Exterior Pairs and Up Step Statistics on Dyck Paths
aa r X i v : . [ m a t h . C O ] O c t EXTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS
SEN-PENG EU AND TUNG-SHAN FU
Abstract.
Let C n be the set of Dyck paths of length n . In this paper, by a new automor-phism of ordered trees, we prove that the statistic ‘number of exterior pairs’, introducedby A. Denise and R. Simion, on the set C n is equidistributed with the statistic ‘numberof up steps at height h with h ≡ m ≥
3, we prove that the twostatistics ‘number of up steps at height h with h ≡ m )’ and ‘number of up stepsat height h with h ≡ m − m )’ on the set C n are ‘almost equidistributed’. Bothresults are proved combinatorially. Introduction
Let C n denote the set of lattice paths, called Dyck paths of length n , in the plane Z × Z from the origin to the point (2 n,
0) using up step (1 ,
1) and down step (1 , −
1) that neverpass below the x -axis. Let U and D denote an up step and a down step, respectively. In[3], Denise and Simion introduced and investigated the two statistics ‘pyramid weight’ and‘number of exterior pairs’ on the set C n . A pyramid in a Dyck path is a section of the form U h D h , a succession of h up steps followed immediately by h down steps, where h is called the height of the pyramid. The pyramid is maximal if it is not contained in a higher pyramid.The pyramid weight of a Dyck path is the sum of the heights of its maximal pyramids. An exterior pair in a Dyck path is a pair consisting of an up step and its matching down stepwhich do not belong to any pyramid. For example, the path shown in Figure 1 containsthree maximal pyramids with a total weight of 4 and two exterior pairs. Figure 1.
A Dyck path with three maximal pyramids and two exterior pairs.
Since a Dyck path in C n with a pyramid weight of k contains n − k exterior pairs, bothof the statistics are essentially equidistributed on the set C n . However, they seem to be‘isolated’ from other statistics in the sense that so far there are no known statistics thatshare the same distribution with them. In the first part of this work, we discover one andestablish an explicit connection with the statistic ‘number of exterior pairs’.For a Dyck path, an up step that rises from the line y = h − y = h is saidto be at height h . It is well known [7] that the number of paths in C n with k up steps at Key words and phrases.
Dyck paths, exterior pairs, ordered trees, planted trees, continued fractions.Partially supported by National Science Council, Taiwan under grants 98-2115-M-390-002 (S.-P. Eu) and97-2115-M-251-001 (T.-S. Fu). even height is enumerated by the
Narayana number N n,k = 1 n (cid:18) nk (cid:19)(cid:18) nk + 1 (cid:19) , for 0 ≤ k ≤ n −
1. Note that P n − k =0 N n,k = n +1 (cid:0) nn (cid:1) = |C n | is the n th Catalan number. Weconsider the number g ( c ;3) n,k of the paths in C n with k up steps at height h such that h ≡ c (mod 3), for some c ∈ { , , } . For example, the initial values of g ( c ;3) n,k are shown in Figure2. n \ k n \ k n \ k g (0;3) n,k g (1;3) n,k g (2;3) n,k Figure 2.
The distribution of Dyck paths with respect to g ( c ;3) n,k . To our surprise, the distribution g (0;3) n,k , shown in Figure 2, coincides with the distributionof the statistic ‘number of exterior pairs’ on the set C n (cf. [3, Figure 2.4]). In additionto an algebraic proof by the method of generating functions (see Example 3.2), one of themain results in this paper is a bijective proof of the equidistribution of these two statistics(Theorem 1.1), which is established by a recursive construction. To our knowledge, it is notequivalent to any previously known bijection on the set C n . Theorem 1.1.
