Positional Marked Patterns in Permutations
aa r X i v : . [ m a t h . C O ] F e b Positional Marked Patterns in Permutations
Jeffrey B. Remmel
Department of MathematicsUniversity of California, San DiegoLa Jolla, CA 92093-0112. USA [email protected]
Sittipong Thamrongpairoj
Department of MathematicsUniversity of California, San DiegoLa Jolla, CA 92093-0112. USA [email protected]
February 9, 2021
The notion of mesh patterns was first introduced by Br¨and´en and Claesson [2]. Manyauthors have further studied this notion. In particular, the notion of marked mesh patternwas introduced by ´Ulfarsson [4], and the study of the distributions of quadrant marked meshpatterns in permutations was initiated by Kitaev and Remmel in [3].Here, we recall the definition of quadrant marked mesh pattern. Let N = { , , , . . . } denote the set of natural numbers and S n denote the symmetric group of permutations of1 , . . . , n , then we consider the graph of σ, G ( σ ), to be the set of points ( i, σ i ), we are interestedin the points that lie in the four quadrants I, II, III, IV of that coordinate system. For any a, b, c, d ∈ N and any σ = σ . . . σ n ∈ S n , we say that σ i matches the simple marked meshpattern M M P ( a, b, c, d ) if in G ( σ ) relative to the coordinate system which has the point( i, σ i ) as its origin, there are at least a points in quadrant I, at least b points in quadrant II,at least c points in quadrant III, and at least d points in quadrant IV. We let mmp ( a,b,c,d ) ( σ )denote the number of i such that σ i matches the marked mesh pattern M M P ( a, b, c, d ) in σ .For example, let σ = 647913258 and consider simple marked mesh pattern M M P (2 , , , i, σ i ) in the graph of σ such that there are at least 2 points to the topright of ( i, σ i ) and at least 2 points to the bottom left of ( i, σ i ). The graph of σ and anillustration of M M P (1 , , ,
0) are shown in Figure 1. As we see in Figure 2, at point (3 , , M M P (2 , , ,
0) in σ . At point (6 , , M M P (2 , , ,
0) in σ . In this case, only 7 matches M M P (2 , , ,
0) in σ , so mmp (2 , , , ( σ ) = 1.We studied a generalization of quadrant marked mesh patterns. As a motivation, con-sider a statistic mmp (1 , , , mmp (1 , , ,
0) for anypermutation σ is to count the number of elements in σ that can be the midpoint of thepattern 123 in σ . One can ask similar questions but for different patterns and positions. Forexample, one may consider the number of elements in σ that can be the starting point ofpattern 2314. 1 ≥ M M P (2 , , ,
0) in σ but 3 does notHere, we briefly define the notion of positional marked patterns (pmp) which enumeratethe described statistics. A positional marked pattern of length k is a permutation of [ k ] withone of the elements underlined. Given any positional marked pattern τ , let π ( τ ) denote theelement in S k obtained from τ by removing the underline, and let u ( τ ) be the position ofthe underlined element in τ . Given σ = σ σ . . . σ n ∈ S n and positional marked pattern τ ,we say that σ is τ -match at position l if σ contains a pattern π ( τ ) in such a way that l -thelement in σ plays the role of the underlined elements in τ . Let pmp τ ( σ ) denote the numberof position l such that σ is τ -match at position l . We will carefully define positional markedpatterns in Section 2.Recall that, in classical patterns, two patterns τ and τ are Wilf-equivalent if the numberof σ in S n avoiding τ is the same as that of τ for all n ∈ N . Here, we adopt the vocabularyfrom the classical definition to our definition. Given two positional marked patterns τ and τ , we say that τ and τ are pmp -Wilf equivalent if pmp τ and pmp τ has the samedistribution over S n for all n . Our main goal is to classify pmp -Wilf equivalent classes forpositional marked patterns. 2he chapter outlines as follows: In section 2, we give a precise definition and somepreliminary results on positional marked patterns. In section 3, we study positional markedpatterns of length 3. We prove that there are only 2 pmp -Wilf equivalent classes for positionalmarked patterns of length 3. The result follows from the following theorem. We will provein Section 3 that there are only 2 pmp -Wilf equivalent classes for positional marked patternsof length 3. The result follows from the following theorem. Theorem 1.
Two positional marked patterns and are pmp -Wilf equivalent, and twopositional marked patterns and are pmp -Wilf equivalent.
In Section 23, we prove some non-trivial equivalences of pairs of positional marked pat-terns of length 4 as well as providing numerical results for every patterns of length 4. InSection 5, we prove the following theorem, which gives an equivalence of a pair of positionalmarked patterns of arbitrary length.
Theorem 2.
Given two positional marked patterns P = 12 p . . . p l − and P = 21 p . . . p l − .Then, P and P are pmp-Wilf equivalent. In Section 6, we discuss further research possibilities, as well as precise connection betweenpositional marked patterns and quadrant marked mesh patterns.
Definition 3.
Let S ∗ k denote a set of permutations of [ k ] with one of the elements underlined.Given any τ ∈ S ∗ k , let π ( τ ) denote a element in S k obtained from τ by removing the underline,and let u ( τ ) be the position of the underlined element in τ . We shall call an element in S ∗ k a positional marked pattern (pmp). For example, an element in S ∗ looks like τ = 1432. In this case, π ( τ ) = 1432 and u ( τ ) = 2. Definition 4.
Given a word w where alphabets are taken from Z ≥ , a reduction of w , denotedby red ( w ) , is the word obtained from w by replacing the i -th smallest alphabet by i . Inparticular, if w is a word with distinct alphabets of length k , then red ( w ) ∈ S k . For example, if w = 25725, then red ( w ) = 12312. Here, we are ready to define a statisticson S n that we mainly focus on in this paper. Definition 5.
Given σ = σ σ . . . σ n ∈ S n and τ ∈ S ∗ k , we say that σ is τ -match at position l if there is a subsequence σ i σ i . . . σ i k in σ such that1. i l = u ( τ ) red ( σ i σ i . . . σ i k ) = π ( τ ) In other words, σ contains a pattern π ( τ ) in such a way that l -th element in σ plays therole of the underlined elements in τ . Let pmp τ ( σ ) denote the number of position l such that σ is τ -match at position l . σ = 26481573 and τ = 1432. Then σ is τ -match at positions 2 and 4, aswe find subsequences 2653 and 2853 respectively. Thus, pmp τ ( σ ) = 2. We are interested inthe generating function P n,τ ( x ) = X σ ∈ S n x pmp τ ( σ ) . We say that two positional marked patterns τ and τ are pmp -Wilf equivalent if P n,τ ( x ) = P n,τ ( x ) for all n . Note that, by looking atthe constant terms of generating functions, if τ and τ are pmp -Wilf equivalent, then π ( τ )and π ( τ ) are Wilf-equivalent. Thus, one might think of pmp -Wilf equivalent as a strongerversion of Wilf-equivalent. In this paper, we classify equivalence classes of S ∗ and S ∗ .We associate a positional marked pattern with a permutation matrix-like diagram. Givenany τ ∈ S ∗ k , the diagram associated to τ is a k by k array with the following filling: For thecell at i -th row and j -th column (i) the cell is filled with ◦ if τ i = j and u ( τ ) = i , (ii) the cellis filled with × if τ i = j and u ( τ ) = i , or (iii) the cell is empty otherwise. By convention, wecount rows and columns of an array from left to right and from bottom to top. For example,if τ = 1432, then the corresponding diagram is ◦ × ×× Similarly, we associate σ ∈ S n with an n by n diagram with the cell at i -th row and j -column is (i) filled with × if σ i = j and (ii) empty otherwise. For example, the diagramcorresponding to π = 26481573 is × ×× ×× ×× × We can visualize pmp matching using diagrams. Given a permutation σ and a pmp τ , σ is τ -match at position i if when replacing × in the i -th column of the diagram of σ by ◦ ,then it contains a subdiagram τ . For example, σ = 26481573 is τ -match at position 2 and 4by looking at subdiagrams marked in red below. × ×◦ ×× ×× × ◦ ×× ×× ×× × pmp by symmetry. Given two pmp τ , τ suchthat the diagram of τ can be obtained from by applying a series of rotations and reflectionsto the diagram of τ , then we can construct a map θ : S n → S n by applying the same seriesof rotations and reflections to elements in S n . The map will obviously have the propertythat pmp τ ( σ ) = pmp τ ( θ ( σ )) for all σ ∈ S n , and so τ and τ are pmp -Wilf equivalent. Weproved the following lemma. Lemma 6.
