<s>[INST] Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\ 
-a-n+1 & 
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\ 
c & 
\end{array} ; 1\right] = \dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$
I give negative parameters to the $\Gamma$ function.
What other identity can I use?
I'm trying to find a closed form to this: $\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
Wolfram Mathematica answered this as a closed form: $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
But I would like to have a manual solution with proof.
This is how I got that hypergeometric series:
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
$\dfrac{t_{i+1}}{t_{i}} = \frac{\binom{a+i+1-1}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \frac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \frac{(a+i) (i-n)}{ (i-a-n+1)(i+1)}$
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\ 
-a-n+1 & 
\end{array} ; 1\right]$
UPDATE
Thanks to David H, I got closer to the solution.
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\ 
-a-n+1 & 
\end{array} ; 1\right]$
$\lim\limits_{\epsilon \to0} \frac{\Gamma (-2 a-2 \epsilon +1) \Gamma (-a-n-\epsilon +1)}{\Gamma (-a-\epsilon +1) \Gamma (-2 a-n-2 \epsilon +1)} = \frac{4^{-a} \Gamma \left(\frac{1}{2}-a\right) \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
As you can see this result is close to the expected $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$ formula. But the $\binom{a+n-1}{n}$ factor is still missing and I don't really understand why.
 [/INST] We may try to keep it simple.
Suppose we seek to evaluate
$$\sum_{k=0}^n {a-1+k\choose k} {a-1+n-k\choose n-k}.$$
It is immediately apparent that this is a convolution of two ordinary generating functions.
To see what they are re-write the sum as
$$\sum_{k=0}^n {a-1+k\choose a-1} {a-1+n-k\choose a-1}.$$
By the Newton binomial this is a convolution of
$$\frac{1}{(1-z)^a}$$
with itself, giving
$$\frac{1}{(1-z)^{2a}}.$$
Extracting coefficients we obtain
$${n+2a-1\choose 2a-1} = {n+2a-1\choose n}.$$
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