Hadamard products, lattice paths, and skew tableaux
A Dissertation
Presented to
The Faculty of the Graduate School of Arts and Sciences
Brandeis University
Department of Mathematics
Ira M. Gessel, Advisor
In Partial Fulfillment
of the Requirements for the Degree
Doctor of Philosophy
by
Jong Hyun Kim
May, 2011
This dissertation, directed and approved by Jong Hyun Kim’s committee, has been
accepted and approved by the Faculty of Brandeis University in partial fulfillment of
the requirements for the degree of:
DOCTOR OF PHILOSOPHY
Adam Jaffe, Dean of Arts and Sciences
Dissertation Committee:
Ira M. Gessel, Dept. of Mathematics, Chair.
Kiyoshi Igusa, Dept. of Mathematics
Lionel Levine, Dept. of Mathematics, Massachusetts Institute of Technology
c Copyright by
Jong Hyun Kim
2011
Acknowledgments
There are so many people who have supported and helped me to become who I
am today. First and foremost, I would like to express my heartfelt gratitude to my
advisor, Ira M. Gessel, for his teaching and generosity from the first time I met him. I
feel very lucky to have been one of his students. I am grateful as well to Kiyoshi Igusa
and Lionel Levine for serving on my dissertation committee. I thank the faculty, my
colleagues, and the staff from the department of mathematics at Brandeis University.
I would also like to thank my undergraduate advisor, Suh-Ryung Kim, for encouraging
me to continue my studies in mathematics. Lastly, I am deeply indebted to my family,
friends, and my brothers and sisters in Christ from Han-Young Presbyterian Church,
Yesim Presbyterian Church, and Korean Church of Boston. I am grateful for them,
especially for my parents and for my soul mate, for their continuous love and prayers.
“O Lord, our Lord, how majestic is your name in all the earth! You have set
your glory above the heavens. Out of the mouth of babies and infants, you have
established strength because of your foes, to still the enemy and the avenger. When
I look at your heavens, the work of your fingers, the moon and the stars, which you
have set in place, what is man that you are mindful of him, and the son of man that
you care for him? Yet you have made him a little lower than the heavenly beings and
crowned him with glory and honor. You have given him dominion over the works of
your hands; you have put all things under his feet, all sheep and oxen, and also the
beasts of the field, the birds of the heavens, and the fish of the sea, whatever passes
along the paths of the seas. O Lord, our Lord, how majestic is your name in all the
earth!” Psalm 8.
iv
Abstract
Hadamard products, lattice paths, and skew tableaux
A dissertation presented to the Faculty of the
Graduate School of Arts and Sciences of Brandeis
University, Waltham, Massachusetts
by Jong Hyun Kim
This thesis concerns the computation of Hadamard products, the enumeration of
lattice paths, and the enumeration of standard Young tableaux of skew shape.
The first chapter is about Hadamard products and tilings. Shapiro gave a combinatorial proof of a bilinear generating function for Chebyshev polynomials equivalent
to the formula
1
1
1 − x2
∗
=
,
1 − ax − x2 1 − bx − x2
1 − abx − (2 + a2 + b2 )x2 − abx3 + x4
where ∗ denotes the Hadamard product. In a similar way, by considering tilings of a
2 × n rectangle with 1 × 1 and 1 × 2 bricks in the top row, and 1 × 1 and 1 × n bricks
in the bottom row, we find an explicit formula for the Hadamard product
xm
1
∗
.
1 − ax − x2 1 − bx − xn
The second chapter deals with lattice path enumerations using redundant generating functions. A redundant generating function is a generating function having
terms which are not part of the solution of the original problem. We use redundant
generating functions to study two path problems. In the first application we explain
a surprising occurrence of Catalan numbers in counting paths that stay below the
line y = 2x. In the second application we prove a conjecture of Niederhausen and
Sullivan.
v
Finally, the third chapter is the enumeration of three-rowed standard Young
tableaux of skew shape in terms of Motzkin numbers. The enumeration of standard Young tableaux (SYTs) of shape λ can be easily computed by the hook-length
formula. In 1981, Amitai Regev proved that the number of SYTs having at most
three rows with n entries equals the nth Motzkin number Mn . In 2006, Regev conjectured that the total number of SYTs of skew shape λ/(2, 1) over all partitions λ
having at most three parts with n entries is the difference of two Motzkin numbers,
Mn−1 − Mn−3 . Ekhad and Zeilberger proved Regev’s conjecture using a computer
program. In his paper [3], S.-P. Eu found a bijection between Motzkin paths and
SYTs of skew shape with at most three rows to prove Regev’s conjecture, and Eu
also indirectly showed that for the fixed µ = (µ1 , µ2 ) the number of SYTs of skew
shape λ/µ over all partitions λ having at most three parts can be expressed as a linear
combination of Motzkin numbers. In this chapter, we will find an explicit formula
for the generating function for the general case: for each partition µ having at most
three parts the generating function gives a formula for the coefficients of the linear
combination of Motzkin numbers. We will also show that these generating functions
are unexpectedly related to the Chebyshev polynomials of the second kind.
vi
Contents
List of Tables
viii
List of Figures
ix
Chapter 1. Hadamard Products and Tilings
1
1.1. Introduction
1
1.2. Hadamard products
3
Chapter 2. Redundant generating functions in lattice path enumeration
19
2.1. Introduction
19
2.2. The ballot problem
20
2.3. Variations of the ballot problem
24
2.4. A conjecture of Niederhausen and Sullivan
30
2.5. Main Theorem of Chapter 2
32
Chapter 3. On the enumeration of three-rowed standard Young tableaux of
skew shape in terms of Motzkin numbers
50
3.1. Introduction
50
3.2. About standard Young tableaux having at most three rows
52
Bibliography
69
vii
List of Tables
1
The values of B̃(m, n)
22
2
The values of C2 (m, n)
24
3
The values of D2 (m, n) for 0 ≤ n ≤ 2m
26
4
The values of D2 (m, n)
28
5
The values of S(m, n)
31
6
The values of S 0 (m, n)
31
7
The values of P10 (m, n)
34
8
The values of P00 (m, n)
46
1
The values of rk,i
61
viii
List of Figures
2.1 The values of D2 (m, n)
29
2.2 Decomposition for a path from (0, 0) to (m, 0) when T = {−1, 2}
32
ix
CHAPTER 1
Hadamard Products and Tilings
1.1. Introduction
The Fibonacci numbers are defined by F0 = 0, F1 = 1, and for n ≥ 2, Fn =
Fn−1 + Fn−2 . It is convenient to write fn for Fn+1 so that fn is the number of ways
to tile a 1 × n strip with 1 × 1 square bricks and 1 × 2 rectangular bricks [1]. The
number of tilings of a 1 × n strip with k bricks is the coefficient of xn in (x + x2 )k , so
we know that the generating function [4] for Fibonacci numbers is
∞
X
f n xn =
n=0
1
.
1 − x − x2
We now define the polynomial fn (a) by
∞
X
1
fn (a)xn .
=
2
1 − ax − x
n=0
(1)
Then we have that fn (1) = fn and fn (a) can be interpreted as the sum of the weights of
tilings of a 1×n strip with 1×1 square bricks weighted by a and 1×2 rectangular bricks
weighted by 1. By applying the geometric series and binomial series to (1−ax−x2 )−1
we obtain that
n
bX
2c
n − k n−2k
fn (a) =
a
.
k
k=0
(2)
So, f0 (a) = 1, f1 (a) = a, f2 (a) = 1 + a2 , f3 (a) = 2a + a3 , f4 (a) = 1 + 3a2 + a4 , etc.
1
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
Louis W. Shapiro [17] gave a combinatorial proof of a bilinear generating function
for Chebyshev polynomials equivalent to
∞
X
fn (a)fn (b)x
n=0
2n
1 − x4
=
.
1 − abx2 − (2 + a2 + b2 )x4 − abx6 + x8
The Hadamard product G∗H of the power series G(x) =
P
k≥0
P
k≥0
(3)
g(k)xk and H(x) =
h(k)xk is defined by
G∗H =
X
g(k)h(k)xk .
k≥0
If G(x) and H(x) are rational power series, then so is the Hadamard product G ∗ H
[18, p. 207].
Using the notation of Hadamard product, we can rewrite (3) as
1
1
1 − x2
∗
=
.
1 − ax − x2 1 − bx − x2
1 − abx − (2 + a2 + b2 )x2 − abx3 + x4
(4)
In this chapter, I will review Shapiro’s proof of (4), and extend this approach to find
an explicit formula for the Hadamard product
1
xm
∗
.
1 − ax − x2 1 − bx − xn
The MacMahon operator Ω≥ is defined on formal Laurent series by
Ω≥
∞
X
an x n =
n=−∞
∞
X
an .
n=0
We assume that the above sum (5) converges in an appropriate sense.
2
(5)
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
We can express the MacMahon operator Ω≥ in terms of Hadamard products. Let
P
P
G(x) = i≥0 g(i)xi and H(x) = j≥0 h(j)xj be power series. Then
G(x)H(x−1 ) =
XX
g(i)h(j)xi−j .
i≥0 j≥0
Since
P
i
(
i≥0
j≤i h(j))x = H(x)/(1 − x), we have
P
−1
Ω≥ G(x)H(x ) =
∞ X
i
X
i=0
H(x) g(i)h(j) = G(x) ∗
.
1 − x x=1
j=0
(6)
In [11], G.-N. Han used computer algebra to show that
Ω≥
1 + z2y
1
=
.
(1 − zx − zx2 )(1 − y/x − y/x2 )
(1 − 2z)(1 − 3zy − z 2 y − zy 2 )
We will derive Han’s result from our formula for 1/(1 − ax − x2 ) ∗ 1/(1 − bx − x3 ).
1.2. Hadamard products
Now we review Shapiro’s [17] proof of a formula for the Hadamard product
∞
X
1
1
∗
=
fk (a)fk (b)xk .
2
2
1 − ax − x
1 − bx − x
k=0
(7)
We consider (7) as counting pairs of tilings. The coefficient fk (a)fk (b) of xk counts
tilings of a 2×k rectangle with 1×1 square bricks weighted by a and 1×2 rectangular
bricks weighted by 1 in the top row, and 1 × 1 square bricks weighted by b and 1 × 2
rectangular bricks weighted by 1 in the bottom row, as in the following figure:
a
a
b b b
a a a
b
b b
3
a a
b
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
The vertical line segments passing from top to bottom serve to factor these tilings into
tilings of smaller length. For example, the following figure shows the factorization of
the above figure.
a
b
a
b b
a a
a
a
b
b
b
a
b
Let us define a prime block to be a tiling that cannot be factored any further without
cutting it through the middle of some brick. So these prime blocks can be classified
as follows:
The prime block of length 1:
The prime blocks of length 2:
,
,
The prime blocks of length 2k + 1 ≥ 3, together with the result of interchanging the
two rows:
···
···
The prime blocks of length 2k ≥ 4, together with the result of interchanging the two
rows:
···
···
4
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
Thus the generating function P2 (x) for the weighted prime blocks of the Hadamard
product is
2
2
2
P2 (x) = abx + (1 + a + b )x +
∞
X
2abx
2k+1
+
k=1
∞
X
(a2 + b2 )x2k
k=2
3
2
2abx
(a + b2 )x4
+
1 − x2
1 − x2
abx + (1 + a2 + b2 )x2 + abx3 − x4
=
.
1 − x2
= abx + (1 + a2 + b2 )x2 +
Since any tiling can be factored uniquely as a sequence of prime blocks [8, p. 1027–
1030], we have (1 − ax − x2 )−1 ∗ (1 − bx − x2 )−1 = 1/(1 − P2 (x)). So we obtain the
following explicit formula:
1
1 − x2
1
∗
=
,
1 − ax − x2 1 − bx − x2
1 − abx − (2 + a2 + b2 )x2 − abx3 + x4
(8)
which is equivalent to Shapiro’s result. Letting a = 1 and b = 1 in equation (8), we
have [18, p. 251]
∞
X
n=0
fn2 xn =
1−x
.
1 − 2x − 2x2 + x3
As noted by Shapiro [17], (8) can be written as an identity for Chebyshev polynomials. The Chebyshev polynomials of the second kind Un (a) (n ≥ 0) can be defined
by the generating function
∞
X
1
=
Un (a)z n .
1 − 2az + z 2
n=0
5
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
By substituting −2ai for a and iz for x in equation (1) we have the relation Un (a) =
in fn (−2ai), and from identity (2) we have
n
bX
2c
n−k
Un (a) =
(−1)k (2a)n−2k .
k
k=0
By replacing a, b, and x with −2ai, −2bi, and −z respectively in equation (8) we
can obtain the Chebyshev polynomial identity
∞
X
Un (a)Un (b)z n =
n=0
1 − z2
.
1 − 4abz − (2 − 4a2 − 4b2 )z 2 − 4abz 3 + z 4
We now want to prove an identity which we will use later.
Lemma 1.2.1. For m ≥ −1 and n ≥ −1,
fm (a)fn+1 (a) − fm+1 (a)fn (a) = (−1)min(m,n+1) f|m−n|−1 (a),
(9)
where f−1 (a) = 0.
