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Claim. On 2026-04-23 Liam Price posted in the thread of Problem 1194 a note in which GPT-5.4 Pro claims a partial solution. The site's curator summarized it the same day. Write $A=\{x_1<x_2<\cdots\}$. If $a_n\ll n^c$ for every $n$, then $x_k-x_{k-1}\gg x_k^{1/c}$, so $x_k\gg k^{c/(c-1)}$ and $|A\cap[1,x]|\ll x^{1-1/c}$, and the counting argument in the site's remarks then gives $a_n\gg n^{c/(2c-2)}$ infinitely often. Balancing the exponents at $c=3/2$ gives the result as the curator states it: $a_n\gg n^{3/2}$ for infinitely many $n$.

Submission note. Posted to the site's forum by Liam Price on 23 April 2026:

Although not specified here, I would assume this result is known. Regardless, GPT-5.4 pro claims a partial solution here.

Covers. A lower bound $a_n\gg n^{3/2}$ along infinitely many $n$. It does not determine how fast $a_n/n$ must grow.

Standing. Price wrote that they assumed the result was already known. The same day the curator posted in the thread a stronger bound that GPT Pro found at their request, which the site's remarks credit to GPT-5.4 Pro. No acceptance evidence is recorded, so the claim stays claimed.

Depends on. No page of this wiki.