Pattern Avoidance of Generalized Permutations

Transcription

1 Pattern Avoidance of Generalized Permutations arxiv: v1 [math.co] 17 Apr 2018 Zhousheng Mei, Suijie Wang College of Mathematics and Econometrics Hunan University, Changsha, China. Abstract In this paper, we study pattern avoidances of generalized permutations and show that the number of all generalized permutations avoiding π is independent of the choice of π S 3, which extends the classic results on permutations avoiding π S 3. Extending both Dyck path and Riordan path, we introduce the Catalan- Riordan path which turns out to be a combinatorial interpretation of the difference array of Catalan numbers. As applications, we interpret Riordan numbers in two ways, via semistandard Young tableaux of two rows and generalized permutations avoiding π S 3. Analogous to Lewis s method, we establish a bijection from generalized permutations to rectangular semistandard Young tableaux which will recover several known results in the literature. Keywords: Generalized permutations, Pattern avoidances, Riordan numbers, Young tableaux, RSK correspondence 1 Introduction A composition of a positive integer N is a sequence α = (α 1,α 2,...,α n ) of nonnegative integers such that α 1 + +α n = N. Denote by [α] = {1 α 1,2 α 2,...,n αn } the multiset with multiplicity α i of i [n]. Given two compositions α and β of N, a generalized permutation of α β is a one-to-one correspondence from multisets [α] to [β]. Denote by S αβ the set of all generalized permutations of α β. Each generalized permutation τ S αβ can be written uniquely in two-rowed array ( ) n n τ = τ 11 τ 1α1 τ 21 τ 2α2 τ n1... τ nαn such that τ i1 τ i2 τ iαi for all 1 i n and {τ 11,τ 12,...,τ nαn } = [β] as multisets. Without causing confusions, we abbreviate τ = τ 11 τ 12...τ nαn and call the 1

2 subsequence τ(i) = τ i1 τ i2...τ iαi the i-th block of τ. For convenience, write continuous repetitions of equal elements as exponential notations, e.g., the composition (1, 2, 2, 2, 1) is abbreviated (1,2 3,1). When α = β = (1 n ), S (1 n )(1 n ) = S n is the set of all permutations on [n] = {1,2,...,n}. As classical counting problems, pattern avoidances of permutations have been well studied, see [9, 14, 17] etc.. In this paper, we consider the pattern avoidances of generalized permutations. Definition 1.1. For τ = τ 11 τ 12...τ nαn S αβ and π S m, τ avoids the pattern π if no index sets 1 i 1 < i 2 < < i m n and j 1,j 2,...,j m satisfy that the subsequence τ i1 j 1 τ i2 j 2...τ imj m is order isomorphic to π, i.e., τ isj s < τ itj t iff π s < π t. Let S αβ (π) denote the set of generalized permutations in S αβ which avoid the pattern π. ( ) For example, for α = (1,2,2,1),β = (3,2,1), and τ = S αβ, we have τ S αβ (132) and τ / S αβ (231), ( ) since the subsequence of τ is of pattern 231. In 1973, Knuth [6] established a celebrated result on pattern avoidances of permutations. Theorem 1.2. [6] The cardinality S (1 n )(1 n )(π) = C n = 1 ( ) 2n n+1 n is independent of the choice of π S 3, where C n is the n-th Catalan number. In 1985, Simion and Schmidt [14] determined the number S (1 n )(1 n )(T) of permutations simultaneously avoiding any given set T of patterns in S 3. In 2001, Albert, Aldred, Atkinson, Handley, and Holton [1] calculated S (1 n )β(t) for any subset T S 3. In 2006, using the results in [1, 2], Savage and Wilf [13] observed that Theorem 1.3. [13] If β = (β 1,...,β m ) is a composition of N, then the cardinality of S (1 N )β(π) is independent of the choice of π S 3 and the order of entries of β. In 2007, Myers [11] gave a bijection proof on Theorem 1.3 by extending the construction of Simion and Schmidt [14] for permutations. Let λ be a partition of N and µ a composition of N. The number K λµ of SSYTs (semistandard Young tableaux) of shape λ and type µ is called the Kostka number, which is is independent of the order of entries of µ and Schur-concave on µ, i.e., K λµ K λµ if µ dominates µ for any two partitions µ and µ. Below we consider generalized permutations avoiding π S 3 and a parallel result as above, whose proof is given in Section 2. 2