Proof. It is enough to show that (9) holds for the case m ≥ n ≥ −1. Fix
m > n > 0 and let A be the set of tilings of a 1 × m strip and a 1 × (n + 1) strip with
1 × 1 square bricks weighted by a and 1 × 2 rectangular bricks weighted by 1. Then
there are fm (a)fn+1 (a) weighted tilings in A. Similarly, let B be the set of tilings of
a 1 × (m + 1) strip and a 1 × n strip with 1 × 1 square bricks weighted by a and 1 × 2
rectangular bricks weighted by 1. Then there are fm+1 (a)fn (a) weighted tilings in B.
Now we will find a bijection from a subset A to the set B if n is odd, a bijection from
a subset of B to the set A if n is even that proves (9). Let us consider a tiling in
A drawn in two rows so that the top row is a 1 × m strip and the bottom row is a
6
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
1 × (n + 1) strip indented m − n spaces, as follows:
···
←−
m−n −→
···
Let us find the rightmost vertical line segment, if there is one, that passes through
both strips without cutting through the middle of some brick. We call the part of the
tiling to the right of this line the tail of the tiling. In the following figure the tail is
separated.
···
···
Switching the two rows of the tail of this tiling produces the following tiling in B:
···
←−
m−n −→
···
where the top row is a 1 × (m + 1) strip and the bottom row is a 1 × n strip indented
m − n spaces.
When n is odd, this tail switching pairs up every element of A with every element
of B except for the tilings in A of the form:
······
←− m−n −→
···
···
where the top row is a 1 × m strip, the bottom row is a 1 × (n + 1) strip indented
m − n spaces,
······
represents any strip of length m − n − 1 tiled with
1 × 1 square bricks and 1 × 2 rectangular bricks, and every other brick is 1 × 2. In this
case, tail switching cannot be applied to the tiling. So there are fm−n−1 (a) weighted
7
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
tilings in the set A which cannot matched with those in the set B by tail switching.
Therefore we have fm (a)fn+1 (a) − fm+1 (a)fn (a) = fm−n−1 (a).
When n is even, tail switching pairs up every element of A with every element of
B except for the tilings in B of the form:
······
←− m−n −→
···
···
where the top row is a 1 × (m + 1) strip, the bottom row is a 1 × n strip indented
m − n spaces,
······
represents any strip of length m − n − 1 tiled with
1 × 1 square bricks and 1 × 2 rectangular bricks, and every other brick is 1 × 2. In this
case, tail switching cannot be applied to the tiling. So there are fm−n−1 (a) weighted
tilings in the set B which cannot matched with those in the set A by tail switching.
Therefore we have fm (a)fn+1 (a) − fm+1 (a)fn (a) = −fm−n−1 (a).
In the case m > n = 0, the definition of the tail must be modified slightly. We
leave the details to the reader. Now suppose n > m. Let D(m, n) = fm (a)fn+1 (a) −
fm+1 (a)fn (a). Then D(m, n) = −D(n, m) = (−1)m fn−m−1 (a). This is equivalent to
the desired formula. In the other cases in which m or n is −1 or m = n, we can easily
see that equation (9) is true because f−1 (a) = 0.
A special case of identity (9) for m = n + 1 and a = 1 is Cassini’s Fibonacci
2
identity fn+1
− fn+2 fn = (−1)n+1 which was proved in the same way in [1, p. 8] and
[22].
Next, we can use this combinatorial method to obtain an explicit formula for
Hadamard product (1 − ax − x2 )−1 ∗ (1 − bx − xn )−1 .
8
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
Theorem 1. For n ≥ 2, the Hadamard product
1
1
∗
2
1 − ax − x
1 − bx − xn
is equal to
1 − fn−2 xn
1 − abx − b2 x2 − (fn + fn−2 )xn − (2bfn−1 − abfn−2 )xn+1 + (−1)n x2n
where fn represents fn (a) and f−1 = 0.
Proof. We now consider the Hadamard product
1
1
∗
2
1 − ax − x
1 − bx − xn
(10)
as counting tilings, using 1 × n rectangular bricks instead of 1 × 2 rectangular bricks
in the bottom row. In this setting a prime block cannot have a 1 × 1 square brick in
the bottom row anywhere except at the beginning or end. The possible prime blocks
can be classified as follows:
The prime block of length 1:
The prime blocks of length n:
NNNNN
The prime blocks of length nk (k ≥ 2):
NNNN
NNN
NNN
9
···
···
NNN
NNNN
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
The prime blocks of length nk + 1 (k ≥ 1):
NNNN
NNN
NNN
NNN
NNN
···
···
NNN
···
···
NNN
NNN
NNN
NNNN
The prime blocks of length nk + 2 (k ≥ 0):
NNN
where
NNN
NNN
···
···
NNN
NNN
NNN NNNN
NNNNN
,
, and
represent any strips of length n − 2, n − 1, and n
respectively tiled with 1 × 1 square bricks and 1 × 2 rectangular bricks.
Thus the generating function Pn (x) for the weighted prime blocks of the Hadamard
product (10) is
n
Pn (x) = abx + fn (a)x +
∞
X
2bfn−1 (a)fn−2 (a)k−1 xnk+1
k=1
+
∞
X
2
k nk+2
b fn−2 (a) x
k=0
+
∞
X
fn−1 (a)2 fn−2 (a)k−2 xnk
k=2
2bfn−1 (a)xn+1
b2 x 2
fn−1 (a)2 x2n
+
+
1 − fn−2 (a)xn 1 − fn−2 (a)xn 1 − fn−2 (a)xn
abx + b2 x2 + fn (a)xn + 2bfn−1 (a) − abfn−2 (a) xn+1 + (−1)n−1 x2n
=
1 − fn−2 (a)xn
= abx + fn (a)xn +
where we have used the identity fn−1 (a)2 − fn (a)fn−2 (a) = (−1)n−1 obtained by
substituting n − 1 for m and n − 2 for n in identity (9) of Lemma 1.2.1. Since
any tiling can be factored uniquely as a sequence of prime blocks, we obtain that
(1 − ax − x2 )−1 ∗ (1 − bx − xn )−1 = 1/(1 − Pn (x)). This is equivalent to the desired
formula.
10
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
Note that Theorem 1 also holds for n = 1. The polynomials fn (a) + fn−2 (a) in
Theorem 1 are Lucas polynomials.
Using (6) we can prove Han’s result
Ω≥
1
1 + z2y
=
,
(1 − zx − zx2 )(1 − y/x − y/x2 )
(1 − 2z)(1 − 3zy − z 2 y − zy 2 )
(11)
by computing the Hadamard product 1/(1 − zx − zx2 ) ∗ 1/((1 − yx − yx2 )(1 − x))
and then setting x = 1. Substituting z 1/2 for a and z 1/2 x for x in 1/(1 − ax − x2 )
gives 1/(1 − zx − zx2 ), and substituting (y + 1)/(−y)1/3 for b and (−y)1/3 x for x in
1/(1 − bx − x3 ) gives 1/(1 − (y + 1)x + yx3 ). Making these substitutions in Theorem
1 and using the fact that if f (x) ∗ g(x) = h(x) then f (αx) ∗ g(βx) = h(αβx), we have
that the Hadamard product
1
1
∗
1 − zx − zx2 1 − (y + 1)x + yx3
is equal to
1 + z 2 yx3
.
1 − (zy + z)x − z(y + 1)2 x2 + (3z 2 y + z 3 y)x3 + (z 3 y + 2z 2 y)(y + 1)x4 − z 3 y 2 x6
Then setting x = 1 gives Han’s result (11), which he proved in a more complicated
way.
We now modify the above setting to obtain a formula for the Hadamard product
1/(1 − ax − x2 ) ∗ xm /(1 − bx − x2 ).
Theorem 2. For m ≥ 0, the Hadamard product
1
xm
∗
1 − ax − x2 1 − bx − x2
11
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
is equal to
fm (a)xm + bfm−1 (a)xm+1 − fm−2 (a)xm+2
1 − abx − (2 + a2 + b2 )x2 − abx3 + x4
where f−1 (a) = 0 and f−2 (a) = 1.
Proof. When m = 0, the formula reduces to (8). When m ≥ 1, we consider the
Hadamard product
xm
1
∗
1 − ax − x2 1 − bx − x2
as counting tilings. We modify the tilings of a 2 × k rectangle so that the bottom row
starts with a 1 × m rectangular brick to account for the factor xm in xm /(1 − bx − x2 ).
In this setting the first block where the bottom row starts with a 1 × m rectangular
brick will be different from all the others, but the following blocks can be built up
from a sequence of prime blocks which are exactly the same as the prime blocks in
the Hadamard product (1 − ax − x2 )−1 ∗ (1 − bx − x2 )−1 . The first blocks can be
classified as follows:
The first blocks of length m:
NNNNN
The first blocks of length m + 2k (k ≥ 1):
NNNN
···
···
The first blocks of length m + 2k + 1 (k ≥ 0):
NNNN
···
···
12
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
where
NNNN
and
NNNNN
represent any strips of length m − 1 and m respectively
tiled with 1 × 1 square bricks and 1 × 2 rectangular bricks. So the generating function
Qm,2 (x) where m ≥ 2 for weighted first blocks is
m
Qm,2 (x) = fm (a)x +
∞
X
afm−1 (a)x
m+2k
k=1
+
∞
X
bfm−1 (a)xm+2k+1
k=0
m+2
afm−1 (a)x
bfm−1 (a)xm+1
+
1 − x2
1 − x2
fm (a)xm + bfm−1 (a)xm+1 − fm−2 (a)xm+2
.
=
1 − x2
= fm (a)xm +
Since any tiling can be factored uniquely as a first block followed by a sequence of
prime blocks, we have 1/(1 − ax − x2 ) ∗ xm /(1 − bx − x2 ) = Qm,2 (x)/(1 − P2 (x)) where
m ≥ 1. This is equivalent to the desired formula.
Now we can generalize the previous theorem by computing an explicit formula for
the Hadamard product 1/(1 − ax − x2 ) ∗ xm /(1 − bx − xn ).
Theorem 3. For m ≥ 0 and n ≥ 2, the Hadamard product
xm
1
∗
1 − ax − x2 1 − bx − xn
is equal to
fm xm + bfm−1 xm+1 + (−1)min(m−1,n−1) f|m−n+1|−1 xm+n
1 − abx − b2 x2 − (fn + fn−2 )xn − (2bfn−1 − abfn−2 )xn+1 + (−1)n x2n
where fn represents fn (a) and f−1 = 0.
Proof. When m = 0, Theorem 3 reduces to Theorem 1. Let us consider the
Hadamard product
1
xm
∗
1 − ax − x2 1 − bx − xn
13
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
as counting pairs of tilings where m ≥ 1 and n ≥ 2. We slightly modify the above
tiling by using 1 × n rectangular bricks instead of 1 × 2 rectangular bricks in the
bottom row. In this setting, the first block where the bottom row starts with a 1 × m
rectangular brick will be different from all the others, but the following blocks can
be built up from a sequence of prime blocks whick are exactly the same as the prime
blocks in the Hadamard product (1 − ax − x2 )−1 ∗ (1 − bx − xn )−1 . The possible first
blocks can be classified as follows:
The first blocks of length m:
LLLLLL
The first blocks of length m + nk (k ≥ 1):
LLLLL
NNN
NNN
···
···
NNN
NNNN
The first blocks of length m + nk + 1 (k ≥ 0):
LLLLL
where
NNN
NNN
···
···
NNN
NNN
NNN NNNN LLLLL
LLLLLL
,
,
, and
represent any strips of length n−2,
n − 1, m − 1, and m respectively tiled with 1 × 1 square bricks and 1 × 2 rectangular
bricks.
So the generating function Qm,n (x) where m ≥ 1 and n ≥ 2 for the weighted first
blocks is
m
Qm,n (x) = fm (a)x +
∞
X
fm−1 (a)fn−1 (a)fn−2 (a)k−1 xm+nk
k=1
14
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
+
∞
X
bfm−1 (a)fn−2 (a)k xm+nk+1
k=0
fm−1 (a)fn−1 (a)xm+n bfm−1 (a)xm+1
+
1 − fn−2 (a)xn
1 − fn−2 (a)xn
fm (a)xm + bfm−1 (a)xm+1 + fm−1 (a)fn−1 (a) − fm (a)fn−2 (a) xm+n
= fm (a)xm +
=
=
1 − fn−2 (a)xn
fm (a)xm + bfm−1 (a)xm+1 + (−1)min(m−1,n−1) f|m−n+1|−1 (a)xm+n
1 − fn−2 (a)xn
where we use the identity
fm−1 (a)fn−1 (a) − fm (a)fn−2 (a) = (−1)min(m−1,n−1) f|m−n+1|−1 (a)
obtained by substituting m − 1 for m and n − 2 for n in identity (9) of Lemma 1.2.1.
Since any tiling can be factored uniquely as a first block followed by a sequence of
prime blocks, we have 1/(1 − ax − x2 ) ∗ xm /(1 − bx − xn ) = Qm,n (x)/(1 − Pn (x)) where
m ≥ 1 and n ≥ 2. This is equivalent to the desired formula.
