Congruences involving Bernoulli and Euler numbers Zhi-Hong Sun

Transcription

1 The aer will aear in Journal of Nuber Theory. Congruences involving Bernoulli Euler nubers Zhi-Hong Sun Deartent of Matheatics, Huaiyin Teachers College, Huaian, Jiangsu 300, PR China Received January 006; revised August 006 Counicated by David Goss Abstract. Let [x] be P the integral art of x. Let > 5 be a rie. In the aer [/4] we ainly deterine x= x od P, [/4] od 3, = od 3 P = od in ters of Euler Bernoulli nubers. For exale, we have [/4] X x= 8E x 3 4E B 3 od, where E n is the nth Euler nuber B n is the nth Bernoulli nuber. MSC: Priary B68, Secondary A07. Keywords: Congruence, Bernoulli nuber, Bernoulli olynoial, Euler nuber. Introduction. The Bernoulli nubers {B n } Bernoulli olynoials {B n x} are defined by B 0 =, n n B = 0 n B n x = =0 n =0 n B x n n 0. The Euler nubers {E n } Euler olynoials {E n x} are defined by e t e t + = n=0 which are equivalent to see [MOS] E n t n n! t < π e xt E 0 =, E n = 0, E n x = n n r=0 n r=0 e t + = n=0 E n x tn n! n E r = 0 n r n x n r E r. r t < π, E-ail: hyzhsun@ublic.hy.js.cn URL: htt://

2 Let [x] be the integral art of x. For a given rie let Z denote the set of rational integers those rational nubers whose denoinator is not divisible by. For a Z with a 0 od, as usual we define the Ferat quotient q a = a /. In the aer we establish soe congruences involving Bernoulli Euler nubers. In articular, in Z we have q q + E 4 E q 3 od 3, 4 << [ 4 ] [ 4 ] < 4 + = = + 3q + 3q E 3 od 3, 4 q 8 q + q B 3 od 3, q 7 B 3 od 3, q + 3 q B od, where is a rie greater than 5. In addition to the above notation, we also use throughout this aer the following notation: Z the set of integers, N the set of ositive integers, {x} the fractional art of x, ϕn Euler s totient function.. Basic Leas. We begin with a useful identity involving Bernoulli olynoials. Lea.. Let, N, r Z with 0. Then x = { r } { r B B + x=0 x rod }. In the case = Lea. is well nown. See [MOS] [IR]. Lea. was established by the author in 99. A roof is given in [S4], a generalization was ublished by the author s brother Z.W. Sun [Su]. Fro [S, Lea.3] [IR, Proosition 5..4,. 38] we have Lea.. Suose that, N with >. If x, y Z, then B x Z B x B y/ Z. If is an odd rie such that, then B x/ Z. Lea.3 [MOS]. Let x y be variables n N. Then i B n+ = 0. ii B n x = n B n x. iii B n x + y = n Bn r yx r. r=0 iv E n x = n n B n x+ B n x. n r

3 Lea.4 [MOS], [GS]. Let n N. Then 3 B n = B n = n n B n, B n = B n = 3 3n n B n 5 B n = B n 6 6 = n 3 3 n 6 n B n. Lea.5. For any nonnegative integer n we have E n = 4 n+ B n+ 4 n +. Proof. It is well nown that E n = n E n. Thus alying Lea.3 we see that E n = n E n+ = n n+ n + = 4n+ B n+ B n+ n This roves the lea. Fro [S3, Corollary 3. Theore 4.] we have: 3 B n+ B n+ 4 4 = 4n+ n+ n + B. 4 Lea.6. Let be an odd rie, x Z, b N with b. Then B +b x + b B bx b B +b x + b B +bx + b B bx b od for b od for b >. Lea.7. Let > 3 be a rie, r Z, N with < 3. Then x= x rod x B +{ r ϕ3 } B ϕ 3 +{ r } ϕ B { r } + B { r } + { r } + + B od 3. 3

