古典類型的不相交劃分與m-可除的不相交劃分的循環篩選現象

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(1)國立高雄大學應用數學系碩士論文. 古典類型的不相交劃分與 m-可除的不相交劃分的循環篩選現象 The cyclic sieving phenomenon for noncrossing partitions and m-divisible noncrossing partitions of classical types. 研究生：劉誠澤撰指導教授：鄭斯恩. 中華民國 99 年 6 月.

(2) 致謝感謝指導教授鄭斯恩教授這兩年來的指導與陪伴，以及給我足夠的空間去運用。感謝老師的一席話，「自己決定畢業時間」。在學業上，感謝老師不厭其煩的提醒與叮嚀，每每讓遇到瓶頸的我頓時開通；在生活上，感謝老師的歡樂個性與詼諧言語，總能讓大家開懷愉悅；在感情上，感謝老師的經驗分享與寶貴建議，讓我更清楚此事不可急進。感謝口試委員傅東山教授與潘業忠教授提供的寶貴建議，讓我獲益良多。感謝游森棚教授的教導與鼓勵，使我省悟了一些事。接著，我要感謝所屬的大家庭，感謝系上的老師們，感謝系上三位親切的大姐姐，感謝系上的學長姐，感謝班上的同學，感謝系上的學弟妹，感謝系上提供的資源，以及感謝這兩年所經歷的晴天與雨天。最後，我要感謝關心我的家人，無時無刻的鼓勵與無形的支持是我最大的後盾，讓我無後顧之憂得以完成學業，我永遠愛你們。. 劉誠澤撰.

(3) The cyclic sieving phenomenon for noncrossing partitions and m-divisible noncrossing partitions of classical types. By Cheng-Tse Liu Advisor Szu-En Cheng. Department of Applied Mathematics, National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. June 2010.

(4) Contents 1 Introduction 1.1 Definitions and notations . . . . . . . . . . . . . . . . . . . . . 1.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Noncrossing partitions of classical 2.1 Noncrossing partitions of type A 2.2 Noncrossing partitions of type B 2.3 Noncrossing partitions of type D. types 8 . . . . . . . . . . . . . . . . 8 . . . . . . . . . . . . . . . . 10 . . . . . . . . . . . . . . . . 16. 3 m-Divisible noncrossing partitions of 3.1 m-Divisible noncrossing partitions of 3.2 m-Divisible noncrossing partitions of 3.3 m-Divisible noncrossing partitions of 4 Remarks. 1 1 4. classical type A . type B . type D .. types 26 . . . . . . . . . 26 . . . . . . . . . 36 . . . . . . . . . 42 43. i.

(5) 古典類型的不相交劃分與 m-可除的不相交劃分的循環篩選現象指導教授：鄭斯恩教授國立高雄大學應用數學系. 學生：劉誠澤國立高雄大學應用數學系. 摘要. 在文獻[12]中，Reiner、Stanton 與 White 介紹了循環篩選現象。在循環群的作用之下，我們利用軌道計數定理計算古典類型的不相交劃分與 m-可除的不相交劃分的等價類個數，並且找到自然的 q-類比去建立循環篩選現象。關鍵字：循環篩選現象、不相交劃分、m-可除的不相交劃分、循環群. ii.

(6) The cyclic sieving phenomenon for noncrossing partitions and m-divisible noncrossing partitions of classical types. Advisor: Professor Szu-En Cheng Department of Applied Mathematics National University of Kaohsiung. Student: Cheng-Tse Liu Department of Applied Mathematics National University of Kaohsiung. ABSTRACT. The cyclic sieving phenomenon was introduced by Reiner, Stanton, and White [12]. We use the Orbit-counting Lemma to count the number of orbits of noncrossing partitions and m-divisible noncrossing partitions of classical types, and find the natural q-analogues to exhibit the cyclic sieving phenomenon under cyclic group actions.. Keywords: cyclic sieving phenomenon, noncrossing partitions, m-divisible noncrossing partitions, cyclic group. iii.

(7) 1. Introduction. 1.1. Definitions and notations. We will mostly follow the definitions and notations introduced in [1, 10, 11]. We first define set partitions for the classical reflection groups of types A, B and D. An An−1 -partition π is a partition of the set [n] = {1, 2, . . . , n}, and π is said to be noncrossing if i < j < k < l with i, k in the same block B of π and j, l together in a block B ′ of π, then B = B ′ . We use a model to describe a partition π of [n]: placing the numbers 1, 2, . . . , n clockwise around a circle in order, and for every block B of π, drawing the convex hull of {i|i ∈ B}. Hence π is noncrossing if and only if the convex hulls of different blocks of π are disjoint from each other. The set of noncrossing An−1 -partitions will be denoted by NC A (n) = {π : π is a noncrossing An−1 -partition}, and the set of noncrossing An−1 -partitions with k blocks will be denoted by NC A (n, k) = {π : π is a noncrossing An−1 -partition with k blocks}. See Figure 1 for two elements of NC A (n). 11. 12. 11. 1. 10. 2. 9. 6. 2. 9. 4 7. 1. 10 3. 8. 12. 3. 8. 5. 4 7. 6. 5. Figure 1: Two elements of NC A (n) for n = 12 with blocks {2}, {3, 4}, {6, 8}, {7}, {10, 11, 12}, {1, 5, 9} and {1, 3, 4}, {2}, {5, 12}, {6, 11}, {7, 8}, {9, 10}. A Bn -partition is a partition π of [±n] = {1, 2, . . . , n, −1, −2, . . . , −n} into blocks with the property that for any block B of π, its negative −B is also a block of π, and there is at most one block, called the zero block, 1.

(8) containing both +i and −i for some i. We also use the same model as An−1 partitions: placing the numbers 1, 2, . . . , n, −1, −2, . . . , −n clockwise around a circle in order, and for every block B of π, drawing the convex hull of {i|i ∈ B}. Hence π is noncrossing if and only if the convex hulls of different blocks of π are disjoint from each other. The set of noncrossing Bn -partitions will be denoted by NC B (n) = {π : π is a noncrossing Bn -partition}. We define the statistic nz(π) to be k if π has 2k blocks other than the zero block. The set of noncrossing Bn -partitions with nz(π) = k will be denoted by NC B (n, k) = {π : π is a noncrossing Bn -partition with 2k non-zero blocks}. See Figure 2 for two elements of NC B (n). −5. −5. 1. −4. −4. 2. −3. 3. −2. 4 −1. 2. −3. 3. −2. 1. 5. 4 −1. 5. Figure 2: Two elements of NC B (n) for n = 5 with blocks ±{2, 3}, ±{4}, {1, 5, −1, −5} and ±{1, 2, −5}, ±{3, −4}. A Dn -partition is a Bn -partition π with the property that the zero block of π, if present, does not consist of a single pair {i, −i}. We use a model to describe a Dn -partition π: placing the number 1, 2, · · · , n−1, −1, −2, · · · , −(n− 1) clockwise around a circle in this order, and label its centroid with both n and −n. Given a Dn -partition π and a block B of π, let ρ(B) denote the convex hull of the set of points labeled with the elements of B. Two distinct blocks B and B ′ of π are said to cross if ρ(B) and ρ(B ′ ) do not coincide and one of them contains a point of the other in its relative interior. Observe that the case ρ(B) = ρ(B ′ ), which we have allowed, can occur only when B and B ′ are the singletons {n} and {−n}, and that if π has a zero block B, then B and the block containing n cross unless {n, −n} ⊆ B. The set of 2.

(9) noncrossing Dn -partitions will be denoted by NC D (n) = {π : π is a noncrossing Dn -partition}, and the set of noncrossing Dn -partitions with nz(π) = k will be denoted by NC D (n, k) = {π : π is a noncrossing Dn -partition with 2k non-zero blocks}. See Figure 3 for two elements of NC D (n). −5 −6 1 −4 7 −7. 6. −7. 7. 3. −2. 4 −1. 1 2. −3. 3. −2. −6. −4. 2. −3. −5. 5. 4 −1. 6. 5. Figure 3: Two elements of NC D (n) for n = 7 with blocks ±{1, 6}, ±{2, 3}, ±{4, 5}, ±{7} and ±{1, 4, 5, 7}, ±{2, 3}, ±{6}. We follow the definition of the cyclic sieving phenomenon from Reiner, Stanton, and White [12]. Let X be a finite set, with an action of a cyclic group C of order n. Let X(q) be a polynomial in q having nonnegative integer coefficients, with the property that X(1) = |X|. Fix an isomorphism ω of C with the complex n-th roots of unity, that is, an embedding ω : C ֒→ C× . Definition 1.1 [12] A triple (X, X(q), C) as above is said to exhibit the cyclic sieving phenomenon, if for every c ∈ C, [X(q)]q=ω(c) = |{x ∈ X : c(x) = x}|.. (1). This thesis is organized as follows. In Section 2 we count the number of orbits of NC A (n, k) (resp. NC B (n, k) and NC D (n, k)) under the action of the cyclic group G of order n (resp. 2n and 2(n − 1)) and find the natural q-analogue for |NC A (n, k)| (resp. |NC B (n, k)| and |NC D (n, k)|) to exhibit the cyclic sieving phenomenon. In Section 3 we count the number of orbits A B of NC(m) (n, k) (resp. NC(m) (n, k)) under the action of the cyclic group G A of order mn (resp. 2mn) and find the natural q-analogue for |NC(m) (n, k)| B (resp. |NC(m) (n, k)|) to exhibit the cyclic sieving phenomenon. In Section 4 we conclude with a few remarks. 3.

