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Let $m,k$ be an positive integers with $k\le m$. I am trying to prove $$\sum_{j=0}^k{\frac{1}{2}\choose k-j}\frac{2^{2j}(m+j)!}{(m-j)!(2j)!}=\frac{P(n,k)}{(2k)!}$$ where $n=2m+1$ and $P(n,k)=\prod_{i=0}^{k-1} (n^2-(2i)^2)$. We can use Pochhammer's symbol and write the left side as $$(-1)^k\sum_{j=0}^k\frac{(-1/2)_{k-j}(m+1)_j(-m)_j}{(1/2)_j(k-j)!j!}$$ We can also write the right hand side as $$(-1)^k\frac{(-m-\frac{1}{2})_k(m+\frac{1}{2})_k}{(\frac{1}{2})_kk!}.$$ So I think the identity must follow from some identities in hypergeometric functions. Does anyone see a way to do it?

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  • $\begingroup$ Note: Recently duplicated on MathOverflow and treated there. $\endgroup$
    – ccorn
    May 6 '13 at 3:04
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Treating both sides as coefficients of $x^k$ in a power series with variable $x$, the left-hand side turns into a power-series product, one factor representing $\sqrt{1-x}$. We end up with a special case of the known hypergeometric identity $$(1-x)^{a+b-c} {}_2F_1(a,b;c;x) = {}_2F_1(c-a,c-b;c;x)$$ with $a = m+1$, $b = -m$, $c = 1/2$. Cf. (34) of this MathWorld entry.

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  • $\begingroup$ This formula can be found in: Abramowitz, M. and Stegun, I. A. (Eds.). "Hypergeometric Functions." Ch. 15 inHandbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 555-566, 1972. The formula is 15.3.3 on page 559. $\endgroup$
    – TCL
    May 6 '13 at 15:31

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