How to calculate $ \sum_{n=1}^\infty\frac{3^{\omega(n)}\Omega(n)}{n^s} $? For $s\in\mathbb C$ with say $\Re s>1$, how to write
$$
\sum_{n=1}^\infty\frac{3^{\omega(n)}\Omega(n)}{n^s}
$$
in terms of the Riemann Zeta function (where $\omega$ is the number of prime factors without multiplicity, and $\Omega$ is the number of prime factors with multiplicity)?
I tried the following: denote by $\Pi$ the indicator function of the prime powers, then $\Omega = 1*\Pi$, i.e. the convolution of the function constant equal to $1$ and $\Pi$. By changing the order of summation, we then see
$$
\sum_{n=1}^\infty\frac{3^{\omega(n)}\Omega(n)}{n^s}=\sum_{d=1}^\infty \frac{\Pi(d)}{d^s}\sum_{n=1}^{\infty}\frac{3^{\omega(nd)}}{n^s}
$$
Now I am able to show that if $d=p^l$ for some prime $p$, then
$$
\sum_{n=1}^{\infty}\frac{3^{\omega(nd)}}{n^s}=\frac{3}{1+2p^{-s}}\sum_{n=1}^{\infty}\frac{3^{\omega(n)}}{n^s},
$$
but even for $\sum_{n=1}^{\infty}\frac{3^{\omega(n)}}{n^s}$ I am not sure how to proceed. Note that solar we have:
$$
\sum_{n=1}^\infty\frac{3^{\omega(n)}\Omega(n)}{n^s}=\left(\sum_{n=1}^{\infty}\frac{3^{\omega(n)}}{n^s}\right)\left(\sum_{p \text{ prime, }l\geq 1}\frac{3}{p^{ls}(1+2p^{-s})}\right)=\left(\sum_{n=1}^{\infty}\frac{3^{\omega(n)}}{n^s}\right)\left(\sum_{p \text{ prime}}\frac{3}{(p^s-1)(1+2p^{-s})}\right).
$$
Is this the way to proceed or am I completely on the wrong path? Thanks in advance!
 A: 
$$F(s):=\sum^\infty_{n=1}\frac{3^{\omega(n)}\Omega(n)}{n^s}=\zeta(s)\prod_p\left(1+\frac2{p^{s}}\right)\sum_p\left(\frac1{p^s-1}+\frac2{p^s+2}\right) $$

Derivation:
$$F(s):=\sum^\infty_{n=1}\frac{3^{\omega(n)}\Omega(n)}{n^s}=\frac{\partial}{\partial\alpha}\underbrace{\sum^\infty_{n=1}\frac{3^{\omega(n)}e^{\alpha\Omega(n)}}{n^s}}_{:=g(\alpha,s)}\bigg\vert_{\alpha=0}$$
Let $\displaystyle{f(n)=\frac{3^{\omega(n)}e^{\alpha\Omega(n)}}{n^s}}$.
For the moment, consider the set of integers $S_N=\{k\,|\,k=p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N},a_{(\cdot)}\ge1\}$
Summing $f(n)$ over $S_N$ gives
$$\begin{align}
\sum_{n\in S_N}f(n)
&=\sum^\infty_{a_N=1}\cdots\sum^\infty_{a_1=1}\frac{3^N\exp\alpha(a_1+\cdots+a_N)}{(p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N})^s} \\
&=3^N\prod^N_{j=1}\sum^\infty_{a_j=1}\left(\frac{e^\alpha}{p_j^s}\right)^{a_j} \\
&=3^N\prod^N_{j=1}\underbrace{\frac1{p_j^s\cdot e^{-\alpha}-1}}_{E_j} \\
&=3^N\prod^N_{j=1}{E_j}
\end{align}
$$ 
Now, consider another set of integers $S^*_N=\{k\,|\,k=p_1^{a_1}\cdots p_N^{a_N},a_{(\cdot)}\color{red}{\ge0}\}$.
Summing $f(k)$ over all the elements in $S^∗_N$ with $a_\alpha=0$ and other indexes non-zero gives $\displaystyle{\frac1{3E_{\alpha}}\cdot 3^N\prod^N_{j=1}{E_j}}$.
Similarly, with only two zero indexes we get $\displaystyle{\frac1{9E_{\alpha}E_\beta}\cdot 3^N\prod^N_{j=1}{E_j}}$, and $\displaystyle{\frac1{27E_{\alpha}E_\beta E_\gamma}\cdot 3^N\prod^N_{j=1}{E_j}}$ for three. So on and so on.
Therefore, summing all these gives
$$\sum_{n\in S^*_N}f(n)=\left(1+\frac1{3E_1}\right)\cdots\left(1+\frac1{3E_N}\right)3^N\prod^N_{j=1}{E_j}=\prod^N_{j=1}(1+3E_j)$$
Taking $N\to\infty$,
$$g(\alpha,s)=\prod_p\left(1+\frac3{p^s\cdot e^{-\alpha}-1}\right)$$
By logarithmic differentiation,
$$\frac1{g(0,s)}\frac{\partial}{\partial\alpha}g(\alpha,s)\bigg\vert_{\alpha=0}=\sum_p\frac{\frac{3p^s}{(p^s-1)^2}}{1+\frac3{p^s-1}}$$
$$\begin{align}
F(s)
&=\prod_p\left(1+\frac3{p^s-1}\right)\sum_p\frac{3p^s}{p^{2s}+p^s-2} \\
&=\prod_p\left(\frac{p^s+2}{p^s-1}\right)\sum_p\frac{3p^s}{p^{2s}+p^s-2} \\
&=\prod_p\left(\frac{1+2p^{-s}}{1-p^{-s}}\right)\sum_p\frac{3p^s}{p^{2s}+p^s-2} \\
&=\zeta(s)\prod_p\left(1+\frac2{p^{s}}\right)\sum_p\frac{3p^s}{p^{2s}+p^s-2} \\
&=\zeta(s)\prod_p\left(1+\frac2{p^{s}}\right)\sum_p\left(\frac1{p^s-1}+\frac2{p^s+2}\right)
\end{align}
$$
Since the summation term resembles the prime zeta function, I suspect that $\text{Re }s=0$ is the natural boundary of $F(s)$. 
Moreover, it seems like that the pole at $s=1$ is of order $3$. If so, it would be interesting to calculate the residue there (as well as the coefficients of $(s-1)^{-3}$ and $(s-1)^{-2}$).
