Apéry Numbers, Franel Numbers and Binary Quadratic Forms

Transcription

1 A tal given at Tsinghua University (April 12, 2013) and Hong Kong University of Science and Technology (May 2, 2013) Apéry Numbers, Franel Numbers and Binary Quadratic Forms Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun May 2, 2013

2 Abstract The Apéry numbers and the Franel numbers are given by and A n = n ( ) n 2 ( ) n + 2 (n = 0, 1, 2,...) f n = n ( ) n 3 (n = 0, 1, 2,...) respectively, they play important roles in number theory and combinatorics. In this tal we will give a survey of the recent developments of congruences involving Apéry numbers, Franel numbers and representations of primes by certain binary quadratic forms. 2 / 39

3 Part I. Introduction to Apéry numbers and Franel numbers 3 / 39

4 Apéry Numbers In 1978 Apéry proved that ζ(3) = n=1 1/n3 is irrational! During his proof he used the sequence {B n /A n } n=1 of rational numbers to approximate ζ(3), where A 0 = 1, A 1 = 5, B 0 = 0, B 1 = 6, and both {A n } n 0 and {B n } n 0 satisfy the recurrence (n + 1) 3 u n+1 = (2n + 1)(17n n + 5)u n n 3 u n 1 (n = 1, 2,...). In fact, A n = n ( ) n 2 ( ) n + 2 = n ( ) n + 2 ( ) and these numbers are called Apéry numbers. 4 / 39

5 Beuers Conjecture Dedeind eta function in the theory of modular forms: η(τ) = q 1/24 (1 q n ) with q = e 2πiτ n=1 Note that q < 1 if τ H = {z C : Im(z) > 0}. Beuers Conjecture (1985). For any prime p > 3 we have A ()/2 a(p) (mod p 2 ), where a(n) (n = 1, 2, 3,...) are given by η 4 (2τ)η 4 (4τ) = q (1 q 2n ) 4 (1 q 4n ) 4 = a(n)q n. n=1 n=1 5 / 39

6 An equivalent form of Beuers conjecture A Simple Observation. Let p = 2n + 1 be an odd prime. Then ( )( ) ( )( ) n n + n n 1 ( 1) = ( )( ) (p 1)/2 ( p 1)/2 = ( ) 1/2 2 ( ( 2 ) ) 2 ( 2 ) 2 = ( 4) = 16 (mod p 2 ). Thus Beuers conjecture has the following equivalent form: ()/2 ( 2 ) a(p) (mod p2 ). 6 / 39

7 Ahlgren and Ono s Proof of the Beuers conjecture Key steps in S. Ahlgren and Ken Ono s proof [2000]. (i) For an odd prime p let N(p) denote the number of F p -points of the following Calabi-Yau threefold x + 1 x + y + 1 y + z + 1 z + w + 1 w = 0. Then a(p) = p 3 2p 2 7 N(p). (ii) For any positive integer n we have n ( ) n 2 ( ) n + 2 (1 + 2H n+ + 2H n 4H ) = 0, =1 where H = 0<j 1/j. T. Kilbourn [Acta Arith. 123(2006)]: For any odd prime p we have ( 2 ) a(p) (mod p3 ). 7 / 39

8 Some combinatorial identities Comparing the coefficients of x n in the expansions of one finds the nown identity n ( ) n 2 n = For n = 1, 3, 5,..., clearly (x + 1) n (x + 1) n = (x + 1) 2n n ( n 3 ( 1) ) = Dixon s Identity 2n ( )( ) n n = n ( ) 2n. n n ( n 3 ( 1) ) n = 0. ( ) 2n 3 ( 2n ( 1) = ( 1) n n )( 3n n ). 8 / 39

9 Franel numbers In 1894 J. Franel introduced the Franel numbers n ( ) n 3 f n = (n = 0, 1, 2,...) and noted the recurrence relation (n + 1) 2 f n+1 = (7n(n + 1) + 2)f n + 8n 2 f n 1 (n = 1, 2, 3,...). In 2008 D. Callan gave a combinatorial interpretation of the Franel numbers. V. Strehl s Identity: A n = n ( n )( n + Barrucand s Identity: n ( ) n f = g n where g n := ) f. n ( n ) 2 ( 2 ). 9 / 39