3 Theorem 1.4. Let α = (α 1,...,α n ) and β = (β 1,...,β m ) be two compositions of N and denote (α,β) = (α 1,...,α n,β 1,...,β m ). The cardinality of S αβ (π) is the Kostka number K (N,N)(α,β), i.e., S αβ (π) = S βα (π) = K (N,N)(α,β) which is independent of the choice of π S 3 and the order of entries of α and β, and Schur-concave on both α and β if α and β are partitions. Recall that the n-th Catalan number C n is the number of Dyck paths from (0,0) to (2n,0), and the number of SYTs (standard Young tableaux) of shape (n 2 ), i.e., K (n 2 )(1 2n ) = S (1 n )(1 n )(π) = C n. The n-th Riordan number R n is the number of Riordan paths, i.e., Motzkin paths from (0,0) to (n,0) with no horizontal steps at level 0. In Section 3, we introduce the Catalan-Riordan paths which extend both Dyck and Riordan paths, which turns out to be a combinatorial interpretation on the difference array of Catalan numbers. As applications, we obtain two new combinatorial interpretations for Riordan numbers, K (n 2 )(2 n ) = R n and S (2 n )(2 n )(π) = R 2n. In 2011, Lewis [7] established a bijection between block-ascending permutations and rectangular SYTs to enumerate two classes of block-ascending permutations via the celebrated hook length formula. In Section 4, Lewis s construction will be extended to any pair of SSYTs. Theorem 1.5. Let α = (α 1,...,α n ) and β = (β 1,...,β m ) be two compositions of N. There is a bijection between the set S αβ of all generalized permutations of α β and the set of rectangular SSYTs of shape (N m ) and type (α,n β), i.e., S αβ = λ NK λα K λβ = K (N m )(α,n β), where (α,n β) = (α 1,...,α n,n β 1,...,N β m ). In Section 5, we present some applications of above results which contain the main results of Lewis [7], Mei and Wang [10], and Chen [5]. 2 Proof of Theorem 1.4 Lemma 2.1. [18] Kostka numbers K λµ are Schur-concave on µ. Namely, if λ, µ, and ν are partitions of N, µ = (µ 1,µ 2,...) and ν = (ν 1,ν 2,...) satisfy that µ 1 + +µ i ν 1 + +ν i for all i 1, then K λµ K λν. 3

4 With the above result, Theorem 1.4 is an easy consequence of three lemmas below. If α = (α 1,α 2,...,α n ) is a composition, let ᾱ = (α n,α n 1,...,α 1 ) be the reverse of α. For β = (β 1,β 2,...,β m ) and τ S αβ, define τ Sᾱβ and τ S α β as follows. The i-th block of τ is the (n+1 i)-th block of τ, i.e., τ ij = τ (n+1 i)j for all 1 j α n+1 i. The i-th block of τ is defined by τ ij +τ i(α i +1 j) = m+1 for all 1 j α i. ( ) E.g., α = (1,1,2,1),β = (2,2,1), and τ = S αβ, then ᾱ = (1,2,1,1), β = (1,2,2), τ = ( Below is a quick fact. Lemma 2.2. ) Sᾱβ and τ = S αβ (123) = Sᾱβ (321), ( S αβ (213) = Sᾱβ (312) = Sᾱ β(132) = S α β(231). ) S α β. Given two compositions α = (α 1,...,α n ) and β = (β 1,...,β m ) of N. Let P and Q be SSYTs of the same shape λ = (λ 1,λ 2,...) and of types α and β respectively. Denote by P Q a rectangular SSYT of shape (N,N) and type (α, β) whose (i,j)-entry is P Q(i,j) = { P(i,j) if j λi ; n+m+1 Q(3 i,n +1 j) else. Namely, P Q is obtained by rotating Q by 180, replacing each entry i of Q with n + m + 1 i, and jointing the resulting diagram with P. E.g., If P = and Q = , then P Q = For any τ S αβ (321), if τ RSK (P,Q), then P and Q have at most two rows. It is clear that the composition of and RSK correspondence τ RSK (P,Q) P Q gives a bijection from S αβ (321) to the set of all rectangular SSYTs of shape (N,N) and type (β,ᾱ). Thus, S αβ (321) = K (N,N)(β,ᾱ) = K (N,N)(α,β). Lemma 2.3. For two compositions α and β of N, the cardinality S αβ (321) = S βα (321) = K (N,N)(α,β) is independent of the order of the entries of α and β. 4

