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