

The questionRohan Pandey, arXiv:2601.03293v1, Conjecture 4.1 (Parity Conjecture), PDF page 4:

Why it mattersIt allows an exact graph counterexample to address the published biconditional without assuming a transfer-matrix formula or approximate root calculations.

The ideaUse vertices Bool x Fin n, representing ui and vi, and the undirected, loop-free closure of the source's outer-successor, inner-k-step, and equal-index spoke relations, with indices modulo n.

Our contributionFor all integers n ≥ 2k+1, the independence polynomial I(GP(n,k),x) has only real roots if and only if k is even.

Verification details
Release source recordRefuted in Lean

arXiv

Pandey's Parity Conjecture 4.1

Rohan Pandey, arXiv:2601.03293v1, Conjecture 4.1 (Parity Conjecture), PDF page 4:

For all integers n ≥ 2k+1, the independence polynomial I(GP(n,k),x) has only real roots if and only if k is even.

Definition 2.1 supplies n >= 3 and 1 <= k < n/2. Precisely, for every n,k : Nat with 3 <= n, 1 <= k, and 2*k < n, the assertion is (forall z : Complex, eval₂ (Int.castRingHom Complex) z (independencePolynomial (gp n k) univ) = 0 -> z.im = 0) <-> Even k. The polynomial is the existing integer independent-set generating polynomial on all vertices. No experimental-range restriction is added.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/GeneralizedPetersen/ParityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2025Two indexed wreath eigenvalues coincideThe wreath eigenvalues indexed by three and five coincide at fourteen and five.Refuted

The questionJan Petr and Pavel Turek, The wreath matrix, arXiv:2501.07269v2, Question 5.5, page 16, ask:

Why it mattersTheorem 1.5 assigns multiplicities to indexed components.

Our contributionThe wreath eigenvalues indexed by three and five coincide at fourteen and five.

Verification details
Release source recordRefuted in Lean

Open problem

Two indexed wreath eigenvalues coincide

Jan Petr and Pavel Turek, The wreath matrix, arXiv:2501.07269v2, Question 5.5, page 16, ask:

Are the eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_k$ from Theorem 1.5 all distinct?

The source and its formulas appear in the Library note. The preregistration is #19058. The answer is negative: at $(n,k)=(14,5)$, the values indexed by $3$ and $5$ are both $17107200$.

Statement echo

| Source clause | Lean counterpart | Scope | | --- | --- | --- | | Positive integers $k<n/2$ | Natural parameters, $0<k$ and $2k<n$ in claim | Exactly the positive strict-half range | | $g=\gcd(n,k)$ | Nat.gcd; Nat.Coprime n k in claim | Coprime restriction; a counterexample here refutes the unrestricted question | | Both cases of $\lambda_1$ in Theorem 1.5 | lambdaOne | Literal rational formula; natural division and factorial arguments | | Formula for $2\leq l\leq k$ when $g=1$ | lam | Index one uses lambdaOne; other indices use the published finite sum | | All indexed values are distinct | claim quantifies over two distinct members of Finset.Icc 1 k | The conclusion compares the formula values | | Negative answer | result proves the negation of claim | Witness $(14,5,3,5)$ |

The formulas are total outside the admissible range, without a spectral assertion there. The factorial arguments and the sum exponent use natural subtraction; the factor $n-2k-1$ uses rational subtraction. The source's identification of the formulas with the spectrum is a literature premise, not a Lean theorem in this module.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/SpectralGraph/PetrTurekWreathEigenvaluesRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026Pham–Vu Question 3.9: diameter of graphs with well-covered powersCartesian products with a sufficiently large clique preserve well-covered powers and increase the diameter by one.Refuted

The questionMy Hanh Pham and Thanh Vu, On graphs whose powers are well-covered, arXiv:2610.09300v1 (7 October 2026), Question 3.9:

Why it mattersThe diameter bound would sharply restrict the geometry of graphs whose every positive power is well-covered.

The ideaIts distance is dH((u,a),(v,b)) = dG(u,v) + 1{a≠b}.

Our contributionCartesian products with a sufficiently large clique preserve well-covered powers and increase the diameter by one.

Verification details
Release source recordRefuted in Lean

Open problem

Pham–Vu Question 3.9: diameter of graphs with well-covered powers

My Hanh Pham and Thanh Vu, On graphs whose powers are well-covered, arXiv:2610.09300v1 (7 October 2026), Question 3.9:

Question 3.9. Let G be a simple connected graph such that G^d is well-covered for all d ≥ 1. Is it true that diam(G) ≤ 3? Furthermore, if diam(G) = 3, are there only finitely many such graphs without true twins?

Definition 2.2 puts an edge between distinct vertices of G^d precisely when their G-distance is at most d. Definition 2.3 defines well-coveredness by equal cardinalities of minimal vertex covers. Complementation identifies these with maximal independent sets. Corollary 3.5 defines true twins as adjacent vertices with equal closed neighbourhoods.

This dossier settles the first clause. The closed Lean claim quantifies over finite vertex types, simple graphs, their decidable adjacency, connectedness, and every positive power, and asserts diameter at most three. The settling statement is literally result : ¬ claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/WellCoveredPowerCliqueProduct.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Sahbi's Conjecture 4.1: bipartite equality for all positive kBipartite equality for all positive k is false.Refuted

The questionRafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 4.1:

Why it mattersAn equality asserted for every positive k must hold at two; the bipartite two-star counterexample tests this necessary specialization.

The ideaThe imported two-star refutation uses a ten-vertex bipartite graph whose cross-edge matrix is

Our contributionBipartite equality for all positive k is false.

Verification details
Release source recordRefuted in Lean

Open problem

Sahbi's Conjecture 4.1: bipartite equality for all positive k

Rafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 4.1:

For every bipartite graph G and every integer k ≥ 1, βk(G) = SFk(G).

A set I is k-independent when every vertex of the induced graph G[I] has degree less than k. The maximum such cardinality is βₖ(G). A set S is k-star-forming when, for every outside vertex v, G[S ∪ {v}] contains a copy of K₁,ₖ containing v. The copy need not be induced. SFₖ(G) is the maximum cardinality of an inclusion-minimal k-star-forming set.

The Lean claim quantifies over every n and every simple graph G on Fin n with decidable adjacency, and every natural k, assumes 1 ≤ k and G.Colorable 2, and asserts beta G k = SF G k. Every finite simple graph can be relabelled onto Fin n; all defining adjacency, subset and cardinality conditions are invariant under this bijection. Positive integers k correspond exactly to naturals with 1 ≤ k. The theorem result : ¬ claim refutes the quoted statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/StarFormingBipartiteEqualityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2026Sahbi's Conjecture 7.2: triangle-free equality at twoFor every triangle-free graph G, β2(G) = SF2(G).Refuted

The questionRafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 7.2:

Why it mattersTriangle-free graphs include every bipartite graph, so the two-star bipartite counterexample also tests this broader equality.

The ideaThe imported two-star refutation uses a ten-vertex bipartite graph whose cross-edge matrix is

Our contributionFor every triangle-free graph G, β2(G) = SF2(G).

Verification details
Release source recordRefuted in Lean

Open problem

Sahbi's Conjecture 7.2: triangle-free equality at two

Rafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 7.2:

For every triangle-free graph G, β2(G) = SF2(G).

A set I is k-independent when every vertex of the induced graph G[I] has degree less than k. The maximum such cardinality is βₖ(G). A set S is k-star-forming when, for every outside vertex v, G[S ∪ {v}] contains a copy of K₁,ₖ containing v. The copy need not be induced. SFₖ(G) is the maximum cardinality of an inclusion-minimal k-star-forming set.

The Lean claim quantifies over every n and every simple graph G on Fin n with decidable adjacency, assumes G.CliqueFree 3, and asserts beta G 2 = SF G 2. Every finite simple graph can be relabelled onto Fin n; all defining adjacency, subset and cardinality conditions are invariant under this bijection. Positive integers k correspond exactly to naturals with 1 ≤ k. The theorem result : ¬ claim refutes the quoted statement.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/StarFormingTriangleFreeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryConjecture refutedFirst preprint 2026Sharma's primitive-polynomial Conjecture 4.2A degree-41 finite-field certificate refutes Sharma's full primitive-polynomial conjecture.Refuted

The questionThe target is the literal Conjecture 4.2 in Avnish K.

Why it mattersThe source strengthens its existential Conjecture 4.1 to every primitive lambda when the base field has cardinality p squared.

The ideaThis is a strict positive divisor bound below the claimed order, giving the contradiction and the full claim negation.

Our contributionA degree-41 finite-field certificate refutes Sharma's full primitive-polynomial conjecture.

Verification details
Release source recordRefuted in Lean

arXiv

Sharma's primitive-polynomial Conjecture 4.2

The target is the literal Conjecture 4.2 in Avnish K. Sharma's Primitive Polynomials of the Form g(x)+λ over Finite Fields: Non-Existence Results and Conjectures, arXiv:2608.07262v2. For every odd prime p, every finite field K with characteristic p and |K|=p^2, and every \lambda\in K satisfying the source multiplicative-generator condition \operatorname{ord}(\lambda)=|K|-1, the source claim says that X^p+X+\lambda is the monic minimal polynomial of a multiplicative generator in an extension L/K of degree p. “Primitive” here means multiplicative order; it does not mean Polynomial.IsPrimitive.

The primary statement, printed page 16, reads (mathematical typography transliterated):

Let p be an odd prime. Then, for every primitive element λ∈F_(p²), the polynomial x^p+x+λ is primitive over F_(p²).

The negated statement proved here is [ \neg\bigl[\forall p\text{ odd prime},\ \forall K\text{ finite field of characteristic }p\text{ with }|K|=p^2,\ \forall\lambda\in K, \ \operatorname{ord}(\lambda)=|K|-1\ \Longrightarrow \exists L/K,\alpha\in L:\ [L:K]=p\ \land \operatorname{minpoly}_K(\alpha)=X^p+X+\lambda\ \land \operatorname{ord}(\alpha)=|L|-1\bigr]. ] The extension is a finite field extension. The minimal polynomial is monic. The existential Conjecture 4.1 is not the target.

The Lean definition fullClaim preserves all of these quantifiers and the source conclusion. The candidate coefficient is the source leading coefficient, so the public result is exactly Not fullClaim after unfolding the coefficient parameter.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/ArithUnits/SharmaPrimitivePolynomialRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025The improved entanglement-swapping product bound fails in dimension fourThe improved product bound for entanglement swapping conjectured by Starke, Basso, Celeri and Maziero fails for two identical partially entangled ququarts: the average is 723/625, whereas the proposed bound is 507/625.Refuted

The questionMaziero, “Entanglement swapping for partially entangled qudits and the role of quantum complementarity”, arXiv:2508.00813v2, p. 8, Eq. (59), conjecture:

Why it mattersIt distinguishes the proposed dimension-dependent factor from the paper's unscaled product bound and its exact low-dimensional results.

The ideaA local have inside result establishes, for every positive dimension and every complex coefficient pair,

Our contributionThe improved product bound for entanglement swapping conjectured by Starke, Basso, Celeri and Maziero fails for two identical partially entangled ququarts: the average is 723/625, whereas the proposed bound is 507/625.

Verification details
Release source recordRefuted in Lean

Open problem

The improved entanglement-swapping product bound fails in dimension four

D. S. Starke, M. L. W. Basso, L. C. Céleri and J. Maziero, “Entanglement swapping for partially entangled qudits and the role of quantum complementarity”, arXiv:2508.00813v2, p. 8, Eq. (59), conjecture:

Therefore, we conjecture that the improved upper bound has the form $\big\langle E_{l_1}\big(|\hat{\phi}{pq}^{AB}\rangle\big)\big\rangle \le \frac{E{l_1}(|\xi\rangle_{AC})E_{l_1}(|\eta\rangle_{C'B})}{d-1}.$

For every dimension $d\ge2$, the inputs are $|\xi\rangle_{AC}=\sum_jc_j|jj\rangle$ and $|\eta\rangle_{C'B}=\sum_kb_k|kk\rangle$, with $\sum_j|c_j|^2=\sum_k|b_k|^2=1$; the paper's second coefficient $d_k$ is renamed $b_k$. The measure is the unscaled ordered sum $E_{l_1}(a)=\sum_{j\ne k}|a_ja_k|$.

The Bell outcome $(p,q)$ gives $\phi_{pq}=d^{-1/2}\sum_kc_{p\oplus k}b_k\bar\omega^{qk}|p\oplus k,k\rangle$, where $\omega=\exp(2\pi i/d)$ and $\oplus$ is addition modulo $d$. Its probability is its squared Hilbert norm; its normalized state is obtained by dividing by that norm. The average includes all $d^2$ outcomes, with zero-probability outcomes contributing zero. Each conditional state is in Schmidt form, since $k\mapsto p\oplus k$ is a bijection.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/QuditSwappingProductBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    A distinct released result depends on this result’s module in the recorded proof graph. This supports reuse, not field-wide foundational status.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2026Taylor-coefficient weak majorization for Hermitian matricesRefutation of unrestricted Hermitian Taylor majorization.Refuted

The questionZhang, Weak majorization inequalities for the cubic and quartic coefficients of e^{(A+B)t} versus e^{At}e^{Bt}, arXiv:2601.07286v1, Section 5, Conjecture 5.5:

Why it mattersThe question asks whether the cubic and quartic spectral comparisons extend to every higher Taylor coefficient for arbitrary Hermitian inputs.

Our contributionRefutation of unrestricted Hermitian Taylor majorization.

Verification details
Release source recordRefuted in Lean

Open problem

Taylor-coefficient weak majorization for Hermitian matrices

T. Zhang, Weak majorization inequalities for the cubic and quartic coefficients of e^{(A+B)t} versus e^{At}e^{Bt}, arXiv:2601.07286v1, Section 5, Conjecture 5.5:

For A, B ∈ Hₙ and every k ≥ 5, λ(H^k) ≺_w λ(R_k).

Here $H_n$ consists of all complex Hermitian $n\times n$ matrices, $H=A+B$,

$$Q_k=\sum_{p=0}^k\binom{k}{p}A^pB^{k-p},\qquad R_k=\operatorname{Re}Q_k=\frac{Q_k+Q_k^*}{2}.$$

The list $\lambda$ is nonincreasing, and $x\prec_w y$ means $\sum_{i=1}^r x_i\le\sum_{i=1}^r y_i$ for every $1\le r\le n$. The paper proves the comparisons for $k=3,4$. Its conjecture imposes no positivity assumption on $A$ or $B$.

Formal reading

All Lean names below have prefix D5.S3.Quantum.Matrix.TaylorMajorization; the Pauli matrices are from D5.S3.Quantum.FiniteDimensional.

| Source object | Lean name | Correspondence | | --- | --- | --- | | $H_n$, $n\ge1$ | Matrix (Fin n) (Fin n) ℂ, Matrix.IsHermitian | Defs.claim explicitly requires 1 ≤ n and both Hermitian hypotheses. | | $Q_k$ | Defs.taylorQ | The ordered binomial sum, with the $A$ powers on the left. | | $R_k$ | Defs.taylorR | Complex Hermitian symmetrization; Defs.taylorR_isHermitian supplies its Hermitian proof. | | $\lambda$, $\prec_w$ | Defs.weakMaj | Mathlib's nonincreasing eigenvalues₀, compared by all initial sums up to $n$. The extra $r=0$ sum is automatically zero. | | All dimensions, inputs, orders and partial sums | Defs.claim | Every natural $n\ge1$, Hermitian $A,B$, natural $k\ge5$ and $1\le r\le n$. | | $\Delta_k=\operatorname{Tr}R_k-\operatorname{Tr}(A+B)^k$ | Defs.defect | Real parts of traces; both compared matrices are Hermitian, so their traces are real. | | $Z=\operatorname{diag}(1,-1)$, $X=\left[\begin{smallmatrix}0&1\1&0\end{smallmatrix}\right]$ | qubitZ, qubitX | The existing two-dimensional Pauli matrices. | | Refutation | Result.result | Exactly ¬ Defs.claim, without additional assumptions. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/TaylorMajorization/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 2/4 · 30% weight

    An explicitly identified research question in the original source.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2018Three weak measurements of projections below minus one eighthFor three sequential weak measurements of projection observables on a pure state, without post-selection, the mean product of the pointer positions can be -1/6, below -1/8.Refuted

The questionBranciard (arXiv:1805.09364, quant-ph; Quantum 3, 194 (2019)) study sequential weak measurements without post-selection.

Why it mattersFor two projection observables the paper's example attains −1/8, and its Appendix shows that this is "the most anomalous value obtainable"; the conjecture asserts that longer sequences of projections do not go further.

The ideaTake d = 3, the state e3, the projection P1 onto (1, 1, −1), and P2 = I − v v, P3 = I − w w with v = (0, 1, −1)/√2, w = (1, 0, −1)/√2; all three have rational entries.

Our contributionFor three sequential weak measurements of projection observables on a pure state, without post-selection, the mean product of the pointer positions can be -1/6, below -1/8.

Verification details
Release source recordRefuted in Lean

Open problem

Three weak measurements of projections below minus one eighth

A. A. Abbott, R. Silva, J. Wechs, N. Brunner and C. Branciard (arXiv:1805.09364, quant-ph; Quantum 3, 194 (2019)) study sequential weak measurements without post-selection. For n observables A_1, …, A_n and a state ρ, the mean product of the pointer positions in the weak regime is 2^{−(n−1)} Tr[{A_1, {A_2, …, {A_{n−1}, A_n}…}} ρ]. For two projection observables (eigenvalues 0 and 1) the paper proves that this is at least −1/8, and states:

Interestingly, by numerically minimising the mean product of the pointer positions for sequences of up to 5 projection observables, we were unable to obtain a value smaller than −1/8, and we conjecture that this is in fact the case for all n.

Issue #11377 fixes the reading. A projection observable is an orthogonal projection P = P² = P* on ℂ^d of any rank, as in the paper's two-observable theorem. The formal claim is the conjecture for every dimension, every number of projections and every unit vector; the result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/SequentialWeakPointerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
  • Prior research · 1/4 · 5% weight

    The motivation identifies an earlier mathematical result or partial case.

    Source 1
Quantum informationCounterexample foundFirst preprint 2015GHZ phase tangents refute a rank-only lower bound for normalized Pauli Fisher informationGHZ phase tangents refute a uniform lower bound for mean Pauli Fisher information.Refuted

The questionGuţă, Statistically efficient tomography of low rank states with incomplete measurements, New J.

Why it mattersThe bound would provide the missing spectral input for extending its concentration argument from uniformly random orthonormal bases to product Pauli settings.

The ideaAt that site each outcome vector has zero overlap with one of the two computational branches, so its directional readout vanishes.

Our contributionGHZ phase tangents refute a uniform lower bound for mean Pauli Fisher information.

Verification details
Release source recordRefuted in Lean

Open problem

GHZ phase tangents refute a rank-only lower bound for normalized Pauli Fisher information

A. Acharya, T. Kypraios and M. Guţă, Statistically efficient tomography of low rank states with incomplete measurements, New J. Phys. 18 (2016), 043018; arXiv:1510.03229v2, Section 4, subsection “Pauli settings”, Efficient-Ion-Tomography.tex, line 360:

While the full dependence of $\lambda_{min}$ on $r$ and $d$ is unclear, we conjecture that for any fixed rank, $\lambda_{min}$ is larger than a fixed constant for all states of rank $r$, of arbitrarily many ions.

Here $d=2^n$, $\bar I$ averages the Fisher information over all $3^n$ product Pauli settings, and $\lambda_{min}=\lambda_{min}(G^{-1/2}\bar I G^{-1/2})$, where $G$ is the Frobenius metric of the local rank-$r$ parametrisation. The measurement model, metric and Fisher matrix are in the same source, lines 114–121, 157–176 and 178–180; see also the literature note.

Issue #24270 fixes the tangent reading. For complex matrices on $n$ qubits, let $u_{s,o}$ be the product Pauli eigenvector, $p_\rho(o\mid s)=\operatorname{Re}\langle u_{s,o}|\rho|u_{s,o}\rangle$, and $\bar I_\rho(v)=3^{-n}\sum_s\sum_o(\operatorname{Re}\langle u_{s,o}|v|u_{s,o}\rangle)^2/p_\rho(o\mid s)$, with zero-probability terms assigned zero.

IsRankTangent r ρ v means that there exists a curve $\gamma:\mathbb R\to M_{2^n}(\mathbb C)$ with $\gamma(0)=\rho$, derivative $v$ at zero, and $\gamma(t)$ a positive semidefinite trace-one matrix of rank $r$ for every real $t$. Differentiability is required only at zero.

claim asserts that for every natural rank $r\ge1$ there exists a real $c>0$ such that, for every $n\in\mathbb N$, every rank-$r$ density matrix $\rho$ and every IsRankTangent r ρ v, one has $c\operatorname{Re}\operatorname{Tr}(v^2)\le\bar I_\rho(v)$. The quantifier over $n$ includes zero. The module proves result : ¬ claim; its witnesses have rank one and $n>0$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Tomography/PauliFisherGHZRefutation.result

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Quantum informationCounterexample foundFirst preprint 2020Compatible quantum channels need not be Jordan compatibleTwo compatible channels on four-dimensional complex matrices need not be Jordan compatible.Refuted

The questionSikora, Jordan products of quantum channels and their compatibility, Nat.

Why it mattersFor a fixed pair of channels, Jordan compatibility is a semidefinite feasibility condition on A, linear in A.

The ideaLet P = [[0,0,1,0],[0,0,0,1],[1,0,0,0],[0,1,0,0]] and Q = diag(1, 1, −1, −1).

Our contributionTwo compatible channels on four-dimensional complex matrices need not be Jordan compatible.

Scope & next questionJordan compatibility implies compatibility (Proposition 26), and for invertible channels the converse holds (Theorem 30). - The conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

Compatible quantum channels need not be Jordan compatible

M. Girard, M. Plávala and J. Sikora, Jordan products of quantum channels and their compatibility, Nat. Commun. 12, 2129 (2021), arXiv:2009.03279v1, study when two channels can be implemented jointly.

  • Jordan compatibility. For a Hermitian A on X ⊗ X_1 ⊗ X_2 with Tr_{X_1}(A) = Tr_{X_2}(A) = J(1_{L(X)}), the generalized Jordan product Φ_1 ⊙_A Φ_2 has Choi matrix (1 ⊗ Φ_1 ⊗ Φ_2)(A) (Definition 25). The channels are Jordan compatible if this product is completely positive for some such A (Definition 27). Jordan compatibility implies compatibility (Proposition 26), and for invertible channels the converse holds (Theorem 30).
  • The conjecture. After Theorem 30: "the question of whether compatibility of channels is equivalent to Jordan compatibility remains open. We conjecture that such a pair of channels does not exists."

The verbatim statements are in the literature note.

Issue #17171 fixes the reading: for every d, every compatible pair of channels on M_d is Jordan compatible. The conjecture is refuted at d = 4.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumChannels/JordanCompatibilityRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2014The multilayer percolation determinant can have three zeros in [0, 1]Three layers with degree identically 65 give three distinct interior zeros of the multilayer percolation determinant, refuting its conjectured uniqueness.Refuted

The questionBasu, Spanning connectivity in a multilayer network and its relationship to site-bond percolation, arXiv:1402.7057v4 (Phys.

Why it mattersIn the source, qc(M) is the site-occupation threshold above which M independently site-percolated copies of a configuration-model graph have a giant spanning cluster.

The idea1. The matrix. Take M = 3 and the point-mass degree distribution p = δ₆₅, so z = 65 and C = 64.

Our contributionThree layers with degree identically 65 give three distinct interior zeros of the multilayer percolation determinant, refuting its conjectured uniqueness.

Verification details
Release source recordRefuted in Lean

Open problem

The multilayer percolation determinant can have three zeros in [0, 1]

S. Guha, D. Towsley, P. Nain, Ç. Çapar, A. Swami and P. Basu, Spanning connectivity in a multilayer network and its relationship to site-bond percolation, arXiv:1402.7057v4 (Phys. Rev. E 93, 062310):

Numerically, it is easy to verify that $\det (\bA (q))$ has a unique zero in $[0,1]$ for all $M \ge 1$ as long as $C > 1$. We conjecture that this is always true.

Here A(q) is the source's M × M matrix:

  • A_ij = δ_ij − C (binom(M, j) − binom(M − i, j)) q^j (1 − q)^{M−j} for i, j ∈ {1, …, M};
  • C = Σ_k (k − 1) k p_k / z for a degree distribution p with mean degree z.

The definitions are quoted in the literature note.

Issue #19823 fixes the reading: for every M ≥ 1 and every finitely supported degree distribution with C > 1, det A(q) would have exactly one zero in [0, 1]. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/Percolation/MultilayerDeterminantRootNonuniqueness.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2020The Harrow channel pair's parallel Chernoff exponent is not oneFor every positive number of parallel uses, the product state with qubits zero and one gives an error exponent at least two, excluding convergence to one.Refuted

The questionWinter, Usefulness of adaptive strategies in asymptotic quantum channel discrimination, arXiv:2011.06569v3 [quant-ph] (Phys.

Why it mattersExample 5 exhibits a separation between adaptive and parallel discrimination: the source gives a two-use adaptive strategy with zero error and a finite parallel upper bound.

The ideaThe theorem amplifiedoutput (n : ℕ) identifies the stabilized channel difference on this density matrix with the diagonal matrix whose entries are difference n c; entries at reference strings other than the fixed reference n vanish.

Our contributionFor every positive number of parallel uses, the product state with qubits zero and one gives an error exponent at least two, excluding convergence to one.

Scope & next questionThe source's general question about conditions for optimality of non-adaptive protocols remains open.

Verification details
Release source recordRefuted in Lean

Open problem

The Harrow channel pair's parallel Chernoff exponent is not one

F. Salek, M. Hayashi and A. Winter, Usefulness of adaptive strategies in asymptotic quantum channel discrimination, arXiv:2011.06569v3 [quant-ph] (Phys. Rev. A 105, 022419 (2022)), study the two channels in Example 5 from two qubits $A\otimes C$ to one qubit $B$. The source text is the TeX member cq-channel-PRXQ-revised.14jan22mh+aw.tex of arXiv v3, md5 f950feede0e95a6a2aa6ab8f74ee23d0.

For these channels, the source defines the parallel error probability at lines 1305–1306 by $P_{e,\mathbb P}^{(n)}=\frac12\left(1-\frac12\lVert\mathcal M^{\otimes n}-\overline{\mathcal M}^{\otimes n}\rVert_\diamond\right)$ and the non-adaptive Chernoff exponent at lines 1308–1315 by $C^{\mathbb P}(\mathcal M,\overline{\mathcal M})=\lim_{n\to\infty}-\frac1n\log P_{e,\mathbb P}^{(n)}$, with the value $+\infty$ allowed. Example 5 gives $\mathcal M(\rho\otimes\gamma)=|0\rangle\langle0|\langle0|\gamma|0\rangle+|0\rangle\langle0|\langle1|\gamma|1\rangle\langle0|\rho|0\rangle+\frac12I\langle1|\gamma|1\rangle\langle1|\rho|1\rangle$ and $\overline{\mathcal M}(\rho\otimes\gamma)=|+\rangle\langle+|\langle0|\gamma|0\rangle+|1\rangle\langle1|\langle1|\gamma|1\rangle\langle+|\rho|+\rangle+\frac12I\langle1|\gamma|1\rangle\langle-|\rho|-\rangle$, extended by linearity (lines 1406–1417).

After the upper bound $C^{\mathbb P}\le 13.83$, the source states at lines 1509–1514:

Note that a lower bound is the Chernoff bound of the two pure output states $|0\rangle\langle0| = \mathcal{M}(|00\rangle\langle00|)$ and $|+\rangle\langle+| = \overline{\mathcal{M}}(|00\rangle\langle00|)$, which is $\log 2 = 1$, so $C^{\mathbb{P}}(\mathcal{M},\overline{\mathcal{M}}) \geq 1$. It seems reasonable to conjecture that this is optimal, but we do not have at present a proof of it.

