Dirichlet series of GCD

That's a fun problem! Just what I needed on a slow day at the office!

If we write $n=2^a3^b7^cm$ , where $\gcd(m,2016)=1$ , then the sum can be written as

$\left(\sum_{a=0}^{\infty}\frac{\gcd(2^a,2016)}{4^a}\right)\left(\sum_{b=0}^{\infty}\frac{\gcd(3^b,2016)}{9^b}\right)\left(\sum_{c=0}^{\infty}\frac{\gcd(7^c,2016)}{49^c}\right)\left(\sum_{\gcd(m,2016)=1}\frac{1}{m^2}\right)$

This is not as bad as it looks. We know that the last term is $\frac{16\pi^2}{147}$ , and the others are easy to evaluate as they turn into geometric series after a few terms, for example, $\frac{1}{9^0}+\frac{3}{9^1}+\frac{9}{9^2}+\frac{9}{9^3}+..+\frac{9}{9^k}+..$ $=1+\frac{1}{3}+\frac{1}{8}$ $=\frac{35}{24}$ . We find

$\frac{95}{48}\frac{35}{24}\frac{55}{48}\frac{16\pi^2}{147}=\frac{26125\pi^2}{72576}$

and the answer is $\boxed{98701}$ .

There are fancier ways to solve this, but why use a sledge hammer to crack a nut ;)

Indeed, a nice problem. We can generalize Otto's solution as follows: Note that $\sum_{a \ge 0} \frac{p^{\mathrm{min}(a,n)}}{p^{2a}} \; = \; \displaystyle \sum_{a=0}^{n-1}p^{-a} + \sum_{a \ge n} p^{n-2a} \; = \; \displaystyle \frac{1+ p^{-1} - p^{-n-1}}{1 - p^{-2}}$ for any prime $p$ and integer $n \ge 1$ . If $k \in \mathbb{N}$ has prime factorization $k\,=\, p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}$ , then $\sum_{n \ge 1} \frac{\mathrm{gcd}(n,k)}{n^2} \; = \; \prod_{j=1}^m \left(\sum_{a \ge 0} \frac{p_j^{\mathrm{min}(a,a_j)}}{p_j^{2a}}\right) \sum_{\mathrm{gcd}(n,k) = 1} \frac{1}{n^2}$ and hence $\begin{array}{rcl} \displaystyle \sum_{n \ge 1} \frac{\mathrm{gcd}(n,k)}{n^2} & = & \displaystyle \prod_{j=1}^m \frac{1 + p_j^{-1} - p_j^{-a_j-1}}{1 - p_j^{-2}} \sum_{\mathrm{gcd}(n,k)=1} \frac{1}{n^2} \\ & = & \displaystyle \prod_{j=1}^m \frac{1 + p_j^{-1} - p_j^{-a_j-1}}{1 - p_j^{-2}} \times \prod_{j=1}^m \big(1 - p_j^{-2}\big) \zeta(2) \\ & = & \displaystyle \zeta(2) \prod_{j=1}^m\big(1 + p_j^{-1} - p_j^{-a_j-1}\big) \end{array}$ giving the sum $\tfrac{26125}{72576}\pi^2$ , and the answer $\boxed{98701}$ , when $k = 2016$ .