Dyck paths with a first return decomposition constrained by height

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1 Dyck paths with a first return decomposition constrained by height Jean-Luc Baril, Sergey Kirgizov and Armen Petrossian LEI UMR-CNRS 6306, Université de Bourgogne B.P , 1078 DIJON-Cede France {barjl,sergey.kirgizov,armen.petrossian}@u-bourgogne.fr October 16, 017 Abstract We study the enumeration of Dyck paths having a first return decomposition with special properties based on a height constraint. We ehibit new restricted sets of Dyck paths counted by the Motzkin numbers, and we give a constructive bijection between these objects and Motzkin paths. As a byproduct, we provide a generating function for the number of Motzkin paths of height k with a flat resp. with no flats) at the maimal height. Keywords: Enumeration, Dyck and Motzkin paths, first return decomposition, statistics, height, peak. 1 Introduction and notations A Dyck path of semilength n 0 is a lattice path starting at 0,0), ending at n,0), and never going below the -ais, consisting of up steps U = 1,1) and down steps D = 1, 1). Let D n, n 0, be the set of all Dyck paths of semilength n, and let D = n 0 D n. The cardinality of D n is given by the nth Catalan number, which is the general term n+1 1 n n) of the sequence A in the On-line Encyclopedia of Integer Sequences of N.J.A. Sloane [17]. A large number of various classes of combinatorial objects are enumerated by the Catalan numbers such as planar trees, Young tableau, stack sortable permutations, Dyck paths, and so on. A list of over 60 types of such combinatorial classes has been compiled by Stanley [19]. 1

2 In combinatorics, many papers deal with Dyck paths. Most of them consist in enumerating Dyck paths according to several parameters, such as length, number of peaks or valleys, number of double rises, number of returns to the -ais see for instance [, 8 16, 18]). Other studies investigate restricted classes of Dyck paths avoiding some patterns or having a specific structure. For instance, Barcucci et al. [1] consider non-decreasing Dyck paths which are those having a non-decreasing sequence of heights of valleys see also [6,7]), and it is well known [5] that Dyck paths avoiding the triple rise UUU are enumerated by the Motzkin numbers see A in [17]). Anynon-emptyDyckpathP D hasauniquefirstreturndecomposition [8] of the form P = UαDβ where α and β are two Dyck paths in D. See Figure 1 for an illustration of this decomposition. α β Figure 1: First return decomposition UαDβ of a Dyck path P D. Based on this decomposition, we construct a new collection of subsets of D as follows. Given a function s : D N, called statistic, and a comparison operator on N for instance or >), the set D s, is the union of the empty Dyck path with all Dyck paths P having a first return decomposition P = UαDβ satisfying the conditions: { α,β D s,, 1) suαd) sβ). For n 0, we denote by D s, the set of Dyck paths of semilength n in D s,. Thus, we have D s, = n D s, n 0 n. For instance, if the operator is = and s is a constant statistic i.e., sp) = 0 for any P D), then we obviously have Dn s, = D n for n 0. If s is the number of returns i.e., sp) is the number of down steps D that return the path P to the -ais) and suαd) sβ) is suαd) sβ), then it is straightforward to see that Dn s, consists of Dyck paths built over the grammar S ǫ USD USDUSD. So, the generating function S) for the cardinalities of Dn s,, n 0, satisfies the functional equation S) = 1+S)+ S). The solution of this equation is the well-known generating function for the Motzkin numbers A in [17]).

