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The paper will appear in Discrete Mathematics.CONGRUENCES INVOLVING BERNOULLI POLYNOMIALS
Zhi-Hong Sun
Department of Mathematics, Huaiyin Teachers College,Huaian, Jiangsu 223001, P.R. China
E-mail: hyzhsun@public.hy.js.cnHomepage: http://www.hytc.edu.cn/xsjl/szh
Let {Bn(x)} be the Bernoulli polynomials. In the paper we establish some
congruences for Bj(x) (mod pn), where p is an odd prime and x is a rational p-integer.
Such congruences are concerned with the properties ofp-regular functions, the congru-ences for h(
sp) (mod p) (s = 3, 5, 8, 12) and the sum
kr (mod m)
p
k
, where h(d) is
the class number of the quadratic field (
d) of discriminant d and p-regular func-tions are those functions f such that f(k) (k = 0, 1, . . . ) are rational p-integers and n
k=0
n
k
(1)kf(k) 0 (mod pn) for n = 1, 2, 3, . . . We also establish many congru-ences for Euler numbers.
MSC: Primary 11B68, Secondary 11A07, 11R29.
Keywords: Congruence, Bernoulli polynomial, p-regular function, class number, Eulernumber
1. Introduction.
The Bernoulli numbers{Bn} and Bernoulli polynomials{Bn(x)} are defined by
B0= 1,n1k=0
n
k
Bk = 0 (n2) and Bn(x) =
nk=0
n
k
Bkx
nk (n0).
The Euler numbers{En} and Euler polynomials{En(x)} are defined by
2et
e2t + 1=
n=0
Entn
n! (|t|<
2) and
2ext
et + 1=
n=0
En(x)tn
n! (|t|< ),
which are equivalent to (see [MOS])
E0 = 1, E2n1= 0,n
r=0
2n
2r
E2r = 0 (n1)
and
(1.1) En(x) +n
r=0
n
r
Er(x) = 2x
n (n0).1
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It is well known that([MOS])
(1.2)
En(x) = 1
2n
nr=0
n
r
(2x 1)nrEr
= 2n+ 1
Bn+1(x) 2n+1Bn+1x2.LetZ and N be the set of integers and the set of positive integers respectively. Let
[x] be the integral part ofxand{x} be the fractional part ofx. Ifm, sN and pisan odd prime not dividing m, in Section 2 we show that
(1)s mp
p1k=1
ksp(mod m)
p
k
Bp1 (s1)pm Bp1 spm (modp) if 2|m,12
(1)[ (s1)pm ]Ep2
(s1)pm
(1)[ spm]Ep2 spm (mod p) if 2m.For a discriminant d let h(d) be the class number of the quadratic field Q(
d) (Q
is the set of rational numbers). Ifp >3 is a prime of the form 4m+3, it is well knownthat (cf. [IR])
(1.3) h(p) 2Bp+12
(mod p).
Ifp is a prime of the form 4m+ 1, according to [Er] we have
(1.4) 2h(4p)Ep12
(mod p).
Let ( an) be the Kronecker symbol. For odd primes p, in Section 3 we establish thefollowing congruences:
h(8p)Ep12
14
(mod p);
h(3p) 4
p
3Bp+1
2 1
3(mod p) forp1 (mod 4);
h(12p)8p3
Bp+12
112
(mod p) for p7, 11, 23 (mod 24);
h(5p) 8Bp+12
15
(mod p) forp11, 19 (mod 20).
For m N let Zm be the set of rational numbers whose denominator is coprimetom. For a prime p, in [S5] the author introduced the notion ofp-regular functions.If f(k) Zp for any nonnegative integers k and
nk=0
nk
(1)kf(k) 0 (mod pn)
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for all n N, then f is called a p-regular function. Iff is a p-regular function andk,m,n,tN, in Section 4 we show that
(1.5) f(ktpm1)n1
r=0
(1)n1rk 1 rn
1
r
k
rf(rtpm1) (mod pmn),
which was annouced by the author in [S5, (2.4)]. We also show that
(1.6) f(kpm1)(1 kpm1)f(0) +kpm1f(1) (mod pm+1) for p >2.
Let p be a prime, x Zp and let b be a nonnegative integer. Lettp be theleast nonnegative residue oft modulopand x = (x+ xp)/p. From [S4, Theorem3.1] we know that f(k) = p(pBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)) is a pregularfunction. If p 1 b, in [S5] the author showed that f(k) = (Bk(p1)+b(x)
pk(p1)+b1
Bk(p1)+b(x
))/(k(p 1) + b) is also a pregular function. Using such re-sults in [S4, S5] and (1.5), in Section 5 we obtain general congruences for pBk(ps)+b(x),pBk(ps)+b, (mod p
sn), where k,n, s N, is Eulers totient function and is aDirichlet character modulo a positive integer. As a consequence of (1.6), if 2|b and
p 1b, we have
Bk(ps)+b
k(ps) +b (1 kps1)(1 pb1) Bb
b +kps1
Bp1+bp 1 +b (mod p
s+1).
In Section 6 we establish some congruences fornk=0
nk(1)
kpBk(p1)+b(x) mod-
ulopn+1
, where pis an odd prime,nN,xZp andb is a nonnegative integer.Let p be an odd prime and b {0, 2, 4, . . . }. In Section 7 we show that f(k) =(1(1) p12 pk(p1)+b)Ek(p1)+b is apregular function. Using this and (1.5) we givecongruences for Ek(pm)+b (mod p
mn), where k, mN. By (1.6) we have
Ek(pm)+b(1 kpm1)(1 (1)p12 pb)Eb+kp
m1Ep1+b (mod pm+1).
We also show that f(k) =E2k+b is a 2regular function and
E2mkt+bn1r=0
(1)n1rk 1 rn 1 rkrE2mrt+b (mod 2mn+n),
wherek,m,n,tN and N is given by 21 n
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Theorem 2.1. Letp, mN andk, rZ withk0. Then
p1x=0
xr(mod m)
xk = mk
k+ 1
Bk+1
pm
+r p
m
Bk+1
rm
and
p1x=0
xr(mod m)
(1)xrm xk =mk
2
(1)[ rpm ]Ek
pm
+r p
m
(1)[ rm ]Ek
rm
.
Proof. For any real number t and nonnegative integer n it is well known that (cf.[MOS])
(2.1) Bn(t+ 1) Bn(t) =ntn1 (n= 0) and En(t+ 1) +En(t) = 2tn.
Hence, for xZ we have
Bk+1
x+ 1m
+r x 1
m
Bk+1
xm
+r x
m
=
Bk+1
x+1m +
rxm
1m Bk+1 xm + rxm = 0 if m x r,Bk+1
x+1m +
m1m
Bk+1 xm= (k+ 1) xmk ifm|x r.Thus
Bk+1
pm
+
r pm
Bk+1
rm
=
p1x=0
Bk+1
x+ 1m
+r x 1
m
Bk+1
xm
+r x
m
=k+ 1
mk
p1x=0
xr(mod m)
xk.
Similarly, ifxZ, by (2.1) we have
(1)[ rx1m ]Ekx+ 1
m +
r x 1m
(1)[ rxm ]Ek
xm
+r x
m
=
(1)[ rxm ]Ekx+1m + rxm} 1m Ek xm + rxm = 0ifmx r,
(1) rxm 1Ekx+1m +
m1m
(1) rxm Ek xm=(1) rxm 2( xm)kifm|x r.
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Thus
(1)[ rpm ]Ekp
m+r p
m
(1)[ rm ]Ek
rm
=
p1
x=0(1)[ rx1m ]Ek
x+ 1
m +
r x 1
m (1)[ rxm ]Ek
xm
+r x
m
= 2mk
p1x=0
xr(mod m)
(1)xrm xk.
This completes the proof.
Corollary 2.1. Let p be an odd prime and k {0, 1, . . . , p2}. Let r Z andmN withpm. Then
p1x=0
xr(mod m)
xk mk
k+ 1
Bk+1
r pm
Bk+1
rm
(mod p)
and
p1x=0
xr(mod m)
(1)xrm xk
mk
2(1)
[ rpm
]Ekr pm
(1)[ rm
]Ekrm(mod p).
Proof. Ifx1, x2 Zp and x1 x2 (mod p), by [S5, Lemma 3.1] and [S3, Lemma3.3] we have
(2.2) Bk+1(x1) Bk+1(x2)
k+ 1 x1 x2
p pBk0 (mod p)
and
(2.3) Ek(x1)Ek(x2) (mod p).
