Combinatorics of Riordan arrays with identical A and Z sequences

Discrete Mathematics 312 (2012) 2040–2049 Contents lists available at SciVerse ScienceDirect Discrete Mathematics journal homepage: www.elsevier.com/locate/disc Combinatorics of Riordan arrays with identical A and Z sequences✩ Gi-Sang Cheon a,∗ , Hana Kim a , Louis W. Shapiro b a Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea b Department of Mathematics, Howard University, Washington, DC 20059, USA article abstract info Article history: Received 5 June 2011 Received in revised form 16 March 2012 Accepted 16 March 2012 Available online 12 April 2012 In theory, Riordan arrays can have any A-sequence and any Z -sequence. For examples of combinatorial interest they tend to be related. Here we look at the case that they are identical or nearly so. We provide a combinatorial interpretation in terms of weighted Łukasiewicz paths and then look at several large classes of examples. © 2012 Elsevier B.V. All rights reserved. Keywords: Riordan array Łukasiewicz path Dyck path Consistent Riordan array 1. Introduction We begin by briefly describing the concept of the Riordan group. Let Fn for n = 0, 1, 2, . . . be the set of formal power series defined by Fn = {f (z ) = fn z n + fn+1 z n+1 + fn+2 z n+2 + · · · |fn ̸= 0, fi ∈ C}. A Riordan array denoted (g (z ), f (z )) is an infinite lower triangular matrix such that the generating function of the kth column for k = 0, 1, 2, . . . is g (z )f k (z ) where g (z ) ∈ F0 , g (0) = 1 and f (z ) ∈ F1 . Given a Riordan array (g (z ), f (z )) and column vector h = [h0 , h1 , h2 , . . .]T , then the product of (g (z ), f (z )) and h gives a column vector whose generating function is g (z ) · h(f (z )). Simply we write (g (z ), f (z ))h(z ) = g (z ) · h(f (z ))  j where h(z ) = j≥0 hj z . This property is called the fundamental theorem for Riordan arrays and this leads to the matrix multiplication for Riordan arrays: (g (z ), f (z ))(h(z ), ℓ(z )) = (g (z ) · h(f (z )), ℓ(f (z ))). The set of all Riordan arrays forms a group under the above operation of a matrix multiplication. This is the Riordan group introduced by Shapiro et al. [11]. One subgroup of the Riordan group that we will encounter is the Bell subgroup which consists of the matrices of the form (g (z ), zg (z )). A Riordan array R = (g (z ), f (z )) = [rn,k ]n,k≥0 can be characterized [5,6,14] by two sequences A = (a0 , a1 , . . .) and Z = (z0 , z1 , . . .) such that rn+1,0 =  j≥0 zj rn,j and rn+1,k+1 =  j≥ 0 aj rn,k+j ✩ This work was supported by the National Research Foundation of Korea Grant funded by the Korean Government (NRF-2011-0003187). ∗ Corresponding author. E-mail addresses: [email protected] (G.-S. Cheon), [email protected] (H. Kim), [email protected] (L.W. Shapiro). 0012-365X/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2012.03.023 (1) G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 2041 for n, k ≥ 0. If A(z ) and Z (z ) are the generating functions for the A- and Z -sequences respectively, then it follows that g (z ) = 1 and 1 − zZ (f (z )) (2) f (z ) = zA(f (z )). Riordan arrays arise in the enumeration of lattice paths, e.g. Dyck paths, Motzkin paths, and Schröder paths and so on; see [3,10,11,14]. In particular, a connection between Riordan arrays and ‘unweighted’ lattice paths having privileged access to the main diagonal has been studied in [6] by means of the A-matrix, which includes the concept of A- and Z -sequences. As suggested by the referee, the concept of consistent Riordan arrays is also related to the concept of lattice paths having unprivileged access to the main diagonal in [6]. The concept of the A- and Z -sequences is often useful for enumeration problems such as deriving combinatorial identities or recursion formulas for entries of a Riordan array. In general, Riordan