Given τ , τ ∈ S ∗ k , such that τ can be obtained by applying series of reflectionsand rotation to τ . Then τ and τ are pmp -Wilf equivalent. Lemma 6 reduces the problem tremendously. There are 3! · pmp of length 3.However, with Lemma 6, there are at most 4 equivalence classes, which are represented by123 , , , ××◦ ×◦× × ×◦ ◦ ×× S ∗ pmp -Wilf equivalent, and 123 and 132 are pmp -Wilf equivalent. and In this section, we will prove the following theorem:
Theorem 7.
Two pmp and are pmp -Wilf equivalent.
For our convenience, we let τ = 123 and τ = 132. We will first prove by showing that twogenerating functions satisfy a same recursive formula. We later will construct a bijection from S n to itself that maps pmp τ to pmp τ , that is, a map θ such that pmp τ ( σ ) = pmp τ ( θ ( σ )).Given any permutation σ = σ σ . . . σ n ∈ S n . We will look at the position where thelast ascent occurs. In particular, σ is either has no ascent or there is k such that σ k <σ k +1 > σ k +2 > . . . > σ n . In that case, we say that the last ascent of σ is at position k . Let P n,τ ,k ( x ) = X x pmp τ ( σ ) where the sum is over all permutation in S n with the last ascent isat position k . Then, we have P n,τ ( x ) = 1 + n − X k =1 P n,τ ,k ( x ).Given any permutation σ = σ σ . . . σ n ∈ S n . Similar to τ , for τ , we will look at theposition where the last descent occurs. σ is either 1 2 . . . ( n − n , or there is k such that σ k > σ k +1 < σ k +2 < . . . < σ n . We say that the last descent of σ is at position k . Let P n,τ ,k ( x ) = X x pmp τ ( σ ) where ths sum is over all permutation in S n with the last descentis at position k . Then, we have P n,τ ( x ) = 1 + n − X k =1 P n,τ ,k First, we derive recursive formula for P n,τ ,k ( x ).5 emma 8. Let τ = 123 and let P n,τ ,k ( x ) = X x pmp τ ( σ ) where the sum is over all σ ∈ S n with the last ascent of σ is at position k . Then, P n,τ ,k ( x ) satisfies the following recursiveformula: P n,τ ,k ( x ) = ( k − xP n − ,τ ,k − ( x ) + 1 + k X l =1 P n − ,τ ,l ( x ) where P n,τ , ( x ) = P n,τ ,n ( x ) = 0 by convention.Proof. We derive a recursive formula of P n,τ ,k ( x ) by looking at position of 1 in σ where thelast ascent is at position k . So, σ has a following form: σ = σ σ . . . σ k − σ k < σ k +1 > σ k +2 > . . . > σ n Let t be the position of 1. We have 3 cases:1. 1 ≤ t ≤ k −
1. In this case, 1 , σ k , σ k +1 form a pattern τ . Thus, σ is τ -match atpostion t . Moreover, 1 does not inflence pmp -match at other positions. Thus, we canremove 1 from σ and reduce other elements. The last ascent will be at position k − x ( k − P n − ,τ ,k − ( x ).2. t = k . In this case, σ is not τ -match at position k . We can remove 1 from σ andreduce other elements. The remaining permutation will either has no ascent, or thelast ascent appears at some position between 1 and k −
1. Thus, this case contributes1 + k − X l =1 P n − ,τ ,l ( x )3. t = n . In this case, σ is not τ -match at postion n . We can remove 1 from σ and reduceother elements. The remaining permutation will have the last ascent at position k .Thus, this case contributes P n − ,τ ,k ( x )In total, we have P n,τ ,k ( x ) = ( k − xP n − ,τ ,k − ( x ) + 1 + k − X l =1 P n − ,τ ,l ( x ) ! + P n − ,τ ,k ( x )= ( k − xP n − ,τ ,k − ( x ) + 1 + k X l =1 P n − ,τ ,l ( x )Note that, the first case does not exist when k = 1, but the formula is correct since P n − ,τ , ( x ) = 0. Also, the third case does not exist when k = n −
1. However, the formulais also correct since P n,τ ,n ( x ) = 0.Here, we prove a similar result for τ . 6 emma 9. Let τ = 132 and let P n,τ ,k ( x ) = X x pmp τ ( σ ) where the sum is over all σ ∈ S n with the last descent of σ is at position k . Then, P n,τ ,k ( x ) satisfies the following recursiveformula: P n,τ ,k ( x ) = ( k − xP n − ,τ ,k − ( x ) + 1 + k X l =1 P n − ,τ ,l ( x ) where P n,τ , ( x ) = P n,τ ,n ( x ) = 0 by convention.Proof. We use the same strategy as in the case for τ . We derive a recursive formula for P n,τ ,k ( x ) by looking at the position of 1 in σ where the last descent is at position k . So, σ has a following form: σ = σ σ . . . σ k − σ k > σ k +1 < σ k +2 < . . . < σ n Let t be the position of 1. We have 2 cases:1. 1 ≤ t ≤ k −
1. In this case, 1 , σ k , σ k +1 form a pattern τ . Thus, σ is τ -match atpostion t . Moreover, 1 does not inflence τ -match at other positions. Thus, we canremove 1 from σ and reduce other elements. The last descent will be at position k − x ( k − P n − ,τ ,k − ( x ).2. t = k + 1. In this case, σ is not τ -match at position k + 1. We can remove 1 from σ and reduce other elements. The remaining permutation will either has no descent, orthe last ascent appears at some position between 1 and k . Thus, this case contributes1 + k X l =1 P n − ,τ ,l ( x )In total, we have P n,τ ,k ( x ) = ( k − xP n − ,τ ,k − ( x ) + 1 + k X l =1 P n − ,τ ,l ( x )Note that, the first case does not exist when k = 1, but the formula is correct since P n − ,τ , ( x ) = 0. Also, for the second case, if k = n −
1, the last descent cannot be atposition k = n −
1. However, the formula is still correct since P n − ,τ ,n ( x ) = 0.One can check that P ,τ , ( x ) = 1 = P ,τ , ( x ). Since P n,τ ,k ( x ) and P n,τ ,k ( x ) satisfy thesame recursive formula and have the same initial values, we proved that P n,τ ,k ( x ) = P n,τ ,k ( x )for all n, k such that 1 ≤ k ≤ n −
1, and hence P n,τ ( x ) = P n,τ ( x ) for all n ≥
1. Therefore,we prove Theorem 7.To conclude this section, we give a bijection θ : S n → S n such that pmp τ ( σ ) = pmp τ ( θ ( σ )). The bijection is constructed based on the recursive formula. We start with θ (1) = 1, θ (12) = 21 and θ (21) = 12. In general, θ will have following properties7. σ has no ascent if and only if θ ( σ ) has no descent.2. The last ascent of σ is at the same position as the last descent of θ ( σ ).3. pmp τ ( σ ) = pmp τ ( θ ( σ ))By observation, θ satisfies the properties for S and S . For S n +1 where n ≥
2, we definethe map recursively. Given σ ∈ S n , we have σ ′ = θ ( σ ) ∈ S n , where the last ascent of σ is atthe same position as the last descent of σ ′ . Let k be the position. We decompose σ and σ ′ at k . σ = σ σ . . . σ k σ k +1 . . . σ n σ ′ = σ ′ σ ′ . . . σ ′ k σ ′ k +1 . . . σ ′ n If σ has no ascent, and so σ ′ has no descent, we decompose σ and σ ′ by having the “firstpart” empty, or, equivalently, set k = 0.Here, we obtain an element ˆ σ ∈ S n +1 by increase every element in σ by 1 and insert 1 atsome position, then θ (ˆ σ ) will be obtained from increasing elements in σ ′ by 1 and insert 1at some position based on the position of 1 in ˆ σ . Suppose we insert 1 at position r from theleft in σ , then the position r ′ of 1 inserted in σ ′ is r ′ = r r ≤ kr + 1 k + 1 ≤ r ≤ nk + 1 r = n + 1The corresponding positions of 1 can be viewed from the diagram below: σ σ σ σ k σ k +1 σ k +2 σ k +3 σ n σ ′ σ ′ σ ′ σ ′ k σ ′ k +1 σ ′ k +2 σ ′ k +3 σ ′ n Corresponding positions of 1By observation, the position of the last ascent of ˆ σ is the same as the position of the lastdescent of ˆ σ . Also, pmp τ will increase by 1 if and only if 1 is inserted at the first k positions,which is the same condition for pmp τ to increase by 1. Thus, θ satisfies the conditions.As an example, we start with θ (12) = 21. Here, the diagram for inserting 1 looks like1 22 18uppose we insert 1 to 12 at the last position, so we get 231. According to the diagram,we should insert 1 to 21 at the second position, so we get 312. Thus, θ (231) = 312. Now thediagram looks like 2 3 13 1 2Here, we insert 1 to 231 at the first position, so we get 1342. We should also insert 1 to312 at the first position, so we get 1423. Thus, θ (1342) = 1423.Below are enumerations of P n, ( x ) for the first few n : P , ( x ) = 1 P , ( x ) = 2 P , ( x ) = 5 + xP , ( x ) = 14 + 8 x + 2 x P , ( x ) = 42 + 47 x + 25 x + 6 x P , ( x ) = 132 + 244 x + 216 x + 104 x + 24 x P , ( x ) = 429 + 1186 x + 1568 x + 1199 x + 538 x + 120 x P , ( x ) = 1430 + 5536 x + 10232 x + 11264 x + 7814 x + 3324 x + 720 x P n, ( x ) | x k Even though we do not know a formula for P n, ( x ) in general, we can still explain someof the coeffients. For example, P n, ( x ) | x = C n , the n -th Catalan number. This is obvioussince a permutation is 123 avoiding if and only if it is not 123-match at any position. Thereare other coefficients that have a nice formula. Theorem 10.
