Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Source. Theorem 4, p. 8, 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 4 (p. 8). "Let f(n)f(n) be as in Theorem 3. If nn is sufficiently large, then f(n)<n4/9f(n)<n^{4/9}."

Here, as in Theorem 3, S(n)S(n) is the sum of the prime factors of nn counted with multiplicity and f(n)f(n) is the least number with ∑i=1f(n)S(n+i)>n\sum_{i=1}^{f(n)}S(n+i)>n.

Read depth. Claims checked: the statement was read clause by clause on the page image on 2026-10-08, and the deduction from Lemma 3 on p. 8 was followed. Nothing here is independently reviewed.

Proof sketch

P. 8. With g(n)=n4/9g(n)=n^{4/9}, each prime p>n5/9+δp>n^{5/9+\delta} dividing an integer of (n,n+g(n)](n,n+g(n)] adds at least n5/9+δn^{5/9+\delta} to ∑i≤g(n)S(n+i)\sum_{i\le g(n)}S(n+i), and Lemma 3 gives ≫n4/9\gg n^{4/9} such primes. The sum is therefore ≫n1+δ\gg n^{1+\delta}, which exceeds nn for large nn, so f(n)≤g(n)f(n)\le g(n).

Dependencies

Lemma 3.

Bears on

  • Problem 420: the paper derives the companion bound Corollary 3 on K(n)K(n) from the same lemma; see that page for the relation.