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Source. Hughes, arXiv:2609.10902v1, Remark 6 (pp. 4–5); read on the page image.

Statement

$h(n!)\gg(\log n)^2$ as $n\to\infty$.

Proof sketch

Let $T=\tau(n!)$ and $k=h(n!)$. If $k\le T/2$, every $1\le m\le n!$ is a subset sum of at most $k$ of the $T$ divisors, so $n!\le\sum_{i\le k}\binom Ti\le(eT/k)^k$. By Chebyshev's bound $\pi(x)\ll x/\log x$, $\log T=\sum_{p\le n}\log(v_p(n!)+1)\ll n/\log n$ (primes $p\le\sqrt n$ contribute $O(\sqrt n\log n)$; primes $p>\sqrt n$ in $(n/2^{r+1},n/2^r]$ have $v_p(n!)=\lfloor n/p\rfloor<2^{r+1}$ and number $O(n/(2^r\log n))$). Since $\log n!\asymp n\log n$, the counting bound gives $n\log n\ll k\log(eT/k)\le k(1+\log T)\ll kn/\log n$, so $k\gg(\log n)^2$. If instead $k>T/2$, then $k>n/2$, because each of $1,\dots,n$ divides $n!$ and so $T\ge n$; this again gives $k\gg(\log n)^2$.

The remark ends (p. 5) by noting that the lower bound $(\log n)^2$ and the upper bound $n/\log n$ are still far apart, and recalls Erdős's questions whether $h(n!)<n^{o(1)}$ and whether even $h(n!)<(\log n)^{O(1)}$ ([ErGr80], pp. 37–38).

Reconstruction

An author-recorded reconstruction of the argument, not an independent review, is filed as the Remark 6 reconstruction.

Bears on

• Problem 18: the third question cannot be answered with an exponent below $2$; the second and third questions are restated.