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THE CATALAN NUMBERS
AND THEIR APPLICATIONS
1, 2, 5, 14, 42, 132, …
Christen Mirando
Western New England University
IN THIS PRESENTATION…
 History
of the Catalan Numbers
 Examples/Applications






Euler’s triangulation of convex polygons
Catalan’s parenthesization problem
Mountain ranges on a graph
Hankel matrix
Relationship to computer science
Pascal’s Triangle
 Formulas



Recursive
Explicit
Generating Function
THE FIBONACCI SEQUENCE
0, 1, 1, 2, 3, 5, 8, 13…
… is a set of numbers with which most people are
familiar. These numbers appear in several areas of
mathematics, in nature and in many other places. One
set of numbers people may not be as familiar with is
the set of Catalan numbers. This set of numbers, like
the Fibonacci numbers, has many unexpected
occurrences and beautiful relationships in
mathematics.
THE FIRST FEW CATALAN NUMBERS
n
Cn
0
1
1
1
2
2
3
5
4
14
5
42
6
132
7
429
8
1,430
9
4,862
10
16,796
11
58,786
12
208,012
HISTORY OF THE CATALAN NUMBERS

Eugene Charles Catalan (1814-1894) was a Belgian mathematician
who “discovered” the Catalan Numbers in 1838 while studying wellformed sequences of parentheses.


Although this set of numbers is named after him, he was not the first to
discover it.
Leonhard Euler (1707-1783) discovered this set of numbers while
studying triangulations of convex polygons.
Published a recursive formula in 1761.
 Worked with Hungarian mathematician, Johann von Segner (1704-1777) ,
to develop a second order recurrence relation.


Gabriel Lame (1795-1870), was a French mathematician who used
Euler’s recursive formula to find a explicit formula in 1838.

Although Euler and Catalan are given credit for this sequence, there
is speculation that a Chinese mathematician Antu Ming (1692-1763)
discovered them in the 1730’s through geometric models.
EULER’S TRIANGULATION OF A CONVEX N-GON

Let Tn be the number of ways a convex n-gon can be divided into n-2
triangles by drawing n-3 nonintersecting diagonals where n ≥ 3.
When n = 4, Tn = 2
When n = 5, Tn = 5
When n = 6, Tn = 14
EULER’S RECURSIVE FORMULA
Euler let Tn be the number of triangulations of a convex n-gon
Tn = 2610…(4n – 10)
(n – 1)!
n≥3
T3 = 2 = 1, T4 = 26 = 2, and T5 = 2610 = 5
2!
3!
4!
These look like the Catalan numbers, except the subscript of Tn is shifted two places to the
right, so Cn = Tn+2. Substituting n + 2 into Tn yields:
Cn = 2610…(4n – 2)
(n + 1)!
Cn =
4n – 2
n+1

2610…(4n – 6)
n!
Cn = 4n – 2  Tn+1
n+1
= 4n – 2
n+1
 Cn – 1
CATALAN’S PARENTHESIZATION PROBLEM
 Suppose
we are given n pairs of left and right parentheses.
Find the number of correctly parenthesized sequences, Pn,
that can be formed.
Correct: () ()
Incorrect: ())(
()
When n = 1, Pn = 1
()(), (())
When n = 2, Pn = 2
()()(), (())(), ()(()), (()()), ((()))
When n = 3, Pn = 5
(((()))), ((()())), ((())()), (()(())),
(()()()), (())(()), ()()()(), ()(())(),
()()(()), ()(()()), ()((())), (())()(),
(()())(), ((()))()
When n = 4, Pn = 14
SINCE |PN| = |TN| = CN, A BIJECTION
MUST EXIST BETWEEN TN AND PN

In 1961, H.G. Forder found a solution to this problem. He let each
diagonal spanning adjacent sides be labeled with the parenthesized
concatenation of the labels of the sides.
c
c
b
d
c
b
d
b
d
(ab)
a
e
(ab)
a
e
c
b
a
a
e
c
((ab)c)
(ab)
(de)
d
(de)
b
((ab)c)
(ab)
e
a
d
(de)
e
((ab)c)(de))
Since n = 6, there are 14 different ways to triangulate a hexagon
CLASSIC PROBLEM – MOUNTAIN RANGES
Find the number of mountain ranges that can be drawn with n
upstrokes and n downstrokes. That is, find the number of different
paths, Mn, we can choose from the origin to the lattice point (2n, 0).
 Can touch the x-axis but cannot cross it.
 From the point (x, y) we can climb up to the point (x+1, y +1) or
climb down to the point (x+1, y −1).


