c 2009 Society for Industrial and Applied Mathematics

Transcription

1 SIAM J. DISCRETE MATH. Vol. 23, No. 2, pp c 2009 Society for Industrial and Applied Mathematics CROSSINGS AND NESTINGS OF TWO EDGES IN SET PARTITIONS SVETLANA POZNANOVIĆ AND CATHERINE YAN Abstract. Let π and λ be two set partitions with the same number of blocks. Assume π is a partition of [n]. For any integer l, m 0, let T (π, l) be the set of partitions of [n + l] whose restrictions to the last n elements are isomorphic to π, andt (π, l, m) the subset of T (π, l) consisting of those partitions with exactly m blocks. Similarly define T (λ, l) andt (λ, l, m). We prove that if the statistic cr (ne), the number of crossings (nestings) of two edges, coincides on the sets T (π, l) and T (λ, l) forl =0, 1, then it coincides on T (π, l, m) andt (λ, l, m) for all l, m 0. These results extend the ones obtained by Klazar on the distribution of crossings and nestings for matchings. Key words. set partitions, crossings, nestings AMS subject classifications. 05A18, 05A15 DOI / Introduction and statement of main result. In a recent paper [5], Klazar studied distributions of the numbers of crossings and nestings of two edges in (perfect) matchings. All matchings form an infinite tree T rooted at the empty matching, in which the children of a matching M are the matchings obtained from M by adding to M in all possible ways a new first edge. Given two matchings M and N on [2n], Klazar decided when the number of crossings (nestings) have identical distribution on the levels of the two subtrees of T rooted at M and N. In the last section of [5], Klazar raised the question as to apply the method to other structures besides matchings. In the present paper we consider set partitions, which have a natural graphic representation by a set of arcs. We establish the Klazar-type results to the distribution of crossings and nestings of two edges in set partitions. Our approach follows that of Klazar for matchings [5], but is not a straightforward generalization. The structure of set partitions is more complicated than that of matchings. For example, partitions of [n] may have a different number of blocks, while every matching of [2n] has exactly n blocks. To get the results, we first defined an infinite tree T (Π) on the set of all set partitions, which, when restricted to matchings, is different than the one introduced by Klazar. We state our main result in Theorem 1.1, whose proof and applications are given in sections 2 and 3. Section 4 is devoted to the enumeration of the crossing/nesting similarity classes. Though the ideas of the proofs are similar to those in [5], in many places we have to supply our own argument to fit in the different structure, and use a variety of combinatorial structures, in particular, Motzkin paths, Charlier diagrams, and binary sequences. We also analyze the joint generating function of the statistics cr and ne over partitions rooted at π, and derive a continued fraction expansion for general π. Received by the editors October 12, 2007; accepted for publication January 31, 2009; published electronically April 9, Department of Mathematics, Texas A&M University, College Station, TX (spoznan@ math.tamu.edu). Department of Mathematics, Texas A&M University, College Station, TX and Center of Combinatorics, LPMC-TJKLC, Nankai University, Tianjin , People s Republic of China (cyan@math.tamu.edu). This author was supported in part by NSF grants DMS and DMS

2 788 SVETLANA POZNANOVIĆ AND CATHERINE YAN We begin by introducing some necessary notations. A (set) partition of [n] = {1, 2,...,n} is a collection of disjoint nonempty subsets of [n], called blocks, whose union is [n]. A matching of [2n] is a partition of [2n] inn two-element blocks, which we also call edges. If a partition π has k blocks, we write π = k. A partition π is often represented as a graph on the vertex set [n], drawn on a horizontal line in the increasing order from left to right, whose edge set consists of arcs connecting the elements of each block in numerical order. We write an arc e as a pair (i, j) with i<j. For a partition π of [n], we say that the arcs (i 1,j 1 )and(i 2,j 2 ) form a crossing if i 1 < i 2 < j 1 < j 2, and they form a nesting if i 1 < i 2 < j 2 < j 1. By cr(π) (respectively, ne(π)), we denote the number of crossings (respectively, nestings) of π. The distribution of the statistics cr and ne on matchings has been studied in a number of articles, including [2, 5, 6, 7, 8], to list a few. The symmetry of cr and ne for set partitions was established in [4]. In this paper we investigate the distribution of the statistics cr(π) and ne(π) over the partitions of [n] with a prefixed restriction to the last k elements. Denote by Π n the set of all partitions of [n], and by Π n,k the set of partitions of [n] withk blocks. For n =0,Π 0 contains the empty partition. Let Π = n=0 Π n = n=0 k n Π n,k. We define the tree T (Π) of partitions as a rooted tree whose nodes are partitions such that we have the following: 1. The root is the empty partition. 2. The partition π of [n + 1] is a child of λ, a partition of [n], if and only if the restriction of π on {2,...,n+1} is order-isomorphic to λ. See Figure 1 for an illustration of T (Π). Fig. 1. ThetreeofpartitionsT (Π). Observe that if λ is a partition of [n] with λ = k, thenλ has k + 1 children in T (Π). Let B 1,...,B k be the blocks of λ ordered in increasing order with respect to their minimal elements. For a set S, lets +1 = {a +1 : a S}. Wedenotethe children of λ by λ 0,λ 1,...,λ k as follows: λ 0 is a partition of [n +1] with k +1 blocks, λ (0) = {{1},B 1 +1,...,B k +1};