For n ≥ , the degree of P n, ( x ) is n − , and P n, ( x ) | x n − = ( n − Proof.
For any σ ∈ S n , it is clear that σ is not 123-match at position n or n −
1. Thus, pmp ( σ ) ≤ n −
2. For any σ ∈ S n , we claim that pmp ( σ ) = n − σ n − = n − σ n = n . The coverse obviously true since σ would be 123-match atposition i for 1 ≤ i ≤ n −
2. Suppose that pmp ( σ ) = n −
2. Let p i be the position of i in σ . That is σ p i = i . Note that n − n cannot be a starting point of the pattern 123, so σ is not 123-match at positions p n − and p n . However, σ is 123-match at every position except n − n . Thus, { p n − , p n } = { n − , n } . Consider position p n − . σ must be 123-matchat position p n − . That is n − σ . However, the onlyway for n − n − , n − , n form a pattern 123. Thatis n − n . So, p n − = n − p n = n , which means σ n − = n − σ n = n . 9hus, we know that all σ such that pmp ( σ ) = n − σ of the form: σ = σ σ . . . σ n − σ n − n − n There are ( n − σ ’s, so P n, ( x ) | x n − = ( n − Theorem 11.
Given any positive integers n, k such that n ≥ k , and any τ ∈ S ∗ k , the degreeof P n,τ ( x ) is n − k + 1 , and P n,τ ( x ) | x n − k +1 = ( n − k + 1)! .Proof. Given k ≤ n and τ ∈ S ∗ k . Suppose τ has the form τ = τ τ . . . τ l . . . τ k − τ k That is u ( τ ) = l and π ( τ ) = τ τ . . . τ k . Note that there are l − τ and there are k − l numbers to the right of the underlined numberin τ . Given σ ∈ S n . Suppose σ is τ -match at position i . Then, there must be at least l − i in σ and there must be at least k − l to the right of position i in σ . Thus, l ≤ i ≤ n − k + l . Therefore, pmp τ ( σ ) ≤ n − k + 1. Thus, the degree of P n,τ ( x )is at most n − k + 1.Here, we count the number of σ ∈ S n such that pmp τ ( x ) = n − k + 1. We shall prove thefollowing claim: Claim 12.
Given σ = σ . . . σ n ∈ S n , pmp τ ( σ ) = n − k + 1 if and only if all of the followingshold:(1) { σ , σ , . . . , σ l − , σ n − k + l +1 , . . . , σ n } = { , , . . . , τ l − , n − k + τ l + 1 , . . . , n } (2) red ( σ σ . . . σ l − σ n − k + l +1 . . . σ n ) = red ( τ τ . . . τ l − τ l +1 . . . τ k ) That is, the first l − numbers and last k − l numbers in σ is a rearrangement of { , , . . . , τ l − , n − k + τ l + 1 , . . . , n } in the way that they have the same relative order as τ τ . . . τ l − τ l +1 . . . τ k .Proof. (of the claim)We first prove the converse. Suppose σ satisfies (1) and (2). We want to show that σ is τ -match at any position i when l ≤ i ≤ n − k + l . Given any such position i . Consider thefollowing subsequence σ σ . . . σ l − σ i σ n − k + l +1 . . . σ n Note that σ i is the τ l -th smallest number in the subsequence, and τ l is the τ l -th smallestnumber in subsequence τ τ . . . τ k . Thus, we can insert σ i to σ . . . σ l − σ l +1 . . . σ n and τ l to τ . . . τ l − τ l +1 τ k . So, σ σ . . . σ l − σ i σ n − k + l +1 . . . σ n have the same relative order as τ . . . τ k .However, τ . . . τ k = π ( τ ) is a permutation, so red ( σ σ . . . σ l − σ i σ n − k + l +1 . . . σ n ) = π ( τ ).So, σ is τ -match at position i .Here, we prove the forward direction. Suppose pmp τ ( σ ) = n − k + 1, then σ is τ -match at all positions i when l ≤ i ≤ n − k + l , and σ is not τ -match at all posi-tions j ∈ { , , . . . , l − , n − k + l + 1 , . . . , n } . Note that, in order for σ to match at10ostion i , there must be at least τ l − σ i and there must be at least k − τ l numbers greater than σ i . Therefore τ l ≤ σ i ≤ n − k + τ l . In other words, σ is not τ -match at positions of 1 , , . . . , τ l − n − k + τ l , n − k + τ l +1 , . . . , n . There are k − { p , p , . . . , p τ l − , p n − k + τ l +1 , . . . , p n } = { , , . . . , l − , n − k + l + 1 , . . . , n } , or in other words, { σ , σ , . . . , σ l − , σ n − k + l +1 , . . . , σ n } = { , , . . . , τ l − , n − k + τ l + 1 , . . . , n } . Thus, we prove (1).To prove (2), we only need to show that there is at most one rearrangement of { , , . . . , τ l − , n − k + τ l + 1 , . . . , n } such that pmp τ ( σ ) = n − k + 1. Then, by converse direction, weknow that the rearrangement in (2) make pmp τ ( σ ) = n − k + 1. In that case, we concludethat the unique rearrangement that make pmp τ ( σ ) = n − k + 1 exists and must be (2).Let’s call numbers appearing at positions 1 , , . . . , l − left part of σ , while call thenumbers appearing at positions n − k + l + 1 , n − k + l + 2 , . . . , n the right part of σ . Then, σ has a following structure: 1 , , . . . , τ l − , n − k + τ l + 1 , . . . nσ = left part right partWe shall prove that ways to rearrange 1 , , . . . , τ l − , n − k + τ l + 1 , . . . n in left and rightpart of σ so that pmp τ ( σ ) = n − k + 1 is unique if exist.Consider the number τ l in σ . We know that σ is τ -match at position p τ l . Then, numbers1 , , . . . , τ l − σ must involve in τ -match at position p τ l since we need τ l − τ l . Therefore, for each number i less than τ l , we can determine whether i is inthe left part or right part of σ based on the relative postion of i and τ l in τ . Also, considerthe number n − k + τ l . σ must be τ -match at position p n − k + τ l . By the same reasoning, foreach number n − k + τ l + i , we determine whether n − k + τ l + i is in the left part or rightpart of σ based on the relative position of τ l + i and τ l in τ . Thus, we determine both leftand right part as sets.Now, consider position l in σ . σ must be τ -match at position l . Thus, numbers in first l − σ must involve in τ -match at postion l , and so, the first l − σ must have the same relative order as first l − τ . By the same reasoning, byconsidering position n − k + l , the last k − l numbers in σ have the same relative order asthe last k − l numbers in τ . Therefore, we completely determine both left and right part of σ . Thus, there is at most one way to rearrange { , , . . . , τ l − , n − k + τ l + 1 , . . . , n } .11ince the rearrangement in (2) makes pmp ( σ ) = n − k + 1, σ , then σ must satisfies (2).As an example, if τ = 164352, any σ ∈ S with pmp τ ( σ ) = 9 − σ = 1 9 7 σ σ σ σ n − k + 1)! ways to rearrange the “middle” part of σ . Thus, P n,τ ( x ) | x n − k +1 =( n − k + 1)!Another coefficient we can describe is P n, ( x ) | x . The sequence are A029760 and A139262on OEIS. The sequence A139262 counts the sum of all inversion of all elements in S n (132),the set of 132-avoiders in S n . Theorem 13. P n, ( x ) | x = X σ ∈ S n (132) inv ( σ ) Proof.
