The Chu-Vandermonde theorem states that

$$ \,_{2}F_{1}\left(\left.-n, a \atop c\right|1\right) = \frac{(c-a)_n}{(c)_n} $$

where the LHS denotes a terminating (due to the negative upper index, $-n$) hypergeometric series with base 1, and $(c)_n$ denotes a rising factorial.

I'm wondering if an equivalent closed form is possible in the following case

$$ \,_{3}F_{2}\left(\left.r-n, q-s, q-t \atop q-n, -s-t+n+q+1\right|1\right) = \sum^{\infty}_{k=0} \frac{(r-n)_k(q-s)_k(q-t)_k}{(q-n)_k(-s-t+n+q+1)_kk!} $$

which terminates when $k>s-q$, since $q-s$ is always negative.

I ask partly because Michael Hirschhorn appears to allude to a generalisation to his approach for proving binomial identities involving $\,_{2}F_{1}$ series to cases involving $\,_{p}F_{q}$ series in this paper (top of page 2).

I have searched this recently collected list of $\,_{3}F_{2}$ identities, but without any luck.

To give context to my question, I am trying to find a closed form for the following sum of binomials

$$ \sum_{k} \binom{s-q}{k} \binom{n-s}{t-q-k} \binom{n-q-k}{r-q} $$

which, using Hirschhorn's approach, can be re-arranged into the $\,_{3}F_{2}$ given above, multiplied by $\binom{n-q}{r-q}\binom{n-s}{t-q}$.

Any help is greatly appreciated.

  • $\begingroup$ I would suggest to try to understand how the first identity follows from the Mellin-Barnes integral representation for $_2F_1$ and then to think if similar simplifications are possible for $_3F_2$. $\endgroup$ – Start wearing purple Jul 9 '13 at 10:05
  • $\begingroup$ Thanks - I considered trying to apply Euler's integral transform, but this requires that the upper indices of the series are non-negative, which is not the case for my $\,_{3}F_{2}$. $\endgroup$ – matk Jul 9 '13 at 10:59
  • $\begingroup$ I insist on Mellin-Barnes, since Chu-Vandermonde identity looks like two ways of contracting the contour therein. $\endgroup$ – Start wearing purple Jul 9 '13 at 11:02
  • $\begingroup$ The Mellin-Barnes also requires non-negative upper indices according to your link. $\endgroup$ – matk Jul 9 '13 at 11:05
  • 1
    $\begingroup$ I am not saying it is straightforward. The formula for negative integers should be obtained in a limit where both left and right side go to infinity. $\endgroup$ – Start wearing purple Jul 9 '13 at 11:10

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