Download Volume 1 (December 2009)

Macalester Journal of
Catalan Numbers
Volume 1
December 2009
Published by:
Department of Mathematics, Statistics and Computer Science
Macalester College
1600 Grand Avenue
Saint Paul, MN 55105
2009 All rights reserved
Cover art: Yenee Soh
Editor: Andrew Beveridge
Table of Contents
For each Catalan problem below, we enumerate the 5 elements for the n = 3 case. Moreover, each
list appears in the same order. In other words, the natural bijection between any pair maps the
first element of one list to the first element of the other list, and so on.
Introduction to the Catalan Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Andrew Beveridge
1 − 1 − 1−
1 − 11 − −
11 − −1−
11 − 1 − −
111 − −−
Non-Crossing Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Mo Liu
Non-Decreasing Sequences of Positive Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Matt Kusner
123 122 113 112 111
Non-Crossing Muraski Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Colin Keeley
(13)(24)-Free Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Michael Kapernaros
(12)(34)(56) (12)(36)(45) (14)(23)(56) (16)(23)(45) (16)(25)(34)
Non-Intersecting Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
James Leonard
Standard Young Tableaux of Shape (n, n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Jack Shirek
135
246
134
256
125
346
124
356
123
456
Stacking Coins in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Tom Woodward
Sequences Below the Sequence 012 . . . (n − 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Congcong Nie
000 001 010 011 012
Upper Triangular Ferrers Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Katie Agnew
Pairs of Non-Crossing Lattice Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Drew Van Denover
Partially Connected Polygon Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
Jeanmarie Youngblood
Plane Binary Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
David Klock
Non-Intersecting Convex Hulls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
Jorge Banuelos
Dyck Paths with No Peaks at Height Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Yenee Soh
Pairs of Internally Disjoint Lattice Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Matthew Hurni
Dyck Paths with Odd Maximal Descents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
Emily Merrill
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
An Introduction to the Catalan Numbers
Andrew Beveridge∗
1
Preface
Welcome to the inaugural volume of the Macalester Journal of Catalan Numbers!
This journal collects the work of the students enrolled in Math 379: Combinatorics
during Fall 2009 at Macalester College. The 16 students were assigned problems
drawn from the exhaustive list of Catalan exercises in Richard Stanley’s Enumerative Combinatorics, Volume 2. Each student produced an article proving that the
counting problem was enumerated by the Catalan numbers. Future Combinatorics classes at Macalester will tackle other Catalan problems, to be collected in
later volumes of this journal.
The articles are arranged loosely in terms of difficulty. The early articles are
typically straight forward arguments, either showing that a sequence satisfies the
Catalan recurrence or presenting a simple bijection to a known Catalan problem.
Later articles are a bit more involved, requiring multiple steps, more creative
bijections, or the heavy machinery of generating functions. Like the class of Catalan problems, these articles are diverse, yet unified by common themes. We hope
that the variety of the articles reflects the beauty and flexibility of the Catalan
numbers.
We thank Tom Halverson for suggesting that we collect these articles into
a journal, and for contributing a lecture on the Catalan numbers during the
semester.
∗ Department
MN 55105.
of Mathematics, Statistics and Computer Science, Macalester College, Saint Paul
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Macalester Journal of Catalan Numbers
2
The Catalan Numbers
The sequence of Catalan numbers is among the most well known sequences in
combinatorics. The Catalan sequence {Cn }n≥0 begins
1, 1, 2, 5, 14, 42, 132, 429, 1340, 4862, . . .
The formula for the nth Catalan number is
1
2n
Cn =
.
n+1 n
The Catalan numbers are a combinatorial chameleon. This remarkable sequence of numbers arises in many independent settings. In particular, it is the
solution to many counting problems. The objective of this journal is to explore
the variety of Catalan problems in enumerative combinatorics.
The key to the flexibility of the Catalan numbers lies in their recurrence relation. Indeed, we have
Cn =
n −1
∑ Ck Cn−k−1 for n ≥ 1.
k =0
This recurrence relation is perfectly suited for problems that have a particular
type of natural recurrence. We will discuss this recurrence further in the next
section.
Generating functions are a powerful tool in combinatorics, and investigation
of Catalan numbers is no exception. The generating function for the Catalan
numbers is
√
∞
1
−
1 − 4x
C ( x ) = ∑ Cn x n =
.
2x
n =0
3
Balanced sequences of 1’s and −1’s
We consider one of the classic counting problems whose solution is given by the
Catalan numbers. This particular formulation is quite useful when proving other
Catalan results. Many of the articles in this journal create a bijection from their
counting problem to this one.
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Macalester Journal of Catalan Numbers
Suppose that we have n copies of the symbol “1” and n copies of the symbol
“−1.” How many ways are there to arrange them in a sequence s1 , s2 , . . . , s2n so
j
that the partial sums satisfy ∑i=1 si ≥ 0 for 1 ≤ j ≤ 2n? In other words, as we
move from left to right in such a “valid” sequence, we always encounter at least as
many 1’s as −1’s. Using “−00 to denote “ − 100 , we enumerate all valid sequences
for 0 ≤ n ≤ 4:
n
valid sequences
0
∅
1
1−
2
1 − 1−, 11 − −
3
1 − 1 − 1−, 1 − 11 − −, 11 − −1−, 11 − 1 − −, 111 − −−
4
1 − 1 − 1 − 1−, 1 − 1 − 11 − −, 1 − 11 − −1−, 1 − 11 − 1 − −,
1 − 111 − −−, 11 − −1 − 1−, 11 − −11 − −, 11 − 1 − −1−,
11 − 1 − 1 − −, 11 − 11 − −−, 111 − − − 1−, 111 − −1 − −
111 − 1 − −−, 1111 − − − −
Table 3.1: An enumeration of sequences of balanced 1’s and −1’s.
Let An denote the number of valid sequences of length 2n. For 0 ≤ n ≤ 4, the
values of An are
1, 1, 2, 5, 14,
matching the initial the Catalan numbers. Here we opt for the convention that
there is one way to create a sequence of length 0. Continuing our enumeration
for n = 5 and n = 6 would be quite a task: there are a total 42 and 132 valid
sequences, respectively!
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Macalester Journal of Catalan Numbers
To successfully enumerate all valid sequences for n = 5, we would surely
have to develop a systematic approach. After some trial and error, inspiration
would strike. Let us find the first time that we have seen as many −1’s as 1’s. In
mathematical terms, let 2m be the first index such that ∑2m
i =1 si = 0. (This index
must be even since we must have equal numbers of 1’s and −1’s.) This index
naturally partitions our sequence into two smaller valid sequences: the first with
length 2m and the second with length 2n − 2m. Note that when m = n the second
sequence has length 0. Let us reorganize our valid sequences for n = 4 according
to this index m, where we use ∅ to denote a sequence of length 0:
m=1
1−
1−
1−
1−
1−
1 − 1 − 1−
1 − 11 − −
11 − −1−
11 − 1 − −
111 − −−
m=2
11 − − 1 − 1−
11 − − 11 − −
m=3
m=4
11 − 1 − − 1−
111 − −− 1−
∅
∅
∅
∅
∅
11 − 1 − 1 − −
11 − 11 − −−
111 − −1 − −
111 − 1 − −−
1111 − − − −
Table 3.2: Valid sequences of balanced 1’s and −1’s, partitioned by the
length 2m of the first valid subsequence.
We make one final observation. The first 2m elements are a special type of
balanced sequence. Namely, the index 2m is the first time that we have seen an
equal number of 1’s and −1’s. (This property does not necessarily hold for the
latter sequence of length 2n − 2m.) Such an indecomposable sequence must be
of the form 1, s1 , s2 , · · · , s2m−2 , −1 where s1 , s2 , . . . , s2m−2 is any valid sequence
of length 2m − 2. Setting k := m − 1, we incorporate this observation into our
previous table by highlighting the embedded sequence of length 2k:
k=0
1
1
1
1
1
∅
∅
∅
∅
∅
−
−
−
−
−
1 − 1 − 1−
1−11− −
1 1 − −1−
11−1− −
1 1 1 − −−
k=1
1 1− −
1 1− −
1 − 1−
11− −
k=2
1 1−1− −
1 11− − −
k=3
1−
1−
1
1
1
1
1
1 − 1 − 1−
1 − 11 − −
1 1 − −1 −
11 − 1 − −
1 1 1 − −−
−
−
−
−
−
∅
∅
∅
∅
∅
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Macalester Journal of Catalan Numbers
Table 3.3: Valid sequences of balanced 1’s and −1’s, where the embedded sequence
of length 2k within the complete subsequence has been highlighted.
In summary, we can naturally partition any valid sequence into two valid
subsequences. The first sequence, of length 2k + 2, is indecomposable. The second
valid sequence, which has length 2n − 2k − 2, has no additional restrictions. An
indecomposable sequence of length 2k + 2 corresponds to a valid sequence of
length 2k.
Every valid sequence of length 2n is built from a valid sequence of length
2k and a valid sequence of length 2n − 2 − 2k. In other words, we can create
a bijection from pairs pairs of valid sequences of lengths 2k and 2n − 2k − 2 to
valid sequences of length 2n. Indeed, given two valid sequences s1 , s2 , . . . , s2k and
t1 , t2 , . . . t2n−2−2k , the sequence
1, s1 , s2 , . . . , s2k , −1, t1 , t2 , . . . , t2n−2−2k
is a valid sequence of length 2n.
Since our choices for the first and second sequences are independent, our
counting problem satisfies the Catalan recurrence:
An =
n −1
∑ Ak An−k−1 for n ≥ 1.
k =0
Our initial values also match the Catalan numbers, so we can conclude that these
sequence are identical, { An }n≥0 = {Cn }n≥0 . In other words, the number of valid
sequences of length 2n is equal to the nth Catalan number, Cn .
The form of our systematic partition is the hallmark of Catalan problems.
Every instance of size n can be decomposed into two independent substructures,
of size k and n − 1 − k respectively, where 0 ≤ k ≤ n − 1. Why do these numbers
only add up to n − 1? The answer is that the final “missing” index actually plays
an essential role. It acts as the border between the first and second substructures.
This self-partitioning property unifies all Catalan problems.
4
Dyck Paths
We describe another classic Catalan problem that is referenced by articles in the
journal. Dyck paths are lattice paths from (0, 0) to (2n, 2n) that use steps (1, 1)
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Macalester Journal of Catalan Numbers
and (1, −1) which never falling below the x-axis. We enumerate all Dyck paths
for 0 ≤ n ≤ 4.
n
Dyck paths of length 2n
0
∅
1
2
3
4
Table 4.1: An enumeration of Dyck paths.
There is a simple bijection from Dyck paths to balanced sequences of 1’s and
−1’s. Namely, map the step (1, 1) to the symbol 1 and map the step (1, −1) to
the symbol −1. Finding a bijection to a known Catalan problem is a particularly
satisfying form of proof.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Non-Crossing Arcs
Mo Liu∗
1
An Introduction
Consider n + 1 points on a horizontal line in the plane. Connect the points with
n arcs such that
1. the arcs do not pass below any point
2. the graph thus formed is a tree
3. no two arcs intersect in their interiors (i.e., the arcs are noncrossing)
4. at every vertex, all the arcs exit in the same direction (left or right)
We first enumerate some small examples. Look at the graphs below:
n=0
n=1
∗
Mo is from Beijing, China. He has been in the United States for one year. He likes jumping
up and down on his bed, crawling up and down stairs and standing upside down doing counting
problems.
Macalester Journal of Catalan Numbers
8
n=2
n=3
n=4
We show the number of such structures is governed by the Catalan Numbers.
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Macalester Journal of Catalan Numbers
2
The Catalan Numbers and Recurrence Relations
Define An is the number of ways of drawing n + 1 such points connected by arcs.
