在Dyck路徑、不相交劃分，與2-Motzkin路徑上的統計量

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(1)國立高雄大學應用數學系碩士論文. Some Statistics in Dyck Paths, Non-crossing Partitions, and 2-Motzkin Paths 在 Dyck 路徑、路徑、不相交劃分，不相交劃分，與 2-Motzkin 路徑上的路徑上的統計量. 研究生：王佩錡撰指導教授：鄭斯恩教授. 中華民國 104 年 1 月.

(2) 致謝首先，感謝我的指導教授. 鄭斯恩教授，在這三年半來，教授不僅在學術研究. 上提供專業的指導，更當我因工作、婚姻、學業無法兼顧而萌生休學念頭時，適時地開導我，讓我得以堅持下去。感謝老師用那嚴謹的態度，仔細耐心地給予指導，讓我的論文得以順利完成。在此致上萬分的感謝。同時，也感謝口試委員張惠蘭教授與李渭天教授，提供了寶貴的意見，使論文更臻完善。不但拓展我的視野，更加深我的知識與見聞，在此由衷地感謝！此外，感謝應數系上的老師們，傳授許多專業知能，增長了我的學識。也感謝在研究期間，共同勉勵打氣的同學及學弟妹們，不藏私地提供我許多意見與資訊。特別感謝學弟邱奕彰，提供許多 Latex 技術上的指導。也要感謝高雄大學，提供一個良好、僻靜、舒適的研究空間，讓我得以在此做研究，更謝謝應用數學系親切的行政同仁，在這段期間提供許多協助與資源。同時，要感謝立德國中的校長與同仁，在我在職進修期間給予協助、支持與鼓勵。更特別感謝英語科丁韋安老師，在英文方面提供我許多寶貴的意見與協助，使我的文章更趨於通順。最後，感謝我的父母，當學業與工作蠟燭兩頭燒時，給予我最大的鼓勵與支持，更感謝我的先生，婚後默默地包容與陪伴。真心感謝一路相伴支持的人，僅將本論文獻給所有關心我的人，以表達我無限的感激。. 王佩錡謹誌高雄大學 2015.01.16.

(3) Some Statistics in Dyck Paths, Non-crossing Partitions, and 2-Motzkin Paths. by Pei-Chi Wang Advisor Dr. Szu-En Cheng. Department of Applied Mathematics National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. November 2014.

(4) Contents Abstract (in Chinese). ii. Abstract (in English). iii. 1 Introduction 1.1 Definitions and notations . . . . . . . . . . 1.1.1 Dyck paths . . . . . . . . . . . . . 1.1.2 Non-crossing partitions . . . . . . . 1.1.3 Motzkin paths and 2-Motzkin paths 1.2 Structure . . . . . . . . . . . . . . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. 2 Bijection between non-crossing partitions and Dyck paths. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. 1 1 1 2 4 5 5. 3 The generating functions of some statistics in Dyck paths 10 3.1 The statistics edes, odes, easc, and oasc . . . . . . . . . . . . . . . . . 10 3.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4 Bijection between 2-Motzkin paths and Dyck paths 20 4.1 The statistics emaj and ecomaj . . . . . . . . . . . . . . . . . . . . . 23 4.2 The statistics omaj and ocomaj . . . . . . . . . . . . . . . . . . . . . 25 5 Conclusion Remark. 27. References. 28. i.

(5) 在 Dyck路路徑、不相交劃分，與 2-Motzkin路路徑上的統計量指導教授：鄭斯恩教授國立高雄大學應用數學系. 學生：王佩錡國立高雄大學應用數學系. 摘要本論文主要利用兩個對射函數來得到在 Dyck 路徑上四個統計量 edes、odes、easc，和 oasc 的生成函數，其中一個是有關不相交劃分和 Dyck 路徑上的對射，而另一個是有關 2-Motzkin 路徑和 Dyck 路徑上的對射。接著我們進一步推導出 Dyck 路徑上統計量 emaj、omaj、ecomaj，和 ocomaj 的計數多項式，並連同討論在不相交劃分及 2-Motzkin 路徑上的相關結果。. 關鍵字: Dyck 路徑、不相交劃分、2-Motzkin 路徑、統計量. ii.

(6) Some Statistics in Dyck Paths, Non-crossing Partitions, and 2-Motzkin Paths Advisor: Dr. Szu-En Cheng Department of Applied Mathematics National University of Kaohsiung. Student: Pei-Chi Wang Department of Applied Mathematics National University of Kaohsiung. Abstract In this study, we use two bijections: one is between non-crossing partitions and Dyck paths, and the other is between 2-Motzkin paths and Dyck paths. We obtain the generating function of statistics edes, odes, easc, and oasc in Dyck paths. Moreover, we derive the enumerating polynomials of the statistics emaj, omaj, ecomaj, and ocomaj in Dyck paths. The corresponding results for non-crossing partitions, and 2-Motzkin paths are also discussed.. Keywords: Dyck path, non-crossing partition, 2-Motzkin path, statistic.. iii.

(7) 1. Introduction. In [8] F¨ urlinger and Hofbauer showed that the generating function for the major index on Dyck paths is a natural q-analogue of the Catalan numbers. They also refined the major index to two statistics: α and β, and found some formulas. In [14, 15] Sapounakis, Tasoulas, and Tsikouras dealt with the statistics “number of occurrences of τ at even (or odd) height”, for some strings τ in Dyck paths. In this thesis, we will investigate some refinements of the statistics: the descent number, the ascent number, the major index, and the comajor index on Dyck paths.. 1.1 1.1.1. Definitions and notations Dyck paths. A Dyck path of semilength n is a path on the plane from (0,0) to (2n,0) consisting of up steps u=(1,1) and down steps d=(1,−1), such that the path does not go below the x-axis. The set of Dyck paths of semilength n is denoted by D(n). The number of 2n 1 [9, A000108]. Dyck paths of semilength n is called the Catalan number Cn = n+1 n Let ω = ω1 ω2 . . . ω2n be a Dyck path of semilength n. We define the descent set, the descent number, and the major index of ω by X Des(ω) := {i : ωi ωi+1 = du}, des(ω) := | Des(ω)|, and maj(ω) := i. i∈Des(ω). Analogously, we define the ascent set, the ascent number, and the comajor index of ω by X Asc(ω) := {i : ωi ωi+1 = ud}, asc(ω) := | Asc(ω)|, and comaj(ω) := i. i∈Asc(ω). Example 1.1. Let ω = uuuduudddduududd ∈ D(8) as shown in the figure below. By the definitions, we have Des(ω) = {4, 10, 13}, Asc(ω) = {3, 6, 12, 14}, maj(ω) = 4 + 10 + 13 = 27, and comaj(ω) = 3 + 6 + 12 + 14 = 35. We can easily find that comaj(ω) − maj(ω) = 8. 6 3. 4. 5. 7 8. 2. 12. 9 10 11. 1. 13 14 15 16. In a Dyck path, a peak is an occurrence of ud, and a valley is an occurrence of du. By the location (respectively, height) of a peak or of a valley, we mean the 1.

(8) x-coordinate (respectively, y-coordinate) of the intersection point of its two steps. Thus, the elements of Des(ω) (respectively, Asc(ω)) are the locations of the valleys (respectively, peaks). We call a peak or valley is even (respectively, odd ) if its location (or, equivalently, its height) is even (respectively, odd). We define the even descent set, the even descent number, and the even major index of ω by eDes(ω):= {i : i ∈ X Des(ω) and i is even}, edes(ω):= | eDes(ω)|, and emaj(ω) := i. i∈eDes(ω). The others are defined similarly. For ω ∈ D(n), n ≥ 1, we can uniquely decompose ω in the form ω = uµdν, where µ ∈ D(k − 1), ν ∈ D(n − k), and (2k, 0) is the first touch point (except for (0, 0)) at the x-axis for some 1 ≤ k ≤ n. See Figure 1. This is so called the first return decomposition [10, 11, 15]. µ ∈ D(k − 1) ν ∈ D(n − k) 2k. 1. Figure 1: The first return decomposition of a nonempty Dyck path.. The number of Dyck paths of semilength n with k valleys is the Narayana number n Nn,k = n1 nk k+1 [4][9, A001263], and the sum of Nn,k for 0 ≤ k < n is the n-th Catalan numbers Cn . 1.1.2. Non-crossing partitions. A partition of [n] := {1, 2, ..., n} is a collection of disjoint nonempty subsets of [n], called blocks, whose union is [n]. A partition of [n] with k blocks is denoted by π = B1 /B2 /.../Bk , where the blocks are ordered in the increasing order of their minimum elements and within each block, the elements are written in the numerical order. A partition π = B1 /B2 /.../Bk of [n] is non-crossing if for all 1 ≤ a < b < c < d ≤ n, it does not have a, c ∈ Bi and b, d ∈ Bj for i 6= j; thus any two blocks Bi , Bj of π do not cross. Let N C(n) denote the set of non-crossing partitions of [n], and let N C(n, k) denote the set of non-crossing partitions of [n] with k blocks. A block B is called a singleton if |B| = 1.. 2.

