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“Catalan Numbers and Pascal’s Triangle”

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## “Catalan Numbers and Pascal’s Triangle”

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**“Catalan Numbers and Pascal’s Triangle”**Jim Olsen and Allison McGannWestern Illinois University http://www.wiu.edu/users/mfjro1/wiu/ IMACC Robert Allerton Park, IL March 27, 2010**Catalan Numbers are:**1, 1, 2, 5, 14, 42, 132, 429, 1430, … Notation: C1 = 1 C2 = 2 C3 = 5 C4 = 14 Also, C0 = 1**Pascal’s Triangle**1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 The nth entry in row r is (n and r start with 0) Example:**Catalan Numbers in Pascal’s Triangle**by subtraction 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 2 4 6 15 20 Characterization #18 of Pascal’s Triangle The difference of the middle number in an even rowand its neighbor is a Catalan Number.**Catalan Numbers in Pascal’s Triangle**by division 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 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.**Formulas**Problems Difference Quotient Product Recursive Summation**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**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**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**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**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**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**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**Formulas**Problems 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**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**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 Recursive n pair of parentheses Summation Trivalent trees n+1 branches**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.**Induction Step**is a formula for the number of triangulations of an h+2-gon, for all h<k. Inductive hypothesis: Assume is a formula for the number of triangulations of a k+2-gon Prove**Consider a k+2-gon.**Here are a couple of triangulations.**…**P2 Pi P1 Label two consecutive vertices X and Y. This leaves k additional points. We label these points P0, P1, P2,… Pk-1. Consider a k+2-gon. P0 X Y Every triangulation of the polygon will include a triangle with the points X, Y, and Pi.**…**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.**…**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**For example, k=6**Octagon, i=3 Pi =P3 P2 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 X Y (Only one triangulation on each side is shown.) 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).**…**P2 We let the point Pi go from P0 to Pk-1. We see that The number of triangulations of a k+2-gon is Pi P1 P0 X Y 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.)**Formulas**Problems 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**C**A B C D E B D E A**C**A B C D E B D E A**C**A B C D E B D (CD) E A**C**A B C D E B D (CD) (B(CD)) E A**C**A B C D E B D (CD) (B(CD)) (A(B(CD))) E A (A(B(CD)))E)**C**A B C D E B D (CD) (B(CD)) (A(B(CD))) E A (A(B(CD)))E)**C**A B C D E B D E A**C**A B C D E B D E A**C**A B C D E B D (AB) E A**C**A B C D E B D (DE) (AB) E A**C**A B C D E B D (DE) (AB) ((AB)C) E A (((AB)C)(DE))**C**A B C D E B D (DE) (AB) ((AB)C) E A (((AB)C)(DE))**Formulas**Problems 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**Formulas**Problems 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**Formulas**Problems solves Difference 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**For further**Investigation: 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**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**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**Thank You**Jim Olsen and Allison McGannWestern Illinois University http://www.wiu.edu/users/mfjro1/wiu/ JR-Olsen@wiu.edu