Table Of ContentSUPERCONGRUENCES RELATED TO F (1)
3 2
INVOLVING HARMONIC NUMBERS
ROBERTO TAURASO
Abstract. Weshowvarioussupercongruencesfortruncatedserieswhichinvolvecentral
7
binomial coefficients and harmonic numbers. The corresponding infinite series are also
1
evaluated.
0
2
n
a
J 1. Introduction
0
3 In 1997, Van Hamme [21] established the p-adic analogs of several Ramanujan type
series. For one of them, the series labeled (H.1),
]
T
N ∞ 2k 3 1, 1, 1 π
k = F 2 2 2;1 = , (1)
. 64k 3 2 1,1 Γ4(3)
h k=0 (cid:0) (cid:1) (cid:20) (cid:21) 4
t X
a the modulo p2 congruence (H.2) for the truncated version has been recently improved by
m
Long and Ramakrishna in [9, Theorem 3],
[
29v3 Xkp−=10 (cid:0)62k4k(cid:1)k3 ≡p3 ( −−Γp1624pΓ(cid:0)4p41(cid:1)41 iiff pp ≡≡44 13,. (2)
7 where p isany prime greater than 3 (we use the n(cid:0)ota(cid:1)tiona m b to meana b (mod m)).
0 ≡ ≡
In this paper we will investigate the series and the corresponding partial sums where
0
. the terms have one of the following forms
1
0 3 3 (2) 3 3 (2)
2k H 2k H 2k O 2k O
7 k, k , k , k .
1 k 64k k 64k k 64k k 64k
: (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
v
(r) (r)
i Here H denotes the k-th generalized harmonic number of order r and O is the sum
X k k
with odd denominators,
r
a
k k
1 1
(r) (r)
H = and O =
k jr k (2j 1)r
j=1 j=1 −
X X
(1) (1)
where we adopt the convention that H = H and O = O .
k k k k
Date: January 31, 2017.
2010 Mathematics Subject Classification. 11A07,33C20,11S80,33B15,11B65.
Key words and phrases. Supercongruences, hypergeometric series, harmonic numbers, p-adic Gamma
function.
1
2 ROBERTOTAURASO
The main results are presented in Section 3 (evaluations of the infinite series) and
Section 5 (congruences for the truncated series). For example we show that
∞ 2k 3 O π2 p−1 2k 3 O 0 if p 4 1,
k k ≡
= , and
Xk=1(cid:18)k (cid:19) 64k 6Γ4(34) Xk=1(cid:18)k (cid:19) 64k ≡p2 −1p2Γ4p 14 if p ≡4 3.
The correspondence between the right-hand sides of the infinite serie(cid:0)s(cid:1)and the finite sum
is particularly striking for the appearance of the classic Gamma function and the p-adic
analog.
2. Similar results of lower degree
Before dealing with the main issue, we are going to take a look to similar sums already
in the literature, where the central binomial coefficient is raised to a power less than 3.
Assume that p is a prime greater than 3. For n 1, we have
≥
n−1
2k H 2n 2n(H 2)
k n−1
= − +2.