Note that Theorem 3 also holds for n = 1. In Theorem 3, there are some special
cases: when b = 0, we have that
X
fm+nk (a)xm+nk =
k≥0
fm (a)xm + (−1)min(m−1,n−1) f|m−n+1|−1 (a)xm+n
.
1 − (fn (a) + fn−2 (a))xn + (−1)n x2n
In particular, when n = 1 and b = 0, we have
X
k≥m
fk (a)xk =
fm (a)xm + (−1)min(m−1,0) fm−1 (a)xm+1
.
1 − ax − x2
Using a similar method, we can also compute an explicit formula for the Hadamard
product xm /(1 − ax − x2 ) ∗ 1/(1 − xn ).
15
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
Theorem 4. For m ≥ 1 and n ≥ 2, the Hadamard product
xm
1
∗
2
1 − ax − x
1 − xn
is equal to
fn−r (a)x(q+1)n + (−1)n−r−1 f|r−1|−1 (a)x(q+2)n
1 − (fn (a) + fn−2 (a))xn + (−1)n x2n
if m = qn + r for some positive integers q and r with 0 < r < n, and is equal to
xm − fn−2 (a)xm+n
1 − (fn (a) + fn−2 (a))xn + (−1)n x2n
if m = qn for some positive integer q.
Proof. Let us consider the Hadamard product
1
xm
∗
2
1 − ax − x
1 − xn
as counting pairs of tilings. We modify the tilings of a 2 × k rectangle in the proof of
Theorem 3 so that the top row starts with a 1 × m rectangular brick to account for
the factor xm in xm /(1 − ax − x2 ). In this setting the first block where the top row
starts with a 1 × m rectangular brick will be different from all the others, but the
following blocks can be built up from a sequence of prime blocks which are exactly
the same as the prime blocks in the Hadamard product 1/(1 − ax − x2 ) ∗ 1/(1 − xn ).
For the case m = qn + r where 0 < r < n, the possible first blocks can be classified
as follows:
The first blocks of length (q + 1)n:
LLL
16
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
The first blocks of length (q + k)n (k ≥ 2):
LL
where
···
···
NNNN
NNNNN
LL LLL NNNN
NNNNN
,
,
, and
represent any strips of length n − r − 1,
n − r, n − 2, and n − 1 respectively tiled with 1 × 1 square bricks and 1 × 2 rectangular
bricks.
So, the generating function Rm,n (x) for the weighted first blocks is
Rm,n (x) = fn−r (a)x
(q+1)n
+
∞
X
fn−1−r (a)fn−1 (a)fn−2 (a)k−2 x(q+k)n
k=2
fn−1−r (a)fn−1 (a)x(q+2)n
1 − fn−2 (a)xn
(q+1)n
fn−r (a)x
+ fn−1 (a)fn−r−1 (a) − fn−2 (a)fn−r (a) x(q+2)n
= fn−r (a)x(q+1)n +
=
=
1 − fn−2 (a)xn
fn−r (a)x(q+1)n + (−1)min(n−r−1) f|r−1|−1 (a)x(q+2)n
1 − fn−2 (a)xn
where we use the identity
fn−1 (a)fn−r−1 (a) − fn−2 (a)fn−r (a) = (−1)min(n−r−1,n−1) f|r−1|−1 (a)
obtained by substituting n − 2 for m and n − r − 1 for n in identity (9) of Lemma
1.2.1. Since any tiling can be factored uniquely as a first block followed by a sequence
of prime blocks, we have xm /(1 − ax − x2 ) ∗ 1/(1 − xn ) = Rm,n (x)/(1 − Pn (x)) where
b = 0. This is equivalent to the desired formula.
For the case m = qn, we have only one first block of length m:
17
CHAPTER 1. HADAMARD PRODUCTS AND TILINGS
So the generating function Rm,n (x) for the weighted first blocks is Rm,n (x) = xm .
Therefore we have xm /(1 − ax − x2 ) ∗ 1/(1 − xn ) = xm /(1 − Pn (x)) where b = 0. This
is equivalent to the desired formula
Note that Theorem 4 also holds for n = 1.
18
CHAPTER 2
Redundant generating functions in lattice path enumeration
2.1. Introduction
One method for solving recurrences with boundary conditions is extending the
region in which the recurrence is satisfied. Then the generating function for the
extended recurrence in the new region may become simpler. Following MacMahon
[12, pp. 128], we call the generating function a redundant generating function, since
it contains some terms which are not part of the solution of the original problem.
This method is especially useful in lattice path counting problems. In this chapter
we first review the redundant generating function for ballot numbers. Then we study
variations of the ballot problem. In one of these variations, we explain a surprising
occurrence of Catalan numbers. When we define the number D20 (m−n, n) for m, n ≥ 0
by
∞
X
m,n=0
D20 (m
∞
−1
X
n+1
n+1
− n, n)x t = 1 +
(−1) Cn t
1 − x(1 + t)
,
m n
(12)
n=0
where Cn is the nth Catalan number [19, pp. 219–229], we will see that D20 (m, n)
is the number of lattice paths from (0, 0) to (m, n) (where 0 ≤ n ≤ 2m), with unit
up (↑) steps (0, 1) and unit right (→) steps (1, 0), that never cross the line y = 2x.
There is no combinatorial significance for D20 (m, n) where 2m < n or n < 0.
19
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Next, we prove a conjecture [15] of Niederhausen and Sullivan using a redundant
generating function. When we define the number S 0 (m, n) for m, n ≥ 0 by
∞
X
m,n=0
0
m n
S (m, n)x t =
3 + t − p(1 + t)2 + 4t3 2
−1
1 − x(1 + t + t + t )
,
2
3
(13)
the conjecture says that for 2m ≥ n, S 0 (m, n) is the number of paths from (0, 0) to
(4m − n − 1, 2m − n + 1), with up (%) steps (1, 1) and down (&) steps (1, −1),
that avoid four consecutive up (%) steps and never go below the x-axis. There is no
combinatorial meaning for S 0 (m, n) where 2m < n. Finally, we will find a redundant
generating function for the numbers counting lattice paths with arbitrary given up
steps, but one down step. In Theorem 6 a redundant generating function is given
that generalizes the previous two redundant generating functions (12) and (13).
2.2. The ballot problem
Let us consider the ballot problem [14, pp. 1–8] that asks about the number
B(m, n) of lattice paths from (1, 0) to (m, n) (where m > n), with unit up (↑) steps
(0, 1) and unit right (→) steps (1, 0), that stay below the line x = y. The number
B(m, n) can easily be computed by the recurrence
B(m, n) = B(m − 1, n) + B(m, n − 1) for 0 ≤ n < m, (m, n) 6= (1, 0)
(14)
with the initial condition B(1, 0) = 1 and the boundary conditions B(m, −1) = 0 and
B(m, n) = 0 for 0 ≤ m ≤ n. The recurrence (14) holds since a path ending at (m, n)
can be obtained from either a path ending at (m − 1, n) followed by a unit right (→)
step (1, 0) or a path ending at (m, n − 1) followed by a unit up (↑) step (0, 1).
20
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Let c(x) be the Catalan number generating function
c(x) =
∞
X
n
Cn x =
n=0
∞
X
n=0
√
1
2n n 1 − 1 − 4x
x =
.
n+1 n
2x
X
Now we want to find the generating function
B(m, n)xm y n . From the well-
m≥n≥0
known fact that B(n + 1, n) is the nth Catalan number Cn , we can use a variation of
(14) which is
B(m, n) − B(m − 1, n) − B(m, n − 1) =





1
if (m, n) = (1, 0)
−Cm−1 if m = n




0
(15)
otherwise.
Then the recurrence (15) is valid for all m, n ≥ 0, and is also easy to see combinatorially. Multiplying the recurrence (15) by xm y n and summing on m and n, we
get
(1 − x − y)
X
B(m, n)xm y n = x − xyc(xy).
m≥n≥0
So, we have
X
B(m, n)xm y n =
m≥n≥0
x 1 − yc(xy)
1−x−y
.
This is equivalent to
1+
X
B(m, n)xm y n =
m≥n≥0
1
.
1 − xc(xy)
(16)
Identity (16) can also be proved combinatorially by using the prime decomposition
for Dyck paths [8, pp. 1027–1030].
Instead of taking B(m, n) = 0 for m < n, we could try to define B(m, n) everywhere in the first quadrant so that the recurrence (14) is satisfied as much as possible.
21
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
6
5
4
3
2
1
0
n/m
−1 −5 −14 −28 −42 −42
−1 −4 −9 −14 −14
0
−1 −3 −5 −5
0
14
−1 −2 −2
0
5
14
−1 −1
0
2
5
9
−1
0
1
2
3
4
0
1
1
1
1
1
0
1
2
3
4
5
Table 1. The values of B̃(m, n)
0
42
42
28
14
5
1
6
We can do this by rewriting the recurrence (14) as
B(m − 1, n) = B(m, n) − B(m, n − 1) for m < n,
with B(m, n) = 0 for all m < 0. So let us define B̃(m, n) to be B(m, n) for m ≥ n ≥ 0,
and by
B̃(m − 1, n) = B̃(m, n) − B̃(m, n − 1) for m < n,
(17)
with B̃(m, n) = 0 for all m < 0. Table 1 shows the values of B̃(m, n) for the
first quadrant. It is easy to see that B̃(m, n) can be extended to the whole first
quadrant so that the recurrence is satisfied everywhere except (1, 0) and (0, n) for
n ∈ P = {1, 2, 3, . . .}. In fact, the recurrence (17) is satisfied everywhere except at
(1, 0) and (0, 1). To see this, we define B 0 (m, n) by B 0 (1, 0) = 1, B 0 (0, 1) = −1,
B 0 (m, −1) = 0, and B 0 (−1, m) = 0 for all m ∈ N = {0, 1, 2, . . .} with the recurrence
for all (m, n) ∈ N2 , (m, n) 6= (1, 0) or (m, n) 6= (0, 1),
B 0 (m, n) = B 0 (m − 1, n) + B 0 (m, n − 1).
Then it is easy to show by induction that B 0 (m, n) = −B 0 (n, m) for all m, n ∈ N, so
we have B 0 (m, m) = 0 for all m ∈ N. Therefore, the values of B̃(m, n) and B 0 (m, n)
for all (m, n) ∈ N2 coincide.
22
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
From this fact, we can assume that the recurrence for B 0 (m, n) for the first quadrant is


1 if (m, n) = (1, 0)



B 0 (m, n) − B 0 (m − 1, n) − B 0 (m, n − 1) =
−1 if (m, n) = (0, 1)




0 otherwise,
(18)
with the boundary conditions B 0 (m, n) = 0 for m < 0 or n < 0.
In terms of generating functions, the recurrence (18) is equivalent to the equation
(1 − x − y)
X
B 0 (m, n)xm y n = x − y,
m,n≥0
which is the same as
X
B 0 (m, n)xm y n =
m,n≥0
x−y
.
1−x−y
(19)
Expanding in powers in x and y and equating coefficients of xm y n in equation (19),
we have the ballot number formula:
m+n−1
m+n−1
m−n m+n
B (m, n) =
−
=
,
m−1
m
m+n
m
0
which satisfies the recurrence (14) with the initial and boundary conditions. That is,
B(m, n) = B 0 (m, n) for all 0 ≤ n ≤ m.
Also, we can see that the redundant generating function (x − y)/(1 − x − y) for
B 0 (m, n) is much simpler than the generating function x 1 − yc(xy) /(1 − x − y) for
B(m, n).
23
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
4
0
0 12 30
3
2
0 3 7 12 18
0 1 2 3 4 5 6
1
0
0 1 1 1 1 1 1 1 1
n/m 0 1 2 3 4 5 6 7 8
Table 2. The values of C2 (m, n)
2.3. Variations of the ballot problem
Now let us consider [7] the line x = py, where p ∈ P, as the boundary line in the
ballot problem instead of the boundary line x = y. Then this problem asks about
the number Cp (m, n) of lattice paths from (1, 0) to (m, n), (where m > pn) with unit
up (↑) steps (0, 1) and unit right (→) steps (1, 0), that never touch the line x = py.
Table 2 shows the values of C2 (m, n) in the case p = 2.
If we extend Cp (m, n) to the first quadrant then the recurrence seems to be satisfied everywhere except at (1, 0) and (0, 1). More precisely, let us define values of
Cp0 (m, n) by the recurrence
Cp0 (m, n) = Cp0 (m − 1, n) + Cp0 (m, n − 1),
and the initial conditions Cp0 (1, 0) = 1, Cp0 (0, 1) = −p, and the boundary conditions
Cp0 (m, −1) = 0 and Cp0 (−1, m) = 0 for all m ∈ N. This is equivalent to


1 if (m, n) = (1, 0)



Cp0 (m, n) − Cp0 (m − 1, n) − Cp0 (m, n − 1) =
−p if (m, n) = (0, 1)




0 otherwise,
(20)
with the boundary conditions Cp0 (m, n) = 0 for m < 0 or n < 0. The recurrence (20)
is valid for all m, n ≥ 0.