(10) 1.2. Background. Lemma 1.2 (Orbit-counting Lemma) [3] Let G be a permutation group on a set X. For each element g ∈ G, let fix(g) denote the number of points x ∈ X fixed by g. The number of orbits of a permutation group G is equal to the average number of fixed points of its elements, viz. 1 X fix(g). |G| g∈G. (2). Lemma 1.3 [3] The cyclic group of order n contains, for each divisor d of n, φ(d) elements of order d, where φ(d) is the Euler φ function. Each has n/d cycles of length d. Lemma 1.4 We have φ(2d) =. . φ(d) 2φ(d). if d is odd, if d is even.. (3). Proof. Let d = p1 k1 p2 k2 · · · pr kr be odd, where pi is a prime number for 1 ≤ i ≤ r and kj > 0 is an integer for 1 ≤ j ≤ r. Then 1 1 1 φ(2d) = 2d(1 − )(1 − ) . . . (1 − ) 2 p1 pr 1 1 = d(1 − ) . . . (1 − ) = φ(d). p1 pr Let d = p1 k1 p2 k2 · · · pr kr 2kr+1 be even, where pi is a prime number for 1 ≤ i ≤ r and kj > 0 is an integer for 1 ≤ j ≤ r + 1. Then φ(2d) = 2d(1 −. 1 1 1 1 )(1 − ) · · · (1 − )(1 − ) = 2φ(d). p1 p2 pr 2 . Proposition 1.5 [11] Let G be the cyclic group of order n and G act on NC A (n) by rotations. If g ∈ G is of order d > 1, then there is a bijection between {π ∈ NC A (n) : π is fixed by g} and NC B ( nd ). Thus, fix(g) is the number of NC B ( nd ).. 4.

(11) As in Proposition 1.5, we call the block of π central if it is corresponded to the zero block of π ′ ∈ NC B ( nd ). We will use a few constructions from [1, 11]. Setting PnB = {(L, R) : L, R ⊆ [n], |L| = |R|}, Pn−1 ={(L, R, ε) : L, R ⊆ [n − 1], |R| = |L| + 1, ε = S B B ±1}, and PnD = Pn−1 Pn−1 . Note that Pn−1 and Pn−1 are disjoint. B B B Recall that a map τ : Pn → NC (n) is constructed in [11] as follows. Given x = (L, R) ∈ PnB , place a left parenthesis before each occurrence of i and −i in the infinite cyclic sequence . . . , −1, −2, . . . , −n, 1, 2, . . . , n, . . . for i ∈ L and a right parenthesis after each occurrence of i and −i for i ∈ R. To read off a partition of NC B (n), create a block for each of the strings of integers inside the lowest level matching pairs of parentheses, and the integers they enclose and continue similarly with the remaining parenthesization until all parentheses have been removed. The remaining integers, if any, form the zero block of τ B (x). To reverse the bijection, given π ∈ NC B (n), find a nonzero block which forms a contiguous sequence of values in the above infinite cyclic order (there will always be such blocks since π is noncrossing, unless π has only a zero block). For such a contiguous sequence i1 , i2 , . . . , ik , put the absolute values |i1 | in L and |ik | in R. Then remove this block of π and remove this contiguous sequence of values wherever it occurs in the cyclic order. Repeat this process until there is nothing left but the zero block of π. Proposition 1.6 [11] The map τ B is a bijection from the set PnB to NC B (n) in which |L| = |R| is mapped to the statistic nz(π). Therefore |NC B (n, k)| = n 2 and |NC B (n)| = 2n . n k. Lemma 1.7 [10] There is a bijection between X {(L, f ) : L ⊆ [n], |L| = k, f (l) ≤ n} and NC B (n, k), l∈L. where f is a function from L to positive integer. Proof. Given (L, f ) and the infinite cyclic sequence . . . , 1, 2, . . . , n, −1, −2, . . . , −n, 1, 2, . . . We place a left parenthesis before each occurrence of i and −i for each element i in L. We choose an element i∗ ∈ L, if none of the f (i∗ ) − 1 integers on the sequence immediately to its right are in L or −L. We claim that such 5.

(12) elements exist. For i ∈ L, let g(i) − 1 be the number of integers strictly between i and the next element of L or −L to its right. We can see that the sum of g(i) is n. Hence we have f (i) ≤ g(i) for at least one i ∈ L, which means that it is i∗ . Let i∗ and the f (i∗ ) − 1 integers immediately to its right from a block B and let −B be another block. Then we remove from the sequence the blocks B and −B. Continue similarly, until all elements of L are removed. The remaining elements (if any) are placed into the zero block. This creates a π ∈ NC B (n, k). To reverse the bijection, given a π ∈ NC B (n, k) with non-zero blocks B1 , B2 , . . . Bk , −B1 , −B2 , . . . − Bk . We use the same method as above that the inverse of τ B to construct L. Let li ∈ L be the integer that is chosen from Bi (or −Bi ), and define f by mapping li to the size of Bi . Hence we get the (L, f ). Because of the sum of the sizes of B1 , B2 , . . . , Bk must be less P than or equal to n, i.e. l∈L f (l) ≤ n. . Theorem 1.8 [10, 12] The number of π ∈ NC B (n, k) with no zero block is 2 n n−1 n k (4) = k k−1 k n. and the number of π ∈ NC B (n, k) which has a zero block is 2 n n−1 n n−k = . k k k n. (5). Recall that a map τ D : PnD → NC D (n) is constructed in [1] as follows. If B x ∈ Pn−1 , then π is the partition obtained from τ B (x) by adding n and −n to the zero block of τ B (x), if such a block exists, and by adding the singletons B {n} and {−n} to τ B (x) otherwise. Suppose that x is not in Pn−1 , say, x = (L, R, ǫ). We parenthesize the infinite cyclic sequence . . . , −1, −2, . . . , −(n − 1), 1, 2, . . . , n − 1, −1, −2, . . . as in the type B case and form blocks of π with the same procedure, until a right parenthesis remains after each occurrence of i and −i for a unique i ∈ [n−1]. Then let B and −B be blocks of π, where B consists of the integers in {−i − 1, . . . , −n + 1, 1, 2, . . . , i} which have not been removed from the infinite sequence together with n or −n, if ǫ = 1 or ǫ = −1, respectively. Proposition 1.9 [1] The map τ D is a bijection from the set PnD to NC D (n). Moreover, for any x ∈ PnD , the number of pairs {B, −B} of non-zero blocks of τ D (x) is equal to 6.

(13)   |R| |R| + 1  |R|. B if x ∈ Pn−1 and τ B (x) has a zero block, B if x ∈ Pn−1 and τ B (x) has no zero block, if x ∈ Pn−1 .. Therefore |NC D (n, k)|= 2n−2 + 2 2n−2 . n−1 n. 2 k−1 n−1 n−1 2 n−1−k + n−1 +2 n−1 k n−1 k−1 n−1 k−1 k. (6). and |NC D (n)| =. Proposition 1.10 [14] Let m, n, and k be nonnegative integers, and r be an integer. Then m X m n m+n = . k k+r m+r. (7). k=0. We use the following notations and lemmas: 1 − qn . 1−q n Y 1 − qi [n]q ! = [n]! = [n][n − 1] · · · [1] = if n ≥ 1. 1−q i=1  [n]!  if 0 ≤ k ≤ n, n n [n − k]![k]! = =  k q k 0 if k < 0 or k > n.. [n]q = [n] = 1 + q + q 2 + · · · + q n−1 =. Theorem 1.11 (q-Binomial Theorem) [3] For n ≥ 1, n−1 Y. i. (1 + q x) =. i=0. n X k=0. k n k ( ) q 2 x . k. (8). Theorem 1.12 (q-Lucas Theorem) [5, 9] Let m, k, and d be positive integers. Let ω be any primitive d-th root of unity. If m = ad+b and k = rd+s, where 0 ≤ b, s ≤ d − 1, then a b m = . (9) k q=ω r s q=ω The following Lemma is an easy consequence of L’Hospital’s Rule. 7.

(14) Lemma 1.13 Let m1 and m2 be positive integers. Let ω be any primitive d-th root of unity. If m1 ≡ m2 (mod d), then  m 1 if m1 ≡ m2 ≡ 0 (mod d), [m1 ]  m 2 lim = (10) q→ω [m2 ]  1 if m1 ≡ m2 6≡ 0 (mod d).. 2. 2.1. Noncrossing partitions of classical types Noncrossing partitions of type A. Recall that NC A (n, k) is the set of noncrossing partitions of [n] with k blocks. The number of NC A (n, k) is 1 n n A (1) N (n, k) = Nar (An−1 , n − k) = , n k−1 k n n where Nar (1) (An−1 , k) = n1 n−k−1 is the type A Narayana number (see k A [2, 4]). The sum of N (n, k) for 1 ≤ k ≤ n is the type A Catalan number 1 2n A C (n) = . n+1 n Lemma 2.1 [10] Let the cyclic group G = h(1, 2, . . . , n)i of order n act on NC A (n, k) by cyclic rotations. If g ∈ G is of order d > 1. The number of π ∈ NC A (n, k) fixed by g is  n 2 k  d   k   n   d n 2 fix(g) = n−k+1 d   k−1  n   d   0. if d|k, if d|k − 1, otherwise.. Moreover, the number of π ∈ NC A (n) fixed by g is 2n fix(g) = nd . d. 8. (11). (12).