k 4k n 4n
k=1(cid:18) (cid:19) (cid:18) (cid:19)
X
Thus, by n = p, we obtain (see [19, (1.10)] for the modulo p3 version)
p−1
2k H 1
k 2 2p+4p2q (2) 6p3q2(2) p3B ,
k 4k ≡p4 − p − p − 3 p−3
k=1(cid:18) (cid:19)
X
where q (a) = ap−1−1 is the Fermat quotient and we used the Wolstenholme’s theorem
p p
2p 2, and the congruences
p ≡p3
(cid:0) (cid:1) H 1p2B , 4p−1 1+2pq (2)+p2q2(2) (3)
p−1 ≡p3 −3 p−3 ≡p3 p p
(for the first one we can refer to [14, Theorem 5.1 (a)]). Moreover, the identity
n−1 2k H(2) 2n 2nH(2) n−1 2k
k = n−1 2 k
k 4k n 4n − k4k
k=1(cid:18) (cid:19) (cid:18) (cid:19) k=1 (cid:0) (cid:1)
X X
implies (see [19, (1.11)] for the modulo p version)
p−1 (2)
2k H 4 1
k 4q (2)+2pq2(2) p2q3(2) p2B ,
k 4k ≡p3 − p p − 3 p − 2 p−3
k=1(cid:18) (cid:19)
X
where we employed the congruence established in [20, Theorem 1.1],
n−1 2k
kk4k ≡p3,−Hp−21
k=1 (cid:0) (cid:1)
X
and
2 2 7
Hp(2−)1 ≡p2 3pBp−3, Hp−21 ≡p3 −2qp(2)+pqp2(2)− 3p2qp3(2)− 12p2Bp−3 (4)
3
given in [14, Theorem 5.1 (a)]) and [14, Theorem 5.2 (c)] respectively.
As regards the squared case, the identities [12, (2.4) and (2.8)]
n
n n+k
( 1)kH = 2( 1)nH ,
k n
k k − −
k=1(cid:18) (cid:19)(cid:18) (cid:19)
X
n n n+k n ( 1)k
( 1)kH(2) = 2( 1)n+1 − ,
k k − k − k2
k=1(cid:18) (cid:19)(cid:18) (cid:19) k=1
X X
and the congruence for 0 k n = (p 1)/2 (note that p divides 2k for n < k < p)
≤ ≤ − k
n n+k 2k k ((2j 1)2 p2) (cid:0) (cid:1)2k 2
( 1)k = j=1 − − k (5)
k k − k 4k(2k)! ≡p2 16k
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)Q (cid:0) (cid:1)
imply [19, Theorem 4.1] (see also [17, Theorems 1.1 and 1.2] for a more general p2-
congruence)
p−1 2
2k H
k 16kk ≡p2 (−1)p+21(4qp(2)−2pqp2(2)),
k=1(cid:18) (cid:19)
X
p−1 2 (2)
2k H
k 8E +4E ,
k 16k ≡p2 − p−3 2p−4
k=1(cid:18) (cid:19)
X
where we also used
(2) 7 (2) p−1 14
Hp−21 ≡p2 3pBp−3, H⌊p4⌋ ≡p2 (−1) 2 (8Ep−3−4E2p−4)+ 3 pBp−3 (6)
given in [14, Corollary 5.2], [15, Corollary 3.8] and
n ( 1)k 1 (2) (2) p−1
−k2 = 2H⌊p4⌋ −Hp−21 ≡p2 (−1) 2 (8Ep−3 −4E2p−4). (7)
k=1
X
3. Evaluations of the infinite series
The generalized hypergeometric function is defined as
a ,a , ,a ∞ (a ) (a ) (a ) zk
1 2 r 1 k 2 k r k
F ··· ;z = ···
r s
b ,b , ,b (b ) (b ) (b ) · k!
(cid:20) 1 2 ··· s (cid:21) k=0 1 k 2 k··· s k
X
where (x) = x(x+1) (x+k 1) for k 1 and (x) = 1 is the Pochhammer symbol
k 0
··· − ≥
and a , b and z are complex numbers with none of the b being negative integers or zero.
i j j
We recall some well-known hypergeometric identities:
i) Dixon’s theorem [1, p.13]
a,b,c Γ(1+ a)Γ(1+a b)Γ(1+a c)Γ(1+ a b c)
F ;1 = 2 − − 2 − − , (8)
3 2 1+a b,1+a c Γ(1+a)Γ(1+ a b)Γ(1+ a c)Γ(1+a b c)
(cid:20) − − (cid:21) 2 − 2 − − −
4 ROBERTOTAURASO
ii) Whipple’s theorem [1, p.16]
a,1 a,c π21−2cΓ(e)Γ(1+2c e)
F − ;1 = − . (9)
3 2 e,1+2c e Γ(a+e)Γ(1−a+e)Γ(1+c a+e)Γ(1+c 1−a+e)
(cid:20) − (cid:21) 2 2 − 2 − 2
In the next theorem we evaluate four specific series.
