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Moments, Narayana numbers, and the cut and paste for lattice paths
Journal of Statistical Planning and Inference 135 (2005) 229 – 244 www.elsevier.com/locate/jspi
Moments, Narayana numbers, and the cut and paste for lattice paths Robert A. Sulanke Department of Mathematics, Boise State University, Boise, ID, USA Received 20 September 2002; received in revised form 18 December 2003; accepted 7 February 2005 Available online 23 March 2005
Abstract Let U(n) denote the set of unrestricted lattice paths that run from (0, 0) to (n, 0) with permitted steps (1, 1), (1, −1), and perhaps a horizontal step. Let E(n + 2) denote the set of paths in U(n + 2) that run strictly above the horizontal axis except initially and finally. First we review the cut-and-paste bijection which relates points under paths of E(n + 2) to points on paths of U(n). We apply it to obtain area and enumeration results for paths, some involving the Narayana distribution. We extend the cut-and-paste bijection to a formula relating factorial moments for the paths of E(n+2) to factorial moments for the paths of U(n). © 2005 Published by Elsevier B.V. MSC: 05A15 Keywords: Lattice path moments; Catalan numbers; Narayana distribution; Schröder numbers
1. Introduction Consider lattice paths in the integer plane which we represent as concatenations of the directed steps U := (1, 1), D := (1, −1), and, perhaps, H := (h, 0) where h is a fixed positive integer. For a given set of steps, let U(n) denote the set of all unrestricted paths running from (0, 0) to (n, 0). Let C(n) denote the set of paths in U(n) constrained never to pass beneath the horizontal axis. Let E(n) denote the set of paths in C(n) that are elevated strictly above the horizontal axis except at their initial and final points. For any set of paths L with unit-weighted steps, |L| will denote its cardinality. For example, for the step set E-mail address: [email protected]. 0378-3758/$ - see front matter © 2005 Published by Elsevier B.V. doi:10.1016/j.jspi.2005.02.016
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S = {U, D} and for n 0, we have that |U(2n)| is the central binomial coefficient ( 2n n ), C(2n) is the set of Dyck paths of length 2n, and |C(2n)| = |E(2n + 2)| is a Catalan number 1/(n + 1)( 2n n ). For any step set S and any path P running from (0, 0) to (n, 0), let (0, p0 ), (1, p1 ), (2, p2 ), . . . , (x, px ), . . . , (n, pn )
(1)
denote all integer lattice points traced by P, not just the end points of its steps. Our principal result relates sums of factorial moments for constrained paths to sums of factorial moments for restricted paths. These factorial moments are expressed in terms of binomial coefficients. Specifically, for the steps U, D, and H, and for any real r, n n qx + r px + r + 2 . (2) = qx px P ∈C(n) x=0
Q∈U(n) x=0
This identity was motivated by the following known identities for the usual first and second moments, which later will be seen to be simple consequences of identity (2). For the first moment, the sum of the areas of the regions bounded by the elevated paths of E(n + 2) and the horizontal axis is equal to the total number of x-intercepts on the paths of U(n). When the horizontal steps are disallowed, this yields that the sum of the areas under the paths of E(2n + 2) is 4n . For the second moment we have
n
P ∈E(n+2) x=0
px2 =
n
1 = (n + 1)|U(n)|.
Q∈U(n) x=0
This paper is a continuation of the paper of Pergola et al. (2003) which introduces the cut-and-paste bijective method relating lattice points under elevated paths of E(n + 2) to lattice points on the unrestricted paths of U(n). Section 2 reviews this method and give some additional notation. Section 3 gives examples obtained by restricting the domain and codomain of the cut-and-paste bijection. In particular, the cut-and-paste delivers the Narayana numbers. Section 4, which can be read immediately after Section 2, proves and mildly extends the result of identity (2) to paths with weighted steps. Section 5 extends a result of Chapman (1999) concerning a recurrence for factorial moments. The paper concludes with further examples, mainly for paths enumerated by the Schröder numbers.
2. Background 2.1. The cut-and-paste method Here we re-define the cut-and-paste method, which was presented with more detail, including its invertibility, in Pergola et al. (2003). Let S = {U, D, H } be the step set. First we need the notion of a dot. Given a path P ∈ E(n + 2), given a lattice point (x, y) lying strictly under P but weakly above the horizontal axis, and given an integer k, a dot is a triple, [P , (x, y), k]. The index k distinguishes a dot as there may be more than one dot at a point
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θ ε λ
(x,y)
ρ
(x’,k)
Fig. 1. For k = 3, the dot [P , (27, 3), 3] in DOTS(40) and its image (Q, (25, 3)) in PPATHS (38).
(x, y) under a given path; moreover, the index k has a role in the formula of the cut-and-paste bijection (see Fig. 1). With the notation of (1), the domain for the cut-and-paste bijection is DOTS(n + 2) := {[P , (x, y), k] : P ∈ E(n + 2), 0 < x < n + 2, 0 y < px , −px + y < k < px − y}. One should view this domain as being partitioned into many triangular arrays of dots, with one array corresponding to each lattice point on the trace of each path in E(n + 2). For each trace point (x, px ), it is easy to see that each array has (px )2 dots. For the path set E(n + 2) there will be a total of (n + 1)|E(n + 2)| nonempty triangular arrays of dots. For example, for P = U U DU U DDD, if x = 5, then p5 = 3 and the corresponding array of dots appears as [P , (5, 2), 0], [P , (5, 1), −1] [P , (5, 1), 0] [P , (5, 1), 1], [P , (5, 0), −2] [P , (5, 0), −1] [P , (5, 0), 0] [P , (5, 0), 1]
[P , (5, 0), 2].
The codomain for the proposed bijection is a set of pointed paths, where each path Q is unrestricted and is marked by a distinguished lattice point (x, qx ) on its trace. Hence the codomain is PPATHS(n) := {(Q, (x, qx )) : Q ∈ U(n), 0 x n}. We now define the cut-and-paste bijection
: DOTS(n + 2) → PPATHS(n)
(3)
to satisfy the following: First assume k 0. Each [P , (x, y), k] ∈ DOTS(n + 2) determines four subpaths of P (see Fig. 1): • Let = (1 , 2 ) be the nearest point on P to the left of (x, y) such that 2 = px − k − 1. (This indicates the role k plays in defining the bijection.)
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Table 1 The listing of the cut-and-paste bijection where P = U U U DDD and P
= U U DU DD
([P , (1, 0), 0]) = (U U DD, (0, 0)) ([P , (2, 0), −1]) = (DDU U , (1, −1)) ([P , (2, 0), 0]) = (U DDU , (0, 0)) ([P , (2, 0), 1]) = (U U DD, (1, 1)) ([P , (2, 1), 0]) = (DU U D, (2, 0)) ([P , (3, 0), −2]) = (DDU U , (2, −2)) ([P , (3, 0), −1]) = (DU U D, (1, −1)) ([P , (3, 0), 0]) = (DDU U , (0, 0)) ([P , (3, 0), 1]) = (U DDU , (1, 1)) ([P , (3, 0), 2]) = (U U DD, (2, 2)) ([P , (3, 1), −1]) = (U DDU , (3, −1)) ([P , (3, 1), 0]) = (DU DU , (2, 0)) ([P , (3, 1), 1]) = (DU U D, (3, 1)) ([P , (3, 2), 0]) = (DDU U , (4, 0)) ([P , (4, 0), −1]) = (DDU U , (3, −1))
• • • • • •
([P , (4, 0), 0]) = (U DDU , (2, 0)) ([P , (4, 0), 1]) = (U U DD, (3, 1)) ([P , (4, 1), 0]) = (DU U D, (4, 0)) ([P , (5, 0), 0]) = (U U DD, (4, 0)) ([P
, (1, 0), 0]) = (U DU D, (0, 0)) ([P
, (2, 0), −1]) = (DU DU , (1, −1)) ([P
, (2, 0), 0]) = (DU DU , (0, 0)) ([P
, (2, 0), 1]) = (U DU D, (1, 1)) ([P
, (2, 1), 0]) = (U DDU , (4, 0)) ([P
, (3, 0), 0]) = (U DU D, (2, 0)) ([P
, (4, 0), −1]) = (DU DU , (3, −1)) ([P
, (4, 0), 0]) = (DU U D, (0, 0)) ([P
, (4, 0), 1]) = (U DU D, (3, 1)) ([P
, (4, 1), 0]) = (DU DU , (4, 0)) ([P
, (5, 0), 0]) = (U DU D, (4, 0))
Let = (1 , 2 ) be the nearest point on P to the left of (x, y) such that 2 = y. Let = (1 , 2 ) be the nearest point on P to right of (x, y) such that 2 = y. Let L1 be that subpath of P running from (0, 0) to . Let L2 be that subpath of P running from to . Let R1 be that subpath of P running from to . Let R2 be that subpath of P running from to (n + 2, 0).
