Two examples of unbalanced Wilf-equivalence

Abstract

We prove that the set of patterns {1324,3416725} is Wilf-equivalent to the pattern 1234 and that the set of patterns {2143,3142,246135} is Wilf-equivalent to the set of patterns {2413,3142}. These are the first known unbalanced Wilf-equivalences for classical patterns between finite sets of patterns. A pattern is an equivalence class of sequences under order-isomorphism. Two sequences �1 and �2 over totally ordered alphabets are order-isomorphic if, for any pair of positions i and j, we have �1(i) < �1(j) if and only if �2(i) < �2(j). We identify a pattern with its canonical representative, in which the kth smallest letter is k. We say that a permutation � contains a patternifhas a subsequence order-isomorphic to �, otherwise we say that � avoids �. We denote the set of permutations of length n by Sn, the set of permutations in Sn avoiding patternby Avn(�), and the set of permutations in Sn avoiding every pattern in a setby Avn(�). We say that two (sets of) patterns � ' and � '' are Wilf-equivalent, denoted � ' ∼ � '' , if |Avn(� ' )| = |Avn(� '' )| for all n ∈ N. We call a Wilf-equivalence unbalanced if the two sets of patterns do not contain the same number of patterns of each length. We will sometimes talk about the type of an unbalanced Wilf-equivalence, defined by the lengths of the patterns: for example, the two Wilf-equivalences proved in this paper have types (4) ∼ (4,7) and (4,4) ∼ (4,4,6). In Section 1, we prove that {1324,3416725} ∼ 1234. In Section 2, we prove that {2143,3142,246135} ∼ {2413,3142}. In Section 3, we conjecture a few other unbalanced Wilf-equivalences.