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THÔNG TIN TÀI LIỆU
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Value-Peaks of Permutations Pierre Bouchard D´epartement de math´ematiques Universit´e du Qu´ebec `a Montr´eal Case postale 8888, Succu rsale Centre-ville Montr´eal (Qu´ebec), Canada H3C 3P8 bouchard.pierre@uqam.ca Hungyung Chang ∗ Department of Applied Mathematics National Sun Yat-sen University Kaohsiung, Taiwan 80424 changhy@math.nsysu.edu.tw Jun Ma † Institute of Mathematics Academia Sinica Taipei, Taiwan majun904@sjtu.edu.cn Jean Yeh Department of Mathematics National Taiwan Univers ity Taipei, Taiwan jean.yh@ms45.url.com.tw Yeong-Nan Yeh ‡ Institute of Mathematics Academia Sinica Taipei, Taiwan mayeh@math.sinica.edu.tw Submitted: Nov 29, 2009; Accepted: Mar 16, 2010; Published: Mar 29, 2010 Mathematics Subject Classification: 05A15 Abstract In th is paper, we f ocus on a “local property” of permutations: value-peak. A permutation σ has a value-peak σ(i) if σ(i − 1) < σ(i) > σ(i + 1) for some i ∈ [2, n − 1]. Define V P(σ) as the set of value-peaks of the permutation σ. For any S ⊆ [3, n], define V P n (S) such that V P (σ) = S. Let P n = {S | V P n (S) = ∅}. we make the set P n into a poset P n by defining S T if S ⊆ T as sets. We prove that the poset P n is a simplicial complex on the set [3, n] and study some of its properties. We give enumerative formulae of permutations in th e set V P n (S). ∗ Partially supported by NSC 98-211 5-M-110 -009 † Corresponding author ‡ Partially supported by NSC 96-211 5-M-001 -005 the electronic journal of combinatorics 17 (2010), #R46 1 1 Introduction Let [m, n] := {m, m + 1, ··· , n}. If m > n, then [m, n] = ∅. Let [n] := [1, n] and S n be the set of all the permutations on the set [n]. We write permutations of S n in the form σ = (σ(1)σ(2) ···σ(n)). Fix a permutation σ in S n . For every i ∈ [n − 1], if σ(i) > σ(i + 1), then we say that i is a position-descent of σ. Define the position-descent set o f a permutation σ, denoted by PD(σ), as P D(σ) = {i ∈ [n − 1] | σ(i) > σ( i + 1)}. Given a set S ⊆ [n − 1], suppose P D(σ) = S for some σ ∈ S n . We easily obtain the increasing and decreasing intervals of σ from the set S. The permutation σ is a function from the set [n] to itself. Since the monotonic property of a function is a global property of the function, the position-descent set of a permutation gives a “global property” of the permutation. We say a permutation σ ∈ S n has a value-descent σ(i) if σ(i) > σ(i + 1) for some i ∈ [n −1]. Define the value-descent set of a permuta t io n σ, denoted by V D(σ), as V D(σ) = {σ(i) | σ(i) > σ(i + 1)}. The value-descent set of a permutatio n is different from its position-descent set. Let S ⊆ [2, n]. Suppose V D(σ) = S for some σ ∈ S n . We only have t hat k is larger than its immediate right neighbour in the permutation σ for any k ∈ S and do not obtain the increasing and decreasing intervals o f σ from t he set S. So the value-descent set of a permutation gives a “local property” of the permutat io n. For any S ⊆ [2, n], define a set V D n (S) as V D n (S) = {σ ∈ S n | V D(σ) = S} and