Binomial coefficients and k-regular sequences

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1 Binomial coefficients and k-regular sequences Eric Rowland Hofstra University New York Combinatorics Seminar CUNY Graduate Center, Eric Rowland Binomial coefficients and k-regular sequences / 27

2 Valuations of binomial coefficients Pascal s triangle: For this talk: p is a prime Let ν p (n) denote the exponent of the highest power of p dividing n. Example: ν 3 (18) = 2. Theorem (Kummer 1852) ν p ( ( n m) ) = number of carries involved in adding m to n m in base p. Eric Rowland Binomial coefficients and k-regular sequences / 27

3 Valuations of binomial coefficients 2-, 3-, 5-, and 7-adic valuations: Eric Rowland Binomial coefficients and k-regular sequences / 27

4 Odd binomial coefficients Main theme: Arithmetic information about binomial coefficients reflects the base-p representations of integers Glaisher (1899) counted odd binomial coefficients: 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16,... θ 2,0 (n) = 2 n 1 Definition θ p,α (n) := {m : 0 m n and ν p ( ( n m) ) = α}. n d := number of occurrences of d in the base-p representation of n. Eric Rowland Binomial coefficients and k-regular sequences / 27

5 Derivation from Kummer s theorem Glaisher s result θ 2,0 (n) = 2 n 1 follows from Kummer s theorem. Theorem (Kummer) ν p ( ( n m) ) = number of carries involved in adding m to n m in base p. Example n = 25. How many m satisfy ν 2 ( ( 25 m) ) = 0? θ 2,0 (25) = = 8. n = 25 = m = 00 2 Eric Rowland Binomial coefficients and k-regular sequences / 27

6 Binomial coefficients not divisible by p Number of binomial coefficients with 3-adic valuation 0: 1, 2, 3, 2, 4, 6, 3, 6, 9, 2, 4, 6, 4, 8, 12, 6,... θ 3,0 (n) = 2 n 1 3 n 2 Theorem (Fine 1947) Write n = n l n 1 n 0 in base p. Then θ p,0 (n) = (n 0 + 1) (n l + 1) = 1 n 0 2 n 1 3 n 2 p n p 1. Eric Rowland Binomial coefficients and k-regular sequences / 27

7 Prime powers? Carlitz found a recurrence involving θ p,α (n) and a secondary quantity ψ p,α (n) := {m : 0 m n and νp ((m + 1) ( n m) ) = α}. Theorem (Carlitz 1967) θ p,α (pn + d) = (d + 1)θ p,α (n) + (p d 1)ψ p,α 1 (n 1) { (d + 1)θ p,α (n) + (p d 1)ψ p,α 1 (n 1) if 0 d p 2 ψ p,α (pn + d) = pψ p,α 1 (n) if d = p 1. Is there a better formulation of this recurrence? Eric Rowland Binomial coefficients and k-regular sequences / 27

8 k-regular sequences

9 Constant-recursive sequences Fibonacci recurrence: Matrix form: F (n + 2) = F(n + 1) + F(n) [ ] F(n + 1) = F(n + 2) [ ] [ 0 1 F(n) 1 1 F(n + 1) Matrix product: F(n) = [ 1 0 ] [ ] n [ ] ] Characterizations of constant-recursive sequences over Q: s(n) is determined by a linear recurrence in s(n + i) (along with finitely many initial conditions) {s(n + i) n 0 : i 0} is finite-dimensional s(n) = u M n v for some matrix M and vectors u, v generating function n 0 s(n)x n is rational Eric Rowland Binomial coefficients and k-regular sequences / 27

10 k-regularity Definition Let k 2. A sequence s(n) n 0 is k-regular if the vector space generated by is finite-dimensional. {s(k e n + i) n 0 : e 0 and 0 i k e 1} Characterizations of k-regularity (Allouche & Shallit 1992): s(n) is determined by finitely many linear recurrences in s(k e n + i) (along with finitely many initial conditions) s(n) = u M(n 0 ) M(n 1 ) M(n l ) v for some M(d) and vectors u, v generating function in k non-commuting variables is rational Eric Rowland Binomial coefficients and k-regular sequences / 27

11 Examples of k-regular sequences ν p (n) ν p (F(n)) ν 2 (n) n 1 : 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4,... k-automatic sequences (e.g., the Thue Morse sequence 0, 1, 1, 0, 1, 0, 0, 1,... ) polynomial and quasi-polynomial sequences sums and products of k-regular sequences A k-regular sequence reflects the base-k representation of n, so many nested sequences are k-regular. How to guess a recurrence? Eric Rowland Binomial coefficients and k-regular sequences / 27

