Omega

n=12ω(n)n2\large \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^2}

Let ω(n)\omega(n) denote the number of distinct prime divisors of nn.

If the series above can be expressed as ab\frac{a}{b}, where aa and bb are coprime positive integers, find a+ba+b.

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