(9) We can express non-crossing partitions graphically by plotting 1, 2, . . . , n on the real line and joining successive elements of the same block by arcs, and draw circles at singletons. See Figure 2.. 1. 2. 3. 4. 5. 6. 7. Figure 2: Illustration of the non-crossing partition π = 157/234/6 ∈ N C(7).. Note that, in general, most people only draw a point at a singleton, but we draw a circle at a singleton because it is convenient to label the interval later. For π ∈ N C(n, k), we define Fs(π) := {f1 , f2 , . . . , fk }, the set of the smallest elements of the blocks, Ls(π) := {l1 , l2 , . . . , lk }, the set of the largest elements of the blocks, Ir(π) := {i : 1 ≤ i ≤ n, i 6∈ Fs(π) ∪ Ls(π)}, the set of the interiors of the blocks of π, and Sg(π) := {i : 1 ≤ i ≤ n, i ∈ Fs(π) ∩ Ls(π)}, the union of the singletons of π, where fi = min{a : a ∈ Bi } is the smallest element, and li = max{a : a ∈ Bi } is the largest element, respectively, of the ith block of π. We also define the number of and the sum of the elements of each set by X X nir(π) := | Ir(π)|, nsg(π) := | Sg(π)|, mir(π) := i, msg(π) := i. i∈Ir(π). i∈Sg(π). If we represent non-crossing partitions graphically, we could see that there are four kinds of points, which are related to Fs, Ls, Ir, and Sg (illustrated in Figure 3).. a c b a ∈ Fs(π) \ Sg(π) b ∈ Ls(π) \ Sg(π) c ∈ Ir(π). d d ∈ Sg(π). Figure 3: Illustration of four kinds of points of non-crossing partitions.. We will use the following encoding as in [16]. Given a non-crossing partition π ∈ N C(n), we label each interval [j, j+1], 1 ≤ j < n, with λj ∈ {b, e, l, r} as follows:. 3.

(10)   b, if j j + 1 and j is not the largest element in its block;        e, if j j + 1 and j + 1 is not the smallest element in its block; λj = l, if j j + 1, j is the largest element in its block,     and j + 1 is the smallest element in its block;     r, if j ∼ j + 1, where j ∼ j + 1 means that j and j + 1 are in the same block. The set of indices i for which λi = b (e, l, r, respectively) will be denoted by B(π) (E(π), L(π), R(π), respectively). And we define the number of and the sum of the elements of them by nb(π) := | B(π)|, ne(π) := | E(π)|, nl(π) := | L(π)|, nr(π) := | R(π)|, X X X X mb(π) := i, me(π) := i, ml(π) := i, mr(π) := i. i∈B(π). i∈E(π). i∈L(π). i∈R(π). The four kinds of intervals are illustrated in Figure 4.. e. b. r. l. Figure 4: Illustration of intervals labeled b, e, l, r.. 1.1.3. Motzkin paths and 2-Motzkin paths. A Motzkin path of length n is a path on the plane from (0,0) to (n,0) consisting of up steps u=(1,1), down steps d=(1,−1), or horizontal steps h=(1,0) that never goes below the x-axis. The set of Motzkin paths of length n is denoted by M(n). The number of Motzkin paths of length n is called the Motzkin number Mn = bn/2c X n Ck (see [9, A001006]). The first few terms of the Motzkin numbers {Mn } 2k k=0 for n ≥ 0 are 1, 1, 2, 4, 9, 21, 51, 127, 323, 835, · · · . A 2-Motzkin path of length n is a Motzkin path, where the horizontal steps can be of two kinds: straight s or wavy w (see Figure 5). The set of 2-Motzkin paths of length n is denoted by T M(n). It is known that the number of 2-Motzkin paths of length n is given by the Catalan number Cn+1 (see, for example [6]).. u. s. d. w. Figure 5: Four types of steps in 2-Motzkin paths.. 4.

(11) 1.2. Structure. The rest of the thesis is organized as follows. In Section 2, we introduce a bijection φ between non-crossing partitions and Dyck paths. In Section 3, we calculate the generating function of the statistics edes, odes, easc, and oasc in Dyck paths. Then we use it to obtain the generating functions for some interesting quantities. In Section 4, we introduce a bijection ψ between 2-Motzkin paths and Dyck paths, and use these two bijections φ and ψ to get the enumerating polynomials of the statistics emaj, omaj, ecomaj, and ocomaj in Dyck paths. Finally, we summarize our results and list some open questions.. 2. Bijection between non-crossing partitions and Dyck paths. Now we introduce a bijection between non-crossing partitions and Dyck paths. Define a mapping φ : N C(n) → D(n) by φ(π) = θ(1)θ(2) . . . θ(n), where    uu, if i ∈ Fs(π) \ Sg(π);   dd, if i ∈ Ls(π) \ Sg(π); θ(i)=  du, if i ∈ Ir(π);    ud, if i ∈ Sg(π). The mapping φ is a well-known bijection. Since we cannot find a proof in the literature, we take this opportunity to give a proof here. Theorem 2.1. The mapping φ : N C(n) → D(n) is a bijection. Proof. For π = B1 /B2 / · · · /Bk ∈ N C(n), let Bi = {fi , gi , . . . , li }, where fi < gi < · · · < li , and bi = |Bi | for all i = 1, 2, . . . , k. (Bi = {fi } for |Bi | = 1.) Then Bi will generate a Dyck path θ(fi )θ(gi ) · · · θ(li ) = u(ud)bi −1 d ∈ D(bi ). We now claim that φ(π) = θ(1)θ(2) · · · θ(n) is a Dyck path of semilength n. First, let D(1) = θ(f1 = 1)θ(g1 ) · · · θ(l1 ) = u(ud)b1 −1 d ∈ D(b1 ) be the Dyck path generated by B1 . Then, for 1 < j ≤ k, we insert the Dyck path θ(fj )θ(gj ) · · · θ(lj ) = u(ud)bj −1 d generated by Bj into D(j−1) at the position after θ(mj−1 ) to produce a Dyck path P S D(j) ∈ D( ji=1 bi ), where mj−1 = max{t ∈ j−1 i=1 Bi | t < fj }. Note that we have Sj−1 mj−1 + 1 = fj , and the elements of i=1 Bi are ordered increasingly to produce D(j) by the non-crossing property. Hence, we obtain D(k) = θ(1)θ(2) · · · θ(n) which is a Dyck path of semilength n. 5.

(12) For the inverse mapping φ−1 : D(n) → N C(n), given ω = ω1 ω2 . . . ω2n ∈ D(n), we parenthesize the sequence 1, 2, . . . , n as follows.  (i, put a left parenthesis before i if ω2i−1 ω2i =uu,     i), put a right parenthesis after i if ω 2i−1 ω2i =dd,  i , put no parenthesis on i if ω2i−1 ω2i =du,    (i), put a pair of parentheses enclosing i if ω2i−1 ω2i =ud. Then create a block of φ−1 (ω) for each of the consecutive strings inside lowest level parethesis pairs (i.e. paretheses which pair each other and enclose no others.) Now remove these lowest level parenthesis pairs and all the numbers they enclose, and continue with the remaining parenthesization. Clearly the partition φ−1 (ω) is noncrossing. Let us see an example of the bijection φ. Example 2.2. (a) π = 146/23/5 ∈ N C(6) is mapping to ω = uuuuddduuddd ∈ D(6). (b) ω = uuuuddduuddd ∈ D(6) is leading to the parenthesization (1(23)4(5)6), then we get π = 146/23/5. Illustrate in Figure 6. π=146/23/5 Fs(π) \ Sg(π) = {1, 2}, Ls(π) \ Sg(π) = {3, 6}, Ir(π) = {4}, Sg(π) = {5}.. ω=uuuuddduuddd =uu/ uu/ dd/ du/ ud/ dd. ω1 ω2 ω3 ω4 ω5 ω6 ω7 ω8 ω9 ω10ω11ω12. (1. ω=u u u u d d d u u d d d. (2. 3). 4. (5). 6). π = 146/23/5. (b) φ−1. (a) φ. Figure 6: Example of the bijection φ. We could describe the bijection φ by the card arrangements as well, which is similar to the technique used in [17]. Let us redefine φ−1 : D(n) → N C(n) for ω = ω1 ω1 · · · ω2n by φ−1 (ω) = λ1 λ2 . . . λn , where    use the Uj card if ω2i−1 ω2i = uu and the height of ω2i−1 is 2j;   use the D card if ω j 2i−1 ω2i = dd and the height of ω2i−1 is 2j; λi =  use the Tj card if ω2i−1 ω2i = du and the height of ω2i−1 is 2j;    use the Kj card if ω2i−1 ω2i = ud and the height of ω2i−1 is 2j. 6.

