- Email: [email protected]
Proof of two conjectural supercongruences involving Catalan–Larcombe–French numbers
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.1 (1-9) Journal of Number Theory ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Proof of two conjectural supercongruences involving Catalan–Larcombe–French numbers ✩ Guo-Shuai Mao Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 13 October 2016 Received in revised form 27 January 2017 Accepted 6 March 2017 Available online xxxx Communicated by D. Wan
a b s t r a c t In this paper we mainly prove the following two conjectures of Z.-W. Sun [10]: For any odd prime p, we have p−1 k=0
p−1
MSC: primary 11B65, 11B68 secondary 05A10, 11A07 Keywords: Central binomial coefficients Supercongruence Harmonic numbers
Pk ≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 ), 8k
k=0
where Pn =
Pk ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ), 16k
n k=0
2k 2 2(n−k) 2 k n−k n k
is the n-th Catalan–Larcombe–
French number, En are the Euler numbers which are defined by
E0 = 1, En = −
n/2
k=1
n En−2k (n ≥ 1). 2k
© 2017 Published by Elsevier Inc.
✩
This research was supported by the Natural Science Foundation (Grant No. 11571162) of China. E-mail address: [email protected].
http://dx.doi.org/10.1016/j.jnt.2017.03.017 0022-314X/© 2017 Published by Elsevier Inc.
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.2 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
2
1. Introduction The Catalan–Larcombe–French numbers P0 , P1 , P2 , . . . (cf. [2]) are given by
Pn =
n
2k2 2(n−k)2 k
n/2
nn−k
= 2n
k
k=0
k=0
n 2k
2 2k 4n−2k , k
which arose from the theory of elliptic integrals (see [4]). It is known that (n + 1)Pn+1 = (24n(n + 1) + 8)Pn − 128n2 Pn−1 for all n ∈ Z+ . The sequence (Pn )n≥0 is also related to the theory of modular forms. See D. Zagier [13]. Let Sn = Pn /2n . Zagier noted that n/2
Sn =
k=0
2k k
2
n n−2k . 4 2k
(1.1)
Recently Z.-W. Sun obtained the following identity 2 n 2k k Sn = (−4)n−k k n−k
(1.2)
k=0
n 1 2k 2(n − k) k = (−4)k . (−2)n k n−k n−k k=0
The Bernoulli numbers {Bn } are defined by
B0 = 1,
n−1 k=0
n Bk = 0 (n ≥ 2). k
In [11] Z.-W. Sun showed that
(p−3)/2
k=0
2k2 k
(2k +
1)16k
≡ −2qp (2) − pqp (2)2 +
where qp (2) denotes the Fermat quotient (2p−1 − 1)/p. The Harmonic numbers Hn are defined by
Hn =
n 1 , k
k=1
and H0 = 0.
5 2 p Bp−3 (mod p3 ), 12
(1.3)
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.3 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
3
Z.-W. Sun [12] also did some research on Harmonic numbers Hn and H2n , such as (p−1)/2
2k k
k=0
2
Hk ≡ 2(−1)(p−1)/2 H(p−1)/2 (mod p2 ) 16k
and (p−1)/2
k=0
2k k
2
3 H2k ≡ (−1)(p−1)/2 H(p−1)/2 + pEp−3 (mod p2 ). 16k 2
In [8] Z.-W. Sun proved that 2k2
(p−1)/2
k
16k
k=0
≡ (−1)(p−1)/2 + p2 Ep−3 (mod p3 ).
(1.4)
Motivated by the above work, we mainly obtain the following results. Theorem 1.1. Let p > 3 be a prime. Then
(p−3)/2
k=0
2k2
Hk 7 ≡ 4qp (2)2 +2(−1)(p−1)/2 (E2p−4 −2Ep−3 )+ pBp−3 (mod p2 ), (1.5) k (2k + 1)16 12 k
and
(p−3)/2
k=0
2k2
H2k ≡ −2(−1)(p−1)/2 Ep−3 (mod p). (2k + 1)16k k
(1.6)
Remark 1.1. Our approach to Theorem 1.1 is somewhat unique in the sense that it depends heavily on some special combinatorial identities. The reason why we research these two congruences is to prove the following theorem. The congruence (1.6) was conjectured by Z.-W. Sun in 2012. He also listed the equivalent one as a conjecture in [11]:
(p−1)/2
k=1
2k2 k
k16k
H2k ≡ 4(−1)(p−1)/2 Ep−3 (mod p).
