Dyck路徑與不相交分割間的一些結果

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(1)國立高雄大學應用數學系碩士論文. Some results about relations between Dyck paths and noncrossing partitions Dyck 路徑與不相交分割間的一些結果. 研究生：邱奕璋撰指導教授：鄭斯恩. 中華民國 103 年 11 月.

(2) Some results about relations between Dyck paths and non-crossing partitions. by Yi-Chang Chiu Advisor Szu-En Cheng. Department of Applied Mathematics, National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. Date(Nov 2014).

(3) 致謝轉眼，在高雄大學應用數學系已是七年，中間所有大大小小的人事物，都使我更加完整。感謝指導教授鄭斯恩老師，若不是有您，我就無法在這兩年繼續沉浸在我最有興趣的科目與方向，這段時間感謝您不斷忍受我的惰性與過度隨興。不需要別人提醒，謝謝您不欲人知並滿滿的愛。感謝游森棚教官，何其幸運在人生中對的時間遇到您，終身感恩。感謝張惠蘭老師與蔡宜霖、蔡一慈兩位同學這兩年的砥礪與照顧，使得大學延誤的那一年變得充滿意義。感謝鄭益新，不管公事私事都有太多的事情受你幫助，幸好有你。感謝應數系的千惠姐、雅鳳姐、佩君姐，校內的行政事務與生活上的大小煩惱都多虧有妳們三位應數系的大家長在。感謝應數系的每位師長，是你們對自身專業的熱情與驕傲感染了我，讓我期望自己能變得更好；更因為你們無私的關心與付出，使得應數系成了一個家。最後謝謝人生，讓我遇見了王瓊誼。. 邱奕璋撰 i.

(4) Contents 1 Introduction. 3. 2 q-Narayana numbers and non-crossing partitions. 6. 3 Recursions for the statistics rb, ls, rs, and lb on non-crossing partitions 10 4 Connections on rs and area 18 4.1 Area of a Dyck path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.2 Area of a 2-Motzkin path . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5 Conclusions. 26. References. 28. ii.

(5) Dyck路徑與不相交分割間的一些結果指導教授：鄭斯恩教授國立高雄大學應用數學系. 學生：邱奕璋國立高雄大學應用數學系. 摘要在不相交分割上有統計量ls，rb，rs，lb，bk及rank，在Dyck路徑上有統計量area及rank，在2-Motzkin路徑上有統計量area。我們在不相交分割上得到了 Narayana數最明顯的q-類比。對於所有不相交分割我們定義了兩個有趣的分解方法，並推導出幾個雙變數分布函數的遞迴式。我們也證明了不相交分割上的一組統計量組合2rs + n − bk會和Dyck路徑上的area等分布。最後藉著不相交分割與2-Motzkin路徑之間的連結，我們得到了一個Catalan數的(q, x)-細分。. 關鍵字：不相交分割、Dyck路徑、對射、分解. 1.

(6) Some results about relations between Dyck paths and non-crossing partitions Advisor: Szu-En Cheng Department of Applied Mathematics National University of Kaohsiung. Student: Yi-Chang Chiu Department of Applied Mathematics National University of Kaohsiung. Abstract There are statistics, ls, rb, rs, lb, bk, and rank on non-crossing partitions, area and rank on Dyck paths, and area on 2-Motzkin paths. We give the most obvious qanalogue of the Narayana numbers based on non-crossing partitions. We also define two interesting decompositions of non-crossing partitions, and obtain recursions of joint distributions of statistics on non-crossing partitions. We show that 2rs+n−bk on non-crossing partitions and area on Dyck paths are equidistributed. Finally, we get an interesting (q, x)-refinement of the Catalan numbers through the connection between non-crossing partitions and 2-Motzkin paths.. Keywords: non-crossing partitions, Dyck paths, bijections, decompositions.. 2.

(7) 1. Introduction. A partition of [n] := {1, 2, . . . , n} is a collection of disjoint nonempty subsets of [n], called blocks, whose union is [n]. A partition π with k blocks is written as π = B1 /B2 / . . . /Bk where the blocks are ordered in increasing order of their minimum elements. A partition π = B1 /B2 / . . . /Bk of [n] is non-crossing if whenever a quadruple of elements 1 ≤ a < b < c < d ≤ n satisfies a, c ∈ Bi and b, d ∈ Bj for some 1 ≤ i, j ≤ k, then in fact i = j; thus, the blocks do not “cross”. The set of all non-crossing partitions of [n] is denoted by N C(n), and the set of non-crossing partitions of [n] into k blocks is denoted by N C(n, k). In Section 3, we need the notation N C([i, i + j]) to denote the set of all non-crossing partitions of {i, i + 1, . . . , i + j} for all integer i > 0 and j ≥ 0. Otherwise, N C([i, i + j]) is empty. We can represent non-crossing partitions graphically, as illustrated in Figure 1, by plotting 1, 2, . . . , n on a horizontal line and joining successive elements of the same block by arcs.. 1. 2. 3. 4. 5. 6. Figure 1: The non-crossing partition π = 1 3 4 / 2 / 5 6 ∈ N C(6, 3). A partition π of [n] also can be represented by its restricted growth f unction [4] (“RG function”), w : [n] → [n], w(i) = the index of the block of π which contains i. In this thesis the restricted growth function associated with π will be the word w = w1 w2 . . . wn , where wi = w(i). Given a partition π = B1 /B2 / . . . /Bk of [n], let w = w1 w2 . . . wn be its RG function. Simion [7] defined four statistics on partitions: ls(π) := lb(π) :=. n X i=1 n X. #{wj | wj < wi , j < i}, #{wj | wj > wi , j < i},. i=1. rs(π) := and. rb(π) :=. n X i=1 n X. #{wj | wj < wi , j > i}, #{wj | wj > wi , j > i}.. i=1. Example 1.1. If n = 8 and π = 1 5 6 7 / 2 3 / 4 / 8, then w = 1 2 2 3 1 1 1 4, ls(π) = 0 + 1 + 1 + 2 + 0 + 0 + 0 + 3 = 7, lb(π) = 0 + 0 + 0 + 0 + 2 + 2 + 2 + 0 = 6, rs(π) = 0 + 1 + 1 + 1 + 0 + 0 + 0 + 0 = 3, and rb(π) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 0 = 11. For a partition π = B1 /B2 / . . . /Bk , let bi := |Bi | , fi := min{a | a ∈ Bi } , and li := max{a | a ∈ Bi }. Simion has proved the following results. 3.

