All Random Solved Random Open
Let $r\geq 2$ and suppose that $A\subseteq\{1,\ldots,N\}$ is such that, for any $m$, there are at most $r$ solutions to $m=pa$ where $p$ is prime and $a\in A$. Give the best possible upper bound for \[\sum_{n\in A}\frac{1}{n}.\]
Erdős observed that \[\sum_{n\in A}\frac{1}{n}\sum_{p\leq N}\frac{1}{p}\leq r\sum_{m\leq N^2}\frac{1}{m}\ll r\log N,\] and hence \[\sum_{n\in A}\frac{1}{n} \ll r\frac{\log N}{\log\log N}.\] See also [536] and [537].