(15) Theorem 2.2 [10] The number of orbits of NC A (n, k) under the action of the cyclic group G = h(1, 2, . . . , n)i of order n is   2 n 2 n X X 1 k n − k + 1 d d N A (n, k) +  , (13) φ(d) + φ(d) k k−1  n n n d d d|(n,k) d>1. d|(n,k−1) d>1. or equivalently  n 2 1X k φ(d) kd + n n d d|(n,k). X. d|(n,k−1). φ(d). . n 2 n d k−1 d. . − k + 1 − N A (n, k). (14) n. Moreover, the number of orbits of NC A (n) is   2n X  1 d C A (n) + , φ(d) n  n d. (15). d|n d>1. or equivalently.   2n X 1 φ(d) nd  − C A (n). n d. (16). d|n. There is a q-analogue of the Narayana number (see [4, 12]), 1 n n A N (n, k; q) = q k(k−1) , [n] k k − 1 and the sum of N A (n, k; q) for 1 ≤ k ≤ n is the q-analogue of the Catalan number 1 2n A C (n; q) = . [n + 1] n Lemma 2.3 [12] If d|n, d ≥ 2, and ω is any primitive d-th root of unity, then  2 n  k  d  if d|k,  k  n   d  2 n N A (n, k; q) q=ω = (17) n−k+1 d  if d|k − 1,  k−1  n  d     0 otherwise. 9.

(16) Moreover, A. C (n; q). . q=ω. 2n = nd .. (18). d. Theorem 2.4 [12] Let X be the set NC A (n, k) and X(q) = N A (n, k; q). Let the cyclic group G = h(1, 2, . . . , n)i of order n act on X by cyclic rotations. Then (X,X(q),G) exhibits the cyclic sieving phenomenon. Moreover, let X be the set NC A (n) and X(q) = C A (n; q). Then (X,X(q),G) exhibits the cyclic sieving phenomenon. Proof. Let ω be any primitive d-th root of unity and g ∈ G be of order d > 1. By Lemma 2.1 and 2.3, we can see that N A (n, k; q) q=ω is equal to the number of π ∈ NC A (n, k) fixed by g. Moreover, C A (n; q) q=ω is equal to the number of π ∈ NC A (n) fixed by g. . 2.2. Noncrossing partitions of type B. The number of NC B (n, k) is 2 n N (n, k) = Nar (Bn , n − k) = , k n where Nar (1) (Bn , k) = nk n−k is the type B Narayana number (see [2, 11]). B The sum of N (n, k) for 0 ≤ k ≤ n is the type B Catalan number 2n B C (n) = . n B. (1). Lemma 2.5 Let the cyclic group G = h(1, 2, . . . , n, −1, −2, . . . , −n)i of order 2n act on NC B (n, k) by cyclic rotations. If g ∈ G is of order d ≥ 1. The number of π ∈ NC B (n, k) fixed by g is  n 2   d  if d is odd and d|k,  k   d   2n 2 fix(g) = (19) d  if d is even and d|2k,  2k   d     0 otherwise. 10.

(17) Moreover, the number of π ∈ NC B (n) fixed by g is  2n  d  if d is odd,   n d fix(g) = 4n   d  if d is even.  2n. (20). d. Proof. In the literature [11], we know that π ∈ NC A (2n) which is invariant under 180◦ rotation is isomorphic to NC B (n) by identifying their circular diagrams in the following way: relabel the numbers 1, 2, . . . , n, n + 1, n + 2, . . . , 2n as 1, 2, . . . , n, −1, −2, . . . , −n. Therefore, we can apply the bijection in Proposition 1.5 for NC B (n). Let g ∈ G be of order d ≥ 1. As d is even, , 2k ). π ∈ NC B (n, k) which is fixed by g is corresponded to π ′ ∈ NC B ( 2n d d B As d is odd, π ∈ NC (n, k) which is fixed by g and invariant under 180◦ rotation simultaneously is corresponded to π ′ ∈ NC B ( nd , kd ). See Figure 4 for examples of order d = 8 and d = 3. Note that 2k must be divided by d. E. G. F. Figure 4: Examples of order d = 8 and d = 3, we can see that E = F and E = G. The number of π ∈ NC B (n, k) fixed by g is |{π ∈ NC B (n, k) : π is fixed by g}|  n 2  n k  B d  if d is odd,   |NC ( d , d )| = k d = 2n 2   2n 2k  B d  if d is even.  |NC ( , )| = 2k d d d. Moreover, the number of π ∈ NC B (n) fixed by g is. 11.

(18) 2n  n  B d    |NC ( d )| = n d fix(g) = 4n  2n  d   |NC B ( )| = 2n d d. if d is odd, if d is even. . Theorem 2.6 The number of orbits of NC B (n, k) under the action of the cyclic group G = h(1, 2, . . . , n, −1, −2, . . . , −n)i of order 2n is n 2 1 X φ(d) kd . n d. (21). d|(n,k). Moreover, the number of orbits of NC B (n) is 2n 1X φ(d) nd . n d. (22). d|n. Proof. By the Orbit-counting Lemma, Lemma 1.3, Lemma 1.4, and Lemma 2.5, the number of orbits of NC B (n, k) is . 1  X φ(d)  2n d|(2n,2k) is odd. =. d. 1  X φ(d)  2n d|(n,k) is odd. =. d. 1  X φ(d)  2n d|(n,k) d is odd. n 2 d k d. n 2 d k d. n 2 d k d. +. X. φ(d). d|(2n,2k) d is even. +. X. φ(2d). d|(n,k). +. X. n 2 1 X = φ(d) kd . n d. d 2k d. n 2. φ(d). d|(n,k) d is odd. 2n 2 d k d. .  .  . n 2 d k d. . +. X. d|(n,k) d is even. 2φ(d). n 2 d k d.   . d|(n,k). Moreover, using the similar method, the number of orbits of NC B (n) is 12.

(19) . 1  X φ(d)  2n d|2n d is odd. 1X φ(d) = n d|n. 2n d n d. +. X. φ(d). d|2n d is even. 4n d 2n d. 2n d . n.   . d. We give a table of the number of orbits of NC (n, k) and NC (n) under rotations for 1 ≤ n ≤ 8 and 0 ≤ k ≤ 8 (see Table 1). They are appeared in OEIS [13, A123610 and A123611]. B. nk 1 2 3 4 5 6 7 8. 0 1 1 1 1 1 1 1 1. 1 2 3 4 5 6 7 8 1 2 1 3 3 1 4 10 4 1 5 20 20 5 1 6 39 68 39 6 1 7 63 175 175 63 7 1 8 100 392 618 392 100 8 1. B. 2 4 8 20 52 160 492 1620. Table 1: The number of orbits of NC B (n, k) and NC B (n) under rotations. We have the following useful lemmas for the q-analogue of type B. Lemma 2.7 Let n be a nonnegative integer, . n 2 X 2n n 2 = q 2k . n q2 k q2 k=0. Proof. Substituting n by 2n in (8), we have. 13. (23).

(20) 2n X k=0. 2n−1 Y k 2n k ( ) q 2 x = (1 + q i x) k i=0 =. n−1 Y. (1 + q i x). i=0. n−1 Y. (1 + q i (q n x)). i=0. ! ! X n j k n n q (2) xj q (2 ) (q n x)k . j k j=0 k=0. n X. =. Comparing coefficients of xn on both sides, we have X n n n−k k 2n n n nk ( ) ( ) ( ) q 2 = q 2 q 2 q n n − k k k=0 n X k n n (n−k = q 2 )+(2)+nk . n−k k k=0. Hence. X n 2 2n n 2 = qk . n k k=0. Substituting q by q 2 in the above equation, we get the formula (23).. . 2 n 2n 2k 2 B Let N (n, k; q) = q and C (n; q) = . k q2 n q2 B. Lemma 2.8 If d|2n and ω is any primitive d-th root of unity, then  n 2   d  if d is odd and d|k,  k   d   2n 2 N B (n, k; q) q=ω = d  if d is even and d|2k,  2k   d     0 otherwise.. (24). Moreover,. B. C (n; q). . q=ω.  2n  d    n d = 4n   d   2n d. 14. if d is odd, (25) if d is even..

(21) Proof. Note that ω 2 is a primitive d-th root of unity if d is odd, and ω 2 is a primitive d2 -th root of unity if d is even. By Theorem 1.12 and Lemma 1.13, we have ! 2 n 2 N B (n, k; q) q=ω = q 2k k q2 q=ω 2 ! n 2 = qk k q=ω 2  n 2 n 2 2  2k d  ω = kd if d is odd and d|k,   k   d d    2n 2  2n 2 2k 2 d d = ω = 2k if d is even and d|2k, 2k   d d    2  2  [n] n − 1 2   q k = 0 otherwise.  lim2 2 q→ω [k] k−1 Moreover,. C B (n; q). . q=ω. =. ! 2n n q2. = q=ω. . 2n n q=ω 2.  2n  d    n d = 4n   d   2n. if d is odd, if d is even.. d. . Theorem 2.9 Let X be the set NC B (n, k) and X(q)=N B (n, k; q). Let the cyclic group G = h(1, 2, . . . , n, −1, −2, . . . , −n)i of order 2n act on X by cyclic rotations. Then (X, X(q), G) exhibits the cyclic sieving phenomenon. Moreover, let X be the set NC B (n) and X(q)=C B (n; q), then (X, X(q), G) exhibits the cyclic sieving phenomenon. Proof. Let ω be any primitive d-th root of unity and g ∈ G be of order d ≥ 1. By Lemma 2.5 and 2.8, we can see that N B (n, k; q) q=ω is equal to the number of π ∈ NC B (n, k) fixed by g. Moreover, C B (n; q) q=ω is equal to the number of π ∈ NC B (n) fixed by g. . 15.