3 In this paper, we focus on the sets D h, where the statistic h is the maimal height of a Dyck path, i.e., hp) is the maimal ordinate reached by the path P. In Section, we deal with the case in which operator is a strict inequality >. We prove that the cardinalities of the sets Dn h,>, n 0, are given by the sequence A in [17]. This sequence corresponds to the coefficients of the series lim k P k ) where P k ) is a polynomial recursively defined by P 0 ) =, P 1 ) =, P k ) = P k 1 ) + P k ) if k is even, and P k ) = P k 1 ) P k ) if k is odd. In Section 3, we focus on the set D h, where huαd) hβ) the operator is ). Using generating functions, we prove that the cardinalities of the Dn h,, n 0, are given by the Motzkin numbers A in [17]). Moreover, we give a constructive one-to-one correspondence φ between Dyck paths in Dn h, and Motzkin paths of length n. Also, we show how φ transforms peaks UD into peaks UD and flats F in Motzkin paths. Finally, we deduce generating function for the total number of peaks in Dn h,. Table 1 presents the two main enumerative results of Dn h, obtained in Sections and 3. -constraint Sequence OEIS a n,1 n 9 huαd) > hβ) A ,1,,3,6,1,4,50,107 huαd) hβ) Motzkin A ,,4,9,1,51,17,33,835 Table 1: Cardinalities of D h, n according to the -constraint. Enumeration of D h,> n In this section, we enumerate the set Dn h,> of Dyck paths of semilength n 0 with a first return decomposition satisfying huαd) > hβ) where h is the maimal height of a Dyck path. For instance, we have D h,> 1 = {UD}, D h,> = {UUDD}, and D h,> 3 = {UUDDUD,UUUDDD} see Figure ). Let A k ) = n 0 a n,k n resp. B k ) = n 0 b n,k n ) be the generating function where the coefficient a n,k resp. b n,k ) is the number of Dyck paths in Dn h,> with a maimal height equal to k resp. of at most k). So, we have B k ) = k A i ) and the generating function B) for the set D h,> i=0 3

4 Figure : The two Dyck paths in D h,> 3. is given by B) = lim k + B k). Any Dyck path of height k in D h,> is either empty, or of the form UαDβ where α is a Dyck path in D h,> of height k 1 and β D h,> such that hβ) k 1. So, we deduce easily the following functional equations: { A 0 ) = B 0 ) = 1, Theorem 1 We have for k 0, A k ) = A k 1 ) B k 1 ) for k 1. B k ) = P k) where P k is the polynomial recursively defined by P 0 ) =, P 1 ) =, P k ) = P k 1 )+P k ) and P k+1 ) = P k ) P k 1 ). As consequence, we have for k 1 A k ) = P k) P k ), and B) is generating function of the sequence A in [17]. Proof. We proceed by induction on k. For k = 0, the property holds since P 0 ) = = B 0 ). Assuming the property for 0 i k 1, we prove it for k. Taking into account that A k ) = B k ) B k 1 ) in equation ), we obtain B k ) = B k 1 )+ B k 1 ) B k 1 ) B k ) ). With the recurrence hypothesis, we have B k ) = P k )+P k ) P k ) P k 4 ) ) = P k )+P k )P k 3 ) = P k )+P k 1 ) = P k ). ) 4

5 An induction on k completes the proof and the epression of A k ) is deduced from A k ) = B k ) B k 1 ). For instance, we have B ) = , B 3 ) = , and the first ten terms of B) are We refer to Table for an overview of the coefficients a n,k for 1 n 10 and 1 k 9. Notice that the family of sets D h,>, n 1, seems to be the first eample of combinatorial objects enumerated by the sequence A in [17]. k\n Table : Number a n,k of Dyck paths of height k in D h,> n, 1 n 10 and 1 k 9. 3 Enumeration of D h, n 3.1 Enumeration using generating functions In this section, we enumerate the set Dn h, of Dyck paths of semilength n 0 with a first return decomposition satisfying huαd) hβ) where α,β D h, and h is the maimal height of a Dyck path. For instance, we have D h, 1 = {UD}, D h, = {UDUD,UUDD}, and D h, 3 consists of the four Dyck paths UDUDUD,UUDDUD,UUDUDD and UUUDDD see Figure 3). 5

