EQUIDISTRIBUTION AND SIGN-BALANCE ON 132-AVOIDING PERMUTATIONS. Toufik Mansour 1,2

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1 Séminaire Lotharingien de Combinatoire 5 (2004), Article B5e EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS Toufik Mansour,2 Department of Mathematics, Chalmers University of Technology S Göteborg, Sweden toufik@mathchalmersse Abstract Let R n be the set of all permutations of length n which avoid 32 In this paper we study the statistics last descent ( ldes ), first descent ( fdes ), last rise ( lris ), first rise ( fris ) on the set R n In particular, we prove that the bistatistic ( fris, lris ) on the set of all permutations R n \{n 2} the bistatistic ( n ldes, n fdes ) on the set of all permutations of R n \{2 n} are equidistributed Furthermore, we consider the case of sign balance for these statistics on the set of all permutations R n, we give a combinatorial interpretation for some of these statistics Introduction Let S n denote the set of permutations of {,, n}, written in one-line notation, suppose that π S n σ S k We say that a subsequence of π has type σ whenever it has the same pairwise comparisons as σ For example, the subsequence of the permutation has type 2435 We say that π avoids σ (or π is σ-avoiding) whenever π contains no subsequence of type σ For example, the permutation avoids , but it has 2589 as a subsequence, so it does not avoid 234 We denote the set of σ-avoiding permutations in S n by S n (σ) We define R n = S n (32) For π R n, we define the following statistics: () ldes(π) = last descent of π = max{ i n π i > π i+ } where ldes(2 n) = 0, (2) fdes(π) = first descent of π = min{ i n π i > π i+ } where fdes(2 n) = 0, (3) lris(π) = last rise of π = max{ i n π i < π i+ } where lris(n 2) = 0, (4) fris(π) = first rise of π = min{ i n π i < π i+ } where fris(n 2) = 0, (5) lind(π) = π (n) = the index of the letter n in π, (6) find(π) = π () = the index of the letter in π Key words phrases 32-avoiding permutations, Dyck paths, equidistribution of statistics, signbalance Research financed by EC s IHRP Programme, within the Research Training Network Algebraic Combinatorics in Europe, grant HPRN-CT Current address: Department of Mathematics, University of Haifa, 3905 Haifa, Israel toufik@mathhaifaacil

2 2 TOUFIK MANSOUR Foata Schützenberger [FS, Theorem ] proved that the major index the inversion number are equidistributed on S n whose inverse has a prescribed descent set Recently, Adin Roichman [AR, Theorem ] gave an analogue for this result for S n (32) They proved that the statistics ldes lind are equidistributed on the set S n (32) Note that these statistics are identical on the set R n (To prove that, let s = π We may assume that s >, otherwise π equals the identity of S n, which has no descents Clearly, π s > π s Furthermore, we have = π s < π s+ < < π n since π avoids 32 Therefore, s is exactly the last descent of π) The Catalan sequence is the sequence (C n ) = (,, 2, 5, 4, 42, 32, 429, 430, ), where C n = n+( 2n ) n is called the nth Catalan number The generating function for the Catalan numbers is denoted by C(t) = 4t The Catalan numbers provide 2t a complete answer to the problem of counting certain properties of more than 00 different combinatorial structures (see [S, page 29 Exercise 69]) The structures that are useful for us in the present paper are Dyck paths (see [S]) 32-avoiding permutations (see [Kn]) A Dyck path is a path in the plane integer lattice Z 2 consisting of up-steps U = (, ) down-steps D = (, ) which never passes below the x-axis We denote the set of Dyck paths of length 2n by P n A point on the Dyck path is called a peak if it is immediately preceded by an up-step immediately followed by a down-step For p P n, we define lpeak(p) = the height (the y-coordinate) of the last peak of p Recently, Adin Roichman [AR, Theorem 2] proved that the statistics ldes on the set S n (32), ldes on the set P n, lind are equidistributed on the set P n In this paper we prove the following analogue of this result Theorem For all n, we have p P n q n lpeak(p) = q ldes(π) In addition, we study the sign-- find sign-- lind enumerators for R n (see [AR, Theorem 3] for S n (32)), we prove the following result Theorem 2 For all n, π R 2n sign(π)q ldes(π) = ( q) π R 2n sign(π)q ldes(π) = π R 2n+ sign(π)q find(π), q 2(find(π) ) The paper is organized as follows In Section 2, we study the statistics ldes, fdes, lris, fris on the set R n In particular, we prove that the bistatistic ( fris, lris ) on the set R n \{n 2} the bistatistic ( n ldes, n fdes ) on the set R n \{2 n} are equidistributed In Section 3, we consider the case of sign balance of some statistics on the set R n Finally, in Section 4, we give a combinatorial interpretation (using Dyck paths) for some of these statistics

