1. The Catalan Numbers: Sequence of Integer
Pai Sukanya Suksak
December 16, 2010
1. Introduction
Catalan Numbers: Sequence of Integer
Figure 1 The central binomial coefficients of Pascal’s Triangle
As shown in figure 1, the middle numbers in Pascal’s triangle are following;
1, 2, 6, 20, 70,…
They can be divided by
1, 2, 3, 4, 5,…
then, we obtain the sequence of following numbers
1, 1, 2, 5, 14,…
⎛ 2n ⎞
Since each middle number is derided from the binomial coefficient, ⎜ ⎟ , then we
⎝ n⎠
divide each number by (n + 1). These numbers are called Catalan numbers. The formula
of Catalan numbers will be presented in Theorem 1.[1]
2. General Knowledge of the Catalan Numbers
Catalan numbers, which appear in many counting problem such as monotonic paths,
Dyck words, dividing polygons, and noncrossing partitions, form a sequence of a
sequence of natural numbers.
Historical Information
Catalan Numbers were previously discussed by the Chinese mathematician Antu
Ming in 1730, but his work was not well-known in the Western world because it was
introduced only in China. Then, in 1751, the Swiss mathematician Leonhard Euler found

2. Catalan numbers while he was studying the triangulations of convex polygons. Until
1838, the Belgian mathematician Eugene Charles Catalan discovered Catalan numbers
while he was studying the sequence of parentheses. Even though Catalan numbers were
discovered long time ago, they were named just after Eugene Charles Catalan introduced
them to the Western world.[2]
In fact, there are various ways to define the Catalan numbers, we are going to
introduce one of them. Kreher and Stinson (1956) define the Catalan numbers in term of
totally balanced sequence. [3]
Definition 1. Let n be a positive integer, and let a = [a1, a2, …, a2n-1, a2n] ∈ (Ζ2)2n. The
sequence is said to be totally balanced sequence if the following two properties are
satisfied:
a) a contains n 0s and n 1s, and
b) for any I, 1 ≤ I ≤ 2n, it hold that
| {j : 1 ≤ j ≤ I, ai = 0}| ≥ | {j : 1 ≤ j ≤ I, ai = 1} |.
Now, we are ready to define the Catalan numbers using the defintion of the totally
balance sequence.
Definition. Let Cn denote the set of all totally balanced sequences in (Ζ2)2n. Cn is referred
to a Catalan family of order n. The Catalan number Cn is defined to be
Cn = |Cn|.
By the definition of the Catalan numbers above, the first fourth terms of the
Catalan families Cn and the Catalan numbers Cn in Table 1.
Table 1 shows the Catalan families Cn and the Catalan numbers Cn for 1 ≤ n ≤ 4.
N Cn Cn
1 01 1
2 0011 0101 2
3 000111 001011 001101 010011 010101 3
4 00001111 00010111 00011011 00011101 00100111 4
00101011 00101101 00110011 00110101 01000111
01001011 01001101 01010011 01010101
(Kreher and Stinson, 1956)
In the first theorem, we are going to state a formula of the Catalan number that is
commonly used.
3. Formula of the Catalan Number and Its Proof
Theorem. For any integer n ≥ 1, the Catalan number Cn is given in term of binomial
coefficients by

3. 1 ⎛ 2n ⎞ (2n)!
Cn = ⎜ n ⎟ = (n + 1)!n! , for n ≥ 1.
n + 1⎝ ⎠
(a)
For instance, the first tenth Catalan numbers are C1 = 1, C2 = 2, C3 = 5, C4 =
14, C5 = 42, C6 = 132, C7 = 429, C8 = 1430, C9 = 4862 and C10 = 16796.
Proof. To derive the formula, we are going to prove by using the triangulation
definition. We are going to apply Euler’s Triangulation problem. Consider the ways of
triangulating n-gon, where 3 ≤ n ≤ 5.[2]
Figure 2 Triangulations of an n-gon, where 3 ≤ n ≤ 5.
One way of triangulating a triangle Two ways of triangulating a square
http://www.toulouse.ca/EdgeGuarding/MobileGuards.html
Five ways of triangulating a pentagon
Euler used an inductive argument to establish the following formula.
2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 10)
An = , n ≥ 3.
( n − 1)!
However, we need to extend the formula of Euler that was published in 1761 to
include the case n = 0, 1, and 2 because the original formula makes sense only for n ≥ 3.
Then we have
2 ⋅ 6 ⋅10 ⋅ ...⋅ (4k + 2)
Ak+3 = , k ≥ 0.
( k + 2 )!
Then, let Cn = Ak+2. Thus,
2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 2)
Cn = , n ≥ 1.
( n + 1)!

