p-regular functions and congruences for Bernoulli and Euler numbers

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1 p-regular functions and congruences for Bernoulli and Euler numbers Zhi-Hong Sun( Huaiyin Normal University Huaian, Jiangsu , PR China Notation: Z the set of integers, N the set of positive integers, [x] the greatest integer not exceeding x, {x} the fractional part of x, ( m a the Jacobi symbol, Z p the set of rational p adic integers. [S1] Z.H. Sun, Congruences for Bernoulli numbers and Bernoulli polynomials, Discrete Math. 163(1997, [S2] Z.H. Sun, Congruences concerning Bernoulli numbers and Bernoulli polynomials, Discrete Appl. Math. 105(2000,

2 [S3] Z.H. Sun, Congruences involving Bernoulli polynomials, Discrete Math. 308(2008, [S4] Z.H. Sun, Congruences involving Bernoulli and Euler numbers, J. Number Theory 128(2008, [S5] Z.H. Sun, Euler numbers modulo 2 n, Bull. Austral. Math. Soc. 82(2010, [S6] Z.H. Sun, Congruences for sequences similar to Euler numbers, J. Number Theory 132(2012, [S7] Z.H. Sun, Identities and congruences for a new sequence, Int. J. Number Theory 8(2012, [S8] Z.H. Sun, Some properties of a sequence analogous to Euler numbers, Bull. Austral. Math. Soc., to appear. 2

3 [S9] Z.H. Sun and L.L. Wang, An extension of Stern s congruences, Int. J. Number Theory 9(2013, Definition of {B n }, {E n } and {U n } The Bernoulli numbers B 0, B 1, B 2,... are given by B 0 = 1, n 1 k=0 ( n k Bk = 0 (n 2. The first few Bernoulli numbers are given below: B 0 = 1, B 1 = 1 2, B 2 = 1 6, B 4 = 1 30, B 6 = 1 42, B 8 = 1 30, B 10 = 5 66, B 12 = , B 14 = 7 6, B 16 = Basic properties of {B n }: B 2n+1 = 0 for n 1. 3

4 von Staudt-Clausen Theorem (1844: B 2n + p 1 2n 1 p Z, where p runs over all distinct primes satisfying p 1 2n. von Staudt-Clausen: pb k(p 1 1 (mod p (k 1. Kummer (1850: If p is an odd prime and p 1 b, then B k(p 1+b k(p 1 + b B b b (mod p. The Euler numbers {E n } are given by E 2n 1 = 0, E 0 = 1, n r=0 ( 2n E2r = 0 (n 1. The first few Euler numbers are shown below: E 0 = 1, E 2 = 1, E 4 = 5, E 6 = 61, E 8 = 1385, E 10 = 50521, E 12 = r 4

5 The sequence {U n } is defined by U 0 = 1 and U n = 2 [n/2] k=1 Clearly U 2n 1 = 0 for n p-regular functions ( n 2k Un 2k (n 1. Definition 2.1 Let p be a prime. If f(0, f(1, f(2,... are all algebraic numbers which are integral for p, and for n = 1, 2, 3,..., n k=0 ( n k ( 1 k f(k 0 (mod p n, we call that f is a p-regular function. Example 2.1 Let p be a prime, b {0, 1, 2,...} and m N with p m. Then f(k = m k(p 1+b and g(k = m k(p 1+b 1 are p-regular functions. (by Fermat s little theorem and the binomial theorem 5

6 Example 2.2 ([S1] Let p be an odd prime and b {0, 1, 2,...}. Then f(k = p(p p k(p 1+b B k(p 1+b is a p-regular function. Example 2.3 ([S2] Let p be an odd prime, b N and p 1 b. Then f(k = (1 p k(p 1+b 1 B k(p 1+b k(p 1 + b is a p-regular function. Example 2.4 ([S3] Let p be an odd prime and b {0, 2, 4,...}. Then f(k = ( 1 ( 1 p 1 2 p k(p 1+b E k(p 1+b is a p-regular function. Example 2.5 ([S6] Let p be an odd prime and b {0, 2, 4,...}. Then f(k = ( 1 ( p 3 pk(p 1+b U k(p 1+b is a p-regular function. 6