(22) 2.3. Noncrossing partitions of type D. The number of NC D (n, k) is N D (n, k) =N ar(1) (Dn , n − k) 2 2 n−1 n−1−k n−1 k−1 n−1 n−1 + +2 k n−1 k−1 n−1 k−1 k n n−1 n−2 = + k k k−2 n n−2 n−1 = + , k k k−1 =. n−1 n−2 n where Nar (1) (Dn , k) = nk n−k + k n−k is the type D Narayana number D (see [1, 2]). The sum of N (n, k) for 0 ≤ k ≤ n is the type D Catalan number 2n − 1 2n − 2 2n 2n − 2 2n − 2 2n − 2 C (n) = + = − = +2 n n−2 n n−1 n−1 n 3n − 2 2n − 2 . = n n−1 D. Note that 2 2 n−1 n−1−k n−1 k−1 n−1 n−1 + +2 k n−1 k−1 n−1 k−1 k n−1 n−2 n−1 n−2 n−1 n−1 = + +2 k k k−1 k−2 k−1 k n−1 n−2 n−1 n−1 n−2 n−1 = + + + k k k−1 k−1 k−2 k n−1 n−2 n−2 n−2 n−1 n−2 n−1 = + + + + k k k−1 k−2 k−1 k−2 k n−1 n−1 n−2 n−1 = + + k k−1 k−2 k n n−2 n−1 = + . k k−2 k. |N C D (n, k)| =. Lemma 2.10 Let the cyclic group G = h(1, 2, . . . , n − 1, −1, −2, . . . , −(n − 1))(n, −n)i of order 2n − 2 act on NC D (n, k) by cyclic rotations. If g ∈ G is of order d ≥ 1. The number of π ∈ NC D (n, k) fixed by g is. 16.

(23)  2 2 n−1 n−1−k n−1 k−1    +   k n−1 k−1 n−1           2 2    n−1 k−1 n−1 n−1−k   +   n−1 k−1 n−1 k      n−1 2   n−1−k  d   k  n−1 d fix(g) = n−1 2   k−1  d   k−1  n−1   d    2n−2 2   n−1−k  d   2k  n−1   d    2 2n−2   k−1  d   2k−2  n−1   d    0. n−1 n−1 +2 if d = 1 k−1 k or (d = 2 and n is even), if d = 2 and n is odd, if d > 2 is odd and d|k, if d > 2 is odd and d|k − 1, if d > 2 is even and d|2k, if d > 2 is even and d|2k − 2, otherwise. (26). Moreover, the number of π ∈ NC D (n) fixed by g is. fix(g) =.  2n − 2 2n − 2   +2   n − 1 n      2n−2         . if d = 1 or (d = 2 and n is even),. d n−1 d 4n−4 d 2n−2 d. if d > 2 is odd, if d > 2 is even or (d = 2 and n is odd). (27). Proof. We use the similar method as in Lemma 2.5. By Proposition 1.9, we separate all π ∈ NC D (n, k) into three cases as following. (i) π ∈ NC D (n, k) with the singletons {n} and {−n} and has no zero block; (ii) π ∈ NC D (n, k) which has a zero block; (iii) π ∈ NC D (n, k) with one of B, −B contains n and the other contains −n, and B 6= {n} or {−n}. 17.

(24) Let g ∈ G be of order d ≥ 1. Case (i). There is a bijection that maps π to π ′ ∈ NC B (n − 1, k − 1) with no zero block by removing the singletons {n} and {−n}. In this way, we can use Proposition 1.5 to count the number of π ′ ∈ NC B (n − 1, k − 1) with no zero block and fixed by g. Then |{π ∈ NC D (n, k) : π is fixed by g and has no zero block}| =|{π ′ ∈ NC B (n − 1, k − 1) : π ′ is fixed by g and has no zero block}|  ′′ B n−1 k−1 ′′  if d is odd and d|k − 1,  |{π ∈ NC ( d , d ) : π has no zero block}| = ′′ B 2n−2 2k−2 ′′   |{π ∈ NC ( d , d ) : π has no zero block}| if d is even and d|2k − 2,  n−1 2  k−1  d    k−1 n − 1 d = 2n−2 2   k−1  d   2k−2 n−1 d. if d is odd and d|k − 1,. if d is even and d|2k − 2.. Case (ii). There is a bijection that maps π to π ′ ∈ NC B (n−1, k) which has a zero block by removing {n} and {−n}. In this way, we can use Proposition 1.5 to count the number of π ′ ∈ NC B (n − 1, k) which has a zero block and fixed by g. Then |{π ∈ NC D (n, k) : π is fixed by g and has a zero block}| =|{π ′ ∈ NC B (n − 1, k) : π ′ is fixed by g and has a zero block}|  ′′ B n−1 k ′′  if d is odd and d|k,  |{π ∈ NC ( d , d ) : π has a zero block}| = ′′ B 2n−2 2k ′′  if d is even and d|2k,  |{π ∈ NC ( d , d ) : π has a zero block}|  n−1 2  n−1−k  d  if d is odd and d|k,  k  n−1 d = 2n−2 2   n−1−k  d  if d is even and d|2k.  2k n − 1 d. Case (iii). Let π ∈ NC D (n, k) in case (iii) and fixed by g. Since there are just two blocks containing the centroid of π, d must be 1 or 2. The n−1 number of π which is fixed by the identity is 2 n−1 . For d = 2, k−1 k let B = {x1 , x2 , . . . , n} be the block of π containing n. Then B becomes 18.

(25) hgi. d=1 1. NC D (3) −2. 1 −2 −3 3. −1. d=4 g 1 −2. 1 −2 3 −3. −3 3 2 −1. d=2 g2. 2 −1. d=4 g3 1 −2 −3. 3 −3 2 −1. 1 3. 2 −1. Figure 5: An example for NC D (3) with blocks ±{1, 2, 3}. hgi=h(1, 2, −1, −2)(3, −3)i act on NC D (3) by rotations.. 2 Let. {−x1 , −x2 , . . . , (−1)n−1 n} after the action of g. Also, {−x1 , −x2 , . . . , (−1)n−1 n} = −B only when n is even. See Figure 5 for NC D (3). Hence  n−1 n−1   2 if n is even, k−1 k fix(g) =   0 if n is odd. This establishes the formula (26). Moveover, we separate all π ∈ NC D (n) into two cases as following. (i′ ). π ∈ NC D (n) with the singletons {n} and {−n} and has no zero block, or π ∈ NC D (n) which has a zero block; (ii′ ) π ∈ NC D (n) with one of B, −B contains n and the other contains −n, and B 6= {n} or {−n}. Case (i′ ). There is a bijection that maps π to π ′ ∈ NC B (n − 1) by removing {n} and {−n}. In this way, we can use Proposition 1.5 to count the number of π ′ ∈ NC B (n − 1) fixed by g. Then |{π ∈ NC D (n) in case (i′ ) : π is fixed by g}| =|{π ′ ∈ NC B (n − 1) : π ′ is fixed by g}| 2n−2  n−1  B d  |NC ( )| = n−1 if d is odd,   d d = 4n−4  2n − 2  B d  )| = 2n−2 if d is even.  |NC ( d d 19.

(26) Case (ii′ ). As in case (iii), the number of π in case (ii′ ) and fixed by g is  2n − 2   2 if d = 1 or (d = 2 and n is even), n fix(g) =   0 otherwise.. This establishes the formula (27).. . Theorem 2.11 The number of orbits of NC D (n, k) under the action of the cyclic group G = h(1, 2, . . . , n − 1, −1, −2, . . . , −(n − 1))(n, −n)i of order 2n − 2 is . 1  D n−1 n−1 N (n, k) + (χe (n) − 1) n−1 k−1 k +. X. φ(d). d|(n−1,k) d>1. n−1 2 d k d. n−1−k + n−1. X. d|(n−1,k−1) d>1. φ(d). n−1 2 d k−1 d. . k − 1 , n−1. (28). or equivalently  1  n−1. X. d|(n−1,k). n−1 2 n−1−k φ(d) kd + n−1 d. X. d|(n−1,k−1).  n−1 2 k − 1 d  + (1 + χe (n))N A (n − 1, k). φ(d) k−1 n−1 d (29). Moreover, the number of orbits of NC D (n) is   X 2n−2  1  d C D (n) + (χe (n) − 1) 2n − 2 + , φ(d) n−1   n−1 n d. (30). d|n−1 d>1. or equivalently. . 1  n−1. X. d|n−1.  2n−2  + (1 + χe (n))C A (n − 1), φ(d) d n−1 d. where χe (n) is 1 if n is even and 0 if n is odd. 20. (31).

