Inversion Formulae on Permutations Avoiding 31 Pingge Chen College of Mathematics and Econometrics Hunan University Changsha, P. R. China. chenpingge@hnu.edu.cn Suijie Wang College of Mathematics and Econometrics Hunan University Changsha, P. R. China. wangsuijie@hnu.edu.cn Zhousheng Mei College of Mathematics and Econometrics Hunan University Changsha, P. R. China. zhousheng@hnu.edu.cn Submitted: July 9, 015; Accepted: Nov 4, 015; Published: Nov 13, 015 Abstract We will study the inversion statistic of 31-avoiding permutations, and obtain that the number of 31-avoiding permutations on [n] with m inversions is given by S n,m (31) = b m (b) ). l(b) where the sum runs over all compositions b = (b 1,b,...,b k ) of m, i.e., m = b 1 +b + +b k and b i 1, l(b) = k is the length of b, and (b) := b 1 + b b 1 + + b k b k 1 + b k. We obtain a new bijection from 31-avoiding permutations to Dyck paths which establishes a relation on inversion number of 31-avoiding permutations and valley height of Dyck paths. Keywords: pattern avoidance; Catalan number; Dyck path; generating function Supported by NSFC 11401196 & 11571097 the electronic journal of combinatorics (4) (015), #P4.8 1
1 Introduction Let S n denote the permutation group on [n] = {1,,...,n}. Write σ S n in the form σ = σ 1 σ σ n. For m n, if σ S n and π = π 1 π m S m, we say that σ contains the pattern π if there is an index subsequence 1 i 1 < i < < i m n such that σ ij < σ ik iff π j < π k for 1 j,k m, that is, σ has a subsequence which is order isomorphic to π. Otherwise, σ avoids the pattern π, or say, σ is π-avoiding. We denote by S n (π) the set of all permutations σ S n that are π-avoiding, i.e., S n (π) = {σ S n σ avoids the pattern π}. For example, the permutation 4153 avoids the pattern 31, but contains the pattern 13 since its subsequence 153 is order isomorphic to 13, hence 4153 S 5 (31) and 4153 / S 5 (13). In 1970 s, Knuth [1, 13] obtained a well known result on permutations avoiding patterns, that is for any π S 3, S n (π) = C n = 1 ( ) n, n+1 n where C n is the n-th Catalan number which counts the number of Dyck paths of length n. In past decades, various articles considered the bijections between 31-avoiding permutations and Dyck paths, see [4, 7, 10, 11, 14, 15, 17, 18, 19, 1, ]. In this paper, we will study the inversion distribution of 31-avoiding permutations. For σ = σ 1 σ σ n S n (π), we define the inversion set Inv(σ) to be Inv(σ) = {(σ i,σ j ) i < j and σ i > σ j }, and denote by inv(σ) = #Inv(σ), called the inversion number of σ, where the hash sign denotes cardinality. The generating function I n (π,q) of the inversion numbers is I n (π,q) = σ S n(π) q inv(σ). for σ S n (π). This generating function was first introduced and explored in [8, 0] and some recurrence formulae have been obtained for π S 3 and π 31. Conjecture 3. of [8] states that, for all n 1, n I n (31,q) = I n 1 (31,q)+ q i+1 I i (31,q)I n i 1 (31,q). (1) i=0 Soon afterwards a bijective proof of the recursive formula (1) was obtained by Szu-En Cheng et al. [6]. There are some other works on inversions of restricted permutations, see [1, 3, 5, 9, 15, 16]. In 014, M. Barnabei, F. Bonetti, S. Elizalde and M. Silimbani [] studied the distribution of descents and major indexes of 31-avoiding involutions. the electronic journal of combinatorics (4) (015), #P4.8