10 Connection with modular forms Don Zagier (2009) investigated what integer sequence {u n } satisfies u 1 = 0, u 0 = 1, and the Apéry-lie recurrence relation ( + 1) 2 u +1 = (A 2 + A + B)u + C 2 u 1 ( = 1, 2, 3,...). When (A, B, C) = (7, 2, 8), u n is just the Franel number f n, and Zagier noted that n=0 ( η(τ) 3 η(6τ) 9 ) n f n η(2τ) 3 η(3τ) 9 = η(2τ)η(3τ)6 η(τ) 2 η(6τ) 3 for any complex number τ with Im(τ) > 0, where η(τ) := e πiτ/12 (1 e 2πinτ ). n=1 10 / 39

11 Supercongruences involving Franel numbers Wolstenholme s Congruence: For any prime p > 3, we have =1 1 0 (mod p2 ) and 1 0 (mod p). 2 Sula-Granville Congruence: For any prime p > 3 we have =1 2 2 ( 2 ) 2 1 (mod p). Theorem (Z. W. Sun, 2011). For any prime p > 3, we have =1 ( 1) f 0 (mod p 2 ), ( p ) ( 1) f 3 (mod p 2 ), p =1 ( 1) f 0 (mod p), 2 ( 1) f 2 ( p ) 3 3 (mod p 2 ). 11 / 39

12 Part II. Connections to Binary Quadratic Forms 12 / 39

13 Gauss congruence Gauss Congruence. Let p 1 (mod 4) be a prime and write p = x 2 + y 2 with x 1 (mod 4) and y 0 (mod 2). Then ( ) (p 1)/2 2x (mod p). (p 1)/4 Further Refinement of Gauss Result (Chowla, Dwor and Evans, 1986): ( ) (p 1)/ ( 2x p ) (p 1)/4 2 2x (mod p 2 ). It follows that ( ) (p 1)/2 2 2 (4x 2 2p) (mod p 2 ). (p 1)/4 13 / 39

14 Determining x mod p 2 with p = x 2 + y 2 and 4 x 1 Z. W. Sun [Acta Arith. 156(2012)]: Let p 1 (mod 4) be a prime. Write p = x 2 + y 2 with x 1 (mod 4) and y 0 (mod 2). Then ( 1) ()/4 x ( 16) ( ) 2 2 ( ) 2 2 (mod p 2 ). Z. W. Sun [Finite Fields Appl. 22(2013)]: For any prime p 3 (mod 4), we have ( 2 ) 2 8 ( 2 ) 2 ( 16) ( 1)(p+1)/4 2p ( (p+1)/2 (p+1)/4 ) (mod p 2 ). 14 / 39

15 Yeung s result on ( ) ()/2 ()/3 mod p 2 For a prime p and an integer a 0 (mod p), the Fermat quotient q p (a) := a 1 p Z. K. M. Yeung [J. Number Theory 33(1989)]: Let p 1 (mod 3) be a prime and write p = x 2 + 3y 2 with x 1 (mod 3). Then we have ( ) (p 1)/2 (p 1)/3 ( 2x p ) ( 1 2 2x 3 p q p(2) + 3 ) 4 p q p(3) (mod p 2 ). Remar. Yeung s result is an analogue of Gauss congruence but it is less elegant since the right-hand side contains the unpleasant expression p q p(2) p q p(3). 15 / 39

16 A conjecture on Apéry numbers Conjecture (Z. W. Sun, 2010). For any odd prime p, we have also, A ( 1) A { 4x 2 2p (mod p 2 ) if p = x 2 + 2y 2, 0 (mod p 2 ) if p 5, 7 (mod 8); { 4x 2 2p (mod p 2 ) if p = x 2 + 3y 2, 0 (mod p 2 ) if p 2 (mod 3). Remar. In 2011 I proved the mod p version of both congruences and that ( 1) A 0 (mod p 2 ) for any prime p 2 (mod 3). 16 / 39