(13) The height of ωi is defined to be the y-coordinate of the initial point of ωi , and the cards are described in Figure 7. Then λ1 λ2 . . . λn would be a graphically representation of a non-crossing partition. Example 2.3. Let ω = uuuuddduuddd ∈ D(6). φ−1. λi : U0. (1 (2 3) 4 (5) 6) We have φ−1 (ω) = 146/23/5 ∈ N C(6). 1. U1 D2 T1 K1 D1. 2. 3. U cards :. 4. 5. 6. i+1 i. U0. U1. U2. D cards :. Ui i i-1. D1. D2. D3. T cards :. Di i. T1. T2. T3. K cards :. i. Ti i. K0. K1. K2. i. Ki. Figure 7: Four kinds of cards. Now we go back to the definition of the bijection φ and discuss the locations of the valleys in the Dyck path ω = ω1 ω2 · · · ω2n = φ(π) for a non-crossing partition π ∈ N C(n). We consider when the valleys of ω occur at even and odd locations. An even valley ω2i ω2i+1 = du occurs in ω if and only if i ∈ Ls(π) and i + 1 ∈ Fs(π) if and only if [i, i + 1] is labelled by l. An odd valley ω2i−1 ω2i = du occurs in ω if and only if i ∈ Ir(π). See Figure 8.. 7.

(14) odd valley:. even valley:. i i+1 l interval. ω2i ω2i+1. across two groups. i Ir points. ω2i−1 ω2i. in one group. Figure 8: Illustration of the formation of valleys. Hence we have the following relations between non-crossing partitions and Dyck paths. π ∈ N C(n) ω = φ(π) ∈ D(n) i ∈ L(π). ←→. 2i ∈ eDes(ω). (1). i ∈ Ir(π). ←→. 2i − 1 ∈ oDes(ω). (2). Also, edes(ω) = nl(π), odes(ω) = nir(π), and des(ω) = nl(π) + nir(π).. (3). Thus we have the following theorem. Theorem 2.4. For π ∈ N C(n), ω = φ(π), we have maj(ω) = 2 ml(π) + 2 mir(π) − nir(π). Proof. Using (1) and (2), we have X X X maj(ω) = i= 2i + (2i − 1) i∈Des(ω). X. =. 2i +. i∈L(π). i∈L(π). i∈Ir(π). X. X. 2i −. i∈Ir(π). 1 = 2 ml(π) + 2 mir(π) − nir(π).. i∈Ir(π). Example 2.5. Let π = 15/2/34/678 and ω = φ(π), then L(π) = {2, 5} and Ir(π) = {7}. By (1), (2) and Theorem 2.4, we have Des(ω) = {4, 10, 13} and maj(ω) = 4 + 10 + 13 = 27. See the figure below. 6 3 2. interval: point:. 1. 2. l. 3. 4. 5. l. 6. 7. 4. 5. 7 8. 8. Des(ω) = {4, 10, 13}. Ir. 8. 12. 9. 1. 10 11. 13 14 15 16.

(15) Similarly, we discuss the locations of the peaks in ω. We consider that the peaks of ω occur at even and odd locations. An even peak ω2i ω2i+1 = ud occurs in ω if S S and only if i ∈ (Fs(π) \ Sg(π)) Ir(π) and i + 1 ∈ (Ls(π) \ Sg(π)) Ir(π) if and only if [i, i + 1] is labelled by r. An odd peak ω2i−1 ω2i = ud occurs in ω if and only if i ∈ Sg(π). See Figure 9. Hence we have the following relations between non-crossing partitions and Dyck paths. π ∈ N C(n) ω = φ(π) ∈ D(n) i ∈ R(π). ←→. 2i ∈ eAsc(ω). (4). i ∈ Sg(π). ←→. 2i − 1 ∈ oAsc(ω). (5). easc(ω) = nr(π), oasc(ω) = nsg(π), and asc(ω) = nr(π) + nsg(π).. (6). Also,. odd peak:. even peak:. i i+1. r interval. ω2i ω2i+1. across two groups. i ω2i−1 ω2i. in one group. Sg points. Figure 9: Illustration of the formation of peaks.. Thus, we have the following theorem. Theorem 2.6. For π ∈ N C(n), ω = φ(π), we have comaj(ω) = 2 mr(π) + 2 msg(π) − nsg(π). Proof. Using (4) and (5), we have X X X comaj(ω) = i= 2i + (2i − 1) i∈Asc(ω). =. X i∈R(π). 2i +. i∈R(π). X i∈Sg(π). i∈Sg(π). 2i −. X. 1. i∈Sg(π). = 2 mr(π) + 2 msg(π) − nsg(π). Note that it is easy to see that maj(ω) = comaj(ω) − n for all ω ∈ D(n).. 9.

(16) Example 2.7. Let π = 15/2/34/678 and ω = φ(π), then R(π) = {3, 6, 7} and Sg(π) = {2}. By (4), (5) and Theorem 2.6, we have Asc(ω) = {3, 6, 12, 14} and comaj(ω) = 3 + 6 + 12 + 14 = 35 (maj(ω) = 35 − 8 = 27). See the figure below. 6 3. 4. 7. 5. 2. interval: point:. 3. 1. 2. 3. r. 4. 5. 6. r. 7. r. 1. 8. 8 12. 9 10 11. 13 14 15 16. Asc(ω) = {3, 6, 12, 14}. Sg. The generating functions of some statistics in Dyck paths. 3.1. The statistics edes, odes, easc, and oasc. It is clear that edes(∅) = odes(∅) = easc(∅) = oasc(∅) = 0. We can easily get the following relations. Lemma 3.1. Let ω ∈ D(n), n ≥ 1, and ω = uµdν be the first return decomposition of ω, where µ ∈ D(k − 1) and ν ∈ D(n − k) for some 1 ≤ k ≤ n. Then we have ( odes(µ) + 1 + edes(ν) if 1 6 k < n; (7) edes(ω) = odes(µ) if k = n; odes(ω) = edes(µ) + odes(ν). if 1 6 k 6 n;. (8). easc(ω) = oasc(µ) + easc(ν) ( easc(µ) + oasc(ν) oasc(ω) = 1 + oasc(ν). if 1 6 k 6 n;. (9). if 1 < k 6 n; if k = 1.. (10). Proof. Let µ+ = uµd ∈ D(k). Since even valleys of µ+ are odd valleys of µ, we have edes(µ+ ) = odes(µ). And there is an even valley at location 2k if ν 6= ∅. Even valleys of ν remain even in ω, thus we complete the proof of Equation (7). The others can be proved similarly. Now we encode edes, odes, easc, and oasc by α, β, γ, and δ, respectively, and write down the generating function of the enumerating polynomials according to these statistics by X X G(z) = G(α, β, γ, δ; z) := αedes(ω) β odes(ω) γ easc(ω) δ oasc(ω) z n . n≥0 ω∈D(n). 10.

