Congruences for Franel numbers

0.21 MB·English

by  Zhi-Wei Sun

Preprint, arXiv:1112.1034
CONGRUENCES FOR FRANEL NUMBERS
Zhi-Wei Sun
2
1 Department of Mathematics, Nanjing University
0
2 Nanjing 210093, People’s Republic of China
[email protected]
n
a http://math.nju.edu.cn/∼zwsun
J

6
2 pAlbaystirmapcotr.taTntheroFlersaninelbnoutmhbceormsbgiinvaetnorbicysfannd=nuPmnkb=e0r(cid:0)tnkh(cid:1)e3or(yn. I=n 0th,1is,2p,a.p.e.)r
]
T we initiate the systematic investigation of fundamental congruences for Franel
N numbers. Let p>3 be a prime. We mainlyestablish the followingcongruences:
.
h
at p−1(−1)kf ≡ p (mod p2), p−1(−1)kkf ≡−2 p (mod p2),
m X k (cid:16)3(cid:17) X k 3 (cid:16)3(cid:17)
k=0 k=0
0 [ pX−1 (−k1)kfk ≡0 (mod p2), pX−1 (−k12)kfk ≡0 (mod p).
1 k=1 k=1
v
4 We also pose several conjectural congruences.
3
0
1
.
2
1. Introduction
1
1
1 It is well known that
:
v
n 2
i n 2n
X = (n = 0,1,2,...)
(cid:18)k(cid:19) (cid:18)n (cid:19)
r kX=0
a
and central binomial coefficients play important roles in mathematics. A fa-
mous theorem of J. Wolstenholme [W] asserts that
1 2p 2p−1
= ≡ 1 (mod p3) for any prime p > 3.
2(cid:18)p(cid:19) (cid:18)p−1 (cid:19)
The reader may consult [S11a], [S11b], [ST1] and [ST2] for recent work on
congruences involving central binomial coefficients.
2010MathematicsSubjectClassification.Primary11A07,11B65;Secondary05A10,11B68,
11E25.
Keywords. Franel numbers, Ap´ery numbers, binomial coefficients, congruences.
Supported by the National Natural Science Foundation (grant 11171140) of China.
1
2 ZHI-WEI SUN
In 1895 J. Franel [F] noted that the numbers
n 3
n
f = (n = 0,1,2,...) (1.1)
n
(cid:18)k(cid:19)
X
k=0
(cf. [Sl, A000172]) satisfy the recurrence relation:
(n+1)2f = (7n(n+1)+2)f +8n2f (n = 1,2,3,...).
n+1 n n−1
SuchnumbersarenowcalledFranelnumbers. Forcombinatorialinterpretations
of Franel numbers and Barrucand’s identity
n n 2
n n 2k
f = g with g = , (1.2)
k n n
(cid:18)k(cid:19) (cid:18)k(cid:19) (cid:18)k (cid:19)
X X
k=0 k=0
the reader may consult D. Callan [C]. Recall that Ap´ery numbers given by
n 2 2 n 2 2
n n+k n+k 2k
A = = (n = 0,1,2,...)
n
(cid:18)k(cid:19) (cid:18) k (cid:19) (cid:18) 2k (cid:19) (cid:18)k (cid:19)
X X
k=0 k=0
were introduced by Ap´ery [Ap] (see also [Po]) in his famous proof of the irra-
tionality of ζ(3) = ∞ 1/n3, and they can be expressed in terms of Franel
n=1
numbers as follows:P
n
n n+k
A = f (1.3)
n k
(cid:18)k(cid:19)(cid:18) k (cid:19)
X
k=0
(see V. Strehl [St92]).
The Franel numbers are also related to the theory of modular forms. Let η
be the Dedkind eta function given by
∞
η(τ) := eπiτ/12 (1−e2πinτ)
nY=1
with Im(τ) > 0. It is known that
∞ η(τ)3η(6τ)9 n η(2τ)η(3τ)6
f =
n(cid:18)η(2τ)3η(3τ)9(cid:19) η(τ)2η(6τ)3
nX=0
for any complex number τ with Im(τ) > 0. (See, e.g., D. Zagier [Z].)