To prove the theorem, we construct sets whose cardinality are X σ ∈ S n (132) inv ( σ ). Definition 14.
Let
IM S n (132) be the set of 132-avoiding σ that a pair of elements causingan inversion are marked with ∗ . As an example,
IM S (132) contains 8 elements, which are ∗ ∗ ∗ ∗
1, 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
1, and 3 ∗ ∗
1. It is easy to see that | IM S n (132) | = X σ ∈ S n (132) inv ( σ ) Definition 15.
Given σ ∈ S n , i, j ∈ Z , ≤ i ≤ n , ≤ j ≤ n + 1 . Let θ ( σ, i, j ) be anelement in S n +1 obtained from σ by1. increases every number greater than or equal to i by 12. insert i at the j -th position from the left. (The first position is in front of the firstelement, and the n + 1 -th position is behind the last element) For example, to find θ (2143 , , θ (2143 , ,
2) = 23154.Here, we are ready to define a map Φ :
IM S n (132) → { σ ∈ S n +1 | pmp ( σ ) = 1 } Definition 16.
Given σ ∈ IM S n (132) . Let σ be the underlying permutation of σ . Let j bethe position of the first * in σ , and let i be the number underneath the second * in σ . Define Φ( σ ) = θ ( σ, i, j ) Lemma 17.
Given σ ∈ IM S n (132) and ˆ σ = Φ( σ ) , then mmp (ˆ σ ) = 1 . Thus, Φ is a mapfrom IM S n (132) to { σ ∈ S n +1 | pmp ( σ ) = 1 } roof. Let σ ∈ IM S n (132) has a form σ = σ . . . σ k − ∗ σ k σ k +1 . . . σ l − ∗ σ l σ l +1 . . . σ n and σ is the underlying permutation. Then, Φ( σ ) = θ ( σ, σ l , k ) looks likeΦ( σ ) = σ ′ . . . σ ′ k − σ l σ ′ k σ ′ k +1 . . . σ ′ l − σ ′ l σ ′ l +1 . . . σ ′ n where σ ′ i = (cid:26) σ i + 1 for σ i ≥ σ l σ i for σ i < σ l Note that ˆ σ = σ ′ l = σ l + 1. Also, since σ k and σ l cause an inversion, so σ k > σ l , and thus σ ′ k = σ k + 1. Therefore, σ l < σ ′ l < σ ′ k . Equvialently, σ l , σ ′ k , σ ′ l form a pattern 132 in ˆ σ . Thus, mmp (ˆ σ ) ≥
1. To prove that pmp (ˆ σ ) = 1, we need to show that ˆ σ does not contain apattern 132 that starts at and element other than σ l .Suppose otherwise. That is, ˆ σ contains a pattern 132 where the starting element is σ ′ t for some t . We have several cases:1. If the pattern does not involve σ l , then there must be u, v such that σ ′ t , σ ′ u , σ ′ v form apattern 132. It is easy to see that σ t , σ u , σ v also form a pattern 132 in σ , which is acontradiction since σ is a 132-avoider.2. If the pattern involves σ l , then t < k and σ ′ t < σ l . In this case, it is easy to see that σ ′ t , σ ′ k , σ ′ l also form a pattern 132. Thus, σ t , σ k , σ l form a pattern 132 in σ , which isagain a contradiction.Thus, ˆ σ does not contain the pattern 132 that starts at any elements except σ l . So, pmp (ˆ σ ) = 1.The map Φ is clearly an injection. Let σ = σ . . . σ n be in the image of Φ, and let σ i bethe starting point of the pattern 132. Then, i determines the first starred element, and σ i determines the second starred element. To prove that the map Φ is a surjective, we needanother lemma: Lemma 18.
Given σ = σ . . . σ n ∈ S n such that pmp ( σ ) = 1 . Suppose the pattern 132 in σ starts at σ t , then the following must be true:(1) σ t +1 ≥ σ t (2) σ t + 1 is on the right of σ t (3) σ t +1 = σ t + 1 That is, the graph of σ should look like t σ t +1 σ t + 1 Proof.
Suppose (1) is not true. There must be u, v such that σ t , σ u , σ v form a pattern 132.Then, σ t +1 , σ u , σ v also form a pattern 132. So, pmp ( σ ) ≥ u, v such that σ t , σ u , σ v form a pattern 132. Then, σ t + 1 , σ u , σ v also form a pattern 132. So, pmp ( σ ) ≥ u, v such that σ t , σ u , σ v form a pattern 132. Then, σ t +1 , σ u , σ v also form a pattern 132. So, pmp ( σ ) ≥ Lemma 19.
The map
Φ :
IM S n (132) → { σ ∈ S n +1 | σ ) = 1 } is surjective.Proof. Given any σ ∈ { σ ∈ S n +1 | σ ) = 1 } . Let σ t be the starting poing of the pattern132 in σ . By lemma 18, we know that σ t , σ t +1 , σ t + 1 form a pattern 132. Let ˜ σ be apermutation obtained from σ by removing σ t and reducing. Let ˜ σ ∗ be ˜ σ with * marked onelements corresponding to σ t +1 and σ t + 1 before reducing ( σ t +1 − , σ t after reducing). ˜ σ ∗ is an element is IM S n (132) since σ t +1 , σ t + 1 form an inversion. Clearly, Φ(˜ σ ∗ ) = σ For example, consider σ = 785269314 ∈ S with pmp ( σ ) = 1, with only possiblestarting point of the patter 132 is 2. So, in this example, σ t , σ t +1 , σ t + 1 are 2,6,3. Then,removing 2 and reducing give ˜ σ = 67458213. Then, put * at elements corresponding to 6,3before reducing, which are 5,2 after reducing. So, ˜ σ ∗ = 674 ∗ ∗ and By observing diagrams of 132 and 132, we see that those two patterns are equivalent bylemma 6. Thus, we will instead prove that 123 and 132 are equivalent. ◦ ×× × ◦×
Diagrams of 132 and 132In this section, we prove the following theorem
Theorem 20.
Two generalized pmp and are pmp -Wilf equivalent. τ = 123 and τ = 132. We will first prove by showingthat two generating functions satisfy the same recursive formula. We later will constructa bijection from S n to itself that maps an pmp τ to pmp τ , that is, a map θ such that pmp τ ( σ ) = pmp τ ( θ ( σ )).Given any permutation σ ∈ S n . We will look at the position of 1 in σ . Let P n,τ ,k ( x ) = X x pmp τ ( σ ) , where the sum is over all permutation in S n with 1 at position k . Then, wehave P n,τ ( x ) = n X k =1 P n,τ ,k ( x ).Here, we derive a recursive formula for P n,τ ,k ( x ) Lemma 21.
Let τ = 123 and let P n,τ ,k = x pmp τ ( σ ) where the sum is over all σ ∈ S n with1 at position k . Then, P n,τ ,k ( x ) satisfies the following recursive formula: P n,τ ,k ( x ) = k − X l =1 P n − ,τ ,l ( x ) ! + ( x ( n − k −
1) + 1) P n − ,τ ,k ( x ) where P n,τ ,N ( x ) = 0 for N > n by convention.Proof.
