# Hypergeometric function 3F2 with unit argument

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):
$${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= \sum _{k=0}^n \frac{(-1)^{k} \binom{n}{k} \left(\prod _{j=1}^k (-b+j-n)\right)\left(\prod _{j=1}^k (a-b+j-1)\right) }{ \left(\prod _{j=1}^k (-a+j-n)\right)\left(\prod _{j=1}^k (b+j)\right) }$$
• Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway. – K. Keeper Jan 25 at 13:39
• Thanks for clarifying. For $n=3$ you get $$\frac{(b+2) \left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2\right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising. – James Arathoon Jan 25 at 14:48