The size of a graph is the number of points. We have A0 = 1, A1 = 1, A2 = 2,
A3 = 5, A4 = 14. We know that the Catalan numbers start with 1, 1, 2, 5, 14, 42....
We show that An is the nth Catalan number.
1
2n
th
The formula for the n Catalan number is Cn =
. There is also a
n+1 n
recurrence ralation for the Catalan numbers:
Cn = C0 Cn−1 + C1 Cn−2 + ... + Cn−2 C1 + Cn−1 C0
We prove that the sequence of An also satisfies this recurrence:
An = A0 An−1 + A1 An−2 + ... + An−2 A1 + An−1 A0
3
Proof
Take n = 4 as an example. A0 = 1 is simply true from the first graph drawn.
When n ≥ 1, we can see that the leftmost point is always connected to the rightmost point. Indeed, the arcs connected to the leftmost point can only exit to the
right and the arcs connected to the rightmost point can only exit to the left. If
these two points are not connected, there must be another point connecting both
. In this case, the middle point is conof them, which would be like
nected to two different directions of arcs, which is not allowed, so the leftmost
and the rightmost points are always connected. As the outmost arc exists for every graph, we can just remove it and focus on inner arcs.
We partition the graphs in the following four catagories:
1. P0 and P1 are disconnected:
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
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Macalester Journal of Catalan Numbers
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
The graph is split into two components: the first component with a size of 1
and the second component with a size of 4, so the number of ways to create
graphs of this type is A1−1 A4−1 = A0 A3 .
2. P1 and P2 are disconnected:
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
The graph is split into two components: the first component with a size of 2
and the second component with a size of 3, so the number of ways to create
graphs of this type is A2−1 A3−1 = A1 A2 .
3. P2 and P3 are disconnected:
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
The graph is split into two components: the first component with a size of 3
and the second component with a size of 2, so the number of ways to create
graphs of this type is A3−1 A2−1 = A2 A1 .
4. P3 and P4 are disconnected:
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
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Macalester Journal of Catalan Numbers
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
P0
P1
P2
P3
P4
The graph is split into two components: the first component with a size of 4
and the second component with a size of 1, so the number of ways to create
graphs of this type is A4−1 A1−1 = A3 A0 .
So the number of ways of drawing 5 points connected by arcs is A0 A3 +
A1 A2 + A2 A1 + A3 A0 = 14.
Now consider the general case. Suppose we have a valid graph of size n + 1.
Removing the arc connecting P0 , Pn+1 disconnects the graph into 2 intervals:
(P0 , P1 , P2 , ..., Pk ), (Pk+1 , Pk+2 , ..., Pn+1 ). The number of different graphs is Ak An−k−1 .
We sum over for all possible k, and we get the recurrence relation
An =
n −1
∑ ( Ak An−1−k ) = A0 An−1 + A1 An−2 + ... + An−2 A1 + An−1 A0
k =0
Since the An satisfies the Catalan recurrence and A0 = C0 = 1, the numbers are
equal for all n.
Macalester Journal of Catalan Numbers
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Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Non-Decreasing Sequences of Positive Integers
Matthew Kusner∗
1
Introduction
Consider sequences of integers, a1 a2 · · · an , such that,
1 ≤ a1 ≤ · · · ≤ a n
and
ai ≤ i
We define Sn as the set containing all of the sequences of length n that satisfy
these two constraints. We will prove that |Sn | = Cn , the nth Catalan number.
2
Enumeration for n ≥ 4
We list the sets Sn for 0 ≤ n ≤ 4.
n
0
1
2
3
4
∗ Matthew
Sn
∅
1
11 12
111 112 113 122 123
1111 1112 1113 1114
1124 1134 1224 1234
1123 1223 1133 1122
1222 1233
Cn
1
1
2
5
14
is a junior majoring in mathematics and computer science. Born in Cleveland, OH
he currently calls home the city of Iowa City, IA. If Matthew had the opportunity to be endowed
with the skills of any basketball player he would choose Carmelo Anthony for his smooth moves
and steady jump shot.
14
Macalester Journal of Catalan Numbers
3
Proof
We will construct a bijection from our problem to a known Catalan problem.
Consider the set Tn of all northeast lattice paths which do not go above the main
diagonal for a n x n grid. For example, here are the elements of T3 :
T3
It is vital to note that there must be three horizontal sections of each path. In other
words, a path must move along the bottom edge of three different squares within
the lattice path. These horizontal movements may occur at any height below the
main diagonal of the lattice. A path in Tn must begin with a horizontal move to
the right. We will define movements along the bottom of the lattice as having a
height of one. Thus, it is the case that h0 = 1. In general, the height of the ith
horizontal movement of a path in Tn is equal to:
hi = v + 1
where v is the total number of vertical movements that occurred before the horizontal movement.
We now define our function f : Tn → Sn and show that it is a bijection. For
t ∈ Tn , define f (t) to be:
f ( t ) = h0 h1 h2 · · · h n
For example, the corresponding heights for T3 are 111, 112, 113, 122, and 123.
First, we show that f (t) ∈ Sn . Consider a lattice path t ∈ Tn with a sequence
of n horizontal movement heights h0 h1 h2 · · · hn . It must be the case that:
1 ≤ h1 ≤ h2 ≤ · · · ≤ h n
(1)
because no downward movements are allowed. Thus, the horizontal movement
heights may only increase. As well, we have:
hi ≤ i
(2)
Macalester Journal of Catalan Numbers
15
Otherwise the point (i, hi ) in our path is above the main diagonal. Therefore,
f ( t ) ∈ Sn .
We claim that f is onto. Indeed, the horizontal movement heights of any
t ∈ Tn are always nondecreasing and positive. Therefore, given a sequence s ∈ Sn
there exists at least one t ∈ Tn whose horizontal movement heights correspond to
the numbers in s.
The function f must also be one-to-one. Given two lattice paths t1 , t2 ∈ T,
such that t1 6= t2 , there is a specific point at which the paths diverge. Say this
occurs at position j. It is then the case that h j of t1 is not equal to h j of t2 . As such,
f ( t1 ) 6 = f ( t2 ).
Thus, f is indeed a bijection. We know that the Catalan numbers count the
number of northeast lattice paths that do not go above the main diagonal. Therefore they also count the number of nondecreasing sequences of positive integers.
Macalester Journal of Catalan Numbers
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Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Non-Crossing Muraski Diagrams
Colin Keeley∗
1
Introduction
Noncrossing Muraski diagrams consist of n equally spaced vertical lines. The
lines are then connected to each other from the top of one to the top of another or
we can choose not to connect them at all. The horizontal lines are not allowed to
cross each other. This can be achieved by varying the heights of the vertical lines.
We show that the number of noncrossing Muraski diagrams of size n is equal to
the n Catalan number, Cn .
2
Examples for 0 ≤ n ≤ 4
2.1
n = 0, C0 = 1
∅
2.2
n = 1, C1 = 1
C1 = C0 C0
∗ Colin,
class of 2011, is an economics and mathematics major. He plays on the basketball team
and is from Kildeer, IL.
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Macalester Journal of Catalan Numbers
2.3
n = 2, C2 = 2
C2 = C0 C1 + C1 C0
2.4
n = 3, C3 = 5
C3 = C0 C2 + C1 C1 + C2 C0
2.5
n = 4, C4 = 14
C3 C0 : Do not connect 1st line to any other line.
Macalester Journal of Catalan Numbers
19
C0 C3 : Connect 1st line to 2nd line
C1 C2 : Connect 1st line to 3rd line.
C2 C1 : Connect 1st line to 4th line.
C4 = C3 C0 + C1 C2 + C2 C1 + C0 C3
3
General Solution
For n lines we connect the 1st line to the kth line. This divides the diagram
up into 2 separate problems because no line from 1 to k can be connected to any
line that is after k without crossing another line. The number of diagrams on the
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Macalester Journal of Catalan Numbers
left side up to k is Ck because it is equal to only having the lines from 1 to k. The
number of diagrams on the right side of k is Cn−1−k because it is equivalent to
only being able to use the lines from k+1 to n so the number of available lines is
n − 1 − k.
Consider k = 1. We begin by not connecting the 1st line to any other line and
then we find all the possible diagrams of the other n − 1 lines. By the product
principle, the total number of diagrams is C0 Cn−1 . When k = 2, we connect the
1st line to the 2nd line and find all the possible diagrams of the other n − 2 lines.
By the product principle, the total number of diagrams is C1 Cn−2 . We continue
this method from k = 0 to when k = n − 1 and add all the values together to get
Cn .
Therefore we have:
C0 = 1
Cn =
n −1
∑ Ck C(n−1)−k for n ≥ 0.
k =0
So the number of noncrossing Muraski diagrams satisfies the Catalan recurrence.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
(13)(24)-Free Involutions
Michael Kapernaros∗
1
Problem Statement
We seek to show that the number of fixed-point-free involutions of [2n] such that
if i < j < k < l and w(i ) = k, then w( j) 6= l is equivalent to the series known as
the Catalan Numbers, generated recursively as follows:
Cn = Σnk=0 (Ck ∗ Cn−k−1 )
C0 = 1
The Catalan Numbers are defined for n ≥ 0. Recall that an involution of [m]
is an m-permutation w such that w2 is the identity permutation.
2
Valid involutions through n=4
The following subsections enumerate the involutions that satisfy the conditions
of the problem statement for n = 1, 2, 3, 4.
n=1
n=2
n=3
n=4
∗ Michael,
Essex, CT.
(12)
(12)(34), (14)(23)
(12)(34)(56), (12)(36)(45), (14)(23)(56),
(16)(23)(45), (16)(25)(34)
(12)(34)(56)(78), (12)(34)(58)(67), (12)(36)(45)(78)
(12)(38)(45)(67), (12)(38)(47)(56), (14)(23)(56)(78)
(14)(23)(58)(67), (16)(78)(23)(45), (16)(78)(25)(43)
(18)(23)(45)(67), (18)(23)(47)(56), (18)(25)(34)(67)
(18)(27)(34)(56), (18)(27)(36)(45)
class of 2010, is a Mathematics major and an avid Scrabble player. He hails from
Macalester Journal of Catalan Numbers
3
22
Proof
Let Vn be the set of fixed-point free involutions of [2n] such that if i < j < k < l
and w(i ) = k, then w( j) 6= l. Let |V |n = Vn . We show that the following recursion
holds:
Vn = Σnk=0 (Vk ∗ Vn−k−1 )
Since all these are involutions and fixed-point-free, they cannot contain anything but 2-cycles.
We will partition Vn into two sets, as follows:
An = {w ∈ Vn |w(1) = 2m + 1, m ≥ 1}
Bn = {w ∈ Vn |w(1) = 2m, m ≥ 1}
An
Note that any remaining number less than 2m + 1 cannot map to a number greater
than 2m + 1, or we break the conditions of the problem. Thus, all numbers
{2, 3, ..., 2m} must map to another number within their own set. However, there
are 2m − 1 such numbers. Thus, there are no possible combinations of 2-cycles
within these. Thus, An = ∅.
Bn
Any remaining number less than 2m cannot map to a number greater than 2m, or
we break the conditions of the problem. Thus,w must map the set {2, 3, ..., 2m − 1}
must map to itself. However, this is simply a reduced form of our problem: there
are Vm−1 such legal mappings. (It is obvious that shifting the numbers to the
right will not affect their position relative to the others).
Similarly, w must map the set {2m + 1, 2m + 2, ..., 2n} to itself. Again, this is a
reduced form of the problem and there are Vn−m legal mappings.
Thus, since any legal mapping from the set {2, 3, ..., 2m − 1} can pair with any
legal mapping from the set {2m + 1, 2m + 2, ..., 2n},
Vn = Σnm=1 (Vm−1 ∗ Vn−m )
23
Macalester Journal of Catalan Numbers
Reindexing, k = m − 1,
Vn = Σnk=0 (Vk ∗ Vn−k−1 ),
V0 = 1.