3 EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS 3 2 Equidistribution of statistics In this section, we study the statistics ldes, fdes, lris, fris on the set R n by using the block decomposition approach of 32-avoiding permutations in S n (see [MV]) First of all, let us describe the block decomposition of a permutation π R n Let n π = (π, n, π ) R n such that π j = n π avoids 32 if only if π is a permutation of the numbers n j +, n j + 2,, n, π is a permutation of the numbers, 2,, n j, both π π avoid 32 This representation is called the block decomposition of π Theorem 2 The statistic fris on the set R n \{n 2} the statistic n ldes on the set R n \{2 n} are equidistributed, that is, for n, we have q fris(π) = q n ldes(π) Moreover, \{n2} \{2n} q fris(π) x n = qx + qxc(x) q ldes(π) = C(qx) Proof Let F n (q) = q ldes(π) for n 0 Using the block decomposition of π R n, we may derive a recurrence for F n (q) Namely, by making use of the fact that S m (32) = C m = m+( 2m m ) (see [Kn]), we get for n, n 2 (2) F n (q) = F n (q) + q q j C j F n j (q) Multiplying both sides by x n, summing over all n, finally using F 0 (q) =, we obtain that F (x; q) = + q ldes(π) x n = C(xq) n Let now H n (q) = q fris(π) for n 0 Again, using the block decomposition of π, we may derive a recurrence for H n (q) First of all, the contribution of the permutations π with π = n gives + q(h n (q) ) Secondly, the contribution of the permutations π with π 2 = n gives C n 2 q Finally, for π j = n with j 3, we get as contribution (H j (q) )C n j whenever π (n ) (n + j) in the decomposition π = (π, n, π ), q j C n j for π = (n ) (n + j) Hence, for n, we have n n H n (q) = + q(h n (q) ) + C n 2 q + C n j (H j (q) ) + q j C n j, j=2 j=2

4 4 TOUFIK MANSOUR or, equivalently, n n (22) H n (q) = + q(h n (q) ) + C n j (H j (q) ) + q j C n j If we write K n (q) for q n (H n (q ) ) +, then it is easy to see that j=2 n 2 K n (q) = K n (q) + q q j C j K n j (q) Since K 0 (q) = K (q) =, an induction on n together with Equation (2) gives for n 0, q n (H n (/q) ) + = F n (q) The rest follows now easily Theorem 22 The statistic lris on the set R n \{n 2} the statistic n fdes on the set R n \{2 n} are equidistributed, that is, for n, we have q lris(π) = q n fdes(π) Moreover, \{n2} \{2n} q fdes(π) x n = q + q()2 C(x) ()(q) q lris(π) x n = + x(c(xq) )C(xq) Proof Let L n (q) = q fdes(π) for n 0 Let π = (π, n, π ) R n such that π j = n Using this block decomposition of π, we may derive a recurrence for L n (q) Namely, the contribution of the permutations with j = gives C n q, the contribution for j = 2 gives C n 2 q 2, the contribution for n j 3 π (n + j) (n ) gives C n j (L j (q) ), the contribution for n j 3 π = (n + j) (n ) gives q j C n j, the contribution for j = n π = (n + j) (n )n gives Hence, for n, we have n n L n (q) = C n q + C n 2 q 2 + C n j (L j (q) ) + q j C n j +, or, equivalently, n 2 n (23) L n (q) = + q q j C n j + C n j (L j (q) ) j=3 Multiplying by x n, summing over n, finally using L 0 (q) = L (q) =, we obtain that L n (q)x n = q + q()2 C(x) ()(q) j=3 j=