The formal proposition is claim, namely $\operatorname{Tendsto};\operatorname{exponent};\operatorname{atTop};(\operatorname{nhds}(1:\mathrm{EReal}))$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/HarrowParallelChernoffExponentRefutation.result

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Quantum informationCounterexample foundFirst preprint 2015The balanced-cut gradient bound of Wölk and Gühne fails for eight qubitsFour singlets within the two halves of an eight-qubit chain have zero squared total spin and gradient second moment 240, exceeding the proposed balanced-cut bound 48.Refuted

The questionGühne, Characterizing the width of entanglement, arXiv:1507.07226v2, New J.

Why it mattersA violation of the bound would certify entanglement across the middle of the chain from two collective measurements: the total spin and a spin component in a magnetic gradient.

The ideaIn each half, place singlets on the half-sites (1,4) and (2,3), so the state is Φ = χ ⊗ χ.

Our contributionFour singlets within the two halves of an eight-qubit chain have zero squared total spin and gradient second moment 240, exceeding the proposed balanced-cut bound 48.

Scope & next questionFor N = 4 each half has only the adjacent pairing, which is why the case N = 4 holds. - Repair.

Verification details
Release source recordRefuted in Lean

Open problem

The balanced-cut gradient bound of Wölk and Gühne fails for eight qubits

S. Wölk and O. Gühne, Characterizing the width of entanglement, arXiv:1507.07226v2, New J. Phys. 18 (2016) 123024, Appendix D. The conjecture states that every state of N spins with entanglement only within the subgroups {1, …, N/2} and {N/2+1, …, N} satisfies

$\langle B^2 \rangle \leq \langle B^2\rangle_- + \frac{\langle B\rangle_\textrm {cl}-\langle B^2\rangle_-}{2N},\langle J^2\rangle$

with ⟨B²⟩₋ = 6N. The gradient operator and the reference values are quoted in the literature note. The printed numerator has ⟨B⟩_cl; the preceding sentence and the endpoint values fix it as ⟨B²⟩_cl.

Issue #19442 fixes the reading:

  • for every even N = 2m ≥ 4 and every density matrix that is separable across the two halves;
  • the coefficient of ⟨J²⟩ is (⟨B²⟩_cl − 6N)/(2N).

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuantumBounds/WidthGradientBalancedCutRefutation.result

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Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2013λ_f = min{1, 1/[4f(0)]} fails for a regular symmetric operator monotone functionA regular symmetric operator monotone function has a skew-information integrand below both endpoint values, refuting the proposed formula for its least value.Refuted

The questionOh, Robertson-Schrödinger uncertainty relation refined by skew information, arXiv:1303.6404v1, Appendix 2:

Why it mattersThe constant λf enters the source's refinement of the Robertson-Schrödinger uncertainty relation by f-skew information: a larger λf gives a stronger relation.

The ideaLet f(x) = (2/9)(1 + x) + (10/3)·x/(11x + 1) + (10/3)·x/(x + 11).

Our contributionA regular symmetric operator monotone function has a skew-information integrand below both endpoint values, refuting the proposed formula for its least value.

Verification details
Release source recordRefuted in Lean

arXiv

λ_f = min{1, 1/[4f(0)]} fails for a regular symmetric operator monotone function

S. Yu and C. H. Oh, Robertson-Schrödinger uncertainty relation refined by skew information, arXiv:1303.6404v1, Appendix 2:

Numerical evidences show that $\lambda_f=1$ whenever $4f(0)\le 1$ and we conjecture that $\lambda_f=\min{1,1/[4f(0)]}$ always holds.

Here λ_f is the minimum over x ≥ 0 of F(x) = (1 + x − f(0)(1 − x)²/(2f(x)))/(2f(x)), for f nonnegative on [0, ∞), operator monotone on Hermitian matrices, symmetric (x f(1/x) = f(x)), normalized (f(1) = 1) and regular (f(0) > 0). The definitions are quoted in the literature note.

Issue #19664 fixes the reading. For every such f, min{1, 1/(4f(0))} would be the least value of F on [0, ∞). The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SkewInformationLambdaRefutation.result

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Quantum informationCounterexample foundFirst preprint 2018Collective dephasing of two qutrits admits nonstationary states with constant positive negativityA nonstationary two-qutrit density matrix has constant positive negativity under Ali's collective dephasing.Refuted

The questionAli, Freezing dynamics of entanglement and nonlocality for qutrit-qutrit (3 ⊗ 3) quantum systems, arXiv:1810.01123v1, published in Mod.

Why it mattersIn the source the feature means that the state changes at every instant while its entanglement stays constant.

The idea1. Evolution. The channel multiplies each entry ρ{ij} by Σk Dk(i) Dk(j).

Our contributionA nonstationary two-qutrit density matrix has constant positive negativity under Ali's collective dephasing.

Verification details
Release source recordRefuted in Lean

Open problem

Collective dephasing of two qutrits admits nonstationary states with constant positive negativity

M. Ali, Freezing dynamics of entanglement and nonlocality for qutrit-qutrit (3 ⊗ 3) quantum systems, arXiv:1810.01123v1, published in Mod. Phys. Lett. A 34, 1950102 (2019). The conjecture is in the Conclusions:

We have found no evidence for time-invariant entanglement for qutrit-qutrit states and based on our results, we conjecture that under current dynamics, this feature of time-invariant entanglement may not exist for qutrit-qutrit systems.

The feature, the Kraus operators of the collective dephasing and the entanglement measure are quoted in the literature note.

Issue #19382 fixes the reading:

  • Hypotheses: a rate Γ > 0 and a density matrix ρ₀ whose negativity is positive and constant along Φ_{Γ,t} for all t ≥ 0.
  • Conclusion: ρ₀ is stationary.

The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Decoherence/QutritCollectiveDephasingInvariantNegativityRefutation.result

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Quantum informationCounterexample foundFirst preprint 2020The quantum crystal potential is not a Yangian combination on the bosonic statesThe three-box bosonic identity-class state contradicts the proposed expression of the negative-flip potential as a linear combination of affine Yangian generators.Refuted

The questionReffert, Quantum crystals, Kagome lattice, and plane partitions fermion-boson duality (Phys.

Why it mattersAn identification of the crystal Hamiltonian with Yangian generators, Eq. (51), would make the quantum crystal an integrable system governed by the affine Yangian of gl̂(1).

The idea2. The potential. As single-layer plane partitions, the row, hook and column have 1, 2 and 1 removable boxes.

Our contributionThe three-box bosonic identity-class state contradicts the proposed expression of the negative-flip potential as a linear combination of affine Yangian generators.

Verification details
Release source recordRefuted in Lean

Open problem

The quantum crystal potential is not a Yangian combination on the bosonic states

T. Araujo, D. Orlando and S. Reffert, Quantum crystals, Kagome lattice, and plane partitions fermion-boson duality (Phys. Rev. D 103, 026020 (2021); arXiv:2005.09103), study the quantum crystal melting Hamiltonian H = −J(Γ + Π) + V(√q O₊ + O₋/√q) on plane-partition states.

  • Γ and Π add and remove boxes in all possible ways.
  • O₊ and O₋ count the boxes that can be added and removed.
  • A fermion-boson correspondence, Eq. (46), defines bosonic states |Λ(N K)⟩⟩ = Σ_Λ χ_Λ[C(K)]|Λ⟩ through characters of symmetric groups; it preserves the number of boxes.

Section VI conjectures: "The growth operators Γ, Π and O±(Γ, Π) are linear combinations of the generators of the algebra Y[gl̂(1)], and the states |N K⟩⟩ fulfill the Y[gl̂(1)] MacMahon representation."

  • In that representation e_j adds and f_j removes a box on the bosonic states, Eq. (49), and φ_j acts diagonally on them, Eq. (50).
  • The authors conclude that "O±(Π, Γ) ∈ B since they can be written as linear combinations of φ_j".

The verbatim statements are in the literature note.

Issue #18238 fixes the reading: the consequences of the conjecture at the three-box bosonic state of the identity class of S₃. The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/QuantumCrystalYangianRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2019The quasidistillable candidates are not two-qutrit negativity maximizers for fixed marginalsAt fixed two-qutrit marginals diag(49,47,4)/100 and diag(94,5,1)/100, a density matrix has greater negativity than the candidate of Baio, Chruscinski, Horodecki, Messina and Sarbicki.Refuted

The questionSarbicki, Bounds on the entanglement of two-qutrit systems from fixed marginals (Phys.

Why it mattersA state maximizing entanglement for given marginals gives an upper bound on the entanglement that local measurements alone allow.

The ideaTake λ = (47, 4)/100 and μ = (5, 1)/100, inside the region of Eq. (27): ρA = diag(49, 47, 4)/100 and ρB = diag(94, 5, 1)/100.

Our contributionAt fixed two-qutrit marginals diag(49,47,4)/100 and diag(94,5,1)/100, a density matrix has greater negativity than the candidate of Baio, Chruscinski, Horodecki, Messina and Sarbicki.

Verification details
Release source recordRefuted in Lean

Open problem

The quasidistillable candidates are not two-qutrit negativity maximizers for fixed marginals

G. Baio, D. Chruściński, P. Horodecki, A. Messina and G. Sarbicki, Bounds on the entanglement of two-qutrit systems from fixed marginals (Phys. Rev. A 99, 062312 (2019); arXiv:1904.07650), study the largest entanglement of a bipartite state whose marginals are fixed.

  • For two qubits the maximum of the negativity N(ρ) = (‖ρ^Γ‖₁ − 1)/2 is attained by an explicit state (12).
  • For two qutrits with ordered marginals ρ_A = diag(1−λ₁−λ₂, λ₁, λ₂) and ρ_B = diag(1−μ₁−μ₂, μ₁, μ₂) (Eq. (24)), they maximize the negativity within the circulant family (22). They obtain three quasidistillable states, Eqs. (27)–(29).
  • From numerical optimization they conclude: "all the above results strongly support our conjecture that our quasidistillable states (27, 28, 29) are legitimate candidates for two-qutrits MEMS with respect to fixed marginals".

The verbatim statements are in the literature note.

Issue #18345 fixes the reading: in the region of Eq. (27), the state (27) maximizes the negativity over all two-qutrit states with the marginals (24). The conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/FixedMarginalNegativityRefutation.result

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Quantum informationCounterexample foundFirst preprint 2014The unitary scaling landscape has a nonglobal local minimumA unitary matrix of order three has a nonglobal local scaling minimum.Refuted

The questionDe Baerdemacker, Scaling a unitary matrix, arXiv:1401.7883v2, Open Syst.

Why it mattersThe landscape governs the authors' iterative algorithm for scaling a unitary matrix to one with all line sums 1, which yields a decomposition of U(n) into diagonal unitary factors and unitary factors whose rows and columns all sum to 1.

The ideaTake n = 3 and A = (1/34)[[−2, 24, 24], [24, 1+17i, 1−17i], [24, 1−17i, 1+17i]].

Our contributionA unitary matrix of order three has a nonglobal local scaling minimum.

Verification details
Release source recordRefuted in Lean

Open problem

The unitary scaling landscape has a nonglobal local minimum

A. De Vos and S. De Baerdemacker, Scaling a unitary matrix, arXiv:1401.7883v2, Open Syst. Inf. Dyn. 21 (2014) 1450013, Appendix B:

We conjecture that all these stationary points are either maxima or saddle points or global minima. In other words: we conjecture that no local minima exist. Moreover, we conjecture that the global minima satisfy~$\Psi=0$.

The landscape Ψ(λ, ρ) = n² − |sum(B)|² with B_{jk} = e^{i(λ_j+ρ_k)} A_{jk} is quoted in the literature note.

Issue #19521 fixes the reading: for every unitary A, every local minimum of Ψ_A is a global minimum. This first sentence of the conjecture is refuted. The second sentence, that global minima have Ψ = 0, is the existence of a scaling, attributed in the v2 comment to Idel and Wolf.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Matrix/UnitaryScalingLandscapeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2014. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021A nonfactorizable doubled triangleA doubled triangle with simultaneous exchange of edge copies has no local factorization.Refuted

The questionNetzer, Polynomial decompositions with invariance and positivity inspired by tensors, arXiv:2109.06680v2; Linear Algebra and its Applications 698 (2024), 537–588, DOI 10.1016/j.laa.2024.05.025, ask in Question 38:

Why it mattersFactorizability supplies local normalization factors for the source's conversion from separable decompositions over local sum-of-squares polynomials to invariant sum-of-squares decompositions.

The ideaThe vertex action is trivial. facetAction fixes copies under the identity and swaps the two copies of every edge simultaneously under the other element. trianglecompatible establishes compatibility for precisely this complex and these two actions.

Our contributionA doubled triangle with simultaneous exchange of edge copies has no local factorization.

Verification details
Release source recordRefuted in Lean

Open problem

A nonfactorizable doubled triangle

G. De las Cuevas, A. Klingler and T. Netzer, Polynomial decompositions with invariance and positivity inspired by tensors, arXiv:2109.06680v2; Linear Algebra and its Applications 698 (2024), 537–588, DOI 10.1016/j.laa.2024.05.025, ask in Question 38:

Are there non-factorizable $(\Omega,G)$ structures?

This is the verbatim question at lines 1546–1549 of the v2 TeX source Manuscript_final.tex (md5 e601e0a3202f352cba4bf6d4d13e7db4). Question 38 is also on PDF page 28; the conclusion repeats the existence question at TeX line 2035. Definitions and further quotations are in the literature note.

The source's definitions at lines 318–332, 425–444 and 1515–1522 give the following reading. A weighted simplicial complex on $[n]={0,\ldots,n}$ assigns natural weights $\Omega(S)$ to vertex subsets, with $\Omega(S)\mid\Omega(T)$ whenever $S\subseteq T$, and nonzero singleton weights. Each maximal nonzero-weight subset $F$ is a facet with $\Omega(F)$ distinct copies. Write $\widetilde{\mathcal F}$ for these multifacets and $\widetilde{\mathcal F}_i$ for those containing $i$. A finite group $G$ acts on vertices preserving weights and on multifacets compatibly with collapse to their underlying facets. Its action on labelings is inverse pullback, ${}^g\gamma(x)=\gamma(g^{-1}x)$.

For each finite index set $\mathcal I$ and labeling $\alpha:\widetilde{\mathcal F}\to\mathcal I$, define $K_\alpha=\left|\left{\gamma:\widetilde{\mathcal F}\to\mathcal I:\ \forall i\in[n],\ \exists g_i\in G,\ g_i i=i\ \text{and}\ ({}^{g_i}\gamma)|i=\alpha|i\right}\right|$. Factorizability requires, for every finite $\mathcal I$, strictly positive real factors $C^{[i]}\beta$, equivariant under $C^{[gi]}{{}^g\beta}=C^{[i]}\beta$, satisfying $\prod{i\in[n]}C^{[i]}_{\alpha|i}=K\alpha^{-1}$ for every global labeling $\alpha$.

Issue #21655 fixes the formal claim as universal factorizability: for every natural $n$, every such weight function, every finite group and every compatible pair of actions, the structure is factorizable. result : ¬ claim refutes that universal assertion and answers the quoted existence question affirmatively.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/TensorNetworks/NonFactorizableDoubledTriangle.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2019Polygon and three-time Leggett–Garg inequalities are insufficient at six timesSix-time pair probabilities satisfy the polygon and three-time inequalities without admitting a joint distribution.Refuted

The questionMawby, Fine's theorem for Leggett-Garg tests with an arbitrary number of measurement times, Phys.

Why it mattersIts failure separates positivity and compatibility of the individual snapshots, together with the listed correlation inequalities, from consistency of the full family with a joint probability.

Our contributionSix-time pair probabilities satisfy the polygon and three-time inequalities without admitting a joint distribution.

Verification details
Release source recordRefuted in Lean

Open problem

Polygon and three-time Leggett–Garg inequalities are insufficient at six times

J. J. Halliwell and C. Mawby, Fine's theorem for Leggett-Garg tests with an arbitrary number of measurement times, Phys. Rev. A 100, 042103 (2019), arXiv:1906.04865v2, Section 4.B, “Sufficient Conditions: A Conjecture”, state:

A set of sufficient conditions for the existence of a joint probability matching all possible pairwise probabilities consists of the $n$-gon equalities Eq.(\ref{neweq}) together with all possible three-time LG inequalities of the form

$1 + s_i s_j C_{ij} + s_i s_k C_{ik} + s_j s_k C_{jk} \ge 0,$

where $i<j<k$.

This quotation is from the downloaded v2 e-print member LGn_paper_revised.tex, lines 554–559, md5 9f186d04357875722ffef09fcdb26481. The word “equalities” is in the source; the referenced Eq. (neweq), lines 494–512, states the inequalities $n+2\sum_{i<j}s_is_jC_{ij}\ge1$ for odd $n$ and $\ge0$ for even $n$, for all signs $s_i\in{\pm1}$. Section 4.B assumes nonnegative pairwise probabilities, and the shared-time compatibility condition is Eq. (cons), lines 150–153.

Issue #23609 fixes the assertion: for every $n\ge3$, every family of nonnegative normalized pair tables compatible on shared times has a matching joint probability whenever its correlators satisfy all full-set $n$-gon inequalities and all three-time LG inequalities. This assertion is refuted by a six-time family. The source and its pairwise setting are documented in the literature note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/LeggettGargSixTimeSufficiencyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    Year in the primary source attribution: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedFirst preprint 2012Ishikawa and Koutschan's Pfaffian conjecture for the Motzkin triangle, part (i)At k = n = 2 the Pfaffian of the matrix of the second column of the Motzkin triangle is -8, while the printed product in part (i) of Ishikawa and Koutschan's Conjecture conj.gen is 8.Refuted

The questionIshikawa and Koutschan write 𝓜^{(k)}i for the number of Motzkin paths from (0,0) to (i−1,k−1), with steps U = (1,1), H = (1,0), D = (1,−1) that never run below the horizontal axis, and define the Pfaffian of a skew-symmetric 2n × 2n matrix as the signed sum over the partitions of [2n] into two-element subsets.

Why it mattersThe formal statement makes no claim about a sign-corrected version of the conjecture.

The ideaThe column 𝓜^{(2)}1, …, 𝓜^{(2)}5 is 0, 1, 2, 5, 12, so the upper entries of the 4 × 4 matrix are a₁₂ = 0, a₁₃ = 2, a₁₄ = 6, a₂₃ = 2, a₂₄ = 10, a₃₄ = 12, and

Our contributionAt k = n = 2 the Pfaffian of the matrix of the second column of the Motzkin triangle is -8, while the printed product in part (i) of Ishikawa and Koutschan's Conjecture conj.gen is 8.

Verification details
Release source recordRefuted in Lean

Open problem

Ishikawa and Koutschan's Pfaffian conjecture for the Motzkin triangle, part (i)

Ishikawa and Koutschan write 𝓜^{(k)}_i for the number of Motzkin paths from (0,0) to (i−1,k−1), with steps U = (1,1), H = (1,0), D = (1,−1) that never run below the horizontal axis, and define the Pfaffian of a skew-symmetric 2n × 2n matrix as the signed sum over the partitions of [2n] into two-element subsets. Part (i) of their Conjecture conj.gen states, for positive integers n and k:

Pf((j−i)𝓜^{(k)}{i+j−2}){1≤i,j≤2n} equals ∏_{i=0}^{m−1}∏_{j=0}^{k−1}(4ki+2j+k) if m = n/k is an integer, and it equals (∏_{j=1}^{⌊k/2⌋} 1/(2j−k))(∏_{i=0}^{m−1}∏_{j=1}^{k}(4ki+2j−k)) if k is odd and m = (n+⌊k/2⌋)/k is an integer. The Pfaffian is zero in all other cases.

At k = 1 this is the paper's Theorem thm.pfMotz. Issue #10039 fixes the readings: the Pfaffian is the source's matching sum over the rationals; the value 𝓜^{(k)}_0 enters only the diagonal entry at i = j = 1, where it is multiplied by j − i = 0; the first case is read as k ∣ n and the second as k odd with k ∣ n + ⌊k/2⌋, the first taking precedence (both hold only at k = 1, where the two products agree); part (ii) is not settled.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/IshikawaKoutschanMotzkinPfaffianRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 2/4 · 10% weight

    First preprint year: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2017Monogamy of the qutrit CGLMP inequalityKumari, Ghose and Mann (arXiv:1704.06516, Phys.Refuted

The questionMann, Sufficient condition for nonexistence of symmetric extension of qudits using Bell inequalities, Phys.

Why it mattersWith the paper's Theorem 3, Eq. (35) was offered as a test for the absence of symmetric extensions of two-qutrit states.

Our contributionKumari, Ghose and Mann (arXiv:1704.06516, Phys.

Verification details
Release source recordRefuted in Lean

Open problem

Monogamy of the qutrit CGLMP inequality

M. Kumari, S. Ghose and R. B. Mann, Sufficient condition for nonexistence of symmetric extension of qudits using Bell inequalities, Phys. Rev. A 96, 012128 (2017), arXiv:1704.06516v2, Eq. (35):

The CGLMP inequality for 2-qutrit states is monogamous, that is, if $\rho_{ABC}$ is any 3-qutrit state such that $\rho_{AB}$, $\rho_{BC}$ and $\rho_{AC}$ are its three 2-qutrit RDMs, at most one of these violates the CGLMP inequality.

Here $\mathcal B_{CGLMP}(\rho)$ is the maximum of the CGLMP expression $\mathcal I_3$ (Eq. (CGLMP1)) over the angles of $A_k=U_{\text{FT}}U(\vec\phi_k)$ and $B_l=U^*{\text{FT}}U(\vec\varphi_l)$, with $U(\vec\phi)=\mathrm{diag}(e^{-i\phi(j)})$ and a computational-basis measurement after each operator (Eq. (CGLMP2)). A state violates the inequality when $\mathcal B{CGLMP}>2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/CglmpMonogamyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2021Optimality of cat pairs for metrology with noisy measurementsKołodyński, Quantum metrology with imperfect measurements, Nature Communications 13, 6971 (2022), arXiv:2109.01160v2.Refuted

The questionKołodyński, Quantum metrology with imperfect measurements, Nature Communications 13, 6971 (2022), arXiv:2109.01160v2.

Our contributionKołodyński, Quantum metrology with imperfect measurements, Nature Communications 13, 6971 (2022), arXiv:2109.01160v2.

Verification details
Release source recordRefuted in Lean

Open problem

Optimality of cat pairs for metrology with noisy measurements

Y. L. Len, T. Gefen, A. Retzker and J. Kołodyński, Quantum metrology with imperfect measurements, Nature Communications 13, 6971 (2022), arXiv:2109.01160v2. An imperfect measurement multiplies the quantum Fisher information by

$\gamma_\cM=\max_{\ket{\xi},\ket{\xi_{\perp}}} \sum_x \frac{\mathrm{Re}!\left{\langle\xi_{\perp}|M_{x}|\xi\rangle\right}^{2}}{\langle\xi|M_{x}|\xi\rangle}$

and for $N$ probes measured independently $M_{\xvec}=M_{x_1}\otimes\cdots\otimes M_{x_N}$. The supplement conjectures:

for any classical noise channel, $M_{x}=\sum_{i}p!\left(x|i\right)\Pi_{i}$ that is applied independently on each of the $N$ probes, the optimal $|\zeta^{N} \rangle,|\zeta_{\perp}^{N}\rangle$ take the form of ``cat states": $|\zeta^{N} \rangle =\cos\left(\theta\right)|j\rangle^{\otimes N}+\sin\left(\theta\right)|k\rangle^{\otimes N}$, $|\zeta_{\perp}^{N} \rangle =-\sin\left(\theta\right)|j\rangle^{\otimes N}+\cos\left(\theta\right)|k\rangle^{\otimes N}$

Here $\zeta,\zeta_\perp$ are the images of $(\psi\pm\psi_\perp)/\sqrt2$ under the encoding unitary, so the optimized pair is $\xi=(\zeta+\zeta_\perp)/\sqrt2$, $\xi_\perp=(\zeta-\zeta_\perp)/\sqrt2$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/NoisyMeasurementCatOptimalityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2019A directed graph with democracy coefficient above oneA weakly connected directed graph on six vertices with twelve unweighted arcs has forward hierarchical levels (227, -991, -991, 329, 767, 659)/2694 and forward democracy coefficient 901/898, which is larger than 1.Refuted

The questionJarvis (arXiv:1908.04358, physics.soc-ph; Sci.

Why it mattersThe democracy coefficient measures how much the vertices that drive a network are themselves influenced, and the conjectured bound η ≤ 1 would make 1 the value of perfectly democratic, balanced networks; the paper derives from it the bound η(G) ≤ 1 − m/n relating the coefficient to its influenced subgraph.

The idea2. g = (227, −991, −991, 329, 767, 659)/2694 satisfies Mᵀ(M g − d) = 0, so ‖M x − d‖² = ‖M g − d‖² + ‖M (x − g)‖² for every x.

Our contributionA weakly connected directed graph on six vertices with twelve unweighted arcs has forward hierarchical levels (227, -991, -991, 329, 767, 659)/2694 and forward democracy coefficient 901/898, which is larger than 1.

Verification details
Release source recordRefuted in Lean

Open problem

A directed graph with democracy coefficient above one

G. Moutsinas, C. Shuaib, W. Guo and S. Jarvis (arXiv:1908.04358, physics.soc-ph; Sci. Rep. 11, 13943 (2021)) define hierarchical levels of the vertices of a weighted simple directed graph. With in-degrees d_j = Σ_i a_ij and M = (diag(d) − A)ᵀ, the forward levels g are the minimum-norm minimizer of ‖M x − d‖₂, and the forward democracy coefficient is η_f = 1 − Mean(g_j − g_i), the mean over the arcs i → j weighted by a_ij; the backward coefficient η_b uses out-degrees. The paper proves η_f, η_b ≥ 0 for weakly connected graphs and η_f = η_b = 1 for balanced graphs, and states:

Let $G$ be a weakly connected directed graph. Then the following are true: $\eta_f(G) \le 1$ and $\eta_b(G) \le 1$; $\eta_f(G)=\eta_b(G)=1$ if and only if the graph is balanced.

Issue #11800 fixes the reading: weighted simple directed graphs with antiparallel arcs allowed, weak connectivity of the underlying graph, the levels as in the paper's Definition 3.1 without assuming uniqueness, and the first bullet's forward inequality as the target. The result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/HierarchyDemocracyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedPosed in 2018Refutation of the A126762 least-witness exponent-shift conjectureThe printed exponent-shift conjecture for OEIS A126762 fails at n = 363.Refuted

The questiona(n) is the least k > n such that the remainder when n^k is divided by k is n.

Why it mattersThis first-tier OEIS conjecture was printed in 2018 and remains labelled as a conjecture in the current entry.

The ideaThe least k > 363 satisfying 363^k ≡ 363 (mod k) is 366: the residues of 363^k modulo k are respectively 1, 333, and 363 at k = 364, 365, 366.

Our contributionThe printed exponent-shift conjecture for OEIS A126762 fails at n = 363.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A126762 least-witness exponent-shift conjecture

OEIS A126762, NAME (verbatim):

a(n) is the least k > n such that the remainder when n^k is divided by k is n.

COMMENT (verbatim; Thomas Ordowski, Aug 03 2018):

a(n) is the smallest number k > n such that n^k == n (mod k). Conjecture: a(n) is the smallest number k > n such that n^(k-1) == 1 (mod k). Thus a(n) is coprime to n. - Thomas Ordowski, Aug 03 2018

The published b-file line 363 366 already contradicts the coprimality consequence, since gcd(363, 366) = 3.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/OrdowskiLeastWitnessRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedPosed in 2012Refutation of the A175406 floor formula conjectureThe greatest integer k such that (1+1/n)^k <= 2Refuted

The questionThe greatest integer k such that (1+1/n)^k <= 2

Why it mattersThe entry presents a universal floor formula for the greatest admissible exponent.

