GPT 6 Astra Ultra attempt on Math-737

Erdős Problem #267: independent continuation

Date: 23 September 2026. Source versions read in full: 267.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

The question asks whether $\sum_k1/F_{n_k}$ is irrational whenever $n_{k+1}/n_k\ge c>1$. The supplied product-denominator proof covers $c\ge2$ and the condition $n_{N+1}-\sum_{j\le N}n_j+N\to\infty$. I retain the exact Binet prefactor and weaken the required limiting condition to one along a subsequence.

2) WORK: A SHARPER PRODUCT CRITERION

Set $\varphi=(1+\sqrt5)/2$ and $\kappa=\log(\sqrt5)/\log\varphi$. For every strictly increasing sequence of positive indices, irrationality follows if \[ \limsup_{N\to\infty} \left(n_{N+1}-\sum_{j\le N}n_j+\kappa N\right)=+\infty. \tag{1} \] Binet's identity gives \[ Q_N:=\prod_{j\le N}F_{n_j} =5^{-N/2}\varphi^{\sum_{j\le N}n_j} \prod_{j\le N}(1-(-1)^{n_j}\varphi^{-2n_j}). \] The final product is positive and bounded above independently of $N$: its factors are positive, and the sum of the absolute deviations from one is at most $\sum_{m\ge1}\varphi^{-2m}<\infty$. Also \[ R_N:=\sum_{j>N}1/F_{n_j}\le C\varphi^{-n_{N+1}}, \] using $F_m\ge\varphi^{m-2}$ and $n_{N+j}\ge n_{N+1}+j-1$. Thus $Q_NR_N\le C'\varphi^{-E_N}$ with $E_N$ the expression in $1$. If the full sum were $p/q$, then $qQ_NR_N$ would be a positive integer. Along a subsequence with $E_N\to\infty$ it instead tends to zero, a contradiction.

The exact constant satisfies $\kappa>1$, so this is a modest strengthening of the supplied sufficient criterion. It is meaningful for index sequences whose first omitted index is only linearly below the previous-index sum. For indices asymptotic to $c^N$ with fixed $1<c<2$, however, \[ n_{N+1}-\sum_{j\le N}n_j \sim n_{N+1}\frac{c-2}{c-1}, \] which is negative on an exponential scale; no linear correction in $N$ can repair the product-denominator method in that regime.

3) VERIFICATION AND FINAL

All product factors, including $n_1=1$, are positive. An infinite subsequence is enough because the alleged rational denominator $q$ is fixed. The exact Binet identity, rather than a new Fibonacci divisibility claim, supplies the improvement. UNRESOLVED. No proof for every $1<c<2$ is obtained. The known $c\ge2$ range was verified in the indexed official record: https://www.erdosproblems.com/267.

8) COMPLETION ESTIMATE

COMPLETION: 15% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.