In this paper we study congruences for Franel numbers systematically. As
usual, for an odd prime p and integer a, (a) denotes the Legendre symbol, and
p
q (a) stands for the Fermat quotient (ap−1 −1)/p if p ∤ a.
p
Now we state our main results.
CONGRUENCES FOR FRANEL NUMBERS 3
Theorem 1.1. Let p > 3 be a prime. For any p-adic integer r we have
p−1 p−1 2
k +r 2k k +r
(−1)k f ≡ (mod p2). (1.4)
k
(cid:18) k (cid:19) (cid:18)k (cid:19)(cid:18) k (cid:19)
X X
k=0 k=0
In particular,
p−1
p
(−1)kf ≡ (mod p2), (1.5)
k
3
X (cid:16) (cid:17)
k=0
p−1
2 p
(−1)kkf ≡− (mod p2), (1.6)
k
3 3
X (cid:16) (cid:17)
k=0
p−1
10 p
(−1)kk2f ≡ (mod p2), (1.7)
k
27 3
X (cid:16) (cid:17)
k=0
and
p−1 2k f p−1 2k 3
k k ≡ k (mod p2). (1.8)
(cid:0)(−4(cid:1))k (cid:0)16(cid:1)k
X X
k=0 k=0
We also have
p−1
(−1)k
f ≡0 (mod p2), (1.9)
k
k
X
k=1
p−1 (−1)k
f ≡0 (mod p), (1.10)
k2 k
X
k=1
p−1
(−1)k
f ≡3q (2)+3pq (2)2 (mod p2), (1.11)
k−1 p p
k
X
k=1
and
p−1
f
(3k +1) k ≡ p2 −2p3q (2)+4p4q (2)2 (mod p5). (1.12)
8k p p
X
k=0
Remark 1.1. Fix a prime p > 3. As f ≡ (−8)kf (mod p) for all k =
k p−1−k
0,... ,p−1 by [JV, Lemma 2.6], (1.11) implies that
p−1 p−1 p−1
f (−1)k (−1)p−k
k
≡ f = f ≡ 3q (2) (mod p).
k8k k p−1−k p−k k−1 p
X X X
k=1 k=1 k=1
4 ZHI-WEI SUN
Concerning (1.8) the author [S11b, Conj. 5.2(ii)] conjectured that
p−1 2k 3 4x2 −2p (mod p2) if p ≡ 1 (mod 3) & p = x2 +3y2 (x,y ∈ Z),
k ≡
(cid:0)16(cid:1)k (cid:26) 0 (mod p2) if p ≡ 2 (mod 3).
X
k=0
We also have many such conjectures for p−1 2k f /mk mod p2. (1.10) can
k=0 k k
be extended as P (cid:0) (cid:1)
p−1
(−1)kr
(r)
f ≡ 0 (mod p), (1.13)
kr−1 k
X
k=1
where r is any positive integer and f(r) := k k r. Note that f(2) = 2k
k j=0 j k k
and p−1 2k /k ≡ 0 (mod p2) by [ST1]. P (cid:0) (cid:1) (cid:0) (cid:1)
k=1 k
P (cid:0) (cid:1)
Let p > 3 be a prime. Similar to (1.5)-(1.7), we are also able to show that
p−1 p−1
10 p 14 p
(−1)kk3f ≡ − (mod p2) and (−1)kk4f ≡ − (mod p2).
k 81 3 k 35 3
X (cid:16) (cid:17) X (cid:16) (cid:17)
k=0 k=0
In general, for any positive integer r there should be an odd integer a such
r
that
p−1
2a p
(−1)kkrf ≡ r (mod p2).
k 32r−1 3
X (cid:16) (cid:17)
k=0
For example, if p > 5 then
p−1 p−1
322 p 2030 p
(−1)kk5f ≡ (mod p2), (−1)kk6f ≡ − (mod p2).
k 37 3 k 39 3
X (cid:16) (cid:17) X (cid:16) (cid:17)
k=0 k=0
The Ap´ery polynomials introduced in [S11c] are those
n 2 2
n n+k
A (x) = xk (n = 0,1,2,...).
