I was looking into this sequence. And I'm particularly interested in the asymptotic behavior of the following series (which is stated on the site) $$\sum_{k=1}^n \frac{1}{a(k)} \sim \frac{3(\log n)^2}{2\pi^2} + \cdots, $$ where $a(k) = \min\{m \in \mathbb{N}: k \mid m^2 \}$. But I couldn't find any further details about it on the site, except for the author Vaclav Kotesovec, but with no success finding any further details about the series. Can you provide me with some relevant references, please?

## 1 Answer

I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ then $a(k) = ab$, so $$\begin{align*} \sum_{k\leq x}\frac1{a(k)} &= \sum_{a b^2 \leq x} \frac{\mu^2(a)}{ab} = \sum_{a \leq x} \frac{\mu^2(a)}{a} \sum_{b^2 \leq x/a}\frac1b = \sum_{a \leq x} \frac{\mu^2(a) H_{\sqrt{x/a}}}{a}\\ &= \sum_{a \leq x} \frac{\mu^2(a) \left(\frac12\log(x) - \frac12\log(a) + \gamma + O(\sqrt{a/x})\right)}{a}\\ &= \left(\frac12\log(x) + \gamma\right) \sum_{a \leq x}\frac{\mu^2(a)}a - \frac12 \sum_{a \leq x}\frac{\mu^2(a)\log(a)}{a} + O(1). \end{align*}$$

We have $$\sum_{a \leq x}{\mu^2(a)} = \frac{6x}{\pi^2} + O(\sqrt x),$$ so by partial summation $$\sum_{a \leq x}\frac{\mu^2(a)}a = \frac6{\pi^2}\log(x) + C + O\left(\frac1 {\sqrt x}\right),$$ and $$\sum_{a \leq x}\frac{\mu^2(a) \log(a)}a = \frac{3}{\pi^2} \log^2(x) - \frac6{\pi^2} \log(x) + C_2 + O\left(\frac{\log(x)}{\sqrt x}\right),$$ which gives the rough result. The more precise result can likely be found by directly looking at the sums (without passing through $\sum_{a \leq x}\mu^2(a)$) and using more terms for the harmonic numbers.