For months, I've been trying to find a closed form for a generating function for this three-variable sequence
$$a(G,M,T)=(-1)^{G+T+M} 2^{M-1} \frac{(M-1)!}{(2M-2)!} \binom{M}{T} {}_3F_2(M+1, -\frac{G}{2},\frac{1-G}{2};M-\frac{1}{2},M-T+1;1)$$
where $G\geq0, T\geq0, M\geq2$ and $G\geq 2T-2M$.
If the generating function is $$f(x,y,z)=\sum_{G,M,T} a(G,M,T) x^G y^T z^M$$ what I have so far is this $$f(x,y,z)= -\frac{2 \sqrt{\pi}}{1+x} \sum_{M=2}^\infty \frac{\left(\frac{z}{2}(y-1)\right)^M}{\Gamma(M-\frac{1}{2})} {}_2F_1\left( \frac{1}{2},1;M-\frac{1}{2},\frac{x^2(1-y)}{(1+x)^2} \right)$$ and this $$ f(x,y,z) = \sum_{k=0}^{\infty} 2^{k-\frac{1}{2}} (1+x)^{-1-2k} (x-xy)^k (z(y-1))^{\frac{3}{2}-k} \left( - \Gamma\left(k+\frac{1}{2}\right) + \Gamma\left(k+\frac{1}{2}, \frac{z}{2}(y-1)\right)\right) $$ without any luck summing the remaining series in both cases.
In the first case, I can use the form given to me in this answer but only the first part sums out to a closed form.
Similarly, in the second case, the part with the gamma function can be summed out relatively easily, but the part with the incomplete gamma function becomes problematic.
I've tried everything I know using Mathematica, Maple and I digged through endless papers with various hypergeometric identities, but to no avail, I could really use some help.
Why do I even believe this can be written in closed form? I can't give you a satisfying answer, I guess I believe it enough to spend months of work on it, but I might be wrong.
I only see an integral representation (like the first one here but incomplete).
The first expression for $f(x,y,z)$ contains a sum of the form (after $M=n+2$) $$F(u,v)=\sum_{n=0}^\infty{}_2F_1\left(\frac12,1;n+\frac32;u\right)\frac{v^n\sqrt\pi}{\Gamma(n+3/2)}.$$ Substituting the integral representation $${}_2F_1\left(\frac12,1;n+\frac32;u\right)=\frac{\Gamma(n+3/2)}{n!\sqrt\pi}\int_0^1 t^{-1/2}(1-t)^n(1-ut)^{-1}\,dt$$ here, we get $$F(u,v)=\int_0^1 t^{-1/2}(1-ut)^{-1} e^{v(1-t)}\,dt.$$
A number of expansions can be obtained from here. Say, the geometric series $(1-ut)^{-1}$ results in $e^v$ times a series with incomplete gamma functions (not the one you have though). Another one comes from $(1-ut)^{-1}e^{-vt}=\sum_{n=0}^\infty(ut)^n\sum_{k=0}^n(-v/u)^k/k!$.