The ideaThe proof uses one certified estimate chain. (i) Real.hasSumlogoneaddinv at a = 1 gives the 36-term lower sum S36 and the geometric upper tail S36 + 3⁻⁷² for log 2. (ii) At n₀ = 1121626023352383, the same series gives a lower bound from its first two positive terms for log(1 + 1/n₀), while Real.loglesuboneofpos gives the upper bound log(1 + x) ≤ x. (iii) Exact rational normalization yields ⌊(n₀ + 1/2) log 2⌋₊ = M with M = 777451915729368, and proves (1 + 1/n₀)^M > 2 while (1 + 1/n₀)^(M−1) ≤ 2.

Our contributionThe greatest integer k such that (1+1/n)^k <= 2

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A175406 floor formula conjecture

OEIS A175406 %N (verbatim):

The greatest integer k such that (1+1/n)^k <= 2

OEIS A175406 %C (verbatim):

The sequence of first differences consists of zeros and ones, with no two consecutive zeros and no more than three consecutive ones.

OEIS A175406 %F (verbatim):

a(n) = n log 2 + O(1). Conjecture: a(n) = floor((n + 1/2) log 2). - Charles R Greathouse IV, Apr 03 2012

OEIS A175406 %A (verbatim):

Zak Seidov, May 01 2010

The literal claim formalized here is ∀ n : ℕ, 1 ≤ n → a n = ⌊((n : ℝ) + 1 / 2) * Real.log 2⌋₊. The definition of a uses the literal natural supremum sSup {k : ℕ | (1 + 1 / (n : ℝ)) ^ k ≤ 2}; in the natural conditionally complete order, an unbounded set has sSup = 0. The result refutes this universal claim at one explicit positive index.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/GreathouseLogTwoFloorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2012. Historical context is capped; this does not establish sustained unsuccessful research.

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Discrete structuresConjecture refutedPosed in 2014Refutation of the A245786 integrality-implies-gcd-record conjectureThe seventh listed A245786 member is not an A216793 gcd record.Refuted

The questionNumbers n such that k(n) = (n/tau(n) + sigma(n)/n) is an integer.

Why it mattersThis first-tier OEIS conjecture was printed in 2014 and remains a universal subsequence claim linking an integrality condition to record values of gcd(sigma(n), n).

The ideaMultiplicativity gives tau(N) = 288 and sigma(N) = 919636480, hence N/288 + 919636480/N = 957958.

Our contributionThe seventh listed A245786 member is not an A216793 gcd record.

Scope & next questionIt asserts no corrected statement and no claim about any other input.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A245786 integrality-implies-gcd-record conjecture

OEIS A245786, NAME (verbatim):

Numbers n such that k(n) = (n/tau(n) + sigma(n)/n) is an integer.

COMMENT conjecture line (verbatim; Jaroslav Krizek, Aug 15 2014):

Conjecture: Subsequence of A216793.

OEIS A216793, NAME (verbatim; Michel Marcus, Sep 16 2012):

Numbers n such that gcd(sigma(n), n) > gcd(sigma(m), m) for all m < n.

The literal claim is ∀ n ≥ 1, IsMember n → IsRecord n, where IsMember is the A245786 integrality condition and IsRecord is the A216793 strict gcd-record condition.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/KrizekIntegralityGcdRecordRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2014. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresConjecture refutedPosed in 2018Refutation of the A319927 odd non-unitary power-sum conjectureThe sequence member 9216 refutes Ianakiev's power-sum divisibility conjecture.Refuted

The questionNumbers k such that the sum of the squares of the odd non-unitary divisors of k divides the sum of the squares of the non-unitary divisors of k.

Why it mattersIanakiev's 2018 comment states a universal divisibility conjecture for every nonnegative power at every sequence member.

The ideaFinite evaluation gives O(2,9216)=9 and S(2,9216)=41243877, so 9216 satisfies the guarded sequence-membership predicate.

Our contributionThe sequence member 9216 refutes Ianakiev's power-sum divisibility conjecture.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A319927 odd non-unitary power-sum conjecture

OEIS A319927, NAME (verbatim):

Numbers k such that the sum of the squares of the odd non-unitary divisors of k divides the sum of the squares of the non-unitary divisors of k.

COMMENT conjecture line (verbatim; Ivan N. Ianakiev, Oct 02 2018):

Conjecture: For any nonnegative integer power p the sum of the p-th powers of the odd non-unitary divisors of a(n) divides the sum of the p-th powers of the non-unitary divisors of a(n).

Peter Munn's COMMENT (verbatim):

The start of this sequence, including the 58 terms currently shown in the data section, is consistent with a definition "nonsquarefree numbers not divisible by 4", but some larger terms are divisible by 4: for example, a(1305) = 9216 = 2^10 * 3^2. - Peter Munn, Sep 21 2020

The PARI membership test (verbatim) is:

isok(n) = my(suo = suo2(n)); if (suo, (su2(n) % suo) == 0);

The AUTHOR line is:

Ivan N. Ianakiev, Oct 02 2018

The first terms printed in the DATA section are 9, 18, 25, 27, 45, 49, 50, 54, 63, 75, ....

For natural numbers, define nonunitaryDivisors(n) = Nat.divisors(n).filter(d => 1 < gcd(d,n/d)), where n/d is natural-number division and is exact for every divisor in Nat.divisors(n). Define S(k,n) as the sum of d^k over those divisors, and O(k,n) as the corresponding sum after the further filter d % 2 = 1. The source program's nonzero guard is retained in member(n) := 0 < n and 0 < O(2,n) and O(2,n) divides S(2,n). The literal refuted statement is ∀ n : ℕ, member n → ∀ k : ℕ, O k n ∣ S k n.

The result does not assert the 2022 remark that p*a(n) is a term, and it does not assert anything about A034444 or A048105.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/IanakievOddNonunitaryPowerSumRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Dated conjecture statement: 2018. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2021Two-qubit conditional domains need not be C*-subalgebrasA two-qubit density matrix with invertible marginals has a conditional domain that is not closed under multiplication.Refuted

The questionParzygnat, Conditional Distributions for Quantum Systems (QPL 2021, EPTCS 343, 1–13; arXiv:2102.01529), constructs quantum conditionals by categorical Bayesian inversion.

Why it mattersOn a conditional domain the conditional is a positive map, so inference by conditioning is valid there.

The ideaIn the basis (00, 01, 10, 11), let ν = (1/32)·[[11, 1, 1, 0], [1, 7, 0, −1], [1, 0, 9, −1], [0, −1, −1, 5]].

Our contributionA two-qubit density matrix with invertible marginals has a conditional domain that is not closed under multiplication.

Verification details
Release source recordRefuted in Lean

Open problem

Two-qubit conditional domains need not be C*-subalgebras

A. J. Parzygnat, Conditional Distributions for Quantum Systems (QPL 2021, EPTCS 343, 1–13; arXiv:2102.01529), constructs quantum conditionals by categorical Bayesian inversion. For a density matrix ν on M_m ⊗ M_n with invertible marginals ρ = tr_B ν and σ = tr_A ν (Corollary 4.5), the conditional domains are:

  • B_ν = {B : tr_B(ν(1 ⊗ B)) commutes with ρ};
  • A_ν = {A : tr_A((A ⊗ 1)ν) commutes with σ} (Definition 5.5).

They are the largest operator systems on which the conditionals are positive (Lemma 5.8, Corollary 5.13).

  • After Example 5.10 the author asks whether the domains are always C*-subalgebras, and answers no with an unrecorded 9 × 9 density matrix of rank 2.
  • Footnote 13 adds: "Also, I could not find a 4 × 4 density matrix ν for which the conditional domains are not C∗-subalgebras, and I suspect that this may always be the case. I hope to resolve this in future work."

The verbatim statements are in the literature note.

Issue #18167 fixes the reading: for every two-qubit density matrix with invertible marginals, both conditional domains are C*-subalgebras of M₂(ℂ). This suspicion is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/ConditionalDomainSubalgebraRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2021. Historical context is capped; this does not establish sustained unsuccessful research.

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Quantum informationCounterexample foundFirst preprint 2019The quantum coherence fraction can increase under incoherent operationsA four-dimensional density matrix and two diagonal Kraus matrices give a strict increase of the quantum coherence fraction.Refuted

The questionA 100, 032324 (2019), arXiv:1906.09789v1, Section V, CF.tex, lines 422–423:

Why it mattersThe fraction optimizes the overlap with a single maximally coherent state, making it a natural measure of proximity to that target set.

Our contributionA four-dimensional density matrix and two diagonal Kraus matrices give a strict increase of the quantum coherence fraction.

Verification details
Release source recordRefuted in Lean

Open problem

The quantum coherence fraction can increase under incoherent operations

Y. Yao, D. Li and C. P. Sun, Quantum coherence fraction, Phys. Rev. A 100, 032324 (2019), arXiv:1906.09789v1, Section V, CF.tex, lines 422–423:

First, similar to the behavior of fully entangled fraction \cite{Badziag2000}, we conjecture that $C_{\mathcal{F}}(\rho)$ may not satisfy the monotonicity under general incoherent operations.

The source fixes the computational basis as the incoherent basis. An incoherent operation has a Kraus representation whose individual branches map diagonal density matrices to diagonal matrices; the normalized nonzero branches are again incoherent states (Section II, lines 122–131). Its quantum coherence fraction is $C_{\mathcal F}(\rho)=\max_{|\phi\rangle\in\mathcal{MCS}}\langle\phi|\rho|\phi\rangle$, where the maximally coherent unit vectors have coordinates of modulus $1/\sqrt d$ (lines 141–150). The definitions and source locator are in the literature note.

Issue #22563 fixes claim as the universal monotonicity assertion: for every $d,m\in\mathbb N$, every complex $d\times d$ matrix $\rho$ that is positive semidefinite with trace one, and every family $K:\mathrm{Fin},m\to M_d(\mathbb C)$ satisfying $\sum_\alpha K_\alpha^\dagger K_\alpha=I$ and preserving diagonality under each sandwich $X\mapsto K_\alpha X K_\alpha^\dagger$ for every complex diagonal matrix $X$, one has $C_{\mathcal F}(\sum_\alpha K_\alpha\rho K_\alpha^\dagger)\le C_{\mathcal F}(\rho)$. Here the module defines $C_{\mathcal F}(A)=\sup{\operatorname{Re}(z^\dagger Az)/d: z\in\mathbb C^d,\ |z_j|=1\text{ for every }j}$ using the real sSup.

The formal resolution is refuted, because result : ¬ claim negates universal monotonicity. This confirms the source's conjecture of non-monotonicity.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Measurement/CoherenceFractionIncoherentNonmonotone.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    Year in the primary source attribution: 2019. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2017Yu's mixed-state skew-coherence polygamy conjecture is falseA mixed state on two by three levels violates skew-coherence polygamy.Refuted

The questionChang-shui Yu, Quantum coherence via skew information and its polygamy, Phys.

Why it mattersThe question concerns the distribution of a single coherence measure between a composite state and its marginals, with all coherences evaluated in compatible reference bases.

The ideaAll matrix certificates below concern only the module's fixed rho, tau, beta and their explicitly defined candidate roots.

Our contributionA mixed state on two by three levels violates skew-coherence polygamy.

Scope & next questionThe numerical observations in Appendix D cannot justify a universal dimension threshold; a characterization of the mixed states satisfying the unmodified inequality remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Yu's mixed-state skew-coherence polygamy conjecture is false

Chang-shui Yu, Quantum coherence via skew information and its polygamy, Phys. Rev. A 95, 042337 (2017), arXiv:1704.04871v1, Discussion and Conclusions, coherenceLy.tex, lines 997–999:

However, we would like to conjecture that it could hold for the bipartite mixed states with the dimension $N \geq 6$.

Appendix D, “The conjecture,” lines 1396–1397, repeats:

In this sense, we would like to conjecture that this relation should be satisfied in $(N\geq 6)$-dimensional systems.

The quotations are checked against the v1 TeX member coherenceLy.tex, md5 026b6fb07e6d276a7bd743cc480af58c. Lemma 1, lines 170–178, defines the reference-basis coherence by $C(\rho)=1-\sum_k\langle k|\sqrt\rho|k\rangle^2$, with the positive square root. Write $Q(\rho)=1-C(\rho)$. Theorem 2, lines 543–550, states $Q(|\Psi\rangle_{AB})\le Q(\rho_A)Q(\rho_B)$ for every bipartite pure state and its two reduced density matrices. The conjecture asks for the same inequality for arbitrary bipartite mixed states of total dimension $N\ge6$; the $2\otimes3$, $3\otimes3$, $3\otimes4$ and $4\otimes4$ figure captions identify $N$ as the total dimension.

Issue #23835 fixes the reading: for all natural $m,n$ with $mn\ge6$ and every positive semidefinite complex matrix $\rho$ of trace one on $\mathbb C^m\otimes\mathbb C^n$, in the product reference basis,

$Q(\rho)\le Q(\operatorname{Tr}_B\rho)Q(\operatorname{Tr}_A\rho)$.

This is claim; result : ¬ claim refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/SkewCoherencePolygamyMixedRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2017. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023Glauber annealing from a thermal state need not preserve neighbourhood passivityA Petersen-graph Glauber annealing run violates neighbourhood passivity.Refuted

The questionDahlsten, Bound on annealing performance from stochastic thermodynamics, with application to simulated annealing, arXiv:2311.10424v1, Phys.

Why it mattersThe paper uses neighbourhood passivity as the mechanism behind the activity bound A^D ≤ 1/2, which enters its bound on annealing performance.

The idea1. Witness. The Petersen graph on ten states, which is connected and 3-regular.

Our contributionA Petersen-graph Glauber annealing run violates neighbourhood passivity.

Verification details
Release source recordRefuted in Lean

Open problem

Glauber annealing from a thermal state need not preserve neighbourhood passivity

Y. Luo, Y.-Z. Zhen, X. Liu, D. Ebler and O. Dahlsten, Bound on annealing performance from stochastic thermodynamics, with application to simulated annealing, arXiv:2311.10424v1, Phys. Rev. E 108, 054119 (2023), bound the activity A^D of simulated annealing with Glauber acceptance (Eqs. (30)–(36)).

  • Neighbourhood passivity (§IV.B, Eq. (41)): P_i = Σ_{j∈N(i)} (γ′_i − γ′_j)(p_i − p_j)/(γ′_i + γ′_j), where primes denote the next step.
  • Its role. If P_i ≥ 0 for all i, then A^D ≤ 1/2 (Eq. (40)), which would support the conjecture A^D(k) ≤ 1/2 of Eq. (39).
  • The expectation: "as the system initially staying in a thermal state evolves towards a new thermal state in SA, we expect the passivity at least in the neighbourhoods to be preserved."

The verbatim statements are in the literature note.

Issue #17931 fixes the reading: for every run from a Gibbs state on a connected regular neighbourhood with distinct energies and a strictly cooling schedule, P_i(k) ≥ 0 for every step k and state i. This expectation is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/GlauberNeighbourhoodPassivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Analysis & dynamicsCounterexample foundFirst preprint 2023The cell-order dynamics of two-dimensional granular systems can have two physical steady statesStrictly positive merging and splitting rates admit two distinct positive normalized stationary cell-order distributions at C = 10.Refuted

The questionBlumenfeld, Steady states of two-dimensional granular systems are unique, stable, and sometimes satisfy detailed balance, J.

Why it mattersThe paper connects uniqueness to experiments in which sheared granular packings always settle into detailed-balance steady states.

The ideaTake C = 10, with its sixteen processes ordered (3,3), (3,4), (3,5), (3,6), (3,7), (3,8), (3,9), (4,4), (4,5), (4,6), (4,7), (4,8), (5,5), (5,6), (5,7), (6,6).

Our contributionStrictly positive merging and splitting rates admit two distinct positive normalized stationary cell-order distributions at C = 10.

Verification details
Release source recordRefuted in Lean

Open problem

The cell-order dynamics of two-dimensional granular systems can have two physical steady states

A. D. C. Myhill and R. Blumenfeld, Steady states of two-dimensional granular systems are unique, stable, and sometimes satisfy detailed balance, J. Phys. A 56, 345001 (2023), arXiv:2306.10526v2, study the evolution of the cell-order distribution Q_3, …, Q_C of a two-dimensional granular packing.

  • The dynamics. Cells of orders i and j merge into a cell of order i + j − 2 at rate p_{i,j}, and split back at rate q_{i,j}; the net flux of the process is η_{i,j} = p_{i,j}Q_iQ_j − q_{i,j}Q_{i+j−2} (§I.A). Eq. (1) gives Q̇_k as the gain from processes producing k-cells, minus the loss from processes consuming them, plus Q_k Σ η, the renormalization for the changing number of cells.
  • What is proved there. A detailed-balance steady state (every η_{i,j} = 0) is unique when it exists; for C = 6 and C = 7 no other steady state exists when it does.
  • The conjecture. "Based on these investigations, we conjecture that, for any choice of constant rates, there is only one physical solution regardless of the upper order, C" (§I.D), where a physical solution has every Q_k real and positive (§III).

The verbatim statements are in the literature note.

Issue #16957 fixes the reading:

  • every C ≥ 3, every assignment of strictly positive constant rates to the processes 3 ≤ i ≤ j with i + j − 2 ≤ C, and normalized positive states;
  • stationarity is the vanishing of the full right-hand side of Eq. (1), read with the flux of §I.A;
  • the printed loss term of Eq. (1) is read as η_{k,i−k+2}, whose splitting factor is the product cell Q_i, as in Eq. (2).

The conjecture is refuted at C = 10.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/GranularCellOrderSteadyStateRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2022Compatible projective tests do not imply compatible quantum channelsEvery pair of projective tests is compatible for a qubit channel that has no joint extension with itself.Refuted

The questionQing-Hua Zhang and Ion Nechita, A Fisher information-based incompatibility criterion for quantum channels, Entropy 24(6), 805 (2022), DOI 10.3390/e24060805, arXiv:2204.09963v1, Section 4, conjecture immediately following Lemma lem:channel-POVM-compatible (article.tex, lines 263–265):

Why it mattersA basis measurement is a classical test of a channel's output.

Our contributionEvery pair of projective tests is compatible for a qubit channel that has no joint extension with itself.

Verification details
Release source recordRefuted in Lean

Open problem

Compatible projective tests do not imply compatible quantum channels

Qing-Hua Zhang and Ion Nechita, A Fisher information-based incompatibility criterion for quantum channels, Entropy 24(6), 805 (2022), DOI 10.3390/e24060805, arXiv:2204.09963v1, Section 4, conjecture immediately following Lemma lem:channel-POVM-compatible (article.tex, lines 263–265):

Consider two quantum channels $\Phi, \Psi: \mathcal L(H_d) \to \mathcal L(H_d)$ such that, for all orthonormal bases $\mathbf e = (e_1, \ldots, e_d)$, $\mathbf f = (f_1, \ldots, f_d)$ of $\mathbb C^d$, the POVMs

$\Big[ \Phi^(|e_i\rangle\langle e_i|) \Big]_{i=1}^d \quad \text{ and } \quad \Big[ \Psi^(|f_j\rangle\langle f_j|) \Big]_{j=1}^d$

are compatible. Then, $\Phi$ and $\Psi$ are compatible channels.

The source's \ketbra macro is expanded in the displayed quotation. The source archive member article.tex has MD5 614f0961ea15b65fb0b83ac83bd3fffb.

Here channels are completely positive, trace-preserving complex linear maps. Two POVMs are compatible when a positive normalized family indexed by both outcome sets has them as its marginals. Two channels are compatible when a single channel to the tensor product of their outputs has them as its partial-trace marginals on every input matrix. These definitions are in the source at lines 124–146 and 179–199 and in the literature note.

Issue #21608 fixes the reading: for every positive dimension and every pair of channels on that space, compatibility of the two induced POVMs for every pair of orthonormal bases implies channel compatibility. This is refuted by two copies of the qubit depolarizing channel $\Phi_{7/10}$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/ProjectiveTestChannelCompatibilityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Quantum informationCounterexample foundFirst preprint 2023Word-trace domination does not control fractional Schatten normsDomination of every word trace does not order Schatten norms of non-integer order.Refuted

The questionKnill, Towards a resolution of the spin alignment problem, arXiv:2307.06894v3, Commun.

Why it mattersThe proposed implication would extend the paper's use of the overlap lemma from integer Schatten orders to every real order at least one.

The ideaThere is no positivity or nonempty-word hypothesis.

Our contributionDomination of every word trace does not order Schatten norms of non-integer order.

Scope & next questionWhether those additional constraints permit a different extension remains open here; the original and strong spin alignment conjectures are not settled by result.

Verification details
Release source recordRefuted in Lean

Open problem

Word-trace domination does not control fractional Schatten norms

M. A. Alhejji and E. Knill, Towards a resolution of the spin alignment problem, arXiv:2307.06894v3, Commun. Math. Phys. 405, 119 (2024), Section VI, Concluding remarks, item 1 (PDF p. 30; TeX main.tex, line 759), ask:

(Schatten norms of non-integer order) We used Cor. IV.6 to resolve the spin alignment problem for Schatten norms of integer order. It is of interest to know if a similar approach can be used in cases where the order of the norm is not an integer. Suppose that $A_{0}, A_{1}, B_{0}, B_{1} \geq 0$ satisfy $\lambda(A_{0}) = \lambda(B_{0})$, $\lambda(A_{1}) = \lambda(B_{1})$. For a string $s \in {0, 1}^{*}$, define the ordered products $\Pi_s(A_{0}, A_{1}) := \Pi_{i \in [|s|]} A_{s(i)}$ and $\Pi_s(B_{0}, B_{1}) := \Pi_{i \in [|s|]} B_{s(i)}$. If for all strings $s$, $\operatorname{tr}(\Pi_s(A_{0}, A_{1})) \geq |\operatorname{tr}(\Pi_s(B_{0}, B_{1}))|$, does it hold that $\lVert A_{0} + A_{1} \rVert_{p} \geq \lVert B_{0} + B_{1} \rVert_{p}$ for all $p \in [1,\infty)$? If so, then the overlap lemma may be used to prove the spin alignment conjecture for all Schatten norms, and by extension for the von Neumann entropy as well.

The quotation follows the PDF, with mathematical notation typeset in Markdown. The source defines $\lambda$ using eigenvalues with multiplicity (main.tex, line 117), and defines $\lVert R\rVert_p=(\operatorname{tr}|R|^p)^{1/p}$ with $|R|=\sqrt{R^*R}$ (lines 274–278). See the literature note.

Issue #21843 fixes the reading: for every natural size $d$, four positive semidefinite complex $d\times d$ matrices with $A_0$ and $B_0$ having equal characteristic polynomials, and likewise $A_1$ and $B_1$, satisfy the norm comparison for every real $p\ge1$ whenever every ordered-word trace of the $A$ pair is real and dominates the absolute value of the corresponding $B$ trace. The universal implication is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Schatten/FractionalSchattenAlignmentRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

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    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023Majorization need not survive normalization in a common charge sectorNormalization inside a common charge sector can reverse a majorization order.Refuted

The questionTonni, Probing RG flows, symmetry resolution and quench dynamics through the capacity of entanglement, arXiv:2301.02117v1, JHEP 03 (2023) 175, Section 7, “Symmetry-resolved related issues”, ask (CDraft-Final.tex, line 4140):

Why it mattersThe question connects a comparison of whole entanglement spectra to comparisons conditioned on a measured charge.

The idea1. State hypotheses. rstate and sstate establish nonnegativity of every entry and total mass one for these respective vectors.

Our contributionNormalization inside a common charge sector can reverse a majorization order.

Verification details
Release source recordRefuted in Lean

Open problem

Majorization need not survive normalization in a common charge sector

R. Arias, G. Di Giulio, E. Keski-Vakkuri and E. Tonni, Probing RG flows, symmetry resolution and quench dynamics through the capacity of entanglement, arXiv:2301.02117v1, JHEP 03 (2023) 175, Section 7, “Symmetry-resolved related issues”, ask (CDraft-Final.tex, line 4140):

For instance, given two density matrices such that $\rho\succ \sigma$, we may ask whether this majorization order survives once we decompose $\rho$ and $\sigma$ according to (\ref{eq:decompositionrho}). One can try to diagnose this aspect using $M_A^{(n)}(q)$ in (\ref{SR-M_n}), given that these quantities are entanglement monotones also in a fixed charge sector. We leave this analysis for future investigations.

The referenced equation is the normalized charge-block decomposition of Section 6, equation (6.1), TeX lines 3550–3569: $\rho_A=\bigoplus_q p(q)\rho_A(q)$, with $p(q)=\operatorname{Tr}(\Pi_q\rho_A)$ and $\operatorname{Tr}\rho_A(q)=1$. Section 6 obtains this decomposition for a reduced state commuting with its subsystem charge; the registered comparison uses one common subsystem charge operator. See the literature note.

Issue #22533 fixes the sectorwise reading: does $\rho\succ\sigma$ imply $\rho(q)\succ\sigma(q)$ for every charge whose weight is positive in both states? This implication is refuted already for finite diagonal states.

The formal claim quantifies over every $n\in\mathbb N$, charge map $c:\mathrm{Fin},n\to\mathbb Z$ and real vectors $r,s:\mathrm{Fin},n\to\mathbb R$ with nonnegative entries and $\sum_i r_i=\sum_i s_i=1$. Put $p_r(q)=\sum_{i:c(i)=q}r_i$, the module's reused pushforward, and let $\operatorname{sec}_q(r)$ be the descending list of $r_i/p_r(q)$ for all indices with $c(i)=q$, the module's sectorSpectrum. Zero eigenvalues retain their multiplicities. The claim says that $\operatorname{desc}(r)\succ\operatorname{desc}(s)$ implies $\operatorname{sec}_q(r)\succ\operatorname{sec}_q(s)$ for every integer $q$ with $p_r(q)>0$ and $p_s(q)>0$.

For equal-length descending lists $x,y$, the reused Majorizes requires every prefix sum of $x$ to be at least the corresponding prefix sum of $y$ for $1\le k\le |x|$, and equal total sums. This is the direction used in the source's definition of majorization (TeX line 830).

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/ChargeSectorMajorizationRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2024Refutation of the regular elementary symmetric image tableConjecture 16's d=5 column is false at n=4: the actual difference is 2, not 1.Refuted

The questionBallantine, Beck, Merca and Sagan, arXiv:2409.11268v3, Section 5, Conjecture 16 (printed page 19), state:

Why it mattersThe conjecture is a universal residue-class table for four regular-image count differences.

The ideaEnumerating their degree-two and degree-three elementary symmetric images gives

Our contributionConjecture 16's d=5 column is false at n=4: the actual difference is 2, not 1.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the regular elementary symmetric image table

Ballantine, Beck, Merca and Sagan, arXiv:2409.11268v3, Section 5, Conjecture 16 (printed page 19), state:

The value of r_{d,2}(n) - r_{d,3}(n) for d = 2, 3, 4, and 5 are shown in the columns of the following table. These values depend on the congruence class of n modulo 2, 6, 4, and 10, respectively. The first column of the table gives the congruence class for n.

Here pre_k(lambda) is the partition whose parts are the products indexed by the k-element position subsets of the parts of lambda. The set ImP_k(n) is the image of the partitions of n having at least k parts, with equal images counted once. The value r_{d,k}(n) counts the members of that image having no part divisible by d.

For d = 5 and n congruent to 4 modulo 10, the printed table gives 3 floor(n/10) + 1. Thus its entry at n = 4 is one.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Recurrence/Partitions/BallantineRegularSymmetricImageRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2022A two-element T1-groupoid whose reduct is not a reduced T1-groupoidA first tetrahedral groupoid can have a reduct that is not a reduced first tetrahedral groupoid.Refuted

The questionLeshkov, Set-theoretical solutions of simplex equations, arXiv:2206.08906v1, Section 9.2 (Мат. труды 27(1) (2024), pp. 51–53; Siberian Adv.

Our contributionA first tetrahedral groupoid can have a reduct that is not a reduced first tetrahedral groupoid.