17 Apéry polynomials Define Apéry polynomials by A n (x) := n ( ) n 2 ( ) n + 2 x (n = 0, 1, 2,...). Z. W. Sun [J. Number Theory 132(2012)]. Let p be an odd prime. Then ( 1) A (x) ( 2 ) 3 16 x (mod p 2 ). Also, for any p-adic integer x 0 (mod p) we have A (x) ( ) ( x 4 ),,, p (256x) (mod p). 17 / 39

18 A reduction Let ε {±1}. Then m=0 ε m A m (x) = m=0 = = = n where n = (p 1)/2. ε m ( 2 ( 2 ( 2 ( 2 m ( m + 2 ) 2 x m= ) 2 x r=0 ) 2 ε x ) 2 ( ) 2 2 x ( ) m + 2 ε m 2 r=0 ) 2 ε x r=0 ( ) 2 + r 2 ε +r r ( ) p 1 2 p 2 ε r r ε r ( 2(n ) p r ) 2 (mod p 2 ) 18 / 39

19 An auxiliary theorem Theorem (Z. W. Sun [J. N. Number Theory 132(2012)]) Let p be an odd prime and let x be any p-adic integer. (i) If x 2 (mod p) with {0,..., (p 1)/2}, then we have ( ) x 2 ( ) x ( 1) r ( 1) (mod p 2 ). r r=0 (ii) If x (mod p) with {0,..., p 1}, then ( ) x 2 ( ) 2x (mod p 2 ). r r=0 It is interesting to compare parts (i)-(ii) with the nown identities 2n ( ) 2n 2 ( ) 2n ( 1) = ( 1) n n and n ( ) n 2 = ( ) 2n. n 19 / 39

20 Setch of the proof of the first part of the auxiliary theorem Define ( ) 2 + py 2 f (y) := ( 1) r for N. r r=0 We want to prove that f (y) ( 1) ( 2 + py ) (mod p 2 ) for any p-adic integer y and {0, 1,..., (p 1)/2}. Applying the Zeilberger algorithm via Mathematica 7, we find that where (py )f +1 (y) + 4(py )f (y) = (p(y 1) ( ) 3)2 F (y) py (py )(py ) 2, p 1 F (y) = p+2p 2 +17py+20py 6p 2 y+5p 2 y / 39

21 Setch of the proof of the first part of the auxiliary theorem It follows that py f (y) 4(py ) f +1(y) (mod p 2 ) for = 0,..., p 3 2. If 0 < (p 1)/2 and f +1 (y) ( 1) +1 ( 2( + 1) + py + 1 ) (mod p 2 ), then ( ) py ( + 1) + py f (y) 4(py ) ( 1) = ( 10 (py ) 2 ( ) ( ) 2 + py 2 + py ( 1) 4( + 1)(py + + 1) (mod p 2 ). 21 / 39

22 Consequences of the auxiliary theorem Corollary Let p be an odd prime. (i) (Conjectured by Rodriguez-Villegas and proved by Mortenson) We have ( ) 1/2 2 ( ) 1 p (mod p 2 ). (ii) Let a n := n ( n 2C ) for n = 0, 1, 2,..., where C denotes the Catalan number ( ) ( 2 /( + 1) = 2 ) ( 2 +1). Then, for any odd prime p we have a a 0 (mod p 2 ). 22 / 39

23 A general conjecture on supercongruences Conjecture (Z. W. Sun). Let p be an odd prime and let n be a positive integer. Suppose that x is a p-adic integer with x 2 (mod p) for some {1,..., (p + 1)/(2n + 1) }. Then we have ( ) x 2n+1 ( 1) r 0 (mod p 2 ). r r=0 We proved this for n = 1 via the Zeilberger algorithm (see Z. W. Sun [J. Number Theory 132(2012)]). 23 / 39

24 A mod p 3 conjecture Conjecture (Z. W. Sun [J. Number Theory 132(2012)]). Let p > 3 be a prime. If p 1 (mod 3), then ( 1) A If p 1, 3 (mod 8), then A ( 2 ) 3 16 (mod p 3 ). ( 4 ),,, 256 (mod p 3 ). Remar. It was conjectured by Rodriguez-Villegas and proved by E. Mortenson and Z. W. Sun that for any odd prime p we have ( 4 ) {,,, 4x 2 2p (mod p 2 ) if p 1, 3 (mod 8) & p = x 2 + 2y 2, (mod p 2 ) if p 5, 7 (mod 8). 24 / 39