(8) Lemma 1.1. [7] (i) Let π = B1 /B2 / . . . /Bk be a partition of [n]. Then ls(π) =. k X. (i − 1)bi ,. i=1. and. lb(π) = k · n + k −. k X i=1. k X fi − (i · bi ). i=1. (ii) Let π = B1 /B2 / . . . /Bk ∈ N C(n, k). Then rb(π) =. and. rs(π) =. k X i=1 k X. (fi − 1) , li −. i=1. k X. fi − n + k.. i=1. It is easy to see that for π ∈ N C(n, k), we have ls(π) + rb(π) + lb(π) = (k − 1)n.. (1). Note that while f1 < f2 < . . . < fk , the entries of (l1 , l2 , . . . , lk ) are not necessarily in increasing order. The increasingly order of elements of {l1 , l2 , . . . , lk } is denoted by l10 < l20 < . . . < lk0 . For a partition π = B1 /B2 / . . . /Bk of [n], we define its b-vector, f -vector, l-vector, and l0 -vector to be. and. b(π) = (b1 , b2 , . . . , bk ), f (π) = (f1 , f2 , . . . , fk ), l(π) = (l1 , l2 , . . . , lk ), l0 (π) = (l10 , l20 , . . . , lk0 ), respectively.. There are two helpful lemmas to describe non-crossing partitions. Lemma 1.2. [7] Let 1 = f1 < f2 < . . . < fk ≤ n and 1 ≤ l10 < l20 < . . . < lk0 = n be two integer sequences. If (f1 , f2 , . . . , fk ) and (l10 , l20 , . . . , lk0 ) are the f -vector and l0 -vector of some partition of [n], then fi ≤ li0 for all i = 1, 2, . . . , k, and fi+1 ≤ li0 + 1 for all i = 1, 2, . . . , k − 1. Moreover, if these conditions are satisfied, then there is a unique non-crossing partition of [n] having the given f -vector and l0 -vector. Lemma 1.3. [7] Let 1 = f1 < f2 < . . . < fk ≤ n and 1 ≤ b1 , b2 , . . . , bk ≤ n. Then a non-crossing partition π ∈ N C(n, k) exists with these as its f - and b-vectors if and only if fi + bi + bi+1 + . . . + bk − 1 ≤ n for all i = 1, 2, . . . , k, with equality for i = 1. If these inequalities hold, then π ∈ N C(n, k) is uniquely determined by the vectors f and b. A Dyck path of length 2n is a lattice path from (0, 0) to (2n, 0) using up steps U = (1, 1) and down steps D = (1, −1), which never goes below the x-axis. The set of all Dyck paths 4.

(9) of length 2n is denoted by Dn . It is well known that the cardinality of the set Dn is the n-th Catalan number 2n 1 . Cn = n+1 n Let pi stand for the i-th step in a Dyck path p. We call the subword pi pi+1 = DU (resp. pi pi+1 = U D) a valley (resp. peak). The set of Dyck paths of length 2n with k valleys is denoted by Dn,k . It is well known that the cardinality of Dn,k is the Narayana number 1 n n Nn,k = . n k+1 k Definition 1.1. Let p be any Dyck path of length 2n, then the descent set and major index of p are. and. Des(p) := {i | pi pi+1 = DU }, X maj(p) := i. i∈Des(p). F¨ urlinger and Hofbauer [3] refined the major index by the following definition. Definition 1.2. Given p ∈ Dn , let α(p) :=. X. #{j | pj = U, j ≤ i},. i∈Des(p). and. β(p) :=. X. #{j | pj = D, j ≤ i}.. i∈Des(p). For p ∈ Dn , obviously α(p)+β(p) = maj(p). Note that if p = U u1 Dd1 U u2 Dd2 . . . U uk Ddk ∈ Dn,k−1 , then and. α(p) = u1 + (u1 + u2 ) + . . . + (u1 + u2 + . . . + uk−1 ), β(p) = d1 + (d1 + d2 ) + . . . + (d1 + d2 + . . . + dk−1 ).. (2) (3). For convenience, we define its u-vector and d-vector to be u(p) = (u1 , u2 , . . . , uk ) and d(p) = (d1 , d2 , . . . , dk ). The standard q-analogue for integer n ≥ 1 is [n]q = 1 + q + q 2 + . . . + q n−1 , and [n]q ! = [n]q [n − 1]q · · · [1]q for n ≥ 1 with [0]q ! = 1. The q-binomial coefficient is n [n]q ! = . k q [k]q ![n − k]q ! In the next section, we give the most obvious q-analogue of the Narayana numbers based on non-crossing partitions. In Section 3, we define joint distributions for the number of blocks, and one of rb, ls, rs and lb of non-crossing partitions. And we use two decompositions on non-crossing partitions to get recursions for these distributions. In Section 4.1, we will prove that the statistic area on Dyck paths and a combination of some statistics on non-crossing partitions are equally distributed. In Section 4.2, we define the joint distribution of statistics rs and rank on non-crossing partitions and prove a similar result as in [1]. We finish with Section 5, where we give some problems we are interested in. 5.

(10) 2. q-Narayana numbers and non-crossing partitions. In [3], F¨ urlinger and Hofbauer gave the obvious q-analogue of the Narayana numbers and the Catalan numbers based on Dyck paths. Lemma 2.1. [3] We have X. q. maj(p). p∈Dn,k. and. X p∈Dn. q maj(p). 1 n n k2 +k = q , [n]q k + 1 q k q 1 2n = . n+1 n q. In [10], Zhao and Zhong gave the same q-Narayana numbers by using the statistics ls and rb on non-crossing partitions. In our work, we use two bijections between N C(n, k) and Dn,k−1 to count α and β, then get some formulas for the q-Narayana numbers. Define a map φ1 from N C(n, k) to Dn,k−1 as follows: Given π ∈ N C(n, k) with 0 , and di = fi+1 − fi f (π) = (f1 , f2 , . . . , fk ) and l0 (π) = (l10 , l20 , . . . , lk0 ), let ui = li0 − li−1 0 for all i = 1, 2, . . . , k, where l0 = 0 and fk+1 = n + 1. Then let (u1 , u2 , . . . , uk ) and (d1 , d2 , . . . , dk ) be the u-vector and d-vector of φ1 (π). Obviously u1 + u2 + . . . + ui = li0 , and d1 +d2 +. . .+di = fi+1 −1 for all i = 1, 2, . . . k. Thus by Lemma 1.2, u1 +u2 +. . .+ui ≥ d1 + d2 + . . . + di for all i = 1, 2, . . . , k − 1, with equality for i = k. Hence φ1 (π) ∈ Dn,k−1 . Proposition 2.2. φ1 is bijective. Proof. It is easy to see that φ1 is one-to-one. Given p ∈ Dn,k−1 with u(p) = (u1 , u2 , . . . , uk ) and d(p) = (d1 , d2 , . . . , dk ), let li0 = u1 + u2 + . . . + ui and fi = d1 + d2 + . . . + di−1 + 1 for all i = 1, 2, . . . , k. Then li0 ≥ fi for all i = 1, 2, . . . , k. Similarly li0 + 1 ≥ fi+1 for all i = 1, 2, . . . , k − 1. By Lemma 1.2, we can uniquely construct a non-crossing partition, φ−1 (p), with f (φ−1 (p)) = (f1 , f2 , . . . fk ), and l0 (φ−1 (p)) = (l10 , l20 , . . . , lk0 ). Hence φ1 is bijective. Example 2.1. Let π = 1 5 6 7 / 2 3 / 4 / 8 ∈ N C(8, 4), then (u1 , u2 , u3 , u4 ) = (3, 1, 3, 1) and (d1 , d2 , d3 , d4 ) = (1, 2, 4, 1). Hence, φ1 (π) = U 3 D1 U 1 D2 U 3 D4 U 1 D1 ∈ D8,3 .. Figure 2: φ1 (π) of Example 2.1.. 6.

(11) Theorem 2.3. Let π ∈ N C(n, k) and p = φ1 (π) ∈ Dn,k−1 . Then α(p) = rs(π) + rb(π), and β(p) = rb(π). Proof. Let π ∈ N C(n, k), and p = φ1 (π) = U u1 Dd1 U u2 Dd2 . . . U ui Ddi . . . U uk Ddk . Then α(p) = u1 + (u1 + u2 ) + . . . + (u1 + u2 + . . . + uk−1 ) 0 = l10 + l20 + . . . + lk−1 =. k X. by Equation (2). li0 − n. i=1. =. k X. li0 −. i=1. k X. ! fi − n + k. +. i=1. k X. ! fi − k. i=1. = rs(π) + rb(π). by Lemma 1.1,. and β(p) = d1 + (d1 + d2 ) + . . . + (d1 + d2 + . . . + dk−1 ) = (f2 − 1) + (f3 − 1) + . . . + (fk − 1) =. by Equation (3). k X (fi − 1) i=1. = rb(π). by Lemma 1.1.. Corollary 2.4. We have X π∈ N C(n,k). q. 2rb(π)+rs(π). n 1 n 2 q k −k . = [n]q k q k − 1 q. In order to get another formula, we need the following definition. Definition 2.1. For p ∈ Dn,k−1 with u(p) = (u1 , u2 , . . . , uk ) and d(p) = (d1 , d2 , . . . , dk ), we define the reverse of p, called prev , to be the Dyck path with u(prev ) = (dk , dk−1 , . . . , d1 ) and d(prev ) = (uk , uk−1 , . . . , u1 ). By Definition 2.1, we get. and. α(prev ) = dk + (dk + dk−1 ) + . . . + (dk + dk−1 + . . . + d2 ) = (k − 1)n − β(p), (4) β(prev ) = uk + (uk + uk−1 ) + . . . + (uk + uk−1 + . . . + u2 ) = (k − 1)n − α(p). (5). Example 2.2. Let p = U 3 D1 U 1 D2 U 3 D4 U 1 D1 ∈ D8,3 ,. 7.