(27) Proof. We use the similar method as in Theorem 2.6. By the Orbitcounting Lemma and Lemma 2.10, the number of orbits of NC D (n, k) is . 2 2 ! n−1 n−1−k n−1 k−1 n−1 n−1 + +2 k n−1 k−1 n−1 k−1 k. 1   φ(1) 2n − 2. 2 ! 2 n−1 n−1−k n−1 k−1 n−1 n−1 + + 2χe (n) + φ(2) n−1 k−1 n−1 k−1 k k X. +. d|(2n−2,2k) d>1 is odd. X. +. n−1 2 n−1−k φ(d) kd + n−1 d. d|(2n−2,2k−2) d>1 is odd. . 1  = 2 2n − 2 +2. n−1 2 k−1 d + φ(d) k−1 n −1 d. X. d|(2n−2,2k) d>2 is even. X. d|(2n−2,2k−2) d>2 is even. 2n−2 2 n−1−k d φ(d) 2k n−1 d  2n−2 2 k − 1 d φ(d) 2k−2  n −1 d. ! 2 2 n−1 n−1−k n−1 k−1 n−1 n−1 + + 2(1 + χe (n)) k n−1 k−1 n−1 k−1 k. X. d|(n−1,k) d>1. n−1 2 n−1−k φ(d) kd +2 n−1 d. X. d|(n−1,k−1) d>1.  n−1 2 k − 1 d φ(d) k−1  n −1 d. . 1  D n−1 n−1 =  N (n, k) + (χe (n) − 1) n−1 k−1 k +. X. n−1 2 n−1−k φ(d) kd + n−1 d. X. n−1 2 n−1−k φ(d) kd + n−1 d. d|(n−1,k) d>1.  1  = n−1. d|(n−1,k). X.  n−1 2 k − 1 d φ(d) k−1  n −1 d. X.  n−1 2 k − 1 d  + (1 + χe (n))N A (n − 1, k). φ(d) k−1 n − 1 d. d|(n−1,k−1) d>1. d|(n−1,k−1). Moreover, the number of orbits of NC D (n) is. 21.

(28) .  X 2n−2  1  d  C D (n) + (χe (n) − 1) 2n − 2 + φ(d) n−1  n−1 n d. d|n−1 d>1. =. 1 n−1. .  2n−2 X  φ(d) d  + (1 + χe (n))C A (n − 1). d|n−1. n−1 d. We give a table of the number of orbits of NC (n, k) and NC (n) under rotations for 2 ≤ n ≤ 8 and 0 ≤ k ≤ 8 (see Table 2). They are not appeared in [13]. D. nk 2 3 4 5 6 7 8. 0 1 1 1 1 1 1 1. 1 2 2 4 4 6 6 8. 2 3 4 5 6 7 1 2 1 8 4 1 12 12 4 1 33 56 33 6 1 42 97 97 42 6 1 88 328 500 328 88 8. D. 8. 1. 4 6 18 34 136 292 1350. Table 2: The number of orbits of NC D (n, k) and NC D (n) under rotations.. Lemma 2.12 Let n be a positive integer, ! 2 2 n X 2 2 [n − 1 − k] [k − 1] n−1 n−1 2n − 2 q 2(k−1)2 q 2k(k+1) q + q = . 2 2 k [n − 1] k − 1 [n − 1] n − 1 q q 2 2 2 q q q k=0 (32). Proof. First,. 22.

(29) ! 2 2 n X n − 1 [n − 1 − k] k(k+1) n − 1 [k − 1] (k−1)2 q + q k [n − 1] k − 1 [n − 1] k=0 2 2 n n X n − 1 [n − 1 − k] k(k+1) X n − 1 [k − 1] (k−1)2 = q + q k k − 1 [n − 1] [n − 1] k=0 k=0 2 2 n−1 n−1 X X [k] k2 n−1 n − 1 [n − 1 − k] k(k+1) = q + q k k [n − 1] [n − 1] k=0 k=0 2 2 n−1 k X n−1 q = (q k [n − 1 − k] + [k]) k [n − 1] k=0 2 n−1 X n−1 2n − 2 k2 = q = . k n−1 k=0. Substituting q by q 2 in the above equation, we get (32).. . Let N D (n, k; q)= . n−1 k. 2. q2. 2 [k − 1]q2 2(k−1)2 [n − 1 − k]q2 2k(k+1) n−1 n−1 n−1 q + q + (1 + q n ) q 2k(k−1) k − 1 q2 [n − 1]q2 k − 1 q2 k q2 [n − 1]q2 (33). 2n − 2 2n − 2 n and C (n; q) = + (1 + q ) . n − 1 q2 n 2 q D. . Lemma 2.13 If d|2n − 2 and ω is any primitive d-th root of unity, then. 23.

(30) N D (n, k; q) q=ω  2 2 n−1 n−1−k n−1 k−1 n−1 n−1    + + 2 if d = 1   k n−1 k−1 n−1 k−1 k       or (d = 2 and n is even),       2 2    n−1 n−1−k n−1 k−1   + if d = 2 and n is odd,   k n−1 k−1 n−1     n−1 2    n−1−k  d  if d > 2 is odd and d|k,  k n−1 d = n−1 2    k−1  d  k−1 if d > 2 is odd and d|k − 1,   n−1  d    2n−2 2    n−1−k  d  if d > 2 is even and d|2k,  2k  n−1   d   2n−2 2    k−1  d  if d > 2 is even and d|2k − 2,  2k−2  n−1   d    0 otherwise. (34) Moreover,. C D (n; q). . q=ω.  2n − 2 2n − 2   +2 if d = 1 or (d = 2 and n is even),   n−1 n      2n−2 d if d > 2 is odd, = n−1   d   4n−4    d   2n−2 if d > 2 is even or (d = 2 and n is odd). d. (35). Proof. We separate (33) into three parts and use the similar method as in Lemma 2.8 to calculate. Part 1.. 24.

(31) . n−1 k. 2. q2. [n − 1 − k]q2 2k(k+1) q [n − 1]q2. !. Part 2.. . n−1 k−1. 2. q2. [k − 1]q2 2(k−1)2 q [n − 1]q2. !. q=ω. Part 3.. q=ω.  n−1 2  n−1−k  d  if d is odd and d|k,  k  n−1   d  2n−2 2 = n−1−k d  if d is even and d|2k,  2k  n−1   d    0 if d ∤ 2k..  n−1 2  k−1  d  if d is odd and d|k − 1,  k−1  n−1   d  2n−2 2 = k−1 d  if d is even and d|2k − 2,  2k−2  n−1   d    0 if d ∤ 2k − 2.. ! n − 1 n − 1 (1 + q n ) q 2k(k−1) k − 1 q2 k q2 q=ω ! 2 [n − 1] 2 n − 2 n − 2 q = (1 + q n ) q 2k(k−1) [k − 1]q2 [k]q2 k − 2 q2 k − 1 q2 q=ω  n−1 n−1   2 if d = 1 or (d = 2 and n is even), k−1 k =   0 otherwise.. This establishes the formula (34). Moreover, ! ! 2n − 2 2n − 2 C D (n; q) q=ω = + (1 + q n ) n − 1 q2 n q2 q=ω. By Lemma 2.8, we have . 2n − 2 n−1. ! q2. q=ω.  2n−2  d  if d is odd,   n−1 d = 4n−4   d  if d is even.  2n−2 d. 25. . q=ω.

(32) By Theorem 1.12 and Lemma 1.13, we have  2n − 2   2 if d = 1 or (d = 2 and n is even),   ! n  2n − 2 (1 + q n ) = !  n 2 [2n − 2] 2n − 3 q  n q 2 q=ω  = 0 otherwise.   (1 + q ) [n]q2 n − 1 q2 q=ω. This establishes the formula (35).. . Theorem 2.14 Let X be the set NC D (n, k) and X(q)=N D (n, k; q). Let the cyclic group G = h(1, 2, . . . , n − 1, −1, −2, . . . , −(n − 1))(n, −n)i of order 2n − 2 act on X by cyclic rotations. Then (X, X(q), G) exhibits the cyclic sieving phenomenon. Moreover, let X be the set NC D (n) and X(q)=C D (n; q), then (X, X(q), G) exhibits the cyclic sieving phenomenon. Proof. Let ω be any primitive d-th root of unity and g ∈ G be of order d ≥ 1. By Lemma 2.10 and 2.13, we can see that N D (n, k; q) q=ω is equal to the number of π ∈ NC D (n, k) fixed by g. Moreover, C D (n; q) q=ω is equal to the number of π ∈ NC D (n) fixed by g. . 3 3.1. m-Divisible noncrossing partitions of classical types m-Divisible noncrossing partitions of type A. Let m be a positive integer and we say that π ∈ NC A (mn) is a m-divisible noncrossing partition of type A if the size of each block of π is divisible by A m. Let NC(m) (n) denote the subset of NC A (mn) consisting of m-divisible A noncrossing partitions. The set of π ∈ NC(m) (n) with k blocks will be A A denoted by NC(m) (n, k). See Figure 6 for two elements of NC(2) (n). Now we follow an idea from [8, 10] for using the block size to describe a block. We define a function f from L to m-divisible positive integers to A construct a π ∈ NC(m) (n, k) with a marked block B ∗ . Lemma 3.1 There is a bijection between {(L, f ) : L ⊆ [mn], |L| = k − 1,. X. f (l) < mn}. l∈L m|f (l) A A and ∗ NC(m) (n, k) = {(π, B ∗ ) : π ∈ NC(m) (n, k) and B ∗ is a block of π}.. 26.

