$$x!/(x+1)! = 1/(x+1)$$
$$x!/(x+2)! = 1/(x+1)(x+2)$$
$$x!/(x+3)! = 1/(x+1)(x+2)(x+3)$$
$$ \vdots $$
in general $x!/(x+i)! = (x!/(x+i-1)!)(1/(x+i))$
but what would be the equation for $x!/(x+1/3)!$ ??
I tried expanding the first equations (1/polynomial) and trying to apply the pattern to non-integers. It didn´t work so much.
For example, in the polynomials formed by $(x+1)(x+2)\cdots(x+n)$ the first term will always be $x^n$, the second will be $(n(n+1)/2)x^n/x$ since it is the sum of choosing $(n-1)$ "$x$´s" so that you will end adding $1x^m+2x^m+3x^m+\cdots$, $m=n-1$. And so I approximated $x!/(x+3/2)!$ to $1/(x^1x^.5+(15/8)x^.5)$.
(Having trouble writing the equations if someone correct them that would be really nice), but I wasn´t able to anything else.
I also tried using the gamma function in the form of a convergent infinite sum, but nothing useful came out of that.
Ant thoughts would really appreciated.
One thing that comes to mind is the triplication formula $$ \Gamma(3z) ~=~\frac{3^{3z-\frac{1}{2}}}{2\pi}\Gamma(z)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right) \tag{6.1.19} $$ for the Euler Gamma function, cf. Abramowitz and Stegun, eq. (6.1.19), p. 256. If we define the notation $$z!~:=~\Gamma(z+1),$$ then eq. (6.1.19) becomes $$ (3z+2)! ~=~\frac{3^{3z+\frac{5}{2}}}{2\pi}z!\left(z+\frac{1}{3}\right)!\left(z+\frac{2}{3}\right)!~. $$ OP's expression then becomes $$\frac{z!}{\left(z+\frac{1}{3}\right)!} ~=~\frac{3^{3z+\frac{5}{2}}}{2\pi}\frac{(z!)^2}{(3z+2)! } \left(z+\frac{2}{3}\right)!~.$$
The reciprocal $$\frac{\left(z+\frac{1}{3}\right)!}{z!}~=:~(z\!+\!1)_{1/3}$$ is a generalized Pochhammer symbol.