3 CROSSINGS AND NESTINGS IN PARTITIONS 789 for 1 i k, λ i is a partition of [n +1]withk blocks, λ (i) = {{1} (B i +1),B 1 +1,...,B i 1 +1,B i+1 +1,...,B k +1}. For a partition λ, let T (λ) denote the subtree of T (Π) rooted at λ, and let T (λ, l) be the set of all partitions at the lth level of T (λ). T (λ, l, m) isthesetofall partitions on the lth level of T (λ) withm blocks. Note that T (λ, l, m) if and only if k m k + l. Let G be an abelian group and α, β two elements in G. Consider the statistics s α,β :Π G given by s α,β (λ) =cr(λ)α + ne(λ)β. In [5], Klazar defines a tree of matchings and shows that for two matchings M and N, ifthestatistics α,β coincides at the first two levels of T (M) andt (N), then it coincides at all levels, and similarly for the pair of statistics s α,β,s β,α. In this article we prove that in the tree of partitions defined above, the same results hold. Precisely, we present the following theorem. Theorem 1.1. Let λ, π T(Π) be two nonempty partitions, and s α,β (T ) be the multiset containing {s α,β (t) :t T }. We have the following: (a) If s α,β (T (λ, l)) = s α,β (T (π, l)) for l =0, 1, then s α,β (T (λ, l, m)) = s α,β (T (π, l, m)) for all l, m 0. (b) If s α,β (T (λ, l)) = s β,α (T (π, l)) for l =0, 1, then s α,β (T (λ, l, m)) = s β,α (T (π, l, m)) for all l, m 0. In other words, if the statistic s α,β coincides on the first two levels of the trees T (λ) and T (π), then it coincides on T (λ, l, m) and T (π, l, m) on all levels, and similarly for the pair of statistics s α,β,s β,α. Note that the conditions of Theorem 1.1 imply that λ and π have the same number of blocks. But they are not necessarily partitions of the same [n]. At the end of the introduction we would like to point out the major differences between the structure of crossings and nestings of set partitions and that of matchings. 1. The tree of partitions T (Π) and the tree of matchings are different. In T (Π), children of a partition π are obtained by adding a new vertex, instead of adding a first edge. Hence Klazar s tree of matchings is not a subposet of T (Π). The definition of T (Π) allows us to define the analogous operators R α,β,i, as in [5, section 2]. Since some descendants of π are obtained by adding isolated points, we need to introduce an extra operator M (see Definition 2.2) and supply some new arguments to work with our structure and M. 2. The type of a matching is encoded by a Dyck path, while for set partitions the corresponding structure is restricted bicolored Motzkin paths (RBM) (cf. section 3). 3. The enumeration of crossing/nesting similarity classes is different. A crossingsimilarity class is determined by a value cr(m) (cr(π)) and a composition (a 1,a 2,...,a m )ofn. For matchings, cr(m) can be any integer between 0 and 1 + a 2 +2a 3 + +(m 1)a m. But for partitions the possible value of cr(π) depends only on m, nota i s. In matchings there is a bijection between the set of nesting sequences of matchings of [2n] and the set of Dyck paths D(n). There is no analogous result between set partitions and restricted bicolored Motzkin paths. 4. For matchings every nesting-similarity class is a subset of a crossing-similarity class. This is not true for set partitions. 2. Proof of Theorem 1.1. Throughout this article we will generally adapt Klazar s notation on multisets. Formally a multiset is a pair (A, m), where A is a set, called the underlying set, and m : A N is a mapping that determines the

4 790 SVETLANA POZNANOVIĆ AND CATHERINE YAN multiplicities of the elements of A. We often write multisets by repeating the elements according to their multiplicities. For a map f : X Y and Z X, letf(z) denote the multiset whose underlying set is {f(z) :z Z} and in which each element y appears with multiplicity equal to the cardinality of the set {z : z Z and f(z) =y}. S(X) denotes the set of all finite multisets with elements in the set X. Any function f : X S(Y ) naturally extends to f : S(X) S(Y )byf(z) = z Z {f(z)}, where is the union of multisets (the multiplicities of elements are added). For each b i =minb i of λ define u i (λ) to be the number of edges (p, q) such that p<b i <qand define v i (λ) to be the number of edges (p, q) such that p<q<b i. They satisfy the obvious recursive relations (2.1) (2.2) (2.3) (2.4) { u i (λ 0 0 if i =1, )= u i 1 (λ) if 2 i k +1, { v i (λ 0 0 if i =1, )= v i 1 (λ) if 2 i k +1, 0 if i =1, u i (λ j )= u i 1 (λ)+1 if2 i j, u i (λ) if j +1 i k, 0 if i =1, v i (λ j )= v i 1 (λ) if 2 i j, v i (λ)+1 ifj +1 i k for j =1,...,k,wherek = λ 1. For the statistics s α,β :Π G defined by s α,β (λ) =cr(λ)α + ne(λ)β, wehavethat (2.5) (2.6) s α,β (λ 0 )=s α,β (λ 1 )=s α,β (λ), s α,β (λ j )=s α,β (λ)+u j (λ)α + v j (λ)β, j 1. For simplicity, we will write λ ij for (λ i ) j. Lemma 2.1. For λ 1, (2.7) s α,β (λ 0j )) = { s α,β (λ) if j =0, 1, s α,β (λ j 1 ) if j 2, and for i 1, (2.8) s α,β (λ i ) if j =0, 1, s α,β (λ ij )= s α,β (λ i )+s α,β (λ j 1 ) s α,β (λ 1 )+α if 2 j i, s α,β (λ i )+s α,β (λ j ) s α,β (λ 1 )+β if j i +1. Proof. We first show (2.8). The first line in (2.8) follows directly from (2.5). For

5 CROSSINGS AND NESTINGS IN PARTITIONS 791 the other two, s α,β (λ ij )=s α,β (λ i )+u j (λ i )α + v j (λ i )β { s α,β (λ i )+u j 1 (λ)α + α + v j 1 (λ)β if 2 j i, = s α,β (λ i )+u j (λ)α + v j (λ)β + β if j i +1 { s α,β (λ i )+s α,β (λ j 1 ) s α,β (λ 1 )+α if 2 j i, = s α,β (λ i )+s α,β (λ j ) s α,β (λ 1 )+β if j i +1. The first and third equalities follow from (2.6) and the second one follows from (2.3) and (2.4). Similarly, (2.7) follows from (2.1), (2.2), (2.5), and (2.6). To each partition λ with k blocks (k 1), we associate a sequence seq α,β (λ) :=s α,β (λ 1 )s α,β (λ 2 )...s α,β (λ k ). The sequence seq α,β (λ) encodes the information about the distribution of s α,β on the children of λ in T (Π), in which s α,β (λ 1 ) plays a special role when we analyze the change of seq α,β (λ) below. This is due to the fact that s α,β (λ 1 ) carries information about λ andtwochildrenofλ, namely,λ 0 and λ 1. For an abelian group G, letg l denote the set of finite sequences of length l over G, andg = l 1 G l.ifu = x 1x 2...x k G and y G, thenweusetheconvention that the sequence (x 1 + y)(x 2 + y)...(x k + y) is denoted by x 1 x 2...x k + y. Definition 2.2. For α, β G and i 1, define R α,β,i : G l G l,(i l) by setting R α,β,i (x 1 x 2...x l )=x i (x 1...x i 1 +(x i x 1 + α))(x i+1...x l +(x i x 1 + β)) and R α,β : G S(G ) by setting R α,β (x 1 x 2...x l )={R α,β,i (x 1 x 2...x l ):1 i l}. In addition, define M : G G by setting Lemma 2.1 immediately implies that M(x 1 x 2...x l )=x 1 x 1 x 2...x l. seq α,β (λ 0 )=M(seq α,β (λ)), seq α,β (λ i )=R α,β,i (seq α,β (λ)) for 1 i λ. For l 0, let E α,β (λ, l, m) ={seq α,β (μ) :μ T(λ, l, m)}, the multiset of sequences seq α,β (μ) associated to partitions μ T(λ, l, m). Then for l 1, (2.9) E α,β (λ, l, m) =R α,β (E α,β (λ, l 1,m)) M(E α,β (λ, l 1,m 1)). Next, we define an auxiliary function f which reflects the change of the statistic s α,β along T (Π). Then we prove two general properties of f and use these properties to prove Theorem 1.1. For an integer r 0andγ G, the function f : G S(G) is defined by f r γ (x 1x 2...x l ):={x a1 + x a2 + + x ar (r 1)x 1 + γ :1<a 1 <a 2 < <a r l}.

