How can we show this relationship between the sum of divisors function and the sum $p^{m}+2p^{m-1}+3p^{m-2}+\ldots+(m+1)$?

Question

The sum of divisors function is commonly denoted by $\sigma(n)$. Now let us introduce a recursive definition of divisor functions:

• $r_{n,1}=\sigma(n)$
• $r_{n,2}=\sum_{d|n}r_{d,1}$
• $r_{n,3}=\sum_{d|n}r_{d,2}$
• $\ldots$

Note that these are simplified Dirichlet products so that the functions are multiplicative.

My question:

How can we show that if $p^{m}$ divides $n$ and $p^{m+1}$ does not divide $n$, then $p^{m}+2p^{m-1}+3p^{m-2}+\ldots+(m+1)$ divides $r_{n,2}$?

A first idea:

Formulated differently, $\frac{\frac{p^{m+2}-1}{p-1}-(m+2)}{p-1}$ divides $r_{n,2}$ and if $n$ is squarefree, we have $\prod_{p|n}(p+2)=r_{n,2}$.

Answer 1

Didn't you say the function was multiplicative? Because then $r_{p^m,2}\mid r_{n,2}$, and $$ r_{p^m,2} = \sum_{k=0}^m\sigma(p^k)=\sum_{k=0}^m\sum_{j=0}^kp^j=\sum_{j=0}^mp^j\sum_{k=j}^m1 = \sum_{j=0}^m(m-j+1)p^j. $$