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Discrete Mathematics 313 (2013) 2614–2625

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Discrete Mathematics

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Partial permutations avoiding pairs of patternsNoah ArbesfeldDepartment of Mathematics, Columbia University, New York, NY 10027, United States

a r t i c l e i n f o

Article history:Received 1 January 2013Received in revised form 1 August 2013Accepted 2 August 2013Available online 25 August 2013

Keywords:Pattern avoidancePartial permutationsWilf-equivalence

a b s t r a c t

We continue the study of pattern avoidance in partial permutations initiated by Claesson,Jelínek, Jelínkova, and Kitaev. We extend previous definitions of shape-Wilf-equivalenceand ⋆-Wilf-equivalence to sets of patterns, and determine new shape-Wilf-equivalencesand shape-⋆-Wilf-equivalences between pairs of patterns of length 3. Using these results,we find infinite families of Wilf-equivalences and ⋆-Wilf-equivalences between pairs ofpatterns. We also classify pairs of patterns of length up to 4 up to ⋆-Wilf-equivalence.

© 2013 Elsevier B.V. All rights reserved.

1. Introduction

We continue the study of pattern avoidance in partial permutations initiated by Claesson, Jelínek, Jelínkova, and Kitaevin [6]. We begin by recalling concepts from classical pattern avoidance. A permutation p of length n is an ordered list p1 · · · pnof {1, . . . , n} in which each element appears exactly once. Let Sn denote the set of permutations of length n. Given twopermutations p = p1 · · · pn and q = q1 · · · qm, the permutation p contains q if there exist indices 1 ≤ i1 < · · · < im ≤ n suchthat pij < pij′ if and only if qj < qj′ for all j, j′; in such a case the substring pi1 · · · pim is an occurrence of q. In this setting, qis called a pattern. If p does not contain q, the permutation p avoids the pattern q. If a permutation p avoids each pattern ina set A of patterns then p avoids the set A. Let Sn(A) denote the set of permutations in Sn avoiding A; two sets of patterns Aand B areWilf-equivalent if |Sn(A)| = |Sn(B)| for all n.

A central goal of the study of pattern avoidance is to classify sets of patterns up to Wilf-equivalence. For singleton setsof patterns, the following results are known. First, a pattern p = p1 · · · pn is trivially Wilf-equivalent to its reversal pn · · · p1,its complement (n + 1 − p1) · · · (n + 1 − pn), and its inverse q1 · · · qn, where qi satisfies pqi = i. Beyond these trivial Wilf-equivalences, Backelin, West, and Xin prove in [2] that the patterns 1 · · · n ⊕ x and n · · · 1 ⊕ x are Wilf-equivalent for anyx, and Stankova and West prove in [14] that the patterns 231 ⊕ x and 312 ⊕ x are Wilf-equivalent for any x, where, for twopermutations p and qwith |p| = m and |q| = n, the direct sum p⊕q is the permutation p1 · · · pm(q1 +m) · · · (qn +m). Thesefamilies ofWilf-equivalences were proven by showing that the patterns in question satisfy a stronger equivalence conditioncalled shape-Wilf-equivalence, a condition on the number of fillings of Ferrers diagrams that avoid permutation matrices.The only other knownWilf-equivalence is between 2314 and 2413, shown by Stankova in [13].

In [6], the study of pattern avoidance is expanded to the setting of partial permutations. We recall the necessarydefinitions from that paper. Introduce a symbol �, called a hole. Then, a partial permutation of {1, . . . , n} is an orderedlist π1 · · · πm of elements of {1, . . . , n, �} in which each of {1, . . . , n} appears exactly once. Then Skn is the set of partialpermutations of {1, . . . , n− k} with exactly k holes; for example S13 = {12�, 1 � 2, 21�, 2 � 1, �12, �21}. Let π = π1 · · · πn

be a partial permutation in Skn whose nonhole elements areπi1 , . . . , πin−k with i1 < · · · < in−k; then, a standard permutationp ∈ Sn is an extension of the partial permutation π1 · · · πn if the following statement holds: πij < πij′ if and only if pij < pij′for all j, j′. For example, the extensions of 1 � 32 are 2143, 1243, 1342, and 1432.

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N. Arbesfeld / Discrete Mathematics 313 (2013) 2614–2625 2615

Then, a partial permutation π avoids a pattern p if each extension of π avoids p, and a partial permutation π avoids a setA of patterns if π avoids each element of A. In this sense, a partial permutation π can be thought of as encoding the set of itsextensions. We let Skn(A) denote the set of all partial permutations in Skn that avoid A, and let skn(A) = |Skn(A)|. When writingan expression of the form Skn(A) we omit the brackets around the set A.

This notion of pattern avoidance for partial permutations enables us to extend the notion of Wilf-equivalence to thesetting of partial permutations. Two sets A and B of patterns are k-Wilf-equivalent if skn(A) = skn(B) for all n, and A and Bare ⋆-Wilf-equivalent if they are k-Wilf-equivalent for k ≥ 0. In particular, 0-Wilf-equivalence is the same as the standardnotion of Wilf-equivalence.

In [6], the authors find that results from pattern avoidance have analogues which hold in the setting of partialpermutations. They find ⋆-Wilf-equivalences between patterns by proving that the patterns in question satisfy a conditioncalled shape-⋆-Wilf-equivalence, an extension of shape-Wilf-equivalence that imposes a condition on the number ofso-called partial fillings of diagrams that avoid permutation matrices. Namely, the authors show that the shape-Wilf-equivalence between 1 · · · n ⊕ x and n · · · 1 ⊕ x and the shape-Wilf-equivalence between 231 ⊕ x with 312 ⊕ x extend toshape-⋆-Wilf-equivalences. So, all known Wilf-equivalences extend to ⋆-Wilf-equivalences with the following exceptions:a pattern is not necessary ⋆-Wilf-equivalent to its inverse, and the patterns 2314 and 2413 are not ⋆-Wilf-equivalent. Weremark that the condition shape-Wilf-equivalencewas recently extended in [3] to the setting of vincular patterns, a differentgeneralization of pattern avoidance.

In the present work, we find ⋆-Wilf-equivalences between pairs of patterns, with a focus on determining which resultsfrom Wilf-equivalence can be extended to ⋆-Wilf-equivalences. A set of patterns is trivially ⋆-Wilf-equivalent to the setsformed by taking the complements and reversals of the set’s patterns; we seek nontrivial ⋆-Wilf-equivalences.