5 EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS 5 Now, let P n (q) = q lris(π) for n 0 Again, we may derive a recurrence for P n (q) by using the above block decomposition of π Namely, the contribution for j = n gives C n q n, the contribution for j = n gives C n 2 q n 2, the contribution for j n 2 π (n j) 2 gives C j q j (P n j (q) ), the contribution for n 2 j π = (n j) 2 gives q j C j Hence, for n, we have n 2 n 2 P n (q) = C n q n + C n 2 q n 2 + C j q j (P n j (q) ) + q j C j, or, equivalently, n n (24) P n (q) = q j C j + q C j q j (P n j (q) ) j= Multiplying by x n, summing over n, using finally P 0 (q) = P (q) =, we obtain that P n (q)x n + x(c(xq) )C(xq) = Using Equations (23) (24) together with an induction on n, we conclude that q n (P n (q ) ) + = L n (q), this completes the proof More generally, we prove that the bistatistics ( fris, lris ) on the set R n \{n 2} ( n ldes, n fdes ) on the set R n \{2 n} are equidistributed Theorem 23 The bistatistic ( fris, lris ) on the set R n \{n 2} the bistatistic ( n ldes, n fdes ) on the set R n \{2 n} are equidistributed, that is, for n, we have p n ldes(π) q n fdes(π) = p fris(π) q lris(π) Moreover, \{2n} \{n2} p fris(π) q lris(π) x n = + x( p + p( 2qx + pq2x2 )C(qx)) ()( pqx)( pqxc(qx)) Proof Let A n (p, q) = p fris(π) q lris(π) for n A 0 (p, q) = Using the block decomposition π = (π, n, π ) R n where π j = n once more, we derive a recurrence for A n (p, q) if n Namely, the contribution of the permutations with π = (n ) (n + j) π = (n j) gives n (pq)j The contribution of the permutations where π = (n ) (n + j) π (n j) gives n pq(a n (p, q) ) + q j p j (P n j (q) ) j=2 (P n (q) is the polynomial appearing in the proof of Theorem 22) The contribution of the permutations where π (n ) (n + j) π = (n j) gives n j= qj (H j (p) ) (H n (p) is the polynomial appearing in the proof of Theorem 2) Finally, the contribution of the permutations where π (n ) (n+ j) j=

6 6 TOUFIK MANSOUR π (n j) equals n j= qj (H j (p) )(P n j (q) ) Hence, for n, we have n n A n (p, q) = pq(a n (p, q) ) + (pq) j + q (pq) j (P n j (q) ) n n + q j (H j (p) ) + q q j (H j (p) )(P n j (q) ) Let A(x; p, q), P (x; q), H(x; p) be the generating functions for the sequences A n (p, q), P n (q), H n (p), respectively, that is, j= A(x; p, q) = A n (p, q)x n, P (x; q) = P n (q)x n, H(x; p) = H n (p)x n Multiplying the above recurrence by x n, summing over n, finally using A 0 (p, q) = H 0 (p) = P 0 (q) =, we obtain that x ( pqx)a(x; p, q) = + ()( pqx) + x ( H(qx; p) ) qx + qx ( P (x; q) ) ( qx P (x; q) ) pqx pqx ( + xq H(qx; p) ) ( P (x; q) ) qx From Theorem 2 we have while Theorem 22 yields H(x; q) = qxc(x) + qx, Hence, P (x; q) = ( + x(c(qx) )C(qx)) A(x; p, q) = + x( p + p( 2qx + pq2 x 2 )C(qx)) ()( pqx)( pqxc(qx)) We now turn to the computation of the generating function B(x; p, q) = p n ldes(π) q n fdes(π) x n Using the arguments in the proof of the formula for A(x; p, q), we get that B(x; p, q) = A(x; p, q) + pqx