Theorem 1.2. For any odd prime p, we have p−1 Pk k=0
and
8k
≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 )
(1.7)
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.4 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
4
p−1 Pk ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ). k 16 k=0
(1.8)
We are going to prove Theorem 1.1 in Section 2, and prove Theorem 1.2 in Section 3. 2. Proof of Theorem 1.1 Lemma 2.1. [12, Lemma 4.2] Let p = 2n + 1 be an odd prime, and let k ∈ {0, . . . , n}. Then n 2k k
k
/(−4)k
≡1−p
k j=1
1 (mod p2 ) 2j − 1
(2.1)
and 2k2 n n+k k (−1) ≡ k k (mod p2 ). 16 k k
(2.2)
For any positive integer n, we have the following identities n k=0
nn+k k
(−1)k 2 (−1)k Hk = 2k + 1 2n + 1 k n
k
(2.3)
k=1
and n 2k2 4n k = , 2n (2k + 1)(−4)k 16k (2n + 1) n k=0 k=0 n
n2k k
k
(2.4)
which can be easily proved by induction. Remark 2.1. The above two identities are found by the package Sigma via software Mathematica. The reader may consult [6] to see how to produce such identities. Lemma 2.2. [7, Theorem 3.2] For any prime p > 3, we have 1≤k
and
1 ≡ − 3qp (2) + p k
3 qp (2)2 + (−1)(p−1)/2 (E2p−4 − 2Ep−3 ) 2
7 − p2 qp (2)3 + Bp−3 (mod p3 ) 12
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.5 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
≡ qp (2) − p
p/4
5
1 qp (2)2 + (−1)(p−1)/2 (E2p−4 − 2Ep−3 ) 2
1 + p2 qp (2)3 (mod p3 ). 3 Lemma 2.3. Let p > 3 be a prime. Then
p−1 ≡ (−1)(p−1)/2 4p−1 (mod p3 ). (p − 1)/2
Remark 2.2. Lemma 2.3 is a famous congruence of Morley [5]. It has been utilized by Guo [1] to prove a supercongruence conjectured by Z.-H. Sun [9]. Proof of Theorem 1.1. For any prime p > 3, taking n = (p − 1)/2 in (2.3) and in view of Lemma 2.2 and Lemma 2.3 we can get that
(p−3)/2
(p−1)/2(p−1)/2+k k
k
2k + 1
k=0
=−
≡−
≡
(−1)k Hk
H(p−1)/2
p−1 (p−1)/2
(−1)(p−1)/2
p H(p−1)/2
p−1 (p−1)/2
(−1)(p−1)/2
p
1≤k
(2qp (2) − pqp (2)2 +
+
2 p
−qp (2) + p
≡ 4qp (2)2 + 2(−1)
(p−1)/2 2 (−1)k p k k=1 ⎛ 1 2 + ⎝ − p k
+
p−1 2
2 2 3 3 p qp (2)
+
7 2 12 p Bp−3 )(1
1≤k
⎞ 1⎠ k
+ 2pqp (2) + p2 qp (2)2 )
p
p−1 1 1 qp (2)2 + (−1) 2 (E2p−4 − 2Ep−3 ) − p2 qp (2)3 2 3
(E2p−4 − 2Ep−3 ) +
7 pBp−3 (mod p2 ). 12
Then with (2.2) we can deduce that
(p−3)/2
k=0
(p−1)/2(p−1)/2+k (p−3)/2 2k2 (−1)k Hk Hk k k k ≡ (mod p2 ), 2k + 1 (2k + 1)16k k=0
and so we get the desired (1.5). By using Lemma 2.3, (1.4) and noting that 2p−1 = 1 +pqp (2), 4p−1 = (2p−1 −1 +1)2 = (1 + pqp (2))2 = 1 + 2pqp (2) + p2 qp (2)2 , we can deduce that
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.6 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
6
(p−1)/22k
(p−3)/2
k
k
(2k + 1)(−4)k
k=0
(p−1)/2 2k2 2p−1 k + p−1 p (p−1)/2 k=0 16k
=−
p−1 (p−1)/2 p(−4)(p−1)/2
≡−
p−1 (−1)(p−1)/2 2p−1 + (−1) 2 + p2 Ep−3 p−1 p p2
≡−
2p−1 1 + p−1 + (−1)(p−1)/2 pEp−3 p p2
≡ − 2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3 (mod p2 ). So we have
(p−3)/2
(p−1)/22k k
k
(2k + 1)(−4)k
k=0
≡ −2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3 (mod p2 ).