7 Lemma 2.1 Let n 1 and k 0 be integers. For any function f, Proof. find f(k = n 1 n 1 r=0 k ( 1 n 1 r( k 1 r n 1 r r ( k r f(r + ( k ( 1 r ( r ( 1 s f(s. r=n r s=0 s As m j=0 ( 1 j( x j = ( 1 m ( x 1 m, we ( k r r=0 n 1 = = = = ( 1 r n 1 r s=0 ( k ( r r s=0 r=s n 1 ( k n 1 s=0 n 1 s s s ( r s ( 1 s f(s ( 1 r s f(s ( k s ( 1 r s f(s r s s=0 r=s n 1 ( k n 1 s f(s r=0 j=0 ( k s ( 1 j ( 1 n 1 r( k 1 r n 1 r j ( k r f(r. 7

8 By the binomial inversion formula, f(k = k ( k r=0 r ( 1 r Thus the result follows. r ( r s=0 s ( 1 s f(s. Theorem 2.1 Let p be a prime, n N, k {0, 1, 2,...} and let f be a p-regular function. Then f(k n 1 r=0 ( 1 n 1 r( k 1 r n 1 r ( k r f(r (mod p n. Example: Let p be an odd prime, k {0, 1, 2,...}, n, b N and p 1 b. Then (1 p k(p 1+b 1 B k(p 1+b k(p 1 + b n 1 r=0 ( 1 n 1 r( k 1 r n 1 r ( k (1 p r(p 1+b 1 B r(p 1+b r(p 1 + b (mod pn. r 8

9 Using the properties of Stirling numbers we deduce that: Theorem 2.2 Let p be a prime. Then f is a p-regular function if and only if for each positive integer n, there are a 0, a 1,..., a n 1 {0, 1,..., p n 1} such that f(k a n 1 k n a 1 k + a 0 (mod p n for every k = 0, 1, 2,.... Moreover, we may assume a s s!/p s Z p for s = 0, 1,..., n 1. If p n and f is a p-regular function, then a 0,..., a n 1 are unique. Example: For k N, B 4k+2 4k k k 3 700k k 1042 (mod 5 5, E 4k 750k k 2 620k (mod 5 5 (k > 1. 9

10 Lemma 2.2. Let p be a prime. Let f be a p-regular function. Suppose m, n N and t Z with t 0. Then n r=0 ( n r ( 1 r f(p m 1 rt 0 (mod p mn. Moreover, if A k = p k k r=0 ( k r ( 1 r f(r, then n r=0 ( n r ( 1 r f(p m 1 rt p mn t n A n (mod p mn+1 if p > 2 or m = 1, 2 mn t n n ( n Ar+n (mod 2 mn+1 r=0 r if p = 2 and m 2. 10

11 Theorem 2.3. Let p be a prime, k, m, n, t N, and let f be a p-regular function. Then f(ktp m 1 n 1 r=0 ( 1 n 1 r( k 1 r n 1 r ( k r f(rtp m 1 (mod p mn. Moreover, setting A s = p s s r=0 ( s r ( 1 r f(r we then have n 1 f(ktp m 1 ( 1 n 1 r( k 1 r ( k f(rtp m 1 r=0 n 1 r r p mn( k ( t n A n (mod p mn+1 n if p > 2 or m = 1, 2 mn( k ( t n n ( n Ar+n (mod 2 mn+1 n r=0 r if p = 2 and m 2. From Theorem 2.3 we deduce: 11

12 Theorem 2.4. Let p be a prime, k, m, t N. Let f be a p-regular function. Then (i ([S2] f(kp m 1 f(0 (mod p m. (ii f(ktp m 1 kf(tp m 1 (k 1f(0 (mod p 2m. (iii We have f(ktp m 1 k(k 1 f(2tp m 1 k(k 2f(tp m 1 2 (k 1(k 2 + f(0 (mod p 3m. 2 (iv We have f(kp m 1 f(0 k(f(0 f(1p m 1 (mod p m+1 if p > 2 or m = 1, f(0 2 m 2 k(f(2 4f(1 + 3f(0(mod 2 m+1 if p = 2 and m 2. 12