24
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
To show that Cp0 (m, n) = Cp (m, n) for m ≥ pn, it is enough to show that
Cp0 (pn, n) = 0 for all n ∈ N. Multiplying the recurrence (20) by xm y n and summing on m and n, we get
X
Cp0 (m, n)xm y n =
m,n≥0
x − py
.
1−x−y
(21)
Expanding in powers of x and y and equating coefficients of xm y n in equation (21),
we have the formula: for all m, n ∈ N
Cp0 (m, n)
m+n−1
m+n−1
m − pn m + n
.
=
−p
=
m+n
m
m−1
m
From the formula for Cp0 (m, n), it is easy to see Cp0 (pn, n) = 0. Therefore the values
of Cp0 (m, n) are the same as the values of Cp (m, n) = 0 since Cp0 (pn, n) = 0 and
Cp (pn, n) = 0 for all n ∈ N.
At this point it might be natural to ask what happens if we use the line y = px
(where p ∈ P) as the boundary line instead of the line y = x. For convenience’ sake
we allow a lattice path to touch the line y = px. Let Dp (m, n) be the number of
lattice paths from (0, 0) to (m, n) (where n ≤ pm), with unit up (↑) steps (0, 1) and
unit right (→) steps (1, 0), that never cross the line y = px. Let us consider the case
p = 2. Since the slope of the boundary line is 2, we require D2 (m, 2m + 1) = 0 and
D2 (m, 2m + 2) = 0 for all m ∈ N. Then the number D2 (m, n) can easily be computed
by the recurrence
D2 (m, n) = D2 (m − 1, n) + D2 (m, n − 1) for 0 ≤ n ≤ 2m, (m, n) 6= (0, 0),
with the initial condition D2 (0, 0) = 1 and the boundary conditions
D2 (m, 2m + 1) = 0, D2 (m, 2m + 2) = 0, and D2 (m, −1) = 0 for all m ∈ N.
25
(22)
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
5
4
3
2
1
0
n/m
Table 3.
0 12
0 3 12
0 3 9
0 1 3 6
0 1 2 3
1 1 1 1
0 1 2 3
The values of
43 108 228 431
31 65 120 203
19 34 55 83
10 15 21 28
4
5
6
7
1
1
1
1
4
5
6
7
D2 (m, n) for 0 ≤ n ≤ 2m
Table 3 shows the values of D2 (m, n). From the table we can observe that the
numbers on the line y = 2x are the same as the numbers on the line x − 1 = 2y in
Table 2, but the other numbers off the line y = 2x are different from the numbers in
Table 2. We can prove that the numbers on the line y = 2x in Table 3 are the same
as the numbers on the line x − 1 = 2y in Table 2 by finding a bijection between the
set of lattice paths from (0, 0) to (pn, n), with unit up (↑) steps (0, 1) and unit right
(→) steps (1, 0), that never cross the line x = py and the set of lattice paths from
(0, 0) to (n, pn), with unit up (↑) steps (0, 1) and unit right (→) steps (1, 0), that
never cross the line y = px.
Now let us present the bijection. When we are given a lattice path P from (0, 0)
to (pn, n), first reverse it to obtain a path P r from (pn, n) to (0, 0) consisting of down
(↓) steps and left (←) steps and change a down (↓) step with a right (→) step and
a left (←) step with a up (↑) step. Then we have a lattice path P 0 from (0, 0) to
(n, pn). In the same way, we can transform the lattice path P 0 from (0, 0) to (n, pn)
into the lattice path P from (0, 0) to (pn, n). Since the path P does not cross the line
y = px, we know that for each i, the number of left steps among the first i steps in
the path P r is always less than or equal to p times the number of down steps among
the first i steps. So the corresponding path P 0 to the path P is a path that never
26
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
crosses the line y = px, and vice versa. Therefore we have
Dp (n, pn) = Cp (pn + 1, n) for n ≥ 0 and p ≥ 1.
From the path transformation, we can also derive the number of lattice paths
starting at a point that is below the line y = px and ending on the line y = px,
with unit up (↑) steps (0, 1) and unit right (→) steps (1, 0), that never cross the line
y = px. That is, the number of lattice paths from (i, j) to (n, pn) (where j ≤ pi)
having n − i unit right (→) steps and pn − j unit up (↑) steps, that never cross the
line y = px is Cp pn − j + 1, n − i .
2.3.1. Another way to get Catalan numbers. If we compute, as before, the
values of D2 (m, n) using the recurrence
D2 (m − 1, n) = D2 (m, n) − D2 (m, n − 1)
(23)
with the boundary conditions (22) extended to the first quadrant, then we do not
find anything interesting. So let us extend these values of D2 (m, n) satisfying the
recurrence (23) to the region { (x, y) ∈ Z2 | y ≥ −x, y ≥ 0 }. There is no combinatorial
significance for D2 (m, n) where n > 2m.
Table 4 shows the values of D2 (m, n). Surprisingly, we find that the number
D2 (−n − 1, n + 1) is equal to (−1)n+1 Cn for n ≥ 0 on the boundary line y = −x in
the table. Note that the numbers on the line x = −2 in Table 4 are (−1)n Mn , where
Mn is the nth coefficient of the Motzkin number generating function
m(x) =
∞
X
n
Mn x =
1−x−
n=0
√
1 − 2x − 3x2
.
2x2
(We will deal with the Motzkin number generating function in Chapter 3.)
27
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
6
42
28 19
−14 −9
5
4
5
3
2
1
0
n/m −6 −5 −4
Table 4.
13
9
6
3
0
−6 −4 −3 −3 −3
3
2
1
0
0
−2 −1 −1 −1
0
1
0
0
1
−1
0
1
1
1
−3 −2 −1
0
1
The values of D2 (m, n)
0 12 55
0 12 43
3 12 31
3 9 19
3 6 10
2 3 4
1 1 1
2 3 4
Before we find the redundant generating function for D2 (m, n) (We will give it
after Theorem 6) let us prove that we have the Catalan numbers on the boundary
line y = −x in Table 4. Since D2 (n, 2n) = C2 (2n + 1, n) for all n ∈ N, we have
3n
1
.
D2 (n, 2n) =
2n + 1 n
So, by the recurrence (23) with D2 (n, 2n + 1) = 0, we have for all n ∈ N
1
3n
D2 (n − 1, 2n + 1) = −
.
(2n + 1) n
For each n ∈ N, let us define an to be D2 (n−1, 2n+1) and bn to be D2 (−n−1, n+1).
Now let us find a formula for bn in terms of ai in Table 4. As shown in Figure 2.1,
we can see
D2 (m − 1, n) = D2 (m, n) − D2 (m, n − 1).
By iterating this recurrence, we can easily derive bn =
P∞
i=0 (−1)
n−2i
cni ai , where cni
is the number of paths from (i − 1, 2i + 1) to (−n − 1, n + 1) with unit left (←) steps
(−1, 0) and left up (-) steps (−1, 1). That is,
3i + n − 2i
n+i
cni =
=
.
3i
3i
28
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
The formula bn = (−1)n+1 Cn , where Cn is the nth Catalan number, is a consequence of the following
lemma.
+ −9 + −6 + −4 + −3 + −3 + −3 a
−14 b4 2
I–
@
I–
@
I–
@
I–
@
I–
@
@
@
@
@
@
+
+
+
2
1 + 0
G3
5 3
I–
@
I–
@
I–
@
I – @
@
@
@
@
+
+
−1 −1+ −1 a
−2 b2 1
I–
@
I–
@
@ @
+ 0
b1 1 I – @
@ −1 b0 a0
Figure 2.1. The values of D2 (m, n)
Lemma 2.3.1. For n ≥ 0,
∞ X
n+i
i=0
3i
3i
1
2n
1
=
.
2i + 1 i
n+1 n
Proof. Since the binomial coefficient
n+i
3i
is nonzero only for 0 ≤ i ≤ bn/2c,
the summation i on the left is in that range. So, for n ≥ 0
∞ X
n+i
i=0
3i
bn/2c
X
3i
1
(3i)!
1
(n + i)!
=
·
·
2i + 1 i
(3i)! (n − 2i)! 2i + 1 i! (2i)!
i=0
bn/2c
=
X
i=0
bn/2c
(n + i)!
(n − 2i)! i! (2i + 1)!
(n + i)!
n!
·
(n − 2i)! i! n! (2i + 1)!
i=0
∞ 1 X n+i
n+1
=
n + 1 i=0
n
n − 2i
1
2n
=
,
n+1 n
=
X
29
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
where the second last equality is only valid for 0 ≤ i ≤ bn/2c and the last equality
n+1
is derived from the fact that the coefficient of xn in (1 + x)/(1 − x2 )
is the same
as the coefficient of xn in 1/(1 − x)n+1 .
Note that in the paper [21] Sun proved Lemma 2.3.1 combinatorially by using
binary and ternary trees. Another proof of Lemma 2.3.1 is in the paper [13].
So we have the following result.
Theorem 5. For n ≥ 0, the number D2 (−n − 1, n + 1) is equal to (−1)n+1 Cn , where
Cn is the nth Catalan number.
Note that there is no combinatorial significance for D2 (−n − 1, n + 1) for n ≥ 0.
2.4. A conjecture of Niederhausen and Sullivan
A generalized Dyck path is a path starting at (0, 0), with up (%) steps (1, 1)
and down (&) steps (1, −1), that never goes below the x-axis. Now let us count the
number of generalized Dyck paths without four consecutive up (%) steps. Define
S(m, n) to be the number of such generalized Dyck paths from (0, 0) to (m, n) that
end with a down (&) step (1, −1). Since a path starts at (0, 0), we set the initial
condition S(0, 0) = 1. It is obvious that S(2, 0) = 1, S(3, 1) = 1, and S(4, 2) = 1.
But we have S(5, 3) = 0 because we don’t allow four consecutive up steps. Since
a path cannot have four consecutive up (%) steps, such a generalized Dyck path
ending at (m, n) can be obtained uniquely from a generalized Dyck path ending at
(m − i − 1, n − i + 1) followed by i consecutive up (%) steps for 0 ≤ i ≤ 3 and a down
(&) step. So, when we define S(m, n) = 0 for n < 0 we have the recurrence relation
S(m, n) =
2
X
S(m − i − 2, n − i) for m, n ∈ N, (m, n) 6= (0, 0).
i=−1
30
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
6
5
4
3
2
1
3
1
3
1
0
1
1
2
5
n/m 0 1 2 3 4 5 6
Table 5. The
1
3
1
2
6
9
10
8
19
28
33
32
23
116
101
68
13
36
104
7 8 9 10 11 12
values of S(m, n)
205
13
2
1 0 5 32 112 297
6
5
−1 −1 0 8 33 90 200
1
0 2 10 28 61 115
4
3
−1
0 3 9 19 34 55
2
0
1 3 6 10 15 21
1
0
1 2 3 4
5
6
0
1
1 1 1 1
1
1
n/m
0
1 2 3 4
5
6
0
Table 6. The values of S (m, n)
Table 5 shows the values of S(m, n) with the 0’s omitted.
In the paper [15], Niederhausen and Sullivan conjectured that the number S 0 (m, n)
for m, n ∈ N defined by
∞
X
S 0 (m, n)xm tn =
m,n=0
3 + t − p(1 + t)2 + 4t3 2
−1
1 − x(1 + t + t2 + t3 )
is equal to S(4m − n, 2m − n) in the case 2m ≥ n. That is, in the case 2m ≥ n ≥ 0,
S 0 (m, n) is the number of generalized Dyck paths from (0, 0) to (4m − n, 2m − n)
without four consecutive up steps that end with a down (&) step (1, −1). Note that
there is no combinatorial meaning for S 0 (m, n) when 2m < n. Table 6 shows the
values of S 0 (m, n).
31
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
2.5. Main Theorem of Chapter 2
Now we will prove a generalization of the Niederhausen and Sullivan conjecture
involving more general paths. To do this, let us define a step set T to be a subset of
the set N ∪ {−1}. We assume that the step set T has −1 as an element and that K
is the largest element of T . Now let us consider a path from (0, 0) to (m, 0) (where
m ∈ N), with steps (1, i) (where i ∈ T ), that never goes below the x-axis. Put a
weight of ci x on each step (1, i), where ci is an arbitrary weight for all i ∈ T except
c−1 = 1 and cK = 1, and let f (x) be the generating function for weighted paths from
(0, 0) to (m, 0), with steps (1, i) (where i ∈ T ), that never go below the x-axis.
r
r
@r
@r
@ r
r
r
r
@r
@ r @r
r
r
r
@r @ r @ r @ r
@ r
@r
r
r @ r
r
@
@ r @r
@ r
@ r @r
@r
@ r
@ r
@r
r
Figure 2.2. Decomposition for a path from (0, 0) to (m, 0) when T = {−1, 2}
As shown in Figure 2.2, any such path starting with a step (1, i), where i ≥ 0, can
be decomposed uniquely [8, pp. 1027–1030] by cutting it right before the first points
at height j for j = i − 1, i − 2, . . . , 0. This gives the functional equation for f :
f (x) =
X
ci xf (x)
i+1
,
(24)
i∈T
where the term for i = −1 corresponds to the empty path.