Some of these subpaths may be empty. Here P = L1 L2 R1 R2 . Let L1 R1 be the path obtained by deleting the first and last steps from the concatenation L1 R1 . (When y = 0, L1 has zero length, and L1 R1 denotes R1 with its first and last steps missing.) Define
([L1 L2 R1 R2 , (x, y), k]) = (R2 L1 R1 L2 , (x , k)), where x = x + n + 1 − 1 − 1 − 1. The point which is directly above (x, y) on R1 is moved along with R1 under the map and becomes (x , k). If k < 0, let REFL(Q) denote the reflection of the path Q about the x-axis and define
([P , (x, y), k]) = (REFL(Q), (x , k)), where ([P , (x, y), |k|]) = (Q, (x , |k|)). Table 1 completely lists the cut-and-paste bijection for step set S = {U, D} and for n = 4. There E(6) = {P , P
} where P = U U U DDD and P
= U U DU DD. 2.2. Additional background and notation In this paper, n will denote an arbitrary nonnegative integer. Usually, U, D, and H := (h, 0) will denote unit-weighted steps, while Ut and Hs will denote steps weighted by the
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indeterminates t and s. When the steps are weighted, the weight of a path P, denoted by |P |, is the product of the weights of its steps. The weight of a path set L, denoted |L|, is the sum of the weights of its paths. To obtain the cut-and-paste bijection for weighted paths, one needs only to account for the weights of the steps deleted from E(n + 2) under to insure that is weight preserving. For notational brevity, one can allow h to be either a positive integer or ‘∞’. Effectively, {U, D, (∞, 0)} = {U, D}, and the power z∞ will make no contribution to any power series. In the following, whenever we embellish the notation for the step set, ‘S’, we then embellish the corresponding ‘E’, ‘C’, ‘U’, and generating functions with the same mark. generating functions: c(z) := For any stepnset S = {Ut 2, D, Hs },consider the following n+2 , and u(z) := , e(z) := tz c(z) = | E (n + 2)|z | U (n)|zn . | C (n)|z n0 n0 n0 From decompositions of paths sets we have, c(z) = 1 + szh c(z) + tz2 c(z)2 ,
(4)
u(z) = 1 + szh u(z) + 2tz2 c(z)u(z).
(5)
To see (5) note that (i) every path in U(n) either has zero length; (ii) begins with H; or (iii) begins with U followed by a constrained path or begins with D followed by the reflection about the x-axis of a constrained path, and then later returns to the horizontal axis for the first time. Identity (4) follows in a similar manner. Solving these yields e(z) = tz2 c(z) = 1 − szh − (1 − szh )2 − 4tz2 2, u(z) = 1 (1 − szh )2 − 4tz2 .
(6)
We remark that Example 6.3.8 of Stanley (1999) extends easily to an alternative derivation of (6). n The Narayana numbers, N (n, k), are defined so N (n, k) = 1/n( nk )( k−1 ) and their generating function is n
N (n, k)t k zn =
1 − (1 − t)z −
1 − 2(1 + t)z + (1 − t)2 z2
2z.
n 0 k=1
See sequence A001263 in Sloane. Good introductions to the enumeration of lattice paths with respect to generating functions and results concerning the Narayana numbers are given by Deutsch (1999) and Sulanke (2000, 2002). We note that the binomial coefficient is defined in terms of a rising factorial: ( r+k k )= r+k ) = 1, and ( ) = 0 if otherwise. (r + 1)(r + 2) · · · (r + k)/k! for positive integer k, ( r+0 0 k We denote the truth value of any statement A by (A) so that (A) = 1 if A is true, and = 0 if otherwise.
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3. Some examples resulting from the cut-and-paste bijection Here we illustrate uses of the cut-and-paste bijection by restricting its domain and codomain to prove some known results, mainly concerning the Narayana distribution and the large Schröder numbers. Other such examples appear in Pergola et al. (2003).
3.1. Cardinality results analogous to the cycle lemma For the step set S = {Ut , D, Hs } let E(n, m) and U(n, m) denote those subsets of E(n) and U(n), respectively, where each path has m U-steps. Proposition 1. For m 0, (m + 1)|E(n + 2, m + 1)| = t|U(n, m)|. When S = {U, D}, this formula reduces to (n + 1)|E(2n + 2)| = |U(2n)|, and thus the cut-and-paste bijection explains the factor (n + 1). The paper (Pergola et al., 2003) compares this explanation with that given by the classical cycle lemma of Dvoretzky and Motzkin (1947). Proof of Proposition 1. For each P ∈ E(n + 2, m + 1), form m + 1 dots, [P , (x, 0), k], by selecting the points (x, 0) to lie on the x-axis directly below the final points of the U steps. Specifically, apply the restricted bijection
: {[P , (x, 0), 0] : P ∈ E(n + 2, m + 1) and (x, 0) below a final point of a U step} → {(Q, (0, 0)) : Q ∈ E(n, m)}. We see that the weight of the domain is (m + 1)|E(n + 2, m + 1)| while the weight of the codomain is t|U(n, m)|, where the factor t corresponds to the deletion of a U step in the cut-and-paste bijection. We define the elevated large Schröder paths to be the paths in E(n + 2) having S = {U, D, (2, 0)}. We take the large Schröder numbers to be defined in terms of the cardinality of these path sets. Specifically, (|E(2n + 2)|)n 0 = (1, 2, 6, 22, 90, 394, . . .). (Sequence A006318 of Sloane.) Here E(n, m) and U(n, m) denote those subsets of E(n) and U(n), respectively, where each path has m U-steps. Proposition 1 with some elementary counting shows that large Schröder numbers can be formulated as |E(2n + 2)| =
m0
=
m0
|E(2n + 2, m + 1)|, 1 (m + n)! |U(2n, m)| = . (m + 1) (m + 1)!m!(n − m)! m0
(7)
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3.2. The Narayana distribution in terms of oddly positioned up steps = {Uot , Ue , D}, where Uot denotes an up step of weight t that Consider the step set S must be oddly positioned on any path, Ue denotes a unit-weighted up step that must be evenly positioned on any path, and D denotes a unit weighted down step. (On any path running from (0, 0), the abscissa of the final point of an oddly positioned step will be odd.) As we will see, for t = 2, (| E(2n + 2)|)n 1 = (2, 6, 22, 90, 394, . . .) is the sequence of the large Schröder numbers with missing initial term. Consider also the step set S∗ = {Ut , D, (1, 0)t+1 }. It is known that for t = 1, (|E∗ (n + 1)|)n 1 = (1, 2, 5, 14, 42, . . .) is the sequence of the Catalan numbers with first term missing. For t = 2, (|E∗ (n + 1)|)n 1 = (2, 6, 22, 90, 394, . . .) is the sequence of the large Schröder numbers with first term missing. Proposition 2. For n 1, | E(2n + 2)| = |E∗ (n + 1)| =
n 1 n n ti, n i i−1
(8)
i=1
n ) is a Narayana number. For t = 2, the right-hand side of (8) is a known where (1/n)( ni )( i−1 formula for the large Schröder numbers.
Proof. In order to see that the weights of E∗ (n+1) and E(2n+2) agree, first let E# (n+1, k) # denote the set of elevated paths using step set S ={Ut , D, (1, 0)} and having k (1, 0) steps. Let E# (n + 1, b, r) denote the set of all replications of paths of E# (n + 1) having b blue (1, 0) steps, r red (1, 0) steps, and no other (1, 0) steps. If the blue steps are weighted by t and the red steps by 1, then
t b |E# (n + 1, b, r)| =
k b+r=k
k
b
tb
k |E# (n + 1, k)| b
(1 + t)k |E# (n + 1, k)| = |E∗ (n + 1)|. = k
On the other hand, if we apply the morphism to each path in E # (n + 1, b, r) having the step replacement rules Ut → Uot Ue ,
D → DD,
(1, 0)blue → Uot D,
we can establish a bijection which yields | E(2n + 2)| =
k b+r=k
t b |E# (n + 1, b, r)|.
and
(1, 0)red → DU e ,
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By Proposition 1, |E∗ (n + 2)| =
|E∗ (n + 2, m)|
m
=
t|U∗ (n, m)|/(m + 1) = t
m
m
t m (t + 1)n−2m n! . (m + 1)!m!(n − 2m)!
Applying the binomial theorem and then the Chu–Vandermonde convolution yields the proposition. 3.3. The Narayana distribution in terms of peaks For this application of the cut-and-paste method we count elevated paths with respect to the number of bicolored peaks. Again we will derive the formula for the Narayana numbers and the large Schröder numbers. On any path, a ‘right-hand turn’ or a ‘peak’ is
= {U, D}, let
the intermediate vertex of a consecutive UD pair. For S E(n, b, r) denote the set of elevated paths using the steps U and D and having b blue peaks, r red peaks, and no others. Proposition 3. For 1 b + r n, |
E(2n + 2, b, r)| =
1 b+r
n−1 b+r −1
n b+r −1
b+r b
.
