Int. J. Number Theory 12(2016), no. 2, 527–539.
TWO CONGRUENCES INVOLVING HARMONIC
NUMBERS WITH APPLICATIONS
GUO-SHUAI MAO AND ZHI-WEI SUN
P
Abstract. The harmonic numbers Hn = 0<k6n 1/k (n = 0, 1, 2, . . .)
play important roles in mathematics. Let p > 3 be a prime. With
helps of some combinatorial identities, we establish the following
two new congruences:
p−1 2k
X
1
1 p
k
Bp−2
Hk ≡
(mod p)
k
3 3
3
k=1
and
p−1
X
k=1
2k
k
k
H2k
7 p
≡
Bp−2
12 3
1
(mod p),
3
where Bn (x) denotes the Bernoulli polynomial of degree n. As
Pp−1
Pp−1
an application, we determine n=1 gn and n=1 hn modulo p3 ,
where
n 2
n 2 X
X
n
n
2k
Ck
and hn =
gn =
k
k
k
k=0
k=0
with Ck = 2k
k /(k + 1).
1. Introduction
For n ∈ N = {0, 1, 2, . . .}, define
X 1
X 1
and Hn(2) :=
.
Hn :=
k
k2
0<k6n
0<k6n
Those Hn with n ∈ N are the classical harmonic numbers, and those
(2)
Hn with n ∈ N are called the second-order harmonic numbers.
Let p > 3 be a prime. By a classical result of J. Wolstenholme [W],
we have
(2)
Hp−1 ≡ 0 (mod p2 ) and Hp−1 ≡ 0 (mod p),
Key words and phrases. Central binomial coefficients, congruences, harmonic
numbers, Bernoulli polynomials.
2010 Mathematics Subject Classification. Primary 11B65, 11B68; Secondary
05A10, 11A07.
The second author is the corresponding author. This research was supported by
the Natural Science Foundation of China (grant 11571162).
1
2
GUO-SHUAI MAO AND ZHI-WEI SUN
which imply that
1 2p
2p − 1
=
≡ 1 (mod p3 ).
2 p
p−1
Z.-W. Sun [S12a] established some fundamental congruences
involving
Pp−1
k
harmonic numbers; for example, he showedPthat
k=1 Hk /(k2 ) ≡
∞
0 (mod p) motivated by the known identity k=1 Hk /(k2k ) = π 2 /12.
Let p > 3 be a prime. By Sun and R. Tauraso [ST11, (1.9)], and
Sun [S12c, (2.9)], we have
p−1 X
2k
p
≡
(mod p2 )
k
3
k=0
and
p−1
p
X
2k
k p−1
(−1)
≡ 3p−1
(mod p2 )
k
k
3
k=0
respectively, where (−) denotes the Legendre symbol. Hence
p−1 p 1 − 3p−1
X
2k
(mod p)
Hk ≡
k
3
p
k=0
since
Y p−1
p
(−1)
=
1−
≡ 1 − pHk (mod p2 )
k
j
0<j6k
k
for all k = 0, 1, . . . , p − 1. In 2010 Sun and Tauraso [ST10] proved that
p−1 2k
X
8
k
≡ p2 Bp−3 (mod p3 ),
k
9
k=1
where B0 , B1 , B2 , . . . are the Bernoulli numbers given by
∞
X xn
x
=
Bn
(0 < |x| < 2π).
ex − 1 n=0
n!
In 2011 Sun [S11] showed that
(p−1)/2 2k
X
−1 8
k
≡−
pEp−3 (mod p2 ),
k
p
3
k=1
where ( p· ) denotes the Legendre symbol and Ep−3 stands for the (p−3)th Euler number.
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
3
Recall that the Bernoulli polynomials are given by
n X
n
Bn (x) =
Bk xn−k (n = 0, 1, 2, . . .).
k
k=0
Motivated by the above work, we mainly obtain the following result in
this paper.
Theorem 1.1. Let p > 3 be a prime. Then
p−1 2k
X
1 p
1
k
Hk ≡
Bp−2
(mod p)
k
3
3
3
k=1
(1.1)
and
p−1
X
k=1
2k
k
k
H2k
7 p
≡
Bp−2
12 3
1
(mod p).
