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Statement

For $n\ge1$ the paper sets (p. 83)

$$f(n)=\sum_{p\nmid\binom{2n}{n},\ p\le n}\frac1p ,$$

the sum over primes $p\le n$ that do not divide $\binom{2n}{n}$.

Theorem 2 (p. 86).

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

The introduction (p. 83) adds that the authors "cannot decide if $f(n)$ is unbounded".

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 definition of $f$ and the constant $c_0$ on p. 83, Theorem 2 on p. 86, its proof on pp. 86--87. The edition is identified on the source card.

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

Proof pointer

Pages 86--87. Exchanging the order of summation writes $\sum_{n\le x}f(n)$ as $\sum_{p\le x}A(p;x)/p$, where $A(p;x)$ counts the $k$ with $p\le k<x$ and $p\nmid\binom{2k}{k}$. By the digit criterion (1) of p. 84 (see Theorem 1), a prime with exactly $r$ base-$p$ digits below $x$ leaves about $x/2^r$ such $k$; so $A(p;x)=x/2^r+o(x)$ for $x^{1/(r+1)+\epsilon}<p<x^{1/r-\epsilon}$, small primes ($p\le x^{1/s}$ with $s$ large) contribute a negligible amount because $A(p;x)$ decays geometrically in $r$, and the thin ranges near the endpoints $x^{1/r}$ contribute $O(\epsilon)$ by (4) of p. 86. Mertens' estimate over $x^{1/(r+1)}<p\le x^{1/r}$ gives $\log(1+1/r)$, and $\sum_{r\ge1}2^{-r}\log(1+1/r)=\sum_{r\ge2}2^{-r}\log r=c_0$.

Dependencies

The digit criterion (1) of the same paper (p. 84); Mertens' theorem on $\sum1/p$.

Bears on

• Problem 377: the problem asks whether $f(n)$ is bounded by an absolute constant for all $n$. Theorem 2 shows only that $f$ is bounded on average, with mean $c_0$; it does not decide the question, which the authors say on p. 83 they cannot decide.