arrays can have any A-sequence and any Z -sequence, independently. In this paper, we consider Riordan arrays such that the A- and Z -sequences are identical or nearly so. We begin with a Dyck path from (0, 0) to (2n, 0) using the up step U = (1, 1) and the down step D = (1, −1) as its possible steps with the additional constraint that the path cannot go below the x-axis. If we recount the partial Dyck paths that end at (2n − k, k), we get the familiar Catalan triangle matrix: 1 1 2 (C (z ), zC (z )) =  5  1 2 5 14 14 1 3 9       1 4 1 ··· √ where C (z ) = (1 − 1 − 4z )/2z. We note that here both the A- and Z -sequence are (1, 1, 1, . . .) i.e., A(z ) = Z (z ). A Riordan array (g (z ), f (z )) is said to be consistent if A(z ) = Z (z ). It follows from (2) that (g (z ), f (z )) is consistent if and only if g (z ) = 1−1f (z ) or g (z ) = 1 + g (z )f (z ) where g ′ (0) ̸= 0. Hence a consistent Riordan array can be expressed as either 1 ) or ( 1−1f (z ) , f (z )). In particular, ( 1−1f (z ) , f (z ))−1 = (1 − z , f¯ (z )) where f¯ (z ) is the compositional inverse of f (z ) (g (z ), g (gz()− z) i.e., f (f¯ (z )) = f¯ (f (z )) = z. One purpose of this paper is to show that every consistent Riordan array with nonnegative integer coefficients has a certain combinatorial interpretation. Specifically, in Section 2 we first provide a combinatorial interpretation in terms of weighted Łukasiewicz paths. Then several large classes of examples such as k-Dyck paths, colored Schröder paths, colored Motzkin paths, and ordered trees are discussed in Section 3. 2. Combinatorial interpretations for consistent Riordan arrays Throughout this paper, a Łukasiewicz path (or simply L-path) is a lattice path that starts at the origin, cannot go below the x-axis and  hasas possible steps S−r = (1, −r )(r ≥ −1). It is known that the number of L-paths from (0, 0) to (n, k) is ηn,k := k+1 n+1 2n−k n , which is the (n, k)-entry of the Catalan triangle (C (z ), zC (z )) above. Let us consider an L-path P with weights. The weight of the step S−r is denoted by ω(S−r ) and the weight of P is defined as the product of weights of the steps used and is denoted by ω(P ). Theorem 2.1. Let R = [rn,k ]n,k≥0 be a Riordan array with A = (a0 , a1 , . . .). Then R is consistent if and only if rn,k is the sum of weights of weighted L-paths from the origin to the point (n, k) using the weights ω(S−r ) =  ar ar +1 if S−r touches the x-axis, otherwise (3) where r ≥ 0 and ω(S1 ) = a0 .  n Proof. Let A(z ) = Z (z ) = n≥0 an z . Let us define σ (n, k) to be the sum of weights of weighted L-paths from (0, 0) to (n, k) using the weights given by (3) where σ (0, 0) := 1. It is enough to consider the final steps of the paths that go to the point (n, k). If k = 0, the final step S−r (r = 0, 1, 2, . . .) meets the x-axis. Hence from (3) we have σ (n + 1, 0) =  j≥ 0 ω(S−j )σ (n, j) =  aj σ (n, j). j≥ 0 Let k ≥ 1. We note that when k = 1 the paths coming from the x-axis have S1 as their final step. Otherwise, the final steps of paths do not meet the x-axis. Since ω(S1 ) = a0 for all k ≥ 1, from (3) we have σ (n + 1, k) = ω(S1 )σ (n, k − 1) + = a0 σ (n, k − 1) +   j≥ 0 j≥ 0 ω(S−j )σ (n, k + j) aj+1 σ (n, k + j) =  j≥ 0 aj σ (n, k + j − 1). 2042 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 By substituting aj for zj in (1), we see that rn,k and σ (n, k) satisfy the same recurrence relation with the same initial condition r0,0 = σ (0, 0) = 1. It is easy to show that the converse holds for k ≥ 0. Hence the proof is complete.  