Several congruences involving harmonic numbers

WANG Yunpeng，YANG Jizhen

(1.Department of Mathematics and Physics，Luoyang Institute of Science and Technology，Luoyang，Henan 471022，China;2.Department of Mathematics，Shanghai Normal University，Shanghai 200234，China;3.Department of Mathematics，Luoyang Normal College，Luoyang，Henan 471934，China)

Abstract: Harmonic numbers play important roles in mathematics. Let p>5be a prime. We establish the following congruences：

Key words： Bernoulli numbers; congruences; harmonic numbers

1 Introduction

Forα∈，the generalized harmonic numbers are defined byforn∈.Whenα=1，they reduce to the well-known harmonic numbersH0= 0andn∈.In 1862，Wolstenholme[1]proved that ifp>3 is a prime，then

(1)

Using these congruences，Sun obtained a series of modulopcongruences involving harmonic numbers. Wang[7，9-10]generalized some results of Sun[8]and established congruences for

In this paper，we use the method of Wang[10]and establish congruences involving harmonic numbers and Bernoulli numbers．

2 Lemmas

In this section，we first state some basic facts which will be used very often．

(2)

which implies that

(3)

(4)

Then we have

(5)

Lemma1Letp>3be a prime. Then

S(0，n)≡0 (modp)，

(6)

S(1，n)≡-S(2，n) (modp)，

(7)

(8)

Lemma 2([9]，Theorem 3.1) Letp>3be a prime and0

H(m，1)≡

(9)

Remark1By Lemma 2，we can get

(10)

Lemma 3([9]，Theorem 4.1) Letp>3 be a prime andm=1，2，…，(p-3)/2. Then

H(2m+1，4)≡

(modp2).

(11)

Lemma 4Letp>3be a prime and0

(12)

(13)

(14)

ProofThe first congruence appeared as the corollary 3.3 of literature [9]，and the second congruence is (3.3)of literature [10] and the third is the corollary 3.7 of Wang[10]．

Lemma5Letp>3 be a prime.Then

(15)

(16)

ProofThe first congruence appeared as (4.5) of literature [9] and the second congruence is the corollary 4.2 of literature [9]．

Lemma6Letp>5 be a prime.Then

H(0，4)≡2Bp-3-24 (modp)，

(17)

H(1，4)≡12-Bp-3(modp)，

(18)

(19)

(20)

Proof(18)，(19) and (20) appeared as (2.19)，(2.9) and (2.12) of Wang[10]respectively．

Observe that

(21)

Combining (18) and (21)，we obtain (17)．

Lemma7([10]，Lemma 2.4) Letp>3 be a prime，l=1，2，…，p-2 andt=l-1，…，p-3+l. Then

(22)

Lemma8([10]，Theorem 3.1) Letp>3be a prime. Then

(23)

(24)

(25)

(26)

Remark2Combining (7)，(23) and (24)，we can get

(27)

(28)

Lemma9Letp>3be a prime. Then

(29)

(30)

(31)

(32)

(33)

ProofThe first two congruences are (3.26) and (3.29) of Wang[13]，respectively.Combining (7) and (30)，we obtain (31)．

By (8)，we have

(34)

Writing (23) and (29) in (34)，we obtain (32)．

Note that

(35)

Writing (32) in(35)，we obtain (33)．

3 Main results

Theorem1Letp>5be a prime. Then

(36)

Observe thatH0=0 andHp-1≡0 (modp). Therefore

which implies that

H(4，4)≡-2H(3，4)-2H(2，4)-

(37)

Note thatH(1，0)≡0 (modp)．

Writing (10)，(13)，(16)，(17)，(18)，(19) and (20) in (37)，we finally obtain (36)．

Theorem2Letp>5be a prime. Then

(38)

(39)

(40)

ProofLetm，nbe nonnegative integers. With the help of (4) and by the binomial theorem，we have

S(2m，n)≡

S(2m-r，n-r)) (modp).

(41)

Observe thatH(-1，0)≡0 (modp). Therefore

H(2-r，3-r)) (modp).

(42)

By Lemma 7，we have

(43)

Using(12)，(15)，(23)，(25) and(43)，we can get

The proof of (38) is completed．

Observe that

which implies that

Then we have

(44)

Takingm=6，n=1 in (44)，we have

(45)

Note that

(46)

Takingm=4in Lemma 2 andt=5，l=2 in Lemma 7，we can obtain

(47)

Writing(23)，(27)，(29)，(32)，(46) and(47) in(45)，we obtain

(48)

Again takingm=6，n=2 in (44)，we have

(49)

Recall that

(50)

and by Lemma 2 and Lemma 7，we have

H(3，1)≡0 (modp) andS(4，0)≡0 (modp).

(51)

Similarly，takingm=6，n=3 in (44)，we can get

(52)

Note thatH(1，0)≡0 (modp). By (7) and Lemma 7，we can get

(53)

Writing (13)，(16)，(26)，(29)，(30)，(31)，(38)，(39)and (53) in (52)，we finally obtain (40)．

Theorem3Letp>5be a prime. Then

(54)

ProofTakingm=2，n=4 in Lemma 3，we have

Writing(10)，(13)，(16)，(36) and (40)，we finally arrive at (54)．