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Let $(k,l)$ be coprime positive integers, $(k,l)=1$. Let $p,q$ be two real number satisfying $0<p<1$ and $0<q<1$. Consider the following summation: \begin{equation} {\sum_{n_1,n_2=0}^{+\infty}}' p^{n_1} q^{n_2}, \end{equation} where the summation is constrained to be $ln_1 - n_2 \equiv 0, \mod k$. For $l = 1$ it is straightforward. Is there a closed formula for the summation when $l > 0$?

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    $\begingroup$ Notice that $$ \sum_{m,n=0}^{\infty} ' p^m q^n = \frac{1}{(1-p^k)(1-q^k)} \sum_{\substack{0 \leq m, n < k \\ n\equiv lm\pmod{k}}} p^m q^n. $$ I am not sure if the latter sum can be simplified. $\endgroup$ – Sangchul Lee Aug 1 '17 at 15:53
  • $\begingroup$ Isn't the sum $$\sum_{n_1=0}^\infty \sum_{i=0}^\infty p^{n_1} q^{l n_1 + ik}\text{?}$$ If yes, I don't see the need for $k$ and $l$ to be coprime. $\endgroup$ – Idéophage Aug 1 '17 at 16:04
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    $\begingroup$ @Idéophage No, in your summation $i$ does not start from $0$. It can start from some negative number which makes $ln_1+ ik$ positive. $\endgroup$ – Kevin Ye Aug 1 '17 at 16:10
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$\newcommand{\C}{\mathbb{C}}$This is not an answer but we can obtain a different expression with a more simple finite sum. Let $r$ be a primitive $k$th root of the unity. Your sum is then

$$\frac{1}{k} \sum_{n=0}^{k-1} \frac{1}{(1-pr^{-nl}) (1 - q r^n)}\text{.}$$

(I checked it numerically.)


First, we work in $\C[[X,Y]]$. As noted by Sangchul Lee in the comments, the problem reduces to compute $$f(X,Y) := \sum_{\substack{0 \leq m, n < k \\ n\equiv lm\pmod{k}}} X^m Y^n \text{.}$$

The idea is to interpolate $f$ from its values on $\{X^k = Y^k = 1\}$, where it becomes simply $$\sum_{n=0}^{k-1} X^n Y^{nl} = \sum_{n=0}^{k-1} (X Y^l)^n \text{.}$$ Let $\alpha$ and $\beta$ be integers modulo $k$ such that $X = r^\alpha$ and $Y = r^\beta$. Then $f(X,Y)$ is $k$ if $XY^l = 1$ and $0$ otherwise. That is to say, $f(X,Y) = k$ when $\alpha + l \beta = 0$.

The interpolation gives $$\begin{align*} f(X,Y) &= \sum_{\substack{\beta=0\\\alpha=-l\beta}}^{k-1}k \times \frac{1-X^k}{k (1 - X r^{-\alpha})} \times \frac{1-Y^k}{k (1 - Y r^{-\beta})}\\ &= \sum_{n=0}^{k-1} \frac{(1-X^k)(1-Y^k)}{k (1-X r^{-l n}) (1-Y r^n)} \text{.} \end{align*}$$

Dividing by $(1-X^k)(1-Y^k)$, we get the result.

Question: can we interpret the fact that the product appearing in the interpolation simplifies?

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