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,…

Since each middle number is derided from the binomial coefficient, 2nn

⎛⎝⎜

⎞⎠⎟

, then we

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

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

00101011 00101101 00110011 00110101 01000111 01001011 01001101 01010011 01010101

4

(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

Five ways of triangulating a pentagon

Cn =1

n +12nn

⎛⎝⎜

⎞⎠⎟=

(2n)!(n +1)!n!

, for 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.

http://www.toulouse.ca/EdgeGuarding/MobileGuards.html

Euler used an inductive argument to establish the following formula.

An = 2 ⋅6 ⋅10 ⋅ ... ⋅ (4n −10)n −1( )! , n ≥ 3.

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

Ak+3 = 2 ⋅6 ⋅10 ⋅ ... ⋅ (4k + 2)k + 2( )! , k ≥ 0.

Then, let Cn = Ak+2. Thus,

Cn = 2 ⋅6 ⋅10 ⋅ ... ⋅ (4n − 2)n +1( )! , n ≥ 1.

One way of triangulating a triangle Two ways of triangulating a square

The expression of Cn can be rewritten as

Cn = (4n − 2) ⋅2 ⋅6 ⋅10 ⋅ ... ⋅ (4n − 6)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;

C0 = 1 and Cn = (4n − 2) ⋅Cn−1

n +1( )

To show the recursive formula implying the formula (a) of the Catalan number, it can be processed algebraically.

Cn = (4n − 2) ⋅Cn−1

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( )!

= 2n (2n)!2n n! n +1( )!

= (2n)!n! n +1( )!

= 1n +1

2nn

⎛⎝⎜

⎞⎠⎟

Thus, we obtain the formula (a) of the Catalan number by using the recursion relation as we wish.

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

Cp (n) =1

(p −1)(n +1)pnn

⎛⎝⎜

⎞⎠⎟

, can be defined recursively by

Cp (0) = 1, and Cp (n) = (−1)k−1(p −1)(n − k) +1

k⎛⎝⎜

⎞⎠⎟Cp (n − k)

k=1

( p−1)n+1p

∑ , n ≥ 1.

In particular, p = 2 corresponds to the usual Catalan numbers C2(n) satisfy the linear recursion. i.e.,

C2 (n) = (−1)k−1(n − k) +1

k⎛⎝⎜

⎞⎠⎟C2 (n − k)

k=1

n+12

∑ , n ≥ 1.

Theorem. The prime p | Cp(n) if and only if n ≠pk −1p −1

for all integers k ≥ 0. In

particular, C2(n) is odd if and only if n is a Mersenne number 2k – 1 where some integers k. [5]

Proof. We are going to prove that if the prime p | Cp(n) then n ≠pk −1p −1

.

Let Cp(n) be the generalized Catalan number, for any p ≥ 2, i.e.

Cp (n) = (−1)k−1(p −1)(n − k) +1

k⎛⎝⎜

⎞⎠⎟Cp (n − k)

k=1

( p−1)n+1p

∑ .

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 pr ≤ n ≤ pr+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.

For the term of modulo p, the summation of the term n – k is the form pN −1p −1

. However,

since pr ≤ n ≤ pr+1 , the only non-zero term modulo p, is the one corresponding to the

index k for which (p – 1)(n – k) = pr – 1 if n ≤pr+1 −1p −1

(respectively,

(p −1)(n − k) = pr+1 −1 if n >pr+1 −1p −1

). This term is, to within sign,

pr

n − pr −1p −1

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟Cp

pr −1p −1

⎛⎝⎜

⎞⎠⎟

if n ≤pr+1 −1p −1

(respectively, (p −1)(n − k) = pr+1 −1 if

n > pr+1 −1p −1

). As the binomial coefficient pr

s⎛⎝⎜

⎞⎠⎟

is a multiple of p if and only if 0 < s <

pr, the above term is a multiple of p if and only if 0 < n − pr −1p −1

< pr if

n ≤ pr+1 −1p −1

(respectively, pr+1 < (p −1)n +1 < pr+2 if n >pr+1 −1p −1

). This is equivalent to

pr < (p −1)n +1 < pr+1 if n ≤pr+1 −1p −1

(respectively, pr+1 < (p −1)n +1 < pr+2 if

n > pr+1 −1p −1

). This implies that (p – 1)n + 1 is not a power of p. Therefore, we can

conclude that if the prime p | Cp(n) then n ≠pk −1p −1

.

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

Cn be the nth Catalan number, C(x) can be expressed as following; C(x) = 1− 1− 4x2x

.[6]

Proof. We are going to show that the generating function of the Catalan number, C,

satisfies the equation, C(x) = 1− 1− 4x2x

.

Based on the evident recurrence relation, we can express the nth Catalan number Cn as shown,

Ck = CiCji+ j= k−1∑ , k ≥ 1, and C0 = 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) = Ckxk

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.,

C(x) = 1± 1− 4x2x

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

C(x) = 1− 1− 4x2x

(b).

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.

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]

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]

Theorem. Given the formula of the kth Catalan number, Ck, as following

C0 = 1, and Ck = 1k2kk −1

⎛⎝⎜

⎞⎠⎟

, 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.,

2kk

⎛⎝⎜

⎞⎠⎟−2kk +1

⎛⎝⎜

⎞⎠⎟=1k2kk −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.

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

P to Q is (a + b) − (c + d)

a − d⎛⎝⎜

⎞⎠⎟

, and hence we have the number of good paths from P to Q as

shown below (a + b) − (c + d)

a − c⎛⎝⎜

⎞⎠⎟−(a + b) − (c + d)

a − d⎛⎝⎜

⎞⎠⎟

.

Figure  7  

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 −1a −1

⎛⎝⎜

⎞⎠⎟−

a + b −1a

⎛⎝⎜

⎞⎠⎟=a − ba + b

a + ba

⎛⎝⎜

⎞⎠⎟

(1).

Thus, a − ba + b

is the probability that a path from (0,0) to a,b( ) proceeds first to (1,0), then

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

2kk

⎛⎝⎜

⎞⎠⎟−

2kk +1

⎛⎝⎜

⎞⎠⎟=1k

2kk −1

⎛⎝⎜

⎞⎠⎟

.

Here, we obtain the formula of the kth Catalan number, Ck, as we wish.

C0 = 1, and Ck = 1k2kk −1

⎛⎝⎜

⎞⎠⎟

, k ≥ 1.

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GENERATION, ENUMERATION, ANS SEARCH”. CRC Press, Florida. [4] Koshy, T, & Salmassi, M. (2006). “Parity and Primality of Catalan Numbers”. The

College Mathematics Journal 37, 1, pp. 52-53. [5] Sury, B. (2009). “Generalized Catalan Numbers: Linear Recursion and Divisibility”,

Journal of Integer Sequences 12. [6] Hilton and J. Pedersen. (1991). “Catalan numbers, their generalization and their uses”,

Math. Intelligencer 13, pp. 64–75. [7] Cofman, Judita (08/01/1997). "Catalan Numbers for the Classroom?". Elemente der

Mathematik (0013-6018), 52 (3), p. 108. [8] Stanley, R. (1944). “ENUMERATIVE COMBENATORICS”. Wadsworth &

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