We derive a recursive formula for P n,τ ,k ( x ) by looking at the position of 2 in σ wherethe position of 1 is k . σ has a following form σ = σ σ . . . σ k − σ k +1 . . . σ n Let l be position of 2. We have 3 cases:1. 1 ≤ l ≤ k −
1. In this case, we can remove 1 from σ and reduce other elements withouteffecting τ -match at other position. To see this, first note that if σ is not τ -matchat a particular position, removing 1 and reducing will not change it. Also, if σ is τ -match at some position before 1, then removing 1 will not change τ -match at theposition, since 1 can only serve as 1 in 123. However, 1 appears after the position that τ -match occur. Thus, σ is τ -match at the position without considering 1. Lastly, if σ is τ -match at some position t after 1. Note that 2 appears before 1 in σ . Thus, 2can serve as 1 in 123. Therefore, σ is still τ -match at position t when not considering1. So, we can remove 1 and reduce without effecting τ -match.After removing 1 and reducing, the position of 1 will be at l . Thus, this case contribute k − X l =1 P n − ,τ ,k ( x ).2. k + 1 ≤ l ≤ n −
1. In this case, σ is τ -match at position l since 1 2 σ n form a pattern τ . Morever, by the same reason as in the first case, removing 2 and reducing will noteffect τ -match at other position.After removing 2 and reducing, the position of 1 is still at l . Thus, this case contribute x ( n − k − P n − ,τ ,k ( x )3. l = n . In this case, σ is not τ -match at position l , and removing 2 will not effect τ -match at other position. Thus, this case contribute P n − ,τ ,k ( x )15n total, we have P n,τ ,k ( x ) = k − X l =1 P n − ,τ ,l ( x ) ! + x ( n − k − P n − ,τ ,k ( x ) + P n − ,τ ,k ( x )= k − X l =1 P n − ,τ ,l ( x ) ! + ( x ( n − k −
1) + 1) P n − ,τ ,k ( x )Note that, the first case vanishes when k = 1, but the formula is correct since it wouldcontribute an empty summation. The second case vanishes when n − ≤ k ≤ n , but theformula is still correct since n − k − k = n − P n − ,τ ,n ( x ) = 0 for k = n . Thelast case also vanishes when k = n , but the formula is still correct as P n − ,τ ,k ( x ) = 0.Here, we prove a similar result of τ = 132. Lemma 22.
Let τ = 132 and let P n,τ ,k ( x ) = x pmp τ ( σ ) where the sum is over all σ ∈ S n with 1 at position k . Then, P n,τ ,k ( x ) satisfies the following recursive formula: P n,τ ,k ( x ) = k − X l =1 P n − ,τ ,l ( x ) ! + x ( n − k − P n − ,τ ,k ( x ) Proof.
We derive a recursive formula for P n,τ ,k ( x ) by looking at the position of 2 in σ wherethe position of 1 is k . σ has a following form σ = σ σ . . . σ k − σ k +1 . . . σ n Let l be position of 2. We have 3 cases:1. 1 ≤ l ≤ k −
1. In this case, by the same reason as the first case in the proof of Lemma21, we can remove 1 from σ and reduce other elements without effecting τ -match atother positions.After removing 1 and reducing, the position of 1 will be at l . Thus, this case contribute k − X l =1 P n − ,τ ,k ( x ).2. l = k + 1. In this case, σ is not τ -match at position l , and removing 2 will not effect τ -match at other position by the same reason as above. Thus, this case contribute P n − ,τ ,k ( x )3. k + 2 ≤ l ≤ n . In this case, σ is τ -match at position l since 1 σ k +1 τ . Moreover, by the same reason as above, removing 2 and reducing will not effect τ -match at other position.After removing 2 and reducing, the position of 1 is still at l . Thus, this case contribute x ( n − k − P n − ,τ ,k ( x ) 16n total, we have P n,τ ,k ( x ) = k − X l =1 P n − ,τ ,l ( x ) ! + x ( n − k − P n − ,τ ,k ( x ) + P n − ,τ ,k ( x )= k − X l =1 P n − ,τ ,l ( x ) ! + ( x ( n − k −
1) + 1) P n − ,τ ,k ( x )Note that the first case vanishes when k = 1, but the formula is still correct as itcontributes an empty summation. The second case vanishes when k = n . The formula isstill correct as P n − ,τ ,n ( x ) = 0. The last case vanishes when k = n − , n . When k = n − n − k − k = n , the formula is correct as P n − ,τ ,n ( x ) = 0. It is easy to see that P ,τ , ( x ) = 1 = P ,τ , ( x ). Since, P n,τ ,k ( x ) and P n,τ ,k ( x ) satisfy thesame recursive formula and have the same initial values, we prove that P n,τ ,k ( x ) = P n,τ ,k ( x )for all n, k such that 1 ≤ k ≤ n , and thus P n,τ ( x ) = P n,τ ( x ) for all n ≥
1. Hence, we proveTheorem 20.We also construct a bijection θ : S n → S n such that pmp τ ( σ ) = pmp τ ( θ ( σ )) based onthe recursive formula. We define θ (1) = 1 , θ (12) = 12 and θ (21) = 21. In general, θ willsatisfies the following properties1. The position of 1 in σ is the same as the position of 1 in θ ( σ ).2. pmp τ ( σ ) = pmp τ ( θ ( σ )) θ satisfies the properties for S and S . For S n +1 where n ≥
2, we define the maprecursively. Given σ ∈ S n , we have σ ′ = θ ( σ ). We know that the position of 1 in σ is thesame as the position of 1 in σ ′ . Let k be the position.Here, we obtain an element ˆ σ ∈ S n +1 by applying one of the following:1. Increase every element by 1, and insert 1 at or after position k + 1.2. Increase every element except 1 by 1, and insert 2 at or after position k + 1.Then, we obtain θ (ˆ σ ) by applying similar action to σ ′ based on an action applied to σ :1. If 1 was inserted to σ , then increase every element in σ ′ by 1 and insert 1 at the sameposition as inserted in σ .2. If 2 was inserted to σ at position r , then increase every element in σ ′ except 1 by 1and insert 2 at position r ′ where r ′ = (cid:26) r + 1 k + 1 ≤ r ≤ nk + 1 r = n + 117orresponding positions when inserting 1 or 2 can be viewed from the diagram below: σ σ σ k − σ k +1 σ k +2 σ n − σ n σ ′ σ ′ σ ′ k − σ ′ k +1 σ ′ k +2 σ ′ n − σ ′ n Corresponding positions when inserting 1 σ σ σ k − σ k +1 σ k +2 σ n − σ n σ ′ σ ′ σ ′ k − σ ′ k +1 σ ′ k +2 σ ′ n − σ ′ n Corresponding positions when inserting 2The position of 1 in ˆ σ and θ (ˆ σ ) will be the same. Also, pmp τ ( σ ) would increase by 1 ifand only if 2 was inserted at a non-last position, while pmp τ ( σ ′ ) will increase by 1 if andonly if 2 was inserted at any position but k + 1. Thus, pmp τ (ˆ σ ) = pmp τ ( θ (ˆ σ )). Also, it isnot hard to see that every σ ∈ S n can be obtained by inserting 1 or 2 repeatedly in a uniqueway. Therefore, θ is a bijection.As an example, we start with θ (12) = 12. Say, we would like to insert 2, then the diagramlooks like 1 21 2Suppose we insert 2 to the preimage 12 at the last position, so we get 132. According tothe diagram, we should insert 2 to the image 12 at the first position, so we get 123. Thus, θ (132) = 123. Here, suppose we would like to insert 1, then the diagram looks like:1 3 21 2 318uppose we insert 1 to 132 at the first position, so we get 2143. According to the diagram,we insert 1 to 123 at the same position, so we get 2134. Thus, θ (2143) = 2134.Now, suppose we want to insert 2. The diagram looks like:2 1 3 42 1 4 3Suppose we insert 2 to 2134 at the second position, so we get 31245. According to thediagram, we insert 2 to 2143 at the last position, so we get 31542. Thus, θ (31245) = 31542.Below are enumerations of P n, ( x ) for the first few n . P , ( x ) = 1 P , ( x ) = 2 P , ( x ) = 5 + xP , ( x ) = 14 + 8 x + 2 x P , ( x ) = 42 + 46 x + 26 x + 6 x P , ( x ) = 132 + 232 x + 220 x + 112 x + 24 x P , ( x ) = 429 + 1093 x + 1527 x + 1275 x + 596 x + 120 x P , ( x ) = 1430 + 4944 x + 9436 x + 11384 x + 8638 x + 3768 x + 720 x Remark
