Compute the Dirichlet inverse of $f(n)=\frac{1}{1+|\mu(n)|}$, where $\mu(n)$ is the Möbius function Let for integers $n\geq 1$ the arithmetical function defined by $$f(n)=\frac{1}{1+|\mu(n)|},$$
where $\mu(n)$ is the Möbius function. 
Note that $f(1)=\frac{1}{2}\neq 0$, and $f(n)$  isn't multiplicative.
I am interested in 

Question. Can you get in a closed-form the Dirichlet inverse of previous function $f(n)$? Thanks in advance.

I've written a little dictionary but I'm stuck about how use it succesfully: $f(1)=1/2$; if $p$ is a prime $f(p)=1/2$, and for $a>1$ we've $f(p^a)=1$; if $1<n$ is square-free then $f(n)=1/2$ but if isn't then $f(n)=1$. 
My computation were humbles (and it isn't necessary if you answer  my Question prove by mathematical induction yours statements, I believe that it is neccesary nested induction but with your closed-form I could do it): we've 
$$f^{-1}(1)=2,$$
if $p$ is a prime number 
$$f^{-1}(p)=-\frac{1}{f(1)}f(p)f^{-1}(1)=-2\frac{1}{2}2=-2,$$ 
and we can write for prime powers 
$$f^{-1} \left( p^a \right) =-2 \sum_{k=0}^{a-1} f \left( p^{a-k} \right) f ^{-1}\left( p^{k} \right),$$
when $a>1$.
Thus $f ^{-1}\left( p^{2} \right)=-2$ and $f ^{-1}\left( p^{3} \right)=2,\ldots$
 A: It's easy to verify that your arthmetic funtion is
$$
f(n)=1-|\mu(n)|+\frac{|\mu(n)|}{2}=1-\frac{|\mu(n)|}{2}.
$$
Taking Dirichlet series of both sides, we get
$$
D(f,s)=\zeta(s)-\frac{\zeta(s)}{2\zeta(2s)}.
$$
Thus, we have
$$
D(f^{-1},s)=\frac{1}{D(f,s)}=\frac{1}{\zeta(s)}\frac{2\zeta(2s)}{2\zeta(2s)-1}.
$$
Considering the form of $D(f^{-1},s)$, especially the tricky part $\frac{1}{2\zeta(2s)-1},$ I guess it's pretty hard or even impossible to obtain simple closed forms for all $f^{-1}(n)$.
A: In response to the last comment about the author wanting an exact closed-form formula for the inverse, there are a couple of ways to go about this. The first one can be obtained from the Dirichlet series for $D(f^{-1}, s)$ in the previous solution. In particular, by applying the integral representation in Perron's formula you can find an expression for the summatory function $F^{-1}(x) := \sum_{n \leq x}^{\prime} f^{-1}(n)$. Then by taking the backwards difference of the summatory function you arrive at an exact formula for the function itself. Note that it may not be simple to express the result you obtain from Perron's formula as an easy arithmetic function. In fact, based on the formula given above it should depend on sums over the non-trivial zeros of the zeta function $\pm$ error term series. 
The second method can be done exactly via a known recurrence relation for the Dirichlet inverse which is also given in Chapter 2 of Apostol's book. Notice that $f(n) = 1 / (1+\chi_{\operatorname{sq}}(n))$ where $\chi_{\operatorname{sq}}(n)$ denotes the characteristic function of the squarefree numbers. Namely, $f^{-1}(1) = 1 / f(1) = 2$ and for $n > 1$ we have that 
$$f^{-1}(n) = -2 \sum_{\substack{d|n \\ d>1}} \frac{f^{-1}(n/d)}{1+\chi_{\operatorname{sq}}(d)}.$$ 
This formula can also be used to find the expression for the Dirichlet series in the previous post by noting that $D(\chi_{\operatorname{sq}}, s) = \zeta(s) / \zeta(2s)$ and that $1-\chi_{\operatorname{sq}}(n)$ is the characteristic function of the non-squarefree integers (split the infinite sum up into two pieces and then solve for $D(f^{-1}, s)$ at the end).