(17) Using Lemma 3.1, we can derive the following theorem. Theorem 3.2. The generating function G(z) = G(α, β, γ, δ; z) has the explicit form of h 1 1 + (2β − αδ − βγ)z + (αγ − 2α + 1)(βδ − 1)z 2 G(z) = 2 2(α − αβδ)z + 2βz i p − (αβγδz 2 − βδz 2 − αγz 2 + z 2 − αδz − βγz + 1)2 − 4z 2 . (11) P Proof. Let G(n; α, β, γ, δ) = ω∈D(n) αedes(ω) β odes(ω) γ easc(ω) δ oasc(ω) . We can get that G(0; α, β, γ, δ) = 1 and G(1; α, β, γ, δ) = δ. For n ≥ 2, by Lemma 3.1, we have G(n; α, β, γ, δ) X = αedes(ν)+1 β odes(ν) γ easc(ν) δ 1+oasc(ν) ν∈D(n−1). +. n−1 X. . X. α. odes(µ)+edes(ν)+1 edes(µ)+odes(ν) oasc(µ)+easc(ν) easc(µ)+oasc(ν). β. γ. δ. . k=2 (µ,ν)∈D(k−1)×D(n−k). X. +. αodes(µ) β edes(µ) γ oasc(µ) δ easc(µ). µ∈D(n−1). X. = αδ. αedes(ν) β odes(ν) γ easc(ν) δ oasc(ν). ν∈D(n−1). +α. n−1 X k=2. +. X. α. odes(µ) edes(µ) oasc(µ) easc(µ). β. γ. δ. µ∈D(k−1). X. X. α. edes(ν) odes(ν) easc(ν) oasc(ν). β. γ. ν∈D(n−k). αodes(µ) β edes(µ) γ oasc(µ) δ easc(µ). µ∈D(n−1). = αδG(n − 1; α, β, γ, δ) + α. n−1 X. G(k − 1; β, α, δ, γ) · G(n − k; α, β, γ, δ). k=2. + G(n − 1; β, α, δ, γ). Let G = G(α, β, γ, δ; z) = G = 1 + δz + αδz. X. P. n≥0. G(n; α, β, γ, δ)z n . Then. G(n − 1; α, β, γ, δ)z n−1. n≥2. + αz. n−1 XX. G(k − 1; β, α, δ, γ)z k−1 · G(n − k; α, β, γ, δ)z n−k. n≥2 k=2. +z. X. G(n − 1; β, α, δ, γ)z n−1. n≥2. 11. δ. .

(18) = 1 + δz + αδz G(α, β, γ, δ; z) − 1 + αz G(β, α, δ, γ; z) − 1 G(α, β, γ, δ; z) − 1 + z(G(β, α, δ, γ; z) − 1) = 1 + δz + αδzG − αδz + αz G(β, α, δ, γ; z) · G − G − G(β, α, δ, γ; z) + 1 + zG(β, α, δ, γ; z) − z. Hence, G 1 + αz − αδz − αzG(β, α, δ, γ; z) = 1 − z + αz + δz − αδz + (z − αz)G(β, α, δ, γ; z).. (12). Solving Equation (12) for G, we get G =G(α, β, γ, δ; z) =. 1 − (1 − α − δ + αδ)z + (z − αz)G(β, α, δ, γ; z) . 1 + αz − αδz − αzG(β, α, δ, γ; z). (13). Interchanging variables α and β in Equation (13), we obtain G(β, α, δ, γ; z) =. 1 − (1 − β − γ + βγ)z + (z − βz)G(α, β, γ, δ; z) . 1 + βz − βγz − βzG(α, β, γ, δ; z). (14). Substituting Equation (14) into Equation (12), we have 1 − (1 − β − γ + βγ)z + (z − βz)G G(1 + αz − αδz) − αzG 1 + βz − βγz − βzG 1 − (1 − β − γ + βγ)z + (z − βz)G = 1 − (1 − α − δ + αδ)z + (z − αz) . 1 + βz − βγz − βzG ⇒ G(1 + αz − αδz)(1 + βz − βγz) − G2 (1 + αz − αδz)βz − αzG[1 − (1 − β − γ + βγ)z] − αz(z − βz)G2 = [1 − (1 − α − δ + αδ)z](1 + βz − βγz) − βz[1 − (1 − α − δ + αδ)z]G + (z − αz)[1 − (1 − β − γ + βγ)z] + (z − αz)(z − βz)G. h ⇒ G2 α(1 − β)z 2 + βz(1 + αz − αδz) − G (1 + αz − αδz)(1 + βz − βγz) i − αz[1 − (1 − β − γ + βγ)z] + βz[1 − (1 − α − δ + αδ)z] − (1 − α)(1 − β)z 2 h i + [1 − (1 − α − δ + αδ)z](1 + βz − βγz) + (1 − α)z[1 − (1 − β − γ + βγ)z] = 0.. (15). Solving Equation (15) for G = G(α, β, γ, δ; z), we get the desired result.. 12.

(19) 3.2. Applications. By substituting 0, 1, t, or −t into G(α, β, γ, δ; z), we get some interesting cases. (1) α = 1, β = 1, γ = 1, δ = 1 Setting α = β = γ = δ = 1 in Equation (11), we have √ 1 − 1 − 4z X = Cn z n , G(1, 1, 1, 1; z) = 2z n≥0 where Cn are the Catalan numbers. The first few terms are 1 + z + 2z 2 + 5z 3 + 14z 4 + 42z 5 + 132z 6 + 429z 7 + · · · . (2) α = t, β = 1, γ = 1, δ = 1 The generating function of the enumerating polynomials for statistic edes is G(t, 1, 1, 1; z) =. X X. tedes(ω) z n =. 1 + (1 − t)z −. n≥0 ω∈D(n). p. 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . 2z. (16). The first few terms are 1 + z + (1 + t)z 2 + (2 + 2t + t2 )z 3 + (4 + 6t + 3t2 + t3 )z 4 + (9 + 16t + 12t2 + 4t3 + t4 )z 5 +(21+45t+40t2 +20t3 +5t4 +t5 )z 6 +(51+126t+135t2 +80t3 +30t4 +6t5 +t6 )z 7 +· · · . Let Tn,k = [tk z n ]G(t, 1, 1, 1; z). Then Tn,k is the number of Dyck paths of semilength n with k even valleys. It is also the number of Dyck paths of semilength n with k peaks at even height (see [9, A091869]). n\k 0 1 2 3 4 5 6 7. 0 1 1 1 2 4 9 21 51. 1. 2. 3. 1 2 1 6 3 1 16 12 4 45 40 20 126 135 80. 4. 5. 6. 1 5 30. 1 6. 1. In the case of t = 0 in Equation (16), we get the generating function for the number of Dyck paths of semilength n with no even valleys: √ X 1 + z − 1 − 2z − 3z 2 =1+ Mn−1 z n , G(0, 1, 1, 1; z) = 2z n≥1 13.

(20) where Mn are the Motzkin numbers. Since nl(π) = edes(φ(π)) for all π ∈ N C(n), we have p X X 1 + (1 − t)z − 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 nl(π) n t z = . 2z n≥0 π∈N C(n). (3) α = 1, β = t, γ = 1, δ = 1 The generating function of the enumerating polynomials for statistic odes is X X G(1, t, 1, 1; z) = todes(ω) z n n≥0 ω∈D(n). =. 1 − (1 − t)z −. p 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . 2z[t + (1 − t)z]. (17). The first few terms are 1+z+2z 2 +(4+t)z 3 +(9+4t+t2 )z 4 +(21+15t+5t2 +t3 )z 5 +(51+50t+24t2 +6t3 +t4 )z 6 +(127 + 161t + 98t2 + 35t3 + 7t4 + t5 )z 7 + · · · . Let Tn,k = [tk z n ]G(1, t, 1, 1; z). Then Tn,k is the number of Dyck paths of semilength n with k odd valleys. It is also the number of Dyck paths of semilength n with k uuu’s (triple rise) steps (see [9, A092107]). n\k 0 1 2 3 4 5 6 7. 0 1 1 1 2 4 1 9 4 21 15 51 50 127 161. 2. 3. 1 5 1 24 6 98 35. 4. 5. 1 7. 1. In the case of t = 0 in Equation (17), we get the generating function for the number of Dyck paths of semilength n with no odd √ valleys: 1 − z − 1 − 2z − 3z 2 X G(1, 0, 1, 1; z) = = Mn z n . 2z 2 n≥0 Since nir(π) = odes(φ(π)) for all π ∈ N C(n), we have X. X. t. nir(π) n. z =. 1 − (1 − t)z −. n≥0 π∈N C(n). 14. p 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . 2z[t + (1 − t)z].