12 Guessing a constant-recursive sequence {s(n + i) n 0 : i 0} is finite-dimensional. s(n) = 2 n + n: s(n) : 1, 3, 6, 11, 20, 37,... basis element! s(n + 1) : 3, 6, 11, 20, 37, 70,... basis element! s(n + 2) : 6, 11, 20, 37, 70, 135,... basis element! s(n + 3) : 11, 20, 37, 70, 135, 264,... = 2s(n) 5s(n + 1) + 4s(n + 2) Matrix form: s(n + 1) s(n) s(n + 2) = s(n + 1) s(n + 3) s(n + 2) Eric Rowland Binomial coefficients and k-regular sequences / 27

13 Guessing a 2-regular sequence s(n) = θ 2,1 (n) = number of binomial coefficients ( ) n m with ν2 ( ( n m) ) = 1: s(n) : 0, 0, 1, 0, 1, 2, 2, 0,... basis element! s(2n + 0) : 0, 1, 1, 2, 1, 4, 2, 4,... basis element! s(2n + 1) : 0, 0, 2, 0, 2, 4, 4, 0,... = 2s(n) s(4n + 0) : 0, 1, 1, 2, 1, 4, 2, 4,... = s(2n) s(4n + 2) : 1, 2, 4, 4, 4, 8, 8, 8,... basis element! s(8n + 2) : 1, 4, 4, 8, 4, 12, 8, 16,... = 2s(n) + 2s(2n) + s(4n + 2) s(8n + 6) : 2, 4, 8, 8, 8, 16, 16, 16,... = 2s(4n + 2) Matrix form: s(2n) s(n) s(n) s(4n) = s(2n) = M(0) s(2n) s(8n + 2) s(4n + 2) s(4n + 2) s(2n + 1) s(n) s(n) s(4n + 2) = s(2n) = M(1) s(2n) s(8n + 6) s(4n + 2) s(4n + 2) Eric Rowland Binomial coefficients and k-regular sequences / 27

14 An implementation in Mathematica IntegerSequences is available from Eric Rowland Binomial coefficients and k-regular sequences / 27

15 Sequences of polynomials

16 Fibonacci numbers Combinatorial interpretation: F(n) = # compositions of n 1 using 1, 2. n = 5: Refinement: F (n, x) := compositions λ of n 1 using 1, 2 x λ 1 The coefficient of x α is the number of compositions with α 1s. n F (n, x) x 3 x n F (n, x) 4 x 3 + 2x 5 1x 4 + 3x x 5 + 4x 3 + 3x 7 x 6 + 5x 4 + 6x In particular, F(n, 1) = F(n). Eric Rowland Binomial coefficients and k-regular sequences / 27

17 Fibonacci polynomials Recurrence: F(n + 2, x) = x F (n + 1, x) + F(n, x) Matrix form: [ ] F(n + 1, x) = F(n + 2, x) [ ] [ ] 0 1 F(n, x) 1 x F(n + 1, x) Matrix product: F(n, x) = [ 1 0 ] [ ] n [ ] x 1 Eric Rowland Binomial coefficients and k-regular sequences / 27

18 Generating function Define n T p (n, x) := x νp(( nm)) = θ p,α (n)x α. m=0 α 0 θ p,α (n) is the number of binomial coefficients with p-adic valuation α. p = 2: n T 2 (n, x) x x 2 + x x x 2 + 2x n T 2 (n, x) 8 4x 3 + 2x 2 + x x 2 + 2x x 3 + x 2 + 4x x x 3 + 5x 2 + 2x x 2 + 4x x 3 + 2x 2 + 4x In particular, T p (n, 1) = n + 1. Eric Rowland Binomial coefficients and k-regular sequences / 27

19 Guessing matrices for T p (n, x) p = 2: p = 3: M 3 (0) = p = 5: M 5 (0) = General p: [ ] 0 1 M 2 (0) = 2x 2x + 1 [ ] 0 1 3x 3x + 1 [ ] 0 1 5x 5x + 1 M p (d) = [ M 3 (1) = [ 3 2 M 5 (1) = [ 5 4 dp p 1 (d 1)px + dp p M 2 (1) = x x ] ] [ ] x [ ] 3 0 M 3 (2) = 3x + 3 x [ ] 5 0 M 5 (4) = 15x + 5 x p 1 d p 1 (p d)x + p 1 d p 1 But this matrix isn t unique... There are many bases. Eric Rowland Binomial coefficients and k-regular sequences / 27 ].

