If we take a function $f$, we define
$$
\widetilde{f}(n)=\frac{1}{d(n)} \sum_{m∣n} f(m),
$$
where $d(n)$ is the divisor function.
Given $f(n) =2^{\Omega(n)}$, show that $$\widetilde{f}(n) = \frac{2^{a+1} - 1}{a + 1}$$
Note: $\Omega(n)$ is the total number of prime factors.
I think this has to do with the Dirichlet series and/or Euler product but I am having trouble putting it all together. Any help would greatly appreciated!
From the fundamental theorem of arithmetic, we can take $n$ which has $l$ prime factors as $$n = p_1^{k_1}\cdot p_2^{k_2} \cdot \dots p_l^{k_l} \vert \text{ each } p_i \text{ is prime and } k_i >0$$ Then $$\begin{align}\widetilde{f}(n)&=\frac{1}{d(n)} \sum_{m∣n} f(m) \\&= \frac{1}{\displaystyle \prod_{i=1}^l (k_i+1)}\sum_{m∣n}2^{\Omega(m)} \\&= \frac{\displaystyle\sum_{i=0}^l\left((\#\{m : (m|n) \land (\Omega(m)=i) \})\cdot 2^i\right)}{\displaystyle \prod_{i=1}^l (k_i+1)}\\ &= \frac{\displaystyle\sum_{i=0}^l\left(\left(\sum_{\mathbb{S}\in\mathcal{P}(\{k_1, k_2, \dots k_l\}) \land |S|=i}\left( \prod_{k_j\in\mathbb{S}} k_j\right)\right)\cdot 2^i\right)}{\displaystyle \prod_{i=1}^l (k_i+1)}\\&= \frac{\displaystyle\prod_{i=1}^l\left(2k_i+1\right)}{\displaystyle \prod_{i=1}^l (k_i+1)} = \prod_{i=1}^l\left(2-\frac{1}{k_i+1}\right)\end{align}$$
I don't think you can reduce this further to the form you have written down $\widetilde{f}(n) = \frac{2^{a+1} - 1}{a + 1}$ for any significant $a$, but yeah there will always exist an $a$ for which this is true (unless $n=1$).