Write $f(n)$ for the quotient of $n$ by its largest squarefree divisor. In other words, $f$ is a multiplicative function with $f(p^k) = p^{k-1}$ for all $k \geq 1$.


What, if anything, is known about the asymptotics of
$$(1/X) \sum\_{n=1}^X f(n)$$
or about the behavior of the Dirichlet series
$$\sum\_{n=1}^\infty f(n) n^{-s}$$
near $s=1$?


I call this function "strange" because it feels to me like it has an infinite-order pole at s=1 (am I right?) and this kind of behavior is not what I expect from the mild-mannered arithmetic functions I usually encounter, and not something I know how to interpret.


The function $f$ [appears in OEIS](https://oeis.org/A003557) but the discussion there does not seem to lead to any computation of the desired asymptotic average, though [a paper of Finch linked there](https://arxiv.org/abs/math/0605019) references on p.7 a 1965 theorem of Schwarz on a similar problem.



Asymptotic behavior of a "strange" arithmetic function
Let $k(n)=\prod\_{p \mid n}p$ be the kernel of the integer, so that $f(n)=n/k(n)$. As indicated in p. 7 of Finch's article,
$$\sum\_{n\le x} \frac{1}{k(n)} = \exp\left( \left( \frac{8\log x}{\log \log x}\right)^{1/2}(1+o(1))\right)$$
was shown by de Bruijn in "On the number of integers $\le x$ whose prime factors divide $n$", [Illinois J. Math. 6 (1962) 137–141](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-6/issue-1/On-the-number-of-integers-leqq-x-whose-prime-factors/10.1215/ijm/1255631814.full). This solved a question of Erdős.
The proof went by first showing that $F(s)=\sum\_n \frac{1}{k(n)} n^{-s}$ behaves likes $s^{-1} (\log (s^{-1}))^{-1}$ as $s \to 0^+$, and then applying a Tauberian theorem.


De Bruijn and van Lint showed that
$$\sum\_{n \le x} \frac{n}{k(n)} = o\left( x \sum\_{n \le x} \frac{1}{k(n)}\right)$$
in "On the number of integers $\le x$ whose prime factors divide $n$", [Acta Arith. 8 (1963) 349–356](https://eudml.org/doc/207305). See their paper for the motivation. Combining with the result of de Bruijn, this gives an upper bound for $\sum\_{n \le x} n/k(n)$ agreeing with Will Sawin's heuristic result from the comments.


You might also be interested in their paper "On the asymptotic behaviour of some Dirichlet series with a complicated singularity" ([Nieuw Archief voor Wiskunde, 3/11, 68-75, 1963](https://research.tue.nl/en/publications/on-the-asymptotic-behaviour-of-some-dirichlet-series-with-a-compl)).


Wolfgang K. Schwarz in "Einige Anwendungen Tauberscher Sätze in der Zahlentheorie. B" ([J. Reine Angew. Math. 219, 157-179 (1965)](https://eudml.org/doc/150692)) determined the asymptotics of $\sum\_{n \le x} \frac{1}{k(n)}$ up to $(1+o(1))$ in terms of an implicit quantity (this is in the first page of his paper, but see Satz III.17 for a more explicit result that is less accurate). He also recovers the $o(1)$ result of de Bruijn and van Lint (Satz III.16).


There might be an earlier reference for a lower bound for $\sum\_{n \le x} n/k(n)$, but the one I can find follows from the the 113-page paper "Sur la répartition du noyau d’un entier" by Robert and Tenenbaum ([Indag. Math., New Ser. 24, No. 4, 802-914 (2013)](https://www.sciencedirect.com/science/article/pii/S0019357713000578)). They denote $\sum\_{n \le x} n/k(n)$ by $K\_2(n)$ and estimate it in Théorème 4.4, which gives
$$\sum\_{n \le x} \frac{n}{k(n)} \sim x \sqrt{\frac{2}{\log x \log \log x}}\sum\_{n \le x} \frac{1}{k(n)}.$$
Combining this with Schwarz's work gives an asymptotic formula for your sum.