For ≤ k ≤ n − , there is a bijection Π : C n → C n such that a path π ∈ C n with k exterior pairs is carried to the corresponding path Π( π ) containing k up steps atheight h with h ≡ (mod 3). Recall that a path in C n with k up steps at even height contains n − k up steps at oddheight and that N n,k = N n,n − − k (0 ≤ k ≤ n − C n . Specifically, the number of paths in C n with k steps at evenheight equals the number of paths with k + 1 up steps at odd height. (However, the one-to-one correspondence between the two sets is not apparent.) Moreover, as one has noticedin Figure 2 that g (0;3) n,k = g (2;3) n,k +1 for k ≥
1, the two statistics ‘number of up steps at height h with h ≡ h with h ≡ C n .Motivated by this fact, for an integer m ≥ R ⊆ { , , . . . , m − } we study theenumeration of the paths in C n with k up steps at height h such that h ≡ c (mod m ) and c ∈ R . Let g ( R ; m ) n,k denote this number and let G ( R ; m ) be the generating function for g ( R ; m ) n,k ,where G ( R ; m ) = G ( R ; m ) ( x, y ) = X n ≥ X k ≥ g ( R ; m ) n,k y k x n . XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 3
We shall show that G ( R ; m ) satisfies an equation that is expressible in terms of continuedfractions (Theorem 3.1), which is equivalent to a quadratic equation in G ( R ; m ) . If R is asingleton, say R = { c } , we write g ( c ; m ) n,k and G ( c ; m ) instead. The other main result in thispaper is to prove combinatorially that the two statistics ‘number of up steps at height h with h ≡ m − m )’ and ‘number of up steps at height h with h ≡ m )’ arealmost equidistributed, i.e., g (0; m ) n,k = g ( m − m ) n,k +1 , for k ≥
1, and g (0; m ) n, = g ( m − m ) n, + g ( m − m ) n, (see Theorem 1.2). Theorem 1.2.
For m ≥ , the following equation holds. (1) G ( m − m ) − y · G (0; m ) = (1 − y ) U m − ( √ x ) √ xU m − ( √ x ) , where U n ( x ) denotes the n th Chebyshev polynomial of the second kind, U n (cos θ ) = sin(( n +1) θ )sin θ . We remark that U m − ( √ x ) / ( √ xU m − ( √ x )), a polynomial in x , is a generating functionfor the number of paths in C n of height at most m −
2, as pointed out by Krattenthaler [6,Theorem 2] (see also [1] and [8]). Note that in Eq. (1) the terms with y i vanish, for i ≥ Proof of Theorem 1.1
In this section, we shall establish the bijection requested in Theorem 1.1. A block of aDyck path is a section beginning with an up step whose starting point is on the x -axis andending with the first down step that returns to the x -axis afterward. Dyck paths that haveexactly one block are called primitive . We remark that the requested bijection is establishedfor primitive Dyck paths first and then for ordinary ones in a block-by-block manner. Infact, the bijection is constructed in terms of ordered trees.An ordered tree is an unlabeled rooted tree where the order of the subtrees of a vertexis significant. Let T n denote the set of ordered trees with n edges. There is a well-knownbijection Λ : C n → T n between Dyck paths and ordered trees [4], i.e., traverse the tree fromthe root in preorder, to each edge passed on the way down there corresponds an up stepand to each edge passed on the way up there corresponds a down step. For example, Figure3 shows a Dyck path of length 14 with 2 blocks and the corresponding ordered tree. Figure 3.
A Dyck path and the corresponding ordered tree.
For an ordered tree T and two vertices u, v ∈ T , we say that v is a descendant of u if u is contained in the path from the root to v . If also u and v are adjacent, then v is called a child of u . A vertex with no children is called a leaf . By a planted (ordered) tree we meanan ordered tree whose root has only one child. (We will speak of planted trees withoutincluding the word ‘ordered’.) Let τ ( uv ) denote the planted subtree of T consisting of theedge uv and the descendants of v , and let T − τ ( uv ) denote the remaining part of T when SEN-PENG EU AND TUNG-SHAN FU τ ( uv ) is removed. In this case, the edge uv is called the planting stalk of τ ( uv ). It is easyto see that the Dyck path corresponding to a planted tree is primitive.The level of edge uv ∈ T is defined to be the distance from the root to the end vertex v .The height of T is the highest level of the edges of T . The edge uv is said to be exterior if τ ( uv ) contains at least two leaves. One can check that the exterior edges of T are inone-to-one correspondence with the exterior pairs of the corresponding Dyck path Λ − ( T ).Moreover, the edges at level h in T are in one-to-one correspondence with the up steps atheight h in Λ − ( T ). Hence, under the bijection Λ, the following result leads to the bijectionΠ = Λ − ◦ Φ ◦ Λ requested in Theorem 1.1.
Theorem 2.1.
For ≤ k ≤ n − , there is a bijection Φ : T n → T n such that a tree T ∈ T n with k exterior edges is carried to the corresponding tree Φ( T ) containing k edges at level h with h ≡ (mod 3). Our strategy is to decompose an ordered tree (from the root) into planted subtrees, findthe corresponding trees of the planted subtrees, and then merge them (from their roots)together. In the following, we focus the construction of Φ on planted trees.2.1.