6 792 SVETLANA POZNANOVIĆ AND CATHERINE YAN In particular, f0 0 (x 1 x 2...x l )={x 1 }, f0 1 (x 1x 2...x l )={x 2,...,x l }. Lemma 2.3. Let X, Y S(G ) be two multisets such that fγ r(x) =f γ r (Y ) for every r 0 and γ G. Then (a) fγ(m(x)) r = fγ(m(y r )), (b) fγ(r r α,β (X)) = fγ r (R α,β (Y )), (c) fγ r(r α,β(x)) = fγ r(r β,α(y )) for every r 0 and γ G. Proof. (a)theelementsinfγ r(m(x)) have the form y a 1 + y a2 + + y ar (r 1)y 1 + γ for some y 1 y 2...y l+1 M(X), where y 1 y 2...y l+1 = x 1 x 1 x 2...x l for some x 1 x 2...x l X. Forr =0, f 0 γ (M(X)) = {y 1 + γ : y 1 y 2...y l+1 M(X)} = {x 1 + γ : x 1 x 2...x l X} = f 0 γ (X). Hence f 0 γ (X) =f 0 γ (Y ) implies f 0 γ (M(X)) = f 0 γ (M(Y )). For r 1, divide the multiset f r γ (M(X)) into two disjoint multisets, and A = {y a1 + y a2 + + y ar (r 1)y 1 + γ : y 1 y 2...y l+1 M(X), a 1 =2} B = {y a1 + y a2 + + y ar (r 1)y 1 + γ : y 1 y 2...y l+1 M(X), a 1 > 2}. The elements of A can be written as y a1 + y a2 + + y ar (r 1)y 1 + γ = x 1 + y a2 + + y ar (r 1)x 1 + γ = y a2 + + y ar (r 2)x 1 + γ = x a x ar 1 (r 2)x 1 + γ. Since a 2 1 >a 1 1 = 1, the multiset A is equal to fγ r 1 (X). The elements in B can be written as y a1 + y a2 + + y ar (r 1)y 1 + γ = x a1 1 + x a x ar 1 (r 1)x 1 + γ. Since a 1 3, the indices on the right-hand side run through all the increasing r-tuples 1 <a 1 1 <a 2 1 < <a r 1 l. Therefore, B is equal to f r γ (X). So, (2.10) fγ(m(x)) r = fγ r 1 (X) fγ(x). r By assumption we have fγ(m(x)) r = fγ r 1 (X) fγ r (X) =fγ r 1 (Y ) fγ(y r )=fγ r (M(Y )). (c) We will prove only (c) because the proof of (b) is similar and easier. Since fγ r(x) is a translation of f 0 r (X) byγ, it is enough to prove the result for γ =0only. The elements of f0 r (R α,β (X)) have the form y a1 + y a2 + + y ar (r 1)y 1,where y 1 y 2...y l R α,β (X) isequaltox i (x 1...x i 1 + x i x 1 + α)(x i+1...x l + x i x 1 + β) for some x 1 x 2...x l X and i [l].

7 For 0 t r, let CROSSINGS AND NESTINGS IN PARTITIONS 793 C t,α,β (X) ={y a1 + y a2 + + y ar (r 1)y 1 : y 1 y 2...y l R α,β,i (X) anda t i<a t+1 for some i [l]}. An element y a1 + y a2 + + y ar (r 1)y 1 C t,α,β (X) isequalto x a1 1 + x at 1 + t(x i x 1 + α)+x at x ar +(r t)(x i x 1 + β) (r 1)x i = x a1 1 + x at 1 + x i + x at x ar rx 1 + tα +(r t)β. (2.11) Again, we consider two cases, according to the value of a 1. By (2.11), the submultiset of C t,α,β (X) fora 1 > 2isequaltof r+1 tα+(r t)β (X), and for a 1 = 2 the corresponding submultiset is equal to ftα+(r t)β r (X). Therefore, (2.12) C t,α,β (X) =f r+1 tα+(r t)β (X) f r tα+(r t)β (X). Similarly, (2.13) So, C t,β,α (Y )=f r+1 tβ+(r t)α (Y ) f tβ+(r t)α r (Y ). f r 0 (R α,β (X)) = = = = = r C t,α,β (X) r r r f r+1 tα+(r t)β (X) r f r+1 tα+(r t)β (Y ) r f r+1 (r t)α+tβ (Y ) r r C t,β,α (Y ) = f r 0 (R β,α(y )). f r tα+(r t)β (X) f r tα+(r t)β (Y ) f r (r t)α+tβ (Y ) The second and fifth equalities follow from (2.12) and (2.13), respectively. The third equality follows from the assumption of the lemma, while the fourth equality is just a reordering of the unions. Lemma 2.4. If X, Y S(G ) are one-element sets such that f0 0(X) =f 0 0 (Y ) and f0 1 (X) =f0 1 (Y ), thenfγ r (X) =fγ(y r ) for every r 0 and γ G. Proof. We need to prove that if u, v G are two sequences beginning with the same term and having equal numbers of occurrences of each g G, thenfγ r(u) =f γ r(v) for every r 0andγ G. It suffices to prove the statement for γ = 0, because fγ(u) r is a translation of f0 r(u) byγ. Let u = u 1...u l and v = v 1...v l. Since u 1 = v 1,it suffices to prove that the multisets {u a1 +u a2 + +u ar :1<a 1 <a 2 < <a r l}

