“Catalan Numbers and Pascal’s Triangle”

“Catalan Numbers and Pascal’s Triangle”Jim Olsen and Allison McGann

Western Illinois University

http://www.wiu.edu/users/mfjro1/wiu/

IMACCRobert Allerton Park, IL

March 27, 2010

Catalan Numbers are:

1, 2, 5, 14, 42, 132, 429, 1430, …

Notation:C1 = 1C2 = 2C3 = 5C4 = 14Also, C0 = 1

1,

Pascal’s Triangle 11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

The nth entry in row r is or r n

rC

n

(n and r start with 0)Example:

6=20

3

Catalan Numbers in Pascal’s Triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

by subtraction

1 21 2 1 1C

4 6 2 6 4 2C

15 203 20 15 5C

Characterization #18 of Pascal’s TriangleThe difference of the middle number in an even row

and its neighbor is a Catalan Number.

Catalan Numbers in Pascal’s Triangle

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

by division

21

1 2 12

C

62

1 6 23

C

20 31 20 54

C

Characterization #19 of Pascal’s TriangleThe middle number in an even row, divided by half

the row number plus 1, is a Catalan Number.

ProblemsFormulasDifference

Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

3

6 620 15 5

3 2C

3

61 1 20 534 4

C

32 6 10 52 3 4

C

3 210 52 54 2

C C

3 0 2 1 1 2 0

1 2 1 1 2 1 5C C C C C C C

n +1’s and n –1’sProblems

Find the number of ways we can order 2n numbers from a list made up of n +1’s and n -1’s such that the sum (from the beginning) at any point is 0.

Example: n = 31, 1, 1, -1, -1, -11, 1, -1, -1, 1, -11, 1, -1, 1, -1, -11, -1, 1, 1, -1, -11, -1, 1, -1, 1, -1

5 ways C3 = 5

n +1’s and n –1’sProblemsFind the number of ways a rook in

the lower left corner of an (n+1) by (n+1) chess board can get to the upper right corner, without ever crossing the diagonal.Example: n = 3

5 ways C3 = 5

Rook movesn+1 by n+1 board

n +1’s and n –1’s


Triangulationsn+2 - gon

ProblemsFind the number of ways to triangulate an (n+2)-gon

Example: n = 3

5 ways C3 = 5




n pair ofparentheses

ProblemsFind the number of ways to put n sets of parentheses in a list of (n+1) letters (to indicate the n multiplications).

Example: n = 3(((ab)c)d)((ab)(cd))(((a(bc))d)(a((bc)d))(a(b(cd)))

5 ways C3 = 5





ProblemsThe number of trivalent rooted trees with n trivalent vertices (it has n+1 branches).

Example: n = 3

5 ways C3 = 5

A B C D A B C D

A B C D

A B C D

A B C D

Trivalent trees n+1 branches





ProblemsFormulas Difference

Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

…But WHY?Now we will show the connections

•Why are the formulas equivalent?•Why are the problems equivalent?•Why do the formulas solve the problems?

First, a few easy ones.







Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

equivalent

Left as an exercise (in simplifying factorials).







Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

equivalent

“obvious”







Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

equivalent

+1 corresponds to a rook move to the right.

–1 corresponds to a rook move up.








Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

solves

Prove the Summation formula solves the Triangulations of an n+2-gon Problem

Prove:

is a formula for the

number of triangulations of an n+2-gonLet 0 1C

Proof by induction on n.n=1: 1 1

1 1 1 0 00

1 1 1i ii

C C C C C

This is a formula for the number of triangulations of a 3-gon, because there is 1 triangulation of a triangle.

1

10

n

n i n ii

C C C

Induction StepInductive hypothesis:Assume is a formula for the

number of triangulations of an h+2-gon, for

all h<k.

is a formula for the

number of triangulations of a k+2-gon

Prove

1

10

h

h i h ii

C C C

1

10

k

k i k ii

C C C

Consider a k+2-gon.

Here are a couple of triangulations.

Consider a k+2-gon.

X Y

Label two consecutive vertices X and Y.This leaves k additional points. We label these points P0, P1, P2,… Pk-1.

Every triangulation of the polygon will include a triangle with the points X, Y, and Pi.