In fact, the equivalence of 123 and 213 follows from a general result fromTheorem 27. S ∗ · pmp in S ∗ . However, there are at most 16 equivalent classes by Lemma6. All the representatives are listed below: 19 ××◦ ××◦× × ××◦ × ×◦× ◦ ××× × × ×◦ ◦ × ×× × ◦ ×× × ×◦ × ×× ×◦ ×◦ ×× ◦× ×× ×× ◦× × ×◦ × ×× ×◦ ×◦ ×× S ∗ . We then will show 4 following patterns are equivalent: 1234 , , , Here, for each pattern τ ∈ S ∗ , we enumerate polynomials P n,τ ( x ) for small n . We also grouppatterns together if they seem to provide the same polynomials. τ = 1234 , , P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 158 x + 43 x + 6 x P ,τ ( x ) = 2761 + 1466 x + 619 x + 170 x + 24 x P ,τ ( x ) = 15767 + 12864 x + 7598 x + 3121 x + 850 x + 120 x .1.2 τ = 1234 , , , P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 157 x + 44 x + 6 x P ,τ ( x ) = 2761 + 1439 x + 638 x + 178 x + 24 x P ,τ ( x ) = 15767 + 12420 x + 7764 x + 3341 x + 908 x + 120 x τ = 1432 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 157 x + 44 x + 6 x P ,τ ( x ) = 2761 + 1438 x + 640 x + 177 x + 24 x P ,τ ( x ) = 15767 + 12393 x + 7809 x + 3332 x + 899 x + 120 x τ = 1432 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 156 x + 45 x + 6 x P ,τ ( x ) = 2761 + 1415 x + 655 x + 185 x + 24 x P ,τ ( x ) = 15767 + 12058 x + 7895 x + 3524 x + 956 x + 120 x .1.5 τ = 1342 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 512 + 160 x + 42 x + 6 x P ,τ ( x ) = 2740 + 1500 x + 614 x + 162 x + 24 x P ,τ ( x ) = 15485 + 13207 x + 7700 x + 3016 x + 792 x + 120 x τ = 1342 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 512 + 158 x + 44 x + 6 x P ,τ ( x ) = 2740 + 1451 x + 646 x + 179 x + 24 x P ,τ ( x ) = 15485 + 12455 x + 7912 x + 3427 x + 921 x + 120 x τ = 1342 , P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 512 + 158 x + 44 x + 6 x P ,τ ( x ) = 2740 + 1454 x + 644 x + 178 x + 24 x P ,τ ( x ) = 15485 + 12533 x + 7897 x + 3377 x + 908 x + 120 x .1.8 τ = 1342 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 512 + 159 x + 43 x + 6 x P ,τ ( x ) = 2740 + 1475 x + 629 x + 172 x + 24 x P ,τ ( x ) = 15485 + 12817 x + 7781 x + 3244 x + 873 x + 120 x τ = 1324 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 158 x + 43 x + 6 x P ,τ ( x ) = 2762 + 1464 x + 620 x + 170 x + 24 x P ,τ ( x ) = 15793 + 12820 x + 7608 x + 3129 x + 850 x + 120 x τ = 1324 P ,τ ( x ) = 1 P ,τ ( x ) = 2 P ,τ ( x ) = 6 P ,τ ( x ) = 23 + xP ,τ ( x ) = 103 + 15 x + 2 x P ,τ ( x ) = 513 + 156 x + 45 x + 6 x P ,τ ( x ) = 2762 + 1414 x + 654 x + 186 x + 24 x P ,τ ( x ) = 15793 + 12041 x + 7861 x + 3539 x + 966 x + 120 x .2 Equivalence of , , and Here, we prove an equivalence of four patterns: 1234 , , ××× ◦ ◦ ××× Diagrams of 2134 and 1243The equivalence of 1234 and 2134 as well as the equivalence of 2134 and 2143 follow froma more general Theorem 27. Thus, in this section, we only prove the equivalence of 1234and 1243.
Theorem 23.
Two pmp and are pmp-Wilf equivalent.
For our convenient, let τ = 1234. Let P n,τ ,s ( x ) = X σ x pmp τ ( σ ) , where the sum is overall σ ∈ S n with the last ascent of σ is at position s . Let P n,τ ,s,t ( x ) = X σ x pmp τ ( σ ) , wherethe sum is over all σ ∈ S n with the last ascent of σ is at position s and the 1 is at position t . All feasible values of s are 1 , , . . . , n −
1. All feasible values of t are 1 , , . . . , s − , n for s = n −
1, and 1 , , . . . , n − s = n −
1. Thus, P n,τ ( x ) = 1 + n − X s =1 P n,τ ,s ( x )Also, P n,τ ,s ( x ) = s − X t =1 P n,τ ,s,t ( x ) ! + P n,τ ,s,n ( x )for s = n −
1, and P n,τ ,n − ( x ) = n − X t =1 P n,τ ,n − ,t ( x )Here, we derive a recursive formula for P n,τ ,s,t ( x ). Lemma 24. P n,τ ,s,t ( x ) satisfies the following recursive formulas: P n,τ ,s,t ( x ) = t − X i =1 P n − ,τ ,s − ,i − ( x )+( s − − t ) xP n − ,τ ,s − ,t ( x )+ s − X j = t P n − ,τ ,j,t ( x )+ P n − ,τ ,s,t ( x ) for ≤ t < s ≤ n − , . n,τ ,s,s ( x ) = 1 + s − X i =1 P n − ,τ ,i ( x ) for ≤ s ≤ n − , and P n,τ ,s,n ( x ) = P n − ,τ ,s ( x ) for ≤ s < n − .By convention, Let P n,τ ,s,t ( x ) = 0 for infeasible value of s, t .Proof. First, consider the formula for P n,τ ,s,t ( x ) when 1 ≤ t < s ≤ n − . Let l be theposition of 2 in σ , where σ ∈ S n with the last ascent of σ is at position s and 1 is at position t . We have 4 cases:1. 1 ≤ l ≤ t −
1. In this case, we can remove 1 from σ without effecting τ -match at otherpositions. That is because σ is not τ -match at the position of 2, and if σ is τ -matchat other positions and matching involves 1, it can still be τ -match by using 2 insteadof 1.After removing and reducing, the position of 1 is l , and the position of the last ascentis s −
1. Thus, this case contributes t − X i =1 P n − ,τ ,s − ,i ( x )2. t + 1 ≤ l ≤ s −
1. In this case, σ is τ -match at position l since 1 2 σ s σ s +1 form thepattern τ . By the same reasoning as the first case, we can remove 2 without effecting τ -match at any other positions.After removing and reducing, the position of 1 stays at t and the position of the lastascent is s −
1. Thus, this case contribute x ( s − − t ) P n − ,τ ,s − ,t ( x ).3. l = s . In this case, we can remove 2 without effecting τ -match at other positions.After removing and reducing, the position of 1 stays at t and the last ascent appearsat some position between t and s −
1. Thus, this case contributes s − X j = t P n − ,τ ,j,t ( x )4. l = n . In this case, we can remove 2 without effecting τ -match at other positions.After removing and reducing, the position of 1 stays at t and the position of the lastascent stays at s . Thus, this case contributes P n − ,τ ,s,t ( x )In total, we have P n,τ ,s,t ( x ) = t − X i =1 P n − ,τ ,s − ,i − ( x )+( s − − t ) xP n − ,τ ,s − ,t ( x )+ s − X j = t P n − ,τ ,j,t ( x )+ P n − ,τ ,s,t ( x )Note that, the first case vanishes if t = 1, but the formula is consistent as it contributesan empty summation. The second case vanishes when t = s −
1, but the formula is still25orrect as P n − ,τ ,s − ,s ( x ) = 0. The last case vanishes when s = n −
1, but the formula iscorrect as P n − ,τ ,n − ,t ( x ) = 0. Thus, we prove the first formula.Here, we derive a recursive formula for P n,τ ,s,s ( x ) when 1 ≤ s ≤ n −
1. Given any σ ∈ S n such that the position of 1 and the position of the last ascent is s . Removing 1 does noteffect τ -matching at other positions. After removing and reducing, the permutation eitherhas no ascent, or the last ascent is at some position between 1 and s −
1. Thus, we have P n,τ ,s,s ( x ) = 1 + s − X i =1 P n − ,τ ,i ( x )Lastly, we derive a formula for P n,τ ,s,n ( x ) when 1 ≤ s < n −
1. Given any σ such that theposition of last secent is s and 1 is at the last position. Removing 1 will not effect τ -matchat other positions. The position of last ascent is still at s . Thus, we have P n,τ ,s,n ( x ) = P n − ,τ ,s ( x )We shall prove similar result for 1243. Let τ = 1243. Let P n,τ ,s ( x ) = X σ x pmp τ ( σ ) ,where the sum is over all σ ∈ S n with the last descent of σ is at position s . Let P n,τ ,s,t ( x ) = X σ x pmp τ ( σ ) , where the sum is over all σ ∈ S n with the last ascent of σ is at position s andthe 1 is at position t . All feasible values of s are t = 1 , , . . . , s − s + 1. Note that,when s = 1, the only possible value for t is 2. Thus, we have P n,τ ( x ) = 1 + n − X s =1 P n,τ ,s ( x )Also, P n,τ ,s ( x ) = s − X t =1 P n,τ ,s,t ( x ) ! + P n,τ ,s +1 ,n ( x )Here, we derive a recursive formula for P n,τ ,s ( x ). Lemma 25. P n,τ ,s,t ( x ) satisfies the following recursive formulas: P n,τ ,s,t ( x ) = t − X i =1 P n − ,τ ,s − ,i − ( x )+( s − − t ) xP n − ,τ ,s − ,t ( x )+ s − X j = t P n − ,τ ,j,t ( x )+ P n − ,τ ,s,t ( x ) for ≤ t < s ≤ n − , . P n,τ ,s,s +1 ( x ) = 1 + s X i =1 P n − ,τ ,i ( x ) for ≤ s ≤ n − , andBy convention, Let P n,τ ,s,t ( x ) = 0 for infeasible value of s, t . roof. First, consider the formula for P n,τ ,s,t ( x ) when 1 ≤ t < s ≤ n −