8 794 SVETLANA POZNANOVIĆ AND CATHERINE YAN and {v a1 + v a2 + + v ar :1<a 1 <a 2 < <a r l} are equal. That is clear because {u 2,...,u l } and {v 2,...,v l } are equal as multisets. ProofofTheorem1.1. We prove (b); the proof of (a) is similar. First, we prove by induction on l that (2.14) f r γ (E α,β (λ, l, m)) = f r γ(e β,α (π, l, m)) for every r 0andγ G. Before we proceed with the induction, it is useful to observe that the assumption of Theorem 1.1(b) is equivalent to s α,β (T (λ, l)) = s β,α (T (π, l)) for l =0, 1 (2.15) s α,β (T (λ, l, m)) = s β,α (T (π, l, m)) for l =0, 1andk m k + l, where k = λ. One direction is clear, the other one follows from the following equalities: s α,β (T (λ, 1,k+1))=s α,β (T (λ, 0,k)) = s α,β (T (λ, 0)), s α,β (T (λ, 1,k)) = s α,β (T (λ, 1))\s α,β (T (λ, 0)), where \ is the difference of multisets. For the same reason the assumption of part (a) is equivalent to s α,β (T (λ, l, m)) = s α,β (T (π, l, m)) for l =0, 1andk m k + l. Now we show (2.14). For l = 0 we need to show fγ r(e α,β(λ, 0,k)) = fγ r(e β,α(π, 0,k)). By Lemma 2.4 we only need to check that f0 0(X) =f 0 0(Y )andf 0 1(X) =f 0 1 (Y )for X = {seq α,β (λ)} and Y = {seq β,α (π)}. This follows from (2.15), because f0 0 (X) = s α,β (T (λ, 0,k)) and f0 1(X) =s α,β(t (λ, 1,k)). Suppose fγ(e r α,β (λ, s, m)) = fγ r (E β,α (π, s, m)) for all 0 s<land all m. Then using (2.9), the induction hypothesis, and Lemma 2.3, we have fγ r (E α,β(λ, l, m)) = fγ r (R α,β(e α,β (λ, l 1,m))) fγ r (M(E α,β(λ, l 1,m 1))) = fγ(r r β,α (E β,α (π, l 1,m))) fγ(m(e r β,α (π, l 1,m 1))) = fγ(e r β,α (π, l, m)), and the induction is completed. This proves (2.14). Now s α,β (T (λ, l, m)) = f 0 0 (E α,β(λ, l, m)) = f 0 0 (E β,α(π, l, m)) = s β,α (T (π, l, m)). 3. Applications and examples. As a direct corollary, we obtain a result of Kasraoui and Zeng [4, eq. (1.6)]. Corollary 3.1. The joint distribution of crossings and nestings of partitions is symmetric, i.e., p cr(π) q ne(π) = p ne(π) q cr(π). π Π n π Π n Proof. LetG =(Z Z, +),α=(1, 0), and β =(0, 1). The result follows from the second part of Theorem 1.1 for λ = π = {{1}}. For a partition λ we say that two edges form an alignment if they form neither a crossing nor a nesting. The total number of alignments in λ is denoted by

9 CROSSINGS AND NESTINGS IN PARTITIONS 795 al(λ). A stronger result of Kasraoui and Zeng [4, eq. (1.4)] can also be derived from Theorem 1.1. Corollary 3.2. p cr(π) q ne(π) t al(π) = p ne(π) q cr(π) t al(π). π Π n π Π n Proof. Again we use G =(Z Z, +), α =(1, 0), β =(0, 1), and λ = π = {{1}}. ( Any partition μ Π n with k blocks has n k edges. Hence cr(μ)+ne(μ)+al(μ) = n k ) 2. The result follows from the second part of Theorem 1.1. Corollary 3.3. Let λ and π be two partitions of [n] with the same number of blocks k. If the statistic al is equidistributed on the first two levels of T (λ) and T (π), it is equidistributed on T (λ, l, m) and T (π, l, m) for all l, m 0. Proof. Againweusetheidentitycr(μ)+ne(μ)+al(μ)= ( ) n k 2, which holds for any partition μ Π n with k blocks. Moreover, al(λ) =al(λ 0 ). Therefore the condition that the statistic al is equidistributed on the first two levels of T (λ) andt (π) implies that the statistic cr + ne is equidistributed on T (λ, l, m) andt (π, l, m) for all l =0, 1 and all m. In other words, if we set G = Z and α = β = 1, then the assumption of Theorem 1.1 is satisfied, and hence cr + ne is equidistributed on T (λ, l, m) and T (π, l, m) for all l, m 0. This, in return, implies that al is equidistributed on T (λ, l, m) andt (π, l, m) for all l, m 0. Example 3.4. Let λ = {{1, 2, 5}, {3, 4}} and π = {{1, 2, 4}, {3, 5}}. There are as many partitions on [n] withm crossings and l nestings which, restricted to the last five points, form a partition isomorphic to λ as there are partitions of [n] with l crossings and m nestings which, restricted to the last five points, form a partition isomorphic to π. Proof. Set G = (Z Z, +),α = (1, 0) and β = (0, 1), s α,β =(cr, ne). The claim follows from part (b) of Theorem 1.1 since s α,β (λ) = (0, 1) = s β,α (π) and s α,β (T (λ, 1)) = {(0, 1), (0, 1), (1, 2)} = s β,α (T (π, 1)) Example 3.5. Consider the two partitions λ = {{1, 7}, {2, 6}, {3, 4}, {5, 8}} and π = {{1, 8}, {2, 4}, {3, 6}, {5, 7}}. There are as many partitions on [n]withmcrossings and l nestings which, restricted to the last eight points, form a partition isomorphic to λ as there are ones which, restricted to the last eight points, form a partition isomorphic to π. Proof. Again set G =(Z Z, +),α=(1, 0), and β =(0, 1). Then s α,β =(cr, ne). The claim follows from part (a) of Theorem 1.1 since and s α,β (λ) =(2, 3) = s α,β (π) s α,β (T (λ, 1)) = {(2, 3), (2, 3), (3, 3), (4, 3), (4, 4)} = s α,β (T (π, 1)). 4. Number of crossing- and nesting-similarity classes. In this section we consider equivalence relations cr and ne on set partitions in the same way Klazar defines them on matchings [5]. We determine the number of crossing-similarity classes in Π n,k. For ne, we find a recurrence relation for the number of nesting-similarity classes in Π n,k, and compute the total number of such classes in Π n. Define an equivalence relation cr on Π n : λ cr π if and only if cr(t (λ, l, m)) = cr(t (π, l, m)) for all l, m 0. The relation cr partitions Π n,k into equivalence classes. Theorem 1.1 implies that λ cr π if and only if cr(λ) =cr(π)andf 1 0 (seq 1,0(λ))