use vd n (S) to denote the number of permutations in the set V D n (S), i.e., vd n (S) = |V D n (S)|. In a joint work [1], Chang, Ma and Yeh derive an explicit formula for vd n (S). In this paper, we are interested in another “local property” of permutations: value- peak. A permutation σ has a value-peak σ(i) if σ(i − 1) < σ(i) > σ(i + 1) for some i ∈ [2, n−1]. Define V P (σ ) as the set of value-p eaks of σ, i.e., V P (σ) = {σ(i) | σ(i −1) < σ(i) > σ (i + 1)}. For example, the value-peak set of σ = (48362517) is {5, 6 , 8}. Since σ has no value-peaks when n 2, we may always suppose that n 3 . For any S ⊆ [n], define a set V P n (S) as V P n (S) = {σ ∈ S n | V P (σ) = S}. Obviously, if {1, 2, }∩ S = ∅ then V P n (S) = ∅. Example 1.1 V P 5 ({4, 5}) = { 14253, 14352, 24153, 34152, 24351, 34251, 15243, 15342, 25143, 35142, 25341, 35241 }. Suppose S = {i 1 , i 2 , ··· , i k }, where i 1 < i 2 < ··· < i k . We prove the necessary and sufficient conditions for V P n (S) = ∅ are i j 2j + 1 for all j ∈ [k]. Let P n = {S | V P n (S) = ∅}. We make the set P n into a poset P n by defining S T if S ⊆ T as sets. Fig. 1 shows the Hasse diagrams of P 3 , P 4 and P 5 {3} {4} {3} {3} {5} {4} {3} {3,5} {4,5} fff Fig.1. the Hasse diagrams of P 3 , P 4 and P 5 . the electronic journal of combinatorics 17 (2010), #R46 2 In the next section we prove that P n is a simplicial complex on the vertex set [3, n] and derive some properties of P n . Then we turn to enumerative problems for permutations by value-peak set. Let vp n (S) denote the number of permutations in the set V P n (S), i.e., vp n (S) = |V P n (S)|. For the cases with |S| = 0, 1, 2, we derive explicit formulae f or vp n (S). For general n 3, we derive the following recurrence relation. Let n 3 and S ⊆ [3, n]. Suppose V P n (S) = ∅ and let r = max S if S = ∅, 1 otherwise. For any 0 k n − r − 1, we have vp n (S ∪[n−k +1, n]) = 2(k +1 )vp n−1 (S ∪[n−k, n−1]) +k(k +1)vp n−2 (S ∪[n−k, n−2]). For any S ⊆ [3, n], we write the set S in the form S = m i=1 [r i − k i + 1, r i ] such that r i r i+1 −k i+1 −1 for all i ∈ [m−1]. For example, let n = 12 and S = {3, 4, 8, 10, 11, 12}. Then S = [3, 4] ∪[8, 8] ∪[10, 1 2]. We have r 1 = 4, k 1 = 2, r 2 = 8, k 2 = 1, r 3 = 12, k 3 = 3. Define the type of the set S, denoted type(S), as (r k 1 1 , r k 2 2 , . . . , r k m m ). We co nclude with a formula for the number of permutations in terms of the type of S. The paper is organised as f ollows. In Section 2, we give the necessary and sufficient conditions for V P n (S) = ∅. We prove the poset P n is a simplicial complex on the set [3, n] and study its some properties. In Section 3, we investig ate enumerative problems o f permutations in the sets V P n (S). In the Appendix, we list vp n (S) for 1 n 8 obtained by computer searches. 