Super congruences involving binomial coefficients and new series for famous constants

Transcription

1 Tal at the 5th Pacific Rim Conf. on Math. (Stanford Univ., 2010 Super congruences involving binomial coefficients and new series for famous constants Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun July 2, 2010

2 Abstract A p-adic congruence is called a super congruence if it not only holds mod p but also happens to hold modulo a higher power of p. The topic of super congruences is related to the p-adic Gamma function, Gauss and Jacobi sums, hypergeometric series, modular forms, Calabi-Yau manifolds, representations of p by certain quadratic forms, and some sophisticated combinatorial identities involving harmonic numbers. Recently the speaer formulated many conjectures on super congruences and revealed that super congruences are related to Euler numbers and series with summation related to π and other constants. In this tal we will analyze few typical conjectures of the speaer and introduce related progress.

3 Part A. Previous Wor by Others

4 Rational p-adic integers Let p be a prime. For m Z define the p-adic valuation (or order ν p (m = sup{a N = {0, 1, 2,...} : p a m}. For x = m/n with m Z and n Z + = {1, 2, 3,...} define the p-adic valuation ν p (x of x by The p-adic norm is given by ν p (x = ν p (m ν p (n. x p = 1 p νp(x. Those x Q with x p 1 (i.e., ν p (x 0 are called rational p-adic integers or p-integers, they form a ring. An Example for Congruences involving p-integers: = 3 (mod 32.

5 Classical congruences for central binomial coefficients A central binomial coefficient has the form ( 2 ( = 0, 1, 2,.... If p = 2n + 1 is an odd prime, then ( 2 = (2! p (mod p for every = (! 2 2,..., p 1. Wolstenholme s Congruence. For any prime p > 3 we have H = =1 1 0 (mod p2 and ( 2p 1 = 1 ( 2p 1 (mod p 3. p 1 2 p Remar. In 1900 Glaiser proved that for any prime p > 3 we have ( 2p p 1 3 p3 B p 3 (mod p 4.

6 Classical congruences for central binomial coefficients Morley s Congruence. For any prime p > 3 we have ( p 1 ( 1 (/2 4 (mod p 3. (p 1/2 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 (mod p 2. (p 1/4 2 2x It follows that ( (p 1/2 2 2 (4x 2 2p (mod p 2. (p 1/4

7 Beuers Conjecture for 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 u n+1 = (2n + 1(17n n + 5u n n 3 u n 1 (n = 1, 2,.... In fact, A(n = n =0 ( n 2 ( n + and these numbers are called Apéry numbers. Dedeind eta function in the theory of modular forms: 2 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}.

8 Beuers Conjecture (1985. For any prime p > 3 we have ( p 1 A a(p (mod p 2, 2 where a(n (n = 1, 2, 3,... are given by η 4 (2τη 4 (4τ = q (1 q 2n 4 (1 q 4n 4 = n=1 a(nq n. n=1 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 ( a(p (mod p2.

9 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 Then x + 1 x + y + 1 y + z + 1 z + w + 1 w = 0. 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 ( a(p (mod p3. =0

10 Gaussian hypergeometric series The rising factorial (or Pochhammer symbol: Note that (1 n = n!. (a n = a(a + 1 (a + n 1 = Classical Gaussian hypergeometric series: Γ(a + n. Γ(a r+1f r (α 0,..., α r ; β 1,..., β r x = n=0 (α 0 n (α 1 n (α r n x n (β 1 n (β r n n!, where 0 α 0 α 1 α r < 1 and 0 β 1 β r < 1.