Theorem 1. We have that
∞ 2k 3H 2π(π 3ln2) ∞ 2k 3 O π2
k k
= − , = , (10)
k 64k 3Γ4(3) k 64k 6Γ4(3)
k=1(cid:18) (cid:19) 4 k=1(cid:18) (cid:19) 4
X X
∞ 2k 3H(2) π(12G π2) ∞ 2k 3O(2) π(π2 8G)
k = − , k = − . (11)
k 64k 3Γ4(3) k 64k 8Γ4(3)
k=1(cid:18) (cid:19) 4 k=1(cid:18) (cid:19) 4
X X
where G = ∞ (−1)k is the Catalan’s constant.
k=0 (2k+1)2
Proof. Let P
k−1
1
(r)
H (x) = .
k (x+j)r
j=0
X
Then
d d
(r) (r+1)
((x) ) = (x) H (x) and H (x) = rH (x).
dx k k · k dx k − k
(cid:16) (cid:17)
For (10), let a = b = 1/2 in (8), then
∂ 1, 1,c ∞ 2k 32O +H
F 2 2 ;1 = k k.
∂c 3 2 1, 3 c k 64k
(cid:18) (cid:20) 2 − (cid:21)(cid:19)(cid:12)c=1 k=1(cid:18) (cid:19)
(cid:12) 2 X
(cid:12)
By setting b = c = 1/2 in (8), we get
(cid:12)
∂ a, 1, 1 ∞ 2k 32O 2H
F 2 2 ;1 = k − k.
∂a 3 2 1 +a, 1 +a k 64k
(cid:18) (cid:20)2 2 (cid:21)(cid:19)(cid:12)a=1 k=1(cid:18) (cid:19)
(cid:12) 2 X
(cid:12)
On the other hand, by differentiating the right-hand side of (8) and (9) and by using
(cid:12)
d d
(Γ(x)) = Γ(x) Ψ(x) and Ψ(r)(x) = Ψ(r+1)(x)
dx · dx
where Ψ(r) is the polygamma function of order r (w(cid:0)ith Ψ(0)(cid:1)= Ψ), we obtain
∂ 1, 1,c 1, 1, 1 1 3 1
F 2 2 ;1 = F 2 2 2;1 Ψ(1) Ψ +Ψ +Ψ ,
∂c 3 2 1, 3 c 3 2 1,1 · − − 4 4 2
(cid:18) (cid:20) 2 − (cid:21)(cid:19)(cid:12)c=1 (cid:20) (cid:21) (cid:18) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
(cid:12) 2
(cid:12)
and
(cid:12)
∂ a, 1, 1 1, 1, 1 3
F 2 2 ;1 = F 2 2 2;1 2Ψ(1) 2Ψ 2ln(2) .
∂a 3 2 1 +a, 1 +a 3 2 1,1 · − 4 −
(cid:18) (cid:20)2 2 (cid:21)(cid:19)(cid:12)a=1 (cid:20) (cid:21) (cid:18) (cid:18) (cid:19) (cid:19)
(cid:12) 2
(cid:12)
(cid:12)
5
By considering a suitable linear combination of the previous two identities, the special
values
1 1 π 3 π
Ψ Ψ(1) = ln4, Ψ Ψ(1) = ln8 , Ψ Ψ(1) = ln8+ .
2 − − 4 − − − 2 4 − − 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
yield immediately (10).
Let a = c = 1/2 in (9), then
∂2 1, 1, 1 ∞ 2k 32H(2)
F 2 2 2 ;1 = k .