(2.5)
It is known that n−1 k=0
2k2
2k2 k n−1 2k2 n−1 H2k Hk 1 1 k k . = + k k k (2k + 1)16 2 (2k + 1)16 (2k + 1)16 j=1 2j − 1 k
k=0
k=0
So in view of (2.1) we have
p
k=0
≡
2k2
n−1
(2k +
1)16k
j=1
2k2
n−1 k=0
k
k
k
(2k + 1)16k
−
1 2j − 1
n−1 k=0
n2k k
k
(2k + 1)(−4)k
(mod p2 ).
Hence p
n−1 k=0
2k2
H2k k (2k + 1)16k
2k2 n2k n−1 2k2 n−1 n−1 Hk p k k k k ≡ + − (mod p2 ). 2 (2k + 1)16k (2k + 1)16k (2k + 1)(−4)k k=0
k=0
k=0
By (1.3), (1.5), (2.5) and in view of [7, (3.1)] E2p−4 ≡ Ep−3 (mod p), we can get that
p
n−1 k=0
2k2
H2k k (2k + 1)16k
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.7 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
7
p 4qp (2)2 − 2(−1)(p−1)/2 Ep−3 + −2qp (2) − pqp (2)2 2 − −2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3
≡
≡ −2(−1)(p−1)/2 pEp−3 (mod p2 ). Hence n−1 k=0
2k2
H2k k ≡ −2(−1)(p−1)/2 Ep−3 (mod p). (2k + 1)16k
Therefore we complete the proof of (1.6). 2 3. Proof of Theorem 1.2 p−l 2 Proof of (1.7). For any prime p > 3, note that p−1 = Π2j l=1 l ≡ 1 − pH2j (mod p ) 2j for all 0 ≤ j ≤ (p − 3)/2. Then with (1.1) and Lemma 2.3 we have k 2j 2 p−1 p−1 k/2 2j Pk j = j 8k 16 j=0 k=0
k=0
(p−1)/2
=
j=0
2j 2 j
16j
2 p−1 (p−1)/2 4p−1
(p−1)/2 2j 2 p−1 k p j = 16j 2j + 1 2j j=0
k=2j
≡
+p
≡4
+p
j
(2j + 1)16j
j=0
2j 2
(p−3)/2 p−1
2j 2
(p−3)/2
j=0
j
(2j + 1)16j
(1 − pH2j )
(p−3)/2
−p
2
j=0
2j 2 j
H2j
(mod p3 ).
(2j + 1)16j
So by (1.3) we have p−1 Pk k=0
8k
≡ 1 + 2pqp (2) + p2 qp (2)2 + p(−2qp (2) − pqp (2)) 2j 2
(p−3)/2
− p2
j=0
j
H2j
(2j + 1)16j
(p−3)/2
≡ 1 − p2
j=0
2j 2 j
H2j
(2j + 1)16j
(mod p3 ).
Then with (1.6) we can easily get that p−1 Pk k=0
8k
≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 ).
(1.7) in the case of p = 3 can be verified directly. So we complete the proof of (1.7). 2
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.8 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
8
Proof of (1.8). First we know that in [8] Z.-W. Sun proved that p−1
2k2 k
k=0
16k
≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ).
By taking m = n and x = 1/2 in [3, (1.1)] we can get that n n+k 1 = 2n . k 2k
k=0
With those two results and (1.2) we can obtain 2 p−1 p−1 p−1 n k Pn Sn 1 2k (−4)n−k = = n n n 16 8 8 k n − k n=0 n=0 n=0 k=0
n
p−1
1 = n (−2) n=0 =
p−1 k=0
2k2 k
2k k
k=0
p−1−k
(−4)k
n=0
p−1 k p−1 2k2 k 1 n−k k = n − k (−4)k (−4)k (−2)n
2
k=0
k n
(−2)n+k
=
p−1 k=0
n (p−1)/2 2k2 k k 1 k = + − 8k n=0 n 2 k=0
(p−1)/2
=
k=0
It is known that
2k2 k 1 k + 8k 2
2k2 k
≡
k
8k
n=0
k=(p+1)/2
k
k=(p+1)/2
k
n
(−2)n
2k2
p−1
2k2
p−1
n=k
k
p−1−k
8k
p−1−k
8k
n=0
p−1−k n=0
n k 1 − 2 n
n k 1 − . n 2
≡ 0 (mod p2 ) for each k = (p + 1)/2, . . . , p − 1, and
p−1−k n=0
2k2
n p−1−k −k + n − 1 1 n k 1 = − 2 2 n n n=0
p−1−k
n 1 p−1−k+n 1 = 2p−1−k ≡ k (mod p). 2 2 n
k=0
So we have p−1 k=(p+1)/2
Hence
2k2 k
8k
p−1−k n=0
k n
1 − 2
n ≡
p−1 k=(p+1)/2
2k2 k
16k
(mod p3 ).
