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Claim. Nat Sothanaphan's post of 10 January 2026 in the discussion thread of Problem 401 claims that the inequality cannot hold for all sufficiently large $n$. For $r\ge2$ and every $n=p_{r+1}^k-1$, where $p_{r+1}$ is the $(r+1)$-th prime, any $a_1,a_2$ with $a_1!a_2!\mid n!\,2^n3^n\cdots p_r^n$ have $a_1+a_2\le n+2p_{r+1}$, so no $f(r)\to\infty$ gives $a_1+a_2>n+f(r)\log n$ for every large $n$. The argument is the ChatGPT conversation linked from the post, which uses Kummer's theorem; the post says that the poster checked it by hand and had pointed ChatGPT to Problems 728 and 729 and to Kummer's theorem.

Submission note. Posted to the erdosproblems.com forum by Nat Sothanaphan on 10 January 2026:

I am happy to announce that the "all $n$" version is false!

Specifically, ChatGPT shows that taking $n = p_{r+1}^k - 1$ we must have $a_1+a_2-n \le 2p_{r+1}$ for $r \ge 2$. Thus in fact $\omega(r) = 0$ for all $r \ge 2$. I have checked the argument manually; it is pretty readable. It's not autonomous because I gave a hint to look at #728, #729 and to use Kummer. ChatGPT took $27+13=40$ minutes.

Why it is rejected. It answers a variant, not the Statement. Erdős and Graham's source leaves the quantifier on $n$ open, and the erdosproblems.com wording when the post was made asked for the inequality for all sufficiently large $n$; the erdosproblems.com curator wrote in the thread the next day that this wording was a misreading of the source and that the reading with infinitely many $n$ was intended. Problem 401 states that reading. A bound on $a_1+a_2$ along the sparse sequence $n=p_{r+1}^k-1$ refutes the inequality for all large $n$ but says nothing about the other $n$, and the problem asks only for infinitely many, so its mathematics answers the variant and does not settle this problem; the value disproved is its answer to that variant.

Standing. Rejected, as answering a variant. The thread records the poster's own check of the argument and no other review, and this corpus has not checked it.

Depends on. No page of this wiki.