6 Figure 3: The four Dyck paths in D h, 3. Let C k ) = n 0 c n,k n resp. D k ) = n 0 d n,k n ) be the generating function where the coefficient c n,k resp. d n,k ) is the number of Dyck paths in Dn h, with a maimal height equal to k resp. of at most k). So, we have D k ) = k C i ) and the generating function D) for the set Dn h, i=0 is given by D) = lim k + D k). Any Dyck path of height k in D h, is either empty, or of the form UαDβ where α resp. β) is a Dyck path in D h, of height k 1 resp. of height at most k). So, we deduce the following functional equations: { C 0 ) = D 0 ) = 1, C k ) = C k 1 ) D k ) for k 1. Theorem provides recursive epressions for the two generating functions D k ) and C k ). As a consequence, D) can be epressed as an infinite product of terms 1 1 C k ). Theorem We have D 0 ) = C 0 ) = 1, C 1 ) = 1 and D k ) = k 1 i=0 1 Ci ) ) 1 for k 1, 3) and C k ) = k 1 i=1 D) = C 1 ) k 1 Ci ) ) for k, k i 1 Ci ) ) 1. i=0 Proof. Since we have C k ) = D k ) D k 1 ), equation 3) implies D k ) = D k 1) 1 C k 1 ), 6

7 and starting from D 0 ) = 1, a straightforward induction on k provides D k ) = k 1 1 Ci ) ) 1. Moreover, from equation 3) and C1 ) = 1, we deduce i=0 C k ) C k 1 ) = D k) = C 1 ) An induction on k completes the proof. k 1 i=1 1 Ci ) ) 1. Lemma 1 For k 1, we have D k 1 ) = D k) 1 D k )+1). Proof. We proceed by induction on k. Since D 0 ) = 1 and D 1 ) = 1 1, it is easy to check that D 0 ) = D 1) 1 D 1 )+1). Assuming D i 1 ) = D i) 1 D i )+1) for 1 i k, we prove the result for i = k 1. From equation 3) and the recurrence hypothesis on D k ) we obtain D k ) = D k 1 ) 1 D k 1 ) D k 1) 1 D k 1 )+1 Isolating D k 1 ), we obtain D k 1 ) = D k) 1 induction. D k )+1) ). ) which completes the Theorem 3 The sets D h, n, n 0, are enumerated by the Motzkin numbers. Proof. Lemma1induces D k )D k 1 )+D k 1 ) = D k ) 1for k 1. By taking the limits when k converges to infinity, one gets D) + 1)D)+1 = 0, which is the functional equation of the generating function of Motzkin numbers. Now, we will show how D k ), k 0, can be epressed as a closed form. For this, let us define the function 1 f : u u For n 1, we denote by f n the function recursively defined by f n u) = f f n 1 u) ) anchored with f 0 u) = u. A simple calculation using Maple for instance) proves that the map f satisfies Remark 1. 7

8 Remark 1 If X = Y 1 Y+1) ) Y f 1 Y X) then the map f satisfies = fy) 1 fy) fx) ). k\n Table 3: Number c n,k of Dyck paths of height k in D h, n, 1 n 10 and 1 k 9. Theorem 4 For k 0, we have D k ) = f k 4 D k mod 4 ) ) with the initial cases D 0 ) = 1, D 1 ) = 1 1, D 1 ) = 1 and D 3 ) = Proof. We proceed by induction on k. Since we have D 0 ) = f 0 D 0 ) ), D 1 ) = f 0 D 1 ) ), D ) = f 0 D ) ) and D 3 ) = f 0 D 3 ) ), the basic case holds. Assuming the result for 0 i k, we prove it for k +1. D From equation 3) we have D k+1 ) = k ) ). Using the 1 D k ) D k 1 ) recurrence hypothesis for D k ) and D k 1 ), we obtain for k 4: 8