(12) prev =. p=. α(p) = 3 + 4 + 7 = 14,. α(prev ) = 1 + 5 + 7 = 13 = 3 · 8 − β(p),. and β(p) = 1 + 3 + 7 = 11.. and β(prev ) = 1 + 4 + 5 = 10 = 3 · 8 − α(p).. Theorem 2.5. Let π ∈ N C(n, k) and p = φ1 (π) ∈ Dn,k−1 . Then α(prev ) = lb(π) + ls(π), and β(prev ) = lb(π) + ls(π) − rs(π). Proof. Let π ∈ N C(n, k) and p = φ1 (π). Then α(prev ) = (k − 1)n − β(p) by Equation (4) = (k − 1)n − rb(π) by Theorem 2.3 = (ls(π) + rb(π) + lb(π)) − rb(π) by Equation (1) = lb(π) + ls(π), and β(prev ) = (k − 1)n − α(p) by Equation (5) = (k − 1)n − (rs(π) + rb(π)) by Theorem 2.3 = lb(π) + ls(π) − rs(π) by Equation (1).. Corollary 2.6. We have X. q. 2lb(π)+2ls(π)−rs(π). π∈ N C(n,k). 1 n n 2 = q k −k . [n]q k q k − 1 q. There is a famous bijective map ζ from Dn,k−1 to N C(n, k), which is similar to Prodinger’s in [6], as follows [2, 10]: Given p ∈ Dn,k−1 , we label its up steps by 1 to n going from left to right. Next, assign to each down step the same label as its corresponding up step, that is the first up step to the left on the same horizontal line. Then we get ζ(p) ∈ N C(n, k) whose blocks are constituted by the labels of each sequence of consecutive down steps. Example 2.3. Let p ∈ D6,2 , 3. 4. 3. 2. p= 1. 6. ζ. 5. 4 2. 6 1. We get ζ(p) = 1 2 4 / 3 / 5 6 ∈ N C(6, 3). 8. 5.

(13) For a path p ∈ Dn,k−1 with u(p) = (u1 , u2 , . . . , uk ), we have l0 (ζ(p)) = (u1 , u1 + u2 , . . . , u1 + u2 + . . . + uk ). Note that l0 (ζ(p)) cannot tell us the order of blocks of ζ(p). In order to overcome this problem, we need the following simple bijection θ from N C(n, k) to N C(n, k). Given π ∈ N C(n, k), we define θ(π) by declaring that i and j are in the same block of θ(π) if and only if n + 1 − i and n + 1 − j are in the same block of π for all i, j ∈ [n]. Example 2.4. Let π ∈ N C(6, 3). π= 1. 2. 3. 4. 5. θ(π) = 6. 1. 2. 3. 4. 5. 6. Now we define a bijection φ2 from Dn,k−1 to N C(n, k) by φ2 = θ ◦ ζ. An equivalent way to state φ2 is : Given p ∈ Dn,k−1 , we label its up steps by n to 1 going from left to right. Next, assign to each down step the same label as its corresponding up step, that is the first up step to the left on the same horizontal line. Then we get φ2 (w) ∈ N C(n, k) whose blocks are constituted by the labels of each sequence of consecutive down steps. Example 2.5. Let p ∈ D6,2 , 4. 3. 4. 5. p=. 1. 6. φ2. 2. 3 5. 1 6. 2. We get φ2 (p) = 1 2 / 3 5 6 / 4 ∈ N C(6, 3). For a path p ∈ Dn,k−1 with u(p) = (u1 , u2 , . . . , uk ), and d(p) = (d1 , d2 , . . . , dk ), we have f (φ2 (p)) = (n + 1 − (u1 + u2 + . . . + uk ), n + 1 − (u1 + u2 + . . . + uk−1 ), . . . , n + 1 − u1 ), and b(φ2 (p)) = (dk , dk−1 , . . . , d1 ). Theorem 2.7. Let p ∈ Dn,k−1 and π = φ2 (p) ∈ N C(n, k). Then α(p) = lb(π) + ls(π), β(p) = ls(π), α(prev ) = rb(π) + lb(π), and β(prev ) = rb(π). Proof. Let p = U u1 Dd1 U u2 Dd2 . . . U ui Ddi . . . U uk Ddk ∈ Dn,k−1 and π = φ2 (p) ∈ N C(n, k) with f (π) = (f1 , f2 , . . . , fk ), and b(π) = (b1 , b2 , . . . , bk ). Then α(p) =u1 + (u1 + u2 ) + . . . + (u1 + u2 + . . . + uk−1 ) by Equation (2) =(n + 1 − fk ) + (n + 1 − fk−1 ) + . . . + (n + 1 − f2 ) k X =(k − 1)n − (fi − 1) i=2. 9.

(14) =(k − 1)n −. k X. (fi − 1). i=1. =(k − 1)n − rb(π) by Lemma 1.1 =lb(π) + ls(π) by Equation (1), and β(p) =d1 + (d1 + d2 ) + . . . + (d1 + . . . + dk−1 ) by Equation (3) =bk + (bk + bk−1 ) + . . . + (bk + bk−1 + . . . + b2 ) =. k X. (i − 1)bi. i=1. =ls(π). by Lemma 1.1.. Also, by Equation (4), and Equation (5), we obtain α(prev ) =(k − 1)n − β(p) = rb(π) + lb(π). by Equation (1),. and β(prev ) =(k − 1)n − α(p) = rb(π). by Equation (1).. Corollary 2.8. We have X π∈ N C(n,k). q. lb(π)+2ls(π). =. X. q. 2rb(π)+lb(π). π∈ N C(n,k). n 1 n 2 q k −k . = [n]q k q k − 1 q. Remark 2.9. Given π ∈ N C(n, k), according to Theorem 2.7, we have lb(π) + 2ls(π) = maj(φ−1 2 (π)). In fact, Zhao and Zhong have given the same result (with Equation (1)) in [10].. 3. Recursions for the statistics rb, ls, rs, and lb on non-crossing partitions. Let bk(π) denote the number of blocks of π ∈ N C(n). In this section, we will derive recursions for the statistics rb, ls, rs, lb, and bk on non-crossing partitions. Our derivations are based on two decompositions of π ∈ N C(n). The first decomposition: Let π = B1 /B2 / . . . ∈ N C(n) where the last element of B1 is i. When i ≥ 2, let π = π1 iπ2 , π1 ∈ N C([1, i − 1]), and π2 ∈ N C([i + 1, n]). When i = 1, let π = 1π2 , π2 ∈ N C([2, n]). 10.