In Section 2, we recall the definition of shape-Wilf-equivalence, and find shape-Wilf-equivalences among pairs ofpatterns of length 3. Namely, we prove the following.

Theorem 1.1. The following seven pairs of patterns of length 3 are shape-Wilf-equivalent:

{123, 213}, {132, 231}, {132, 312}, {213, 231}, {213, 312}, {231, 321} and {312, 321}.

We remark that a complete classification of pairs of patterns of length 3 up to shape-Wilf-equivalence was obtainedindependently by Bloom and Elizalde in [4]. The authors also enumerate matchings and set partitions that avoid pairs ofpatterns of length 3 for most such pairs of patterns.

In Section 3, we extend the definition of shape-⋆-Wilf-equivalence from [6] to include pairs of patterns and generalize aresult from [6, Section 3] to sets of patterns, enabling us to find infinite families ofWilf-equivalences and ⋆-Wilf-equivalencesbetween pairs of patterns.

In Section 4, we classify pairs of patterns of length 3 up to shape-⋆-Wilf-equivalence. Namely, we prove the following.

Theorem 1.2. The shape-⋆-Wilf-equivalences among pairs of patterns of length 3 are as follows:

• {123, 213} and {231, 321},• {132, 231} and {213, 312},• {132, 312} and {213, 231}.

By Proposition 3.1, we deduce the following corollary.

Corollary 1.3. Given a set {pi} of patterns, the following sets of patterns are ⋆-Wilf-equivalent:

• {132 ⊕ pi, 231 ⊕ pi} and {213 ⊕ pi, 312 ⊕ pi},• {132 ⊕ pi, 312 ⊕ pi} and {213 ⊕ pi, 231 ⊕ pi},• {123 ⊕ pi, 213 ⊕ pi} and {231 ⊕ pi, 321 ⊕ pi}.

Whereas all known shape-Wilf-equivalent (single) patterns are shown in [6] to be shape-⋆-Wilf-equivalent, we deduceas a consequence of Theorems 1.1 and 1.2 that there exist shape-Wilf-equivalent sets of patterns that are not shape-⋆-Wilf-equivalent.

In Section 5, we determine the ⋆-Wilf-equivalence classes among pairs of short patterns A = {p, q} for p and q of lengthat most 4. When p or q has length 2 the classification is straightforward.

The classification of pairs of patterns {p, q} with |p| = |q| = 3 up to Wilf-equivalence is stated in [11]. Up to trivial⋆-Wilf-equivalence, there are six such pairs, distributed across three 0-Wilf-equivalence classes. Among these six pairs,there is only one nontrivial ⋆-Wilf-equivalence. Namely, we have the following.

Theorem 1.4. Among pairs of patterns {p, q} with |p| = |q| = 3, the only nontrivial ⋆-Wilf-equivalence is

• {123, 132} and {132, 231}.

The classification up to Wilf-equivalence of pairs of patterns {p, q} with |p| = 3, |q| = 4, and p not contained in q, islisted in [1,16]. Up to trivial ⋆-Wilf-equivalence, there are 24 such pairs (note that some inverses to the pairs listed in [1,16]must also be considered).

2616 N. Arbesfeld / Discrete Mathematics 313 (2013) 2614–2625

Theorem 1.5. Among pairs of patterns {p, q} with |p| = 3 and |q| = 4 with p not contained in q, the nontrivial ⋆-Wilf-equivalences are

• {321, 1342} and {321, 1423},• {321, 2341}, {132, 1234}, {132, 3241}, {132, 3124}, {132, 2134}, and {132, 3412},• {321, 3412}, {321, 3142}, {321, 2413}, {132, 2341}, and {132, 2314}.

The classification of pairs of patterns of length 4 up to Wilf-equivalence is completed in [10], building on work donein [5,7–9,13,15]. In the present work we find that most nontrivially Wilf-equivalent pairs of patterns of length 4 are not⋆-Wilf-equivalent. Namely, the following theorem gives the only nontrivial ⋆-Wilf-equivalences.

Theorem 1.6. Among pairs of patterns {p, q} with |p| = |q| = 4, the nontrivial ⋆-Wilf-equivalence classes are

• {4123, 3214} and {1234, 2143},• {1324, 2314} and {2134, 3124},• {1324, 3124}, {2413, 4213}, and {2134, 2314},• {1234, 2134}, {1342, 2341}, {2314, 3214}, and {1243, 2143}.

We conclude the introduction by mentioning that the present work emerged from an investigation into the questionsposed at the end of [6]. Based on computations, we provide the following conjectures. The first is an analogue of theStanley–Wilf theorem in the setting of partial permutations.

Conjecture 1.7. If p is a pattern then limn→∞ns1n(p) exists and is finite.

In particular, [6, Theorem8.1] reduces this conjecture to the statement that s1m(p)s1n(p) ≤ s1m+n(p) for allm and n. Preliminarycomputations suggest that limn→∞

nskn(p) exists and is finite for all patterns p and all integers k.

We also state the following conjecture.

Conjecture 1.8. The patterns k-Wilf-equivalent to the permutation 12 · · · (k + 3) are exactly the layered permutations in Sk+3.

It would also be interesting to investigate which permutations are ⋆-Wilf-equivalent to their inverses, as this is the one ofthe few cases in which Wilf-equivalent permutations are not ⋆-Wilf-equivalent.

2. Shape-Wilf-equivalences for pairs of patterns of length 3

We begin by recalling the definition of shape-Wilf-equivalence.Let D = (λ1, . . . , λn) with λ1 ≥ · · · ≥ λn ≥ 1 denote a bottom-left justified array of boxes, with λi boxes in the i-th row

from the bottom. Such an array D is a Ferrers diagram. We number the rows of D from bottom to top and the columns fromleft to right (the so-called French notation). We allow the possibility that D is an empty list, in which case D is the emptyFerrers diagram. Note that our definition of a Ferrers diagram differs from that given in [6] as we exclude the possibility ofcolumns of zero height; the inclusion of such diagrams does not require significant modification of our proofs.

A transversal of D is a filling of each cell of D with either a 0 or a 1 such that each row and column of D contains exactlyone 1. By convention, we say that the empty diagram has exactly one transversal. A Ferrers diagram D = (λ1, . . . , λn) isfillable if λ1 = n and λi > n − i for i = 1, . . . , n or if D is empty; in other words, a Ferrers diagram is fillable if and only if ithas a transversal.

