Catalan Numbers Richard P. Stanley Catalan Numbers – p. 1 An OEIS entry OEIS: Online Encylopedia of Integer Sequences (Neil Sloane). See http://oeis.org. A database of over 270,000 sequences of integers. Catalan Numbers – p. 2 An OEIS entry OEIS: Online Encylopedia of Integer Sequences (Neil Sloane). See http://oeis.org. A database of over 270,000 sequences of integers. A000108: 1, 1, 2, 5, 14, 42, 132, 429,... C0 = 1, C1 = 2, C2 = 3, C3 = 5, C4 = 14,... Cn is a Catalan number. Catalan Numbers – p. 2 An OEIS entry OEIS: Online Encylopedia of Integer Sequences (Neil Sloane). See http://oeis.org. A database of over 270,000 sequences of integers. A000108: 1, 1, 2, 5, 14, 42, 132, 429,... C0 = 1, C1 = 2, C2 = 3, C3 = 5, C4 = 14,... Cn is a Catalan number. COMMENTS. This is probably the longest entry in OEIS, and rightly so. Catalan Numbers – p. 2 Catalan monograph R. Stanley, Catalan Numbers, Cambridge University Press, 2015. Catalan Numbers – p. 3 Catalan monograph R. Stanley, Catalan Numbers, Cambridge University Press, 2015. Includes 214 combinatorial interpretations of Cn and 68 additional problems. Catalan Numbers – p. 3 History Sharabiin Myangat, also known as Minggatu, Ming’antu ( ), and Jing An (c. 1692–c. 1763): a Mongolian astronomer, mathematician, and topographic scientist who worked at the Qing court in China. Catalan Numbers – p. 4 History Sharabiin Myangat, also known as Minggatu, Ming’antu ( ), and Jing An (c. 1692–c. 1763): a Mongolian astronomer, mathematician, and topographic scientist who worked at the Qing court in China. Typical result (1730’s): ∞ Cn 1 2n+1 sin(2α) = 2sin α − sin α − 4n 1 n=1 − X Catalan Numbers – p. 4 History Sharabiin Myangat, also known as Minggatu, Ming’antu ( ), and Jing An (c. 1692–c. 1763): a Mongolian astronomer, mathematician, and topographic scientist who worked at the Qing court in China. Typical result (1730’s): ∞ Cn 1 2n+1 sin(2α) = 2sin α − sin α − 4n 1 n=1 − X No combinatorics, no further work in China. Catalan Numbers – p. 4 Manuscript of Ming Antu Catalan Numbers – p. 5 Manuscript of Ming Antu Catalan Numbers – p. 5 Manuscript of Ming Antu Catalan Numbers – p. 5 Manuscript of Ming Antu Catalan Numbers – p. 5 More history, via Igor Pak Euler (1751): conjectured formula for number Cn of triangulations of a convex (n + 2)-gon (definition of Catalan numbers). In other words, draw n 1 noncrossing diagonals of a convex polygon− with n + 2 sides. Catalan Numbers – p. 6 Completion of proof Goldbach and Segner (1758–1759): helped Euler complete the proof, in pieces. Lamé (1838): first self-contained, complete proof. Catalan Numbers – p. 7 Catalan Eugène Charles Catalan (1838): wrote Cn in (2n)! the form n!(n+1)! and showed it counted (nonassociative) bracketings (or parenthesizations) of a string of n + 1 letters. Catalan Numbers – p. 8 Catalan Eugène Charles Catalan (1838): wrote Cn in (2n)! the form n!(n+1)! and showed it counted (nonassociative) bracketings (or parenthesizations) of a string of n + 1 letters. Born in 1814 in Bruges (now in Belgium, then under Dutch rule). Studied in France and worked in France and Liège, Belgium. Died in Liège in 1894. Catalan Numbers – p. 8 Why “Catalan numbers”? John Riordan (1948): introduced the term “Catalan number” in Math Reviews. Catalan Numbers – p. 9 Why “Catalan numbers”? John Riordan (1948): introduced the term “Catalan number” in Math Reviews. Riordan (1964): used the term again in Math. Reviews. Catalan Numbers – p. 9 Why “Catalan numbers”? John Riordan (1948): introduced the term “Catalan number” in Math Reviews. Riordan (1964): used the term again in Math. Reviews. Riordan (1968): used the term in his book Combinatorial Identities. Finally caught on. Catalan Numbers – p. 9 Why “Catalan numbers”? John Riordan (1948): introduced the term “Catalan number” in Math Reviews. Riordan (1964): used the term again in Math. Reviews. Riordan (1968): used the term in his book Combinatorial Identities. Finally caught on. Martin Gardner (1976): used the term in his Mathematical Games column in Scientific American. Real popularity began. Catalan Numbers – p. 9 The primary recurrence n Cn+1 = CkCn k, C0 = 1 − Xk=0 Catalan Numbers – p. 10 The primary recurrence