On Arithmetic Properties of Bell Numbers, Delannoy Numbers and Schröder Numbers

Transcription

1 A tal given at the Institute of Mathematics, Academia Sinica (Taiwan (Taipei; July 6, 2011 On Arithmetic Properties of Bell Numbers, Delannoy Numbers and Schröder Numbers Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun July 6, 2011

2 Abstract Bell numbers, Delannoy numbers and Schröder numbers are important quantities arising from enumerative combinatorics. Surprisingly they also have nice number-theoretic properties. We will tal about congruences related to Bell numbers, Delannoy numbers and Schroder numbers.

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 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.

5 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

6 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,.... 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.

7 Recent results on ( 2 and 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 =1 ( 2 0 (mod p for p > 3. Sun & R. Tauraso [Int. JNT 2011, Adv. in Appl. Math. 2010]. p a 1 ( ( 2 p a (mod p 2, 3 p a 1 =1 pa 3( 3 C 1 2 ( 2 (mod p 2, 8 9 p2 B p 3 (mod p 3 for p > 3.

8 Determination of ( 2 /m mod p 2 Let p be an odd prime. If p/2 < < p then ( 2 = (2! 0 (mod p. (! 2 Thus (/2 ( 2 m where m is an integer with p m. ( 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 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,....

9 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: ]. (/2 ( 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..

10 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 ( 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.

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

12 Sun and Zagier s results on Bell numbers For n = 1, 2, 3,... the nth derangement number D n is the number of permutations σ of {1,..., n} with σ(i i for all i = 1,..., n; in addition, D 0 := 1. It is well nown that D n n n! = ( 1! for all n N. Theorem (Sun and D. Zagier, Bull. Austral. Math. Soc., 84(2011 (i For every positive integer n we have =1 B ( n ( 1n 1 D n 1 (mod p for any prime p not dividing n. (ii Let p be any prime. Then for all n = 1,..., p 1 we have B n ( 1 D 1 ( n (mod p. =1

13 Part (i implies part (ii Part (i implies part (ii. For, n {1,..., p 1} with n, as p 1 n we have ( m n 0 (mod p. m=1 Thus, with the help of part (i, if n {1,..., p 1} then B n =1 B m=1 ( m n = m=1 ( m n ( m n ( 1 m 1 D m 1 (mod p. m=1 =1 B ( m

14 A direct proof of B p 2 (mod p Note that B n = n S(n,, where S(n, (a Stirling number of the second ind is the number of ways to partition {1,..., n} into nonempty sets. It is well nown that S(n, = 1! ( ( 1 j j j=0 j n for all n Z + and N. Clearly, S(n, 0 = 0 and S(n, 1 = S(n, n = 1 for any n Z +. Let p be a prime. With the help of Fermat s little theorem, if 1 < < p then S(p, S(1, = 0 (mod p. Therefore B p S(p, 1 + S(p, p = 2 (mod p.

15 Prove part (i by induction For any prime p, ( p 1 ( 1 B B = B p B 0 1 =1 =1 So the desired result holds when n = 1. Now fix n Z + and suppose that =1 for every prime p n. B ( n ( 1n 1 D n 1 (mod p (mod p. Let p be any prime not dividing n + 1. Recall the easy identity D n = nd n 1 + ( 1 n. If p n, then D n ( 1 n (mod p and hence =1 B ( n 1 =1 B ( 1 1 ( 1n D n (mod p.

16 Prove part (i by induction Now suppose that p n. Observe that =1 B ( n = =1 p 2 = l=0 p 2 = l=0 1 l=0 ( 1 l Bl ( n = B l l ( n l+1 r=1 B l l ( n l+1 r=1 p 2 l B l ( n l+1 l=0 l=0 B l ( n l+1 r=1 p 2 l=0 ( l+r 1 r 1 ( n r 1 ( l 1 r 1 n r 1 ( ( n B l ( n l ( p 1 l r 1 =l+1 n (r 1 ( 1 l ( n l l 1 n l (mod p.