13 Example: E kϕ(p m +b (1 kpm 1 (1 ( 1 p 1 2 p b E b + kp m 1 E p 1+b (mod p m+1, U kϕ(p m +b ( 1 ( p 3 pb U b (mod p m, where ϕ is Euler s totient function. Lemma 2.3. For n = 0, 1, 2,... and any two functions f and g we have n ( n k k=0 n = s=0 ( 1 k f(kg(k ( n s s r=0 ( s ( 1 r F (n s + r G(s, r where F (m = ( m m k=0 k ( 1 k f(k and G(m = ( mk=0 m k ( 1 k g(k. Proof. We first claim that n ( n r=0 r ( 1 r f(r + m = m ( m r=0 r ( 1 r F (r + n. 13

14 Clearly the assertion holds for m = 0. Now assume that it is true for m = k. It is easily seen that n r=0 = = = ( n r n s=0 k ( k s=0 k+1 s=0 ( 1 r f(r + k + 1 ( n ( 1 s n+1 f(k + s ( n + 1 ( 1 s f(k + s s s s ( 1 s F (n + s s=0 k ( k s=0 ( k + 1 ( 1 s F (n + s. s So the assertion is true by induction. s ( 1 s F (n s From the binomial inversion formula we know that g(k = k s=0 ( k s ( 1 s G(s. Thus, by the above assertion we have 14

15 n k=0 ( n k n ( n = = = = = k=0 n ( 1 k f(kg(k k s=0 k=s n ( n s ( 1 k f(k n ( n ( k s=0 k=s n ( n n s s k ( k s=0 s ( 1 s G(s ( 1 k s f(k G(s k s n s=0 r=0 n ( n s ( s s=0 s r=0 r which completes the proof. ( n s ( 1 k s f(k G(s k s ( n s ( 1 r f(r + s G(s r ( 1 r F (n s + r G(s, Theorem 2.5 (Product Theorem. Let p be a prime. If f and g are p-regular functions, then f g is also a p-regular function. 15

16 3. p-regular functions involving Bernoulli polynomials and generalized Bernoulli numbers The Bernoulli polynomial B n (x is given by B n (x = n k=0 ( n k Bk x n k. For x Z p let x p denote the unique number n {0, 1,..., p 1} such that x n (mod p. Theorem 3.1 Let p be a prime and let b be a nonnegative integer. (i ([S2, 2000], [Young, 2001] If p 1 b, x Z p and x = (x + x p /p, then f(k = B k(p 1+b (x pk(p 1+b 1 B k(p 1+b (x k(p 1 + b is a p regular function. 16

17 (ii ([S2, (3.1, Theorem 3.1 and Remark 3.1] If a, b N and p a, then f(k = (1 p k(p 1+b 1 (a k(p 1+b 1 B k(p 1+b k(p 1 + b is a p-regular function. Let χ be a primitive Dirichlet character of conductor m. The generalized Bernoulli number B n,χ is defined by m r=1 χ(rte rt e mt 1 = n=0 B n,χ t n n!. Let χ 0 be the trivial character. It is well known that B 1,χ0 = 1 2, B n,χ 0 = B n (n 1, B n,χ = m n 1 m r=1 χ(rb n( r m If χ is nontrivial and n N, then clearly m r=1 χ(r = 0 and so. 17

18 B n,χ n m = mn 1 r=1 = m n 1 m r=1 ( Bn ( r χ(r m B n n χ(r B n( m r B n. n + B n n When p is a prime with p m, by [S1, Lemma 2.3] we have (B n ( r m B n/n Z p. Thus B n,χ /n is congruent to an algebraic integer modulo p. Theorem 3.2. Let p be a prime and let b be a nonnegative integer. (i ([Young, 1999], [Fox, 2002], [S2, 2000] If b, m N, p m and χ is a nontrivial primitive Dirichlet character of conductor m, then f(k = (1 χ(pp k(p 1+b 1 B k(p 1+b,χ k(p 1 + b is a p regular function. 18

