# 6-j symbols and hypergeometric series

What’s the correct formula for $$_{4}F_{3}(a,b,c,d;e,f,g;1)$$ where $$a+b+c+d-e-f-g=-1$$?

The Wolfram Alpha formula involves $$6j$$ symbols and makes no sense for some specific cases. For example, $$_{4}F_{3}(5/4,1/4,1/4,1/4; 1,1,1; 1)$$ is a finite number, but the $$6j$$ symbol formula gives zero.

• This is link to the formula I’m having issue with: functions.wolfram.com/HypergeometricFunctions/Hypergeometric4F3/… Aug 13, 2021 at 20:18
• In the $6j$ formula from your link, one divides by $\Gamma(1-b_1)$, where $b_1=1$. It might be because of this Mathematica gives zero.
– Nemo
Aug 14, 2021 at 7:47

There is no simple formula for the balanced $${}_4F_3$$. It is only directly related to $$6j$$-symbols when it terminates (is a finite sum) and that is not the case for your parameters.