Part of Passing through a stack $k$ times with reversals by (Toufik Mansour, Howard Skogman and Rebecca Smith) 06 March 2019
The generating function M_2^U(x) (latex):
Let $t_0=C(x)$, $t_1=C(xC(x))$, and $t_2=C(xC(x)C(xC(x)))$. We have
\begin{align*}
M^U_2(x)&=\frac{2x^2\left(t_1t_0^2(2x-1)(t_0-2)(xt_0+x-1)^2(xt_0t_1+xt_0-1)t_2+\sum_{i=0}^4H_i(x)x^i\right)}{(xt_0t_1+xt_0-1)(xt_0+t_1-1)(t_0^2+xt_0-3t_0-2x+2)(xt_0+x-1)^2(2x-1)}\\
&=10x^5+160x^6+1636x^7+13704x^8+102876x^9+722772x^{10}+4867904x^{11}+\cdots,
\end{align*}
where
\begin{align*}
H_0(x)&=(2-t_0)(t_0^2+2t_0t_1-2t_0-t_1+1),\\
H_1(x)&=-2t_0^4t_1^2+3t_0^4t_1+6t_0^3t_1^2+3t_0^4-5t_0^2t_1^2-12t_0^3-9t_0^2t_1+2t_0t_1^2+11t_0^2-7t_0t_1+3t_0+4t_1-4,\\
H_2(x)&=2t_0^5t_1^2-6t_0^5t_1-2t_0^4t_1^2-3t_0^5-7t_0^3t_1^2+12t_0^4+14t_0^3t_1+7t_0^2t_1^2-3t_0^3+18t_0^2t_1-4t_0t_1^2-18t_0^2\\
&+3t_0t_1+5t_0-2t_1+2,\\
H_3(x)&=t_0\bigl(3t_0^5t_1-2t_0^4t_1^2+t_0^5+6t_0^4t_1+3t_0^3t_1^2-4t_0^4-14t_0^3t_1+4t_0^2t_1^2-6t_0^3-15t_0^2t_1-3t_0t_1^2+15t_0^2\\
&-12t_0t_1+2t_1^2+6t_0-4\bigr),\\
H_4(x)&=-t_0^2(4t_0^4t_1-3t_0^3t_1-3t_0^3-10t_0^2t_1+2t_0^2-t_0t_1+7t_0-2t_1-2).
\end{align*}