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Oct 15 at 13:40 comment added GH from MO @DesideriusSeverus Let $\omega(N)$ be the number of distinct prime divisors of $N$, and let $p_k$ denote the $k$-th prime. Then $\sum_{p\mid N}p^{-1}\leq\sum_{k=1}^{\omega(N)}p_k^{-1}\leq\log\log p_{\omega(N)}+O(1)$, while $\log N\geq\sum_{k=1}^{\omega(N)}\log p_k\gg p_{\omega(N)}$. Hence really $\sum_{p\mid N}p^{-1}\leq\log\log\log N+O(1)$.
Oct 15 at 7:16 comment added Desiderius Severus @GHfromMO This would be most welcome, I indeed naively used Mertens' bound, dropping the condition that $p\mid N$. Is that clear? (I am not getting it by e.g. partial summation)
Sep 27 at 14:58 comment added GH from MO @DesideriusSeverus The $\log\log N$ in your remark is really $\log\log\log N$.
Sep 27 at 8:44 comment added Desiderius Severus Thanks for your answer. Even though $B_f$ is bounded independently of $f$, there are two other dependencies (a priori upon the level) that bother me: the extra sums over $p \mid N$ are of size roughly bounded by $\sum_{p\mid N} p^{-1} \ll \log \log N$ which is already too large; and the remaining $O$-error term has also a hidden dependence on $f$.
Sep 27 at 6:32 history edited 2734364041 CC BY-SA 4.0
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Sep 27 at 5:59 history answered 2734364041 CC BY-SA 4.0