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!