5 Lemma 2.4. For any two compositions α and β of N, we have S αβ (213) = S αβ (123). Proof. Suppose α = (α 1,α 2,...,α n ) and β = (β 1,β 2,...,β m ) are compositions of N. The result is trivial for n = 1. For n 2, we will show that S αβ (213) and S αβ (123) have the same recursive formula S αβ (π) = n ( Sα (k)β (π) S α (k)β (π) ) for π = 123 or 213, (1) k=1 where S αβ (π) = 0 if some entry of (α,β) is negative, α 0 = 0, and for 1 k n, α (k) = (α 0 +α 1 + +α k 1,α k 1,α k+1,...,α n ), α (k) = (α 0 +α 1 + +α k 1 1,α k 1,α k+1,...,α n ), β = (β 1,...,β m 1,β m 1), β = (β 1,...,β m 1,β m 2). Given τ = τ 11 τ 12...τ 1α1...τ n1...τ n(αn 1)τ nαn S αβ (123), let k [n] be the minimum index such that τ kαk = m. If k = 1, let τ be the generalized permutation obtained from τ by removing τ 1α1. Then τ S α (1)β and avoids the pattern 123. It gives a bijection between {τ S αβ (123) τ 1α1 = m} and S α (1)β (123). Note S α (1)β (123) = 0 by assumptions. Thus #{τ S αβ (123) τ 1α1 = m} = S α (1)β (123) S α (1)β (123). If k 2, then τ iαi < τ kαk = m for all 1 i < k. Since τ avoids the pattern 123, the subsequence τ 11 τ 12...τ 1α1...τ (k 1)1...τ (k 1)αk 1 of the first k 1 blocks avoids the pattern 12, which implies τ (k 1)1 τ (k 1)αk 1 τ 21 τ 2α2 τ 11 τ 1α1 < m = τ kαk. Let γ = (γ 1,...,γ n k+2 ) = (α 1 + +α k 1,α k,...,α n ). The sequence τ = τ (k 1)1...τ (k 1)αk 1...τ 21...τ 2α2 τ 11...τ 1α1 τ k1...τ kαk...τ n1...τ nαn becomes a generalized permutation in X = {σ S γβ (123) σ 1γ1 < σ 2γ2 = m}. Indeed, the map τ τ defines a bijection in this situation, i.e., for k 2 X = #{τ S αβ (123) τ iαi < τ kαk = m for 1 i < k}. 5

6 Let X = {σ S γβ (123) σ 2γ2 = m} and X = {σ S γβ (123) σ 1γ1 = σ 2γ2 = m}. Obviously, X = X X. Similarasbefore, wecanconstruct two bijections, between X and S γ (2)β (123) by removing σ 2γ 2 from σ X, and between X and S γ (2)β (123) by removing σ 1γ1,σ 2γ2 from σ X. It follows that in this case, Notice Hence X = S γ (2)β (123) S γ (2)β (123). S γ (2)β (123) = S α (k)β (123) and S γ (2)β (123) = S α (k)β (123). #{τ S αβ (123) τ iαi < τ kαk = m for 1 i < k} = S α (k)β (123) S α (k)β (123). So far, we have proved that S αβ (123) satisfies the recursive formula (1). For S αβ (213), the proof is similar. For τ S αβ (213), let k [n] be the minimum index such that τ kαk = m. If k = 1, the proof is the same as before. If k 2, then τ iαi < τ kαk = m for all 1 i < k. Since τ avoids the pattern 213, the subsequence τ 11 τ 12...τ (k 1)1...τ (k 1)αk 1 avoids the pattern 21, which implies τ 11 τ 1α1 τ 21 τ (k 1)1 τ (k 1)αk 1 < m = τ kαk. Then τ can be regarded as a generalized permutation in X. Applying the same arguments as before, we can obtain that S αβ (213) also has the recursive formula (1). The proof completes by induction on N. 3 Catalan-Riordan Paths In this section, we will introduce the Catalan-Riordan path, which extends the concepts of Dyck and Riordan paths. As applications, we obtain two interpretations of Riordan numbers on SSYTs and pattern avoidance respectively, which are new in our knowledge. A Dyck path of length 2n is a lattice path in the xy-plane from (0,0) to (2n,0) with the step set {(1,1),(1, 1)} and never going below the x-axis. The number of Dyck paths of length 2n is the n-th Catalan number C n. A Riordan path of length n is a lattice path in the xy-plane from (0,0) to (n,0) with the step set {(1,1),(1,0),(1, 1)} that has no step (1,0) on the x-axis and never goes below the x-axis. The number of Riordan paths of length n is the n-th Riordan number R n (or called ring number). See OEIS [15, A005043] for more investigations and combinatorial interpretations of Riordan numbers. In 1999, Bernhart [3] studied the difference array of Catalan numbers and found a formula on the relation of Catalan and Riordan numbers. 6