Planted trees.
Let P n ⊆ T n be the set of planted trees with n edges. By a bouquet of size k ( k ≥
1) we mean a planted tree such that there are k − h are colored red if h ≡ black otherwise. Nowwe establish a bijection φ : P n → P n such that the exterior edges of T ∈ P n are transformedto the red edges in φ ( T ).2.2. The map φ . Given a T ∈ P n , let uv be the planting stalk of T . If T contains noexterior edges then T is a path of length n and we define φ ( T ) to be a bouquet of size n .Otherwise, T contains at least one exterior edge. Note that the planting stalk uv itself isone of the exterior edges of T . Let w , . . . , w r be the children of v , for some r ≥
1. Unlessspecified, these children are placed in numeric order of the subscripts from left to right.The tree φ ( T ) is recursively constructed with respect to uv according to the following threecases.Case 1. Edge vw r is an exterior edge of T . For 1 ≤ j ≤ r , we first construct the plantedsubtrees T j = φ ( τ ( vw j )). In particular, in T r we find the rightmost edge, say xz , at level3. Then φ ( T ) is obtained from T r by adding an edge xy (emanating from vertex x ) to theright of xz and adding T , . . . , T r − under the edge xy (i.e., merges the roots of T , . . . , T r − with y ). Note that the red edge xy is created in replacement of the planting stalk uv of T .Case 2. Edge vw r is not an exterior edge but vw r − is an exterior edge. Then τ ( vw r ) isa path of a certain length, say t ( t ≥ ≤ j ≤ r −
1, we first construct the plantedsubtrees T j = φ ( τ ( vw j )). In particular, let pq be the planting stalk of T r − . Then φ ( T ) isobtained from T r − by adding a path qxy of length 2 such that the edge qx is the right mostedge at level 2 (emanating from vertex q ), and then adding t − qz , . . . , qz t − (emanating from vertex q ) to the right of qx and adding T , . . . , T r − under the edge xy .Note that the planting stalk uv of T is replaced by the red edge xy and that the subtree τ ( vw r ) of T is replaced by the edges { qx, qz , . . . , qz t − } .Case 3. Neither vw r − nor vw r is an exterior edge. Then τ ( vw r − ) and τ ( vw r ) are pathsof certain lengths. Let the lengths of τ ( vw r − ) and τ ( vw r ) be t and t , respectively. For XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 5 ≤ j ≤ r −
2, we first construct the planted subtrees T j = φ ( τ ( vw j )). To construct the tree φ ( T ), we create a path pqxy of length 3, where vertex p is the root. Next, add t − qz , . . . , qz t − to the left of the edge qx and add t − qz ′ , . . . , qz ′ t − to the rightof the edge qx . Then add T , . . . , T r − under the edge xy . Note that the planting stalk uv of T is replaced by the red edge xy and that the subtree τ ( vw r − ) (resp. τ ( vw r )) of T isreplaced by the edges { pq, qz , . . . , qz t − } (resp. { qx, qz ′ , . . . , qz ′ t − } ). Example 2.2.
Let T be the tree on the left of Figure 4. Note that the edges uv and ve are exterior edges. To construct φ ( T ), we need to form the subtrees T = φ ( τ ( vc )) , T = φ ( τ ( vd )) and T = φ ( τ ( ve )). By Case 3 of the algorithm, T is a path pqxz of length 3,along with an edge qh on the right of qx . Since ve is an exterior edge of T , by Case 1, φ ( T )is obtained from T by adding the edge xy and adding T = yc and T = yd under the edge xy , as shown on the right of Figure 4. Note that the planting stalk uv of T is transformedto the red edge xy , the rightmost one at level 3 in φ ( T ). hxz ydc qpc d heuv gf Figure 4.
A planted tree with and its corresponding tree.
Example 2.3.
Let T be the tree on the left of Figure 5. Note that the edges uv and vd are exterior edges. To construct φ ( T ), we need to form the subtrees T = φ ( τ ( vc )) and T = φ ( τ ( vd )). By Case 3 of the algorithm, T is a path pqab of length 3. Since τ ( ve ) isa path of length 2, by Case 2, φ ( T ) is obtained from T by adding a path qxy of length 2,along with the edge qz , and then adding T = yc under the edge xy , as shown on the rightof Figure 5. Note that the planting stalk uv of T is transformed to the red edge xy , therightmost one at level 3 in φ ( T ). c edvu hf g zyxcqpab Figure 5.