(33) 11. 12. 11. 1. 10. 2. 9. 6. 2. 9. 4 7. 1. 10 3. 8. 12. 3. 8. 5. 4 7. 6. 5. A Figure 6: Two elements of NC(2) (n) for n = 6 with blocks {2, 3}, {5, 6}, {8, 9}, {11, 12}, {1, 4, 7, 10} and {1, 4}, {2, 3}, {5, 6, 7, 8}, {9, 10}, {11, 12}.. Proof. Given (L, f ) and the infinite cyclic sequence . . . , 1, 2, . . . , mn − 1, mn, . . . We place a left parenthesis before each occurrence of l for each element l in L. We choose an element l∗ ∈ L, if none of the f (l∗ ) − 1 integers on the sequence immediately to its right are in L. We claim that such elements exist. For l ∈ L, let g(l) − 1 be the number of integers strictly between l and the next element of L to its right. We can see that the sum of g(l) is mn. Hence we have f (l) ≤ g(l) for at least one l ∈ L, which means that it is l∗ . Let l∗ and the f (l∗ ) − 1 integers immediately to its right from a block B. Then we remove from the sequence the block B. Continue similarly, until all elements of L are removed. The remaining elements are placed into the A block B ∗ . This creates a (π, B ∗ ) ∈ ∗ NC(m) (n, k). ∗ A To reverse the bijection, given (π, B ) ∈ ∗ NC(m) (n, k). Let B1 , B2 , . . ., ∗ Bk−1 , B be the blocks of π and s be the minimum element of B ∗ . For 1 ≤ i ≤ k − 1, choose li as the first element of Bi when we traverse the cycle from s clockwise. Let L = {l1 , l2 , . . . , lk−1 }, and define f by mapping li to the size of Bi . Because of the sum of the sizes of B1 , B2 , . . . , Bk−1 must be P less than mn, i.e. l∈L f (l) < mn. Hence we get the (L, f ). m|f (l). A Theorem 3.2 The number of NC(m) (n, k) is. 1 mn n = Nar (An−1 , n − k) = , (36) n k−1 k n mn where Nar (m) (An−1 , k) = n1 n−k−1 is the type A Fuss-Narayana number k A (see [2]). Moreover, the number of NC(m) (n) is the type A Fuss-Catalan A N(m) (n, k). (m). 27.

(34) number A C(m) (n). (m + 1)n 1 = . mn + 1 n. (37). mn Proof. Let L and f be as in Lemma 3.1. Since there are k−1 ways to n−1 choose L and for a fixed L there are k−1 choices for f . Hence |(L, f )| = n−1 mn A A . And |∗ NC(m) (n, k)| = k|NC(m) (n, k)|, we have k−1 k−1 A |NC(m) (n, k)|. 1 mn n−1 1 mn n = = . k k−1 k−1 n k−1 k. Moreover, substituting m by n, n by mn, and r by −1 in (7), we obtain A |NC(m) (n)|. =. n X. A |NC(m) (n, k)|. k=1. n X 1 mn n = n k−1 k k=1 1 (m + 1)n 1 (m + 1)n = = . n n−1 mn + 1 n We give a table of the number of and for 1 ≤ n ≤ 8 and 1 ≤ k ≤ 8 (see Table 3). They are appeared in OEIS [13, A120986 and A001764]. A NC(2) (n, k). nk 1 2 3 4 5 6 7 8. 1 1 1 1 1 1 1 1 1. 2. 3. 4. 5. 6. A NC(2) (n). 7. 8. 2 6 5 12 28 14 20 90 120 42 30 220 550 495 132 42 455 1820 3003 2002 429 56 840 4900 12740 15288 8008 1430. A |NC(2) (n)| 1 3 12 55 273 1428 7752 43263. A A Table 3: The number of NC(2) (n, k) and NC(2) (n).. 28.

(35) Let G be a cyclic group of order mn, and g ∈ G. In order to count A the number of π ∈ NC(m) (n, k) which are fixed by g, we need to define the following sets. Let N be a positive integer, and nz B(m) (N, k) = {π ∈ NC B (N, k) : π has no zero block, and the size of. each block of π is divisible by m}, z B(m) (N, k) = {π ∈ NC B (N, k) : π has a zero block, and the size of each. non-zero block of π is divisible by m}, B(m) (N, k) = {π ∈ NC B (N, k) : the size of each non-zero block of π is divisible by m}, and B(m) (N) = {π ∈ NC B (N) : the size of each non-zero block of π is divisible by m}. nz z For counting the number of B(m) (N, k) and B(m) (N, k), we have the following useful lemma. nz Lemma 3.3 The number of B(m) (N, k) is  N −1  m  N if m|N, k−1 k   0 if m ∤ N, z and the number of B(m) (N, k) is  N N −1  m    k k  N ⌊N ⌋   m  k k. (38). if m|N, (39) if m ∤ N.. Moreover, the number of B(m) (N, k) is N N ⌊m⌋ , k k. (40). and the number of B(m) (N) is (m+1)N ⌊ m ⌋ . N ⌊m ⌋ 29. (41).

(36) Proof. By Lemma 1.7, there is a bijection between X {(L, f ) : L ⊆ [N], |L| = k, f (l) = N} l∈L m|f (l). P nz and B(m) (N, k). Note that N must be divided by m. Also, l∈L f (l) = N m|f (l) P is equivalent to l∈L f ′ (l) = N/m, where f (l) = mf ′ (l). Hence nz |B(m) (N, k)|. N N −1 m = |(L, f )| = . k k−1. Again, by Lemma 1.7, there is a bijection between X {(L, f ) : L ⊆ [N], |L| = k, f (l) < N} l∈L m|f (l). z and B(m) (N, k). Since. Hence. P. l∈L m|f (l). f (l) < N is equivalent to.  P N  l∈L f ′ (l) < m  P f ′ (l) ≤ ⌊ N ⌋ l∈L m. if m|N, if m ∤ N..  N N −1  m    k k z |B(m) (N, k)| = |(L, f )| =  N ⌊N ⌋   m  k k. if m|N, if m ∤ N.. Moreover, the number of B(m) (N, k) is  N N N N − 1 N − 1 N  m m m  + = if m|N,   k k−1 k k k k N  N ⌊m⌋    if m ∤ N, k k N N ⌊m⌋ = , k k and the number of B(m) (N) is 30.

(37) N. |B(m) (N)| =. ⌊m⌋ X. |B(m) (N, k)|. k=0 N. =. ⌊ m ⌋ X N ⌊N ⌋ m. k=0. k. k. =. . ⌋ ⌊ (m+1)N m . N ⌊m ⌋ . Lemma 3.4 Let the cyclic group G = h(1, 2, . . . , mn)i of order mn act on A NC(m) (n, k) by cyclic rotations. If g ∈ G is of order d > 1. The number of A π ∈ NC(m) (n, k) fixed by g is  mn n −1 d d   if d|n and d|k,  k k  − 1  d d   mn n   −1   d d  if d|n and d|k − 1,  k−1 k−1 d d fix(g) = (42) mn n   ⌊ ⌋  d d  if d ∤ n and d|k − 1,  k−1 k−1    d d     otherwise.  0. A Moreover, the number of π ∈ NC(m) (n) fixed by g is. (m+1)n ⌊ d ⌋ fix(g) = . ⌊ nd ⌋. (43). Proof. We use the same method as in Lemma 2.1. We separate all π ∈ A NC(m) (n, k) which are fixed by g into two cases. Let g ∈ G be of order d > 1. Case 1. π has no central block. Then k must be divided by d and

(38)

(39) A

(40) {π ∈ NC(m) (n, k) : π is fixed by g and has no central block}

(41)

(42)

(43)

(44) nz mn k

(45) =

(46)

(47) B(m) ( , )

(48) d d

(49)  mn n −1  d d  if d|n, k k −1 = d d   0 if d ∤ n. 31.

(50) Case 2. π has a central block. Then k − 1 must be divided by d and

(51)

(52)

(53) {π ∈ NC A (n, k) : π is fixed by g and has a central block}

(54) (m)

(55)

(56)

(57) z mn k − 1

(58) =

(59)

(60) B(m) ( , )

(61) d d

(62)  mn n −1  d d  if d|n,  k−1  k−1 =. d. d. mn n  ⌊d⌋  d   k−1 k−1 d. if d ∤ n.. d. A Moreover, the number of π ∈ NC(m) (n) fixed by g is. mn )| = fix(g) = |B(m) ( d. . ⌊ (m+1)n ⌋ d . ⌊ nd ⌋ . A Theorem 3.5 The number of orbits of NC(m) (n, k) under the action of the cyclic group G = h(1, 2, . . . , mn)i of order mn is   mn n X X −1 mn k − 1  1  z d d N A (n, k) + φ(d) + φ(d)|B ( , )| (m) (m) k k , mn  d d − 1 d d d|(mn,k−1) d|(mn,k) d>1. d>1. (44) or equivalently   mn n X X − 1 1  mn k − 1  z A d φ(d) kd + φ(d)|B(m) ( , )| − N(m) (n, k), k mn d d −1 d d d|(mn,k) d|(mn,k−1) (45). z where B(m) (N, k) is as in Lemma 3.3. Moreover, the number of orbits of A NC(m) (n) is   (m+1)n X 1  ⌊ d ⌋  A C(m) , (n) + φ(d) (46)   mn ⌊ nd ⌋ d|mn d>1. 32.

(63) or equivalently . 1  mn. X. d|mn.  (m+1)n ⌊ d ⌋  A φ(d) − C(m) (n). n ⌊d⌋. (47). A Proof. The number of π ∈ NC(m) (n, k) which is fixed by the identity is A N(m) (n, k). By the Orbit-counting Lemma and Lemma 3.4, the number of A orbits of NC(m) (n, k) is. . X 1  A N(m) (n, k) + φ(d) mn  d|(mn,k) d>1. =. . 1  X φ(d) mn d|(mn,k). mn n d k d. d k d. mn n d k d. d k d. . X −1 mn k − 1  z + φ(d)|B(m) ( , )|  d d −1 d|(mn,k−1) . d>1. . X −1 mn k − 1  z A , )| − N(m) (n, k). + φ(d)|B(m) ( d d −1 d|(mn,k−1) . A Moreover, the number of orbits of NC(m) (n) is. .  (m+1)n X ⌊ d ⌋  1  C A (n) +  φ(d) (m)   mn ⌊ nd ⌋ d|mn d>1. =. . 1  mn. X. d|mn.  (m+1)n ⌊ d ⌋  A φ(d) − C(m) (n). ⌊ nd ⌋. We give a table of the number of orbits of and under rotations for 1 ≤ n ≤ 8 and 1 ≤ k ≤ 8 (see Table 4). They are not appeared in OEIS [13]. A NC(2) (n, k). A NC(2) (n). Lemma 3.6 Let n be a nonnegative integer, X n 1 1 (m + 1)n mn n k(k−1) = q . n [mn + 1] [n] k − 1 k k=0 Proof. Substituting n by (m + 1)n in (8), we have 33. (48).

