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How do I express $\displaystyle\sum_{r+s+t=n}a_{r,s,t}$ explicitly as nested sums, where $r,s,t,n$ are positive integers? That is to say, how can I express it in the following way...

$$\sum_{r+s+t=n}a_{r,s,t}=\sum_{r=r_0}^{R}\sum_{s=s_0}^S\sum_{t=t_0}^Ta_{r,s,t},$$ where $r_0,s_0,t_0,R,S,T$ are to be found. I went through the following possible combinations:

$$\begin{array}{l|l|&l} r&s&t\\ \hline 0&0&5\\ 0&1&4\\ 0&2&3\\ 0&3&2\\ 0&4&1\\ 0&5&0\\ \hline 1&0&4\\ 1&1&3\\ 1&2&2\\ 1&3&1\\ 1&4&0\\ \hline 2&0&3\\ 2&1&2\\ 2&2&1\\ 2&3&0\\ \hline 3&0&2\\ 3&1&1\\ 3&2&0\\ \hline 4&0&1\\ 4&1&0\\ \hline 5&0&0\\ \end{array} $$

So I should be able to do something like:

$$\sum_{r=0}^5\sum_{s=0}^{5-r}\sum_0^{t=5-r-s}a_{r,s,t},$$

where the strange indexing in the most inner sum denotes counting down - not sure if I can just write it the usual way without having to denote this in a special way?

I think this is right. Can it be generalised for more variables, e.g. $r+s+t+u=n$, $r+s+t+u+v=n$, etc.? Perhaps there's a standard way to do this and my brain's just not working right now.

pshmath0
  • 10,565

1 Answers1

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Your main idea is correct but you messed up at the last element. The correct answer is:

$$\sum_{r=0}^5 \sum_{s=0}^{5-r} a_{r, s, 5 - r - s}$$

This is because $r + s + t = n$ rearranged gives $t = n - r - s$. In other words, $t$ is fully determined by $r$ and $s$.

orlp
  • 10,508