Proof. Here the restricted cut-and-paste bijection is
E(2n + 2, 0, i), (x, 0) is below a peak of P } : {[P , (x, 0), 0] : P ∈
→ {(Q, (0, 0)) : Q ∈ U(2n), Q begins with a D step and has i − 1 right-hand turns} with ([P , (x, 0), 0]) = [L1 R1 L2 , (0, 0)] (L1 has no length). If we tilt each image path counterclockwise by 45◦ , one can check that in the tilted path there would be ( n−1 i−1 ) ways to n ) ways to choose the ordinates for the intermediate vertices of choose the abscissae and ( i−1 E(2n+2, 0, i)|= the right-hand turns, where these turns uniquely determine the path. Thus, |
n (1/ i)( n−1 )( ) by Proposition 1. Now, allowing b of the peaks to be colored blue, while i−1 i−1 b+r the remainder are red, yields the factor ( b ). When we disallow blue peaks, the proposition shows that
E(2n + 2, 0, r) is a Narayana number. When we do not restrict the coloring or the number of peaks, we see that the number
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E(2n + 2) with independently bicolored peaks is the large Schröder number: of paths in
E(2n + 2, b, r)| = E(2n + 2, b, i − b)| |
|
b
r
=
b
i
i
b
|
E(2n + 2, b, i − b)|
n n 1 n−1 = 2i . i i−1 i−1
(9)
i=1
Now let us return to the large Schröder paths, considered in (7), which used the step set S = {U, D, (2, 0)}. In that notation, E(2n, n + 1 − j ) will be the set of elevated paths E(2n + 2, b, r) and having j steps of the (2, 0) type. Here is a simple matching between r
E(2n + 2, b, r), transform each UD pair with a blue E(2n + 2, n + 1 − b): for each P ∈ r
intermediate vertex into a (2, 0) step, and then remove the color red from the remaining UD pairs, in order to obtain the image of P. Hence the number of paths in E(2n + 2) is also counted by large Schröder numbers of (9). 3.4. The total heights of the peaks Here, for S = {Ut , D, Hs }, we will sum the heights of the peaks over all paths in the set of constrained weighted paths C(n). In Section 6 we will consider the sum of the heights of all lattice points on the traces of the paths on E(n + 2), i.e., total area under the paths. Here we have the following generating functional result. Proposition 4. For step set S = {Ut , D, Hs }, ((x, px ) is a peak)px zn = tz2 u(z)2 . n 0 P ∈C(n) 0
Consequently, for the Dyck step set S = {U, D}, the power series for the sum of heights of the peaks on constrained paths of C(n) is z2 u(z)2 = z2 (1 − 4z2 )−1 , whose coefficients are powers of 4. In other words, for n 1, the sum of the heights of the peaks of the Dyck paths of length 2n is 4n−1 . Proof of the Proposition 4. Considering how the dots are arrayed under each lattice point on a path of E(n + 2), we wish to find the sum of the weights of the dots [P , (x, y), 1] (note, k = 1) where (x, y) is under a peak of P ∈ E(n + 2). From the definition of the cut-and-paste bijection one can check that the appropriate restricted bijection is
: {[P , (x, y), 1] ∈ DOTS(n + 2) : (x, px ) is a peak} → {(P , (x, 1)) ∈ PPATHS(n) : px−1 = px+1 = 0}. Here, each dot in the restricted domain is mapped to a path in U(n) having a marked UD with intermediate vertex of ordinate 1. Each such marked path results from the concatenation of three paths, namely, an unrestricted path from (0, 0) to the marked UD, the marked UD
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itself, and a following unrestricted path ending at (n, 0). Hence, with tz2 corresponding to the marked UD, we have the result. 4. Relating moments for constrained paths to those for unrestricted paths For this paper, our principal consequence of the cut-and-paste bijection is Proposition 5. For step set S = {U, D, H }, for n 0, and for real r, n px + r + 2 = px
P ∈C(n) x=0
=
n+1 px + r + 1 px − 1
P ∈E(n+2) x=1 n Q∈U(n) x=0
qx + r qx
.
(10)
Proof. The first equality is immediate from a simple shift. To prove the second equality, we assign the value ( k+r k ) to each dot [P , (x, y), k] and also to its image. The cut-and-paste bijection yields trivially
[P ,(x,y),k]∈ DOTS(n+2)
k+r k
=
(Q,(x,k))∈ PPATHS(n)
k+r k
=
(Q,(x,y))∈ PPATHS(n)
y+r y
.
(11)
The left-hand side of (11) becomes
n+1 p x −1
P ∈E(n+2) x=1 y=0
px −y−1 k=0
k+r k
=
n+1 p x −1 px − y + r px − y − 1
P ∈E(n+2) x=1 y=0
=
n+1 px + r + 1 . px − 1
P ∈E(n+2) x=1
This identity with (11) yields (10).
More generally, for step set S = {Ut , D, Hs } and for any nonnegative integer m, a similar computation shows P ∈C(n)
n px + r + 2 − m |P | = px − m x=0
=
t
P ∈E(n+2)
Q∈U(n)
−1
n+1 px + r − m + 1 |P | px − m − 1 x=1
n qx + r − m . |Q| qx − m x=0
(12)
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Example. This example of (12) concerns the sum of the usual second moments of the heights of the trace points of the paths of E(n + 2). Specifically, we have P ∈E(n+2)
|P |
n+1 x=1
px2 = =
P ∈E(n+2)
Q∈U(n)
=
|P |
t|Q| t|Q|
Q∈U(n)
=
Q∈U(n)
x=1
n qx qx − 1 + t|Q| qx qx − 1
Q∈U(n)
=
n+1 px px + 1 + px − 2 px − 1
t|Q|
x=0 n x=0 n x=0 n
[(qx 0) + (qx 1)] [(qx 0) + (qx < 0)] 1 = (n + 1)t|U(n)|.
x=0
Of course, referring directly to the cut-and-paste map in (3), if we count the domain, which consists of triangular arrays of px2 dots, one array for each point (x, px ) on the trace of a path, and if we count the codomain, we find the same result in the unweighted case. Example. Here we count two ‘Catalan structures’ not in the original catalog of Exercise 6.19 of Stanley (1999). (i) For step set {U, D} and for n 0, the total number of intercepts of the horizontal axis by the Dyck paths (i.e., constrained paths) running from (0, 0) to (2n, 0) is 1/(n+2)( 2n+2 n+1 ). (ii) For step set {U, D} and for n 0, assign the values 1, −2, and 1, respectively, to the points of ordinate 0, 1, and 2 on the unrestricted paths running from (0, 0) to (2n, 0). The sum of these values over all these paths is the Catalan number of (i). These two results follow from q − 3 2 x zn = zn (−1)qx qx qx n 0 Q∈U(n) x n 0 Q∈U(n) x p − 1 x zn = px x n 0 P ∈C(n) = (px = 0)zn n 0 P ∈C(n) x 2 −2
= c(z) = z (c(z) − 1). To see the next to the last identity, notice that P ∈C(n) x (px = 0) counts the set of intercept-marked paths formed from constrained paths of C(n). Since each intercept is realized as the concatenation of a constrained path from (0, 0) to the intercept and a constrained path from the intercept to (n, 0), P ∈C(n) x (px = 0) = i |C(i)||C(n − i)|.
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5. Recurrences for moments We will give a recurrence for binomial moments for constrained paths generalizing the recurrence given by Chapman (1999) for Dyck paths (see also Shapiro et al., 1983; Sulanke, 2000). We will then use Proposition 5 to convert that recurrence into one for moments of unrestricted paths. For path set S = {Ut , D, Hs }, we define the generating functions n px + r − m + 2 n |P | cr (z) := z , px − m n 0 P ∈C(n)
ur (z) :=
n 0 Q∈U(n)
x=0
n qx + r − m n z . |Q| qx − m x=0
Proposition 6. For real r and nonnegative m, (1 − tz2 c(z)2 )cr (z) = cr−1 (z),
(13)
(1 − tz2 c(z)2 )ur (z) = ur−1 (z),
(14)
cr (z) = 21 (1 + (1 − szh )u(z))cr−1 (z).
