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Is there some absolute constant $C>0$ such that \[\sum_{p\leq n}1_{p\nmid \binom{2n}{2}}\frac{1}{p}\leq C\] for all $n$?
A question of Erdős, 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 $f(m)=\gamma_0+o(1)$. They also prove that, for all large $n$, \[f(n) \leq c\log\log n\] for some constant $c<1$. (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 Erdős, Graham, Ruzsa, and Straus say there is 'no doubt' this latter claim is true.