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Source. Theorem 2, p. 4, of P. Erdős, S. W. Graham, A. Ivić and C. Pomerance, On the number of divisors of n!, Analytic Number Theory (Progress in Mathematics), Birkhäuser Boston (1996), 337--355, doi:10.1007/978-1-4612-4086-0_19, read in the authors' manuscript named on the source card; pages here are that manuscript's printed pages 1--16, and the published pagination was not compared.

Statement

Theorem 2 (p. 4). "Let $P(n)$ denote the largest prime factor of $n$. Then

$$\frac{d(n!)}{d((n-1)!)}=1+\frac{P(n)}{n}+O\Bigl(\frac1{n^{1/2}}\Bigr)."$$

Here $d(m)$ is the number of positive divisors of $m$. Since $n/P(n)$ is an integer, the main term $P(n)/n$ is always of the form $1/m$ with $m$ a natural number.

Read depth. Claims checked: the statement was read clause by clause on the page image on 2026-10-08, and the proof on p. 4 was followed step by step. Nothing here is independently reviewed.

Proof sketch

P. 4. Write $p=P(n)$. If $p\le n^{1/2}$, a short case split (on whether the largest prime factor of $n/p$ is at most $n^{1/3}$) gives $S(n)\ll n^{1/2}$, and Lemma 1 puts the ratio within $O(n^{-1/2})$ of $1$, while $P(n)/n\le n^{-1/2}$. If $p>n^{1/2}$, write $n=mp$; then $p$ divides $n$ exactly once and its exponent rises from $m-1$ to $m$, contributing the factor $(m+1)/m=1+P(n)/n$, and the remaining primes, those dividing $m$, contribute a factor between $1$ and $\exp(S(m)/n)\le1+2m/n\le1+2n^{-1/2}$ by the argument of Lemma 1.

Dependencies

Lemma 1 and its display (5).

Bears on

• Problem 419: with $n+1$ in place of $n$ the theorem reads $\tau((n+1)!)/\tau(n!)=1+P(n+1)/(n+1)+O(n^{-1/2})$, the problem's ratio; Corollary 1 reads the set of limit points off this formula.