Cubefree

Beautiful problem!

This is the Euler product, for all primes $p$ , of $\sum_{a=0}^{2}\frac{(-2)^{2\omega(p^a)-\Omega(p^a)}}{p^{2a}}=1-\frac{2}{p^2}+\frac{1}{p^4}=\left(1-\frac{1}{p^2}\right)^2$ . The product of these is $\zeta(2)^{-2}=\frac{36}{1\pi^4}$ , so that the answer is $\boxed{41}$ .

The sum is equal to $\sum_{n \ge 1} \frac{X(n)}{n^2}$ where $X$ is the multiplicative function such that $X(p^n) \; =\; \left\{ \begin{array}{lll} 1 & \qquad & n = 0 \\ -2 && n = 1 \\ 1 && n = 2 \\ 0 && n \ge 3 \end{array} \right.$ for any prime $p$ . Thus $(1 \star X)(p^n) \; = \; \mu(n) \; = \; \left\{ \begin{array}{lll} 1 &\qquad& n = 0 \\ -1 && n = 1 \\ 0 && n \ge 2 \end{array} \right.$ and hence $X \,=\, \mu \star \mu$ . Thus $\sum_{n\ ge 1} \frac{X(n)}{n^2} \; = \; \left(\sum_{n\ ge 1} \frac{\mu(n)}{n^2}\right)^2 \; = \; \big(\zeta(2)^{-1}\big)^2 \; = \; \zeta(2)^{-2} \; = \; \frac{36}{\pi^4}$ which makes the answer $\boxed{41}$ .