1. Let l be theposition of 2 in σ , where σ ∈ S n with the last descent of σ is at position s and 1 is atposition t . we have 4 cases:1. 1 ≤ t ≤ l −
1. In this case, σ is not τ -match at position l . We can remove 1 from σ without effecting τ -match at other positions. That is because if σ is τ -match at otherpositions and the matching involves 1, it is still τ -match by using 2 instead of 1.After removing and reducing, the position of 1 is l and the position of the last descentis s −
1. Thus, this case contributes t − X i =1 P n − ,τ ,s − ,i ( x )2. t + 1 ≤ l ≤ s −
1. In this case, σ is τ -match at position l since 1 2 σ s σ s +1 form thepattern τ . By the same reasoning as the first case, we can remove 2 without effecting τ -match at any other positions.After removing and reducing, the position of 1 stays at t and the position of the lastascent is s −
1. Thus, this case contribute x ( s − − t ) P n − ,τ ,s − ,t ( x ).3. l = s . In this case, we can remove 2 without effecting τ -match at other positions.After removing and reducing, the position of 1 stays at t and the last descent appearsat some position between t + 1 and s . Thus, this case contributes s − X j = t P n − ,τ ,j,t ( x )4. l = n . In this case, we can remove 2 without effecting τ -match at other positions.After removing and reducing, the position of 1 stays at t and the position of the lastascent stays at s . Thus, this case contributes P n − ,τ ,s,t ( x )In total, we have P n,τ ,s,t ( x ) = t − X i =1 P n − ,τ ,s − ,i − ( x )+( s − − t ) xP n − ,τ ,s − ,t ( x )+ s X j = t +1 P n − ,τ ,j,t ( x )+ P n − ,τ ,t − ,t ( x )Note that, the first case vanishes if t = 1, but the formula is consistent as it contributesan empty summation. The second case vanishes when t = s −
1, but the formula is stillcorrect as P n − ,τ ,s − ,s ( x ) = 0. The last case vanishes when s = n −
1, but the formula iscorrect as P n − ,τ ,n − ,t ( x ) = 0. Thus, we prove the first formula.Next, we derive the recursive formula for P n,τ ,s,s +1 ( x ) for 1 ≤ s ≤ n −
1. Given any σ ∈ S n such that the position of the last descent is s and the position of 1 is s + 1. Removing1 will not effect τ -matching at other positions. Afer removing and reducing, the remainingpermutation either has no descent, or the last desent is at some position between 1 and s .Thus, we have P n,τ ,s,s +1 ( x ) = 1 + s X i =1 P n − ,τ ,i ( x )Thus, we prove the second formula. 27ere, we state another lemma proving equality P n,τ ,s,t ( x ) and P n,τ ,s,t ( x ). Lemma 26. P n,τ ,s,t ( x ) = P n,τ ,s,t ( x ) for ≤ t < s ≤ n − , and P n,τ ,s,s ( x ) + P n,τ ,s,n ( x ) = P n,τ ,s,s +1 ( x ) for ≤ s ≤ n − Proof.
One can check that both equations are satisfied for small n . Then, we proceed byinduction. For the first equation, consider P n,τ ,s,t ( x ) − P n,τ ,s,t ( x ), for 1 ≤ t < s ≤ n − P n,τ ,s,t ( x ) and P n,τ ,s,t ( x ) as well as inductive hypothesis, wehave P n,τ ,s,t ( x ) − P n,τ ,s,t ( x ) = P n − ,τ ,t,t ( x ) − P n − ,τ ,t − ,t ( x )Using recursive formula and inductive hypothesis, P n − ,τ ,t,t ( x ) = P n − ,τ ,t − ,t ( x ), thus weprove the first equation. For the second equation: P n,τ ,s,s ( x ) + P n,τ ,s,n ( x ) = 1 + s X i =1 P n − ,τ ,i ( x )= 1 + s X i =1 P n − ,τ ,i ( x )= P n,τ ,s,s +1 ( x )Thus, we prove the second equation. As a consequence, P n,τ ,s ( x ) = P n,τ ,s ( x ) for all1 ≤ s ≤ n −
1, and P n,τ ( x ) = P n,τ ( x ) and so, τ and τ are pmp -Wilf equivalent.As a consequence, P n,τ ,s ( x ) = P n,τ ,s ( x ) for all 1 ≤ s ≤ n − P n,τ ( x ) = P n,τ ( x ), and so τ and τ are pmp -Wilf equivalent. Hence, we prove Theorem 23. In this section, we present the equivalence of two positional marked pattern 12 p . . . p n and21 p . . . p n . Theorem 27.
Given two pmp P = 12 p . . . p l − and P = 21 p . . . p l − . Then, P and P are pmp-Wilf equivalent. We apply the technique introduced in [1] to prove Theorem 27. To prove the theorem,we need to give several definitions. Let τ = red ( p p . . . p l − ).28 efinition 28. Given π ∈ S n . Consider the diagram of π . For each cell ( i, j ) in thediagram of π , the cell is dominant if there is an occurrence of τ in the diagram of π whenonly considering row i + 1 , i + 2 , . . . , n and columns j + 1 , j + 2 , . . . , n . A cell is non-dominant if it is not dominant. It is clear that given any dominant cell, every cell to the left and below the dominant cellis also dominant. Thus, the collection of dominant cells form a Ferrers board.Let
N D ( π ) = { ( i, j ) ∈ [ n ] × [ n ] | cell ( i, j ) is non-dominant and contains ×} . Note that,if a cell is dominant, then one could find an occurrence of τ above and to the right of the cellsuch that all × involved are in non-dominant cell. If not, then every occurence of τ aboveand to the right of the cell ( i, j ) contains × in a dominant cell. Pick a copy T of τ in whicha dominant cell ( i ′ , j ′ ) containing × is the rightmost among all dominant cells containing × in all copies of τ above and to the right of ( i, j ). Since cell ( i ′ , j ′ ) is also dominant, one canfind a copy T of τ above and to the right of ( i ′ , j ′ ), which is also above and to the right ofcell ( i, j ). Thus, this copy will also contain a dominant cell containing × , contradicting tothe fact that the T contains the rightmost dominant cell.Thus, if N D ( π ) is known, one can recover the set of dominant cells in the diagram of π completely. Given any Q ⊆ [ n ] × [ n ], Let S Qn = { π ∈ S n | N D ( π ) = Q } . We shall prove that X π ∈ S Qn x pmp P ( π ) = X π ∈ S Qn x pmp P ( π ) Here, we analyse the set S Qn . First, we only need to consider those Q such that S Qn = ∅ .Given any such Q , we can obtain an element in S Qn by filling × in the dominant part ofthe diagram until every row and column contains precisely one × . This gives all elementsin S Qn since filling × in the dominant part does not alter whether a cell is dominant ornon-dominant.To fill × in dominant cells, we start will all dominant cells and eliminate all rows andcolumns that already contain × from the set Q . The remaining cells form a Ferrers board.Let λ ( Q ) denote the shape of the Ferrers board obtained from the process above. Definition 29.
Given any Ferrers board of shape λ = ( λ , λ , . . . , λ k ) with λ ≥ λ ≥ . . . ≥ λ k , a filling of λ is an assignment of × in the Ferrers board of shape λ such that every rowand columns contain precisely one × . In order for λ to have a filling, the number of rows of λ must be the same as the numberof columns in λ . More specifically, if λ = ( λ , λ , . . . , λ k ), λ has a filling if and only if( k − i + 1) ≤ λ i ≤ k for all k . Let S λ denote the set of fillings of λ .Given any π ∈ S λ , and p ∈ S k , we say that π contains p if π contains a subdiagram of p .Note that, in order for π to contain p , the entire diagram of p has to present as a subdiagramof π including cells not containing × . For example, the filling below contains p = 12 butdoes not contain q = 21. × × ×× S λ similar to S n . Given any π ∈ S λ and p ∈ S ∗ k , we say that π is p -match at position t if when replacing × in the t -th column of π of by ◦ , then it contains a subdiagram p . For example, the diagram above is 12-match atposition 3 as shown below × × ×◦ Let pmp p ( π ) denote the number of position t such that π is p -match at position t . Here,we state a lemma which will be our main tool to prove Theorem 27. Lemma 30.