7 Theorem 3.1. [3] If C n and R n arethe n-th Catalanand Riordannumbers respectively, then n ( ) n n ( ) n C n = R i and R n = ( 1) n i C i. i i In 2008, Chen, Deng, and Yang [4] obtained an interpretation of Riordan numbers via pattern avoidances. Theorem 3.2. [4] The number of derangements on [n] that avoid both 321 and is the n-th Riordan number R n. In 2010, Regev [12] gave an interpretation of Riordan numbers via SYTs. Theorem 3.3. [12] For any nonnegative integer n, the n-th Riordan number R n is the number of SYTs of shape (k,k,1 n 2k ) for all k 0, i.e., R n = 1 n+1 k=0 n ( )( ) n+1 n k 1. k k 1 Definition 3.4 (Catalan-Riordan Path). For 0 k n, let CR(n,k) be the set of lattice paths of length n+k in the xy-plane from (0,0) to (n+k,0) such that (1) the first 2k steps have the step set {(1,1),(1, 1)}; (2) the last n k steps have the step set {(1,2),(1,0),(1, 2)}, and no step (1,0) on the x-axis; (3) never goes below the x-axis. Members of CR(n, k) are called Catalan-Riordan paths of size (n, k), and the cardinality of CR(n, k) is called the (n, k)-th Catalan-Riordan number, denoted CR(n, k). Note that CR(n,n) is the set of Dyck paths from (0,0) to (2n,0). When k = 0, replacing all steps (1,2) and (1, 2) with steps (1,1) and (1, 1) respectively, CR(n,0) becomes the set of Riordan paths from (0,0) to (n,0). Namely, CR(n,0) = R n and CR(n,n) = C n. Theorem 3.5. There is a bijection between CR(n,k) and the set of SSYTs of shape (n,n) and type (1 2k,2 n k ), i.e., CR(n,k) = K (n,n)(1 2k,2 n k ). In particular, C n = K (n 2 )(1 2n ) and R n = K (n 2 )(2 n ). 7

8 Proof. Let T = (t ij ) be a SSYT of shape (n 2 ) and type (1 2k,2 n k ). For 1 s n+k, suppose the number of s in the first row of T is y s more than the second row, i.e., y s = #{j [n] t 1j = s} #{j [n] t 2j = s}. Obviously, y s = 1, 1 if 1 s 2k and y s = 2,0, 2 otherwise. We construct a lattice path L starting at (0,0) such that the s-th step of L is (1,y s ) for all 1 s n+k. We claim that this construction defines a map from the set of SSYTs of shape (n,n) and type (1 2k,2 n k ) to CR(n,k). Note that all entries t ij with t ij s appearing in T form a SSYT whose second row is no longer than the first row, which implies y 1 + +y s 0, i.e., the path L never goes below the x-axis. To prove the path L has no step (1,0) on the x-axis, suppose L goes to the x-axis at step s 2k, i.e., y y s = 0. From the definition of y s, the entries t ij with t ij s appearing in T form a SSYT whose two rows have the same length. It forces that both s+1 must lie in the first row of T, i.e., y s+1 = 2. Since two rows of T have the same length, it follows by similar arguments that the path L ends at (n+k,0). One can easily obtain the inverse map by inverting the above constructions and prove that it is well-defined. Corollary 3.6 (Recursive Formula). For 1 k n, CR(n,k) = CR(n,k 1)+CR(n 1,k 1). Proof. From Theorem 3.5, CR(n,k) = K (n,n)(1 2k,2 n k ). Let T n,k = {T T is a SSYT of shape (n,n) and type (1 2k,2 n k )} = {T = (t ij ) T n,k t 12 = 2} {T = (t ij ) T n,k t 21 = 2}. For T = (t ij ) T n,k with t 12 = 2, replacing t 12 = 1, one obtains a SSYT of shape (n,n) and type (2,0,1 2(k 1),2 n k ). Thus #{T = (t ij ) T n,k t 12 = 2} = K (n,n)(1 2(k 1),2 n (k 1) ). For T = (t ij ) T n,k with t 21 = 2, removing t 11 and t 21 from T, one obtains a SSYT of shape (n 1,n 1) and type (1 2(k 1),2 n k ). Thus By Theorem 3.5, the result holds. #{T = (t ij ) T n,k t 21 = 2} = K (n 1,n 1)(1 2(k 1),2 n k ). Indeed, similar as above arguments, if α = (α 1,...,α m ) is a composition of 2n, we can obtain a recursive formula for K (n 2 )α, K (n 2 )α = min{α 1,α 2 } K (n 2 )(α 1 +α 2 2i,α 3,...,α m). 8

9 Let D(f n ) = f n f n 1 be the difference of the sequence f n, and D k (f n ) = D(D k 1 (f n )) the k-th difference of f n for k 2. It is easily seen that CR(n,k) = D k (C n ), which gives an combinatorial interpretation on the difference array of Catalan numbers, see OEIS [15, A059346]. Corollary 3.7. If 0 k n, then K (n,n)(1 2k,2 n k ) = CR(n,k) = k ( ) k n k ( ) n k R n i = ( 1) i C n i. i i In particular, C n = n n ) ( Ri and R i n = n ( 1)i( n i) Ci. Proof. From Corollary 3.6, we have for any 1 l k n, CR(n,k) = l ( ) l CR(n i,k l). i Recall that CR(n,0) = R n, taking l = k in above equation, we have CR(n,k) = ) Rn i. Since CR(n,n) = C n, we have k ( k i C n = l ( ) l CR(n i,n l) for all n 1. i Via the inclusion-exclusion principle, we have CR(n,n l) = l ( ) l ( 1) i C n i. i Let α = (1 m 1,2 n 1 ) and β = (1 m 2,2 n 2 ) be two compositions of n, and assume m 1 + m 2 = 2k. Applying Theorem 1.4, Theorem 3.5, and Corollary 3.7, we have for any π S 3, S αβ (π) = K (n,n)(α,β) = K (n,n)(1 2k,2 n k ) = CR(n,k) = k ( ) k R n i. i Analogous to Catalan numbers, below are two interpretations on Riordan numbers, one via SSYTs of two rows and the other via π-avoidance of generalized permutations with π S 3. 9

