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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 $k$.

  • 0
    What is $\sigma_0 (k^2)$?2012-04-26
  • 0
    @hkju: http://en.wikipedia.org/wiki/Divisor_function2012-04-26

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