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As the title states, is there any simplification to $\sum\limits_{n=k}^N \binom{n}{k}^2$? I found this which sums the squares of the "rows" of a pascal triangle, but here I'm trying to sum the diagonals. There is also this mathSE question from a few years ago with no answers.

Update: I shall clarify that the answer is not a central binomial coefficient, nor using the hockey-stick identity. This is directly seen as $\binom{3}{3}^2 + \binom{4}{3}^2 + \binom{5}{3}^3 + \binom{6}{3}^2 = 517$ is not a central binomial coefficient.

metamorphy
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Gareth Ma
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  • Have you considered writing this as $\sum_{n=k}^N\binom{n}{n-k}^2$ and substituting $n\mapsto n-k$? Then you get $\sum_{n=0}^{N-k}\binom{n+k}k^2$ but that might complicate things... – PinkyWay Apr 17 '22 at 14:05
  • Don't use \limits in title. https://math.meta.stackexchange.com/questions/9687/guidelines-for-good-use-of-mathjax-in-question-titles –  Apr 17 '22 at 17:42
  • @Yooo Ah, I saw someone else add \limits so I rejected the edit request to remove it again. – Gareth Ma Apr 17 '22 at 17:43
  • How do you feel about approximations? I highly suspect there is no closed form for this sum, or even nontrivial simplifications. One way to see this is to recognize this sum as hypergeometric. There's a "well known" algorithm (see, for instance, the book A=B) which will compute a closed form for such a sum if one exists, and this algorithm is built into sage. Despite this, sage is unable to simplify this sum. – HallaSurvivor Apr 17 '22 at 19:07
  • @HallaSurvivor I don't know what to expect, I am simply trying to generalise the hockey-stick identity. Although I don't know anything about hypergeometric functions, maybe you can write an answer explaining what you said that may help others? Thanks for your help in advance. – Gareth Ma Apr 17 '22 at 19:10
  • Update: it might be here. – Gareth Ma Apr 17 '22 at 19:22
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    Does this answer your question? Sum of square binomial coefficients – Leucippus Apr 19 '22 at 00:03
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    @Leucippus: No, we have a variable upper limit here. (On the other hand, the vague any simplification makes it hard to answer...) – metamorphy Apr 19 '22 at 02:50
  • @metamorphy Yeah I understand the vagueness of the question. I will think more deeply about it and see what I really want to ask :” seeing HallaSurvivor’s comment suggests there’s no closed form (as expected). – Gareth Ma Apr 19 '22 at 07:45
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    For fixed $k=1,2,3$ etc., this sum has a closed form as a sum of finitely many hypergeometric terms for obvious reasons. The dependence on $k$ is somewhat difficult to quantify. Surely , fixing $N$ large enough perhaps, one can ask if the sum $\sum_{n=k}^{N} \binom{n}{k}^2$ is hypergeometric in $k$. That is what the algorithm must have tried : finding a WZ mate, which likely will not have succeeded. We need to see what kind of dependence on $k$ (specifically) is desired by OP here, because fixing $k$, the dependence on $N$ is actually hypergeometric : not really difficult to prove this! – Sarvesh Ravichandran Iyer Apr 19 '22 at 12:03

1 Answers1

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Let $T(n,k)=\sum_{j=k}^{n}{j \choose k}^2$.

Computing some values and then searching OEIS you can find your function at OEIS A110197. I suppose there isn't a closed form for the general case, because there isn't any there. However, there are links to special cases sequences and corresponding formulas:

$$T(n,0)=n+1$$ $$T(n,1)=n(n+1)(2n+1)/6$$ $$T(n+2,2)=n(n+1)(n+2)(3n^2 + 6n + 1)/60$$

and other cases up to $k=5$.

Fabius Wiesner
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