Even at this point there are two ways to develop examples. We can start with an A = Z -sequence and use it to put appropriate weights on L-paths or we can use any g (z ) together with g (z ) − 1 as our leftmost columns to develop the rest of the array. To compute the row sums of any Riordan array we multiply by the column vector [1, 1, 1, . . .]T which has the generating function 1/(1 − z ). It is convenient to switch freely between a sequence, the sequence as a column vector, and its generating function. In the case of consistent arrays we get  g (z ), g (z ) − 1 g (z )  1 = g (z ) · 1−z  1 1−( g (z )−1 g (z ) )  = g 2 (z ). (4) g (z )−1 It follows that the row sums of a consistent array (g (z ), g (z ) ) are the elements of the k = 1 column of the Riordan matrix (g (z ), zg (z )) except for a zero as the initial term. By Theorem 2.1, g 2 (z ) represents the generating function for the sum of weights of L-paths ending at some point on the line x = n using the weights given by (3) where (a0 , a1 , . . .) is the sequence with the generating function h(zz ) for the 1 solution h(z ) of g (h(z )) = 1− . Incidentally, g 2 (z ) is also the generating function counting all points on the x-axis for all z such weighted L-paths. For an integer k, a Riordan array (g , f ) is said to be k-consistent if A(z ) = Z (z ) − k i.e., A = (a0 , a1 , . . .) and Z = (k + a0 , a1 , . . .). It follows from (2) that (g (z ), f (z )) is k-consistent if and only if g (z ) = 1 1 − kz − f (z ) or f (z ) = (1 − kz )g (z ) − 1 . g (z ) (5) For instance, let us consider the k-consistent array (g (z ), f (z )) with A = (1, 1, 1, . . .) and Z = (k + 1, 1, 1, . . .). Then we have g (z ) = C (z ) and f (z ) = zC (z ). 1 − kzC (z ) In particular, we obtain the k-consistent Riordan arrays for k = −1, 0, 1, 2 respectively: (F (z ), zC (z )), (C (z ), zC (z )), (C 2 (z ), zC (z )), (B(z )C (z ), zC (z )) where C (z ), F (z ) and B(z ) are the generating functions for the Catalan numbers, the Fine numbers (A000957, [13]) [2] and the central binomial coefficients, respectively. A similar combinatorial interpretation as in Theorem 2.1 is given for a k-consistent Riordan array. Also see [8]. (k) ( k) Theorem 2.2. Let R = [ri,j ]i,j≥0 be a k-consistent Riordan array with A = (a0 , a1 , . . .). Then ri,j is the sum of weights of weighted L-paths from the origin to the point (i, j) using the weights ω(S−r ) =  kδr ,0 + ar ar +1 if S−r touches the x-axis, otherwise where r ≥ 0, ω(S1 ) = a0 and δr ,0 denotes the Kronecker delta. The following theorem shows that 0-consistent and 1-consistent Riordan arrays may have different combinatorial interpretations by means of lattice paths. Theorem 2.3. Let gi,j be the number of lattice paths from (0, 0) to ((m + 1)i − j, mj) using the step set {(1, 1), (1, −m), (m + 1, 0)} for a nonnegative integer m, where the paths cannot go below the x-axis. Then the number array [gi,j ]i,j≥0 is a 1-consistent  i Riordan array given by (G(z ), zGm (z )), where G(z ) = i≥0 gi,0 z . However, if we do not allow flat steps on the x-axis then the array is a 0-consistent Riordan array given by ( 1−zG1m (z ) , zGm (z )), where we refer to the step (m + 1, 0) as a flat step. Proof. Let us consider a lattice path  whose first return to the x-axis is at ((m + 1)ℓ, 0) (ℓ ≥ 1). Since gi,0 counts lattice paths i m+1 ending at ((m + 1)i, 0) and G(z ) = (z ) as illustrated in Fig. 1. i≥0 gi,0 z , we have G(z ) = 1 + zG(z ) + zG Let j ≥ 1. Since gi,j counts lattice paths ending at ((m + 1)i − j, mj) and the jth column generating function is a similar diagram to that in Fig. 1 shows that i j mj+1 gi,j = [z ]z G Since zGm (z ) = i m  i i≥0 j gi,j z = z G mj+1 (z ). Thus j (z ) = [z ]G(z ) · (zG (z )) . (1−z )G(z )−1 , G(z ) it follows from (5) that [gi,j ]i,j≥0 = (G(z ), zGm (z )) is a 1-consistent Riordan array.  