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First, consider how many $n$-dimensional vector of non-negative integers $(x_1,x_2,\cdots,x_n)$ are there whose sum of all entries satisfies $x_1+x_2+\cdots+x_n=m$?

For example for $n=2,m=2$, there are $(2,0),(1,1),(0,2)$, so $f(2,2)=3$.

For $n=3,m=2$ there are $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(0,1,1),(1,0,1)$, so $f(3,2)=6$.

How about for general $m,n$? What I know is $f(n,m)\le$ n+m choose n by How many $k-$dimensional non-negative integer arrays $(x_1,\cdots,x_k)$ satisfies $x_1+x_2+\cdots+x_k\le n$

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Then, I would like a closed form solution for $$\sum_{m=0}^Mf(n,m)$$

ZHU
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  • You can search the site for weak compositions and read Wikipedia about bit – Ross Millikan Apr 04 '18 at 02:42
  • I changed the question, it this okay now? – ZHU Apr 04 '18 at 03:00
  • This seems a strange thing to be interested in. Where does it come from? The sum of the first term is $\frac 12M(M+1)$. The second is the sum of some of a row of Pascal's triangle. I am not aware of a nice form, though you can use the normal approximation. – Ross Millikan Apr 04 '18 at 03:06
  • It comes from trying to derive algorithm complexity. How $N$ scales is more important here. – ZHU Apr 04 '18 at 03:07
  • @RossMillikan made a mistake, just edited it. – ZHU Apr 04 '18 at 03:14
  • We don't have any $N$ in the problem. As $n$ gets large you should be able to put an upper bound on it based on the fact that the last term is the largest. – Ross Millikan Apr 04 '18 at 03:16
  • How about just $\sum_{m=0}^Mf(m,N)$? Should it be $m+N$ choose $N$? Is this larger than $M^2$? – ZHU Apr 04 '18 at 03:23
  • You didn't look at the link I gave you. $f(m,N)={N+m-1 \choose m-1}$. For large $N$ it is of order $N^{m-1}$ with a constant out front. – Ross Millikan Apr 04 '18 at 03:31

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