10 796 SVETLANA POZNANOVIĆ AND CATHERINE YAN = f0 1(seq 1,0(π)). Define crseq(λ) = seq 1,0 (λ) cr(λ). For the upcoming computations it is useful to observe that λ cr π if and only if cr(λ) = cr(π) and f0 1(crseq(λ)) = f 0 1 (crseq(π)), i.e., λ and π are equivalent if and only if they have the same number of crossings and their sequences crseq(λ) andcrseq(π) areequalas multisets. Denote the multiset consisting of the elements of crseq(λ) bycrset(λ). Similarly, define λ ne π if and only if ne(t (λ, l, m)) = ne(t (π, l, m)) for all l, m 0. Again, from Theorem 1.1 we have that λ ne π if and only if ne(λ) =ne(π) and f0 1(seq 0,1(λ)) = f0 1(seq 0,1(π)). Since the sequence seq 0,1 (λ) is nondecreasing, λ ne π if and only if ne(λ) =ne(π) andseq 0,1 (λ) ne(λ) =seq 0,1 (π) ne(π). With the notation at the beginning of section 2, seq 0,1 (λ) ne(λ) =v 1...v k.denotethis sequence by neseq(λ). AMotzkinpathM =(s 1,...,s n ) is a path from (0, 0) to (n, 0) consisting of steps s i {(1, 1), (1, 0), (1, 1)} which does not go below the x-axis. We say that the step s i is of height l if its left endpoint is at the line y = l. A restricted bicolored Motzkin path is a Motzkin path with each horizontal step colored red or blue which does not have a blue horizontal step of height 0. We will denote the steps (1, 1), (1, 1), red (1, 0), and blue (1, 0) by NE (northeast), SE (southeast), RE (red east), and BE (blue east), respectively. The set of all restricted bicolored Motzkin paths of length n is denoted by RBM n. A Charlier diagram of length n is a pair h =(M,ξ), where M =(s 1,...,s n ) RBM n and ξ =(ξ 1,...,ξ n ) is a sequence of integers such that ξ i =1ifs i isaneorrestep,and1 ξ i l if s i isaseorbestepofheight l. Γ n will denote the set of Charlier diagrams of length n. It is well known that partitions are in one-to-one correspondence with Charlier diagrams. Here we use two maps described in [4], which are based on similar constructions in [3, 9]. For our purpose, we reformulate the maps Φ r, Φ l :Γ n Π n as follows. Given (M,ξ) Γ n,constructλ Π n step by step. The path M =(s 1,...,s n ) determines the type of λ: i [n] is - a minimal but not a maximal element of a block of λ (opener) if and only if s i is a NE step; - a maximal but not a minimal element of a block of λ (closer) if and only if s i is a SE step; - both a minimal and a maximal element of a bock of λ (singleton) if and only if s i is a RE step; - neither a minimal nor a maximal element of a block of λ (transient) if and only if s i is a BE step. To draw the edges in Φ r ((M,ξ)), we process the closers and transients one-by-one from left to right. Each time we connect the vertex i that we are processing to the ξ i th available opener or transient to the left of i, where the openers and transients are ranked from right to left. If we rank the openers and transients from left to right, we get Φ l ((M,ξ)). It can be readily checked that Φ r and Φ l are well defined. Moreover, we have the following proposition. Proposition 4.1. The maps Φ r, Φ l :Γ n Π n are bijections. The proof can be found in [3, 4] and their references. Example 4.2. If (M,ξ) is the Charlier diagram in Figure 2, then Φ r ((M,ξ)) = {{1, 7, 10}, {2, 4, 6, 8}, {3}, {5, 9}, {11, 12}}, Φ l ((M,ξ)) = {{1, 4, 6, 7, 9}, {2, 10}, {3}, {5, 8}, {11, 12}}. For M RBM n let d i be the number of NE and RE steps that start at height i (i 0). The profile of M is the sequence pr(m) =(d 0,...,d l ), where l =max{i :

11 CROSSINGS AND NESTINGS IN PARTITIONS 797 blue blue red blue i ξ i Fig. 2. A Charlier diagram. d i 0}. Note that this implies that d i 1foreachi = 0,...,l, and that the path M is of height l or l +1. The semitype of M =(s 1,...,s i ) is the sequence st(m) =(ɛ 1,...,ɛ n ), where ɛ i =0ifs i is a NE or RE step, and ɛ i =1ifs i is a SE or BE step. For example, if M is the path in Figure 2, then pr(m) =(2, 1, 2), and st(m) =(0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1). Let λ Π n and Φ 1 r (λ) = (M,ξ r ), Φ 1 l (λ) = (M,ξ l ). Define ϕ(λ) = M RBM n. Note that for a given λ, ϕ(λ) can be easily constructed using the four steps above. The next lemma gives the relation between a partition and its corresponding restricted bicolored Motzkin path and Charlier diagram. Lemma 4.3. Let Φ r, Φ l,andϕ be the maps defined above and Φ 1 r (λ) =(M,ξr ), Φ 1 l (λ) =(M,ξ l ). (a) The number of blocks of λ is equal to the total number of NE and RE steps of M. (b) cr(λ) = n i=1 (ξr i 1), ne(λ) = n i=1 (ξl i 1). (c) pr(m) =(d 0,...,d l ) if and only if crset(λ) ={0 d0,...,l d l }. (d) neseq(λ) =v 1...v k if and only if the zeros in st(m) =(ɛ 1,...,ɛ n ) are in the positions v 1 +1,v 2 +2,...,v k + k. Proof. (a) The result follows from the fact that the number of blocks of λ is equal to the total number of openers and singletons. (b) Denote by E be the set of arcs of λ. For e =(i, j) E let c e = {(p, q) E : i<p<j<q}. Then cr(λ) = e E c e. Similarly, if n e = {(p, q) E : p< i<j<q}, thenne(λ) = e E n e. From the definitions of Φ r and Φ l it follows that c (i,j) = ξj r 1andn (i,j) = ξj l 1. Hence the claim. (c) Using the notation introduced at the beginning of section 2, we have crseq(λ) = (u 1,...,u k ), where u i is the number of edges (p, q) such that p < b i < q. Here b i =minb i,thatis,b i is the ith opener or singleton from left to right. But the step in M which corresponds to b i,isofheighth if and only if u i = h. (d) It follows directly from the definitions of neseq(λ) andst(m). A composition of k is an ordered tuple (d 0,...,d l ) of positive integers whose sum is k. Lemma 4.4. Let l 0,k 1, andn k. (a) If λ Π n,k and crset(λ) ={0 d0,...,l d l },then(d 0,...,d l ) is a composition of k into l +1 parts, where l n k and 0 cr(λ) (n k 1)l l(l 1) 2. (b) Given a composition (d 0,...,d l ) of k into l +1 n k +1 parts and an integer c such that 0 c (n k 1)l l(l 1) 2,thereexistsλ Π n,k with crset(λ) ={0 d0,...,l d l } and cr(λ) =c.

