Evaluating the sum of $kl^2$ over $p,q,k,l$ such that $pk +ql = n$

Question

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!

Answer 1

Using elementary methods, for p,q,k,l strictly positive, I find : $\frac{1}{24}\sigma_4(n) + \frac{1}{2}\sigma_3(n) - \frac{1}{24} (12n+1)\sigma_2(n)$

The asymmetry is not a problem if you replace $k$ with $k+l$, which is greater than $l$.

Edit: Sorry @alex-m, you're right, I'll try to summarize.

The idea is first to use the Dirichlet transform of the summand; If we have to sum $f(n)$ from 1 to infinity we consider $\sum {\frac {f(n)}{n^s}}$ with $s$ real but large enough for the sum to converge. Then if we can show that $\sum {\frac {f(n)}{n^s}}$ is equal to some other transform $\sum {\frac {g(n)}{n^s}}$ for all $s$ large enough, it is not difficult to prove that $f(n)=g(n)$ for all $n$.
Typically here we'll use the well-known Dirichlet transform $\sum {\frac {\sigma_a(n)}{n^s}} = \zeta(s) \zeta(s-a)$ and we'll seek a way to transform the wanted original Dirichlet sum into a combination of such known sums.

Here the wanted $f(n)$ is $\sum_{\substack{p, q, k, l \in \mathbb{N} \\ k > l \\ pk + ql = n}}k l^2$
As I said the $k>l$ specification can be ignored if the sum becomes $\sum_{\substack{p, q, k, l \in \mathbb{N} \\ p(k+l) + ql = n}}(k+l)l^2$
The Dirichlet transform $\sum {\frac {f(n)}{n^s}}$ to consider will be simply $\sum_{p, q, k, l \in \mathbb{N}} \frac {(k+l)l^2} {(p(k+l) + ql)^s}$

The next step consists in finding an approaching sum, which can be transformed in at least two different ways, bringing the desired sum. Here this approaching sum will be $ A(s) = \sum_{p, q, k, l \in \mathbb{N}} \frac {k l^2} {(pk + ql)^s}$

I recall that $s$ will be large enough for all the considered sums to converge.
We use a "diagonalization" scheme to transform $A(s)$ in two ways; since we have multiple indices, we split the sum in three parts : $ \sum_{a,b >0} h(a,b) = \sum_{a>0} h(a,a) + \sum_{a,b >0} h(a+b,b) + \sum_{a,b >0} h(a,b+a)$

Here we "diagonalize" $A(s)$ first over $k$ and $l$, and second over $p$ and $q$.
In the process we get the desired $\sum_{p, q, k, l \in \mathbb{N}} \frac {(k+l)l^2} {(p(k+l) + ql)^s}$ and other sums easy to manage, which I don't detail but bring the $\zeta(s)$ function.
At the end we get the desired result for $f(n)$.

Answer 2

Sums of this type can be calculated using standard tools from analytic number theory (Kloosterman sums + Poisson summation). This sum with additional restrictions arose in Frobenius problem. For precise statement see Lemma 8 (page 828) in the article On the Frobenius problem for three arguments (2016) by I. Vorob'ev.

In my earlier work The solution of Arnold's problem on the weak asymptotics of Frobenius numbers with three arguments (2009) the same tools used for calculation of more simple sums $$\sum_{\substack{p, q, k, l \in \mathbb{N} \\ pk + ql = n\\\text{+additional restrictions}}}k $$ but this technique allows to replace the summand $k$ by more or less arbitrary smooth function of $p,q,k,l.$

If you need sharper error terms then you should apply more advanced tools like cancellation of Kloosterman sums, Kuznetsov trace formula and averaging over spectrum of Laplace operator from An Additive Divisor Problem by J. M. Deshouillers and H. Iwaniec.

Answer 3

This will probably get a real answer soon and this is more of a comment, but as you are describing it, the generating function for sums of the form $k^{2t} l^{2s}$ is the Fourier series of a product of Eisenstein series (the wikipedia article has a lot of identities) so your problem seems to be more the parity than the asymmetry.