7 EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS 7 Hence, as requested \{2n} p n ldes(π) q n fdes(π) x n = B(x; p, q) pqx = A(x; p, q) = p fris(π) q lris(π) x n, \{n2} 3 Sign Balance on R n We denote the set of all even (respectively odd) permutations π R n by R + n (respectively R n ) We define e n = R + n, o n = R n, m n = e n o n for all n 0 (Simion Schmidt [SimSch] proved that m n = C (n )/2 if n is odd m n = 0 otherwise) In this section we study the sign-balance of R n with respect to certain statistics Theorem 3 We have M n (q)x n = sign(π)q ldes (π) = + x x2qc(x2q2 ) 2 C(x 2 q 2 ) Proof Let F ± n (q) = π R ± n qldes(π), let π = (π, n, π ) R n such that π j+ = n, 0 j n Then, for this block decomposition of π, we have sign(π) = ( ) (j+)(n j ) sign(π ) sign(π ) = ( ) (j+)(n ) sign(π ) sign(π ), or, equivalently, sign(π) = Therefore, for n, we obtain F ± 2n+(q) = F ± 2n(q) + F ± 2n(q) = F ± 2n (q) + Hence, for all n, we have { sign(π ) sign(π ), if n is odd, ( ) j+ sign(π ) sign(π ), if n is even 2n M 2n+ (q) = M 2n (q) + q j+ (e j F ± 2n j (q) + o jf 2n j (q)),2,4,,2n 2 + M 2n (q) = M 2n (q) j=,3,,2n 3 2n q j+ (e j F 2n j (q) + o jf ± 2n j (q)) q j+ (e j F ± 2n j (q) + o jf 2n j (q) q j+ m j M 2n j (q),,2,4,,2n 2 + q j+ m j M 2n j (q) j=,3,,2n 3 q j+ m j M 2n j (q)

8 8 TOUFIK MANSOUR Let M(x; q) = M n(q)x n m(x) = m nx n Using [M2, Lemma 23], we get that ( xqm(xq))m(x; q) ( + x + xqm( xq))m( x; q) ( + xqm( xq))m(x; q) + ( + x xqm(xq))m( x; q) = xq(m(xq) + m( xq)), = 2 xq(m(xq) m( xq)) Hence, solving the above two equations for the variables M(x; q) M( x; q), using the fact that m(x) = + xc(x 2 ) (see [SimSch]), we obtain the desired result As a corollary of the above theorem, we have the following result concerning the sign-- ldes statistic on R n Corollary 32 For all n, we have π R 2n sign(π)( ) ldes(π) = 2C n Moreover, π R 2n sign(π)( ) ldes(π) = C n + sign(π)( ) ldes(π) x n = ( + x + x 2 C(x 2 ))C(x 2 ) n More generally, Theorem 3 yields the following result Corollary 33 For all n, we have π R 2n sign(π)q ldes(π) = ( q) Theorem 34 We have π R 2n sign(π)q ldes(π) = N n (q)x n = sign(π)q fdes(π) π R 2n+ sign(π)q ldes(π), q 2(ldes(π)) = + x ( q)(q + 2q) q(2 )( + xq 2q)C(x 2 ) ()( 2 q 2 ) Proof Let L ± n (q) = π R ± n qfdes(π), let π = (π, n, π ) R n such that π j+ = n, 0 j n Then, for this block decomposition of π, we have sign(π) = ( ) (j+)(n j ) sign(π ) sign(π ) = ( ) (j+)(n ) sign(π ) sign(π ), or, equivalently, sign(π) = { sign(π ) sign(π ), if n is odd, ( ) j+ sign(π ) sign(π ), if n is even