A planted tree with and its corresponding tree.
Example 2.4.
Let T be the tree on the left of Figure 6. To construct φ ( T ), we need toform the subtrees T = φ ( τ ( vc )) and T = φ ( τ ( vd )), which have been shown in Example2.2 and Example 2.3, respectively. Since neither ve nor vf is an exterior edge, by Case 3 SEN-PENG EU AND TUNG-SHAN FU of the algorithm, we create a path pqxy of length 3, along with the edge qg attached tothe left of qx and with the edges qh, qi attached to the right of qx . As shown on the rightof Figure 6, the tree φ ( T ) is then obtained by adding T = τ ( yc ) and T = τ ( yd ) underthe edge xy . Note that the planting stalk uv of T is transformed to the red edge xy , theunique one at level 3 in φ ( T ), and the previously constructed red edges in T = φ ( τ ( vc ))and T = φ ( τ ( vd )) are transformed to red edges in φ ( T ) by shifting them from level 3 tolevel 6. pqg h ixyc duvc d e fhig Figure 6.
A planted tree with and its corresponding tree.
From the construction of φ , we observe that the planting stalk of T is transformed to therightmost red edge at level 3 in φ ( T ), and that the other red edges recursively constructedso far (in T j ) are transformed to red edges in φ ( T ), either by shifting from level 3 i to level3 i + 3 or by remaining at level 3 i (as the ones in T r of Case 1 or in T r − of Case 2), i ≥ φ ( T ) equals the number of exterior edges in T .2.3. Finding φ − . Indeed the map φ − can be recursively constructed by reversing thesteps involved in the construction of φ . To be more precise, we describe the constructionbelow.Given a T ∈ P n , if T contains no red edges then T is a bouquet of size n and we define φ − ( T ) to be a path of length n . Otherwise, T contains at least one red edge. Let xy be the rightmost red edge at level 3 of T , and let pqxy be the path from the root p to y .Let w , . . . , w d be the children of y , for some d ( d ≥ φ − ( T ) is recursivelyconstructed with respect to xy according to the following three cases.Case 1. Vertex x has more than one child. Let Q = T − τ ( xy ). For 1 ≤ j ≤ d , wefirst construct the planted subtrees T j = φ − ( τ ( yw j )) and T d +1 = φ − ( Q ). Then φ − ( T ) isrecovered by adding the subtrees T , . . . , T d +1 under a new edge, say uv . Note that the rededge xy of T is replaced by the planting stalk uv of φ − ( T ).Case 2. Vertex x has only one child and there is another path P of length at least 2starting from q . Since xy is the rightmost red edge at level 3 of T , the path P must beon the left of the edge qx . Note that there might be some edges, say qz , . . . , qz t ( t ≥ qx . Let Q = T − τ ( qx ) − { qz , . . . , qz t } . For 1 ≤ j ≤ d , form the plantedsubtrees T j = φ − ( τ ( yw j )). Let T d +1 = φ − ( Q ) and let T d +2 be a path of length t + 1. Then φ − ( T ) is recovered by adding the subtrees T , . . . , T d +2 under a new edge uv . Note thatthe planting stalk uv of φ − ( T ) replaces the red edge xy of T , and the path T d +2 ⊆ φ − ( T )replaces the edges { qx, qz , . . . , qz t } ⊆ T . XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 7
Case 3.
Vertex x has only one child and there are no other paths of length at least 2starting from q . In this case xy is the unique red edge at level 3 in T , and there mightbe some edges emanating from q on either side of the edge qx . Suppose that there are t (resp. t ) edges on the left (resp. right) of qx . For 1 ≤ j ≤ d , form the planted subtrees T j = φ − ( τ ( yw j )). Let T d +1 and T d +2 be two paths of length t + 1 and t + 1, respectively.Then φ − ( T ) is recovered by adding the subtrees T , . . . , T d +2 under a new edge uv .From the construction of φ − , we observe that the rightmost red edge at level 3 in T is transformed to the planting stalk of φ − ( T ), and that the exterior edges recursivelyconstructed so far (in T j ) remain exterior edges in φ − ( T ). Hence the number of exterioredges in φ − ( T ) equals the number of red edges in T .We have established the following bijection. Proposition 2.5.