3
(1.2)
Remark 1.1. Another motivation of Theorem 1.1 is our desire to prove
Theorem 1.2 in this section. Why (1.1) and (1.2) should involve the
value of Bp−2 (x) at 1/3 ? There are no intuitive reasons, but a slight
indication comes from the following congruence in [ST11, (1.5)] for any
prime p > 3:
p−1 p−1
X
X
2k
(−1)k p − k
p
≡
+ 2p
(mod p3 ).
k
3
k
3
k=0
k=1
There are no closed forms for the sums in (1.1) and (1.2). Our approach
to Theorem 1.1 is somewhat unique in the sense that it depends heavily
on some special combinatorial identities.
Clearly Theorem 1.1 has the following consequence.
Corollary 1.1. For any prime p > 3 we have
p−1 2k
X
k
(4H2k − 7Hk ) ≡ 0 (mod p).
k
k=1
(1.3)
Recall that Hp−1 ≡ 0 (mod p2 ) for any prime p > 3. So our following
conjecture is much stronger than Corollary 1.1.
Conjecture 1.1. For any prime p > 3 we have
p−1 2k
X
Hp−1 278 3
k
(4H2k − 7Hk ) ≡ −14
+
p Bp−5 (mod p4 ).
k
p
15
k=1
4
GUO-SHUAI MAO AND ZHI-WEI SUN
Note that Theorem 1.1 and Corollary 1.1 are related
second
P∞ to the
3
author’s
following
conjectural
formulas
for
ζ(3)
=
1/n
and
K=
n=1
P∞ k
2
k=1 ( 3 )/k (cf. [S14a]):
∞
∞
X
X
H2k + 2Hk
H2k + 17Hk
5
5√
ζ(3)
and
3πK.
=
=
2k
2k
2
2
3
2
k
k
k
k
k=1
k=1
Pn
3
The Franel numbers fn = k=0 nk (n = 0, 1, 2, . . .) play important
roles in combinatoricsP
and number
theory. In 1975 P. Barrucand [B]
obtained the identity nk=0 nk fk = gn , where
n 2 X
n
2k
gn :=
.
(1.4)
k
k
k=0
The sequences (fn )n>0 and (gn )n>0 are two of the five sporadic sequences (cf. D. Zagier [Z, Section 4]) which are integral solutions of
certain Apéry-like recurrence equations and closely related to the theory of modular forms. In 2013, Sun [S13] revealed some unexpected
connections between those numbers fn , gn and representations of primes
p ≡ 1 (mod 3) in the form x2 + 3y 2 with x, y ∈ Z. For any prime p > 3,
Sun [S14b] and [S12b, (1.15)] showed that
p−1
X
n=1
gn ≡
p−1
X
hn ≡ 0 (mod p2 ),
n=1
where
hn :=
n 2
X
n
k=0
k
Ck
(1.5)
2k
and Ck refers to the Catalan number 2k
/(k + 1) = 2k
− k+1
. The
k
k
numbers h0 , h1 , h2 , . . . appeared naturally in the second author’s study
of Apéry polynomials (cf. [S12b]).
Applying Theorem 1.1, we deduce the following result.
Theorem 1.2. Let p > 3 be a prime. Then
p−1
p−1
X
1 X
5 p
1
(2)
g
≡
g
H
≡
B
(mod p),
k
k
p−2
k
p2 k=1
8
3
3
k=1
(1.6)
and
p−1
p−1
X
1 X
3 p
1
(2)
hk ≡
hk Hk ≡
Bp−2
(mod p).
2
p k=1
4
3
3
k=1
(1.7)
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
5
We are going to prove Theorems 1.1 and 1.2 in Sections 2 and 3
respectively. Our proofs make use of some sophisticated combinatorial
identities.
2. Proof of Theorem 1.1
Lemma 2.1. For any n ∈ N, we have
n X
x
y
x+y
=
,
k
n
−
k
n
k=0
n 2
X
n
2n
Hk =
(2Hn − H2n ),
k
n
k=0
bn/2c n
X
X n 2k 2k
k n
n
(−1)
=(−1)
.
k
k
2k
k
k=0
k=0
(2.1)
(2.2)
(2.3)
The three identities (2.1)-(2.3) are known, see, e.g., [G, (3.1), (3.125)
and (3.86)].