Motivated by [, 6], in this paper we will study the inversion distribution of 31- avoiding permutations. As the main result, we give an explicit formula counting the number of 31-avoiding permutations with the fixed inversion number. We also find a bijection between 31-avoiding permutations and Dyck paths, which is new to the best of our knowledge. From this bijection, we show that the inversion number of 31-avoiding permutations and the valley-sum of Dyck paths are equally distributed. Inversions of Permutations Avoiding 31 For 1 k n, let Sn k (31) be the collection of 31-avoiding permutations of [n] and containing 1 k as a subsequence, S k n(31) = {σ S n (31) σ 1 (1) < σ 1 () < < σ 1 (k)}. More precisely, if σ = σ 1 σ σ n S k n(31) and σ i1 = 1,σ i =,...,σ ik = k, then i 1 < i < < i k. Obviously, we have S n (31) = Sn 1 (31) S n (31) Sn n (31) = {id}. For 1 k n, let In k (31,q) be the generating function defined by In k (31,q) = q inv(σ). σ Sn k(31) Then we have I n (31,q) = In 1(31,q) and In n (31,q) = 1 for all n 1. Lemma 1. For 1 k n, we have I k n(31,q) = I k+1 n (31,q)+ q i In+i k 1(31,q). i Proof. Given σ Sn k(31) with 1 k n 1, consider the position of σ 1 (k + 1). Assuming σ 1 (0) = 0, we have either σ 1 (k) < σ 1 (k + 1), or σ 1 (i) < σ 1 (k + 1) < σ 1 (i+1) for some i k 1. (i): If σ Sn k(31) and σ 1 (k) < σ 1 (k+1), it follows that σ Sn k+1 (31). So this case contributes a term In k+1 (31, q) to the generating function In(31,q). k (ii): If σ Sn(31) k and σ 1 (i) < σ 1 (k +1) < σ 1 (i+1) for some i k 1, since σ avoids the pattern 31, it forces that σ 1 (j) > σ 1 (k + 1) for all j k +. Otherwise, we have σ 1 (j) < σ 1 (k+1) < σ 1 (i+1) which is obviously a contradiction. It implies that σ = σ 1 σ σ n satisfies σ 1 = 1,σ =,...,σ i = i,σ i+1 = k + 1. Denote by σ = σ i+ σ i+3 σ n. Then σ is a permutation of {i+1,...,k,k +,...,n} satisfying σ 1 (i+1) < σ 1 (i+) < < σ 1 (k) and inv(σ) = k i+inv( σ). It implies that case (ii)contributes a term q k i I k i n i 1 (31,q) to Ik n (31,q) for 0 i k 1. Changing the index i to k i, the proof will be complete by combining (i)and (ii). the electronic journal of combinatorics (4) (015), #P4.8 3
In the sequel, we always denote by δ : R {0,1} a function such that δ(u,v) = { 0, u = v; 1, otherwise. In order to characterize the generating function I n (31,q) as a counting function of lattice points in a lattice polytope, we introduce the following lemma. Lemma. Assuming x 0 = 0, for all 1 t n, we have 1 t t In+1 1 (31,q) = I x 1 =0 x =x 1 t+1 xt n+1 x t (31,q) x t=x t 1 q δ(x i,x i 1 )(i+1 x i ) Proof. The statement is true for t = 1 by Lemma 1. To use induction on t, suppose the above equality holds for t. From Lemma 1, we have I t+1 xt n+1 x t (31,q) = t+1 I t+ xt+1 n+1 x t+1 (31,q) q δ(x t+1,x t)(t+ x t+1 ). x t+1 =x t Using above formula to substitute the term I t+1 xt n+1 x t (31,q) in the formula of this Lemma, we can easily conclude that the equality holds for the case t+1. Let Ω n be a convex lattice polytope defined by Ω n = {(x 1...,x n ) Z n 0 x 1 x i i for all 1 i n}. Recall that I n+1 (31,q) = In+1 1 (31,q) and In+1 xn n (31,q) = 1. From above lemma by taking t = n we can easily obtain Proposition 3. Assuming x 0 = 0, we have I n+1 (31,q) = x Ω n n q δ(x i,x i 1 )(i+1 x i ). In the following we will give a more explicit interpretation about this formula. Let inv k (σ) be the number of inversions of σ whose first element is k, i.e, inv k (σ) = #{i (k,i) Inv(σ)} It is obvious that inv k (σ) k 1. From the definition of I n+1 (31,q) and Proposition 3, we have q inv(σ) = q n δ(x i,x i 1 )(i+1 x i ) x Ω n σ S n+1 (31) Below we recursively define a map ϕ : S n+1 (31) Ω n, ϕ(σ) = (x 1,...,x n ) = x, () the electronic journal of combinatorics (4) (015), #P4.8 4