For ≤ k ≤ n − , there is a bijection φ : P n → P n such that aplanted tree T ∈ P n with k exterior edges is carried to the corresponding planted tree φ ( T ) containing k edges at level h with h ≡ (mod 3). Now we are able to establish the bijection Φ requested in Theorem 2.1 as well as inTheorem 1.1.Given an ordered tree T ∈ T n with k exterior edges, let u be the root of T and let v , . . . , v r be the children of u , for some r ≥
1. Then T can be decomposed into r plantedsubtrees T = τ ( uv ) ∪ · · · ∪ τ ( uv r ) . Suppose that τ ( uv i ) contains k i exterior edges, where k + · · · + k r = k . Making use of thebijection φ in Proposition 2.5, we find the corresponding planted subtrees T i = φ ( τ ( uv i ))(1 ≤ i ≤ r ), where T i contains k i red edges. Then the corresponding tree Φ( T ) = T ∪· · ·∪ T k ,obtained by merging the roots of T , . . . , T k , contains k red edges, i.e., k edges at level h with h ≡ Example 2.6.
Given the Dyck path π , shown on the left of Figure 3, with 2 blocks and4 exterior steps, we find the corresponding ordered tree T = Λ( π ), shown on the right ofFigure 3, and decompose T into two planted subtrees T = T ∪ T . Following Examples 2.2and 2.3, we construct the trees φ ( T ) and φ ( T ), respectively. Then the corresponding treeΦ( T ) is obtained by merging the roots of φ ( T ) and φ ( T ), shown on the right of Figure 7.Note that Φ( T ) contains 4 red edges. Hence, by Λ − , we obtain the corresponding Dyckpath Π( π ) = Λ − (Φ(Λ( π ))), shown on the left of Figure 7, which contains 4 up steps atheight h with h ≡ Figure 7.
A Dyck path and the corresponding ordered tree.
SEN-PENG EU AND TUNG-SHAN FU Generating functions
In this section, for m ≥ R ⊆ { , , . . . , m − } ( R = ∅ ), we study the generatingfunction G ( R ; m ) for Dyck paths counted according to length and number of up steps atheight h such that h ≡ c (mod m ) and c ∈ R . Let λ be a boolean function defined by λ (true) = 1 and λ (false) = 0. By abuse of notation, let R − i = { c ′ : c − i + m ≡ c ′ (mod m ), c ∈ R } . Theorem 3.1.
For m ≥ and a nonempty set R ⊆ { , , . . . , m − } , the generatingfunction G ( R ; m ) satisfies the equation G ( R ; m ) = 11 − xy λ (1 ∈ R ) − xy λ (2 ∈ R ) . . . − xy λ ( m − ∈ R ) − xy λ (0 ∈ R ) G ( R ; m ) . Proof.
For 0 ≤ i ≤ m −
1, we enumerate the paths π ∈ C n with respect to the number ofup steps at height h with h ≡ c (mod m ) and c ∈ R − i . By the first-return decomposition of Dyck paths, a non-trivial path π ∈ C n has a factorization π = U µ D ν , where µ and ν areDyck paths of certain lengths (possibly empty). We observe that y marks the first step U if 1 ∈ R − i . Moreover, the other up steps in the first block U µ D that satisfy the heightconstrain are the up steps in µ at height h with h ≡ c − m (mod m ). Hence G ( R − i ; m ) satisfies the following equation G ( R − i ; m ) = 1 + xy λ (1 ∈ R − i ) G ( R − i − m ) G ( R − i ; m ) . Hence we have G ( R − i ; m ) = 11 − xy λ (1 ∈ R − i ) G ( R − i − m ) . By iterative substitution and the fact R − m = R , the assertion follows. (cid:3) Example 3.2.
Take m = 3 and R = { } , we have G (0;3) = 11 − x − x − xyG (0;3) , which is equivalent to xy (1 − x )( G (0;3) ) − (1 − x + xy ) G (0;3) + (1 − x ) = 0 . Solving this equation yields G (0;3) = 1 − x + xy − p (1 − xy ) − x (1 − x )(1 − xy )2 xy (1 − x ) , which coincides with the generating function for Dyck paths counted by length and numberof exterior pairs (cf. [3, Theorem 2.3]). XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 9 A bijective proof of Theorem 1.2
Let A ( m − m ) n,j ⊆ C n (resp. A (0; m ) n,j ⊆ C n ) be the set of paths containing exactly j up stepsat height h with h ≡ m − h ≡
0) (mod m ). In this section, we shall prove Theorem1.2 by establishing the following bijection. Theorem 4.1.