Proof of Theorem 1.1. By (2.2),
k 2
X
k
j=1
j
Hj =
2k
(2Hk − H2k ) for each k = 1, . . . , p − 1.
k
Therefore
p−1
X
k=1
2k
k
k
H2k = 2
p−1
X
k=1
2k
k
k
Hk −
p−1
k 2
X
1X k
k=1
k
j=1
j
Hj .
Observe that
p−1
k 2
X
1X k
k=1
k
j=1
j
p−1
p−1 X
Hj X k
k−1
Hj =
j k=j j
j−1
j=1
(2.4)
6
GUO-SHUAI MAO AND ZHI-WEI SUN
and
p−1 X
k
k−1
j
k=j
j−1
p−1−j
X i + j i + j − 1 p−1−j
X −j − 1−j =
=
i
i
i
i
i=0
i=0
p−1−j X p − j p − 1 − j X p − j p − 1 − j p−1−j
=
≡
i
p−1−j−i
i
i
i=0
i=0
2p − 2j − 1
=
(mod p)
p−1−j
with the help of the Chu-Vandermonde identity (2.1). Thus
p−1
p−1
k 2
X
X
1X k
Hj 2p − 2j − 1
Hj ≡
k
j
j
p−1−j
j=1
j=1
k=1
X
p−1
p−1
X
Hj −2j − 1
Hj p + j − 1
≡
=
(−1)j
j
p
−
1
−
j
j
2j
j=1
j=1
=
p−1
X
Hj
j=1
j
p(−1)j Y 2
·
(p − i2 )
j (2j)!(p + j) i=1
p−1
p−1
X
X
Hj
Hj
≡p
2j ≡ p
2j (mod p).
2
2
j
j
j
j
j=1
j=(p+1)/2
By [S11, Lemma 2.1],
2j
2(p − j)
p+1
j
≡ 2p (mod p2 ) for all j =
, . . . , p − 1.
j
p−j
2
(Tauraso [T] contains a similar technique.) Therefore
p−1
k 2
X
1X k
Hj
k j=1 j
k=1
(p−1)/2 2k
p−1
1 X Hj 2(p − j)
1 X k Hp−k
≡
=
(mod p).
2
j
p−j
2 k=1
p−k
j=(p+1)/2
Since
Hp−k = Hp−1 −
1
1
≡ Hk−1 = Hk − (mod p)
p−j
k
0<j<k
X
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
7
for all k = 1, . . . , p − 1, from the above we obtain
p−1
p−1 2k
p−1 2k
k 2
X
1X k
1X k
1X k
Hj ≡ −
Hk +
(mod p).
k j=1 j
2 k=1 k
2 k=1 k 2
k=1
Combining this with (2.4) we get
p−1 2k
p−1 2k
p−1 2k
X
5X k
1X k
k
H2k ≡
Hk −
(mod p).
2
k
2
k
2
k
k=1
k=1
k=1
(2.5)
For each k = 1, . . . , p − 1, clearly
p
p
p Y p−j
=
≡ (−1)k−1 (1 − pHk−1 ) (mod p3 ).
k 0<j<k j
k
k
Thus
p−1
p−1 2k
X
X
2k
k p
k
(1 − pHk−1 ) (mod p3 ).
(−1)
≡ −p
k
k
k
k=1
k=1
On the other hand, by (2.3) we have
(p−1)/2 p
X
X
2k
p
2k
k p
p
(−1)
=(−1)
k
k
2k
k
k=0
k=0
(p−1)/2
X 1 − pH2k−1 2k ≡−1+p
(mod p3 ).