∂e2 3 2 e,2 e k 64k
(cid:18) (cid:20) − (cid:21)(cid:19)(cid:12)e=1 k=1(cid:18) (cid:19)
(cid:12) X
Moreover, for c = 1/2, e = 1 in (9), we find (cid:12)
(cid:12)
∂2 a,1 a, 1 ∞ 2k 3 8O(2)
F − 2;1 = − k .
∂a2 3 2 1,1 k 64k
(cid:18) (cid:20) (cid:21)(cid:19)(cid:12)(cid:12)a=21 Xk=1(cid:18) (cid:19)
On the right-hand side, we have (cid:12)
(cid:12)
∂2 1, 1, 1 1, 1, 1 π2 3
F 2 2 2 ;1 = F 2 2 2;1 Ψ ,
∂e2 3 2 e,2 e 3 2 1,1 · 3 − 1 4
(cid:18) (cid:20) − (cid:21)(cid:19)(cid:12)e=1 (cid:20) (cid:21) (cid:18) (cid:18) (cid:19)(cid:19)
(cid:12)
and (cid:12)
(cid:12)
∂2 a,1 a, 1 1, 1, 1 1 1 1 3
F − 2;1 = F 2 2 2;1 Ψ Ψ π2 .
∂a2 3 2 1,1 3 2 1,1 · 2 1 4 − 2 1 4 −
(cid:18) (cid:20) (cid:21)(cid:19)(cid:12)a=1 (cid:20) (cid:21) (cid:18) (cid:18) (cid:19) (cid:18) (cid:19) (cid:19)
2
(cid:12)
As before, by combining the res(cid:12)ults and by using the special values
(cid:12)
1 1
Ψ = π2 8G
1
2 ± 4 ∓
(cid:18) (cid:19)
(cid:3)
the conclusion (11) easily follows.
4. Congruences for the truncated series - Preliminary results
If n is an odd integer, by replacing k with (n k) is easy to see that
−
n 3 n 3
n n
( 1)k = 0 and ( 1)k H H = 0. (12)
k n−k
− k − k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
The next lemma follows from [2, Theorem 1].
Lemma 1. For any non-negative odd integer n = 2m+1, we have
n 3
n c
( 1)k H = m, (13)
k
− k − 6
k=0 (cid:18) (cid:19)
X
n 3
n c
( 1)k (3H2 +H(2)) = m (H 4H H +2H ). (14)
− k k k 2 m − 2m+1 − 3m+2 6m+4
k=0 (cid:18) (cid:19)
X
6 ROBERTOTAURASO
( 1)m(6m+3)!(m!)3
where c = − .
m
(3m+1)!((2m+1)!)3
The next lemma establishes some identities involving the harmonic numbers that we
will need later on.
Lemma 2. For any non-negative integer n, we have
2 n (−1)k
n
k=1 k2 if n 0,
n n n+k 2k Hk(2) = (cid:18)n2(cid:19) · P 4n ≡2 (15)
Xk=0(cid:18)k(cid:19)(cid:18) k (cid:19)(cid:18)k (cid:19)(−4)k n−1 −2 −4n−1 if n 1.
n−1 · n2 ≡2
(cid:18) 2 (cid:19)
Moreover, for any even integer,
n 2
n n+k 2k H n H
k n
= , (16)
k k k ( 4)k n 4n
k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) − (cid:18)2(cid:19)
X
n 2
n n+k 2k H n H
2k n
= . (17)
k k k ( 4)k n 2 4n
k=0(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) − (cid:18)2(cid:19) ·
X
Proof. For n = 2m, let
−2
2m 2m+k 2k 2m
F(m,k) = ( 4)2m−k
k k k m −
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
then by Wilf-Zeilberger method we find
2(4m+3)k2 2m+1 2m+k 2k 2m −2
G(m,k) = ( 4)2m−k
− (2m+1)3 k 1 k k m −
(cid:18) − (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
such that
F(m+1,k) F(m,k) = G(m,k +1) G(m,k).
