Restricted 132-Dumont permutations

a r X i v :m a t h /0209379v 2 [m a t h .C O ] 26 N o v 2002RESTRICTED 132-DUMONT PERMUTATIONSToufik Mansour1Department of Mathematics,Chalmers University of Technology,S-41296G¨o teborg,Swedentoufik@math.chalmers.seAbstractA permutation πis said to be Dumont permutations of the first kind if each even integer in πmust be followed by a smaller integer,and each odd integer is either followed by a larger integer or is the last element of π(see,for example,[22]).In [7]Dumont showed that certain classes of permutations on n letters are counted by the Genocchi numbers.In particular,Dumont showed that the (n +1)st Genocchi number is the number of Dummont permutations of the first kind on 2n letters.In this paper we study the number of Dumont permutations of the first kind on n letters avoiding the pattern 132and avoiding (or containing exactly once)an arbitrary pattern on k letters.In several interesting cases the generating function depends only on k .Keywords:Dumont permutations,restricted permutations,generating functions.1.IntroductionClassical patterns .Let α∈S n and τ∈S k be two permutations.We say that αcontains τif there exists a subsequence 1≤i 1<i 2<···<i k ≤n such that (αi 1,...,αi k )is order-isomorphic to τ;in such a context τis usually called a pattern .We say that αavoids τ,or is τ-avoiding ,if such a subsequence does not exist.The set of all τ-avoiding permutations in S n is denoted S n (τ).For an arbitrary finite collection of patterns T ,we say that αavoids T if αavoids every τ∈T ;the corresponding subset of S n is denoted S n (T ).While the case of permutations avoiding a single pattern has attracted much attention,the case of multiple pattern avoidance remains less investigated.In particular,it is natural,as the next step,to consider permutations avoiding pairs of patterns τ1,τ2.This problem was solved completely for τ1,τ2∈S 3(see [19]),and for τ1∈S 3and τ2∈S 4(see [21]).Several recent papers [6,9,13,14,15,16]deal with the case τ1∈S 3,τ2∈S k for various pairs τ1,τ2.Another natural question is to study permutations avoiding τ1and containing τ2exactly t times.Such a problem for certain τ1,τ2∈S 3and t =1was investigated in [17],and for certain τ1∈S 3,τ2∈S k in [1,9,13,18].The tools involved in these papers include generating trees,continued fractions,Chebyshev polynomials,and Dyck words.Also,the tools involved in these papers include many classical sequences,for example sequence of Catalan numbers,Fibonacci numbers,and Pell numbers.2RESTRICTED132-DUMONT PERMUTATIONSWe denote the n th Catalan number by C n=11−4x.1−x2Q r−2(x)We denote the solution of Recurrence1.1with Q0(x)=0and Q1(x)=1by F r(x),and we denote the solution of Recurrence1.1with Q0(x)=Q1(x)=1by G r(x).For example,F2(x)=1+x2, F3(x)=1+x4,and G3(x)=1−x2+x41−x2r−2m=r−1−j(1−x2F m(x))and G r(x)=1+r−1 j=1x2jRESTRICTED132-DUMONT PERMUTATIONS3 ing Recurrence1.1it is easy to see that√x)=1+x+F4((3·2n−2−1)x n,n≥2where f n is the n th Fibonacci number.Organization of the paper.In this paper we use generating function techniques to study those Dumont permutations in S n(n≥0)which avoid132and avoid(or contain exactly once)an arbitrary pattern on k letters.In several interesting cases the generating function depends only on k.The paper is organized as follows.The case of Dumont permutations avoiding both132andτis treated in Section2.We present a simple structure for any Dumont permutation avoiding132.This structure can be obtained explicitly for several interesting