2 The Simplicial Complex P n In this section, we give the necessary and sufficient conditions for V P n (S) = ∅ for any n 3 and S ⊆ [n]. We show P n is a simplicial complex o n the set [3, n] and study some properties of P n . Theorem 2.1 Let n 3. Suppose S = {i 1 , i 2 , ··· , i k } is a subset of [n], where i 1 < i 2 < ··· < i k . Then the necessary and sufficient conditions for V P n (S) = ∅ are i j 2j + 1 for all j ∈ [k]. Proof. Suppose V P n (S) = ∅ and let σ ∈ V P n (S). For any j ∈ [k], all the integers i 1 , i 2 , ··· , i j are a value-peak of σ. Then i j − j j + 1, hence, i j 2j + 1. Conversely, suppose i j 2j + 1 for all j ∈ [k]. Suppose [n] \ S = {a 1 , a 2 , ··· , a n−k } with a 1 < a 2 < ··· < a n−k . Let σ be the permutation in S n defined by σ(2j) = i j for 1 j k, σ(2j − 1) = a j for 1 j k + 1, σ(j) = a j for 2k + 2 j n. Obviously, V P (σ) = S and V P n (S) = ∅. Corollary 2.1 Let n 3 and S ⊆ [n]. Sup pose V P n (S) = ∅. We have |S| ⌊ n−1 2 ⌋. the electronic journal of combinatorics 17 (2010), #R46 3 Proof. Suppose S = {i 1 , i 2 , ··· , i k } with i 1 < i 2 < ··· < i k . Since V P n (S) = ∅, Theorem 2.1 tells us that n i k 2k + 1. Hence k ⌊ n−1 2 ⌋. Corollary 2.2 Let n 3 and S ⊆ [n]. Suppose V P n (S) = ∅. Then for |S| < ⌊ n−1 2 ⌋, we have V P n+1 (S ∪ {n + 1}) = ∅; for |S| = ⌊ n−1 2 ⌋, we have V P n+1 (S ∪ {n + 1}) = ∅ if n is even; otherwis e , V P n+1 (S ∪ {n + 1}) = ∅. Proof. Let k = |S| . k < ⌊ n−1 2 ⌋ implies 2 (k + 1) + 1 2⌊ n−1 2 ⌋ + 1 < n + 1. So, V P n+1 (S ∪ {n + 1}) = ∅ when |S| < ⌊ n−1 2 ⌋. For the case with k = ⌊ n−1 2 ⌋, we have 2(k + 1) + 1 = n + 1 if n is even, n + 2 if n is odd. By Theorem 2.1, V P n+1 (S ∪{n + 1}) = ∅ if n is even; otherwise, V P n+1 (S ∪{n + 1}) = ∅. Following [3], define a simplicial complex ∆ on a vertex set V as a collection o f subsets of V satisfying: (1) If x ∈ V , then {x} ∈ ∆, and (2) if S ∈ ∆ and T ⊆ S, then T ∈ ∆. Theorem 2.2 Let n 3. Then P n is a simplicial complex on the set [3, n]. Proof. Obviously, ∅ ∈ P n . For any 3 x n, Theorem 2.1 implies {x} ∈ P n . Let T be a subset of [n] such that V P n (T ) = ∅. Note that V P n (S) = ∅ for any T ⊆ S. Thus given an S ∈ P n , we have T ∈ P n for all T ⊆ S. Hence, P n is a simplicial complex on the set [3, n]. If P and Q are posets, then the direct product of P and Q is the poset P × Q on the set {(x, y) | x ∈ P and y ∈ Q} such that (x, y) (x ′ , y ′ ) in P × Q if x x ′ in P and y y ′ in Q. Recall that the poset n is formed by the set [n] with its usual order. By Corollary 2.2, we obtain a method to construct the poset P n+1 from P n . Theorem 2.3 P n+1 ∼ = 2 × P n if n is even; P n+1 ∼ = (2 × P n ) \ ({1} × P n,⌊ n−1 2 ⌋−1 ) if n is odd. Now, we derive some properties of the simplicial complex P n . By Theorem 2.3, it is easy to obtain the M¨obius function of the poset P n . Corollary 2.3 Let µ n = µ P n be the M¨obius function of the poset P n . Then µ n (S, T ) = (−1) |T |−|S| for a ny S T in P n . Proof. Obviously, µ 3 (∅, {3}) = −1. By induction for n, we assume µ n (S, T ) = (−1) |T |−|S| for any S T in P n . By Theorem 2.3, it follows that µ n+1 (S, T ) = µ n (S \ {n + 1}, T \ {n + 1}) if n + 1 ∈ S ∩ T, µ n (S, T ) if n + 1 /∈ S ∪ T, −µ n (S, T \ {n + 1}) if n + 1 /∈ S and n + 1 ∈ T the electronic journal of combinatorics 17 (2010), #R46 4 for any S ≺ T . Simple computations show that µ n+1 (S, T ) = (−1) |T |−|S| . For every S ∈ P n , we call the element S a face of P n and the dimension o f S is defined to b e |S| − 1, denoted dim(S). In particular, the void set ∅ is always a face of P n of dimension −1, i.e., dim(∅) = −1. Also define the dimension of P n by dim(P n ) = max S∈P n (dim(S)). Theorem 2.4 dim(P n ) = ⌊ n−1 2 ⌋ − 1. Proof. Taking S = {3, 5, ··· , 2⌊ n−1 2 ⌋ + 1}, by Theorem 2.1, we have S ∈ P n . From Corollary 2.1 it follows that the dimension of P n is ⌊ n−1 2 ⌋ − 1. Define P n,i as the set of all the faces o f dimension i in P n , i.e., P n,i = {S ∈ P n | |S| = i + 1} for any −1 i ⌊ n−1 2 ⌋ − 1. Let p n,i = |P n,i |. The sequence (p n,−1 , p n,0 , . . . , p n,⌊ 1 2 (n−1)⌋−1 ) is called the f-vector of the simplicial complex P n . Define the f-polynomial of P n as P n (x) = ⌊ 1 2 (n−1)⌋ i=0 p n,i−1 x ⌊ 1 2 (n−1)⌋−i . To study the f-vector of P n , we introduce the concept of left factors of Dyck path. An n-Dyck path is a lat t ice path in the first quadrant starting at (0, 0) and ending at (2n, 0) with only two kinds of steps—rise step: U = (1, 1) and fall step: D = (1, −1). We can also consider an n-Dyck path P as a word of 2n letters using only U and D. Let L = w 1 w 2 ···w n be a word, where w j ∈ {U, D} and n 0. If there is anot her word R which consists of U and D such that LR forms a Dyck path, then L is called an n-left factor of Dyck paths. Let L n denote the set of all n-left factors of Dyck paths. For any i 0, let L n,i denote the set of all n-left factors of Dyck paths from (0, 0) to (n, n − 2i). It is well known that |L n |, the cardinality of L n , equals the nth central binomial number b n = n ⌊ n 2 ⌋ and |L n,i | = n−2i+1 i n i−1 (see Cori and Viennot [2]). In the following lemma, we give a bijection φ from the set P n to the set L n−1 . Lemma 2.1 There is a bijection φ between the set P n and the set L n−1 for any n 3. Furthermore, the number of elements in P n is n−1 ⌊ n−1 2 ⌋ . Proof. For any S ∈ P n , we construct a word φ(S) = w 1 w 2 ···w n−1 as follows: w i = D if i + 1 ∈ S U if i + 1 /∈ S for any i ∈ [n − 1]. Theorem 2.1 implies φ(S) is an (n − 1)-left factor of a Dyck path. Conversely, for any an n-left factor w 1 w 2 ···w n−1 of a Dyck path, let S = {i+1 | w i = D}. Then V P n (S) = ∅. Thus the mapping φ is a bijection. Note that the number of (n−1)-left factors of Dyck paths is n−1 ⌊ n−1 2 ⌋ . Hence, |P n | = n−1 ⌊ n−1 2 ⌋ . Corollary 2.4 Let n 3. There is a bijection between the set P n,i and the set L n−1,i+1 for a ny −1 i ⌊ n−1 2 ⌋ − 1. Furthermore, we have p n,i = 1 if i = −1, n−2i−2 i+1 n−1 i if 0 i ⌊ n−1 2 ⌋ − 1. the electronic journal of combinatorics 17 (2010), #R46 5 Proof. We just consider the case with i 0. For any S ∈ P n,i , since |S| = i + 1, the number of the letter D in the word φ(S) is i + 1. Hence, φ(S) is a left factor of a Dyck path from (0, 0) to (n − 1, n − 2i − 3). So, φ(S) ∈ L n−1,i+1 . Hence, p n,i = |L n−1,i+1 | = n−2i−2 i+1 n−1 i . Corollary 2.5 Let n 3. The sequence (p n,−1 , p n,0 , . . . , p n,⌊ 1 2 (n−1)⌋−1 ) satisfies the follow- ing recurrence relation: for a ny even integer n, p n+1,i = p n,i if i = −1, p n,i−1 + p n,i if i = 0, 1, ··· , n 2 − 2, p n,i−1 if i = n 2 − 1; for a ny odd integer n, p n+1,i = p n,i if i = −1, p