Let $n \in \mathbb{N}$. Consider $(x,y,z)\in \mathbb{N}^3$ and admissible triple if $x \geq 4, y \geq 4, z \geq 4$ and $x+y+z=n$.
Also, let $\gamma(x,y,z)=2$ if $x,y$ and $z$ are all distinct, $\gamma(x,y,z)=1$ if two of $x,y,z$ are equal to eachother and $\gamma(x,y,z)=\frac{1}{3}$ if $x=y=z$.
For each admissible triple $(x,y,z)$, consider the product $\Gamma(x,y,z)=\gamma(x,y,z)(c_{x-1}-2)(c_{y-1}-2)(c_{z-1}-2)$ where $c_n$ is the $n$-th Catalan number.
Let $p(n) = \sum_{\text{admissible } (x,y,z)} \Gamma(x,y,z)$
Looking for alternate ways of describing $p(n)$, closed form expression would be great, recursive definition would be cool also.
Some calculations that I've done by hand:
$p(12)=\frac{1}{3}(c_3-2)^3$
$p(13)=(c_4-2)(c_3-2)^2$
$p(14)=(c_5-2)(c_3-2)^2+(c_4-2)^2(c_3-2)$
$p(15)=(c_6-2)(c_3-2)^2+2(c_5-2)(c_4-2)(c_3-2)+\frac{1}{3}(c_4-2)^3$
I'm very poor at this type of counting thing.. Help appreciated!!