I've come across a formula for $\mu_{2}(n)$ and am not sure if this formula is equal to Apostol's generalized Mobius function or Popovici's function.
I am not very knowledgeable about the work of these 2 people, so I would like some help to figure out if what I have is the same as what they've created. I'm more confident that what I found is not equal to Apostol's, but not sure about Popovici's function.
Perhaps they're not the same if there is more than one way to compute the below, that is, the sum of the reciprocals of the cube-free numbers (raised to a power greater than 1):
$$\sum_{k=1}^{\infty}\frac{|\mu_{2}(k)|}{k^{s}}=\frac{\zeta(s)^2}{\zeta(2s)^2}.$$
Now, here's what I have (my version of $\mu_{2}$):
$$\mu_{2}(n)=\begin{cases} 1, & \text{if $n$=1} \\ (-2)^{k_{1}}, & \text{if $n$ is cube-free with $k_{1}$ single prime factors} \\ 0, & \text{if $n$ is not cube-free}. \end{cases}$$
As for the 2 aforementioned results:
Popovici (1963) defined a generalised Mobius function $\mu_{k}=\mu\cdot ...\cdot \mu$ to be the $k$-fold Dirichlet convolution of the Mobius function with itself. It is thus again a multiplicative function with:
$\mu_{k}(p^a) = (-1)^a \binom{k}{a}$
For Apostol, please click link (too much trouble to type his equations here, he has them on the very first page): http://emis.ams.org/journals/AUSM/C1-2/math2-4.pdf
The Popovici function $\mu_2$ is multiplicative with $\mu_2(1)=\mu_2(p^0)=\dots=1$, $\mu_2(p)=-2$, $\mu_2(p^2)=1$, and $\mu_2(p^a)=0$ for $a\ge3$. Therefore, $\mu_2(n)=(-2)^r$ if $n$ is cubefree and there are $r$ primes $p$ such that $p$ divides $n$ but $p^2$ does not divide $n$, while $\mu_2(n)=0$ if $n$ is not cubefree. This agrees with the formula for $\mu_2$ in the question, provided "single prime factors" means primes $p$ such that $p$ divides $n$ but $p^2$ does not divide $n$.