(15) π=. B1. .... .... 1. i π1. or. n. i+1. ... 1. n. 2. π2. π2. Figure 3: Illustration to the first decomposition: π = π1 iπ2 , π1 ∈ N C([1, i − 1]), and π2 ∈ N C([i + 1, n]) (or π = 1π2 , π2 ∈ N C([2, n])). Proposition 3.1. Let π = B1 /B2 / . . . ∈ N C(n). Suppose that the last element of B1 is i ≥ 2, we decompose π as π = π1 iπ2 , with π1 ∈ N C([1, i − 1]) and π2 ∈ N C([i + 1, n]); suppose that i = 1, we decompose π as π = 1π2 , with π2 ∈ N C([2, n]). Then when i ≥ 2, rb(π) = rb(π1 ) + rb(π2 ) + i · bk(π2 ), ls(π) = ls(π1 ) + ls(π2 ) + (n − i)bk(π1 ), lb(π) = lb(π1 ) + lb(π2 ) + (bk(π1 ) − 1), and bk(π) = bk(π1 ) + bk(π2 ); when i = 1, rb(π) = rb(π2 ) + bk(π2 ), ls(π) = ls(π2 ) + (n − 1), lb(π) = lb(π2 ), and bk(π) = bk(π2 ) + 1. Proof. The number of blocks is easy to compute. Note that w(i) = 1. When i ≥ 2, w(j) < w(k) for all integers j and k such that 1 ≤ j ≤ i and i + 1 ≤ k ≤ n. Hence rb(π) = rb(π1 ) + rb(π2 ) + i · bk(π2 ), ls(π) = ls(π1 ) + ls(π2 ) + (n − i)bk(π1 ), and lb(π) = lb(π1 ) + lb(π2 ) + (bk(π1 ) − 1). When i = 1, these results can be computed by the same argument. Simion [7] proved that X q rb(π) = π∈N C(n,k). X. X. q ls(π) , and. π∈N C(n,k). q rs(π) =. π∈N C(n,k). X. q lb(π) .. π∈N C(n,k). Hence, X π∈N C(n). q. rb(π) bk(π). t. =. n X. X. q. rb(π) k. t =. k=1 π∈N C(n,k). n X. X. q ls(π) tk =. k=1 π∈N C(n,k). X. q ls(π) tbk(π) .. π∈N C(n). Similarly, X. q rs(π) tbk(π) =. π∈N C(n). X. q lb(π) tbk(π) .. π∈N C(n). Now we discuss the generating functions that track the number of blocks, and one of rb, ls, rs and lb of non-crossing partitions. Define these functions: 11.

(16) .  X. N C(n; q, t) :=. X. q rb(π) tbk(π) =. q ls(π) tbk(π)  ,. π∈N C(n). π∈N C(n). .  and. X. N C ∗ (n; q, t) :=. X. q rs(π) tbk(π) =. q lb(π) tbk(π)  .. π∈N C(n). π∈N C(n). By Proposition 3.1, we can prove that these generating functions satisfy some Catalanstyle recurrences. Theorem 3.2. For n ≥ 1, we have N C(n; q, t) = tN C(n − 1; q, qt) +. n X. N C(i − 1; q, t)N C(n − i; q, q i t),. i=2. =q. n−1. tN C(n − 1; q, t) +. n X. N C(i − 1; q, q n−i t)N C(n − i; q, t),. i=2. and N C ∗ (n; q, t) = tN C ∗ (n − 1; q, t) +. n 1X. q. N C ∗ (i − 1; q, qt)N C ∗ (n − i; q, t).. (6). i=2. Proof. For n ≥ 1, in light of Proposition 3.1, we express these polynomials as follows. N C(n; q, t) X = q rb(π) tbk(π) π∈N C(n).  X. =. q. rb(π2 )+bk(π2 ) bk(π2 )+1. t. . n X  +. X. i=2 π1 ∈N C([1,i−1]) π2 ∈N C([i+1,n]). π2 ∈N C[2,n].  q rb(π1 )+rb(π2 )+i·bk(π2 ) tbk(π1 )+bk(π2 )  .  X. =. q. rb(π2 )+bk(π2 ) bk(π2 )+1. t. =t. n X  +. X. i=2 π1 ∈N C(i−1) π2 ∈N C(n−i). π2 ∈N C(n−1). X. . q rb(π2 ) (qt)bk(π2 ) + n X. .  X.  i=2. π2 ∈N C(n−1). = tN C(n − 1; q, qt) +. n X.  q rb(π1 )+rb(π2 )+i·bk(π2 ) tbk(π1 )+bk(π2 )  . q rb(π1 ) tbk(π1 )  . π1 ∈N C(i−1). N C(i − 1; q, t)N C(n − i; q, q i t),. i=2. N C(n; q, t) X = q ls(π) tbk(π) π∈N C(n). 12.  X π2 ∈N C(n−i). q rb(π2 ) (q i t)bk(π2 ) .

(17)  X. =. π2 ∈N C[2,n]. . n X q ls(π2 )+(n−1) tbk(π2 )+1 +  . X. i=2 π1 ∈N C([1,i−1]) π2 ∈N C([i+1,n]).  q ls(π1 )+ls(π2 )+(n−i)bk(π1 ) tbk(π1 )+bk(π2 )  .  X. =. q. ls(π2 )+(n−1) bk(π2 )+1. t. n X  +. X. i=2 π1 ∈N C(i−1) π2 ∈N C(n−i). π2 ∈N C(n−1). X. = q n−1 t. . q ls(π2 ) tbk(π2 ) +. n X. = q n−1 tN C(n − 1; q, t) +. n X. . .  X. X. q ls(π1 ) (q n−i t)bk(π1 )  .  i=2. π2 ∈N C(n−1).  q ls(π1 )+ls(π2 )+(n−i)bk(π1 ) tbk(π1 )+bk(π2 )  . q ls(π2 ) tbk(π2 ) . π2 ∈N C(n−i). π1 ∈N C(i−1). N C(i − 1; q, q n−i t)N C(n − i; q, t),. i=2. and. N C ∗ (n; q, t) X = q lb(π) tbk(π) π∈N C(n).  X. =. q. lb(π2 ) bk(π2 )+1. t. . n X  +. X. i=2 π1 ∈N C([1,i−1]) π2 ∈N C([i+1,n]). π2 ∈N C[2,n].  q lb(π1 )+lb(π2 )+(bk(π1 )−1) tbk(π1 )+bk(π2 )  .  X. =. q. lb(π2 ) bk(π2 )+1. t. =t. n X  +. X. i=2 π1 ∈N C(i−1) π2 ∈N C(n−i). π2 ∈N C(n−1). X. . q lb(π2 ) tbk(π2 ) +. π2 ∈N C(n−1). = tN C ∗ (n − 1; q, t) +. 1 q. n X. n 1X. q.  q lb(π1 )+lb(π2 )+(bk(π1 )−1) tbk(π1 )+bk(π2 )  . .  X. q lb(π1 ) (qt)bk(π1 )  .  i=2. π1 ∈N C(i−1).  X. q lb(π2 ) tbk(π2 ) . π2 ∈N C(n−i). N C ∗ (i − 1; q, qt)N C ∗ (n − i; q, t).. i=2. So far, we use the first decomposition to get recursions for the distributions of rb, ls and lb. To compute the statistic rs, we need the second decomposition as follows: Let π = B1 /B2 / . . . ∈ N C(n). When |B1 | ≥ 2 and the second element of B1 is i, let π = 1π1 π2 such that i is the smallest element of π2 , π1 ∈ N C([2, i − 1]), and π2 ∈ N C([i, n]). When |B1 | = 1, let π = 1π2 , π2 ∈ N C([2, n]). Proposition 3.3. Let π = B1 /B2 / . . . ∈ N C(n). Suppose that |B1 | ≥ 2 and the second element of B1 is i, we decompose π as π = 1π1 π2 , with π1 ∈ N C([2, i − 1]) and π2 ∈ 13.