n,i−1 + p n,i if i = 0, 1, ··· , n−3 2 , with initial conditions (p 3,−1 , p 3,0 ) = (1, 1). Proof. First, we consider the case of an even integer n . It is easy to see p n+1,−1 = p n,−1 = 1. For any S ∈ P n+1, 1 2 n−1 , Corollary 2.2 tells us n + 1 ∈ S. Note that S ∈ P n+1, 1 2 n−1 if and only if S \ {n + 1} ∈ P n, 1 2 n−2 . Hence, p n+1, 1 2 n−1 = p n, 1 2 n−2 . For every i ∈ {0, 1, . . . , 1 2 n − 2}, it is easy to see P n,i ⊆ P n+1,i . For any S ∈ P n+1,i with n + 1 ∈ S, S \ {n + 1} can be viewed as an element of P n,i−1 . Conversely, for any S ∈ P n,i−1 , Corollary 2.2 implies S ∪ {n + 1} ∈ P n+1,i . Hence, p n+1,i = p n,i−1 + p n,i . Similarly, we can consider the case of an odd integer n. Theorem 2.5 Let n 3. (1) The f-polynomial P n (x) of the simplicial complex P n satisfies the following recur- rence relation: x ε(n) P n+1 (x) = (1 + x)P n (x) − ε(n) 2 n + 1 n − 1 n−1 2 for any n, w here ε(n) = 0 if n is eve n; ε(n) = 1 otherwise, with ini tial condition P 3 (x) = x + 1. (2) Let P (x, y) = n3 P n (x)y n . Then P (x, y) = (1 + y + xy)[1 + x −C(y 2 )] x −(x + 1) 2 y 2 −1 y 2 , where C(y) = 1− √ 1−4y 2y . the electronic journal of combinatorics 17 (2010), #R46 6 Proof. (1) Obviously, P 3 (x) = x + 1. Given an odd integer n, we suppose n = 2i + 1 with i 1. Corollar y 2.5 implies xP 2i+2 (x) = (1 + x)P 2i+1 (x) − 1 (i+1) 2i i . Similarly, given an even integer n, we suppose n = 2i with i 2. By Corollary 2.5, we have P 2i+1 (x) = (1 + x)P 2i (x). (2) Let P odd (x, y) = i1 P 2i+1 (x)y 2i+1 and P even (x, y) = i2 P 2i (x)y 2i . We have P odd (x, y) = (x + 1)y 3 + (x + 1)yP even (x, y) and P(x, y) = P odd (x, y) + P even (x, y). It is easy to check xP 2i+3 (x) = (1 + x) 2 P 2i+1 (x) − 1 i+1 2i i (x + 1). So, P odd (x, y) satisfies the following equation xP odd (x, y) = (x + 1) 2 y 2 P odd (x, y) + (x + 1)y 3 [1 + x −C(y 2 )], where C(y) = 1− √ 1−4y 2y . Equivalently, P odd (x, y) = (x + 1)y 3 [1 + x −C(y 2 )] x −(x + 1) 2 y 2 . Hence P(x, y) = (1 + y + xy)[1 + x −C(y 2 )] x −(x + 1) 2 y 2 − 1 y 2 . Let H n (x) = P n (x−1) = ⌊ 1 2 (n−1)⌋ i=0 h n,i x ⌊ 1 2 (n−1)⌋−i . The polynomial H n (x) and the sequence (h n,0 ,h n,1 ,···, h n,⌊ 1 2 (n−1)⌋ ) are called the h-polynomial and the h-vector of P n respectively. Corollary 2.6 Let n 3. (1) The h-polynomial H n (x) of the simplicial complex P n satisfies the recurrence relation: (x −1) ε(n) H n+1 (x) = xH n (x) − ε(n) 2 (n + 1) n − 1 n−1 2 for any n, w here ε(n) = 0 if n is eve n; ε(n) = 1 otherwise, with ini tial condition H 3 (x) = x. (2) Let H (x, y) = n3 H n (x)y n . We have H (x, y) = (1 + xy)[x − C(y 2 )] x −1 −x 2 y 2 −1 y 2 . Proof. (1) Since H n (x) = P n (x − 1), by Theorem 2.5, we easily obtain H n+1 (x) = xH n (x) if n is even, and (x − 1)H n+1 (x) = xH n (x) − 2 n+1 n−1 n−1 2 if n is odd, with initial condition H 3 (x) = x. (2) Since H (x, y) = P(x − 1, y), we have H (x, y) = (1 + xy)[x − C(y 2 )] x −1 −x 2 y 2 − 1 y 2 . Corollary 2.7 Let the sequence (h n,0 , h n,1 , ··· , h n,⌊ 1 2 (n−1)⌋ ) be the h-v ector of P n . Then h n,i satisfies the following recurrence relation: h n+1,i = h n,0 if i = 0, h n,i + ε(n)h n+1,i−1 if 1 i ⌊ n 2 ⌋ − 1, ε(n)c ⌊ n 2 ⌋ if i = ⌊ n 2 ⌋, the electronic journal of combinatorics 17 (2010), #R46 7 where c m = 1 m+1 2m m and ε(n) = 0 if n is even; otherw i se, ε(n) = 1, w i th initial conditions (h 3,0 , h 3,1 ) = (1, 0). Equivalently, h n,i = ⌊ n 2 ⌋ − i ⌊ n 2 ⌋ + i ⌊ n 2 ⌋ + i ⌊ n 2 ⌋ . Proof. The recurrence relations a re obtained by comparing coefficients on both sides of the identity in 2.6 (1). Consider t n,i = ⌊ n 2 ⌋−i ⌊ n 2 ⌋+i ⌊ n 2 ⌋+i ⌊ n 2 ⌋ . Note that t n,i and h n,i satisfy the same recurrence relations and (t 3,0 , t 3,1 ) = (1, 0). Hence, h n,i = t n,i = ⌊ n 2 ⌋ − i ⌊ n 2 ⌋ + i ⌊ n 2 ⌋ + i ⌊ n 2 ⌋ . Remark 2.1 Let n 3. The number of left factors of the Dyck path from (0, 0) to (⌊ n 2 ⌋ + i −1 , ⌊ n 2 ⌋ − i −1) equals ⌊ n 2 ⌋−i ⌊ n 2 ⌋+i ⌊ n 2 ⌋+i ⌊ n 2 ⌋ for a ny 0 i ⌊ n−1 2 ⌋. Define the reduced Euler characteristic of P n by ˜χ(P n ) = ⌊ 1 2 (n−1)⌋ i=0 (−1) i−1 p n,i−1 . Corollary 2.8 For any n 3, ˜χ(P n ) = 0 if n is odd, 2(−1) n 2 n n−2 1 2 (n−2) if n is even. Proof. Clearly, P 3 (−1) = 0. Theorem 2.5 tells us P n+1 (−1) = 0 if n is even, 2 n+1 n−1 1 2 (n−1) if n is odd for any n 4. Since ˜χ(P n ) = (−1) ⌊ n−1 2 ⌋−1 P n (−1), we have ˜χ(P n ) = 0 if n is odd, 2(−1) n 2 −2 n n−2 1 2 (n−2) if n is even. Let P be a finite post. Define Z(P, i) to be the number of multichains x 1 x 2 ··· x i−1 in P for any i 2. Z(P, i) is called the zeta polyno mial of P . We state Propo sition 3.11.1a and Proposition 3.14.2 in [3] as the following lemma. Lemma 2.2 [3] Suppose P is a poset. (1) Let d i be the number of c hains x 1 < x 2 < ··· < x i−1 in P . Then Z(P, i) = j2 d j i−2 j− 2 . the electronic journal of combinatorics 17 (2010), #R46 8 (2) If P is simplicial and graded, then Z(P, x + 1) is the rank-gene rating function of P. Corollary 2.9 Let n 3 and i 2. Then (1) Z(P n , i) = (i −1) ⌊ n−1 2 ⌋ P n ( 1 i−1 ) for any i 2, (2) Z(P n , i) satisfies the recurrence relations: Z(P n+1 , i) = iZ(P n , i) −ε(n) 2(i −1) 1 2 (n+1) n + 1 n − 1 1 2 (n − 1) , where ε(n) = 0 if n is even; ε(n) = 1 otherwise, w i th initial condition Z(P 3 , i) = i. (3) Let Z(x, y) = n3 Z(P n , x)y n . We have Z(x, y) = (1 + xy)[x − (x −1)C(y 2 (x − 1))] 1 − x 2 y 2 − 1 y 2 . Proof. (1) Let P n (x) be the f-polynomial of P n . We have the rank-generating function of P n is x ⌊ 1 2 (n−1)⌋ P n ( 1 x ). Lemma 2.2(2) implies that Z(P n , i) = (i −1) ⌊ n−1 2 ⌋ P n ( 1 i−1 ). (2) The recurrence relations for Z(P n , i) follow from Theorem 2.5. (3) Note that Z(P n , x + 1) = x ⌊ n−1 2 ⌋ P n ( 1 x ) = ( √ x) n−2+ε(n) P n ( 1 x ). By the proof of Theorem 2.5, we have P odd (x, y) = (x+1)y 3 [1+x−C(y 2 )] x−(x+1) 2 y 2 and P even (x, y) = P odd (x,y)−(x+1)y 3 (x+1)y . Then Z(x + 1, y) = n3 ( √ x) n−2+ε(n) P n ( 1 x )y n = 1 x P even ( 1 x , y √ x) + 1 √ x P odd ( 1 x , y √ x) = (1 + y + xy)[1 + x −xC(y 2 x)] 1 −(x + 1) 2 y 2 −1 y 2 . Let d P n ,i be the number of chains S n,1 ≺ S n,2 ≺ ··· ≺ S n,i of P n . Theorem 2.6 For any i 1, d P n ,i = n d 1 , d 2 , ··· , d i+1 2d i+1 −n n , where the sum is over all (d 1 , ··· , d i+1 ) s uch that i+1 k=1 d k = n, d 1 0, d k 1 for all 2 k i and d i+1 n − ⌊ n−1 2 ⌋. the electronic