10 Theorem 3.8. (1) The number of SSYTs of shape (n 2 ) and type (2 n ) is K (n 2 )(2 n ) = R n. (2) For any π S 3 and k,n Z 0 with 0 k n, if n 1,m 1,n 2,m 2 Z 0 satisfy m 1 +2n 1 = m 2 +2n 2 = n and m 1 +m 2 = 2k, then In particular, S (1 m 1,2 n 1)(1 m 2,2 n 2) (π) = k ( ) k R n i. i S (2 n )(2 n )(π) = R 2n and S (1,2 n )(1,2 n )(π) = R 2n +R 2n+1. 4 Extension of Lewis s Construction Let α = (α 1,...,α n ) and β = (β 1,...,β m ) be two compositions of N. In the rest of this paper, m and n can be any integer m and n with m m and n n respectively only if we take α i = β j = 0 for n < i n and m < j m. For any d N, denote S d (α,β) = {(P,Q) sh(p) = sh(q) (d m ),type(p) = α,type(q) = β}, T d (α,β) = {T sh(t) = (d m ),type(t) = (α,d β)}, where sh(p) and type(p) are the shape and type of P, d β = (d β m,...,d β 1 ). Define a map θ : S d (α,β) T d (α,β) such that if P = (P ij ) and Q = (Q ij ) are two SSYTs of the same shape λ = (λ 1,λ 2,...) with λ 1 d and of types α and β respectively, then θ(p,q) = T = (T ij ) is defined as follows, for 1 k m, { Pij if 1 i l(λ), 1 j λ i ; T ij = n+k if k i m, λ i k+1 (m k +1) < j λ i k (m k), where λ 0 (k) = d and λ i (k) is the maximal j such that Q ij k, i.e., λ i (k) = #{j Q ij k}. Indeed, from the proof of Theorem 4.3, we will see that the map θ can be realized by the following algorithm which extends those constructions by Lewis [7], Mei and Wang [10]. 10

11 Algorithm 4.1. We start with a Young diagram of shape (d m ) whose upper-left corner of shape λ is filled with P. Next we fill the remaining empty boxes of shape (d m )/λ by m steps. At the first step, we fill the number n+m into each box of the rightmost d β 1 boxes of the bottom row. Generally at the i-th step for 2 i m, for those remaining empty boxes of the diagram (d m ) after step i 1, the bottom one of those d β i columns where Q contains no i is filled by the entry m+n+1 i. After the m-th step, we obtain θ(p,q). Example 4.2. Let P and Q be two SSYTs with sh(p) = sh(q) = (4,3,2), type(p) = (2,2,3,2), and type(q) = (3,2,1,1,2) as follows, P = and Q = Taking d = 5, then we fill the Young diagram of shape (5 5 ) step by step as follows, On the other hand, by the definition of λ i (k), we have = θ(p,q). λ 1 (1) = λ 1 (2) = λ 1 (3) = λ 1 (4) = 3,λ 1 (5) = 4, λ 2 (1) = 0,λ 2 (2) = λ 2 (3) = 2,λ 2 (4) = λ 2 (5) = 3, λ 3 (1) = λ 3 (2) = 0,λ 3 (3) = λ 3 (4) = 1,λ 3 (5) = 2, λ 0 (k) = 5 and λ 4 (k) = λ 5 (k) = 0. By the definition of θ, we have T ij = 4+k iff λ i k+1 (5 k +1) < j λ i k (5 k). So T ij = 5 iff λ i (5) < j λ i 1 (4) = T 15 = T 33 = T 41 = 5, T ij = 6 iff λ i 1 (4) < j λ i 2 (3) = T 51 = T 42 = T 24 = T 25 = 6, T ij = 7 iff λ i 2 (3) < j λ i 3 (2) = T 52 = T 43 = T 34 = T 35 = 7, T ij = 8 iff λ i 3 (2) < j λ i 4 (1) = T 53 = T 44 = T 45 = 8, T ij = 9 iff λ i 4 (1) < j λ i 5 (0) = T 54 = T 55 = 9. 11