(15)
Proof. We prove (13) in the form cr (z) = cr−1 (z) + tz2 c(z)2 cr (z)
(16)
and only for m = 0 and s = t = 1. The binomial coefficients satisfy n px + r + 2 px P ∈C(n) x=0
n n px + r + 1 px + r + 1 = + . px px − 1 P ∈C(n) x=0
(17)
P ∈C(n) x=0
Consider the rightmost double sum. Each P ∈ C(n) either lies entirely on the x-axis and makes no contribution to that sum, or it has a factor which is an elevated path. We will consider the contribution to the sum by the horizontal translations of all Q = U Q D belonging to 0 n n−2 E(n +2). Each time a translation of some Q, where Q ∈ E(n +2), appears as a factor in some concatenation forming a path of C(n), it is preceded by a (perhaps void) constrained path and followed by a (perhaps void) constrained path; thus translations of this Q makes i+i =n−n −2 |C(i)||C(i )| appearances in the paths of C(n). The moment ) is contribution of Q to P ∈C(n) nx=0 ( pxp+r+1 x −1
+1 n
x=1
qx + r + 1 qx − 1
n qx + r + 2 = qx
x=0
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times the frequency of its appearance. (Here we are keeping the notation of (1) for (x, px ), (x, qx ), and (x, qx ) on P, Q, and Q , respectively.) Hence, n px + r + 1 px − 1
P ∈C(n) x=0
=
n−2
n =0 q ∈C(n ) i+i =n−n −2
n qx + r + 2 |C(i)||C(i )| qx
(18)
x=0
and the corresponding generating function is z2 c(z)2 cr (z). The identities (17) and (18) yield (16) for m = 0 and s = t = 1. The proof for cases other than m = 0 and s = t = 1 are similar. An application of Proposition 5 to (13) yields (14). A straightforward computation yields (15). 6. Further examples Here we consider formulas concerning the total area under a set of elevated paths. The cut-and-paste method is only manifested in the proofs of Propositions 7 and 9 through Proposition 5. 6.1. Areas and intercepts Take S = {U, D, H }. By the trapezoid rule, the total area bounded between the horizontal axis and the paths of E(n of + 2)pxis the middle formula of (10) in the form with r = −1. Since, in general, ( y−1 ) = P ∈E(n+2) x px = P ∈E(n+2) x ( px −1 ), i.e., y (y = 0), the right-hand side of (10) becomes Q∈U(n) x (qx = 0) which is the total number of intercepts of the horizontal axis by the paths of U(n). More generally, using (12), we have Proposition 7. For S = {U, D, H } and for m 0, the total area of the regions under the paths of E(n + 2) and above the horizontal line y = m is equal to the number of intercepts of the same line by the paths of U(n). Next we give some further results concerning area, intercepts, and the generating function u(z)2 , whose formula is obtained from (6). Proposition 8. For S = {U, D, H }, the generating function for the number of intercepts of the horizontal axis by the step end points, including (0, 0), of the paths of U(n) satisfies ((x, 0) is a step end point on Q)zn = u(z)2 . n 0 Q∈U(n) 0 x n
Proof. This proposition follows by observing that each intercept contributing to the inner summations results from the concatenation of two paths (perhaps empty),
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specifically, an unrestricted path from (0, 0) to the intercept and an unrestricted path from the intercept to (n, 0). Proposition 9. For S = {U, D, H }, the generating function for the number of intercepts of the horizontal axis by lattice points on the traces of paths of U(n) satisfies (19) (qx = 0)zn = (1 + (h − 1)zh )u(z)2 . n 0 Q∈U(n) 0 x n
Equivalently, the generating function for the total area under the paths of E(n + 2) satisfies px zn+2 = z2 (1 + (h − 1)zh )u(z)2 . n 0 P ∈E(n+2) 1 x n+1
Proof. Consider the terms of the inner sum of the left-hand side of (19). All intercepts that are end points of steps are counted by the generating function u(z)2 , as in the proof of Proposition 8. All intercepts that are interior lattice points of some horizontal step can be collected in a group of size h − 1. Since each such group results from the concatenation of three paths, specifically, an unrestricted path from (0, 0) to horizontal step containing interior intercepts, the horizontal step itself, and an unrestricted path from the horizontal step to (n, 0), we see that the generating function counting the intercepts which are interior to horizontal steps is u(z)(h − 1)zh u(z). The second identity follows by the initial remarks of this subsection. 6.2. Two comparable models Here we examine some comparable, yet different, models whose area results are derivable from the cut-and-paste bijection, by way of Proposition 9. Notice the appearance of powers of 4, which also appeared in Section 3.4. In both cases below we will consider instances of the following recurrence: If a(n) is defined so that n a(n)zn = u(z)2 , then a(n) = 4ta(n − 2) + 2sa(n − h) − s 2 a(n − 2h)
(20)
which follows the comparison of coefficients and formula (6). Case for s = t + 1 and h = 1: Here the step set is S∗ = {Ut , D, (1, 0)t+1 }, as in Section 3.2. For t = 1 and 2, u∗ (z) is the generating function for the central binomial coefficients and the central Delannoy numbers, respectively. (See Sequence A001850 of Sloane and Section 6.5 of Stanley (1999).) The weighted area under the elevated paths of E∗ (n + 2), satisfies
n 0 P ∈E∗ (n+2)
x
px zn+2 = tz2 u∗ (z)2 =
tz2 ((t + 1)z − 1)2 − 4tz2
.
For t = 1, z2 u∗ (z) = z2 /(1 − 4z), a generating function whose coefficients are powers of 4. We remark that there is a simple bijection from E∗ (n + 2) to a set of parallelogram polyominoes of perimeter 2n + 4 which is area preserving as discussed in Sulanke (1999). Further attention to this result appears in Del Lungo et al. (2001).
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For t = 2, 2z2 u∗ (z)2 = 2z2 /(1 − 6z + z2 ) = 2z2 + 12z3 + 70z4 + 408z5 . . . , where the coefficients are double the square–triangular numbers, or, equivalently, every other Pell number (see sequences A000129, A001109, and A001542 in Sloane). By (20), for t = 2, u∗ (z)2 = n a ∗ (n)zn satisfies a ∗ (n) = 6a ∗ (n − 1) − a ∗ (n − 2)
(21)
subject to a ∗ (2) = 2 and a ∗ (3) = 12. This recurrence in terms of the sum of the areas of zebras (i.e., column-bicolored parallelogram polyominoes) was a principal topic of Sulanke (1999). Case for t = 1: Here S = {U, D, (0, h)s }. For h = ∞, e(z) is a generating function for the Catalan numbers. For s = 1 and h = 1, e(z) is the generating function for the Motzkin numbers. For s = 1 and h = 2, C(n) are the usual large Schröder paths with e(z) being the generating function for the large Schröder numbers, as noted after formula (9). For h = ∞, u(z) gives the central binomial coefficients. For s = h = 1, u(z) gives the central trinomial numbers. For s = 1 and h = 2, u(z) gives the central Delannoy numbers. By Proposition 9 the area under the elevated paths of E(n + 2), satisfies
n 0 P ∈E(n+2)
x
px zn+2 =
1 + (h − 1)zh (szh − 1)2 − 4z2
.
For h = ∞, the coefficients of this power series are powers of 4 (see Woan et al., 1997). For s = 1 and h = 2, this power series becomes z2 (1 + z2 ) = 1z2 + 7z4 + 41z6 + 239z8 + 1393z10 . . . , 1 − 6z2 + z4 where the coefficients correspond to pairwise sums of consecutive square-triangular numbers, or equivalently to every other pairwise sum of consecutive Pell numbers, as noted in Bonin et al. (1993) (see sequence A002315 in Sloane). For E(2n + 2), we remark that the square–triangular numbers give both the sums of the ordinates of the points of the path restricted to be end points of steps and the sums of the ordinates of the points which are the mid points of steps. By (20), for s = 1 and h = 2, u(z)2 = n a(n)zn satisfies a(n) = 6a(n − 2) − a(n − 4)
(22)
subject to a(2) = 1 and a(4) = 7, as noted in Bonin et al. (1993). Compare this recurrence to (21). For recent considerations of (22) and other references, see Barcucci and Rinaldi (2001). For a bijective approach to the recurrences for the cardinality, area, and the second moments for large Schröder paths see Sulanke (1998).
Acknowledgements The author thanks Simone Rinaldi for conjecturing Proposition 7, which in turn motivated many of the results of this paper. He again thanks Lou Shapiro for guiding him to Proposition 8. He also thanks the referee for his useful suggestions.
