Exploring Divisor Function 4

Here is another proof of the Mark's result

$(a,b)$ denotes $\text{gcd}(a,b)$

Say $F(k) = \displaystyle \sum_{n = 1}^{\infty} \dfrac{d(kn)}{n^2}$

$F(k) = \displaystyle \sum_{m=1}^{\infty} \sum_{n:m|(kn)} \dfrac{1}{n^2}$

$= \displaystyle \sum_{m=1}^{\infty} \sum_{\frac{m}{(m,k)} | n} \dfrac{1}{n^2}$

$=\displaystyle \dfrac{\pi^2}{6} \sum_{m = 1}^{\infty} \dfrac{(m,k)^2}{m^2}$

$= \displaystyle \dfrac{\pi^2}{6} \sum_{g|k} \sum_{m:(m,k)=g} \dfrac{g^2}{m^2}$

$= \displaystyle \dfrac{\pi^2}{6} \sum_{g|k} \sum_{z:(z,\frac{k}{g})=1} \dfrac{g^2}{g^2 z^2}$

Now, using inclusion-exclusion,

$\displaystyle \sum_{(m,n) = 1} \dfrac{1}{n^2} = \dfrac{\pi^2}{6} \sum_{l|m} \dfrac{\mu(l)}{l^2}$

So, $F(k) = \displaystyle \dfrac{\pi^4}{36} \sum_{g|k} \sum_{l|\frac{k}{g}} \dfrac{\mu(g)}{g^2}$

$\implies F(k) = \dfrac{\pi^4}{36} \displaystyle \sum_{g|k} \dfrac{\mu(g)}{g^2} d \left(\frac{k}{g} \right)$

Let $k = \displaystyle \prod_{i = 1}^{m} {p_i}^{a_i}$

Then expanding the sum, we have:

$F(k) = \displaystyle \dfrac{\pi^4}{36} \left((a_1+1)(a_2+1) \dots (a_m+1)\right) \left ( 1 - \sum_{i=1}^{m} \dfrac{a_i}{(a_i+1)p_i^2} + \sum_{1<=i<j<=m}\dfrac{a_i}{(a_i+1)p_i^2}\dfrac{a_j}{(a_j+1)p_j^2} \dots \right)$

$= \displaystyle \dfrac{\pi^4}{36} \prod_{i=1}^{m}(a_i +1 -a_i{ p_i}^{-2})$

In the first of this series, I showed that $\sum_{n \ge 1} \frac{d(Nn)}{n^2} \; = \; \tfrac{1}{36} \pi^4 \prod_{j=1}^m \big(a_j + 1 - a_jp_j^{-2}\big)$ where $N = p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}$ is the prime factorization of $N$ . With $N = 2016 = 2^5\times3^2\times7$ , the sum is equal to $\tfrac{1}{36}\pi^4\big(6 - \tfrac{5}{4}\big)\big(3 - \tfrac{2}{9}\big)\big(2 - \tfrac{1}{49}\big) \; = \; \tfrac{46075}{63504}\pi^4 \;,$ making the answer $\boxed{109583}$ .