Let $$f(n) = 2^{\omega(n)}\mu^2(n)$$, where $\omega(n)$ is number of distinct prime divisors of $n$ and $\mu(n)$ is Moebius function.
I want to simplify it.
As long as $$ \mu^2(n)=\sum_{d^2|n}\mu(d) $$ and $$ 2^{\omega(n)}=\sum_{d|n}\mu^2(d) $$ so $$ f(n)=\sum_{d|n}\mu^2(d)\sum_{d^2|n}\mu(d) $$ And I can't simplify it further. Is it possible?
Thanks.
NB. By "to simplify" I mean "to express in terms of divisors on $n$"
The best way to "simplify" $f(n)$ is to consider that $f(n)$ is a multiplicative function (since both $2^{\omega(n)}$ and $\mu^2(n)$ are multiplicative functions) whose values on $p^k$ is $2$ if $k=1$, $0$ otherwise. This gives, for instance (assuming $\Re(s)>1$): $$ \sum_{n\geq 1}\frac{f(n)}{n^s}=\prod_{p}\left(1+\frac{2}{p^s}\right)\leq\prod_{p}\left(1+\frac{1}{p^s}\right)^2=\left(\frac{\zeta(s)}{\zeta(2s)}\right)^2.$$