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.9 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
9
p−1 p−1 2k2 Pk k ≡ ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ). k 16k 16 k=0 k=0
Therefore we finish the proof of (1.8). 2 Acknowledgments The author would like to thank Prof. Z.-W. Sun and Prof. Victor J.-W. Guo for some helpful comments. References [1] V.J.W. Guo, Proof of a supercongruence conjectured by Z.-H. Sun, Integral Transforms Spec. Funct. 25 (2014) 1009–1015. [2] F. Jarvis, H.A. Verrill, Supercongruences for the Catalan–Larcombe–French numbers, Ramanujan J. 22 (2010) 171–186. [3] Tom H. Koornwinder, Michael J. Schlosser, On an identity by Chaundy and Bullard. I, Indag. Math. (N.S.) 19 (2) (2008) 239–261. [4] P. Larcombe, D. French, On the ‘other’ Catalan numbers: a historical formulation re-examined, Congr. Numer. 143 (2000) 33–64. [5] F. Morley, Note on the congruence 24n ≡ (−1)n (2n)!/(n!)2 , where 2n + 1 is a prime, Ann. Math. 9 (1895) 168–170. [6] R. Osburn, C. Schneider, Gaussian hypergeometric series and supercongruences, Math. Comp. 78 (2009) 275–292. [7] Z.-H. Sun, Congruences involving Bernoulli and Euler numbers, J. Number Theory 128 (2) (2008) 280–312. [8] Z.W. Sun, Super congruences and Euler numbers, Sci. China Math. 54 (12) (2011) 2509–2535. [9] Z.-H. Sun, Some conjectures on congruences, preprint, arXiv:1103.5384v5, 2013. [10] Z.W. Sun, Number theory: arithmetic in Shangri-La, in: S. Kanemitsu, H. Li, J. Liu (Eds.), Proc. 6th China–Japan Seminar, Shanghai, August 15–17, 2001, World Sci., Singapore, 2013, pp. 244–258. [11] Z.W. Sun, p-Adic congruences motivated by series, J. Number Theory 134 (1) (2014) 181–196. [12] Z.W. Sun, A new series for π 3 and related congruences, Internat. J. Math. 26 (8) (2015) 1550055 (23 pages). [13] D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: From Neolithic Scots to John McKay, in: CRM Proc. Lecture Notes, vol. 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349–366.
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.1 (1-9) Journal of Number Theory ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Proof of two conjectural supercongruences involving Catalan–Larcombe–French numbers ✩ Guo-Shuai Mao Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 13 October 2016 Received in revised form 27 January 2017 Accepted 6 March 2017 Available online xxxx Communicated by D. Wan
a b s t r a c t In this paper we mainly prove the following two conjectures of Z.-W. Sun [10]: For any odd prime p, we have p−1 k=0
p−1
MSC: primary 11B65, 11B68 secondary 05A10, 11A07 Keywords: Central binomial coefficients Supercongruence Harmonic numbers
Pk ≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 ), 8k
k=0
where Pn =
Pk ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ), 16k
n k=0
2k 2 2(n−k) 2 k n−k n k
is the n-th Catalan–Larcombe–
French number, En are the Euler numbers which are defined by
E0 = 1, En = −
n/2
k=1
n En−2k (n ≥ 1). 2k
© 2017 Published by Elsevier Inc.
✩
This research was supported by the Natural Science Foundation (Grant No. 11571162) of China. E-mail address: [email protected].
http://dx.doi.org/10.1016/j.jnt.2017.03.017 0022-314X/© 2017 Published by Elsevier Inc.
