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

Kitaev, Sergey and Remmel, Jeffrey (2010) Enumerating (2+2)-free posets by the number of minimal elements and other statistics. In: 22nd International Conference on Formal Power Series & Algebraic Combinatorics, 2010-08-02 - 2010-08-06, San Francisco State University.

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Abstract

A 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. In a recent paper, Bousquet-Mélou et al. found, using so called ascent sequences, the generating function for the number of (2+2)-free posets: P(t)=∑n≥0 ∏i=1n ( 1-(1-t)i). We extend this result by finding 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. We also show that in a special case when only minimal elements are of interest, our rather involved generating function can be rewritten in the form P(t,z)=∑n,k ≥0 pn,k tn zk = 1+ ∑n ≥0zt(1-zt)n+1∏i=1n (1-(1-t)i) where pn,k equals the number of (2+2)-free posets of size n with k minimal elements. Résumé. Un poset sera dit (2+2)-libre s'il ne contient aucun sous-poset isomorphe à 2+2, l'union disjointe de deux chaînes à deux éléments. Dans un article récent, Bousquet-Mélou et al. ont trouvé, à l'aide de ``suites de montées'', la fonction génératrice des nombres de posets (2+2)-libres: c'est P(t)=∑n≥0 ∏i=1n ( 1-(1-t)i). Nous étendons ce résultat en trouvant la fonction génératrice des posets (2+2)-libres rendant compte de quatre statistiques, dont le nombre d'éléments minimaux du poset. Nous montrons aussi que lorsqu'on ne s'intéresse qu'au nombre d'éléments minimaux, notre fonction génératrice assez compliquée peut être simplifiée en P(t,z)=∑n,k ≥0 pn,k tn zk = 1+ ∑n ≥0 zt(1-zt)n+1∏i=1n (1-(1-t)i), où pn,k est le nombre de posets (2+2)-libres de taille n avec k éléments minimaux.