Arithmetic Properties of Combinatorial Quantities

Transcription

1 A tal given at the National Center for Theoretical Sciences (Hsinchu, Taiwan; August 4, 2010 Arithmetic Properties of Combinatorial Quantities Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun August 4, 2010

2 Abstract In combinatorics there are many combinatorial quantities arising from enumeration problems, e.g., Catalan numbers, central trinomial coefficients, Motzin numbers, central Delannoy numbers, Bell numbers and harmonic numbers. In this tal we introduce various conjectures and results on =0 a /m modulo powers of p, where a is the th certain combinatorial quantity, p is a prime and m is an integer not dividing by p.

3 Congruences for p-integers Let p be a prime. A rational number is called a p-integer (or a rational p-adic integer if it can be written in the form a/b with a, b Z and (b, p = 1. All p-integers form a ring R p which is a subring of the ring Z p of all p-adic integers. For a p-integer a/b, an integer c and a nonnegative integer n if a/b = c + p n q for some q R p (equivalently, a bc (mod p n, then we write a b c (mod pn. An Example for Congruences involving p-integers: = 3 (mod 32.

4 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

5 Classical congruences for central binomial coefficients A central binomial coefficient has the form ( 2 ( = 0, 1, 2,.... 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, where B n denotes the nth Bernoulli number.

6 Classical congruences for central binomial coefficients Morley s Congruence. For any prime p > 3 we have ( ( p (mod p 3. (p 1/2 p 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 A theorem of Stienstra and Beuers J. Stienstra and F. Beuers [Math. Ann. 27(1985]: [q p ]q (1 q 4n 6 = = n=1 { 4x 2 2p if p 1 (mod 4 & p = x 2 + y 2 with 2 x & 2 y, 0 if p 3 (mod 4; [q p ]q = (1 q n 2 (1 q 2n (1 q 4n (1 q 8n 2 n=1 { 4x 2 2p if p 1, 3 (mod 8 & p = x 2 + 2y 2 with x, y Z, 0 if p 5, 7 (mod 8; [q p ]q (1 q 2n 3 (1 q 6n 3 n=1 { 4x 2 2p if p 1 (mod 3 & p = x 2 + 3y 2 with x, y Z, 0 if p 2 (mod 3.

8 Catalan numbers For n N = {0, 1, 2,...}, the nth Catalan number is given by C n = 1 ( ( ( 2n 2n 2n =. n + 1 n n n + 1 Recursion. C 0 = 1 and C n+1 = n=0 n C C n (n = 0, 1, 2,.... =0 Generating Function. C n x n = 1 1 4x. 2x Combinatorial Interpretations. The Catalan numbers arise in many enumeration problems. For example, C n is the number of binary parenthesizations of a string of n + 1 letters, and it is also the number of ways to triangulate a convex (n + 2-gon into n triangles by n 1 diagonals that do not intersect in their interiors.

9 Recent results on =0 ( 2 and =0 C mod p 2 Let p be a prime and let a Z + = {1, 2, 3,...}. 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 [Int. JNT, Adv. in Appl. Math. 2010]. p a 1 ( ( 2 p a (mod p 2, 3 =0 p a 1 =0 =1 pa 3( 3 C 1 2 ( 2 (mod p 2, 8 9 p2 B p 3 (mod p 3 for p > 3.

10 Determination of =0 ( 2 /m mod p 2 Let p be an odd prime. If p/2 < < p then ( 2 = (2! 0 (mod p. (! 2 Thus (/2 =0 ( 2 m where m is an integer with p m. =0 ( 2 m (mod p, Sun [Sci. China Math. 2010]: Let p be an odd prime and let a, m Z with a > 0 and p m. Then p a 1 ( 2 ( m 2 ( 4m m 2 4m + u p ( m 2 (mod 4m p2, p =0 m p a p a 1 where ( is the Jacobi symbol and {u n } n 0 is the Lucas sequence given by u 0 = 0, u 1 = 1, and u n+1 = (m 2u n u n 1 (n = 1, 2, 3,....

