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[29]:= 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[29]= (-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[35]:= 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[35]= -(((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[33]:= 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[33]= 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[36]:= 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[36]= 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.