Wenow recall the notion of pattern avoidance for transversals of diagrams. The permutationmatrix associated to a patternp = p1 · · · pn is the transversal of the square diagram (n, . . . , n) with a 1 in the ith column from the left and pi-th row fromthe bottom for i = 1, . . . , n; when there is no confusion we also let p denote the permutation matrix of the pattern p. Atransversal F of Ferrers diagram D contains a pattern p if there is an n by n submatrix of F that is equal to the permutationmatrix p; i.e., if one may choose i1 < i2 < · · · < in and j1 < j2 < · · · < jn such that for all 1 ≤ l, k ≤ n there is a cell inthe il-th row and jk-th column of F , and the entry found in this cell is the same as the entry found in the l-th row and k-thcolumn of the matrix p. A transversal F avoids p if it does not contain p, and avoids a set A if it avoids each element of A.

Given a Ferrers diagram D and a set A of patterns, let SD(A) denote the set of transversals of D that avoid each element ofA and let sD(A) = |SD(A)|. Two sets of patterns A and B are shape-Wilf-equivalent if sD(A) = sD(B) for any diagram D. Notethat two sets A and B are Wilf-equivalent if sD(A) = sD(B) for all square diagrams D; in particular, shape-Wilf-equivalenceis a stronger condition than Wilf-equivalence.

Up to the present work, the following shape-Wilf-equivalences between patterns are known: in [2] it is shown that12 · · · n ⊕ p and n · · · 21 ⊕ p are shape-Wilf-equivalent for any pattern p, and in [14] it is shown that 312 ⊕ p and 231 ⊕ pare shape-Wilf-equivalent for any pattern p.

We now prove Theorem 1.1. First, given a fillable Ferrers diagram D = (λ1, . . . , λn), set e(D) to be the number of indicesi for which λi > n + 1 − i. As D is fillable, we know that 0 ≤ e(D) < n. In other words, e(D) is the number of rows of Dwhose length exceeds the length of the corresponding row of the so-called staircase Ferrers diagram, (n, n − 1, . . . , 1). Forexample, e((4, 4, 3, 1)) = 2.


Fig. 1. A Ferrers diagram D with n = 7 and λ4 = 4. The top and right parts (in the language of Case 1) are labelled. The remaining boxes comprise thebottom part.

Proposition 2.1. Let D be a fillable Ferrers diagram with e(D) ≥ 0. Then,

sD(123, 213) = sD(132, 231) = sD(132, 312) = sD(231, 321) = sD(312, 321) = 2e(D).

Proof. Our argument is similar to that deduced independently in [4]. Let D = (λ1, . . . , λn) be a nonempty fillable Ferrersdiagram. For each pair A listed in the theorem, we use strong induction on the number of rows n of D to count sD(A). Thebase case n = 1 is clear. So suppose n > 1. We first compute sD(132, 231). Consider filling in D starting from the top rowto obtain a {132, 231}-avoiding transversal. If λn = 1, then there is only one choice for the 1 in the uppermost row. Then,if D′ denotes the Ferrers diagram obtained by removing the column and row containing this 1, by induction there are 2e(D′)

ways of filling in the remainder of D to obtain a transversal that avoids {132, 231}. As λn = 1, one has e(D) = e(D′) so thatsD(132, 231) = 2e(D).

If λn > 1, suppose the 1 in the top row is placed in the kth box from the left. If 1 < k < λn, then the 3 by 3 subdiagramof F containing the 1s in the 1st, kth, and λnth columns contains 132 or 231. So k = 1 or k = λ1; in either case the 1in the uppermost row cannot be part of an occurrence of 132 or 231 regardless of how the rest of the diagram is filledin. If D′ denotes the Ferrers diagram obtained by removing the column and row containing this 1, then by the inductivehypothesis there are 2e(D′) ways of filling in the remainder of D to obtain a transversal that avoids 132 and 231. HencesD(132, 231) = 2 · 2e(D′)

= 2e(D), completing the proof for {132, 231}.Similar reasoning can be used to compute sD(123, 213) and sD(312, 321). Reflecting D about its bottom-left to upper-

right diagonal (in other words, sending D = (λ1, . . . , λn) to (λ′

1, . . . , λ′n) where λ′

i is the largest j for which λj ≥ i) leavese(D) unchanged. The computation then follows for the pairs {132, 312} and {231, 321}. �

Proposition 2.2. Let D be a fillable Ferrers diagram with e(D) ≥ 0. Then,

sD(213, 231) = sD(213, 312) = 2e(D).

Proof. We compute sD(213, 312), using a different method from that in [4]. We consider two cases:Case 1: There is some k > 1 for which λk = n−k+1. For this case only we introduce the following notation: the uppermostn− k+ 1 rows comprise the top part of D, denoted by Dt , the rightmost k− 1 columns comprise the right part of D, denotedby Dr , and the remaining (k − 1) × (n − k + 1) rectangle is the bottom part of D (see Fig. 1). Then, in a transversal of D,the top part must contain 1 exactly n − k + 1 times. But the top part consists of exactly n − k + 1 columns; it followsthat in any transversal of D the bottom part must consist of all zeros and the ones in the bottommost k − 1 rows must bein the right part of F . Thus a {213, 312}-avoiding transversal of D is given by a {213, 312}-avoiding transversal of Dt and a{213, 312}-avoiding transversal of Dr . As e(Dt) + e(Dr) = e(D), by induction we have

sD(213, 312) = sDt (213, 312) · sDr (213, 312) = 2e(Dt )2e(Dr ) = 2e(D),

completing Case 1.Case 2: For all k > 1, we have λk > n − k + 1. Fix some i ∈ {1, . . . , n}. We count all {213, 312}-avoiding transversals of Dwith a 1 in the i-th cell from the left in the bottom row. Let Di denote the Ferrers diagram obtained by removing the bottomrow and the i-th column from the left.