(18) π=. ... 1. ... i−1. 2. i. π1. or. n. ... 1. π2. 2. n. π2. Figure 4: Illustration to the second decomposition: π = 1π1 π2 , π1 ∈ N C([2, i − 1]), and π2 ∈ N C([i, n]) (or π2 ∈ N C([2, n])). N C([i, n]). Suppose that |B1 | = 1, we decompose π as π = 1π2 , with π2 ∈ N C([2, n]). Then when i ≥ 2, rb(π) = rb(π1 ) + rb(π2 ) + (i − 2)(bk(π2 ) − 1) + (bk(π1 ) + bk(π2 ) − 1), rs(π) = rs(π1 ) + rs(π2 ) + (i − 2), and bk(π) = bk(π1 ) + bk(π2 ); when i = 1, rb(π) = rb(π2 ) + bk(π2 ), rs(π) = rs(π2 ), and bk(π) = bk(π2 ) + 1. Proof. The number of blocks is easy to compute. When i ≥ 2, w(j) < w(k) for all integer j, k such that 1 < j < i and i < k ≤ n except w(k) = 1. Moreover, w(1) < w(l) for all integers 2 ≤ l ≤ n except w(l) = 1. Hence rb(π) = rb(π1 ) + rb(π2 ) + (i − 2)(bk(π2 ) − 1) + (bk(π1 ) + bk(π2 ) − 1). Also, w(j) > w(i) for all integers 2 ≤ j < i. Hence rs(π) = rs(π1 ) + rs(π2 ) + (i − 2). When i = 1, these results can be computed by the same argument. We also prove recursions for N C(n; q, t) and N C ∗ (n; q, t) by Proposition 3.3. Theorem 3.4. For n ≥ 1, we have n X 1 N C(n; q, t) = tN C(n − 1; q, qt) + N C(i − 2; q, qt)N C(n − i + 1; q, q i−1 t), i−1 q i=2. and N C ∗ (n; q, t) = tN C ∗ (n − 1; q, t) +. n X. q i−2 N C ∗ (i − 2; q, t)N C ∗ (n − i + 1; q, t). (7). i=2. Proof. For n ≥ 1, in light of Proposition 3.3, we express these polynomials as follows. N C(n; q, t) X = q rb(π) tbk(π) π∈N C(n). =. X. q rb(π2 )+bk(π2 ) tbk(π2 )+1. π2 ∈N C[2,n]. 14.

(19) . . n X + . X. i=2 π1 ∈N C([2,i−1]) π2 ∈N C([i,n]). X. =.  q rb(π1 )+rb(π2 )+(i−2)(bk(π2 )−1)+(bk(π1 )+bk(π2 )−1) tbk(π1 )+bk(π2 )  . q rb(π2 )+bk(π2 ) tbk(π2 )+1. π2 ∈N C(n−1). . . n X +  i=2. =t. X π1 ∈N C(i−2) π2 ∈N C(n−i+1). X.  q rb(π1 )+rb(π2 )+(i−2)(bk(π2 )−1)+(bk(π1 )+bk(π2 )−1) tbk(π1 )+bk(π2 )  . q rb(π2 ) (qt)bk(π2 ). π2 ∈N C(n−1).  n X 1  + i−1 q i=2. .  X. q rb(π1 ) (qt)bk(π1 )  . π1 ∈N C(i−2). = tN C(n − 1; q, qt) +. X. q rb(π2 ) (q i−1 t)bk(π2 ) . π2 ∈N C(n−i+1). n X 1 N C(i − 2; q, qt)N C(n − i + 1; q, q i−1 t), i−1 q i=2. and. N C ∗ (n; q, t) X = q rs(π) tbk(π) π∈N C(n).  X. =. π2 ∈N C[2,n]. . n X q rs(π2 ) tbk(π2 )+1 +  . X. i=2 π1 ∈N C([2,i−1]) π2 ∈N C([i,n]).  q rs(π1 )+rs(π2 )+(i−2) tbk(π1 )+bk(π2 )  .  X. =. q. rs(π2 ) bk(π2 )+1. t. =t. n X  + i=2. π2 ∈N C(n−1). X. . q rs(π2 ) tbk(π2 ) +. i=2. π2 ∈N C(n−1). = tN C ∗ (n − 1; q, t) +. n X. n X. X π1 ∈N C(i−2) π2 ∈N C(n−i+1).  q rs(π1 )+rs(π2 )+(i−2) tbk(π1 )+bk(π2 )  .  q i−2 .  X π1 ∈N C(i−2). q rs(π1 ) tbk(π1 )  .  X. q rs(π2 ) tbk(π2 ) . π2 ∈N C(n−i+1). q i−2 N C ∗ (i − 2; q, t)N C ∗ (n − i + 1; q, t).. i=2. Now, we try to derive some continued fraction expressions for the ordinary generating. 15.

(20) function N C ∗ (q, t, z) =. X. N C ∗ (n; q, t)z n .. n≥0. Proposition 3.5. The ordinary generating function N C ∗ (q, t, z) has the following continued fraction expression: N C ∗ (q, t, z) 1. = 1+. q −1 z. ,. q −1 z. − tz −. q −1 z. 1 + q −1 z − qtz −. q −1 z. 1 + q −1 z − q 2 tz −. 1 + q −1 z − q 3 tz −. q −1 z .. .. i.e., for k ≥ 1, the third term of level k is q k−1 tz. Proof. By Equation (6), we get N C ∗ (q, t, z) X N C ∗ (k; q, t)z k = k≥0. = N C ∗ (0; q, t) +. X. N C ∗ (k; q, t)z k. k≥1. =1+. X. tN C ∗ (k − 1; q, t) + q −1. = 1 + tz. ! N C ∗ (i − 1; q, qt)N C ∗ (k − i; q, t) z k. i=2. k≥1. X. k X. ∗. N C (k − 1; q, t)z. k−1. −1. +q z. k XX. N C ∗ (i − 1; q, qt)z i−1 N C ∗ (k − i; q, t)z k−i. k≥1 i=2. k≥1 ∗. −1. ∗. = 1 + tzN C (q, t, z) + q z (N C (q, qt, z) − 1) N C ∗ (q, t, z). From here, we have N C ∗ (q, t, z) 1 1 + q −1 z − tz − q −1 zN C ∗ (q, qt, z) 1 = q −1 z 1 + q −1 z − tz − 1 + q −1 z − qtz − q −1 zN C ∗ (q, q 2 t, z) 1 = q −1 z 1 + q −1 z − tz − q −1 z 1 + q −1 z − qtz − 1 + q −1 z − q 2 tz − q −1 zN C ∗ (q, q 3 t, z) =. 16.

(21) .. . 1. = 1+. q −1 z. .. q −1 z. − tz −. q −1 z. 1 + q −1 z − qtz −. q −1 z. 1 + q −1 z − q 2 tz −. 1 + q −1 z − q 3 tz −. q −1 z .. .. Proposition 3.6. The ordinary generating function N C ∗ (q, t, z) has the following continued fraction expression: 1. N C ∗ (q, t, z) =. ,. tz. 1− 1−. z qtz 1− qz 1− q 2 tz 1− q2z 1− q 3 tz 1− q3z 1− q 4 tz 1− q4z 1− q 5 tz 1− .. .. i.e., for k ≥ 1, the numerator of level 2k − 1 is q k−1 tz and the numerator of level 2k is q k−1 z. Proof. By Equation (7), we get N C ∗ (q, t, z) X = N C ∗ (k; q, t)z k k≥0. = N C ∗ (0; q, t) +. X. N C ∗ (k; q, t)z k. k≥1. =1+. X. tN C ∗ (k − 1; q, t) +. = 1 + tz. ! q i−2 N C ∗ (i − 2; q, t)N C ∗ (k − i + 1; q, t) z k. i=2. k≥1. X. k X. ∗. N C (k − 1; q, t)z. k≥1 ∗. ∗. k−1. +z. k XX. N C ∗ (i − 2; q, t)(qz)i−2 N C ∗ (k − i + 1; q, t)z k−i+1. k≥1 i=2 ∗. = 1 + tzN C (q, t, z) + zN C (q, t, qz) (N C (q, t, z) − 1) . 17.