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References Barcucci, E., Rinaldi, S., 2001. Some linear recurrences and their combinatorial interpretation by means of regular languages. Theoret. Comput. Sci. 255 (1–2), 679–686. Bonin, J., Shapiro, L., Simion, R., 1993. Some q-analogues of the Schröder numbers arising from combinatorial statistics on lattice paths. J. Statist. Plann. Inference 34, 35–55. Chapman, R., 1999. Moments of Dyck paths. Discrete Math. 204, 113–117. De Lungo, A., Nivat, M., Pinzani, R., Rinaldi, S., 2001. The area of parallelogram polyominoes and a tiling game, fun with algorithms, 2. Proceedings in Informatics 10, 85–101. Deutsch, E., 1999. Dyck path enumeration. Discrete Math. 204, 167–200. Dvoretzky, N., Motzkin, Th., 1947. A problem of arrangements. Duke Math. J. 14, 305–313. Pergola, E., Pinzani, R., Rinaldi, S., Sulanke, R.A., 2003. Lattice path moments by cut and paste. Adv. Appl. Math. 30 (1–2), 208–218. Shapiro, L., Woan, W.-J., Getu, S., 1983. Runs, slides, and moments. SIAM J. Discrete Math. 4, 459–466. Sloane, N.J.A., 2002. On-line Encyclopedia of Integer Sequences. http://www.research.att.com/ ∼njas/sequences/index.html Stanley, R.P., 1999. Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge. Sulanke, R.A., 1998. Bijective recurrences concerning Schröder paths. Electron. J. Combin. 5 (1), R47. Sulanke, R.A., 1999. Three recurrences for parallelogram polyominoes. J. Difference Equations Appl. 5, 155–176. Sulanke, R.A., 2000. Recurrences for moments of generalized Motzkin paths, J. of Integer Seq. 3, Article 00.1.1. Sulanke, R.A., 2002. The Narayana distribution. Special issue on lattice path combinatorics and applications (Vienna, 1998). J. Statist. Plann. Inference 101 (1–2), 311–326. Woan, W.-J., Shapiro, L., Rogers, D.G., 1997. The Catalan numbers, the Lebesgue integral and 4n−2 . Amer. Math. Monthly 104, 926–931.
Moments, Narayana numbers, and the cut and paste for lattice paths Robert A. Sulanke Department of Mathematics, Boise State University, Boise, ID, USA Received 20 September 2002; received in revised form 18 December 2003; accepted 7 February 2005 Available online 23 March 2005
Abstract Let U(n) denote the set of unrestricted lattice paths that run from (0, 0) to (n, 0) with permitted steps (1, 1), (1, −1), and perhaps a horizontal step. Let E(n + 2) denote the set of paths in U(n + 2) that run strictly above the horizontal axis except initially and finally. First we review the cut-and-paste bijection which relates points under paths of E(n + 2) to points on paths of U(n). We apply it to obtain area and enumeration results for paths, some involving the Narayana distribution. We extend the cut-and-paste bijection to a formula relating factorial moments for the paths of E(n+2) to factorial moments for the paths of U(n). © 2005 Published by Elsevier B.V. MSC: 05A15 Keywords: Lattice path moments; Catalan numbers; Narayana distribution; Schröder numbers
1. Introduction Consider lattice paths in the integer plane which we represent as concatenations of the directed steps U := (1, 1), D := (1, −1), and, perhaps, H := (h, 0) where h is a fixed positive integer. For a given set of steps, let U(n) denote the set of all unrestricted paths running from (0, 0) to (n, 0). Let C(n) denote the set of paths in U(n) constrained never to pass beneath the horizontal axis. Let E(n) denote the set of paths in C(n) that are elevated strictly above the horizontal axis except at their initial and final points. For any set of paths L with unit-weighted steps, |L| will denote its cardinality. For example, for the step set E-mail address: [email protected]. 0378-3758/$ - see front matter © 2005 Published by Elsevier B.V. doi:10.1016/j.jspi.2005.02.016
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S = {U, D} and for n 0, we have that |U(2n)| is the central binomial coefficient ( 2n n ), C(2n) is the set of Dyck paths of length 2n, and |C(2n)| = |E(2n + 2)| is a Catalan number 1/(n + 1)( 2n n ). For any step set S and any path P running from (0, 0) to (n, 0), let (0, p0 ), (1, p1 ), (2, p2 ), . . . , (x, px ), . . . , (n, pn )
(1)
denote all integer lattice points traced by P, not just the end points of its steps. Our principal result relates sums of factorial moments for constrained paths to sums of factorial moments for restricted paths. These factorial moments are expressed in terms of binomial coefficients. Specifically, for the steps U, D, and H, and for any real r, n n qx + r px + r + 2 . (2) = qx px P ∈C(n) x=0
Q∈U(n) x=0
This identity was motivated by the following known identities for the usual first and second moments, which later will be seen to be simple consequences of identity (2). For the first moment, the sum of the areas of the regions bounded by the elevated paths of E(n + 2) and the horizontal axis is equal to the total number of x-intercepts on the paths of U(n). When the horizontal steps are disallowed, this yields that the sum of the areas under the paths of E(2n + 2) is 4n . For the second moment we have
n
P ∈E(n+2) x=0
px2 =
n
1 = (n + 1)|U(n)|.
Q∈U(n) x=0
This paper is a continuation of the paper of Pergola et al. (2003) which introduces the cut-and-paste bijective method relating lattice points under elevated paths of E(n + 2) to lattice points on the unrestricted paths of U(n). Section 2 reviews this method and give some additional notation. Section 3 gives examples obtained by restricting the domain and codomain of the cut-and-paste bijection. In particular, the cut-and-paste delivers the Narayana numbers. Section 4, which can be read immediately after Section 2, proves and mildly extends the result of identity (2) to paths with weighted steps. Section 5 extends a result of Chapman (1999) concerning a recurrence for factorial moments. The paper concludes with further examples, mainly for paths enumerated by the Schröder numbers.
2. Background 2.1. The cut-and-paste method Here we re-define the cut-and-paste method, which was presented with more detail, including its invertibility, in Pergola et al. (2003). Let S = {U, D, H } be the step set. First we need the notion of a dot. Given a path P ∈ E(n + 2), given a lattice point (x, y) lying strictly under P but weakly above the horizontal axis, and given an integer k, a dot is a triple, [P , (x, y), k]. The index k distinguishes a dot as there may be more than one dot at a point
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θ ε λ
(x,y)
ρ
(x’,k)
Fig. 1. For k = 3, the dot [P , (27, 3), 3] in DOTS(40) and its image (Q, (25, 3)) in PPATHS (38).
(x, y) under a given path; moreover, the index k has a role in the formula of the cut-and-paste bijection (see Fig. 1). With the notation of (1), the domain for the cut-and-paste bijection is DOTS(n + 2) := {[P , (x, y), k] : P ∈ E(n + 2), 0 < x < n + 2, 0 y < px , −px + y < k < px − y}. One should view this domain as being partitioned into many triangular arrays of dots, with one array corresponding to each lattice point on the trace of each path in E(n + 2). For each trace point (x, px ), it is easy to see that each array has (px )2 dots. For the path set E(n + 2) there will be a total of (n + 1)|E(n + 2)| nonempty triangular arrays of dots. For example, for P = U U DU U DDD, if x = 5, then p5 = 3 and the corresponding array of dots appears as [P , (5, 2), 0], [P , (5, 1), −1] [P , (5, 1), 0] [P , (5, 1), 1], [P , (5, 0), −2] [P , (5, 0), −1] [P , (5, 0), 0] [P , (5, 0), 1]
[P , (5, 0), 2].
The codomain for the proposed bijection is a set of pointed paths, where each path Q is unrestricted and is marked by a distinguished lattice point (x, qx ) on its trace. Hence the codomain is PPATHS(n) := {(Q, (x, qx )) : Q ∈ U(n), 0 x n}. We now define the cut-and-paste bijection
: DOTS(n + 2) → PPATHS(n)
(3)
to satisfy the following: First assume k 0. Each [P , (x, y), k] ∈ DOTS(n + 2) determines four subpaths of P (see Fig. 1): • Let = (1 , 2 ) be the nearest point on P to the left of (x, y) such that 2 = px − k − 1. (This indicates the role k plays in defining the bijection.)
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Table 1 The listing of the cut-and-paste bijection where P = U U U DDD and P
= U U DU DD
([P , (1, 0), 0]) = (U U DD, (0, 0)) ([P , (2, 0), −1]) = (DDU U , (1, −1)) ([P , (2, 0), 0]) = (U DDU , (0, 0)) ([P , (2, 0), 1]) = (U U DD, (1, 1)) ([P , (2, 1), 0]) = (DU U D, (2, 0)) ([P , (3, 0), −2]) = (DDU U , (2, −2)) ([P , (3, 0), −1]) = (DU U D, (1, −1)) ([P , (3, 0), 0]) = (DDU U , (0, 0)) ([P , (3, 0), 1]) = (U DDU , (1, 1)) ([P , (3, 0), 2]) = (U U DD, (2, 2)) ([P , (3, 1), −1]) = (U DDU , (3, −1)) ([P , (3, 1), 0]) = (DU DU , (2, 0)) ([P , (3, 1), 1]) = (DU U D, (3, 1)) ([P , (3, 2), 0]) = (DDU U , (4, 0)) ([P , (4, 0), −1]) = (DDU U , (3, −1))
• • • • • •
([P , (4, 0), 0]) = (U DDU , (2, 0)) ([P , (4, 0), 1]) = (U U DD, (3, 1)) ([P , (4, 1), 0]) = (DU U D, (4, 0)) ([P , (5, 0), 0]) = (U U DD, (4, 0)) ([P
, (1, 0), 0]) = (U DU D, (0, 0)) ([P
, (2, 0), −1]) = (DU DU , (1, −1)) ([P
, (2, 0), 0]) = (DU DU , (0, 0)) ([P
, (2, 0), 1]) = (U DU D, (1, 1)) ([P
, (2, 1), 0]) = (U DDU , (4, 0)) ([P
, (3, 0), 0]) = (U DU D, (2, 0)) ([P
, (4, 0), −1]) = (DU DU , (3, −1)) ([P
, (4, 0), 0]) = (DU U D, (0, 0)) ([P
, (4, 0), 1]) = (U DU D, (3, 1)) ([P
, (4, 1), 0]) = (DU DU , (4, 0)) ([P
, (5, 0), 0]) = (U DU D, (4, 0))
Let = (1 , 2 ) be the nearest point on P to the left of (x, y) such that 2 = y. Let = (1 , 2 ) be the nearest point on P to right of (x, y) such that 2 = y. Let L1 be that subpath of P running from (0, 0) to . Let L2 be that subpath of P running from to . Let R1 be that subpath of P running from to . Let R2 be that subpath of P running from to (n + 2, 0).