For this case only we introduce the following notation: the rightmost n− i columns comprise the right part ofDi, denotedbyDi

r , the rows ofDi lying entirely aboveDir comprise the top part ofDi, and the remaining squares ofDi comprise the bottom


Fig. 2. The Ferrers diagram Di , where D is a diagram with n = 7, λk > n − k + 1 for k > 1, and i = 3. The row and column containing the 1 in the bottomrow are removed, and the top and right parts (in the language of Case 2) are labelled. The bottom part is comprised of the remaining squares.

part of Di; the bottom part consists of the squares of the rightmost i − 1 columns lying below the top part (see Fig. 2). Atransversal of D with a 1 in the ith cell from the left in the bottom row is given by a transversal of Di. If this transversal ofDi contains a 1 in the bottom part, then, in the resulting transversal of Di, this 1 in bottom part of Di, the 1 in the bottomrow of D, and the 1 in the (i + 1)th column of D will form an occurrence of 213 or 312. Therefore, the bottom part mustconsist entirely of 0s. It follows that a partial transversal of Di that induces {213, 312}-avoiding transversal of D is given bya {213, 312}-avoiding transversal of Di

t and a {213, 312}-avoiding transversal of Dir , so that

sD(213, 312) =

ni=1

sDit(213, 312) · sDi

r(213, 312).

We compute the right-hand side of this sum. When 1 < i < n, the right part Dir has n − i columns. So, as λn−i+2 > i − 1,

the right part Dir has a transversal if and only if λn−i+2 = i. When i = n, the diagram Di

r is empty and therefore has onetransversal.

Now, suppose λn−i+2 = i for some i. Then, e(Dir) is the number of j such that 1 ≤ j ≤ n − i satisfying λj+1 > n − j + 1

(for i = n this number is 0). Also, by assumption, e(Dit) = i − 2 for i > 1 and e(Di

t) = 0 for i = 1.So, let i1 = 1 and 1 < i2 < · · · < im = n be the indices i for which λn−i+2 = i. Then e(D

ijr ) = (n − ij) − (m − j). So, for

1 < j ≤ m, we have by induction

sDij (213, 312) = sDijt(213, 312) · s

Dijr(213, 312) = 2ij−22n−ij−(m−j)

= 2n−m+j−2,

and also

sD1(213, 312) = sD1r(213, 312) = 2n−m.

We then compute

sD(213, 312) =

ni=1

sDi(213, 312) = 2n−m+

mj=2

2n−m+j−2= 2n−1

completing the induction.Reflecting about the diagonal we conclude that the same computation holds for the pair {213, 231}. �

From Propositions 2.1 and 2.2 we deduce Theorem 1.1. Counting sD(A) for all pairs A of patterns of length 3 and for alldiagrams D with at most five rows, we see that the each of the six pairs {123, 132}, {123, 231}, {123, 312}, {123, 321},{123, 321} and {213, 321} belong to their own shape-Wilf-equivalence class. In an earlier version of this work, weconjectured that the remaining two pairs, {132, 213} and {231, 312}, belong to the same shape-Wilf-equivalence class asthe pairs in Theorem 1.1. This conjecture was recently resolved in [4], where a complete classification of pairs of patterns oflength 3 up to shape-Wilf-equivalence is given.

We conclude this section with a partial generalization of Theorem 1.1 analogous to [6, Lemma 5.6]. We introduce arefinement of pattern-avoiding fillings. Let A be a set of patterns; let p be a pattern, and let D be a Ferrers diagram with krows of maximum length. Then, define SD(A)p to be the set of all A-avoiding transversals whose bottom k rows avoid p.


Fig. 3. A substitution into a �-column of a partial transversal of a Ferrers diagram.

Proposition 2.3. If D is a fillable Ferrers diagram with k rows of maximum length, then

|SD(132, 231)12| = |SD(213, 312)21| = 2e(D)−k+1.

Proof. For square D, the result is clear. For non-square D, the desired result follows from inducting on the number of rowsof D of sub-maximum length and using methods similar to those used in Propositions 2.1 and 2.2 to compute sD(132, 231)and sD(213, 312). �

We will use this result in Section 4.

3. An extension of a result of Claesson et al. to sets of patterns

We recall from [6] the definition of a partial transversal of a Ferrers diagram and state a straightforward extensionof [6, Proposition 3.1] that will enable us to find infinite families of ⋆-Wilf-equivalent pairs. Given a Ferrers diagramD = (λ1, . . . , λn), a partial transversal of D is a filling of each box with a 0, 1, or � that satisfies the following two conditions:(1) each column either consists entirely of holes (called a �-column) or consists entirely of 0s and 1s, and (2) each row andeach column that is not a �-column contains exactly one 1.

We now recall from [6] the notion of an extension of a partial transversal of D. Let F be a filling of a diagram D with a�-column c of height d. Then, a substitution into c is defined as follows: first, some number 0 ≤ k ≤ d is chosen; then a rowis inserted between row k−1 and k that is at most as long as row k−1 and at least as long as row k (at this stepwe take rows0 and n + 1 to have lengths λ1 and 1 respectively). Finally, a 1 is placed in column c of this new row, while the remainingentries of this row are filled with 0s, and each remaining entry of column c is filled with a 0. The result of a substitutioninto a partial transversal of a diagram is a partial transversal of a new diagram with one fewer �-column (see Fig. 3). If a(non-partial) transversal F ′ can be obtained by a series of substitutions from a partial transversal F of a Ferrers diagram D,then F ′ is said to be an extension of F . A partial transversal F of a Ferrers diagram D is said to avoid a set of patterns A if eachextension of F avoids each element of A.

Given a set A of patterns, a Ferrers diagram D, and a set C of columns of D, let |SD,C (A)| be the number of A-avoidingpartial transversals of D whose �-columns are the elements of C . Two sets A and B of patterns are said to be shape-⋆-Wilf-equivalent if, for any Ferrers diagram D and any set C of columns of D, we have |SD,C (A)| = |SD,C (B)|. If two sets of patternsare shape-⋆-Wilf-equivalent, then the two sets are shape-Wilf-equivalent, and, as long as the minimum length of a patternin each set is the same, the two sets are also ⋆-Wilf-equivalent (this additional condition can be dropped if columns of zeroheight are allowed). In [6] it is shown that {12 · · · n} and {n · · · 21} are shape-⋆-Wilf-equivalent, and that {312} and {231}are shape-⋆-Wilf-equivalent.

We now state a generalization of [6, Proposition 3.1] that will enable us to find infinite families of ⋆-Wilf-equivalentpatterns.

Proposition 3.1. Suppose the sets {pi}i∈I and {qj}j∈J are shape-⋆-Wilf-equivalent sets of patterns, and let {xk} be an arbitrary setof patterns. Then, the two sets {pi ⊕ xk}i∈I,k∈K and {qj ⊕ xk}j∈J,k∈K are shape-⋆-Wilf-equivalent.

