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I want to sum over all the possible combinations of two numbers that give the same result under a certain formula. Specifically, in this case, sum over all the possible combinations of non-negative integers $i,j\in\mathbb{N}_0$ that together with some constant natural number $n\in\mathbb{N}$ give the same value for the formula $k=n\left(i+j\right)+j$.

I tried the following notation:

$$\sum_{k=n\left(i+j\right)+j}2^k$$ But, it seems somewhat ambiguous and unclear, it's not explicit how many and what variables the sum goes over and which are fixed. How do I indicate exactly and clearly my intentions? Maybe something like:

$${\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}}_{s.t\ n\left(i+j\right)+j=k}2^k$$

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    $\begingroup$ I sometimes write $\sum_{k \in I} \text{blah}$ and then describe the index set $I$ explicitly. $\endgroup$ – copper.hat Jan 30 at 19:45
  • $\begingroup$ Your link just points to this same question. $\endgroup$ – RobPratt Jan 30 at 21:31
  • $\begingroup$ @RobPratt, sorry, my bad. This is related to my question math.stackexchange.com/questions/3526286/… $\endgroup$ – SIMEL Jan 30 at 21:35
  • $\begingroup$ Is $k$ a fix number or not? If it is, then $2^k$ is also a constant and it could be taken out of the summation. If it is not, why is not your sum infinite? $\endgroup$ – J.-E. Pin Jan 31 at 8:17
  • $\begingroup$ @J.-E.Pin, yes, it's a fixed number and it can be taken out of the sum and put $1$ inside. $\endgroup$ – SIMEL Jan 31 at 8:26
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" I want to sum (what ?) over all the possible combinations of two numbers .."

the what shall be a function $f \, : \, \mathbb {Z}^2 \to \mathbb X$ so you write $$ S(C) = \sum\limits_{(i,j)\, \in \,C} {f(i,j)} $$ where $C$ is a domain in the plane $i,j$ defined by certain conditions.

In the example you give, $$ C = \left\{ {(i,j):\left\{ \matrix{ 0 \le i,j \hfill \cr n\left( {i + j} \right) + j = k \hfill \cr} \right.} \right\} $$ where $n$ and $k$ are considered as given constants (or parameters), and do not vary while taking the sum.

So by writing $$ S(C) = \sum\limits_{(i,j)\, \in \,C} {2^{\,k} } $$ you are actually telling $$ S(C) = \sum\limits_{(i,j)\, \in \,C} {2^{\,k} } = \sum\limits_{(i,j)\, \in \,C} {2^{\,k} \cdot 1} = 2^{\,k} \sum\limits_{(i,j)\,} {{\bf 1}_{\left\{ {(i,j)\, \in \,C} \right\}} } = 2^{\,k} \left| C \right| $$ i.e., that $f(i,j)=1$, so that when summed over $C$ i it is equivalent to summing the corresponding indicator function over the whole plane, and finally to give the size of $C$, which is the number of non-negative solutions to the diophantine equation $n\left( {i + j} \right) + j = k$.

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Here's one option, using \substack: $$\sum_{\substack{i\ge 0,\ j\ge 0:\\n\left(i+j\right)+j=k}} 2^k$$

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  • $\begingroup$ The first option says something else entirely, the condition is nowhere to be found and gives a completely different answer (specifically $\infty$). The second one is still messy and not clear. $\endgroup$ – SIMEL Jan 30 at 20:42
  • $\begingroup$ So $k$ is fixed? $\endgroup$ – RobPratt Jan 30 at 21:26
  • $\begingroup$ Yes, I want to know the number of different combinations for a certain $k$ and $n$, for example, for $n=3$, for $k=1,2$ the number is $0$, for $k=4$ it's $1$, and for $k=12$ it's $2$ $\endgroup$ – SIMEL Jan 30 at 21:28
  • $\begingroup$ OK, I updated my answer to remove the first option. $\endgroup$ – RobPratt Jan 30 at 21:30
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Following your answer to my comment, I would suggest to write $$ 2^k \bigl|\bigl\{(i,j) \in {\Bbb N}^2 \mid ni + (n+1)j = k\bigr\}\bigr| $$

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