− −
(2)
Let S(m) = F(m,k)H then, by summation by parts (see [5] for a similar ap-
k≥1 k
proach), we have
P
(2)
S(m+1) S(m) = (G(m,k +1) G(m,k))H
− − k
k≥0
X
G(m,k +1) G(m,k)
= =
− (k +1)2 − k2
k≥0 k≥1
X X
1 1
= + .
−(2m+1)2 (2m+2)2
(cid:3)
The other identities can be obtained in a similar way.
7
The Morita’s p-adic Gamma function Γ is defined as the continuous extension to the
p
set of all p-adic integers Z of the sequence
p
n ( 1)n k
→ −
0≤k<n
Y
(k,p)=1
where p is an odd prime and n > 1 is an integer (see [13, Chapter 7] for a detailed
introduction to Γ ). If x Z then Γ (0) = 1 and
p p p
∈
xΓ (x) if x = 1,
p p
Γ (x+1) = − | |
p
Γ (x) if x < 1,
( p p
− | |
where denotes the p-adic norm. By [9, Theorem 14], for all a,b Z ,
p p
|·| ∈
Γ (a+bp) Γ (a)(1+G (a)bp) (18)
p p2 p 1
≡
where G (a) = Γ′(a)/Γ (a) Z . Moreover
1 p p ∈ p
Γ (x)Γ (1 x) = ( 1)sp(x) (19)
p p
− −
where s (x) is the integer in 1,2,...,p such that s (x) x. The above formula is the
p p p
{ } ≡
p-adic analog of the classic reflection formula for the classic Gamma function
π
Γ(x)Γ(1 x) = .
− sin(πx)
Lemma 3. For any prime p > 3,
2m 2 Γ4 1 if p 1,
m − p 4 ≡4 (20)
(cid:0)16m(cid:1) ≡p2 16Γ−(cid:0)4 (cid:1)1 (1+2p) if p 3,
p 4 ≡4
where m = p/4 . Moreover if p 3 then(cid:0) (cid:1)
4
⌊ ⌋ ≡
p 1
c Γ4 . (21)
m ≡p2 2 p 4
(cid:18) (cid:19)
Proof. We start with (21). Since p 3, we have that m = (p 3)/4 and
4
≡ −
( 1)m(2m+p)!(m!)3 ( 1)m+1p(m!)3
c = − −
m p2
(3m+1)!((2m+1)!)3 ≡ 2(3m+2)!((2m+1)!)2
( 1)m+1pΓ3 p+1 ( 1)m+1pΓ3 1 p 1
= − p 4 − p 4 Γ4
2Γ 3p+3 Γ2 p+1 ≡p2 2Γ 3 Γ2 1 ≡p2 2 p 4
p 4 p(cid:0) 2 (cid:1) p 4 p (cid:0)2 (cid:1) (cid:18) (cid:19)
where, by (19), (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
1 1 3
Γ2p 2 = (−1)p+21 = 1, and Γp 4 Γp 4 = (−1)p+41 = (−1)m+1. (22)
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
8 ROBERTOTAURASO
As regards (20), we consider only the case p 3 since the other case can be handled
4
≡
similarly. Then
2m 2 1 2 Γ2(1) Γ2 1 +m Γ2 1 + p
m = 2 m = p p 2 = p −4 4 .