11 On =0 ( 3 /m mod p In 2009 Sun determined ( 3 =0 /m mod p where p > 3 is a prime and m is an integer not divisible by p. Some particular congruences (Sun, 2009: ( 3 1 (mod p, =0 =0 =0 8 3 (( p 4 5 ( 3 { 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.

12 On =0 ( 2 2/16 modulo p 2 A Conjecture of Rodriguez-Villegas proved by E. Mortenson. If p is an odd prime, then =0 ( ( 1 = ( 1 (/2 (mod p 2. p Remar. (a By Stirling s formula, n! ( n n 2πn as n +. e It follows that ( π. (b Mortenson proof involves Gauss and Jacobi sums and the p-adic Gamma function. In fact, now there are elementary proofs.

13 Euler numbers and some congruences mod p 3 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..

14 Some congruences related to 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.

15 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.

16 A conjecture involving Euler numbers Conjecture (Sun, Let p > 3 be a prime. Then and (/2 =1 4 (2 1 ( 2 (/2 E p 3 + ( 1 (/2 1 (mod p E p 3 (mod p. (2 1 ( 2 =1 Remar. Let p > 3 be a prime. The speaer has shown that (/2 =2 (/2 =0 (/2 =1 4 2( 2 4 ( 1 2( 2 4 ( + 1 ( 2 ( 1 (/2 4E p 3 (mod p, 8E p ( 1 p (mod p, ( 1 (4 2E p 3 2 (mod p. p

17 On =0 ( 2 3 mod p 2 Conjecture (Z. W. Sun, 2009: 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. In number theory it is nown that if p is a prime with ( p 7 = 1 (i.e., p 1, 2, 4 (mod 7 then there are unique positive integers x and y such that p = x 2 + 7y 2.

18 The speaer s 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.

19 Conjecture involving x y 2 Conjecture (Sun, Let p > 5 be a prime with p 23, 29. =0 ( 6 ( 3 3,, ( { ( p (x 2 2p (mod p 2 if ( p 163 = 1 & 4p = x y 2, 0 (mod p 2 if ( p 163 = 1. Remar. It is well nown that the only imaginary quadratic fields with class number one are those Q( d with d = 1, 2, 3, 7, 11, 19, 43, 67, 163. For each of the 9 values of d we have corresponding conjectures similar to the above one.

20 Conjecture for Q( d with class number two Let d > 0 be a squarefree integer. It is nown that Q( d has class number two if and only if d is among 5, 6, 10, 13, 15, 22, 35, 37, 51, 58, 91, 115, 123, 187, 235, 267, 403, 427. Except for d = 35, 91, 115, 187, 235, 403, 427 we have found explicit conjectures involving x 2 + dy 2. Conjecture for Q( 15 (Sun. Let p > 3 be a prime. Then ( 2 2 ( 3 ( 27 =0 4x 2 2p (mod p 2 if p 1, 4 (mod 15 & p = x y 2, 20x 2 2p (mod p 2 if p 2, 8 (mod 15 & p = 5x 2 + 3y 2, 0 (mod p 2 if ( p 15 = 1. Also, for any a Z + we have p 1 a 1 ( ( ( 3 p a p a ( 27 4 (mod p 2. 3 =0

21 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,

22 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.

23 Harmonic numbers Stirling numbers of the first ind: s(n, = {σ S n : σ has exactly cycles} Harmonic numbers: H n = 0< n 1 = s(n + 1, 2 n! (n = 0, 1, 2,.... It is nown that =1 H 2 = π2 12 (S. W. Coffman, 1987 and =1 H 2 2 = π4 (D. Borwein and J. M. Borwein, 1995.