We give a short sketch of the proof, which is a straightforward modification of the proof of [6, Proposition 3.1]. Namely,given a set {xk} of permutation matrices, define M({xk}) to be the union of the partial subfillings M(xk), where M(xk) is asdefined in the proof of [6, Proposition 3.1]. In the proof of [6, Proposition 3.1], replacing each instance of P with {pi}, eachinstance of Q with {qj}, each instance of X with {xk} and each instance ofM(X) withM({xk}) yields the desired result.

A similar modification to [2, Proposition 2.3] implies that Proposition 3.1 holds when the symbol ⋆ is removed. Hence,we also have the following corollary to Theorem 1.1.

Corollary 3.2. For any set of patterns {xk}, the following sets of patterns are Wilf-equivalent:

{123 ⊕ xk, 213 ⊕ xk}, {132 ⊕ xk, 231 ⊕ xk}, {132 ⊕ xk, 312 ⊕ xk}, {213 ⊕ xk, 231 ⊕ xk},{213 ⊕ xk, 312 ⊕ xk}, {231 ⊕ xk, 321 ⊕ xk}, {312 ⊕ xk, 321 ⊕ xk}.


Fig. 4. A Ferrers diagram with �-column chosen. The left, right, top, middle, bottom, lower, and upper parts are labelled.

4. Shape-⋆-Wilf-equivalences among pairs of patterns of length 3

In this section, we prove Theorem 1.2. Before proving the theorem, we introduce terminology to be used in this sectiononly. Let D be a Ferrers diagram with one �-column chosen. The left part is comprised of the columns strictly to the left ofthe �-column, and the right part, denoted by Dr , is comprised of the columns strictly to the right of the �-column. The lowerpart is the portion of the left part lying in rows that intersect the �-column. The lower part is divided into the bottom part,the portion of the lower part lying in rows that intersect the right part, and themiddle part, the portion lying in rows whoserightmost cell lies in the �-column. The top part is comprised of the rows that lie strictly above the �-column and the upperpart, denoted by Du, is the union of the top and middle parts (see Fig. 4). Note that this terminology is different from thatused in the proofs of Theorem 1.1 and [6, Theorem 5.1].

We first show that the pairs listed in Theorem 1.2 are shape-⋆-Wilf-equivalent. We consider each shape-⋆-Wilf-equivalence separately.

Proposition 4.1. The pairs {132, 312} and {213, 231} are shape-⋆-Wilf-equivalent.

Proof. Let D be a Ferrers diagram with columns chosen to be �-columns. We show that there is an equal number of partialtransversals with the given �-columns that avoid each pair. If D has two or more �-columns chosen, then any partial fillingeither contains all permutations of length 3 or none of them. When D has zero �-columns chosen the result follows fromTheorem 1.1. It remains to consider the case where a partial transversal F has only one �-column.

We begin with characterizations of {132, 312}-avoiding transversals and {213, 231}-avoiding transversals. Thesecharacterizations and their proofs are analogous to [6, Obs. 5.2, 5.3].

A partial transversal F of Dwith one �-column avoids {132, 312} if and only if F satisfies the following four conditions:

(A1) Any 2 by 2 subfilling of F containing two 1s lies entirely in the left part.(A2) The subfilling induced by the left part of F avoids {132, 312}.(A3) The subfilling induced by the right part of F avoids 12 and 21.(A4) The subfilling induced by the lower part of F avoids 12 and 21.

On the other hand, a partial transversal F of D with one �-column avoids {213, 231} if and only if F satisfies the followingfour conditions:

(A1) Any 2 by 2 subfilling of F containing two 1s lies entirely in the left part.(A2′) The subfilling induced by the left part of F avoids {213, 231}.(A3) The subfilling induced by the right part of F avoids 12 and 21.(A4) The subfilling induced by the lower part of F avoids 12 and 21.

Now, let D be a Ferrers diagram, with a �-column chosen such that there is at least one {132, 312}-avoiding partialtransversal. Furthermore, suppose D has k columns in the right part. Then the right part of D must have at least k rows.But (A1) implies that no partial transversal of D can have a 1 in the bottom part, so that D must have exactly k rows. Fromthe proof of [6, Fact 5.4], which characterizes 12-avoiding and 21-avoiding transversals ofD, we deduce that (A3) implies thatDr must be the staircase Ferrers diagram (k, k− 1, . . . , 1); call this diagram Ek. Condition (A4) then implies that the middlepart consists of at most one row. The same conclusions hold if D is a Ferrers diagram with at least one {213, 231}-avoidingpartial transversal and k columns in the right part.


Conversely, if the right part of D is Ek for some k, and the middle part of D has at most one row, then a {132, 312}-avoiding partial transversal of D is given by a {132, 312}-avoiding transversal of Du and the unique transversal of Ek. By thesame reasoning, a {213, 231}-avoiding partial transversal of such a D is given by a {213, 321}-avoiding transversal of Du andthe unique transversal of Ek. By Theorem 1.1, we have sDu(132, 312) = sDu(213, 231), so that {132, 312} and {213, 231} areshape-⋆-Wilf-equivalent. �


Proof. By Theorem 1.1, the pairs {132, 231} and {213, 312} are shape-Wilf-equivalent. Again it remains to consider partialtransversals of Dwith one �-column chosen. A partial transversal F of Dwith one �-column avoids {132, 231} if and only ifF satisfies the following four conditions:

(B1) Any 2 by 2 subfilling of F containing two 1s lies entirely in the left part or entirely in the right part.(B2) The subfilling induced by the left part of F avoids {132, 231}.(B3) The subfilling induced by the right part of F avoids 21.(B4) The subfilling induced by the lower part of F avoids 12.

On the other hand, a partial transversal F of D with one �-column avoids {213, 312} if and only if F satisfies the followingfour conditions:

(B1) Any 2 by 2 subfilling of F containing two 1s lies entirely in the left part or entirely in the right part.(B2′) The subfilling induced by the left part of F avoids {213, 312}.(B3′) The subfilling induced by the right part of F avoids 12.(B4′) The subfilling induced by the lower part of F avoids 21.

