Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$? 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!
 A: 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 $\tau(n)$.
A: 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:
$$\mathcal{A}_n(x) = \frac{1}{1-x}\mathcal{A}_n(\frac{x}{x-1}).$$