4 Proof. Fro Leas.,.3iii Euler s theore we see that x= x rod x x=0 x rod x ϕ3 = ϕ3 { r } { r ϕ 3 B ϕ B ϕ 3 + } { ϕ3 + = ϕ3 ϕ 3 + j { r } ϕ 3 + j j B ϕ 3 + j j=0 { r B ϕ 3 + }} { = ϕ3 B ϕ 3 +{ r } B ϕ 3 +{ r } ϕ ϕ3 + ϕ 3 + j=3 { r } B ϕ 3 ϕ 3 3 j j j B ϕ 3 + j j 3 + B ϕ 3 { r } { r }} od 3. As < 3 we have ϕ 3 so B ϕ 3 { r } Z by Lea.. For j 4 we have j 3 /j 0 od. Thus j 3 j B ϕ 3 + j{ r } Z for j 3. Hence, by the above we obtain x= x rod x B ϕ 3 +{ r } B ϕ 3 +{ r } ϕ ϕ3 Fro Lea.6 we see that } B ϕ 3 B ϕ 3 { r ϕ 3 { r } ϕ 3 B ϕ 3 + B ϕ 3 { r } od 3. = B { r } + + B { r } { r } od = B { r } + + B { r } B { r } od. 4

5 Thus B ϕ 3 B ϕ 3 { r } { r } + { r } + B od { r } B { r } B Now utting all the above together we obtain the result. od. Lea.8. Let be an odd rie, a Z, a 0 od, n N > n +. Then a ϕn n s n s q a s od n. s s= Proof. As > n + we see that s /s! Z s /s! 0 od for s. Thus, a ϕn n = + q a n n = n n s= = q a + q a + = n s= n s= n s= n s s q a s n n s + s q a s s! s + s q a s s! s s q a s od n. s As > n + we see that s n /s Z for s n +. Thus for s n + we have s s s q a s = s s n s q a s n 0 od n. Now utting the above together we obtain the result. Lea.9. Let be an odd rie {0,,..., }. Then 3 6 i= { { i + i= 3 3 i= i i= 5 i i= i i= } i i + i= } i 3 od 4.

6 Proof. For = 0,, it is easy to verify the result. Now assue 3. Clearly =! = { + + 3! + i<j 3 i<j<l i j i<j<l ijl + i j l + i= i<j i ij i= } +! i + od 4. Observe that i= = i i + i= i<j ij i= 3 = i ijl = 6 i,j,l = 6 i<j<l i<j<l ijl + 3 j= ijl + 3 j i= i= j= j i i 3 i j + i= i= i 3 i 3 + i 3. i= We then have i<j ij = { i= i i= } i i<j<l ijl = { 6 i= 3 3 i i= i i= i + i= } i 3. Now utting all the above together we obtain the result. We rear that the congruence for od 3 was given by Leher in [L,. 360]. 3. Congruences for x x< x = 3, 4, 6 [/4] x= x. 6

7 Theore 3.. Let > 3 be a rie. Then i ii iii = od 3 = od 4 = od 6 q 3 4 q q B 3 od q 3 8 q + 4 q 3 9 B 3 od 3. 3 q + 4 q 3 6 q + 8 q q 3 + q B 3 od 3. Proof. Note that B n 0 = B n B n+ = 0 for n N. Taing = r = in Lea.7 we see that if N, then = od B B { } 3 3 B 3 od 3. As B n x = B n x, for = 3, 4, 6 we have B n { } = B n. Since B od 3 by [S, Corollary 4.], using Leas.4,.8 Euler s theore we see that B B 3 = = B B 3 q 3 q q 3 3 od Siilarly, we have B B 4 = B 4 = B 3 q q + 3 q 3 od

8 B B 6 = 3 3 B 6 = B q q + 3 q Now cobining all the above we obtain the result. Corollary 3.. Let > 3 be a rie. Then = od 4 q 3 q q 3 3 od 3. 4 q 8 q + q B 3 od 3. Proof. Using [S3, Rear 5.3] we now that = od 4 = = + = od 4 q q + 3 q 3 4 B 3 od 3. Thus alying Theore 3.ii we deduce the result. Rear 3. Let {3, 4, 6}. In 938 E. Leher [L] obtained the congruences for od. Using the forulas for, in [S] the author = od gave congruences for = r od od. Corollary 3.. Let > 3 be a rie. Then = od 3 r od 3 q + q q 4 q q q B 3 od 3. 8

9 Proof. Clearly = od 3 = = +3 od 6 = = od 3 = = +3 od 6 = od 6. Thus aealing to Theore 3. we obtain the result. Theore 3.. Let > 3 be a rie. Then 4 << < 4 3 3q + q + E 4 E 3 q B 3 od 3 q q + E 4 E q 3 od 3. Proof. Taing =, r = 0 = 4 in Lea.7 we find x= 4 x x B { 4 } B { } 9 B 3 od 3. 4 As B n 3 4 = n B n 4 we then have < 4 = 4 x= 4 x x B 4 B B 3 { 4 } B od 3. 4 Fro the roof of Theore 3. we now that B { 4 } B B 4 B B 4 3q 3 q + q 3 od 3. 9

