Problem 374

For any $m\in \mathbb{N}$ , let $F(m)$ be the minimal $k\geq 2$ (if it exists) such that there are $a_1<\cdots <a_k=m$ with $a_1!\cdots a_k!$ a square. Let $D_k=\{ m : F(m)=k\}$ . What is the order of growth of $\lvert D_k\cap\{1,\ldots,n\}\rvert$ for $3\leq k\leq 6$ ? For example, is it true that $\lvert D_6\cap \{1,\ldots,n\}\rvert \gg n$ ?

Source: erdosproblems.com/374

Number theory

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ClaimedAnswered

A full solution has been claimed but not yet accepted. Settled in another form, for example when its parts resolve differently or the question is open-ended.

OPEN, the site's label (problem page, discussion thread and proof-claims tab read 2026-10-07). The site's proof-claims tab carries three proof claims. Zeraoulia's partial claim of 2026-07-27, the bound $D_6(x)=\omega(x/\log x)$ with certified computations, determines no order of growth and does not reach $D_6(x)\gg x$ , so it settles no instance and has no claim page. Hartley and Olson (claim page (Hartley and Olson, 2026)) and Yudin (claim page (Yudin, 2026)) each claim the full determination. Hartley and Olson's forum submission covered the positive lower density of $D_6$ , and their revised paper adds the orders of growth of $D_3$ through $D_5$ . The derived standing departs from the label: the two pending full claims agree, so the standing is claimed, answered; neither claim is accepted, and the site's curator has credited neither.

Problem 374

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Proof claims

Discussion