19 (ii ([S2, Lemma 8.1(b] If m N, p m and χ is a nontrivial Dirichlet character of conductor m, then f(k = (1 χ(pp k(p 1+b 1 pb k(p 1+b,χ is a p regular function. Definition 3.1 For a 0 define {E n (a } by [n/2] k=0 ( n 2k a 2k E (a n 2k = (1 an (n = 0, 1, 2,.... Clearly E (1 n = E n. Theorem 3.3 ([S6]. Let a be a nonzero integer and b {0, 1, 2,...}. Then f(k = E (a 2k+b is a 2 regular function. 19

20 Theorem 3.4 ([S6]. Let p be an odd prime and let b be a nonnegative integer. Then f 2 (k = (1 ( 1 p 1 2 b+[p 1 4 ] p k(p 1+b E (2 k(p 1+b and f 3 (k = (1 ( 1 [p+1 6 ] ( p 3 b+1 p k(p 1+b E (3 k(p 1+b are p-regular functions. 4. Congruences for p 1 x=1 1 x k (mod p3 and p 1 2 x=1 1 x k (mod p3 20

21 Theorem 4.1 ([S2] Let p be a prime greater than 3. (a If k {1, 2,..., p 4}, then (b (c (d x=1 x=1 x=1 x=1 1 x k k(k + 1 B p 2 k 2 p 2 k p2 (mod p 3 ( if k is odd, B2p 2 k k 2p 2 k 2 B p 1 k p 1 k if k is even. p (mod p 3 1 x p 3 (1 2 3B p+1p 4 3 p2 (mod p 3. 1 x p 2 (2 + pb p 1p p2 (mod p 3. 1 x p 1 pb 2p 2 3pB p 1 + 3(p 1 (mod p 3. 21

22 Proof. For m Z it is clear that 1 m + 2 m + + (p 1 m = B m+1(p B m+1 m + 1 = 1 m+1 ( m + 1 Bm+1 r p r m + 1 r r=1 = pb m + p2 2 mb m 1 + m+1 r=3 ( m r 1 pbm+1 r pr 4 r p 3. Since pb m+1 r, pr 4 r (4.1 Z p for r 3 we have 1 m +2 m + +(p 1 m pb m + p2 2 mb m 1 (mod p 3. Let k {1, 2,..., p 1}. From (4.1 and Euler s theorem we see that 22

23 x=1 p 1 1 x k x ϕ(p3 k x=1 pb ϕ(p 3 k + p2 2 ( ϕ(p 3 k B ϕ(p 3 k 1 pb ϕ(p 3 k k 2 p2 B ϕ(p 3 k 1 = pb ϕ(p 3 k (mod p3 k 2 p2 B ϕ(p 3 k 1 (mod p3 if k is even, if k is odd. For k {1, 2,..., p 2} we see that B ϕ(p 3 k ϕ(p 3 k = B (p 2 1(p 1+p 1 k (p 2 1(p 1 + p 1 k (p 2 1 B 2p 2 k 2p 2 k (p2 2(1 p p 2 k B p 1 k p 1 k B 2p 2 k 2p 2 k + 2(1 pp 2 k B p 1 k p 1 k (mod p2. 23

24 Thus, (4.2 pb ϕ(p 3 k ( kp B 2p 2 k 2p 2 k + 2(1 pp 2 k B p 1 k p 1 k kp( B 2p 2 k 2p 2 k 2 B p 1 k p 1 k (mod p3 if k < p 3, (p 3p( B p+1 p + 1 2(1 pb 2 2 (mod p3 if k = p 3. When k {1, 2,..., p 3}, it follows from Kummer s congruences that B ϕ(p 3 k 1 ϕ(p 3 k 1 = Thus, (4.3 B (p 2 1(p 1+p 2 k (p 2 1(p 1 + p 2 k B p 2 k p 2 k (mod p. k 2 p2 B ϕ(p 3 k 1 k 2 p2 ( k 1 B p 2 k p 2 k (mod p3. 24

