Convolution product for $2^{\Omega(n)}$

Question

How can I write the multiplicative function $2^{\Omega(n)}$ as a Dirichlet product of two multiplicative functions? That's because I have to find an estimate for $\sum_{n\leq x}2^{\Omega(n)}$.

Answer 1

Well, the Dirichlet series of your function has an obvious Euler product:$$ \sum^\infty_{n=1}\frac{2^{\Omega(n)}}{n^s}=\prod_{p\,prime}\frac{1}{1-\frac{2}{p^s}},$$ but I can't see how to write that as a product of just two factors in a meaningful way.
The RHS has a lot of poles due to the many solutions of $p^s=2$ in complex $s$, making $s=0$ an essential singularity, so I don't think there is a connection to well-investigated functions.
I assume that you mean the arithmetic function usually denoted by that symbol, http://oeis.org/wiki/Omega(n),_number_of_prime_factors_of_n_(with_multiplicity).

Answer 2

One of possible Dirichlet products of two multiplicative functions is $$2^{\Omega(n)} = 2^{\Omega(n)-\omega(n)}*1(n)$$ where $1(n)$ is the constant function (defined by $1(n) = 1$) and $\omega(n)$ counts each distinct prime factor.