2k
k
k=1
Therefore
p−1
X
2k
k
2p
p
−p
1 − pHk +
−
+1
p
k
k
k=1
(p−1)/2
X 1 − p(H2k − 1/(2k)) 2k ≡−1+p
(mod p3 )
2k
k
k=1
2p
Since p ≡ 2 (mod p3 ) by Wolstenholme’s theorem, and
p−1 2k
X
k
≡ 0 (mod p2 )
k
k=1
(2.6)
(2.7)
by [ST10], from (2.6) we get
(p−1)/2 2k
(p−1)/2 2k
(p−1)/2 2k
p−1 2k
X
1 X k
1 X k
5 X k
k
Hk ≡
−
H2k +
(mod p).
k
2p k=1 k
2 k=1 k
4 k=1 k 2
k=1
(2.8)
8
GUO-SHUAI MAO AND ZHI-WEI SUN
(Note that
Clearly,
Pp−1
k=1
2k
k
2 P(p−1)/2
/k ≡ k=1
2k
k
2
/k (mod p).).
2p−2k
pH2p−2k = p
X 1
+ 1 ≡ 1 (mod p)
j
j=1
j6=p
for all k = 1, . . . , (p − 1)/2. So we have
(p−1)/2
X
k=1
(p−1)/2
p−1
X
X pH2p−2k
pH2j
≡
=
2k
2k
2
2
2 2(p−j)
k k
k k
(p
−
j)
p−j
k=1
j=(p+1)/2
2j
p−1
1 X
j
H2j (mod p)
≡
2
j
1
j=(p+1)/2
with the help of [S11, Lemma 2.1]. By [S11, (1.2) and (1.3)],
(p−1)/2
(p−1)/2 2k
X
1 X k
2
≡ 0 (mod p).
+
2
p k=1 k
k 2k
k
k=1
Therefore
p−1
X
k=1
2k
k
k
(p−1)/2
X
H2k ≡
k=1
(p−1)/2 2k
1 X k
H2k −
(mod p).
k
p k=1 k
2k
k
Combining this with (2.8) we get
p−1 2k
p−1 2k
p−1 2k
X
X
X
5
1
k
k
k
Hk ≡
−
H2k (mod p).
2
k
4
k
2
k
k=1
k=1
k=1
(2.9)
(2.5) and (2.9) together imply that
p−1 2k
p−1 2k
p−1 2k
p−1 2k
X
X
2X k
7X k
k
k
Hk ≡
(mod p) and
H2k ≡
(mod p).
k
3 k=1 k 2
k
6 k=1 k 2
k=1
k=1
It is known that
p−1
X
k=1
2k
k
k2
1 p
≡
Bp−2
2 3
1
(mod p)
3
(cf. [MT]). So we get the desired (1.1) and (1.2).
Remark 2.1. In [S11] the second author proved that
(p−1)/2 2k
(p−1)/2
X
1 X k
2
−1 8
≡
≡
Ep−3 (mod p).
−
2k
2
p k=1 k
p
3
k
k
k=1
(2.10)
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
9
3. Proof of Theorem 1.2
Lemma 3.1. For any nonnegative integers m and n, we have
n X
x+k
k=0
m
n+x+1
x
=
−
.
m+1
m+1
(3.1)
and
n 2 X
n
x+k
k=0
k
2n
2
x
=
.
n
(3.2)
Remark 3.1. Both (3.1) and (3.2) can be found in [G, (1.48) and (6.30)].
Lemma 3.2. For any nonnegative integer n, we have
n 2 X
n
x+k
k=0
k
2n + 1
1
=
(4n + 2)
2 n
X
x
2k
(2x − 3k)
.
2n
k
k
n k=0
(3.3)
Remark 3.2. One might wonder how we find (3.3). In fact, we use the
2 x+k P
software package Sigma to find a recurrence for un = nk=0 nk 2n+1
and then obtain (3.3) by solving the recurrence for un via Sigma.
Proof of Lemma 3.2. Let F (x) and G(x) denote the left-hand side and
the right-hand side of (3.3). With the help of (3.2), we see that
2
x
F (x + 1) − F (x) =
.
n
Applying the Zeilberger algorithm (see, e.g., [PWZ, pp. 101-119]) via
Mathematica, we find that
2
x
G(x + 1) − G(x) =
n
for all x = 0, 1, 2, . . . .
So, by induction F (x) = G(x) for all x ∈ N. As F (x) and G(x) are
polynomials in x of degree 2n + 1, we have the desired (3.3).