n = 0, Mn = 1
n = 1, Mn = 1
n = 2, Mn = 2
n = 3, Mn = 5
For n = 4, there are 14 possible
mountain ranges.
HANKEL MATRIX
is a nxn matrix whose (i, j) entry is the Catalan number Ci+j-2.
These matrices always have determinant 1.
(1,1) (1,2)
(2,1) (2,2)
C(1+1-2) C(1+2-2)
C(2+1-2) C(2+2-2)
C0 C1
C1 C2
1 1
1 2
(1)(2) – (1)(1) = 2 – 1 = 1
Now for a 3x3 matrix:
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)
C(1+1-2) C(1+2-2) C(1+3-2)
C(2+1-2) C(2+2-2) C(2+3-2)
C(3+1-2) C(3+2-2) C(3+3-2)
C0 C1 C2
C1 C2 C3
C2 C3 C4
1 1 2
1 2 5
2 5 14
Det = [(2)(14) – (5)(5)] – [(1)(14) – (2)(5)] + 2[(1)(5) – (2)(2)]
= [28 – 25 – 14 + 10 + 10 - 8]
=1
APPLICATION TO COMPUTER SCIENCE
 Stacks,
Pushing, Popping
Using a stack, determine the number of different ways of the ordered
list of integers, 1, 2, 3, ..., n can be pushed onto an empty stack and then
popped resulting in an empty stack.
For n = 1, there is only one way to push/pop data.
For n = 2, there are two ways to push/pop data.
Output: 12
empty stack
1
push 1
pop 1
2
push 2
pop 2
Output: 21
1
empty stack
push 1
2
1
push 2
1
pop 2
pop 1
When n = 3, there are 6 permutations, but only 5
satisfy the stack structure.
1
empty
stack
push 1
2
pop 1
push 2
3
pop 2
push 3
pop 3
output:
123
pop 2
output:
132
pop 3
output:
213
pop 1
output:
231
3
1
empty
stack
push 1
pop 1
2
2
2
push 2
push 3
pop 3
2
empty
stack
1
1
1
push 1
push 2
pop 2
3
pop 1
push 3
3
empty
stack
2
2
2
1
1
1
1
1
push 1
push 2
push 3
pop 3
pop 2
3
empty
stack
2
2
2
1
1
1
1
1
push 1
push 2
push 3
pop 3
pop 2
pop 1
Notice that the permutation, 312, cannot be made. It would have
to start with the steps shown below, but there is no way to pop 1
before popping 2.
3
empty
stack
2
2
2
1
1
1
1
push 1
push 2
push 3
pop 3
output:
321
EXPLICIT FORMULA FROM EULER’S
RECURSIVE FORMULA
Cn =
4n – 2
n+1
 Cn – 1
= (4n – 2)(4n – 6) 
Cn – 2
(n + 1)n
= (4n – 2)(4n – 6)(4n-10) 
Cn – 3
(n + 1)n(n – 1)
This continues on until the last term is C0
= (4n – 2)(4n – 6)(4n-10)…62
(n + 1)n(n – 1)… 32
 C0
2 is factored out from each term in the numerator , and the denominator
becomes (n + 1)! Also, remember C0 = 1
= (2n – 1)(2n – 3)(2n – 5)… 31
 2n
(n+1)!
In order to make the numerator (2n)!, terms are inserted in red
= (2n)(2n – 1)(2n - 2)(2n – 3)(2n - 4)(2n – 5)… 321
 2n
(2n)(2n -2)(2n - 4) … (2) (n+1)!
Note, what is inserted in red, (2n)(2n – 2)(2n-4)…2 = 2n n! by factoring out a 2 from
all n terms. We can substitute this in the denominator.
= (2n)!
 2n
2n n! (n+1)!
= (2n)!
n! (n+1)!
=
(2n)!
(n + 1)! n!
Note:
=
(2n)!
(n + 1) n! n!
2n = (2n)!
n
n! (2n – n)!
=
1
2n
(n + 1) n
= (2n)!
n! n!
This is the explicit formula
from Euler’s recursive
formula.
PASCAL'S TRIANGLE AND THE CATALAN
NUMBERS

