How to express this series in closed form? $\sum_{i=1}^{\infty}\frac{(3i)!}{(i!)^3}x^{i}$
Motive of the generating function is to evaluate the number of the paths from the $(0,0,0)$ to $(n,n,n)$ not passing through $(i,i,i)$ ($1\leq i\leq n-1)$. The answer is coefficient of $x^n$ in $\sum_{k=1}^{n}(-1)^{k-1}\left(\sum_{i=1}^{n}\frac{(3i)!}{(i!)^3}x^{i}\right)^{k}.$