gpt 6 astra ultra attempt on Math-1014

3-accessibility of Fibonacci numbers

Definitions and mathematical statement

Define $F_1=F_2=1$ and $F_{j+2}=F_{j+1}+F_j$ for $j\ge1$, and put $\mathcal F=\{F_j:j\ge1\}\subseteq\mathbb N$. This is a set, so the two initial occurrences of $1$ have no multiplicity significance. In particular, $0$ is not an allowed gap.

Given a set $S\subseteq\mathbb N$, an $S$-difference sequence of length $k$ is a strictly increasing sequence $x_1<\cdots<x_k$ of positive integers such that $x_{i+1}-x_i\in S$ for every $1\le i<k$. For a colouring $\chi:\mathbb N\to[r]$, the sequence is monochromatic if $\chi(x_1)=\cdots=\chi(x_k)$. The set $S$ is called $r$-accessible when for every such colouring and every integer $k\ge2$, a monochromatic $S$-difference sequence of length $k$ exists.

Is $\mathcal F$ three-accessible? Written with its quantifiers, the target is \[ \forall\chi:\mathbb N\to\{1,2,3\}\quad\forall k\ge2\quad \exists x_1<\cdots<x_k: \ \chi(x_i)=\chi(x_1),\ \ x_{i+1}-x_i\in\mathcal F. \] The gap may vary from one step to the next, and Fibonacci values may be reused as gaps. The terms $x_i$ themselves need not be Fibonacci numbers. The length is arbitrary but finite; the statement does not require one infinite monochromatic sequence. A negative answer must provide a three-colouring and a fixed finite length which no monochromatic Fibonacci-difference sequence reaches.

Short English statement

Does every three-colouring of the positive integers contain arbitrarily long monochromatic increasing sequences with Fibonacci gaps?

Research attempt

Scope and process.

This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.

A material literature update.

The historical source asks whether Fibonacci numbers are three-accessible. On 27 September 2026, William J. Wesley posted a primary preprint [R1] claiming a negative answer, using an aperiodic colouring coded by an irrational rotation and a finite directed graph. This source was checked on 1 October 2026. Thus the historical "open" metadata must not be read as a fresh assertion that no resolution has been announced. We do not claim authorship of that result or a complete independent verification of its finite certificate. The independent work below explains rigorously why periodic counterexample searches cannot succeed.

Every modulus divides some positive Fibonacci number.

Include $F_0=0,F_1=1$ for this argument only; the allowed gap remains positive. For any $q\ge1$, the recurrence on pairs modulo $q$ is \[ T(u,v)=(v,u+v),\qquad T^{-1}(u,v)=(v-u,u). \] It is a permutation of the finite set $(\mathbb Z/q\mathbb Z)^2$. Therefore the orbit of $(0,1)$ returns to $(0,1)$ after some positive number $h$ of steps, with $h\le q^2$. This is a return to the starting pair, not just an eventual repeat elsewhere, because $T$ is invertible. Its first coordinate gives $F_h\equiv0\pmod q$. Also $F_h>0$, since $h\ge1$. In the case $q=1$ one may simply take the gap $F_1=1$.

Periodic and eventually periodic colourings cannot obstruct.

Suppose $\chi(n+q)=\chi(n)$ for all $n\ge N$, with any finite number of colours. Choose a positive Fibonacci multiple $D=F_h$ of $q$ as above and any $a\ge N$. Then \[ a,\ a+D,\ a+2D,\ldots \] is an infinite increasing monochromatic sequence. Every consecutive gap is the allowed positive Fibonacci number $D$, reused legally. In particular this colouring contains a sequence of every finite length requested in the definition. The result is stronger than the three-colour statement within the eventual-periodicity subcase.

This also invalidates a common attempted construction: a finite colouring of residues modulo $q$ might forbid all monochromatic gaps in a tested finite Fibonacci list, but the first positive Fibonacci multiple of $q$ creates a monochromatic arithmetic chain automatically. Testing only Fibonacci numbers below a numerical cutoff can therefore misclassify every genuinely periodic proposal.

A correct finite-search reformulation.

Fix $k\ge2$. For each $N$, consider all three-colourings of $[N]$ containing no monochromatic Fibonacci-difference sequence of length $k$. Restriction from $[N+1]$ to $[N]$ preserves this property. If such colourings exist for every $N$, their restriction tree has at most three children per vertex and nodes at every depth. An infinite branch exists: recursively choose a child with descendants at arbitrarily large depths, which exists because there are only finitely many children. The resulting infinite colouring avoids length $k$, since any forbidden finite sequence would lie in some initial interval. Conversely an infinite avoiding colouring supplies all finite restrictions. This is a rigorous compactness equivalence, not an inference from one large finite computation.

Verification and relationship with the recent claim.

The divisibility lemma uses only the recurrence, so duplicate initial values $F_1=F_2=1$ cause no difficulty. The positive chain starts after the eventual-periodic threshold and never returns below it. In the compactness argument $k$ must be fixed while $N$ grows: allowing the avoided length to grow with $N$ would not prove non-accessibility. An irrational floor coding as in [R1] need not be periodic in $n$, so the periodic-colouring theorem does not contradict its claimed counterexample. The missing independent verification here is the all-Fibonacci increment bound and the finite graph certificate for that aperiodic construction, or an alternative explicit aperiodic counterexample with a proved uniform length bound.

Final status of this attempt.

UNRESOLVED as an independent full verification. The eventual- periodicity theorem and compactness reduction are proved. A negative resolution of the historical question is reported in [R1]; this file neither certifies that complete proof nor represents the conjecture as unchallenged by current literature.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3043 (OPG-1825), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, 3-accessibility of Fibonacci numbers. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/3_accessibility_of_fibonacci_numbers.

• R1 William J. Wesley, The Fibonacci numbers are not 3-accessible, arXiv:2609.33744v1, submitted 27 September 2026. Primary preprint; abstract and proof structure checked 1 October 2026. https://arxiv.org/html/2609.33744v1.