journal of combinatorics 17 (2010), #R46 9 Proof. Let i 1 and S n,1 ≺ S n,2 ≺ ··· ≺ S n,i be a chain of P n . Suppose |S n,k | = j k for any k ∈ [i]. Then 0 j 1 < j 2 < ··· < j i ⌊ n−1 2 ⌋. There are p n,j i −1 ways to obtain the set S n,i . Given S n,k with k 2, there are j k j k−1 ways to form the subset S n,k−1 ⊆ S n,k . Hence, d P n ,i = 0=j 0 j 1 <j 2 <···<j i ⌊ n−1 2 ⌋ i−1 k=0 j k+1 j k p n,j i −1 = n d 1 , d 2 , ··· , d i+1 2d i+1 − n n , where the sum is over all (d 1 , ··· , d i+1 ) such that i+1 k=1 d k = n, d 1 0, d k 1 for all 2 k i and d i+1 n − ⌊ n−1 2 ⌋. Corollary 2.10 For any n 3, P n (x) = ⌊ n−1 2 ⌋+2 i=2 x ⌊ n−1 2 ⌋+2−i (i − 2)! i−2 j= 1 (1 −jx) n d 1 , d 2 , ··· , d i 2d i −n n where the second sum is over all (d 1 , ··· , d i ) such that i k=1 d k = n, d 1 0, d k 1 for all 2 k i − 1 and d i n − ⌊ n−1 2 ⌋. Proof. Lemma 2.2(1) implies Z(P n , i) = ⌊ n−1 2 ⌋+2 j= 2 d P n ,j−1 i−2 j− 2 . By Corollary 2.9, we have P n 1 i − 1 = 1 i − 1 ⌊ n−1 2 ⌋ ⌊ n−1 2 ⌋+2 j= 2 d P n ,j−1 i − 2 j − 2 for any i 2. Not e that x ⌊ n−1 2 ⌋ ⌊ n−1 2 ⌋+2 j= 2 d P n ,j−1 1 x − 1 j − 2 = ⌊ n−1 2 ⌋+2 j= 2 x ⌊ n−1 2 ⌋+2−j (j − 2)! j− 2 k=1 (1 −kx)d P n ,j−1 is a polynomial. Hence, P n (x) = ⌊ n−1 2 ⌋+2 j= 2 x ⌊ n−1 2 ⌋+2−j (j−2)! j− 2 k=1 (1 − kx)d P n ,j−1 . 3 Enumerations for Permutations in the Set V P n (S) In this section, we will consider enumerative problems of permutations in the set V P n (S). Let vp n (S) denote the number of permutations in the set V P n (S), i.e., vp n (S) = |V P n (S)|. First, we need the following lemma . the electronic journal of combinatorics 17 (2010), #R46 10 [...]... i, j ∈ [3, n] and i < j Proof (1) For any σ ∈ Sn , suppose the position of the integer 1 is i+1, i.e., σ −1 (1) = i+1 Then σ ∈ V Pn (∅) if and only if σ satisfies σ(1) > · · · > σ(i + 1) < · · · < σ(n) For each integer j = 1, the position of j has two possibilities at the left or right of the integer 1 Hence, vpn (∅) = 2n−1 (2) By Lemma 3.1(2), we first consider the number of permutations in the set... σ −1 (s)} Then, for k = 1, 2, 3, the subsequences of σ, that are determined by elements from Tk (σ), correspond to a permutation in V P|Tk (σ)| (∅) The subsequences of σ, that are determined by elements from T4 (σ) and T6 (σ), are decreasing The subsequences of σ, that are determined by elements from T5 (σ) and T7 (σ), are increasing So, the number of permutations under this subcase is (T1 ,T2 ,T3... also consider a value-peak path P from (0, 0) to (n, k) as a word of n − k letters using only H and R Let Pn,k be the set of all the value-peak paths from (0, 0) to (n, k) Let the electronic journal of combinatorics 17 (2010), #R46 15 i be a nonegative integer and P = e1 e2 · · · en−k ∈ Pn,k For every j ∈ [n − k], define the weight wi (ej ) of the step ej as follows: if the step ej connects a vertex (x,... (S) = 2n−m vpm (S) for any n m Proof (1) It is easy to see ((n + 1)σ(1) · · · σ(n)) ∈ V Pn+1 (S) and (σ(1) · · · σ(n)(n + 1)) ∈ V Pn+1 (S) for any σ = (σ(1) · · · σ(n)) ∈ V Pn (S) Conversely, for any σ ∈ V Pn+1 (S), the position of the integer n + 1 is 1 or n + 1, i.e., σ −1 (n + 1) = 1 or n + 1, since n + 1 ∈ S / Hence, vpn+1 (S) = 2vpn (S) (2) Iterating the identity of Lemma 3.1(1), we obtain vpn (S)... 