Question

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

Answer 1

Well it's not pretty but here it is:

\[\prod_{p} \frac{2+p}{1+p} \]

Answer 2

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*} \)