Vn and Cn satisfy the same recurrence, and V0 =C0 . Thus, Vn = Cn for all n.
Macalester Journal of Catalan Numbers
24
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Non-Intersecting Arcs
James Leonard∗
1
Introduction
How many ways are there to connect 2n points in the plane lying on a horizontal
line by n nonintersecting arcs, each arc connecting two of the points and lying
above the points? We show that the answer is the n’th Catalan number.
2
Examples
Here are the ways to solve the problem up to n = 4.
n=1
n=2
n=3
∗ James,
class of 2010, is a Mathematics major.
Macalester Journal of Catalan Numbers
26
n=4
3
Proof
To prove that these are counted by the Catalan numbers, we make a bijection
from the problem of counting the number of ways to make n pairs of balanced
parentheses. A set of parentheses are balanced if one can take pairs of left and
right parentheses such that each left parenthesis is to the left of each right paren-
Macalester Journal of Catalan Numbers
27
thesis. To make our bijection, we start with a set of points and arcs. Then we
we turn each left endpoint into a left parenthesis and each right endpoint into a
right parenthesis. The parentheses are balanced because we can take the pairs to
be the pairs of endpoints that are connected by arcs.
Now we must show the other direction. To do this I will first show that there
is at least one way to do this, and then I will show that there is at most one way
to do this. Since our pairs of parentheses are balanced, we can find corresponding pairs such that each left parenthesis is to the left of each right parenthesis.
Now, when we turn the parentheses into dots and connect the corresponding
pairs, there may or may not be intersections. Since the number of arcs is finite,
the number of intersections is finite. Suppose we have an intersection that looks
like this.
Then we can reconnect the dots such that the two arcs don’t intersect.
Since the resulting arcs have been shortened, no new intersections are made, but
at least one intersection is removed. Therefore we can repeat this process until
there are no intersections.
Now to show that there is at most one way to connect the dots, we go from
left to right such that each time we reach a right parenthesis we connect it to the
rightmost of the unused left parentheses that is to the left of the right parenthesis.
If at any point we don’t follow this method, we are bound to fail. Suppose we
have a choice of left parentheses and we don’t choose the rightmost one. Then
we have something like this.
Then all the unpaired points inside must be left parentheses, because we have
dealt with all the right parentheses. In order to connect any of these left parentheses, we must cross over an existing arc which is not allowed. Therefore we must
follow this method, and since we have shown there is at least one way to do this,
this must be the only way. Therefore our mapping is a bijection.
Macalester Journal of Catalan Numbers
28
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Standard Young Tableaux of Shape (n, n).
Jack Shirek∗
1
Standard Young Tableaux
A Standard Young Tableaux, Tn of the form (n, n) consists of two rows of length
n, such that the numbers are increasing in each row and column. We enumerate
the cases up through n = 4:
n
0
1
2
3
Tn
∗
1
2
12 13
34 24
123 124 125 134 135
456 356 346 256 246
1234 1235 1236 1237 1245 1246 1247
5678 4678 4578 4568 3678 3578 3568
4
1256 1257 1345 1346 1347 1356 1357
3478 3468 2678 2578 2568 2478 2468
We see that at n = 0, there is one possible standard young tableaux diagram;
at n = 1 there is one; at n = 2 there are 2; at n = 3, 5 possibilities; and at n = 4,
we see 14 variations. The number of variations of standard young tableaux of size
(n, n) appear to be equal to the Catalan numbers at term Cn . We will prove this
in the subsequent section.
∗ Jack
is a Junior majoring in mathematics at Macalester college. Hailing from Minnetonka,
Minnesota he loves cheese, Scrabble, and all manner of challenging puzzles. Jack, a pharmacy
technician at HealthPartners, also enjoys playing with his dogs, video games, and the company
of his amazing friends.
Macalester Journal of Catalan Numbers
2
30
A Bijection to Balanced Parentheses
In order to prove that our counting problem is simply another guise of the Catalan numbers, we will create a bijection between our problem and the balanced
parentheses problem. We know the balanced parentheses problem is counted by
the Catalan numbers.
Instances of the balanced parentheses problem consist of sequences of n (’s
and n )’s such that each open parentheses has a matching closed parentheses,
which comes after it. Additionally, as you move from left to right, you never see
ore )’s than (’s. We’ll call the set of all 2n balanced parentheses, Pn . We list the
first few terms:
n
0
1
2
3
4
Pn
*
()
()(), (())
((())), (()()), (())(), ()(()), ()()()
(((()))), ((()())), ((())()), ((()))(), (())(()), (())()(), (()(()))
(()()()), (()())(), ()((())), ()(())(), ()(()()), ()()(()), ()()()()
Proof:
Let Tn be the set of possible Standard Young Tableaux of size (n,n) and let
t ∈ Tn . Define f : Tn → Pn as f (t) = p = p1 p2 p3 . . . p2n , where pi = ( for all i in
the top row of the Standard Young Tableaux diagram and pi =) for all i in the
12
bottom row. For example: for t =
we have f (t) = (()).
34
Let t ∈ Tn and f (t) be the sequence of n open and n closed parentheses. We
claim that f (t) ∈ Pn . The rows of t are increasing to the right and the columns
increasing downward, therefore t must always begin its first row with one and
end its second with 2n. This ensures that f (t) always has an open parentheses as
the first term of its sequence and a closed parentheses as its last. In other words,
p1 = ( and p2n = ).
Similarly, because the rows are always increasing, as are the columns, we see
that the ith open parentheses must always come before the ith closed parentheses.
t . . . t1n
Let t = 11
. The ith open parentheses occurs at t1i while the ith closed
t21 . . . t2n
parentheses occurs at t2i and t1i < t2i for 1 ≤ i ≤ n.
Macalester Journal of Catalan Numbers
31
We prove that f is a bijection by shoing that f is invertible. Define g : Pn → Tn
t . . . t an
as g( p) = t = a1
, where t a1 = 1 and tbn = 2n and a1 ≤ a2 ≤ . . . ≤ an are
tb1 . . . tbn
the indices of open parentheses and b1 ≤ b2 ≤ . . . ≤ bn are the indices of closed
parentheses. The mapping g simply reverses the mapping of f , so g is the inverse
function for f . For example, given the sequence p = (()()) we can reconstruct the
Standard Young Tableaux diagram, by simply reversing the steps taken to create
124
the sequence p. Thus, g( p) =
356
Since f is a bijection, we have | Tn | = | Pn | for all n.
Macalester Journal of Catalan Numbers
32
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Stacking Coins in the Plane
Tom Woodward∗
1
Problem Description
Let COI Nn be the number of ways to stack coins in the plane with the bottom
row consisting of n consecutive coins. In order for a stacking to be valid, if row i
consists of j coins then row i + 1 can only have k coins where 0 ≤ k ≤ j − 1. We
will show that that COI Nn is equal to the nth Catalan number.
2
Some Examples
Below is a table showing the all possible stackings of coins for 0 ≤ n ≤ 4. Note
that for n = 0 there is one way to stack the coins, namely, it is the empty stacking
in which the bottom row consists of 0 coins.
n = 0: *
n = 1:
n = 2:
n = 3:
n = 4:
1 way
1 way
,
2 ways
,
,
,
,
,
,
,
,
5 ways
,
,
,
,
,
,
,
, 14 ways
,
Table 1: A table of possible stackings for n ≤ 4.
∗ Tom,
class of 2010, is a mathematics major. He hails from Lincoln, NE.
34
Macalester Journal of Catalan Numbers
3
Proof that the Catalan numbers are the solution
Let’s find a recurrence for COI Nn . Our recurrence will be based on the location
of the first gap in the second row of coins. Suppose this gap appears between the
kth and (k + 1)st coins. If there is no gap, then take k = n. We treat the kth coin as
special and therefore we color it black. The black coin is crucial in that it divides
our problem into two pieces. This will become more clear as we proceed through
the proof. Below are the stackings for n = 4 arranged in increasing order of k.
,
,
,
,
,
,
,
,
,
,
,
,
,
Without loss of generality, assume we want to build our stacks one coin at a
time, starting at the bottom left corner. We begin by placing a single coin (k = 1).
This essentially splits the problem into two parts. To the left of (and above) this
coin, we place all possible stackings for k − 1 = 1 − 1 = 0 coins. To the right of
the first coin we place all valid stackings for n − k = n − 1 coins. So we have a
total of COI N0 · COI Nn−1 possibilites here.
we place 0 coins here
we place every stacking for n − 1 coins here
It is clear that this will give a valid stacking of n because the bottom row contains
n − 1 + 1 = n coins.
Now let’s consider starting with 2 coins. To the left of and above the second
th
(k ) coin, we can place any valid stacking for k − 1 = 2 − 1 = 1 coins. To the
right of the second coin we place all of the possible stackings for n − k = n − 2
coins. In this case we have COI N1 · COI Nn−2 total stackings.
we can place any stacking for n − 2 coins here
35
Macalester Journal of Catalan Numbers
Consider k = 3 wherein we place a third coin. We can choose any valid
stacking for n = 2, which again is k − 1, to the left and above the newly placed
coin and any valid stacking for n − k = n − 3 coins to the right. Here we have
COI N2 · COI Nn−3 stackings.
we place every stacking for n − 3 coins here
every stacking for n − 3
If k = 4 and we place a fourth coin, we can choose any valid stacking for
n = 3, which, yet again, is k − 1, to the left and above the fourth (kth ) coin and
any valid stacking for n − k = n − 4 to the right of this coin. Here we have
COI N3 · COI Nn−4 stackings.
we place every stacking for n − 4 coins to the right of each of these 5 stackings
Now it might be obvious what happens after we have placed k coins on the
bottom row. We place all valid stackings for k − 1 above and to the left of the kth
coin and all valid stackings for n − k to the right.
we place every stacking for k − 1 here
we place every stacking for n − k here
Finally, when k = n we place all possible stackings for n − 1 above and to the
left of the nth coin and nothing to the right, giving a total of COI Nn−1 · COI N0
stackings.
36
Macalester Journal of Catalan Numbers
If we add up all possible stackings as outlined above, we obtain the following
recurrence relation
COI N0 = 1
COI Nn = COI N0 · COI Nn−1 + COI N1 · COI Nn−2 + · · · + COI Nn−1 · COI N0
n
=
∑ COI Nk−1 COI Nn−k
k =1
n≥1
Also, recall that the Catalan numbers satisfy the recurrence relation below
C0 = 1
Cn = C0 Cn−1 + C1 Cn−2 + C2 Cn−3 + . . . + Cn−1 C0
n
=
∑ Ck−1 Cn−k
k =1
n≥1
Notice that these two recurrences are the same. We saw in Table 1 that for the first
five values of n, COI Nn = Cn . But all we really need here is that COI N0 = C0 .
Because this is true, and both sequences follow the same recurrence relation, it
follows by induction that they are the same.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Sequences Below the Sequence 012 . . . (n − 1)
Congcong Nie∗
1
The Interesting Sequence w
Consider sequences w= w1 w2 w3 . . . wn , such that w1 = 0 and each digit of w, wi
has the following property,
0 ≤ wi+1 ≤ wi + 1.
We show that the number of such sequences is equal to the nth Catalan number
Cn .
2
Examples for n ≤ 4
wn
w0
w1
w2
w3
w4
∗ Congcong,
Sequences
*
0
00 01
000 001 010 011 012
0000 0001 0010 0011 0012
0100 0101
0110 0111
0112 0120 0121 0122 0123
Count
1
1
2
5
14
class of 2011, was born and raised in Beijing, and she likes spacing out, bubble
tea, collecting cards, making colorful posters, reading in a comfy chair in a cloudy afternoon, and
random things. She <3s Combinatorics as much as she <3s Macalester. :-)
38
Macalester Journal of Catalan Numbers
3
Proof: Catalan numbers are the answers
We prove the number of sequences is Cn by providing a bijection to a well known
Catalan problem: balanced sequences of n 1’s and n −1’s. Note that we’ll never
see more −1’s than 1’s at any digit. To prove this, we define b=b1 b2 . . . bn , where
bi = w i − w i + 1 + 1
for 1 ≤ i ≤ n
where we set wn+1 = 0 for convenience.
