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An unlabeled poset is said to be (2+2)-free if it does not contain an inducedsubposet 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 recentpaper, Bousquet-M\'elou et al.\cite{BCDK} found, using so called ascentsequences, 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 generatingfunction for (2+2)-free posets when four statistics are taken into account, oneof 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 $k$ minimalelements, 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 bederived from the first one by a substitution. On the other hand, $P(t)$ caneasily be obtained from $P(t,z)$ thus providing an alternative proof for theenumeration result in \cite{BCDK}. Moreover, we conjecture a simpler form ofwriting $P(t,z)$. Our enumeration results are extended to certain restrictedpermutations and to regular linearized chord diagrams through bijections in\cite{BCDK,cdk}. Finally, we define a subset of ascent sequences counted by theCatalan numbers and we discuss its relations with (2+2)- and (3+1)-free posets.

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