(21) (4) α = 1, β = 1, γ = t, δ = 1 The generating function of the enumerating polynomials for statistic easc is G(1, 1, t, 1; z) =. X X. t. easc(ω) n. z =. 1 + (1 − t)z −. n≥0 ω∈D(n). p 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . 2z. This is the same as G(t, 1, 1, 1; z) in Equation (16). So we have X X. X X. tedes(ω) z n =. n≥0 ω∈D(n). teasc(ω) z n .. n≥0 ω∈D(n). Since nr(π) = easc(φ(π)) for all π ∈ N C(n), we have p X X 1 + (1 − t)z − 1 + 2(1 + t)z + (−3 + 2t + t2 )z 2 nr(π) n t z = , 2z n≥0 π∈N C(n) X X X X and tnl(π) z n = tnr(π) z n . n≥0 π∈N C(n). n≥0 π∈N C(n). (5) α = 1, β = 1, γ = 1, δ = t The generating function of the enumerating polynomials for statistic oasc is G(1, 1, 1, t; z) =. X X. toasc(ω) z n =. 1 + (1 − t)z −. n≥0 ω∈D(n). p 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . (18) 2z[1 + (1 − t)z]. The first few terms are 1 + tz + (1 + t2 )z 2 + (1 + 3t + t3 )z 3 + (3 + 4t + 6t2 + t4 )z 4 + (6 + 15t + 10t2 + 10t3 + t5 )z 5 +(15+36t+45t2 +20t3 +15t4 +t6 )z 6 +(36+105t+126t2 +105t3 +35t4 +21t5 +t7 )z 7 +· · · . Let Tn,k = [tk z n ]G(1, 1, 1, t; z). Then Tn,k is the number of Dyck paths of semilength n with k odd peaks (see [9, A091867]). n\k 0 1 2 3 4 5 6 7. 0 1 0 1 1 3 6 15 36. 1. 2. 3. 4. 1 0 1 3 0 1 4 6 0 1 15 10 10 0 36 45 20 15 105 126 105 35. 5. 6. 7. 1 0 21. 1 0. 1. 15.

(22) In the case of t = 0 in Equation (18), we get the generating function for the number of Dyck paths of semilength n with no odd peaks: √ 1 + z − 1 − 2z − 3z 2 X G(1, 1, 1, 0; z) = = Rn z n , (19) 2z(1 + z) n≥0 where Rn are the Riordan numbers or ring numbers (see [9, A005043]). The first few terms of Riordan numbers {Rn } for n ≥ 0 are 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, · · · . If we add each two consecutive Riordan numbers, we will get the Motzkin numbers. That is, Mn = Rn + Rn+1 for n ≥ 0. Since nsg(π) = oasc(φ(π)) for all π ∈ N C(n), we have X. X. t. nsg(π) n. z =. 1 + (1 − t)z −. n≥0 π∈N C(n). p 1 − 2(1 + t)z + (−3 + 2t + t2 )z 2 . 2z[1 + (1 − t)z]. Note that Equation (19) is the generating function for the number of non-crossing partitions of [n] with no singletons as well. (6) α = t, β = t, γ = 1, δ = 1 The generating function of the enumerating polynomials for statistic des is q 2 1 − (1 − t)2 z 2 − 1 − 2tz + (1 − t)2 z 2 − 4z 2 . G(t, t, 1, 1; z) = 2tz 1 + (1 − t)z The first few terms are 1 + z + (1 + t)z 2 + (1 + 3t + t2 )z 3 + (1 + 6t + 6t2 + t3 )z 4 + (1 + 10t + 20t2 + 10t3 + t4 )z 5 + (1 + 15t + 50t2 + 50t3 + 15t4 + t5 )z 6 + (1 + 21t + 105t2 + 175t3 + 105t4 + 21t5 + t6 )z 7 + · · · . Let Tn,k = [tk z n ]G(t, t, 1, 1; z). Then Tn,k is the number of Dyck paths of similength n with k valleys. It is the Narayana number Nn,k (see [9, A001263]). n\k 0 1 2 3 4 5 6 7. 0 1 1 1 1 1 1 1 1. 1. 2. 3. 4. 1 3 6 10 15 21. 1 6 1 20 10 1 50 50 15 105 175 105. 5. 6. 1 21. 1 16.

(23) (7) α = 1, β = 1, γ = t, δ = t The generating function of the enumerating polynomials for statistic asc is 2 2 q 1 − 2tz + (1 − t)2 z 2 − 4z 2 1 + (1 − t)z − . G(1, 1, t, t; z) = 2z 1 + (1 − t)z The first few terms are 1 + tz + (t + t2 )z 2 + (t + 3t2 + t3 )z 3 + (t + 6t2 + 6t3 + t4 )z 4 + (t + 10t2 + 20t3 + 10t4 + t5 )z 5 + (t + 15t2 + 50t3 + 50t4 + 15t5 + t6 )z 6 + (t + 21t2 + 105t3 + 175t4 + 105t5 + 21t6 + t7 )z 7 + · · · . Let Tn,k = [tk z n ]G(1, 1, t, t; z). Then Tn,k is the number of Dyck paths of semilength n with k peaks (see [9, A090181]). It is the Narayana number Nn,k−1 for 1 ≤ k ≤ n. n\k 0 1 2 3 4 5 6 7. 0 1 0 0 0 0 0 0 0. 1. 2. 1 1 1 1 1 1 1. 1 3 6 10 15 21. 3. 4. 5. 6. 7. 1 6 1 20 10 1 50 50 15 1 105 175 105 21 1 Note that G(1, 1, t, t; z) = t G(t, t, 1, 1; z) − 1 + 1 because asc(ω) = des(ω) + 1 for ω ∈ D(n), n ≥ 1. (8) α = t, β = 1, γ = t, δ = 1 Setting α = γ = t, β = δ = 1 in Equation (11), we have p 1 + 2(1 − t)z − 2z (1 − 2tz)2 − 4z 2 . G(t, 1, t, 1; z) = 2z The first few terms are 1 + z + 2tz 2 + (1 + 4t2 )z 3 + (6t + 8t3 )z 4 + (2 + 24t2 + 16t4 )z 5 + (20t + 80t3 + 32t5 )z 6 + (5 + 120t2 + 240t4 + 64t6 )z 7 + · · · . Let Tn,k = [tk z n ]G(t, 1, t, 1; z). Then Tn,k is the number of Dyck paths of similength n with a total of k even peaks and even valleys (see [9, A121448]).. 17.

(24) n\k 0 1 2 3 4 5 6 7. 0 1 1 0 1 0 2 0 5. 1. 2. 3. 4. 5. 2 0 6 0 20 0. 4 0 24 0 120. 8 0 80 0. 16 0 32 240 0. 6. 64. (9) α = 1, β = t, γ = 1, δ = t Setting α = γ = 1, β = δ = t in Equation (11), we have p 1 − (1 − 2tz)2 − 4z 2 . G(1, t, 1, t; z) = 2z t + (1 − t2 )z The first few terms are 1 + tz + (1 + t2 )z 2 + (4t + t3 )z 3 + (2 + 11t2 + t4 )z 4 + (15t + 26t3 + t5 )z 5 + (5 + 69t2 + 57t4 + t6 )z 6 + (56t + 252t3 + 120t5 + t7 )z 7 + · · · . Let Tn,k = [tk z n ]G(1, t, 1, t; z). Then Tn,k is the number of Dyck paths of similength n with a total of k odd peaks and odd valleys. It is also the number of 2-Motzkin paths of length n with k horizontal steps having no wavy horizontal steps on the x-axis (see [9, A126222]). n\k 0 1 2 3 4 5 6 7. 0 1 0 1 0 2 0 5 0. 1. 2. 3. 1 0 4 0 15 0 56. 1 0 1 11 0 0 26 69 0 0 252. 4. 5. 6. 7. 1 0 57 0. 1 0 120. 1 0. 1. (10) α = t, β = −t, γ = 1, δ = 1 Setting α = t, β = −t, γ = δ = 1 in Equation (11), we have q 2 2 2 1 − 2tz + (t − 1)z − 1 + (1 − t2 )z 2 ) − 4z 2 G(t, −t, 1, 1; z) = . 2tz − 1 + (1 + t)z The first few terms are 1+z +(1+t)z 2 +(1+t+t2 )z 3 +(1+2t+2t2 +t3 )z 4 +(1+2t+ 4t2 +2t3 +t4 )z 5 +(1+3t+6t2 +6t3 +3t4 +t5 )z 6 +(1+3t+9t2 +9t3 +9t4 +3t5 +t6 )z 7 +· · · . 18.