Now let us consider a path Q from (0, h) to (m, 0) (where h, m ∈ N), with steps
(1, i) (where i ∈ T ), that never goes below the x-axis. Decomposing the path Q,
as before, into the first time it reaches height i for i = h − 1, h − 2, . . . , 0, we can
deduce that the generating function for such weighted paths from (0, h) to (m, 0) is
xh f (x)h+1 .
32
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
For m ≥ 0 and n ≥ 0, we define Ph (m, n) to be the sum of the weights of lattice
paths from (0, h) to (m, n), with steps (1, i) where i ∈ T , that never go below the
x-axis. Then we have the initial condition Ph (0, h) = 1 and Ph (m, n) = 0 for n < 0
because paths cannot go below the x-axis. Since such paths have steps (1, i) (where
i ∈ T ) we have the recurrence
Ph (m + 1, n) =
X
ci Ph (m, n − i) for m, n ≥ 0.
(25)
i∈T
Then we can easily deduce
h
x f (x)
h+1
=
∞
X
m
Ph (m, 0)x and f (x) =
m=0
∞
X
P0 (m, 0)xm .
m=0
Also, we find the generating function for Ph (m, n):
h+1 −1
t
xf
(x)
P
P
Ph (m, n)xm tn =
−
,
i
1 − i∈T (ci t x) 1 − i∈T (ci ti x)
m,n=0
∞
X
th
(26)
where the first term on the right side in (26) counts the sum of the weights of all
lattice paths start at (0, h), and the second term counts the sum of the weights of all
lattice paths start at (0, h) and go below the x-axis.
Now we want to extend the region of definition of Ph (m, n) to all integer values
of n. That is, for m ≥ 0 and n < 0, we want the recurrence (25) to hold below the
x-axis. So, for m ≥ 0 let us define Ph0 (m, n) to be Ph (m, n) for n ≥ −K, and by
Ph0 (m, n) = Ph0 (m + 1, n + K) −
X
ci Ph0 (m, n + K − i) for n < −K.
(27)
i∈T −{K}
Note that Ph0 (m, n) in (27) holds for all m ∈ N and n ∈ Z.
Table 7 shows the numbers P10 (m, n) where T = {−1, 1, 2} and ci = 1 for all i ∈ T .
In this table once we know the values of P10 (0, n) in the first column, we can figure
33
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
4
0
0
2
4
12
32
72
194
0
1
1
6
9
27
60
137
3
2
0
1
2
3
10
16
44
93
1
0
2
2
5
12
21
56
1
0
0
1
0
2
2
5
12
21
0
0
0
0
0
0
0
0
−1
−2
0
0
0
0
0
0
0
0
−3
0 −1
0 −2
−2
−5 −12
−21
−4
0
1
0
2
2
5
12
21
−1 −1 −2 −4
−7 −17 −33
−77
−5
−6
2
2
4
8
14
34
66
154
−3 −5 −8 −17 −33 −72 −155 −345
−7
−8
6
10
18
35
74
155
342
762
−9 −13 −20 −39 −76 −160 −344 −753 −1696
−10
26
43
82 167
348
758 1670
3759
n/m
0
1
2
3
4
5
6
7
Table 7. The values of P10 (m, n)
out the values of P10 (m, n) for m > 0 by the recurrence that corresponds to the step
set T = {−1, 1, 2}. For example, P10 (1, n) can be computed by summing P10 (0, n − i)
for all i ∈ T . Similarly, P10 (m + 1, n) can be computed by summing P10 (m, n − i) for
all i ∈ T and m ≥ 0. So, the values of Ph0 (m, n) are determined by the values of
Ph0 (0, n). Therefore, we will find the generating function for Ph0 (0, n) for n < 0.
In the case T = {−1, 0, 1, 2} with ci = 1 for all i ∈ T , we will prove P00 (m, n) =
S(2m + n, n) (where m, n ∈ N) which is related to the Niederhausen and Sullivan
conjecture. Now we present a direct way to see the relation between the numbers
S(m, n) and P00 (m, n) by giving a bijection between the set A of generalized Dyck
paths from (0, 0) to (2m + n, n) (where n ≥ 0) with no 4 consecutive up steps that
end a down step and the set B of paths from (0, 0) to (m, n) with steps (1, −1), (1, 0),
(1, 1), and (1, 2) that never go below the x-axis. Then the number P00 (m, n) becomes
34
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
the cardinality of the set B. Then the following lemma gives us the bijection between
the sets A and B.
Lemma 2.5.1. For m, n ∈ N, the number S(2m + n, n) is equal to the number
P00 (m, n).
Proof. Let us be given a generalized Dyck path ending at (2m + n, n) that ends
a down (&) step (1, −1). Then since an up step (%) of the path is (1, 1) and a down
step (&) is (1, −1), we can compute that the path has m + n up steps and m down
steps.
Then it is enough to prove this lemma by presenting a bijection from the set A
to the set B. Now let us decompose the path at every down step of the path. Then
we have the following four kinds of subpaths: &, %&, %%&, and %%%&. Now
let us convert each subpath & to a (1, −1) step, %& to a (1, 0) step, %%& to a
(1, 1) step, and %%%& to a (1, 2) step. Then we can get a path with steps (1, −1),
(1, 0), (1, 1), and (1, 2) that never goes below the x-axis. Similarly, when we are given
a path with steps (1, −1), (1, 0), (1, 1), and (1, 2) that never goes below the x-axis,
we can have a generalized Dyck path ending at height at least 0 by converting each
(1, −1) step to &, each (1, 0) step to %&, each (1, 1) step to %%&, and each (1, 2)
step to %%%&.
From the recurrence (27) for Ph0 (m, n), we have the following lemma.
Lemma 2.5.2. Let T ⊂ N ∪ {−1} be a step set and let K be the largest element of T .
We assume that −1 ∈ T , K 6= −1, cK = 1, and c−1 = 1. Then we have for h, m ∈ N,
Ph0 (m, −K − 1) = −Ph0 (m, 0).
35
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Proof. By the recurrence (27) we have
X
Ph0 (m, −K − 1) = Ph0 (m + 1, −1) −
ci Ph0 (m, −1 − i)
i∈T −{K}
= Ph0 (m + 1, −1) − Ph0 (m, 0)
= −Ph0 (m, 0),
where the second and third equalities hold because Ph0 (m, n) = 0 for −K ≤ n ≤
−1.
Now we are trying to find the values of weighted paths with a step set T . For any
h, K ∈ N, let us define A(x) to be the generating function for −Ph0 (m, 0):
A(x) = −
∞
X
Ph0 (m, 0)xm = −xh f (x)h+1 ,
(28)
m=0
and, for j ∈ Z, let Aj (x) be the generating function for Ph0 (m, −K − 1 − j):
Aj (x) =
∞
X
Ph0 (m, −K − 1 − j)xm .
m=0
Then Aj (x) = 0 for −K ≤ j ≤ −1 since Ph0 (m, n) = 0 for −K ≤ n ≤ −1, and by
Lemma 2.5.2 we know A0 (x) = A(x) and A−K−1 (x) = −A(x).
Multiplying both sides of the recurrence (27) by xm and summing on m gives
∞
X
m=0
Ph0 (m, n)xm
=
∞
X
Ph0 (m + 1, n + K)xm −
m=0
X
ci
i∈T −{K}
∞
X
Ph0 (m, n + K − i)xm . (29)
m=0
By setting n = −K − 1 − j, where j ≥ 0, in identity (29) we have
Aj (x) =
1
(Aj−K (x) − Aj−K (0)) −
x
36
X
i∈T −{K}
ci Ai+j−K (x).
(30)
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Let A(x, t) :=
P∞
Aj (x)tj . Multiplying both sides of identity (30) by tj and sum-
j=0
ming on j gives
1
A(x, t) =
x
X
0≤j≤K−1
K
−t
∞
X
X
Aj−K (x)tj + tK A(x, t) −
Aj−K (0)tj
0≤j≤K−1
j
Aj (0)t
j=0
−
X
X
ci
j
K−i
Ai+j−K (x)t + t
A(x, t) .
0≤j≤K−i−1
i∈T −{K}
(31)
The fifth term on the right side in (31) is A(x) from the following simplification:
−
X
i∈T −{K}
ci
X
X
Ai+j−K (x)tj = −
0≤j≤K−i−1
Aj−K−1 (x)tj
0≤j≤K
X
−
i∈T −{K,−1}
X
=−
ci
X
Ai+j−K (x)tj
0≤j≤K−i−1
Aj−K−1 (x)tj
0≤j≤K
=A(x),
where the second and third equalities hold because Aj (x) = 0 for −K ≤ j ≤ −1 and
A−K−1 (x) = −A(x).
Therefore, collecting terms in (31) gives
tK
A(x, t) 1 −
+
x
X
K−i
ci t
i∈T −{K}
∞
tK X
= A(x) −
Aj (0)tj .
x j=0
(32)
To figure out Ph0 (0, n), we need to determine A(0, t). To do this, we work in the
ring of Laurent series C((x))[[t]]. We cannot set x = 0 directly in (32) but we can
use the following lemma to find the constant term in x. Let us define CTx to be the
constant term of a power series in x.
37
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Lemma 2.5.3. Let H(x) be a power series in x. Let α and β be power series in t with
no constant term. Then in C((x))[[t]] the constant term in x of H(x)/(1 − αx−1 − β)
is H α/(1 − β) /(1 − β).
Proof. Let H(x) =
CTx
P∞
n=0
an xn . Then
H(x)
H(x)
= CTx
−1
1 − αx − β
(1 − β) 1 − αx−1 /(1 − β)
H(x)
1
CTx
1−β
1 − αx−1 /(1 − β)
∞ X
α m −m
1
CTx H(x)
x
=
1−β
1−β
m=0
=
∞
∞ X
X
1
α m −m
n
=
CTx
an x
x
1−β
1−β
n=0
m=0
α 1
=
H
.
1−β
1−β
Now let us find the constant term in x in A(x, t) by Lemma 2.5.3. Dividing
P
identity (32) by 1 − tK /x + i∈T −{K} ci tK−i gives
A(x, t) =
K
t
+
1−
x
A(x)
X
−
K−i
ci t
i∈T −{K}
38
∞
tK X
Aj (0)tj
x j=0
tK
+
1−
x
X
i∈T −{K}
ci t
.
K−i
(33)
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Since the second term on the right-hand side of (33) contains only negative powers
of x, we get the constant term in x in A(x, t) as follows:
CTx A(x, t) = CTx
=
1−
tK
x
+
A(x)
P
K−i
i∈T −{K} ci t
1
P
1+
=−
i∈T −{K} ci t
K−i
A
tK
1+
thK
(1 +
P
K−i )h+1
i∈T −{K} ci t
P
f
K−i
i∈T −{K} ci t
tK
1+
h+1
P
K−i
i∈T −{K} ci t
,
where the second equality follows from Lemma 2.5.3 and the last equality follows
from (28).
To simplify the constant term A(0, t) that we computed above, let us define a
power series g by
X
1 tK f
, where B(t) =
ci tK−i .
g(t) =
B(t) B(t)
i∈T
Then the constant term in x in A(x, t) becomes
hK
−t
h+1
g(t)
= A(0, t) =
∞
X
Aj (0)tj
j=0
=
∞
X
Ph0 (0, −K − 1 − j)tj .
j=0
Since Ph0 (0, n) = 0 for n ≥ −K except for Ph0 (0, h) = 1, we have
X
n∈Z
Ph0 (0, h − n)tn =
X
Ph0 (0, h − n)tn = 1 − g(t)h+1 t(h+1)(K+1) .
n∈N
39
(34)
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Finally, from the above generating function for Ph0 (0, h − n), we can find the
generating function for Ph0 (m, n). To do this, we rewrite the recurrence (27) as
Ph0 (m + 1, n + K) =
X
ci Ph0 (m, n + K − i).
i∈T
Then replacing n with Km + h − n gives
X
Ph0 m + 1, K(m + 1) + h − n =
ci Ph0 m, K(m + 1) + h − n − i .
i∈T
Next, multiplying both sides by tn and summing on n gives
∞
X
Ph0
∞
X
n
Ph0 (m, Km + h − n)tn .
m + 1, K(m + 1) + h − n t = B(t)
n=0
n=0
Finally, we conclude
∞
X
Ph0 (m, Km
n
m
+ h − n)t = B(t)
n=0
∞
X
Ph0 (0, h − n)tn
n=0
= B(t)m 1 − g(t)h+1 t(h+1)(K+1) .
We summarize with the following theorem.
Theorem 6. Let T ⊂ N ∪ {−1} be a step set and let K be the largest element of T .
