Let $r_0 := 1$, and for $n \geq 1$ let $r_n$ be the number of ways to do the following:
(i) divide $n$ into an even number of recipes,
(ii) bookmark one recipe from each group, AND
(iii) choose half of the bookmarked recipes and THEN decide the order in which we will cook them.
Let $R(x)$ be the exponential generating function (EGF) of $r_n$, $n \geq 0$. Find a closed form formula for $R(x)$.
Here is my work: Let $k = 2m$ be a fixed integer. Then the number of ways to split the $n$ recipes into $k$ groups is the coefficient of $\frac{x^n}{n!}$ of $A(x)^k$ where $A(x) = \frac{1}{1-x} - 1 - \frac{x}{1-x}$. Thus the EGF of doing (i) and then (ii) is, after we define $b_n := n$, $G(x) = B(A(x)) := \displaystyle\sum_{m=1}^\infty b_{2m}\frac{(A(x)^{2m}}{(2m)!} = \displaystyle\sum_{m=1}^\infty 2m \frac{(A(x)^{2m}}{(2m)!} = A(x) \sin A(x)$. For the task (iii), the EGF is $C(x) := \displaystyle\sum_{m = 1}^\infty \binom{2m}{m}m! \frac{x^{2m}}{(2m)!} = e^{x^2} - 1$. Thus, the formula should be $C(B(A(x))) = e^{(A(x)sinA(x))^2} - 1$.
Am I correct in my reasoning?