25 Arithmetic means involving Apéry numbers Theorem. Let n be a positive integer. (i) (Z. W. Sun [J. Number Theory 132(2012)]) We have n 1 (2 + 1)A 0 (mod n). For any prime p > 3, we have (2 + 1)A p p4 B p 3 (mod p 5 ) where B 0, B 1, B 2,... are Bernoulli numbers. (ii) (Conjectured by Z. W. Sun proved by V.J.W. Guo and J. Zeng) n 1 (2 + 1)( 1) A 0 (mod n). 25 / 39

26 Connection between p = x 2 + 3y 2 and Franel numbers Z.W. Sun [J. Number Theory 133(2013)]: Let p > 3 be a prime. When p 1 (mod 3) and p = x 2 + 3y 2 with x, y Z and x 1 (mod 3), we have f 2 If p 2 (mod 3), then f ( 4) 2x p 2x (mod p 2 ). f 2 2 f ( 4) 3p ( (p+1)/2 (p+1)/6 ) (mod p 2 ). Conjecture (Z. W. Sun): For any prime p = x 2 + 3y 2 with x 1 (mod 3), we have x 1 (3 + 4) f f (3 + 2) 2 ( 4) (mod p 2 ). 26 / 39

27 Two auxiliary identities MacMahon s Identity: n In particular, ( ) n 3 z = f n = A New Identity: f n = n/2 n/2 ( n + 3 ( n + 3 n ( n )( 2 )( 2 )( 2 )( ) 3 z (1 + z) n 2. )( ) 3 2 n 2. )( ) 3 ( 4) n. This can be proved by obtaining the recurrence relation for the right-hand side via the Zeilberger algorithm. 27 / 39

28 Three more identities needed Shi-Chieh Chu s Identity: A Simple Identity: r (mod 6) n m n= ( ) n = ( ) n ( 1) = ( ) m n = An Identity for Combinatorial Sums: If p > 3 is odd, then ( ) p = δ ( ) r ( 1) (p+1 2r)/6 3 ()/2 + 1, 3 2 where δ r taes 1 or 0 according as 3 p + r or not. 28 / 39

29 Some congruences needed Lemma 1. Let p > 3 be a prime and let ε = ( p 3 ). Then and (p ε)/3 =1 ( 2(p ε)/3 (p ε)/3 1 2 ε q p(3) (mod p) ( (p ε)/2 (p ε)/3 ) 2 2(p ε)/3 ) ( 1 3 ) 4 p q p(3) (mod p 2 ). Lemma 2. Let p 1 (mod 3) be a prime. Then ( ) ( ) p + 2(p 1)/3 2(p 1)/3 (mod p 2 ) (p 1)/3 (p 1)/3 and ()/2 = q p(2) (mod p). 29 / 39

30 Conjecture involving g n = n Recall that n ( n ) 2 ( 2 ) ( n ) f = g n where g n = n ( n ) 2 ( 2 ). Conjecture (Z. W. Sun): Let p > 3 be a prime. When p 1 (mod 3) and p = x 2 + 3y 2 with x, y Z and x 1 (mod 3), we have and g 3 g ( 3) 2x p 2x (mod p 2 ) x ( + 1) g 3 g ( + 1) ( 3) (mod p 2 ). If p 2 (mod 3), then 2 g 3 g ( 3) 3p ( (p+1)/2 (p+1)/6 ) (mod p 2 ). 30 / 39

31 On g /(±3) modulo p Let m be 3 or 3. Then m 1 {2, 4}. Observe that g n m n = 1 n ( ) n f ( ) n 1 m n f = m m n n=0 n=0 = f m f m j=0 j=0 ( f m m m 1 ( + j j ( p 1 j ) = n= ) 1 m j = f m j=0 ) ( 1 ) j = m So we obtain g /(±3) modulo p since f 2 f ( 4) f (m 1) (mod p). ( 1 j ) 1 ( m) j ( f m 1 1 ) m { 2x (mod p) if p = x 2 + 3y 2 (3 x 1), 0 (mod p) if p 2 (mod 3). 31 / 39