10 By Leas.5.6 we have 3. E 4 = 4 3 B B 4 = E 3 od. Observe that a s = + q a s + sq a od for a, s Z with a. We then have 4 q 4 od 4 q 4 od. Hence 4 B B 4 = E 4 4 E 3 4 q 4E 4 q 4E 3 = E 4 E 3 q 4E 4 E 3 E 4 E 3 od. On the other h, by Lea.4 we have B 3 4 = B 3 = Thus cobining the above we obtain < B 3 8B 3 od. 3q + 3 q q 3 + E 4 E B 3 od 3. Fro [S3, Theore 5.] we now that < Observe that q + q 3 q 3 7 B 3 od 3. 4 << = < < 4. We then obtain the reaining result. Rear 3. For any rie > 3, the congruence [ 4 ] = 3q od was first established by Lerch see [D], a sile roof concerning the forula for was given by the author in [S]. 4 0

11 Corollary 3.3. Let > 3 be a rie. Then < 4 3q + 3 q E 3 od 4 << q q + E 3 od. Lea 3.. Let > 5 be a rie, r Z,, N, < < 3. Then x= x r od x B { r } B { r } + + B { r B { r } B { r } } od. Proof. Fro Lea.6 we see that if x Z, then B ϕ 3 +x ϕ 3 + = B x B x B + x + od B x B x B x + B x od. This together with Lea.7 gives the result. Putting r = 0, in Lea 3. noting that B n+ = 0n we deduce the following result. Theore 3.3. Let > 5 be a rie,, N, < < 3. Then x= x od x B { } B B { } B od if, B { } B { } + + B od if.

12 [/] x= x B B { } B { } B B { } + + B { } od if, + B { } + + B { } od if. Corollary 3.4. Let > 5 be a rie {3, 5,..., 4}. Then i ii iii iv v [/3] x= [/4] x= [/6] x= 6 <x< 4 4 <x< 3 x 3 3 B od, x B od, x 3 B od, x 3 B od, x 3 + B od. Proof. Fro Lea.6 Theore 3.3 we see that for = 3, 4, 6, [/] x= x B B { } = B B od. Now alying Lea.4 we deduce i-iii. iv follows fro ii iii, v follows fro i ii. Theore 3.4. Let > 5 be a rie {3, 5,..., 4}. Then i ii iii x= x od 3 x= x od 4 x= x od 6 x 3 B 3 B od, x B B od, x 3 B 6 B od.

13 Proof. Let {3, 4, 6}. As B n x = B n x, we see that B n { } = B n. Hence, alying Theore 3.3 we have 3. x= x od x B B B B od. By Lea.4 we have B n 3 B n = 3 n 3B n. Note that 3 3 od. By the above we obtain 3 x= x od 3 x B 3 3 B B = 3 + B B 3 B B = B B + 3B B 3 3 B od. B + 3B This roves i. Now we consider ii. Fro Lea.4 we now that B n 4 B n = 4n n B n. Observe that s = + q s + sq od. Using 3. we see that 4 x= x od 4 x 4 B B 4q q B q q B = + B q B + B B B B od. 3

14 This roves ii. Finally we consider iii. As B n 6 = n 3 n B n, by 3. we have 6 x= x od 6 x 3 B 3 B } B { q 3 q 3 { } q 3 B q q 3 B q q 3 B q 3 = 3 B B 3 q + 3 q 3 B B 3 B B od. This roves iii hence the theore is roved. Corollary 3.5. Let > 5 be a rie {3, 5,..., 4}. Then x= x od 3 x= x od 4 x= x od 6 x 3 3 B od, x B od, x 3 6 B od. Corollary 3.6. Let > 5 be a rie {3, 5,..., 4}. Then x= x od 4 x 3 + B B od. 4

