# Given the Dirichlet generating function for the divisor sum case, what is the Dirichlet generating function for the Möbius function case?

From a search in the OEIS I found from the comment by Antonio G. Astudillo that:

$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}d^{c}\right) \left(\sum\limits_{d|n}d^{z}\right)}{n^{\text{s}}}=\frac{\zeta (s) \zeta (s-c) \zeta (s-z) \zeta (-c+s-z)}{\zeta (-c+2 s-z)}$$

What is then the Dirichlet generating function for:

$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}\mu(d) d^{c}\right) \left(\sum\limits_{d|n}\mu(d)d^{z}\right)}{n^{\text{s}}}=?$$

where $\mu(d)$ is the Möbius function of a divisor $d$. Above $\zeta(s)$ is of course the Riemann zeta function.

First let us try to evaluate

$$L_1(s) = \sum_{n\ge 1} \frac{1}{n^s} q_{a,b}(n) = \sum_{n\ge 1} \frac{1}{n^s} \left(\sum_{d|n} d^a\right) \left(\sum_{d|n} d^b\right).$$

With the prime factorization of $n$ being

$$n = \prod_p p^v$$

we get for

$$q_{a,b}(n) = \prod_p \frac{p^{av+a}-1}{p^a-1} \frac{p^{bv+b}-1}{p^b-1}.$$

This yields for the Euler product

$$L_1(s) = \prod_p \left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}} \frac{p^{av+a}-1}{p^a-1} \frac{p^{bv+b}-1}{p^b-1}\right) \\ = \prod_p \left(1 + \frac{1}{p^a-1} \frac{1}{p^b-1} \sum_{v\ge 1} \frac{1}{p^{vs}} (p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right) \\ = \prod_p \left(\frac{1}{p^a-1} \frac{1}{p^b-1} \sum_{v\ge 0} \frac{1}{p^{vs}} (p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right) \\ = \prod_p \left(\frac{1}{p^a-1} \frac{1}{p^b-1} \\ \times \left(\frac{p^{a+b}}{1-1/p^{s-(a+b)}} - \frac{p^{a}}{1-1/p^{s-a}} - \frac{p^{b}}{1-1/p^{s-b}} + \frac{1}{1-1/p^s} \right)\right) \\ = \prod_p \left(\frac{1}{p^a-1} \frac{1}{p^b-1} \frac{1}{1-1/p^{s-(a+b)}} \frac{1}{1-1/p^{s-a}} \frac{1}{1-1/p^{s-b}} \frac{1}{1-1/p^s} \\ \times \left(p^{a+b}-p^a-p^b+1 - (p^{2a+2b} - p^{2a+b} - p^{a+2b} + p^{a+b})/p^{2s} \right)\right) \\ = \prod_p \left( \frac{1}{1-1/p^{s-(a+b)}} \frac{1}{1-1/p^{s-a}} \frac{1}{1-1/p^{s-b}} \frac{1}{1-1/p^s} \left(1-p^{a+b}/p^{2s}\right)\right).$$

We now obtain by inspection that

$$L_1(s) = \frac{\zeta(s-(a+b))\zeta(s-a)\zeta(s-b)\zeta(s)} {\zeta(2s-(a+b))}$$

as claimed. Continuing we introduce

$$L_2(s) = \sum_{n\ge 1} \frac{1}{n^s} r_{a,b}(n) = \sum_{n\ge 1} \frac{1}{n^s} \left(\sum_{d|n} \mu(d) d^a\right) \left(\sum_{d|n} \mu(d) d^b\right).$$

This time we have

$$r_{a,b}(n) = \prod_p (1-p^a) (1-p^b).$$

We get the Euler product

$$L_2(s) = \prod_p \left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}} (1-p^a)(1-p^b)\right) \\= \prod_p \left(1 + (1-p^a)(1-p^b) \frac{1/p^s}{1-1/p^s}\right) \\ = \zeta(s) \prod_p (1 - 1/p^s + 1/p^s - 1/p^{s-a} - 1/p^{s-b} + 1/p^{s-(a+b)}) \\ = \zeta(s) \prod_p \left(1 + \frac{p^{a+b}-p^a-p^b}{p^s}\right) \\ = \zeta(s) \sum_{n\ge 1} \frac{1}{n^s} \mu(n)^2 \prod_{p|n} (p^{a+b}-p^a-p^b).$$

The inner product runs over genuine prime divisors excluding one.

This seems to be the best we can do at this point. What we have shown here is that

$$\sum_{d|n} \mu(d)^2 \prod_{p|d} (p^{a+b}-p^a-p^b) = \left(\sum_{d|n} \mu(d) d^a\right) \left(\sum_{d|n} \mu(d) d^b\right).$$