cases,including classical patterns and generalized patterns.This allows us tofind explicitly some statistics on Dumont permutations which avoid132. The case of avoiding132and containing another patternτexactly once is treated in Section3.Again, wefind explicitly the generating function for several interesting cases ofτ,including classical patterns and generalized patterns.Most of the explicit solutions obtained in Sections2-4involve the generating functions F k(x)and G k(x).2.Dumont permutations which avoid132and another patternLet D is the set of all Dumont permutations of all sizes including the empty permutation.Let Dτ(n) denote the number of Dumont permutations in S n(132,τ),and let Dτ(x)= n≥0Dτ(n)x n be the corresponding generating function.In this section we describe a method for enumerating Dumont permutations which avoid132and another pattern and we use our method to enumerate Dτ(n)for variousτ.We begin with an observation concerning the structure of the Dumnot permutations of the first kind avoiding132which holds immediately from definitions.Proposition2.1.For anyπ∈D n(132)such thatπj=n,there holds one of the following assertions:(1)if n is odd number thenπ=(π′,n),whereπ′∈D n−1(132);(2)if n is even number thenπ=(π′,n,π′′)such thatπ′is a Dumont permutation on the numbersn−j+1,n−j+2,...,n−1,π′′is nonempty Dumont permutation on the numbers1,2,···,n−j, and j=1,2,4,...,n−2(the minimal element ofπ′cannot be even number).2.1.τ=∅.As a corollary of Proposition2.1wefind an explicit formula for the number of132-avoiding Dumont permutations in S n.Theorem2.2.The generating function for the number of132-avoiding Dumont permutations in S n is given by(1+x)C(x2).In other words,the number of132-avoiding Dumont permutations in S n is given by C[n/2],which is the[n/2]th Catalan number.Proof.By Proposition2.1,we have two possibilities for block decomposition of an arbitraryπ∈D n(132).Let us write an equation for D∅(x).The contribution of thefirst decomposition above equalsD∅(2n+1)x2n+1=x n≥0D∅(2n)x2n,n≥04RESTRICTED 132-DUMONT PERMUTATIONSequivalently,(2.1)D ∅(x )−D ∅(−x )=x (D ∅(x )+D ∅(−x )).The contribution of the second decomposition above equalsn ≥1D ∅(2n )x2n=n ≥1D ∅(2n −1)x2n+n ≥1n j =0D ∅(2j +1)D ∅(2n +2−2j )x 2n ,equivalently,(2.2)D ∅(x )+D ∅(−x )−2==x (D ∅(x )−D ∅(−x ))+x2(D 12...k (x )+D 12...k (−x ))and B k (x )=12(D 12...(k −1)(x )−D 12...(k −1)(−x ))(D 12...k (x )+D 12...k (−x )−2).The rest is easy to check by the definitions of A k and B k .Example 2.5.Theorem2.4,for k =3,yields D 123(x )=1+x +x 4−x 5(1+x )(1−x 2−x 4).Inother words,the number of 132-avoiding Dumont permutation in S n (1234)is f n/2+2+f n/2−2if n is even number,otherwise 2for all n ≥2,where f n is the n th Fibonacci number.As an extension of Theorem 2.4,let us defineA (x 1,x 2,x 3,...)=π∈Dj ≥1x 12...j (π)j,where τ(π)is the number of occurrences of τin π.LetA (1)(x 1,x 2,x 3,...)=12(A (x 1,x 2,x 3,...)−A (−x 1,x 2,x 3,...)).Using the same arguments as in the proof of Theorem 2.4,we obtain the following.RESTRICTED 132-DUMONT PERMUTATIONS 5Theorem 2.6.We haveA(1)(x 1,x 2,x 3,...)=1+x 21A(1)(x 1x 2,x 2x 3,x 3x 4, (1)x2...=C (x 2).Hence,we have D ∅(x )=(1+x )C (x 2)(see Theorem2.2).Another application of Theorem 2.6is to the number of right to left maxima.Let π∈S n ,πi is aright to left maxima if πi >πj for all i <j .We denote the number of right to left maxima of πby rlm (π).Proposition 5of [3]provedlrm (π)=j ≥112...j (π)(−1)j −1.Therefore,π∈Dx |π|y rlm (π)=A (xy,y −1,y,y −1,...)together with Theorem 2.6and A (1)(x,1,1,...)