25 Combining the above we get x=1 1 x k kp ( B2p 2 k 2p 2 k 2 B p 1 k p 1 k if k {2, 4,..., p 5}, ( 1 2 3B p+1p 4 3 p2 (mod p 3 if k = p 3, k(k + 1 B p 2 k p 2 k p2 (mod p 3 2 if k {1, 3,..., p 4}. This proves parts (a and (b. (mod p 3 Now consider parts (c and (d. Note that pb r(p 1 1 (mod p for r 1. From the above and [S1, Corollary 4.2] we see that x=1 1 2 p xp 2 2 p2 B ϕ(p 3 (p 1 p 2 2 p((p2 1pB p 1 (p 2 2(p 1 p 2 2 p(pb p p p(pb p p2 (mod p 3 25

26 and x=1 1 x p 1 pb ϕ(p 3 (p 1 ( p 2 1 pb2p 2 (p 2 1(p 2 3pB p 1 2 (p 1 + ( p (1 3p2 2 pb 2p 2 (3 4p 2 pb p 1 + (3 5p2 (p 1 2 pb 2p 2 3pB p 1 + 3(p 1 (mod p 3. This concludes the proof. One can similarly prove that 26

27 Theorem 4.2 Let p > 5 be a prime and k {1, 2,..., p 5}. Then 1 x=1 x k k ( B3p 3 k 3p 3 k 3 B 2p 2 k 2p 2 k + 3 B p 1 k p 1 k ( k + 2p 3 B p 3 k if 2 k, ( k p 3 k (mod p4 B 2p 3 k ( 2p 3 k 2 B p 2 k p 2 k p2 (mod p 4 if 2 k. p 27

28 Theorem 4.3 ([S2]. Let p > 5 be a prime. p 1 2 x=1 (a If k {2, 4,..., p 5}, then 1 x k k(2k+1 1 p( B 2p 2 k 2 2p 2 k 2 B p 1 k p 1 k (mod p3. (b If k {3, 5,, p 4}, then p 1 2 x=1 1 x k (2k 2(2 B p k p k B 2p 1 k 2p 1 k (mod p2. p 1 2 x=1 (c If q p (2 = (2 p 1 1/p, then 1 x 2q p (2 + pq 2 p (2 2 3 p2 q 3 p ( p2 B p 3 (mod p 3. 28

29 Theorem 4.4 ([S4]. Let p > 3 be a prime and q p (a = (a p 1 1/p. Then k=1 k p (mod 3 1 k 1 2 q p(3 1 4 pq p( p2 q p (3 3 p2 81 B p 3(mod p 3 k=1 k p (mod 4 1 k 3 4 q p(2 3 8 pq p( p2 q p (2 3 k=1 k p (mod 6 p2 192 B p 3 (mod p 3, 1 k ( q p( q p(3 p 6 q p( q p(3 2 ( + p q p( q p( B p 3 (mod p 3. 29

30 5. A congruence for (p 1! (mod p 3 Let p be a prime greater than 3. The classical Wilson s theorem states that (p 1! 1 (mod p. In 1900 J.W.L.Glaisher showed that (p 1! pb p 1 p (mod p 2. Here we give a congruence for (p 1! modulo p 3. Theorem 5.1 ([S2]. For any prime p > 3 we have (p 1! pb ( 2 2p 2 2p 2 pb p 1 pbp 1 p 1 1 (mod p 3. 2 p 1 The proof is based on the following Newton s formula. 30

31 Newton s formula: Suppose that x 1, x 2,..., x n are complex numbers. If S m = x m 1 + xm xm n, A m = x i1 x i2 x im, 1 i 1 <i 2 < <i m n for k = 0, 1,..., n we have S k A 1 S k 1 + A 2 S k ( 1 k 1 A k 1 S 1 + ( 1 k ka k = Congruences involving Bernoulli and Euler polynomials The Euler polynomials {E n (x} are given by E n (x + n r=0 It is known that E n (x = 1 2 n n r=0 = 2 n + 1 ( n r ( n Er (x = 2x n (n 0. (2x 1 n r E r r ( ( B n+1 (x 2 n+1 x B n

32 Also, x=0 x=0 x k = B k+1(p B k+1, k + 1 ( 1 x x k = ( 1p E k (p E k (0. 2 Theorem 6.1 ([S3]. Let p, m N and k, r Z with k 0. Then x k x=0 x r(mod m ( = mk k + 1 B k+1 ( p m + { r p m } B k+1 ({ r m } 32

