RESEARCH / Focused target
Mane's negative-index factorization
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Zeros/NegativeIndexFibonacciFactorization.resultAbsent from this graph
Problem
The question
S. R. Mane, Identically vanishing k-generalized Fibonacci polynomials, arXiv:2507.11596v4, p. 19, Section 6, Conjecture 6.4, states:
For k β₯ 2 and n = βsk, where s β [2, k + 1] (the pattern fails for s β₯ k + 2), π_{βsk,k}(x) = βx(x^k + 1)^{sβ2}(x^k + s). (6.4) Hence π_{βsk,k}(x) has roots x^k = βs.
The recurrence (1.2)β(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 $-sk$.
Motivation
Our foothold
Preregistration #14688
selects MANE-2, the full factorization identity. The settling declaration is
D5/S3/Zeros/NegativeIndexFibonacciFactorization.result : claim.
It provides the exact restricted attainment range related to Conjecture 6.6.
Gap
Missing bridges
This is a tier-1 externally published named conjecture. The source and literature readings recorded in #14688 inspect arXiv versions, author papers, classical root papers and citation indexes; no settlement was found in that scope. Finite searches do not establish exhaustive absence of prior work.
Route
Proposed approach
Put $t=X^k$, $A(z)=\sum_{m\ge0}a_k(m)z^m$ and $B(z)=z^k((1+t)-Xz)$. The recurrence gives
$$ A(z)(1-B(z))=1-tz^k. $$
A finite geometric sum gives the coefficients below degree $kL$ without an analytic convergence assumption. At degree $sk+1$, with $s\le k+1$, only $B^s$ contributes; at degree $(s-1)k+1$, only $B^{s-1}$ contributes. Their coefficients are $-sX(1+t)^{s-1}$ and $-(s-1)X(1+t)^{s-2}$. Subtracting $t$ times the latter gives $-X(1+t)^{s-2}(t+s)$.
The public surface consists of claim and result; the private definitions
and all sixteen private theorems have live consumers. Every theorem is
bind-only after expansion through same-delivery helpers to frozen and
pinned Mathlib declarations. Admission is
open-problem-resolution (#14688; Proved), with no escape witness.
Utility is none: this is a universal polynomial identity, with no
fixed-parameter numerical certificate in the delivered Lean module.
Information-escape registration is paused under CLAUDE.md Β§3.9.
Falsifier
What would falsify this route
A mismatch between the recurrence, index convention, quantifier range or polynomial factors and the source invalidates fidelity. A failing in-range instance refutes the claim. A prior published settlement invalidates the literature gap. The computations below are checks, not proof premises.
Evidence
Evidence to collect
The Lean module and its corresponding Blueprint Scribe establish the exact
MANE-2 claim. The settling Scribe node has OpenProblemResolutionClaim(Proved)
and cites Library/Zeros/mane2026vanishing.md. The axiom closure of every
public declaration is contained in
${\mathrm{propext},\mathrm{Classical.choice},\mathrm{Quot.sound}}$.
Triage
Scope assessment
What the settlement shows
Proved in this module β the full permitted range. result establishes
(6.4) for every $k\ge2$ and $2\le s\le k+1$. Its private A_identity,
A_trunc, G_coeff_diag and a_factor implement the coefficient argument.
Proved by the following paper argument β why the pattern stops at $s=k+1$.
The $r$th summand $B^r$ has degree support in $[kr,(k+1)r]$.
For $r<s$, its largest possible endpoint is $(k+1)(s-1)$, which is
strictly below $sk+1$ exactly when $s\le k+1$. For $r>s$, its lower
endpoint is at least $k(s+1)>sk+1$ for $k\ge2$.
At $s=k+2$, the $r=s-1=k+1$ upper endpoint equals $sk+1$;
its leading coefficient is $(-X)^{k+1}$. The shifted numerator contribution
at degree $(s-1)k+1$ still has only the diagonal summand, because
$s-1=k+1$. Consequently an additional nonzero term $(-X)^{k+1}$ appears.
This is a paper derivation of the sharp next boundary, not an additional
Lean theorem. The kernel-checked window exclusion used within the permitted
range is G_coeff_diag.
Proved by the following paper argument β roots and multiplicities.
Over $\mathbb C$, the factors $x$, $x^k+1$ and $x^k+s$ have disjoint root
sets for $s\ge2$. Every root of either binomial is nonzero; its derivative
$kx^{k-1}$ is nonzero since $k\ge2$. Thus every root with $x^k=-s$ has
multiplicity one, every root with $x^k=-1$ has multiplicity $s-2$
(and is absent at $s=2$), and zero has multiplicity one.
Complex $k$th roots of $-s$ exist and have $|x|^k=s$; all other nonzero
roots have $|x|^k=1$. Hence $\zeta_{-sk,k}=s$ throughout the proved range,
including the restricted attainment range of Conjecture 6.6 for $k\ge3$.
These are algebraic consequences of result proved here on paper; no
separate Lean multiplicity or supremum theorem is delivered.