n
(cid:18)k(cid:19) (cid:18) k (cid:19)
X
k=0
Here we define
n 2
n 2k
g (x) = xk
n
(cid:18)k(cid:19) (cid:18)k (cid:19)
X
k=0
for all n = 0,1,2,.... Note that g (1) = g .
n n
CONGRUENCES FOR FRANEL NUMBERS 5
Theorem 1.2. Let p > 3 be a prime. Then
p−1 p−1
xk
g (x) ≡ p (mod p2). (1.14)
k
2k +1
X X
k=0 k=0
Consequently,
p−1
g ≡0 (mod p2), (1.15)
k
X
k=1
p−1
−1
g (−1) ≡ (mod p2), (1.16)
k
(cid:18) p (cid:19)
X
k=0
p−1
p
g (−3) ≡ (mod p2). (1.17)
k
3
X (cid:16) (cid:17)
k=0
We also have
p−1
g (x)
k
≡0 (mod p), (1.18)
k
X
k=1
p−1
g p
k−1
≡− 2q (3) (mod p), (1.19)
p
k 3
X (cid:16) (cid:17)
k=1
p−1
3
kg ≡− (mod p2), (1.20)
k
4
X
k=1
p−1
g (−1)
k
≡0 (mod p) if p > 5. (1.21)
k2
X
k=1
Moreover,
n−1 n−1 2
1 n−1
(4k+3)g = C (1.22)
3n2 k (cid:18) k (cid:19) k
X X
k=0 k=0
for all n = 1,2,3,..., where C denotes the Catalan number 2k /(k + 1) =
k k
2k − 2k . (cid:0) (cid:1)
k k+1
(cid:0) (cid:1) (cid:0) (cid:1)
Remark 1.2. Let p > 3 be a prime. By [JV, Lemma 2.7], g ≡ (p)9kg
k 3 p−1−k
(mod p) for all k = 0,... ,p−1. So (1.17) implies that
p−1 p−1 p−1
g p g p g
k p−1−k k−1
≡ = ≡ 2q (3) (mod p).
k9k 3 k 3 p−k p
X (cid:16) (cid:17)X (cid:16) (cid:17)X
k=1 k=1 k=1
6 ZHI-WEI SUN
We also introduce the polynomials
n 2 n
n 2k n k 2k
f (x) = xk = xk (n = 0,1,2,...)
n
(cid:18)k(cid:19) (cid:18)n(cid:19) (cid:18)k(cid:19)(cid:18)n−k(cid:19)(cid:18) k (cid:19)
X X
k=0 k=0
which play a central role in our proof of Theorem 1.1.
We are going to show Theorem 1.1 in the next section and investigate in
Section 3 connections among the polynomials A (x), f (x) and g (x). In Sec-
n n n
tion 4 we will prove Theorem 1.2. In Section 5 we shall raise some conjectures
for further research.
2. Proof of Theorem 1.1
Lemma 2.1. For any nonnegative integer n, the integer f (1) coincides with
n
the Franel number f .
n
Remark 2.1. This is a known result due to V. Strehl [St94].
Lemma 2.2. For any nonnegative integer k we have
2k 2
l k x+l x+k
(−1)l = .
(cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) l (cid:19) (cid:18) k (cid:19)
X
l=k
Proof. Observe that
2k
l k x+l
(−1)l
(cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) l (cid:19)
X
l=k
2k
l k −x−1
=
(cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) l (cid:19)
X
l=k
2k
−x−1 −x−1−k k
=
(cid:18) k (cid:19) (cid:18) l−k (cid:19)(cid:18)l−k(cid:19)
X
l=k
k 2 2
−x−1 −x−1−k k −x−1 x+k
= = =
(cid:18) k (cid:19) (cid:18) j (cid:19)(cid:18)k −j(cid:19) (cid:18) k (cid:19) (cid:18) k (cid:19)
X
j=0
with the help of the Chu-Vandemonde identity (cf. (3.1) of [G, p.22]). We are
done. (cid:3)
Lemma 2.3. For each positive integer m we have
n−1
2k 2n
P (k) = nm for all n = 1,2,3,... ,
m
(cid:18)k (cid:19) (cid:18)n (cid:19)
X
k=0
CONGRUENCES FOR FRANEL NUMBERS 7
where
P (x) := 2(2x+1)(x+1)m−1 −xm.
m
Proof. The desired result follows immediately by induction on n. (cid:3)
Remark 2.2. The author thanks Prof. Qing-Hu Hou at Nankai Univ. for his
comments on the author’s original version of Lemma 2.3.