9 D k+1 ) = = f k ) 4 D k mod 4 ) 1 f k 4 D k mod 4 ) ) f k 1 ) ) 4 D k 1 mod 4 ) f f k 4 ) ) 4 D k 4 mod 4 ) ) 1 f f k 4 4 D k 4 mod 4 )) f The recurrence hypothesis implies D k+1 ) = and using Remark 1, we deduce D k+1 ) = f fd k 4 )) 1 fd k 4 )) fd k 5 )) D k 4 ) 1 D k 4 ) D k 5 ) ) ) f k 5 4 D k 5 mod 4 )) ), = fd k 3 )) = f 1+ k 3 4 D k 3 mod 4 )) = f k+1 4 D k+1 mod 4 )). )). The induction is completed. For instance, we have D 13 ) = andthe first terms of its Taylor epansion are Notice that, by taking limits in Theorem 4, we retrieve the continued fraction of the Motzkin numbers P. Barry [3]) D) =

10 3. A constructive bijection In this section, we ehibit a constructive bijection between Dn h, and the set M n of Motzkin paths of length n, i.e., lattice paths starting at 0,0), ending at n,0), never going below the -ais, consisting of up-steps U = 1,1), down steps D = 1, 1) and flat steps F = 1,0). We set M = n 0 M n. Let us define recursively the map φ from D h, to M as follows. For P D h,, we set ǫ if P = ǫ, φp) = φα)f if P = αud, φα)φγ)uφβ)d if P = αuuβdγd. Due to the recursive definition, the image by φ of a Dyck path of semilength n is a Motzkin path of length n. For instance, the images of UDUDUD,UUDDUD,UUDUDD,UUUDDD,UUUUDDDDUUUDDUDD are respectively FFF, UDF, FUD, UFD and UUDDFUFD. We refer to Figure 4 for an illustration of this mapping. α φα) α β γ φα) φγ) φβ) Figure 4: Illustration of the bijection between D h, n and M n. Moreover, we easily deduce the two following facts. Fact 1 If α,β D h, n and αβ D h, n, then we have φαβ) = φα)φβ). Fact We say that a peak UD is odd whenever the maimal sub-path of the form U k D that contains it has an odd number of U-steps, i.e., k is odd. Then, φ transforms peaks resp. odd peaks) of paths of D h, into peaks and flats resp. flats) of Motzkin paths. M n defined above is a bijection satisfy- Theorem 5 The map φ : Dn h, ing for any P D h, n, hp) hφp)) = 10.

11 Proof. We proceed by induction on n. Obviously, for n = 1, we have φud) = F and hf) = 0 = hud). For k n, we assume that φ is a bijection from D h, to M k such that hφp)) = hp) k for any P Dh, k, and we prove the result for n+1. Using the enumerative result of Theorem 3, it suffices to prove that φ is surjective from D h, n+1 to M n+1. So, let M be a Motzkin path in M n+1. We distinguish two cases: i) M = σf with σ M n, and ii) M = σuπd where σ and π are two Motzkin paths in M. i) Using the recurrence hypothesis, there is P Dn h, such that σ = φp) and hσ) = hp). So, the Dyck path PUD belongs to Dh, n+1 and φpud) = σf which proves that M belongs to the image by φ of D h, n+1. Moreover we have hφpud)) = hσf) = hσ) = hp) = hpud). ii) We suppose M = σuπd. Let us define the longest suffi σ s of σ possibly empty) such that σ s M and hφ 1 σ s )) 1 + hφ 1 π)) σ s eists since the empty path ǫ satisfies the inequality, and the recurrence hypothesis ensuresthe eistence and the uniqueness of φ 1 σ s ) and φ 1 π)). Let σ r be the Motzkin path obtained from σ by deleting the suffi σ s, and let S Dn h, resp. R Dn h, ) such that φs) = σ s and hσ s ) = hs) resp. φr) = σ r and hσ r ) = hr) ). Also there is T Dh, n such that φt) = π with hπ) = ht). Duetothemaimality ofσ s,σ r iseitheremptyor1+ht) < hφ 1 σ r σ s )). Using Fact 1 we obtain hφ 1 σ r σ s )) = hrs) = hr), and the last inequality can be written as 1+hT) < hr). - If σ r = ǫ then the condition hs) 1+hT) implies that UUTDSD belongstodn h, andwehaveφuutdsd) = φs)uφt)d = σ s UπD = σu πd. Moreover, we have hφu U T DSD)) = hφs)u φt)d) = ma{φs),1+φt)} = ma{ hs),1+ ht) }. WithhS) 1+hT), we deduce hφuutdsd)) = 1+ ht) = ht)+ = huutdsd) as desired. - If σ r ǫ then we have hs) 1 + ht) < hr) which implies that the Dyck path RUUTDSD belongs to D h, n+1. Moreover, we have hφruutdsd)) = hφr)φs)uφt)d) = ma{φr),φs),1 + φt)} = ma{ hr), hs),1 + ht) }. From hs) 1 + ht) < hr), we obtain hs)+1 ht)+ hr) which induces that hφruutdsd)) = hr) = hruutdsd) as desired. 11