15 Proof. Fro [S3, Theores 5. 5.] we see that x= x x = x= x / x= x B B od. Thus x= x od 4 x + x= x od 4 Now alying Theore 3.4 we deduce the result. x B B od. Theore 3.5. Let > 5 be a rie {, 4,..., 5}. Then x< 4 x 4 E E + + B od + 4 <x< + x 4 E E + B od. + Proof. As B n 3 4 = n B n 4, utting = 4 in Theore 3.3 we see that [/4] x= B 4 x B 4 + Fro Leas.4.5 we have 4 + B od. 4 B 4 4 = E 4 q 4 4E od, B 4 = 4 E 4 4 q 4E od, B = 4 4 B + B od. 5

16 As B 4 / B 4 / od, we see that E E od so [/4] x= x 4 { } q 4E q 4E B 4 E E + + B od. + By [S3, Corollary 5.a] we have Note that result. 4 <x< x< = x x< x= x + B od. + x x< 4 x. By the above we obtain the reaining Corollary 3.7 Leher [L, 0]. Let > 5 be a rie {, 4,..., 5}. Then [/4] x 4 E od. Corollary 3.8. Let > 5 be a rie. Then [/4] x= 8E 3 4E x B 3 od. Theore 3.6. Let > 5 be a rie {3, 5,..., 4}. Then x< 4 x B B 4 E od 4 <x< x 3 + B B + 4 E od. 6

17 Proof. As B n 3 4 = n B n 4, utting = 4 in Theore 3.3 we see that x< 4 x B B 4 B B B od According to Leas.4.5 we have B n B n = + n 4n B n B n+ 4 4 Thus, x< 4 x + 4 B + B E q 4q B + q q B = + B B B + + q B + B B Fro [S3, Theore 5.b] we have x B B x< Now cobining the above we deduce the result. << 3 4 = n + 4 n+ E n E E E od. od. Theore 3.7. Let > 5 be a rie. Then q + q + 6E 3 3E 4 3 q 3 + 4B 3 od 3 7

18 3 4 << Proof. It is clear that + < 4 3 3q q + 6E 3 3E 4 + q B 3 od << = By Corollaries we have < 4 < 4 = < 4 = + = < 4 < < 4 3 9B 3 od < od 3. Thus < 4 < 4 4 E 3 E B 3 od << 9B 3 4 E 3 E B 3 = 4 E 3 E B 3 od 3. Hence alying Theore 3. we obtain 3 4 << 4 E 3 E B 3 + 3q 3 q E 3 E 4 + q B 3 3 = 3q q + 6E 3 3E 4 + q B 3 od 3. 8

19 Fro [L,. 353] or [S3, Theore 5.a] we have By [S3, Theore 5.c], < < 3 B 3 od 3. q + q 3 q 3 7 B 3 od 3. Thus << = < < Observing that q q + 3 q B 3 od 3. << 3 4 = << 3 4 << alying the above we obtain the reaining result. Corollary 3.9. Let > 5 be a rie. Then << 3 4 q + q + 3 E 3 od 3 4 << 3q 3 q 3 E 3 od. Theore 3.8. Let > 5 be a rie. Then [ 4 ] [ 4 ] + 3q + 3q E 3 od 3 [ 4 ] [ 4 ] + 3q + 3q E 3 E q 3 3 q E B 3 od 4. 9

20 Proof. Fro Theore 3. we have [/4] i= 3 i 3q + q + E 4 E 3 By Corollaries we have q B 3 od 3. [/4] i= [/4] i= Fro this we deduce [/4] i= [/4] i i= i 3q + 3 q E 3 od 8E 3 4E i B 3 od. i 9q 8E 3 4E 4 + 9q q E B 3 od. By Corollary 3.4 we have [/4] i= [/4] i= 3 [/4] 3 i i= i 3 9B 3 od. Thus, [/4] i i= [/4] i + i= 3q 3 3 3q 4 E 3 + 9B 3 = 6 9 q q E 3 3B 3 od. i 3 Now utting all the above together with Lea.9 the fact E 4 E 3 od yields the result. Rear 3.3 The congruence [ ] 4 [ + 4 ] 3q od was nown to Leher. See [L, 5]. For any rie > 3 we recall the Legendre sybol { if od 3, = 3 if od 3. 0