Thus the result follows from Theorem 2.1.Remark 2.1 Putting k = p 2 in Corollary 2.1 and then applying Fermats littletheorem we see that ifpis an odd prime not dividing m, then
(2.4)
p1x=1
xr(mod m)
1
x 1
m
Bp1
r pm
Bp1
rm
(mod p)
5
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and
(2.5)
p1x=1
xr(mod m)
(1)xrm 1x
12m
(1)[ rpm ]Ep2r p
m
(1)[ rm ]Ep2rm
(mod p).
Here (2.4) and (2.5) are due to my brother Z.W. Sun. See [Su2, Theorem 2.1]. Inspiredby his work, the author established Theorem 2.1 and Corollary 2.1 in 1991.
Corollary 2.2. Letp be an odd prime. Letk {0, 1, . . . , p 2} andm, sN withpm. Then
(1)kk+ 1
Bk+1
(s 1)pm
Bk+1
spm
(s1)p
m
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Corollary 2.3. Letp be a prime.
(i) (Karpinski[K, UW]) Ifp3 (mod 8), then(p3)/4x=1
xp
= 0.
(ii) (Karpinski[K, UW]) Ifp5 (mod 8), then[p/6]
x=1 xp = 0.
(iii) (Berndt[B, UW]) Ifp5 (mod 24), then(p5)/12
x=1
xp
= 0.
Proof. By Corollary 2.2 and the known fact B2n+1 = 0, for mN with pm wehave
(2.8)
[p/m]x=1
xp
[p/m]x=1
xp12 (1)
p12
p+12
B p+1
2 B p+1
2
pm
2B p+1
2
pm
(mod p) ifp1 (mod 4),
2Bp+12 + 2B
p+12 p
m
(mod p) ifp3 (mod 4).It is well known that B2n(
34 ) = B2n(
14 ) = ( 122n1)B2n/24n1. Thus, if p
3 (mod 8), by (2.8) we see that
p34
x=1
xp
2Bp+1
2+ 2B p+1
2
34
=
1
2p1
1 2p12 Bp+12
2Bp+12
1 2
p
2
B p+1
2= 0 (mod p).
Asp34 p34x=1 xp p34 , we must have(p3)/4x=1 (xp ) = 0. This proves (i).Now we consider (ii). For n {0, 1, 2, . . . } and m N it is well known that (cf.
[IR], [MOS])
(2.9) Bn(1 x) = (1)nBn(x) andm1k=0
Bn
x+
k
m
= m1nBn(mx).
Thus
Bp+12
12n
+B p+1
2
12n
+1
2
= 2
p12 Bp+1
2
1n
and so
(2.10) B p+12
12n
2
p
Bp+1
2
1n
(1) p+12 Bp+1
2
n 12n
(mod p).
Since p5 (mod 8), taking n= 3 in (2.10) we find
(2.11) B p+12
16
Bp+1
2
13
+B p+1
2
13
= 0 (mod p).
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This together with (2.8) and (2.9) yields
[p/6]x=1
xp
2Bp+1
2
p6
=2
p3
Bp+1
2
16
0 (modp).
As|[p/6]x=1 xp| [p6 ] we have[p/6]x=1 xp= 0. This proves (ii).Finally we consider (iii). Assume p5 (mod 24). By (2.10) and (2.11) we have
Bp+12
112
2
p
Bp+1
2
16
+B p+1
2
512
B p+1
2
512
(mod p).
On the other hand, by (2.9) we have
B p+12
112
+B p+1
2
512
= 3
p12 Bp+1
2
14
Bp+1
2
912
3
p
B p+1
2
14
(1) p+12 B p+1
2
14
= 0 (mod p).Thus B p+1
2
112
B p+12
512
0 (mod p). Now applying (2.8) we see that[p/12]x=1
xp
2Bp+1
2
p12
=2Bp+1
2
512
0 (mod p).
This yields (iii) and so the corollary is proved.
Corollary 2.4. Supposep, q,mN, nZ, gcd(p, m) = 1 andqm. ForrZ letAr(m, p) be the least positive solution of the congruencepxr (mod m). Then
r: Ar
(m, p)
q, rZ,
n
r
p
1
n= pq+nm
n
m.
Proof. Using Theorem 2.1 we see thatr: Ar(m, p)q, rZ,nrp 1 n=
qx=1
p1nr=n
rpx (mod m)
1 =
qx=1
p1s=0
spx+n (mod m)
1
=
qx=1
B1 p
m+px+n p
m
B1
px+nm
=
qx=1
pm
+
p(x 1) +nm
px+nm
=pq
m+ n
m
pq+n
m
=
pq+n
m
pq+nm
n
m n
m
=pq+n
m
n
m
.
This proves the corollary.8
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Theorem 2.2. Letm, sN and letp be an odd prime not dividingm. Then
(1)s mp
p1k=1
ksp(mod m)
p
k
(s1)pm
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and
(1)[ (s1)pm ]Ep2 (s 1)p
m
(1)[ spm]Ep2
spm
2
(s1)pm
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Proof. From [UW, p. 58] we know that
(3.2) h(8p) = 2p1a=1
a1(mod 4)
8pa
.
Thus applying Corollary 2.1 in the case r= 1, m = 4 and k= p12 we see that
h(8p) = 2p1a=0
a1(mod 4)
2a
ap
2
p1a=0
a1(mod 4)
(1) a14 a p12
4p12
(1)[ 1p4 ]Ep12
1 p4
Ep1
2
14
(mod p).
Since E2n(0) = 22n+1 (B2n+1 22n+1B2n+1) = 0 by (1.2), we see that
Ep12
1 p
4
=
E2n(0) = 0 if p = 4n+ 1,E2n1(
12 ) = 2
12nE2n1= 0 if p = 4n 1.Thus
h(8p)4 p12 Ep12
14
Ep1
2
14
(mod p).
LetSn= 4nEn(
14 ). Now we show that Sn= S
n forn0. By (1.1) we have
4nSn+n
k=0
n
k
4kSk = 2 4n and so Sn+
nk=0
n
k
4nkSk = 2.
That is, Sn = 1n1
k=0
nk
22n2k1Sk. Since S
0 = S0 = 1 we see that S
n = Sn.
That is,
(3.3) Sn= 4nEn
14
.
HenceSp12
= 4p12 Ep1
2( 14 )h(8p) (mod p).This proves the theorem.
Corollary 3.1. Letp be an odd prime. ThenpSp12
.
Proof. From (3.2) we have 1 < h(8p)< p. Thus the result follows from Theorem
3.1.
Remark 3.1Since Sn = 4nEn(
14 ), by (1.2) and the binomial inversion formula we
have
(3.4) Sn=
nr=0
n
r
(1)nr2rEr and
nr=0
n
r
Sr = 2
nEn.
11
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Theorem 3.2. Letp be a prime greater than3.(i) Ifp1 (mod 4), then
h(3p)4Bp+1
2
13
(mod p) ifp1 (mod 12),
4Bp+12 13 (mod p) ifp5 (mod 12).
(ii) Ifp3 (mod 4), then
h(12p)
8B p+12
112
(mod p) ifp7 (mod 24),
8Bp+12
112
(mod p) ifp11 (mod 12),
8B p+12
112
+ 8Bp+1
2(mod p) ifp19 (mod 24)
and
h(5p)8B p+1
2( 15 ) (mod p) ifp11, 19 (mod 20),
8Bp+12
( 15 ) + 4Bp+12(mod p) ifp
3, 7 (mod 20).
.
Proof. We first assume p1 (mod 4). From [UW, p. 40] or [B] we have
h(3p) = 2[p/3]x=1
px
.
Thus applying (2.8), (2.9) and the quadratic reciprocity law we see that
h(
3p) = 2
[p/3]
x=1
x
p
4Bp+12
p
3=
4p
3Bp+12
1
3 (mod p).
This proves (i).Now let us consider (ii). Assume p3 (mod 4). From [UW, pp. 3-5] we have
h(12p) =
4 p
12
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Thus
h(12p)
8B p+12
512
B p+12
16
(mod p) ifp7 (mod 24),
8
B p+12
112
B p+1
2 16
(mod p) ifp11 (mod 12),
8B p+12 13 B p+12 112 (mod p) ifp19 (mod 24).By (2.10) we have
Bp+12
112
2
p
Bp+1
2
16
B p+1
2
512
(mod p).
Thus, if p 7 (mod 24), then h(12p) 8(B p+12
16
B p+12
512
) 8B p+1
2
112
(mod p). It is well known that ([GS])
B2n1
3= 312n
1
2 B
2n and B
2n1
6= (212n
1)(312n
1)
2 B
2n.
Thus
B p+12
13
=
1
2
3
p12 1
Bp+1
2 1
2
3p
1
Bp+1
2(mod p)
and
Bp+12
16
=
(2p12 1)(3 p12 1)
2 Bp+1
2 1
2
2p
1
3p
1
Bp+1
2(modp).
