Enumerating (2+2)-free posets by the number of minimal elements and other statistics

TL;DR

This paper extends the enumeration of (2+2)-free posets by including additional statistics, especially the number of minimal elements, providing new generating functions and conjectures, with implications for related combinatorial structures.

Contribution

It derives new generating functions for (2+2)-free posets considering multiple statistics, including minimal elements, and offers an alternative proof for existing enumeration results.

Findings

Derived generating function for (2+2)-free posets with statistics

Established relation between p_{n,k} and enumeration with minimal elements

Connected enumeration results to permutations and chord diagrams

Abstract

An unlabeled poset is said to be (2+2)-free if it does not contain an induced subposet that is isomorphic to 2+2, the union of two disjoint 2-element chains. Let $p_n$ denote the number of (2+2)-free posets of size $n$. In a recent paper, Bousquet-M\'elou et al.\cite{BCDK} found, using so called ascent sequences, the generating function for the number of (2+2)-free posets of size $n$: $P(t)=\sum_{n \geq 0} p_n t^n = \sum_{n\geq 0} \prod_{i=1}^{n} (1-(1-t)^i)$. We extend this result in two ways. First, we find the generating function for (2+2)-free posets when four statistics are taken into account, one of which is the number of minimal elements in a poset. Second, we show that if $p_{n,k}$ equals the number of (2+2)-free posets of size $n$ with $k$ minimal elements, then $P(t,z)=\sum_{n,k \geq 0} p_{n,k} t^n z^k = 1+ \sum_{n \geq 0} \frac{zt}{(1-zt)^{n+1}} \prod_{i=1}^n (1-(1-t)^i)$. The second result cannot be derived from the first one by a substitution. On the other hand, $P(t)$ can easily be obtained from $P(t,z)$ thus providing an alternative proof for the enumeration result in \cite{BCDK}. Moreover, we conjecture a simpler form of writing $P(t,z)$. Our enumeration results are extended to certain restricted permutations and to regular linearized chord diagrams through bijections in \cite{BCDK,cdk}. Finally, we define a subset of ascent sequences counted by the Catalan numbers and we discuss its relations with (2+2)- and (3+1)-free posets.