11 Legendre symbols Let p be an odd prime and a Z. The Legendre symbol ( a p is given by ( a 0 if p a, = 1 if p a and x p 2 a (mod p for some x Z, 1 if p a and x 2 a (mod p for no x Z. It is well nown that ( ab p = ( a p ( b p for any a, b Z. Also, ( { 1 = ( 1 (/2 1 if p 1 (mod 4, = p 1 if p 1 (mod 4; ( { 2 = ( 1 (p2 1/8 1 if p ±1 (mod 8, = p 1 if p ±3 (mod 8. The Law of Quadratic Reciprocity: If p and q are distinct odd primes, then ( ( p q = ( 1 2 q 1 2. q p

12 Conjectures of Rodriguez-Villegas In 2001 Rodriguez-Villegas conjectured 22 congruences which relate truncated hypergeometric series to the number of F p -points of some family of Calabi-Yau manifolds. Here we list some of them. ( 2 2 ( 1 16 = ( 1 (/2 (mod p 2, p =0 ( 2 ( 3 ( p ( 3 27 = (mod p 2, 3 p =0 ( 2 ( 4 ( 2 2 (mod p 2, p =0 =0 ( 6 3 =0 64 ( ( 2 3 ( 1 p 64 [q p ]q (mod p 2, (1 q 4n 6 (mod p 2. n=1

13 Where the denominators and ( come from? p By Stirling s formula, n! ( n n 2πn as n + e It follows that ( π, ( (π 3/2, ( ( π, ( ( , 2 2π ( ( π.

14 Progress on Rodriguez-Villegas conjectures The congruences we list have been confirmed, see, E. Motenson, J. Number Theory 99(2003; Trans. AMS 355(2003; Proc. AMS 133(2005. Many of the 22 conjectures remain open.

15 Ramanujan s series for 1/π Here are 5 of the 17 Ramanujan series recorded by him in 1914: =0 ( 2 3 ( 1 (4 + 1 (1/23 (1 3 = (4 + 1 ( 64 = 2 π, = ( 4 =0 =0 = (1/23 ( ( 8 (1/23 ( (1/23 (1 3 (1/2 (1/4 (3/4 (1 3 =0 ( 2 3 = ( = 4 π, =0 ( 2 3 = (6 + 1 ( 512 = 2 2 π, =0 ( 2 3 = ( = 16 π, =0 ( 4,,, = ( ( 1024 = 8 π. =0 Remar. The first one was actually proved by G. Bauer in 1859.

16 Hamme s Conjectures L. Van Hamme [1997] conjectured the p-adic analogues of the above first 4 identities and W. Zudilin [JNT, 2009] obtained the p-adic analogue of the last identity. (/2 =0 (/2 =0 (/2 =0 ( 2 3 (4 + 1 ( 64 ( 1 p (mod p 3, p ( 2 3 ( 1 ( p (mod p 4, p ( 2 3 ( 2 (6 + 1 ( 512 p (mod p 3, p (/2 ( 2 3 ( =0 ( 4,,, ( ( 1024 =0 ( 1 5p (mod p 4, p ( 1 3p (mod p 3. p

17 Progress on Hamme s conjectures The first of the above congruence was proved by E. Mortenson [Proc. AMS 136(2008] and the second one was recently shown by Ling Long, while the last was confirmed by Zudilin via the WZ method. The third and the fourth remain open. The p-adic Gamma function plays an important role in Hamme s formulation of those conjectures. It is defined in the following way: Γ p (n := ( 1 n (n = 1, 2, 3,... 1<<n p and Γ p (x = lim n x Γ p (n for any p adic integer x.

18 Some Series for π D. V. Chudnovsy and G. V. Chudnovsy (1987: ( ( ( = ,, 2π =0 This yielded the record for the calculation of π during D. Zeilberger (1993: 21 8 =1 3( 2 3 = ζ(2 = π2 6. T. Amdeberhan and D. Zeilberger (1997: ( 1 ( = 2ζ(3. =1 5( 2 A Conjecture of J. Guillera (2003: ( =1 7( 2 7 = π4 8.

19 Part B. My Results and Conjectures

20 Some Joint Wor H. Pan and Z. W. Sun [Discrete Math. 2006]. ( ( 2 p d (mod p (d = 0,..., p, + d 3 =0 =1 ( 2 0 (mod p for p > 3. Sun & R. Tauraso [arxiv: , Adv. in Appl. Math.]. p a 1 ( ( 2 p a (mod p 2, 3 =0 =1 ( p2 B p 3 (mod p 3 for p > 3, L. L. Zhao, H. Pan and Z. W. Sun [Proc. AMS, 2010] ( (mod p.