We create our sequence S(b) of 1’s and −1’s as follows. For each i that 1 ≤
i ≤ n, write a single 1 followed by bi −1’s. The sequence S(b) will be a balanced
sequence.
3.1
Numerical Examples
Consider the 4-digits sequence w1 w2 w3 w4 = 0122. We calculate,
b1 = w1 − w2 + 1 = 0 − 1 + 1 = 0
b2 = w2 − w3 + 1 = 1 − 2 + 2 = 0
b3 = w3 − w4 + 1 = 2 − 2 + 1 = 1
b4 = w4 − w5 + 1 = 2 − 0 + 1 = 3
The resulting sequence is 111(−1)1(−1)(−1)(−1). This is a balanced sequence
of four 1’s and four −1’s.
This algorithm works in the reverse direction too. For example, consider the
sequence 1(−1)11(−1)1(−1)(−1), we know that each 1 marks the start of a wi .
Recall that w1 = 0. Since the first 1 is followed by a −1, we have b1 = 1 so that
b1 = w1 − w2 + 1 = 1
Thus w2 = w1 = 0. Since nothing follows w2 , b2 = 0. Therefore,
b2 = w2 − w3 + 1 = 0
which means that w3 = w2 + 1 = 1. Similarly, we can get
b3 = w3 − w4 + 1 = 1,
39
Macalester Journal of Catalan Numbers
so that w4 = w3 = 1.
The corresponding 4-digit sequence is 0011. In the next section, we show that
this is a bijection.
3.2
The Proof
Given a sequence w = w1 w2 . . . wn , let b = b1 b2 . . . bm . The total number of 1’s in
S(b) is n. Again, for convenience we define wn+1 = 0
The total number of −1’s is,
n
∑ bi
i =1
n
=
∑ ( wi − wi +1 + 1 )
i =1
= w1 − w n +1 + n
= 0 − 0 + n = n.
Therefore S(b) has length of 2n with equal numbers of 1’s and −1’s. Meanwhile,
i
∑ bk
k =1
i
=
∑ ( wi − wi +1 + 1 )
k =1
= w1 − w i +1 + i ≤ i
For 1 ≤ i ≤ n, the number of −1’s does not exceed the number of previously
seen 1’s. This algorithm works in the reverse direction too. Given the values of
b1 b2 . . . bn , we can get below n + 2 equations, in n + 1 unknowns w1 w2 w3 . . . wn+1 ,
w1 = 0
bi = w i − w i + 1 + 1
w n +1 = 0
1≤i≤n
However, the equation for bn is redundant.
bn = n −
n −1
∑ bk = w n + 1
k =1
40
Macalester Journal of Catalan Numbers
n −1
n −1
∑ bk
=
∑ bk
= − w n + ( n − 1)
k =1
n −1
k =1
∑ ( w k − w k +1 + 1 )
k =1
Therefore,
bn = n − (−wn + (n − 1)) = 1 − wn
b.
In other words, we have a unique solution w1 , w2 , . . . , wn+1 for each sequence
We now must show the solution w1 , w2 , . . . , wn is a valid sequence. First, we
have,
w i + 1 = w i − bi + 1 ≤ w i + 1
We know that
k
∑ bi
≤ k
∑ ( wi − wi +1 + 1 )
≤ k
i =1
k
i =1
− w k +1 + k ≤ k
Thus we have wk+1 ≥ 0. For 1 ≤ k ≤ n
In summary, the solution both unique and valid, so the mapping is invertible.
Therefore the number of valid sequences of length n is the nth Catalan number
Cn .
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Upper Triangular Ferrers Diagrams
Katie Agnew∗
1
Introduction
A Ferrers Diagram is a collection of left-justifed boxes where each row has no
more boxes than the row above. Let Fn be the set of Ferrers Diagrams that fit in
the shape (n − 1, n − 2, ..., 1). We show that |Fn | = Cn , the nth Catalan number.
2
Examples
The sets Fn for 1 ≤ n ≤ 4 are as follows. In the table below, we give the shape
(n − 1, n − 2, ..., 1), and then list the elements of Fn . Note that we include the
“empty diagram” consisting of no boxes as a member of each Fn .
∗ Katie
is Mathematics major and Environmental Studies minor at Macalester College. She was
born and raised in Minnesota, plays rugby, and enjoys a good blizzard.
42
Macalester Journal of Catalan Numbers
n
(n − 1, n − 2, . . . , 1)
Fn
1
∅
∅
2
3
4
3
Bijection
We will create a bijection from Fn to another set that is counted by the Catalan
numbers. Let Sn be the sequences 1 ≤ a1 ≤ ... ≤ an of integers with ai ≤ i. For
example, S3 = {111, 112, 113, 122, 123}. In his paper in this journal, Matt Kusner
proves that the size of the set Sn is the nth Catalan number.
We now define f : Fn → Sn . Given F ∈ Fn , we always consider F to have
n − 1 rows by allowing a row to have 0 boxes. We create an (n − 1) tuple as
follows: starting from the bottom row, we create an (n − 1)-tuple whose entries
are the number of elements in each row. So, for F3
43
Macalester Journal of Catalan Numbers
we would get:
00 01 02 11 12.
Next, we prepend a 0 to this number to get a string of length n. In our example,
we get
000 001 002 011 012.
Finally, we add 1 to each digit of the number. The resulting sequences for our
example are the elements of S3 :
111 112 113 122 123.
This mapping is invertible. We simply follow these three steps in reverse. For
example, if we begin with S3 , and then subtract 1 from each digit of the numbers
(111 in this case), we get: 000 001 002 011 012. Removing the leading 0
gives 00 01 02 11 12. Which corresponds to F3 .
Now, we will show that this holds for any value of n. Given F ∈ Fn , we can
represent F with the (n − 1)-tuple ( f 1 , f 2 , ..., f n−1 ) where f i is the length of the
(n − i )th row. Since F is a Ferrers Diagram, we have f i ≤ i and f i ≤ f i+1 .
Our mapping φ : Fn → Sn is given by φ( F ) = s where s = (s1 , ..., sn ) =
(1, f 1 + 1, ..., f n−1 + 1). We show that this mapping is well defined by showing
that s ∈ S. We have
s1 = 1,
si = 1 + f i−1 ≤ 1 + (i − 1) = i for 2 ≤ i ≤ n,
si = 1 + f i−1 ≤ 1 + f i = si+1 for 2 ≤ i ≤ n − 1,
sn = 1 + f n−1 ≤ 1 + (n − 1) = n.
Therefore s ∈ Sn . Because φ is a bijection from Fn to Sn , the size of Fn is the nth
Catalan Number.
Macalester Journal of Catalan Numbers
44
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Pairs of Non-Crossing Lattice Paths
Drew Van Denover∗
1
Introduction
Our task is to count the set of unordered pairs of lattice paths with n − 1 steps
each. Both paths must start at the origin and end at the same point, using only
steps of (1,0) or (0,1) such that one path never crosses the other. For example, the
possible pairs when n = 2 are:
The leftmost paths begin at (0,0) and end at (0,1). The paths in the rightmost
pair begin at (0,0) and end at (1,0). In both cases, the paths are identical, so they
technically overlap one another. They are drawn slightly apart so they can be
easily distinguished.
We prove that the number of pairs of lattice paths of this type is given by the
nth Catalan number, Cn . We demonstrate a bijection between the set of these
paths and the set of balanced groupings of 2n parentheses, where a "balanced"
grouping contains equal numbers of open and closed parentheses arranged such
that if you begin at the left and move to the right, you have never passed more
closed parentheses than open ones. The existence of such a bijection proves that
both sets have the same cardinality. Since the number of balanced groupings of
parentheses is known to be the Catalan numbers, the sequence Cn must also describe the number of possible pairs of lattice paths that meet our conditions.
∗ Drew,
class of 2011, is a Mathematics and Philosophy major from Denver, Colorado.
46
Macalester Journal of Catalan Numbers
2
Solutions for n ≤ 4
n = 0:
∅
n = 1:
..
n = 2:
n = 3:
n = 4:
47
Macalester Journal of Catalan Numbers
3
Defining the Mapping
We define a function f : A → B, where A is the set of our lattice pairs of
length n − 1 and B is the set of balanced groupings of 2n parentheses. First, we
separate the two lattice paths and describe each one as strings of U’s and R’s,
which represent steps up and right, respectively. For each step of (0,1) we mark
U, and for each step of (1,0) we mark R. If the two paths are identical, then we
can arbitrarily label one the "top" path and the other the "bottom." If they are not
identical, there is a first step where they disagree. In that case, we call the path
with behavior (0,1) at that step the top path and the path with (1,0) the bottom.
We then write the string for the top path above the string for the bottom path
to create a 2 x (n − 1) matrix, where each row describes the behavior of one of
the paths, and the ith column describes the directions of both paths at the ith step
for 1 ≤ i ≤ n − 1. For example:
R
R
→
→
→
U R
R U
Next, we replace each column of that matrix with a particular string of two
parentheses based on the column’s contents. The columns map to pairs of parentheses as follows:
U
U
U
R
R
R
R
U
⇐⇒ ( )
⇐⇒ ( (
⇐⇒ ) (
(1)
⇐⇒ ) )
Using this key, we can convert our example into a string of parentheses like this:
→
→
→
R
R
U
R
R
→)((())
U
Finally, we add an open parenthesis to the beginning of the string, and a closed
parenthesis to the end. The end result is a balanced grouping of 2n parentheses:
→
→
→
R U
R R
R
→)((())→()((()))
U
Macalester Journal of Catalan Numbers
4
48
The Mapping is Well-defined
The result of the function f outlined above must be a balanced group of parentheses because of two properties of the 2 x n − 1 matrix:
U
R
Claim 1: There must be equal numbers of
steps and
steps.
R
U
Let x be the vertical distance between the two paths after a certain number of
steps. How does each of the four possible path behaviors affect this value? Let yt
represent the y-coordinate of the top path, and yb represent the y-coordinate of
the bottom path. Then, each at step ∆x = ∆yt − ∆yb . There are four cases:
–
–
–
–
U
: Both paths move up one unit. Thus, ∆yt = ∆yb = 1, and ∆x = 0.
U
Movements of this type do not change the vertical distance between the
paths.
R
: Neither path moves in the vertical direction. ∆yt = ∆yb = 0, and
R
∆x = 0. This behavior does not change the vertical distance either.
U
: The top path moves one unit upwards, but the bottom path does not.
R
So ∆yt = 1 and ∆yb = 0, and ∆x = 1. These steps increase the vertical
distance between the paths by one unit.
R
: The opposite is true. ∆yt = 0 and ∆yt = 1, so ∆x = −1. This behavior
U
reduces the vertical distance between the paths by one unit.
U
R
and b steps of
, x = a − b. However, in
R
U
order to belong to the set of our pairs of paths, both paths must end the same
point. This means that the total value of x after all steps have been completed
must be 0. So x = a − b = 0, and a = b. In summary, all of our pairs of paths
U
R
must have equal numbers of
steps and
steps.
R
U
So, for a path with a steps of
Macalester Journal of Catalan Numbers
Claim 2: There can never be more
49
R
U
steps than
steps as we move from
U
R
left to right.
If the two paths are at the same point after n steps, the n + 1th step cannot be
R
, or else the paths will cross (which our lattice paths cannot). Our paths can
U
R
move in a
step without crossing only if the paths are in different positions after
U
R
the previous step. The paths will only be in different positions if the
step is
U
U
preceded by an unmatched
step, since that is the only other type of behavior
R
capable of separating the paths.