21 Theore 3.9. Let be a rie greater than 5. Then [/6] [/3] i 5 3 B 6 od. = = [/6] ii q 3 q 3 + q q 3 3 B 6 od. = [/3] iii 3 q q B 6 od. = [/3] iv B 6 od. = v We have [ 6 ] [ 6 ] = q + 3 q 3 + q + 3q q q 3 6 B od [ 3 ] [ 3 ] + 3 q q 3 60 B od Proof. Taing = = 3, 6 in Theore 3.3 using Lea.6 we see that [/6] = [/3] = B { 6 } B { 3 } B = 3 6 B = 3 3 B od 3 6 B od. 3 3 By Raabe s theore cf. [S, Lea.] we have B 6 +B 6 + = B 3. Thus B = 3 B B = 3 + B 5B od Hence i holds. Suose {3, 6}. Taing =, r = 0 = 3, 6 in Lea.7 using Lea.6 we see that [ ] = = x= x 0 od = B ϕ 3 B ϕ 3 ϕ 3 x B { ϕ3 } B ϕ 3 ϕ B od. B { }

22 Fro the roof of Theore 3. we have B ϕ 3 B ϕ 3 ϕ 3 { 3 q q 3 od if = 3, q q + 3 q q 3 od if = 6. Hence ii iii follow fro the above the fact B 3 5 B 6 od. Now we consider iv. Since < 3 = < 3 = by [S3, Corollary 5.], using i we see that < 3 = < 3 = 3 << = 3 << < 3 < 3 < = < 3 << 3 = < 3 << B 6 0 od 3 << od. On the other h, noting that = 0 od cf. [L], [S3] using iii Theore 3.i we see that = 3 + = = 3 [ 3 ] = = 0 od 3 = od 3 = od q q 3 30 = B od B 3 6 q 3 4 q 3 Thus iv is true. Finally we consider v. By Lea.9, for = 3, 6 we have [ ] [ ] [ ] = + [ ] = [ ] = od 3.

23 Thus aealing to i-iii we obtain [ 6 ] [ 6 ] q 3 q 3 + q q 3 + q 3 q 3 B od B 3 6 [ 3 ] [ 3 ] 3 q q 3 B q 3 B od This yields v hence the theore is roved. Rear 3.4 Let > 5 be a rie. The congruences [/3] [/6] q 3 q 3 od, 3 q 3 od, = = [/3] + 3 [/3] q 3 od, [/6] + q + 3 [/6] q 3 od were nown to Leher [L], the congruence [/3] = due to Schwindt cf. [R, L]. In [S], using the forulas for roved that [ 3 ] = 0 od. [/6] 5 = r od 3 Corollary 3.0. Let > 5 be a rie. Then [/6] [/6] i q 3 q 3 + q + 34 q 3 od. ii = [/3] = iii We have = [/3] = 3 q q 3 od. od is the author [ 6 ] [ 6 ] 0 [ 3 ] [ 3 ] 9 + q 7 q 3 + q + 3q q q 3 od 3. 3

24 Corollary 3.. Let > 5 be a rie. Then i ii iii iv < 3 < 3 < 3 3 < < << 3 < 3 6 << 3 5 [ 6 ] = od, 9 q q 3 od, 4q + q od, 4q 9 q 3 + q q 3 od. 4. Congruences for = Let > 3 be a rie. For n N let G n x = =. = x n. Then G n x Z [x]. In [Gl] Glaisher showed that 4. G = = q od. In 004 Granville [Gr] roved the following Sula s conjecture: 4. G = = q od. In the section we deterine G od 3 G od. Lea 4.. Let be an odd rie. In Z [x] we have G x + x x = x + r= x r r r s= s od. Proof. Fro Lea.9 we now that s= s od for. Thus r= x r r r s= x r s r r r= = r 4 r= r r x r od. r

25 Since 0 tr dt = r, setting y = xt we see that r r= = 0 r= x x r r r r x r t r dt = r y y = dy = y y = = = x + x 0 x y x = x x = = x. Now cobining the above we obtain the result. Lea 4.. Let > 3 be a rie. Then xt xt dt t y y dy G q r r + r s + 3 q 3 B 3 od. r= s= Proof. Fro [S3, Rear 5.3] we now that + = = = = = q q + 3 q 3 B 3 od 3. Thus utting x = in Lea 4. alying the above gives the result. Lea 4.3. Let > 3 be a rie n N. Then Proof. Clearly we have x G n = x ng n+ x n x G n /x G n x od. = i= = x x i i n = = x n = + nx n n n x n + n n = = n ng n+ x + G n x od. + nx n+ This yields the result. 5

