Closed formula for a multiple Dirichlet convolution of the 1-function with the identity

Question

For two multiplicative arithmetic functions $f,g$ the Dirichlet convolution is defined by $(f\ast g) (n)=\sum\limits_{ab=n}f(a)g(b)$. Convoluting any arithmetic function with the $1$-function ($1(n)=1$ for all $n\in\mathbb{N}$) gives a summation function $(1\ast f)(n)=\sum\limits_{d|n}f(d)$. Given the prime factorization of $n$, there is a closed formula for the sum of divisors ($1\ast id)(n)$ of $n$. Is there anything like this if I consider functions of the form $1\ast 1\ast\ldots\ast 1\ast id$ (i.d. the sums of divisors of divisors, etc.)?

Answer 1

If I understand correctly you wish to compute a closed form for the coefficients $q_n$ of the Dirichlet series $$[1/n^s] \zeta(s)^k\zeta(s-1)$$ where $k\ge 1.$ These are given by $$q_n = \sum_{a_1 a_2 \cdots a_k \times d = n} d.$$

Suppose that $n=p^v,$ a prime power. Then by stars-and-bars we get the value $$q_n = \sum_{w=0}^v {w+k-1\choose k-1} p^{v-w}.$$

Finally using the fact that $q_n$ is multiplicative we obtain the formula $$q_n = \prod_{p|n} \sum_{w=0}^v {w+k-1\choose k-1} p^{v-w}$$ where the product ranges over the prime powers $p^v$ in the prime factorization of $n.$

This formula is documented at the OEIS e.g. at links OEIS A007430 and OEIS A007429.