4. The expression of Cn can be rewritten as
(4n − 2) ⋅ 2 ⋅ 6 ⋅10 ⋅ ...⋅ (4n − 6)
Cn =
( n + 1) n!
(4n − 2) ⋅ Cn −1
=
( n + 1)
When n = 1, we will have C0 = C1 = 1. Thus, we can define C0 = 1 and Cn recursively as
following;
(4n − 2) ⋅ Cn −1
C0 = 1 and Cn =
( n + 1)
To show the recursive formula implying the formula (a) of the Catalan number, it can be
processed algebraically.
(4n − 2) ⋅ Cn −1
Cn =
( n + 1)
(4n − 2) ⋅ (4n − 6) ⋅ Cn − 2
=
( n + 1) n
(4n − 2) ⋅ (4n − 6) ⋅ ( 4n − 10 ) ⋅ Cn − 3
=
( n + 1) n(n − 1)
(4n − 2) ⋅ (4n − 6) ⋅ ( 4n − 10 ) ⋅... ⋅ 6 ⋅ 2 ⋅ C0
=
( n + 1) n(n − 1) ⋅ ⋅ ⋅ 3 ⋅ 2
⎡(2n − 1) ⋅ (2n − 3) ⋅ ( 2n − 5 ) ⋅... ⋅ 3 ⋅1⎤ ⋅ 2 n
= ⎣ ⎦
( n + 1)!
2 n (2n)!
=
2 n n!( n + 1)!
(2n)!
=
n!( n + 1)!
1 ⎛ 2n ⎞
= 
n + 1⎜ n ⎟
⎝ ⎠
Thus, we obtain the formula (a) of the Catalan number by using the recursion
relation as we wish.

5. 4. Properties of the Catalan numbers
4.1 Parity of Catalan Numbers
Definition. Let r be the remainder, then the parity of an integer is defined as its attribute
of being even or odd, which means 1) let b = 2 and a be an integer, if the remainder, r,
such that r = 0, the integer a is called even and has the form a = 2q; 2) if the remainder r,
such that r = 1, the integer a is called odd and has the form a = 2q + 1.
Two given integers, s and t, have the same parity, if they are both even, or both
odd; but if one of them is even, and the other is odd, then s and t are said to be of different
parity. [4]
Table 2 The first 18 Catalan Numbers (Koshy and Salmassi, 2006)
n 0 1 2 3 4 5 6 7 8
Cn 1 1 2 5 14 42 132 429 1430
n 9 10 11 12 13 14 15 16 17
Cn 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790
As shown in Table 2, for n ≤ 17, the Catalan numbers, Cn, are odd when n = 0, 1,
3, 7, and 15. All of those numbers that have the same form which is 2m – 1 when m > 0
(i.e. the 2nd and 3rd Catalan numbers) are prime. The numbers of the form 2m – 1 where m
is an integer are known as Mersenne numbers.[4]
In order to prove the next theorem, we are first going to introduce one of the
generalized Catalan numbers formulas. [5]
Lemma. For any p ≥ 2, the generalized Catalan number Cp (n) , such that
1 ⎛ pn ⎞
C p (n) = , can be defined recursively by
( p − 1)(n + 1) ⎜ n ⎟
⎝ ⎠
( p −1)n +1
p
⎛ ( p − 1)(n − k) + 1⎞
Cp (0) = 1, and C p (n) = ∑ (−1)k −1 ⎜
⎝ k ⎟ C p (n − k) ,
⎠
n ≥ 1.
k =1
In particular, p = 2 corresponds to the usual Catalan numbers C2(n) satisfy the
linear recursion. i.e.,
n +1
2
⎛ (n − k) + 1⎞
C2 (n) = ∑ (−1)k −1 ⎜ ⎟ C2 (n − k) , n ≥ 1.
k =1 ⎝ k ⎠
pk − 1
Theorem. The prime p | Cp(n) if and only if n ≠ for all integers k ≥ 0. In
p −1
particular, C2(n) is odd if and only if n is a Mersenne number 2k – 1 where some integers
k. [5]

