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

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!

• Out of curiosity, is there even a closed form for sum of $k^2 l^2?$ – Igor Rivin Feb 21 '18 at 0:40
• (nice question, by the way) – Igor Rivin Feb 21 '18 at 0:41
• If you change the title to something more descriptive, you'll probably get a better response. – Greg Martin Feb 21 '18 at 5:42
• Can you give more backgraund concerning your question? Do you need exactly this sum or sum with additional restrictions on $p,q,k,l$? – Alexey Ustinov Feb 27 '18 at 6:26
• @AlexeyUstinov For this case, I needed this exact sum. It comes up while counting square tiled surfaces. – metallicmural99 Feb 27 '18 at 12:40

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 $f(n)$.

• The OP is trying to simplify a sum, and is looking for suggestions about how to do this. You only provide a closed-form result, instead of an approach; in particular, we have no way of checking whether your result is correct, if you do not provide us with its proof. – Alex M. Feb 21 '18 at 16:33
• @AlexM.: At least numerically, the two expressions match for all $n<100$. I'd be very surprised if they are not equal. – Max Alekseyev Feb 21 '18 at 19:58
• @MaxAlekseyev: Agreed, but I believe that the OP was looking not just for a closed-form result, but rather for a method of obtaining it. Notice that the initial version of this answer (that my comment addressed) was very brief, containing only the part above the later edit; in the meantime, its author has siginificantly improved it, rendering my initial comment somewhat useless. The answer as it stands now is very good. – Alex M. Feb 21 '18 at 20:11
• Thank you @rosab. Most of this is very new to me so I will take some time to digest this. For beginners, what are the $h(a, b)$ functions? – metallicmural99 Feb 22 '18 at 13:35
• @metallicmural99 The line you refer to is here to explain a general fact: if you sum an infinite number of positive terms $h(a,b)$ (without convergence concerns) using two indices $a$ and $b$ each varying from 1 to infinity, you may split the summation into three parts: $a=b$ (the "diagonal"), $a>b$ and $a<b$. Next I apply that to $A(s) = \sum_{p, q, k, l \in \mathbb{N}} \frac {k l^2} {(pk + ql)^s}$ with $a=k$ and $b=l$, and also with $a=p$ and $b=q$. The resulting partial summations have to be considered separately. Sorry I didn't detail everything. – rosab Feb 22 '18 at 14:14

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.

• It seems they want money in exchange for the article... – Pietro Majer Feb 22 '18 at 9:50
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.
• Thank you for the answer. I am having trouble deducing the generating functions for the sums of the form $k^{2t}l^{2s}$ from the wikipedia article you linked. I think I see how to get the ones when $k$ and $l$ have odd exponents, but don't quite get the ones for even exponents.. – metallicmural99 Feb 21 '18 at 1:35
• I think that the asymmetry mentioned is the $k>l$ inequality, and that seems to be a bit tricky regardless of parity.. – Vladimir Dotsenko Feb 21 '18 at 17:10