1)vpn−1(S ∪ [n − k, n − 1]) + k(k + 1)vpn−2 (S ∪ [n − k, n − 2]) Proof For any σ ∈ V Pn (S ∪ [n − k + 1, n]), we consider the following four cases Case 1 There are no integers i ∈ [n−k +1, n] such that the position of i is beside n−k in σ, i.e., |σ −1 (i) − σ −1 (n − k)| = 1 Then σ −1 (n − k) = 1 or n since the permutation σ the electronic journal of combinatorics 17 (2010), #R46 14 has not a value-peak n −... positions of n − k and n in σ Clearly, τ ∈ V Pn (S ∪ [n − k, n − 1]) Lemma 3.1 (1) tells us vpn (S ∪ [n − k, n − 1]) = 2vpn−1(S ∪ [n − k, n − 1]) Hence, the number of permutations under this case is 2 · vpn−1 (S ∪ [n − k, n − 1]) Case 2 There are exactly two integers j, m ∈ [n−k + 1, n] such that |σ −1 (j) −σ −1 (n− k)| = 1 and |σ −1 (m) − σ −1 (n − k)| = 1 Deleting j and m, we obtain a subsequence τ of σ... the number of permutations under this case is k(k − 1)vpn−2(S ∪ [n − k, n − 2]) Case 3 There is exactly one integer j ∈ [n−k+1, n] such that |σ −1 (j)−σ −1 (n−k)| = 1 Then there are k ways to form the set {j} Let τ be the subsequence of σ obtained by deleting j There are the following two subcases Subcase 3.1 σ −1 (n − k) = 1 and n Then φσ (τ ) ∈ V Pn−1 (S ∪ [n − k, n − 1]) Hence, the number of permutations... relation Fixing a set S, let the weight of the vertex (n, k) be vpn (S ∪ [n − k + 1, n]) It is easy to see we can obtain the recurrence relation for vpn (S) by Fig 2 So we introduce the concept of value-peak path in the plane Z × Z as follows A value-peak path is a lattice path in the first quadrant starting at (0, 0) and ending at (n, k) with only two kinds of steps—horizon step H = (1, 0) and rise... electronic journal of combinatorics 17 (2010), #R46 11 We discuss the following two subcases Subcase 1 T3 (σ) = ∅ Let T6 (σ) = {σ(k) | i < σ(k) < j, k > σ −1 (j)} Then T6 (σ) = ∅ since σ must have a 6 value-peak j and k=4 Tk (σ) = [i + 1, j − 1] For k = 1, 2, 6, the subsequences of σ, that are determined by elements from Tk (σ), correspond to a permutation in V P|Tk (σ)| (∅) The subsequences of σ, that are... m=0 k where the sum is over all (k + 1)-tuples (t0 , t1 , · · · , tk ) such that and tm k m=0 m=0 n−2k 0 It is easy to see the sum is the coefficient of x 1 1−2(i+1+m)x tm = n − r − 2k in the power series This completes the proof the electronic journal of combinatorics 17 (2010), #R46 16 Lemma 3.5 Let n 3 and S ⊆ [3, n] Then (1) vpn ([n − k + 1, n]) = w(0; n − 1, k) (2) Suppose S = ∅ and V Pn (S) = . is called an n-left factor of Dyck paths. Let L n denote the set of all n-left factors of Dyck paths. For any i 0, let L n,i denote the set of all n-left factors of Dyck paths from (0, 0) to. intervals of σ from the set S. The permutation σ is a function from the set [n] to itself. Since the monotonic property of a function is a global property of the function, the position-descent set of. 3. Define the type of the set S, denoted type(S), as (r k 1 1 , r k 2 2 , . . . , r k m m ). We co nclude with a formula for the number of permutations in terms of the type of S. The paper is