# Omega

**Number Theory**Level 5

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

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.