RESEARCH / Focused target
First occurrences in Kimberling's array
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S1/Recurrence/KimberlingArrayFirstOccurrence.resultAbsent from this graph
Problem
The question
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.
Motivation
Our foothold
The source proves that each positive occurrence count is attained infinitely many times. First occurrences identify the smallest representatives of these classes and relate their growth to odd-indexed Lucas numbers.
Gap
Missing bridges
Proposition 10 establishes attainment at L_{4v−5}; Conjecture 15 asks for universal minimality. The gap is a bound on the smallest common value of two arbitrary positive rows, followed by a finite count argument. The source supplied with the task is v1; no claim of exhaustive literature priority is made.
Route
Proposed approach
Consecutive 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.
For 1 ≤ i < j, suppose N belongs to both rows and N < L_{2i+1}. Choose positive multipliers k,l and put d = j−i ≥ 2 and u = k−lφ^d. The floor bounds give k < φ^{i+1}, lφ^d < φ^{i+1} and 0 < |u| < φ^{−i}. Its conjugate satisfies |u′| < φ^{i+1}(1+φ^{−2d}), so its nonzero integral norm has absolute value less than φ(1+φ^{−4}) < 2. Hence the norm is ±1.
The unit classification in ℤ[φ] and the negative golden coefficient imply u = ψ^r = (−1)^r φ^{−r}, with ψ = −φ^{−1} and r ≥ i+1. Coefficient comparison gives lF_d = F_r and d ≤ r. If r ≥ i+2, the conjugate equation k = φ^r+lψ^d contradicts k < φ^{i+1}. Thus r = i+1. For even d, the same equations give lφ^j ≥ φ^{2i+1} > L_{2i+1}. For odd d, let H = lL_j and ε = lφ^{−j}, where 0 < ε < φ^{−1}. Then
$$ l\varphi^j=H-(-1)^r\varepsilon,\qquad k\varphi^i=H+(-1)^r(\varphi^{-1}-\varepsilon). $$
These multiples straddle the integer H, contradicting their common floor. Therefore every common value satisfies L_{2i+1} ≤ N.
List any v containing rows in increasing order. Their gaps are at least two, so the penultimate index is at least 2v−3. Applying the pair bound and Lucas monotonicity gives the stronger conclusion
$$ v\ge2\ \land\ a(N)\ge v\quad\Longrightarrow\quad L_{4v-5}\le N. $$
At m = 4v−5, the rows 1,3,…,2v−3 and m contain L_m. The lower odd-row multiplier is L_{m−i}, and the own-row multiplier is one. These are v distinct rows. An additional row would force L_{4v−1} ≤ L_{4v−5}, contradicting strict Lucas increase. This proves attainment and minimality.
Falsifier
What would falsify this route
A counterexample would be a positive N below L_{4v−5} with exactly v occurrences, or a target Lucas number with a different occurrence count. The stronger lower bound excludes the former, and the explicit v-row construction plus the next threshold excludes the latter. The proof must retain strict floor bounds, both interval-gap parities, the integer norm condition, and complete coverage of the finite occurrence count.
Evidence
Evidence to collect
D5/S1/Recurrence/KimberlingArrayFirstOccurrence.result has the exact
closed type claim. The companion module proves the golden unit
classification. Both modules compile with the pinned Lean toolchain.
The settling theorem's axiom closure is propext, Classical.choice,
Quot.sound; there is no sorry, new axiom or native_decide.
Two independent exact-integer checks through N = 1,000,000 reproduce the first occurrences 4, 29, 199, 1364, 9349, 64079, 439204 for v = 2,…,8. The forward enumeration visits 28 rows and 1,618,022 entries. The inverse membership checker verifies 720,519 containing-row pairs with no violation of the pair lower bound. These finite computations support the definitions; the universal conclusion follows from the Lean proof.
Literature: arXiv:2610.05776v1, Proposition 10 and Conjecture 15; sequence records OEIS A128440 and OEIS A358359. The proposition's exact row set for odd m ≥ 3 is {i odd : 1 ≤ i ≤ (m−1)/2} ∪ {m}. The formal attainment argument uses its explicit positive memberships and the stronger lower bound to exclude any further rows.
Triage
Scope assessment
theorem: Conjecture 15 holds for every v ≥ 2. The stronger implication
from a(N) ≥ v to L_{4v−5} ≤ N is also proved.
computed: the two exact-integer enumerations agree on all first
occurrences available through N = 1,000,000.
open: no even-indexed Lucas membership classification is proved here.
The source's Proposition 14 proves existence and positivity of the
densities d_v; exact density values or additional bounds are not settled
by this first-occurrence theorem.
What the settlement shows
- [proved: D5/S1/Recurrence/KimberlingArrayFirstOccurrence.pair_lower_bound] The norm-one constraint on an overlapping pair forces an odd Lucas threshold. The parity cases exclude every smaller common value.
- [proved: D5/S1/Recurrence/KimberlingArrayFirstOccurrence.occurrence_lower_bound] The bound applies to at least v occurrences and does not require a positive N hypothesis. Separation of consecutive rows is what turns a pair estimate into a threshold for arbitrarily many occurrences.
- [proved: D5/S1/Recurrence/KimberlingArrayFirstOccurrence.lucas_attainment] The threshold is sharp for each v ≥ 2. The explicit odd-row witnesses attain it, and the next threshold excludes any extra row.
- [proved: D5/S1/Recurrence/KimberlingArrayFirstOccurrence.result] The representative used in the source's Theorem 13 is optimal as the first occurrence. Its infinite-recurrence argument and Proposition 14's density argument remain separate results of the source.
- [computed:
python3 /Users/macstudio/omega-op/s3688/t6-work/lucas_scan.pyandpython3 /Users/macstudio/omega-op/s3688/t6-work/carmichael/check_inverse.py] The controls through one million agree with the universal threshold. - [open] Exact densities and an even-indexed Lucas membership description require separate proofs; this settlement addresses the odd-indexed first-occurrence family.
ASSUMED-UNVERIFIED
Unverified assumptions
The supplied v1 source and task brief establish the target's wording; remote issue metadata, subsequent literature and exhaustive priority have not been checked in this offline task. The OEIS pages are cited as supplied sequence references and were not fetched. Scoped Lean and Scribe checks are separate from repository freezing, complete admission and official acceptance, which are not performed by this delivery.