i≥0 gi,j z i , 2043 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 Fig. 1. A lattice path with a first return at ((m + 1)i, 0). Now let gi∗,j be the number of lattice paths ending at ((m + 1)i − j, mj) with no flat step (m + 1, 0) on the x-axis, and let g (z ) =  i≥0 m gi∗,0 z i . Since every lattice path with no flat step on the x-axis starts with the step (1, 1), we have g (z ) = 1 + zg (z )G (z ). For j ≥ 1, by a similar argument to the previous case we have g (z ) = 1 , 1−zGm (z ) it follows from (5) that [gi∗,j ]i,j≥0 =( 1 1−zGm (z ) m  i≥0 , zG (z )) is a 0-consistent array. gi∗,j z i = g (z ) · (zGm (z ))j . Since  For instance, let m = 1. Since the step set is {(1, 1), (1, −1), (2, 0)}, gi,0 counts the number of Schröder paths from (0, 0) √  1−6z +z 2 1−z − i to (2i, 0), and G(z ) = r (z ) = i≥0 gi,0 z = 2z the 1-consistent and 0-consistent Riordan arrays start as 1 2 6 (r (z ), zr (z )) =  22  90 1 4 16 68 1 6 30 1  1 8 ··· is the generating function for the large Schröder numbers. Thus 1 3 (s(z ), zr (z )) =  11  45   ,   1 1 3 11 45 1 5 23 ···  1 7      1 1 respectively, where s(z ) = 1−zr is the generating function for the small Schröder numbers. In fact, the number of Schröder (z ) paths from (0, 0) to (5, 1) is 16 = [z 3 ]r (z ) · (zr (z ))1 . Eliminating the paths which do have a flat step on the x-axis leaves 11 = [z 3 ]s(z ) · (zr (z ))1 . Here are three more examples of consistent arrays (g (z ), f (z )). Familiar sequences show up but in unfamiliar arrays.  1 1  2  , zm(z ) =  5 1 − zm(z ) 13  1 35 1 2 5 13 35 1 3 9 26  1 4 14 ··· ··· 1 1   3  1 , z (1 + 2zm(z )) =  √ 7 2 1 − 2z − 3z 19  51 √ 1−z − 1−2z −3z 2 1 3 7 19 51 1 5 1    ,    1 5 15 45 1 7 27 ··· ··· 1 1 1  (1 + zm(z ), zR(z )) =  2 4  9  1 9 1    ,    1 1 2 4 9  1 1 3 6 1 1 4 ··· ··· 1 1 1    ,    (6) 1+zm(z ) where m(z ) = and R(z ) = 1+z are generating functions for the Motzkin numbers and Riordan numbers 2z 2 respectively. Many questions now present themselves. First might be to give combinatorial interpretations to all the entries in these arrays, not just the two left hand columns with the generating functions g (z ) and g (z )f (z ). Even the entries in the next column with generating function g (z )f 2 (z ) leads to some sequences that were not listed in Sloane’s EIS [13]. This is answered in part in the next section starting with L-paths using the ai as weights and then converting these weighted paths to k-Dyck paths. A further bijection leads to 2-paths which are just Dyck paths. Also there are generalizations of several of these examples in the next section. One of these are the α -Motzkin paths with A(z ) = 1 + α z + z 2 and another is the α -Schröder z = 1 + (α + 1)z + (α + 1)z 2 + (α + 1)z 3 + · · ·. paths with A(z ) = 11+α −z Remark. We can think about more of a combinatorial approach and the key is that g (z ) = 1−1f (z ) so that the kth column of our (g (z ), f (z )) represents trees (or paths) consisting of a green tree on the left and then k planted red trees on the right. We get as many examples as we want. For instance if f (z ) = z + z 2 then we get a Fibonacci consistent array and each planted 2044 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 Fig. 2. The five ordered trees with 4 edges, 2 planted red trees. subtree is just one or two edges with no branching other than at the root. The first few entries are as follows:  1 1 − z − z2 , z + z2  1 1 2  = 3 5  1 2 3 5 8 8  1 3 5 8 1 4 8 ··· ··· 1 5        1 and the five ordered trees which are counted by the (4, 2)-entry are illustrated in Fig. 2. 3. Consistent Riordan arrays with special A-sequences In this section, we will look at other combinatorial interpretations for the consistent Riordan arrays for some special A-sequences. 