A poset is (3+1)-free if it does not contain the disjoint union of chains of length 3 and 1 as an induced subposet. These posets play a central role in the (3+1)-free conjecture of Stanley and Stembridge. Lewis and Zhang have enumerated (3+1)-free posets in the graded case by decomposing them into bipartite graphs, but until now the general enumeration problem has remained open. We give a finer decomposition into bipartite graphs which applies to all (3+1)-free posets and obtain generating functions which count (3+1)-free posets with labelled or unlabelled vertices. Using this decomposition, we obtain a decomposition of the automorphism group and asymptotics for the number of (3+1)-free posets.
- Pub Date:
- March 2013
- Mathematics - Combinatorics;
- 28 pages, 5 figures. New version includes substantial changes to clarify the construction of skeleta and the enumeration. An extended abstract of this paper appears as arXiv:1212.5356