(22) From here, we have N C ∗ (q, t, z) 1 − zN C ∗ (q, t, qz) = 1 − tz − zN C ∗ (q, t, qz) 1 = tz 1− 1 − zN C ∗ (q, t, qz) 1 = tz 1− z 1− qtz 1− 1 − qzN C ∗ (q, t, q 2 z) 1 = tz 1− z 1− qtz 1− qz 1− q 2 tz 1− 1 − q 2 zN C ∗ (q, t, q 3 z) .. . 1 = tz 1− z 1− qtz 1− qz 1− q 2 tz 1− q2z 1− q 3 tz 1− q3z 1− q 4 tz 1− q4z 1− q 5 tz 1− .... 4 4.1. .. Connections on rs and area Area of a Dyck path. √ √ We define the area of a Dyck path p, denote area(p), to be the number of 2 × 2 diamonds that can fit below the path and above the x-axis. For convenience, we call a point on (x, y) is odd if x is odd. 18.

(23) We describe a bijection φ3 , due to Stump [9], from Dn to N C(n) as follows [1]: Given a path p ∈ Dn , label the odd points of the path by 1 to n going from left to right. We form the partition φ3 (p) = π ∈ N C(n) by declaring that i and i0 are in the same block of π if and only if: 1. i and i0 are on the line y = j, and 2. the segments of p between i and i0 does not pass below the line y = j. Example 4.1. Let a Dyck path p ∈ D13 , 3. 4. 5. 7. 2 p=. 6. 8. 1. 11 9. 10. 12. 13. We get φ3 (p) = 1 9 / 2 6 8 / 3 4 5 / 7 / 10 12 13 / 11 ∈ N C(13, 6). Proposition 4.1. Let p ∈ Dn and π = φ3 (p) ∈ N C(n, k), then area(p) = 2rs(π) + n − k. Proof. By the operation of φ3 , each block Bi of π generates a zigzag diamond path graphically (see Figure 5). In the zigzag diamond path of Bi , the number of diamonds on the upper row is li − fi , and the number of diamonds on the bottom row is li − fi + 1. Hence, the area(p) can be counted easily by summing the diamonds of all the zigzag diamond paths, except those with bottom row on the x-axis. That is X X area(p) = (2li − 2fi + 1) + (li − fi ) Bi with elements on y=j, j≥3. =. k X. =. X. (2li − 2fi + 1) −. i=1 k X. Bi with elements on y=1. Bi with elements on y=1. (2li − 2fi + 1) − n. i=1. =2. =2. (li − fi + 1). k X i=1 k X i=1. li − li −. k X i=1 k X. ! fi. +k−n !. fi − n + k. +n−k. i=1. = 2rs(π) + n − k. by Lemma 1.1.. 19.

(24) 3 2 p=. 4. 5. 7. B3. B4. 6. 8. B2. 1. 11 9. B1. 10. B6. B5 12. 13. Figure 5: A Dyck path p to illustrate Proposition 4.1. The colored region is the zigzag diamond path generated by B2 of φ3 (p). Proposition 4.1 links Dyck paths and non-crossing partitions by the statistics area and rs. Define the rank of a non-crossing partition π ∈ N C(n) to be n − bk(π). The rank of a Dyck path p is defined by the rank of φ3 (p). In [1], Blanco and Petersen did some work about the generating functions of joint distribution of rank and area on Dyck paths X Dyck(n; q, x) = q area(p) xrank(p) , p∈Dn. and Dyck(q, x, z) =. X. Dyck(n; q, x)z n .. n≥0. They gave the following recursion and continued fraction expression.. Dyck(n; q, x) =. n−1 X. 1 (qx)k Dyck(k; q, )Dyck(n − 1 − k; q, x), x k=0 1. and Dyck(q, x, z) = 1−. , z qxz 1− q2z 1− q 3 xz 1− q4z 1− q 5 xz 1− .. .. i.e., for k ≥ 1, the numerator of level 2k − 1 is q 2k−2 z and the numerator of level 2k is q 2k−1 xz. Actually, by Proposition 4.1, we have X n n ∗ 2 1 q x N C n; q , = q 2rs(π)+n−bk(π) xrank(π) = Dyck(n; q, x), (8) qx π∈N C(n) ∗ 2 1 and N C q , , qxz = Dyck(q, x, z). (9) qx Hence, we can get two recursions for Dyck(n; q, x) and two continued fraction expressions for Dyck(q, x, z). 20.

(25) Theorem 4.2. For n ≥ 1, we have n. x X 2(i−1) Dyck(n; q, x) = Dyck(n − 1; q, x) + q Dyck(i − 1; q, q −2 x)Dyck(n − i; q, x), q i=2 = Dyck(n − 1; q, x) + qx. n X. q 2(i−2) Dyck(i − 2; q, x)Dyck(n − i + 1; q, x).. i=2. Proof. For n ≥ 1, we have Dyck(n; q, x) n n ∗ 2 1 = q x N C n; q , qx = q n xn. ! n X 1 1 1 1 q N C ∗ (n − 1; q 2 , ) + 2 N C ∗ (i − 1; q 2 , )N C ∗ (n − i; q 2 , ) qx qx q i=2 x qx. by Equation (6) 1 = q n−1 xn−1 N C ∗ (n − 1; q 2 , ) qx n x X 2(i−1) i−1 −2 i−1 1 n−i n−i ∗ 2 1 ∗ 2 + ) q x N C (n − i; q , ) q q (q x) N C (i − 1; q , q i=2 q · q −2 x qx n. = Dyck(n − 1; q, x) +. x X 2(i−1) q Dyck(i − 1; q, q −2 x)Dyck(n − i; q, x); q i=2. and Dyck(n; q, x) n n ∗ 2 1 = q x N C n; q , qx = q n xn. ! n X 1 1 1 1 N C ∗ (n − 1; q 2 , ) + q 2(i−2) N C ∗ (i − 2; q 2 , )N C ∗ (n − i + 1; q 2 , ) qx qx qx qx i=2. by Equation (7) 1 = q n−1 xn−1 N C ∗ (n − 1; q 2 , ) qx n X 2(i−2) i−2 i−2 ∗ 2 1 n−i+1 n−i+1 ∗ 2 1 + qx q q x N C (i − 2; q , ) q x N C (n − i + 1; q , ) qx qx i=2 = Dyck(n − 1; q, x) + qx. n X. q 2(i−2) Dyck(i − 2; q, x)Dyck(n − i + 1; q, x).. i=2. 21.

(26) Proposition 4.3. The ordinary generating function Dyck(q, x, z) has the following continued fraction expression: 1. Dyck(q, x, z) = 1+. q −1 xz. ,. q −1 xz. −z−. q −1 xz. 1 + q −1 xz − q 2 z −. 1 + q −1 xz − q 4 z −. q −1 xz .. .. i.e., for k ≥ 1, the third term of level k is q 2(k−1) z. Proof. By Proposition 3.5, we get Dyck(q, x, z) = N C ∗ (q 2 ,. 1 , qxz) qx 1. =. .. q −1 xz. 1 + q −1 xz − z − 1+. q −1 xz. −. q2z. q −1 xz. − 1+. q −1 xz. −. q4z. q −1 xz − .. .. Remark that, by Proposition 3.6, we can obtain the same continued fraction expression for Dyck(q, x, z) in [1]. Proposition 4.4. The ordinary generating function Dyck(q, x, z) has the following continued fraction expression: 1. Dyck(q, x, z) = 1−. , z qxz 1− q2z 1− q 3 xz 1− q4z 1− q 5 xz 1− .. .. i.e., for k ≥ 1, the numerator of level 2k − 1 is q 2k−2 z and the numerator of level 2k is q 2k−1 xz.. 22.