Computed β $s=k+2$. For $k=2,\ldots,7$, exact symbolic recurrence calculations give six of six matches for
$$ \mathcal F_{-k(k+2),k}(x)=-xQ_k(x^k),\qquad Q_k(t)=(t+k+2)(1+t)^k+(-1)^kt. $$
The required orchestrator script python3 /tmp/op-mane/check64.py, exit $0$,
gives 27 of 27 in-range matches and six of six failures of (6.4) at
$s=k+2$, for $k=2,\ldots,7$. The supplemental self-contained script
python3 /tmp/op-mane/stageb/boundary64.py, exit $0$, additionally checks
the displayed $Q_k$ identity in six cases and the root-factor multiplicities
in all 27 in-range cases. These computations establish only their finite
scope; the uniform boundary argument is the paper proof above, with its
Lean formalization open here.
Open β the source's broader simplicity sentence. The source says βAll such roots appear to be simpleβ for $j\in[2,k+1]$ with $x^k=-j$ at the additional negative indices discussed on p. 19. The in-range $j=s$ roots are simple by the paper argument above and have exact symbolic multiplicity one in all 27 tested cases. Simplicity at other indices is open here. The $x^k=-1$ roots are outside $j>1$ and can have multiplicity $s-2$; they do not refute that sentence.
Open β neighbouring questions and dependence. Conjecture 6.5, the universal upper-bound clause and asymptotic branch claims of Conjecture 6.6 are not settled here. The companion settlement #14674 refutes unrestricted attainment. This identity supplies attainment only in the stated restricted range. Any source use of (6.4) within that range can use the proved identity; extrapolations beyond it gain no premise from this result. In particular, the p. 20 inference of unbounded growth at fixed $k$ is not justified by (6.4): its parameter $s$ is bounded by $k+1$, and the $s=1$ case cited there is outside the conjecture's $s\ge2$ range.
Reproducible symbolic checks
The scripts require Python 3 and SymPy. Save the following exact UTF-8 bytes at the named command paths, or run equivalent local paths. The computations are exact polynomial expansions and gcd/division checks, with no numerical tolerance.
Command: python3 /tmp/op-mane/check64.py. Exit code: $0$.
SHA-256: e7ee83f451689b57f35da62752803542cf7acab552b81e0c6980ffe2eeb6e214.
import sympy as sp
x=sp.symbols('x')
def F(k, nmin):
vals={1:sp.Integer(1)}
for n in range(-(k-2),1): vals[n]=sp.Integer(0)
n=1
while n-k>=nmin:
vals[n-k]=sp.expand(vals[n]-sum(x**(k-j)*vals[n-j] for j in range(1,k)))
n-=1
return vals
ok=0; bad=0
for k in range(2,8):
v=F(k,-k*(k+2))
for s in range(2,k+2):
lhs=v[-s*k]; rhs=sp.expand(-x*(x**k+1)**(s-2)*(x**k+s))
if sp.expand(lhs-rhs)==0: ok+=1
else: bad+=1; print("FAIL",k,s)
# s=k+2 fails?
s=k+2; print(k, "s=k+2 matches pattern:", sp.expand(v[-s*k]+x*(x**k+1)**(s-2)*(x**k+s))==0)
print("pattern holds", ok, "fails", bad)
Command: python3 /tmp/op-mane/stageb/boundary64.py. Exit code: $0$.
SHA-256: 535eb1b77cb46fcff894c432957caac480507c9d1e764d9e75bfff399f4b4e63.
import sympy as sp
x=sp.symbols('x')
def F(k, nmin):
vals={1:sp.Integer(1)}
for n in range(-(k-2),1): vals[n]=sp.Integer(0)
n=1
while n-k>=nmin:
vals[n-k]=sp.expand(vals[n]-sum(x**(k-j)*vals[n-j] for j in range(1,k)))
n-=1
return vals
ok=0; bad=0
for k in range(2,8):
v=F(k,-k*(k+2))
for s in range(2,k+2):
lhs=v[-s*k]; rhs=sp.expand(-x*(x**k+1)**(s-2)*(x**k+s))
if sp.expand(lhs-rhs)==0: ok+=1
else: bad+=1; print("FAIL",k,s)
# s=k+2 fails?
s=k+2; print(k, "s=k+2 matches pattern:", sp.expand(v[-s*k]+x*(x**k+1)**(s-2)*(x**k+s))==0)
print("pattern holds", ok, "fails", bad)
q_matches=0; root_matches=0
def multiplicity(p,q):
count=0
while True:
quo,rem=sp.div(p,q,x)
if rem!=0:return count
p=quo;count+=1
for k in range(2,8):
vals=F(k,-k*(k+2)); s=k+2
q=(x**k+k+2)*(1+x**k)**k+(-1)**k*x**k
assert sp.expand(vals[-k*s]+x*q)==0
q_matches+=1
for s in range(2,k+2):
f=vals[-k*s]
assert multiplicity(f,x)==1
assert multiplicity(f,x**k+s)==1
assert multiplicity(f,x**k+1)==s-2
assert sp.gcd(x**k+s,sp.diff(x**k+s,x))==1
assert sp.gcd(x**k+1,sp.diff(x**k+1,x))==1
root_matches+=1
print('Boundary Q_k identity matches:',q_matches,'/ 6')
print('Root-factor multiplicities match:',root_matches,'/ 27')
ASSUMED-UNVERIFIED
Unverified assumptions
Exhaustive absence of a prior settlement is not established by the finite literature search. The general root-multiplicity, amplitude and sharp-boundary consequences have paper proofs here but no separate Lean declarations. The broader simplicity assertion and neighbouring conjectures identified above remain open here.