Exploring Divisor Function 1

Let $N = p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}$ be the prime factorization of $N$ , and let $A(N) \; = \; \big\{ n \in \mathbb{N} \, \big| \, (n,N) = 1 \big\}$ be the collection of positive integers which are coprime to $N$ . Then $\begin{array}{rcl} \displaystyle \sum_{n \ge 1} \frac{d(Nn)}{n^2} & = & \displaystyle \sum_{b_1 \ge 0} \sum_{b_2 \ge 0} \cdots \sum_{b_m \ge 0} \sum_{n \in A(N)} \frac{d\big(p_1^{b_1+a_1}p_2^{b_2+a_2} \cdots p_m^{b_m+a_m} n\big)}{p_1^{2b_1} p_2^{2b_2} \cdots p_m^{2b_m} n^2} \\ & = & \displaystyle \sum_{b_1 \ge 0} \frac{b_1+a_1+1}{p_1^{2b_1}} \sum_{b_2 \ge 0}\frac{b_2+a_2+1}{p_2^{2b_2}} \cdots \sum_{b_m \ge 0} \frac{b_m + a_m + 1}{p_m^{2b_m}} \sum_{n \in A(N)} \frac{d(n)}{n^2} \\ & = & \displaystyle\left(\prod_{j=1}^m f_{a_j + 1}\big(p_j^{-2}\big) \right) \sum_{n \in A(N)} \frac{d(n)}{n^2} \end{array}$ where $f_k(x) \; = \; \sum_{j \ge 0} (j+k)x^j \; = \; \big(k - (k-1)x\big)(1-x)^{-2} \qquad \qquad |x| < 1 \;.$ A similar calculation shows us that $\zeta(2)^2 \; = \; \sum_{n\ ge 1} \frac{d(n)}{n^2} \; = \; \left(\prod_{j=1}^m f_1\big(p_j^{-2}\big) \right) \sum_{n \in A(N)} \frac{d(n)}{n^2}$ and hence that $\sum_{n \ge 1} \frac{d(Nn)}{n^2} \; = \; \prod_{j=1}^m \frac{f_{a_j+1}(p_j^{-2})}{f_1(p_j^{-2})} \zeta(2)^2 \; = \; \tfrac{1}{36} \pi^4 \prod_{j=1}^m \big(a_j + 1 - a_jp_j^{-2}\big)$ With $N = 2017$ , the sum is equal to $\tfrac{1}{36}\pi^4\big(2 - \tfrac{1}{2017^2}\big) \; = \; \tfrac{8136577}{146458404}\pi^4 \;,$ making the answer $\boxed{154594985}$ .

Recall that if $f$ is a completely multiplicative function, $f*f=fd(n)$

Let $F(n)=\begin{cases}0 \text{ if } n \text{ is a multiple of prime }p\\1 \text{ otherwise}\end{cases}$

Notice that $F$ is completely multiplicative. Hence, through Dirichlet Convolution,

$\left[\sum_{n=1}^{\infty} \frac{F(n)}{n^s}\right]^{2}=\sum_{n=1}^{\infty} \frac{d(n)F(n)}{n^s}=\sum_{n=1}^{\infty} \frac{d(n)}{n^s}-\sum_{n=1}^{\infty} \frac{d(pn)}{(pn)^s}$

$\sum_{n=1}^{\infty} \frac{F(n)}{n^s}=\sum_{n=1}^{\infty} \frac{1}{n^s}-\sum_{n=1}^{\infty} \frac{1}{(pn)^s}=\left(1-\frac{1}{p^s}\right)\zeta(s)$

Using the fact that $1*1=d$ and hence $\sum_{n=1}^{\infty} \frac{d(n)}{n^s}=\zeta(s)^2$

$\sum_{n=1}^{\infty} \frac{d(pn)}{(pn)^s}=\zeta(s)^2-\left(1-\frac{1}{p^s}\right)^2\zeta(s)^2=\zeta(s)^2\left(1-\left(1-\frac{1}{p^s}\right)^2\right)$

Substituting $p=2017$ and $s=2$ gives

$\sum_{n=1}^{\infty} \frac{d(2017n)}{n^2}=2017^2\zeta(2)^2\left(1-\left(1-\frac{1}{2017^2}\right)^2\right)=\boxed{\frac{8136577\pi ^4}{146458404}}$