Closed form for $\sum_{k=1}^n \sigma(k^2)$

Question

I know that there is this formula for the sum of sigma function : $$\sum_{k = 1}^{n} k \lfloor \tfrac{n}{k} \rfloor = \sum_{k = 1}^{n} \sigma(k), $$ I was wondering if there were a known formula for : $$\sum_{k=1}^n \sigma(k^2)$$

Answer 1

$$\sum_{k=1}^n \sigma(k) = \sum_{k=1}^n \sum_{d | k} d =\sum_{d=1}^{n} d \sum_{k \le n, d | k} 1=\sum_{d=1}^{n} d\lfloor n/d \rfloor $$ Let $$h(d)\ = \ \inf\ \{m, \ d | m^2\} \ = \ d \prod_{p^{r} \| d, r \text{ is odd}} p$$ then $d | k^2$ iff $h(d) | k$ so that $$\sum_{k=1}^n \sigma(k^2) = \sum_{k=1}^n \sum_{d | k^2} d =\sum_{d=1}^{n^2} d \sum_{k^2 \le n^2, d | k^2} 1 =\sum_{d=1}^{n^2} d \lfloor n / h(d) \rfloor$$

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Note that $h(d)$ is a multiplicative function : $$H(s)=\sum_{d=1}^\infty \frac{d^{-s}}{h(d)}= \prod_p (1+\sum_{k=1}^\infty \frac{p^{-sk}}{h(p^k)}) \\= \prod_p (1+\sum_{k=1}^\infty p^{-2sk}+\frac{p^{-s(2k-1)}}{p})=\prod_p (1+\frac{1+p^{s-1}}{p^{2s}-1})$$ It has a dominating pole of order $1$ at $s=1/2$

From which we can estimate $$\sum_{d < x} \frac{d}{h(d)} \sim A x^{3/2}, \qquad\quad\sum_{k=1}^n \sigma(k^2) \sim n\sum_{d < n^2} \frac{d}{h(d)} \sim A n^4 $$