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.