TITLE: Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?
QUESTION [5 upvotes]: This question is related to the last question about van der Pol's identity for the sum of divisors.
In Touchard (1953) 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:
$$\sum_{k=0}^n\sigma(k)\sigma(n-k)=\tfrac5{12}\sigma_3(n)-\tfrac12n\sigma(n)$$
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!

REPLY [9 votes]: All these identities can indeed be proved essentially trivially using modular forms and quasi-modular forms (those involving $E_2$), and the fact that the dimension
of such spaces is $1$ for weight 4,6,8,10,14, and $2$ for weight 12, in which case the identities involve also the Ramanujan $\tau$ function. Explicitly,
sums $\sum_{1\le k\le n-1}k^a\sigma_b(k)\sigma_c(n-k)$ with $b$ and $c$ odd positive integers ($\sigma_b(k)=\sum_{d\mid k}d^b$) have weight
$w=b+c+2+2a$, so if $w=4,6,8,10,14$ you will obtain identities involving only $\sigma$, and if $w=12$ also $\tau(n)$.