Ifp
11 (mod 12), then 3p = 1 and so B p+12 16 0 (mod p). Hence h(12p)8Bp+12
112
(mod p). If p 19 (mod 24), then 3p =1 and so Bp+12 13
Bp+12
(mod p). Thus h(12p)8(B p+12
112
+B p+1
2) (mod p).
Finally we consider h(5p) (mod p). From [UW, p. 40] or [B] we have
h(5p) = 2
p
5
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From (2.9) we see that
Bp+12
+ 2Bp+12
15
+ 2B p+1
2
25
=
4k=0
B p+12
k5
= 5
p12 Bp+1
2
and so
B p+12
15
+B p+1
2
25
1
2
p5
1
B p+1
2(mod p).
Thus
h(5p) 4p
5
2Bp+1
2
15
+
1
2
1
p5
Bp+1
2
=
8Bp+12
15
(mod p) ifp11, 19 (mod 20),
8Bp+12
15
+ 4Bp+1
2(mod p) ifp3, 7 (mod 20).
The proof is now complete.When d is a negative discriminant, it is known that 1 h(d) < p. Thus, from
Theorem 3.2 we deduce the following result.
Corollary 3.2. Letp be a prime.(i) Ifp1 (mod 4), thenBp+1
2( 13 )0 (mod p).
(ii) Ifp7, 11, 23 (mod 24), thenBp+12
( 112 )0 (mod p).(iii) Ifp11, 19 (mod 20), thenBp+1
2( 15 )0 (mod p).
Remark 3.2Forn= 0, 1, . . . it is well known that nk=0 nk 1nk+1 Bk(x) = xn. From
this we deduce that ifmN and an= mnBn( 1m), thennk=0 n+1k mnkak =n + 1.4. p-regular functions.
For a prime p, in [S5] the author introduced the notion ofp-regular functions. Iff(k) is a complex number congruent to an algebraic integer modulo p for any givennonnegative integer k and
nk=0
nk
(1)kf(k) 0 (mod pn) for all n N, then f
is called a p-regular function. Iff and g are p-regular functions, in [S5] the authorshowed that f g is also a p-regular function. Thus we see thatp-regular functionsform a ring. In the section we discuss further properties ofp-regular functions.
Suppose nN andk {0, 1, . . . , n}. Let s(n, k) be the unsigned Stirling numberof the first kind and S(n, k) be the Stirling number of the second kind defined by
x(x 1) (x n+ 1) =n
k=0
(1)nks(n, k)xk
and
xn =
nk=0
S(n, k)x(x 1) (x k+ 1).14
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For our convenience we also define s(n, k) = S(n, k) = 0 for k > n. FormN it iswell known that
(4.1)n
r=0
n
r
(1)nrrm =n!S(m, n)
In particular, taking m = n we have the following Eulers identity
(4.2)n
r=0
n
r
(1)nrrn =n!.
Lemma 4.1. Letx, d be variables, m, nN and iZ withi0. Thenn
r=0
n
r
(1)nr
rx+d
m
ri
=
n!
m!
mj=ni
mk=j
kj
(1)mk
s(m, k)dkj
S(i+j, n)xj
.
In particular we have
nr=0
n
r
(1)nr
rx
m
ri =
n!
m!
mj=ni
(1)mjs(m, j)S(i+j, n)xj .
Proof. Since
m!rx+d
m = (rx+d)(rx+d 1) (rx+d m+ 1)=
mk=0
(1)mks(m, k)(rx+d)k
=
mk=0
(1)mks(m, k)k
j=0
k
j
(rx)jdkj
=mj=0
mk=j
k
j
(1)mks(m, k)dkj
rjxj ,
we have
nr=0
n
r
(1)nr
rx+d
m
ri
= 1
m!
mj=0
mk=j
k
j
(1)mks(m, k)dkj
xj
nr=0
n
r
(1)nrri+j .
Now applying (4.1) we obtain the result.15
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Lemma 4.2. Letp be a prime andm, nN. Thenm!s(n, m)
n! pnm Zp and m!S(n, m)
n! pnm Zp.
Moreover, ifm < n, we have
m!s(n, m)
n! pnm m!S(n, m)
n! pnm 0 (mod p) forp > 2
andm!s(n, m)
n! 2nm
m
n m
(mod 2).
Proof. It is well known that
(ex 1)mm!
=
n=m
S(n, m)xn
n!.
Thus, applying the multinomial theorem we see that
(ex 1)m = k=1
xk
k!
m=
n=m
k1+k2++kn=mk1+2k2++nkn=n
m!
k1!k2! kn!n
r=1
1
r!kr
xn
and so
(4.3) S(n, m) = k1+k2++kn=mk1+2k2++nkn=n
n!
1!k1k1!2!k2k2!
n!knkn!.
Hence
m!S(n, m)
n! pnm =
k1+k2++kn=mk1+2k2++nkn=n
(k1+k2+ +kn)!k1!k2! kn!
nr=1
pr1r!
kr.
From [S5, pp. 196-197] we also have
(4.4) s(n, m) = k1+k2++kn=mk1+2k2++nkn=n
n!
1k1k1!2k2k2! nknkn!
and
m!s(n, m)
n! pnm =
k1+k2++kn=mk1+2k2++nkn=n
(k1+k2+ +kn)!k1!k2! kn!
nr=1
pr1r
kr.
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It is known that (k1+ +kn)!/(k1! kn!)Z. ForrN we know that ifp r!(thatis p | r! but p+1 r!), then =i=1 rpi rp. Thus pr1/r, pr1/r!Zp. For
p >2 we see that pr1/rpr1/r!0 (mod p) for r >1. Hence the result followsfrom the above. For p = 2 we see that 2r1/r0 (mod 2) forr >2. Thus
m!s(n, m)
n! 2nm
k1+k2=mk1+2k2=n
(k1+k2)!
k1!k2! = m
n m
(mod 2).
Summarizing the above we prove the lemma.From Lemma 4.1 we have the following identities, which are generalizations of
Eulers identity.
Theorem 4.1. Letx, d be variables andm, nN.(i) Ifmn, then
n
r=0
n
r
(1)nr
rx+d
m
rnm =
n!
m!xm.
In particular, whenm= n we haven
r=0
n
r
(1)nr
rx+d
n
= xn.
(ii) Ifmn+ 1, thenn
r=0
n
r
(1)nr
rx+d
m
rn+1m =
n!
m!
n(n+ 1)2
xm m(m 1 2d)2
xm1
.
In particular, whenm= n+ 1 we haven
r=0
nr
(1)nrrx+d
n+ 1
=
d+
n(x
1)
2
xn
.
Proof. Observe thats(m, m) = 1 and S(n, n) = 1. Putting i = n m in Lemma4.1 we obtain (i). By (4.3) and (4.4) we have
s(n, n 1) =S(n, n 1) =n(n 1)/2 for n= 2, 3, 4, . . .Thus applying Lemma 4.1 we see that ifmn+ 1, then
nr=0
n
r
(1)nr
rx+d
m
rn+1m
= n!
m!
m
j=m1
m
k=j
kj(1)
mks(m, k)dkjS(n+ 1 m+j, n)xj
= n!
m!
S(n+ 1, n)xm +
mk=m1
k
m 1
(1)mks(m, k)dk(m1)xm1
= n!
m!
n(n+ 1)2
xm +
dm m(m 1)2
xm1
.
This yields (ii) and so the theorem is proved.17
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Corollary 4.1. Let p be an odd prime, m Z and d {0, 1, . . . , p1}. Thenmp m (mod p) and
mp mp
p1k=1
1
k
km+dp
+m
dk=1
1
k (mod p).
Proof. From Theorem 4.1(i) we have
mp =
pk=0
p
k
(1)pk
km+d
p
=
mp+d
p
+
p1k=1
p
k
(1)pk
km+d
p
.
Asp1
k=11k 0 (modp), we see that
mp+d
p = (mp+d)(mp+d 1) (mp+d p+ 1)
p!
= mp
p (mp+ 1) (mp+d)((m 1)p+d+ 1) ((m 1)p+p 1)
(p 1)!
m
1 +mpd
k=1
1
k+ (m 1)p
p1k=d+1
1
k
m
1 +mpd
k=1
1
k (m 1)p
dk=1
1
k
=m1 +pd
k=1
1
k (mod p2).
Letrkbe the least nonnegative residue ofkm+dmodulop. Fork {1, 2, . . . , p1}we see that
p
k
=
p(p 1) (p k+ 1)k!
(1)k1
k p(mod p2).
Thus,p1k=1
p
k
(1)pk
km+d
p
p1
k=1
p
k(km+d)(km+d 1) (km+d p+ 1)
p!