3.1. k-Dyck path interpretation A k-Dyck path for k ∈ N is a lattice path from (0, 0) to (kn, 0) using the steps U = (1, 1),  Dk =  (1, 1 − k) which is not kn allowed to go below the x-axis. It is well known that the number of k-Dyck paths is (k−11)n+1 n , n ≥ 0, the sequence of k-ary numbers. Its generating function Bk (z ) satisfies the functional equation Bk (z ) = 1 + zBkk (z ). It can be shown [4] that the following identity is valid for all real numbers r: Brk (z ) =  n ≥0 r kn + r  kn + r n  zn. (7) (k)  Let Bk = [bi,j ]i,j≥0 = (Bk (z ), 1 − 1/Bk (z )). Then Bk is the consistent array with A(z ) = Z (z ) = 1/(1 − z )k−1 = j≥ 0  j+k−2 j  z j . Further, Bk can be factored as Bk = (Bk (z ), zBk (z ))(1, zBk−1 (z )). (k) By Theorem 2.1, bi,j is the sum of weights of L-paths from the origin to the point (i, j) using the weights   k+r −2    r  ω(S−r ) =   k + r −1   r +1 if S−r touches the x-axis, (8) otherwise where ω(S1 ) = 1. It follows from (7) and Bk (z ) = 1 + zBkk (z ) that (k)  bi,j = [z i ]Bk (z ) 1 − 1 Bk (z ) j = [z i−j ]B(kk−1)j+1 (z ) = (k − 1)j + 1 (k − 1)i + 1  ki − j i−j  . (9) For k = 2 there is a classic bijection [15] from L-paths to Dyck paths. Replace each S−r by UDr +1 . In fact, in this case, since   2+r −2   =1  r  ω(S−r ) =   2+r −1   =1 r +1 if S−r touches the x-axis, otherwise, the regular unweighted L-paths are counted by B2 (z ) = C (z ) as expected. In the following theorem, for any k ≥ 2 we obtain a bijection between weighted L-paths and sets of partial k-Dyck paths which have no restriction on the end points. G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 2045 Fig. 3. The L-path with ω(L) = 6 and the corresponding six 3-Dyck paths from (0, 0) to (14, 2). Fig. 4. The composition of 3-Dyck path D by disjoint subpaths. Theorem 3.1. Let k ≥ 2. There is a bijection between the set of weighted L-paths ending at the point (i, j) using the weights given by (8) and the collection of sets of partial k-Dyck paths ending at the point (ki − j, (k − 1)j). Proof.  Let  L be the set of weighted L-paths from the origin to (i, j) using the weights (8) with |L| = η, where η := ηi,j = j+1 i+1 2i−j i . We first define the rule Ω1 changing a weighted L-path into a set of partial k-Dyck paths. (i) For a L ∈ L, we replace S1 with the steps U k−1 =  U · · · U. k−1 (ii) For r ≥ 0 if S−r meets the x-axis then  we replace S−r with (k + r ) steps UPk+r −2 Dk where Pk+r −2 is a path consisting of (k − 2) U’s and r Dk ’s. This yields k+r −2 r k-Dyck paths. (iii) If S−r does not meet the x-axis then we replace S−r with (k + r ) steps UPk∗+r −1 where Pk∗+r −1 is a path consisting of (k − 2) U’s and (r + 1) Dk ’s. It is easy to show that all the partial k-Dyck paths obtained from the L-path L ∈ L ending at (i, j) have the same end point (ki − j, (k − 1)j). After Ω1 is applied the number of such partial k-Dyck paths coincides with the weight of L. Let us illustrate with a small example that will show the ideas with less clutter. For k = 3, each L-path with the weights given by (8) now corresponds to the set of partial 3-Dyck paths. For instance we start with a L-path L with ω(L) = 6 ending at (5, 1). Then by the rule Ω1 , we obtain a set of 6 partial 3-Dyck paths ending at (14, 2), see Fig. 3. Let D(L) be the set of all partial k-Dyck paths obtained from L ∈ L by the rule Ω1 and D be the set of all partial k-Dyck paths ending at (ki − j, (k − 1)j). We claim that D is completely partitioned into η sets D(L) where L ∈ L and |L| = η. To see this, we first consider the inverse of the rule Ω1 . Given a partial k-Dyck path D ∈ D of length ki − j, let D = s1 s2 · · · ski−j where sℓ ∈ {U , Dk }. Then divide D into disjoint subpaths P1 , P2 , . . . starting with U and containing exactly k − 1 U’s, see Fig. 4. Replace Pℓ with S1−r where r is the number of down steps in Pℓ . Finally, assign a weight on each step S1−r according to (8). Then we obtain the weighted L-path L(D) corresponding to the k-Dyck path D ∈ D , see Fig. 5. As with the previous figure, it can easily be shown that the resulting weighted L-path L(D) ends at the point (i, j). ′ Let us now define the relation ∼ on the set D by D ∼ D′ if and only if L(D) = L(D ) . It is easy to show that ∼ is an equivalence relation on D . Thus D(L) coincides with the equivalence class containing a k-Dyck path D such that L = L(D) ,  and so D = L∈L D(L). Hence a map ϕ : L → {D(L)|L ∈ L} defined by ϕ(L) = D(L) gives a bijection. This completes the proof.  