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

Question

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.