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

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!

Answer 1

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: $$A_1(n):=\sum_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n).$$ $$A_3(n):=\sum_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$

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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: $$\mathcal{A}_n(x) = \frac{1}{1-x}\mathcal{A}_n(\frac{x}{x-1}).$$

Answer 2

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 $\tau(n)$.