For the Dyck paths in C n of height at least m − , the following results hold. (i) For j ≥ , there is a bijection Ψ j between A ( m − m ) n,j and A (0; m ) n,j − . (ii) For j = 1 , there is a bijection Ψ between A ( m − m ) n, and the set B ⊆ A (0; m ) n, , where B consists of the paths that contain no up steps at height h with h ≡ (mod m )and contain at least one up step at height h ′ with h ′ ≡ m − (mod m ). Fix an integer m ≥
2. Given a π ∈ C n of height at least m −
1, we cut π into segmentsby lines of the form L i : y = mi − i ≥ ω ⊆ π are classified into thefollowing categories.(S1) Segment ω begins with an up step starting from a line L i , for some i ≥
1, endswith the first down step returning to the line L i afterward, and never touches theline L i +1 . We call such a segment an above-block on L i .(S2) Segment ω begins with a down step starting from a line L i , for some i ≥
1, endswith the first up step reaching the line L i afterward, and never touches the line L i − . We call such a segment an under-block on L i .(S3) Segment ω is called an upward link if ω begins with an up step starting from a line L i , for some i ≥
1, and ends with the first up step reaching the line L i +1 afterward.(S4) Segment ω is called a downward link if ω begins with a down step starting froma line L i , for some i ≥
2, and ends with the first down step returning to the line L i − afterward.(S5) The segment from the origin to the first up step that reaches the line L is calledthe initial segment of π . The segment starting from the last down step that leavesthe line L to the endpoint of π is called the terminal segment of π . Example 4.2.
Take m = 3. The Dyck path π shown in Figure 8(a) is decomposed intonine segments π = ω · · · ω , where ω = [ O, A ] is the initial segment, ω = [ H, I ] is theterminal segment, ω = [ A, B ], ω = [ D, E ], and ω = [ G, H ] are above-blocks, ω = [ B, C ]and ω = [ E, F ] are under-blocks, ω = [ C, D ] is an upward link, and ω = [ F, G ] is adownward link.We have the following immediate observations.
Lemma 4.3.
According to the above decomposition of π ∈ C n with respect to lines of theform L i : y = mi − ( i ≥ ), the following facts hold. (i) An above-block ω contains a unique up step (i.e., the first step of ω ) at height h with h ≡ (mod m ), and contains no up steps at height h ′ with h ′ ≡ m − (mod m ). (ii) An under-block ω contains a unique up step (i.e., the last step of ω ) at height h with h ≡ m − (mod m ), and contains no up steps at height h ′ with h ′ ≡ (mod m ). (iii) The first (resp. last) step of an upward link ω is the unique up step at height h with h ≡ (resp. with h ≡ m − ) (mod m ) contained in ω . L AO B C F (a)(b)
A B CO D E FD E G HHG IL I L L L L Figure 8.
Decomposition of a Dyck path by lines of the form y = 3 i − i ≥ (iv) The last step of the initial segment of π is the unique up step at height m − contained in ω . (v) A downward link and the terminal segment of π contain no up steps at height h with h ≡ or m − (mod m ). For the above-blocks and under-blocks ω on some line L i , we define an operation Γ on ω by letting Γ( ω ) be the segment obtained from ω by reflecting ω about the line L i . Note thatΓ( ω ) is an under-block (resp. above-block) on L i if ω is an above-block (resp. under-block)on L i . Making use of this operation, we define an involution Ω : C n → C n as follows. The involution Ω . Given a π ∈ C n , if the height of π is less than m −
1, then we defineΩ( π ) = π . Otherwise, the path π has a factorization π = ω · · · ω d ( d ≥ standard form , with respect to lines of the form L i : y = mi − i ≥ ω isthe initial segment, ω d is the terminal segment, and each ω r is a segment in one of thefour categories (S1)–(S4), for 2 ≤ r ≤ d −
1. The map Ω is defined by carrying π toΩ( π ) = ω b ω · · · b ω d − ω d , where b ω r = (cid:26) Γ( ω r ) if ω r is an above-block or an under-block ω r if ω r is an upward link or a downward link , for 2 ≤ r ≤ d −
1. It is obvious that Ω is an involution.
Example 4.4.
