1. “Catalan Numbers and Pascal’s Triangle”
Jim Olsen and Allison McGann Western Illinois University
IMACC
Robert Allerton Park, IL
March 27, 2010

2. Catalan Numbers are: 1,
1, 2, 5, 14, 42, 132, 429, 1430, … Notation: C1 = 1 C2 = 2 C3 = 5 C4 = 14 Also, C0 = 1

3. Pascal’s Triangle 1
1 1
The nth entry in row r is
(n and r start with 0)
Example:

4. Catalan Numbers in Pascal’s Triangle
by subtraction 1
1 1
1 2
4 6
15 20
Characterization #18 of Pascal’s Triangle
The difference of the middle number in an even row and its neighbor is a Catalan Number.

5. Catalan Numbers in Pascal’s Triangle
by division 1
1 1 2 6 20
Characterization #19 of Pascal’s Triangle
The middle number in an even row, divided by half the row number plus 1, is a Catalan Number.

6. Formulas
Problems
Difference
Quotient
Product
Recursive
Summation

7. Problems
n +1’s and n –1’s
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 = 3
1, 1, 1, -1, -1, -1
1, 1, -1, -1, 1, -1
1, 1, -1, 1, -1, -1
1, -1, 1, 1, -1, -1
1, -1, 1, -1, 1, -1
5 ways  C3 = 5

8. Find 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
Problems
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
5 ways  C3 = 5

9. Problems
n +1’s and n –1’s
Find the number of ways to triangulate an (n+2)-gon
Example: n = 3
5 ways  C3 = 5
Rook moves
n+1 by n+1 board
Triangulations
n+2 - gon

10. Problems
Find 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
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Triangulations
n+2 - gon
n pair of
parentheses

11. Problems
The number of trivalent rooted trees with n trivalent vertices (it has n+1 branches).
Example: n = 3
5 ways  C3 = 5
n +1’s and n –1’s
Rook moves
n+1 by n+1 board A B C D A B C D A B C D
Triangulations
n+2 - gon
n pair of
parentheses A B C D A B C D
Trivalent trees
n+1 branches

12. Now we will show the connections
Formulas
Problems
Difference
n +1’s and n –1’s
…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.
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

13. Left as an exercise (in simplifying factorials).
Formulas
Problems
Difference
n +1’s and n –1’s
equivalent
Rook moves
n+1 by n+1 board
Quotient
Left as an exercise (in simplifying factorials).
Triangulations
n+2 - gon
Product
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

14. equivalent “obvious” Formulas Problems n +1’s and n –1’s Rook moves
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
equivalent
Recursive
n pair of
parentheses
“obvious”
Summation
Trivalent trees
n+1 branches

15. –1 corresponds to a rook move up.
Formulas
Problems
Difference
n +1’s and n –1’s
equivalent
Rook moves
n+1 by n+1 board
Quotient
+1 corresponds to a rook move to the right.
–1 corresponds to a rook move up.
Triangulations
n+2 - gon
Product
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

16. solves Formulas Problems n +1’s and n –1’s Rook moves n+1 by n+1 board
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
solves
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

17. Prove the Summation formula solves the Triangulations of an n+2-gon Problem
is a formula for the number of triangulations of an n+2-gon
Prove:
Let
Proof by induction on n.
n=1:
This is a formula for the number of triangulations of a 3-gon, because there is 1 triangulation of a triangle.

18. Induction Step Inductive 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

19. Consider a k+2-gon.
Here are a couple of triangulations.

20. …
Consider a k+2-gon. P2 Pi P1
Label two consecutive vertices X and Y.
This leaves k additional points. We label these points P0, P1, P2,… Pk-1. P0 X Y
Every triangulation of the polygon will include a triangle with the points X, Y, and Pi.

21. … P2
Every triangulation of the polygon will include a triangle with the points X, Y, and Pi. Pi P1 P0 X Y
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.