2046 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 Fig. 5. The weighted L-path L(D) corresponding to D in Fig. 4. By Theorems 2.1 and 3.1, we have the following corollary. (k) (k) Corollary 3.2. Let bi,j be the same as in (9). Then bi,j counts the number of partial k-Dyck paths from (0, 0) to (ki − j, (k − 1)j). Further,     n  2 kn + 2 (k − 1)j + 1 kn − j = . (k − 1)n + 1 n − j kn + 2 n j= 0 Proof. The identity (10) immediately follows from (4) and (7). (10)  3.2. Colored Schröder path interpretation Rogers [10] studied two kinds of Schröder polynomials rn (α) and sn (α) obtained from the weighted Schröder paths. Specifically for an integer α ≥ 0, rn (α) is the sum of weights of weighted Schröder paths ending at (2n, 0) with the steps U = (1, 1), D = (1, −1) and F = (2, 0) weighted by 1, 1 and α respectively, and sn (α) is the sum of weights of the same weighted Schröder paths with no flat steps on the x-axis. It is known that R(α) = 1 + α zR(α) + z (R(α))2 where R(α) =  n ≥0 r n n and S (α) = 1 − α zS (α) + z (1 + α)(S (α))2 (α)z and S (α) =  n≥0 sn (11) n (α)z . Theorem 3.3. For any complex number α , the Riordan array (S (α), zR(α)) is a consistent array with A = (1, α + 1, α + 1, . . .). Proof. Since R(α) = (1 + α)S (α) − α , it follows from (11) that zR(α) = z ((1 + α)S (α) − α) = z zS (α)((1 + α)S (α) − α) zS (α) = S (α) − 1 S (α) . 1+α zR(α) z Hence (S (α), zR(α)) is consistent. In addition, zR(α) = z 1−zR(α) in (11) implies that A(z ) = 11+α = 1 + (α + 1)z + (α + −z 2 1)z + · · ·.  (α) (α) Let (S (α), zR(α)) = [sn,k ]n,k≥0 . By Theorem 2.1, sn,k can be viewed as the sum of weights of L-paths from the origin to the point (n, k) using the weights ω(S−r ) =  1 α+1 if r = 1 or if the flat step S0 is on the x-axis, otherwise. (12) A weighted Schröder path is said to be α -Schröder path if its flat step has weight α , which can be thought of as α different colors. We denote the flat step with color ℓ by Fℓ (also see [10]). Theorem 3.4. Let α ≥ 1. There is a bijection between the set of weighted L-paths ending at the point (n, k) using the weights given by (12) and the collection of sets of partial α -Schöder paths ending at the point (2n − k, k) with no flat steps on the x-axis. Proof. We proceed by a similar method to Theorem 3.1. Let L be the set of weighted L-paths from the origin to (n, k) using the weights (12) with |L| = η. We begin with the rule Ω2 changing a weighted L-path to a partial α -Schröder path. (i) For an L ∈ L, if S0 meets the x-axis then we replace S0 with UD and if S0 does not meet the x-axis then we replace S0 with one of {UD, F1 , . . . , Fα }. (ii) We replace S1 with U and replace S−r (r ≥ 1) with one of {UDr +1 , F1 Dr , . . . , Fα Dr }. It can be shown that every partial α -Schröder path obtained from L by the rule Ω2 has the same end point (2n − k, k). The set of such α -Schröder paths is denoted by A(L). Obviously, |A(L)| = ω(L). Let S be the set of all partial α -Schröder paths ending at (2n − k, k). Let us now show that S is completely partitioned into the sets A(L) where L ∈ L and |L| = η. First we consider the inverse rule of Ω2 . Given a partial α -Schröder path s ∈ S of length 2n − k, let s = s1 s2 · · · s2n−k where sℓ ∈ {U , D, F1 , . . . , Fα }. Then divide s into disjoint subpaths P1 , P2 , . . . starting with a step which is not D and containing exactly one of {U , F1 , . . . , Fα }. Replace 2047 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 Pℓ with S1−r (resp. S−r ) if Pℓ begins with U (resp. Fi ), where r is the number of D’s in Pℓ . Finally, assign a weight on each step S−r by ω(S−r ) in (12). Then we obtain the weighted L-path L(s) ending at (n, k) corresponding to the α -Schröder path s ∈ S ending at (2n − k, k). ′ Let us now define the relation ∼ on the set S by s ∼ s′ if and only if L(s) = L(s ) . It is easy to show that ∼ is an equivalence (s) relation  on S . Thus A(L) coincides with the equivalence class containing a α -Schröder path s such that L = L , and so S = A ( L ) . Hence a map φ : L → { A ( L )| L ∈ L } defined by φ( L ) = A ( L ) is indeed a bijection. Thus the proof is L∈L completed.  