=p
p1k=1
1
k km+d rk
p 1
(p 1)!p1i=0i=rk
(km+d i)
pp1k=1
1
k km+d rk
p =p
p1k=1
1
k
km+dp
(modp2).
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Now putting all the above together we obtain the result.Remark 4.1 In the case d = 0, Corollary 4.1 was first found by Lerch [Ler]. For adifferent proof of Lerchs result, see [S5].
Theorem 4.2. Letp be a prime. Let f be ap-regular function. Supposem, n Nandd, tZ withd, t0. Then
nr=0
n
r
(1)rf(pm1rt+d)0 (mod pmn).
Moreover, ifAk =pkk
r=0
kr
(1)rf(r), then
n
r=0
n
r
(1)rf(pm1rt+d)
pmntnAn (mod pmn+1) ifp >2 orm= 1,
2mntnn
r=0
nr
Ar+n (mod 2
mn+1) ifp= 2 andm2.
Proof. Since f is a p-regular function, we have AkZp for k0. Set
a0=A0 and ai = (1)inr=i
s(r, i)pr
r!Ar for i= 1, 2, . . . , n.
Aspr/r!
Zp and Ar
Zp we have a0, . . . , an
Zp. From [S5, p. 197] we have
f(k)ni=0
aiki (mod pn+1) for k= 0, 1, 2, . . . .
Thus applying (4.1) and (4.2) we see that
nr=0
n
r
(1)rf(rt+d)
nr=0
n
r
(1)r
ni=0
ai(rt+d)i
=
nr=0
nr
(1)r(antnrn +bn1rn1 + +b1r+b0)
=an(t)nn! = (1)ns(n, n)pn
n!An (t)nn!
=pntnAn (mod pn+1),
whereb0, b1, . . . , bn1Zp. Thus the result is true form= 1.19
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Now assume m2. By the binomial inversion formula we have f(k) =ks=0 ks(p)sAs.Thus
n
r=0
n
r
(1)rf(pm1rt)
=
nr=0
n
r
(1)r
pm1rtk=0
pm1rt
k
(p)kAk
=
pm1ntk=0
(p)kAkn
r=0
n
r
(1)r
pm1rt
k
=
pm1ntk=n
(p)kAk (1)n n!k!
kj=n
(1)kjs(k, j)S(j, n)pm1tj (by Lemma 4.1)
=
pm1ntk=n
(p)n(1)kAkk
j=n(1)kj
s(k, j)j!
k! pkj
S(j, n)n!
j! pjn pm1tj
=Antnpmn +
pm1ntk=n+1
(p)n(1)kAk (1)kns(k, n)n!
k! pkn p(m1)ntn
+k
j=n+1
(1)kjs(k, j)j!k!
pkj S(j, n)n!j!
pjn (pm1t)j
.
By Lemma 4.2, for j, k, nN we haves(k, j)j!
k! pkj Zp and S(j, n)n!
j! pjn Zp.Hence, by the above, Lemma 4.2 and the fact (m 1)(n + 1 ) + nmn + 1 we obtain
nr=0
n
r
(1)rf(pm1rt)
pmntn
An+
pm1ntk=n+1
s(k, n)n!
k! pknAk
pmntnAn (mod p
mn+1) ifp >2,
2mntn2m
1ntk=n
nkn
Ak = 2
mntnn
r=0
nr
Ar+n (mod 2
mn+1) ifp = 2.
Thus the result holds for d= 0.Now suppose g(r) =f(r+d). By the previous argument,
nr=0
n
r
(1)rg(r)pnAn (mod pn+1).
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Thus g is also a p-regular function. Note thatn
r=0
n
r
(1)rf(pm1rt+d) =
nr=0
n
r
(1)rg(pm1rt).
By the above we see that the result is also true for d >0. The proof is now complete.
Theorem 4.3. Let p be a prime, k,m,n,t N and d {0, 1, 2, . . . }. Let f be ap-regular function. Then
f(ktpm1 +d)n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(rtpm1 +d) (mod pmn).
Moreover, settingAs = pss
r=0
sr
(1)rf(r) we then have
f(ktpm1 +d) n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(rtpm1 +d)
pmnkn(t)
nAn (mod pmn+1) ifp >2 orm= 1,
2mnkn
(t)nnr=0 nrAr+n (mod 2mn+1) ifp= 2 andm2.
Proof. From [S4, Lemma 2.1] we know that for any function F,
(4.5)
F(k) =
n1r=0
(1)n1r
k 1 rn 1 r
k
r
F(r)
+k
r=n
k
r
(1)r
rs=0
r
s
(1)sF(s),
where the second sum vanishes when k < n.Now taking F(k) =f(ktpm1 +d) we obtain
f(ktpm1 +d) =n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(rtpm1 +d)
+k
r=n
k
r
(1)r
rs=0
r
s
(1)sf(stpm1 +d).
By Theorem 4.2 we havek
r=n
k
r
(1)r
rs=0
r
s
(1)sf(stpm1 +d)
(1)n
kn
ns=0
ns
(1)sf(stpm1 +d)
kn
pmn(t)nAn (modpmn+1) ifp >2 or m = 1,
kn
2mn(t)nnr=0 nrAr+n (mod 2mn+1) ifp = 2 and m2.
Now combining the above we prove the theorem.Puttingn = 1, 2, 3 and d= 0 in Theorem 4.3 we deduce the following result.
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Corollary 4.2. Letp be a prime, k,m, tN. Letf be ap-regular function. Then(i) ([S5, Corollary 2.1]) f(kpm1)f(0) (mod pm).(ii) f(ktpm1)kf(tpm1) (k 1)f(0) (mod p2m).(iii) We have
f(ktpm1)k(k 1)2
f(2tpm1) k(k 2)f(tpm1)
+(k 1)(k 2)
2 f(0) (mod p3m).
(iv) We have
f(kpm1)
f(0) k(f(0) f(1))pm1 (mod pm+1) ifp > 2 orm= 1,f(0) 2m2k(f(2) 4f(1) + 3f(0)) (mod 2m+1) ifp= 2 andm2.
Theorem 4.4. Letp be a prime and letf be ap-regular function. LetnN.(i) Ford, xZp andm {0, 1, . . . , n 1} we have
nk=0
n
k
(1)k
kx+d
m
f(k)0 (mod pnm).
(ii) We have
n
k=1
n
k(1)kf(k 1) f(pn1 1) (mod pn).
Proof. From [S5, Theorem 2.1] we know that there are a0, a1, . . . , anm1 Zsuch that
f(k)anm1knm1 + +a1k+a0 (mod pnm) for k= 0, 1, 2, . . .
Thus applying Lemma 4.1 and (4.1) we have
n
k=0
n
k
(1)k
kx+d
m
f(k)
n
k=0
n
k
(1)k
kx+d
m
nm1i=0
aiki
=nm1i=0
ai
nk=0
n
k
(1)k
kx+d
m
ki = 0 (mod pnm).
This proves (i).22
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Now we consider (ii). By [S5, Theorem 2.1] there are a0, a1, . . . , an1 Zp suchthats!as/p
s Zp (s= 0, 1, . . . , n 1) and
f(k)an1kn1 + +a1k+a0 (mod pn) for k= 0, 1, 2, . . .
Note that ps1
/s!Zp for sN. We then have a1 an10 (modp). Letan1(k 1)n1 + +a1(k 1) +a0 = bn1kn1 + +b1k+b0.
Then clearlyb1 bn10 (mod p) and
f(k 1)bn1kn1 + +b1k+b0 (mod pn) for k= 1, 2, 3, . . .
Thusf(pn1 1)bn1(pn1)n1 + +b1pn1 +b0b0 (mod pn).
Hence, applying (4.1) we haven
k=1
n
k
(1)kf(k 1)
nk=1
n
k
(1)k(bn1kn1 + +b1k+b0)
=
n1i=1
bi
nk=0
n
k
(1)kki +b0
nk=1
n
k
(1)k
=b0 f(pn1 1) (mod pn).
So the theorem is proved.
5. Congruences for pBk(pm)+b(x) and pBk(pm)+b, (mod pmn).For given prime p and t Zp we recall thattp denotes the least nonnegative
residue oftmodulo p.
Theorem 5.1. Letp be a prime, and k,m,n,t,bZ withm, n1 andk,b, t0.LetxZp andx = (x+ xp)/p. Then
pBkt(pm)+b(x) pkt(pm)+bBkt(pm)+b(x
)
n1
r=0(1)n1r
k 1 rn 1 r
k
r
pBrt(pm)+b(x) prt(p
m)+bBrt(pm)+b(x)
(b,n,p)kn
(t)npmn1 (mod pmn) ifp >2 orm= 1,
0 (mod 2mn) ifp= 2 andm2,
where
(b,n,p) =
1 ifp= 2 andn {1, 2, 4, 6, . . . }or ifp >2, p 1|b andp 1|n,
0 otherwise.23
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Proof. From [S4, Theorem 3.1] we know that
nk=0
n
k
(1)k
pBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)
pn1(b,n,p) (mod pn).