9 EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS 9 Therefore, for n, we obtain L ± 2n+(q) = L ± 2n(q) = 2n q j+ L ± 2n j + (L± j (q) ɛ± )e 2n j + (L j (q) ɛ )o 2n j,2,4,,2n j=,3,,2n 3 j=,3,,2n 3 q j+ L 2n j (q) + (L j (q) ɛ )e 2n j + (L pm j (q) ɛ ± )o 2n j q j+ L ± 2n j (L ± j ɛ± )e 2n j + (L j ɛ )o 2n j, where ɛ + = ɛ = 0 Hence, for all n, we have N 2n+ (q) = + M 2n (q) = 2n q j+ m 2n j +,2,,2n 2,2,,2n 2 2n q j+ m 2n j + (N j (q) )m 2n j, j=,3,,2n 3 (N j (q) )m 2n j + q j+ m 2n j j=,3,,2n (N j (q) )m 2n j Let N(x; q) = N n(q)x n m(x) = m nx n Using [M2, Lemma 23], we get that x (N(x; q) N( x; q)) = 2 + xq ( m(x) 2 2 q + m( x) ) + xq 2 2 q 2 + x [( N(x; q) ) ( m(x) + N( x, q) ) ] m( x), 2 + x (N(x; q) + N( x, q)) = 2 xq (m(x) m( x)) 2 2( 2 q 2 ) x 2 q 2 + (m(x) + m( x) 2) 2( 2 q 2 ) x ( N(x; q) + N( x; q) 2 ) (m(x) m( x)) x ( N(x; q) N( x; q) 2x ) (m(x) + m( x)) 4 2 Hence, solving the above two equations for the variables N(x; q) N( x; q), using the fact that m(x) = + xc(x 2 ) (see [SimSch]), we get the desired result 4 Dyck paths Following [Kr], we define a bijection Φ between permutations in S n (32) Dyck paths from the origin to the point (2n, 0) Let π = π π n be a 32-avoiding permutation We read the permutation π from left to right successively generate a Dyck

10 0 TOUFIK MANSOUR path When π j is read, then in the path we adjoin as many up-steps as necessary, followed by a down-step from height h j + to height h j (measured from the x-axis), where h j is the number of elements in π j+, π j+2,, π n which are larger than π j For example, if π = S 6, then the corresponding Dyck path is the one shown in Figure y x Figure The Dyck path Φ(53426) Namely, the first element to be read is 5 There is one element in 3426 which is larger than 5, therefore the path starts with two up-steps followed by a down-step, thus reaching height Next 3 is read There are 2 elements in 426 which are larger than 3, therefore the path continues with two up-steps followed by a down-step, thus reaching height 2 Etc Conversely, given a Dyck path starting at the origin returning to the x-axis, the obvious inverse of the bijection Φ produces a 32-avoiding permutation Theorem 4 For all n 0, we have p P n q n lpeak(p) = q ldes(π) Proof We prove that n + lpeak(φ(π)) = fdes(π) for any π R n Let π = (π,, π ) R n such that π = j Since π avoids 32, the letters of π are increasing (ie, π a < π b for all a < b) Therefore, by definition of Φ, the Dyck path p = Φ(π) satisfies the equation lpeak(p) = π + = n + j, where π is the number of letters in π Remark 42 The relation n + peak(φ(π)) = fdes(π) for π R n can be read off immediately from the permutation diagram, see [R] Acknowledgment The author thanks the anonymous referee, whose suggestions improved both the content exposition of this paper The final version of this paper was written during the author s stay at the University of Haifa, Israel, he would like to express his gratitude to the University of Haifa for the support References [AR] Ron M Adin Yuval Roichman, Equidistribution sign-balance on 32-avoiding permutations, preprint, 2003, <

11 EQUIDISTRIBUTION AND SIGN-BALANCE ON 32-AVOIDING PERMUTATIONS [FS] D Foata M-P Schützenberger, Major index inversion number of permutations, Math Nachr 83 (978), [Kn] D Knuth, The Art of Computer Programming, vol, Addison Wesley, Reading, MA, 968 [Kr] C Krattenthaler, Permutations with restricted patterns Dyck paths, Adv Appl Math 27 (200), [M] T Mansour, Counting peaks at height k in a Dyck path, Journal Integer of Sequences 5 (2002), Article 02 [M2] T Mansour, Restricted 32-alternating permutations Chebyshev polynomials, Annals of Combinatorics 7:2 (2003), [MV] T Mansour A Vainshtein, Restricted permutations Chebyshev polynomials, Séminaire Lotharingien de Combinatoire 47 (2002), Article B47c [SimSch] R Simion F Schmidt, Restricted permutations, European J Combin 6 (985), [R] A Reifegerste, On the diagram of 32-avoiding permutations, European J Combin 24 (2003), [S] R P Stanley, Enumerative Combinatorics, vol, Wadsworth Brooks/Cole, Pacific Grove, CA, 986, xi pages; second printing, Cambridge University Press, Cambridge, 996