12 Theorem 4.3. The map θ is a bijection. Proof. First we prove that θ is well-defined. Suppose θ(p,q) = T = (T ij ) for P,Q S d (α,β). We need to show that T is a SSYT of shape (d m ) and type (α,d β). By the definition of λ i (k), we have Q (i+1)(λi (k)+1) Q i(λi (k)+1) k + 1 Q (i+1)λi+1 (k+1) which implies λ i (k)+1 > λ i+1 (k +1), i.e., Given k i m, since λ i (k) λ i+1 (k +1). {j T ij = n+k} = {j λ i k+1 (m k +1) < j λ i k (m k)}, {j T ij = n+k +1} = {j λ i k (m k) < j λ i k 1 (m k 1)}, (2)... we have {j T ij n+k} = {j j > λ i k+1 (m k +1)}. Thus for 1 i m k +1, Taking k = 1, we obtain {j T (i+k 1)j n+k} = {j λ i (m k +1) < j d}. {(i,j) T ij n+1} = {(i,j) λ i (m) = λ i < j d}, which implies that the shape of T is (d m ). To prove that the type of T is (α,d β), since λ i (k)isthemaximal j such thatq ij k, it followsthat Q ij = k iff λ i (k 1) < j λ i (k). Thus for 1 i m k +1, By (2), {j Q ij = m k +1} = {j λ i (m k) < j λ i (m k +1)}. {j T (i+k 1)j = n+k} = {j λ i (m k +1) < j λ i 1 (m k)}. Then for 1 i m k +1, we have #{j Q ij = m k +1}+#{j T (i+k 1)j = n+k} = λ i 1 (m k) λ i (m k). Notethatλ i (k) = 0fori > k. Then m i=1 (λ i 1(m k) λ i (m k)) = d. Fori > m k+1, there is no j [d] such that Q ij = m k+1 and T (i+k 1)j = n+k. If each pair (i,j) with T (i+k 1)j = n+k for 1 i m k+1 is substituted by the pair (i,j) with T ij = n+k for k i m, then #{(i,j) T ij = n+k}+#{(i,j) Q ij = m k +1} = d. 12

13 So T is of shape (α,d β). For any k i m, it follows by (2) that in each row of T, the boxes with entry n+k+1 are strictly right of the boxes with entry n+k, i.e., T is weakly increasing across each row. It remains to show that T is strictly increasing down each column. Suppose there exist i and j such that T ij = n + k and T (i+1)j = n + k with k > k. Since λ i (k) λ i+1 (k +1), we have Via the definition of T ij, λ i k +1(m k ) λ i k+1 (m k) λ i k+1 (m k +1). j {j λ i k+1 (m k+1) < j λ i k (m k)} {j λ i k +2(m k +1) < j λ i k +1(m k )}, which is a contradiction. Thus T T d (α,β) and the map θ is well-defined. To prove θ is a bijection, it is enough to show that for any T T d (α,β), there is a unique pair (P,Q) S d (α,β) such that θ(p,q) = T. Given T T d (α,β), define P = (P ij ) as P ij = T ij if 1 T ij n. It is obvious that P is uniquely determined by T. Since T is of shape (d m ) and type (α,d β), we have sh(p) (d m ) and type(p) = α. Denote by λ = (λ 1,λ 2,...) the shape of P and assume λ i = 0 if i > l(λ). For i [m] and s [m+n], let µ i (s) = #{j T ij s}. If i [m], µ i (n) = λ i and µ i (n + k) = d for i k. Let µ i (s) = 0 if i > m. Now we construct Q = (Q ij ) as follows, for k [m], Q ij = m k +1 if 1 i m k +1, µ k+i (n+k) < j µ k+i 1 (n+k 1). To prove that Q is a SSYT of shape λ and type β, the arguments are similar as before. Since T (i+1)(µi (s)+1) > T i(µi (s)+1) s+1 T (i+1)µi+1 (s+1), we have µ i (s)+1 > µ i+1 (s+1), i.e., µ i (s) µ i+1 (s+1). Since {j Q ij m} = = m {j Q ij = m k +1} k=1 m {j µ k+i (n+k) < j µ k+i 1 (n+k 1)} k=1 = {j 1 j λ i }, it follows that Q has shape λ. To prove type(q) = β, since µ i (s) is the maximal j such that T ij s, it follows that T ij = s iff µ i (s 1) < j µ i (s). Thus for k i m, {j T ij = n+k} = {j µ i (n+k 1) < j µ i (n+k)}. 13