33 and x=0 x r(mod m = mk 2 ( 1 x r m x k ( ( 1 [r p ( 1 [ r m ] E k ({ r m ( { m ] p r p E k m + m }. } Proof. For any real number t and nonnegative integer n it is well known that B n (t + 1 B n (t = nt n 1 (n 0, E n (t E n (t = 2t n. Hence, for x Z we have ( { } ( { x + 1 r x 1 x r x B k+1 m + B k+1 m m + { m 0 if m x r, = (k + 1( x m k if m x r. } 33

34 Thus ( { p r p B k+1 m + m = x=0 } ( B k+1 ( x + 1 m + ( { x r x B k+1 m + m = k + 1 m k x=0 x r(mod m ({ r B k+1 m { r x 1 x k. } m } } Similarly, if x Z, then ( { } ( 1 [r x 1 m ] x + 1 r x 1 E k m + ( { m} ( 1 [r x m ] x r x E k m + { m 0 if m x r, = ( 1 r x m 2( x m k if m x r. 34

35 Thus ( 1 [r p = ( { m ] p r p E k m + m ( 1 [ r m ] E k ({ r m x=0 } } { ( 1 [r x 1 m ] E k ( x + 1 m + ( { ( 1 [r x m ] x r x E k m + m = 2 m k x=0 x r(mod m This completes the proof. ( 1 x r m x k. { r x 1 m }} } Corollary 6.1 Let p be an odd prime and k {0, 1,..., p 2}. Let r Z and m N with p m. Then x k x=0 x r(mod m mk ( k + 1 B k+1 ({ r p m } B k+1 ({ r m } (mod p 35

36 and x=0 x r(mod m mk 2 ( 1 x r m x k ( ( 1 [r p ( 1 [ r m ] E k ({ r m ({ m ] r p E k m } } (mod p. Proof. If x 1, x 2 Z p and x 1 x 2 (mod p, then B k+1 (x 1 B k+1 (x 2 k+1 0 (mod p and E k (x 1 E k (x 2 (mod p. Thus the result follows from Theorem 6.1. In the case k = p 2, Corollary 6.1 is due to Zhi-Wei Sun. Inspired by Zhi-Wei Sun s work, I established Theorem 6.1 and Corollary

37 Corollary 6.2 Let p be an odd prime. Let k {0, 1,..., p 2} and m, s N with p m. Then and ( 1 k k + 1 ( ( 1 [(s 1p } ({ (s 1p B k+1 m r k (mod p (s 1p m <r sp m } B k+1 ({ sp m } ({ ({ m ] (s 1p E k ( 1 [sp m ] sp E k m m 2( 1 k 1 ( 1 r r k (mod p. (s 1p m <r sp m } 37

38 Theorem 6.2. Let m, s N and let p be an odd prime not dividing m. Then ( 1 sm p k=1 k sp(mod m ( p k (s 1p m <k<sp m ( 1 km (s 1p B p 1 ({ m } B p 1({ sp }(mod pif 2 m, m 1 (( 1[(s 1p m ] (s 1p E p 2 ({ 2 m } ( 1 [sp m ] E p 2 ({ sp } (mod p if 2 m. m k Corollary 6.3. Let m, n N and let p be an odd prime not dividing m. (i If 2 m, then ({ np B p 1 m m p n s=1 } B p 1 ( 1 s 1 k=1 k sp(mod m ( p k (mod p. 38

39 (ii If 2 m, then ({ ( 1 [np m ] np E p 2 m 2m p n s=1 ( 1 s 1 } + 2p 2 p k=1 k sp(mod m ( p k (mod p. In the cases m = 3, 4, 5, 6, 8, 9 the formulae for p were given by me (published k=0 k r (mod m ( p k in , in the case m = 10 the formula was published by Z.H.Sun and Z.W.Sun in 1992, and for the case m = 12, the formula was given by Zhi-Wei Sun (published in