20 Which basis is best? Can we get integer coefficients? Can we get non-negative integer coefficients? (allows a bijective proof) For each 2 2 invertible matrix S with integer entries j, compute S 1 M p (d)s. T p (n, x) = [ 1 0 ] M p (n 0 ) M p (n 1 ) M p (n l ) [ ] 1 1 Simplest matrix (maximizing monomial entries): [ ] d + 1 p d 1 d x (p d) x Eric Rowland Binomial coefficients and k-regular sequences / 27

21 Matrix product Let [ ] d + 1 p d 1 M p (d) :=. d x (p d) x Theorem (Rowland 2018) Write n = n l n 1 n 0 in base p. Then T p (n, x) := n x νp(( m)) n = [ 1 0 ] [ ] 1 M p (n 0 ) M p (n 1 ) M p (n l ). 0 m=0 Setting x = 0 gives θ p,0 (n) = (n 0 + 1) (n l + 1) as a special case: [ ] [ ] [ ] θp,0 (pn + d) d + 1 p d 1 θp,0 (n) =, or simply θ p,0 (pn + d) = (d + 1) θ p,0 (n). Eric Rowland Binomial coefficients and k-regular sequences / 27

22 Comparing recurrences Carlitz recurrence: θ p,α (pn + d) = (d + 1)θ p,α (n) + (p d 1)ψ p,α 1 (n 1) { (d + 1)θ p,α (n) + (p d 1)ψ p,α 1 (n 1) if 0 d p 2 ψ p,α (pn + d) = pψ p,α 1 (n) if d = p 1. Carlitz has ψ p,α (pn + d) on the left but ψ p,α 1 (n 1) on the right. Recurrence leading to matrix product: θ p,α (pn + d) = (d + 1)θ p,α (n) + (p d 1)ψ p,α 1 (n 1) ψ p,α (pn + d 1) = dθ p,α (n) + (p d)ψ p,α 1 (n 1). [ ] d + 1 p d 1 M p (d) = d x (p d) x Eric Rowland Binomial coefficients and k-regular sequences / 27

23 Multinomial coefficients For a k-tuple m = (m 1, m 2,..., m k ) of non-negative integers, define total m := m 1 + m m k and mult m := (total m)! m 1! m 2! m k!. Theorem (Rowland 2018) Let k 1, and let e = [ ] Z k. Write n = n l n 1 n 0 in base p. Then x νp(mult m) = e M p,k (n 0 ) M p,k (n 1 ) M p,k (n l ) e. m N k total m=n M p,k (d) is a k k matrix... Eric Rowland Binomial coefficients and k-regular sequences / 27

24 Multinomial coefficients Let c p,k (n) be the coefficient of x n in (1 + x + x x p 1 ) k. p = 5: For each d {0,..., p 1}, let M p,k (d) be the k k matrix whose (i, j) entry is c p,k (p (j 1) + d (i 1)) x i 1. Example Let p = 5 and k = 3; the matrices M 5,3 (0),..., M 5,3 (4) are x 10x, x 18x 6x, 3x 19x 3x, 0 10x 2 15x x 2 10x 2 x 2 18x 2 6x x 18x x, 10x 15x 0. 3x 2 19x 2 3x 2 6x 2 18x 2 x 2 Eric Rowland Binomial coefficients and k-regular sequences / 27

25 Sketch of proof Lemma Let n 0. Let k 1. Let 0 i k 1. Let d {0,..., p 1}. Let m N k with total m = pn + d i. Define j = n total m/p. Then total(m mod p) = pj + d i, 0 j k 1, and ( ) (pn + d)! ν p (mult m) + ν p (pn + d i)! = ν p (mult m/p ) + ν p ( n! (n j)! ) + j. Eric Rowland Binomial coefficients and k-regular sequences / 27

26 Sketch of proof For d {0,..., p 1}, 0 i k 1, and α 0, show that β(m) := ( m/p, m mod p) is a bijection from the set { ( )} (pn + d)! A = m N k : total m = pn + d i and ν p (mult m) = α ν p (pn + d i)! to the set B = k 1 j=0 ( { ( ) } n! c N k : total c = n j and ν p (mult c) = α ν p j (n j)! { d {0,..., p 1} k : total d = pj + d i } ). The lemma implies that if m A then β(m) B. Eric Rowland Binomial coefficients and k-regular sequences / 27

27 Unexplored territory Do generalizations of binomial coefficients have analogous products? Fibonomial coefficients q-binomial coefficients Carlitz binomial coefficients word binomial coefficients ( ) u v other hypergeometric terms coefficients in other rational series ( n ) m = n! coefficients in (1 + x + x x p 1 ) k : m!(n m)! ( n+m m ) = [x n y m ] 1 1 x y Eric Rowland Binomial coefficients and k-regular sequences / 27

A MATRIX GENERALIZATION OF A THEOREM OF FINE

A MATRIX GENERALIZATION OF A THEOREM OF FINE ERIC ROWLAND To Jeff Shallit on his 60th birthday! Abstract. In 1947 Nathan Fine gave a beautiful product for the number of binomial coefficients ( n, for m