Verification details
Release source recordRefuted in Lean

Open problem

A two-element T1-groupoid whose reduct is not a reduced T1-groupoid

V. Bardakov, B. Chuzinov, I. Emel'yanenkov, M. Ivanov, T. Kozlovskaya and V. Leshkov, Set-theoretical solutions of simplex equations, arXiv:2206.08906v1, Section 9.2 (Мат. труды 27(1) (2024), pp. 51–53; Siberian Adv. Math. 34(1) (2024) 1–40), define a first tetrahedral 4-groupoid ($T_1$-groupoid) $(X,,\circ,\triangleleft,\triangleright)$ by four identities and a reduced $T_1$-groupoid $(X,,\circ)$ by three identities, both giving elementary 1-solutions of the tetrahedron equation. They write: "It is not clear if there is a $T_1$-groupoid such that forgetting about $x\triangleleft y$ and $x\triangleright y$ will not yield a reduced $T_1$-groupoid." The verbatim statements are in the literature note.

Issue #14643 reads the question as whether the universal statement "the reduct $(X,*,\circ)$ of every $T_1$-groupoid is a reduced $T_1$-groupoid" fails.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/VertexModels/TetrahedralGroupoidReductRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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Analysis & dynamicsCounterexample foundFirst preprint 2021The averaged-kernel characterization failsWe suspect that the property that σ minimizes IK is equivalent to the fact that UK^{σ^{n−2}} is conditionally positive definite, i.e. the two-input energy I{UK^{σ^{n−2}}} is minimized by σ.Refuted

The questionZ. 301 (2022), Section 5, paragraph before Theorem 5.3, verbatim:

Why it mattersAveraging detects a quadratic variation at the uniform measure.

The ideaFor x on the unit sphere in dimension d, define

Our contributionWe suspect that the property that σ minimizes IK is equivalent to the fact that UK^{σ^{n−2}} is conditionally positive definite, i.e. the two-input energy I{UK^{σ^{n−2}}} is minimized by σ.

Verification details
Release source recordRefuted in Lean

Open problem

The averaged-kernel characterization fails

D. Bilyk, D. Ferizović, A. Glazyrin, R. Matzke, J. Park and O. Vlasiuk, Potential theory with multivariate kernels, Math. Z. 301 (2022), Section 5, paragraph before Theorem 5.3, verbatim:

We suspect that the property that σ minimizes I_K is equivalent to the fact that U_K^{σ^{n−2}} is conditionally positive definite, i.e. the two-input energy I_{U_K^{σ^{n−2}}} is minimized by σ. This conjecture is supported by all the examples known to us.

The setting is the unit sphere in real Euclidean space, normalized surface measure σ, and continuous symmetric rotationally invariant kernels with n ≥ 3. Definitions 2.1–2.2 give the measure and energy conditions; (5.1) gives invariance.

| Source object | Formal reading | |---|---| | S^{d−1} | Defs.S d, the unit sphere subtype in EuclideanSpace ℝ (Fin d) with its Borel structure | | σ | Defs.sigma d, radial spherical measure of volume scaled to mass one | | K | (Fin (m+3) → Defs.S d) → ℝ, where n=m+3 | | Continuity, symmetry, rotational invariance | Defs.IsAdmissible; the counterexample is invariant under the full orthogonal group and hence under the special orthogonal group | | I_K(μ) | Defs.energy K μ, the integral against μ in every input | | σ minimizes I_K | Defs.IsMinimizer K (Defs.sigma d), over all Borel probability measures | | U_K^{σ^{n−2}} | Defs.avgKernel K, integrating the first n−2 inputs | | Conditional positive definiteness | Defs.IsCondPosDef, the quadratic form on differences of two finite positive measures of equal total mass; Jordan decomposition covers all finite signed measures of mass zero | | Universal equivalence | Defs.claim, for all d≥2 and n≥3; its orthogonally invariant subclass already contains the refuting family |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MultivariateEnergy/Result.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

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Quantum informationCounterexample foundFirst preprint 2024The Bu–Gu–Jaffe mixed-state quantum inverse sumset bound is falseThe equal mixture of two levels in dimension seven violates the proposed mixed-state quantum inverse sumset bound.Refuted

The questionJaffe, Quantum Ruzsa Divergence to Quantify Magic, IEEE Transactions on Information Theory 71(4) (2025) 2726–2740, DOI 10.1109/TIT.2025.3543276; arXiv:2401.14385v3, Remark 45, equation (58).

Why it mattersThe quantum inverse sumset problem asks how the entropy increase under self-convolution controls distance to the mean state.

Our contributionThe equal mixture of two levels in dimension seven violates the proposed mixed-state quantum inverse sumset bound.

Scope & next questionThis refutation does not settle that separate conjecture, the qubit bounds or the Rényi versions.

Verification details
Release source recordRefuted in Lean

Open problem

The Bu–Gu–Jaffe mixed-state quantum inverse sumset bound is false

K. Bu, W. Gu and A. Jaffe, Quantum Ruzsa Divergence to Quantify Magic, IEEE Transactions on Information Theory 71(4) (2025) 2726–2740, DOI 10.1109/TIT.2025.3543276; arXiv:2401.14385v3, Remark 45, equation (58). The v3 source IEEE_Final_BGJ.tex, line 2106, states verbatim:

In addition, we also expect that the property (2) in Theorem \ref{thm:QIST} holds, which has the following form

The reference is to Theorem 44. Its proposed mixed-state inequality, at lines 2107–2109, is

$D(\rho|\mathcal M(\rho))\leq\dfrac{\log R(\rho)}{\log R(\rho)-\log[1+\lambda(R(\rho)\operatorname{Tr}[\rho^2]-1)]}\log C.\qquad(58)$

Here $R(\rho)=\operatorname{rank}\mathcal M(\rho)$ and $\lambda=(1-MG(\rho))^2$. The characteristic function is $\Xi_\rho(p,q)=\operatorname{Tr}[\rho w(-p,-q)]$; the mean state retains exactly its coefficients of modulus one. The magic gap $MG$ is one minus the largest modulus among the nonzero coefficients whose modulus is not one, with $MG=0$ if that finite set is empty. The source definitions and their locations are in the literature note.

Issue #24984 fixes the single-qudit reading encoded by claim: for every nonzero prime dimension $d$, every nonzero $s,t\in\mathbb Z_d$ with $s^2+t^2=1$, every three density matrices $\rho,\sigma,\tau$ with $\sigma=\mathcal M(\rho)$ and $\tau=\rho\boxtimes_{s,t}\rho$, and every real $C$, require (58) whenever

$1<\delta_q[\rho]=\exp(S(\tau)-S(\rho))\leq C,\qquad A=1+\lambda(R(\rho)\operatorname{Tr}[\rho^2]-1)>0,\qquad\log R(\rho)-\log A>0.$

Density matrices are positive semidefinite complex matrices of trace one. Entropy and relative entropy are the natural-logarithm trace expressions $S(\rho)=-\operatorname{Tr}(\rho\log\rho)$ and $D(\rho|\sigma)=\operatorname{Tr}[\rho(\log\rho-\log\sigma)]$. The two positivity assumptions restrict to a positive logarithm argument and denominator. This restriction of the expected universal bound is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Magic/MixedStateInverseSumsetRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2023A non-semi-positive semi-nonnegative qutrit map can signal information backflowA semi-nonnegative qutrit map that is not semi-positive doubles trace distance between two states in a common Kraus context.Refuted

The questionSP maps are the intermediate maps of CP-indivisible, non-Markovian evolutions. - For qubits, the non-SP SN maps send only one pure state to a state, so they "signal no information backflow". - The Outlook conjectures: "We conjecture that non-SP SNTP maps still signal no information backflow and are not uniquely defined even for higher dimensional H."

Why it mattersInformation backflow, an increase of distinguishability, is the operational signature of non-Markovian evolution in the trace-distance sense that the source cites.

The idea1. The map. On 3 × 3 matrices X = [[A, u], [v, c]], with A of size 2 × 2, let Ψ(X) = [[2A − (tr A)1/2 + c·1, u], [v, −c]].

Our contributionA semi-nonnegative qutrit map that is not semi-positive doubles trace distance between two states in a common Kraus context.

Verification details
Release source recordRefuted in Lean

arXiv

A non-semi-positive semi-nonnegative qutrit map can signal information backflow

N. Cao, M. Fitzsimmons, Z. Mann, R. Pereira and R. Laflamme, Quantum Maps Between CPTP and HPTP, arXiv:2308.01894, classify the Hermiticity-preserving trace-preserving (HPTP) maps that can describe the evolution of an open system.

  • A map is semi-nonnegative (SN) if it sends some density matrix to a density matrix (Definition 2), equivalently if Ψ ∘ Φ = Ξ for some CPTP Φ and Ξ (Theorem 2).
  • It is semi-positive (SP) if it sends some invertible density matrix to an invertible one (Definition 1). SP maps are the intermediate maps of CP-indivisible, non-Markovian evolutions.
  • For qubits, the non-SP SN maps send only one pure state to a state, so they "signal no information backflow".
  • The Outlook conjectures: "We conjecture that non-SP SNTP maps still signal no information backflow and are not uniquely defined even for higher dimensional H."

The verbatim statements are in the literature note.

Issue #18048 fixes the reading of the first conjunct: in every context Ψ ∘ Φ = Ξ of a non-SP SNTP map, the intermediate map does not increase the trace distance of two states. This conjunct, and with it the conjecture, is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/QuantumChannels/SemiNonnegativeBackflowRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Geometry & algebraCounterexample foundFirst preprint 2022Locality tensor products need not be locality vector spacesA locality structure on real five-space refutes both conjectures for componentwise and four-way locality.Refuted

The questionPaycha, Tensor products and the Milnor-Moore theorem in the locality setup, arXiv:2205.14616v1 (European Journal of Mathematics 11, 41, 2025), Conjectural statement 6.16, states:

Why it mattersLocality tensor products are used to build tensor algebras, enveloping algebras and the locality version of the Milnor–Moore theorem.

The ideaFour-way relatedness implies componentwise relatedness, so the same obstruction excludes four-way support comparisons.

Our contributionA locality structure on real five-space refutes both conjectures for componentwise and four-way locality.

Verification details
Release source recordRefuted in Lean

Open problem

Locality tensor products need not be locality vector spaces

P. J. Clavier, L. Foissy, D. A. López and S. Paycha, Tensor products and the Milnor-Moore theorem in the locality setup, arXiv:2205.14616v1 (European Journal of Mathematics 11, 41, 2025), Conjectural statement 6.16, states:

Given a locality vector space $( {E},\top)$ and $V_1$, $V_2$ two of its subspaces, the subspace $I_{\rm bil} {(V_1, V_2)} {\cap\mathbb{K}(V_1\times_\top V_2)}\subset \mathbb{K}(V_1\times_\top V_2)$ is locality compatible with $\top_{V_1\times_\top V_2}$.

The locator is the downloaded TeX Clavier_Foissy_Lopez_Paycha_arxiv.tex, lines 1984–1987, label conj:main1, md5 1f6041a82cd80114573aa776694ccf20. Quotations expand typesetting macros without changing the wording.

The paper's Notations, Clavier_Foissy_Lopez_Paycha_arxiv.tex, line 321, specifies that $\mathbb K$ is a commutative field of characteristic zero. Both conjectures below, under both generator-relation readings, quantify over fields of characteristic zero. The thesis states no characteristic restriction; refuting its assertion even on characteristic-zero fields also refutes the unrestricted assertion.

The tensor-product conjecture is stated explicitly by P. J. Clavier, TRAPs, Generalisations of MZVs, Locality and Resurgence for Quantum Field Theories, arXiv:2506.09493v1 (2025), chapter LOC.tex, lines 1797–1799, label conj:loc_tensor, md5 41f5e772e40d5eaf8583e8903293dbbb:

Let $(E,\top)$ be any locality vector space, $V$ and $W$ any two locality vector subspaces of $E$. Then $(V\otimes_\top W,\overline{\top})$ is a locality vector space.

For a field $\mathbb K$, a locality vector space is a $\mathbb K$-vector space $E$ with a symmetric relation $r$, with $r(0,0)$, such that the polar $U^r={x:\forall u\in U,\ r(u,x)}$ of every subset $U\subseteq E$ is a linear subspace. For subspaces $V,W\subseteq E$, put $F=\mathbb K(V\times_\top W)$, the free span of pairs $(v,w)\in V\times W$ satisfying $r(v,w)$. In the ambient free span on all pairs, let $I_{\rm bil}(V,W)$ be the span of the two additivity and two scalar-multiplication relations, and put $I=F\cap I_{\rm bil}(V,W)$.

The module uses two generator relations on pairs $(v,w)$ and $(v',w')$:

  • Prod.RProd r r: $r(v,v')\wedge r(w,w')$, as in the 2022 paper, Eq. eq:locreltoptimes, TeX lines 447–449;
  • fourWay: $r(v,w)\wedge r(v,v')\wedge r(v,w')\wedge r(w,v')\wedge r(w,w')\wedge r(v',w')$, all six pairs among the four entries, as in the thesis's locality cartesian power, Eq. eq:lprod, LOC.tex lines 219–221, and its description of tensor locality through $E^{\times_\top4}$, lines 743–747.

The sources disagree: the paper's Remark at lines 1972–1975 identifies its componentwise relation, when $V=W$, with $V^{\times_\top4}$, whereas the thesis's cartesian-power definition requires all six pairs to be related. The two generator relations differ in general, even on local pairs; the computed example in Triage distinguishes them.

For either generator relation $g$, write $s_g(a,b)$ when every pair in the support of $a\in F$ is $g$-related to every pair in the support of $b\in F$. The relation on $F/I=V\otimes_\top W$ is existential: two classes are related precisely when they have $s_g$-related representatives. Locality compatibility of $I$ means that for all $x,y,z\in F$ and $w\in I$, if $s_g(x,y)$ and $s_g(x+w,z)$, there is $w'\in I$ with both $s_g(x+w',y)$ and $s_g(x+w',z)$.

The problem is tracked in issue #21552, including comment 6396883196. The module defines TensorConjecture g as the tensor-product assertion and PairConjecture g as the compatibility assertion for the chosen generator relation, each universally quantified over every field $\mathbb K$ of characteristic zero, every $\mathbb K$-vector space $E$ in Type, every locality relation and every pair of subspaces, without a finite-dimensionality hypothesis. The module defines claim := TensorConjecture (fun _ r => Prod.RProd r r) ∨ TensorConjecture fourWay ∨ PairConjecture (fun _ r => Prod.RProd r r) ∨ PairConjecture fourWay and proves result : ¬ claim, refuting both universal assertions for each generator relation.

The definitions and source locators are recorded in the paper note and the thesis note.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/HomologicalAlgebra/LocalityTensorProductRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    First preprint year: 2022. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2023The activity bound on cross-correlation asymmetry fails for general networksA sixteen-state network contradicts the proposed general-network activity bound on cross-correlation asymmetry.Refuted

The questionGu, Thermodynamic bounds on the asymmetry of cross-correlations with dynamical activity and entropy production, Phys.

Why it mattersEq. (8) would bound a measurable asymmetry of correlations by the total activity and the dissipation of an arbitrary network, giving an experimental lower bound on entropy production in terms of activity.

The idea- the complete graph on {0, 6, …, 15} with all rates 1.

Our contributionA sixteen-state network contradicts the proposed general-network activity bound on cross-correlation asymmetry.

Verification details
Release source recordRefuted in Lean

Open problem

The activity bound on cross-correlation asymmetry fails for general networks

J. Gu, Thermodynamic bounds on the asymmetry of cross-correlations with dynamical activity and entropy production, Phys. Rev. E 109, L042101 (2024), arXiv:2312.00354, bounds the normalized asymmetry χ_ab = α_ab/√(D_a D_b) of the cross-correlation of two observables in a Markov jump process on N states.

  • Eq. (8) bounds it by the dynamical activity A, the entropy production σ and the flux scale ratio γ: |χ_ab| ≤ γ/tan(π/N) · √(σ/(2A)).
  • The author proves Eq. (8) for unicyclic networks and the weaker Eq. (9) in general, and writes: "We conjecture that Eq. (8) is valid for arbitrary networks". The evidence is 10⁶ random samples on three four-state topologies.

The verbatim statements are in the literature note.

Issue #18028 fixes the reading: Eq. (8) holds for every N ≥ 3, every stationary network with bidirectional support and a positive flux, and all observables with D_a D_b > 0. This conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/CrossCorrelationActivityBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

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Analysis & dynamicsCounterexample foundFirst preprint 2023Gu's pseudo-coarse-grained dissipation-time trade-off is falseA positive three-state cyclic chain violates the proposed two-state lower bound on non-adiabatic entropy production.Refuted

The questionJie Gu, Speed limit, dissipation bound and dissipation-time trade-off in thermal relaxation processes, Phys.

Why it mattersThe proposed bound uses only two endpoints, an elapsed time and the total rate, while allowing dynamics without detailed balance.

The ideaAll vectors and constants in the construction are fixed, except where a lemma's real parameter and hypotheses are stated explicitly.

Our contributionA positive three-state cyclic chain violates the proposed two-state lower bound on non-adiabatic entropy production.

Scope & next questionWhether detailed balance suffices for Eq. (13), or which additional compatibility condition on the binary dynamics suffices for general multistate processes, remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Gu's pseudo-coarse-grained dissipation-time trade-off is false

Jie Gu, Speed limit, dissipation bound and dissipation-time trade-off in thermal relaxation processes, Phys. Rev. E 108, L052103 (2023), DOI 10.1103/PhysRevE.108.L052103, arXiv:2304.08752v2, states the following conjecture in the section “Trade-off between dissipation and time,” Eq. (13). The verbatim sentence and equation are from relaxation.tex, lines 328 and 331 (the equation's mathematical content is rendered inline):

Let us state the third main result: for $N$-state dynamics evolving from $\boldsymbol{p}^\text{(i)}$ to $\boldsymbol{p}^\text{(f)}$ during $\tau$ with an total rate $\mathcal{W}$, the conjecture is that the entropy production $\Sigma$ is bounded from below by (referred to as trade-off bound)

$\Sigma_\text{na} \ge \Sigma_{\Omega_i}(\mathcal W \tau) \quad i =1,2,\ldots,S_N^{(2)}.$

The displayed inequality concerns non-adiabatic entropy production, despite the sentence's use of $\Sigma$. Lines 333–337 define each binary comparison by a partition into two nonempty sets, the aggregate endpoints, and the original process's total rate and elapsed time. The source conventions and the two-state steady-state formula are also recorded in the literature note.

Issue #21842 fixes the following restricted reading, encoded by claim. For every natural number $n$, rate array $a:\mathrm{Fin},n\to\mathrm{Fin},n\to\mathbb R$, vector $u$, real curve $P:\mathbb R\to\mathbb R^n$, real time $\tau$, and finite subset $\Omega$ of the states, put

$W_{ij}=a_{ij}$ for $i\ne j$, and $W_{jj}=-\sum_{i\ne j}a_{ij}$.

Assume all of the following:

  • $a_{ij}>0$ whenever $i\ne j$;
  • $u_i>0$ for every state, $\sum_i u_i=1$, and $Wu=0$;
  • $P$ has derivative $WP(t)$ at every real $t$;
  • $\tau>0$, every coordinate of $P(0)$ is positive, $\sum_iP_i(0)=1$, and every coordinate of $P(\tau)$ is positive;
  • $\Omega$ and its complement are both nonempty.

Define $D(v|w)=\sum_i v_i\log(v_i/w_i)$, $d(z|b)=D((z,1-z)|(b,1-b))$, and

$x=\sum_{i\in\Omega}P_i(0)$, $y=\sum_{i\in\Omega}P_i(\tau)$, $k=-\operatorname{Tr}W$, $s=x+(y-x)/(1-e^{-k\tau})$.

Whenever $0<s<1$, the proposed inequality is

$d(x|s)-d(y|s)\le D(P(0)|u)-D(P(\tau)|u)$.

The module proves result : ¬ claim. Thus even this restriction to positive off-diagonal rates, positive endpoints and an interior fitted binary steady state is false.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/PseudoCoarseGrainedTradeoffRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    Year in the primary source attribution: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationConjecture refutedFirst preprint 2023Guo's partial-negativity subadditivity conjectureAn equal mixture of two orthogonal maximally entangled states on a 5 × 5 system has reduced function 1/2, whereas both marginals have reduced function 1/5.Refuted

Why it mattersEquation (24) is used as an assumption for later constructions in the source.

Our contributionAn equal mixture of two orthogonal maximally entangled states on a 5 × 5 system has reduced function 1/2, whereas both marginals have reduced function 1/5.

Scope & next questionThe nearest surviving question recorded here is the universal claim for local dimensions at most four, which remains open.

Verification details
Release source recordRefuted in Lean

Open problem

Guo's partial-negativity subadditivity conjecture

Y. Guo, Complete Genuine Multipartite Entanglement Monotone, Results in Physics 57 (2024), 107430, arXiv:2301.00334v3, states the reduced function $\hat h(\rho)=\sqrt{\delta_1\delta_2}$, where $\delta_1$ and $\delta_2$ are the two largest eigenvalues of $\rho$, and conjectures in equation (24) that $\hat h(\rho_{AB})\leq\hat h(\rho_A)+\hat h(\rho_B)$ for every bipartite density matrix.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | two largest eigenvalues counted with multiplicity | hHat uses positions 0 and 1 of the decreasing eigenvalues₀ list | | every pair of local dimensions | claim quantifies dA dB : ℕ | | every bipartite state | claim quantifies ρ : Matrix (Fin dA × Fin dB) (Fin dA × Fin dB) ℂ and assumes IsDensity ρ | | reduced states are the two partial traces | ρ.traceRight and partialTraceFirst ρ in claim | | subadditivity inequality (24) | claim concludes hHat ρ ≤ hHat ρ.traceRight + hHat (partialTraceFirst ρ) | | refutation | result : ¬ claim |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/PartialNegativity/GuoHatSubadditivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundFirst preprint 2023Refutation of the rational Q-system infinite-root primitive countHou--Jiang--Miao's root-of-unity primitive-count formula is negative at the admissible triple (26,11,1), so it cannot be a natural-number count.Refuted

The questionClosed Chains, arXiv:2310.14966v3, SciPost Phys. 16, 129 (2024), Appendix C, conjecture (C.2), page 35:

Why it mattersThe conjecture supplies the infinite-root primitive counts used by the recursive counting algorithm in Appendix C.

The ideaLet formula(L,M,n) be the integer-valued right-hand side of (C.2).

Our contributionHou--Jiang--Miao's root-of-unity primitive-count formula is negative at the admissible triple (26,11,1), so it cannot be a natural-number count.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of the rational Q-system infinite-root primitive count

J. Hou, Y. Jiang and Y. Miao, Rational Q-systems at Root of Unity I. Closed Chains, arXiv:2310.14966v3, SciPost Phys. 16, 129 (2024), Appendix C, conjecture (C.2), page 35:

Before introducing the algorithm, we make the following conjecture for the number of primitive states with infinite Bethe root(s) by observing the numerical results: N^{pri}{±∞}(L, M, n_±) = binom(L, M − n_±) − sum{x=0}^{M−n_±−1} binom(L,x), (C.2), when n_± = n_+ = n_− is a solution to (3.15). When there is no solution to (3.15), N^{pri}_{±∞}(L, M, n_±) = 0.

Appendix C specializes to no twist κ = 1, η = iπ/3, even L and M ≤ L/2. Equations (3.15)–(3.17) imply L ≡ 2(M − n) (mod 6) and 0 ≤ n ≤ 2: η = iπ/3 gives ℓ₁ = 1 and ℓ₂ = 3. For states with infinite roots, n ≥ 1; n ≤ M records that these roots are among the M Bethe roots.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/RationalQSystemInfinityCountRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023QTD squared can exceed base-2 quantum Jensen–Shannon divergenceA full-rank commuting qubit pair has quantum triangular discrimination squared greater than base-2 quantum Jensen–Shannon divergence.Refuted

The questionYupan Liu, Quantum state testing beyond the polarizing regime and quantum triangular discrimination, arXiv:2303.01952v5, §1.3, “Improved inequalities on the quantum triangular discrimination”, printed p. 7, states:

Why it mattersThe proposed lower inequality strengthens the factor-½ lower bound of Theorem 3.4.

The ideaBoth matrices are positive semidefinite with unit trace.

Our contributionA full-rank commuting qubit pair has quantum triangular discrimination squared greater than base-2 quantum Jensen–Shannon divergence.

Verification details
Release source recordRefuted in Lean

Open problem

QTD squared can exceed base-2 quantum Jensen–Shannon divergence

Yupan Liu, Quantum state testing beyond the polarizing regime and quantum triangular discrimination, arXiv:2303.01952v5, §1.3, “Improved inequalities on the quantum triangular discrimination”, printed p. 7, states:

We observe that Theorem 1.4(ii) is not tight. Numerical simulations indicate that the tight bound is QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁) for any states ρ₀ and ρ₁.

After Theorem 3.4:

Numerical simulations suggest that the tight bound is QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁), while we only managed to prove a slightly weaker bound ½QTD²(ρ₀, ρ₁) ≤ QJS₂(ρ₀, ρ₁) ≤ QTD(ρ₀, ρ₁) in Theorem 3.4.

The paper was published in Computational Complexity 34 (2025), article 11. Definition 3.1 specifies the inverse square root on the support. Definition 2.10 defines QJS by the entropy of the average minus the average entropy; QJS₂ uses base-2 logarithms.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/QuantumTriangularJensenShannonRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsCounterexample foundFirst preprint 2023A three-point law exceeds the Meer–Weeks two-sided kurtosis upper limitA three-point probability law exceeds the proposed two-sided kurtosis upper limit.Refuted

The questionThe quotation is checked against the v2 TeX source, md5 4452c28a26c7a76c87642af46fe0c8a9.

Why it mattersThe proposed bounds would let the support, mean and variance constrain the fourth standardized moment of any bounded distribution by the extrema obtained from two-point distributions.

Our contributionA three-point probability law exceeds the proposed two-sided kurtosis upper limit.

Scope & next questionNo general replacement for asymmetric intervals is formalized here; that question remains open in this dossier.

Verification details
Release source recordRefuted in Lean

Open problem

A three-point law exceeds the Meer–Weeks two-sided kurtosis upper limit

David J. Meer and Eric R. Weeks, Bounded distributions place limits on skewness and larger moments, PLOS ONE 19, e0297862 (2024), DOI 10.1371/journal.pone.0297862; arXiv:2308.05006v2, subsection “Kurtosis $D_4$”, plos_latex_template.tex, line 467:

While we have not proven that the bidisperse distribution sets the limits for $D_4$ for all other distributions, this is suggestive that Inequalities \ref{D4mostgeneral} are indeed limits for the kurtosis for any distribution.

The quotation is checked against the v2 TeX source, md5 4452c28a26c7a76c87642af46fe0c8a9. The referenced inequalities are at lines 461–464; their definitions are at lines 437–460. The moment conventions are at lines 228–236 and 290–293. See the literature note.

Issue #21539 fixes the following universal reading. For every real interval $L<U$ and every probability law $P$ on $\mathbb R$ with $L\le X\le U$ almost surely, mean $\mu=\mathbb E X>0$ and variance $v=\mathbb E[(X-\mu)^2]>0$, put

$\sigma=\sqrt v$, $\delta=\sigma/\mu$, $\delta_{\min}=|1-L/\mu|$, $\delta_{\max}=|1-U/\mu|$, $\delta_1=\min(\delta_{\min},\delta_{\max})$, $\delta_2=\max(\delta_{\min},\delta_{\max})$, $\delta'=\sqrt{\delta_1\delta_2}$,

and $D_4=\mathbb E[(X-\mu)^4]/v^2$, $D_{4,m}(\delta)=(\delta/\delta_m-\delta_m/\delta)^2+1$ for $m=1,2$. The assertion requires both:

  • if $\delta<\delta_1$, then $1\le D_4\le D_{4,2}(\delta)$;
  • if $\delta_1\le\delta<\delta'$, then $D_{4,1}(\delta)\le D_4\le D_{4,2}(\delta)$.

This reading is refuted. It makes no assertion for $\delta\ge\delta'$; the positive-mean restriction suffices for the refutation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Analytic/MomentMinors/BoundedKurtosisLimitRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Year in the primary source attribution: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Number theoryCounterexample foundPosed in 2023The permutation question for A359482The greedy decimal sequence A359482 omits every single-digit successor and therefore omits 2.Refuted

The questionIs this sequence a permutation of the integers > 0?

