Mobius and Divisor-Counting function

Question

While looking at the function $$f(x)=\sum_{k|x} \mu(k)*\sigma_0(k)$$ where $\mu(x)$ is the mobius function and $\sigma_0(x)$ is the number of divisors of $x$, I noticed that $f(x)$ is either $1$ or $-1$ and also that if $x$ is a power of a prime number then $f(x)=-1$. At first i thought that $f(x)=-1$ if and only if x is a power of a prime number but this pattern breaks with $x=30$. Can somebody explain to me when $f(x)$ is equal to $1$ and $-1$ and if those are the only numbers that can occure?

Answer 1

Assume that $n=\prod_{k=1}^{\omega(n)} p_k^{\alpha_k}$. For any $d\mid n$, $\mu(d)$ equals $\pm 1$ iff $d$ is squarefree, and in such a case $\sigma_0(d)$ equals $2^{\omega(d)}$. It follows that $$ \sum_{d\mid n}\mu(d)\sigma_0(d) = \sum_{E\subset\{1,\ldots,\omega(n)\}}(-2)^{|E|}=\sum_{k=0}^{\omega(n)}\binom{\omega(n)}{k}(-2)^k=(1-2)^{\omega(n)}=(-1)^{\omega(n)}. $$