These two properties taken in conjunction demonstrate that the function f
will always generate a balanced set of parentheses. That is, f is well-defined. To
test whether a group of parentheses is balanced, we keep track of a sum as we
move from left to right, adding 1 each time we pass an open parentheses, and
subtracting 1 each time we pass a closed one. The grouping is balanced as long
as that sum is never negative and the total sum is zero.
The first parenthesis in the output from f will always be open, so our sum
after one character is 1. Some combination of "( )", ") (", "( (", and ") )" follow.
The groupings "( )" and ") (" each contain one open parenthesis and one closed,
so they have no net effect on our sum when we pass them. We could add any
number of either of these pairings without risking an invalid grouping.
"( (" adds 2 to the sum, and ") )" subtracts 2. These could potentially cause the
sum to become negative if we pass a ") )" pair before a "( (" pair. However, we
know this is impossible. We demonstrated above that there are always at least
U
R
as many
steps as
steps at any point, so there will always be at least as
R
U
many "( (" pairings as ") )" pairings as we move from left to right. Thus, we will
only subtract 2 from the sum when we have already added 2, and the sum will
never become negative.
We also know that there are equal numbers of "( (" pairings and ") )" pairings,
so the +2s and −2s they generate will ultimately cancel each other out. After
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Macalester Journal of Catalan Numbers
adding any number of "( )" pairings, any number of ") (" pairings, and an equal
number of "( (" and ") )" pairings, the net sum is 1 (thanks to the first unmatched
open parentheses). Then, the last parentheses will always be closed, taking the
sum back down to 0, indicating that the grouping is balanced. Clearly, any combination of steps that corresponds to a valid pair of lattice paths will necessarily
generate a balanced grouping of parentheses.
5
Defining the Inverse Mapping
We define a function g : B → A which reverses the function f . Given a
balanced grouping of 2n parentheses, we can follow the same steps in reverse
to produce a unique lattice pair. To demonstrate, we will undo the string of
parentheses we just created.
First, we strip the first and last parentheses from the string, leaving 2(n − 1)
parentheses remaining:
()((()))→)((())
Now, we take the first two parentheses in the string and replace them with a 2
x 1 matrix containing U’s and R’s, using the mappings defined in (1). We repeat
this process for the next two parentheses and so on, until the entire string has
been replaced. Note that there must be an even number of parentheses because
the string contained 2(n − 1) parentheses when we began replacing them. After
replacing them all, we combine the 2 x 1 columns into an 2 x n − 1 matrix. In our
example:
()((()))→)((())→
R
R
U
R
R
U
Finally, we can use the rows of the resulting matrix to reconstruct each path
individually and combine them:
()((()))→)((())→
R
R
U
R
R
U
We have arrived at our original pair of lattices.
→
→
→
Macalester Journal of Catalan Numbers
6
51
Conclusion
We have defined a function f : A → B to transform our lattice pairs into a
balanced grouping of parentheses, as well the function g : B → A. We have also
shown that g ◦ f : A → A, which is the identity mapping id A . That is, the function
g is the inverse function for f . Therefore, f is a bijection and | A| = | B|. Since
Cn counts balanced groupings of 2n parentheses, it must also count our pairs of
lattices with length n − 1.
Macalester Journal of Catalan Numbers
52
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Partially Connected Polygon Graphs
Jeanmarie Youngblood∗
1
Problem
How many ways are there to connect some pairs of vertices of an (n − 1)−gon
with non-crossing edges and then circle a subset of the remaining vertices?
This problem is in effect asking how many graphs of an (n − 1)−gon have
vertices of maximum degree 1 (and non-crossing edges); each vertex may connect
to exactly one other vertex, connect to itself (in other words be circled), or remain
isolated. For consistency, we will orient each polygon such that the (n − 1)th vertex is in the bottom right corner, and label v1 through vn−2 clockwise, beginning
to the immediate left of vn−1 . We will show via a recurrence relation that the
solution to this problem for any (n − 1)−gon is the nth Catalan number Cn .
2
Examples
Here are the solutions for 1 ≤ n ≤ 4:
n = 1: 0−gon
∅
=1 way = C1
n = 2: 1−gon
= 2 ways = C2
∗ Jeanmarie,
class of 2012, has been trying to find a closed formula for primes since her fourth
grade teacher told her she would be on the front page of the New York Times if she did. She can
eat more ice cream than you. She hails from Brooklyn, NY.
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Macalester Journal of Catalan Numbers
n = 3: 2−gon
= 5 ways = C3
n = 4: 3−gon
= 14 ways = C4
= 14 ways = C4
For each of these values of 1 ≤ n ≤ 4, the number of possible desired graphs
is the corresponding Catalan number Cn . We will now show that these graphs
satisfy the recurrence relation for the Catalan numbers.
3
Recurrence Relation
The recurrence relation for the Catalan numbers is
C0 = 1
Cn+1 =
n
∑ Ci Cn−i for n ≥ 0.
i =0
To show that the number of possible graphs for each (n − 1)−gon in fact correlates to Cn , we must partition each set of graphs such that the parts correspond
to C0 Cn−1 , C1 Cn−2 ,. . . , Cn−1 C0 , according to the recurrence relation. Before considering the general case, we explain how to partition these graphs when n = 4.
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Macalester Journal of Catalan Numbers
3.1
For the case n = 4
In this case, we must partition the 14 triangles as follows:
– C0 C3 = 1 ∗ 5: 5 triangles
– C1 C2 = 1 ∗ 2: 2 triangles
– C2 C1 = 2 ∗ 1: 2 triangles
– C3 C0 = 5 ∗ 1: 5 triangles
The characteristics of vertex v3 provide the basis for creating the partition.
Observe that v3 must either share an edge with v1 , v2 , or itself (meaning circled),
or remain isolated.
Consider the graphs in which v3 is circled or isolated as special cases. In each
case, the number of possible graphs of the remaining 2−gon can be drawn C3
ways (as shown in the example where n = 2). Because C0 = 1, these two sets of
graphs can be counted by C0 C3 and C3 C0 respectively.
In all other cases, (when an edge is drawn from v3 to another vertex), the edge
divides the remaining vertices into two distinct groups A and B. Because there
are only three vertices in this case, one group will contain the remaining vertex
and the other will contain no vertices.
Number of of possible graphs when
– v3 remains isolated: 1∗[number of possible graphs with 2 vertices]= 1 ∗ 5 =
C0 C3 .
v2
v2
v2
v2
v2
v1
v3
v1
v3
v1
v3
v1
v3
v1
v3
– v3 connects to v1 : [Side A: number possible graphs with 0 vertices]∗[Side B:
number possible graphs with 1 vertex] = 1 ∗ 2 = C1 C2 .
v2
v2
v1
B
A
v3
v1
B
A
v3
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Macalester Journal of Catalan Numbers
– v3 connects to v2 : [Side A: number possible graphs with 1 vertex]∗[Side B:
number possible graphs with 0 vertices]= 1 ∗ 2 = C2 C1 .
v2
v2
v1
A B
v3
v1
A B
v3
– v3 is circled: [number of possible graphs with 2 vertices]∗1 = 5 ∗ 1 = C3 C0 .
v2
v2
v2
v2
v2
v1
v3
v1
v3
v1
v3
v1
v3
v1
v3
This shows that by classifying parts according to which vertex connects to v3 ,
we partition the set of graphs of a 3−gon such that each part corresponds to a
term in the recurrence relation summation.
3.2
Extention to any (n − 1)−gon:
The partition for an (n − 1)−gon depends on the state of vn−1 . There are three
cases:
– When vn−1 connects to vi , 1 ≤ i ≤ n − 2:
The vertices on either side of the edge between vn−1 and vi (1 ≤ i ≤ n − 2)
form two separate sets of points A and B, where | A| = i − 1 and | B| =
(n − 1) − (i − 1) − 2 = n − i − 2 (The 2 being subtracted corresponds to
the two vertices vn−1 and vi ). More specifically, A = {1, 2, . . . , i − 1}, and
B = {i + 1, i + 2, . . . , vn−2 }.
By the recurrence relation, the ways to draw acceptable graphs is Ci for
the vertices in A and Cn−i−1 for those in B. Because these processes are
independent of one another, multiplying them gives the number of ways to
do both at once. So, the number of desired graphs of an (n − 1)−gon is
Ci Cn−i−1 for all 0 < i < n − 1.
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Macalester Journal of Catalan Numbers
vi
v4
A
v3
v2
v n −4
B
v1
v n −1
v n −3
v n −2
– When vn−1 is circled:
n − 2 vertices remain. We can therefore draw the rest of the graph in Cn−1 =
1 ∗ Cn−1 = C0 Cn−1 ways.
– vn−1 is isolated:
As in the previous case, we can draw the rest of the graph in Cn−1 = Cn−1 ∗
1 = Cn−1 C0 ways.
Summing up the number of graphs for each i gives C0 Cn−1 + C1 Cn − 2 + ... +
Cn−2 C1 + Cn−1 C0 , which we know is equal to Cn . In conclusion,
C0 = 1
Cn+1 =
n
∑ Ci Cn−i for n ≥ 0.
i =0
so the number of our graphs satisfies the Catalan recurrence.
Therefore, the Catalan numbers count the ways to connect some vertices of an
(n − 1)−gon and circle a subset of the remaining vertices.
Macalester Journal of Catalan Numbers
58
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Plane Binary Trees
David Klock∗
1
Problem: Plane Binary Trees with n Internal Nodes
We wish to determine how many different structures are possible for a plane
binary tree with 2n+1 nodes, or, equivalently, n internal (non-leaf) nodes. We
prove that this number is equivalent to Cn , the nth Catalan number.
2
Solution: Catalan Numbers
2.1
Proof Idea
We already know that the number of valid groupings of 2n parentheses is Cn . We
will show that the parentheses problem is equivalent to the binary tree problem
by defining a bijection from the set of all binary trees B to the set of all groupings
of parentheses P, represented as strings of parentheses. Our bijection will be
based on the following three rules:
1. A node is represented by a pair of parentheses.
2. If a node has a left child, that child is represented as the first pair of parentheses inside the pair representing the parent node-that is, (( )) represents a
parent and a left child.
3. If a node has a right child, that child is represented as the first pair of parentheses after the pair representing the parent node-that is, ( )( ) represents a
parent and a right child.
∗ David,
class of 2010, is a double major in Political Science and Computer Science, with a
minor in Mathematics.
Macalester Journal of Catalan Numbers
60
Note that these rules are recursive. For example, if we have the string of
parentheses (( )( )), this represents a parent with a left child which, in turn, has a
right child. (It does not represent a parent with two left children–this is why the
rules emphasize the word “first”.)
For example, if we have the binary tree
we would represent it as “((( ))( ))( )”.
Intuitively, these rules will lead to a unique, and valid, representation of a
binary tree as a string of parentheses. Each “(” has a matching “)”. Invalid
strings like “)()(” do not have a chance to occur. Moreover, any given binary tree
has one unambiguous parenthetical representation. Below, we prove rigorously
that these rules indeed lead to a bijection.
2.2
n=1
n=2
n=3
Enumeration of Trees for n ≤ 4
Macalester Journal of Catalan Numbers
61
n=4
2.3
The Bijection
We now explain exactly how our bijection maps B onto P. The problem concerns
strings and nodes. Because these are concepts treated in computer science, it will
be useful for us to think of the bijection as a computer procedure. Let us call it
WriteTree(tree).
Macalester Journal of Catalan Numbers
62
We begin by defining a subprocedure, WriteNode(node, string), which WriteTree will use to do the actual representation. WriteNode converts a node (including its children, recursively) into a parenthesized representation. node is the node
to be converted, and string is the string to write the parenthesized representation
out to.
