$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$ The average power of an integer $m$ is given by $$ \ap(m):=\log_{\rad(m)}(m)=\frac{\log(m)}{\log(\rad(m))}, $$ where $\rad(m)=\prod_{p|m}p$.

$\ap(m) \ge 2$ if and only if $\rad(m)^2 \le m$ , which is obviously satisfied by the set of powerful integers $P_2$ defined by the relation "$p|m$ implies $p^2|m$". Clearly $$ \sum_{P_2} \frac{1}{m}=\prod_p(1+\frac{1}{p^2}+\frac{1}{p^3}+ \ldots ) =\prod_p\left(1+\frac{1}{p(p-1)}\right)=1.9436\ldots $$ converges. From numerical computation, it seems that we may have $$ \sum_{ap(m) \ge 2} \frac{1}{m} =2.11\ldots\;. $$ Does it actually converge? Is there a similar expression as for $P_2$? Note $\ap(48) \ge 2$ but $48=2^4.3 \notin P_2$

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    $\begingroup$ As an explicit formula: $$\sum_{ap(m) \geq 2}=\sum_{P_2}\frac{1}{m}+\sum_{\beta, \gamma \in S^{(2)}_{*} \\(\beta, \gamma)=1} \frac{1}{\beta\gamma^2} \left(\sum_{\alpha > \beta \\ (\alpha, \beta)=1} \frac{1}{\alpha( \text{rad}(\alpha)^2)}\right)$$, here $S^{(2)}_*$ is the set of all squarefree numbers. $\endgroup$
    – Alapan Das
    Mar 22 at 5:22
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    $\begingroup$ I find it amusing that the right hand side is an explicit formula for the expression on the left hand side (but I have to admit that I don't know the first thing about number theory…)- $\endgroup$
    – Dirk
    Mar 22 at 6:56

1 Answer 1


$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$

I think so. Fix $\rad(m)=p_1\ldots p_k=:P$. Denote by $\Omega$ the set of positive integers with all prime divisors in $\{p_1,\ldots,p_k\}$. Then $m=PQ$ where $Q\in \Omega$ and $Q\geqslant P$. For $s\in (0,1)$ we have $$ \prod_{p|m}(1-p^{s-1})^{-1}=\sum_{n\in \Omega} n^{s-1}\geqslant P^s\sum_{n\in \Omega,n\geqslant P}n^{-1}. $$ Thus $$ \sum_{n\in \Omega,n\geqslant P}n^{-1}\leqslant \prod_{p|m}p^{-s}(1-p^{s-1})^{-1}=\prod_{p|m}(p^s-p^{2s-1})^{-1}. $$ For $s=1/2$ the right hand side reads as $\prod(p^{1/2}-1)^{-1}$. Thus $$ \sum_{m:\ap(m)\geqslant 2} m^{-1}\leqslant \sum_{p_1,\ldots,p_k}\prod_{i=1}^k\frac1{p_i(\sqrt{p_i}-1)}=\prod_p\left(1+\frac1{p(\sqrt{p}-1)}\right)<\infty. $$

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    $\begingroup$ Perhaps it is worthwhile to note that your first inequality uses "Rankin's trick". $\endgroup$
    – GH from MO
    Mar 22 at 16:18
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    $\begingroup$ Ah, I wanted to call it "Chebyshev inequality", but I see that it is indeed called "Rankin trick" in this context. $\endgroup$ Mar 22 at 17:21

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