Recently I obtained the following expression $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1), $$ with $b>a>0$ and $n\in\mathbb{N}$.
My question is: If someone knows a closed form solution to the above expression (either in terms of the gamma function or the rising factorials).
I'm aware of the Saalschütz's theorem which states that $${}_3F_2(-n,a,b; c, 1+a+b-n-c; 1) = \frac{(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}, $$ or the Dixon's identity, however the derived equation (while only based on 3 parameters) does not have the necessary forms. I also tried a lot of identities listed here: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/02/06/ but no success so far.
Alternatively, I can also obtain the equation
$${}_3F_2(-n,-a-1 ,1-b-n; b + 1, 1-a-n; -1), $$
by using few operations in the early stage of the problem, but the list of formulas is much larger for $z=1$.
I'm not really an expert on the hypergeometric functions, so a help from a trained eye in this problematic would be appreciated and very helpfull. From what I have read, the ${}_3F_2(-n,...;...; 1)$ form of the hypergeometric function ${}_3F_2$ frequently appears in many problems.
I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):
$${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= \sum _{k=0}^n \frac{(-1)^{k} \binom{n}{k} \left(\prod _{j=1}^k (-b+j-n)\right)\left(\prod _{j=1}^k (a-b+j-1)\right) }{ \left(\prod _{j=1}^k (-a+j-n)\right)\left(\prod _{j=1}^k (b+j)\right) }$$