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!

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

$$A_2 = \sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$$

Edit (22.08.2024): Thanks to @TatendaKubalalika for pointing to an error in the last equation: The OPN satifies the following equation:

$$8n^4-2n^3-3n^2\sigma_3(n)+24A_2 = 0$$

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!