40 7. Extension of Stern s congruence for Euler numbers For k, m N and b {0, 2, 4,...}. The Stern s congruence states that (7.1 E 2 m k+b E b + 2 m k (mod 2 m+1. In 1875 Stern gave a brief sketch of a proof of (7.1. Then Frobenius amplified Stern s sketch in There are many modern proofs of (7.1. (Ernvall(1979, Wagstaff(2002, Zhi-Wei Sun(2005, Zhi-Hong Sun(2008 Let b {0, 2, 4,...} and k, m N. In [S5, 2010] I showed that E 2 m k+b E b + 2 m k (mod 2 m+2 for m 2, E 2 m k+b E b m k (mod 2 m+3 for m 3, 40

41 and for m 4, E 2 m k+b { Eb m k (mod 2 m+4 if b 0, 6 (mod 8, E b 3 2 m k (mod 2 m+4 if b 2, 4 (mod 8. In [S9], Z.H.Sun and L.L.Wang (IJNT, 2013 established a congruence for E 2 m k+b (mod 2 m+7. In particular, for m 7, E 2 m k+b E b +2 m k(7(b (mod 2 m+7. For a 0 recall that {E n (a } is defined by [n/2] k=0 ( n 2k a 2k E (a n 2k = (1 an (n = 0, 1, 2,.... From [S6] we know that E (a n = (2a n E n ( 1 2a = n k=0 ( n k = [n/2] k=0 ( n 2k 2 k+1 ( 1 2 k+1 B k+1 k + 1 ak. (1 a n 2k a 2k E 2k 41

42 Theorem 7.1([S6, 2012]. Let a be a nonzero integer, k, m N, m 2 and b {0, 1, 2,...}. Then E (a 2 m k+b E(a b 2 m k(a 3 ((b a + 2 m ka 3 (b 1 (mod 2 m+4+3α if 2 α a and α 1, 2 m ka((b m k(b + 1 (mod 2 m+4 if 2 a and 2 b, 2 m k(a 2 1 (mod 2 m+4 if 2 ab. As E n (1 = E n, the theorem is a vast generalization of Stern s congruence. 42

43 8. ( 1 n U 2n and ( 1 n E (a 2n are realizable are two sequences sat- If {a n } n=1 and {b n} n=1 isfying a 1 = b 1, b n +a 1 b n 1 + +a n 1 b 1 = na n (n > 1, we say that (a n, b n is a Newton-Euler pair. If (a n, b n is a Newton-Euler pair and a n Z for all n = 1, 2, 3,..., we say that {b n } is a Newton- Euler sequence. Let {b n } be a Newton-Euler sequence. clearly b n Z for all n = 1, 2, 3,.... Then Z.H. Sun, On the properties of Newton-Euler pairs, J. Number Theory 114(2005, Lemma 8.1. Let {b n } n=1 be a sequence of integers. Then the following statements are equivalent: 43

44 (i {b n } is a Newton-Euler sequence. (ii d n µ(n d b d 0 (mod n for every n N. (iii For any prime p and α, m N with p m we have b mp α b mp α 1 (mod p α. (iv For any n, t N and prime p with p t n we have b n bn p (mod p t. (v There exists a sequence {c n } of integers such that b n = d n dc d for any n N. (vi For any n N we have k 1 +2k 2 + +nk n =n b k 1 1 bk 2 2 bk n n 1 k 1 k 1! 2 k 2 k 2! n k n k n! Z. 44

45 (vii For any n N we have 1 n! b 1 b 2 b 3... b n 1 b 1 b 2... b n 1 2 b b n 2. (n 1 b 1 Z. Let {b n } n=1 be a sequence of nonnegative integers. If there is a set X and a map T : X X such that b n is the number of fixed points of T n, following Puri and Ward we say that {b n } is realizable. Puri and Ward (2001 proved that a sequence {b n } of nonnegative integers is realizable if and only if for any n N, n 1 d n µ(n d b d is a nonnegative integer. Thus, using Möbius inversion formula we see that a sequence {b n } is realizable if and only if there exists a sequence {c n } of nonnegative integers such that b n = d n dc d for any n N. 45