6. pk − 1
Proof. We are going to prove that if the prime p | Cp(n) then n ≠ .
p −1
Let Cp(n) be the generalized Catalan number, for any p ≥ 2, i.e.
( p −1)n +1
p
⎛ ( p − 1)(n − k) + 1⎞
C p (n) = ∑ (−1)k −1 ⎜
⎝ k ⎟ C p (n − k) .
⎠
k =1
To prove the statement, we shall apply induction on n. For the base step, the result holds
for n = 1, since Cp(1) = 1. Then, we assume n > 1 and that the result holds for all number
m, m < n. Let p r ≤ n ≤ p r +1 − 1 . We will consider the right-hand side of Cp(n). We will let
it called modulo p. Then, we apply the induction hypothesis that for m < n, we will have
Cp(m) is a multiple of p if (p – 1) m +1 is not a power of p.
pN − 1
For the term of modulo p, the summation of the term n – k is the form . However,
p −1
since p r ≤ n ≤ p r +1 , the only non-zero term modulo p, is the one corresponding to the
p r +1 − 1
index k for which (p – 1)(n – k) = pr – 1 if n ≤ (respectively,
p −1
p r +1 − 1
( p − 1)(n − k) = p r +1 − 1 if n > ). This term is, to within sign,
p −1
⎛ pr ⎞
⎟ C ⎛ p − 1 ⎞ if n ≤ p − 1 (respectively, ( p − 1)(n − k) = p r +1 − 1 if
r r +1
⎜
⎜ n − p − 1⎟ p ⎜ p − 1 ⎟
r
⎜ ⎝ ⎠ p −1
⎝ p −1 ⎟
⎠
p r +1 − 1 ⎛ pr ⎞
n> ). As the binomial coefficient ⎜ ⎟ is a multiple of p if and only if 0 < s <
p −1 ⎝s ⎠
pr − 1
pr, the above term is a multiple of p if and only if 0 < n − < p r if
p −1
p r +1 − 1 p r +1 − 1
n≤ (respectively, p r +1 < ( p − 1)n + 1 < p r + 2 if n > ). This is equivalent to
p −1 p −1
p r +1 − 1
p r < ( p − 1)n + 1 < p r +1 if n ≤ (respectively, p r +1 < ( p − 1)n + 1 < p r + 2 if
p −1
p r +1 − 1
n> ). This implies that (p – 1)n + 1 is not a power of p. Therefore, we can
p −1
pk − 1
conclude that if the prime p | Cp(n) then n ≠ . 
p −1

7. 5. The Generating Function of Catalan Numbers
The linear recursion of generalized Catalan numbers, Cp(x), p ≥ 2, have been given in the
previous section. In this section, we will provide a proof of the generating function of the
Catalan number
Theorem. Let C be the generating function of the second-ordered Catalan number, and
1 − 1 − 4x
Cn be the nth Catalan number, C(x) can be expressed as following; C(x) = .[6]
2x
Proof. We are going to show that the generating function of the Catalan number, C,
1 − 1 − 4x
satisfies the equation, C(x) = .
2x
Based on the evident recurrence relation, we can express the nth Catalan number Cn as
shown,
Ck = ∑ Ci C j , k ≥ 1, and C0 = 1.
i + j = k −1
A sequence of numbers, a0, a1, a2,…, elements of the sequence can be formed the power
series The generating function for the sequence is shown below.
∞
C(x) = ∑ Ck x k .
k=0
= C0 + C1x + C2x2 + C3x3 +…
then the recurrence relation expression shows that C satisfies
xC2(x) – C(x) +1 = 0, C(0) = 1,
so that C can be solved using Quadratic formula, i.e.,
1 ± 1 − 4x
C(x) =
2x
In fact, we must choose the minus sign to obtain the positive coefficients of the
power of x in the generating function of C(x) because all Catalan numbers are always
positive. Moreover,
Thus, the generating function of the Catalan number is
1 − 1 − 4x
C(x) = (b).
2x
Therefore, we have proved the formula of the generating function of the second-ordered
Catalan numbers as we wish. [6]
We generate the second-ordered Catalan numbers, since they are the usual one that we
mainly discuss in this paper.

8. 5. Interpretations of the Catalan Numbers
There are connections among the combinatorial interpretations which are found in
the triangulation of n-gon with n+2 sides, the construction of the binary tree-diagram
with n+1 leaves, Dyck words with 2n length, and the monotonic paths with n × n square
cells.[7]
A1) The Catalan Numbers and Triangulation of Polygon. The Catalan number Cn is
the number of different ways a polygon with n + 2 sides can be partition into triangle by
drawing nonintersecting diagonals. For example, the following pentagon as shown in
figure 3 demonstrates the case n = 3. The 3rd Catalan number is 5, so there are 5 different
ways to triangulate the pentagon. [7]
Figure 3. Five different partitions of a convex pentagon into triangles. [7]
Figure 4. The construction of the tree-diagram with n+1 leaves, corresponding to the
partitions. [7]
Figure 5. The labeling of the branches of the tree-diagrams. [7]
Figure 6. The codes with length 2n, derived from the labeled tree-diagrams. [7]