Figure 1: ϕ : σ = 31579468 (0,0,,0,1,0,,0,3) x = (0,1,1,4,4,5,5,6) such that x 1 = inv (σ) and for k, { x x k = k 1, if inv k+1 (σ) = 0; k +1 inv k+1 (σ), otherwise. Figure.1 shows an example, where the second vector is ( inv 1 (σ),...,inv 9 (σ) ). Theorem 4. The map ϕ defined above is a bijection. Moreover, if ϕ(σ) = (x 1,...,x n ), then n inv(σ) = δ(x i,x i 1 )(i+1 x i ). Proof. We first show that ϕ is well defined in the sense that if x = ϕ(σ) then x Ω n = {(x 1,...,x n ) Z n 0 x 1 x i i for all 1 i n}. We use induction on i. For i = 1, it is obvious x 1 = inv (σ) 1. Suppose 0 x 1 x i 1 i 1. If inv i+1 (σ) = 0, then x i = x i 1 i by the induction hypothesis. If inv i+1 (σ) 0, then x i = i + 1 inv i+1 (σ) i. It remains to show that if inv i+1 (σ) 0, then x i 1 x i. Let k i be maximal such that inv k (σ) 0, i.e., inv k+1 (σ) = = inv i (σ) = 0. It follows that there exists an inversion (k,l) Inv(σ). Since σ is 31-avoiding, we have σ 1 (k) < σ 1 (i+1), otherwise (i+1,k,l) is a subsequence of σ and of type 31. Hence we obtain inv i+1 (σ) inv k (σ)+i k, and x i = i+1 inv i+1 (σ) k +1 inv k (σ) = x k 1 +1 > x k 1. Note that inv k+1 (σ) = = inv i (σ) = 0. By definitions, we have x k 1 = = x i 1 which provesthatϕiswelldefined. Toprovethemapϕisabijection, notethateachpermutation σ can be uniquely recovered from its inversion vector ( inv 1 (σ),...,inv n+1 (σ) ). Now we construct an inverse map ψ : Ω n S n+1 (31) of ϕ recursively as follows. Given x = (x 1,...,x n ), define ψ(x) = σ = σ 1 σ n+1 such that inv 1 (σ) = 0 and for k n+1, { 0 if x inv k (σ) = k 1 = x k ; k x k 1 otherwise. It is not difficult to see that both ψ ϕ and ϕ ψ are identity map, i.e., ϕ is a bijection. This completes the proof. the electronic journal of combinatorics (4) (015), #P4.8 5
Figure : x = (0,1,1,4,4,5,5,6) D = uuduuddduuduududdd When n = 3, the inversion polynomial of S n (31) is I 3 (31,q) = q 4 +4q 3 +5q +3q+1. Below is the list of the bijection ϕ, q 0 : {134} ϕ {(0,0,0)}; q 1 : {143,134,134} ϕ {(0,0,3),(1,1,1),(0,,)}; q : {134,143,143,314,314} ϕ {(0,,3),(0,0,),(1,1,3),(1,,),(0,1,1)}; q 3 : {341,413,314,413} ϕ {(1,,3),(1,1,),(0,1,3),(0,0,1)}; q 4 : {341} ϕ {(0,1,)}. A Dyck path D is a lattice path from (0,0) to (n,0) in the (x,y)-plane with up-steps (1,1) (abbreviated as u ) and down-steps (1, 1) (abbreviated as d ), such that D never falls below the x-axis. A valley du of the Dyck path D is a down-step followed by an up-step. The height of a valley is defined to be the y-coordinate of its bottom. Denote by D n the set of all Dyck paths of length n. Several bijections between S n (31) and D n have been established in the literature, see [4, 7, 10, 11, 14, 15, 17, 18, 19, 1, ]. Here we will give a new bijection obtained easily from the above theorem. Morever, this bijection will allow to read the inversion number of a permutation as the sum of all valley heights and the number of valleys in the corresponding Dyck path. Indeed, for x = (x 1,...,x n ) Ω n, assuming x 0 = 0 and x n+1 = n+1, we construct a Dyck path D x as follows. By reading i from 1 to n+1, for each i we add an up-step and x i x i 1 down-steps from left to right. Figure. presents an example. It is obvious that this construction gives an bijection from Ω n+1 to D n+1. If all valleys of a Dyck path D have heights a 1,...,a k, denote by v(d) = (a i +1). Combining with Theorem 4, we can easily obtain our first main result. Theorem 5. The map σ D ϕ(σ) is a bijection from S n+1 (31) to D n+1 such that where ϕ is defined in (). inv(σ) = v(d ϕ(σ) ), As an application of Theorem 5, we will give a counting formula on the number of 31-avoiding permutations with a fixed inversion number. For any D D n, we define a the electronic journal of combinatorics (4) (015), #P4.8 6