12 The induction is completed. Corollary 1 Let P be a Dyck path in D h, n, n 1. The Motzkin path φp) contains a flat step F at height hφp)) if and only if hp) is odd. Proof. We proceed by induction on n. For n = 1, we have P = UD and φp) = F and the result holds since hp) = 1 is odd. Assuming the result for i n, we prove it for n + 1. We distinguish two cases: i) P = αud and ii) P = αuuβdγd, where α,β and γ belong to D h,. i) For α ǫ, the recurrence hypothesis means that φα) contains a flat step F at height hφα)) if and only if hα) is odd. Since we have φp) = φα)f and hp) = hα), φp) contains a flat step F a height hφp)) = hφα)) if and only if hp) is odd. ii) If P = αuuβdγd then we have hα) +hβ) 1 + hγ) and φp) = φα)φγ)uφβ)d. - As we have done in i), a flat step F appears at height hφp)) = hφα)) in φα) if and only if hα) = hp) is odd. - φp) contains a flat step at height hφp)) in φβ) if and only if a flat step appears at height hφβ)) in φβ) and hφp)) = hφβ)) + 1, i.e., hp) = hβ) + 1. Using the recurrence hypothesis, this is equivalent to hβ) is odd, which means hp) = hβ) + 1. Since hp) hβ)+, this is equivalent to hp) = hβ)+ and thus hp) is odd. - Let us prove by contradiction that a flat step cannot occur at height hφp)) in φγ). Indeed, this should imply the following: hφp)) = hφγ)) 1+hφβ)). With Theorem 5 we obtain hγ) hβ) 1+. Due to the recurrence hypothesis, hγ) is odd, so 1+hγ) +hβ) huγd) huuβdd) 1, and >, which implies huγd) > huuβdd). This last inequality contradicts P D h,. 1

13 The induction is completed. As a byproduct, we derive in Corollary three generating functions for the number of Motzkin paths of height k according to a constraint on the maimal height of a flat. The two first results do not seem to appear in the literature, while the third one is presented in [4]. See Tables 4 and 5 for numerical data. Corollary The generating function M k ) where the coefficient of n is the number Motzkin paths in M n of height eactly k and where there is a flat F at height k is M k ) = D k+1 ) D k ). The generating function M k ) where the coefficient of n is the number Motzkin paths in M n of height eactly k and where there is no flat F at height k is M k ) = D k ) D k 1 ). The generating function M k ) where the coefficient of n is the number Motzkin paths in M n of height eactly k is M k ) = D k+1 ) D k 1 ). Proof. The proof is directly deduced from Theorem 5 and Corollary 1. For instance, we have M 1 ) = 3, and the coefficient of 1 )1 n ) is given by n Fn + ) where Fn) is the nth Fibonacci number. The first terms of its Taylor epansion are see sequence A in [17]). Also, M 1 ) = 1 )1 ) and the coefficient of n is Fn) 1. The first terms of its Taylor epansion are see sequence A in [17]). Notice that the two sequences defined by M k ) and M k ) do not appear in [17]. We k 0 k 0 leave open the question of finding closed forms for their generating functions. Corollary 3 The bivariate generating function Q, y) where the coefficient of n y k is the number of Dyck paths of Dn h, containing eactly k peaks U D is 1 y+ y 4 y 4 y+ 3 y y+ y y y+1. 13

