Is there a closed formula for the generating function of some trinomial coefficients? We learn in calculus how to obtain a sum of binomial coefficients $\frac{(2d)!}{(d!)^2}$ in terms of a generating function 
$\sum_{d \geq 0} \frac{(2d)!}{(d!)^2} x^d$
by the Taylor series of $(1-4x)^{-1/2}$ at $x=0$. 
My question: is there a way to do this for trinomial coefficients? In particular what is 
$\sum_{d \geq 1} \frac{(3d-1)!}{(d!)^3} x^d =?$
I can't imagine this not being studied before, but can not find a specific answer after a few futile hours of searching. 
 A: What is "a way"? Of course, your question (in even more general form) was asked
centuries ago and gave rise to hypergeometric series, series of the form $\sum_n c_n$
with ratio $c_{n+1}/c_n$ being a rational function of index $n$. The most convenient
form is therefore the hypergeometric $_4F_3$ series expression given in Robert's answer.
Note that the hypergeometric function is not algebraic (i.e., transcendental) in
this case, so that no formula like $(1-4z)^{-1/2}$ can be given;
all algebraic hypergeometric instances are now tabulated thanks to
a fantastic result of Beukers and Heckman.
Dyson's famous 1962
paper Statistical theory of the energy levels of complex systems
originated the study of constant term identities. In particular,
Dyson's ex-conjecture states that for $a_1,\dots,a_n$ nonnegative integers
$$
\text{constant term}
\prod_{1\leq i\neq j\leq n} \biggl(1-\frac{x_i}{x_j}\biggr)^{a_i}
=\frac{(a_1+a_2+\cdots+a_n)!}{a_1!a_2!\cdots a_n!}\,.
$$
This could serve a different basis for other type generating functions.
A: Maple writes this as $2\ x\ {}_4F_3(1,1,4/3,5/3;\ 2,2,2;\ 27 x)$.
A: For what it is worth (probably very little!),
$$\sum_{d\ge 0}\binom{3d}{d,d,d} x^d$$
is the coefficient of $y^0z^0$ in
$$\frac{1}{1-(yz+1/y+1/z)^3x},$$
and the original series is $\frac13$ of the derivative.