32 Conjectures involving n ) 4x ( n In 2011 the author introduced the polynomials S n (x) = n ( n 4x ) (n = 0, 1, 2,...) and posed 13 related conjectures. Here is one of them. Conjecture (Z. W. Sun) For any prime p > 2 we have S n (12) n=0 4x 2 2p (mod p 2 ) if p 1 (mod 12) & p = x 2 + y 2 (3 x), ( xy 3 )4xy (mod p2 ) if p 5 (mod 12) & p = x 2 + y 2, 0 (mod p 2 ) if p 3 (mod 4), (4 + 3)S (12) p Moreover, ( ( )) 3 p (mod p 2 ). 1 n 1 (4 + 3)S (12) Z for all n = 1, 2, 3,.... n 32 / 39

33 Conjectures involving n ( n ) 4x Here is another conjecture. Conjecture (Z. W. Sun). Let p be an odd prime. Then S ( 20) 4x 2 2p (mod p 2 ) if p 1, 9 (mod 20) & p = x 2 + y 2 (5 x), 4xy (mod p 2 ) if p 13, 17 (mod20), p = x 2 + y 2 (5 x y), 0 (mod p 2 ) if p 3 (mod 4). And (6 + 5)S ( 20) p Moreover, ( ) ( p ( )) 5 p (mod p 2 ). 1 n 1 (6 + 5)S ( 20) Z for all n = 1, 2, 3,.... n 33 / 39

34 Part III. More conjectures on Apéry numbers and Franel numbers 34 / 39

35 A conjecture on f n and g n Conjecture [Z. W. Sun, JNT 133(2013)]. For each n = 1, 2, 3,..., 1 n 1 2n 2 (3 + 2)( 1) f Z and 1 n 1 n 2 (4 + 1)g 9 n 1 Z. Moreover, for any prime p > 3 we have (3 + 2)( 1) f 2p 2 (2 p 1) 2 (mod p 5 ), (4 + 1) g 9 p2 2 ( ( p )) 3 p 2 (3 p 3) (mod p 4 ). 3 Remar. The part for Franel numbers has been confirmed by V. J. W. Guo. 35 / 39

36 More conjectures for f n and g n Conjecture (Z. W. Sun) (i) For any integer n > 1, we have n 1 ( )( 1) f 0 (mod (n 1)n 2 ), n 1 ( )( 1) f 0 (mod 4(n 1)n 3 ), n 1 ( )g 0 (mod 3n 3 ). (ii) For each odd prime p we have ( )( 1) f 3p 2 (p 1) 16p 3 q p (2) (mod p 4 ), ( )( 1) f 4p 3 (mod p 4 ). 36 / 39

37 More conjectures for Apéry numbers Conjecture (Z. W. Sun) (i) For any positive integer n, we have n 1 ( )( 1) A 0 (mod n 3 ), n 1 ( )( 1) A 0 (mod n 4 ). (ii) Let p > 3 be a prime. Then ( )A p 3 +2p 4 H 2 5 p8 B p 5 (mod p 9 ), where B 0, B 1, B 2,... are Bernoulli numbers. If p > 5, then ( )( 1) A 2p 4 + 3p 5 + (6p 8)p 5 H 12 5 p9 B p 5 (mod p 10 ). 37 / 39

38 A conjecture involving 3-adic valuations For a rational number a/b, its 3-adic valuation (or 3-adic order) ν 3 (a/b) is defined as ν 3 (a) ν 3 (b), where ν 3 (m) := sup{n N : p n m} for any nonzero integer m. Conjecture (Z. W. Sun) Let n be any positive integer. Then ( n 1 ) ν 3 ( 1) f 2ν 3 (n), ( n 1 ) ν 3 ( 1) f 2ν 3 (n), ( n 1 ) ( n 1 ν 3 (2 +1)( 1) A = 3ν 3 (n) ν 3 (2 +1) 3 ( 1) A ). If n is a positive multiple of 3, then ( n 1 ) ν 3 (2 + 1) 3 ( 1) A = 3ν 3 (n) / 39

39 Than you! 39 / 39