WriteNode(node, string):
1. Write an opening parenthesis, “(”, to string.
2. If node has a left child left, invoke WriteNode(left, string).
3. Write a closing parenthesis, “)”, to string.
4. If node has a right child right, invoke WriteNode(right, string).
Given this procedure, WriteTree(tree) works as follows:
WriteTree(tree):
1. Let string equal an empty string, “ ”.
2. Let node equal the root node of tree.
3. Invoke WriteNode(node, string).
4. Return string as the result.
In other words, WriteTree(tree) takes the root of tree, invokes WriteNode on it,
and returns the resulting string of parentheses.
2.4
The Proof
We claim that this function will produce a bijection from B to P such that the
image of a tree with n nodes is a string with n pairs of parentheses. We do so by
showing that WriteTree is a valid, invertible function.
2.4.1
Validity
For WriteTree to be a valid function, the string of parentheses given by WriteTree(tree) must be within P for every tree in B. A string of parentheses is valid
(within P) if there are an equal number of open and closed parentheses and if,
beginning at the left and moving right, one at no point encounters more closed
parentheses than open parentheses. We observe that:
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63
1. The procedure WriteNode(node, string) writes only two characters–one “(”
and one “)”. No other part of the function writes a character. Thus, the
number of opening parenthesis must equal the number of closing parenthesis.
2. WriteNode(node, string) always writes a “(” first, followed by a matching
“)” at some point later in the string. Thus, at any point in the string, there
must be at least as many opening parentheses as closing parentheses in the
portion of the string coming before that point.
Thus, WriteTree produces strings which are valid members of P. It is, therefore,
a valid function from B into P.
2.4.2
Invertibility
We must also show that WriteTree is invertible. To do so, we define its inverse,
ReadString. We begin by creating a procedure, ReadParens(string, currentNode,
index), which reverses the operation of WriteNode(node, string). The currentNode
in ReadParens(string, currentNode, index) requires some explanation; in essence, if
one expects a pair of parentheses is about to come up in string (by looking ahead),
one creates the node represented by those parentheses in advance, then invokes
ReadParens on that node to process its parentheses and, potentially, add children
to it. The index tracks our position in the string.
ReadParens(string, currentNode, index):
1. Begin reading from index.
2. Read the opening parenthesis representing currentNode, “(”, from string.
3. Increment index.
4. If the character now at index in string is “(”:
(a) Create a new node, left. Make this currentNode’s left child.
(b) Invoke ReadParens(string, left, index).
5. Read the closing parenthesis representing currentNode, “)”, from string.
6. Increment index.
7. If the character now at index in string is “(”:
Macalester Journal of Catalan Numbers
64
(a) Create a new node, right. Make this currentNode’s right child.
(b) Invoke ReadParens(string, right, index).
Given this procedure, ReadString(string) for string a valid string of parentheses
(as defined above) works as follows. (We ignore the edge case where P is empty,
representing a tree with only one node.)
ReadString(string):
1. Create a new node root, which will serve as the root node of our tree.
2. Invoke ReadParens(string, root, 0).
3. Return a binary tree with root as root.
We will prove that ReadString(string) does, in fact, invert WriteTree by induction on n, the number of internal nodes in our tree.
We will make use of the fact that a node with two leaves will be represented
by WriteNode as “(( ))( )”, regardless of where it is in the tree. More generally,
WriteNode will depict any internal node in the form “(L)R”, where L and R are
the representations of the subtrees of the left and right child of the given node.
(When L and R are both leaves, this reduces to “(( ))( )”.)
Our base case in the induction will be that n = 1. In this case, we have a
binary tree tree consisting of the root root and two leaves. WriteTree(tree) will
take root and invoke WriteNode on it to get a string of parentheses. By the note
above, this string will simply be “(( ))( )”. Call this string string. If we invoke
ReadString(string), we will get back tree.
Thus, ReadString inverts WriteTree for this base case.
We now consider the induction case. Suppose that ReadString((WriteTree(tree)))
= tree for all trees tree with n - 1 internal nodes. We show that ReadString(WriteTree(tree))
= tree for tree tree with n internal nodes.
Suppose we have treen , a tree with n internal nodes. Moreover, suppose node is
an internal node with two leaves. At least one such node must exist. If we replace
node with a leaf, we get a tree treen−1 with n - 1 internal nodes. Let f(treen−1 ) =
stringn−1 . Then, by our assumption, ReadString(stringn−1 ) = treen−1 .
Now, let WriteTree(treen ) = stringn . We show that ReadString(stringn ) = treen .
Our proof depends on whether node is a left or a right child of node parent . The
two cases are similar; we consider the case where node is a left child, and omit the
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Macalester Journal of Catalan Numbers
other case.
node parent
right child R (potentially has children)
leaf
The subtree of node parent for treen−1 .
WriteNode represents this as “(( ))R”.
node parent
node
leaf
right child R (potentially has children)
leaf
The subtree of node parent for treen .
WriteNode represents this as “((( ))( ))R”.
Consider how node parent is represented in stringn−1 . To obtain the tree bn−1 , we
replaced node with a leaf. Thus, in treen−1 , node parent has a leaf for its left child. As
observed above, invoking WriteNode on node parent represents it as “(L)R”; since
the left child is a leaf in treen−1 , this reduces to “(( ))R” in stringn−1 .
Now, consider how node parent is represented in stringn . To obtain thetree treen ,
node parent has node as its left child. Now, node itself has two leaves as children;
as noted before, this situation is represented as “(( ))( )”. Thus, WriteNode will
represent node parent in stringn as “((( ))( ))R”.
We can now consider how ReadString acts on stringn−1 and stringn . We show
that ReadString acts the same way on each string except when processing the
modified part of the string, and, moreover, uses the modified part of the string
to create a leaf in treen−1 and the node node in treen . This proves that ReadString
works as expected.
ReadString(stringn−1 ) and ReadString(stringn ) both begin in the same way;
they create a root node root, and invoke ReadParens(root, stringn−1 , index) and
ReadParens(root, stringn , index), respectively. Since ReadParens operates in a linear fashion and only ever looks ahead one character, both functions will create
the same structure until they begin reading the modified part of the string representing node parent : “(( ))R” in stringn−1 , and “((( ))( ))R” in stringn .
At this point, the two functions will differ. ReadString(stringn−1 ) will interpret
Macalester Journal of Catalan Numbers
66
"(( ))R" to create a node parent with a leaf on the left and the subtree R on the right,
as desired. ReadString(stringn ), on the other hand, will interpret "((( ))( ))R" to
create a node parent with the internal node node (with two leaves) as a left child and
the subtree R as a right child. This is also as desired.
From here on, the two functions are parsing the same string. As a result, the
rest of the tree structure they create is identical. Thus, the only difference between
the two trees they generate comes at node parent . At this point, when ReadString
parses stringn−1 , it gives node parent a leaf for a left child. On the other hand, when
ReadString parses stringn , it gives node parent an internal node for a left child. This
is exactly what we want to occur. Thus, the induction step is valid in the case
where node is a left child of node parent . The proof for a right child is similar.
Thus, by induction, ReadString is the inverse of WriteTree when WriteTree
acts on binary trees for which n ≥ 1. (We may define a special case to handle the
trivial situation where n = 0; we ignore it here.) Since WriteTree has an inverse,
it is a bijection between B and P. Since we can define a bijection between the set
of all binary trees with n internal nodes and the set of all groupings of n pairs of
parentheses, the sizes of these sets must be the same. Finally, since the number
of groupings of n pairs of parentheses is Cn , the nth Catalan number, the number
of binary trees with n internal nodes must also be Cn , and our claim is proven.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Non-Intersecting Convex Hulls
Jorge Banuelos∗
1
Counting Problem
Consider the set of partitions B(n) = { B1 , . . . , Bk } of [n] such that if the numbers
1, 2, . . . , n are arranged in order around a circle, then the convex hulls of the
blocks { B1 . . . , Bk } are pairwise disjoint. For example, when n = 3 we have 5
such partitions:
1
3
1
2
3
1
2
3
1
2
3
1
2
3
2
We show the number of such partitions is equal to the nth Catalan number.
2
Bijection with the Balanced Parentheses Problem
We will prove a bijection exists between the set C(n) of partitions of Balanced
Parentheses, which we know has a bijection with the Catalan numbers, and our
given set of non-intersecting convex hulls B(n) by demonstrating that a bijection
exists for both sets with the set A(n)={(i1 , . . . , in ) : ∑nk=1 ik = n and ∑kj=1 i j ≤
k }. We begin by proving a bijection between B(n) and A(n). We then prove a
bijection between C(n) and A(n).
2.1
Non-Intersecting Convex Hulls
We enumerate the set B(n) for n ∈ {1, 2, 3, 4}
∗ Jorge,
class of 2011, is a mathematics major from Zacatecas, Mexico.
68
Macalester Journal of Catalan Numbers
1
n=1
1 partition
n=2
2 partitions
n=3
5 partitions
1 2
1 2
1
1
3
2
3
1
3
3
1
n=4
14 partitions
2
3
1
2
3
2
1
2
2
1
2
4
1
1
2
4
1
2
4
2
4
3
3
3
3
1
1
1
1
2
4
2
4
2
4
2
4
3
3
3
3
1
1
1
1
2
4
2
4
3
3
1
1
2
4
3
2
4
3
2
4
3
2
4
3
Partitions in B(n) : n ∈ {1, 2, 3, 4}
Now, we define a bijection g : B(n) → A(n). If B ∈ B(n) then g( B) = (i1 , i2 , . . . , in )
where the ik are defined as follows. If the element k ∈ [n] is in a z-gon by itself
then g(k ) = 1. If k is the largest element in the z-gon then g(k ) = z, otherwise
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Macalester Journal of Catalan Numbers
g(k ) = 0.
Example 1: g : B 7→ A
1
4
2
7→
(0, 1 , 2 , 1)
3
The inverse function g−1 : A(n) → B(n) simply reverses the process outlined
above. If ik = 1 then g−1 places the element k in a 1-gon by itself, otherwise it is
in a z-gon with other elements. Considering the elements in the domain of g−1
where ik 6= 0, 1 in ascending order, g−1 represents the last element in a ik -gon with
the directly preceding ik elements for which ik = 0, which have not been used in
the creation of another z-gon. g−1 evidently maps each element in its domain to
a unique element in B(n). Since our function g is invertible, it is a bijection from
B(n) to A(n).
2.2
Balanced Parentheses
Now we define an invertible function f : C(n) → A(n). We have f (C ) =
(i1 , i2 , . . . , in ) where the ik are defined as follows. For each partition C ∈ C(n)
number the left parentheses in ascending order with the element k ∈ [n]. For
k < n, let ik be the number of right parentheses between the kth and k + 1th left
parentheses. For k = n, let ik be the number of right parentheses after the nth left
parenthesis.
Example 2: f : C 7→ A
( ) ( ( ( ( ) ) ( ) ) )
7→ (1,0,0,0,2,3)
The function f is one-to-one: it maps each element of C(n) to a different element
of A(n).
We define f −1 : A(n) → C(n) by reversing the above process. First for each
element k ∈ [n] draw a corresponding left parenthesis in ascending order. Then
after each parenthesis draw ik right parentheses. Since ∑kj=1 i j ≤ k : 1 ≤ k ≤ n
Macalester Journal of Catalan Numbers
70
we are assured that a partition of balanced parentheses is created. Furthermore,
each element A evidently maps to a different element C, since the number of
right parenthesis after a certain left parenthesis will not be the same for any 2
distinct elements.
Since the function f : C(n) → A(n) is invertible it follows that a bijection exists
between the sets C(n) and A(n).
