For $n$ and $k$ non-negative integers, let $$F(n,k) = \sum_{i=0}^{n}\binom{n}{i}^k$$ For example, $F(n,0)=n+1$, $F(n,1)=2^n$ and $F(n,2)=\binom{2n}{n}$.
Does there exist a general formula for $F(n,k)$?
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as an experiment, you can compute several values of $F(n,3)$ and check the resulting integer sequence with the database at oeis.org ... – Greg Martin Jun 18 '14 at 22:10
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3It can be proven using the WZ algorithm that the sum of the cubes of binomial coefficients does not possess a closed form. See A = B for more details. – Lucian Jun 18 '14 at 22:27
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@Lucian, thanks - I did not know about this WZ Algorithm. – Nicky Hekster Jun 19 '14 at 08:58
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Related: "Is there a closed form for the sum of the cubes of the binomial coefficients?" – Blue Jun 10 '24 at 07:53
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For $n=3$, they are the Franel numbers (sequence $A000172$ in $OEIS$).
In fact, there is a general formula which involves generalized hypergeometric functions (which are summations up to infinity !!). Look at the pattern $$F(n,3)=\, _3F_2(-n,-n,-n;1,1;-1)$$ $$F(n,4)=\, _4F_3(-n,-n,-n,-n;1,1,1;1)$$ $$F(n,5)=\, _5F_4(-n,-n,-n,-n,-n;1,1,1,1;-1)$$
Claude Leibovici
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