I'm attempting to solve exercise 46(b) in Stanley's Enumerative Combinatorics Volume 1. The problem is stated as follows:
Let $g(n)$ be the number of sublattices of $B_{n}$ that contain $\varnothing$ and $[n]$ and let $f(n)$ be the number of rank $n$ sublattices of the boolean algebra $B_{n}$ (or equivalently the number of partial orders $P$ on $[n]$). Write $$F(x) = \sum_{n \ge 0}f(n)\frac{x^{n}}{n!}$$ $$G(x) = \sum_{n \ge 0}g(n)\frac{x^{n}}{n!}$$ Show that $G(x) = F(e^{x}-1)$
My Attempt: Clearly $$F(e^{x}-1) = \sum_{n \ge 0}f(n)\frac{(e^{x}-1)^{n}}{n!}$$ which is the $e^{e^{x}-1}$ generating function of $f(n)$. It is know that $$e^{e^{x}-1} = \sum_{n \ge 0}b(n)\frac{x^{n}}{n!}$$ where $b(n)$ is the $n$th bell number. It then follows $$F(e^{x}-1) = \sum_{n \ge 0}f(n)b(n)\frac{x^{n}}{n!}$$ so we need show that $f(n)b(n) = g(n)$ for all $n$.
The problem I'm having is that $f(n)b(n) = g(n)$ fails for $n = 2$ so I went wrong somewhere. $f(2) = 3$ and $b(2) = 2$ but $g(2) = 4$. Am I proceeding in the right direction to solve this problem or am I missing a key element in the proof?
As you've noted, it does not follow that $F(e^x - 1) = \sum f(n)b(n) x^n/n!$.
Think combinatorially about what the formula $G(x) = F(e^x - 1)$ tells you. The function $e^x - 1$ is the generating function for non-empty sets. Composing $F$ with $e^x - 1$ tells you to take $[n]$ and partition it into non-empty sets $[n] = S_1 \sqcup \cdots \sqcup S_k$ ($\sqcup =$ disjoint union) then take a sub-lattice of $B_k$ labeled with $S_1, \dots, S_k$.
How might we find such a decomposition? Given a lattice $L$ we want to find $S_1,\dots,S_k$. Note that $k$ must be the rank of $L$ (by definition of $f(k)$). So take a maximal chain $\mathcal C$ in $L$. We can write $\mathcal C$ as
$$ \emptyset \subset S_1 \subset S_1 \sqcup S_2 \subset \dots \subset S_1 \sqcup S_2 \sqcup \dots \sqcup S_k = [n]. $$
I believe this gives you $S_1, \dots, S_k$.