(64) nk 1 2 3 4 5 6 7 8. 1 1 1 1 1 1 1 1 1. 2. 3. 1 1 2 2 3 3 4. 4. 5. 6. 7. 8. 2 4 3 10 12 6 20 49 43 14 34 130 219 143 34 54 312 802 966 504 95. 1 2 4 10 31 130 564 2738. A A Table 4: The number of orbits of NC(2) (n, k) and NC(2) (n) under rotations.. (m+1)n. X k=0. k (m + 1)n ( ) q 2 xk k. (m+1)n−1. = = =. Y. i=0 mn Y. (1 + q i x) i. (1 + q x). i=0 mn+1 X i=0. n−2 Y. (1 + q i (xq mn+1 )). i=0. ! ! X n−1 k mn + 1 n − 1 q( ) xi q (2 ) (xq mn+1 )k . i k k=0 i 2. . Comparing coefficients of xmn on both sides, we have X n mn mn−k+1 k−1 (m + 1)n mn + 1 n − 1 ( ) ( ) ( ) 2 q 2 = q q 2 q (mn+1)(k−1) mn mn − k + 1 k−1 k=0 n X mn + 1 n − 1 (mn−k+1 +(mn+1)(k−1) )+(k−1 2 2 ) = q . k k−1 k=0. Hence. X n 1 1 (m + 1)n mn + 1 n − 1 k(k−1) = q [mn + 1] n [mn + 1] k k−1 k=0 n X 1 mn n k(k−1) = q . [n] k − 1 k k=0. 34.

(65) Let A N(m) (n, k; q). and A C(m) (n; q). 1 mn n k(k−1) = q [n] k − 1 k. 1 (m + 1)n = . [mn + 1] n. Lemma 3.7 If d|mn, d ≥ 2, and ω is any primitive d-th root of unity, then  mn n −1  d d  if d|n and d|k,  k k  −1   d d  mn n    −1  d d  if d|n and d|k − 1, k−1 k−1 A (49) N(m) (n, k; q) q=ω = d d  mn n   ⌊d⌋  d  if d ∤ n and d|k − 1,  k−1 k−1    d d    0 otherwise.. Moreover,. A C(m) (n; q) q=ω. . (m+1)n ⌊ d ⌋ = . ⌊ nd ⌋. (50). Proof. For d|k, using Theorem 1.12 and Lemma 1.13, we have [k] mn n k(k−1) A N(m) (n, k; q) q=ω = q k [mn − k + 1][n] k q=ω  mn n mn n −1 k d  d d d  = k k k n kd −1 = d d d   0. if d|n, if d ∤ n.. For d | k − 1, we have [n − k + 1] mn n A k(k−1) N(m) (n, k; q) q=ω = q k−1 k−1 [n][k] q=ω  n mn n −1 n − k + 1 mn  d d d d   = k−1  k−1 k−1 k−1  n d d d d = mn n mn n n  ⌊d⌋ ⌊d⌋ n − d⌊ d ⌋  d d  =  k−1 k−1 k−1 k−1  0 d d d d q=ω 35. if d|n, if d ∤ n..

(66) For d ∤ k and d ∤ k − 1, we have [mn] mn − 1 n − 1 k(k−1) A N(m) (n, k; q) q=ω = q = 0. k−1 [k][k − 1] k − 2 q=ω. Moreover,. A C(m) (n; q) q=ω. . =. . (m+1)n ⌊ d ⌋ 1 (m + 1)n = . [mn + 1] n ⌊ nd ⌋ q=ω . A A Theorem 3.8 Let X be the set NC(m) (n, k) and X(q)=N(m) (n, k; q). Let the cyclic group G = h(1, 2, . . . , mn)i of order mn act on X by cyclic rotations. Then (X, X(q), G) exhibits the cyclic sieving phenomenon. Moreover, let X A A be the set NC(m) (n) and X(q)=C(m) (n; q). Then (X, X(q), G) exhibits the cyclic sieving phenomenon.. Proof. Let ω be any primitive d-th root of unity and g ∈ G be of order A d > 1. By Lemma 3.4 and 3.7, we can see that N(m) (n, k; q) is equal q=ω A A to the number of π ∈ NC(m) (n, k) fixed by g. Moreover, C(m) (n; q) is q=ω. A equal to the number of π ∈ NC(m) (n) fixed by g.. 3.2. . m-Divisible noncrossing partitions of type B. Let m be a positive integer and we say that π ∈ NC B (mn) is a m-divisible noncrossing partition of type B if the size of each block of π is divisible by m. B Let NC(m) (n) denote the subset of NC B (mn) consisting of m-divisible nonB crossing partitions. The set of π ∈ NC(m) (n) with nz(π)=k will be denoted B B by NC(m) (n, k). See Figure 7 for two elements of NC(2) (n). B Theorem 3.9 The number of NC(m) (n, k) is B N(m) (n, k). = Nar. (m). (Bn , n − k) =. . mn n , k k. (51). mn where Nar (m) (Bn , k) = nk n−k is the type B Fuss-Narayana number. MoreB over, the number of NC(m) (n) is the type B Fuss-Catalan number (m + 1)n B C(m) (n) = . (52) n 36.

(67) −5. −6. −5. 1. −4. 6. 3. −2. 4 −1. 2. −3. 3. −2. 1. −4. 2. −3. −6. 5. 4 −1. 6. 5. B Figure 7: Two elements of NC(2) (n) for n = 3 with blocks ±{3, 4}, ±{5, 6}, {1, 2, −1, −2} and ±{1, 4}, ±{2, 3}, ±{5, 6}.. B B Proof. We use the facts that NC(m) (n, k) = B(m) (mn, k) and NC(m) (n) = B(m) (mn), then apply Lemma 3.3 to get the result. B B We give a table of the number of NC(2) (n, k) and NC(2) (n) for 1 ≤ n ≤ 6 and 0 ≤ k ≤ 6 (see Table 5). They are appeared in OEIS [13, A110608 and A005809].. nk 1 2 3 4 5 6. 0 1 1 1 1 1 1. B 1 2 3 4 5 6 |NC(2) (n)| 2 3 8 6 15 18 45 20 84 32 168 224 70 495 50 450 1200 1050 252 3003 72 990 4400 7425 4752 924 18564. B B Table 5: The number of NC(2) (n, k) and NC(2) (n).. Lemma 3.10 Let the cyclic group G = h(1, 2, . . . , mn, −1, −2, . . . , −mn)i B of order 2mn act on NC(m) (n, k) by cyclic rotations. If g ∈ G is of order B d ≥ 1. The number of π ∈ NC(m) (n, k) fixed by g is. 37.

(68)  mn n ⌊d⌋  d   k k     d d 2mn ⌊ 2n ⌋ fix(g) = d d   2k 2k   d d    0. if d is odd and d|k, if d is even and d|2k,. (53). otherwise.. B Moreover, the number of π ∈ NC(m) (n) fixed by g is.  (m+1)n  ⌊ d ⌋     ⌊ nd ⌋ fix(g) = 2(m+1)n   ⌊ d ⌋    ⌊ 2n ⌋ d. if d is odd, (54) if d is even.. Proof. We use the similar method as in Lemma 2.5. Let g ∈ G be of order d ≥ 1, B |{π ∈ NC(m) (n, k) : π is fixed by g }| mn n  ⌊d⌋ mn k  d  |B ( , )| = if d is odd, (m)  k k  d d d d = 2mn 2n  ⌊d⌋ 2mn 2k  d  , )| = 2k if d is even.  |B(m) ( 2k d d d d. B Note that 2k must be divided by d. Moreover, the number of π ∈ NC(m) (n) fixed by g is.  (m+1)n  mn ⌊ d ⌋     |B(m) ( d )| = ⌊ nd ⌋ fix(g) = 2(m+1)n   ⌊ d ⌋ 2mn   )| =  |B(m) ( d ⌊ 2n ⌋ d. if d is odd, if d is even. . B Theorem 3.11 The number of orbits of NC(m) (n, k) under the action of the cyclic group G = h(1, 2, . . . , mn, −1, −2, . . . , −mn)i of order 2mn is mn n ⌊d⌋ 1 X φ(d) kd . (55) k mn d d d|(mn,k). 38.

(69) B Moreover, the number of orbits of NC(m) (n) is. (m+1)n ⌊ d ⌋ 1 X φ(d) . mn ⌊ nd ⌋. (56). d|mn. Proof. We use the same method as in Theorem 2.6. By the Orbit-counting B Lemma and Lemma 3.10, the number of orbits of NC(m) (n, k) is . 1   2mn. X. φ(d). d|(2mn,2k) d is odd. mn d k d. ⌊ nd ⌋ k d. +. X. φ(d). d|(2mn,2k) d is even. 2mn . mn n ⌊d⌋ 1 X = φ(d) kd . k mn d d. d 2k d. . ⌊ 2n ⌋  d  2k d. d|(mn,k). B Moreover, the number of orbits of NC(m) (n) is. . (m+1)n ⌊ d ⌋ 1  X + φ(d)  2mn ⌊ nd ⌋ d|2mn d is odd. (m+1)n 1 X ⌊ d ⌋ φ(d) . = mn ⌊ nd ⌋.  X ⌊ 2(m+1)n ⌋  d φ(d)  2n ⌊d⌋ d|2mn. d is even. d|mn. We give a table of the number of orbits of and under rotations for 1 ≤ n ≤ 6 and 0 ≤ k ≤ 6 (see Table 6). They are not appeared in OEIS [13]. B NC(2) (n, k). B NC(2) (n). Lemma 3.12 Let n be a nonnegative integer, n X (m + 1)n mn n 2 = q 2k . n k q2 k q2 q2 k=0. Proof. Substituting n by (m + 1)n in (8), we have. 39. (57).