Set f(k) =ppBk(p1)+b(x) pk(p1)+bBk(p1)+b(x)
. Then
nk=0
nk
(1)kf(k)
(b,n,p)pn (mod pn+1).Thusf is ap-regular function. Hence appealing to Theorem4.3 we have
f(ktpm1) n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(rtpm1)
pmnkn
(t)n(b,n,p) (mod pmn+1) ifp >2 or m = 1,
2mn
kn
(t)n
nr=0
nr
(b, n+r, 2) (mod 2mn+1) ifp = 2 and m2.
Note that(b, n+r, 2) =
1 ifn+r {1, 2, 4, 6, . . . },0 ifn+r {3, 5, 7, . . . }.
We then have
nr=0
n
r
(b, n+r, 2)
=
(b, 1, 2) +(b, 2, 2) = 1 + 10 (mod 2) ifn = 1,n
r=02|n+r
nr= 2
n1 0 (mod 2) if n >1.
Thusf(ktpm1)
p
n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(rtpm1)
p
pmn1kn
(t)n(b,n,p) (modpmn) ifp >2 orm= 1,
0 (mod 2mn) ifp= 2 and m2.This is the result.
Corollary 5.1. Let p be a prime, and k,m, b Z with k, m 1 and b 0. LetxZp andx
= (x+ xp)/p. Supposep >2 orm >1. Then
pBk(pm)+b(x)
3 (mod 4) ifp= m = 2, k= 1 andb= 0,
pBb(x) pbBb(x) (mod pm) otherwise.Proof. Putting n= t = 1 in Theorem 5.1 we see that
pBk(pm)+b(x) pk(pm)+bBk(pm)+b(x
)pBb(x) pbBb(x) (mod pm).24
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Ifp = m = 2, k = 1 and b = 0, then pBk(pm)+b(x) = 2B2(x) = 2(x2 x+ 16 )
3 (mod 4). Otherwise, we havek(pm) + bm + 1 and sopk(pm)+bBk(pm)+b(x)0 (mod pm). Thus the result follows from the above.
In the case p >2, Corollary 5.1 has been proved by the author in [S4].Let be a primitive Dirichlet character of conductor m. The generalized Bernoulli
number Bn, is defined bymr=1
(r)tert
emt 1 =n=0
Bn,tn
n!.
Let0 be the trivial character. It is well known that (see [W])
B1,0 = 1
2, Bn,0 =Bn(n= 1) and Bn,= mn1
mr=1
(r)Bn r
m
.
If is nontrivial and nN, then clearlymr=1(r) = 0 and so
Bn,n
=mn1mr=1
(r)Bn( rm ) Bn
n +
Bnn
= mn1
mr=1
(r)Bn( rm) Bn
n .
When p is a prime with pm, by [S4, Lemma 2.3] we have (Bn(rm ) Bn)/n Zp.
Thus Bn,/nis congruent to an algebraic integer modulo p.
Lemma 5.1. Letp be a prime and letb be a nonnegative integer.(i) ([S5, Theorem 3.2], [Y2]) Ifp1b,xZp andx = (x+xp)/p, thenf(k) =
(Bk(p1)+b(x) pk(p1)+b1Bk(p1)+b(x))/(k(p 1) +b) is apregular function.(ii) ([S5, (3.1), Theorem 3.1 and Remark 3.1]) If a, b N andp a, then f(k) =
(1
pk(p1)+b1)(ak(p1)+b
1)Bk(p1)+b/(k(p
1) +b) is ap-regular function.
(iii) ([Y3, Theorem 4.2], [Y1, p. 216], [F], [S5, Lemma 8.1(a)]) If b, m N, p mand is a nontrivial primitive Dirichlet character of conductor m, then f(k) =(1 (p) pk(p1)+b1)Bk(p1)+b,/(k(p 1) +b) is apregular function.
(iv) ([S5, Lemma 8.1(b)]) If m N, p m and is a nontrivial Dirichlet char-acter of conductorm, thenf(k) = (1 (p)pk(p1)+b1)pBk(p1)+b, is apregularfunction.
From Lemma 5.1 and Theorem 4.3 we deduce the following theorem.
Theorem 5.2. Letp be a prime, k,n,s, tN andb {0, 1, 2, . . . }.(i) Ifp 1b, xZp andx = (x+ xp)/p, then
Bktps1(p1)+b(x) pktps1
(p1)+b1Bktps1(p1)+b(x)ktps1(p 1) +b
n1r=0
(1)n1r
k 1 rn 1 r
k
r
Brtps1(p1)+b(x) prtps1(p1)+b1Brtps1(p1)+b(x
)
rtps1(p 1) +b (mod psn).
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(ii) Ifa, bN andpa, then
1 pktps1(p1)+b1aktps1(p1)+b 1 Bktps1(p1)+b
ktps1(p 1) +b
n1r=0
(1)n1rk 1 rn 1 rkr1 prtps1(p1)+b1 artps1(p1)+b 1 Brtps1(p1)+b
rtps1(p 1) +b (mod psn).
(iii) If b, m N, p m and is a nontrivial primitive Dirichlet character ofconductorm, then
(1 (p)pktps1(p1)+b1)Bktps1(p1)+b,ktps1(p
1) +b
n1r=0
(1)n1r
k 1 rn 1 r
k
r
(1 (p)prtps1(p1)+b1)Brtps1(p1)+b,
rtps1(p 1) +b (mod psn).
(iv) IfmN, p m and is a nontrivial Dirichlet character of conductorm, then
1 (p)pktps1(p1)+b1
pBktps1(p1)+b,
n1r=0
(1)n1rk 1 rn 1 rkr 1 (p)prtps1(p1)+b1pBrtps1(p1)+b, (mod psn).
Remark 5.1 Theorem 5.2 can be viewed as generalizations of some congruencesin [S5]. In the case n = 1, Theorem 5.2(i) was given by Eie and Ong [EO], andindependently by the author in [S5, p. 204]. In the case s = t = 1, Theorem 5.2(i)was announced by the author in [S4] and proved in [S5], and Theorem 5.2(iii) (in thecase p 1 b) and Theorem 5.2(iv) were also given in [S5]. When n = 1, Theorem5.2(iii) was given in [W, p. 141].
Combining Lemma 5.1 and Corollary 4.2(iv) we obtain the following result.
Theorem 5.3. Letp be an odd prime, k, sN andb {0, 1, 2, . . . }.(i) Ifp 1b, xZp andx = (x+ xp)/p, then
Bk(ps)+b(x)
k(ps) +b (1 kps1) Bb(x) p
b1Bb(x)
b +kps1
Bp1+b(x)
p 1 +b (mod ps+1).
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(ii) If b, m N, p m and is a nontrivial primitive Dirichlet character ofconductorm, then
Bk(ps)+b,
k(ps) +b(1 kps1)
1 (p)pb1
Bb,
b +kps1
Bp1+b,
p 1 +b (mod ps+1).
(iii) IfmN, p m and is a nontrivial Dirichlet character of conductorm, then(1 (p)pk(ps)+b1)pBk(ps)+b,(1 kps1)1 (p)pb1pBb,
+kps1
1 (p)pp2+bpBp1+b, (mod ps+1).Corollary 5.2. Letp be an odd prime andk,s, bN with2|b andp 1b. Then
Bk(ps)+bk(ps) +b
(1 kps1)(1 pb1) Bbb
+kps1 Bp1+bp
1 +b
(mod ps+1).
Theorem 5.4. Letp be a prime, a, nN andpa.(i) There are integersb0, b1, , bn1 such that
1 pk(p1)1ak(p1) 1 Bk(p1)
k(p 1)bn1kn1 + +b1k+b0 (mod pn) for k= 1, 2, 3, . . .
(ii) Ifp >2 orn >2, then
nk=1
nk
(1)k
(1 pk(p1)1
)(ak(p1)
1)Bk(p1)
k(p 1) 1
a(p
n)
pn (mod pn
).
Proof. Suppose bN. From Lemma 5.1(ii) we know that
f(k) =
1 pk(p1)+b1
ak(p1)+b 1 Bk(p1)+b
k(p 1) +bis a p-regular function. Hence taking b = p 1 and applying [S5, Theorem 2.1] weknow that there exist integers a0, a1, . . . , an1 such that
1 p(k+1)(p1)1a(k+1)(p1)
1 B(k+1)(p1)
(k+ 1)(p 1)an1kn1 + +a1k+a0 (modpn) for k= 0, 1, 2, . . .That is,
1 pk(p1)1ak(p1) 1 Bk(p1)
k(p 1)an1(k 1)n1 + +a1(k 1) +a0 (mod pn) for k= 1, 2, 3, . . .