By Theorems 2.1 and 3.4, we have the following corollary. (α) Corollary 3.5. sn,k counts the number of α -Schröder paths from (0, 0) to (2n − k, k) with no flat steps on the x-axis. 3.3. Colored Motzkin path interpretation A Motzkin path is a lattice path that starts at the origin and ends on the x-axis, cannot go below the x-axis and has as its possible steps U = (1, 1), D = (1, −1) and the flat step F = (1, 0). The number of Motzkin paths ending at the point (n, 0) is denoted Mn and is called the nth Motzkin number. A Motzkin path with no flat steps on the x-axis is called a Riordan path. The number of Riordan paths ending at (n, 0) is called the nth Riordan number Rn . The Motzkin numbers have been generalized in several ways, for example see [9,12]. The Motzkin paths whose steps are all colored have been studied in [7] in connection with Riordan arrays. In this section, we consider so called α -Motzkin path which is the (partial) Motzkin path with flat steps F1 , . . . , Fα of α colors. Let Mn (α) and Rn (α) be the number of α -Motzkin paths and the number of α -Riordan paths ending at (n, 0). The corresponding generating functions will be denoted by M (α) and R(α), respectively. It is known [12] that 1 − αz − M (α) =  (1 − α z )2 − 4z 2 2z 2 = 1 + α zM (α) + z 2 (M (α))2 (13) and R(α) =  (1 − α z )2 − 4z 2 = 1 + z 2 M (α)R(α). 2z (α + z ) 1 + αz − (14) We note that M (1), R(1) and R(2) give the Motzkin numbers, the Riordan numbers (A005043, [13]) and the Fine numbers (A000957, [13]), respectively. Also R(3) is known as the generating function for (A117641, [13]), which counts the number of 3-Motzkin paths with no flat steps on the x-axis. We are now interested in combinatorial interpretations for the two consistent Riordan arrays Mα =  1 1 − zM (α)  , zM (α)  and Rα = 1 1 − α zR(α) (α)  , α zR(α) . (α) Theorem 3.6. For α ≥ 0, let Mα = [mn,k ]n,k≥0 . Then mn,k counts the number of α -Motzkin paths of length n − 1 ending at level ≥k − 1. Proof. Since the 0th column and the 1st column of Mα are the same, it suffices to give a combinatorial interpretation for Mα := (M (α)/(1 − zM (α)), zM (α)). First we count the α -Motzkin paths of length n ending at any level. The generating function for the numbers of those paths is given by M (α) M (α) + z (M (α))2 + z 2 (M (α))3 + · · · = 1 − zM (α) , which is the generating function for the 0th column of Mα . Similarly, the α -Motzkin paths of length n ending at level k or higher are counted by the generating function: z k (M (α))k+1 + z k+1 (M (α))k+2 + · · · = z k (M (α))k+1 1 − zM (α) = M (α) 1 − zM (α) (zM (α))k , (α) which is the generating function for the k-th column of M α . Since Mα is consistent, mn,k counts the number of α -Motzkin paths of length n − 1 ending at level ≥k − 1.  A simple computation shows that M α can be expressed as the matrix decomposition: Mα =  1 , z 1−z 1−z  M (α − 1) 1 − zM (α − 1)  , zM (α − 1) := P Mα−1 . 