17 Prove part (i by induction Thus, applying Fermat s little theorem we get n =1 B ( n l=1 B l ( n 1 l l=1 B l ( 1 l (mod p. Therefore l=1 B l ( n 1 l n =1 This concludes the induction step. B ( n + l=1 B l ( 1 l n( 1 n 1 D n = ( 1 n D n (mod p.

18 A further extension The Touchard polynomial T n (x of degree n is given by T n (x = n S(n, x. Note that T n (1 = B n. Similar to the recursion for Bell numbers, we have the recursion n ( n T n+1 (x = x T (x. Theorem (Sun & Zagier, Bull. Austral. Math. Soc., 84(2011. For every positive integer m, we have ( x m 0<n<p m 1 T n (x ( m n x p for any prime p not dividing m. (m 1! ( x (mod p!

19 Consequences Let p be a prime. The theorem implies the congruence 0<n<p m 1 T n (x ( m n 1 ( x m 1 l=0 (m 1! ( x l (mod p l! for any p-adic integer x not divisible by p, special cases being 0<n<p 0<n<p 0<n<p T n (x ( 2 n x 1 x (mod p for p 2, T n (x ( 3 n x 2 2x + 2 x 2 (mod p for p 3, T n (x ( 4 n x 3 3x 2 + 6x 6 x 3 (mod p for p 2.

20 On central Delannoy numbers D n := n ( ( n n + = 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 > 3 be a prime. Then ( 1 ( 2 D p 2 E p 3 (mod p 3, D 2 p (mod p, p =1 =1 D q p(2 (mod p, D 2 2 2q p(2 2 (mod p, =1 D 2 2 ( 1 p E p 3 (mod p, where q p (2 denotes the Fermat quotient (2 1/p.

21 An auxiliary identity In the proof of the theorem, the following new identity plays an important role. New Identity (Sun. For s = 1, 2 we have n = n ( 1 ( 2n (2 + 1 s = n + 16 n (2n + 1 s( 2n n Via Zeilberger s algorithm we find that both sides satisfy the same recurrence relation..

22 On central Delannoy numbers Theorem (Sun, Let p > 3 be a prime. Then (2 + 1( 1 D p 7 12 p4 B p 3 (mod p 5, (2 + 1D p + 2p 2 q p (2 p 3 q p (2 2 (mod p 4, where B 0, B 1, B 2... are Bernoulli numbers. Conjecture (Sun, For any prime p > 3 we have =1 D q p(2 + p q p (2 2 (mod p 2. Remar. We can show the congruence modulo p.

23 Congruences involving Schröder numbers The nth Schröder number is given by n ( n + n S n = C = ( 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. Theorem (Sun, Let p be an odd prime and let m be an integer not divisible by p. Then =1 ( S m m2 6m m The theorem in the case m = 6 gives that =1 ( m 2 6m + 1 (mod p. p S 0 (mod p for any prime p > 3. 6

24 An observation of van Hammer Let p = 2n + 1 be a prime. As observed by van Hammer, for = 0,..., n we have ( ( ( ( n n + n n 1 ( 1 = ( ( (p 1/2 ( p 1/2 = 1 j=0 = ( 2 j( 1 2 j ( 1 2 (! 2 = j2 p2 4 (! 2 ( ( 1/2 2 ( 2 2 = ( 4 = ( 2 j= (mod p 2 Thus D (/2 n ( 2 2 ( 16 (mod p 2.

25 On D (/2 and S (/2 mod p 2 Theorem 1 (conjectured by Z. W. Sun in 2009 and proved by Z. H. Sun [Proc. AMS, 2011] 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 (/2 ( 2 2 (/2 ( 2 2 ( 8 ( 16 ( 1(/4 2x p (mod p 2. 2x Theorem 2 (Z. W. Sun, 2011 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 (/2 S (/2 ( 1 (/4 2 ( 2 C ( 16 8 ( 2x p x (/2 ( 2 2 ( 16 (mod p 2.

26 A conjecture Conjecture (Sun, 2010 Let p > 3 be a prime. Then and (/2 =1 D S D S 2pH (/2 (mod p 3, =1 { 4x 2 (mod p if p 1 (mod 4 & p = x 2 + 4y 2, 0 (mod p if p 3 (mod 4.

27 Than you!