(25) Let Tn,k = [tk z n ]G(t, −t, 1, 1; z). Then Tn,k for n ≥ 1 is the number of symmetric Dyck paths of semilength n with k valleys (see [9, A088855]). n\k 0 1 2 3 4 5 6 7. 0 1 1 1 1 1 1 1 1. 1. 2. 3. 4. 5. 6. 1 1 2 2 3 3. 1 2 4 6 9. 1 2 6 9. 1 3 9. 1 3. 1. (11) α = 1, β = −t, γ = 1, δ = t Setting α = γ = 1, β = −t, δ = t in Equation (11), we have √ 1 − 2tz − 1 − 4z 2 . G(1, −t, 1, t; z) = 2z − t + (1 + t2 )z The first few terms are 1+tz+(1+t2 )z 2 +(2t+t3 )z 3 +(2+3t2 +t4 )z 4 +(5t+4t3 +t5 )z 5 + (5 + 9t2 + 5t4 + t6 )z 6 + (14t + 14t3 + 6t5 + t7 )z 7 + · · · . Let Tn,k = [tk z n ]G(1, −t, 1, t; z). Then Tn,k is the number of Motzkin paths of length n with k horizontal steps having no horizontal steps at positive heights (see [9, A053121]). n\k 0 1 2 3 4 5 6 7. 0 1 0 1 0 2 0 5 0. 1. 2. 3. 4. 5. 6. 7. 1 0 2 0 5 0 14. 1 0 3 0 9 0. 1 0 4 0 14. 1 0 5 0. 1 0 6. 1 0. 1. 19.

(26) 4. Bijection between 2-Motzkin paths and Dyck paths. Natural q-analogues of the Catalan numbers and the Narayana numbers are given by 2n 1 n n 1 2 and Nn,k (q) = q k +k , respectively, Cn (q) = [n + 1]q n q [n]q k q k + 1 q n. −1 where [n]q := qq−1 = 1 + q + q 2 + q 3 + · · · + q n−1 , [n]q ! := [n]q [n − 1]q · · · [1]q , and n := [n]q !/([k]q ![n − k]q !). k q F¨ urlinger and Hofbauer [8] proved that X X 1 2n 1 n n 2 maj(ω) maj(ω) q = and q = q k +k . [n + 1]q n q [n]q k q k + 1 q ω∈D(n) ω∈D(n). des(ω)=k. In this section, we would like to find some similar results for the statistics emaj, omaj, ecomaj, and ocomaj. To this end, we encode emaj (omaj, ecomaj, ocomaj, respectively) and edes (odes, easc, oasc, respectively) by q and t, and write down the enumerating polynomials according to these statistics by X H1 (n; q, t) := q emaj(ω) tedes(ω) , ω∈D(n). H2 (n; q, t) :=. X. q omaj(ω) todes(ω) ,. ω∈D(n). H3 (n; q, t) :=. X. q ecomaj(ω) teasc(ω) , and. ω∈D(n). H4 (n; q, t) :=. X. q ocomaj(ω) toasc(ω) .. ω∈D(n). Clearly, emaj(∅) = omaj(∅) = ecomaj(∅) = ocomaj(∅) = 0, and H1 (0; q, t) = H2 (0; q, t) = H3 (0; q, t) = H4 (0; q, t) = 1. Using the first return decomposition, we get the following relations. Lemma 4.1. Let ω ∈ D(n), n ≥ 1, and ω = uµdν be the first return decomposition of ω, where µ ∈ D(k − 1) and ν ∈ D(n − k) for some 1 ≤ k ≤ n. Then we have ( emaj(ω) =. omaj(µ) + odes(µ) + 2k + emaj(ν) + 2k · edes(ν) if 1 ≤ k < n; omaj(µ) + odes(µ) if k = n;. omaj(ω) = emaj(µ) + edes(µ) + omaj(ν) + 2k · odes(ν). if 1 ≤ k ≤ n;. ecomaj(ω) = ocomaj(µ) + oasc(µ) + ecomaj(ν) + 2k · easc(ν) if 1 ≤ k ≤ n; ( 1 + ocomaj(ν) + 2 oasc(ν) if k = 1; ocomaj(ω) = ecomaj(µ) + easc(µ) + ocomaj(ν) + 2k · oasc(ν) if 1 < k ≤ n.. 20. (20) (21) (22) (23).

(27) Proof. Let µ+ = uµd ∈ D(k). Since the locations of even valleys of µ+ are the locations of odd valleys of µ plus one, we have emaj(µ+ ) = omaj(µ) + odes(µ). And there is an even valley at the location 2k if ν 6= ∅. Also, the locations of even valleys of ν plus 2k are the locations of even valleys of ω, thus we complete the proof of Equation (20). The others can be proved similarly.. Using Lemma 4.1 and 3.1, we have the following lemma. Lemma 4.2. For n ≥ 1, we have H1 (n; q, t) = H2 (n; q, t) = H3 (n; q, t) = H4 (n; q, t) =. n−1 X k=1 n X k=1 n X k=1 n X. q 2k t · H2 (k − 1; q, qt) · H1 (n − k; q, q 2k t) + H2 (n − 1; q, qt);. (24). H1 (k − 1; q, qt) · H2 (n − k; q, q 2k t);. (25). H4 (k − 1; q, qt) · H3 (n − k; q, q 2k t);. (26). H3 (k − 1; q, qt) · H4 (n − k; q, q 2k t) + qtH4 (n − 1; q, q 2 t).. (27). k=2. Proof. Here we will only prove Equation (24) and (25), and the others can be proved similarly. Using Equation (20) and (7), we have H1 (n; q, t) =. n−1 X. X. q omaj(µ)+odes(µ)+2k+emaj(ν)+2k edes(ν) todes(µ)+1+edes(ν). k=1 (µ,ν)∈D(k−1)×D(n−k). X. +. q omaj(µ)+odes(µ) todes(µ). µ∈D(n−1). =. n−1 X. q 2k t. k=1. X. . µ∈D(k−1). X. +. q omaj(µ) (qt)odes(µ). X. q emaj(ν) (q 2k t)edes(ν). . ν∈D(n−k). q omaj(µ) (qt)odes(µ). µ∈D(n−1). =. n−1 X. q 2k tH2 (k − 1; q, qt) · H1 (n − k; q, q 2k t) + H2 (n − 1; q, qt).. k=1. Using Equation (21) and (8), we have H2 (n; q, t) =. n X. X. q emaj(µ)+edes(µ)+omaj(ν)+2k odes(ν) tedes(µ)+odes(ν). k=1 (µ,ν)∈D(k−1)×D(n−k). 21.

(28) =. n X k=1. =. n X. X. q emaj(µ) (qt)edes(µ). . µ∈D(k−1). X. q omaj(ν) (q 2k t)odes(ν). . ν∈D(n−k). H1 (k − 1; q, qt) · H2 (n − k; q, q 2k t).. k=1. Note that if we let H1 (q, t; z) = then we have H2 (q, t; z) = 1 + z. P. n XX. n≥0. H1 (n; q, t)z n and H2 (q, t; z) =. P. n≥0. H2 (n; q, t)z n ,. H1 (k − 1; q, qt)z k−1 · H2 (n − k; q, q 2k t)z n−k .. n≥1 k=1. However, we cannot get the generating function of H2 (n; q, t) from the above equation. H1 (n; q, t), H3 (n; q, t), and H4 (n; q, t) are also in a similar situation. So we need other tools. Here we establish a bijection between 2-Motzkin paths of length n − 1 and Dyck paths of semilength n to achieve our purpose (see [5, 6]). Define ψ : T M(n − 1) −→ D(n) where σ = σ1 σ2 · · · σn−1 ∈ T M(n − 1) is mapping to ω = ω1 ω2 · · · ωn−1 ∈ D(n) such that ω1 = u, ω2n = d, and  uu if σi = u,     dd if σ = d, i ω2i ω2i+1 =  ud if σi = s,    du if σi = w. We take n = 3 as an example to illustrate the bijection ψ in Figure 10. Let σ = σ1 σ2 · · · σn−1 ∈ T M(n − 1). We define Str(σ) and Wav(σ) as follows: Str(σ) = {i: σi = s}, and Wav(σ) = {i: σi = w}. And we define the number of and the sum of the elements X of them by wav(σ) := | Wav(σ)|, mwav(σ) := i, Xi∈Wav(σ) str(σ) := | Str(σ)|, mstr(σ) := i. i∈Str(σ). According to the bijection ψ, we have the following relations between 2-Motzkin paths and Dyck paths. σ ∈ T M(n − 1) ω = ψ(σ) ∈ D(n) i ∈ Str(σ). ←→. 2i ∈ eAsc(ω). (28). i ∈ Wav(σ). ←→. 2i ∈ eDes(ω). (29). Now we are going to discuss the statistics emaj and ecomaj using the bijection ψ. 22.