16m (1) Γ2 1 · Γ2(1+m) Γ2 1 + p
(cid:0) (cid:1) (cid:0) (cid:1)m ! p 2 p(cid:0) (cid:1) p(cid:0) 4 4 (cid:1)
By (18) and by [7, Lemma 2.4], (cid:0) (cid:1) (cid:0) (cid:1)
1 p 1 p
Γ + Γ 1+(G (1)+H ) ,
p p2 p 1 3m+1
−4 4 ≡ −4 4
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
and
1 p 1 p
Γ + Γ 1+(G (1)+H ) .
p p2 p 1 m
4 4 ≡ 4 4
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
Therefore, since Γ 1 = 4Γ 3 ,
p −4 p −4
2m 2 (cid:0) Γ2(cid:1) 1 (cid:0) (cid:1) p 1
m p −4 1+(H H ) 16Γ−4 (1+2p)
16m ≡p2 Γ2 1 · 3m+1 − m 2 ≡p2 p 4
(cid:0) (cid:1) p(cid:0) 4 (cid:1) (cid:18) (cid:19)
(cid:16) (cid:17)
where we also used (22) a(cid:0)nd(cid:1)
m+1
1 4
H = H H + H +4.
3m+1 p−1 p m p m
− p j ≡ 3p+1 ≡
j=1 −
X
(cid:3)
5. Congruences for the truncated series - Main results
Theorem 2. For any prime p > 3,
p−1 3 Γ4 1 (2q (2) pq2(2)) if p 1,
2k Hk p 4 · p − p ≡4
(23)
Xk=1(cid:18)k (cid:19) 64k ≡p2 −1p(cid:0)2Γ(cid:1)4p 41 if p ≡4 3,
and (cid:0) (cid:1)
p−1 3 (2) Γ4 1 (4E 2E ) if p 1,
2k Hk − p 4 · p−3 − 2p−4 ≡4 (24)
Xk=1(cid:18)k (cid:19) 64k ≡p2 −41Γ4p(cid:0) (cid:1)41 if p ≡4 3.
Proof. For (24), if p 1 then n = (p 1)(cid:0)/2(cid:1)is even and we use (5) and (15). Finally we
4
≡ −
use (20). If p 3 then n = (p 1)/2 = 2m+1 is odd. We have
4
≡ −
2k k−1 −1 k−1
p p 1 p
k = 1 1 = 1 (H H )
( 4)k n − 2j +1 ≡p2 − 2 n j − 2 n − n−k
−(cid:0) (cid:1)k j=0(cid:18) (cid:19) j=0 −
Y X
and therefore(cid:0) (cid:1)
1 2k n p
( 1)k 1 (H H ) . (25)
4k k ≡p2 − k − 2 n − n−k
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
9
Thus, by (12), (16), and (21),
p−1 3 n 3
2k H n 3p
k ( 1)k 1 (H H ) H
k 64k ≡p2 − k − 2 n − n−k k
k=1(cid:18) (cid:19) k=0 (cid:18) (cid:19) (cid:18) (cid:19)
X X
3p c p 1
1 H m Γ4 .
≡p2 − 2 n − 6 ≡p2 −12 p 4
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
As regards (24), we use (5) and (15) with n = (p 1)/2. Then we apply (20) and (7). (cid:3)
−
Theorem 3. For any prime p > 3,
p−1 3 0 if p 1,
2k O 4
k ≡
(26)
Xk=1(cid:18)k (cid:19) 64k ≡p2 −1p2Γ4p 41 if p ≡4 3,
and (cid:0) (cid:1)
p−1 3 (2) 1Γ4 1 E if p 1,
2k Ok 2 p 4 p−3 ≡4 (27)
Xk=1(cid:18)k (cid:19) 64k ≡p −116Γ(cid:0)4p(cid:1)41 if p ≡4 3.
Proof. If n = (p 1)/2 is even then by (5), (16) a(cid:0)nd(cid:1)(17),
−
p−1 3 p−1 2k 3 2 2
2k O H n H 1 n H
k = k H k n n 0.
k 64k 64k 2k − 2 ≡p2 n 2 4n − 2 n 4n ≡p2
k=1(cid:18) (cid:19) k=1 (cid:0) (cid:1) (cid:18) (cid:19) (cid:18)2(cid:19) · (cid:18)2(cid:19)
X X
Assume now that n = (p 1)/2 = 2m+1 is odd. We have that
−
2k
1 p
(2) (2) (2)
H = H H +pH H + H H .
