How can I prove the following identity:
$$\sum_{k=1}^{n}{\sigma_{\ 0} (k^2)} = \sum_{k=1}^{n}{\left\lfloor \frac{n}{k}\right\rfloor \ 2^{\omega(k)}}$$
where $\omega(k)$ is the number of distinct prime divisors of $k$.
How can I prove the following identity:
$$\sum_{k=1}^{n}{\sigma_{\ 0} (k^2)} = \sum_{k=1}^{n}{\left\lfloor \frac{n}{k}\right\rfloor \ 2^{\omega(k)}}$$
where $\omega(k)$ is the number of distinct prime divisors of $k$.