This question is related to the [last question about van der Pol's identity for the sum of divisors](https://mathoverflow.net/questions/372476/van-der-pols-identity-for-the-sum-of-divisors-and-a-quartic-polynomial-equation).
In [Touchard (1953)](https://oeis.org/A000385/a000385.pdf) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$):


$$n^2(n-1) = \frac{6}{\sigma(n)} \sum\_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k)$$


We can evaluate the convolution part with [Ramanujan's identity](https://en.wikipedia.org/wiki/Eisenstein_series#Ramanujan_identities):


$$\sum\_{k=0}^n\sigma(k)\sigma(n-k)=\tfrac5{12}\sigma\_3(n)-\tfrac12n\sigma(n)$$


which for our case reads like this:


$$\sum\_{k=1}^{n-1}\sigma(k)\sigma(n-k)=\tfrac5{12}\sigma\_3(n)-\tfrac12n\sigma(n)+\tfrac{\sigma(n)}{12}$$


Substituting in van der Pol's equation a perfect number $n = \sigma(n)/2$ and making use of Ramanujan's identity, we find that the perfect number $n$ satisfies the following quartic equation:


$$
8n^4-2n^3+3 \sigma\_3(n)n^2+24A\_2 =0
$$


where


$$A\_2 = \sum\_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$$


I asked an expert of convolution identities for $\sigma(n)$ if $A\_2$ can be evaluated and he said, that one could prove a similar formula, like the one of Ramanujan, "simply by considering the first and the second derivative of suitable identities between Eisenstein series".


However I am not very confident with Eisenstein series, so **I am asking the experts for help to help evaluate $A\_2$.**


Thanks for your help!



Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?
Numerical experiments suggest that
$$A\_2(n) := \sum\_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma\_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$
PS. In fact, it directly follows from the quoted Touchard and Ramanujan identities.


A couple of similar identities:
$$A\_1(n):=\sum\_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma\_3(n) - \frac{6n^2-n}{24}\sigma(n).$$
$$A\_3(n):=\sum\_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma\_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$




---


**ADDED.** A recurrent formula for $A\_d(n)$ with an *odd* $d$ can be obtained from the observation:
\begin{split}
A\_d(n) & := \sum\_{k=1}^{n-1} k^d\sigma(k)\sigma(n-k) \\
&= \sum\_{k=1}^{n-1} (n-k)^d\sigma(k)\sigma(n-k) \\
&= \sum\_{i=0}^d \binom{d}{i} n^{d-i} (-1)^i A\_i(n).
\end{split}
implying that
\begin{split}
A\_d(n) &= \frac{1}{2} \sum\_{i=0}^{d-1} \binom{d}{i} n^{d-i} (-1)^i A\_i(n) \\
&=\frac{1}{d+1} \sum\_{i=0}^{d-1} \binom{d+1}{i} n^{d-i} (-1)^i A\_i(n).
\end{split}
However, to use this formula one would need to compute $A\_t(n)$ for even $t<d$ by other means.


It also follows that the generating function:
$$\mathcal{A}\_n(x) := \sum\_{d=0}^{\infty} \frac{A\_d(n)}{n^d}x^d$$
satisfied the functional equation:
$$\mathcal{A}\_n(x) = \frac{1}{1-x}\mathcal{A}\_n(\frac{x}{x-1}).$$


