Asymptotics of $\sum_{k=1}^nd(k^s)$, the sum of the divisor counts of the squares, cubes, etc.

Question

Let $d(n)$ be the number of divisors of $n$. It is well known that $$\sum_{k=1}^nd(k)=\sum_{d=1}^n\left\lfloor\frac{n}{d}\right\rfloor=\Theta(n\log n).$$ Purely based on experimentation, I expect for integral $s\geq1$, in general, we get $$\sum_{k=1}^nd(k^s)=\Theta(n\log^sn),$$ but I see no way of proving or disproving this claim, not even for $s=2$. Are there any known results on the asymptotics of $\sum_{k=1}^nd(k^s)$?

Answer 1

I'll do $s=2$. The higher values of $s$ should be able to be handled similarly.

$\sum_{k \le n} \tau(k^2) = \sum_{k \le n} \sum_{d \mid k^2} 1 = \sum_{d \le n^2} \sum_{\substack{k \le n \\ d \mid k^2}} 1$. This motivates us to write $d=a^2b$ with $b$ square-free. In the following, $b$ is always square-free. $\sum_{a^2b \le n} \sum_{\substack{k \le n \\ a^2b \mid k^2}} 1 = \sum_{a^2b \le n^2} \sum_{\substack{k \le n \\ ab \mid k}} 1 = \sum_{a^2b \le n^2} \lfloor \frac{n}{ab}\rfloor = \sum_{\substack{a^2b \le n^2 \\ ab \le n}} \lfloor \frac{n}{ab}\rfloor$ and since $ab \le n$ implies $a^2b \le n^2$, we get $= \sum_{ab \le n} \left[\frac{n}{ab}+O(1)\right] = \sum_{b \le n} \sum_{a \le \frac{n}{b}} \left[\frac{n}{ab}+O(1)\right] = \sum_{b \le n} \left[\frac{n}{b}\log(\frac{n}{b})+O(\frac{n}{b})\right]$. Recall the sum is over square-free $b$. However, square-free integers are just $\frac{6}{\pi^2}$ density of integers and are uniformly distributed enough, so since you just want a $\Theta$ bound, we can just pretend the (purely analytic and not arithemtic) sum is over all $b \le n$. Then we get $\Theta(n\log^2 n)$, as desired.