Why it mattersThis Tier 1 OEIS question has a local decimal obstruction that can decide the unbounded permutation question.

The ideaFor x>0 and 1<=d<=9 the sum has at least as many digits as x.

Our contributionThe greedy decimal sequence A359482 omits every single-digit successor and therefore omits 2.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The permutation question for A359482

Is this sequence a permutation of the integers > 0?

OEIS A359482 defines the lexicographically earliest sequence of distinct positive terms for which a(n)+a(n+1) is a decimal substring of the concatenation of a(n) and a(n+1). This dossier concerns exactly the permutation question. Hasler's July 3, 2023 comment conjectures a negative answer and, more specifically, that the initial 1 is the only single-digit term.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/SumInConcatenation.not_positive_permutation

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    Dated conjecture statement: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Analysis & dynamicsConjecture refutedFirst preprint 2024A pinned triangle refutes the spanning-tree dimension conjectureA triangle of pinned reversible edges in unit-rate K4 makes the rooted spanning-tree vectors linearly dependent.Refuted

The questionPolettini, Harunari, Dal Cengio and Lecomte, Coplanarity of rooted spanning-tree vectors, arXiv:2407.16093v1, Introduction, conjecture:

Why it mattersThe proposed dimension depends only on the number of pins.

Our contributionA triangle of pinned reversible edges in unit-rate K4 makes the rooted spanning-tree vectors linearly dependent.

Scope & next questionThe source's one-pin and two-pin theorems remain untouched; no other source theorem is claimed to be invalidated.

Verification details
Release source recordRefuted in Lean

Open problem

A pinned triangle refutes the spanning-tree dimension conjecture

Polettini, Harunari, Dal Cengio and Lecomte, Coplanarity of rooted spanning-tree vectors, arXiv:2407.16093v1, Introduction, conjecture:

Finally, we conjecture that the 3ⁿ-vectors representing the deletion-constriction through n reversible edges span a n+1 dimensional space.

The source is recorded in the Library note. The preregistration is #19157. The universal assertion is refuted by unit-rate $K_4$ with the reversible edges $(0,1),(1,2),(2,0)$ pinned.

Statement echo

| Source clause or encoding | Lean counterpart | Fidelity and scope | | --- | --- | --- | | Irreducibility: a path of nonzero weight between any two vertices | Irreducible | Reflexive transitive closure of positive-rate arcs; all rates are nonnegative | | A spanning tree directs every edge toward its root | IsRootedTree | The parent fixes the root; every non-root edge has positive rate; every vertex reaches the root | | An arc belongs to the tree | Contains | Parent-map encoding excludes the auxiliary root loop | | Deletion-constriction through $n$ reversible edges | HasStatus | Neither arc, forward only, or backward only, giving $3^n$ coordinates | | Definition 1: sum of products over the conditioned trees | tau | The literal sum over parent maps with the selected statuses; products exclude the root | | Deletion of all pins | deletePins | Both orientations of every pin receive rate zero | | Dimension $n+1$ | claim | Real finrank of the span of the root-indexed vectors | | Distinct reversible edges | Positive rates in both directions, distinct endpoints and injectivity through Sym2.mk | No parallel duplicate pins | | Irreducibility after deletion | Last hypothesis of claim | A stronger admissibility condition, satisfied by the counterexample; it does not weaken the refutation | | Universal parameters | Natural $N,n$, rates on Fin N, pins indexed by Fin n | Finite labelled graphs; aliases are expanded without changing the source convention |

The root loop encodes a fixed root and contributes no weight. An empty product is one. The defining formulas and all parameters appear in the Blueprint mirror; result displays the entire negated universal assertion.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/PolettiniPinnedTreeSpanRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
  • Historical context · 1/4 · 10% weight

    First preprint year: 2024. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2024The four-qubit Fredkin gate generates more than two ebits across AD:BCThe four-qubit Fredkin gate F4 generates more than two ebits across the cut AD:BC: one product input with auxiliary systems gives 2.00034... ebits.Refuted

The questionA 111, 022407 (2025)) define the entanglement generation K{L:L^c}(U) of a multipartite gate as the supremum, over pure inputs ψi of each party with a local auxiliary system Ri, of the von Neumann entropy in ebits of the reduced state on L of U(ψ1 ⊗ ⋯ ⊗ ψn).

Why it mattersFor the three other cuts the paper determines the entanglement generation exactly, and the conjecture would fix the entangling power of F4 at two ebits.

The ideaThe output amplitude matrix across the cut is M = DA N DB, with DA, DB diagonal and N rational.

Our contributionThe four-qubit Fredkin gate F4 generates more than two ebits across the cut AD:BC: one product input with auxiliary systems gives 2.00034... ebits.

Verification details
Release source recordRefuted in Lean

Open problem

The four-qubit Fredkin gate generates more than two ebits across AD:BC

X. Qiu, Z. Song and L. Chen (arXiv:2410.15253, quant-ph; Phys. Rev. A 111, 022407 (2025)) define the entanglement generation K_{L:L^c}(U) of a multipartite gate as the supremum, over pure inputs ψ_i of each party with a local auxiliary system R_i, of the von Neumann entropy in ebits of the reduced state on L of U(ψ_1 ⊗ ⋯ ⊗ ψ_n). For the four-qubit Fredkin gate F_4 = (|00⟩⟨00| + |01⟩⟨01| + |10⟩⟨10|)_{AB} ⊗ I_{CD} + |11⟩⟨11|_{AB} ⊗ (S_2)_{CD} they prove K_{AD:BC}(F_4) ∈ [2, log_2 5] and state:

Further we conjecture that $K_{AD:BC}(F_4)\leq 2$ by numerical results and thus have the following. The entanglement generation $K_{AD:BC}(F_4)=2$. Hence the entangling power of a four-qubit Fredkin gate $F_4$ is equal to two ebits.

Issue #11460 fixes the reading. Each auxiliary system has any finite dimension d_i ≥ 1, the inputs are unit vectors, F_4 acts as the identity on the auxiliary systems, and the entropy is taken on A R_A D R_D. The formal claim is the first sentence, K_{AD:BC}(F_4) = 2; the result refutes it.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/FredkinEntanglingPowerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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Analysis & dynamicsCounterexample foundFirst preprint 2023Squares do not maximize the mean exit time of lattice clustersA connected lattice disk with 121 vertices has larger mean first exit time than the 11 by 11 square.Refuted

The questionNicosia, Clusters in randomly-coloured spatial networks, arXiv:2308.07880v1 (J.

Why it mattersThe source proposes the mean exit time as a descriptor of the compactness of coloured clusters: lines minimize it, and the square, with the most internal 4-cycles and the smallest surface, is conjectured to maximize it.

Our contributionA connected lattice disk with 121 vertices has larger mean first exit time than the 11 by 11 square.

Verification details
Release source recordRefuted in Lean

Open problem

Squares do not maximize the mean exit time of lattice clusters

S. Rognone and V. Nicosia, Clusters in randomly-coloured spatial networks, arXiv:2308.07880v1 (J. Phys. Complexity, 2026), section on the mean exit time from a cluster:

Characterising clusters with maximal exit time at each fixed $N$ seems to be a harder quest. Here we conjecture that the maximal exit time is obtained for squares with $N$ nodes.

Here:

  • Cluster: a set of nodes of the square lattice, connected through nearest neighbours; the walk moves to each of the four neighbours with probability 1/4.
  • Mean exit time $\tau(S)$: the average over the starting nodes of $S$ of the expected first time the walk is outside $S$.

The definitions are quoted in the literature note. The source's displayed formula averages node-to-node hitting times, while its prose and every worked first-step system use the first-exit time; the settlement uses the first-exit time.

Issue #20034 fixes the reading: for every $m\ge1$ and every connected cluster $S$ of $m^2$ nodes, $\tau(S)\le\tau({0,\dots,m-1}^2)$. This is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/RandomWalks/SquareClusterExitTimeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    Source 1 · Source 2
Quantum informationCounterexample foundFirst preprint 2023Steering does not guarantee atemporalityA two-qubit state is spatially steerable and temporally compatible.Refuted

The questionGu, Causal classification of spatiotemporal quantum correlations, arXiv:2306.09336v2 (Phys.

Why it mattersThe source introduces atemporality as a certificate that observed correlations need a common cause: no temporal process, sending one qubit through a channel, reproduces them.

The ideaLet R = (1/10000)[[5391,0,0,1782],[0,3609,1188,0],[0,1188,401,0],[1782,0,0,599]].

Our contributionA two-qubit state is spatially steerable and temporally compatible.

Scope & next questionThe question remains open for states steerable in both directions, and for Bell-nonlocal states.

Verification details
Release source recordRefuted in Lean

Open problem

Steering does not guarantee atemporality

M. Song, V. Narasimhachar, B. Regula, T. J. Elliott and M. Gu, Causal classification of spatiotemporal quantum correlations, arXiv:2306.09336v2 (Phys. Rev. Lett. 133, 110202), Discussion:

Some entangled states are not atemporal, so is atemporality a strictly stronger notion of quantum correlations? And if so, is it guaranteed by steering or Bell non-locality?

Here:

  • Temporally compatible from A to B: R = (id ⊗ Λ)K with K = {ρ_A ⊗ I/2, S} for a quantum channel Λ, the source's Eq. (3). The reverse direction is symmetric.
  • Atemporal: temporally compatible in neither direction.

The definitions are quoted in the literature note.

Issue #19923 fixes the reading of the steering clause: every two-qubit density operator that is EPR-steerable in at least one direction would be atemporal. This is refuted. The Bell-nonlocality clause is not addressed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/SteeringTemporalCompatibilityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

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    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
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    First preprint year: 2023. Historical context is capped; this does not establish sustained unsuccessful research.

    Source 1 · Source 2
Discrete structuresCounterexample foundSource 2025A stronger cumulative Sturmian progression bound is falseRefute the cumulative Sturmian bound at slope two minus square root three.Refuted

The questionGandhar Joshi, Automatic sequences from algebraic, dynamical, and computational perspectives, Open University PhD thesis, §6.1, p. 118:

Why it mattersIts unrestricted version would order all shorter differences by a convergent denominator.

The ideaThe proof uses the published FibonacciMapFirstStart.openrotationrun to exclude wrapping inside either open symbol interval.

Our contributionRefute the cumulative Sturmian bound at slope two minus square root three.

Verification details
Release source recordRefuted in Lean

Open problem

A stronger cumulative Sturmian progression bound is false

Gandhar Joshi, Automatic sequences from algebraic, dynamical, and computational perspectives, Open University PhD thesis, §6.1, p. 118:

“The data generated by the A(d) formulae in Corollary 4.3.13 heuristically directs to a conjecture that for every n ∈ N, d ≤ qₙ ⟹ A(d) ≤ A(qₙ) (Proposition 4.4.4) could be true for all g(d). As done for the specific case in [32, Corollary 4.17], this possibility for all Sturmians will be assessed in the author’s future work.”

The literal conjecture quantifies over irrational $0<\alpha<1/2$, positive $n$, the convergent denominator $q_n$, and $1\le d\le q_n$. Preregistration: https://github.com/the-omega-institute/trureturing/issues/17837.

Statement echo

| Source clause | Formal counterpart | Fidelity | | --- | --- | --- | | Definition 2.2.3: the letter at $n$ uses ${(n+1)\alpha}$ and intervals $[0,1-\alpha)$, $[1-\alpha,1)$ | lowerMechanicalWord α α; private sturmian_letter_iff | Same indexing and intercept; true is the second letter. | | Definitions 2.6.1–2.6.2: all starts, either letter, positive finite lengths, infinity if unbounded | mapLength | Full nested supremum in $\mathbb N\cup{\infty}$. | | Irrational slope in Chapter 4's $0<\alpha<1/2$ | claim slope hypotheses | Literal domain. | | Every positive $n$, with $q_0=1$ and $q_1=a_1$ | 1 ≤ n, (GenContFract.of α).dens n | Natural index and real denominator, identified with natural $q$. | | Every positive $d\le q_n$ has $A(d)\le A(q_n)$ | The final quantified implication of claim | No additional condition on $g(d)$. | | Refutation | result : ¬ claim | Witness $\alpha=2-\sqrt3$, $n=1$, $q=3$, $d=2$. |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S1/Words/SturmianMapCumulativeRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
  • Historical context

    Year recorded in the source bibliography: 2025.

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    The motivation identifies an earlier mathematical result or partial case.

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Discrete structuresCounterexample foundFirst preprint 2026The unrestricted grid-power zero-forcing conjecture is falseConjecture 12 as printed fails: two initially black vertices force the three-by-two grid at power one, whereas the conjectured value is three.Refuted

The questionAbiad, Flagg, Ghasemi Nezhad and Zeijlemaker, The Zero Forcing Number of Graph Powers, arXiv:2610.03892v1, §3, Conjecture 12 (conj:grid):

Why it mattersThe targeted issue is the literal quantified assertion, not the classical grid formula.

The ideaName vertices A=(0,0), B=(0,1), C=(1,0), D=(1,1), E=(2,0), F=(2,1).

Our contributionConjecture 12 as printed fails: two initially black vertices force the three-by-two grid at power one, whereas the conjectured value is three.

Scope & next questionIt does not settle a corrected b≤ℓ conjecture or assert author intent.

Verification details
Release source recordRefuted in Lean

arXiv

abiad flagg ghasemi nezhad zeijlemaker 2026 grid power zero forcing

Abiad, Flagg, Ghasemi Nezhad and Zeijlemaker, The Zero Forcing Number of Graph Powers, arXiv:2610.03892v1, §3, Conjecture 12 (conj:grid):

Let b≥1, k≥1 and ℓ≥k+1. Then Z((P_b □ P_ℓ)^k)=bk.

The endpoint is refutation of this full universal statement as printed. It does not settle a corrected b≤ℓ conjecture or assert author intent.

Statement echo

| Source clause | Lean counterpart | | --- | --- | | Every natural b,k,ℓ | ∀ b k ell : ℕ in claim | | b≥1, k≥1, ℓ≥k+1 | 1 ≤ b → 1 ≤ k → k + 1 ≤ ell → | | P_n is the n-vertex path | Mathlib pathGraph n on Fin n, adjacency of consecutive values | | Cartesian product | Mathlib pathGraph b □ pathGraph ell, on Fin b × Fin ell | | Distinct vertices at base distance ≤k | graphPower: u ≠ v ∧ G.edist u v ≤ (k : ℕ∞) | | Ordinary unique-white-neighbor forcing | Frozen ZeroForcingThree.Black and IsZeroForcing | | Minimum initial cardinality | Frozen ZeroForcingThree.zeroForcingNumber, an infimum of attainable finite cardinalities | | Equality to bk | zeroForcingNumber (...) = b * k | | Negation of the entire assertion | result : ¬ claim |

All admissible grids are connected and finite: positive path factors are connected, and Cartesian products preserve reachability. Extended graph distance is used directly, so no disconnected-distance collapse or k=1-only power definition affects the universal statement. The coordinate correspondence is (i,j)↦(i+1,j+1). No source hypothesis is dropped or added. Black.force can also derive a vertex already black, which adds no vertex and does not alter the closure; when its target is white, its premises say precisely that the source is black and every other neighbor is black.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ZeroForcing/GridPowerZeroForcingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresCounterexample foundFirst preprint 2026Problem 15.5 answered negatively by a five-element Clifford semigroupA five-element orthodox semigroup with an idempotent full-ordering product has no complete mapping.Refuted

The questionAraújo, Bentz, Cameron, Hendrey, and Kinyon, Complete Mappings of Semigroups, arXiv:2608.25092v1 (2026), define a complete mapping in the abstract on printed page 1:

Why it mattersThe counterexample is the five-element semigroup W with the following multiplication table.

The ideaThe result introduces no intermediate proposition beyond finite computation and normalization.

Our contributionA five-element orthodox semigroup with an idempotent full-ordering product has no complete mapping.

Verification details
Release source recordRefuted in Lean

Open problem

Problem 15.5 answered negatively by a five-element Clifford semigroup

Araújo, Bentz, Cameron, Hendrey, and Kinyon, Complete Mappings of Semigroups, arXiv:2608.25092v1 (2026), define a complete mapping in the abstract on printed page 1:

A complete mapping of a semigroup S is a bijection α : S → S such that the map θ : S → S defined by xθ = x · xα is also a bijection.

Section 2 defines regularity on printed page 6:

An element a of a semigroup S is said to be regular if there exists b ∈ S such that aba = a. If every element of a semigroup is regular, then the semigroup itself is said to be regular.

Section 14 defines E-semigroups and orthodox semigroups on printed page 59:

A semigroup S is said to be an E-semigroup if the set E(S) of idempotents is a subsemigroup of S. A regular E-semigroup is said to be orthodox.

Theorem 14.3 on printed page 59 states:

Theorem 14.3. Let S be a finite E-semigroup with a complete mapping. Then there exists an ordering c1, . . . , cn of all elements of S such that c1 · · · cn is an idempotent.

The following remark on printed page 59 records the zero case:

We remark that for semigroups with zero this conclusion can be uninformative: nothing rules out the possibility that the product of elements given by Theorem 14.3 is zero, even if restricted to the non-zero elements. For instance, for the 3-element commutative idempotent semigroup S = {0, 1, 2} with 1 · 2 = 0, let α = θ = id_S and note that Pα = {{0}, {1}, {2}}.

Section 15 then says on printed page 60:

The converse of Theorem 14.3 leads to the following question. An affirmative answer would generalize the Hall–Paige conjecture.

Problem 15.5 on printed page 60 asks:

Problem 15.5. Let S be an orthodox semigroup with an ordering c1, . . . , cn of all elements of S such that c1 · · · cn is an idempotent. Must S have a complete mapping?

The formal reading makes six decisions: (i) an ordering of all elements is a duplicate-free finite list containing every element; (ii) c₁ ⋯ c_n is the left fold; (iii) idempotent means p * p = p; (iv) orthodox means regular and closed under multiplication of idempotents; (v) a complete mapping is a bijection whose product map is also bijective; and (vi) the question is the universal claim over S : Type, so one counterexample answers No.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/AraujoOrthodoxCompleteMappingRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2025A closest product vector to a q-Dicke state need not be q-symmetricAt q equal to four, no closest normalized product vector to the one-excitation three-qubit q-Dicke state is q-symmetric.Refuted

The questionRelancio, Deformations of the symmetric subspace of qubit chains, J.

Why it mattersThe paper proposes transferring the simplification of geometric entanglement optimization for ordinary symmetric states to q-Dicke states.

The ideaAll vectors and norms below are over the ordinary complex Euclidean spaces.

Our contributionAt q equal to four, no closest normalized product vector to the one-excitation three-qubit q-Dicke state is q-symmetric.

Scope & next questionA general replacement theorem using consistently deformed norms and overlaps remains open here. - Nearby statements that survive.

Verification details
Release source recordRefuted in Lean

Open problem

A closest product vector to a q-Dicke state need not be q-symmetric

A. Ballesteros, I. Gutierrez-Sagredo, J. de Ramón and J. J. Relancio, Deformations of the symmetric subspace of qubit chains, J. Phys. A 58, 405301 (2025), arXiv:2503.23554v3, subsection “Entanglement of q-symmetric states”, TeX member newsubmission__1_.tex, l. 1258:

Therefore, the closest separable state to a $q$-symmetric state for two qubits is also $q$-symmetric. If we conjecture that a similar relation holds for arbitrary $q$-Dicke states, it would mean that we can construct whole new families of states whose multipartite entanglement has an analytic expression.

The optimization in that subsection is the ordinary overlap modulus over product vectors with every local factor normalized (l. 1234–1243). Write $|0\rangle=|\downarrow\rangle$ and $|1\rangle=|\uparrow\rangle$. For $q>0$, the three-qubit vectors at l. 581–582 are

$|D^1_3\rangle_q=(\sqrt q,|001\rangle+|010\rangle+q^{-1/2}|100\rangle)/\sqrt{q+1+q^{-1}}$,

$|D^2_3\rangle_q=(\sqrt q,|011\rangle+|101\rangle+q^{-1/2}|110\rangle)/\sqrt{q+1+q^{-1}}$.

Here $[3]_q=q+1+q^{-1}$ follows from the source's q-number at l. 469, with its undeformed convention at $q=1$. The source identifies $\mathrm{Sym}_q\mathcal H$ with the complex span of $|000\rangle$, $|D^1_3\rangle_q$, $|D^2_3\rangle_q$ and $|111\rangle$ (l. 586). These definitions are recorded in the literature note.

Issue #21970 fixes the reading: for every real $q>0$, there exist unit $a,b,c\in\mathbb C^2$ such that $a\otimes b\otimes c$ maximizes $|\langle D^1_3|_q(a\otimes b\otimes c)\rangle|$ over all unit local factors and belongs to $\mathrm{Sym}_q\mathcal H$. This three-qubit, one-excitation instance of the conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/GeometricMeasure/QDickeClosestProductRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresConjecture refutedFirst preprint 2026Refutation of Beluhov's invariant-kernel dimension conjecture at type (1,3,6)For every proper type 𝐧 of order n, it holds that dim Ω⊠ = ⌊n/2⌋.Refuted

The questionNikolai Beluhov, Diamond Determinants and Somos Sequences, arXiv:2602.24239v2, Section 10, page 24, states:

Why it mattersThe settlement studies the proper type (1,3,6), whose order is 10, and tests the predicted dimension 5 over the independent-parameter field.

Our contributionFor every proper type 𝐧 of order n, it holds that dim Ω⊠ = ⌊n/2⌋.

Scope & next questionThe full type-dependent gauge interpretation, an exact dimension at type (1,3,6), a corrected universal formula, and consequences for other types remain open.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of Beluhov's invariant-kernel dimension conjecture at type (1,3,6)

Nikolai Beluhov, Diamond Determinants and Somos Sequences, arXiv:2602.24239v2, Section 10, page 24, states:

For every proper type 𝐧 of order n, it holds that dim Ω⊠ = ⌊n/2⌋.

The formal claim uses the paper's stated parity-only gauge space, the rational-function field in three independent coefficients, and the kernel of the displayed operator over the constrained monomial subspace.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/Combinatorics/GaleRobinsonKernelDimensionRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2025A type-4c diagonal state refutes the geometric tangle ansatzA type-4c canonical state on the Bloch-norm diagonal refutes the Benedito–Sierra geometric tangle ansatz.Refuted

The questionSierra, Visualizing Three-Qubit Entanglement, arXiv:2505.23638v2, p. 7, Eq. (15), state:

Why it mattersThe diagonal distance vanishes, yet the proposed leading term disagrees with the canonical tangle.

The ideaThe three Pauli vectors are (0,0,−1/2), (1/2,0,0) and (1/2,0,0).

Our contributionA type-4c canonical state on the Bloch-norm diagonal refutes the Benedito–Sierra geometric tangle ansatz.

Verification details
Release source recordRefuted in Lean

arXiv

A type-4c diagonal state refutes the geometric tangle ansatz

A. Benedito and G. Sierra, Visualizing Three-Qubit Entanglement, arXiv:2505.23638v2, p. 7, Eq. (15), state:

“τ(r⃗) = 1 − |r⃗|²/3 − d(r⃗, V_line) · F(r⃗) where |ψ⟩ ∈ GHZ excluding type 5 and F(r⃗) ≥ 0.”

The canonical state is λ₀|000⟩ + λ₁ e^{iφ}|100⟩ + λ₂|101⟩ + λ₃|110⟩ + λ₄|111⟩, with 0 ≤ λⱼ ≤ 1, Σ λⱼ² = 1 and 0 ≤ φ ≤ π (p. 3, Eqs. (3)–(4)). The vector r⃗ = (r_A,r_B,r_C) contains the lengths of the three reduced one-qubit Pauli vectors. V_line = {(t,t,t)} is the main diagonal in ℝ³, and d is its Euclidean distance, defined as the infimum of distances to points on that line. The canonical three-tangle is τ(ψ) = 4|Hdet(t_ijk)| = 4λ₀²λ₄² (p. 5, Eq. (10)).

Issue #11500 preregisters the existence of a nonnegative real function F satisfying the equation for every normalized canonical GHZ state that is not type 5. The formal claim retains exactly this state domain. Type 5 means every λⱼ ≠ 0 and every Jₖ ≠ 0. The J invariants are the Acín invariants cited by the paper, with μⱼ = λⱼ² and Δ = |λ₁λ₄ exp(iφ) − λ₂λ₃|²: J₁ = Δ, J₂ = μ₀μ₂, J₃ = μ₀μ₃, J₄ = μ₀μ₄ and J₅ = μ₀(Δ + μ₂μ₃ − μ₁μ₄). Their canonical definitions are Eq. (23) of arXiv:quant-ph/0003050, the paper's cited classification source. The witness has λ₁ = 0, which excludes it from the full type-5 predicate without evaluating any J invariant.

The complex tensor amplitudes gives all eight coefficients of the state. jointDensity is its outer product, with entries t_ijk conjugate(t_i'j'k'). rhoA, rhoB and rhoC are finite-sum partial traces using the frozen partialTraceLeft and partialTraceRight. The frozen bloch map extracts (2 Re ρ₀₁, −2 Im ρ₀₁, Re ρ₀₀ − Re ρ₁₁). blochLengths takes the Euclidean lengths of these actual vectors. cayley is the full complex hyperdeterminant quartic; tangle is 4 ‖cayley (amplitudes ψ)‖, with no coordinate shortcut in the definition of the claim.

The source and the formal distance definition agree as follows:

| Source definition | Lean definition | | --- | --- | | “the distance from that point to the straight line spanned by the main diagonal is the length of the vector connecting it to a point on the line such that this vector is perpendicular to the line” (arXiv v2, Appendix B, before the displayed distance formula) | V_line : Set (EuclideanSpace ℝ (Fin 3)) := Set.range (fun t : ℝ => WithLp.toLp 2 ![t,t,t]); euclideanVector r := WithLp.toLp 2 ![r.1,r.2.1,r.2.2]; distanceToDiagonal r := Metric.infDist (euclideanVector r) V_line |

The vector embedding and the diagonal set both live in EuclideanSpace ℝ (Fin 3). The local distance step in result applies Metric.infDist_zero_of_mem with range witness t = 1/2.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/ThreeQubitGeometricTangleRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Quantum informationCounterexample foundFirst preprint 2026Subsystem Lanczos positivity for general initial statesThe first squared subsystem Lanczos coefficient can be negative for an entangled two-qubit state.Refuted

The questionLoc, Complexity Inequalities for Quantum Subsystems, arXiv:2606.20790v2, §3.1, printed p. 15 (PDF page 16), state:

Why it mattersThe conjecture supplies the proposed reality of all subsystem Lanczos coefficients.

The ideaIt differentiates the actual matrix exponential, the conjugated density, partial trace and overlap.

Our contributionThe first squared subsystem Lanczos coefficient can be negative for an entangled two-qubit state.

Scope & next questionAny general construction requiring real off-diagonal coefficients needs an additional condition; which broader complexity conclusions can be retained with complex coefficients remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

Subsystem Lanczos positivity for general initial states

P. Caputa, G. Di Giulio and T. Q. Loc, Complexity Inequalities for Quantum Subsystems, arXiv:2606.20790v2, §3.1, printed p. 15 (PDF page 16), state:

At present, however, we are unable to establish the sign of $\left(b_n^{(A)}\right)^2$ in full generality. Based on all the examples discussed in this manuscript, we conjecture that $\left(b_n^{(A)}\right)^2>0$ for every $n$, and hence that all the coefficients $b_n^{(A)}$ are real.