14 By the construction of Q ij, {j Q (i k)j = m k +1} = {j µ i (n+k) < j µ i 1 (n+k 1)}. For k i m+1, we have #{j T ij = n+k}+#{j Q (i k)j = m k +1} = µ i 1 (n+k 1) µ i (n+k 1). Note that µ m+1 (k) = 0, µ i (n+k) = d for i k. Then m+1 i=k (µ i 1(n+k 1) µ i (n+ k 1)) = d. For i < k, there is no j [d] such that Q (i k)j = m k+1 and T ij = n+k. Similar as before, we have #{(i,j) T ij = n+k}+#{(i,j) Q ij = m k +1} = d. This proves that Q is of type β. It remains to show that Q is a SSYT. Note that {j Q ij = m k +1} = {j µ k+i (n+k) < j µ k+i 1 (n+k 1)}, {j Q ij = m k} = {j µ k+i+1 (n+k +1) < j µ k+i (n+k)}. It follows that in the i-th row of Q, the boxes filled with the number m k is strictly left of the boxes filled with the number m k+1, which means Q is weakly increasing across each row. Suppose there exist i and j such that Q ij = m k+1 and Q (i+1)j = m k+1 with k < k. Then we have µ k +i+1(n+k ) < j µ k +i(n+k 1) and µ k+i (n+k) < j µ k+i 1 (n+k 1), in contradiction with µ k +i(n+k 1) µ k +i(n+k ) µ k+i (n+k). Therefore Q is a SSYT and the map θ is a bijection. Indeed, we can obtain (P,Q) from T by inverting Algorithm 4.1. Example 4.4. Let T be a SSYT of shape (5 5 ) and type (α,5 β) = (2,2,3,2,3,4,4,3,2) for α = (2,2,3,2) and β = (2,1,1,2,3). We construct a pair (P,Q) of SSYTs of the same shape from T as follows, T = P = Q =

15 Note that if type(p) = (α 1,...,α n ) and type(q) = (β 1,...,β m ), sh(p) = sh(q) (d m ) iff sh(p) = sh(q) (d n ). By the definition of S d (α,β), we have S d (β,α) = S d (α,β) = K λα K λβ. By Theorem 4.3, we have λ N;λ (d m ) S d (α,β) = T d (α,β) = K (d m )(α,d β) = K (d m )(α,d β), S d (β,α) = T d (β,α) = K (d n )(β,d ᾱ) = K (d n )(β,d α). Corollary 4.5. Let α = (α 1,...,α n ) and β = (β 1,...,β m ) be two compositions of N. Suppose m n. We have K λα K λβ = K (d m )(α,d β) = K (d n )(β,d α). λ N;λ (d m ) Recall that RSK correspondence is a bijection sending each generalized permutation of α β to a pair of SSYT of the same shape and types α and β respectively, i.e., if d = N, S αβ = S N (α,β) = λ NK λα K λβ = K (N m )(α,n β) = K (N n )(β,n α), which proves Theorem Applications In this section, we will give some applications of previous results. Let Sαβ k be the set of generalized permutations τ S αβ whose second row τ 11 τ has no weakly increasing subsequence oflengthk, i.e., no indexsets 1 i 1 i 2 i k andj 1,j 2,...,j k satisfy τ i1 j 1 τ i2 j 2 τ ik j k. When β = (1 α i ), each member of Sαβ k is a block-ascending permutation which was first studied in [7]. Theorem 5.1. [7] If N = kn, there are two bijections between S k+1 (k n )(1 N ) and the set of SYTs of shape (kn ), S k+1 ((k 1) n )(1 N ) and the set of SYTs of shape (kn ). By the hook length formula, we have S k+1 (k n )(1 N ) = Sk+1 ((k 1) n )(1 N ) = f(kn). 15

16 His proof is a revisional RSK correspondence which merges a pair (P,Q) of SYTs obtained by RSK correspondence from a blcok-ascending permutation to a SYT of shape (k n ). Analogous to Lewis s construction, we [10] obtained the following extension. Theorem 5.2. [10] Let α = (α 1,...,α n ) be a composition of N and α i {k 1,k} for all i [n]. There is a bijection between S k+1 and the set of SYTs of shape α(1 N ) (kn ). Then by the hook length formula, the cardinality S k+1 α(1 N ) = f(kn ) is independent of the choice of α i {k 1,k} for all i [n]. Later, Chen [5] proved that for any composition α of N, the cardinality of S k+1 is α(1 N ) independent of the order of entries of α. As byproducts, he gave a direct bijection between S k+1 α(1 N ) andsk+1 forany α (k n )(1 N ) {k 1,k}n without RSK correspondence involved, and proved that the cardinality of S k+1 is Schur-concave, i.e., if α = (α α(1 N ) 1,...,α n ) and β = (β 1,...,β m ) are partitions of N with α 1 + +α i β 1 + +β i for all i 1 under the assumption of α i = β j = 0 for i > n and j > m, S k+1 α(1 N ) Sk+1 β(1 N ). If τ S k+1 αβ and τ RSK (P,Q), it is well known that the first row of sh(p) = sh(q) is no longer than k. Indeed, RSK correspondence gives a bijection between τ S k+1 αβ and S k (β,α). The following result recovers Chen s Schur-concavity [5] of S k+1 on the α(1 N ) partition α and Theorem 5.2 by taking α {k 1,k} n and β = (1 N ). Corollary 5.3. Let α and β be two compositions of N. If k N, then S k+1 αβ = Sk+1 βα = K λα K λβ = K (k m )(α,k β) λ N;λ (k m ) is independent of the order of entries of α and β. Moreover, S k+1 αβ is Schur-concave on partitions α and β. In particular, S k+1 αβ S N(12 (k +1)), where S N (12 (k +1)) is the set of all 12 (k +1)-avoiding permutations on [N]. From Corollary 5.3, taking α = 1 N and k = 2, we have S 3 (1 N )β = K (2 m )(1 N,2 β). Corollary 5.4. Let α = (α 1,...,α n ) be a composition of N with α i > 0. If α {1,2} n, then the number of permutations on the multiset [α] which contain no weakly increasing subsequence of length 3 is the n-th Catalan number, i.e., S 3 (1 N )α = { Cn if α {1,2} n, 0 otherwise. 16