26 Theore 4.. Let > 3 be a rie. Then i ii iii iv = = = = q 7 B 3 od 3. q + 3 q B od. q q od. q od. As Proof. Suose x Z x 0 od. Fro Lea 4. we see that x G /x x + x we see that Thus s= r= x G /x x r r s = s= r s= = s r= s = = x x + = x x r 0 x + r= x s= r s = x x x Hence alying Leas we see that x r= r= s= + = x r r r s x r s= od, od. s= s od. s od. G 3 x x G /x G x { x x x x x + + = x s= s s= = s od. 6 + x }

27 Therefore, 4.4 G 3 x { x x + = x + x x x } od. Taing x = we find G 3 {4q = = } od. = It is well nown that = 0 od cf. [L,.353]. Fro [S3, Theore 5.c] we also now that = + = = q + q 3 q 3 7 B 3 od 3. Thus 4.5 G 3 { q +q 3 q 3 7 B 3 } od. As = 4.6 = = = = = = G 3 + G + G od 3, = by 4.5 we have Naely, G 3 q + q 3 q 3 7 B 3 + G 3 + G + G od G G q q + 3 q B 3 od 3. According to Lea 4. we have G q = + = 7 + r= r r r s= s od.

28 As > 3, by [S3, Corollary 5. Theore 5.] or [L] we have = 3 B 3 od Thus alying 4.3 we see that = 3 B 3 od 3. r= r r hence r s= s = = = s s= = G q s= = s s= s s= = s 8 = 3 4 B 3 od 3 B B 3 od. On the other h, using [S3, Corollaries 5. 5.] we have Hence This yields q G = = = = = = = i= i = 3 + B 3 3 B 3 = B 3 od. 3 B B 3 G B 3 od 3. G = = So i holds. Substituting this into 4.7 gives q 7 B 3 od 3. G q B q 3 od 3. That is, G q + 3 q B 3 od. 8

29 Thus ii is true. By 4.6 we have = G G od. Fro the above we now that Thus = G q od G q od. G G q + q + q q + q q q q od. This roves iii. Fro Lea 4.3 we have G 3 G / G od. Thus, G G q This roves iv hence the theore is roved. Corollary 4.. Let > 3 be a rie. Then od. q = od. Rear 4. Let > 3 be a rie. By [DS, 5] [S3, Corollary 5.b] we have G 3 = = 3 3 q j= This together with 4.5 Corollary 3. yields = j 3 3 q B 3 od. q + q 6 q q 48 B 3 q q + 3 q B 3 4 = 4 + od 3. 9

30 By the roof of Theore 4., we also have = = G / G 3 + G / q 3 q B 3 + q + q B 3 od. References =0 rod q + 3 q B 3 [D] L.E. Dicson, History of the Theory of Nubers, Vol.I, Chelsea Publ., 999, [DS] K. Dilcher L. Sula, The cube of the Ferat quotient, Integers: Electronic J. Cobin. Nuber Theory 6 006, #A4. [Gl] J.W.L. Glaisher, On the residues of the sus of roducts of the first nubers their owers to odulus or 3, Quart. J. Math. Oxford 3 900, [Gr] A. Granville, The square of the Ferat quotient, Integers: Electronic J. Cobin. Nuber Theory 4 004, #A. [GS] A. Granville Z.W. Sun, Values of Bernoulli olynoials, Pacific J. Math , [IR] K. Irel M. Rosen, A Classical Introduction to Modern Nuber Theory second edition, Sringer, New Yor, 990, [L] E. Leher, On congruences involving Bernoulli nubers the quotients of Ferat Wilson, Ann. Math , [MOS] W. Magnus, F. Oberhettinger R.P. Soni, Forulas Theores for the Secial Functions of Matheatical Physics 3rd Edition, Sringer-Verlag, New Yor, 966, [R] P. Ribenboi, Thirteen Lectures on Ferat s Last Theore, Sringer, New Yor, 979, np n [S] Z.H. Sun, Cobinatorial su its alications in nuber theory I, J. Nanjing Univ. Math. Biquarterly 9 99, [S] Z.H. Sun, Congruences for Bernoulli nubers Bernoulli olynoials, Discrete Math , [S3] Z.H. Sun, Congruences concerning Bernoulli nubers Bernoulli olynoials, Discrete Al. Math , [S4] Z.H. Sun, Congruences involving Bernoulli olynoials, Discrete Math. doi:0.06/j.disc [Su] Z.W. Sun, General congruences for Bernoulli olynoials, Discrete Math ,