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Claim. Write $D_k(X)=\lvert D_k\cap[1,X]\rvert$, $s(v)$ for the squarefree part of $v$, $S=\{s(c!):c\ge1\}$ and $\kappa_3=\sum_{a\in S\setminus\{1\}}a^{-1/2}=2.709751\ldots$. Theorem 1.1: for every $\varepsilon>0$, $D_3(X)=\kappa_3\sqrt X+O_\varepsilon(X^{2/5+\varepsilon})$. Theorem 1.2: there is $c^*>0$ such that for $k\in\{5,6\}$ and all large $X$ at least $c^*X$ squarefree $n\in[X/2,X]$ have $F(n)=k$, for $k=5$ separately among even and among odd $n$. Corollary 1.3 collects the orders of growth: $D_2(X)=\lfloor\sqrt X\rfloor-1$, $D_3(X)\sim\kappa_3\sqrt X$, $D_4(X)\asymp D_5(X)\asymp D_6(X)\asymp X$ and $D_k(X)=0$ for $k\ge7$, the lower bound for $D_4$ coming from the identity $n!(n-1)!u!(u-1)!=(vu(n-1)!(u-1)!)^2$ for $n=v^2u$. The corollary answers the question of Problem 374 for every $k$ from $3$ to $6$ and confirms the conjecture of Erdős and Graham that $D_6(X)\gg X$; whether $D_k(X)/X$ converges for $k=4,5,6$ is left open. The source is F. Yudin, Square products of factorials and a conjecture of Erdős and Graham, arXiv:2610.01899, version 1 of 2026-10-01 (Creative Commons Attribution 4.0, by the arXiv record), submitted to the site's proof-claims tab on 2026-10-02. The three-factorial estimate counts representations $n!(n-\ell)!c!=\square$ with $\ell\ge2$ by combining Tao's uniform bound for Pell equations (card) with the larger sieve of Gallagher and Weil's character-sum bound, the exponent $2/5$ coming from balancing the two cases; the consecutive representations $n!(n-1)!c!=\square$ give the main term. For five and six factorials the argument takes $n=ap$ with $p>X^{1-\alpha}$ prime, uses the six-factor identity of Erdős and Graham, and excludes shorter representations by bounding the two factorial blocks through the prime equidistribution estimate of Matomäki, Radziwiłł, Shao, Tao and Teräväinen with the large sieve, after Huxley's theorem on primes in short intervals has given preliminary bounds on the two block lengths (Lemma 4.2). The positive lower density of the surviving $n$ comes from the prime number theorem, summed over cofactors $a$ from a set of positive lower density (Proposition 7.3). An appendix constructs, for every $m\ge2$, perfect $m$th powers that are products of a bounded number of distinct factorials with affine arguments, through a zero-sum theorem for finite abelian groups.

Submission note. Posted to erdosproblems.com as a proof claim by Fedir Yudin (account Fedir) on 2 October 2026, giving "GPT 6 Astra, Claude 5.5 Opus" as the AI used:

I have posted an arXiv preprint claiming a full solution. Writing $D_k(X)=|D_k\cap[1,X]|$, the main results are

$$> D_3(X)=\kappa_3\sqrt X+O_\varepsilon(X^{2/5+\varepsilon}), \qquad \kappa_3=2.709751\ldots, >$$

and

$$> D_5(X)\asymp D_6(X)\asymp X. >$$

Together with the classical result $D_4(X)\asymp X$, these determine all the requested orders of growth. The three-factor estimate builds on Tao’s bound for the interval length, combining a uniform bound for Pell equations with Gallagher’s larger sieve and the Weil bound. For five and six factors, the argument uses integers with a very large prime factor and excludes shorter representations using the prime equidistribution estimate of Matomäki–Radziwiłł–Shao–Tao–Teräväinen. I would welcome comments or corrections, particularly on Section 5, where the equidistribution estimate is used to bound the lengths of the two factorial blocks.

Standing. The claim is pending. The site's proof-claims tab lists it as a full proof claim, the site's label is unchanged, and the curator has not credited the result, so there is no acceptance evidence. The claimant's submission invites corrections, especially to the fifth section, in which the equidistribution estimate controls how long the two blocks of consecutive integers in a representation can be. The claim's tools field names the AI systems GPT 6 Astra and Claude 5.5 Opus; in a comment on the claim of 2026-10-02 the claimant says most of the proof ideas came from GPT, that their own part was to choose the questions, make occasional suggestions, work through the arguments and revise the exposition, and that the appendix came first. A reader's comment of 2026-10-03 observes that the $k=3$ estimate is a small modification of Tao's argument, which split its cases suboptimally. The same conclusions were reached independently by Hartley and Olson, whose paper records the differences between the two arguments and whose Lean development, which the corpus has not built, covers their own proof, not this one; this preprint has no formalization.