Expressing ${_3F_2}$ in terms of gamma functions I have a generalized hypergeometric funtion of the following form
$$\,_3F_2\left(\frac{1}{2},\frac{\beta }{2},\frac{1}{4} (\beta +2 p-2);\frac{\beta +1}{2},\frac{1}{4} (\beta +2 p+5);1\right)$$ where $\beta$ can take integer values $2,1,0,-1,\ldots$ and $p = 0$ or, $\frac{1}{2}$. Is it possible to express this function in terms of gamma-functions?  
 A: Since $\beta$ is an integer, I will rename it $n$. So we have:
$$n=2,1,0,-1,-2,\dots$$
And two functions (for each value of $p$):
$$F_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-2}{4} ;\frac{n +1}{2},\frac{n+5}{4};1\right)$$
$$G_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-1}{4} ;\frac{n +1}{2},\frac{n+6}{4};1\right)$$

Before attempting the simplification for general $n$, I have decided to do some experiments in Mathematica. Here are the results:
$$\begin{array}{c,c,c} n & F_n & G_n \\ 2 & 1 & \frac{\sqrt{\pi } \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{5}{4}\right)}-\frac{4}{3} \\ 1 & \, _3F_2\left(-\frac{1}{4},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1\right) & 1 \\ 0 & 1 & 1 \\ -1 & \infty & \infty \\ -2 & -\frac{1}{3} & \frac{1}{4} \\ -3 & \infty & 0 \\ -4 & -\frac{3}{5} & -\frac{1}{4} \\ -5 & \infty & \infty \\ -6 & -\frac{3}{5} & \infty \\ -7 & \infty & -\frac{1}{6} \\ -8 & -\frac{65}{63} & -\frac{65}{112} \end{array}$$
For all further $n$ the functions either don't exist (which is denoted by $\infty$) or are rational numbers.
Values of $n$ for which $F_n$ diverges are all negative odd numbers (as far as experiments show):
$$n=-1,-3,-5,\dots$$
Values of $n$ for which $G_n$ diverges are:
$$n=-1,-5,-6,-9,-10,-13,-14,-17,-18,-21,-22,\dots$$
Except for $n=-1$ the other numbers seem to have the form $-(4k+1)$ and $-(4k+2)$. Where $k$ is a positive integer.

My attempts at general simplification (using Euler integral formulas) have not been successful so far. I will edit if I have more results.

Some comments.
Hypergeometric functions with negative integer parameters in the numerator are polynomials (which is why we have rational values) while negative integer parameters in the denominator lead to divergence, unless they are compensated by the numerator. By considering different cases for $n$, it's quite easy to see why the functions dirverge when they do.
$F_n$ diverges for odd negative $n$ because for $n=-2k-1$ we have:
$$F_{-2k-1}=\,_3F_2\left(\frac{1}{2},-k-\frac{1 }{2},\frac{-2k-3}{4} ;-k,\frac{-k+2}{2};1\right)$$
So the negative integer in the demoniators of the series will lead to divergence for any $k$.
On the other hand, let's consider $G_n$ for the values of $n$ where it apparently always diverges:
$$G_{-4k-1}=\,_3F_2\left(\frac{1}{2},-2k-\frac{1 }{2},-k-\frac{1}{2} ;-2k,-k+\frac{5}{4};1\right)$$
$$G_{-4k-2}=\,_3F_2\left(\frac{1}{2},-2k-1,-k-\frac{3}{4} ;-2k-\frac{1}{2},-k+1;1\right)$$
Again,  the negative integers in the denominators of the series are not compensated (in the second case since $-2k-1 \neq -k+1$ unless $k=-2$, though we were assuming $k$ to be a positive integer).
