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Statement

With $f(n)$ and $c_0$ as in Theorem 2:

Corollary (p. 89). For every $\epsilon>0$,

$$\lim_{x\to\infty}\frac1x\bigl|\{n\le x:\ |f(n)-c_0|>\epsilon\}\bigr|=0 .$$

The introduction (p. 83) states the same conclusion as: for all but $o(n)$ integers $m\le n$, $f(m)=c_0+o(1)$.

Source. P. Erdős, R. L. Graham, I. Z. Ruzsa and E. G. Straus, On the prime factors of $\binom{2n}{n}$, Math. Comp. 29 (1975), no. 129, 83--92; the unnumbered Corollary on p. 89, announced on p. 83. The edition is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image.

Proof pointer

The paper gives no separate proof. By Theorems 2 and 3 the mean of $(f(n)-c_0)^2$ over $n\le x$ tends to $c_0^2-2c_0^2+c_0^2=0$, and Chebyshev's inequality gives the density statement.

Dependencies

Theorem 2 and Theorem 3.

Bears on

• Problem 377: the problem asks whether $f(n)\le C$ for all $n$. The Corollary bounds $f(n)$ by $c_0+\epsilon$ outside a set of density $0$ and leaves the exceptional $n$ uncontrolled, so it does not decide the problem.