=C (x 2)we getπ∈Dx |π|y rlm (π)=1+xC (x 2)y +n ≥2x 2n −2C n −1(x 2)y n .Corollary 2.7.The generating function for the number of Dumont permutations avoiding 132andhaving exactly k right to left maxima is given by x 2k −2C k −1(x 2)for all k ≥2,and x k C k (x 2)for k =0,1.2.3.A classical pattern τ=2134...k .Similarly as in Theorem 2.4,we obtain the case τ=2134...k .Theorem 2.8.For all k ≥2,D 213...k (x )=G k −1(x )+xG k −2.Example 2.9.Theorem 2.8for k =3,4yields D 213(x )=1+x −x 3(1−x 2)2.6RESTRICTED132-DUMONT PERMUTATIONS2.4.A generalized pattern12-3-···-k.In this subsection we use the notation of generalized pat-terns(see Section1).For example,we write the classical pattern132as1-3-2.By definitions,we get D12(x)=1+x+x2.So,by the same arguments as in the proof of Theorem2.4, together withD12(x)=D1-2(x)=1+x+x2,we obtain the following.Theorem2.10.For all k≥1,D12-3-...-k(x)=D1-2-3-...-k(x)=F k(x)+xF k−1(x).A comparison of Theorem2.4with Theorem2.10suggests that there should exist a bijection between the sets S n(1-3-2,12-3-···-k)and S n(1-3-2,1-2-3-···-k).However,we failed to produce such a bijection,andfinding it remains a challenging open question.Now,let us defineB(x1,x2,x3,...)= π∈D x1(π)1 j≥2x12-3-...-j(π),1whereτ(π)is the number of occurrences ofτinπ.LetA(2)(x1,x2,x3,...)=1(B(x1,x2,x3,...)−B(−x1,x2,x3,...)).2Using the same arguments as those in the proof of Theorem2.4,we getTheorem2.11.x21(1−x2+x2A(2)(x1,x2x3,x3x4,...))A(2)(x1,x2,x3,...)=1+1−4x2yRESTRICTED132-DUMONT PERMUTATIONS7 2.5.A generalized patternτ=21-3-···-k.In this subsection,we use the notation of generalized patterns(see Section1).For example,we write the classical pattern132as1-3-2.By definitions,we get D21(x)=1+x.So,by the same arguments as in the proof of Theorem2.4 together withD21(x)=D2-1(x)=1+x,we obtain the following.Theorem2.13.For all k≥2,D21-3-...-k(x)=D2-1-3-...-k(x)=G k−1(x)+xG k−2(x).A comparison of Theorem2.8with Theorem2.13suggests that there should exist a bijection between the sets S n(1-3-2,21-3-···-k)and S n(1-3-2,2-1-3-···-k).However,we failed to produce such a bijection,andfinding it remains a challenging open question.Now,let us defineC(x1,x2,x3,...)= π∈D x1(π)1 j≥2x21-3-...-j(π),1whereτ(π)is the number of occurrences ofτinπ.LetA(3)(x1,x2,x3,...)=1(C(x1,x2,x3,...)−C(−x1,x2,x3,...)).2Using the same arguments as in the proof of Theorem2.4,we get the following.Theorem2.14.We havex21x2A(3)(x1,x2x3,x3x4,...)A(3)(x1,x2,x3,...)=1+1−x2−x2F k−3(x).8RESTRICTED132-DUMONT PERMUTATIONSing the same arguments as in the proof of Theorem2.2we getD23...k1(x)−D23...k1(−x)=x(D23...k1(x)+D23...k1(−x)),andD23...k1(x)+D23...k1(−x)−2=x(D23...k1(x)−D23...k1(−x))++x1−2x2−x4.In other words,the number of Dumont permutation in S n(132,23451)is given by P[n/2],which is the[n/2]th Pell number for all n≥2.3.Dumont permutations which avoid132and contain another pattern exactly once Let Dτ;r(n)denote the number of Dumont permutations in S n(132)containingτexactly r times, and let Dτ;r(x)= n≥0Dτ;r(n)x n be the corresponding generating function.3.1.A classical patternτ=12...k.Theorem3.1.LetA k(x)=x2(1−x2F k−2(x))2A k−2(x)for all k≥2,where A1(x)=0and A2(x)=x4.Then for all k≥2D12...k;1(x)=A k(x)+xA k−1(x).Proof.By Proposition2.1,we have two possibilities for the block decomposition of an arbitraryπin D n(132).Let us write an equation for D12...k;1(x).The contribution of thefirst decomposition above isn≥0D12...k;1(2n+1)x2n+1=x n≥0D12...(k−1);1(2n)x2n,equivalently(3.1)D12...k;1(x)−D12...k;1(−x)=x(D12...