Can't find a solution of $\sum_{d|n}d^2\mu\left(\frac nd \right)$ using Dirichlet convolution

Question

I searched for a solution of the sum $$\sum_{d|n}d^2\mu\left(\frac nd \right)$$ running over the divisors of $n$ and $\mu$ the Möbius function. In the available textbooks it is not dealt with or I overlooked the proof of a helpful result from which one can derive it. Also the generalization would be interesting $$\sum_{d|n}d^k\mu\left (\frac nd\right)$$

Answer 1

As a Dirichlet convolution, $\cdot^2 * \mu $ is a multiplicative function, and: $$ (\cdot^2*\mu)(p^k) = p^{2k}-p^{2k-2} = p^{2k}\left(1-\frac{1}{p^2}\right)\tag{1} $$ implies: $$ (\cdot^2*\mu)(n)=\sum_{d\mid n}d^2\mu\left(\frac{n}{d}\right) = n^2\prod_{p\mid n}\left(1-\frac{1}{p^2}\right) = n\,\varphi(n)\prod_{p\mid n}\left(1+\frac{1}{p}\right). \tag{2}$$

Answer 2

As an addendum to the excellent answer already provided suppose we seek to evaluate $$f_k(n) = \sum_{d|n} d^k\mu\left(\frac{n}{d}\right).$$

Working with the corresponding Dirichlet series we find that $$L(s) = \sum_{n\ge 1}\frac{1}{n^s} \sum_{d|n} d^k\mu\left(\frac{n}{d}\right) = \left( \sum_{n\ge 1} \frac{n^k}{n^s} \right) \times \frac{1}{\zeta(s)}.$$

This yields $$L(s) = \frac{\zeta(s-k)}{\zeta(s)}.$$

The Euler product here is $$L(s) = \prod_p \frac{1-1/p^s}{1-1/p^{s-k}} = \prod_p \frac{1-1/p^s}{1-p^k/p^{s}}.$$

Expanding the geometric series we get $$L(s) = \prod_p \left(\sum_{q\ge 0} \frac{p^{qk}}{p^{qs}} - \sum_{q\ge 0} \frac{p^{qk}}{p^{(q+1)s}} \right).$$

This is $$L(s) = \prod_p \left(1 + \sum_{q\ge 1} \frac{p^{qk}-p^{(q-1)k}}{p^{qs}} \right).$$

This finally implies that $f_k(1) = 1$ and $$f_k(p^v) = p^{vk}-p^{(v-1)k}.$$

Therefore we have $$f_k(n) = n^k\prod_{p|n} \left(1-\frac{1}{p^k}\right).$$

This is Jordan's totient function.