10
GUO-SHUAI MAO AND ZHI-WEI SUN
Proof of Theorem 1.2. (i) With the help of Lemma 3.1, we have
p−1
X
p−1 2
p−1 X
2k X n
=
k n=k k
k
k
k=0
p−1 k 2 p−1 X
2k X X k
n+j
=
k n=k j=0 j
2k
k=0
k 2 p−1 p−1 X
2k X k X n + j
=
k j=0 j n=k
2k
k=0
k 2 p−1 X
2k X k
p+j
=
.
k
j
2k
+
1
j=0
k=0
gn =
n=0
p−1 n 2 X
X n
2k
n=0 k=0
Thus, by applying Lemma 3.2 we get
p−1
X
p−1
X
2 k
1 X
p
2j
gk =
(2p − 3j)
4k + 2 j=0
j
j
k=0
k=0
2 X
p−1
p−1
X
1X
p
2j
1
1
(2p − 3j)
=
−
2 j=0
2k
+
1
2i + 1
j
j
06i<j
k=0
2 p−1
1X
p
2j
Hp−1
Hj
=
(2p − 3j)
H2p−1 −
− H2j +
.
j
j
2 j=0
2
2
Note that pHp−1 ≡ 0 (mod p3 ) and
pH2p−1
p−1 p−1
X
X
1
2p
1
=1+p
=1 + p
+
p−j p+j
p2 − j 2
j=1
j=1
(2)
≡1 − 2p2 Hp−1 ≡ 1 (mod p3 ).
Therefore
p−1
X
k=0
gk ≡
2 p−1
X
2p − 3j p
2j
j
j
2 p−1
X
2p − 3j p
2j
Hj
+
− H2j (mod p3 )
2
j
j
2
j=0
j=0
2p
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
11
and hence
p−1
X
2 p−1
X
2p − 3j p2 p − 1
2j
gk ≡
· 2
2p
j j−1
j
j=1
k=1
2 p−1
X
2p − 3j p2 p − 1
2j
Hj
+
· 2
− H2j
2
j
j
−
1
j
2
j=1
2
p−1 2j
p−1 2j X j
X
p−1
3p
j
2
≡p
−
j2
2 j=1 j j − 1
j=1
p−1 2j Hj
3 2X j
− H2j (mod p3 ).
− p
2 j=1 j
2
(Note that 2jj H2j is p-adic integral for all j = 1, . . . , p − 1.) Clearly,
2
p−1
≡ (1 − pHj−1 )2 ≡ 1 − 2pHj−1 (mod p2 ).
(3.4)
j−1
Thus
p−1 2k
p−1
p−1 2k
p−1 2k X
X
3 1X k
1
1 X
k
k
gk ≡
−
−2
Hk −
p2 k=1
k2
2 p k=1 k
k
k
k=1
k=1
p−1 2k
p−1 2k
X
3 1X k
k
−
Hk −
H2k
2 2 k=1 k
k
k=1
p−1 2k
p−1 2k
p−1 2k
X
9X k
3X k
k
+
Hk +
H2k (mod p)
≡−2
2
k
4
k
2
k
k=1
k=1
k=1
with the help of (2.7). Now, applying Theorem 1.1 and (2.10) we
immediately get that
p−1
1 X
5 p
1
gk ≡
Bp−2
(mod p).
(3.5)
2
p k=1
8 3
3
(ii) Observe that
p−1
p−1 n 2 X
X
X n
1
2k
(2gn − hn ) =
2−
k
k+1
k
n=0
n=0 k=0
p−1 2
p−1
X
2k + 1 2k X n
.