We assume that −1 ∈ T and K 6= −1. Let Ph (m, n) be the sum of the weights of the
paths from (0, h) to (m, n) (where h, m, n ∈ N), with steps (1, i) (where i ∈ T ), that
never go below the x-axis, where we put a weight of ci x on each step (1, i) (where ci
is an arbitrary weight for all i ∈ T except c−1 = 1 and cK = 1). For each m ≥ 0,
40
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
define Ph0 (m, n) for n ≤ Km + h by
Ph (x, t) =
∞
X
Ph0 (m, Km + h − n)xm tn
m, n=0
−1
h+1 (h+1)(K+1)
= 1 − g(t) t
1 − xB(t)
,
where the power series g(t) is uniquely determined by
X
ci t
K−i
g(t)
i
X
tK+1 g(t)
n−1
= 1,
(35)
n=1
i∈T −{0,−1}
and Ph0 (m, n) = 0 for n > Km + h. Then we have
Ph0 (m, n) = Ph (m, n) for m, n ≥ 0.
Proof. It is enough to show that the power series g defined by (34) satisfies the
functional equation (35) and g is uniquely determined by (35). From the functional
equation (24) for f , we can find the functional equation for g. Substituting tK /B(t)
for x in equation (24) and simplifying gives the functional equation for g
X
B(t)g(t) =
i+1
ci tK g(t)
.
(36)
i∈T
Since B(t) =
X
ci tK−i where c−1 = 1 and cK = 1, our theorem follows from the
i∈T
factorization:
X
i∈T
ci t
K−i
X
i+1
g(t) −
ci tK g(t)
i∈T
= tK+1 +
X
K−i
ci t
g(t) − 1 +
i∈T −{0,−1}
X
i∈T −{0,−1}
41
i+1
ci t g(t)
K
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
= tK+1 g(t) +
X
i∈T −{0,−1}
K+1
= t
g(t) − 1 1 −
X
ci tK−i g(t) − 1 −
ci tK g(t)
i+1
i∈T −{0,−1}
X
i∈T −{0,−1}
K−i
ci t
g(t)
i
X
K+1
t
n−1
g(t)
.
n=1
Now we want to show that the power series g is uniquely determined by (35).
Every term, except i = K and n = 1, on the left side of (35) has a factor tj for j ≥ 1
and the term for i = K and n = 1 is g(t). Then the functional equation (35) may be
written as
g(t) = 1 + tT (t, g(t)), for some polynomial T in t.
Equating coefficients gives a unique power series solution. Therefore the power series
g is uniquely determined by (35).
Note that in Theorem 6 we prefer to take Ph0 (m, Km + h − n) instead of Ph0 (m, n)
as the coefficient of xm tn in Ph (x, t) because we want Ph (x, t) to be a power series,
not a Laurent series. Comparing the generating function (26) for Ph (m, n) with the
generating function for Ph0 (m, n) in Theorem 6, we can see that the degree of equation
(35) that g satisfies is one less than the degree of equation (24) that f satisfies. In
particular, when the largest element of the step set T is 2 (that is, K = 2) we know
that the degree of f is 3 and the degree of the corresponding function g is 2. So,
while the power series g can be easily computed by using the quadratic formula, the
power series f cannot be. We will see more details with examples in the following
subsection.
2.5.1. Corollaries to our main theorem. First let us consider the case T =
{−1, 1} and h = 0. Then we will see that Theorem 6 gives (19) in this case. Since
42
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
we defined B(m, n) to be the number of lattice paths from (1, 0) to (m, n) (where
m > n), with unit up steps (0, 1) and unit right steps (1, 0), that stay below the line
x = y we can conclude
B(m + 1, n) = P00 (m + n, m − n) for m ≥ n.
Since the functional equation for g in Theorem 6 is g(t) = 1, we have
∞
X
P00 (m, m − n)xm tn =
m,n=0
1 − t2
.
1 − x(1 + t2 )
(37)
1−t
.
1 − x(1 + t)
(38)
Replacing t2 with t and n with 2n in (37) gives
∞
X
P00 (m, m − 2n)xm tn =
m,n=0
Next, replacing m with m + n and t with t/x in (38) gives
X
P00 (m + n, m − n)xm tn =
n≥0
m≥−n
x−t
.
x(1 − x − t)
(39)
Since the right side of equation (39) does not have terms xi tj for i < −1, we reduce
the range from m ≥ −n to m ≥ −1 on the left side. That is,
X
P00 (m + n, m − n)xm tn =
n≥0
m≥−1
x−t
.
x(1 − x − t)
Finally, multiplying by x and substituting m + 1 for m in (40) gives
∞
X
P00 (m − 1 + n, m − 1 − n)xm tn =
m,n=0
43
x−t
,
1−x−t
(40)
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
which is the same as (19) because
B(m + 1, n) = P00 (m + n, m − n) and B 0 (m, n) = B(m, n) for 0 ≤ n ≤ m.
A more interesting case is T = {−1, 0, 1, 2} where K = 2. As we mentioned before,
in this case we can easily compute the power series g using the quadratic formula.
Then the functional equation for g in Theorem 6 becomes t3 g(t)2 +(1+c1 t)g(t)−1 = 0,
so
g(t) =
−1 − c1 t +
√
1 + 2c1 t + c1 2 t2 + 4t3
,
2t3
(41)
where the sign of the square root is determined because g is a power series.
Now as a special case we are going to prove the Niederhausen and Sullivan conjecture and give another proof of Theorem 5. For the conjecture we set h = 0, c0 = 1,
and c1 = 1 because the paths start at (0, 0) and each path with steps (1, i) where
i ∈ T = {−1, 0, 1, 2} is counted once. Then by (41) we have the power series
g(t) =
−1 − t +
√
1 + 2t + t2 + 4t3
.
2t3
That is, g(t) = 1 − t + t2 − 2 t3 + 4 t4 − 7 t5 + 13 t6 − 26 t7 + 52 t8 − 104 t9 + 212 t10 + · · · .
So, by Theorem 6 we have the generating function for the values of P00 (m, n)
∞
X
m,n=0
P00 (m, 2m − n)xm tn =
3+t−
√
−1
1 + 2t + t2 + 4t3 2
3
1 − x(1 + t + t + t )
,
2
which is the conjecture [15] of Niederhausen and Sullivan because S 0 (m, n) = P00 (m, 2m−
n) for m, n ≥ 0. By P00 (m, 2m − n) = S(4m − n, 2m − n) where 2m ≥ n, we conclude
S 0 (m, n) = S(4m − n, 2m − n) for 0 ≤ n ≤ 2m. That is, S 0 (m, n) is the number of
paths from (0, 0) to (4m − n − 1, 2m − n + 1), with up (%) steps (1, 1) and down
44
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
(&) steps (1, −1), that avoid four consecutive up (%) steps and never go below the
x-axis.
For another proof of Theorem 5, first let us consider a lattice path R from (0, 0)
to (m, n) (where n ≤ 2m), with unit up (↑) steps (0, 1) and unit right (→) steps
(1, 0), that never crosses the line y = 2x. To apply Theorem 6 we transform the path
R into a path from (0, 0) to (m + n, 2m − n), with up steps (1, 2) and down steps
(1, −1), that never goes below the x-axis. We replace a unit right (→) step (1, 0) with
a step (1, 2) and a unit up (↑) step (0, 1) with a step (1, −1). The transformed path
does not go below the x-axis because when the path R takes a unit right (→) step
(1, 0) the y-coordinate difference between the path R and the line y = 2x increases
by 2, whereas when the path R takes a up (↑) step (0, 1) the y-coordinate difference
decreases by 1. So, we have
D2 (m, n) = P00 (m + n, 2m − n) for 2m ≥ n.
(42)
In Theorem 6 we take the step set T = {−1, 2}. Then we set h = 0, c0 = 0, and
c1 = 0 since paths start at (0, 0) and have the step set T = {−1, 2}. Table 8 shows
the values of P00 (m, n).
We see from Table 4 and Table 8 that
D2 (m, n) = P00 (m + n, 2m − n) for n ≥ 0 and m + n ≥ 0.
(43)
We can prove that (43) holds by using the recurrences (23) and (27). There is no
combinatorial meaning for D2 (m, n) where 2m < n or n < 0.
By (41) we have the power series
g(t) =
−1 +
√
∞
1 + 4t3 X
=
(−1)n Cn t3n .
3
2t
n=0
45
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
3
0
0
0
2
0
0
9
0
0
1
0
0
3
0
0
12
2
1
0
0
1
0
0
3
0
0
1
0
0
1
0
0
3
0
0
−1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
−2
−3 −1
0
0 −1
0
0 −3
0
−4
0
0
0
0
0
0
0
0
−5
0
0 −1
0
0 −3
0
0
1
0
0
1
0
0
3
0
−6
−7
0 −1
0
0 −3
0
0 −12
0
0
2
0
0
6
0
0
−8
−9 −2
0
0 −4
0
0 −15
0
−10
0
3
0
0
9
0
0
36
−11
0
0 −6
0
0 −21
0
0
−12
5
0
0 13
0
0
51
0
−13
0 −9
0
0 −30
0
0 −127
−14
0
0 19
0
0
72
0
0
n/m
0
1
2
3
4
5
6
7
0
Table 8. The values of P0 (m, n)
That is, g(t) = 1 − t3 + 2 t6 − 5 t9 + 14 t12 − 42 t15 + 132 t18 − 429 t21 + 1430 t24 + · · · .
So, by Theorem 6 the generating function for the values of P00 (m, n) is
∞
X
P00 (m, 2m
∞
−1
X
n+1
3(n+1)
3
− n)x t = 1 +
(−1) Cn t
1 − x(1 + t )
.
m n
m,n=0
n=0
Replacing t with t1/3 and 3n with n gives
∞
X
P00 (m, 2m
∞
−1
X
.
− 3n)x t = 1 +
(−1)n+1 Cn tn+1 1 − x(1 + t)
m,n=0
m n
n=0
With D20 (m, n) defined as in (12), we know P00 (m, 2m − 3n) = D20 (m − n, n) for
m, n ∈ N. So, we have
D20 (−n − 1, n + 1) = (−1)n+1 Cn for all n ≥ 0.
46
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
From (43) we conclude
D2 (−n − 1, n + 1) = (−1)n+1 Cn for all n ≥ 0.
Also, by (42) we deduce
D20 (m, n) = D2 (m, n) for 0 ≤ n ≤ 2m.
That is, D20 (m, n) is the number of lattice paths from (0, 0) to (m, n) (where 0 ≤ n ≤
2m), with unit up (↑) steps (0, 1) and unit right (→) steps (1, 0), that never cross the
line y = 2x.
For another application of Theorem 6 we can find a redundant generating function
for Dp (m, n) which was defined in Section 2.3. To do this let us consider a lattice
path S from (0, 0) to (m, n) (where n ≤ pm), with unit up (↑) steps (0, 1) and unit
right (→) steps (1, 0), that never crosses the line y = px. To apply Theorem 6 we
transform the path S into a path from (0, 0) to (m + n, pm − n), with up steps (1, p)
and down steps (1, −1), that never goes below the x-axis. We replace a unit right
(→) step (1, 0) with a step (1, p) and a unit up (↑) step (0, 1) with a step (1, −1).
The transformed path does not go below the x-axis because when the path S takes
a unit right (→) step (1, 0) the y-coordinate difference between the path S and the
line y = px increases by p, whereas when the path S takes a up (↑) step (0, 1) the
y-coordinate difference decreases by 1. So we take the step set T = {−1, p} and
Dp (m, n) = P00 (m + n, pm − n). By Theorem 6, we know that the functional equation
for g(t) is
g(t)
p X
g(t)t
p+1
n=1
47
n−1
= 1,
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
which is equal to
Pp
n=1
p+1
g(t)t
n
= tp+1 . From this equation we see that g(t)tp+1
is a power series in tp+1 , so we may define a power series γ(t) by γ(tp+1 ) = g(t)tp+1 .
Then the equation becomes
p
X
γ(t)n = t.
(44)
n=1
That is, γ(t) is the compositional inverse of t + t2 + · · · + tp . So, by Theorem 6 (where
h = 0) we have the generating function for the values of P00 (m, n) is
∞
X
−1
P00 m, pm − (p + 1)n xm tn = 1 − g(t)tp+1 1 − x(1 + tp+1 ) .
(45)
m,n=0
By Dp (m − n, n) = P00 m, pm − (p + 1)n and Theorem 6 (where h = 0), we can
deduce the following corollary after replacing tp+1 with t and (p + 1)n with n in (45).
Corollary 2.5.1. Let Dp (m, n) be the number of lattice paths from (0, 0) to (m, n)
(where n ≤ pm), with unit up (↑) steps (0, 1) and unit right (→) steps (1, 0), that
never cross the line y = px. Define the number Dp0 (m, n) for m, n ∈ N by
∞
X
Dp0 (m
−1
− n, n)x t = 1 − γ(t) 1 − x 1 + t
,
m n
m, n=0
where the power series γ(t) satisfies equation (44). Then we have
Dp0 (m, n) = Dp (m, n) for 0 ≤ n ≤ pm.
