arithmetic properties of weighted catalan numbers Jason Chen Mentor: Dmitry Kubrak May 20, 2017 MIT PRIMES Conference
background: catalan numbers Definition The Catalan numbers are the sequence of integers C n = 1 ( ) 2n. n + 1 n The first few values are: C 0 = 1 C 3 = 5 C 6 = 132 C 1 = 1 C 4 = 14 C 7 = 429 C 2 = 2 C 5 = 42 C 8 = 1430. 1
background: catalan numbers C n is the number of full triangulations of an (n + 2)-gon. Example C 3 = 5 2
background: catalan numbers C n is the number of ways to pair n sets of brackets. Example C 3 = 5 ((())) (()()) (())() ()(()) ()()() 3
background: catalan numbers Definition A Dyck path of length 2n is a continuous broken line lying in the first quadrant of the plane, starting at the origin (0, 0) and consisting of n up-steps in the direction 1, 1 and n down-steps in the direction 1, 1. Example length 6: length 8: 4
catalan numbers: dyck paths C n is the number of Dyck paths of length 2n. Example C 3 = 5 5
catalan numbers: recurrence The Catalan numbers are known to satisfy the recurrence n C n+1 = C k C n k, C 0 = 1. k=0 6
catalan numbers: recurrence The Catalan numbers are known to satisfy the recurrence n C n+1 = C k C n k, C 0 = 1. k=0 From the recurrence, we get the equation C(x) = x C(x) 2 + 1, where C(x) is the generating function of the Catalan numbers: C 0 + C 1 x + C 2 x 2 + C 3 x 3 +. 6
catalan numbers: generating function Solving C(x) = x C(x) 2 + 1 gives C(x) = 1 1 4x. 2x 7
catalan numbers: generating function Solving C(x) = x C(x) 2 + 1 gives C(x) = 1 1 4x. 2x Another expression for C(x) is the continued fraction: C(x) = 1. x 1 x 1 1 x 1... 7
weighted catalan numbers: dyck paths Suppose we have a sequence of integers b = (b 0, b 1, b 2, b 3,... ) and a Dyck path P, as shown 8
weighted catalan numbers: dyck paths Suppose we have a sequence of integers b = (b 0, b 1, b 2, b 3,... ) and a Dyck path P, as shown b 3 b 2 b 1 b 0 wt(p) = b 2 0 b2 1 b 2 8
weighted catalan numbers Definition The nth weighted Catalan number C b n is the sum of weights over all Dyck paths P of length 2n. Cn b = wt(p). paths P 9
weighted catalan numbers The first few values in terms of b i are: C0 b = 1 C1 b = b 0 C2 b = b2 0 + b 0b 1 C3 b = b3 0 + 2b2 0 b 1 + b 0 b 2 1 + b 0 b 1 b 2 C4 b = b4 0 + 3b3 0 b 1 + 3b 2 0 b2 1 + b 0 b 3 1 + 2b 2 0 b 1b 2 + 2b 0 b 2 1b 2 + b 0 b 1 b 2 2 + b 0b 1 b 2 b 3 C5 b = b 5 0 + 4b4 0 b 1 + 6b 3 0 b2 1 + 3b 3 0 b 1b 2 + 4b 2 0 b3 1 + 2b 2 0 b 1b 2 2 + + b 0 b 1 b 3 2 + 3b 0b 2 1b 2 2 + b 0b 1 b 2 b 2 3 + 3b 0b 3 1b 2 + 2b 0 b 1 b 2 2 b 3 + 2b 0 b 2 1b 2 b 3 + b 0 b 1 b 2 b 3 b 4. 10
weighted catalan numbers: specific cases Example The sequence b = (1, 1, 1,... ) gives the Catalan numbers. 11
weighted catalan numbers: specific cases Example The sequence b = (1, 1, 1,... ) gives the Catalan numbers. Example For an integer a, b = (a, a, a,... ) gives the sequence Cn b = a n C n. 11
weighted catalan numbers: specific cases Example The sequence b = (1, 1, 1,... ) gives the Catalan numbers. Example For an integer a, b = (a, a, a,... ) gives the sequence Cn b = a n C n. Example The sequence b = (1, 2, 3, 4,... ) gives C b n = (2n 1)!!. Similarly, b = (1, 1, 2, 2, 3, 3, 4,... ) gives C b n = n!. 11
weighted catalan numbers: specific cases Example For an integer q, b = (q 0, q 1, q 2, q 3,... ) gives the q-catalan numbers. 12