24 Congruences on harmonic numbers Theorem (Z. W. Sun, 2009: Let p > 3 be a prime. Then When p > 5 we have =1 =1 H 3 =1 H 0 (mod p, 2 2 H 2 4 (mod p, 9 6 (mod p, H 2 2p 2 (mod p2. =1 =1 H 2 0 (mod p. 2

25 Congruences on harmonic numbers of even order Conjecture (Z. W. Sun, 2009 Let m be any positive even integer. If p is a prime with p 1 3m, then =1 H 2,m m where H,m = 0<j 1/j m. 0 (mod p, Remar. Sun proved the conjecture in the case 2p/3 < m < p via Bernoulli numbers. Later his former student Li-Lu Zhao (Hong Kong Univ. fully confirmed the conjecture.

26 Beuers conjecture on Apéry numbers In his proof of the irrationality of ζ(3, Apéry introduced n ( n 2 ( n + 2 A n = (n = 0, 1, 2,.... =0 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 (2τη 4 (4τ = q (1 q 2n 4 (1 q 4n 4 = n=1 a(nq n. n=1

27 Beuers conjecture on Apéry numbers 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 =0 ( a(p (mod p2.

28 Ahlgren and Ono s Proof of the Beuers conjecture S. Ahlgren and Ken Ono [J. Reine Angew. Math. 518(2000]: The Beuers conjecture is true! Outline of their proof. First show that a(p can be expressed as a special value of the Gauss hypergeometric function 4 F 3 (λ defined in terms of Jacobi sums. Then rewrite Jacobi sums in terms of Gauss sums and apply the Gross-Koblitz formula to express Gauss sums in terms of the p-adic Gamma function Γ p (x. Finally use combinatorial properties of Γ p (x and some sophisticated combinatorial identities involving harmonic numbers H n = 0< n 1/.

29 Two ey points in Ahlgren and Ono s Proof Two ey points 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

30 Congruences on Apéry numbers Recall that Apéry numbers are those integers n ( n 2 ( n + 2 A n = (n = 0, 1, 2,.... =0 Theorem (Sun, For any n Z + we have 1 n 1 n 1 ( ( ( ( n 1 n + n + 2 (2 + 1A = Z. n =0 =0 If p > 3 is a prime, then =0 (2 + 1A p (mod p 4. Conjecture (Sun, For any positive integer n we have 1 n 1 (2 + 1( 1 A Z. n =0 If p > 3 is a prime, then ( p (2 + 1( 1 A p (mod p 3. 3 =0

31 Congruences 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.

32 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.

33 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+.

34 Congruences involving Schröder numbers D n := n ( ( n n + = =0 The nth Schröder number is given by n ( n + n S n = C = 2 =0 =0 n ( ( n = ( n ( n + which is the number of lattice paths from (0, 0 to (n, n with steps (1, 0, (0, 1 and (1, 1 that never rise above the line y = x. Conjecture (Sun, 2010 Let p > 3 be a prime. Then and (/2 =1 D S 1 + 4pq p (2 2p 2 q p (2 2 (mod p 3, =0 D S { 4x 2 (mod p if p 1 (mod 4 & p = x 2 + y 2 (2 x, 0 (mod p if p 3 (mod 4.

35 On central trinomial coefficients The nth central trinomial coefficient: T n :=[x n ](1 + x + x 2 n (the coefficient of x n in (1 + x + x 2 n n ( ( n/2 n n ( ( n 2 = =. 2 =0 =0 Theorem (H. Q. Cao and Sun, For any prime p > 3 we have ( p T 3 (mod p 2. 3 Conjecture (Sun, 2010 For any n Z + we have If p is a prime, then n 1 (8 + 5T 2 0 (mod n. =0 ( p (8 + 5T 2 3p (mod p 2. 3 =0

36 Mod p 2 congruences on Motzin numbers The nth Motzin number M n := n/2 =0 ( n C 2 is the number of paths from (0, 0 to (n, 0 in an n n grid using only steps (1, 0, (1, 1 and (1, 1. Conjecture (Sun, Let p > 3 be a prime. Then ( p M 2 (2 6p (mod p 2, 3 =0 ( p M 2 (9p 1 (mod p 2, 3 =0 M T 4 3 =0 ( p + p ( ( p (mod p 2.