12 798 SVETLANA POZNANOVIĆ AND CATHERINE YAN Proof. (a) It is clear that d d l = k. It follows that all the d i s are positive from part (c) of Lemma 4.3. Moreover, λ has at least l openers and, therefore, at least l closers. So, k + l n, i.e., l +1 n k +1. Let c i (respectively, t i ) be the number of SE (respectively, BE) steps at level i, 1 i l +1. Then l+1 i=1 (c i + t i )=n k and c i 1, 1 i l. Using part (b) of Lemma 4.3, we have 0 cr(λ) l l(l 1) i=1 (1 + 0)(i 1) + (n k l)l =(n k 1)l 2. (b) Suppose first that l +1 n k. LetM RBM n consist of d 0 1REsteps followed by an NE step, then d 1 1 RE steps followed by one NE step, etc., d l 1 RE steps followed by an NE step, then n k l 1 BE steps, and l + 1 SE steps. It is not hard to see that indeed M RBM n. The path never crosses the x-axis and all the BE steps, if any, are at height l Also pr(m) =(d 0,...,d l ). Consider all the sequences ξ =(ξ 1,...,ξ n ) such that (M,ξ) is a Charlier diagram. Then (4.1) ξ i =1, 1 i k, 1 ξ i l +1, k+1 i n l 1, 1 ξ n i+1 i, 1 i l +1. Hence (4.2) 0 n (ξ i 1) (n k l 1)l + l +(l 1) + +1=(n k 1)l i=1 l(l 1). 2 In the case l = n k, constructm RBM n similarly: d 0 1 RE steps followed by an NE step, then d 1 1 RE steps followed by one NE step, etc., d l RE steps followed by l SE steps. (Note that, unlike in the case l<n k, the path M is of height l.) All the sequences ξ =(ξ 1,...,ξ n ) such that (M,ξ) is a Charlier diagram satisfy the following properties: (4.3) Hence (4.4) 0 ξ i =1, 1 i k, 1 ξ n i+1 i, 1 i l. n (ξ i 1) (l 1) + +1=(n k 1)l i=1 l(l 1). 2 Because of (4.2) (respectively, (4.4)), for any integer c between 0 and (n k 1)l l(l 1) 2, ξ can be chosen to satisfy the conditions (4.1) (respectively, (4.3)) such that n i=1 (ξ i 1) = c. Since Φ is a bijection, there is λ Π n,k such that Φ(λ) =(M,ξ) and, by part (b) and (c) of Lemma 4.3, cr(λ) =c and crset(λ) ={0 d0,...,l d l }. Theorem 4.5. Let n k 1 and m =min{n k, k 1}. Then (4.5) Π n,k / cr = In particular, if n 2k 1, m ( k 1 l l=0 )[ (n k 1)l l(l 1) 2 ] +1. (4.6) Π n,k / cr =(n k 1)(k 1)2 k 2 +2 k 1 (k 1)(k 2)2 k 4.

13 CROSSINGS AND NESTINGS IN PARTITIONS 799 Proof. Recall that λ cr π if and only if cr(λ) =cr(π) andcrset(λ) =crset(π). Therefore, Π n,k / cr = {(crset(λ),cr(λ)) : λ Π n,k }. Using Lemma 4.4 and the fact that the number of compositions of k into l +1 parts, 0 l k 1, is ( ) k 1 l,we derive (4.5). In particular, when n 2k 1, k 1 ( k 1 Π n,k / cr = l l=0 )[ (n k 1)l l(l 1) 2 ] +1. But k 1 ( ) k 1 =(1+x) k 1 x=1 =2 k 1, l l=0 k 1 ( ) ( ) k 1 d l = (1 + x)k 1 x=1 =(k 1)(1 + x) k 2 x=1 l dx l=0 =(k 1)2 k 2, k 1 ( ) ( ) k 1 d 2 l(l 1) = (1 + x)k 1 l dx2 x=1 l=0 =(k 1)(k 2)(1 + x) k 3 x=1 =(k 1)(k 2)2 k 3, and (4.6) follows. Theorem 4.5 implies that there are many more examples of different partitions λ and π for which the statistic cr has same distribution on the levels of T (λ) andt (π). For example, Π 2k,k > (2k 1)!! 2 ( ) 2k k e while Π2k,k / cr 3k 2 2 k 4. Next we analyze the number of nesting-similarity classes. First we derive a recurrence for the numbers f n,k = Π n,k / ne. Theorem 4.6. Let n k 1. Then (4.7) (4.8) f n,1 =1, f n,k = n 1 r=k 1 ( ) n 2 f r,k 1 +(k 1), k 2. k Proof. Equation (4.7) is clear since Π n,1 =1. Recall that λ ne π if and only if ne(λ) =ne(π) andneseq(λ) =neseq(π). By Lemma 4.3, f n,k is equal to the number of pairs (ɛ, c) such that there exists λ Π n,k with ne(λ) =c and st(ϕ(λ)) = ɛ. It is not hard to see that for a given sequence ɛ =(ɛ 1,...,ɛ n ) {0, 1} n,thereexistsλ Π n,k such that st(ϕ(λ)) = ɛ if and only if ɛ has k zeros and ɛ 1 = 0. Denote the set of all such sequences by Sn,k 0 and denote the set of all ɛ {0, 1} n with k zeros by S n,k. For a sequence ɛ S n,k define a bicolored Motzkin path M = M(ɛ) =(s 1,...,s n ) as follows. For i from n to 1 we have the following: If ɛ i =0ands i is not defined yet, then set s i to be an RE step. If ɛ i =1andthereisj<isuch that ɛ j =0ands j is not defined yet, then set s i to be an SE step and s j0 to be an NE step, where j 0 =min{j : ɛ j = 0 and s j is not defined yet}. If ɛ i = 1 and there is no j<isuch that ɛ j =0ands j has not been defined yet, set s i to be a BE step.