The source evolves a normalized pure state by $U(t)=e^{-\mathrm{i}Ht}$, with a time-independent Hermitian Hamiltonian on a bipartite Hilbert space. Its reduced density and fixed-normalization return amplitude are

$$ \rho_A(t)=\operatorname{Tr}_B(U(t)|\psi\rangle\langle\psi|U(t)^\dagger), \qquad R_A(t)=\frac{\operatorname{Tr}(\rho_A(t)\rho_A(0))} {\operatorname{Tr}(\rho_A(0)^2)}. $$

Equations (3.3) and (3.5) define $\mu_n^{(A)}=\partial_t^nR_A(t)|_{t=0}$ and $(b_1^{(A)})^2=(\mu_1^{(A)})^2-\mu_2^{(A)}$. The finite-dimensional $n=1$ clause quantifies over both subsystem dimensions, every normalized pure state and every Hermitian Hamiltonian. A negative first square disproves the full conjecture, including a nonnegative reading that allows terminating zero coefficients.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Dynamics/SubsystemLanczosPositivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresConjecture refutedFirst preprint 2025Cohen Conjecture 29The printed thresholds in Cohen's Conjectures 28 and 29 fail at n = 11 and n = 12.Refuted

The questionCohen, Journal of Integer Sequences 28 (2025), Article 25.4.7, Section 2.5, Conjecture 29, asserts that there are at least two cousin-prime pairs between consecutive cubes for every natural n > 1 and that the count eventually exceeds every positive integer.

Why it mattersThe printed thresholds are part of the named conjecture, but the count sequence printed immediately before them gives seven cousin-prime pairs for n = 12.

The ideaDefine cousinPairCount n as the number of natural numbers p with n^3 < p, p + 4 < (n+1)^3, prime endpoints, and no prime among p + 1, p + 2, and p + 3.

Our contributionThe printed thresholds in Cohen's Conjectures 28 and 29 fail at n = 11 and n = 12.

Verification details
Release source recordRefuted in Lean

Open problem

Cohen Conjecture 29

Cohen, Journal of Integer Sequences 28 (2025), Article 25.4.7, Section 2.5, Conjecture 29, asserts that there are at least two cousin-prime pairs between consecutive cubes for every natural n > 1 and that the count eventually exceeds every positive integer. Its printed threshold table further states N(8) = N(9) = N(10) = 12: every natural n >= 12 has at least eight, nine, and ten such pairs, respectively.

The formal claim29 retains both general assertions and all ten printed threshold clauses. The theorem result29 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/CohenConsecutiveCubePrimePairThresholdRefutation.result29

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresConjecture refutedFirst preprint 2025Cohen Conjecture 28The printed thresholds in Cohen's Conjectures 28 and 29 fail at n = 11 and n = 12.Refuted

The questionCohen, Journal of Integer Sequences 28 (2025), Article 25.4.7, Section 2.5, Conjecture 28, asserts that there are at least two twin-prime pairs between every pair of consecutive positive cubes and that the count eventually exceeds every positive integer.

Why it mattersThe printed threshold is part of the named conjecture, but the count sequence printed immediately before it gives nine twin-prime pairs for n = 11.

The ideaDefine twinPairCount n as the number of natural numbers p with n^3 < p, p + 2 < (n+1)^3, and both p and p + 2 prime.

Our contributionThe printed thresholds in Cohen's Conjectures 28 and 29 fail at n = 11 and n = 12.

Verification details
Release source recordRefuted in Lean

Open problem

Cohen Conjecture 28

Cohen, Journal of Integer Sequences 28 (2025), Article 25.4.7, Section 2.5, Conjecture 28, asserts that there are at least two twin-prime pairs between every pair of consecutive positive cubes and that the count eventually exceeds every positive integer. Its printed threshold table further states N(10) = 11: for every natural n >= 11, there are at least ten such pairs strictly between n^3 and (n+1)^3.

The formal claim28 retains both general assertions and all seventeen printed threshold clauses. The theorem result28 proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/CohenConsecutiveCubePrimePairThresholdRefutation.result28

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Analysis & dynamicsCounterexample foundFirst preprint 2026The unrestricted multivariate hard-core occupancy bound is falseThe three-vertex star with fugacities (15,2,2) has expected independent-set size 9/8, below the proposed degree-sequence lower bound 259/230.Refuted

The questionDavies, Sandhu, Seo and Tan, arXiv:2605.05149v1, Section 1 after Theorem 1, state:

Why it mattersIt answers the unrestricted multivariate question with the three-vertex star K{1,2} and fugacities (15,2,2).

The ideaUse the existing IndependentPartitionDeletion.configurations and partition: actual independent subsets and their product-weight sum. expectedSize is their cardinality-weighted sum divided by the partition.

Our contributionThe three-vertex star with fugacities (15,2,2) has expected independent-set size 9/8, below the proposed degree-sequence lower bound 259/230.

Verification details
Release source recordRefuted in Lean

Open problem

The unrestricted multivariate hard-core occupancy bound is false

Davies, Sandhu, Seo and Tan, arXiv:2605.05149v1, Section 1 after Theorem 1, state:

Strengthening the conjecture, we believe that the multivariate version should hold for any λ in the positive orthant.

The bound in Theorem 1 is tight by the example of a disjoint union of complete graphs (such that λ is constant on each component), though we believe that the upper bound on the entries of λ can be removed.

The proposed bound is

$$ \mathbb E_{G,\lambda}|I|\geq \sum_{v\in V(G)}\frac{\lambda_v}{1+(d_v+1)\lambda_v}, \qquad \lambda_v>0. $$

Lee and Seo, arXiv:2602.02450v2, Section 5, write:

Having seen the Davies–Kang conjecture, it seems plausible to look for a strengthening of Theorem 1.1 in terms of occupancy fractions.

Their inequality (5.2), at t = 1, is the same bound after multiplication by the vertex count and identification of the sum of vertex marginals with expected independent-set cardinality. Its nonnegative orthant contains the strictly positive witness used here. The formal claim quantifies all natural n, all simple graphs on Fin n, all adjacency-decision instances, and every strictly positive real fugacity vector. Classical decidability supplies such an instance for every graph. The empty graph gives 0 ≤ 0.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/HardCore/MultivariateOccupancyRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationConjecture refutedFirst preprint 2026A mixed-sign Gaussian refutes Galindo–Rowell Conjecture 10.6A mixed-sign Gaussian in dimension fifteen refutes the Pauli-direction rigidity conjecture for cyclic unitary Yang–Baxter operators.Refuted

The questionRowell, Unitary Yang–Baxter Operators: Towards a Classification, arXiv:2608.16865v1, Conjecture 10.6 (§10.2.7), state:

Why it mattersThis supplies a concrete counterexample outside the listed computations.

Our contributionA mixed-sign Gaussian in dimension fifteen refutes the Pauli-direction rigidity conjecture for cyclic unitary Yang–Baxter operators.

Scope & next questiontheorem: this dossier settles Conjecture 10.6 only in the quantified form encoded by claim; it does not settle the paper's other conjectures or its equivalent classification sentence beyond the witness supplied here.

Verification details
Release source recordRefuted in Lean

arXiv

A mixed-sign Gaussian refutes Galindo–Rowell Conjecture 10.6

C. Galindo and E. C. Rowell, Unitary Yang–Baxter Operators: Towards a Classification, arXiv:2608.16865v1, Conjecture 10.6 (§10.2.7), state:

Let $d\geq2$ with $4\nmid d$, let $\alpha\in(\mathbb Z/d\mathbb Z)^\times$, and put $P_\alpha=X\otimes Z^\alpha$, $\tau_\alpha(s,t)=w^{\alpha st}$. If $a=(1,a_1,\ldots,a_{d-1})\in\mathbb T_d^{\mathrm{coef}}$, $R_\alpha(a)$ satisfies the Yang–Baxter equation, and $R_\alpha(a)$ is projectively unitary, then there is one sign $\epsilon\in{+1,-1}$ with $a_{s+t}=a_sa_t\tau_\alpha(s,t)^\epsilon$ for every $s,t\in\mathbb Z/d\mathbb Z$.

The Lean claim is this quantified statement with ZMod d, the displayed braid equation, and Mathlib projective unitarity. The theorem result proves its negation.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Algebra/CyclicYangBaxterPauliRigidityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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Discrete structuresCounterexample foundFirst preprint 2026The p = 29 orientation instances have no solutionGarcia's four normalized AGL(1,29) orientation instances cannot avoid a cycle of length 64 after H15 vertex replacement.Refuted

The questionDaniel Garcia, arXiv:2609.04686v1, §§3 and 5, leaves the order-812 affine-group orientation instance undecided after several CPU-hours.

The ideaThis is the literal vertex-replacement rule; its definition does not select or encode cycles.

Our contributionGarcia's four normalized AGL(1,29) orientation instances cannot avoid a cycle of length 64 after H15 vertex replacement.

Verification details
Release source recordRefuted in Lean

Open problem

garcia 2026 erdos64 agl29 orientation

Daniel Garcia, arXiv:2609.04686v1, §§3 and 5, leaves the order-812 affine-group orientation instance undecided after several CPU-hours. The settlement refutes the solvability of this instance, using the universal obstruction preregistered in issue #14828.

Let $V=\operatorname{AGL}(1,29)=(\mathbb Z/29\mathbb Z)^\times\times\mathbb Z/29\mathbb Z$, with $(x,y)(z,w)=(xz,xw+y)$. For each $a\in{3,8,10,11}$, put $t=(-1,0)$, $g=(a,1)$ and $r=g^{-1}$. Let $B_a$ be the simple undirected Cayley graph joining $x$ to $xt,xg,xr$.

The claim is: some orientation of Garcia's p = 29 instance avoids 64-cycles. Precisely, there exist one of these four multipliers, a function $\sigma:V\to{t,g,r}$, and bijections $\tau_x:{t,g,r}\to{u,v,w}$ with $\tau_x(\sigma(x))=u$, such that the literal $H_{15}$ replacement has no simple cycle of length 64. The closed Lean theorem result : ¬ claim refutes this existence assertion. Every independent assignment of the remaining incident edges to $v,w$ is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ErdosGyarfasGarciaOrientation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2026The p = 31 orientation instances have no solutionGarcia's four normalized AGL(1,31) orientation instances cannot avoid a cycle of length 64 after H15 vertex replacement.Refuted

The questionDaniel Garcia, arXiv:2609.04686v1, §5, leaves the order-930 affine-group orientation instance open.

The ideaEach word closes at the identity with fourteen distinct prefix products.

Our contributionGarcia's four normalized AGL(1,31) orientation instances cannot avoid a cycle of length 64 after H15 vertex replacement.

Verification details
Release source recordRefuted in Lean

Open problem

garcia 2026 erdos64 agl31 orientation

Daniel Garcia, arXiv:2609.04686v1, §5, leaves the order-930 affine-group orientation instance open. The settlement refutes the solvability assertion preregistered in issue #14838.

Let $V=\operatorname{AGL}(1,31)=(\mathbb Z/31\mathbb Z)^\times\times\mathbb Z/31\mathbb Z$, with $(x,y)(z,w)=(xz,xw+y)$. For each $a\in{11,17,22,24}$, put $t=(-1,0)$, $g=(a,1)$ and $r=g^{-1}$. Let $B_a$ be the simple undirected Cayley graph joining $x$ to $xt,xg,xr$.

The claim is: some orientation of Garcia's p = 31 instance avoids 64-cycles. Precisely, there exist one of these four multipliers, a function $\sigma:V\to{t,g,r}$, and bijections $\tau_x:{t,g,r}\to{u,v,w}$ with $\tau_x(\sigma(x))=u$, such that the literal $H_{15}$ replacement has no simple cycle of length 64. The closed Lean theorem result : ¬ claim refutes this existence assertion. Every independent assignment of the remaining incident edges to $v,w$ is included.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/ErdosGyarfasGarciaOrientationThirtyOne.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryCounterexample foundFirst preprint 2026Goel's OQ4: Fibonacci period-to-rank ratios omit fiveThe prime-domain Fibonacci period-to-rank ratios omit every multiple of five, answering Goel's OQ4 in the negative.Refuted

The questionAradhya Goel, Sophie Germain Primes and the Totient of Fibonacci Numbers, arXiv:2604.17847v3, Section 10, OQ4, p. 10, asks:

Our contributionThe prime-domain Fibonacci period-to-rank ratios omit every multiple of five, answering Goel's OQ4 in the negative.

Verification details
Release source recordRefuted in Lean

arXiv

Goel's OQ4: Fibonacci period-to-rank ratios omit five

Aradhya Goel, Sophie Germain Primes and the Totient of Fibonacci Numbers, arXiv:2604.17847v3, Section 10, OQ4, p. 10, asks:

OQ4. Values of $\pi(q)/z(2q+1)$. Is ${\pi(q)/z(2q+1):q>5\text{ with }z(2q+1)\mid\pi(q)}={\text{odd integers}}$?

The versioned TeX source reads verbatim:

\item[OQ4.] \textbf{Values of $\pi(q)/z(2q+1)$.} Is $\{\pi(q)/z(2q+1) : q > 5 \text{ with } z(2q+1) \mid \pi(q)\} = \{\text{odd integers}\}$?

The Introduction defines $z(p)$ at primes as the least positive index $k$ with $p\mid F_k$, and $\pi(n)$ as the period of the Fibonacci residues modulo $n$. The standing domain in Sections 5–7 is Sophie Germain primes: $q$ and $2q+1$ are both prime. The assertion here uses $q>5$ and $z(2q+1)\mid\pi(q)$. The right-hand side is the positive odd integers; reading it as all odd integers also fails because the ratios are positive.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/GoelPisanoRatioRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2025Maximum single-party entropy need not decrease on average under LOCCA local measurement on a four-party pure state increases the average maximum single-party entropy.Refuted

The questionJin, Measure of entanglement and the monogamy relation: a topical review, arXiv:2512.21992v1, Section 9.4, p. 88, state:

Why it mattersThis external named problem has no anchor atom.

The ideaThe entropy calculation applies NoisyGraphStateCoherentInformationRefutation.entropy2, ChenKatoBrandaoCMIRefutation.entropydiagonal, ThreeByThreeModulusEntropy.entropyeqspectrum, and PartialTraceMutualInformation.spectralEntropy, with the functional-calculus carrier correspondence. QuantitativeMultiOrbitWeilNegativeCertificate.finiteComplexEnergy supplies the squared-norm expression.

Our contributionA local measurement on a four-party pure state increases the average maximum single-party entropy.

Verification details
Release source recordRefuted in Lean

Open problem

Maximum single-party entropy need not decrease on average under LOCC

Y. Guo and Z.-X. Jin, Measure of entanglement and the monogamy relation: a topical review, arXiv:2512.21992v1, Section 9.4, p. 88, state:

Note here that we can not prove here E′^(n) is nonincreasing on average under LOCC, but we conjecture that E′^(n) does not increase on average under LOCC for these cases mentioned above. Hereafter we always assume the conjecture is true.

Equation (189) defines $E'^{(n)}(|\psi\rangle)=\max_i h(\rho^{A_i})$, and Table 16 includes the von Neumann entropy $h=S$. Section 2 defines $S(\rho)=-\operatorname{Tr}(\rho\log_2\rho)$. The settlement refutes the unrestricted entropy assertion using four parties of local dimension nine and one two-outcome local projective measurement.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite pure multipartite state | The amplitude argument of claim, indexed by Fin n → Fin d | Every finite party count and common local dimension; the witness has four parties and dimension nine. | | Reduced state of party $i$ | marginal | The literal coordinate partial trace from the binding conventions in #20726. | | $\max_i S(\rho^{A_i})$ | maxMarginalEntropy | The finite real supremum of the literal trace-log expression, divided by $\log 2$. | | One local measurement, an LOCC operation | applyLocal and the Kraus-completeness hypothesis in claim | One coordinate changes, and the sum of adjoint products is the identity. A violation for this subclass refutes LOCC monotonicity. | | Normalized input, probabilities, normalized branches | The finite squared-norm sums, norm-one hypothesis, and inverse-square-root scaling in claim | The binding convention is inlined literally; zero-probability branches contribute zero. Proofs reuse finiteComplexEnergy for that same sum. | | Does not increase on average | The concluding inequality in claim | The maximum is inside the outcome average. | | Refutation | result | Negation of the complete claim; no extra hypotheses. |

The real supremum represents a maximum for the nonempty finite party set in the witness. The claim extends the trace expression to arbitrary matrices, but its counterexample consists of normalized pure states and complete projectors. No positivity or physical-realization gap is used.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/GuoJinMaxMarginalEntropyLOCC.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2025A weight density below Janssen's moment lower bound for the annealed Potts critical parameterA mixture of two uniform densities has a unique positive Potts criticality zero below the proposed general moment lower bound.Refuted

The questionJanssen, Analysis of quantities determining the critical inverse temperature in the annealed Potts model with Pareto vertex weights, arXiv:2508.21409v3, Section 6:

Why it matterstc determines the critical inverse temperature of the annealed Potts model on sparse rank-1 random graphs.

The ideaTake q = 3 and the density of W = A + U, where P(A = 1) = 1331/1340, P(A = 11) = 9/1340, and U is independent of A and uniform on [−1/1000, 1/1000].

Our contributionA mixture of two uniform densities has a unique positive Potts criticality zero below the proposed general moment lower bound.

Verification details
Release source recordRefuted in Lean

arXiv

A weight density below Janssen's moment lower bound for the annealed Potts critical parameter

A. J. E. M. Janssen, Analysis of quantities determining the critical inverse temperature in the annealed Potts model with Pareto vertex weights, arXiv:2508.21409v3, Section 6:

In fact, a limited numerical exploration with some other pdfs (like the exponential $\exp({-}w),\cax_{[0,\infty)}(w)$ and $\cax_{[0,1)}(w)$) shows that there may hold more generally 2,\frac{\mu_3}{\mu_4},{\rm ln}(q-1)<t_c<2,\frac{\mu_0}{\mu_1},{\rm ln}(q-1)~, ~~~~~~q>2~,

Here 𝓕₀ and 𝒦 are defined in Section 1. Under the standing assumption that 𝓕₀'' is first positive and then negative, t_c is the unique positive zero of 𝒦. The verbatim definitions are in the literature note.

Issue #18836 fixes the reading: the lower inequality, for every q > 2 and every probability density on (0, ∞) with finite fourth moment satisfying the standing assumption and the uniqueness of the positive zero of 𝒦. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/AnnealedPottsMomentBoundRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Geometry & algebraCounterexample foundFirst preprint 2026Natural-basis dependence of balance in degenerate evolution algebrasDegenerate natural bases with graph balances two and one.Refuted

The questionKaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, Question 27, page 24, ask:

Why it mattersAn annihilator direction permits a shear of a natural basis without changing any mixed product.

The ideaOn any closed signed path the height increments telescope, so every balance is even.

Our contributionDegenerate natural bases with graph balances two and one.

Scope & next questionThe other questions in Question 27, including a complete invariant system for nondegenerate algebras, remain open here.

Verification details
Release source recordRefuted in Lean

Open problem

Natural-basis dependence of balance in degenerate evolution algebras

Kaygorodov, Martín González and Viruel, Evolution algebras: a survey of results and a list of open problems, arXiv:2610.22248v1, Question 27, page 24, ask:

Is the balance independent of the natural basis for arbitrary (possibly degenerate) evolution algebras?

The source's section 1.1 states: “Unless stated otherwise, algebras are finite-dimensional.” The assertion concerns every field, bilinear evolution multiplication and pair of natural bases. It adds no nondegeneracy, perfectness, characteristic or connectedness assumption. The other two questions about a complete system of graph invariants are outside this settlement. Preregistration: #21625.

Statement echo

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | A field and a finite-dimensional algebra | claim quantifies k, V, field, additive-group, module and finite-dimensional instances, and bilinear μ | No associativity or unit assumption | | Evolution algebra | Existing DerivationAnnihilatorObstruction.IsEvolutionAlgebra | The arbitrary-field natural-basis predicate is reused | | Every two natural bases | ι, κ, their Fintype instances, B, F, and both zero mixed-product laws in claim | Different index types are allowed | | Row square coefficient gives an arrow | associatedGraph | Basis coordinates of the square, with the source's row order | | A path retains actual arrows traversed forwards or backwards | Path.vertex, Path.edge, Path.isEdge, Path.traverses | Opposing arrows remain distinct choices | | A cycle has equal first and last vertices | IsCycle | All finite lengths, including zero | | Forward cardinality minus backward cardinality | pathBalance | Both counts are cast to integers; loops enter both sets | | Gcd of all absolute cycle balances | graphBalance | Absolute integer ideal generator; no length cutoff | | The balances agree | Equality in claim; result proves its negation | The rational four-dimensional instance refutes the universal assertion |

The graph definitions also accept infinite vertex types and the height lemma arbitrary moduli; these stronger helper domains do not restrict the finite-basis assertion. Closing universes at Type supplies a closed proposition and includes the rational counterexample. The zero gcd is zero.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/QuadraticForms/EvolutionAlgebras/DegenerateGraphBalanceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025A qubit refutation of the maximal reverse Pinsker boundA noncommuting qubit pair violates the maximal reverse Pinsker bound.Refuted

The questionRioul, From Classical to Quantum: Explicit Classical Distributions Achieving Maximal Quantum f-Divergence, arXiv:2501.14340v1, §III-C, p. 5:

Why it mattersThe maximal-divergence instance suffices to test the source's universal statement.

Our contributionA noncommuting qubit pair violates the maximal reverse Pinsker bound.

Verification details
Release source recordRefuted in Lean

Open problem

A qubit refutation of the maximal reverse Pinsker bound

D. Lanier, J. Béguinot and O. Rioul, From Classical to Quantum: Explicit Classical Distributions Achieving Maximal Quantum f-Divergence, arXiv:2501.14340v1, §III-C, p. 5:

We conjecture that (49) is in fact not necessary for (50) to hold.

The same paragraph states:

Our bound holds in general for any f-divergence, in particular for the maximal f-divergence.

Section I specifies a convex real generator on the nonnegative half-line with $f(1)=0$. Equation (4) defines

$$ D_f^{\max}(\rho\Vert\sigma) =\operatorname{Tr}\bigl(\sigma^{1/2}f(\sigma^{-1/2}\rho\sigma^{-1/2})\sigma^{1/2}\bigr). $$

Theorem 4 assumes $|\rho-\sigma|\leq\rho+\sigma$ (49), and asserts

$$ D_f(\rho\Vert\sigma)\leq\frac{|\rho-\sigma|_1}{2} \left(\frac{f(m)}{1-m}+\frac{f(M)}{M-1}\right),\tag{50} $$

where $m,M$ are the extreme eigenvalues of $\sigma^{-1/2}\rho\sigma^{-1/2}$. Preregistration: #19943.

Statement echo

| Source clause | Lean counterpart | Fidelity and scope | | --- | --- | --- | | Every finite dimension | claim: natural dimension and Matrix (Fin n) (Fin n) ℂ | Includes the qubit instance. | | Convex real generator for nonnegative arguments, zero at one | ConvexOn ℝ (Set.Ici 0) f, f 1 = 0 | Values outside the half-line do not affect the positive relative matrix. | | States; invertible second state | ρ.PosSemidef, ρ.trace = 1, σ.PosDef, σ.trace = 1 | Positive definiteness expresses an invertible positive state. | | Equation (4) | maxDivergence | Literal relative-matrix expression and real part of the CFC trace. | | Equation (50) | reversePinskerBound | Repository trace norm; real spectral infimum and supremum. The witness has nonzero denominators. | | Remove condition (49) | claim | No absolute-difference order hypothesis. | | Conditional printed assertion | private claimTheorem4 | Adds exactly the matrix-order condition (49). | | Refutation | public result | Negates the complete binding claim. |

The public formulas inline the source relative-matrix expression. The proof uses D5.S3.Resource.LogDet.PathSpectralClassical.relativeCongruence, which has that expression, without introducing a duplicate definition. The library note is Library/Divergence/lanier2025maximal.md.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Divergence/LanierBeguinotRioulReversePinskerRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
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    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Quantum informationCounterexample foundFirst preprint 2026The powered binary Rényi entropy is not monotone in the orderPowered binary Renyi entropy can decrease as its positive order increases.Refuted

The questionLiu, Computational hardness of estimating quantum entropies via binary entropy bounds, arXiv:2601.03734v2, Section 1.4(b):

Why it mattersThe comparison with order 2 (the source's Theorem 3.4) transfers hardness results between Rényi orders for rank-2 states.

The ideaSince √(1/17) + √(16/17) = 5/√17 and (1/17)^{1/4} + (16/17)^{1/4} = 3·17^{−1/4}:

Our contributionPowered binary Renyi entropy can decrease as its positive order increases.

Verification details
Release source recordRefuted in Lean

Open problem

The powered binary Rényi entropy is not monotone in the order

Y. Liu, Computational hardness of estimating quantum entropies via binary entropy bounds, arXiv:2601.03734v2, Section 1.4(b):

Can the inequalities in \Cref{table:Rank2RenyiQEA-hardness} be tightened? For instance, is it possible to prove that $\rbra[\big]{\frac{\Halpha(x)}{\ln(2)}}^{2/\alpha}$ is monotonically non-decreasing in $\alpha$ for all fixed $x\in[0,1]$, as suggested by numerical evidence and as a generalization of \Cref{thm:Renyi-binary-entropy-upper-bound-0leAle2}?

Here H^R_α(x) = ln(x^α + (1−x)^α)/(1−α). The verbatim passages are in the literature note.

Issue #19090 fixes the reading: for every x ∈ [0, 1] and all orders 0 < a < b with a, b ≠ 1, (H_a(x)/ln 2)^{2/a} ≤ (H_b(x)/ln 2)^{2/b}. It is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Entropy/RenyiCombining/BinaryRenyiPowerMonotonicityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

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Discrete structuresCounterexample foundFirst preprint 2026The ordinary Loopy polynomial does not determine the generalized degree polynomialTwo eight-vertex trees, each carrying a single loop, have one common ordinary Loopy polynomial and two different generalized degree polynomials.Refuted

The questionProblem 11.3 of kirillov2026loopy asks exactly:

Why it mattersDisplay (6.10) of the source defines the generalized degree polynomial of Crew as the sum over subsets S of the vertex set of q to the size of S, r to the number of edges inside S, and t to the number of edges with exactly one endpoint in S.

The ideaTwo trees on the vertices 0 through 7, one carrying a loop at vertex 1 and one carrying a loop at vertex 4, are exhibited.

Our contributionTwo eight-vertex trees, each carrying a single loop, have one common ordinary Loopy polynomial and two different generalized degree polynomials.

Verification details
Release source recordRefuted in Lean

arXiv

The ordinary Loopy polynomial does not determine the generalized degree polynomial

Problem 11.3 of kirillov2026loopy asks exactly:

Does L_G determine the generalized degree polynomial GD_G?

The source's conventions admit every finite undirected multigraph, loops and parallel edges included, and Problem 11.3 adds no restriction; the sentences that follow it introduce the companion Problem 11.4 about degree sequences, which Theorem 4.3 settles in the loopless case and Theorem 6.15 settles through the refined polynomial. The pair delivered here carries loops and therefore lies inside the problem's own domain.

In repository coordinates the target quantifies independently over finite vertex sets V W, edge-occurrence lists E F and accumulated-loop functions ell eta. From Valid V E ell, Valid W F eta and loopy V E ell = loopy W F eta it concludes equality of the exponent-triple multisets of the generalized degree polynomial. No looplessness, connectedness, simplicity or equal-order hypothesis is present.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Factorization/Combinatorics/LoopyGeneralizedDegree.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2024The Luiz–Vieira generalized Blanuša closing conjecture is falseThe first generalized Blanuša snark has a [21]-Roman function of weight 108 and every independent [21]-Roman function has weight at least 110.Refuted

The questionVieira, Independent [k]-Roman Domination on Graphs, arXiv:2406.11688v1, Discuss.

Our contributionThe first generalized Blanuša snark has a [21]-Roman function of weight 108 and every independent [21]-Roman function has weight at least 110.

Verification details
Release source recordRefuted in Lean

Open problem

The Luiz–Vieira generalized Blanuša closing conjecture is false

A. G. Luiz and F. A. S. Vieira, Independent [k]-Roman Domination on Graphs, arXiv:2406.11688v1, Discuss. Math. Graph Theory 45 (2025), 1249–1272, Section 6, conjecture that “this property holds for all generalized Blanuša snarks,” where the property is $\gamma_{[kR]}(G)=i_{[kR]}(G)$ for $k\ge 2$. The Lean claim is the literal quantification over $t\in{1,2}$, $i\ge 1$, and $k\ge 2$ in D5/S3/Combinatorics/Graph/Domination/LuizVieiraBlanusaIndependentRoman.claim.