37 Generalized central trinomial coefficients and generalized Motzin numbers Given b, c Z, the generalized central trinomial coefficients T n (b, c :=[x n ](x 2 + bx + c n = [x 0 ](b + x + cx 1 n = n/2 =0 ( n 2 ( 2 b n 2 c = n/2 =0 ( n ( n b n 2 c and we introduce the generalized Motzin numbers n/2 ( n/2 n ( ( n n b M n (b, c := C b n 2 c n 2 c = =0 (n = 0, 1, 2,.... Note that =0 T n = T n (1, 1, M n = M n (1, 1, T n (2, 1 = [x n ](x + 1 2n = and M n (2, 1 = n =0 ( n C 2 n 2 = C n+1. 2 ( 2n, n

38 Continued D n = T n (3, 2, but M n (3, 2 is different from S n. H. S. Wilf observed that T n (b, cx n = n=0 which implies the recursion 1 1 2bx + (b 2 4cx 2 (n+1t n+1 (b, c = (2n+1bT n (b, c+(4c b 2 nt n 1 (b, c (n Z +. Theorem (Sun, Let p be an odd prime and let b, c, m Z with m 0 (mod p. Then ( T (b, c (m b 2 4c m (mod p p and 2c =0 =0 M (b, c m ( (m b (m b 2 ((m b 2 2 4c 4c (mod p. p

39 Continued Theorem (Sun, Let b, c Z. (i For any n Z +, we have n 1 (2 + 1T (b, c 2 (4c b 2 n 1 0 (mod n, =0 and furthermore n 1 b (2 + 1T (b, c 2 (4c b 2 n 1 = nt n (b, ct n 1 (b, c. =0 (ii Suppose that b 2 4c = 1. Then 1 n 1 n ( ( ( n n + 1 b 1 1 (2 + 1T (b, c = Z n 1 2 =0 =1 for all n Z +. If p is a prime not dividing c, then (2 + 1T (b, c p (mod p 2. =0

40 Conjecture on generalized central trinomial coefficients Conjecture (Sun, Let b, c Z. (i For any n Z + we have n 1 (2 + 1T (b, c 2 (b 2 4c n 1 0 (mod n 2. =0 (ii Suppose that b 2 4c = 1. Then n 1 (2 + 1T (b, c m 0 (mod n =0 for all m, n Z +. If p is a prime not dividing c, then ( 2b 1 (2 + 1T (b, c 3 p (mod p 2 p and =0 (2 + 1T (b, c 4 p (mod p 2. =0

41 A new ind of numbers For b, c Z we introduce a new ind of numbers: n ( n 2 ( n 2 n/2 W n (b, c := b n 2 c = =0 =0 ( n 2 Note that W n ( b, c = ( 1 n W n (b, c. Conjecture (Sun, Let p be an odd prime. Then W (1, 1 =0 2 ( 2 2 b n 2 c. { 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. If p 1, 3 (mod 8, then (16 + 3W (1, 1 8p (mod p 2. =0 If p 5, 7 (mod 8 and p 7,then W (1,1 =0 0 (mod p 2.

42 More conjectures Conjecture (Sun, (i Let p > 3 be a prime. Then ( 1 W (1, 1 =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. (ii For any n Z + we have n 1 (6 + 5( 1 W (1, 1 0 (mod n. =0 If p is an odd prime, then (6 + 5( 1 W (1, 1 p =0 ( ( p (mod p 2.

43 More conjectures Conjecture (Sun, Let p be an odd prime. Then =0 W (2, 1 ( 2 { 4x 2 2p (mod p 2 0 (mod p 2 if p 3 (mod 4. If p 1 (mod 4, then if p 1 (mod 4 and p = x 2 + y 2 (2 x, (4 + 3 W (2, 1 ( 2 0 (mod p 2. =0

44 More conjectures Conjecture (Sun, 2010 (i Let p be an odd prime. Then ( 1 W (2, 1 ( 2 p =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. (ii For any n Z + we have n 1 (4 + 3W (2, 1( 2 n 1 0 (mod n. =0 If p is an odd prime, then (4 + 3 W (2, 1 ( 2 =0 ( p 2 ( 2 + p ( 1 p (mod p 2.

