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Statement

With $f(n)$ and $c_0=\sum_{k\ge2}(\log k)/2^k$ as in Theorem 2:

Theorem 3 (p. 87).

$$\lim_{x\to\infty}\frac1x\sum_{n=1}^{x}f^2(n)=c_0^2 .$$

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; Theorem 3 on p. 87, its proof on pp. 87--88. The edition is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page images. The proof was read for its structure only and not re-derived.

Proof pointer

Pages 87--88. The sum $\sum_{n\le x}f^2(n)$ becomes $\sum_{p,q\le x}A(p,q;x)/(pq)$, with $A(p,q;x)$ the number of $k$ with $p,q\le k\le x$ and $\binom{2k}{k}$ coprime to $pq$. The pairs of primes split into three classes by comparison with $x^{1/s}$: both small, one small, both large. The first two classes contribute $o(x)$ as $s\to\infty$. For two large primes with $r$ and $t$ digits, under the spacing condition (5) of p. 88 on the powers $p^i$ and $q^j$, the digit conditions of (1) behave independently and $A(p,q;x)=x/2^{r+t}+o(x)$; this yields $c_0^2x+o(x)$, and the pairs violating (5) contribute $o(1)$ to the normalized sum as $\epsilon\to0$. The argument proves only the upper bound $\sum_{n\le x}f^2(n)\le c_0^2x+o(x)$ (p. 88); the matching lower bound follows from Theorem 2 and the inequality between the arithmetic and the quadratic mean.

Dependencies

Theorem 2 and the digit criterion (1) of the same paper.

Bears on

• Problem 377: together with Theorem 2 it gives the Corollary (p. 89), that $f(n)$ is within any $\epsilon$ of $c_0$ outside a set of $n$ of density $0$; it says nothing about how large $f$ is on that exceptional set, which is what the problem asks.