# Omega

$\large s = \sum_{n=1}^\infty \dfrac{\omega(n)}{n^2}$

Let $$\omega(n)$$ denote the number of distinct prime factors of $$n$$. Compute $$\lceil 1000s\rceil$$.

Details and Assumptions:

• You may use the fact that $$\displaystyle P(2) = \sum_{p \text{ prime}} \dfrac1{p^2} \approx 0.452247$$.

• You may read up Prime zeta function.

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