Up to now we considered a path starting at height 0, that is, h = 0. Now we want
to consider the more general case of a boundary line y = px + h instead of y = px
where h, p ∈ P. For a fixed h, p ∈ P, let Ep,h (m, n) be the number of lattice paths
from (0, 0) to (m, n) (where n ≤ pm + h), with unit up (↑) steps (0, 1) and unit right
(→) steps (1, 0), that never cross the line y = px + h.
48
CHAPTER 2. REDUNDANT GENERATING FUNCTIONS
Applying the same transformation as before, we have
Ep,h (m − n, n) = Ph0 m, pm + h − (p + 1)n with T = {−1, p}.
Therefore, by Theorem 6 we can conclude the following corollary after replacing tp+1
with t and (p + 1)n with n.
0
(m, n) for m, n ∈ N is defined by
Corollary 2.5.2. For h, p ∈ N, the number Ep,h
∞
X
−1
0
Ep,h
(m − n, n)xm tn = 1 − γ(t)h+1 1 − x(1 + t) ,
m, n=0
where the power series γ(t) satisfies equation (44). Then we have
0
Ep,h
(m, n) = Ep,h (m, n) for 0 ≤ n ≤ pm + h.
Note that in the case h = 0, Corollary 2.5.2 reduces to Corollary 2.5.1 because
Ep,0 (m, n) = Dp (m, n).
49
CHAPTER 3
On the enumeration of three-rowed standard Young tableaux
of skew shape in terms of Motzkin numbers
3.1. Introduction
A partition λ of a nonnegative integer n is a weakly decreasing sequence
λ = (λ1 , λ2 , . . . , λl )
such that
Pl
i=1
λi = n and λl > 0. The integers λi are called parts. The number
of parts of λ is the length of λ, denoted l(λ). We denote Par(n) to be the set of all
partitions of n, with Par(0) consisting of the empty partition ∅, and we let
Par := ∪n≥0 Par(n).
If λ ∈ Par(n), then we also write λ ` n or |λ| = n. Any partition λ can be identified
with its Young diagram, which is a collection of boxes arranged in left-justified rows
with the ith row containing λi boxes for 1 ≤ i ≤ l(λ). If λ and µ are partitions such
that µi ≤ λi for all i, we write µ ⊂ λ. Let us assume |λ| = n and µ ⊂ λ throughout
this chapter.
A skew shape λ/µ is a pair of partitions (λ, µ) such that the Young diagram of λ
contains the Young diagram of µ. If µ = ∅, then we assume that λ/µ = λ. A skew
Young diagram of shape λ/µ is a Young diagram of λ with a Young diagram of µ
removed from it. A standard Young tableau (SYT) of skew shape λ/µ is obtained
50
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
by taking a skew Young diagram of shape λ/µ and writing numbers 1, 2, . . . , n in
the n boxes of this diagram such that the numbers increase from left to right in
each row and from top to bottom down in each column. A Young tableau is called
semistandard (SSYT) if the entries weakly increase from left to right in each row and
strictly increase from top to bottom down in each column. The size of a SSYT is the
number of its entries. As shown in Figure below, (a) is a Young diagram of shape
(4, 3, 1), (b) is a SYT of shape (4, 3, 1), and (c) is a SYT of skew shape (4, 3, 1)/(2, 1),
respectively.
Now let us review several definitions for symmetric functions [19]. Let x =
(x1 , x2 , . . .) be a set of indeterminates. For n ∈ N, a homogeneous symmetric function
of degree n over the complex number field C is a formal power series
f (x) =
X
cα x α ,
α
where α ranges over all weak compositions α = (α1 , α2 , . . .) of n, cα ∈ C, xα stands for
the monomial xα1 1 xα2 2 · · · , and f (xω(1) , xω(2) , . . .) = f (x1 , x2 , . . .) for every permutation
ω of the positive integers P. Note that a symmetric function of degree zero is just a
complex number.
The monomial symmetric functions mλ for λ ∈ Par are defined by
mλ :=
X
xα ,
α
where the sum ranges over all distinct permutations α = (α1 , α2 , . . .) of the entries of
the vector λ = (α1 , α2 , . . .).
51
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
The complete homogeneous symmetric functions hλ for λ ∈ Par are defined by the
formulas
hn :=
X
mλ =
X
xi1 · · · xin , for n ≥ 1
(with h0 = 1)
i1 ≤···≤in
λ`n
hλ = hλ1 hλ2 · · ·
if λ = (λ1 , λ2 , . . .).
Let λ/µ be a skew shape. The skew Schur function sλ/µ = sλ/µ (x) of λ/µ in the
variables x = (x1 , x2 , . . .) is the sum of monomials
sλ/µ (x) :=
X
xT =
T
X
xt11 xt22 · · · ,
T
summed over all SSYTs T of skew shape λ/µ where each ti counts the occurrences
of the number i in the SSYT T . If µ = ∅, (that is, λ/µ = λ), then we call sλ (x) the
Schur function of shape λ. For example, the SSYTs of shape (2, 1) with largest part
at most three are given by
.
So
S(2,1) (x1 , x2 , x3 ) = x21 x2 + x1 x22 + x21 x3 + x1 x23 + x22 x3 + x2 x23 + 2x1 x2 x3
= m(2,1) (x1 , x2 , x3 ) + 2m(1,1,1) (x1 , x2 , x3 ).
3.2. About standard Young tableaux having at most three rows
In their paper [10], Gordon and Houten showed that the sum of Schur functions
of skew shape λ/µ, for fixed µ over all partitions λ with at most a fixed number of
rows, can be expressed as a Pfaffian of a matrix of complete homogeneous symmetric
52
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
functions. Before we state their results, let us define the functions h, gi , and fj used
in them.
For i ≥ 0 and j ≥ 1, let us define h, gi , and fj by
h :=
∞
X
hn =
n=0
gi :=
∞
X
∞ X
X
mλ ,
n=0 λ`n
hn hn+i =
∞ X
X
n=0 λ1 `n
n=0
X
mλ1
mλ2 ,
λ2 `n+i
and
fj := g0 + 2(g1 + · · · + gj−1 ) + gj .
Gordon and Houten proved that
X
sλ/µ = Pf(D2m ),
l(λ)≤2m
where Pf denotes the Pfaffian and D2m = (fλi −µj +j−i )1≤i,j≤2m , and

X
sλ/µ = Pf 
l(λ)≤2m+1
0
−H
H
t

,
(46)
D2m
where the matrix is obtained by bordering D2m with a row H of h’s, a column −H t
(transpose of −H) of −h’s, and a zero.
For µ = ∅, Gordon [9] showed that
X
sλ = det(gi−j + gi+j−1 )1≤i,j≤m ,
l(λ)≤2m
53
(47)
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
and
X
sλ = h det(gi−j − gi+j )1≤i,j≤m .
(48)
l(λ)≤2m+1
For the case m = 1, identity (47) reduces to
X
sλ = g0 + g1 .
(49)
l(λ)≤2
We can directly prove identity (49) by considering the two cases of partitions of even
or odd numbers and then by applying Pieri’s rule to each case. For m = 1, identity
(48) reduces to
X
sλ = h(g0 − g2 ),
l(λ)≤3
which can also be proved by applying Pieri’s rule to expand hg0 and hg2 .
To count SYTs of skew shape λ/µ, we need to find the coefficient of x1 x2 · · · xn−|µ|
in sλ/µ . Now let us transform symmetric functions into formal power series by applying the map θ from the algebra of symmetric functions to formal power series in x,
which is defined by

 xr /r!, if λ = (1r ) for some r
θ(mλ ) =
 0,
otherwise
and extended by linearity. Then we can easily show that the map θ is a homomorphism
with the property
θ(hn ) =
54
xn
.
n!
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
Let us apply the linear map θ to gi . For i ≥ 0, we have
θ(gi ) = θ
∞
X
hn hn+i
n=0
=
∞
X
θ(hn )θ(hn+i )
n=0
∞
X
xn xn+i
n! (n + i)!
n=0
2n+i
∞ X
2n + i
x
=
.
n
(2n + i)!
n=0
=
Now we want to convert the generating function θ(gi ) to an ordinary generating
function. To do this, we need the following lemma.
Lemma 3.2.1. Define L to be the linear map from the algebra of formal power series
C[[x]] to itself defined by
xn
n!
7→ xn , extended by linearity. Then we have
x 1
L(e f (x)) =
F
, where F (x) = L f (x) .
1−x
1−x
x
Proof. Let f (x) =
P∞
x
P∞
xn
a
.
Then
F
(x)
=
L
f
(x)
= n=0 an xn and
n
n=0
n!
∞
X
xn an L ex
by linearity
n!
n=0
n+k ∞ ∞
X
X
n+k
x
=
an L
k
(n + k)!
n=0
k=0
∞ ∞
X
X
n+k k
n
an x
x
=
k
n=0
k=0
L(e f (x)) =
∞
X
xn
(1 − x)n+1
n=0
x 1
=
F
.
1−x
1−x
=
an
55
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
Let m(x) the Motzkin number generating function
m(x) =
∞
X
Mn xn =
1−x−
n=0
where Mn =
Pbn/2c
well-known fact
k=0
√
1 − 2x − 3x2
,
2x2
n!/ k!(k + 1)!(n − 2k)! is the nth Motzkin number. From the
∞ X
2n + s n
c(x)s
,
x =√
n
1 − 4x
n=0
(50)
where c(x) is the Catalan number generating function, we can express the Motzkin
number generating function in terms of the Catalan number generating function:
1
c
1−x
x2
(1 − x)2
= m(x).
Let Ψ := L ◦ θ. Then Ψ is a linear map from the algebra of symmetric functions
to the algebra of formal power series C[[x]].
Now let us compute Ψ(gi ) for i ≥ 0.
Ψ(gi ) = L(θ(gi ))
2n+i !
∞ X
2n + i
x
=L
n
(2n + i)!
n=0
∞ X
2n + i 2n+i
x
=
n
n=0
xi c(x2 )i
=√
,
1 − 4x2
where the last equation follows from identity (50).
56
(51)
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
For j, k ≥ 1, let us consider a partition µ = (µ1 , µ2 ) having µ1 = j + k − 2 and
µ2 = j − 1 in equation (46). Then equation (46) reduces to
X
sλ/µ = h(fj + fk − fj+k ).
(52)
l(λ)≤3
Let G3 (µ1 , µ2 ) be the sum (52). That is,
G3 (j + k − 2, j − 1) = h(fj + fk − fj+k ).
(53)
Now let us first consider the case µ = (k − 1, 0), that is, µ = (k − 1). Letting j = 1
in equation (53), we can find the number of SYTs of skew shape λ/(k − 1) over all
partitions λ of n having at most three parts, filled with the numbers 1, 2, . . . , n−k +1.
Since
fn = g0 + 2(g1 + · · · + gn−1 ) + gn for n > 0,
we have
G3 (k − 1, 0) = h(g0 + g1 − gk − gk+1 ).
By equation (51) and Lemma 3.2.1, we have
Ψ(hgk ) = L θ(hgk )
= L θ(h)θ(gk )
= L ex θ(gk )
=
x k x2 k
1
√
c
2
1−x
(1 − x)
1 − 2x − 3x2
xk m(x)k
=√
.
1 − 2x − 3x2
57
(54)
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
To simplify the term that we computed above, let us define αk by
αk := Ψ(hgk ), for all k ≥ 0.
√
Then we have αk = xk m(x)k / 1 − 2x − 3x2 .
To find the number of SYTs of skew shape λ/(k − 1) over all partitions λ of n
having at most three parts, filled with the numbers 1, 2, . . . , n − k + 1, let us apply Ψ
to G3 (k − 1, 0). Then by linearity we deduce
Ψ G3 (k − 1, 0) = α0 + α1 − αk − αk+1
1 + xm(x) − xk m(x)k − xk+1 m(x)k+1
√
1 − 2x − 3x2
1 + xm(x) 1 − xk m(x)k
√
=
1 − 2x − 3x2
m(x)
k
k
=
1
+
xm(x)
1
−
x
m(x)
1 − x2 m(x)2
=
=
m(x)
(1 − xk m(x)k ).
1 − xm(x)
(55)
For example, consider the case k = 1 in (55), that is, µ = ∅. Then we know that
the number of SYTs of shape λ over all partitions λ of n having at most three parts,
filled with the numbers 1, 2, . . . , n, is equal to the nth Motzkin number since
Ψ G3 (0, 0) = m(x).
Consider the case k = 2 in (55), that is, µ = (1). Since the Motzkin number
generating function m(x) satisfies the functional identity
m(x) = 1 + xm(x) + x2 m(x),
58
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
we deduce
Ψ G3 (1, 0) = m(x) 1 + xm(x)
=
m(x) − 1
.
x
This tells us that the number of SYTs of skew shape λ/(1) over all partitions λ of n
having at most three parts, filled with the numbers 1, 2, . . . , n − 1 is equal to the nth
Motzkin number for n ≥ 1.