=
k
+
1
k
k
k=0
n=k
12
GUO-SHUAI MAO AND ZHI-WEI SUN
Similar to the proof in part (i), we have
2 p−1
p−1
k
X
X
(2k + 1)/(k + 1) X
p
2j
(2gn − hn ) =
(2p − 3j)
4k + 2
j
j
n=0
j=0
k=0
2 p−1
p
2j
1
1X
(2p − 3j)
Hp−1 + − Hj
=
2 j=0
j
j
p
and thus
p−1
X
(2gk − hk )
k=1
2 p−1
1X
p2 p − 1
2j
1
− Hj
≡
(2p − 3j) 2
2 j=1
j j−1
j
p
p−1 2j
p−1 2j
X
X
3p
j
j
≡p2
−
(1 − 2pHj−1 ) (1 − pHj ) (by (3.4))
2
j
2
j
j=1
j=1
p−1 2j
p−1 2j X j
X j
3
2
2
≡p
− p
1+p
− 3Hj
j2
2 j=1 j
j
j=1
p−1 2j
p−1 2j
p−1 2j
X
3 X j
9 2X j
j
2
= − 2p
− p
+ p
Hj (mod p3 )
2
j
2 j=1 j
2 j=1 j
j=1
Combining this with (1.1), (2.7) and (2.10), we obtain that
p−1
X
p2 p 1
(2gk − hk ) ≡
Bp−2
(mod p3 ).
2
3
3
k=1
Thus,
p−1
X
3 2 p
Bp−2
hk ≡ p
4
3
k=1
1
(mod p3 )
3
(3.6)
with the help of (3.5).
(iii) By [S14b, Theorem1.1],
p−1
X
gk ≡ p
k=1
Therefore
p−1
X
k=1
2
p−1
X
7
(2)
gk Hk + p3 Bp−3 (mod p4 ).
6
k=1
(2)
gk Hk
p−1
1 X
≡ 2
gk (mod p)
p k=1
and hence (1.6) holds in view of (3.5).
(3.7)
TWO CONGRUENCES INVOLVING HARMONIC NUMBERS
13
From [S14b, Theorem1.1] we know that
p−1
p−1
X
X
p
2 (2)
2 (2)
gk (x) 1 − p Hk
≡
1 − 2p Hk xk (mod p4 ), (3.8)
2k
+
1
k=0
k=0
Pk
2
where gk (x) = j=0 kj 2jj xj . Therefore, the left-hand side of (3.8)
minus the right-hand side of (3.8) can be written as p4 P (x), where P (x)
is a polynomial of degree at most p − 1 with p-adic integer coefficients.
Since
Z 1
n 2 X
n
2k
gn (x)dx for n = 0, 1, 2, . . . ,
hn =
Ck =
k
k
0
k=0
we deduce that
p−1
X
(2)
hk (1 − p2 Hk )
k=0
Z
=
p−1
1X
0
p−1
=
X
k=0
≡
p−1
X
k=0
(2)
gk (x)(1 − p2 Hk )dx
k=0
p
(2)
(1 − 2p2 Hk )
2k + 1
Z
1
k
x dx + p
4
0
Z
1
P (x)dx
0
p−1
X
2p
p
(2)
(2)
(1 − 2p2 Hk ) −
(1 − 2p2 Hk ) (mod p3 ).
2k + 1
k+1
k=0
Combining (3.7) and (3.8) we see that
p−1
X
7 3
(2)
1 + p Bp−3 ≡
gk 1 − p2 Hk
6
k=0
≡
p−1
X
k=0
p (2)
1 − 2p2 Hk
(mod p4 ).
2k + 1
Therefore
p−1
p−1
p−1
(2)
X
X
X
p
Hk
7 3
2 (2)
3
+ 2p
hk (1 − p Hk ) ≡2 + p Bp−3 −
3
k+1
k+1
k=0
k=0
k=0
(2)
≡2 − 1 − pHp−1 + 2p2 Hp−1 ≡ 1 (mod p3 )
which implies that
p−1
X
k=1
(2)
hk (1 − p2 Hk ) ≡ 0 (mod p3 ).
14
GUO-SHUAI MAO AND ZHI-WEI SUN
Combining this (3.6) we obtain the desired (1.7).
So far we have completed the proof of Theorem 1.2.
Acknowledgment. The authors would like to thank the referee for
helpful comments.
References
[B]
[G]
[MT]
[PWZ]
[S11]
[S12a]
[S12b]
[S12c]
[S13]
[S14a]
[S14b]
[ST10]
[ST11]
[T]
[W]
[Z]
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(Guo-Shuai Mao) Department of Mathematics, Nanjing University,
Nanjing 210093, People’s Republic of China
E-mail address: [email protected]
(Zhi-Wei Sun) Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
E-mail address: [email protected]