n Cn+1 = CkCn k, C0 = 1 − Xk=0 e Catalan Numbers – p. 10 Solving the recurrence n Cn+1 = CkCn k, C0 = 1 − Xk=0 n Let y = n 0 Cnx . ≥ MultiplyP recurrence by xn and sum on n 0. ≥ Catalan Numbers – p. 11 Solving the recurrence n Cn+1 = CkCn k, C0 = 1 − Xk=0 n Let y = n 0 Cnx . ≥ MultiplyP recurrence by xn and sum on n 0. ≥ n n n Cn+1x = CkCn k x − n 0 n 0 k=0 ! X≥ X≥ X Catalan Numbers – p. 11 A quadratic equation n n n Cn+1x = CkCn k x − n 0 n 0 k=0 ! X≥ X≥ X Catalan Numbers – p. 12 A quadratic equation n n n Cn+1x = CkCn k x − n 0 n 0 k=0 ! X≥ X≥ X n n Now x n 0 Cn+1x = n 1 Cnx = y 1. ≥ ≥ − P n P n Moreover, k=0 CkCn k is the coefficient of x in − n 2 2 n n 0 Cnx = y , since in general, k=0 akbn k P − is the≥ coefficient of xn in the product P n n P n 0 anx n 0 bnx . ≥ ≥ P P Catalan Numbers – p. 12 A quadratic equation n n n Cn+1x = CkCn k x − n 0 n 0 k=0 ! X≥ X≥ X n n Now x n 0 Cn+1x = n 1 Cnx = y 1. ≥ ≥ − P n P n Moreover, k=0 CkCn k is the coefficient of x in − n 2 2 n n 0 Cnx = y , since in general, k=0 akbn k P − is the≥ coefficient of xn in the product P n n P n 0 anx n 0 bnx . ≥ ≥ P yP1 − = y2 xy2 − y +1=0 ⇒ x ⇒ Catalan Numbers – p. 12 Solving the quadratic equation 1 √1 4x xy2 y +1=0 y = ± − − ⇒ 2x Which sign is correct? Catalan Numbers – p. 13 Solving the quadratic equation 1 √1 4x xy2 y +1=0 y = ± − − ⇒ 2x Which sign is correct? Well, in general (Taylor series) α un (1+u)α = un = α(α 1) (α n+1) . n − ··· − n! n 0 n 0 X≥ X≥ Catalan Numbers – p. 13 Solving the quadratic equation 1 √1 4x xy2 y +1=0 y = ± − − ⇒ 2x Which sign is correct? Well, in general (Taylor series) α un (1+u)α = un = α(α 1) (α n+1) . n − ··· − n! n 0 n 0 X≥ X≥ Let u = 4x, α = 1, to get − 2 √1 4x = 1 2x 2x2 + . − − − ··· Catalan Numbers – p. 13 Which sign? n 1 √1 4x Recall y = n 0 Cnx = ± 2x− . ≥ P Catalan Numbers – p. 14 Which sign? n 1 √1 4x Recall y = n 0 Cnx = ± 2x− . ≥ The plus signP gives 1 + (1 2x 2x2 + ) 1 − − ··· = 1 x + , 2x x − − ··· which makes no sense. The minus sign gives 1 (1 2x 2x2 + ) − − − ··· =1+ x + , 2x ··· which is correct. Catalan Numbers – p. 14 A formula for Cn We get 1 y = (1 √1 4x) 2x − − 1 1/2 = 1 ( 4x)n , 2x − n − n 0 ! X≥ 1 1 3 2n−3 1/2 2 ( 2 )( 2 ) ( 2 ) − − ··· − where n = n! . Catalan Numbers – p. 15 A formula for Cn We get 1 y = (1 √1 4x) 2x − − 1 1/2 = 1 ( 4x)n , 2x − n − n 0 ! X≥ 1 1 3 2n−3 1/2 2 ( 2 )( 2 ) ( 2 ) − − ··· − where n = n! . 1 2n n Simplifies to y = n 0 n+1 n x , so ≥ P1 2n (2n)! Cn = = n + 1 n n! (n + 1)! Catalan Numbers – p. 15 Other combinatorial interpretations Pn := triangulations of convex (n + 2)-gon { } # = C (where #S = number of elements of S) ⇒ Pn n We want other combinatorial interpretations of C , i.e., other sets for which C = # . n Sn n Sn Catalan Numbers – p. 16 Other combinatorial interpretations Pn := triangulations of convex (n + 2)-gon { } # = C (where #S = number of elements of S) ⇒ Pn n We want other combinatorial interpretations of C , i.e., other sets for which C = # . n Sn n Sn bijective proof: show that Cn = # n by giving a bijection S ϕ: Tn → Sn (or ), where we already know # = C . Sn →Tn Tn n Catalan Numbers – p. 16 Bijection Reminder: a bijection ϕ: S T is a function that is one-to-one and onto, that→ is, for every t T there is a unique s S for which ϕ(s)= t. ∈ ∈ Catalan Numbers – p. 17 Bijection Reminder: a bijection ϕ: S T is a function that is one-to-one and onto, that→ is, for every t T there is a unique s S for which ϕ(s)= t. ∈ ∈ If S, T are finite and ϕ: S T is a bijection, then #S = #T (the “best” way→ to prove #S = #T ). Catalan Numbers – p. 17 Binary trees 4. Binary trees with n vertices (each vertex has a left subtree and a right subtree, which may be empty) Catalan Numbers – p. 18 Binary trees 4. Binary trees with n vertices (each vertex has a left subtree and a right subtree, which may be empty) Catalan Numbers – p. 18 Binary trees 4. Binary trees with n vertices (each vertex has a left subtree and a right subtree, which may be empty) Catalan Numbers – p. 18 Bijection with triangulations Catalan Numbers – p. 19 Bijection with triangulations Catalan Numbers – p. 19 Bijection with triangulations Catalan Numbers – p. 19 Bijection with triangulations Catalan Numbers – p.