Now let us find the generating function for Ψ G3 (k − 1, 0) for k ≥ 1. From (55),
we have
∞
X
k
m(x)
(1 − xk m(x)k )y k
1
−
xm(x)
k=1
m(x) 1
1
=
−
1 − xm(x) 1 − y 1 − xym(x)
(α0 + α1 − αk − αk+1 )y =
k=1
∞
X
y2
=−
(1 − y)(x + xy − y + xy 2 )
xy
m(x),
+
(1 − y)(x + xy − y + xy 2 )
(56)
where the last equation follows by rationalizing the denominator. By factoring the
denominator of the first expression in the right side of equation (56), we have
−
y2
y2
=
−
,
(1 − y)(x + xy − y + xy 2 )
x(1 − y)(1 + y − y/x + y 2 )
(57)
so when expanded in powers of y, the function (57) has only negative powers of
x. However, the left side of equation (56) has no negative powers of x. So we can
conclude that the function (57) must be cancelled with negative powers of x from
the second expression in the right side of equation (56). Therefore, we know that the
59
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
generating function for Ψ G3 (k − 1) is the part of
xy
m(x),
(1 − y)(x + xy − y + xy 2 )
consisting of nonnegative powers of x.
Let
R(x, y) :=
xy
.
(1 − y)(x + xy − y + xy 2 )
(58)
By factoring
xy
y
=
,
2
(x + xy − y + xy )
1 + (1 − x−1 )y + y 2
we can see that the coefficient of y n in (58) is a polynomial in 1/x. So there are
P
polynomials rk (x) = i rk,i xi such that the function (58) can be written as
∞
X
y
rk
=
(1 − y) (1 + (1 − x−1 )y + y 2 ) k=1
1
yk .
x
(59)
Now we need to compute the coefficient of xn−(k−1) y k in the expression R(x, y)m(x)
to find the number of SYTs of skew shape λ/(k − 1) over all partitions λ of n having
at most three parts, filled with the numbers 1, 2, . . . , n − (k − 1). By (59), we have
[x
1
y ]R(x, y)m(x) = [x
]rk
m(x)
x
∞
X
1
n−k+1
i
= [x
]
Mi x rk
x
i=0
n−k+1 k
n−k+1
= [x
n−k+1
]
∞
X
Mi xi rk,j x−j
i,j=0
=
X
=
X
Mi rk,i−n+k−1
i
i
60
Mi+n−k+1 rk,i .
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
k\i
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6 7
0
1
0
1
0 −1
1
1
0 −2
1
0
2
1 −3
1
0 −2
3
3 −4
1
1
0 −6
3
6 −5
1
0
3
3 −12
1 10 −6 1
Table 1. The values of rk,i
So, we summarize with the following theorem.
Theorem 7. For k ≥ 1, the number of standard Young tableaux of skew shape
λ/(k − 1) over all partitions λ of n having at most three parts, filled with the numbers
1, 2, . . . , n − k + 1 can be written as the linear combination of Motzkin numbers
X
rk,i Mi+n−k+1 ,
i
where the number Mn is the nth Motzkin number and the coefficients rk,i are defined
by
∞
X
k=1
k
rk (x)y =
∞ X
k−1
X
rk,i xi y k =
k=1 i=0
y
.
(1 − y) (1 + (1 − x)y + y 2 )
Table 1 shows the coefficients rk,i for k from 1 to 8.
3.2.1. Generalization of Theorem 7. Now we want to consider a generalization of Theorem 7; for a partition µ having at most three rows, we want to express
the number of SYTs of skew shape λ/µ over all partitions λ of n having at most three
parts, filled with the numbers 1, 2, . . . , n − |µ|, as a linear combination of Motzkin
numbers. It is enough to consider partitions µ having at most two rows.
61
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
First let us compute Ψ(hfk ) for k > 0. By equation (54) we have
Ψ(hfk ) = L θ(h g0 + 2(g1 + · · · + gk−1 ) + gk
1
1 + 2 xm(x) + · · · + xk−1 m(x)k−1 + xk m(x)k
=√
1 − 2x − 3x2
1
1 − xk+1 m(x)k+1 + xm(x) − xk m(x)k
=√
1 − xm(x)
1 − 2x − 3x2
1 + xm(x) 1 − xk m(x)k
m(x)
=
1 − x2 m(x)2
1 − xm(x)
m(x)
k
k
=
2 1 − x m(x) .
1 − xm(x)
(60)
Let βk := Ψ(hfk ) for k > 0. Then by (60), we can compute the generating function
B(y) for βk as the following:
B(y) =
∞
X
k
βk y =
k=1
=
=
∞
X
m(x)
k=1
1 − xm(x)
k
k
k
2 1 − x m(x) y
1
1
−
2
1 − y 1 − xm(x)y
1 − xm(x)
m(x)
−y(x2 − x2 y)m(x)
y(x − y + 2xy)
+
,
2
(1 − y)(1 − 3x)(x + xy − y + xy ) (1 − y)(1 − 3x)(x + xy − y + xy 2 )
(61)
where the last equation follows by rationalizing the denominator. Then the generating function for Ψ(hfj ) + Ψ(hfk ) − Ψ(hfj+k ) is
∞
X
Ψ(hfj )+Ψ(hfk ) − Ψ(hfj+k ) y j z k
j,k=1
=
∞
X
βj + βk − βj+k y j z k
j,k=1
62
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
∞
X
z
y
=
B(y) +
B(z) −
βj+k y j z k
1−z
1−y
j,k=1
∞
X
y
y n z − yz n
z
B(y) +
B(z) −
βn
=
1−z
1−y
y−z
n=1
y
1 z
B(y) +
B(z) −
zB(y) − yB(z) .
=
1−z
1−y
y−z
(62)
So, plugging equation (61) into (62) gives the generating function for Ψ(hfj ) +
Ψ(hfk ) − Ψ(hfj+k ), which is equal to
yz(1 − yz)m(x)
(1 − y) 1 + (1 − 1/x)y + y 2 (1 − z) 1 + (1 − 1/x)z + z 2
yz(xyz + xy + xz − yz)
.
−
(1 − y) (x + xy − y + xy 2 ) (1 − z) (x + xz − z + xz 2 )
(63)
Let
T (x, y, z) :=
yz(1 − yz)
,
(1 − y) 1 + (1 − 1/x)y + y 2 (1 − z) 1 + (1 − 1/x)z + z 2
(64)
and let
S(x, y) :=
y
.
(1 − y) 1 + (1 − x)y + y 2
Then by equation (59) we have
S(x, y) =
∞
X
rk (x)y k .
k=1
We know that when expanded in powers of y, the second term in (63) has only
negative powers of x. However the left side of equation in (62) has no negative powers
of x. So the first term in (63) must be cancelled with negative powers of x from the
second term in (63). Also, the expression (64) has only negative powers of x. Then
63
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
we simplify the function
T (1/x, y, z) =
yz(1 − yz)
(1 − y) 1 + (1 − x)y + y 2 (1 − z) 1 + (1 − x)z + z 2
= yz(1 − yz)S(x, y)S(x, z)
= yz(1 − yz)
∞
X
rj (x)y
j
j=1
=
∞
X
∞
X
rk (x)z k
k=1
(rj (x)rk (x) − rj−1 (x)rk−1 (x)) y j z k ,
(65)
j,k=1
where r0 (x) = 0.
Define the polynomial rj,k (x) by
T (1/x, y, z) =
X
rj,k (x)y j z k .
(66)
j,k
Then by (65) and (66) we have
rj,k (x) = rj (x)rk (x) − rj−1 (x)rk−1 (x).
Next, let us compute the coefficient of xn−(|µ1 |+|µ2 |) y j z k in the expression T (x, y, z)m(x).
By (66), we have
[x
1
m(x)
y z ] T (x, y, z)m(x) = [x
] rj,k
x
∞
X
1
n−(|µ1 |+|µ2 |)
i
= [x
]
Mi x rj,k
x
i=0
n−(|µ1 |+|µ2 |) j k
n−(|µ1 |+|µ2 |)
= [xn−(|µ1 |+|µ2 |) ]
∞
X
Mi xi ri,j,k x−l
i,l=0
=
X
i
64
Mi ri−n+(|µ1 |+|µ2 |),j,k
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
=
X
Mi+n−(|µ1 |+|µ2 |) ri,j,k .
i
So, we summarize with the following theorem.
Theorem 8. For positive integers µ1 and µ2 with µ1 ≥ µ2 , the number of standard
Young tableaux of skew shape λ/(µ1 , µ2 ) over all partitions λ of n having at most
three parts, filled with the numbers 1, 2, . . . , n − (µ1 + µ2 ), can be written as the linear
combination of Motzkin numbers
X
Mi+n−(|µ1 |+|µ2 |) ri,j,k
i
where the number Mn is the nth Motzkin number, j = µ2 + 1, k = µ1 − µ2 + 1, and
ri,j,k is the coefficient of xi in rj,k (x) = rj (x)rk (x) − rj−1 (x)rk−1 (x) defined in (66).
3.2.2. Focus on the polynomial rn (x). Now we study the polynomial rn (x).
Define polynomials qn (x) by
∞
X
qn (x)z n =
n=0
1
.
1 − xz + z 2
(67)
Then qn (x) = Un (x/2) where Un (x) (defined in Chapter 1) is the Chebyshev polynomial of the second kind.
Let us find a formula for rn (x) in terms of the Chebyshev polynomial of the second
kind. Substituting 1/x for x in (59) gives
∞
X
rn (x)y n =
n=0
y
.
(1 − y) 1 + (1 − x)y + y 2
(68)
By (68), we have
∞
X
n=0
rn (x + 1)y n =
y
.
(1 − y)(1 − xy + y 2 )
65
(69)
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
So, from (67) and (69), we deduce
n−1
X
rn (x + 1) =
Ui (x/2).
i=0
Then we can rewrite (69) as the sum of an even function and an odd function of x:
y
y(1 + y 2 )
=
(1 − y)(1 − xy + y 2 )
(1 − y)(1 − xy + y 2 )(1 + xy + y 2 )
+
xy 2
.
(1 − y)(1 − xy + y 2 )(1 + xy + y 2 )
Let
Pe (x, y) :=
y(1 + y 2 )
(1 − y)(1 − xy + y 2 )(1 + xy + y 2 )
Po (x, y) :=
xy 2
.
(1 − y)(1 − xy + y 2 )(1 + xy + y 2 )
and
Then the following lemma shows that we can express Pe (x, y) and Po (x, y) as the
Chebyshev polynomial of the second kind.
Lemma 3.2.2. The even function Pe (x, y) and the odd function Po (x, y) in y/(1 −
y)(1 − xy + y 2 ) can be expressed as the Chebyshev polynomial of the second kind:
Pe (x, y) = y(1 + y)
∞
X
Un
x 2
2
n=0
y 2n ,
(70)
and
Po (x, y) = (1 + y)
∞
X
Un
n=1
66
x
2
Un−1
x
2
y 2n .
(71)
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
Proof. Let i =
√
−1. Substituting iy 2 for x and −xi for a and b in equation (7)
in Section 1.2 gives us that
Pe (x, y)
1
1
=
∗
2
4
y(1 + y)
1 − xy + y
1 − xy 2 + y 4
∞
∞
x
x
X
X
2n
=
Un
y ∗
Un
y 2n
2
2
n=0
n=0
=
∞
X
Un
x 2
2
n=0
y 2n ,
where ∗ denotes the Hadamard product in y as in Chapter 1. So we have (70).
Similarly, the function Po (x, y)/(1 + y) can be written as
1
y2
Po (x, y)
=
∗
(1 + y)
1 − xy 2 + y 4 1 − xy 2 + y 4
∞
∞
x
x
X
X
2n
Un
Un
y ∗
y 2n+2
=
2
2
n=0
n=0
=
∞
X
Un
n=1
x
2
Un−1
x
2
y 2n ,
where ∗ denotes the Hadamard product in y.
By Lemma 3.2.2 and qn (x) = Un (x/2), the even function Pe of x satisfies that
Pe (x, y) = y(1 + y)
∞
X
qn (x)2 y 2n
n=0
= q0 (x)2 y + q0 (x)2 y 2 + q1 (x)2 y 3 + q1 (x)2 y 4 + · · ·
=
∞
X
qb(n−1)/2c (x)2 y n ,
n=1
67
CHAPTER 3. THREE-ROWED STANDARD YOUNG TABLEAUX OF SKEW SHAPE
where qn (x) is the polynomial in (67) for all n ≥ 0. Also the odd function can be
expressed as
Po (x, y) =
∞
X
qbn/2c (x)qbn/2c−1 (x)y n .
n=2
So the expression (69) can be written as
∞
X
n
y
2
2
=
q
(x)
y
+
q
(x)
+
q
(x)q
(x)
y . (72)
0
b(n−1)/2c
bn/2c
bn/2c−1
(1 − y)(1 − xy + y 2 )
n=1
Therefore, equating coefficients in (72) gives r1 (x + 1) = q0 (x)2 and
rn (x + 1) = qb(n−1)/2c (x)2 + qbn/2c (x)qbn/2c−1 (x), when n ≥ 2.
That is, for all m > 0,
r2m (x + 1) = qm−1 (x) qm−1 (x) + qm (x) ,
and
r2m+1 (x + 1) = qm (x) qm−1 (x) + qm (x) .
68
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