weighted catalan numbers: specific cases Example For an integer q, b = (q 0, q 1, q 2, q 3,... ) gives the q-catalan numbers. Macdonald polynomials Geometry of Hilbert schemes Harmonic analysis Representation theory Mathematical physics Algebraic combinatorics 12
history of weighted catalan numbers Theorem (A. Postnikov, 2000) The number of plane Morse links of order n are the weighted Catalan numbers C b n for the sequence b = (1 2, 3 2, 5 2, 7 2,... ). This sequence is commonly denoted L n. 13
history of weighed catalan numbers Theorem (Kummer) If ν 2 (n) denotes the largest power of 2 dividing n, and s 2 (n) denotes the sum of the binary digits of n, then ν 2 (C n ) = s 2 (n + 1) 1. 14
history of weighed catalan numbers Definition Let be the difference operator, acting on functions f : Z 0 Z by ( f)(x) = f(x + 1) f(x). Theorem (A. Postnikov, 2006) If the sequence b satisfies: b(0) is odd, and 2 n+1 ( n b)(x) for all n 1 and x Z 0, then ν 2 (C b n) = ν 2 (C n ) = s 2 (n + 1) 1. 15
current research: recursion Let S be the shift operator: S : (a 0, a 1, a 2, a 3,... ) (a 1, a 2, a 3,... ). 16
current research: recursion Let S be the shift operator: S : (a 0, a 1, a 2, a 3,... ) (a 1, a 2, a 3,... ). Then we have the recurrence: C b n+1 = b 0 n k=0 C S(b) k C b n k. C S(b) k C b n k 16
current research: generating function Recall the generating function for C n 1 C(x) =. x 1 1 x 1... For Cn, b C b 1 (x) =. b 0 x 1 b 1 x 1 1 b 2x 1... 17
current research We want to study C b n modulo primes p, or more generally, p n. Since C b n is a polynomial in b i, it suffices to consider the residues of b i modulo p. Theorem Let b = (b 0, b 1, b 2,... ) and p be prime. Then C b n is eventually periodic modulo p iff any of b i are congruent to 0 mod p. This is because if any of the b i are 0, then C b (x) is rational, i.e. C b (x) = p(x) q(x) for some polynomials p(x), q(x). 18
current research: results If p = 2, then we can describe the period: Theorem Consider a sequence b F N 2 such that b k = 0, and b i = 1 for i < k. Then C b (x) = p k+1(x) p k+2 (x), where with p 0 (x) = 0 and p 1 (x) = 1. p k+2 (x) = p k+1 (x) + x p k (x), The period of the sequence C b n is equal to the minimal number m such that p k+2 (x) divides x m 1. 19
current research: results It turns out that p 2 k(x) = 1 for all k. In particular, we get: Corollary Let k 1 be a natural number. Consider a sequence b F N 2 such that b 2 k 2 = 0, and b i = 1 for i < 2 k 2. Then C b (x) = p 2 k 1 (x). In particular, it is a polynomial of degree 2 k 1 1 and so the sequence C b n is identically 0 for n 2 k 1. 20
direction of future research What is the period of L n modulo p n? 21
direction of future research What is the period of L n modulo p n? 21
direction of future research What is the period of L n modulo p n? Conjecture (A. Postnikov) The period of L n modulo 3 k+3 is exactly 2 3 k. 21
direction of future research What is the period of L n modulo p n? Conjecture (A. Postnikov) The period of L n modulo 3 k+3 is exactly 2 3 k. What is the period of Cn b modulo p n? 21
direction of future research What is the period of L n modulo p n? Conjecture (A. Postnikov) The period of L n modulo 3 k+3 is exactly 2 3 k. What is the period of Cn b modulo p n? Can we classify sequences for primes greater than 2? 21
acknowledgments MIT PRIMES 22
acknowledgments MIT PRIMES Professor Alexander Postnikov 22
acknowledgments MIT PRIMES Professor Alexander Postnikov Dmitry Kubrak 22
acknowledgments MIT PRIMES Professor Alexander Postnikov Dmitry Kubrak My parents 22
Questions? 22