14 800 SVETLANA POZNANOVIĆ AND CATHERINE YAN Note that we build M backwards, from (n, 0) to (0, 0). Let h i be the height of s i and ne(ɛ) = (h i 1), where the sum is over all the indices i such that ɛ i =1. For example, if ɛ =(0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1), then M(ɛ) =(NE,NE,NE,BE,NE,BE,BE,SE,SE,SE,RE,SE). The sequence of the heights of all the steps of M is (0, 1, 2, 3, 3, 4, 4, 4, 3, 2, 1, 1) and ne(ɛ) =(3 1) + (4 1) + (4 1) + (4 1) + (3 1) + (2 1) + (1 1) = 14. Although clearly M(ɛ) staysabovethex-axis, it is not necessarily a restricted bicolored Motzkin path. The reason is that any 1 in ɛ before the first zero would produce a BE step on the x-axis. Hence, M(ɛ) RBM n if and only if ɛ 1 =0,or, equivalently, ɛ Sn,k 0. We claim that for a fixed ɛ Sn,k 0,thereisλ Π n,k such that st(ϕ(λ)) = ɛ and ne(λ) =c if and only if 0 c ne(ɛ). To show the if part, one can choose a sequence ξ = (ξ 1,...,ξ n )with1 ξ i h i if ɛ i = 1, ξ i = 1 if ɛ i = 0, and n i=1 (ξ i 1) = c. Then Lemma 4.3 implies that Φ l ((M,ξ)) satisfies the requirements. Conversely, suppose λ Π n,k is such that st(ϕ(λ)) = ɛ. LetΦ 1 l (λ) =(M,ξ ). Then the height h i of each BE and SE step of M satisfies h i min{# zeros in (ɛ 1,...,ɛ i 1 ), (# ones in (ɛ i+1,...,ɛ n )) + 1} = h i. Now, by Lemma 4.3, ɛ S 0 n,k ne(λ) = (ξ i 1) (h i 1) (h i 1) = ne(ɛ). The claim is proved. Back to the proof of Theorem 4.6, we have f n,k = (ne(ɛ)+1) = ( (hi 1) + 1) = ) n 1 hi (n k 1)(. k 1 ɛ S 0 n,k ɛ S 0 n,k Set g n,k = hi and g n,k = hi, ɛ S 0 n,k ɛ S n,k where the inner sums are taken over all the indices i such that ɛ i =1. Withthis notation, ( ) n 1 (4.9) f n,k = g n,k (n k 1). k 1 The sequences g n,k and gn,k satisfy the following recurrence relations: ( ) n 2 (4.10) g n,k = g n 1,k 1 + gn 2,k 1 +(n k), k 1 (4.11) g n,k = n r=k g r,k. To see (4.10), note that if ɛ n = 0, then (ɛ 1,...,ɛ n 1 ) Sn 1,k 1 0 and M(ɛ) is M(ɛ 1,...,ɛ n 1 ) with one RE step appended, and if ɛ n =1,then(ɛ 2,...,ɛ n 1 ) S n 2,k 1 and M(ɛ 2,...,ɛ n 1 ) is obtained from M(ɛ) by deleting the first NE and the

15 CROSSINGS AND NESTINGS IN PARTITIONS 801 last SE step. For (4.11), if ɛ 1 = = ɛ r 1 =1andɛ r =0,thenM(ɛ r,...,ɛ n )is obtained from M(ɛ) by deleting the first r 1 BE steps at level 0. Substituting (4.11) into (4.10) gives (4.12) g n,k = n 1 r=k 1 ( ) n 2 g r,k 1 +(n k). k 1 Finally, by substituting g n,k from (4.9) into (4.12) and simplifying, we then obtain (4.8). Corollary 4.7. Π 1 / ne =1, Π 2 / ne =2, Π n / ne =2 n 5 (n 2 5n + 22), n 3. Proof. Denote Π n / ne by F n.usingf n = n k=1 f n,k, (4.7), and (4.8), we get F n =1+F F n 1 + This yields the recurrence relation n ( ) n 2 (k 1) k k=2 = F F n 1 +(n 4)2 n 3 +2, n 2. F n =2F n 1 +(n 3)2 n 4, n 3 with initial values F 1 =1andF 2 = 2, which has the solution F n =2 n 5 (n 2 5n + 22), n 3. The following tables give the number of crossing/nesting-similarity classes on Π n,k for small n and k. crossing-similarity classes n \ k nesting-similarity classes n \ k The two equivalence relations cr and ne on set partitions are not compatible. From the tables it is clear that cr is not a refinement of ne. On the other hand, let π = {{1, 3}, {2, 4}, {5, 6}} and λ = {{1, 3, 6}, {2, 4}, {5}}. It is easy to check that π ne λ, but π cr λ,ascr(π) =1andcr(λ) =2. 5. Generating function for crossings and nestings. In this section we analyze the generating function S π (q, p, z) = l 0 λ T (π,l) q cr(λ) p ne(λ) z l