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.2 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
2
1. Introduction The Catalan–Larcombe–French numbers P0 , P1 , P2 , . . . (cf. [2]) are given by
Pn =
n
2k2 2(n−k)2 k
n/2
nn−k
= 2n
k
k=0
k=0
n 2k
2 2k 4n−2k , k
which arose from the theory of elliptic integrals (see [4]). It is known that (n + 1)Pn+1 = (24n(n + 1) + 8)Pn − 128n2 Pn−1 for all n ∈ Z+ . The sequence (Pn )n≥0 is also related to the theory of modular forms. See D. Zagier [13]. Let Sn = Pn /2n . Zagier noted that n/2
Sn =
k=0
2k k
2
n n−2k . 4 2k
(1.1)
Recently Z.-W. Sun obtained the following identity 2 n 2k k Sn = (−4)n−k k n−k
(1.2)
k=0
n 1 2k 2(n − k) k = (−4)k . (−2)n k n−k n−k k=0
The Bernoulli numbers {Bn } are defined by
B0 = 1,
n−1 k=0
n Bk = 0 (n ≥ 2). k
In [11] Z.-W. Sun showed that
(p−3)/2
k=0
2k2 k
(2k +
1)16k
≡ −2qp (2) − pqp (2)2 +
where qp (2) denotes the Fermat quotient (2p−1 − 1)/p. The Harmonic numbers Hn are defined by
Hn =
n 1 , k
k=1
and H0 = 0.
5 2 p Bp−3 (mod p3 ), 12
(1.3)
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.3 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
3
Z.-W. Sun [12] also did some research on Harmonic numbers Hn and H2n , such as (p−1)/2
2k k
k=0
2
Hk ≡ 2(−1)(p−1)/2 H(p−1)/2 (mod p2 ) 16k
and (p−1)/2
k=0
2k k
2
3 H2k ≡ (−1)(p−1)/2 H(p−1)/2 + pEp−3 (mod p2 ). 16k 2
In [8] Z.-W. Sun proved that 2k2
(p−1)/2
k
16k
k=0
≡ (−1)(p−1)/2 + p2 Ep−3 (mod p3 ).
(1.4)
Motivated by the above work, we mainly obtain the following results. Theorem 1.1. Let p > 3 be a prime. Then
(p−3)/2
k=0
2k2
Hk 7 ≡ 4qp (2)2 +2(−1)(p−1)/2 (E2p−4 −2Ep−3 )+ pBp−3 (mod p2 ), (1.5) k (2k + 1)16 12 k
and
(p−3)/2
k=0
2k2
H2k ≡ −2(−1)(p−1)/2 Ep−3 (mod p). (2k + 1)16k k
(1.6)
Remark 1.1. Our approach to Theorem 1.1 is somewhat unique in the sense that it depends heavily on some special combinatorial identities. The reason why we research these two congruences is to prove the following theorem. The congruence (1.6) was conjectured by Z.-W. Sun in 2012. He also listed the equivalent one as a conjecture in [11]:
(p−1)/2
k=1
2k2 k
k16k
H2k ≡ 4(−1)(p−1)/2 Ep−3 (mod p).
Theorem 1.2. For any odd prime p, we have p−1 Pk k=0
and
8k
≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 )
(1.7)
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.4 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
4
p−1 Pk ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ). k 16 k=0
(1.8)
We are going to prove Theorem 1.1 in Section 2, and prove Theorem 1.2 in Section 3. 2. Proof of Theorem 1.1 Lemma 2.1. [12, Lemma 4.2] Let p = 2n + 1 be an odd prime, and let k ∈ {0, . . . , n}. Then n 2k k
k
/(−4)k
≡1−p
k j=1
1 (mod p2 ) 2j − 1
(2.1)
and 2k2 n n+k k (−1) ≡ k k (mod p2 ). 16 k k
(2.2)
For any positive integer n, we have the following identities n k=0
nn+k k
(−1)k 2 (−1)k Hk = 2k + 1 2n + 1 k n
k
(2.3)
k=1
and n 2k2 4n k = , 2n (2k + 1)(−4)k 16k (2n + 1) n k=0 k=0 n
