<s>[INST] Hypergeometric functions for $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c\geq 0$ The sum is: $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c \geq 0$.
The equivalent hypergeometric function can be found in page 214, Concrete Mathemtics, which is:
$$_3F_2\left(1-a-2n,1-b-2n,-2n;a,b;1\right) (*)$$
This is my approach:
Let $n = max(-a,-b,-c)$.
$$\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k\geq n} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k\geq0} {{a+b} \choose{a+k+n}}{{b+c} \choose{b+k+n}}{{c+a} \choose{c+k+n}}(-1)^{k+n}$$
Now let $$t_k = {{a+b} \choose{a+k+n}} {{b+c} \choose{b+k+n}}{{c+a} \choose{c+k+n}}(-1)^{k+n} = \frac{(a+b)!}{(a+k+n)!(b-k-n)!}\frac{(b+c)!}{(b+k+n)!(c-k-n)!}\frac{(c+a)!}{(c+k+n)!(a-k-n)!}(-1)^{k+n}$$
I would like to evaluate $$\frac{t_{k+1}}{t_k}$$,which is:
$$\frac{(a-k-n)(b-k-n)(c-k-n)(-1)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n = \frac{(k-a+n)(k-b+n)(k-c+n)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n $$
The above fraction doesn't have the form of:
$$ \frac{(k+a_1)(k+a_2)...(k+a_m)z}{(k+b_1)(k+b_2)...(k+b_n)(k+1)}$$
Even if we let n be even, the answer would be:
$$_4F_3\left(n-a,n-b,n-c,1;n+1+a,b+1+n,c+1+n;1\right)$$, not matching $(*)$. Apart from this way, i can't find any other approaches to this problem. Any help would be greatly appreciated.
*Edit: We might assume the biggest number of three is c. So $a \leq b \leq c$. This could work because we will have $k!$ in  $t_k$. (9/4/2021)
 [/INST] The dumbest mistake is in this $(-1)^n$ in:
$$\frac{(a-k-n)(b-k-n)(c-k-n)(-1)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n $$
Clearly, it should be canceled as well! And the idea in the comment works!
Let we assume $a \leq b \leq c$. So, we have $n = -c$!
Now, $$\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k \geq0} {{a+b} \choose{a+k-c}}{{b+c} \choose{b+k-c}}{{c+a} \choose{k}}(-1)^k$$
Repeating the steps, we have:
$$\frac{t_{k+1}}{t_k} = \frac{(b+c-k)(c+a-k)(2c-k)(-1)}{(a-c+k+1)(b-c+k+1)(k+1)} = \frac{(k-(b+c))(k-(c+a))(k-2c)}{(a-c+k+1)(b-c+k+1)(k+1)}$$
And it is: $$_{3} F_{2}\left(\begin{matrix} -b-c,-c-a,-2c\\a-c+1,b-c+1 \end{matrix} ;1\right)$$
Finally, we simply let $a_0 = a-c+1$, $b_0 = b-c+1$ and $c=n$. So, $a=a_0+c-1=$ and $b=b_0+c-1$.
$$_{3} F_{2}\left(\begin{matrix} -b_0-c+1-c,-c-a_0-c+1,-2c\\a_0,b_0 \end{matrix} ;1\right)$$ = $$_{3} F_{2}\left(\begin{matrix} -b_0-2n+1,-a_0-2n+1,-2n\\a_0,b_0 \end{matrix} ;1\right)$$, we're done!
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