(70) nk 1 2 3 4 5 6. 0 1 1 1 1 1 1. 1 2 1 2 2 3 8 4 22 5 46 6 84. 3. 4. 5. 6. 4 28 10 120 106 26 368 623 396 80. 2 5 16 65 304 1558. B B Table 6: The number of orbits of NC(2) (n, k) and NC(2) (n) under rotations.. (m+1)n. X k=0. (m+1)n−1 Y k (m + 1)n k ( ) q 2 x = (1 + q i x) k i=0 =. n−1 Y. (1 + q i x). i=0. =. n X i=0. mn−1 Y. (1 + q i (xq n )). i=0. ! ! X mn i k n mn q (2) xi q (2 ) (xq n )k . i k k=0. Comparing coefficients of xn on both sides, we have X n n n−k k (m + 1)n n mn nk ( ) ( ) ( ) q 2 = q 2 q 2 q n n − k k k=0 n X k n mn (n−k = q 2 )+(2)+nk . k k k=0. Hence . X n (m + 1)n mn n k2 = q . n k k k=0. Substituting q by q 2 in above equation, we get (57). Let. B N(m) (n, k; q). mn n (m + 1)n 2k 2 B = q and C(m) (n; q) = . k q2 k q2 n 2 q. 40. .

(71) Lemma 3.13 If d|2mn and ω is any primitive d-th root of unity, then  mn n ⌊d⌋  d  if d is odd and d|k,  k k    d d  2mn 2n B ⌊d⌋ N(m) (n, k; q) q=ω = (58) d if d is even and d|2k,   2k 2k   d d    0 otherwise.. Moreover,. B C(m) (n; q) q=ω. .   ⌊ (m+1)n ⌋  d    ⌊ nd ⌋ = 2(m+1)n   ⌊ d ⌋    ⌊ 2n ⌋ d. if d is odd, (59) if d is even.. Proof. Using the same method as in Lemma 2.8, we have B N(m) (n, k; q) q=ω.  ! mn n mn n  ⌊ nd ⌋ ⌊d⌋ ⌋ n − d⌊  d d d  = if d is odd and d|k,  k k k k  0   d d d d q 2 q=ω   "  #   2mn 2n  2mn ⌊ 2n ⌋ n ⌊d⌋ n − (d/2)⌊ ⌋ d d d (d/2)   = if d is even and d|2k, = 2k 2k 2k 2k  0  d d d d  q 2 q=ω  !     n 2  [mn]q2 mn − 1  q 2k =0 otherwise.   [k]q2 k − 1 q2 k q2 q=ω. Moreover,. B C(m) (n; q) q=ω. . =. ! (m + 1)n n q2. q=ω.  (m+1)n  ⌊ d ⌋     ⌊ nd ⌋ = 2(m+1)n   ⌊ d ⌋    ⌊ 2n ⌋ d. if d is odd, if d is even. . B B Theorem 3.14 Let X be the set NC(m) (n, k) and X(q)=N(m) (n, k; q). Let the cyclic group G = h(1, 2, . . . , mn, −1, −2, . . . , −mn)i of order 2mn act. 41.

(72) on X by cyclic rotations. Then (X, X(q), G) exhibits the cyclic sieving pheB B nomenon. Moreover, let X be the set NC(m) (n) and X(q)=C(m) (n; q). Then (X, X(q), G) exhibits the cyclic sieving phenomenon. Proof. Let ω be any primitive d-th root of unity and g ∈ G be of order B d ≥ 1. By Lemma 3.10 and 3.13, we can see that N(m) (n, k; q) is equal q=ω B B to the number of π ∈ NC(m) (n, k) fixed by g. Moreover, C(m) (n; q) is q=ω. equal to the number of π ∈. 3.3. B NC(m) (n). fixed by g.. . m-Divisible noncrossing partitions of type D. We will follow the definition of m-divisible noncrossing partitions of type D in [6, 7]. Let π be a partition of {1, 2, . . . , mn, −1, −2, . . . , −mn}. We represent π using the annulus with the integers 1, 2, . . . , mn−m, −1, −2, . . . , −(mn−m) in the outer circle in clockwise order and the integers mn − m + 1, mn − m + 2, . . . , mn, −(mn − m + 1), −(mn − m + 2), . . . , −mn in the inner circle in counter clockwise order. We represent each block B of π as follows. Let B = {a1 , a2 , . . . , au }. We can assume that a1 , a2 , . . . , ai are arranged in the outer circle in the clockwise order and ai+1 , ai+2 , . . . , au are arranged in the inner circle in counter clockwise order for some i. Then we draw curves, which lie inside of the annulus, connecting aj and aj+1 for all j ∈ [u], where au+1 =a1 . If we can draw the curves for all the blocks of π such that they do not intersect, then we call π a noncrossing partition on the (2m(n − 1), 2m) − annulus. A m-divisible noncrossing partition of type D is a noncrossing partition π on the (2m(n − 1), 2m)-annulus satisfying the following conditions: 1. If B ∈ π, then −B ∈ π. 2. For each block B ∈ π with u elements, if a1 , a2 , . . . , au are the elements of B such that a1 , a2 , . . . , ai are arranged in the outer circle in clockwise order and ai+1 , ai+2 , . . . , au are arranged in the inner circle in counter clockwise order, then |aj+1| ≡ |aj | + 1 mod m for all j ∈ [u], where au+1 = a1 . 3. If there is a zero block B of π, i.e. B = −B, then B contains all the integers in the inner circle. 4. If a block B contains only the integers on the inner circle , then B is either {mn − m + 1, mn − m + 2, . . . , mn} or {−(mn − m + 1), −(mn − m + 2), . . . , −mn}. 42.

(73) D The set of m-divisible NC D (mn) will be denoted by NC(m) (n). See Figure D 8 for two elements of NC(2) (n). We are unable to find a cyclic group of D order 2m(n − 1) act on NC(m) (n) for m > 1. If the inner circle is rotated, D then some of π ∈ NC(m) (n) will contradict with the above condition 4 after the rotation. If the outer circle is rotated and the signs of the inner circle D are exchanged, then some of π ∈ NC(m) (n) will contradict with the above condition 2 after the action.. −6. −5 −4 −3. 8. 2 −7. −8. −3. 3. −7. −2. 1. −4. 2 7. −6. −5. 1. −8. 8. 3. 7 −2. 4. 4. 5 5 −1 6 6 D Figure 8: Two elements of NC(2) (n) for n = 4 with blocks ±{3, 4}, ±{1, 2, 5, 6, −7, −8} and ±{1, 4}, ±{2, 3}, ±{5, 6},±{7, 8}. −1. 4. Remarks 1. It would be interesting to find some combinatorial statistics on X in the above theorems such that X(q) is a q-count by a combinatorial statistic. D 2. It would be interesting to find the number of the unlabeled NC(m) (n) diagrams.. References [1] C.A. Athanasiadis and V. Reiner, Noncrossing partition for the group Dn , SIAM J. Discrete Math. 18 (2004), no. 2, 397−417. [2] D. Armstrong, Generalized noncrossing partitions and combinatorics of Coxeter groups, Mem. Amer. Math. Soc. 202 (2009), no. 949. 43.

(74) [3] P.J. Cameron, Combinatorics: topics, techniques, algorithms, Cambridge University Press, Cambridge, 1994. [4] J. F¨ urlinger and J. Hofbauer, q-Catalan Numbers, J. Combin. Theory Ser. A 40 (1985), no. 2, 248−264. [5] V.J.W. Guo and J. Zeng, Some arithmetic properties of the q-Euler numbers and q-Sali´e numbers, European J. Combin. 27 (2006), no. 6, 884−895. [6] J.S. Kim, Chain enumeration of k-divisible noncrossing partitions of classical types, arxiv:0908.2641v2 [math.CO] (2009). [7] C. Krattenthaler and T.W. M¨ uller, Decomposition numbers for finite Coxeter groups and generalised non-crossing partitions, Trans. Amer. Math. Soc. 362 (2010), no. 5, 2723−2787. [8] S.C. Liaw, H.G. Yeh, F.K. Hwang, and G.J. Chang, A simple and direct derivation for the number of noncrossing partitions, Proc. Amer. Math. Soc. 126 (1998), no. 6, 1579−1581. [9] G. Olive, Generalized powers, Amer. Math. Mon. 72 (1965), no. 6, 619−627. [10] C.W. Pei, Cyclic group actions on non-crossing partitions, University of Kaohsiung, master thesis, 2009. [11] V. Reiner, Non-crossing partitions for classical reflection groups, Discrete Math. 177 (1997), no. 1-3, 195−222. [12] V. Reiner, D. Stanton, and D. White, The cyclic sieving phenomenon, J. Combin. Theory Ser. A 108 (2004), no. 1, 17−50. [13] N.J.A Sloane, The on-line encyclopedia of integer sequences, published electronically at http://www.research.att.com/~njas/sequences/ [14] A. Tucker, Applied combinatorics, 5th edition, John Wiley and Sons, 2007.. 44.

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