# Expression simplified for explicit int values, but not with FullSimplify/FunctionExpand and Assuming. What formula does Mathematica use?

I have a $$_4F_3$$ hypergeometric function (Mathematica 11.3)

``HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n, 1 + 2 m + n}, {2 + n,    2 + n, 3/2 + 2 m + n}, z] ``

If I plug in explicit integer values for $$n$$ I get e.g.

``In:= HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n,     1 + 2 m + n}, {2 + n, 2 + n, 3/2 + 2 m + n}, z] /. {n -> 0}  Out= (-1 - 4 m)/(  4 m^2 z) + ((1 + 4 m) HypergeometricPFQ[{1/2, 2 m, 2 m}, {1,      1/2 + 2 m}, z])/(4 m^2 z)  In:= HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n,     1 + 2 m + n}, {2 + n, 2 + n, 3/2 + 2 m + n}, z] /. {n -> 1}  Out= -(((3 + 4 m) (1 + 4 m + 4 m^2 z))/(   3 m^2 (1 + 2 m)^2 z^2)) + ((1 + 4 m) (3 + 4 m) HypergeometricPFQ[{1/     2, 2 m, 2 m}, {1, 1/2 + 2 m}, z])/(3 m^2 (1 + 2 m)^2 z^2) ``

and so on. However, passing $$n$$ being an integer as an assumption and using `FullSimplify` or `FunctionExpand` leads to nothing

``In:= Assuming[{n \[Element] Integers, n >= 0},   FunctionExpand[   HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n, 1 + 2 m + n}, {2 + n,      2 + n, 3/2 + 2 m + n}, z]]]  Out= HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n,    1 + 2 m + n}, {2 + n, 2 + n, 3/2 + 2 m + n}, z] ``

and

``In:= Assuming[{n \[Element] Integers, n >= 0},   FullSimplify[   HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n, 1 + 2 m + n}, {2 + n,      2 + n, 3/2 + 2 m + n}, z]]]  Out= HypergeometricPFQ[{1, 3/2 + n, 1 + 2 m + n,    1 + 2 m + n}, {2 + n, 2 + n, 3/2 + 2 m + n}, z] ``

Ultimately, I want to know what formula does Mathematica use to obtain these simplifications from $$_4F_3$$ to $$_3F_2$$ (I looked at some resources like DLMF, but couldn’t find anything). Also, it would be nice to find a way to get Mathematica to apply whatever formula it is using to the general case with assumptions.