$$\sum\limits_{d\mid n}\mu^2(d)/\sigma(d)=?$$

Question

kainui · Answer

Well it's not pretty but here it is:

$$\prod_{p} \frac{2+p}{1+p} $$

ikram002p · Answer

so same idea as previous question note it's multiplicative function
 
$ \begin{align*}
n&=\prod_{i=1}^{\omega(n)} p_i^{k_i}  \
n_{\omega(n)}&=\prod_{i=1}^{\omega(n)} p_i \
f(n) &=\sum\limits_{d\mid n}\mu^2(d)/\sigma(d)  \ 
 &= \sum\limits_{d\mid n_{\omega(n)}} 1/\sigma(d) \ 
 f(n_{\omega(n)})&=f(p_1p_2....p_{\omega(n)}) \ 
 &=f(p_1)f(p_2) .... f(p_\omega(n))\
 &=(1+\frac{1}{p_1+1})(1+\frac{1}{p_2+1})...(1+\frac{1}{p_{\omega(n)}+1})\
&=\prod_{i=1}^{\omega(n) }\frac{p_i+2}{p_i+1} 

\end{align*} $