(k−1);1(x)+D12...(k−1);1(−x)).The contribution of the second decomposition above isn≥1D12...k;1(2n)x2n= n≥1D12...k;1(2n−1)x2n++ n≥1n j=0D12...(k−1);1(2j+1)D12...k;0(2n+2−2j)x2n++ n≥1n j=0D12...(k−1);0(2j+1)D12...k;1(2n+2−2j)x2n, equivalently(3.2)D12...k;1(x)+D12...k;1(−x)=x(D12...k;1(x)−D12...k;1(−x))++x2(D12...(k−1);0(x)−D12...(k−1);0(−x))(D12...k;1(x)+D12...k;1(−x)).Using Theorem2.4,Equation3.1,Equation3.2,and Definition1.1,we get the desired result.RESTRICTED132-DUMONT PERMUTATIONS9 Example3.2.Theorem3.1for k=3we getD123;1(x)=x5(1+x−x2)(1−x2)(1−x2−x4)2.As an extension of Theorem3.1,let us consider the case r≥1.Theorem2.6,for given k and r,yields an explicit formula for D12...k;r(x).For example,for k=3and r=0,1,2,3,4,we have the following. Theorem3.3.We have(i)D123;0(x)=1+x+x4−x51−x2;(iii)D123;2(x)=x5(1+x2)(1+2x−2x2−x3+x4)(1−x2)2;(v)D123;4(x)=x9(1+x2)(−1−3x+3x2+3x3−3x4−x5+x6)1−x2G k−2(x)A k−1(x)+x4G k−1(x)1−x2.Then,for all k≥2,D213...k;1(x)=A k(x)+xA k−1(x).3.3.A generalized patternsτ=12-3-···-k andτ=21-3-···-k.Similarly to Theorem3.1,we getTheorem3.5.LetA k(x)=x2(1−x2F k−2(x))2A k−2(x)for all k≥4,where A1(x)=x2and A2(x)=2x4.Then,for all k≥2,D12-3-···-k;1(x)=A k(x)+xA k−1(x).As an extension of Theorem3.5,let us consider the case r≥1.Theorem2.11,for given k and r, yields an explicit formula for D12-3-···-k;r(x).For example,for k=3and r=0,1,2,3,4,we have the following.10RESTRICTED132-DUMONT PERMUTATIONSTheorem3.6.We have(i)D12-3;0(x)=1+x+x4−x5(1−x2)2;(iii)D12-3;2(x)=x7(2+2x−6x2−x3+6x4+x5−2x6)(1−x2)4;(v)D12-3;4(x)=x9(5+5x−23x2−7x3+40x4−x5−30x6+5x7+5x8−x9+5x10+x11−2x12)1−x2G k−2(x)A k−1(x)+x4G k−1(x)1−x2,and A4(x)=x6(2−x2)1−x2;(ii)D21-3;1(x)=x3(1+x−x2)(1−x2)2;(iv)D21-3;3(x)=x5(1+x−x2+x3−x4−x5+x6)(1−x2)3.4.Further resultsHere we present three different directions to generalize the results of the previous sections.Thefirst of these directions is to consider one occurrence of the classical pattern132.For example,the following result is true.RESTRICTED132-DUMONT PERMUTATIONS11 Theorem4.1.There does not exist a Dumont permutation containing132(classical pattern)exactly once.Proof.Letπ=(π′,n,π′′)be a Dumont permutation of length n,which contains the pattern132 exactly once.It is easy to see that there does not exist a Dumont permutation where n=0,1,2,3. Suppose n≥4,and let us assume by induction on n that there does not exist a Dumont permutation of length m≤n−1containing132exactly once.To prove this property for n,let us consider the following two cases together using Proposition2.1:n is either an even number,or n is an odd number.(1)Let n be an odd number.Sinceπis a Dumont permutation,we getπ′′=∅,soπcontains132exactly once if and only ifπ′contains132exactly once.(2)Let n be an even number.Sinceπis a Dumont permutation we haveπ′′=∅.Now,let usconsider two cases:either n does not appear in the occurrence of132,or n does.(a)Let the occurrence of132not contain the element n.So,every element ofπ′is greater thanevery element ofπ′′.Therefore,eitherπ′is a Dumont permutation of length m≤n−2that contains132exactly once,orπ′′is a Dumont permutation of length m≤n−1thatcontains132exactly once.