Modulo $$p^2$$ p 2 Congruences Involving Generalized Harmonic Numbers
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Modulo p2 Congruences Involving Generalized Harmonic Numbers Yunpeng Wang1 · Jizhen Yang2 Received: 25 June 2020 / Revised: 9 September 2020 / Accepted: 22 September 2020 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2020
Abstract (n) Let p be an odd prime and Hk = kj=1 1/ j n denote the generalized harmonic num p−1 (n) ber. In this paper, the authors establish a kind of congruences involving k=1 k m Hk (mod p 2 ), where m, n are positive integers. Furthermore, the authors prove a congru p−1 ence for k=1 k p−2 (Hk(2) )2 (mod p 2 ). Keywords Bernoulli numbers · Congruences · Generalized harmonic numbers Mathematics Subject Classification 11A07 · 11B75 · 05A19
1 Introduction For α ∈ N, the generalized harmonic numbers are defined by (α)
H0
= 0 and Hn(α) =
n 1 , for n ∈ Z+ . iα i=1
In 1862, Wolstenholme [13] proved that if p > 3 is a prime, then H p−1 ≡ 0
(2) (mod p 2 ) and H p−1 ≡0
(mod p).
(1.1)
Communicated by Emrah Kilic.
B
Jizhen Yang [email protected]
1
Department of Mathematics and Physics, Luoyang Institute of Science and Technology, Luoyang 471023, People’s Republic of China
2
Department of Mathematics, Luoyang Normal College, Luoyang 471934, People’s Republic of China
123
Y. Wang, J. Yang
From (1.1), it is easy to deduce that 2p − 1 ≡ 1 (mod p 3 ). p−1
(1.2)
The congruence (1.2) is usually called Wolstenholme’s theorem and its extensions are various (see, e.g., [2–4,14]). In 2011, Sun [10] established a kind of congruences p−1 involving k=1 k m Hkn (mod p). Wang [12] generalized some of these congruences p−1 to modulo p 2 type. Sun [10] prove that k=1 k −2 Hk2 ≡ 0 (mod p) and conjectured two modulo p 2 congruences of Euler-type. These congruences were conformed in [8,11], respectively. In addition, Zhao [14] studied a kind of congruences involving multiple harmonic sums, which contained some congruences related to this kind. p−1 (n) In this paper, we focus on the properties of k=1 k m Hk (mod p 2 ) for 0 < m, n < p( p − 1). This is the biggest difference from the earlier work, which allows us to calculate Euler-type congruences by Euler’s theorem. Further, we establish a p−1 (2) congruence for k=1 k p−2 (Hk )2 (mod p 2 ). Our main results are as follows. Theorem 1.1 Let p > 3 be a prime and m, n be two positive integers less than p( p−1) such that m ≡ −1 (mod p) and p − 1 m, n. (i) If m ≤ n and δm,n is the Kronecker delta, then (m + 1)
p−1
(n)
k m Hk
≡
m m+1 p (n+r −m−1) Br H p−1 δm,n − r 2
(mod p 2 ).
r =δm,n
k=1
(ii) If n ≤ m < p − 1 or m, n are of different parity such that n < m and m ≡ n + 1 (mod p − 1), then p−1
k m Hk(n) ≡ −
(n−r ) n B m+1−r H p−1 m r =1
k=1
m +1−r
r
−p
m−n r =0
m Br Bm+1−n−r m +1−r r
(mod p2 ).
(iii) If m, n are two positive integers of same parity such that n < m and m < 2( p−1) and n < p − 1, then p−1
(n)
k m Hk
k=1
≡
1 m p 1+ Bm−n 2 n+1 n
(mod p 2 ).
(iv) Suppose that n ≤ p − 4 and define λm ≡ m − n (mod p − 1) such that λm ∈ {2, . . . , p − 2
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