Erdős Problem #377

declared status 'open'. Formalized: yes. Is there some absolute constant $C>0$ such that$$\sum_{p\leq n}1_{p\nmid \binom{2n}{n}}\frac{1}{p}\leq C$$for all $n$ (where the summation is restricted to primes $p\leq n$)? Current best: A question of Erd\H{o}s, Graham, Ruzsa, and Straus [EGRS75], who proved that if $f(n)$ is the sum in question then$$\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n) = \sum_{k=2}^\infty \frac{\log k}{2^k}=\gamma_0$$and$$\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n)^2 = \gamma_0^2,$$so that for almost all integers $f(m)=\gamma_0+o(1)$. (It is trivial from Mertens estimates that $f(n)\leq (1+o(1))\log\log n$.) A positive answer would imply that$$\sum_{p\leq n}1_{p\mid \binom{2n}{n}}\frac{1}{p}=(1-o(1))\log\log n,$$and Erd\H{o}s, Graham, Ruzsa, and Straus say there is 'no doubt' this latter claim is true. Prize: no. Tags: binomial coefficients, number theory.

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