(b)Let the occurrence of132contain the element n.So,π=(π′,a,n,π′′,a+1,π′′′)(see[16])such thatπp=n andπq=a+1,where every element ofπ′is greater than every elementofπ′′and every element ofπ′′is greater than every element ofπ′′′.Since n is evennumber and maximal inπwe have that a is an odd number,so a+1is an even number.Therefore,by using Proposition2.1we get that p,q are even numbers,(π′,a)is of oddlength,andπ′′is of even length.On the other hand,q=p+1+|π′′|,so q is an oddnumber,a contradiction.Hence,by induction on n we get the desired result.The second direction is to consider more than one additional restriction.For example,the following result is true.Theorem 4.2.Let k≥2.The generating function for the number of Dumont permutations in S n(1-3-2,1-2-3···-k,2-1-3···-k)is given byG k−1(x)+xG k−2(x).A comparison of Theorem4.2with Theorem2.8suggests that there should exist a bijection between the sets S n(1-3-2,2-1-3-···-k)and S n(1-3-2,1-2-3···-k,2-1-3-···-k).However,we failed to produce such a bijection,andfinding it remains an open question.The third direction is to consider another3-letter pattern instead of1-3-2.Theorem4.3.The number of Dumont permutation of the second kind in S n(3-2-1)is the same as the number of Dumont permutation in S n(2-3-1)(or in S n(3-1-2))which is equal to C[n/2].Acknowledgments:The author is grateful to S.Kitaev and the anonymous refree for their careful reading of the manuscript.12RESTRICTED132-DUMONT PERMUTATIONSReferences[1]M.Atkinson,Restricted permutations,Discr.Math.195(1999),27–38.[2]E.Babson and E.Steingr´ımsson,Generalized permutation patterns and a classification of the Mahonianstatistics,S´e minaire Lotharingien de Combinatoire,B44b:18pp,(2000).[3]P.Br¨a nd´e n,A.Claesson,and E.Steingr´ımsson,Continued fractions and increasing subsequences in permuta-tions,Discr.Math.,to appear.[4]A.Claesson,Generalised pattern avoidance,European Journal of Combinatorics,22(2001)961–973.[5]A.Claesson and T.Mansour,Permutations avoiding a pair of generalized patterns of length three with exactlyone dash,preprint CO/0107044.[6]T.Chow and J.West,Forbidden subsequences and Chebyshev polynomials,Discr.Math.204(1999)119–128.[7]D.Dumont,Interpretations combinatoires des nombres de Genocchi,Duke Math.J.41(1974),305–318.[8]S.Kitaev,Multi-avoidance of generalised patterns,to appear in Discrete Mathematics.[9]C.Krattenthaler,Permutations with restricted patterns and Dyck paths,Adv.in Applied Math.27(2001),510–530.[10]T.Mansour,Continued fractions and generalized patterns,European Journal of Combinatorics,23:3(2002),329–344.[11]T.Mansour,Restricted1-3-2permutations and generalized patterns,Annals of Combinatorics6(2002),1–12.[12]T.Mansour,Continued fractions,statistics,and generalized patterns,Ars Combinatoria,to appear(2002),preprint CO/0110040.[13]T.Mansour and A.Vainshtein,Restricted permutations,continued fractions,and Chebyshev polynomialsElectronic bin.7(2000)#R17.[14]T.Mansour and A.Vainshtein,Restricted132-avoiding permutations,Adv.Appl.Math.126(2001),258–269.[15]T.Mansour and A.Vainshtein,Layered restrictions and Chebychev polynomials(2000),Annals of Combina-torics5(2001),451–458.[16]T.Mansour and A.Vainshtein,Restricted permutations and Chebyshev polynomials,S´e minaire Lotharingiende Combinatoire47(2002),Article B47c.[17]A.Robertson,Permutations containing and avoiding123and132patterns,Discrete Mathematics and Theo-retical Computer Science,3(1999)151–154.[18]A.Robertson,H.Wilf,and D.Zeilberger,Permutation patterns and continued fractions,Electronic bin.6(1999)#R38.[19]R.Simion,F.W.Schmidt,Restricted Permutations,Europ.J.of Combinatorics6(1985),383–406.[20]N.J.A.Sloane and S.Plouffe,The Encyclopedia of Integer Sequences,Academic Press,New York(1995).[21]J.West,Generating trees and forbidden subsequences,Discr.Math.157(1996),363–372.[22]————,Available in http://www.theory.csc.uvic.ca/∼cos/inf/perm/GenpcchhiInfo.html.。