2(n−k) p−1 − p j ≡p2 2k 2k ≡p2 2k 4 k − n−k
Xj=1 − (cid:16) (cid:17)
Hence
n 3 n 3
n n
( 1)k H = ( 1)k H
2k 2(n−k)
− k − − n k
k=1 (cid:18) (cid:19) k=0 (cid:18) − (cid:19)
X X
n 3 n 3
n p n
( 1)k H ( 1)k H(2) H(2)
≡p2 − − k 2k − 4 − k k − n−k
Xk=1 (cid:18) (cid:19) Xk=0 (cid:18) (cid:19) (cid:16) (cid:17)
which implies
n 3 n 3 n 3
n p n p n
( 1)k H ( 1)k H(2) H(2) = ( 1)k H(2).
− k 2k ≡p2 −8 − k k − n−k −4 − k k
Xk=1 (cid:18) (cid:19) Xk=0 (cid:18) (cid:19) (cid:16) (cid:17) Xk=0 (cid:18) (cid:19)
(28)
Moreover
k−1 k−1
H 1 1 1 1
k
H = + H (H +H H ). (29)
2k p k p k n−k n
2 2j +1 ≡ 2 − n j ≡ 2 −
!
j=0 j=0 −
X X
10 ROBERTOTAURASO
Consequently by (25), (28), (29),
p−1 3 n 3
2k H n 3p
2k ( 1)k 1 (H H ) H
k 64k ≡p2 − k − 2 n − n−k 2k
k=1(cid:18) (cid:19) k=1 (cid:18) (cid:19) (cid:18) (cid:19)
X X
n 3 n 3
n 3p n
( 1)k H ( 1)k (H H )(H +H H )
p2 2k n n−k k n−k n
≡ − k − 4 − k − −
k=1 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
n 3 n 3
p n 3p n
( 1)k H(2) H ( 1)k H
≡p2 −4 − k k − 2 n − k k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
n 3 n 3
3p n 3p n
+ ( 1)k H2 H ( 1)k H
4 − k n−k − 4 n − k n−k
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X
n 3
p n c p 1
( 1)k 3H2 +H(2) m Γ4 (30)
≡p2 −4 − k k k ≡p2 − 4 ≡p2 −8 p 4
Xk=0 (cid:18) (cid:19) (cid:16) (cid:17) (cid:18) (cid:19)
where in the last step we used (13), (14), and (21) (note that m < 2m+1 < 3m+2 <
p < 6m+4 < 2p). Finally, by (23),
p−1 3 p−1 2k 3
2k O H p 1
k = k H k Γ4
k 64k 64k 2k − 2 ≡p2 −12 p 4
k=1(cid:18) (cid:19) k=1 (cid:0) (cid:1) (cid:18) (cid:19) (cid:18) (cid:19)
X X
and the proof of (26) is complete.
As regards (27), we have by (6)
k−1 k−1 (2) (2) (2)
1 1 H H H
O(2) = = n − n−k n−k (31)
k (2j +1)2 ≡p 4(n j)2 4 ≡p − 4
j=0 j=0 −
X X
where n = (p 1)/2. Then, by (25) and (31),
−
p−1 2k 3O(2) 1 n n 3 ( 1)n+1 n n 3
k ( 1)k H(2) = − ( 1)k H(2)
k 64k ≡p −4 − k n−k 4 − k k
k=1(cid:18) (cid:19) k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
X X X
( 1)n+1 p−1 2k 3H(2)
− k
p
≡ 4 k 64k
k=1(cid:18) (cid:19)
X
(cid:3)
and the desired result follows from (24).
Remark 4. By (23), (26), and (2) for any prime p > 3
p−1 3 p−1 2k 3
2k H H
2k − k q (2) k
k 64k ≡p p 64k
k=1(cid:18) (cid:19) k=0 (cid:0) (cid:1)
X X
