# Omega 2

$\large \sum_{n=1}^\infty \dfrac{\Omega(n)\tau(n)}{n^2}=\dfrac{\pi^a}{b} \sum_{p \text{ prime}}^\infty \dfrac{1}{p^2-1}$

If the equation above holds true for integer constants $$a$$ and $$b$$, find $$a+b$$.

Notations:

• $$\Omega(n)$$ counts the number of prime factors of $$n$$ (with multiplicity)

• $$\tau(n)$$ counts the number of divisors of $$n$$.

×