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On setting
an1(k 1)n1 + +a1(k 1) +a0=bn1kn1 + +b1k+b0we obtain (i).
Now we consider (ii). Supposep >2 or n >2. Since f(k) is a p-regular function,
by Theorem 4.4(ii) we have
nk=1
n
k
(1)k(1 p(k1)(p1)+b1)(a(k1)(p1)+b 1) B(k1)(p1)+b
(k 1)(p 1) +b
(1 p(pn11)(p1)+b1)(a(pn11)(p1)+b 1) B(pn11)(p1)+b(pn1 1)(p 1) +b (mod p
n).
Substitutingbbyp 1 +b we see that for b0,
(5.1)
n
k=1
n
k(1)k(1 pk(p1)+b1)(ak(p1)+b 1) Bk(p1)+b
k(p
1) +b
(1 p(pn)+b1)(a(pn)+b 1)B(pn)+b(pn) +b
(mod pn).
By Corollary 5.1 we have pB(pn)p 1 (mod pn). Thus taking b= 0 in (5.1) andnoting that (pn)n+ 1 we obtain
nk=1
n
k
(1)k(1 pk(p1)1)(ak(p1) 1)Bk(p1)
k(p 1)
(1 p(pn)1)(a(pn) 1) B(pn)(pn)
=(1 p(pn)1) a(pn) 1
pn pB(pn)
p 1 a(p
n) 1pn
(mod pn).
This completes the proof of the theorem.
6. Congruences forn
k=0
nk
(1)kpBk(p1)+b(x) (mod pn+1).
ForaN and bZ we define (a|b) = 1 or 0 according as a|b orab.Lemma 6.1. Letp be an odd prime andnN. Then
n
s=1sn+1 (mod p1)
n
s (p 1|n) (mod p).Proof. Let n0 {1, 2, . . . , p 1} be such that nn0 (modp 1). Since Glaisher
(see [D]) it is well known that
ns=0
sr (mod p1)
n
s
n0s=0
sr (mod p1)
n0s
(mod p) for rZ.
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From [S1] we know that
ns=0
sr (mod p1)
n
s
=
ns=0
snr (mod p1)
n
s
.
Thusn
s=0sn+1 (mod p1)
n
s
=
ns=0
s1 (mod p1)
n
s
n0s=0
sp2 (mod p1)
n0s
=
p 1 1 (mod p) ifn0 = p 1,1 (mod p) ifn0 = p 2,0 (mod p) ifn0 < p 2.
Hence
ns=1
sn+1 (mod p1)
ns
=
ns=0
sn+1 (mod p1)
ns
(p1|n+1) (p1|n) (modp).
This proves the lemma.
Proposition 6.1. Letp be an odd prime, nN andxZp. Letb be a nonnegativeinteger. Then
nk=0
n
k
(1)k
pBk(p1)+b(x) pk(p1)+bBk(p1)+b
x+ xpp
p1j=0
j=xp
(x+j)bnpnBn
(x+j)p (x+j)p(p 1)
+pn(b,n,p) (mod pn+1),
where
(b,n,p) =
(n b)T n ifp 1|b andp 1|n,(n b)T ifp 1b andp 1|n,b n ifp 1|b andp 1|n+ 1,0 otherwise
and
T =p1j=0
j=xp
(x+j)p1+b (x+j)bp
.
Proof. Let
Sn=
nk=0
n
k
(1)k
pBk(p1)+b(x) pk(p1)+bBk(p1)+b
x+ xpp
.
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From [S4, p.157] we know that
Sn=
n(p1)+br=0
prBr
p1j=0
j=xp
nk=0
n
k
(1)k
k(p 1) +b
r
(x+j)k(p1)+br.
By [S5, p.199] we know that for any functionsf and g we have
(6.1)
nk=0
n
k
(1)kf(k)g(k)
=n
s=0
n
s
nsi=0
n s
i
(1)if(i+s)
sj=0
s
j
(1)jg(j).
Now taking f(k) =k(p1)+b
r
and g(k) = ak(p1)+br (a= 0) in (6.1) we obtainn
k=0
n
k
(1)k
k(p 1) +b
r
ak(p1)+br
=n
s=0
n
s
nsi=0
n s
i
(1)i
(i+s)(p 1) +b
r
sj=0
s
j
(1)jaj(p1)+br
=n
s=0
n
s
abr(1 ap1)s
nsi=0
n s
i
(1)i
i(p 1) +s(p 1) +b
r
.
Thus applying the above and Lemma 4.1 we have
Sn=
n(p1)+br=0
prBr
p1j=0
j=xp
ns=0
n
s
(x+j)br
1 (x+j)p1s
nsi=0
n s
i
(1)i
i(p 1) +s(p 1) +b
r
=
p1j=0
j=xp
ns=0
ns1 (x+j)
p1
psn(p1)+b
r=nspr+sBr (x+j)br
nsi=0
n s
i
(1)i
i(p 1) +s(p 1) +b
r
.
SincepBrZp and sopr+sBr0 (mod pn+1) forrn s + 2, by Theorem 4.1 we30
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have
n(p1)+br=ns
(x+j)brpr+sBr
nsi=0
n s
i
(1)i
i(p 1) +s(p 1) +b
r
(x+j)b(ns)pnBnsnsi=0
n s
i
(1)ii(p 1) +s(p 1) +b
n s
+ (x+j)b(ns+1)pn+1Bns+1
nsi=0
n s
i
(1)i
i(p 1) +s(p 1) +b
n s+ 1
= (x+j)b(ns)pnBns (1 p)ns + (x+j)b(ns+1)pn+1Bns+1 (s(p 1) +b+ (n s)(p 2)/2)(1 p)ns
(x+j)b(ns)(1 p)nspnBns+ (x+j)b(ns+1)(b
n)pn+1Bns+1 (mod p
n+1).
Thus,
Snp1j=0
j=xp
ns=0
n
s
1 (x+j)p1p
s(x+j)bn+s(1 p)nspnBns
+ (x+j)bn+s1(b n)pn+1Bns+1
=
p1j=0
j=xp
(x+j)bn(1 p)npnn
s=0
n
s
1 (x+j)p1p
x+j1 p
sBns
+
p1j=0
j=xp
ns=0
n
s
1 (x+j)p1p
s(x+j)bn+s1(b n)pn+1Bns+1
p1j=0
j=xp
(x+j)bn(1 p)npnBn(xj) +p1j=0
j=xp
ns=0
p1|ns+1
n
s
1 (x+j)p1p
s
(x+j)bn+s1(n b)pn (modpn+1),
where
xj = (x+j)p (x+j)
p(p 1) .
In the last step we use the facts
Bn(t) =
ns=0
n
s
tsBns and pBk (p 1|k) (mod p) (k1).
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ForaZ, using Lemma 6.1 and Fermats little theorem we see thatn
s=0sn+1 (mod p1)
n
s
as =
ns=1
sn+1 (mod p1)
n
s
as +(p 1|n+ 1)
an+1n
s=1sn+1 (mod p1)
n
s
+(p 1|n+ 1)
(p 1|n)an+1 +(p 1|n+ 1)
=
an+1 a (modp) ifp 1|n,1 (mod p) ifp 1|n+ 1,0 (mod p) ifp 1n and p 1n+ 1.
We also note that (see [S5, (5.1)])
(6.2)
p1j=0
j=xp
(x+j)b p1r=1
rb (p 1|b) (mod p).
Thus
p1j=0
j=xp
ns=0
p1|ns+1
n
s
1 (x+j)p1p
s(x+j)bn+s1(n b)pn
pn(n b) p1j=0
j=xp
(x+j)bn
s=0sn+1 (mod p1)
ns
1 (x+j)p1
p
s
pn(n b)p1j=0
j=xp
(x+j)b((x+j)p1 1)/p(mod pn+1)
ifp 1|n,
pn(n b)p1
j=0j=xp(x+j)b (p 1|b)(n b)pn (mod pn+1)
ifp 1|n+ 1,0 (mod pn+1) ifp 1n and p 1n+ 1.
On the other hand, fortZp we have Bn(t) BnZp (cf. [S4, Lemma 2.3]) and so
(np)pnBn(xj) npn+1Bn
npn (mod pn+1) ifp 1|n,0 (mod pn+1) ifp 1n.