Statement echo

The source active-neighbour filter appears literally in IsRomanDom; the other source definitions are mirrored by IsIndepRomanDom, weight, gammaRoman, indepRoman, the block and link edge sets, the recursive verts and edges, and snark. The settling declaration is D5/S3/Combinatorics/Graph/Domination/LuizVieiraBlanusaIndependentRoman.result : ¬ claim; it instantiates $t=1$, $i=1$, $k=21$.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Graph/Domination/LuizVieiraBlanusaIndependentRoman.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2024Non-local partial entanglement annihilation is not closed under aligned tensor productsTwo non-local channels with Schmidt-number-two outputs can have an aligned tensor product with a Schmidt-rank-four output.Refuted

The questionMajumdar, On the characterization of partially entanglement breaking and annihilating channels, arXiv:2411.19315v2; J.

Why it mattersThe question tests whether a bound on the internal entanglement dimension of every single-use output remains valid for parallel uses across the aligned split.

The ideaThese are the module's fixed phi and replacement.

Our contributionTwo non-local channels with Schmidt-number-two outputs can have an aligned tensor product with a Schmidt-rank-four output.

Verification details
Release source recordRefuted in Lean

Open problem

Non-local partial entanglement annihilation is not closed under aligned tensor products

B. Mallick, N. Ganguly and A. S. Majumdar, On the characterization of partially entanglement breaking and annihilating channels, arXiv:2411.19315v2; J. Math. Phys. 67, 082201 (2026), ask in Section VI, Conclusions, PDF page 16:

We have also shown that $r$-partially entanglement breaking channels are not closed under tensor products, motivating the investigation of whether a similar property holds for $r$-partially entanglement annihilating channels.

This is the verbatim sentence in main.tex, line 1167, from the v2 source archive (member MD5 02f5d47ee4475c5ba30a52b1e5931b9b). The definitions are in Section IV.A, PDF pages 9–10, main.tex lines 630–632; see also the literature note.

Write $\operatorname{PEA}_{d,r}(E)$ for the module's IsPEA d r E: $E$ is a completely positive, trace-preserving complex-linear map on the matrices of $\mathbb C^d\otimes\mathbb C^d$, cannot be written as a tensor product of two individual CPTP maps, satisfies $r<d$, and sends every positive semidefinite trace-one input to a state of Schmidt number at most $r$.

Issue #21802 fixes the closure assertion: for every $d,r\in\mathbb N$ and every pair of complex-linear bipartite matrix maps $E,F$, $\operatorname{PEA}{d,r}(E)$ and $\operatorname{PEA}{d,r}(F)$ imply $\operatorname{PEA}{d^2,r}(E\otimes{\mathrm{aligned}}F)$. The aligned split groups $B_1B'_1,|,B_2B'_2$ when the two copies are $B_1B_2$ and $B'_1B'_2$. The threshold remains $r$.

This closure assertion is refuted. The source asks whether non-closure occurs; it does not assert closure. Thus refuting the registered assertion answers the source's non-closure question affirmatively.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Entanglement/SchmidtNumber/PartiallyEntanglementAnnihilatingTensorRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryCounterexample foundFirst preprint 2026Refutation of Nathanson's additive h-basis strict inequalityThe strict inequality in Nathanson's Problem 12(2) fails for a three-element integer set.Refuted

The questionProblem 12(2) of arXiv:2605.26425v3 asks whether, for integers h >= 2 and k >= 2, every k-element integer set A with a negative element satisfies ellh(A) < n-flath(k) whenever ellh(A) is defined.

Why it mattersThe source prints A = {-1, 1, 2} as a reflection example, while Problem 12(2) asks whether the strict inequality holds universally.

The ideaAt h = 2, k = 3, and A = {-1, 1, 2}, the double sumset is {-2, 0, 1, 2, 3, 4}, so the covered initial segment reaches 4.

Our contributionThe strict inequality in Nathanson's Problem 12(2) fails for a three-element integer set.

Verification details
Release source recordRefuted in Lean

arXiv

Refutation of Nathanson's additive h-basis strict inequality

Problem 12(2) of arXiv:2605.26425v3 asks whether, for integers h >= 2 and k >= 2, every k-element integer set A with a negative element satisfies ell_h(A) < n-flat_h(k) whenever ell_h(A) is defined.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Arith/NathansonAdditiveHBasisRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Number theoryConjecture refutedRefutation of the A089610 eventual strict-increase conjectureOdd-prime counting rules out eventual strict increase for OEIS A089610.Refuted

The questionThere exists an n1 such that a(n) is < a(n+1) for all n >= n1.

Why it mattersThe entry proposes that the square-interval prime count eventually rises at every step.

The ideaFor every natural k, each prime counted by a(2k) is greater than two and odd.

Our contributionOdd-prime counting rules out eventual strict increase for OEIS A089610.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A089610 eventual strict-increase conjecture

OEIS A089610 %N (verbatim):

Number of primes between n^2 and (n+1/2)^2.

OEIS A089610 %C (verbatim sub-conjecture):

There exists an n_1 such that a(n) is < a(n+1) for all n >= n_1.

OEIS A089610 %A (verbatim):

Cino Hilliard, Dec 30 2003

The literal refuted statement is ∃ N : ℕ, ∀ n : ℕ, N ≤ n → a(n) < a(n+1), where a(n) = ((Finset.Ioc (n*n) (n*n+n)).filter Nat.Prime).card. Since (n+1/2)^2 = n^2+n+1/4, this natural interval selects exactly the integer primes between the endpoints printed in the NAME, as T. D. Noe's comment notes. The separate conjecture that a(n) > 1 after n = 17 and Oppermann's positivity conjecture a(n) > 0 are not claimed or resolved here.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/PrimeGaps/HilliardSquareIntervalPrimeCountEventualIncreaseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundRefutation of the A103585 period-43 commentThe 129th and 172nd entries refute the proposed period 43 of OEIS A103585.Refuted

The questionConsider numbers k such that (A102370(k)-k)/2 = 1; read them mod 4 to get the sequence.

Why it mattersStephan's comment asserts a universal period.

The ideaThe theorem Nat.nthcount therefore gives Nat.nth P 128 = 383, so A(128)=383 mod 4=3.

Our contributionThe 129th and 172nd entries refute the proposed period 43 of OEIS A103585.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the A103585 period-43 comment

OEIS A103585, NAME (verbatim):

Consider numbers k such that (A102370(k)-k)/2 = 1; read them mod 4 to get the sequence.

Ralf Stephan's COMMENT of May 18, 2007 (verbatim):

Sequence appears to have period 43.

For natural numbers, define s(k) := k + Sum_{m in Icc(1,k+1)} (if (k+m) mod 2^m = 0 then 2^m else 0), the finite form of the A102370 formula; define P(k) := (s(k) = k+2); and define A(i) := nth(P,i) mod 4. Thus A is zero-indexed and OEIS a(m) equals A(m-1). The literal refuted statement is for every natural i, A(i+43) = A(i).

The definition of A102370, the %N definition of A103585, any true period, and eventual aperiodicity are not claimed or assessed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/StephanSlopingBinaryPeriodRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundRefutation of Layman's A196226 divisor-power congruencesLayman's three divisor-power congruence conjectures.Refuted

The questionLayman's conjecture comment gives three universal congruences:

Why it mattersThe value m = 690 is an A196226 term above all three thresholds and is not an even semiprime.

The ideaThe first identity gives 1728 mod 690 = 348 = 3 + 690/2, and 690 is even, so membership 690 holds.

Our contributionLayman's three divisor-power congruence conjectures.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of Layman's A196226 divisor-power congruences

OEIS A196226 defines its terms as the natural numbers m for which the sum of divisors of m, reduced modulo m, is 3 + m/2. John W. Layman's conjecture comment gives three universal congruences:

(1) If m>=14 is a term of this sequence, then sigma(2,m) is congruent to 5 + m/2 modulo m.

(2) If m>=22 is a term of this sequence, then sigma(3,m) is congruent to 9 + m/2 modulo m.

(3) If m>=38 is a term of this sequence, then sigma(4,m) is congruent to 17 + m/2 modulo m.

Here sigma(k,m) is the sum of the k-th powers of the positive divisors of m. The formal membership predicate makes the entry's evenness condition explicit:

def membership (m : Nat) : Prop :=
  2 | m /\ ArithmeticFunction.sigma 1 m % m = 3 + m / 2

Natural-number division is used throughout. The closed claim is the disjunction of the three displayed universal statements, so Not claim asserts that every one of the three conjectures is false.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/LaymanDivisorPowerCongruenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundThe literal adjacent difference of OEIS A378495The literal adjacent-column identity with the zero-based index n-4 fails at n=5: the actual cardinality difference is twenty and the absolute sequence value is six.Refuted

The questionConjecture: T(n,n-1) - T(n,n-2) = abs(A238474(n-4)) for n >= 4.

Why it mattersIt is Peter Kagey's revision-59 formula from 2024.

The ideaA derangement on five labels cannot have a 4-cycle, since the unused label would be fixed.

Our contributionThe literal adjacent-column identity with the zero-based index n-4 fails at n=5: the actual cardinality difference is twenty and the absolute sequence value is six.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

The literal adjacent difference of OEIS A378495

OEIS A378495, revision 70, FORMULA, states verbatim:

Conjecture: T(n,n-1) - T(n,n-2) = abs(A238474(n-4)) for n >= 4.

On the source-agreeing columns $2\le k\le n$, $T(n,k)$ counts derangements of $S_n$ having no $k$-cycle; the source explicitly sets $T(n,1)=0$. The triangle has offset 1,5. The dependent A238474 entry, revision 15, has offset 0,1 and formula $b(m)=(-1)^m(m+3)!/(2(m+1))$ for every natural $m$. This is interpreted in $\mathbb Q$, with rational ordinary subtraction between cast cardinalities and rational division in $b$.

Statement echo

| Source clause | Exact formal counterpart | | --- | --- | | Every $n\ge4$ | claim quantifies over every n : Nat with 4 <= n | | Actual derangements with no $k$-cycle | Imported TerminalDifference.T, using permutations of Fin n, no fixed points, and cycle-factor support sizes | | Ordinary difference of adjacent columns | (T n (n-1) : Q) - (T n (n-2) : Q) | | Zero-based A238474 | b (m : Nat) : Q uses the source factorial quotient | | Literal source shift $n-4$ | Exactly abs (b (n-4)), with natural subtraction | | Universal assertion is false | theorem result : Not claim, a closed negation of the literal Prop definition |

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Derangements/AdjacentDifferenceRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedRefutation of the numerator conjecture of OEIS A362534At n = 19 the adiabatic polarization-transfer bound g(19) = 215955/131072 has numerator 215955, while the ratio f(19)/g(19) = 92378/43191 has denominator 43191, so the conjecture of OEIS A362534 fails.Refuted

The questionThe symmetry-constrained upper bounds are given by the function f(n): (1) for even n, f(n) = (2^(1-n))nbinomial(n-1, n/2) (2) for odd n, f(n) = (2^(1-n))nbinomial(n-1, (n-1)/2).

Why it mattersThe conjecture relates the two bounds of the entry for all numbers of X spins.

The idea19 is odd, so f(19) = 2^(-18) · 19 · C(18, 9) and g(19) = 2 (1 − 2^(-19) · C(19, 9)).

Our contributionAt n = 19 the adiabatic polarization-transfer bound g(19) = 215955/131072 has numerator 215955, while the ratio f(19)/g(19) = 92378/43191 has denominator 43191, so the conjecture of OEIS A362534 fails.

Verification details
Release source recordRefuted in Lean

Integer sequences (OEIS)

Refutation of the numerator conjecture of OEIS A362534

OEIS A362534 (Mohamed Sabba, 2023) concerns upper bounds on the transfer of z-magnetization from a group X_n of spin-1/2 nuclei to a single spin-1/2 nucleus A in NMR. Its COMMENTS define

The symmetry-constrained upper bounds are given by the function f(n): (1) for even n, f(n) = (2^(1-n))*n*binomial(n-1, n/2) (2) for odd n, f(n) = (2^(1-n))*n*binomial(n-1, (n-1)/2). The adiabatic bounds are given by the function g(n): (3) for even n, g(n) = 2*(1-(2^(-n))*binomial(n, n/2)) (4) for odd n, g(n) = 2*(1-(2^(-n))*binomial(n, (n-1)/2)).

and close with

Conjecture: the numerator of g(n) is the denominator of f(n)/g(n).

Issue #10062 fixes the readings: f(n) and g(n) are the rational numbers given by these formulas for n ≥ 1 (the entry's offset); numerator and denominator are those of the fraction in lowest terms, as in the entry's own lists g = 1, 1, 5/4, 5/4, 11/8, 11/8, 93/64, … and f/g = 1, 1, 6/5, 6/5, 15/11, 15/11, 140/93, …; the conjecture is asserted for every n ≥ 1.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/SabbaPolarizationTransferRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Analysis & dynamicsCounterexample foundFirst preprint 2025A disordered exclusion ring whose ε² current response vanishes although neither symmetry condition holdsA four-site exclusion ring has zero quadratic current response despite failing both proposed rate-symmetry conditions.Refuted

The questionAkimoto, Symmetry breaking of current response in disordered exclusion processes, arXiv:2512.03316v2, Phys.

Why it mattersThe source treats the inversion-correspondence condition (ii) as a finely tuned, measure-zero case and still counts it as a mechanism for J₂ = 0.

The idea1. Witness. L = 4 and N = 2, with trap-model rates w{i,i+1} = w{i,i−1} = ai, a = (2, 4, 3, 41).

Our contributionA four-site exclusion ring has zero quadratic current response despite failing both proposed rate-symmetry conditions.

Verification details
Release source recordRefuted in Lean

Open problem

A disordered exclusion ring whose ε² current response vanishes although neither symmetry condition holds

I. Sakai and T. Akimoto, Symmetry breaking of current response in disordered exclusion processes, arXiv:2512.03316v2, Phys. Rev. E 113, 044123 (2026), study the stationary current J(ε) of an exclusion process on a ring of L sites with N particles.

  • Model. A particle at site i hops to i + 1 at rate p_i(ε) = (1 + ε) w_{i,i+1}/2 and to i − 1 at rate q_i(ε) = (1 − ε) w_{i,i−1}/2, with ∏ w_{i,i−1}/w_{i,i+1} = 1 (Eq. (1)).
  • Condition (i). The bond-bias ratio r_i(ε) = p_i(ε)/q_{i+1}(ε) does not depend on i (Eq. (5)).
  • Condition (ii). There is a k such that, for every i, p_{k+i}(ε) = q_{k−i}(−ε) or p_{k+i}(ε) = q_{k−i+1}(−ε) (Appendix C).
  • The conjecture. Either condition makes J odd in ε. The authors write (§III.D, p. 3): "We conjecture—and later confirm numerically— that an ε² contribution appears whenever neither condition (i) nor condition (ii) is satisfied."

The verbatim statements are in the literature note.

Issue #17823 fixes the reading:

  • the ε² contribution is the coefficient c₂ of J(ε) = c₁ε + c₂ε² + o(ε²) under the unique stationary law;
  • condition (ii) is read in its weakest per-site form;
  • the claim is restricted to at least two particles and at least two holes, because the source (§III.F) already reports an exactly odd current for a single particle.

This restricted conjecture is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/StatisticalMechanics/DisorderedExclusionQuadraticResponseRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
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    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2023The perp hyperplane refutes the three-qubit hexagon contextuality generalizationThe IIZ perp hyperplane refutes the classical-hexagon contextuality generalization.Refuted

The questionGiorgetti and H. de Boutray, Hexagons govern three-qubit contextuality, Quantum 9 (2025) 1601, DOI 10.22331/q-2025-01-20-1601; arXiv:2312.07738v3, Section 4, main.tex, lines 736–740:

Why it mattersThe conjectured equality would turn a minimization over hidden-variable assignments into a geometric intersection problem.

The ideaThe lower bound on the natural infimum uses lecsInf with Set.rangenonempty; the range contains the violation count of the zero assignment.

Our contributionThe IIZ perp hyperplane refutes the classical-hexagon contextuality generalization.

Scope & next questionWhich further geometric hypotheses guarantee equality remains open here.

Verification details
Release source recordRefuted in Lean

Open problem

The perp hyperplane refutes the three-qubit hexagon contextuality generalization

M. Saniga, F. Holweck, C. Kelleher, A. Muller, A. Giorgetti and H. de Boutray, Hexagons govern three-qubit contextuality, Quantum 9 (2025) 1601, DOI 10.22331/q-2025-01-20-1601; arXiv:2312.07738v3, Section 4, main.tex, lines 736–740:

The message from the above-given observations is more than obvious: the degree of contextuality of a contextual quantum geometry of the three-qubit $\mathcal{W}(5,2)$ whose contexts are i) all the 315 lines of the whole space $\mathcal{W}(5,2)$, ii) all the 45 lines of an elliptic quadric, iii) all the 105 lines of a hyperbolic quadric or iv) all the 15 lines of a doily is equal to the number of lines each of these geometries shares with a classically-embedded hexagon, with the understanding that each shared line corresponds to an unsatisfiable context. A natural generalization of these findings is that this should hold for any contextual sub-geometry of $\mathcal{W}(5,2)$ whose contexts are lines.

The next paragraph begins, “As a particular stance of this conjecture, let us consider the complement of a skew-embedded hexagon” (main.tex, lines 742–744), so the proposed generalization extends beyond the four named families. Definitions and source locators are also in the literature note.

For the formal reading, put $V=(\mathbb F_2^2)^3$. Its nonzero vectors label the canonical three-qubit Pauli observables $P_v$. Write $B$ for the binary symplectic form. Lines are the triples ${u,v,u+v}$ with $u,v$ nonzero, distinct and $B(u,v)=0$. Set $s(\ell)=1$ when $\prod_{q\in\ell}P_q=-I$, and $s(\ell)=0$ otherwise. For a finite family $C$ of these lines, define $d(C)=\min_{f:V\to\mathbb F_2}|{\ell\in C:\sum_{q\in\ell}f(q)\ne s(\ell)}|$.

Issue #24997 fixes claim: for every such $C$ with $d(C)>0$, there exists $H$ satisfying IsClassicalHexagon with $d(C)=|C\cap H|$. The predicate, specified below, includes all the source's classical embeddings. This claim is refuted.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Contextuality/PerpHyperplaneHexagonContextualityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
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  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Quantum informationCounterexample foundFirst preprint 2025A faithful mixed-state counterexample to KD-real SLD exclusionA faithful nine-dimensional mixed state has a KD-real symmetric logarithmic derivative with nonzero quantum Fisher information, refuting the mixed-state exclusion conjecture.Refuted

The questionMatéo Spriet, Christopher Langrenez, Raymond Brummelhuis and Stephan De Bièvre, What Is Special About the Kirkwood–Dirac Distributions?

The ideaThe public result is the designated refutation result (basis=refutes) and is exempt from four-slot escape registration (CLAUDE.md §3.9).

Our contributionA faithful nine-dimensional mixed state has a KD-real symmetric logarithmic derivative with nonzero quantum Fisher information, refuting the mixed-state exclusion conjecture.

Verification details
Release source recordRefuted in Lean

Open problem

A faithful mixed-state counterexample to KD-real SLD exclusion

Matéo Spriet, Christopher Langrenez, Raymond Brummelhuis and Stephan De Bièvre, What Is Special About the Kirkwood–Dirac Distributions? Only They Produce Natural Conditional Expectations, arXiv:2511.01996v2, §6.3, immediately after (6.23), state:

In view of what precedes, it is tempting to conjecture that, if $\hat \rho$ and $\hat X$ are both KD real, the optimal observable $\hat L$ must not be KD-real itself, provided $\rm I_{QF}$ does not vanish.

The conjecture is unnumbered; Proposition 6.5 is the published pure-state theorem. The journal article, DOI 10.3390/sym18061008, retains the conjecture after (168), with the pure-state result numbered Proposition 7. The claim is bound to arXiv v2. Source and literature evidence are in #22488; the source note is sprietlangrenez2026kdreal.

Statement echo

All module-local names below belong to D5.S3.Quantum.Information.KirkwoodDiracSLDRefutation; the four public definitions and result are in its SprietKirkwoodDiracSLD namespace.

| Source clause | Lean counterpart | Fidelity | | --- | --- | --- | | Finite Hilbert space of positive dimension | d, the positivity hypothesis, and complex matrices indexed by Fin d in claim | All positive finite dimensions. | | Two orthonormal CSCO eigenbases | a, b in Matrix.unitaryGroup | Their columns are the eigenvectors; eigenvalue labels play no role. | | Complementarity | Every entry of the overlap matrix is nonzero | No mutual-unbiasedness hypothesis is added. | | Left KD distribution | qLeft | Conjugate overlap times the matrix coefficient. | | Left observable symbol | symbolLeft | Inverse overlap times the matrix coefficient. | | KD-real Hermitian sector | IsKDReal | Hermitian symmetry and zero imaginary symbol at every label pair. | | Density matrix | Reused SLDSpectralEnergy.IsDensity | Positive semidefinite with trace one. | | KD-real state and generator | The two IsKDReal hypotheses on the same bases | No reality or commutativity assumption is added. | | SLD tangent equation and Hermitian solution | Reused FisherDiscord.Defs.IsSLD | The tangent is $i(\rho X-X\rho)=-i[X,\rho]$. | | Existence of the SLD | Existential conclusion in claim | Existence is asserted, rather than assumed. | | Nonvanishing source QFI | Four times reused FisherDiscord.Defs.fisherSLD | The owner is one quarter of $\operatorname{Tr}(\rho L^2)$. | | Every such SLD fails to be KD-real | Universal conclusion over L | result negates the whole literal claim. |

The public definitions retain the literal source expressions. Density, SLD and Fisher information use their existing owners. Each binder or conjunct above either represents the quoted mathematical setting or makes an implicit existence assertion explicit; none narrows the source class. Faithfulness is a property of the counterexample, not a hypothesis restricting claim.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Quantum/Information/KirkwoodDiracSLDRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresCounterexample foundFirst preprint 2024Noninteger parameters do not ensure injectivity of the Szalay–Tóth partition generatorsDistinct partitions have equal power sums at exponent one half and equal entanglement dimensions at base five halves.Refuted

The questionTóth, Alternatives of entanglement depth and metrological entanglement criteria, arXiv:2408.15350v3; Quantum 9, 1718 (2025), Section 4.7, state:

Why it mattersThese generators assign numerical labels to multipartite entanglement properties indexed by integer partitions.

Our contributionDistinct partitions have equal power sums at exponent one half and equal entanglement dimensions at base five halves.

Verification details
Release source recordRefuted in Lean

Open problem

Noninteger parameters do not ensure injectivity of the Szalay–Tóth partition generators

Sz. Szalay and G. Tóth, Alternatives of entanglement depth and metrological entanglement criteria, arXiv:2408.15350v3; Quantum 9, 1718 (2025), Section 4.7, state:

On the other hand, for $q\not\in\field{N}$ the equation becomes transcendental, and we expect the injectivity of $s_q$.

For the case $d\not\in\field{N}$, we expect the injectivity of $\Dim_d$, although the meaning of $\Dim_d$ is not established for noninteger $d$, which is just a parameter then.

These are verbatim TeX quotations from depths.tex, line 2437 and lines 2445–2447, respectively, in the v3 source archive (MD5 bbacd570d915c7fdda31d547fe476fec). The source macros \field{N} and \Dim_d denote $\mathbb N$ and $\operatorname{Dim}_d$.

An integer partition $\xi$ of $n$ is a multiset of positive integers summing to $n$ (depths.tex, lines 462–466). The source defines $s_q(\xi)=\sum_{x\in\xi}x^q$ for real $q$ (Eq. eq:sq, lines 1903–1907), and $\operatorname{Dim}d(\xi)=\sum{x\in\xi}d^x$ (Eq. eq:Dimd, lines 2306–2312). Repetitions contribute with their multiplicities. For the second expectation, $d\ge2$ remains a mathematical parameter; a noninteger value is not a Hilbert-space dimension.

Issue #21841 fixes claim as the disjunction of these two assertions:

  • For every $n\in\mathbb N$ and every real $q$ unequal to every natural number, $s_q$ is injective on the partitions of $n$.
  • For every $n\in\mathbb N$ and every real $d\ge2$ unequal to every natural number, $\operatorname{Dim}_d$ is injective on the partitions of $n$.

The theorem result : ¬ claim refutes both universal assertions.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S3/Combinatorics/Partitions/EntanglementDepthGeneratorInjectivityRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    The dossier identifies an established research line; its workflow tier alone cannot establish foundational importance.

    Source 1
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1
Discrete structuresConjecture refutedFirst preprint 2025Refutation of Zelinsky and Zhang's Conjecture 22The integer 6 refutes a published lower bound for a divisor-weighted logarithmic sum.Refuted

The questionZelinsky and Zhang, Kullback-Leibler divergence and primitive non-deficient numbers, arXiv:2501.04209v2, page 14, Conjecture 22 (verbatim):

Why it mattersThe published universal lower bound has a small exact counterexample.

The ideaAt n = 6, the positivity and exclusion hypotheses hold and Nat.minFac 6 = 2.

Our contributionThe integer 6 refutes a published lower bound for a divisor-weighted logarithmic sum.

Verification details
Release source recordRefuted in Lean

Open problem

Refutation of Zelinsky and Zhang's Conjecture 22

Zelinsky and Zhang, Kullback-Leibler divergence and primitive non-deficient numbers, arXiv:2501.04209v2, page 14, Conjecture 22 (verbatim):

Assume that n is a positive integer and assume that n is not in {1, 12, 24, 30, 36, 48, 60, 72, 120, 180, 240, 360}. Let p be the smallest prime factor of n. Then v(n) >= 1/p^2.

For a positive integer n, the paper defines v(n) = sum_{d | n, d > 1} (1/d) log((tau(n) - 1)/d). The literal claim formalized here quantifies over every positive natural n, uses exactly the twelve published exclusions, and takes p to be Nat.minFac n. No additional restriction on n is imposed.

Proof record

Module present in current Truth release; resolution evidence is pinned below.

D5/S0/Certificates/ZelinskyZhangConjectureTwentyTwoRefutation.result

Frozen module record

Evidence snapshot: b52262a023c4db9bb30a128ebfd80ea4ef4c1a53. The linked mdBook follows upstream development.
About this highlight
Why this order
  • Problem significance · 1/4 · 30% weight

    A released resolution of an explicitly recorded external question.

    Source 1 · Source 2
  • Proof methods · 1/4 · 15% weight

    A verified proof or refutation is available; additional method evidence is scored only when documented.

    Source 1
  • Mathematical advance · 1/4 · 15% weight

    The exact recorded statement is refuted; a single witness does not by itself imply a broader advance.

    Source 1

About this order

Open questions & research directions

Curated problem collections

19 selected questions

Named questions & related foundations

Weil spectra

Paper

Connes–Consani–Moscovici's route to RH: do genuine lowest Weil modes converge to Ξ?

Open question

Our foothold

Finite-scale Weil forms provide a concrete spectral starting point.

What remains

Relate genuine lowest modes to the Xi limit; the explicit model alone is insufficient.

Question & possible approaches

The question

Do suitably normalized Fourier transforms of genuine lowest Weil modes converge locally uniformly to the Riemann Xi function as the physical support grows?

A possible next step

Reuse the existing Weil research lanes. Match operator domains, physical window, normalization and Fourier convention before comparing any two certificates.

Compare one genuine mode and its actual candidate

The question

Can a full-space residual and complement-coercivity certificate bound the Fourier error for the same physical Weil operator and prolate trial?

What remains

A retained matrix residual leaves an exterior term. Domain error and candidate normalization must be paid separately.

A possible next step

Check the existing numerical and proof-script outputs before creating another certificate; assemble prime, pole, Gamma and tail terms with one set of parameters.

What would count as progress

An independently replayable fixed-window bound with every premise identified. It should not be relabelled as a uniform all-window theorem.