45 More conjectures Conjecture (Sun, 2010 (i Let p 2, 5 be a prime. Then we have W (1, 4 =0 4x 2 2p (mod p 2 if p 1, 9 (mod 20 & p = x 2 + 5y 2, 2x 2 2p (mod p 2 if p 3, 7 (mod 20 & 2p = x 2 + 5y 2, 0 (mod p 2 if p 11, 13, 17, 19 (mod 20. (ii For any n Z + we have n 1 ( W (1, 4 0 (mod n. =0 If p is an odd prime, then ( W (1, 4 p =0 ( 10 ( p (mod p 2.

46 More conjectures Conjecture (Sun, 2010 (i For any prime p > 5, we have W (1, 81 =0 4x 2 2p (mod p 2 if p 1, 9, 11, 19 (mod 40 & p = x y 2, 2p 2x 2 (mod p 2 if p 7, 13, 23, 37 (mod 40 & 2p = x y 2, 0 (mod p 2 if ( 10 p = 1. (ii For any n Z + we have n 1 (10 + 9W (1, 81 0 (mod n. =0 If p > 3 is a prime, then (10 + 9W (1, 81 p =0 ( 4 ( p (mod p 2.

47 More conjectures Conjecture (Sun, 2010 (i For any prime p 7, we have ( 1 W (1, 16 =0 { 4x 2 2p (mod p 2 if p 1, 2, 4 (mod 7 & p = x 2 + 7y 2, 0 (mod p 2 if p 3, 5, 6 (mod 7. (ii For all n Z + we have n 1 ( ( 1 W (1, 16 0 (mod n. =0 If p is a prime, then ( ( 1 W (1, 16 p =0 ( ( p (mod p 2.

48 More conjectures Conjecture (Sun, 2010 (i For any prime p > 3, we have W (1, 324 =0 4x 2 2p (mod p 2 if ( 13 p = ( 1 p = 1 & p = x y 2, 2x 2 2p (mod p 2 if ( 13 p = ( 1 p = 1 & 2p = x y 2, 0 (mod p 2 if ( 13 p = 1. (ii For any n Z + we have n 1 ( W (1, (mod n. =0 If p > 3 is a prime, then ( W (1, 324 p =0 ( 130 ( p (mod p 2.

49 Bell numbers For n = 1, 2, 3,..., the nth Bell number b n denotes the number of partitions of a set of cardinality n. In addition, b 0 := 1. Here are values of b 1,..., b 10 : 1, 2, 5, 15, 52, 203, 877, 4140, 21147, Recursion: b n+1 = n =0 ( n b (n = 0, 1, 2,.... Exponential Generating Function: x n b n n! = 1 eex. n=0 Touchard s Congruence: For any prime p and m, n = 0, 1, 2,... we have b p m +n mb n + b n+1 (mod p.

50 A conjecture on Bell numbers Conjecture (Sun, July 17, For any positive integer n there is a unique integer s(n such that =0 In particular, b s(n (mod p for any prime p n. ( n s(2 = 1, s(3 = 2, s(4 = 1, s(5 = 10, s(6 = 43, s(7 = 266, s(8 = 1853, s(9 = 14834, s(10 = Remar. It is easy to see that s(1 = 2. In fact, if p is a prime then ( p 1 ( 1 b b = b p =0 =0 b 0 + b 1 = 2 (mod p (by Touchard s congruence.

51 More Conjectures on Congruences For more conjectures of mine on congruences, see Z. W. Sun, Open Conjectures on Congruences, arxiv: which contains 100 unsolved conjectures raised by me. You are welcome to solve my conjectures!

52 Than you!