17 Let P and Q be two SSYTs of the same shape (n 2 ) and types (2 n ) and (1 2n ) respectively. If (P,Q) RSK τ, it follows from RSK correspondence [16] that τ S (1 2n )(2 n )(321) S n+1 (1 2n )(2 n ). From Theorem 3.4, the number of such pairs (P,Q) is C n R n. Corollary 5.5. The number of permutations on the multiset {1,1,2,2,...,n,n} which avoid the pattern 321 and contain no weakly increasing subsequence of length n is C n R n, i.e., C n R n = S (1 2n )(2 n )(321) S n+1 (1 2n )(2 n ). If α = (α 1,...,α n ) is a composition of N, it is known that the number of permutations on the multiset [α] is ( ) N α 1,...,α n, i.e., ( ) N S (1 )α =. N α 1,...,α n Corollary 5.6. If α is a composition of N and f λ the number of SYTs of shape λ, then ( ) f λ N K λα =. α 1,...,α n λ N From the theory of symmetric functions [8, 16], h λ = µ ( ν K νλ K νµ )m µ, where h λ is the complete symmetric function and m µ the monomial symmetric function. Theorem 1.5 implies the following result. Corollary 5.7. If λ N and the length of λ is l(λ), then h λ = µ NK (N l(λ) )(µ,n λ)m µ, i.e., the transition matrix from the basis (h λ ) to the basis (m λ ) consists of Kostka numbers of rectangular shape. References [1] M. H. Albert, R. E. L. Aldred, M. D. Atkinson, C. Handley, and D. Holton, Permutations of a Multiset Avoiding Permutations of Length 3, European J. Combin. 22 (2001), no. 8,

18 [2] M. D. Atkinson, S. A. Linton, and L. A. Walker, Priority Queues and Multisets, Electron. J. Combin. 2 (1995), #R24. [3] F. R. Bernhart, Catalan, Motzkin, and Riordan Numbers, Discrete Math. 204 (1999), no. 1-3, [4] William Y. C. Chen, Eva Y. P. Deng, and Laura L. M. Yang, Riordan Paths and Degrangements, Discrete Math. 308(2008), no. 11, [5] E. Chen, Schur-Concavity for Avoidance of Increasing Subsequences in Block- Ascending Permutations, Electron. J. Combin. 24 (2017), no. 4, #P4.4. [6] D. E. Knuth, The Art of Computer Programming. Vol. III, Addison-Wesley Reading MA, [7] J. B. Lewis, Pattern Avoidance for Alternating Permutations and Young Tableaux, J. Combin. Theory Ser. A 118 (2011), no. 4, [8] I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd Edition, Oxford University Press, Oxford, [9] P. A. MacMahon, Combinatory Analysis, London Cambridge University Press, Volume I, Section III, Chater [10] Z. S. Mei, S. J. Wang, Pattern Avoidance and Young Tableaux, Electron. J. Combin. 24 (2017), no. 1, #p1.6. [11] A. N. Myers, Pattern Avoidance in Multiset Permutations: Bijective Proof, Ann. Comb. 11 (2007), no. 3-4, [12] A. Regev, Identities for the Number og Standard Young Tableaux in Some (k, l)- Hooks, Séminaire Lotharingien de Combinatoire, 63(3), [13] C. D. Savage and H. S. Wilf, Pattern Avoidance in Compositions and Multiset Permutations, Adv. in Appl. Math. 36 (2006), no. 2, [14] R. Simion and F. W. Schmidt, Restricted Permutations, European J. Combin. 6 (1985), no. 4, [15] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequence, Published electronically at njas/sequences/. [16] R. P. Stanley, Enumerative Combinatoric, vol.2, Cambridge University Press, [17] J. West, Permutations with Forbidden Subsequences and Stack-Sortable Permutations, Ph.D. Thesis, M.I.T., Cambridge, MA,

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