22. … P2
We need to triangulate the points to the right of the triangle, and triangulate the points to the left of the triangle. Pi P1 P0
The number of points to
the right and to the left is
less than k+2, so we know how
many triangulations there are to the right and to the left. We will multiply these, by the fundamental counting principle. X Y
In fact, the number of triangulations to right is Ci and the number of triangulations to left is Ck-1-i

23. Pi =P3 P2 P1 P0 X Y For example, k=6 Octagon, i=3
We need to triangulate the 5 points to the right of triangle XYP3, and triangulate the 4 points to the left of the triangle. P1 P0
(Only one triangulation on each side is shown.) X Y
The number of triangulations to right is C3=5. The number of triangulations to left is C2=2. So, there are 25=10 triangulations which contain this particular triangle (XYP3).

24. We let the point Pi go from P0 to Pk-1. We see that
The number of triangulations of a k+2-gon is X Y P0 P1 P2 Pi …
Therefore, by the principle of strong induction,
is a formula for the number of triangulations of an n+2-gon.
(The formula is valid for all natural numbers.)

25. equivalent Formulas Problems n +1’s and n –1’s Rook moves
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
equivalent
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

26. Triangulation
Expression
Tree
ABCED

27. Triangulation
Expression
Tree
ABCED C A B C D E B D E A

28. Triangulation
Expression
Tree
AB(CD)E C A B C D E B D E A

29. Triangulation
Expression
Tree
A(B(CD))E C A B C D E B D
(CD) E A

30. (A(B(CD)))E Triangulation Expression Tree C A B C D E B D (CD) (B(CD))

31. ((A(B(CD)))E) Triangulation Expression Tree C A B C D E B D (CD)

32. ((A(B(CD)))E) Triangulation Expression Tree C A B C D E B D (CD)

33. >> Second Example <<
Triangulation
Expression
Tree
ABCED
>> Second Example <<

34. Triangulation
Expression
Tree
ABCED

35. Triangulation
Expression
Tree
ABCED C A B C D E B D E A

36. Triangulation
Expression
Tree
(AB)CDE C A B C D E B D E A

37. Triangulation
Expression
Tree
(AB)C(DE) C A B C D E B D
(AB) E A

38. ((AB)C)(DE) Triangulation Expression Tree C A B C D E B D (DE) (AB) E

39. (((AB)C)(DE)) Triangulation Expression Tree C A B C D E B D (DE) (AB)

40. (((AB)C)(DE)) Triangulation Expression Tree C A B C D E B D (DE) (AB)

41. equivalent Formulas Problems n +1’s and n –1’s Rook moves
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Quotient
equivalent
Triangulations
n+2 - gon
Product
See Handout
(Math Induction)
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

42. solves solves Formulas Problems n +1’s and n –1’s Rook moves
Difference
n +1’s and n –1’s
solves
Rook moves
n+1 by n+1 board
Quotient
solves
Triangulations
n+2 - gon
Product
See
Rook Moves Proofs
Handout
Recursive
n pair of
parentheses
Summation
Planted trivalent
Trees n branches

43. solves solves Formulas Problems n +1’s and n –1’s Rook moves
Difference
solves
n +1’s and n –1’s
solves
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
See
Handout on the n +1’s and n -1’s problem
Recursive
n pair of
parentheses
Summation
Planted trivalent
Trees n branches

44. For further Investigation: Solves
Formulas
Problems
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Quotient
Triangulations
n+2 - gon
Product
Solves
Interesting proof (can be found online). Involves building new triangulations from old.
Recursive
n pair of
parentheses
Summation
Trivalent trees
n+1 branches

45. See Martin Gardner article.
For further
Investigation:
Formulas
Problems
Difference
n +1’s and n –1’s
Rook moves
n+1 by n+1 board
Equivalent
Quotient
Triangulations
n+2 - gon
Product
Recursive
n pair of
parentheses
Pretty easy –
See Martin Gardner article.
Summation
Trivalent trees
n+1 branches

46. 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
Wikipedia.com – Catalan Numbers
Catalan Numbers with Applications. Book by Thomas Koshy. Oxford Press pages.
Enumerative Combinatorics ( ) 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) Type in 1, 2, 5, 14, 42

47. Jim Olsen and Allison McGann Western Illinois University
Thank You
Jim Olsen and Allison McGann Western Illinois University