46 J. Arias de Reyna (Acta Arith. 119(2005, showed that {E 2n } is a Newton-Euler sequence and { E 2n } is realizable. In [S6, S7] I proved the following results. Theorem 8.1 ([S7, 2012]. {U 2n } is a Newton- Euler sequence and {( 1 n U 2n } is realizable. Theorem 8.2 ([S6, 2012]. Let a N. Then {( 1 n E (a 2n } is realizable. 9. Congruences involving {U n } In [S7], Z.H. Sun introduced the sequence {U n } as below: U 0 = 1, Since U 1 = 0. n 1. U n = 2 [n/2] k=1 ( n 2k Un 2k (n 1. By induction, U 2n 1 = 0 for 46

47 The first few values of U 2n are shown below: U 2 = 2, U 4 = 22, U 6 = 602, U 8 = 30742, U 10 = , U 12 = Theorem 9.1 For n N we have U 2n = 3 2n E 2n ( 1 3 = 2(22n n 2 2n + 1 = 2 ( 2 2n nB 2n+1( 1 3 2n + 1 B2n+1( 1 6 2n + 1. For d Z with d < 0 and d 0, 1 (mod 4 let h(d denote the class number of the form class group consisting of classes of primitive, integral binary quadratic forms of discriminant d. Theorem 9.2 ([S7]. Let p be a prime of the form 4k + 1. Then ( 1 + 2( 1 p 1 4 h( 3p (mod p U p 1 2 and so p U p

48 Recall that the Fermat quotient q p (a = (a p 1 1/p. Theorem 9.3 ([S7]. Let p be a prime greater than 5. Then (i [p/6] k=1 1 k 2q p (2 3 2 q p(3+p ( q p ( q p(3 2 5p 2 (p 3 U p 3 (mod p 2, (ii [p/3] k=1 1 k 3 2 q p( pq p(3 2 p( p 3 U p 3 (mod p 2, (iii [2p/3] k=1 ( 1 k 1 k 9 k=1 3 k+p 1 k 3p( p 3 U p 3 (mod p 2. 48

49 (iv We have ( 1 [ p 6 ]( p 1 [ p 6 ] ( 1 + p 2q p (2 + 3 ( 2 q p(3 + p 2 q p ( q p (2q p (3 + 3 ( 8 q p(3 2 p 5 U p 3 (mod p 3 3 and ( 1 [ p 3 ]( p 1 [ p 3 ] pq p( p2 q p (3 2 p2 2 ( p 3 U p 3 (mod p 3. Theorem 9.4. Let p > 3 be a prime and k {2, 4,..., p 3}. Then [p/6] x=1 1 x k 6k x=1 6 x p 1 x k 6k (2 k + 1 4(2 k ( p 3 U p 1 k (mod p 49

50 and [p/3] x=1 1 x k 3k x=1 3 x p 1 x k 6 k 4(2 k ( p 3 U p 1 k (mod p Theorem 9.5. Let p > 3 be a prime and k {2, 4,..., p 3}. Then and [p/3] x=1 [ p+3 6 ] x=1 ( 1 x 1 1 x k 3k 2 ( p 3 1 (2x 1 k 3 k 2 k U p 1 k (mod p ( p 3 U p 1 k (mod p. 50

51 By [S7], (9.1 B p 2 ( 1 3 6U p 3 (mod p for any prime p > 3. S. Mattarei and R. Tauraso (Congruences for central binomial sums and finite polylogarithms, J. Number Theory 133(2013, proved that for any prime p > 3, k=0 Thus, (9.2 ( 2k ( p k 3 p2 3 B p 2 k=0 ( 1 3 (mod p 3. ( 2k ( p 2p k 3 2 U p 3 (mod p 3 for any prime p > 3. The congruence k=0 ( 2k ( p 3 k (mod p 2 was found and proved by Z.W. Sun and R. Tauraso (Adv. in Appl. Math. 45(2010,

52 Suppose that p is a prime of the form 3k+1 and so 4p = L 2 +27M 2 for some integers L and M. Assume L 1 (mod 3. From (9.1 and the work of J. B. Cosgrave and K. Dilcher (Mod p 3 analogues of theorems of Gauss and Jacobi on binomial coefficients, Acta Arith. 142(2010, we have (9.3 ( 2(p 1 3 p 1 3 ( L + p L + p2 L 3 (1 + p 2 U p 3 L + p L + p2 ( 1 L 3 LU p 3 (mod p 3. 52