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Thus applying (6.2) we get
p1j=0
j=xp
(x+j)bn (np)pnBn(xj)
p1j=0
j=xp
(x+j)b npn npn(p 1|b) (modpn+1) ifp 1|n,
0 (mod pn+1) ifp 1n.
Hence, by the above and the fact (1 p)n 1 np(mod p2) we obtainp1
j=0j=xp
(x+j)bn(1 p)npnBn(xj) p1
j=0j=xp
(x+j)bnpnBn(xj)
p1j=0
j=xp
(x+j)bn (np)pnBn(xj)
npn (mod pn+1) ifp 1|b and p 1|n,
0 (mod pn+1) ifp 1b or p 1n.
Now combining the above we see that
Snp1j=0
j=xp
(x+j)bnpnBn(xj)
npn + (n b)pnT (mod pn+1) ifp 1|b andp 1|n,pn(n b)T (modpn+1) ifp 1b andp 1|n,pn(b n) (mod pn+1) ifp 1|b andp 1|n+ 1,0 (mod pn+1) otherwise.
This is the result.Remark 6.1 When p= 2, b1 and n2, setting (b,n,p) =b n we can showthat the result of Proposition 6.1 is also true.Theorem 6.1. Letp be a prime greater than3, xZp, nN, n0, 1 (mod p 1)andb {0, 1, 2, . . . }. Letn0 be given bynn0 (modp1)andn0 {2, 3, . . . , p2}.Set
Sn=
nk=0
n
k
(1)k
pBk(p1)+b(x) pk(p1)+bBk(p1)+b
x+ xpp
.
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Then
Sn
nn0
Sn0pn0 + (n+2)b2pn (mod pn+1) ifp 1|b andp 1|n+ 1,
nn0
Sn0pn0pn (mod pn+1) ifp 1b orp 1n+ 1.
Proof. Sincep 1 n we know that Bn/nZp. FortZp, by [S4, Lemma 2.3]we have (Bn(t) Bn)/nZp. Thus
Bn(t)
n =
Bn(t) Bnn
+Bn
n Zp.
Asn0, 1 (mod p 1), by [S5, Corollary 3.1] we have
Bn(t)
n Bn0(t) p
n01Bn0
(t+ tp)/p
n0 Bn0(t)
n0(mod p).
Set xj = ((x+j)p (x+j))/(p(p 1)). Then xj Zp. ThusBn(xj)/n Zp andBn(xj)/nBn0(xj)/n0 (mod p). From Proposition 6.1 and the above we see that
Snpn
p1j=0
j=xp
(x+j)bnBn(xj) + (b n)(p 1|b)(p 1|n+ 1)
np1j=0
j=xp
(x+j)bn0Bn0(xj)
n0+ (b n)(p 1|b)(p 1|n+ 1) (mod p)
and so
Sn0pn0
n0p1j=0
j=xp
(x+j)bn0Bn0(xj)
n0+ (b n0)(p 1|b)(p 1|n+ 1) (mod p).
Thus
Snpn
nn0
Sn0pn0
(b n0)(p 1|b)(p 1|n+ 1)+ (b n)(p 1|b)(p 1|n+ 1)
= n
n0 Sn0
pn0+b
1 n
n0
(p 1|b)(p 1|n+ 1)
nn0
Sn0pn0
+b
1 +n
2
(p 1|b)(p 1|n+ 1) (mod p).
This proves the theorem.34
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Theorem 6.2. Let p be an odd prime, x Zp, b, n Z with n 1 and b 0. Ifp|n andp 1n, then
n
k=0
n
k
(1)k
pBk(p1)+b(x) pk(p1)+bBk(p1)+b
x+ xp
p
bpn (mod pn+1) ifp 1|b andp 1|n+ 1,0 (mod pn+1) ifp 1b orp 1n+ 1.
Proof. As p1 n and p| n, for t Zp we see that Bn(t)/n Zp and soBn(t) =nBn(t)/n0 (modp). Thus the result follows from Proposition 6.1.Theorem 6.3. Letp be an odd prime, nN andb {0, 2, 4, . . . }. Ifp(p 1)| n,then
nk=0
n
k
(1)k(1 pk(p1)+b1)pBk(p1)+b
pn1
2pn (mod pn+1) ifp
1|
b,
0 (mod pn+1) ifp 1b.Proof. From Proposition 6.1 we see that
nk=0
n
k
(1)k(1 pk(p1)+b1)pBk(p1)+b
p1j=1
jbnpnBn
jp jp(p 1)
bT pn (modpn+1),
where
T =
p1j=1
jp1+b jbp
.
For p >3 and mN, from [S5, (5.1)] we havep1j=1
jm pBm+ p2
2mBm1+
p3
6m(m 1)Bm2 (mod p3).
Ifm4 is even, then Bm1 = 0 andpBm2Zp. Thus
(6.3)
p1j=1j
m
pBm (mod p2
) for m= 2, 4, 6, . . .
Hence
T
pBp1+bpBbp (mod p) ifp > 3 and b >0,
pBp1(p1)p (mod p) ifp > 3 and b = 0,
22+b2b
3 = 2b (1)b = 1 3B223 (mod 3) ifp= 3.
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Ifp >3 and b = k(p 1) for some kN, by [S4, Corollary 4.2] we have
(6.4) pBb = pBk(p1)kpBp1 (k 1)(p 1) (mod p2)
and
pBp1+b=pB(k+1)(p1)(k+ 1)pBp1 k(p 1) (mod p2).Thus
T pBp1+bpBbp
pBp1 (p 1)p
(mod p).
Ifp >3 and p 1b, by Kummers congruences we have
Bp1+bp 1 +b
Bbb
(mod p) and so Bp1+b(b 1) Bbb
(mod p).
Thus
T pBp1+bpBbp
b 1b
Bb Bb=Bbb
(mod p).
Summarizing the above we have
(6.5) T
pBp1(p1)p (modp) ifp 1|b,
Bbb (mod p) ifp 1b.
Asp(p 1)|n, from Corollary 5.1 we have pBn(x)p 1 (mod p2) for xZp.Note that jn 1 (mod p2) forj = 1, 2, . . . , p 1. Combining the above we obtain
nk=0
n
k
(1)k(1 pk(p1)+b1)pBk(p1)+b
p1j=1
jbnpn1 pBn jp j
p(p 1)
bT pn
p1j=1
jbpn1(p 1) bT pn (mod pn+1).
From (6.3) and (6.4) we see that
p1j=1
jb
pBb bp1pBp1 ( bp1 1)(p 1) (mod p2)ifp >3, b >0 and p 1|b,
pBb (mod p2) ifp >3 and p 1b,
p 1 (mod p2) ifp >3 and b = 0,1 + (1 + 3)
b2 2 + 3b2 2 + 6b (mod 9) ifp= 3.
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That is,
p1
j=1jb
bp1 (pBp1 (p 1)) +p 1 (mod p2) ifp 1|b,pBb (mod p2) ifp
1b.
Hence
nk=0
n
k
(1)k(1 pk(p1)+b1)pBk(p1)+b
pn1(p 1)p1j=1
jb bT pn
pn1
(b(pBp1 (p 1)) + (p 1)2
) pn1
b(pBp1 (p 1))=pn1(p 1)2 pn1 2pn (mod pn+1) ifp 1|b,
pn1(p 1) pBb bpn (Bbb ) =pn+1Bb0 (mod pn+1) ifp 1b.
This completes the proof.
Theorem 6.4. Letp be a prime greater than3, xZp, nN, n0, 1 (mod p 1)andb {0, 1, 2, . . . }. Letn0 be given bynn0 (modp1)andn0 {2, 3, . . . , p2}.Let
f(k) =pBk(p1)+b(x) pk(p1)+bBk(p1)+b
x+ xpp
.
Then fork= 0, 1, 2, . . . we have
f(k)n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(r) +
n
n0n0
s=0
n0s
(1)sf(s)
pn0
k
n
(p)n
+(p 1|n+ 1)(p 1|b)(n+ 2)b
2 k
n k
n+ 1 (p)n (modpn+1).
Proof. From [S4, Theorem 3.1] we have
mk=0
m
k
(1)kf(k)pm1(p 1|m)(p 1|b) (mod pm) for mN.
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Thus applying [S4, Lemma 2.1], Theorem 6.1 and the above we see that
f(k) n1r=0
(1)n1r
k 1 rn 1 r
k
r
f(r)
=
kr=n
kr
(1)rr
s=0
rs
(1)sf(s)
k
n
(1)n
ns=0
n
s
(1)sf(s) +
k
n+ 1
(1)n+1
n+1s=0
n+ 1
s
(1)sf(s)
k
n
(1)npn
nn0
n0
s=0
n0s
(1)sf(s)
pn0+
(n+ 2)b
2 (p 1|n+ 1)(p 1|b)
+
k
n+ 1
(1)n+1pn(p 1|n+ 1)(p 1|b) (mod pn+1).
