Sum of number of divisors for square n

Question

So I've found that you can find the sum of $\sigma_0(n)$ by using:

$\sum_{i=1}^N \sigma_0(i) = \sum_{i=1}^N \lfloor \frac{N}{i} \rfloor$

or better yet:

$\sum_{i=1}^N \sigma_0(i) = \sum_{i=1}^f 2 * \lfloor \frac{N}{i} \rfloor - f^2$ where $f = \lfloor N^\frac{1}{2} \rfloor$

Is there a similarly useful expression for:

$\sum_{i=1}^N \sigma_0(i^2) = \space?$

Answer 1

$\sum_{i=1}^N \sigma_0(i^2) = \sum_{i=1}^N 2^{\omega(i)}\lfloor \frac{N}{i} \rfloor$

and more generally

$\sum_{i=1}^N \sigma_0(i^k) = \sum_{i=1}^N k^{\omega(i)}\lfloor \frac{N}{i} \rfloor$