tunnel of D to be a horizontal segment between two lattice points of D that intersects D only in these two points, and stays always below D. From Theorem 5, for m 0, there is a bijection S n,m (31) := {σ S n (31) : inv(σ) = m} D n,m := {D D n : v(d) = m}. Theorem 6. For every m 0, S n,m (31) = b m (b) ). l(b) where the sum runs over all compositions b = (b 1,b,...,b k ) of m, denoted b m, i.e., m = b 1 +b + +b k and b i 1, l(b) = k is the length of b, and (b) := b 1 + b b 1 + + b k b k 1 + b k. Proof. It is sufficient to consider D n,m. For any D D n,m, suppose that D has k valleys with heights a 1,a,,a k, then m = v(d) = k (a i + 1). Let l i be the length of the path D located between the i-th and (i+1)-th valley, for i = 0,1,,,k. Then we have l i = a i+1 a i +t i, l i = n, t i 1. i=0 Where t i is the number of tunnels between the i-th and (i+1)-th valley. Let a 0 = 0 and a k+1 = 0 be the heights of the starting point and the terminal point of the Dyck path D, respectively. Write (a) = a i+1 a i. Then i=0 #{D D n all valleys of D have heights a 1,a,...,a k } = #{(l 0,l 1,,l k ) l i = a i+1 a i +t i, l i = n,t i 1}. = #{(t 0,t 1,,t k ) t 0 +t 1 + +t k = n (a),t i 1}. (a) ) 1 = k i=0 So we have D n,m = (a 1 +1)+(a +1)+ +(a k +1)=m a i +1 1 (a) ) 1 k the electronic journal of combinatorics (4) (015), #P4.8 7
Let b i = a i +1 for 1 i k, b 0 = b k+1 = 0, obviously (b) = k i=0 b i+1 b i = (a)+. Hence (b) ) D n,m =. k b 1 +b + +b k =m b i 1 Acknowledgements We would like to express our gratitude to the anonymous referee for many useful comments. References [1] J. Bandlow, Eric S. Egge, and K. Killpatrick. A weight-preserving bijection between Schröer paths and Schröer permutations. Ann. Comb., 6(3-4): 35 48, 00. [] M. Barnabei, F. Bonetti, S. Elizalde, and M. Silimbani. Descent sets on 31-avoiding involutions and hook decompositions of partitions. J. Combin. Theory Ser. A, 18:13 148, 014. [3] Andrew M. Baxter. Refining enumeration schemes to count according to the inversion number. Pure Math. Appl. (PU. M. A.), 1(): 137 160, 010. [4] S. C. Billey, W. Jockusch, and R. P. Stanley. Some combinatorial properties of Schubert polynomials. J. Algebraic Combin., (4): 345 374, 1993. [5] William Y. C. Chen, Yu-Ping Deng, and Laura L.M. Yang. Motzkin paths and reduced decompositions for permutations with forbidden patterns. Electron. J. Combin., 9(), #R15, 003. [6] S. E. Cheng, S. Elizalde, A. Kasraouic, and B. E. Sagan. Inversion polynomials for 31-avoiding permutations. Discrete Math., 313():55 565, 013 [7] A. Claesson and S. Kitaev. Classification of bijections between 31- and 13-avoiding permutations. Sém. Lothar. Combin. 60, Art. B60d, 30 pp, 008. [8] T. Dokos, T. Dwyer, B. P. Johnson, B. E. Sagan, and K. Selsor. Permutation patterns and statistics. Discrete Math., 31(18):760 775, 01. [9] Eric S. Egge. Restricted 341-avoiding involutions, continued fractions, and Chebyshev polynomials. Adv. in Appl. Math., 33(3):451 475, 004. [10] S. Elizalde and E. Deutsch. A Simple and Unusual Bijection for Dyck Paths and its Consequences. Ann. Comb., 7(3):81 97, 003. [11] S. Elizalde and I. Pak. Bijections for refined restricted permutations. J. Combin. Theory Ser. A, 105():07 19, 004. [1] D. E. Knuth. The Art of Computer Programming I: Fundamental Algorithms, Addison-Wesley, Publishing Co., Reading, Mass.-London-Don Mills, Ont, 1969. the electronic journal of combinatorics (4) (015), #P4.8 8
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