Some of these subpaths may be empty. Here P = L1 L2 R1 R2 . Let L1 R1 be the path obtained by deleting the first and last steps from the concatenation L1 R1 . (When y = 0, L1 has zero length, and L1 R1 denotes R1 with its first and last steps missing.) Define
([L1 L2 R1 R2 , (x, y), k]) = (R2 L1 R1 L2 , (x , k)), where x = x + n + 1 − 1 − 1 − 1. The point which is directly above (x, y) on R1 is moved along with R1 under the map and becomes (x , k). If k < 0, let REFL(Q) denote the reflection of the path Q about the x-axis and define
([P , (x, y), k]) = (REFL(Q), (x , k)), where ([P , (x, y), |k|]) = (Q, (x , |k|)). Table 1 completely lists the cut-and-paste bijection for step set S = {U, D} and for n = 4. There E(6) = {P , P
} where P = U U U DDD and P
= U U DU DD. 2.2. Additional background and notation In this paper, n will denote an arbitrary nonnegative integer. Usually, U, D, and H := (h, 0) will denote unit-weighted steps, while Ut and Hs will denote steps weighted by the
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indeterminates t and s. When the steps are weighted, the weight of a path P, denoted by |P |, is the product of the weights of its steps. The weight of a path set L, denoted |L|, is the sum of the weights of its paths. To obtain the cut-and-paste bijection for weighted paths, one needs only to account for the weights of the steps deleted from E(n + 2) under to insure that is weight preserving. For notational brevity, one can allow h to be either a positive integer or ‘∞’. Effectively, {U, D, (∞, 0)} = {U, D}, and the power z∞ will make no contribution to any power series. In the following, whenever we embellish the notation for the step set, ‘S’, we then embellish the corresponding ‘E’, ‘C’, ‘U’, and generating functions with the same mark. generating functions: c(z) := For any stepnset S = {Ut 2, D, Hs },consider the following n+2 , and u(z) := , e(z) := tz c(z) = | E (n + 2)|z | U (n)|zn . | C (n)|z n0 n0 n0 From decompositions of paths sets we have, c(z) = 1 + szh c(z) + tz2 c(z)2 ,
(4)
u(z) = 1 + szh u(z) + 2tz2 c(z)u(z).
(5)
To see (5) note that (i) every path in U(n) either has zero length; (ii) begins with H; or (iii) begins with U followed by a constrained path or begins with D followed by the reflection about the x-axis of a constrained path, and then later returns to the horizontal axis for the first time. Identity (4) follows in a similar manner. Solving these yields e(z) = tz2 c(z) = 1 − szh − (1 − szh )2 − 4tz2 2, u(z) = 1 (1 − szh )2 − 4tz2 .
(6)
We remark that Example 6.3.8 of Stanley (1999) extends easily to an alternative derivation of (6). n The Narayana numbers, N (n, k), are defined so N (n, k) = 1/n( nk )( k−1 ) and their generating function is n
N (n, k)t k zn =
1 − (1 − t)z −
1 − 2(1 + t)z + (1 − t)2 z2
2z.
n 0 k=1
See sequence A001263 in Sloane. Good introductions to the enumeration of lattice paths with respect to generating functions and results concerning the Narayana numbers are given by Deutsch (1999) and Sulanke (2000, 2002). We note that the binomial coefficient is defined in terms of a rising factorial: ( r+k k )= r+k ) = 1, and ( ) = 0 if otherwise. (r + 1)(r + 2) · · · (r + k)/k! for positive integer k, ( r+0 0 k We denote the truth value of any statement A by (A) so that (A) = 1 if A is true, and = 0 if otherwise.
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3. Some examples resulting from the cut-and-paste bijection Here we illustrate uses of the cut-and-paste bijection by restricting its domain and codomain to prove some known results, mainly concerning the Narayana distribution and the large Schröder numbers. Other such examples appear in Pergola et al. (2003).
3.1. Cardinality results analogous to the cycle lemma For the step set S = {Ut , D, Hs } let E(n, m) and U(n, m) denote those subsets of E(n) and U(n), respectively, where each path has m U-steps. Proposition 1. For m 0, (m + 1)|E(n + 2, m + 1)| = t|U(n, m)|. When S = {U, D}, this formula reduces to (n + 1)|E(2n + 2)| = |U(2n)|, and thus the cut-and-paste bijection explains the factor (n + 1). The paper (Pergola et al., 2003) compares this explanation with that given by the classical cycle lemma of Dvoretzky and Motzkin (1947). Proof of Proposition 1. For each P ∈ E(n + 2, m + 1), form m + 1 dots, [P , (x, 0), k], by selecting the points (x, 0) to lie on the x-axis directly below the final points of the U steps. Specifically, apply the restricted bijection
: {[P , (x, 0), 0] : P ∈ E(n + 2, m + 1) and (x, 0) below a final point of a U step} → {(Q, (0, 0)) : Q ∈ E(n, m)}. We see that the weight of the domain is (m + 1)|E(n + 2, m + 1)| while the weight of the codomain is t|U(n, m)|, where the factor t corresponds to the deletion of a U step in the cut-and-paste bijection. We define the elevated large Schröder paths to be the paths in E(n + 2) having S = {U, D, (2, 0)}. We take the large Schröder numbers to be defined in terms of the cardinality of these path sets. Specifically, (|E(2n + 2)|)n 0 = (1, 2, 6, 22, 90, 394, . . .). (Sequence A006318 of Sloane.) Here E(n, m) and U(n, m) denote those subsets of E(n) and U(n), respectively, where each path has m U-steps. Proposition 1 with some elementary counting shows that large Schröder numbers can be formulated as |E(2n + 2)| =
m0
=
m0
|E(2n + 2, m + 1)|, 1 (m + n)! |U(2n, m)| = . (m + 1) (m + 1)!m!(n − m)! m0
(7)
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3.2. The Narayana distribution in terms of oddly positioned up steps = {Uot , Ue , D}, where Uot denotes an up step of weight t that Consider the step set S must be oddly positioned on any path, Ue denotes a unit-weighted up step that must be evenly positioned on any path, and D denotes a unit weighted down step. (On any path running from (0, 0), the abscissa of the final point of an oddly positioned step will be odd.) As we will see, for t = 2, (| E(2n + 2)|)n 1 = (2, 6, 22, 90, 394, . . .) is the sequence of the large Schröder numbers with missing initial term. Consider also the step set S∗ = {Ut , D, (1, 0)t+1 }. It is known that for t = 1, (|E∗ (n + 1)|)n 1 = (1, 2, 5, 14, 42, . . .) is the sequence of the Catalan numbers with first term missing. For t = 2, (|E∗ (n + 1)|)n 1 = (2, 6, 22, 90, 394, . . .) is the sequence of the large Schröder numbers with first term missing. Proposition 2. For n 1, | E(2n + 2)| = |E∗ (n + 1)| =
n 1 n n ti, n i i−1
(8)
i=1
n ) is a Narayana number. For t = 2, the right-hand side of (8) is a known where (1/n)( ni )( i−1 formula for the large Schröder numbers.
Proof. In order to see that the weights of E∗ (n+1) and E(2n+2) agree, first let E# (n+1, k) # denote the set of elevated paths using step set S ={Ut , D, (1, 0)} and having k (1, 0) steps. Let E# (n + 1, b, r) denote the set of all replications of paths of E# (n + 1) having b blue (1, 0) steps, r red (1, 0) steps, and no other (1, 0) steps. If the blue steps are weighted by t and the red steps by 1, then
t b |E# (n + 1, b, r)| =
k b+r=k
k
b
tb
k |E# (n + 1, k)| b
(1 + t)k |E# (n + 1, k)| = |E∗ (n + 1)|. = k
On the other hand, if we apply the morphism to each path in E # (n + 1, b, r) having the step replacement rules Ut → Uot Ue ,
D → DD,
(1, 0)blue → Uot D,
we can establish a bijection which yields | E(2n + 2)| =
k b+r=k
t b |E# (n + 1, b, r)|.
and
(1, 0)red → DU e ,
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By Proposition 1, |E∗ (n + 2)| =
|E∗ (n + 2, m)|
m
=
t|U∗ (n, m)|/(m + 1) = t
m
m
t m (t + 1)n−2m n! . (m + 1)!m!(n − 2m)!
