I have come across the following sum:

$$\sum_{\substack{p, q, k, l \in \mathbb{N} \\ k > l \\ pk + ql = n}}kl^2$$

and I am trying to simplify it, hoping to get a nice formula in terms of $n$ and some arithmetic functions of $n$.

For instance, the following sum can be evaluated as follows:

\begin{align*} \sum_{\substack{p, q, k, l \in \mathbb{N} \\ pk + ql = n}}kl &= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}}\sum_{k | \alpha}k \sum_{l | \beta}l \\ &= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}} \sigma_1(\alpha) \sigma_1(\beta)\\ &= (\sigma_1 \Delta \sigma_1)(n) \end{align*} where $\sigma_1(n)$ is the sum of divisors of $n$ and $\Delta$ is the discrete convolution. Ramanujan has a formula for the discrete convolution of $\sigma_1$ with itself given by $$(\sigma_1 \Delta \sigma_1)(n) = \frac{5}{12}\sigma_3(n) + \frac{1}{12}\sigma_1(n) - \frac{1}{2} n \sigma_1(n)$$ where $\sigma_3(n)$ is the sum of the cubes of the divisors of $n$.

Any thoughts on how one might proceed with the sum in the beginning of the question? The asymmetry between $k$ and $l$ is causing some problems so the same approach as for the second sum does not quite work.

I would be very much grateful for any suggestions. Thanks!