(29) Dyck paths. 2-Motzkin paths T M(2). D(3). bijection ψ. s. s. s. w. u. d. w. s. w. w. Figure 10: Bijection between T M(2) and D(3).. 4.1. The statistics emaj and ecomaj. In this subsection, we compute the enumerating polynomials H1 (n; q, t) and H3 (n; q, t). We will use the q-Binomial Theorem (see, for example [1]). Theorem 4.3. (The q-Binomial Theorem) For n ≥ 0, n n Y X k+1 n k i ( ) (1 + q x) = q 2 x . k q i=1 k=0 Using Theorem 4.3 and relations (28) and (29) between 2-Motzkin paths and Dyck paths, we get the following results. Theorem 4.4. For n ≥ 1, we have X. q. mwav(σ) wav(σ). t. σ∈T M(n−1). and H1 (n; q, t) =. X ω∈D(n). q. n−1 X. k+1 n − 1 ( ) = q 2 Mn−1−k tk , k q k=0. emaj(ω) edes(ω). t. =. n−1 X k=0. q. k2 +k. n−1 Mn−1−k tk . k q2. (30). (31). Proof. Let T M(n − 1, k) be the set of 2-Motzkin paths of length n − 1 with exactly k wavy horizontal steps. Given S = {s1 , s2 , . . . , sk } ⊆ [n − 1], if we choose S to be the positions of the wavy horizontal steps for a σ ∈ T M(n − 1, k), then the 23.

(30) remaining n − 1 − k steps form a Motzkin path of length n − k − 1. So the number of σ ∈ T M(n − 1, k) with Wav(σ) = S is Mn−k−1 . Hence we have X. q. mwav(σ) wav(σ). t. =. n−1 X. σ∈T M(n−1). =. k=0. =. n−1 X. S⊆[n−1] σ∈T M(n−1,k) |S|=k Wav(σ)=S. k=0. X. q. k=0. P. i∈S. i k. t. . S⊆[n−1] |S|=k. .. S⊆[n−1] |S|=k. And by the q-Binomial Theorem, X. t. . n−1 X X P q mwav(σ) tk = q i∈S i tk · Mn−k−1. X. Mn−k−1. q. mwav(σ) k. σ∈T M(n−1,k). k=0. n−1 X X. X. q. mwav(σ) wav(σ). t. =. σ∈T M(n−1). P. n−1 X. q. S⊆[n−1] |S|=k. Mn−k−1. k=0. P. . i∈S. i. k+1 2. = q(. ) n−1 . So we have k q. k+1 n − 1 ( ) q 2 tk k q. n−1 X. k+1 n − 1 ( ) q 2 Mn−k−1 tk . = k q k=0 By (29), we have edes(ψ(σ)) = wav(σ) and emaj(ψ(σ)) = 2 mwav(σ) for all σ ∈ T M(n − 1). Thus X. X. q emaj(ω) tedes(ω) =. q 2 mwav(σ) twav(σ). σ∈T M(n−1). ω∈D(n). =. n−1 X. q. k2 +k. k=0. . n−1 Mn−k−1 tk . k q2. Since, in T M(n − 1), steps s and steps w are interchangeable. Hence, we have the same result for the statistics mstr(σ) and ecomaj(ω). Theorem 4.5. For n ≥ 1, we have X. q. mstr(σ) str(σ). t. σ∈T M(n−1). and H3 (n; q, t) =. X ω∈D(n). q. n−1 X. k+1 n − 1 ( ) = q 2 Mn−k−1 tk , k q k=0. ecomaj(ω) easc(ω). t. =. n−1 X k=0. 24. q. k2 +k. . n−1 Mn−k−1 tk . k q2.

(31) 4.2. The statistics omaj and ocomaj. We use the bijection φ between non-crossing partitions and Dyck paths to compute the enumerating polynomial H4 (n; q, t). Theorem 4.6. For n ≥ 1, we have X. q. t. X. ocomaj(ω) oasc(ω). π∈N C(n). and H4 (n; q, t) =. n X. k+1 n ( ) 2 = q Rn−k tk , k q k=0. msg(π) nsg(π). q. t. ω∈D(n). n X. (32). n = q Rn−k tk . k q2 k=0 k2. (33). d Proof. Let N C(n, k) be the set of non-crossing partitions with exactly k singletons. Given P = {p1 , p2 , . . . , pk } ⊆ [n], if we choose P to be Sg(π) for a d π ∈ N C(n, k), then of the remaining n − k numbers form a non-crossing partid tion without singletons. So the number of π ∈ N C(n, k) with Sg(π) = P is Rn−k , the (n − k)-th Riordan number (see Equation (19)).Thus we have X. q. msg(π) nsg(π). t. n X. =. =. =. n X. X. X. q. d π∈N C(n,k) n X msg(π) k. t. Rn−k. X. q. k=0. P. i∈P. i k. t. . X. =. P ⊆[n] π∈N d C(n,k) |P |=k Sg(π)=P. k=0. q msg(π) tk. . k=0. π∈N C(n) n X. X. q. P. i k. t · Rn−k. i∈P. . P ⊆[n] |P |=k. .. P ⊆[n] |P |=k. k=0. And by the q-Binomial Theorem, X. q. π∈N C(n). msg(π) nsg(π). t. =. n X. P. P ⊆[n] |P |=k. Rn−k. . k=0. q. P. i∈P. i. k+1 2. = q(. k+1 n k ( ) q 2 t k q. ) n . Hence, we have k q n X. k+1 n ( ) = q 2 Rn−k tk . k q k=0. By Equation (5) and (6), we have oasc(φ(π)) = nsg(π) and ocomaj(φ(π)) =. X i∈oAsc(φ(π)). X. (2i − 1) = 2 msg(π) − nsg(π) for all π ∈ N C(n). Thus. i∈Sg(π). X ω∈D(n) n X. q ocomaj(ω) toasc(ω) =. X. q 2 msg(π)−nsg(π) tnsg(π). π∈N C(n). n X 2(k+1 −k n k k2 n ) 2 Rn−k tk . = q Rn−k t = q k k 2 2 q q k=0 k=0 25. i=.

(32) Now the remaining question is to get the enumerating polynomial H2 (n; q, t). We need another decomposition of Dyck paths. For ω ∈ D(n), n ≥ 1, using the touch points of ω at the x-axis, we can uniquely decompose ω in the form + + ω = µ+ 1 µ2 · · · µi = uµ1 duµ2 d · · · uµi d,. where µ+ j = uµj d, µj ∈ D(kj − 1), kj ≥ 1, j = 1, 2, · · · , i, with k1 + k2 + · · · + ki = n for some 1 ≤ i ≤ n. This is so called the prime decomposition [15] (see Figure 11). µ1 ∈ D(k1 − 1). µ2 ∈ D(k2 − 1). µi ∈ D(ki − 1) 2(k1 + k2 + · · · + ki ). 2k1. 1. Figure 11: The prime decomposition of a nonempty Dyck path. Theorem 4.7. For n ≥ 1, we have X H2 (n; q, t) = q omaj(ω) todes(ω) ω∈D(n). =. n X i=1. i Y. X k1 +k2 +···+ki =n kj ≥1 for 1≤j≤i. H1 (kj − 1; q, q 2(k1 +k2 +···+kj−1 )+1 t).. j=1. Proof. Using the prime decomposition, we have X H2 (n; q, t) = q omaj(ω) todes(ω) ω∈D(n). =. n X i=1. X. q. n X. n X. i=1. k1 =1 µ1 ∈D(k1 −1). ·. ·. . . (µ1 ,µ2 ,··· ,µi ) ∈D(k1 −1)×D(k2 −1)×···×D(ki −1). +2(k1 +k2 +···+ki−1 ) odes(µ+ i ). =. . + + + omaj(µ+ 1 )+ omaj(µ2 )+2k1 odes(µ2 ) +···+ omaj(µi ). X. q. . ! + + odes(µ+ 1 )+odes(µ2 )+···+odes(µi ). ·t. emaj(µ1 ). n−k 1 −k2 X. X. k3 =1. µ3 ∈D(k3 −1). edes(µ1 ). (qt). n−k X1. X. q emaj(µ2 ) q 2k1 +1 t. k2 =1 µ2 ∈D(k2 −1). edes(µ3 ) q emaj(µ3 ) q 2(k1 +k2 )+1 t ···. X. edes(µi ) q emaj(µi ) q 2(k1 +k2 +···+ki−1 )+1 t. µi ∈D(n−k1 −k2 −···−ki−1 −1). 26. !!!!. edes(µ2 ).