Suppose D has a �-column chosen such that there is at least one {132, 231}-avoiding partial transversal. Then, (B1) impliesthat if the right partDr has k columns thenDr also has k rows. A {132, 231}-avoiding partial transversal is therefore given by a{132, 231}-avoiding (non-partial) transversal of Du such that themiddle part avoids 12, and a 21-avoiding transversal of Dr ;similarly, a {213, 312}-avoiding partial transversal is given by a {213, 312}-avoiding transversal of Du such that the middlepart avoids 21, and a 12-avoiding transversal of Dr . But, by Proposition 2.3 we have |SDu(132, 231)12| = |SDu(213, 312)21|and by [6, Fact 5.4] we have sDr (21) = sDr (12) = 1. So, the pairs {132, 231} and {231, 312} are shape-⋆-Wilf-equivalent. �


Proof. By Theorem 1.1, the pairs {132, 312} and {213, 231} are shape-Wilf-equivalent. Again it remains to consider partialtransversals of Dwith one �-column chosen. A partial transversal F of Dwith one �-column avoids {123, 213} if and only ifF satisfies the following three conditions:

(C1) A 2 by 2 subfilling of F containing 12 must be contained entirely in the left part.(C2) The subfilling induced by the left part of F avoids {123, 213}.(C3) The subfilling induced by the lower part of F avoids 12 and 21.

A partial transversal F of D with one �-column avoids {231, 321} if and only if F satisfies the following three conditions:

(C1′) A 2 by 2 subfilling of F containing 21 must be contained entirely in the left part.(C2′) The subfilling induced by the left part of F avoids {231, 321}.(C3) The subfilling induced by the lower part of F avoids 12 and 21.

Suppose D has at least one {123, 213}-avoiding transversal and the right part of D has k columns. Then (C3) implies that theright part of D has k or k+ 1 rows. The same holds if D has at least one {231, 321}-avoiding transversal and the right part ofD has k columns.

Suppose first that the right part has k rows. Then, D has a {123, 213}-avoiding partial transversal only if the middle partof D contains at most one row. In this case a {123, 213}-avoiding partial transversal of D is given by a {123, 213}-avoiding(non-partial) transversal of Du and a 12-avoiding transversal of Dr ; similarly a {231, 321}-avoiding partial transversalof D is given by a {231, 321}-avoiding transversal of Du and a 21-avoiding transversal of Dr . By Theorem 1.1 we havesDu(123, 213) = sDu(231, 321); it follows that if the right part ofD has k rows then there are an equal number of {123, 213}-avoiding partial transversals of D and {231, 321}-avoiding partial transversals of D.

Now suppose that the right part ofD = (λ1, . . . , λn) has k+1 rows, that the right part ofD is (λr1, . . . , λ

rk+1), and that the

top part of D is (λt1, . . . , λ

tl ). In particular λr

1 = k. Condition (C1) implies that, in a {123, 231}-avoiding partial transversalof D, the 1s in Dr lie in the bottom k rows. It follows that D has a partial transversal only if the middle part is empty. In thiscase, a transversal of D is given by a {123, 231}-avoiding (non-partial) transversal of (λ1 − (k + 1), λt

1, . . . , λtl ), i.e., the top

part adjoined with the top row from the bottom part, and a 12-avoiding transversal of the bottom k rows of Dr .If the right part of D has k + 1 rows, then in a {231, 321}-avoiding transversal, the middle part must be empty. Our

characterization of {231, 321}-avoiding transversals of D implies that such a transversal can be determined uniquely from a


Fig. 5. A {231, 321}-avoiding transversal of a Ferrers diagram D constructed from a {231, 321}-avoiding transversal F t+and a 21-avoiding transversal F+

r .

21-avoiding transversal F r+of (λr

1 + 1, . . . , λrk+1 + 1) and a {231, 321}-avoiding transversal F t

+of (λ1 − (k+ 1), λt

1, . . . , λtl )

in the followingmanner. First, fill in the top part of Dwith the subfilling induced by all but the bottommost row of F+

t . Then,fill in the right part of Dwith the subfilling induced by all but the leftmost column of F+

r . Finally, if the 1 in the bottommostrow of F+

t is in the i-th column, and the 1 in the leftmost column of F+r is in the j-th row, then place a 1 in the i-th column

and j-th row of D (see Fig. 5).By Theorem 1.1, the diagram (λ1 − (k + 1), λt

1, . . . , λtl ) has an equal number of {123, 231}-avoiding transversals and

{231, 321}-avoiding transversals, and by [6, Fact 5.4], there exists exactly one 12-avoiding transversal of the bottom k rowsof Dr , and there exists exactly one 21-avoiding transversal of (λr

1 + 1, . . . , λrk+1 + 1). So, we deduce the shape-⋆-Wilf-

equivalence of {123, 231} and {231, 321}. �

Computing sD(A) for D with at most five columns and pairs A of patterns of length 3, we see that the list of shape-⋆-Wilf-equivalences in the statement of Theorem 1.2 is exhaustive, completing the proof. This fact that there are no further shape-⋆-Wilf-equivalences among pairs of patterns of length 3 can also be deduced from Corollary 1.3 and Theorem 1.6.

5. Computation of ⋆-Wilf-equivalences among pairs of short patterns

In this section, we prove Theorems 1.4–1.6. Given two strings of integers P and Q , we let PQ denote the concatenation ofthese two strings and we say that P > Q if every element of P is greater than every element of Q . We prove each theoremseparately.

5.1. Proof of Theorem 1.4

We consider pairs {p, q} with |p| = |q| = 3.

Proposition 5.1. The pair {123, 132} is ⋆-Wilf-equivalent to the pair {132, 231}.

Proof. In [11] it is stated that the pairs {123, 132} and {132, 231} are 0-Wilf-equivalent; it remains to show that these twopairs are 1-Wilf-equivalent.

Suppose π ∈ S1n(123, 132). Write π as P �Q . Then, Q is either the string 1 or the empty string, and P must be decreasing.So if n > 1 we have S1n(123, 132) = {(n − 1)(n − 2) · · · 2 � 1, (n − 1)(n − 2) · · · 21�} so that s1n(123, 132) = 2.

Now, suppose P � Q ∈ S1n(132, 231). Then either P or Q is empty. If Q is empty, then P must be decreasing and if P isempty, then Q must be increasing. If n > 1, we have S1n(132, 231) = {(n− 1)(n− 2) · · · 1�, �12 · · · (n− 1)} and s1n(B) = 2.So, the pairs are ⋆-Wilf-equivalent as claimed. �

Computing s1n(p, q) for |p| = |q| = 3 in a similar manner, we see that the Wilf-equivalence given in Theorem 1.4 is the onlynontrivial Wilf-equivalence between pairs of patterns of length 3.