P0

P1

P2

Pi

…

X Y

Every triangulation of the polygon will include a triangle with the points X, Y, and Pi.

P0

P1

P2

Pi

To form an entire triangulation of the polygon containing triangle XYPi, we need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle.

…

X Y

P0

P1

P2

Pi

We need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle.The number of points tothe right and to the left isless than k+2, so we know howmany triangulations there are to the right and to the left. We will multiply these, by the fundamental counting principle.

In fact, the number of triangulations to right is Ci and the number of triangulations to left is Ck-1-i

…

X Y

P0

P1

P2Pi =P3

We need to triangulate the 5 points to the right of triangle XYP3, and triangulate the 4 points to the left of the triangle.

The number of triangulations to right is C3=5. The number of triangulations to left is C2=2. So, there are 25=10 triangulations which contain this particular triangle (XYP3).

For example, k=6Octagon, i=3

(Only one triangulation on each side is shown.)

We let the point Pi go from P0 to Pk-1. We see that

The number of triangulations of a k+2-gon is

Therefore, by the principle of strong induction,

is a formula for the number of triangulations of an n+2-gon.

1

10

k

k i k ii

C C C

1

10

n

n i n ii

C C C

X Y

P0

P1

P2

Pi

…

(The formula is valid for all natural numbers.)






Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

equiv

alent


Triangulation Expression Tree

ABCED


ABCEDA

B

C

D

E

A B C D E


AB(CD)EA

B

C

D

E

A B C D E


A(B(CD))EA

B

C

D

E

A B C D E

(CD)


(A(B(CD)))EA

B

C

D

E

A B C D E

(CD)

(B(CD))


((A(B(CD)))E)A

B

C

D

E

A B C D E

(CD)

(B(CD))

(A(B(CD)))

(A(B(CD)))E)


((A(B(CD)))E)A

B

C

D

E

A B C D E

(CD)

(B(CD))

(A(B(CD)))

(A(B(CD)))E)


ABCED

>> Second Example <<


ABCED


ABCEDA

B

C

D

E

A B C D E


(AB)CDEA

B

C

D

E

A B C D E


(AB)C(DE)A

B

C

D

E

A B C D E

(AB)


((AB)C)(DE)A

B

C

D

E

A B C D E

(AB) (DE)


(((AB)C)(DE))A

B

C

D

E

A B C D E

(AB) (DE)

((AB)

C)

(((AB)C)(DE))


(((AB)C)(DE))A

B

C

D

E

A B C D E

(AB) (DE)

((AB)

C)

(((AB)C)(DE))






Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

equivalent

See Handout(Math Induction)






Planted trivalentTrees n branches


Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

solves

solves

See Rook Moves ProofsHandout





Planted trivalentTrees n branches


Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

solves

solves

See Handout on the n +1’s and n -1’s problem







Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

Solves

For furtherInvestigation:

Interesting proof (can be found online). Involves building new triangulations from old.






Quotient

Product

Recursive

Summation

2 21n

n nC

n n

211n

nC

nn

14 2

1n nnC Cn

1

4 21

n

ni

iCi

1

10

n

n i n ii

C C C

Equiv

alent

For furtherInvestigation:

Pretty easy –See Martin Gardner article.


References• Mathematical Games: Catalan numbers: an integer sequence that

materializes in unexpected places. By Martin Gardner. Scientific American. June 1976, Vol 234, No. 6. pp. 120-125.

• Wikipedia.com – Catalan Numbers• Catalan Numbers with Applications. Book by Thomas Koshy.

Oxford Press. 2009. 422 pages.• Enumerative Combinatorics (http://math.mit.edu/~rstan/ec/ ) two-

volume book and website by Richard Stanley. Includes 66 combinatorial interpretations of Catalan numbers!

• The On-Line Encyclopedia of Integer Sequences (by AT&T) - http://www.research.att.com/~njas/sequences/index.html Type in 1, 2, 5, 14, 42

http://math.mit.edu/~rstan/ec/

http://www.research.att.com/~njas/sequences/index.html

Thank You

Jim Olsen and Allison McGannWestern Illinois University

http://www.wiu.edu/users/mfjro1/wiu/[email protected]

http://www.wiu.edu/users/mfjro1/wiu/

mailto:[email protected]

Documents

“Catalan Numbers and Pascal’s Triangle”