21 My own results Recall that if p/2 < < p then ( 2 = (2! 0 (mod p. (! 2 Thus =0 ( 2 (/2 m =0 where m is an integer with p m. ( 2 m (mod p, In 2009 I [arxiv: , arxiv: , ] determined =0 ( 2 m mod p 2, (/2 =0 in terms of linear recurrences. ( 2 m mod p 2, =0 ( 3 m mod p

22 Some particular congruences due to me =0 =0 =0 ( (( p 4 5 ( 3 1 (mod p, { 2 (mod p if p ±2 (mod 7, 7 1 (mod p otherwise. ( 4 1 (mod p if p 1 (mod 5 & p 11, 5 1/11 (mod p if p 2, 3 (mod 5, 9/11 (mod p if p 4 (mod 5. If p 1 (mod 3 then =0 ( (/3 (mod p.

23 Connection between super congruences and Euler numbers Recall that Euler numbers E 0, E 1,... are given by E 0 = 1, ( n E n = 0 (n = 1, 2, 3, It is nown that E 1 = E 3 = E 5 = = 0 and sec x = ( 1 n x 2n ( E 2n x < π. (2n! 2 n=0 Z. W. Sun [arxiv: ]. =0 (/2 =0 ( 2 2 ( 1 (/2 p 2 E p 3 (mod p 3, ( 2 8 ( 2 + p ( 2 p 2 p 4 E p 3 (mod p 3..

24 Connection between super congruences and Euler numbers Theorem (Sun, For any prime p > 3 we have (/2 =1 (/2 =1 (/2 =0 ( 2 ( 1 (p+1/2 8 3 pe p 3 (mod p 2, 1 2( ( 1 (/ E p 3 (mod p, ( 2 Remar. Note that lim ( 1 (/2 + p 2 E p 3 (mod p 3. ( = 1 π and =1 1 2( 2 = π2 18.

25 Some auxiliary results needed for the proof A Lemma (Sun, (i If p = 2n + 1 is an odd prime, then ( ( n n + ( ( 1 1 p ( (H n+ H n 16 (mod p 4. (ii We have n ( n ( 1 n =0 ( n + ( 1 (H n+ H n = 3 2 Some auxiliary identities: ( n 2 = n + 1 ( n 2n n n =1 2( n =1 =1 =1 n =1 1 2 (Staver, 2( n ( 1 n ( n+ = ( 1 (3 n 1 1 2( =1 n 1 n 1 n 2( = 3 n+ 2( 2 =1 =1 ( 2. n ( 1 (Apéry = (Sun.

26 Six conjectured series for π 2 and other constants Conjecture (Z. W. Sun, 2010: We have =1 =1 (10 38 =1 3( 2 2 ( 3 = π2 2, ( =1 3( 2 2 ( 3 =8π 2, ( ( 4 2 =1 3( 2 (15 4( 27 2 ( 3 3( 2 (5 1( 144 =1 3( 2 ( ( 64 5( 2 4 ( 3 2 ( 4 2 =12π 2, = 27 = 45 2 =1 =1 = 14ζ(3. ( 3 2, ( 3 2,

27 Conjecture involving x 2 + 7y 2 Let p > 3 be a prime. Then =0 ( { 2 3 4x 2 2p (mod p 2 if ( p 7 = 1 & p = x 2 + 7y 2, 0 (mod p 2 if ( p 7 = 1. Moreover, (/2 =0 ( 2 3 ( p + ( 1 32p 3 E p 3 (mod p 4. p Remar. M. Jameson and K. Ono are woring on the first part of this conjecture but they have not yet got a proof.

28 Conjecture involving x y 2 Let p > 3 be a prime. Then ( 2 2 ( 3 64 =0 { x 2 2p (mod p 2 if ( p 11 = 1 & 4p = x y 2 (x, y Z, 0 (mod p 2 if ( p 11 = 1, i.e., p 2, 6, 7, 8, 10 (mod 11. Furthermore, ( =0 (/2 p =1 ( 2 2 ( p p4 B p 3 (mod p 5, ( ( 2 2 ( p 64 3 p2 B p 3 (mod p 3. Remar. It is well-nown that the quadratic field Q( 11 has class number one and hence for any odd prime p with ( p 11 = 1 we can write 4p = x y 2 with x, y Z.