Lemma 2.4. Let p > 3 be a prime.
(i) ([ST, (1.6)]) We have
p−1
2k p
≡ (mod p2).
(cid:18)k (cid:19) 3
X (cid:16) (cid:17)
k=0
(ii) ([S11c]) We have
p−1
(2k +1)A ≡ p (mod p4).
k
X
k=0
Lemma 2.5. Let m be a positive integer. For n = 0,1,... ,m we have
n
x −x m−n x−1 −x
= .
(cid:18)k(cid:19)(cid:18)m−k(cid:19) m (cid:18) n (cid:19)(cid:18)m−n(cid:19)
X
k=0
Remark 2.3. This is a known result due to E. S. Andersen, see, e.g., (3.14) of
[G, p.23].
Lemma 2.6 (Sun [S11b, Lemma 2.1]). Let p be an odd prime. For any k =
1,... ,p−1 we have
2k 2(p−k)
k ≡ (−1)⌊2k/p⌋−12p (mod p2).
(cid:18) k (cid:19)(cid:18) p−k (cid:19)
Recall that harmonic numbers are given by
1
H = (n = 0,1,2,...).
n
k
0<Xk6n
In general. for any positiveinteger m, harmonic numbers of order m aredefined
by
1
H(m) := (n = 0,1,2,...).
n km
0<Xk6n
8 ZHI-WEI SUN
Let p > 3 be a prime. In 1862 J. Wolstenholme [W] proved that
H ≡ 0 (mod p2) and H(2) ≡ 0 (mod p).
p−1 p−1
Note that
(p−1)/2
1 1 1 1
(2) (2)
H = + = H ≡ 0 (mod p).
(p−1)/2 2 (cid:18)k2 (p−k)2(cid:19) 2 p−1
X
k=1
In 1938 E. Lehmer [L] showed that
H ≡ −2q (2)+pq (2)2 (mod p2). (2.1)
(p−1)/2 p p
These fundamental congruences are quite useful. Recently the author [S12a]
proved some further congruences involving harmonic numbers.
Lemma 2.7. Let p > 3 be a prime. Then
f ≡ 1+3pq (2)+3p2q (2)2 (mod p3). (2.2)
p−1 p p
Proof. For any k = 1,... ,p−1, we obviously have
k
p−1 p
(−1)k = 1−
(cid:18) k (cid:19) (cid:18) j(cid:19)
jY=1
p2 2 p2
≡1−pH + = 1−pH + H2 −H(2) (mod p3).
k 2 ij k 2 k k
16Xi<j6k (cid:16) (cid:17)
Thus
p−1 p−1 3 p−1 p2 3
f −1 = ≡ (−1)k 1−pH + H2 −H(2)
p−1 (cid:18) k (cid:19) (cid:18) k 2 k k (cid:19)
X X (cid:16) (cid:17)
k=1 k=1
p−1 p−1 p−1
9 3
=−3p (−1)kH + p2 (−1)kH2 − p2 (−1)kH(2) (mod p3).
k 2 k 2 k
X X X
k=1 k=1 k=1
Clearly
p−1 p−1 k (−1)k p−1 p−1(−1)k p−1 1
(−1)kH = = k=j =
k j P j j
Xk=1 kX=1Xj=1 Xj=1 Xj=1
2|j
1 p
= H ≡ −q (2)+ q (2)2 (mod p2) (by (2.1))
(p−1)/2 p p
2 2
CONGRUENCES FOR FRANEL NUMBERS 9
and
p−1 p−1 p−1(−1)k (p−1)/2 1
(−1)kH(2) = k=j = ≡ 0 (mod p).
