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Claim. Let $\rho=(\sqrt{13}-1)/2=1.3027756\ldots$, the positive root of $\rho(1+\rho)=3$. For every $1<\alpha<\rho$ there is a strictly increasing sequence of integers $2\le a_1<a_2<\cdots$ such that

$$\sum_{n\ge1}\frac{1}{a_n}\in\mathbb{Q},\qquad \sum_{n\ge1}\frac{1}{a_n-1}\in\mathbb{Q},\qquad \lim_{n\to\infty}a_n^{1/\alpha^n}=C_\alpha$$

for some constant $C_\alpha>1$; hence for every $1<\beta<\rho$ such a sequence has $a_n^{1/\beta^n}\to\infty$. This is Theorem 1 of the write-up A residual-state construction for Erdős Problem #265 (ten pages, dated August 2026), the preprint link, pinned to the commit of 2026-08-28 that uploaded it; the forum entry (the discussion link) was submitted the same day under the name Cam by the account donteatllamas, as a partial claim. The write-up's disclosure says that the construction, its checking and the text were developed with substantial assistance from OpenAI's GPT-5.6 Sol; the forum entry names the system as GPT 5.6 high and adds that the AI completed most of the proof on its own and that a human read it over.

Submission note. Posted to erdosproblems.com as a proof claim by Cam (account llamaboy) on 28 August 2026, giving "GPT 5.6 high" as the AI used:

The main point of this proof is to show that we can improve the known bound on beta up to a substantially better value of $1.3027756\ldots$ More specifically, for every

$$> 1<\beta<\frac{\sqrt{13}-1}{2}, >$$

the construction gives a strictly increasing sequence $a_n$ for which both reciprocal sums are rational and $a_n^{1/\beta^n}\to\infty$. The proof is based on two telescoping residuals $x_n,y_n$. The key step is a covering argument showing we can choose each integer $a_n$ so that (x_{n+1}\asymp x_n^\alpha) while preserving the required relation between $x_n$ and $y_n$. AI Disclosure: AI completed most of this proof on its own, but it was read over by a human.

Covers. The growth question of Problem 265 for exponents $1<\beta<\rho$: it shows that doubly exponential growth with any such base is compatible with both sums being rational. It does not settle the problem, which asks for the exact growth threshold; in particular it says nothing about $\limsup a_n^{1/2^n}>1$, the point the site's commentary leaves open and which a later Lean claim asserts to be impossible, since $\rho<2$. The write-up's own Remark 1 says that $\rho$ is a threshold of this construction, not a claimed optimum.

Argument. Two residuals are driven to zero by telescoping, $x_{n+1}=x_n-1/a_n$ and $y_{n+1}=y_n-1/(a_n(a_n-1))$, so that the two sums equal $x_1$ and $y_1$, chosen rational; a one-step lemma shows that, within an invariant family of intervals, each integer $a_n$ can be chosen so that $x_{n+1}\asymp x_n^{\alpha}$ while $y_n$ stays in the interval the next step needs, and the condition $\alpha(1+\alpha)<3$ is what makes the admissible child intervals wider than the spacing of their centers as $x_n\to0$.

Comparison with the literature. Kovač and Tao's Theorem 2.8, recorded on [[problems/irrationality/E0265/claims/2024_11_27_kovac_tao|their accepted partial claim page]] and on the card kovac_2024_several_irrationality_problems_ahmes_series, gives for $d$ consecutive shifts any $1<\beta<((2d+2)/(2d+1))^{1/d}$ (their condition (7.9)); for the two shifts of this problem, after replacing $a_n$ by $a_n+1$ to pass between their pair $1/a_n,\,1/(a_n+1)$ and the problem's pair, this is $\beta<\sqrt{6/5}=1.0954\ldots$. The claim, if correct, raises that bound to $\rho$. The corpus has checked neither statement against its proof.

Standing. Claimed. The site labels the problem OPEN (page last edited 21 January 2026). The corpus has checked the statement and the outline of the proof, not the proof.

Depends on. Nothing in this wiki.