3
Final Result
It follows that since a bijection exists between the sets A(n) and C(n) and A(n)
and B(n) that a bijection must exists between the sets C(n) and B(n). Since the
numbers of elements in the set C(n), of partitions of Balanced Parentheses, is
known to have a bijection with the Catalan numbers, the set B(n) does as well.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Dyck Paths With No Peaks At Height Two
Yenee Soh∗
1
Dyck Paths: An Introduction To The Problem
A mountain range (also called a Dyck path) of length 2n is a path from (0, 0)
to (2n, 0) which uses only steps (1, 1) and (1, −1) and does not go beneath the
x-axis. The resulting number of paths is the nth Catalan number. In this paper,
we consider Dyck paths from (0, 0) to (2n + 2, 0) with no peaks at height two. A
peak occurs when a (1, 1) step is followed by a (−1, −1) step. We prove that the
number of ways to create paths from (0, 0) to (2n + 2, 0) with this restriction is
equal to the nth Catalan number Cn .
2
Enumeration of Examples for n ≤ 4
We enumerate all such Dyck paths for n ≤ 4. From the examples below we can
see that the number of paths equal the Catalan numbers for C0 , C1 C2 , C3 and C4 .
For n = 0, we have 1 possible path:
For n = 1, we have 1 possible path:
For n = 2, we have 2 possible paths:
∗ Yenee
grew up in Seattle, WA and currently lives in Seoul, South Korea. She enjoys traveling and has been to Micronesia, Guam, Japan, the Philippines, Cambodia, Vanuatu, Australia,
Vietnam and Canada. She loves experiencing new cultures, learning languages and meeting new
people.
Macalester Journal of Catalan Numbers
For n = 3, we have 5 possible paths:
For n = 4, we have 14 possible paths:
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Macalester Journal of Catalan Numbers
3
73
Proof: Catalan Numbers Are The Solution
We give a bijection between regular Dyck paths from 0 to 2n and our problem,
Dyck paths of length 2n + 2 with no peaks at height 2. It is known that the
number of regular Dyck paths of length 2n is Cn . By proving a bijection, we
conclude that our problem also results in the Catalan numbers.
Let S be the set of regular Dyck paths created with n upstrokes and n downstrokes that all stay above the x-axis. Let T be the set of Dyck paths from (0, 0) to
(2n + 2, 0) with no peaks at height two.
We now define our function f : S → T. First, starting with a Dyck path s from
S, we obtain a Dyck path t from T using the following steps.
Macalester Journal of Catalan Numbers
74
(1) We attach a Dyck path of length 2 to the left of s to create s∗ , which gives us
the increased length of 2 necessary for T.
(2) A maximal sub-Dyck path refers to a sub-Dyck path with no peaks at height 1
that is followed directly by a sub-Dyck path of height 1. Let p∗ be a maximal
sub-Dyck path of s∗ . To each such p∗ add a (1, 1) step at the beginning and a
(−1, −1) step at the end to create sub-Dyck path p∗∗ with no peaks at height
2. This step produces a Dyck path s∗∗ which may be larger than 2n + 2.
(3) From s∗∗ eliminate each Dyck path of length 2 that is to the immediate left
of each p∗∗ . Removing these peaks will result in a Dyck path of length
2n + 2. By lengthening the height of the sub-Dyck paths to ensure no peaks
at height 2, we have overall increased the length. If there is one maximal
sub-Dyck path, we just get rid of the Dyck path of length 2 in step (1),
since both increased the Dyck path by length 2. If we have more than one
maximal sub-Dyck path, we can always get rid of the Dyck path of length
2, with height 1 right before it. Indeed a peak of height 1 exists between
the maximal sub-Dcyk paths. The result is in T since we have changed the
regular Dyck path of length n to a Dyck path of length 2n + 2 that has no
peaks at height 2 and is a unique element t ∈ T.
We now define the reverse function g : T → S. To obtain s from t, we do the
above steps in reverse as follows:
(1) Let t∗ be a sub-Dyck path of t between two consecutive points on the x-axis
with t∗ having no peaks at height 1. To each t∗ add a Dyck path of length 2
immediately to the left. This step produces a Dyck path s∗∗ .
(2) Let p∗∗ be a maximal sub-Dyck path of s∗∗ . From each such p∗∗ remove the
left-most (1, 1) step and the right-most (−1, −1) step to produce a sub-Dyck
path p∗ . This step produces a Dyck path s∗ of length 2n + 2.
(3) From s∗ , remove the left-most Dyck path of length 2 to produce s.
Through these steps, we regain the Dyck paths of length 2 and of height 1
that we had gotten rid of in step (3) of the original function, and decrease the
height of the maximal sub-Dyck paths to their original height by getting rid of
the left-most (1, 1) step and the right-most (−1, −1) step, and remove the step we
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Macalester Journal of Catalan Numbers
had added in step(1) of the original function. This results in a regular Dyck path
with length 2n.
In conclusion, f is a bijection since the reverse function g gives us back with
the Dyck path we started with.
For example, with n = 3 we have a regular Dyck path with length 6 and add a
Dyck path of length 2 to the left. We will define this as s∗ . This can be illustrated
as the following:
s
s∗
becomes
Next, we ensure that we have no peaks at height two. We would look at the
peaks in s∗ . First we do not have to consider those that have peaks of height one,
since this does not affect our requirements for T. So we would focus on peaks
where the height is greater than one. To eliminate any possibilities of having any
peaks at height 2 we would increase the regular Dyck path by adding a step on
both ends of these paths, creating s∗∗ as illustrated below:
s∗
s∗∗
becomes
By doing the two operations above, we have provided the length 2n requirement and the requirement that there are no peaks of height 2. However, by doing
the second operation we have increased the length to 2n + 4, so we eliminate each
Dyck path of length 2 to the immediate left of each s∗∗ . This gives us a unique
t ∈ T, as illustrated below:
t
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Macalester Journal of Catalan Numbers
It seems that the adding a Dyck path and then removing it is unnecessary, yet
it is an important step when there are multiple maximal sub-Dyck paths where
we increase the height. In this case, the number of Dyck paths of height 1 equals
the number of mountain ranges we increase (so the two cancel out each other).
With the same example, if we go through the second set of steps that define
the reverse function, we start with a t ∈ T that gives us back the s ∈ S that we
started with.
First on the t that we acquired as a result of the first operation, we add a Dyck
path of length 2 immediately to the left to produce Dyck path s∗∗ .
t
s∗∗
becomes
We have regained the Dyck paths of length 2 that we had gotten rid of. Next,
we back step the increased height of the maximal sub-Dyck paths, to produce s∗ ,
as illustrated below.
s∗∗
s∗
becomes
Then we have to get rid of the Dyck path of length 2 added to the left-most of
the Dyck path to end with our original Dyck path.
s∗
s
becomes
As we can see from the example above, the result of the reverse function gives
us back what we started with in the beginning.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Pairs of Internally Disjoint Lattice Paths
Matthew Hurni∗
1
Counting Problem
Let Ln be the set of unordered pairs of lattice paths with n + 1 steps each, starting
at (0, 0), using steps (1, 0) or (0, 1), and ending at the same point. We show | Ln |
is the nth Catalan number.
2
A Bijection with Sequences of 1’s and (-1)’s
Let Sn be the set of balanced sequences of n 1’s and n (-1)’s. This is a well known
Catalan Problem. We will describe a bijection from Ln to Sn . Begin by breaking
the lattice paths down into a grid of squares with sides of length 1. Suppose
this grid has k columns C1 , . . . , Ck . Let ai be the number of squares in column Ci ,
for 1 ≤ i ≤ k, and let bi be the number of rows in common to Ci and Ci+1 , for
1 ≤ i ≤ k − 1. Define a sequence Ω of 1’s and (-1)’s as follows (where exponentiation denotes repitition):
Ω = 1a1 (−1) a1 −b1 +1 1a2 −b1 +1 (−1) a2 −b2 +1 1a3 −b2 +1 ...1ak −bk−1 +1 (−1) ak .
The following table lists the lattice paths along with its corresponding sequence and the values of ( a1 , . . . , ak ) and (b1 , . . . , bk−1 ).
∗ Matt,
class of 2011, is from Little Falls, MN. He is a mathematics major and a physics minor,
and is also a member of the Macalester football team.
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Macalester Journal of Catalan Numbers
n
grid of squares sequence of 1’s and (-1)’s
ai
bi
1
1−
(1)
∅
2
1 − 1−
(1, 1)
(1)
11 − −
(2)
∅
1 − 1 − 1−
(1, 1, 1)
(1, 1)
1 − 11 − −
(1, 2)
(1)
11 − 1 − −
(2, 1)
(1)
11 − −1−
(2, 2)
(2)
111 − −−
(3)
∅
3
1 − 1 − 1 − 1−
4
(1, 1, 1, 1) (1, 1, 1)
1 − 1 − 11 − −
(1, 1, 2)
(1, 1)
1 − 11 − 1 − −
(1, 2, 1)
(1, 1)
1 − 11 − −1−
(1, 2, 2)
(1, 2)
1 − 111 − −−
(1, 3)
(1)
11 − −1 − 1−
(2, 1, 1)
(1, 1)
11 − −11 − −
(2, 2)
(1)
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Macalester Journal of Catalan Numbers
n
4
grid of squares sequence of 1’s and (-1)’s
ai
bi
11 − 1 − −1−
(2, 2, 1) (2, 1)
11 − 1 − 1 − −
(2, 2, 2) (2, 2)
11 − 11 − −−
(2, 3)
(2)
111 − 1 − −−
(3, 1)
(1)
111 − −1 − −
(3, 2)
(2)
111 − − − 1−
(3, 3)
(3)
1111 − − − −
(4)
∅
This mapping is well defined due to the following:
– The total number of 1’s and (-1)’s are the same. All that must be done is to
add up the exponents of each. Indeed, the sum of the exponents of the 1’s
is
a1 + ( a2 − b1 + 1) + ( a3 − b2 + 1) + · · · + ak − bk−1 + 1 =
k
k −1
i =1
j =1
∑ ai − ∑ b j + k
and the sum of the exponents of the (-1)’s is
( a1 − b1 + 1) + ( a2 − b2 + 1) + ( ak−1 − bk−1 + 1) + · · · + ak =
k
k −1
i =1
j =1
∑ ai − ∑ bj + k.
– At any given point in the sequence of 1’s and (-1)’s there are at least as many
1’s as (-1)’s. The most likely point to find more (-1)’s than 1’s occurs at the
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Macalester Journal of Catalan Numbers
end of a sequence of (-1)’s. For 1 ≤ r ≤ k − 1, by the r th pair of 1’s and (-1)’s,
the total number of 1’s we have encountered is:
r
a 1 + ∑ ( a j − b j −1 + 1 ) =
j =2
r
r −1
j =1
j =1
∑ aj −
∑ b j + (r − 1)
and the total number of (-1)’s we have encountered is:
r
r
r
r
r −1
j =1
j =1
j =1
j =1
j =1
∑ ( a j − b j −1 + 1 ) = ∑ a j − ∑ b j + r ≤ ∑ a j − ∑ b j + ( r − 1 )
since br ≥ 1.
3
Bijection: From Squares to 1’s and (-1)’s
We show the mapping is one-one and onto.
3.1
One-One
For any valid shape a unique series of 1’s and (-1)’s exists. For instance, suppose
two shapes are described by a1 , b1 , a2 , b2 , . . . , an and c1 , d1 , c2 , d2 , . . . , cm . If n 6= m
there are a different number of columns. So the resulting sequences are distinct.
Now consider the two sequences a1 , b1 , a2 , b2 , . . . , an and c1 , d1 , c2 , d2 , . . . , cn . If any
ai 6= ci or any bi 6= di then there is obviosly a distinct series of 1’s and (-1)’s
corresponding to that term as a result.
3.2
Onto
Consider a balanced sequence 1r1 (−1)s1 1r2 (−1)s2 · · · 1rk (−1)sk where ri ≥ 1 and
j
j
si ≥ 1. The exponents satisfy ∑ik=1 ri = ∑ik=1 sk = n and ∑i=1 si ≤ ∑i=1 ri for
1 ≤ j ≤ k.