Given any Q ⊆ [ n ] × [ n ] such that S Qn = ∅ . Then X π ∈ S Qn x pmp P ( π ) = X π ∈ S λ ( Q ) x pmp ( π ) and X π ∈ S Qn x pmp P ( π ) = X π ∈ S λ ( Q ) x pmp ( π ) Proof.
Note that there is a natural bijection φ : S λ ( Q ) → S Qn which maps any filling π ∈ λ ( Q )to a permutation with non-dominant part corresponding to Q and dominant part having thesame filling as π . We will show that π is 12-match at a position if and only if φ ( π ) is P -matchat the corresponding position according to the map φ The converse is obvious. For the forward direction, suppose π ∈ S λ ( Q ) is 12-match ata certain position. Thus, the corresponding position in φ ( π ) is 12-match with every cellsinvolved are dominant. Thus, we can find a copy of τ above and to the right of every cellsinvolved in 12-match. Therefore, the 12-match together with τ makes φ ( π ) P -match at theposition corresponding to the 12-match in π .Therefore, φ is a bijection between S λ ( Q ) and S Qn such that pmp ( π ) = pmp P ( φ ( π )).Thus, it proves the first equality of the lemma. The second equation can be proved with theexact same reasoning.As an example, let n = 9, τ = 12, (so P = 1234 and P = 2134), and Q is as below ×× × × Q , we recover dominant and non-dominant cells. We fill • in non-dominant cells. • • • • • • • • ו × • • • • • • •• • • • • • • •• • • × • • • •• • • • •• • • • •• • • × •• •• • We want to fill × into the diagram so that every row and column contains precisely one × , thus we eliminate all empty cells that are in same rows or columns with cell containing × . We fill • in such cells. • • • • • • • • ו × • • • • • • •• • • • • • • •• • • • × • • • •• • • • • •• • • • • •• • • • • • • × •• • • •• • • • Thus, the remaining cells form a Ferrers board λ ( Q ) = (5 , , , , S Q and S λ ( Q ) , we fill cells in λ ( Q )the same way we fill available cells in a diagram Q . For example, below is an example of thecorrespondence: π = × ×× ×× ←→ φ ( π ) = • • • • • • • • ו × • • • • • • •× • • • • • • • •• • • • × • • • •• × • • • • •• × • • • • •• • • • • • • × •• • × • •• • × • • π corresponds to 1234-matching at position4 of φ ( π ). π = × ◦× ×× ←→ φ ( π ) = • • • • • • • • ו × • • • • • • •× • • • • • • • •• • • • × • • • •• ◦ • • • • •• × • • • • •• • • • • • • × •• • × • •• • × • • Here, in order to prove theorem 27, we only need to prove that, given any Ferrers board λ , X π ∈ S λ x pmp ( π ) = X π ∈ S λ x pmp ( π ) . In fact, we find a formula for both polynomials. Lemma 31.
Let λ = ( λ , λ , . . . , λ k ) be a Ferrers board such that ( k − i + 1) ≤ λ i ≤ k forall ≤ i ≤ k . Then, X π ∈ S λ x pmp ( π ) = X π ∈ S λ x pmp ( π ) = k Y i =1 (1 + ( λ i − ( k − i + 1)) x ) Proof.
Given any λ satisfying ( k − i + 1) ≤ λ i ≤ k for all 1 ≤ i ≤ k , let λ be a Ferrersboard obtained from λ by removing the top most row and left most column. That is λ =( λ − , λ − , . . . , λ k − − X π ∈ S λ x pmp ( π ) = (1 + ( λ k − x ) X π ∈ S λ x pmp ( π ) To prove the equation above, we consider the topmost row in λ . The filling is 12-matchat the position of × in the topmost row if and only if × is not in the rightmost possible cell.So, there are λ k − × so that it is 12-match, and 1 position otherwise. Oncethe top row is filled, we consider filling the rest of λ by remove the top row and the columncontaining × . The remaining cells form a shape λ . Thus, the equation above is proved.For the pattern 21, a similar reasoning can also be applied. The filling of λ is 21-matchat the position of × in the topmost row if and only if × is not in the leftmost possible cell.Hence, we have X π ∈ S λ x pmp ( π ) = (1 + ( λ k − x ) X π ∈ S λ x pmp ( π ) Therefore, we proved the lemma.Here, we prove the main theorem: 32 roof. (of Theorem 27) Note that S n is a disjoint union of S Qn for all Q such that S Qn = ∅ .So, we have P n,P ( x ) = X σ ∈ S n x pmp P ( σ ) = X Q X σ ∈ S Qn x pmp P ( σ ) = X Q X σ ∈ S λ ( Q ) x pmp ( σ ) By Lemma 30= X Q X σ ∈ S λ ( Q ) x pmp ( σ ) By Lemma 31= X Q X σ ∈ S Qn x pmp P ( σ ) By Lemma 30= X σ ∈ S n x pmp P ( σ ) = P n,P ( x )Thus, P and P are pmp -Wilf equivalent. Remark
Let P = 21 p p . . . p l − . By applying the same reasoning, we can prove that P and P are pmp -Wilf equivalent by considering the rightmost column of any Ferrers board λ . Thus, we prove that P , P and P are pmp -Wilf equivalent. One way to generalize positional marked pattern is to consider multiple patterns. Given acollection of positional marked patterns Γ and σ ∈ S n , we say that σ is Γ-match at position l if σ is τ -match at position l for some τ ∈ Γ. Given σ ∈ S n , let pmp Γ ( σ ) denote thenumber of position l such that σ is Γ-match at position l . Let P n, Γ ( x ) = X σ ∈ S n x pmp Γ ( σ ) .Given two collections of positional marked patterns Γ and Γ , they are Wilf-equivalent if P n, Γ ( x ) = P n, Γ ( x ). This is a nice generalization to positional marked patterns since theconstant term of P n, Γ ( x ) enumerates the number of σ ∈ S n avoiding every patterns in Γ,which replicates what P n,τ ( x ) provides in single pattern cases. Here, we stated, withoutproof, a result on collection of positional marked patterns similar to Theorem 27. The proofalso follows the same logic as in the proof of Theorem 27. Theorem 32.
Let τ , τ , . . . , τ l be rearrangements of { , , . . . , k } , so that τ i and τ i areelements of S ∗ k . Let Γ = { τ i | ≤ i ≤ l } and Γ = { τ i | ≤ i ≤ l } . Then Γ and Γ areWilf-equivalent. Moreover, define Γ-matching would let us realize positional marked pattern as a refine-ment of marked mesh pattern defined by Remmel and Kitaev [3]. Given a, b, c, d ∈ Z ≥ , andgiven σ ∈ S n , we say σ matchs M M P ( a, b, c, d ) at position l if in the diagram of σ relative33o the coordinate system which has × in the l -th column as its origin there are at least a × ’s in quadrant I, at least b × ’s in quadrant II, at least c × ’s in quadrant III, and at least d × ’s in quadrant IV. As an example, σ = 25417683 matches M M P (3 , , ,
1) at position 3: ×× ×× × ×× ×
IIIIII IVIt is easy to see that, for any a, b, c, d , M M P ( a, b, c, d ) is equivalent to Γ-matching forsome collection of positional marked patterns Γ. For example, to match M M P (2 , , ,
0) atsome position, it is the same as to match Γ = { , } at the same position. Therefore, aswe introduce multiple positional marked patterns, one can realize posional marked patternsas a refinement of marked mesh pattern. References [1] E. Babson and J. West. The permutations 123p4 . . . pm and 321p4 . . . pm are wilf-equivalent.
Graphs and Combinatorics , 16:173:373–380, 2000.[2] P. Br¨and´en and A. Claesson. Mesh patterns and the expansion of permutation statisticsas sums of permutation patterns.
Electronic J. Combin. , 18:2, 2011.[3] S. Kitaev and J. Remmel. Quadrant marked mesh patterns.
Journal of Integer Sequence ,15 issue 4(12.4.7):29, 2012.[4] Henning A. ´Ulfarsson. A unification of permutation patterns related to schubert varieties.