Need help with finding Dirichlet generating function

Question

I am unable to find the following generating function: $$\tag{1}{\sum _{n=1}^{\infty } \frac{2^{-k_n} \mu \big(\frac{n}{2^{k_n}}\big)}{n^s}}$$

$\mu$ is Möbius function, $k_n$ is the highest integer number so that $2^{k_n}$ divides $n$ (maybe someone is more familiar with notation $2^{k_n}\mid\mid n$).

I am able to find generating functions of $\mu \big(\frac{n}{2^{k_n}}\big)$ and $2^{-k_n}$ separately: $${\sum _{n=1}^{\infty } \frac{\mu \big(\frac{n}{2^{k_n}}\big)}{n^s}=\frac{1}{\left(1-2^{-s}\right)^2 \zeta (s)}}$$ $${\sum _{n=1}^{\infty } \frac{2^{-k_n}}{n^s}=\frac{\left(2^s-1\right) \zeta (s)}{2^s-\frac{1}{2}}}$$

Then I tried to use the formula in this answer, that relates term-wise product of arithmetic functions with its generating function, but I could not simplify the integral.

Any ideas how to find the generating function of $(1)$?

Answer 1

Writing $n=mk_n$ where $m$ is odd, \begin{align*} \sum _{n=1}^{\infty } \frac{2^{-k_n} \mu \big(\frac{n}{2^{k_n}}\big)}{n^s} &= \sum_{k=0}^\infty \sum_{\substack{n\ge1 \\ k_n = k}} \frac{2^{-k_n} \mu \big(\frac{n}{2^{k_n}}\big)}{n^s} \\ &= \sum_{k=0}^\infty 2^{-k} \sum_{\substack{m\ge1 \\ m\text{ odd}}} \frac{\mu(m)}{(2^km)^s} \\ &= \sum_{k=0}^\infty 2^{-k(s+1)} \sum_{\substack{m\ge1 \\ m\text{ odd}}} \frac{\mu(m)}{m^s} = \frac1{1-2^{-s-1}} \frac1{\zeta(s)}(1-2^{-s})^{-1}. \end{align*}