16 802 SVETLANA POZNANOVIĆ AND CATHERINE YAN for a given partition π, and derive a continued fraction expansion for S π (q, p, z). For this we work with the group G = Z Z and α =(1, 0),β =(0, 1). Fix a partition π with k blocks. Define E α,β (π, l) = k+l m=k E α,β(π, l, m), i.e., E α,β (π, l) is the multiset of sequences seq α,β (μ) associated to the partitions μ T(π, l). A recurrence analogous to (2.9) holds. Namely, for l 1 (5.1) E α,β (λ, l) =R α,β (E α,β (λ, l 1)) M(E α,β (λ, l 1)). For simplicity we write E l instead of E α,β (π, l) when there is no confusion. Define b l,r to be the generating function of the multiset f r 0 (E l), i.e., b l,r (q, p) = (x,y) f r 0 (E l) q x p y, where (x, y) f r 0 (E l) contributes to the sum above according to its multiplicity in f r 0 (E l ). By convention, let b l,r (q, p) =0iff r 0 (E l )=, oroneofl, r is negative. For simplicity we write b l,r for b l,r (q, p). Note that b l,0 = λ T (π,l) qcr(λ) p ne(λ) and hence (5.2) S π (q, p, z) = l 0 b l,0 z l. By the formulas (5.1), (2.10), and the proof of part (c) of Lemma 2.3, we get f0 r (E l )=f0 r (M(E l 1 )) f0 r (R α,β (E l 1 )) r = f0 r 1 (E l 1 ) f0 r (E l 1 ) f r+1 tα+(r t)β (E l 1) r ftα+(r t)β r (E l 1), which leads to a recurrent relation for b l,r : ( r ) ( r ) b l,r = b l 1,r 1 + b l 1,r + q t p r t b l 1,r+1 + q t p r t b l 1,r. Using the standard notation [r] q,p := qr p r q p, we can write this as the following proposition. Proposition 5.1. b l,r = b l 1,r 1 +(1+[r +1] q,p )b l 1,r +[r +1] q,p b l 1,r+1. If the sequence associated to the partition π is x 1 x 2...x k with x i = u i α+v i β, 1 i k, then (5.3) b 0,0 = q u1 p v1, b 0,r = q ui 1 + +uir (r 1)u1 p vi 1 + +vir (r 1)v1 for r 1. 1<i 1< <i r k In particular, b 0,r =0ifr k. Given l and s, both nonnegative integers, consider the paths from (l, 0) to (0,s) using steps ( 1, 0), ( 1, 1), and ( 1, 1), which do not go below the x-axis. Each step ( 1, 0) (( 1, 1), ( 1, 1), respectively) starting at the line y = r has weight [r +1] q,p (1+[r +1] q,p,1, respectively). The weight w(m) of such a path M is defined

17 CROSSINGS AND NESTINGS IN PARTITIONS 803 to be the product of the weights of its steps. Let c l,s = w(m), where the sum is over all the paths M described above. Then from Proposition 5.1 one has b l,0 = 0 s k 1 c l,s b 0,s. Set a r =[r +1] q,p and c r =[r +1] q,p + 1. By the well-known theory of continued fractions (see [3]), c l,s is equal to the coefficient in front of z l in (5.4) K s (z) :=J /0/ (z)a 0 zj /1/ (z)a 1 z J /s/ (z) = 1 z s (Q s 1(z)J(z) P s 1 (z)), where J /h/ (z) = 1 a h z 2 1 c h z 1 c h+1 z a h+1z 2... and P k(z) Q k (z) is the kth convergent of J(z) :=J /0/ (z). Hence we have the following theorem. Theorem 5.2. Let π be a partition with k blocks whose associated sequence is x 1 x 2...x k,wherex i = u i α + v i β for 1 i k. Then S π (q, p, z) = 0 s k 1 b 0,s K s (z), where b 0,s is given by the formula (5.3), andk s (z) is given by (5.4). In particular, when k = 1, i.e., π is a partition with only one block, then b 0,0 =1 and b l,0 = c l,0 b 0,0 = c l,0. Therefore we have the following corollary. Corollary 5.3. If π =1,then S π (q, p, z) = 1 ([1] q,p +1)z 1 [1] q,p z 2 1 ([2] q,p +1)z [2] q,pz 2 Remark. Corollary 5.3 leads to a continued fraction expansion for the generating function of crossings and nestings over Π: Just taking π to be the partition of {1} and bearing in mind that we are counting the empty partition as well, we get q cr(λ) p ne(λ) z n =1+zS {1} (q, p, z) n 0 λ Π n (5.5) =1+ 1 ([1] q,p +1)z z... [1] q,p z 2 1 ([2] q,p +1)z [2] q,pz

18 804 SVETLANA POZNANOVIĆ AND CATHERINE YAN A different expansion was given in [4], as (5.6) n 0 λ Π n q cr(λ) p ne(λ) z n = 1 z 2. 1 z [2] q,p z 2 1 ([1] q,p +1)z 1 ([2] q,p +1)z [3] q,pz 2... The fractions (5.6) and (5.5) can be transformed into each other by applying twice the following contraction formula for continued fraction, (for example, see [1]): c 0 1 c 1z 1 c 2z... = c (c 1 + c 2 )z c 0 c 1 z c 2 c 3 z 2 1 (c 3 + c 4 )z c 4c 5 z REFERENCES [1] R.J. Clarke, E. Steingrímsson, and J. Zeng, New Euler-Mahonian permutation statistics, Sém. Lothar. Combin., 35 (1995), Art. B35c. [2] W.Y.C. Chen, E.Y.P. Deng, R.R.X. Du, R.P. Stanley, and C.H. Yan, Crossing and nestings of matchings and partitions, Trans. Amer. Math. Soc., 359 (2007), pp [3] P. Flajolet, Combinatorial aspects of continued fractions, Discrete Math., 32 (1980), pp [4] A. Kasraoui and J. Zeng, Distribution of crossings, nestings and alignments of two edges in matchings and partitions, Electron. J. Combin., 13 (2006), Research Paper 33. [5] M. Klazar, On identities concerning the numbers of crossings and nestings of two edges in matchings, SIAM J. Discrete Math., 20 (2006), pp [6] J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), pp [7] M. de Sainte-Catherine, Couplages et Pfaffiens en Combinatoire, Physique et Informatique, Ph.D. thesis, University of Bordeaux I, Talence, France, [8] J. Touchard, Sur un probléme de configurations et sur les fractions continues, Canadian J. Math., 4 (1952), pp [9] X. Viennot, Une théorie combinatoire des polynômes orthogonaux, Notes de cours, UQAM, Montréal, 1983.