29 A conjecture motivated by some series for ζ(3 and ζ(4 Conjecture (Sun, Let p > 7 be a prime and let H = =1 1/ p2 B p 3 /3 (mod p 3. Then and Also, =1 =1 1 4( 2 (/2 =1 ( p 2 H (mod p 2 H p pb p 5 (mod p 2. ( 1 3( 2B 2 p 3 (mod p. Motivation. ( 1 3( = ζ(3 and =1 =1 1 4( 2 = ζ(4.

30 A conjecture on divisibility of binomial coefficients Recall that C n = 1 ( 2n Z for all n N. n + 1 n The author observed that for any, l Z + we have ( ln + 1 n + ln (, ln + 1 for any n N. n In particular, if all prime factors of divides l then (ln + 1 ( n+ln n for every n = 0, 1, 2,.... Conjecture (Sun, Let and l be positive integers. If (ln + 1 ( n+ln n for all sufficiently large positive integers n, then each prime factor of divides l. In other words, if has a prime factor not dividing l then there are infinitely many positive integers n such that (ln + 1 ( n+ln n.

31 Some numerical examples Let and l be positive integers such that not all prime factors of divides l. Define f (, l as the smallest positive integer n such that (ln + 1 ( n+ln n. Via Mathematica we obtained the following data: f (7, 36 = 279, f (10, 192 = 362, f (11, 100 = 1187, f (13, 144 = 2001, f (22, 200 = 6462, f (31, 171 = 1765; f (43, 26 = 640, f (53, 32 = 790, f (67, 56 = 2004, f (73, 61 = 2184, f (74, 62 = 885, f (97, 81 = 2904, f (179, 199 = 28989, f (223, 93 = 13368, f (307, 189 = 31915, f (277, 254 = 36552, f (313, 287 =

32 On Apéry numbers A n := n ( n 2 ( n + 2. =0 Theorem (Sun, n 1 =0 (2 + 1A 0 (mod n for any n Z +. If p > 3 is a prime, then (2 + 1A p (mod p 4. =0 Conjecture (Sun, For any positive integer n we have If p > 3 is a prime, then n 1 n (2 + 1( 1 A. =0 ( p (2 + 1( 1 A p (mod p 3. 3 =0

33 On Apéry numbers Conjecture (Sun, 2010 Let p > 3 be a prime. Then and A =0 { 4x 2 2p (mod p 2 if p 1, 3 (mod 8 and p = x 2 + 2y 2, 0 (mod p 2 if p 5, 7 (mod 8; ( 1 A =0 { 4x 2 2p (mod p 2 if p 1 (mod 3 and p = x 2 + 3y 2, 0 (mod p 2 if p 2 (mod 3.

34 On central Delannoy numbers D n := n =0 ( n ( n + In combinatorics, D n is the number of lattice paths from (0, 0 to (n, n with steps (1, 0, (0, 1 and (1, 1. Theorem (Sun, Let p be an odd prime. Then D =0 When p > 3 we also have ( 1 p. p 2 E p 3 (mod p 3. (2 + 1( 1 D p (mod p 4, =0 (2 + 1D p + 2p 2 q p (2 p 3 q p (2 2 (mod p 4, =0 where q p (2 denotes the Fermat quotient (2 1/p.

35 On central Delannoy numbers Conjecture (Sun, (i For any n Z + we have n 1 (2 + 1D 2 0 (mod n2. =0 If p > 3 is a prime, then (2 + 1D 2 p2 4p 3 q p (2 2p 4 q p (2 2 (mod p 5. =0 (ii Let p be any odd prime. Then =1 D 2 2 ( 1 E p 3 (mod p and p ( 2 D 2 p =0 (mod p. Remar. I can show that n n 1 =0 (2 + 1( 1 D 2 for n Z+.

36 More Conjectures on Congruences For more conjectures of mine on congruences, see Z. W. Sun, Open Conjectures on Congruences, arxiv: You are welcome to solve my conjectures!

37 Than you!