AI 中文总结
本文证明阶乘的不同除数之和的覆盖数上界为(2log2+o(1))n/logn，通过结合递减贪心构造与更锐利的除数间隙估计，改进了先前结果。
AI 中文摘要
对于实用数$N$，设$h(N)$为最小的$k$，使得每个整数$1\le m\le N$都是$N$的至多$k$个不同除数之和。我们证明$h(n!)\le(2\log2+o(1))\\,n/\log n$。这改进了Tenenbaum-Yokota引理4和Yokota 1995年背包笔记中建立的$n/(\log n)^{1/2-\varepsilon}$阶的界。我们将他们的递减贪心构造与Berend-Harmse更锐利的阶乘除数间隙估计相结合。分别在$\sqrt{n!}$以下和以上计数步骤，其中上限范围通过倒数除数处理，保留了间隙指数中的首项系数，并得出显式常数$2\log2$。
英文摘要
For practical $N$ let $h(N)$ be the least $k$ such that every integer $1\le m\le N$ is a sum of at most $k$ distinct divisors of $N$. We prove $h(n!)\le(2\log2+o(1))\,n/\log n$. This improves the bounds of order $n/(\log n)^{1/2-\varepsilon}$ established in Tenenbaum-Yokota's Lemma 4 and Yokota's 1995 knapsack note. We combine their decreasing greedy construction with the sharper factorial divisor-gap estimate of Berend-Harmse. Counting the steps separately below and above $\sqrt{n!}$, with the upper range handled through reciprocal divisors, retains the leading coefficient in the gap exponent and yields the explicit constant $2\log2$.
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