2048 G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049  n  n is the Pascal matrix. By induction we have M α = P α M 0 where P α = [α n−k k ]n,k≥0 and M 0 = Here P = k n , k ≥ 0   [ ⌈(n+nk)/2⌉ ]n,k≥0 . Hence we obtain the explicit form for m(α) n,k as    n  j n (α) . mn+1,k+1 = α n−j j ⌈ j+2k ⌉ j=0 (α) (α) Theorem 3.7. For α ≥ 1, let Rα = [rn,k ]n,k≥0 . Then α1 rn+1,k+1 counts the number of α -Motzkin paths ending at (n, 0) that have at least k flat steps on the x-axis. Proof. Since 1−α1zR(α) = 1 + α zM (α) from (13) and (14), we have Rα = (1 + α zM (α), α zM (α)). Deleting the first row and the first column of Rα yields (α M (α), α zR(α)) = (M (α), α zR(α))(α, z ). (α) Thus (M (α), α zR(α)) = [ α1 rn+1,k+1 ]n,k≥0 . Since M (α)(α zR(α))k is the generating function for the number of α -Motzkin paths ending at (n, 0) that have at least k flat steps on the x-axis, we obtain the desired one.  3.4. Ordered tree interpretation (k) Let us consider the consistent Riordan array Tk = [ti,j ]i,j≥0 with A(z ) = 1 + z + · · · + z k−1 (k ≥ 2). In particular, it is (3) known [1] that (ti,0 )i≥0 = (1, 1, 2, 5, 13, 35, 96, . . .) (A005773, [13]) counts the number of ordered trees with i edges and having nonroot vertices of outdegree at most 2. It also counts the number of directed animals in the first quadrant using unit North and East steps. Deutsch [1] introduced a simple method to obtain the generating function T (z ) for ordered trees with the prescribed sets R, N, and L of root degrees, node degrees, and branch lengths, respectively. In fact, T (z ) = 1 +  P ℓ (z )H ℓ (z ) zℓ and (15) ℓ∈R where P (z ) and H (z ) are the generating functions of all paths and of all trees except the planted trees (ordered trees with the root degree 1) with the prescribed sets R, N, and L respectively, i.e., P (z ) =  ℓ∈L H (z ) = 1 +  P ℓ (z )H ℓ (z ). ℓ∈N ,ℓ̸=1 z 2 C 2 (z ) For regular ordered trees, H (z ) = 1 + z 2 C 2 (z ) + z 3 C 3 (z ) + z 4 C 4 (z ) + · · · = 1 + 1−zC (z ) . ( k) Enumeration of ordered trees with certain prescribed sets R, N, and L gives a new combinatorial interpretation for ti,j . ( k) Theorem 3.8. Let ti,j be the number of ordered trees with i edges having root degree ≥j and nonroot outdegrees ≤k − 1. Then (k) Tk = [ti,j ]i,j≥0 is the consistent Riordan array with A = (1, 1, . . . , 1, 0, . . .) where 1’s appear k times. (k) Proof. Since ti,j = 0 for i, j ≥ 0 such that i < j, Tk is a lower triangular matrix. Let us consider the generating function T (z ) for the number of ordered trees with i ≥ 0 edges having root degree at least 0 and nonroot outdegrees at most k − 1. Thus z we have R = {1, 2, 3, . . .}, N = {1, 2, . . . , k − 1}, and L = {1, 2, 3, . . .}. By (15) we obtain T (z ) = 1−P (1z )H (z ) , P (z ) = 1− z and H (z ) = 1 + (P (z )H (z ))2 + (P (z )H (z ))3 + · · · + (P (z )H (z ))k−1 .  (16) (k) i For j ≥ 1, let Tj (z ) = i≥j ti,j z . In this case, we have R = {j, j + 1, j + 2, . . .}, N = {1, 2, . . . , k − 1} and L = {1, 2, 3, . . .}. Since i ≥ j ≥ 1, there is no empty tree. Hence by (15) we obtain j  1 . (P (z )H (z ))j = T (z ) · 1 − 1 − P (z )H (z ) T (z ) ℓ≥j   It follows Tk is the consistent Riordan array given by T (z ), 1 − T (1z ) . In addition, by a simple computation the A-sequence Tj (z ) = 1  (P (z )H (z ))ℓ = follows from (2) and (16) that 1− 1 T (z ) =z k−1   ℓ=0 1− 1 T (z ) where A(z ) = 1 + z + · · · + z k−1 . ℓ    1 = zA 1 − T (z ) (17) G.-S. Cheon et al. / Discrete Mathematics 312 (2012) 2040–2049 2049 √ Fig. 6. 0 · 1 · 2 trees counted by 1/ 1 − 2z − 3z 2 for n = 0, 1, 2, 3, 4. √ We end this section with a consistent Riordan array (1/ 1 − 2z − 3z 2 , z (1 + 2zm(z ))) observed in (6) and a combinatorial interpretation to accompany it. We note that f (z ) := z (1 + 2zm(z )) = 1 −  1 − 2z − 3z 2 = z + 2z 2 + 2z 3 + 4z 4 + 8z 5 + · · · where m is the generating function for the Motzkin numbers. √ In terms of 0 · 1 · 2 trees, the generating function 1/ 1 − 2z − 3z 2 counts such trees where, along the right most branch any vertex of updegree 2 has the left edge red or green. The first few terms are illustrated in Fig. 6. If the right most branch has length 1 we get the f trees as circled in g Fig. 6. 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