Counting Self-dual posets

Question

An exercise in Stanley's Enumerative Combinatorics (Chapter 3, ex. 8) asked for an example of a finite self-dual poset, (i.e. there is a bijection $f: P\to P$ such that $s\le t \Longleftrightarrow f(s) \ge f(t)$) for which there is no such bijection $f$ satisfying $f(f(t)) = t$ for all $t \in P$.

I'm interested in how many of these such posets exist for a given order. Has there been any result obtained about this? It seems like an interesting problem, but I struggled to find a single one (I would put the Hasse diagram here, but I'm not good enough at $\LaTeX$ to do that) , much less count them.

Answer 1

I posted this same question on MO, and this answer was given there.

There is a 12-element poset $P$ satisfying the stated condition. Let $Q$ be any self-dual poset that does not have $P$ as a connected component. Then the disjoint union $P + Q$ also satisfies the condition. Hence a lower bound for the number of $n$-element posets satisfying the condition is $d(n-12)-d(n-24)$, where $d(m)$ is the number of self-dual $m$-element posets. I suspect that this lower bound is not far from the truth. More specifically, if $f(n)$ is the number of $n$ element posets satisfying the condition, then it shouldn't be too hard to show that $\log f(n)\sim \log(d(n-12)-d(n-24))\sim \log d(n)$. I don't know whether anyone looked at the asymptotics of $d(n)$. Since the total number $g(n)$ of $n$-element posets satisfies $\log g(n)\sim \frac{n^2}{4}\log 2$, we easily get that $\log d(n)$ is asymptotically at least as large as $\frac{n^2}{16}\log 2$. This suggests the question: is $\log d(n)\sim \frac{n^2}{16}\log 2$? If not, what constant should replace $1/16$? Addendum. In fact, since $d(n-12)-d(n-24)<f(n)<d(n)$ and $d(n)>g(n/2)$, it follows that $\log f(n)\sim \log d(n)$.