k P j2 (2i)2
Xk=1 Xj=1 Xi=1
Observe that
p−1 p−1 p−1 2
1
(−1)kH2 = (−1)p−kH2 = (−1)k−1 H −
k p−k (cid:18) p−1 p−j(cid:19)
X X X X
k=1 k=1 k=1 0<j<k
p−1 2
1
≡− (−1)k H −
k
(cid:18) k(cid:19)
X
k=1
p−1 p−1 p−1
(−1)k (−1)k
≡− (−1)kH2 +2 H − (mod p).
k k k k2
X X X
k=1 k=1 k=1
Clearly,
p−1 p−1 (p−1)/2
(−1)k 1+(−1)k 2
≡ = ≡ 0 (mod p),
k2 k2 (2j)2
kX=1 kX=1 Xj=1
and
p−1 (−1)k p−1 H p−1 H q (2)2 q (2)2
k k p p
H = − ≡ − − (mod p)
k
k k k 2 (cid:18) 2 (cid:19)
X X X
k=1 k=1 k=1
2|k 2∤k
by [S12a, Lemma 2.3]. Therefore
p−1 p−1
(−1)k
(−1)kH2 ≡ H ≡ q (2)2 (mod p).
k k k p
X X
k=1 k=1
Combining the above, we finally obtain
p 9
f −1 ≡−3p −q (2)+ q (2)2 + p2q (2)2 (mod p3)
p−1 p p p
2 2
(cid:16) (cid:17)
and hence (2.2) holds. (cid:3)
Proof of Theorem 1.1. (i) Let r be any p-adic integer. Observe that
p−1 p−1 l
l+r l+r l k 2k
(−1)l f (x) = (−1)l xk
l
(cid:18) l (cid:19) (cid:18) l (cid:19) (cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) k (cid:19)
X X X
l=0 l=0 k=0
p−1 min{2k,p−1}
2k l k l+r
= xk (−1)l .
(cid:18)k (cid:19) (cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) l (cid:19)
X X
k=0 l=k
10 ZHI-WEI SUN
If (p−1)/2 < k 6 p−1 and k 6 l 6 2k, then
2k (2k)! l l!
= ≡ 0 (mod p) and = ≡ 0 (mod p).
(cid:18)k (cid:19) (k!)2 (cid:18)k(cid:19) k!(l−k)!
Therefore
p−1 p−1 2k
l+r 2k l k l+r
(−1)l f (x) ≡ xk (−1)l (mod p2).
l
(cid:18) l (cid:19) (cid:18)k (cid:19) (cid:18)k(cid:19)(cid:18)l−k(cid:19)(cid:18) l (cid:19)
X X X
l=0 k=0 l=k
Applying Lemma 2.2 we obtain
p−1 p−1 2
l+r 2k k +r
(−1)l f (x) ≡ xk (mod p2). (2.3)
l
(cid:18) l (cid:19) (cid:18)k (cid:19) (cid:18) k (cid:19)
X X
l=0 k=0
In the case x = 1 this yields (1.4).
Taking r = 0 in (1.4) and applying Lemma 2.4(i), we immediately get (1.5).
By (2.3) with r = 0,1,
p−1
(3(k +1)−1)(−1)kf (x)
k
X
k=0
p−1 p−1
2k 2k
≡ xk 3(k+1)2 −1 = P (k) xk (mod p2)
2
(cid:18)k (cid:19) (cid:18)k (cid:19)
kX=0 (cid:0) (cid:1) kX=0
where P (x) = 2(2x+1)(x+1) −x2 = 3x2 +6x+ 2. Thus, with the help of
2
Lemma 2.3, we have
p−1
(3k +2)(−1)kf ≡ 0 (mod p2) (2.4)
k
X
k=0
and hence (1.6) holds in view of (1.5).
Taking r = 2 in (2.3) we get
p−1 p−1
2k
2 (k2 +3k +2)(−1)kf (x) ≡ xk((k+1)(k +2))2 (mod p2).
k
(cid:18) k (cid:19)
X X
k=0 k=0
In view of (2.4), this yields
p−1 p−1
2k
2 (−1)kk2f ≡ (k2 +3k +2)2 (mod p2).
k
(cid:18) k (cid:19)
X X
k=0 k=0