(27) Proof. By Proposition 3.6, we get Dyck(q, x, z) = N C ∗ (q 2 , 1. = 1−. 4.2. 1 , qxz) qx. . z qxz 1− q2z 1− q 3 xz 1− q4z 1− q 5 xz 1− .. .. Area of a 2-Motzkin path. A Motzkin path of length n is a path starting on (1, 0) and ending on (n + 1, 0) but never going below the x-axis, with possible steps u = (1, 1), d = (1, −1), and s = (1, 0). The set of all Motzkin paths of length n is denoted by Motz(n). A 2-Motzkin path of length n is a path starting on (1, 0) and ending on (n + 1, 0) but never going below the x-axis, with possible steps u = (1, 1), d = (1, −1), and (1, 0), where the level step (1, 0) can be either of two kinds: straight (s) and zigzag (z). The set of all 2-Motzkin paths of length n is denoted by 2-Motz(n). Let λi stand for the i-th step in a path m which is comprised of u, d, s, and z. We denote U (m) = {j | λj = u, λj ∈ m}, D(m) = {j | λj = d, λj ∈ m}, S(m) = {j | λj = s, λj ∈ m}, and Z(m) = {j | λj = z, λj ∈ m}. Note that if m ∈ Motz(n), then Z(m) must be empty. For an m ∈ 2-Motz(n), let area(m) denote the total area of the regions bounded by the path of m and the x-axis. Note that it is not the same as the area of Dyck paths. Define a map φ4 from N C(n) to 2-Motz(n − 1) as in [8]. Given π ∈ N C(n), we convert each interval [j, j + 1], 1 6 j < n with λj ∈ {u, d, s, z} as follows:   u, if j and j + 1 are in different blocks and j is not the largest element in its block;       d, if j and j + 1 are in different blocks and j + 1 is not the smallest element in its block; λj = s, if j and j + 1 are in the same block;    z, if j and j + 1 are in different blocks, j is the largest element in its block,     and j + 1 is the smallest element in its block. Then let φ4 (π) = λ1 λ2 . . . λn−1 . About φ4 , Simion and Ullman gave an observation. Observation 4.1. [8, Observation 2.4] If m = λ1 λ2 . . . λn−1 is the path associated with π ∈ N C(n) as above, then |U (m)| = |D(m)| and |U (λ1 λ2 . . . λi )| ≥ |D(λ1 λ2 . . . λi )| for all i = 1, 2, . . . , n − 1 (in other words, there is a complete parenthesization of the u’s and the d’s).. 23.

(28) For π ∈ N C(n), φ4 (π) will never go below the x-axis and end on the x-axis by Observation 4.1. Hence φ4 (π) ∈ 2-Motz(n − 1). For the inverse mapping, given m = λ1 λ2 . . . λn−1 ∈ 2-Motz(n − 1), first we plot 1, 2, ..., n on a horizontal line with a left parenthesis “(” on the left of 1 and a right parenthesis “)” on the right of n. For each λi = u (resp. d), we put a left parenthesis “(” (resp. a right parenthesis “)”) between i and i + 1. For each λi = z, we put a right parenthesis and a left parenthesis “)(” between i and i + 1. For each λi = s, we do nothing. Since m ∈ 2-Motz(n − 1), the u’s and d’s are completely parenthesized. In addition, inserting “)(” inside a complete parenthesization of pairs of parentheses does not change the completion. So, we now have a complete parenthesization of pairs of parentheses on [n]. Then, we can deduce it as a partition of [n], φ−1 4 (m), by creating a −1 block of φ4 (m) for each of the consecutive integers inside the lowest level parenthesis pairs (i.e., parenthesis which pair each other and enclose no others), and removing these lowest level parenthesis pairs and all the integers they enclose. Repeat this procedure until all parentheses have been removed. Clearly the partition φ−1 4 (m) is non-crossing. Hence we have the following proposition. Proposition 4.5. φ4 is bijective. Example 4.2. Let π = 1 6 7 / 2 3 / 4 5 / 8 14 / 9 / 10 / 11 13 / 12 ∈ N C(14), then φ4 (π) = uszsdszuzzudd. 12 2. φ4 (π) = 1. 3. 4. 5. 9 6. 7. 10. 11. 8. 13 14. Example 4.3. Let m = uudszudzd ∈ 2-Motz(9), then we have (1(2(3)4 5)(6(7)8)(9)10). Firstly, create blocks {3}, {7}, and {9} and remove (3), (7), and (9), then we have (1(2 4 5)(6 8)10). Secondly, create blocks {2, 4, 5} and {6, 8} and remove (2 4 5) and (6 8), then we have (1 10). Finally, create a block {1, 10} and remove (1 10), then we get φ−1 4 (m) = 1 10 / 2 4 5 / 3 / 6 8 / 7 / 9. Lemma 4.6. Let π ∈ N C(n) and m = φ4 (π) = λ1 λ2 . . . λn−1 ∈ 2-Motz(n − 1). For integers 1 ≤ a < b ≤ n − 1, λa and λb are a pair of matched u and d in m if and only if a and b + 1 are successive elements in the same block of π. Proof. Let π ∈ N C(n) and m = φ4 (π) = λ1 λ2 . . . λn−1 ∈ 2-Motz(n − 1). If λa and λb are a pair of matched u and d in m, we can get the result easily by operations of φ−1 4 . For the converse, let a and b + 1 be successive elements in the same block of π, we can derive that λa = u and λb = d immediately. Since a and b + 1 are successive elements in the same block, the blocks of π in {a + 1, . . . , b} are non-crossing. So there is a complete parenthesization of the u’s and the d’s in λa+1 . . . λb−1 . Hence λa and λb are matched in m. Proposition 4.7. Let π ∈ N C(n) and m = φ4 (π) ∈ 2-Motz(n − 1), then area(m) = rs(π). 24.

(29) Proof. Given m = λ1 λ2 . . . λn−1 ∈2-Motz(n − 1), we denote the points on m by (i, y(i)) for all integer 1 ≤ i ≤ n. We will count the area of m column by column. If λi = u, then the area of the region bounded by x = i, x = i + 1, the x-axis, and the segment of m on [i, i + 1] is y(i) + 0.5. Similarly, if λi = d, then the area is y(i) − 0.5. If λi = s or z, then the area is y(i). Hence, X X X area(m) = (y(i) + 0.5) + (y(i) − 0.5) + y(i) λi =u. =. X. λi =d. y(i) +. λi =u. =. n−1 X. X. λi =s or z. X. y(i) +. λi =d. y(i). λi =s or z. y(i).. i=1. Remark that m ∈ 2-Motz(n − 1), hence the u’s and d’s in m are completely parenthesized. For integer i on the line y = y(i), it means that there are y(i) matched pairs of λa = u and λb = d such that a < i and b ≥ i. Translate to non-crossing partition, a and b + 1 are successive elements in the same block of π by Lemma 4.6. So there are y(i) blocks with their first elements less than i and their last elements greater than i. That is y(i) is equal to #{wj | wj < wi , j > i}. Hence n−1 X. y(i) =. i=1. =. n−1 X i=1 n X. #{wj | wj < wi , j > i} #{wj | wj < wi , j > i}. i=1. = rs(π).. Example 4.4 (continued). We have rs(π) = 0+1+1+1+1+0+0+0+1+1+1+2+1+0 = 10. And area(φ4 (π)) = 0.5 + 1 + 1 + 1 + 0.5 + 0 + 0 + 0.5 + 1 + 1 + 1.5 + 1.5 + 0.5 = 10. 12 2. φ4 (π) =. 3. 4. 5. 1. 9 6. 7. 8. 10. 11. 13 14. The rank function for the lattice of non-crossing partitions n of [n] is well understood. 1 n For instance, the Narayana numbers Nn,n−k−1 = n k k+1 count the number of noncrossing partitions of [n] with rank k. One of the more interesting results about this function is the following identity [5]. X X xrank(π) = γn−1,j xj (1 + x)n−1−2j , (10) π∈N C(n). 0≤j≤. (n−1) 2. 25.