Applying the binomial theorem and then the Chu–Vandermonde convolution yields the proposition. 3.3. The Narayana distribution in terms of peaks For this application of the cut-and-paste method we count elevated paths with respect to the number of bicolored peaks. Again we will derive the formula for the Narayana numbers and the large Schröder numbers. On any path, a ‘right-hand turn’ or a ‘peak’ is
= {U, D}, let
the intermediate vertex of a consecutive UD pair. For S E(n, b, r) denote the set of elevated paths using the steps U and D and having b blue peaks, r red peaks, and no others. Proposition 3. For 1 b + r n, |
E(2n + 2, b, r)| =
1 b+r
n−1 b+r −1
n b+r −1
b+r b
.
Proof. Here the restricted cut-and-paste bijection is
E(2n + 2, 0, i), (x, 0) is below a peak of P } : {[P , (x, 0), 0] : P ∈
→ {(Q, (0, 0)) : Q ∈ U(2n), Q begins with a D step and has i − 1 right-hand turns} with ([P , (x, 0), 0]) = [L1 R1 L2 , (0, 0)] (L1 has no length). If we tilt each image path counterclockwise by 45◦ , one can check that in the tilted path there would be ( n−1 i−1 ) ways to n ) ways to choose the ordinates for the intermediate vertices of choose the abscissae and ( i−1 E(2n+2, 0, i)|= the right-hand turns, where these turns uniquely determine the path. Thus, |
n (1/ i)( n−1 )( ) by Proposition 1. Now, allowing b of the peaks to be colored blue, while i−1 i−1 b+r the remainder are red, yields the factor ( b ). When we disallow blue peaks, the proposition shows that
E(2n + 2, 0, r) is a Narayana number. When we do not restrict the coloring or the number of peaks, we see that the number
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E(2n + 2) with independently bicolored peaks is the large Schröder number: of paths in
E(2n + 2, b, r)| = E(2n + 2, b, i − b)| |
|
b
r
=
b
i
i
b
|
E(2n + 2, b, i − b)|
n n 1 n−1 = 2i . i i−1 i−1
(9)
i=1
Now let us return to the large Schröder paths, considered in (7), which used the step set S = {U, D, (2, 0)}. In that notation, E(2n, n + 1 − j ) will be the set of elevated paths E(2n + 2, b, r) and having j steps of the (2, 0) type. Here is a simple matching between r
E(2n + 2, b, r), transform each UD pair with a blue E(2n + 2, n + 1 − b): for each P ∈ r
intermediate vertex into a (2, 0) step, and then remove the color red from the remaining UD pairs, in order to obtain the image of P. Hence the number of paths in E(2n + 2) is also counted by large Schröder numbers of (9). 3.4. The total heights of the peaks Here, for S = {Ut , D, Hs }, we will sum the heights of the peaks over all paths in the set of constrained weighted paths C(n). In Section 6 we will consider the sum of the heights of all lattice points on the traces of the paths on E(n + 2), i.e., total area under the paths. Here we have the following generating functional result. Proposition 4. For step set S = {Ut , D, Hs }, ((x, px ) is a peak)px zn = tz2 u(z)2 . n 0 P ∈C(n) 0
Consequently, for the Dyck step set S = {U, D}, the power series for the sum of heights of the peaks on constrained paths of C(n) is z2 u(z)2 = z2 (1 − 4z2 )−1 , whose coefficients are powers of 4. In other words, for n 1, the sum of the heights of the peaks of the Dyck paths of length 2n is 4n−1 . Proof of the Proposition 4. Considering how the dots are arrayed under each lattice point on a path of E(n + 2), we wish to find the sum of the weights of the dots [P , (x, y), 1] (note, k = 1) where (x, y) is under a peak of P ∈ E(n + 2). From the definition of the cut-and-paste bijection one can check that the appropriate restricted bijection is
: {[P , (x, y), 1] ∈ DOTS(n + 2) : (x, px ) is a peak} → {(P , (x, 1)) ∈ PPATHS(n) : px−1 = px+1 = 0}. Here, each dot in the restricted domain is mapped to a path in U(n) having a marked UD with intermediate vertex of ordinate 1. Each such marked path results from the concatenation of three paths, namely, an unrestricted path from (0, 0) to the marked UD, the marked UD
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itself, and a following unrestricted path ending at (n, 0). Hence, with tz2 corresponding to the marked UD, we have the result. 4. Relating moments for constrained paths to those for unrestricted paths For this paper, our principal consequence of the cut-and-paste bijection is Proposition 5. For step set S = {U, D, H }, for n 0, and for real r, n px + r + 2 = px
P ∈C(n) x=0
=
n+1 px + r + 1 px − 1
P ∈E(n+2) x=1 n Q∈U(n) x=0
qx + r qx
.
(10)
Proof. The first equality is immediate from a simple shift. To prove the second equality, we assign the value ( k+r k ) to each dot [P , (x, y), k] and also to its image. The cut-and-paste bijection yields trivially
[P ,(x,y),k]∈ DOTS(n+2)
k+r k
=
(Q,(x,k))∈ PPATHS(n)
k+r k
=
(Q,(x,y))∈ PPATHS(n)
y+r y
.
(11)
The left-hand side of (11) becomes
n+1 p x −1
P ∈E(n+2) x=1 y=0
px −y−1 k=0
k+r k
=
n+1 p x −1 px − y + r px − y − 1
P ∈E(n+2) x=1 y=0
=
n+1 px + r + 1 . px − 1
P ∈E(n+2) x=1
This identity with (11) yields (10).
More generally, for step set S = {Ut , D, Hs } and for any nonnegative integer m, a similar computation shows P ∈C(n)
n px + r + 2 − m |P | = px − m x=0
=
t
P ∈E(n+2)
Q∈U(n)
−1
n+1 px + r − m + 1 |P | px − m − 1 x=1
n qx + r − m . |Q| qx − m x=0
(12)
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Example. This example of (12) concerns the sum of the usual second moments of the heights of the trace points of the paths of E(n + 2). Specifically, we have P ∈E(n+2)
|P |
n+1 x=1
px2 = =
P ∈E(n+2)
Q∈U(n)
=
|P |
t|Q| t|Q|
Q∈U(n)
=
Q∈U(n)
x=1
n qx qx − 1 + t|Q| qx qx − 1
Q∈U(n)
=
n+1 px px + 1 + px − 2 px − 1
t|Q|
x=0 n x=0 n x=0 n
[(qx 0) + (qx 1)] [(qx 0) + (qx < 0)] 1 = (n + 1)t|U(n)|.
x=0
Of course, referring directly to the cut-and-paste map in (3), if we count the domain, which consists of triangular arrays of px2 dots, one array for each point (x, px ) on the trace of a path, and if we count the codomain, we find the same result in the unweighted case. Example. Here we count two ‘Catalan structures’ not in the original catalog of Exercise 6.19 of Stanley (1999). (i) For step set {U, D} and for n 0, the total number of intercepts of the horizontal axis by the Dyck paths (i.e., constrained paths) running from (0, 0) to (2n, 0) is 1/(n+2)( 2n+2 n+1 ). (ii) For step set {U, D} and for n 0, assign the values 1, −2, and 1, respectively, to the points of ordinate 0, 1, and 2 on the unrestricted paths running from (0, 0) to (2n, 0). The sum of these values over all these paths is the Catalan number of (i). These two results follow from q − 3 2 x zn = zn (−1)qx qx qx n 0 Q∈U(n) x n 0 Q∈U(n) x p − 1 x zn = px x n 0 P ∈C(n) = (px = 0)zn n 0 P ∈C(n) x 2 −2
= c(z) = z (c(z) − 1). To see the next to the last identity, notice that P ∈C(n) x (px = 0) counts the set of intercept-marked paths formed from constrained paths of C(n). Since each intercept is realized as the concatenation of a constrained path from (0, 0) to the intercept and a constrained path from the intercept to (n, 0), P ∈C(n) x (px = 0) = i |C(i)||C(n − i)|.
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5. Recurrences for moments We will give a recurrence for binomial moments for constrained paths generalizing the recurrence given by Chapman (1999) for Dyck paths (see also Shapiro et al., 1983; Sulanke, 2000). We will then use Proposition 5 to convert that recurrence into one for moments of unrestricted paths. For path set S = {Ut , D, Hs }, we define the generating functions n px + r − m + 2 n |P | cr (z) := z , px − m n 0 P ∈C(n)
ur (z) :=
n 0 Q∈U(n)
x=0
n qx + r − m n z . |Q| qx − m x=0
Proposition 6. For real r and nonnegative m, (1 − tz2 c(z)2 )cr (z) = cr−1 (z),
(13)
(1 − tz2 c(z)2 )ur (z) = ur−1 (z),
(14)
cr (z) = 21 (1 + (1 − szh )u(z))cr−1 (z).