We now consider pairs {p, q} with |p| = 3 and |q| = 4 such that p is not contained in q.

Proposition 5.2. The pairs {321, 1342} and {321, 1423} are ⋆-Wilf-equivalent.

Proof. The 0-Wilf-equivalence of these pairs is stated in [1,16]. It remains to prove 1-Wilf-equivalence: we will shows1n(321, 1342) = s1n(321, 1423) = 3.


Suppose n ≥ 3 and π = P � Q ∈ S1n(321, 1342). As π avoids 321, both P and Q must be increasing with P < Q . But Pmust also avoid 123, so that |P| ≤ 2. Thus

S1n(321, 1342) = {�12 · · · (n − 1), 1 � 23 · · · (n − 1), 12 � 3 · · · (n − 1)}.

Now, suppose π = P � Q ∈ S1n(321, 1423). Once again, we have P < Q where P and Q are increasing. But π avoids 1423 sothat either P is empty or |Q | ≤ 1. So

S1n(321, 1423) = {�12 · · · (n − 1), 12 · · · (n − 2) � (n − 1), 12 · · · (n − 1)�},

completing the proof. �

Similar counting arguments can be used to prove the following propositions.

Proposition 5.3. The pairs {321, 2341}, {132, 1234}, {132, 3241}, {132, 3124}, {132, 2134}, and {132, 3412} are ⋆-Wilf-equivalent. In particular, for such a pair {p, q} we have s1n(p, q) = 3.

Proposition 5.4. The pairs {321, 3412}, {321, 3142}, {321, 2413}, {132, 2341}, and {132, 2314} are ⋆-Wilf-equivalent. Inparticular, for such a pair {p, q} we have s1n(p, q) = n.

Computing s1n(p, q) for small n, we see that the list in Theorem 1.5 is exhaustive.


We now prove Theorem 1.6, considering pairs {p, q} with |p| = |q| = 4. We introduce notation that will help us to countpartial permutations. For a pair A, let S{h}

n (A) be the subset of S1n(A) such that πh = � and let s{h}n (A) = |S{h}n (A)|.


Proof. It is shown in [9] that the two pairs are 0-Wilf-equivalent. We now check that {4123, 3214} and {1234, 2143} are1-Wilf-equivalent. We introduce notation which will help us to count partial permutations.

Let n ≥ 2. We claim that for h ∈ {1, . . . , n} we have s{h}n (4123, 3214) = s{h}n (1234, 2143) = 2n−2. We proceed byinduction on n; the base case n = 2 is straightforward. Suppose n > 2. Then, we have s{1}n (4123, 3214) = s0n−1(123, 213) =

2n−2 and s{n}n (4123, 3214) = s0n−1(312, 321) = 2n−2, where the last equality in both of these computations is given in [11].Now suppose h ∈ {1, n} and π ∈ S{h}

n (4123, 3214). Then π1 = 1 or πn = 1. If π1 = 1, regardless of the other entries of π ,this 1 could not be part of an occurrence of either element of {4123, 3214}; the same is true when πn = 1. After the positionof this 1 is chosen, by induction there are 2n−3 ways to fill in the remaining elements of π , so that s{h}n (4123, 3214) = 2n−2.

We now count s{h}n (1234, 2143). Suppose n > 2. We have s{1}n (1234, 2143) = s0n−1(123, 132) = 2n−2, and s{n}n =

s0n−1(123, 213) = 2n−2, where the last equality in both of these computations is given in [11]. Now suppose h ∈ {1, n}and π ∈ S{h}

n (1234, 2143). If h = n − 1 then either π1 = n or π2 = n; this n cannot be part of an occurrence of an elementof {1234, 2143} so by induction there are 2n−3 ways to fill in the remaining elements. If h = n − 1, then we must haveπn−1 = 1 or πn = 1; in either case, this 1 cannot be part of an occurrence of an element of {1234, 2143}. By induction, thereare 2n−3 ways to fill in the remaining elements of π , so that s{h}n (1234, 2143) = 2n−2.

We now check that {4123, 3214} and {1234, 2143} are 2-Wilf-equivalent. Suppose n > 3 and P�Q �R ∈ S2n(4123, 3214).Then, |Q | ≤ 1 and one of |P| or |R| must be 0. If |Q | = 1 and |R| = 0 is empty then π must be 12 · · · (n − 3) � (n − 2)�;

similarly, for each choice of the length of Q , and which of P or R is empty, there is exactly one partial permutation avoiding{4123, 3214} so that s2n(4123, 3214) = 4.

Now, suppose n > 3 and P � Q � R ∈ S2n(1234, 2143). Then both |P| and |R| are at most 1. If |P| = |R| = 1, then theresulting partial permutation must be (n − 2) � (n − 3) · · · 2 � 1; similarly each choice of the lengths of |P| and |R| yieldsone partial permutation avoiding {1234, 3214} so that s2n(1234, 2143) = 4. �


Proof. The claim follows from setting {pi} = {1} in Corollary 1.3. �

Proposition 5.7. The pairs {1324, 3124}, {2413, 4213}, and {2134, 2314} are ⋆-Wilf-equivalent.

Proof. Setting {pi} = 1 in Corollary 1.3 we deduce that the pairs {1324, 3124} and {2134, 2314} are ⋆-Wilf-equivalent.We now show that {1324, 3124} and {2413, 4213} are ⋆-Wilf-equivalent. It is shown in [8] that these pairs are 0-Wilf-equivalent; we check 1-Wilf-equivalence. We claim that for h ∈ {1, . . . , n} we have s{h}n (1324, 3124) = s{h}n (2413, 4213) =

2n−2. We proceed by induction on n; the base case n = 2 is straightforward. We first count s{h}n (1324, 3124). Supposen > 2. We have s{n}n (1324, 3124) = s0n−1(132, 312) = 2n−2, where the last equality is given in [11]. If h = n and


Fig. 6. The structure of a {2413, 4312}-avoiding permutation.