A physical-window rate for the true mode error

The question

Can arithmetic cancellation and complement coercivity control the actual-mode Fourier error along an unbounded sequence of support windows?

What remains

Increasing numerical precision or an exterior cutoff does not reduce the physical error by itself; gaps may deteriorate with the window.

A possible next step

Separate the physical window, finite matrix dimension and numerical tail cutoff. Seek bounds uniform over each target compact set with an explicit vanishing budget.

What would count as progress

A rate for the genuine family with all growth terms retained, or an obstruction to the particular proposed estimate.

Related mathematics in the Library

This is a 2025 source rechecked alongside the June 2026 Suzuki paper. The explicit model limit and convergence of genuine lowest modes are separate. Finite spectral agreement is not an all-scale proof.

Weil spectra

Paper

Does Suzuki's boundary characteristic satisfy the zeta limit?

Open question

Our foothold

Our finite resolvent identities supply nearby operator and spectral constructions.

What remains

Identify the actual boundary operator and prove the stated limiting relation.

Question & possible approaches

The question

Can theta(a) and a nonvanishing normalization make W(a,theta;z) approach z² xi(1/2-iz)/xi'(1/2-iz) in the sense intended by equation (1.12)?

A possible next step

First resolve the analytic meaning of the meromorphic quotient and normalization. Then identify the paper's W with a repository operator construction without supplying zeta zeros as spectral input.

Identify the actual finite-window characteristic

The question

Can the paper's deficiency-space boundary characteristic be identified with a resolvent or Fredholm construction using the genuine screw kernel?

What remains

A finite atomic measure chosen to look like a Clark measure does not establish this identification.

A possible next step

Fix the domain, boundary convention and inner product; derive the relevant boundary function and measure rather than assume their desired expansion.

What would count as progress

An operator-to-characteristic identity on a specified analytic domain with normalization constants retained.

Control the spectral shift without assuming RH

The question

Can the shift needed to define the window Hilbert spaces be controlled as a grows, without assuming positivity of every Weil window?

What remains

The source's global Hilbert-space heuristic assumes RH. Finiteness of prime terms at each fixed a gives no uniform estimate.

A possible next step

Use explicit lower bounds and compare shifted constructions on common test-function cores; track any effect on characteristic zeros and normalization.

What would count as progress

A shift-control lemma with unconditional hypotheses useful for the proposed limit, or a precise failure of a proposed normalization.

Related mathematics in the Library

The limit is conjectural; Corollary 1.6 proves an implication from it. Section 7 assumes RH for its heuristic. That assumption cannot be imported into an unconditional proof of the limit.

Quantum geometry

Paper

Zauner's conjecture: do SIC-POVMs exist in every dimension?

Open question

Our foothold

Weyl displacement identities provide algebraic inputs for symmetric measurements.

What remains

Construct a valid SIC in every dimension, beyond the available conditional identities.

Question & possible approaches

The question

Does every positive dimension d admit d² unit vectors forming an equiangular tight frame, equivalently a SIC-POVM?

A possible next step

Connect the existing displacement operators to a precisely normalized fiducial system, keeping every extra symmetry and number-theoretic premise explicit.

From a verified fiducial to a SIC orbit

The question

Can exact displacement-overlap equations for a unit fiducial imply the d²-vector equiangular tight-frame and POVM identities?

What remains

Trace orthogonality is available but the orbit normalization, distinctness and identity resolution must be derived.

A possible next step

Use the existing displacement composition and trace owners; derive twirling and the orbit Gram entries rather than introducing another Weyl convention.

What would count as progress

A conditional fiducial-to-SIC theorem with d and phase conventions explicit, plus exact low-dimensional checks.

An exact certificate behind a numerical SIC candidate

The question

Can a proposed algebraic fiducial be certified against all required overlap and normalization equations?

What remains

A small floating residual does not establish equality or existence. An imposed order-three symmetry narrows the search class.

A possible next step

Use explicit number-field embeddings and exact polynomial reductions; list which equations and embedding choices are certified.

What would count as progress

A verified instance with its dimension and assumptions, never presented as the universal existence theorem.

Related mathematics in the Library

The survey records the SIC existence conjecture and conditional number-theoretic progress. Neither those assumptions nor the MUB conjecture are identified with an unconditional all-dimension SIC proof.

Quantum geometry

Paper

Can four mutually unbiased bases exist in dimension six?

Open question

Our foothold

Our rank-one contexts and overlap identities describe the relevant basis geometry.

What remains

Four bases in dimension six must satisfy all pairwise compatibility equations at once.

Question & possible approaches

The question

Does a family of four orthonormal bases of complex dimension six exist with every cross-basis squared inner product equal to 1/6?

A possible next step

Establish the exact basis/context correspondence, then express all cross-basis compatibility equations in one coordinate system.

An exact basis-to-projector bridge

The question

Can orthonormal bases modulo individual phases be represented faithfully as labelled complete rank-one projector contexts, preserving the MUB overlap condition?

What remains

Rank-one, completeness, orthogonality, labelling and phase equivalence all need explicit hypotheses.

A possible next step

Formalize both directions and prove trace(P_i Q_j) = |inner(v_i,w_j)|^2, using the repository context definitions.

What would count as progress

A proved correspondence at dimension six that translates the four-basis feasible set without adding a two-design assumption.

A certificate for joint Hadamard compatibility

The question

Can an independently verified order-six Hadamard parameterization support a complete exact exclusion or construction for three Hadamards relative to the identity basis?

What remains

A classification claim alone leaves pairwise and joint compatibility to prove. Numerical near-solutions do not satisfy exact equations.

A possible next step

First check parameterization coverage and exceptional branches, then express every required cross-Gram modulus equation and seek exact certificates.

What would count as progress

An exact solution or branch-covering exclusion with proved coverage. Treat the dossier-mentioned classification preprint as unverified input until checked.

Which projector-coordinate constraint could exclude four bases?

The question

After checking the vector-coordinate obstruction, can a faithful projector formulation impose a dimension-six constraint stronger than m <= d+1?

What remains

The existing elementary projector bound permits four, five, six and seven bases at d=6. No stronger exclusion is supplied by that bound.

A possible next step

Track polynomial degree before and after P=vv*. First reproduce the stated separation, then test exact certificates for a genuinely stronger constraint.

What would count as progress

A scope-correct formalization of the reported obstruction or a new certificate for the four-basis feasible set. Failure of one encoding is not nonexistence.

Related mathematics in the Library

Arithmetic

Paper

Wall–Sun–Sun primes: can a prime have equal Fibonacci periods modulo p and p²?

Open question

Our foothold

Golden-unit orders and their lifts connect Fibonacci periods to local arithmetic.

What remains

Determine whether a prime can retain the same period when lifted to its square.

Question & possible approaches

The question

Does a prime p exist with pi(p) = pi(p²), where pi is the Pisano period? The infinitely-many assertion is a separate stronger question.

A possible next step

Identify recurrence period, matrix order and golden-unit order before studying the first lift defect.

Pisano period as matrix and golden-unit order

The question

For m >= 2, can the pair-recurrence period be proved equal to the multiplicative order of A = [[0,1],[1,1]] over Z/mZ and of its golden-unit representation?

What remains

Matrix order concerns the whole matrix, so equality with the orbit period needs proof; invertibility and indexing must be explicit.

A possible next step

Prove the Fibonacci formula for A^n and compare the return condition (F_n,F_(n+1)) = (0,1) modulo m.

What would count as progress

A reusable equivalence for all m >= 2, with small-modulus conventions stated and checked.

The first-lift defect criterion modulo p²

The question

With r = pi(p) and A^r = I + pB modulo p², is pi(p²) equal to r exactly when B vanishes modulo p, and otherwise equal to pr?

What remains

The argument requires the reduction-order relation and a justified calculation of the order of I+pB, including small primes.

A possible next step

Define B as a residue class, prove the kernel-order dichotomy, and cross-check recurrence and matrix computations independently.

What would count as progress

A proved local lift dichotomy with every prime case covered. This does not settle whether a vanishing defect ever occurs.

Related mathematics in the Library

Next questions from our results

Sequence complexity

Paper

What determines the signed even recurrence of Thue–Morse complexity?

Open question

Our foothold

Our odd and even recurrences use the same run-compressed class count.

What remains

Recover the signed difference and the missing even recursion beyond Equation (11).

Equation (11) is now a released result ↗The absolute difference is covered. The sign and full recursion are separate questions.

Question & possible approaches

The question

For the same run-compressed class count R, what determines the sign of R(4n+2)-R(4n), and can R(4n) and the full sequence be described by an exact recursion?

A possible next step

Start from the released even-complexity theorem and the extrema recurrences. Isolate the information needed for the sign and R(4n), keeping the nonautomaticity formalization as a separate target.

Determine the sign and the remaining even recursion

The question

Can the difference R(4n+2)-R(4n) be characterized for every n >= 1, including its sign when nonzero?

What remains

The released absolute-difference theorem fixes the magnitude only; it leaves the sign and R(4n) outside its stated scope.

A possible next step

Read the released even-complexity proof and its alternation-extrema recurrences, then isolate the additional information needed to recover the signed difference.

What would count as progress

An all-index signed relation for the same class count, or a recurrence that determines the remaining even values.

Formalize the nonautomaticity argument

The question

Can one prove that R(n), for n >= 1, is not k-automatic for any k >= 2?

What remains

The dossier points to a written unboundedness argument, but does not bind a separate kernel-checked nonautomaticity endpoint.

A possible next step

Locate any newer formal endpoint first. Otherwise formalize the written unboundedness argument for positive lengths and connect it to the finite-output definition of k-automaticity.

What would count as progress

A release-bound unboundedness or non-k-automaticity theorem with the sequence definition and quantified bases explicit.

Related mathematics in the Library

The cited paper records these related questions. The released Equation (11) dossier now covers the absolute difference. Its scope note distinguishes the sign and full recursion, and records a written nonautomaticity argument without a separately kernel-checked endpoint. This is a scope update from the release dossier, not a new literature search.

Real-rooted polynomials

Paper

What survives of the Pochhammer interval conjecture in higher degree?

Open question

Our foothold

The exact quadratic classification gives a starting point for higher-degree root geometry.

What remains

Which interval and monotonicity claims survive the counterexample to the original bound?

Question & possible approaches

The question

After the degree-two counterexample to c_(2k)(a) < 2a, is M_(2k)(a) still a single symmetric interval for all k >= 2 and a > 0, and what can be proved about the sequence of its left endpoint radii?

A possible next step

Start with degree four, derive the transformed coefficients exactly, and isolate parameter ranges with certified real-root counts.

A certified degree-four admissible interval

The question

For each positive a, is the set of t for which every root of L_a((X+t)^4) lies in [-1,0] a single interval?

What remains

The quadratic discriminant argument does not classify all four complex roots of the quartic.

A possible next step

Construct L_a((X+t)^4) from the same basis action and use exact root isolation or a Sturm certificate, including repeated roots and endpoints.

What would count as progress

A proved parameter classification or an exact disconnected-set counterexample, with every complex root accounted for.

Compare the degree-two and degree-four radii

The question

With c_(2k)(a) defined independently from the admissible set, does c2(a) <= c4(a) hold for every a > 0?

What remains

Monotonicity was bundled with a false strict bound in the source conjecture. Its validity has to be established separately.

A possible next step

Use the proved c2 closed form as a baseline for the exact quartic endpoint, preserving infimum existence and boundedness obligations.

What would count as progress

A universal comparison or a certified parameter counterexample. Later degree comparisons and the limiting value remain separate targets.

Related mathematics in the Library

Our quadratic result refutes the original strict upper bound. This family asks about the remaining clauses separately. Degree-two root-radius progress in upstream QuadraticRadii addresses other questions and is not an all-even-degree result.

Additive combinatorics

Paper

Does the greedy formula hold for the separate g+1 seed family?

Open question

Our foothold

The two-parameter theorem links accepted blocks to the literal greedy process.

What remains

The separate g+1 seed family falls outside that theorem’s d ≥ 2 hypothesis.

Question & possible approaches

The question

For every g >= 2, does the literal greedy three-sumfree sequence starting 1,g,g+1 have the membership formula in published Conjecture 16, with modulus 10g+3 and its two residue intervals?

A possible next step

Encode the published Conjecture 16 candidate and prove its accepted and forbidden block statements uniformly in g.

Accepted blocks and forbidden sums for g+1

The question

Can the candidate set with exceptions 1,2g+1,6g+1 and residue intervals [g,2g] and [6g+2,7g+1] modulo 10g+3 be characterized by restricted triple sums?

What remains

The single-interval periodic decomposition from Conjecture 17 cannot be reused without handling both new intervals and all cross-sums.

A possible next step

Prove three-sumfreeness and explicit representations of every excluded later value, separating the initial segment from periodic translates.

What would count as progress

A uniform accepted/forbidden partition for every g>=2 that preserves the requirement of three distinct earlier summands.

Connect the candidate to the least-next-entry sequence

The question

Does the new candidate describe every term selected by the literal greedy process from 1,g,g+1?

What remains

A sumfree candidate by itself does not prove that the greedy algorithm generates it.

A possible next step

Adapt the prefix invariant and uniqueness argument to the new accepted/forbidden partition, including the seed boundary.

What would count as progress

An all-parameter membership theorem matching Conjecture 16. Verified samples remain examples, not the universal result.

Related mathematics in the Library

The source gives fixed-parameter checks, distinct from the universal conjecture. Published Conjecture 16 and our solved Conjecture 17 are separate families. Numbering differs in the early arXiv version.

Automata

Paper

Is the 22-state base-4 golden-ratio DFAO minimal?

Open question

Our foothold

Exact golden-ratio digits and Zeckendorf encodings supply arithmetic inputs.

What remains

Does the sparse powers-of-four input language really require all 22 states?

Question & possible approaches

The question

Must every leading-zero-invariant DFAO that produces the correct golden-ratio digit on all Zeckendorf encodings of 4^i have at least 22 states?

A possible next step

Verify the published transition table and exact input/output conventions before searching for a smaller machine.

A sparse-language distinguishability certificate

The question

Can 22 prefixes be found such that every pair has a common suffix making both completed words legal power inputs with different required outputs?

What remains

Each separation witness must constrain both words inside the sparse domain. Full-language Myhill-Nerode witnesses may be irrelevant here.

A possible next step

Build exact labelled prefix trees; search for a 22-clique of pairwise incompatible prefixes and retain both completed power inputs for every edge.

What would count as progress

A checked finite incompatibility certificate plus a proof that it forces 22 distinct states. Failure to find a clique leaves other lower-bound methods open.

A finite, replayable 21-state UNSAT obstruction

The question

Can a finite set of exact power-input digit constraints already exclude all DFAOs with at most 21 states under the chosen conventions?

What remains

A solver verdict needs a proof-checking and encoding-soundness bridge. A SAT candidate still requires all-powers verification.

A possible next step

Encode a bounded labelled sample, export an independently checkable UNSAT certificate, and account for state renaming and leading-zero invariance.

What would count as progress

Every globally correct DFAO must satisfy the sample; a sound finite UNSAT certificate therefore proves the lower bound. Finite sample completeness is unnecessary for this direction.

Related mathematics in the Library

Golden numeration

Paper

Do negative base-phi prefixes obey the trident classification?

Open question

Our foothold

Canonical golden numeration provides exact order and digit representations.

What remains

Connect negative-prefix occurrences to the proposed trident classification.

Question & possible approaches

The question

Is each admissible negative-prefix occurrence sequence one Lucas-parameterized V_F, V_G or V_H component, or an allowed union of three components?

A possible next step

Port the two-sided expansion and the three sequence families with explicit union and multiplicity conventions.

A bounded-state negative-prefix transducer

The question

For each fixed prefix length m, can a finite-state transducer read canonical Zeckendorf input and output the first m negative-position base-phi digits?

What remains

Contractive intuition is insufficient: carry residues and a uniform reachable-state bound for fixed m need proof.

A possible next step

Compute exact Z[phi] residue traces, propose a state invariant, and prove transition closure and output correctness.

What would count as progress

A finite machine with a proof for every canonical input at fixed m. Finite samples alone establish neither the state bound nor correctness.

From negative cylinders to exact return itineraries

The question

Can each negative-prefix cylinder be identified with the paper's F, G or H itinerary, or its allowed three-component interleaving?

What remains

Existing golden-word return theorems apply only after the cylinder/subshift identification is established.

A possible next step

Start with one admissible single-component prefix; derive its gap word and Lucas parameters by exact desubstitution.

What would count as progress

A quantified cylinder-to-itinerary equality and Lucas-parameter recursion. Record any violated gap or return-word invariant.

Use the proved Kimberling criterion as a conversion check

The question

Can the exponent-doubling criterion in Shallit and Vukusic Theorem 5 be transported faithfully to the repository's exact two-sided phi representation?

What remains

The criterion is reported proved externally. Its formalization needs the same canonical digit and symmetry conventions and does not settle the trident conjecture.

A possible next step

Port the symmetry and integer-value predicates, replay the exact algebraic criterion, and use it as a regression case for two-sided digit conversion.

What would count as progress

A faithful theorem-level bridge to a known result, with no claim of a new conjecture resolution.

Related mathematics in the Library

Rewriting games

Paper

Does the Ordered Zeckendorf Long Game Strategy maximize length?

Open question

Our foothold

The rewriting rules and exact sequence encodings provide a concrete game model.

What remains

Prove that the proposed long-game strategy is optimal over every legal play.

Released result ↗

Question & possible approaches

The question

Among legal terminal runs starting from n copies of F_1, does the specified Long Game Strategy attain maximum length?

A possible next step

Define the five ordered moves and resolve whether all switch tie choices are intended to be equally optimal.

Does switch-order freedom preserve LGS length?

The question

Do all allowed switch tie choices lead to the same LGS terminal length from the specified initial states?

What remains

The universal LGS statement depends on this interpretation; normal-form uniqueness does not imply equal run lengths.

A possible next step

Enumerate exact ordered states and all switch tie choices for bounded n; retain differing lengths and full move traces.

What would count as progress

A switch-order independence proof, or an exact counterexample requiring a deterministic or existential reformulation.

Height-preserving local exchanges for LGS

The question

Can each LGS priority choice be shown to attain an argmax successor of the longest-remaining-path Bellman recursion?

What remains

The game DAG needs its own termination proof. Local exchange lemmas may fail even when global optimality survives.

A possible next step

Test swap-first, leftmost-one merge and rightmost-split exchanges against exact state heights; save both suffix witnesses at the first failure.

What would count as progress

Proved height dominance for every reachable strategy state, followed by well-founded induction. A failed exchange refutes that route only.

Related mathematics in the Library

Probability

Paper

Do random Zeckendorf game lengths have a Gaussian limit?

Open question

Our foothold

Exact game paths and arithmetic structure provide objects for a probabilistic model.

What remains

Identify a mechanism that yields a Gaussian limit for random game lengths.

Question & possible approaches

The question

Under a precisely specified random game measure, does the centered, variance-normalized length converge to a Gaussian?

A possible next step

Define both path measures exactly; prioritize the dossier's locally uniform mixing question with explicit quantifiers.

Two exact game-length generating functions

The question

Can exact finite recursions distinguish uniform terminal-path sampling from locally uniform successor sampling on the game DAG?

What remains

Move multiplicities and suffix-path counts affect transition weights. Treating the measures as interchangeable changes the target.

A possible next step

Compute both rational generating functions independently on small value-preserving state DAGs, with the terminal state contributing 1.

What would count as progress

Proved recursions, normalization and moment identities for both measures, including an explicit comparison example.

A uniform regenerative-barrier estimate

The question

Can a canonical digit barrier bound the dependence of separated split indicators uniformly over initial input size?

What remains

Finite carry locality need not survive conditioning on entire random histories. The relevant mixing-CLT summability and variance hypotheses must be checked explicitly.

A possible next step

Search exact history pairs that share a proposed barrier, then measure and bound influence-crossing events under the chosen measure.

What would count as progress

An explicit uniform mixing estimate strong enough for the selected CLT, or a minimal history pair invalidating the proposed barrier.

Related mathematics in the Library

Sequence complexity

Paper

How large is maximum order complexity along polynomial inputs?

Open question

Our foothold

Zeckendorf order and carry structure offer exact tools for polynomial subsequences.

What remains

Control successor behavior well enough to bound maximum order complexity.

Question & possible approaches

The question

For the parity of the Zeckendorf digit sum along n², can maximum order complexity of the first N terms be bounded above by C N^(1/4) for all sufficiently large N?

A possible next step

Implement the exact finite repeated-block criterion before attempting an effective carry-window bound.

An exact successor-collision certificate

The question

Can finite maximum order complexity be certified by equal-length block collisions and their following bits, including degenerate constant-prefix cases?

What remains

Off-by-one conventions and special zero-memory cases must match the source definition.

A possible next step

Cross-check a naive implementation against a suffix-based method; emit equal blocks with unequal successors for every claimed lower bound.

What would count as progress

A proved finite criterion and replayable witnesses, independent of any conjectured asymptotic exponent.

Can bounded carry state force the next parity bit?

The question

For square inputs, can a history of length O(N^(1/4)) determine enough boundary carry information to determine the next Zeckendorf digit-sum parity?

What remains

Unbounded carry propagation and noninjective history-to-state maps are possible obstructions.

A possible next step

Track exact normalization of 2nh+h² at Fibonacci/Lucas scales, then search for equal observed histories with conflicting next bits.

What would count as progress

An explicit uniform boundary-state theorem implying the upper bound, or exact witness pairs falsifying the proposed window.

Related mathematics in the Library

Phase retrieval

Paper

Is real phase retrieval exponentially ill-conditioned at 2M-1 measurements?

Open question

Our foothold

Rank-one measurements and exact overlap geometry connect to phase retrieval.

What remains

A Gaussian asymptotic does not establish the universal deterministic stability claim.

Question & possible approaches

The question

For every full-spark real (2M-1)-by-M matrix, M>1, do universal C>0 and beta in (0,1) bound omega(A) by C times the maximum row norm times beta^M?

A possible next step

Reduce the full-spark omega definition to minimum singular value over square row submatrices, retaining row scaling, then test candidate deterministic inequalities.

An exact square-submatrix formulation of omega

The question

Under full spark at N=2M-1, is omega(A) exactly the minimum of sigma_min(A_T) over M-row sets T?

What remains

The source definition also allows larger row sets; its equivalence needs a rank and Loewner-order argument.

A possible next step

Formalize rectangular row restriction and the variational least singular value. Preserve M>1 and the source's complement-rank convention.

What would count as progress

A reusable equivalence with a row-rescaling test and exact small cases. This is a formalization bridge, not a new resolution.

A deterministic obstruction stronger than a Gaussian example

The question

Can row-subset geometry force an exponentially small least singular value uniformly over every normalized full-spark frame?

What remains

Average-case small-ball estimates and the existence of a badly conditioned example do not imply a worst-case upper bound.

A possible next step

Compare extremal deterministic families using certified spectral intervals; seek a packing, determinant or partition argument with dimension-independent constants.

What would count as progress

A uniform inequality or a counterfamily for a stated effective bound, with numerical probes kept separate from the universal claim.

Related mathematics in the Library

Conjecture 20 is the universal full-spark statement. The lower-case n in the displayed complement-rank formula is read as the column dimension M; the normalized domain is stated explicitly here.

Phase retrieval

Paper

Does injectivity at 4M-5 complex measurements become rare?

Open question

Our foothold

Complex measurement geometry offers a shared language for injectivity.

What remains

Understand the probability of injectivity at the 4M−5 measurement threshold.

Question & possible approaches

The question

For M>=2 and a Haar-random M-plane in C^(4M-5), let p_M be the probability that x modulo global phase maps injectively to |Ax|². Is p_M<1 for each M, and does p_M tend to zero?

A possible next step

Encode the measurement kernel on actual differences xx* - yy*. Define the Grassmannian measure and bad event before relying on dimension counts.

A faithful Hermitian lift of equal-intensity pairs

The question

Is equal intensity equivalent to the linear measurement map annihilating xx* - yy*, with zero difference exactly global-phase equivalence?

What remains

Unnormalized signals and the zero vector need explicit treatment. Arbitrary rank-two matrices are not automatically differences with the required inertia.

A possible next step

Prove the trace identity and represent the difference cone exactly; relate normalized nonzero signals to the rank-one projector predicate.

What would count as progress

A bidirectional statement preserving all signals, not just unit-norm or generic vectors.

From an algebraic witness to positive probability

The question

Can a noninjective-frame witness be extended to a positive-measure set at the critical measurement count?

What remains

A single real-algebraic solution may lie in a measure-zero stratum. Dimension counting alone does not settle transversality or probability.

A possible next step

Analyze the rank-two difference incidence set on a concrete Grassmannian chart and verify a local persistence or measure argument.

What would count as progress

A proved positive-measure bad set for a stated dimension or a uniform probabilistic estimate; retain the distinction between the two conjecture clauses.

Related mathematics in the Library

The statement is recorded in the checked survey. The 11-measurement example in C^4 refutes the older necessity claim, not these two probability clauses. This entry does not assert exhaustive absence of later work.

Fourier and optimization

Paper

Does the expected Lovasz number of random circulants approach sqrt(n)?

Open question

Our foothold

Finite Fourier analysis supplies the characters of the cyclic model.

What remains

Bridge that structure to a probabilistic bound with the conjectured leading constant.

Question & possible approaches

The question

For dense undirected random circulant graphs with independent Bernoulli(1/2) choices of nonzero step pairs {k,-k}, is E theta(G)=(1+o(1)) sqrt(n)?

A possible next step

Derive the cyclic averaging and Fourier-diagonal reduction with the repository character convention; then attach sound primal and dual bounds.

A normalization-safe circulant theta reduction

The question

Can cyclic averaging identify the theta SDP with a finite Fourier positivity problem without changing its optimum?

What remains

The SDP uses trace one while a scaled first-row LP can use x_0=1. Confusing these conventions changes the objective by n.

A possible next step

Prove averaging preserves PSD, edge constraints and objective, then derive eigenvalues using the existing ZMod characters, with sign and scale recorded.

What would count as progress

A bidirectional feasible-point translation and equality of objectives under the chosen normalization.

Certified bounds for a concrete circulant graph

The question

Can explicit primal and dual witnesses give rigorous theta bounds for a supplied circulant graph?

What remains

Solver tolerance and approximate Fourier eigenvalues can falsely certify feasibility. Real algebraic or interval enclosures need a soundness theorem.

A possible next step

Construct exact small-graph witnesses and certified residual margins; test graph-complement and scaling conventions independently.

What would count as progress

Replayable upper and lower bounds with all rounding budgets. These are finite footholds for the expectation conjecture.

Related mathematics in the Library

The survey gives an O(sqrt(n log log n)) upper estimate and the sharper leading-constant conjecture. Its arXiv submission date is 31 March 2026; HTML generation in August is not a new arXiv version.

External classical questions

Arithmetic

OEIS A006577

Does every Collatz orbit reach 1?

Open question

Why it matters

Integer iteration, recurrence relations and termination proofs.

What remains

The source records the universal termination question as open; finite computations do not settle it.

Question & possible approaches

The question

Starting from any positive integer, repeatedly halve even values and replace odd values n by 3n + 1. Must the orbit eventually reach 1?

Arithmetic

OEIS A000043

Are there infinitely many Mersenne primes?

Open question

Why it matters

Prime structure, divisibility and arithmetic sequences.

What remains

OEIS records infinitude as unproved; primality certificates for individual terms do not establish infinitude.

Question & possible approaches

The question

Are there infinitely many prime exponents p for which 2^p - 1 is prime?

Additive combinatorics

Erdős #3

Erdős's conjecture on arithmetic progressions: does a divergent reciprocal sum force long progressions?

Open question

Why it matters

Additive structure, density bounds and sum-free sets.

What remains

The full conjecture remains open. Bloom and Sisask proved the three-term case in 2020; that does not settle arbitrary lengths.

Three-term case provedBloom & Sisask · 2020

Question & possible approaches

The question

If a set A of positive integers has a divergent sum of reciprocals, must it contain arithmetic progressions of every finite length?