14 k\n Table 4: Number of Motzkin paths of length n and height k with a flat at height k, 1 n 10 and 0 k 4. k\n Table 5: Number of Motzkin paths of length n and height k with no flats at height k, 1 n 10 and 1 k 5. Proof. By Fact, the number of peaks UD in Dyck paths of Dn h, is equal to the number of peaks UD and flats F in Motzkin paths in M n. Since a non empty Motzkin path in M is either αf or αuβd where α,β M, the generating function Q, y) satisfies the functional equation Q, y) = 1 + yq,y) + yq,y) + Q,y) 1)Q,y). A simple calculation with Maple for instance) completes the proof. It is interesting to notice that Q,y) 1 = yq,y) where Q,y) is the bivariate generating function where the coefficient of n y k is the number of Motzkin n-paths with k weak valleys see A in [17]). As a consequence, we deduce in Corollary 4 the popularity of peaks in D h, n. Corollary 4 The generating function where the coefficient of n is the total 14

15 number of peaks in all Dyck paths of Dn h, is 1 1+) Proof. Using Corollary 3, the result is given by Q,y) y y=1. ThepopularityofpeaksinDn h,, n 1, isgiven bythesequencea05566 in [17], which is the first difference of the sequence A that counts the directed animals of size n. The first terms are Corollary 5 The bivariate generating function R, y) where the coefficient of n y k is the number of Dyck paths of Dn h, containing eactly k odd peaks UD is 1 y 1 y 4 + y. is also the number of flats on M n. Since a non empty Motzkin path in M is either αf or αuβd where α, β M, the generating function R, y) satisfies the functional Proof. By Fact, the number of odd peaks on D h, n equation R,y) = 1+yR,y) + R,y). A simple calculation with Maple for instance) completes the proof. Corollary 6 The generating function where the coefficient of n is the total number of odd peaks in all Dyck paths of Dn h, is Proof. Using Corollary 5, the result is given by R,y) y y=1. The popularity of odd peaks is given by the sequence A in [17], wherethe nth term is the nthmotzkin numberdivided by n. Thefirst terms are

16 4 Final remarks In this paper, we study the enumeration of Dyck paths having a first return decomposition with special properties based on a height constraint. For future research, it would be interesting to investigate other statistics on Dyck paths such that number of peaks, valleys, zigzag or double rises, etc. More generally, can we etend this study for other combinatorial objects such as Motzkin and Ballot paths or permutations? 5 Acknowledgement We would like to thank the anonymous referees for their very careful reading of this paper and their helpful comments and suggestions. References [1] E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani. Nondecreasing Dyck paths and q-fibonacci numbers. Discrete Math., ), 1-3, [] J.-L. Baril and A. Petrossian. Equivalence classes of Dyck paths modulo some statistics. Discrete Math., ), 4, [3] P. Barry. On sequences with { 1, 0, 1} Hankel transforms. preprint, 01, available electronically at [4] C. Brennan and A. Knopfmacher. The height of two types of generalised Motzkin paths. J. of Statistical Planning and Inference, ), [5] D. Callan. Two bijections for Dyck path parameters. preprint, 004, available electronically at [6] E. Czabarka, R. Florez and L. Junes. Some enumeration on non-decreasing Dyck paths. Electronic Journal of Combinatorics, 1)015). [7] A. Denise and R. Simion. Two combinatorial statistics on Dyck paths. Discrete Math., ), 1-3, [8] E. Deutsch. Dyck path enumeration. Discrete Math., ),

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