π ∈ S{h}n (1324, 3124) then πn must equal 1 or 2, in which case πn cannot be part of an occurrence of either element of

{1324, 3124}. By induction there are 2n−3 ways to fill in the remaining elements of π so that s{h}n (1324, 3124) = 2n−2.Now, suppose n > 2 and that P �Q ∈ s1n(2413, 4213). Then, the numbers comprising the string P must be |P| consecutive

integers. Additionally, we may write the string Q as Q1Q2 where Q1 > P > Q2, the string Q1 avoids {312, 213}, and thestring Q2 is decreasing. Conversely, if the elements of {1, . . . , n − 1} are arranged into three strings P,Q1, and Q2 such thatQ1 > P > Q2, the string P avoids {231, 321}, the string Q1 avoids {312, 213}, and Q2 is decreasing, then P � Q1Q2 avoids{2413, 4213} (see Fig. 6). It is given in [11] that s0k(231, 321) = s0k(312, 213) = 2k−1 for k ≥ 2. So, counting the number ofsuch arrangements P � Q1Q2 when |P| = h − 1 for h ≥ 1, we obtain

s{h}n (2413, 4213) = 2h−1+

n−2k=h

2h−2· 2(n−1)−(k−1)−1

= 2n−2

as desired.We now check that {1324, 3124} and {2413, 4213} are 2-Wilf-equivalent. Suppose n > 3 and P�Q �R ∈ S2n(1324, 3124).

Then |P| + |Q | ≤ 1, so that

S2n(1324, 3124) = {� � (n − 2) · · · 1, (n − 2) � �(n − 3) · · · 1, �(n − 2) � (n − 3) · · · 1}.

Now suppose n > 3 and P � Q � R ∈ S2n(2413, 4213). Then |P| ≤ 1, at least one of |P| or |R| must be 0, and at least one of|Q | or |R| must be 0. So,

S2n(2413, 4213) = {1 � (n − 2) · · · 2�, �(n − 2) · · · 1�, � � (n − 2) · · · 1}

and therefore s2n(1324, 3124) = s2n(2413, 4213) = 3. �

Proposition 5.8. The pairs {1234, 2134}, {1342, 2341}, {2314, 3214}, and {1243, 2143} are ⋆-Wilf-equivalent.

Proof. Setting {pi} = 1 in Corollary 1.3, the pairs {1234, 2134} and {2314, 3214} are ⋆-Wilf-equivalent. By [6, Proposition4.1], the patterns 12 and 21 are shape-⋆-Wilf-equivalent, so setting {pi} = {12, 21} in Proposition 3.1, we deduce that{1234, 1243} and {2134, 2143} are star-Wilf-equivalent. But, taking the reversal and then the complement of each elementof {1234, 1243} we obtain {1234, 2134}, so we deduce that {1234, 1243} and {1243, 2143} are ⋆-Wilf-equivalent.

It remains to show that {2314, 3214} and {1342, 2341} are ⋆-Wilf-equivalent. It is shown in [8] that these pairs are 0-Wilf-equivalent. We check 1-Wilf-equivalence. First we compute s1n(2314, 3214) for n > 2. We have s{1}n (2314, 3214) =

s0n−1(213) = Cn−1, where Ck denotes the kth Catalan number, and s{n}n (2314, 3214) = s0n−1(231, 321) = 2n−2. Now, supposen > 2, thatπ ∈ s1n(2314, 3214) and thatπ = P �Q where P is nonempty. Then, if Q has length k, the pattern Q must consistof k of the numbers {1, . . . , k + 1} or else π will contain an element of {2314, 3214}. Let a ∈ {1, . . . , k + 1} be the numberthat appears in P . Then, eitherπ1 = a orπ2 = a, andQ = Q1Q2 whereQ1 > a > Q2 and bothQ1 andQ2 avoid 213 (see Fig. 7).We count the number of partial permutations in S1n(2314, 3214) that satisfy these conditions.Whenπ2 = � (or equivalentlyk = n−2), wemust have π1 = a so that there are

k+1a=1 Ca−1Ck+1−a = Ck+1 partial permutations. When π2 = �, there are 2

choices for the position of a and 2n−k−3 choices for the remainder of P , for a total of 2·2n−k−3 k+1a=1 Ca−1Ck+1−a = 2n−k−2Ck+1

partial permutations. It follows that s1n(2314, 3214) = Cn−1 +n−2

k=0 2n−k−2Ck+1.

We claim that we also have s1n(1342, 2341) = Cn−1 +n−2

k=0 2n−k−2Ck+1. We proceed by induction; the claim is clear for

n = 2. Suppose n > 2. We have s{1}n (1342, 2341) = s0n−1(231) = Cn−1 and similarly s{n}n (1342, 2341) = Cn−1. Now supposeh ∈ {1, n} and π ∈ S{h}

n (1342, 2341). Then we must have π1 = n − 1 or πn = n − 1. In either case, n − 1 cannot appear in


Fig. 7. The structure of a {2314, 3214}-avoiding permutation P � Q with |Q | ≥ 1 where π1 = a is the element of {1, . . . , |Q | + 1} not appearing in Q .

an occurrence of either element of {1342, 2341}, so if π1 = n − 1 there are s{h−1}n−1 (1342, 2341) ways to fill in the remaining

entries, and if πn = n − 1 there are s{h}n−1(1342, 2341) ways to fill in the remaining entries. So,

s1n(1342, 2341) = s{1}n (1342, 2341) + s{n}n (1342, 2341) + 2s1n−1(1342, 2341) − s{1}n−1(1342, 2341)

− s{n}n−1(1342, 2341)

= 2Cn−1 +

n−3k=0

2n−k−2Ck+1 = Cn−1 +

n−2k=0

2n−k−2Ck+1,

as claimed. The sequence s1n(2314, 3214) is sequence A134635 in [12].We now show 2-Wilf-equivalence. Suppose P � Q � R ∈ S2n(2314, 3214). Then |P| ≤ 1, Q is increasing, R is decreasing,

and R < P < Q so that s2n(2314, 3214) = 2n − 3. Similarly, suppose P � Q � R ∈ S2n(2314, 3214). Then P is decreasing, Qis decreasing, R is increasing, Q < P,Q < R, and at least one of |P| or |R| must be empty. Thus s2n(2314, 3214) = 2n − 3, sothat {2314, 3214} and {1342, 2341} are ⋆-Wilf-equivalent. �

Acknowledgements

This research was conducted at the University of Minnesota Duluth, supported by grants NSF/DMS 106279, and NSAH98230-11-1-0224. I am grateful to Joe Gallian for supervising this research, to Alan Deckelbaum, JonathanWang and RickyLiu for their advice, and to Anders Claesson for sharing code used to work with examples.

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