9. Figure 7. The lattices paths which is called the monotonic path are obtained from the
codes. [7]
A2) The Catalan Numbers and Tree Diagram. There are various applications in metric
trees. For example, the Catalan number, Cn, is the number of full binary trees with n + 1
leaves when a full binary tree is defined as a tree whose every vertex has either two
branches (leaves) or no branch (leaf). [8]
For instance, if we let a full binary tree with 3 + 1 = 4 leaves, then n = 3. The 3rd
Catalan number, C3, is 5. Thus, there are five patterns of full binary trees with 4 leaves as
shown in the Figure 5.
A3) The Catalan Numbers and the Dyck Words. There is an application of the Catalan
number in Computer Science. In fact, the Catalan number, Cn, is the number of Dyck
words of the length 2n. [8] Dyck word is defined as “a string consisting of n X’s and n Y’s
such that no initial segment of the string has more Y’s than X’s”.
Then we construct monotonic path, which is a set of steps begining at the lower
left corner, finish at the upper right corner of the grid with n×n square cells. Beside, each
step stays below the diagonal, y = x. According to the codes. let R(right) stand for X and
L(left) stand for Y. As been seen in Figure 6, given 3 × 3 square cells. Thus, there are
five different monotonic paths along the edge of the given grid. Thus, the Catalan
number, Cn, is the number of different monotonic paths that can occur in n x n grid
square. In fact, monotonic path is similar to the ballot problem that we are going to
discuss.
A4) The Catalan Numbers and the Ballot Problem (Hilton and Pederson).
The ballot problem was introduced in the late nineteen century to determine the
probability of counting of votes when there are two candidates A, and B such that A
receives a votes, and B receives b votes with a > b. Originally, the problem was first
solved by Bertrand, but it became famous in 1887 after the French mathematician Desire
Andre gave the solution for the problem. We will introduce the term of p-good before we
are continuing the discussion.
Definition. A path from P to Q is p-good if it lies entirely below the line y = (p – 1)x.
Thus, a bad path is defined as the complementary set of a set of good paths. [6]
According to the linear recursion of generalized Catalan numbers, the usual
Catalan numbers are defined when p = 2, so we will call 2-good paths for this
interpretation.
The method of solving the ballot problem is to count the number of 2-good paths
from (c,d) to (a,b), where (c,d), (a,b) are any two lattice points below the line y = x. [6]

10. Theorem. Given the formula of the kth Catalan number, Ck, as following
1 ⎛ 2k ⎞
C0 = 1, and Ck = , k ≥ 1.
k ⎜ k − 1⎟
⎝ ⎠
The kth Catalan numbers, Ck, is an expression of the number of good path from (0,-1) to
(k, k-1), i.e.,
⎛ 2k ⎞ ⎛ 2k ⎞ 1 ⎛ 2k ⎞
⎜ k ⎟ − ⎜ k + 1⎟ = k ⎜ k − 1⎟ .
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Proof. Let us assume that both good paths and bad paths exist. As shown in Figure 7., we
have
d < c ≤ b < a. (Note that this condition is satisfied the ballot problem that the loser
receives at least 1 vote.). Let ℘ be bad path, PF, FQ be subpaths from P to Q, and ℘1 be
the path obtained from ℘1 by reflecting with the line y = x.
Figure
7
The total number of paths from P(c,d) to Q(a,b) can be expressed as the binomial
coefficient.
⎛ (a + b) − (c + d)⎞
⎜
⎝ a−c ⎟;
⎠
As we have seen from Figure 7, ℘ = ℘1℘2 . Then, if ℘ = ℘1 ℘2 , where ℘ is a path from
P(d, c) to Q(a,b) . Assume the rule ℘  ℘ be injective correspondence between the set of
bad paths from P to Q and the set of paths from P to Q. Thus, the number of bad paths from
⎛ (a + b) − (c + d)⎞
P to Q is ⎜ ⎟ , and hence we have the number of good paths from P to Q as
⎝ a−d ⎠
shown below
⎛ (a + b) − (c + d)⎞ ⎛ (a + b) − (c + d)⎞
⎜
⎝ a−c ⎟ −⎜
⎠ ⎝ a−d ⎟.
⎠

11. In particular, if the starting point of the path is (1,0) to ( a,b ) , the number of good paths is
given as
⎛ a + b − 1⎞ ⎛ a + b − 1⎞ a − b ⎛ a + b ⎞
⎜ a −1 ⎟ − ⎜
⎝ ⎠ ⎝ ⎠
=
a ⎟ a+b⎜ a ⎟ ⎝ ⎠
(1).
a−b
Thus, is the probability that a path from (0,0) to ( a,b ) proceeds first to (1,0), then
a+b
continues as a good path to ( a,b ) . According to the solution of the ballot problem, if we let
the path starting from (0, -1) to ( k, k − 1) , we can rewrite (1) as
⎛ 2k ⎞ ⎛ 2k ⎞ 1 ⎛ 2k ⎞
⎜ k ⎟ − ⎜ k + 1⎟ = k ⎜ k − 1⎟ .
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Here, we obtain the formula of the kth Catalan number, Ck, as we wish.
1 ⎛ 2k ⎞
C0 = 1, and Ck = , k ≥ 1. 
k ⎜ k − 1⎟
⎝ ⎠
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[3]
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[4]
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[5]
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[6]
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