(33) =. n X i=1. ·. n X. H1 (k1 − 1; q, qt). k1 =1 n−k 1 −k2 X. n−k X1. H1 (k2 − 1; q, q 2k1 +1 t). k2 =1. !!!! i−1 X Pi−1 H1 k3 − 1; q, q 2(k1 +k2 )+1 t · · · H1 n − kj − 1; q, q 2( j=1 kj )+1 t j=1. k3 =1. =. n X n n−k 1 −k2 X X1 n−k X i=1 k1 =1 k2 =1. · · · H1 (k1 − 1; q, qt)H1 (k2 − 1; q, q 2k1 +1 t). k3 =1. · H1 (k3 − 1; q, q 2(k1 +k2 )+1 t) · · · H1 (n −. i−1 X. ! kj − 1; q, q. P 2( i−1 j=1 kj )+1. t). j=1. =. n X i=1. X. H1 (k1 − 1; q, qt)H1 (k2 − 1; q, q 2k1 +1 t). k1 +k2 +···+ki =n kj ≥1 for 1≤j≤i. Pi−1. · H1 (k3 − 1; q, q 2(k1 +k2 )+1 t) · · · H1 (ki − 1; q, q 2( =. n X i=1. 5. X k1 +k2 +···+ki =n kj ≥1 for 1≤j≤i. i Y. j=1. kj )+1. t). H1 (kj − 1; q, q 2(k1 +k2 +···+kj−1 )+1 t).. j=1. Conclusion Remark. We have used bijections φ : N C(n) → D(n) and ψ : T M(n − 1) −→ D(n) to obtain the results shown in Table 1. In the case of t = 0 for the generating functions (2)∼(5) in Table 1, we have the results shown in Table 2. There are some questions we could do in the future. (1) Calculate the generating functions of statistics | Fs |, | Ls |, nb, and ne in noncrossing partitions. (2) Calculate the generating functions of the number of up steps, and the number of down steps in 2-Motzkin paths. (3) Discuss why G(t, −t, 1, 1; z) can get the numbers of symmetric Dyck paths with k peaks. (4) Discuss why G(1, −t, 1, t; z) can get the number of Motzkin paths with k horizontal steps having no horizontal steps at positive heights.. 27.

(34) References [1] Martin Aigner. A Course in Enumertion. Springer Verlag: Berlin Heidelberg, 2007. [2] Frank Reiff Bernhart. Catalan, Motzkin, and Riordan numbers. Discrete Mathematics, 204:73–112, 1999. [3] Sa´ ul A. Blanco and T. Kyle Petersen. Counting Dyck Paths by Area and Rank. Annals of Combinatorics, 18:171–197, 2014. [4] Mikl´os Bona and Bruce Eli Sagan. On divisibility of Narayana numbers by primes. Journal of Integer Sequences, 8, 2005. Article 05.2.4. [5] Marie-Pierre Delest and G´erard Viennot. Algebraic languages and polyominoes enumeration. Theoretical Computer Science, 34:169–206, 1984. [6] Emeric Deutscha and Louis W. Shapiro. A bijection between ordered trees and 2-Motzkin paths and its many consequences. Discrete Mathematics, 256:655– 670, 2002. [7] Robert Donaghey and Louis W. Sharpiro. Motzkin numbers. J. Combin. Theory Ser. A, 23:291–301, 1977. [8] Johannes F¨ urlinger and Josef Hofbauer. q-Catalan numbers. J. Combin. Theory Ser. A, 40:248–264, 1985. [9] OEIS Foundation Inc. http://oeis.org. 2011.. The on-line encyclopedia of integer sequences,. [10] Toufik Mansour. Statistics on Dyck paths. Journal of Integer Sequences, 9, 2006. Article 06.1.5. [11] Toufik Mansour. Combinatorics of Set Partitions. Taylor & Francis Group, LLC, 2013. [12] Toufik Mansour, Eva Yu-Ping Deng, and Rosena Ruo-Xia Du. Dyck paths and restricted permutations. Discrete Applied Mathematics, 154:1593–1605, 2006. [13] Donatella Merlini, Renzo Sprugnoli, and Maria Cecilia Verri. Some statistics on Dyck paths. Journal of Statistical Planning and Inference, 101:211–227, 2002. [14] Aristidis Sapounakis, Ioannis Tasoulas, and Panagiotis Tsikouras. Counting strings in Dyck paths. Discrete Mathematics, 307:2909–2924, 2007. 28.

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(36) 30. √. 2 √. . 1−2tz+(1−t)2 z 2. 2z 1+(1−t)z. . r −. √ 2 2 (1−2tz) −4z . (11) G(1, −t, 1, t; z) =. r. 2z −t+(1+t2 )z. √ 1−2tz− 1−4z 2 . 2. 2. . −4z 2. −4z 2. −4z 2. 1+(1−t2 )z 2 ). 2tz −1+(1+t)z. . 1−2tz+(t2 −1)z 2 −. 2z t+(1−t2 )z. 1−. 1+2(1−t)z−2z (1−2tz)2 −4z 2 2z. 1+(1−t)z. 2tz 1+(1−t)z. . 1−2tz+(1−t)2 z 2. . 2. 1−2(1+t)z+(−3+2t+t2 )z 2 2z[1+(1−t)z]. 1−2(1+t)z+(−3+2t+t2 )z 2 2z. 1−2(1+t)z+(−3+2t+t2 )z 2 2z[t+(1−t)z]. r. 1−(1−t)2 z 2 −. 1+(1−t)z−. √. 1+(1−t)z−. (10) G(t, −t, 1, 1; z) =. (9) G(1, t, 1, t; z) =. (8) G(t, 1, t, 1; z) =. (7) G(1, 1, t, t; z) =. (6) G(t, t, 1, 1; z) =. (5) G(1, 1, 1, t; z) =. (4) G(1, 1, t, 1; z) =. (3) G(1, t, 1, 1; z) =. 1−(1−t)z−. √. G(α, β, γ, δ; z) generating function √ (1) G(1, 1, 1, 1; z) = 1− 2z1−4z √ 1+(1−t)z− 1−2(1+t)z+(−3+2t+t2 )z 2 (2) G(t, 1, 1, 1; z) = 2z. steps having no horizontal steps at positive heights. # Motzkin paths of length n with k horizontal. # symmetric Dyck paths of semilength n with k valleys. and odd valleys # 2-Motzkin paths of length n with k horizontal steps having no wavy horizontal steps on the x-axis. # Dyck paths of semilength n with a total of k odd peaks. # Dyck paths of semilength n with a total of k even peaks and even valleys. # Dyck paths of semilength n with k peaks. # Dyck paths of semilength n with k valleys. # non-crossing partitions of [n] with k elements in the set Sg.. # Dyck paths of semilength n with k odd peaks. # Dyck paths of semilength n with k even peaks # non-crossing partitions of [n] with k elements in the set R. # Dyck paths of semilength n with k uuu’s (triple rise) steps # non-crossing partitions of [n] with k elements in the set Ir. A053121. A088855. A126222. A121448. A090181. A001263. A091867. A091869. A092107. A091869. # Dyck paths of semilength n with k even valleys # non-crossing partitions of [n] with k elements in the set L # Dyck paths of semilength n with k odd valleys. OEIS A000108. Combinatorial meaning of [tk z n ]G # Dyck paths of semilength n. Table 1: The generating functions for cases (1)∼(11).

(37) Table 2: t = 0 for cases (2)∼(5) in Table 1 G(α, β, γ, δ; z) (2) G(0, 1, 1, 1; z) X =1+ Mn−1 z n. Combinatorial meaning of [z n ]G # Dyck paths of semilength n with no even valleys # non-crossing partitions of [n] with no elements in the set L. n≥1. (3) G(1, 0, 1, 1; z) X = Mn z n n≥0. # Dyck paths of semilength n with no odd valleys # Dyck paths of semilength n with no uuu’s(triple rise)steps # non-crossing partitions of [n] with no interiors. (4) G(1, 1, 0, 1; z) X =1+ Mn−1 z n. # Dyck paths of semilength n with no even peaks # non-crossing partitions of [n] with no elements in the set R. n≥1. (5) G(1, 1, 1, 0; z) X = Rn z n. # Dyck paths of semilength n with no odd peaks # non-crossing partitions of [n] with no singletons. n≥0. 31.

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