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Here is a power series, which looks a bit like a Hypergeometric function series, but I don't think that it is. Has anyone any idea what it is? Here $n,p,r$ are integers with $n\ge 0$ and $p\ge r\ge 0$:

$$ f_{n,p,r}(x)\,=\,\sum_{s=0}^p \frac{x^s}{s!}\ \frac{p!\,(2n+p+s+2)!\,(n+r+s+2)!}{(p-s)!\,(2n+r+s+3)!\,(n+s+2)!} $$

Originally this occurred as a $q$-factorial series, but if the $q=1$ case given here could be recognised, it would be a big help.

Thanks for the answer below. A bit before I declare this answered (if I remember how to do that!) I would like to sneak in another question. Is there any sensible way to sum this in the case $x=-1$? The answer is likely zero if $r<p$ and reasonably simple if $r=p$, but how to prove that?

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Isn't it hypergeometric? Maybe I'm messed up, but it looks like a multiple of ${}_3F_3(-p, n+r+3, 2n+p+3; n+3, n+r+2, 2n+r+4; -x)$. (This is from taking the ratio of the $s+1$ and $s$ terms and factoring.) Where did this function come from? It seems plausible that one could say more than just that it's hypergeometric, but I'm not sure what. –  Henry Cohn Nov 6 '13 at 13:01
    
Thanks, that is a more parameter version of the Hypergeometric than I am used to! It comes from looking at the eigenvalues of the Hodge Laplacian in the 1-forms of the noncommutative sphere (guess q=1 is just commutative sphere). The sum, evaluated at x=-1, is used in finding the normalisation of the eigen-1-forms in the Hodge inner product. I am hoping that knowing the classical analogue will help in the quantum case. –  Edwin Beggs Nov 6 '13 at 13:40
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2 Answers 2

up vote 6 down vote accepted

Henry is almost right; the sum can be written as $$\frac { \left( 2n+p+2 \right) !\, \left( n+r+2 \right)!} { \left( 2n+r+3 \right) !\, \left( n+2 \right) !} {}_{3}F_{2}\left({{2n+p+3,n+r+3,-p\,\,\,}\atop {2n+r+4,n+3}}\Bigm |-x\right)$$ For $x=-1$ (no need to assume $r\le p$) the sum can be evaluated by Saalschütz's theorem (also called the Pfaff-Saalschütz theorem), giving $$ \left( -1 \right) ^{p}\frac { \left( 2n+p+2 \right) !\, \left( n+r +2 \right) !\,r!\, \left( n+p \right) !}{ \left( 2+n+p \right) !\, \left( 3+2n+r+p \right) !\, \left( r-p \right) !\,n!}. $$

Saalschütz's theorem is probably the most commonly used hypergeometric summation formula after Vandermonde's theorem (and it has a $q$-analogue). Zeilberger's algorithm can be used to prove any hypergeometric summation formula, but it won't tell you what it is. (Though Saalschütz's theorem can be found in the book $A=B$ in section 3.5, and section 3.6 has some information on how to identify hypergeometric summation formulas.)

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According to Mathematica, for $r=p$, the sum for $x=-1$ gives $$ \frac{(n+1) \left(-\frac{1}{4}\right)^p \left(\frac{1}{2} (2 n+3)\right)! p! (2 n+p+2)!}{(2 n+3)! (n+p+1) \left(\frac{1}{2} (2 n+2 p+3)\right)!} $$ This formula can can be proven using Zeilberger’s algorithm [D. Zeilberger, The method of creative telescoping, J. Symbolic Comput. 11 (1991) 195–204], implemented as a Mathematica package [P. Paule, M. Schorn, A Mathematica version of Zeilberger’s algorithm for proving binomial coefficient identities, J. Symbolic Comput. 20 (5–6) (1995) 673–698], which establishes recurrence relations for $f_{n,p,p}$.

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The canonical reference for this sort of thing is Petkovsek-Zeilberger A=B. –  Igor Rivin Nov 6 '13 at 15:13
    
Thanks! I have been comprehensively answered on all points. Thanks to all who contributed! –  Edwin Beggs Nov 6 '13 at 16:12
    
Don't forget that Wilf is also a co-author of A=B. –  Ira Gessel Nov 7 '13 at 13:31
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