Take m = 3 and the path π shown in Figure 8(a). As shown in Example4.2, π is factorized into the standard form π = ω . . . ω . The corresponding path Ω( π ) = ω b ω . . . b ω ω is shown in Figure 8(b), where b ω = Γ( ω ), b ω = Γ( ω ), b ω = ω , b ω = Γ( ω ), b ω = Γ( ω ), b ω = ω , and ω = Γ( ω ).Let F ( m ) n,j,k ⊆ C n be the set of paths containing j up steps at height h with h ≡ m − m ) and k up steps at height h ′ with h ′ ≡ m ). Proposition 4.5.
For j ≥ and k ≥ , the involution Ω induces a bijection Ω j,k : F ( m ) n,j,k → F ( m ) n,k +1 ,j − . XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 11
Proof.
In particular, for ( j, k ) = (1 , , : F ( m ) n, , → F ( m ) n, , to be an identitymapping, i.e., Ω , ( π ) = π , for π ∈ F ( m ) n, , .For ( j, k ) = (1 , π ∈ F ( m ) n,j,k , we factorize π into the standard form π = ω · · · ω d ( d ≥ L i : y = mi − i ≥ t segmentsamong ω , . . . , ω d − , which are upward links. Since π contains j up steps at height h with h ≡ m − m ) and k up steps at height h ′ with h ′ ≡ m ), by Lemma 4.3,there are j − − t segments µ , . . . , µ j − − t ∈ { ω , . . . , ω d − } that are under-blocks and k − t segments ν , . . . , ν k − t ∈ { ω , . . . , ω d − } that are above-blocks. Under the involutionΩ, the corresponding path Ω( π ) contains j − − t above-blocks b µ , . . . , b µ j − − t and k − t under-blocks b ν , . . . , b ν k − t . Along with the t upward links in Ω( π ) and the initial segment,by Lemma 4.3, Ω( π ) contains k + 1 up steps at height h with h ≡ m − m ) and j − h ′ with h ′ ≡ m ). Hence Ω j,k ( π ) = Ω( π ) ∈ F ( m ) n,k +1 ,j − .It is easy to see that Ω − j,k = Ω | F ( m ) n,k +1 ,j − = Ω k +1 ,j − : F ( m ) n,k +1 ,j − → F ( m ) n,j,k . (cid:3) Example 4.6.
Following Example 4.4, the path π shown in Figure 8(a) contains four upsteps at height h with h ≡ h ′ with h ′ ≡ , ( π ), shown in Figure 8(b), contains five up steps at height h with h ≡ h ′ with h ′ ≡ Proof of Theorem 4.1. (i) For j ≥
2, we have A ( m − m ) n,j = ∪ k ≥ F ( m ) n,j,k and A (0; m ) n,j − = ∪ k ≥ F ( m ) n,k +1 ,j − . It follows from Proposition 4.5 that the map Ψ j : A ( m − m ) n,j → A (0; m ) n,j − is established by the refinement,Ψ j | F ( m ) n,j,k = Ω j,k : F ( m ) n,j,k → F ( m ) n,k +1 ,j − , for k ≥ . (ii) For j = 1, we have A ( m − m ) n, = ∪ k ≥ F ( m ) n, ,k and B = ∪ k ≥ F ( m ) n,k +1 , . It follows fromProposition 4.5 that the map Ψ : A ( m − m ) n, → B is established by the refinement,Ψ | F ( m ) n, ,k = Ω ,k : F ( m ) n, ,k → F ( m ) n,k +1 , , for k ≥ . (cid:3) Now we are able to prove Theorem 1.2. For j ≥
2, by Theorem 4.1(i), we have[ y j x n ] { G ( m − m ) − y · G (0; m ) } = g ( m − m ) n,j − g (0; m ) n,j − = |A ( m − m ) n,j | − |A (0; m ) n,j − | = 0 . For j = 1, by Theorem 4.1(ii), we have g ( m − m ) n, = |A ( m − m ) n, | = |B| , where B consists ofthe paths in C n that contain no up steps at height h with h ≡ m ) and contain atleast one up step at height h ′ with h ′ ≡ m − m ). Hence[ y x n ] { G ( m − m ) − y · G (0; m ) } = g ( m − m ) n, − g (0; m ) n, = |B| − |A (0; m ) n, | = − U m − ( √ x ) √ xU m − ( √ x ) , which is the negative of the number of paths in C n of height at most m −
2. Moreover,[ y x n ] { G ( m − m ) − y · G (0; m ) } = g ( m − m ) n, is also the number of paths in C n of height at most m −
2. This completes the proof of Theorem 1.2. Concluding Notes
Given a positive integer s , an s -ary path of length n is a lattice path from (0 ,
0) to(( s + 1) n, ,
1) and grand down step (1 , − s ), that never passes below the x -axis. When s = 1 it is an ordinary Dyck path. One can consider the s -generalizationof pyramids and exterior pairs on s -ary paths. For example, a pyramid of height k is asuccession of sk up steps followed immediately by k down steps. An exterior down step isa down step that does not belong to any pyramid. Let p ( s ) n,k (resp. e ( s ) n,k ) be the number of s -ary paths of length n with a pyramid weight of k (resp. with k exterior down steps), andlet P and E be the generating functions for p ( s ) n,k and e ( s ) n,k , respectively, where P = P ( x, y ) = X n ≥ X k ≥ p ( s ) n,k y k x n , E = E ( x, y ) = X n ≥ X k ≥ e ( s ) n,k y k x n . Note that E ( x, y ) = P ( xy, y − ) since an s -ary path of length n with a pyramid weight of k contains n − k exterior down steps. Proposition 5.1.