(15)
Proof. We prove (13) in the form cr (z) = cr−1 (z) + tz2 c(z)2 cr (z)
(16)
and only for m = 0 and s = t = 1. The binomial coefficients satisfy n px + r + 2 px P ∈C(n) x=0
n n px + r + 1 px + r + 1 = + . px px − 1 P ∈C(n) x=0
(17)
P ∈C(n) x=0
Consider the rightmost double sum. Each P ∈ C(n) either lies entirely on the x-axis and makes no contribution to that sum, or it has a factor which is an elevated path. We will consider the contribution to the sum by the horizontal translations of all Q = U Q D belonging to 0 n n−2 E(n +2). Each time a translation of some Q, where Q ∈ E(n +2), appears as a factor in some concatenation forming a path of C(n), it is preceded by a (perhaps void) constrained path and followed by a (perhaps void) constrained path; thus translations of this Q makes i+i =n−n −2 |C(i)||C(i )| appearances in the paths of C(n). The moment ) is contribution of Q to P ∈C(n) nx=0 ( pxp+r+1 x −1
+1 n
x=1
qx + r + 1 qx − 1
n qx + r + 2 = qx
x=0
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times the frequency of its appearance. (Here we are keeping the notation of (1) for (x, px ), (x, qx ), and (x, qx ) on P, Q, and Q , respectively.) Hence, n px + r + 1 px − 1
P ∈C(n) x=0
=
n−2
n =0 q ∈C(n ) i+i =n−n −2
n qx + r + 2 |C(i)||C(i )| qx
(18)
x=0
and the corresponding generating function is z2 c(z)2 cr (z). The identities (17) and (18) yield (16) for m = 0 and s = t = 1. The proof for cases other than m = 0 and s = t = 1 are similar. An application of Proposition 5 to (13) yields (14). A straightforward computation yields (15). 6. Further examples Here we consider formulas concerning the total area under a set of elevated paths. The cut-and-paste method is only manifested in the proofs of Propositions 7 and 9 through Proposition 5. 6.1. Areas and intercepts Take S = {U, D, H }. By the trapezoid rule, the total area bounded between the horizontal axis and the paths of E(n of + 2)pxis the middle formula of (10) in the form with r = −1. Since, in general, ( y−1 ) = P ∈E(n+2) x px = P ∈E(n+2) x ( px −1 ), i.e., y (y = 0), the right-hand side of (10) becomes Q∈U(n) x (qx = 0) which is the total number of intercepts of the horizontal axis by the paths of U(n). More generally, using (12), we have Proposition 7. For S = {U, D, H } and for m 0, the total area of the regions under the paths of E(n + 2) and above the horizontal line y = m is equal to the number of intercepts of the same line by the paths of U(n). Next we give some further results concerning area, intercepts, and the generating function u(z)2 , whose formula is obtained from (6). Proposition 8. For S = {U, D, H }, the generating function for the number of intercepts of the horizontal axis by the step end points, including (0, 0), of the paths of U(n) satisfies ((x, 0) is a step end point on Q)zn = u(z)2 . n 0 Q∈U(n) 0 x n
Proof. This proposition follows by observing that each intercept contributing to the inner summations results from the concatenation of two paths (perhaps empty),
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specifically, an unrestricted path from (0, 0) to the intercept and an unrestricted path from the intercept to (n, 0). Proposition 9. For S = {U, D, H }, the generating function for the number of intercepts of the horizontal axis by lattice points on the traces of paths of U(n) satisfies (19) (qx = 0)zn = (1 + (h − 1)zh )u(z)2 . n 0 Q∈U(n) 0 x n
Equivalently, the generating function for the total area under the paths of E(n + 2) satisfies px zn+2 = z2 (1 + (h − 1)zh )u(z)2 . n 0 P ∈E(n+2) 1 x n+1
Proof. Consider the terms of the inner sum of the left-hand side of (19). All intercepts that are end points of steps are counted by the generating function u(z)2 , as in the proof of Proposition 8. All intercepts that are interior lattice points of some horizontal step can be collected in a group of size h − 1. Since each such group results from the concatenation of three paths, specifically, an unrestricted path from (0, 0) to horizontal step containing interior intercepts, the horizontal step itself, and an unrestricted path from the horizontal step to (n, 0), we see that the generating function counting the intercepts which are interior to horizontal steps is u(z)(h − 1)zh u(z). The second identity follows by the initial remarks of this subsection. 6.2. Two comparable models Here we examine some comparable, yet different, models whose area results are derivable from the cut-and-paste bijection, by way of Proposition 9. Notice the appearance of powers of 4, which also appeared in Section 3.4. In both cases below we will consider instances of the following recurrence: If a(n) is defined so that n a(n)zn = u(z)2 , then a(n) = 4ta(n − 2) + 2sa(n − h) − s 2 a(n − 2h)
(20)
which follows the comparison of coefficients and formula (6). Case for s = t + 1 and h = 1: Here the step set is S∗ = {Ut , D, (1, 0)t+1 }, as in Section 3.2. For t = 1 and 2, u∗ (z) is the generating function for the central binomial coefficients and the central Delannoy numbers, respectively. (See Sequence A001850 of Sloane and Section 6.5 of Stanley (1999).) The weighted area under the elevated paths of E∗ (n + 2), satisfies
n 0 P ∈E∗ (n+2)
x
px zn+2 = tz2 u∗ (z)2 =
tz2 ((t + 1)z − 1)2 − 4tz2
.
For t = 1, z2 u∗ (z) = z2 /(1 − 4z), a generating function whose coefficients are powers of 4. We remark that there is a simple bijection from E∗ (n + 2) to a set of parallelogram polyominoes of perimeter 2n + 4 which is area preserving as discussed in Sulanke (1999). Further attention to this result appears in Del Lungo et al. (2001).
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For t = 2, 2z2 u∗ (z)2 = 2z2 /(1 − 6z + z2 ) = 2z2 + 12z3 + 70z4 + 408z5 . . . , where the coefficients are double the square–triangular numbers, or, equivalently, every other Pell number (see sequences A000129, A001109, and A001542 in Sloane). By (20), for t = 2, u∗ (z)2 = n a ∗ (n)zn satisfies a ∗ (n) = 6a ∗ (n − 1) − a ∗ (n − 2)
(21)
subject to a ∗ (2) = 2 and a ∗ (3) = 12. This recurrence in terms of the sum of the areas of zebras (i.e., column-bicolored parallelogram polyominoes) was a principal topic of Sulanke (1999). Case for t = 1: Here S = {U, D, (0, h)s }. For h = ∞, e(z) is a generating function for the Catalan numbers. For s = 1 and h = 1, e(z) is the generating function for the Motzkin numbers. For s = 1 and h = 2, C(n) are the usual large Schröder paths with e(z) being the generating function for the large Schröder numbers, as noted after formula (9). For h = ∞, u(z) gives the central binomial coefficients. For s = h = 1, u(z) gives the central trinomial numbers. For s = 1 and h = 2, u(z) gives the central Delannoy numbers. By Proposition 9 the area under the elevated paths of E(n + 2), satisfies
n 0 P ∈E(n+2)
x
px zn+2 =
1 + (h − 1)zh (szh − 1)2 − 4z2
.
For h = ∞, the coefficients of this power series are powers of 4 (see Woan et al., 1997). For s = 1 and h = 2, this power series becomes z2 (1 + z2 ) = 1z2 + 7z4 + 41z6 + 239z8 + 1393z10 . . . , 1 − 6z2 + z4 where the coefficients correspond to pairwise sums of consecutive square-triangular numbers, or equivalently to every other pairwise sum of consecutive Pell numbers, as noted in Bonin et al. (1993) (see sequence A002315 in Sloane). For E(2n + 2), we remark that the square–triangular numbers give both the sums of the ordinates of the points of the path restricted to be end points of steps and the sums of the ordinates of the points which are the mid points of steps. By (20), for s = 1 and h = 2, u(z)2 = n a(n)zn satisfies a(n) = 6a(n − 2) − a(n − 4)
(22)
subject to a(2) = 1 and a(4) = 7, as noted in Bonin et al. (1993). Compare this recurrence to (21). For recent considerations of (22) and other references, see Barcucci and Rinaldi (2001). For a bijective approach to the recurrences for the cardinality, area, and the second moments for large Schröder paths see Sulanke (1998).
Acknowledgements The author thanks Simone Rinaldi for conjecturing Proposition 7, which in turn motivated many of the results of this paper. He again thanks Lou Shapiro for guiding him to Proposition 8. He also thanks the referee for his useful suggestions.
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