Exploring Divisor Function 3

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 $N = 2048 = 2^{11}$ , the sum is equal to $\tfrac{1}{36}\pi^4\big(12 - \tfrac{11}{4}\big) \; = \; \tfrac{37}{144}\pi^4 \;,$ making the answer $\boxed{185}$ .

When it comes to evaluating for when $k=p^a$ , where p is a prime, a problem arises, since

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

Isn't completely multiplicative, so we can't use the same method as what we did in part 1.

However, we can split the sum up and then sum them all together.

$\displaystyle \sum _{n=1}^{\infty}\frac{d\left(p^an\right)}{(p^a n)^s}=\sum _{n=1}^{\infty}\frac{d\left(pn\right)}{(pn)^s}-\sum _{m=1}^{a-1}\sum _{n \text{ coprime to }p}\frac{d\left(p^mn\right)}{(p^m n)^s}$

Since $d(n)$ is multiplicative, in other words, if a and b are coprime integers, $d(a)\times d(b)=d(ab)$

$\displaystyle \sum _{m=1}^{a-1}\sum _{n \text{ coprime to }p}\frac{d\left(p^mn\right)}{(p^m n)^s}=\sum _{m=1}^{a-1}\frac{d\left(p^m\right)}{p^{ms}}\sum _{n \text{ coprime to }p}\frac{d\left(n\right)}{n^s}=\sum _{m=1}^{a-1}\frac{m+1}{p^{ms}}\sum _{n \text{ coprime to }p}\frac{d\left(n\right)}{n^s}$

Since through part 1,

$\displaystyle \sum _{n=1}^{\infty}\frac{d\left(pn\right)}{\left(pn\right)^s}=\sum _{n=1}^{\infty}\frac{d\left(n\right)}{n^s}-\sum _{n \text{ coprime to }p}\frac{d\left(n\right)}{n^s}=\zeta(s)^2-\left(1-\frac{1}{p^s}\right)^2\zeta(s)^2=\zeta(s)^2\left(1-\left(1-\frac{1}{p^s}\right)^2\right)$

And that

$\displaystyle \sum _{m=1}^{a-1}\frac{m+1}{p^{ms}}=\frac{2p^{s(a+1)}-p^{as}-ap^{2s}+ap^s-p^{2s}}{p^{as}\left(p^s-1\right)^2}$

Putting it all together,

$\displaystyle \sum _{n=1}^{\infty}\frac{d\left(p^an\right)}{(p^a n)^s}=\sum _{n=1}^{\infty}\frac{d\left(pn\right)}{(pn)^s}-\sum _{m=1}^{a-1}\sum _{n \text{ coprime to }p}\frac{d\left(p^mn\right)}{(p^m n)^s}$

$\displaystyle =\zeta(s)^2\left(1-\left(1-\frac{1}{p^s}\right)^2\right)-\frac{2p^{s(a+1)}-p^{as}-ap^{2s}+ap^s-p^{2s}}{p^{as}\left(p^s-1\right)^2}\left(1-\frac{1}{p^s}\right)^2\zeta(s)^2$

Hence

$\displaystyle \sum _{n=1}^{\infty}\frac{d\left(p^an\right)}{ n^s}=p^{sa}\left(\zeta(s)^2\left(1-\left(1-\frac{1}{p^s}\right)^2\right)-\frac{2p^{s(a+1)}-p^{as}-ap^{2s}+ap^s-p^{2s}}{p^{as}\left(p^s-1\right)^2}\left(1-\frac{1}{p^s}\right)^2\zeta(s)^2\right)$

Simplifying gives

$\zeta(s)^2\left(a+1-ap^{-s}\right)$

Substituting p=2, s=2, a=11, gives

$\displaystyle \sum _{n=1}^{\infty}\frac{d\left(2048n\right)}{n^2}=\left(\frac{\pi ^2}{6}\right)^2\left(12-11\cdot 2^{-2}\right)=\boxed{\frac{37\pi ^4}{144}}$

Due to how well the final equation collapsed, I'm pretty sure there is a much better solution out there.