Explicit formula for the Catalan numbers; Cn = 1  2n
n+1
n
row 0
1
1
1
1
1
1
1
1
8
7
15
Example C4:
56
1
6
10
21
28
3
4
5
6
1
3
1
1
2
4
1
10
20
35
70
1
8
(4+1) 4
5
15
35
1
6
21
56
1
7
28
= 1
5
1
8
1
(70) = 14
ANOTHER EXAMPLE, CN = 1  2N
N
2n
n -1
N
-1
is the term directly to the right/left of the CBC .
Now, we divide this term by row number n.
1
1
1
1
1
1
1
1
1
7
8
Example C4:
6
15
1
4
10
20
35
56
1
3
10
21
28
2
4
6
1
3
5
row 0
5
15
35
70
1
1
6
21
56
1
7
28
1
8
1
1  8 = 1 (56) = 14
4 3
4
GENERATING FUNCTION FOR THE
CATALAN NUMBERS

We have given a recursive and an explicit formula for finding the nth
Catalan number, Cn. We will now find the generating function for the
Catalan numbers.

Segner’s Recursive Formula:

Cn = (C0, C1, C2,…, Cn-1)  (Cn-1 ,Cn-1,…,C0 )
= C0Cn-1 + C1Cn-2 +…+ Cn-1C0

Let C(x) = C0 + C1x + C2x2 + C2x2 + … + Cnxn

[C(x)]2 = (C0 + C1x + C2x2 + C2x2 + … + Cnxn) (C0 + C1 + C2x2 + C2x2 + … + Cnxn)
= (C0C0) + (C0C1 + C1C0) x + (C0C2 + C1C1 + C2C0) x2 +…..
= (11) + (11 + 11) x + (12 + 11 + 21) x2 +…..
= 1 + 2x + 5 x2 +…..
= C1x0 + C2x1 + C3x2 +…. + Cn+1xn
C(x) 2 = C1 + C2x + C3x2 +…. + Cn+1xn
C(x) 2 = C1x + C2x2 + C3x3 …. + Cnxn
x
C(x) 2 = (C0 + C1x + C2x2 + C3x3 …. + Cnxn) – C0
x
C(x) 2 = C(x) – C0
x
xC(x) 2 = C(x) – C0
xC(x) 2 - C(x) + 1 = 0
Using the quadratic formula, a = x, b = -1, c = 1
C(x) = 1 +/- (-1)2 – 4(x)(1)
2x
C(x) = 1 - 1 – 4(x)
2x
IF ANY OF THESE PROBLEMS APPEAR:

Number of binary trees with n vertices

Number of full binary trees with n vertices

Number of n multisets {a1, a2, …, an} of elements ai such that a1 + a2 + … + an = 0

Number of ways n semicircles can be arranged such that no two intersect

Number of permutations for the set {1, 2, …, 2n} where f is an involution that
has no fixed points and satisfies the noncrossing condition

Number of ways to arrange 2n soldiers in two rows in ascending height order

Number of tilings of the staircase shape that can be made with n rectangles

Number of paths a rook can take from the upper left-hand corner to the lower
right-hand corner on an nxn chessboard without crossing the main diagonal

Number of ideals in a ring is Cn+1

Number of nilpotent ideals in a ring

….. Many more!
… THE SOLUTION IS THE CATALAN NUMBERS!
ANY QUESTIONS?
REFERENCES








Guy, R.K. Letters to the Editor. Mathematics Magazine 61 (Oct. 1988): 269.
Larcombe, P. “The 18th Century Chinese Discovery of the Catalan Numbers.”
Mathematical Spectrum 32 (1999-2000): 5-6.
Jarvis, F. “Catalan Numbers.” Mathematical Spectrum 36 (2003-2004): 9-12.
Koshy, Thomas. “The Ubiquitous Catalan Numbers.” Mathematics Teacher 100.3
(2006): 184-88.
Gardner, M. “Mathematical Games.” Scientific American 234 (June 1976): 120-125.
Grimaldi, Ralph P. Fibonacci and Catalan Numbers: An Introduction. Hoboken, NJ:
John Wiley & Sons, 2012. Print.
Koshy, Thomas. Catalan Numbers with Applications. Oxford: Oxford UP, 2009.
Print.
Brualdi, Richard A. Introductory Combinatorics. New York: North-Holland, 1977.
Print.