This yields the result.
Corollary 6.1. Letk, nN.(i) Ifn2 (mod 4), then
(5 54k)B4kn1r=0
(1)n1r
k 1 rn 1 r
k
r
(5 54r)B4r
+ 3n
k
n
5n (mod 5n+1)
and
(5 54k+2)B4k+2n1r=0
(1)n1rk 1 rn 1 rkr(5 54r+2)B4r+2 n
k
n
5n (mod 5n+1).
(ii) Ifn3 (mod 4), then
(5 54k)B4kn1r=0
(1)n1r
k 1 rn 1 r
k
r
(5 54r)B4r
+ k
n+ 15n (mod 5n+1)
and
(5 54k+2)B4k+2n1r=0
(1)n1r
k 1 rn 1 r
k
r
(5 54r+2)B4r+2
+n
k
n
5n (mod 5n+1).
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Theorem 7.1. Letp be an odd prime, k,m,n,tN andb {0, 2, 4, . . . }. Then
1 (1) p12 pktpm1(p1)+bEktpm1(p1)+b
n1
r=0
(
1)n1r
k 1 r
n 1 rk
r1
(
1)
p12 prtp
m1(p1)+b Ertpm1(p1)+b (mod pmn).
Puttingn = 1, 2, 3 and t= 1 in Theorem 7.1 we obtain the following result.
Corollary 7.1. Letp be an odd prime, k, mN andb {0, 2, 4, . . . }. Then(i) ([C, p. 131]) Ek(pm)+b
1 (1) p12 pbEb (mod pm).
(ii) Ek(pm)+bkE(pm)+b (k 1)
1 (1) p12 pbEb (mod p2m).(iii) We have
Ek(pm)+b
k(k 1)2
E2(pm)+b
k(k
2)1 (1)
p12 p(p
m)+bE(pm)+b+
(k 1)(k 2)2
1 (1)p12 pbEb (mod p3m).
From Lemma 7.1 and Corollary 4.2(iv) we have:
Theorem 7.2. Letp be an odd prime, k, mN andb {0, 2, 4, . . . }. Then
Ek(pm)+b(1 kpm1)(1 (1)p12 pb)Eb+kp
m1Ep1+b (mod pm+1).
Corollary 7.2. Letp be an odd prime andk, mN. Then
Ek(pm)
kpm1Ep1 (mod pm+1) ifp1 (mod 4),2 +kpm1(Ep1 2) (mod pm+1) ifp3 (mod 4).
From [S5, Theorem 2.1] and Lemma 7.1 we have:
Theorem 7.3. Letp be an odd prime, nN andb {0, 2, 4, . . . }. Then there areintegersa0, a1, . . . , an1 such that
(1 (1) p12 pk(p1)+b)Ek(p1)+ban1kn1 + +a1k+a0 (mod pn)for every k = 0, 1, 2, . . . Moreover, if p n, then a0, a1, . . . , an1 (mod pn) areuniquely determined.
As examples, we have
(1 + 32k)E2k 12k+ 2 (mod 33),(7.2)(1 54k)E4k 750k3 + 1375k2 620k (mod 55),(7.3)(1 54k+2)E4k+21000k3 + 1500k2 + 540k+ 24 (mod 55).(7.4)
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Theorem 7.4. LetnN andb {0, 2, 4, . . . }. Supposen N and2n1 n 1
and thusf(k) =E2k+b is a2regular function.Proof. Suppose nN and 2n1 n n, thus n < nand hence 2n n n+ 1. Therefore, forn 3 we have
nk=0
nk
(1)kE2k+b
0 (mod 2n+1). AsE0 E2 = 1 (1) = 2 and E0 2E2+E4 = 1 2(1) + 5 = 8,applying (7.6) and the above we see that E
bE
b+2 E
0E
2 = 2 (mod 8) and
Eb 2Eb+2+Eb+40 (mod 8).So the result follows.Theorem 7.5. Suppose k,m,n,tN andb {0, 2, 4, . . . }. ForsN letsN begiven by2s1 s
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From Corollary 7.3 and the proof of Theorem 4.2 we know that
nr=0
n
r
(1)rE22m1rt+b
=Antn 2mn +2m1ntr=n+1
(2)n(1)rAr (1)rns(r, n)n!r!
2rn 2(m1)ntn
+r
j=n+1
(1)rjs(r, j)j!r!
2rj S(j, n)n!j!
2jn (2m1t)j
.
By Lemma 4.2, for n + 1jr we haves(r, j)j!
r! 2rj ,
S(j, n)n!
j! 2jn Z2 and s(r, n)n!
r! 2rn
n
r n
(mod 2).
As 2n+1n+1
|Ar forr
n+ 1, by the above we obtain
(7.7)
nr=0
n
r
(1)rE2mrt+b2mnAntn 2mntnen (mod 2mn+n+1n+1)
and so
(7.8)n
r=0
n
r
(1)rE2mrt+b0 (mod 2mn+nn).
Forrn + 1 we havemr+ r rm(n + 1 ) + n + 1 n+1mn + n + 2 n+1.Thus, ifrn+ 1, by (7.8) we have
(7.9)r
s=0
r
s
(1)sE2mst+b0 (mod 2mn+n+2n+1).
By (4.5) we have
E2mkt+b =
n1r=0
(1)n1r
k 1 rn 1 r
k
r
E2mrt+b
+k
r=n
k
r
(1)r
r
s=0
r
s
(1)sE2mst+b.
Hence, applying (7.9) we obtain
(7.10)
E2mkt+bn1r=0
(1)n1r
k 1 rn 1 r
k
r
E2mrt+b
k
n
(1)n
ns=0
n
s
(1)sE2mst+b (mod 2mn+n+2n+1).
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In view of (7.7), we get
E2mkt+bn1r=0
(1)n1r
k 1 rn 1 r
k
r
E2mrt+b
+k
n
(1)n
2mn
tn
en (mod 2mn+n+1n+1
).
Now assume m2. Then (m 1)(n+ 1) +nmn+ 1. From the above we seethat
nr=0
n
r
(1)rE2mrt+b
2mnAntn +2m1ntr=n+1
(2)n(1)rAr(1)rns(r, n)n!
r! 2rn 2(m1)ntn
2mntnAn+ 2m1ntr=n+1
nr n
Ar2mntn n+2
r=n
nr n
Ar
2mntn
en+nen+1+
n
2
en+2
(mod 2mn+n+2n+1).
This together with (7.10) yields the remaining result. Hence the proof is complete.As 2nn |en and n + 1 n+1n n, by Theorem 7.5 we have:
Corollary 7.4. Let k,m,n,t N and b {0, 2, 4, . . . }. Let N be given by21 n
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Corollary 7.6. Letk, mN, m2 andb {0, 2, 4, . . . }. Then
E2mk+bkE2m+b (k 1)Eb+ 22mk(k 1) (mod 22m+2).
Taking m = 2 and b = 0, 2 in Corollary 7.6 we get:
Corollary 7.7. ForkN we have
E4k
4k+ 1 (mod 64) ifk0, 1 (mod 4),4k+ 33 (mod 64) ifk2, 3 (mod 4)
and
E4k+2
4k 1 (mod 64) ifk0, 1 (mod 4),4k 33 (mod 64) ifk2, 3 (mod 4).
Corollary 7.8. Letk, mN, m2 andb {0, 2, 4, . . . }. Letk = 0 or1 accordingas4k 3 or4|k 3. Then
E2mk+b
k
2
E2m+1+b k(k 2)E2m+b+
k 1
2
Eb+ 2
3m+1k (mod 23m+2).
Proof. Observe thate3 = 10, e4 = 104, e5 = 1816 andk3
k (mod 2). Takingn= 3 and t = 1 in Theorem 7.5 we obtain the result.
Taking m = 2, b = 0, 2 in Corollary 7.8 and noting that E8 105 (mod 256),E10 89 (mod 256) we deduce:Corollary 7.9. LetkN andk = 0 or1 according as4k 3 or4|k 3. Then
E4k48k244k+1+128k (mod 256)andE4k+216k276k1+128k (mod 256).
Remark 7.2 Let{Sn} be given by (3.1). From Remark 3.1 we know that (1)kSkis a 2-regular function and hence f(k) = (1)k+bSk+b is also a 2-regular function,where b {0, 1, 2, . . . }. Thus, by Corollary 4.2, for m 2, k 1 and b 0 we haveS2m1k+bSb(mod 2m) andS2m1k+bSb2m2k(Sb+2+4Sb+1+3Sb) (mod 2m+1).
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