Just a bit of Number Theory

For any prime $p$ we have $\sum_{n=0}^\infty \frac{2^{\omega(p^n)}}{p^{ns}} \; = \; 1 + \sum_{n=1}^\infty \frac{2}{p^{ns}} \; = \; 1 + \frac{2p^{-s}}{1-p^{-s}} \; = \; \frac{1+p^{-s}}{1-p^{-s}}$ for $s > 1$ , and hence $\begin{aligned} \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s} & = \; \prod_p \frac{1+p^{-s}}{1-p^{-s}} \; = \; \zeta(s) \times \frac{\zeta(s)}{\zeta(2s)} \; = \; \frac{\zeta(s)^2}{\zeta(2s)} \end{aligned}$ for $s > 1$ , while $\sum_{n=1}^\infty \frac{d(n)}{n^s} \; = \; \zeta(s)^2$ for $s > 1$ , so that $S_s \; = \; \frac{1}{\zeta(2s)} \hspace{2cm} s > 1$ With $s=4$ we obtain $S_4 = \zeta(8)^{-1} = \frac{9450}{\pi^8}$ , making the answer $9450 + 1 + 8 = \boxed{9459}$ .