(30) where the γn−1,j are nonnegative integers. (In fact, γn−1,j =. X m∈Motz(n−1) |U (m)|=j. n−1 1= Cj .) 2j. Now we consider the joint distribution of rs and rank on non-crossing partitions. Proposition 4.7 links N C(n) and 2-Motz(n − 1) by the statistics area and rs. In [1, 8], they proved that for a non-crossing partition π, rank(π) = |U (φ4 (π))| + |S(φ4 (π))|. With those properties, we can prove a result which is similar to Equation (10). There is an onto mapping ρ from 2-Motz(n) to Motz(n) such that U (ρ(m)) = U (m), S D(ρ(m)) = D(m), and S(ρ(m)) = S(m) Z(m) for m ∈ 2-Motz(n). Theorem 4.8. We have X q rs(π) xrank(π) =. γn−1,j (q)xj (1 + x)n−1−2j ,. c 0≤j≤b n−1 2. π∈N C(n). X. where γn−1,j (q) =. X. q area(m) .. m∈Motz(n−1) |U (m)|=j. Proof. By proposition 4.7, we have X q rs(π) xrank(π) = π∈N C(n). X. q area(m) x|U (m)|+|S(m)|. m∈2-Motz(n−1). =. X. q area(m) x|U (m)| x|S(m)| 1|Z(m)|. m∈2-Motz(n−1). =. X. q area(m) x|U (m)| (1 + x)|S(m)|. m∈Motz(n−1). =. X. q area(m) x|U (m)| (1 + x)n−1−2|U (m)|. m∈Motz(n−1). =. X. γn−1,j (q)xj (1 + x)n−1−2j .. c 0≤j≤b n−1 2. 5. Conclusions. The following table describes the connections proved in Section 2 and 4 between noncrossing partitions and Dyck paths. Bijection φ1 φ2 φ3. maj(p) area(p) 2rb(π) + rs(π) (1) 2lb(π) + 2ls(π) − rs(π) lb(π) + 2ls(π) (2) 2rb(π) + lb(π) (3) 2rs(π) + n − bk(π) 26.

(31) Of course we are interested in the following question. Problem 5.1. Can (1), (2), and (3) be expressed by rb, ls, rs, and lb on non-crossing partitions? In fact, we can express (1) by other statistics on non-crossing partitions. Proposition 5.2. Let π ∈ N C(n, k) with f (π) = (f1 , f2 , . . . , fk ), and l0 (π) = (l10 , l20 , . . . , lk0 ). Let p = φ1 (π) ∈ Dn,k−1 . Then area(p) =. k X fi+1 − fi 2. i=1. +. (li0. − fi+1 + 1)(fi+1 − fi ) .. Proof. For π ∈ N C(n, k) with f (π) = (f1 , f2 , . . . , fk ), and l0 (π) = (l10 , l20 , . . . , lk0 ), let p = φ1 (π). First, we construct√k isosceles right triangles with the length of the leg of the i-th triangles is (fi+1 − fi ) 2 for i = 1, 2, . . . , k, where fk+1 = n + 1. Secondly, we put these triangles on the lattice one by one such that the right angle is at the top and the hypotenuse is on the x-axis. Note that the i-th triangle covers fi+12−fi diamonds. √ √ Thirdly, we put a rectangle with area (li0 − fi+1 + 1) 2 × (fi+1 − fi ) 2 on the top right of the right leg of the i-th triangle for i = 1, 2, . . . , k. Summing the diamonds of all the covers, then we get k X fi+1 − fi 0 area(p) = + (li − fi+1 + 1)(fi+1 − fi ) . 2 i=1 Example 5.1. Let π ∈ N C(8, 4) with f (π) = (1, 2, 3, 5) and l0 (π) = (3, 5, 7, 8). Let p = φ1 (π) = U 3 D1 U 2 D1 U 2 D2 U 1 D4 ∈ D8,3 . To illustrate Proposition 5.2, we denote the i-th triangle (resp. rectangle) by Ti (resp. Ri ).. R3. p=. R2 R1 T3 T1. T4. T2. In Section 2, we have deduced that X X q 2rb(π)+rs(π) = q 2lb(π)+2ls(π)−rs(π) = π∈N C(n,k). π∈N C(n,k). X π∈N C(n,k). 1 n n 2 = q k −k . [n]q k q k − 1 q 27. q lb(π)+2ls(π) =. X π∈N C(n,k). q 2rb(π)+lb(π).

(32) X. Also, by Equation (1) we have. q. rb(π)+lb(π)+ls(π). π∈N C(n,k). 1 n n = q (k−1)n . n k k−1. Problem 5.3. Are there any other combinations of rb, ls, rs, and lb on non-crossing partitions with closed-form? We are also interested in the next problem. Problem 5.4. Are there any formulas for   X X q rb(π) = q ls(π)  and π∈N C(n,k). π∈N C(n,k).  X. q rs(π) =. π∈N C(n,k).  X. q lb(π) ?. π∈N C(n,k). In Theorem 4.8, we proved the equidistribution result X X q rs(π) xrank(π) = q area(m) x|U (m)|+|S(m)| . π∈N C(n). m∈2-Motz(n−1). Problem 5.5. Can we find a statistic st on 2-Motz(n − 1) such that   X X X q rb(π) xrank(π) = q ls(π) xrank(π)  = q st(m) x|U (m)|+|S(m)| ? π∈N C(n). π∈N C(n). m∈2-Motz(n−1). References [1] Blanco, S. A., and Petersen, T. K. Counting Dyck paths by area and rank. Ann. Comb. 18 (2014), 171–197. [2] Deutsch, E. Dyck path enumeration. Discrete Math. 204 (1999), 167–202. ¨ rlinger, J., and Hofbauer, J. q-Catalan numbers. J. Combin. Theory Ser. [3] Fu A 40 (1985), 248–264. [4] Milne, S. Restricted growth functions, ranks row matchings of partition lattices, and q-stirling numbers. Adv. in Math. 43 (1982), 173–196. [5] Postnikov, A., Reiner, V., and Williams, L. Faces of generalized permutohedra. Doc. Math. 13 (2008), 207–273. [6] Prodinger, H. A correspondence between ordered trees and noncrossing partitions. Discrete Math. 46 (1983), 205–206. [7] Simion, R. Combinatorial statistics on non-crossong partitions. J. Combin. Theory Ser. A 66 (1994), 270–301. 28.

(33) [8] Simion, R., and Ullman, D. On the structure of the lattice of noncrossing partitions. Discrete Math. 98 (1991), 193–206. [9] Stump, C. More bijective catalan combinatorics on permutations and colored permutations. J. Combin. 4 (2013), 419–447. [10] Zhao, H., and Zhong, Z. Two statistics linking Dyck paths and noncrossing partitions. Electron. J. Combin. 18 (2011), paper 83.. 29.

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