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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 but the discussion there does not seem to lead to any computation of the desired asymptotic average, though a paper of Finch linked there references on p.7 a 1965 theorem of Schwarz on a similar problem.

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    $\begingroup$ $f(n)$ is upper-bounded by the function whose Dirichlet series is $\prod_{k=1}^{\infty} \zeta( ks - (k-1) ) =\prod_p \prod_{k=1}^\infty (1 + p^{k-1- ks} + \dots ) $. (Proof: Ignore the $\dots$ and the cross-terms.) The $\mathbb F_q[t]$-analogue of this Dirichlet series is $\prod_{k=1}^{\infty} 1/ (1-q^{ 1- (ks-(k-1))}) =\prod_{k=1}^{\infty} (1 - (q^{1-s})^k)$ which is just the partition generating function so its $n$'th coefficient is asymptotic to $q^n e^{ C \sqrt{n}}$. $\endgroup$
    – Will Sawin
    Commented Apr 24, 2023 at 1:46
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    $\begingroup$ Thus I would guess that the partial sums of the upper bound are asymptotic to $X e^{C \sqrt{\log X}}$ and guess less confidently that the partial sums of $f(n)$ are asymptotic to $X e^{ C ' \sqrt{\log X}}$ for some $C' \leq C$. $\endgroup$
    – Will Sawin
    Commented Apr 24, 2023 at 1:47
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    $\begingroup$ For lower bounds, a lower bound is $\sum_{\substack{ n_1,\dots, n_k\in \mathbb N \\ n_1 n_2^2 \dots n_k^k \leq X\\ n_1,\dots, n_k \textrm{pairwise coprime}}} n_2 n_3^2 \dots n_k^{k-1}$ which differs by a constant factor from $\sum_{\substack{ n_1,\dots, n_k\in \mathbb N \\ n_1 n_2^2 \dots n_k^k \leq X}} n_2 n_3^2 \dots n_k^{k-1}$ which is asymptotic to $X \log X^{k-1}$ (up to a constant factor) by breaking each variable into dyadic ranges so indeed your function grows faster than any power of log which implies it can't have a finite-order pole at $s=1$. $\endgroup$
    – Will Sawin
    Commented Apr 24, 2023 at 1:49

1 Answer 1

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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. 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. 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).

Wolfgang K. Schwarz in "Einige Anwendungen Tauberscher Sätze in der Zahlentheorie. B" (J. Reine Angew. Math. 219, 157-179 (1965)) 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)). 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.

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    $\begingroup$ Fantastic, thanks, Ofir! $\endgroup$
    – JSE
    Commented Apr 24, 2023 at 15:54
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    $\begingroup$ It's interesting that my guess, while roughly correct, was not quite right: I didn't predict the $\log \log x$ factor in the denominator. I wonder if there is a slick explanation for why it appears as compared to the partition function. I guess one related question is the behavior of $\prod_{k=1}^{\infty} \zeta( ks+1)$ as $s\to 0^+$. $\endgroup$
    – Will Sawin
    Commented Apr 24, 2023 at 18:07
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    $\begingroup$ @WillSawin When you pass to $\prod \zeta(ks-(k-1))$ you lose some: the singularities at $1$ of the generating function of the partition function differs from that of the function field analogue of $k(n)$. Namely, $\log \sum_{n \ge 0} p(n)u^n \sim (\log (u^{-1}))^{-1}$ as $u\to 1^-$ (a quick derivation is given in D. J. Newman's book "Analytic Number Theory") while, based on de Bruijn's work, I expect that $F(u):=\sum_{f \in \mathbb{F}_q[T], \, monic} |k(f)|^{-1}u^{\deg f}$ satisfies $F(u) \sim C_q (\log (u^{-1}))^{-1} \log( 1/\log(u^{-1}))^{-1}$ in the same limit, which is a bit slower. $\endgroup$
    – Ofir Gorodetsky
    Commented Apr 24, 2023 at 19:57

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