The generating functions P and E satisfy respectively the equations P = 1 + x ( P s − − y − xy ) P, E = 1 + x ( yE s + 1 − y − x ) E. Proof.
By the first-return decomposition of s -paths, a nontrivial s -path π has a factorization π = U µ · · · U s µ s Dν , where D is the first (grand) down step that returns to the x -axis, U i is the last up step in the first block β = U µ · · · U s µ s D ⊆ π , which rises from theline y = i − y = i (1 ≤ i ≤ s ), and µ , . . . , µ s , ν are s -ary paths of certainlengths (possibly empty). To enumerate the s -ary paths with respect to pyramid weightand length, we observe that the first down step D is marked y if and only if the first block β is a pyramid, in which case µ = · · · = µ s − = ∅ and µ s is a pyramid of certain length.Hence P satisfies the equation P = 1 + x ( P s − − xy + y − xy ) P. Similarly, if we enumerate the s -ary paths with respect to the number of exterior downsteps and length, then the first down step D is marked y if and only if the first block β isnot a pyramid. Hence E satisfies the equation E = 1 + x ( y ( E s − − x ) + 11 − x ) E, as required. (cid:3) We are interested to know if there is any statistic regarding up steps, which is equidis-tributed with p ( s ) n,k (or e ( s ) n,k ) on the s -ary paths.Theorem 1.2 gives a relation between the two generating functions G ( m − m ) and G (0 ,m ) .It is natural to consider if there is any relation between G ( i ; m ) and G ( j,m ) , for 0 ≤ i, j ≤ m − Conjecture 5.2.
The following relations hold.
XTERIOR PAIRS AND UP STEP STATISTICS ON DYCK PATHS 13 (i)
For m ≥ , we have G ( m − ,m ) − G (1 ,m ) = (1 − t ) U m − ( √ x ) U m − ( √ x ) − y √ xU m − ( √ x ) . (ii) For m ≥ , we have G ( m − ,m ) − G (2 ,m ) = (1 − t ) U m − ( √ x ) U m − ( √ x ) − yU m − ( √ x ) + √ xU m − ( √ x ) . References [1] T. Chow, J. West, Forbidden subsequences and Chebyshev polynomials, Discrete Math. 204 (1999)119–128.[2] E. Deutsch, H. Prodinger, A bijection between directed column-convex polyominoes and ordered treesof height at most three, Theoretical Computer Science 307 (2003) 319–325.[3] A. Denise, R. Simion, Two combinatorial statistics on Dyck paths, Discrete Math. 137 (1995) 155–176.[4] N. Dershowitz, S. Zaks, Applied Tree Enumerations, Lecture Notes in Computer Science, vol. 112,Springer, Berlin, 1981, pp. 180–193.[5] P. Flajolet, Combinatorial aspects of continued fractions, Discrete Math. 306 (2006) 992–1021.[6] C. Krattenthaler, Permutations with restricted patterns and Dyck paths, Adv. Applied Math. 27 (2001)510–530.[7] G. Kreweras, Joint distributions of three descriptive parameters of bridges, Lecture Notes in Mathe-matics, vol. 1234, Springer, Berlin, 1986, pp. 177–191.[8] T. Mansour, A. Vainshtein, Restricted permutations, continued fractions, and Chebyshev polynomials,Electron. J. Combin. 7 (2000) R17.
Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung 811,Taiwan, ROC
E-mail address : [email protected] Mathematics Faculty, National Pingtung Institute of Commerce, Pingtung 900, Taiwan,ROC
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