We claim that a balanced sequence of 1’s and (-1)’s, has a unique solution for
a1 , · · · , ak , b1 , · · · bk−1 . Indeed, the exponents result in 2k equations. However, we
have 2k − 1 unknown variables (the a’s and the b’s). This would be a problem
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Macalester Journal of Catalan Numbers
except for the fact that the last equation is redundant.
We show that the constraint sk = ak is redundant. If the other 2k − 1 equations
are satisfied, then
sk =
k
k −1
k
k −1
i =1
i =1
i =2
i =1
∑ r i − ∑ s i = a 1 + ∑ ( a i − bi − 1 + i ) − ∑ a i − bi + 1 = a k .
So we have 2k − 1 constraints and 2k − 1 unknown variables. So given a set of
r’s and s’s, we are able to obtain the a’s and the b’s.
Now we must show that the a’s and the b’s we’ve obtained lead to a valid shape.
A pair of sequences ( a1 , . . . , ak ),(b1 , . . . , bk−1 ) create a valid shape if and only if
the following conditions hold.
1. a j ≥ 1, 1 ≤ j ≤ k
2. b j ≥ 1, 1 ≤ j ≤ k − 1
3. b j ≤ a j , 1 ≤ j ≤ k − 1
4. b j−1 ≤ a j , 2 ≤ j ≤ k
We now show that the a’s and b’s cooresponding to the balanced sequence
satisfy all 4 conditions.
– For 1 ≤ i ≤ k − 1, we have 1 ≤ si ≤ ai − bi + 1. Therefore bi ≤ ai and
condition 3 holds.
– For 2 ≤ i ≤ k, we have 1 ≤ ri ≤ ai − bi−1 + 1. Therefore bi−1 ≤ ai and
condition 4 holds.
We now show that condition 2 holds. We have
j
∑ si ≤
i =1
j
∑ ri
i =1
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Macalester Journal of Catalan Numbers
j
j
i =1
j
i =2
∑ ( a i − bi + 1 ) ≤ a 1 + ∑ ( a i − bi − 1 + 1 )
j
∑ a i − ∑ bi + j ≤
i =1
i =1
j
j −1
i =1
i =1
∑ a i − ∑ b j + ( j − 1)
− b j ≤ −1
Therefore b j ≥ 1. Finally condition 1 also holds since a1 = r1 ≥ 1 and 1 ≤ b j−1 ≤
a j for 2 ≤ j ≤ r.
In conclusion, our mapping is a bijection, so | Ln | is the nth Catalan number.
Macalester Journal of Catalan Numbers
Volume 1 (December 2009)
Dyck Paths With Odd Maximal Descents
Emily Merrill∗
1
Problem
We will consider a specific group of Dyck paths and prove that they can be enumerated by the Catalan numbers. Dyck paths are lattice paths with steps (1, 1)
and (1, −1) which never fall below the x-axis. We consider Dyck paths from (0, 0)
to (2n + 2, 0) such that any maximal sequence of consecutive steps (1, −1) ending on the x-axis has odd length. Let Ω be the set of all such Dyck paths. Let
ωn be the subset of paths of length 2n + 2, and let zn = |ωn |. As an example,
both figures below are Dyck paths from (0, 0) to (4, 0). However, only the one on
the left is a member of Ω, since the one on the right has two consecutive (1, −1)
steps leading to the x-axis. This proof will demonstrate that zn = cn for all n ≥ 0,
where cn is the nth Catalan number.
2
Enumeration of Solutions for n ≤ 4
We enumerate all members of ω for 0 ≤ n ≤ 4.
z0 = 1
∗ Emily,
class of 2010, is a math and environmental studies major with a geography minor. She
was raised in Illinois, Indiana, and Texas.
Macalester Journal of Catalan Numbers
z1 = 1
z2 = 2
z3 = 5
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Macalester Journal of Catalan Numbers
z4 = 14
85
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Macalester Journal of Catalan Numbers
3
Proof
3.1
Establishing a Generating Function
Let αn be the set of all Dyck paths from (0, 0) to (2n, 0) which never touch the
x-axis between 0 and 2n and have an odd number of consecutive (1, −1) steps at
S
the end. Let A = n≥0 αn . We can see that A is a subset of Ω, and αn is a subset
of ωn for any given n. More specifically, A is the set of all paths in Ω which never
touch the x-axis in between their endpoints.
Now, we will try to construct Ω entirely from A, which will be easier to work
with in the long-term. If we cut up a path in Ω at each point where it touches
the x-axis, each smaller component will be a member of A. Indeed, each of the
smaller paths does not touch the x-axis between its endpoints. Furthermore, the
definition of Ω requires that each sequence of (1, −1) which ends on the x-axis
be of odd length.
In order to write out Ω in terms of A, we must define a notation for appended paths. The path which is composed of some member of αm followed by
some member of αn (connected at their endpoints) will be represented by αm αn .
We must also introduce a way to refer to members of ωn by the number of interp
sections which they have with the x-axis. We will in the future refer to ωn for
paths in ωn which touch the x-axis p times between their endpoints:
For any given n, ωn consists of the union of:
– ωn0 , paths which never touch the x-axis between the two endpoints
– ωn1 , paths which touch the x-axis once between the two endpoints
– ωn2 , paths which touch the x-axis twice between the two endpoints
– et cetera.
In other words,
ωn =
n
[
k =0
ωnk .
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Macalester Journal of Catalan Numbers
Furthermore, we have
ωn0 = αn+1 ,
ωn1 =
ωn2 =
n
[
k =1
n
[
j =1
α k α n − k +1 ,


j
[
k =1

α k α j − k +1  α n − j ,
and so on.
We now develop formulae for an = |αn | and zn = |ωn |. The above equations
lead to analogous equations for the sizes of these sets.
zn =
n
∑ zkn ,
k =0
z0n
= a n +1 ,
z1n =
z2n =
n
∑ a k a n − k +1 ,
k =1
n
j
j =1
k =1
∑ ∑ a k a j − k +1
!
an− j .
We now introduce the generating functions for these sequences:
A( x ) =
∞
∑ ai x i ,
i =0
Z(x) =
∞
∑ zi x i .
i =0
It will be useful to employ the product rule for generating functions. Given two
ordinary generating functions F ( x ) and G ( x ), whose coefficients of x n are f n and
gn , respectively, we can find the coefficients of H ( x ) = F ( x ) G ( x ) in terms of f n
and gn using the so-called product formula:
hn =
n
∑ f i gn −i .
i =0
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Macalester Journal of Catalan Numbers
This sum looks similar to our equation for z1n , with the main difference coming
from the indexing. By this, z1n should be a coefficient of the generating function
A ( x ) A ( x ) = A ( x )2 .
At this point, though, it is important to recall that paths in ωn extend from 0 to
2(n + 1), while paths in αn travel from 0 to 2n. This means that z1n is actually the
(n + 1)th coefficient of A( x )2 . We can express this as: z1n = [ x n+1 ] A( x )2 From our
knowledge of the product rule and the formula for z2n above, we can extrapolate
that z2n = [ x n+1 ] A( x )3 . More generally, zkn = [ x n+1 ] A( x )k+1 . This leads us to
A ( x ) k +1
Z(x) =
.
x
k
This is the generating function for Ωk . We know that the set Ω is the union of
Ωk s, k ≥ 0 (all paths which touch zero times, one time, two times, etc.) Therefore:
Z(x) =
xZ ( x ) =
xZ ( x ) + A( x )0 =
xZ ( x ) + 1 =
∞
∑
k =0
∞
A ( x ) k +1
x
∑ A( x )k
k =1
∞
∑ A( x )k
k =0
∞
∑ A( x )k
k =0
xZ ( x ) + 1 =
3.2
1
.
1 − A( x )
(1)
Relating Generating Functions for Dyck Paths
We now have Z ( x ) in terms of the geometric power series of A( x ). We want to
relate A( x ) to the Catalan numbers, cn , and their generating function, C ( x ). The
Catalan numbers enumerate the general Dyck paths, with cn counting paths from
(0, 0) to (2n, 0).
First, we’ll need to define β n , the set of all Dyck paths which never hit the
x-axis between 0 and 2n and end in an even number of (1, −1) steps down. Next,
we’ll define B( x ) as the generating function for bn = | β n |. The sum ( an + bn )
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Macalester Journal of Catalan Numbers
counts all Dyck paths from (1, 1) to (2n − 1, 1) with one leg added to each side.
In other words, an + bn = cn−1 , so A( x ) + B( x ) = xC ( x ).
We can also prove a few more equalities about A( x ) and B( x ):
A( x ) = x (1 + C ( x ) B( x )),
(2)
B( x ) = xC ( x ) A( x ).
(3)
From the product rule, C ( x ) B( x ) is a series consisting of all paths from the
set counted by B( x ) appended to all paths from the set counted by C ( x ). In
practice, this encompasses all paths which end in an even number of steps down.
If we add 1x0 to that (for paths strictly in B with an empty Catalan path on
the left) and multiply the result by x, we have added a leg to both sides, and
what formerly ended in an even number of steps now ends in an odd number of
steps. These new paths do not touch the x-axis between their endpoints, making
them members of A( x ). In the same way, placing C ( x ) and A( x ) together and
multiplying by x gives us all of the paths counted by B( x ).
Combining (2) and (3) gives:
A( x ) =
3.3
x
1−
x 2 C ( x )2
.
(4)
The Catalan Generating Function
Let C ( x ) be the generating function for the Catalan numbers. Starting from the
recursive formula of the Catalan numbers and using the product formula, we
find that:
!
∞
∑
n =1
cn x n =
∞
n −1
xn
∑ ∑ c k c n −1− k
n =1
C(x) − 1 = x
C(x) − 1 = x
k =0
∞
n −1
n =1
k =0
∞
m
m =0
k =0
2
∑ ∑ c k c n −1− k
∑ ∑ ck cm−k
C ( x ) = 1 + xC ( x ) .
Now, by moving terms around, we see:
!
!
x n −1
xm
Macalester Journal of Catalan Numbers
90
1 = C ( x ) − xC ( x )2
1 = C ( x ) 1 − xC ( x )2
1
1 − xC ( x )
x
xC ( x ) =
1 − xC ( x )
C(x) =
x
1 − xC ( x )
x
1 = 1 + xC ( x ) −
1 − xC ( x )
1 − xC ( x )
x
1 = (1 + xC ( x ))
−
1 − xC ( x ) 1 − xC ( x )
x
(1 + xC ( x )) (1 − xC ( x ))
1=
−
1 − xC ( x )
1 − xC ( x )
2
2
x
1 − x C(x)
−
1=
1 − xC ( x )
1 − xC ( x )
2
2
1 − x C(x) − x
1=
1 − xC ( x )
1 − xC ( x )
1=
.
1 − x 2 C ( x )2 − x
0 = xC ( x ) −
3.4
(5)
Proving that C ( x ) = Z ( x )
Now, we’ll use the relationship among A( x ), C ( x ), and Z ( x ) to prove C ( x ) =
Z ( x ). By (1) and (4), we have:
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Macalester Journal of Catalan Numbers
1 + xZ ( x ) =
1
=
1 − A( x )
1−
1
x
1− x 2 C ( x )2
1 − x 2 C ( x )2
=
1 − x 2 C ( x )2 − x
1 − xC ( x )
= (1 + xC ( x ))
1 − x 2 C ( x )2 − x
= 1 + xC ( x ),
where the last equality follows from (5). Therefore, Z ( x ) = C ( x ) and |ωn | = zn =
cn for n ≥ 0. We have now proven that our variation of the Dyck paths are in fact
enumerated by the Catalan numbers.
Macalester Journal of Catalan Numbers
92