n2k k
k
(2.4)
which can be easily proved by induction. Remark 2.1. The above two identities are found by the package Sigma via software Mathematica. The reader may consult [6] to see how to produce such identities. Lemma 2.2. [7, Theorem 3.2] For any prime p > 3, we have 1≤k
and
1 ≡ − 3qp (2) + p k
3 qp (2)2 + (−1)(p−1)/2 (E2p−4 − 2Ep−3 ) 2
7 − p2 qp (2)3 + Bp−3 (mod p3 ) 12
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.5 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
≡ qp (2) − p
p/4
5
1 qp (2)2 + (−1)(p−1)/2 (E2p−4 − 2Ep−3 ) 2
1 + p2 qp (2)3 (mod p3 ). 3 Lemma 2.3. Let p > 3 be a prime. Then
p−1 ≡ (−1)(p−1)/2 4p−1 (mod p3 ). (p − 1)/2
Remark 2.2. Lemma 2.3 is a famous congruence of Morley [5]. It has been utilized by Guo [1] to prove a supercongruence conjectured by Z.-H. Sun [9]. Proof of Theorem 1.1. For any prime p > 3, taking n = (p − 1)/2 in (2.3) and in view of Lemma 2.2 and Lemma 2.3 we can get that
(p−3)/2
(p−1)/2(p−1)/2+k k
k
2k + 1
k=0
=−
≡−
≡
(−1)k Hk
H(p−1)/2
p−1 (p−1)/2
(−1)(p−1)/2
p H(p−1)/2
p−1 (p−1)/2
(−1)(p−1)/2
p
1≤k
(2qp (2) − pqp (2)2 +
+
2 p
−qp (2) + p
≡ 4qp (2)2 + 2(−1)
(p−1)/2 2 (−1)k p k k=1 ⎛ 1 2 + ⎝ − p k
+
p−1 2
2 2 3 3 p qp (2)
+
7 2 12 p Bp−3 )(1
1≤k
⎞ 1⎠ k
+ 2pqp (2) + p2 qp (2)2 )
p
p−1 1 1 qp (2)2 + (−1) 2 (E2p−4 − 2Ep−3 ) − p2 qp (2)3 2 3
(E2p−4 − 2Ep−3 ) +
7 pBp−3 (mod p2 ). 12
Then with (2.2) we can deduce that
(p−3)/2
k=0
(p−1)/2(p−1)/2+k (p−3)/2 2k2 (−1)k Hk Hk k k k ≡ (mod p2 ), 2k + 1 (2k + 1)16k k=0
and so we get the desired (1.5). By using Lemma 2.3, (1.4) and noting that 2p−1 = 1 +pqp (2), 4p−1 = (2p−1 −1 +1)2 = (1 + pqp (2))2 = 1 + 2pqp (2) + p2 qp (2)2 , we can deduce that
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.6 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
6
(p−1)/22k
(p−3)/2
k
k
(2k + 1)(−4)k
k=0
(p−1)/2 2k2 2p−1 k + p−1 p (p−1)/2 k=0 16k
=−
p−1 (p−1)/2 p(−4)(p−1)/2
≡−
p−1 (−1)(p−1)/2 2p−1 + (−1) 2 + p2 Ep−3 p−1 p p2
≡−
2p−1 1 + p−1 + (−1)(p−1)/2 pEp−3 p p2
≡ − 2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3 (mod p2 ). So we have
(p−3)/2
(p−1)/22k k
k
(2k + 1)(−4)k
k=0
≡ −2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3 (mod p2 ).
(2.5)
It is known that n−1 k=0
2k2
2k2 k n−1 2k2 n−1 H2k Hk 1 1 k k . = + k k k (2k + 1)16 2 (2k + 1)16 (2k + 1)16 j=1 2j − 1 k
k=0
k=0
So in view of (2.1) we have
p
k=0
≡
2k2
n−1
(2k +
1)16k
j=1
2k2
n−1 k=0
k
k
k
(2k + 1)16k
−
1 2j − 1
n−1 k=0
n2k k
k
(2k + 1)(−4)k
(mod p2 ).
Hence p
n−1 k=0
2k2
H2k k (2k + 1)16k
2k2 n2k n−1 2k2 n−1 n−1 Hk p k k k k ≡ + − (mod p2 ). 2 (2k + 1)16k (2k + 1)16k (2k + 1)(−4)k k=0
k=0
k=0
By (1.3), (1.5), (2.5) and in view of [7, (3.1)] E2p−4 ≡ Ep−3 (mod p), we can get that
p
n−1 k=0
2k2
H2k k (2k + 1)16k
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.7 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
7
p 4qp (2)2 − 2(−1)(p−1)/2 Ep−3 + −2qp (2) − pqp (2)2 2 − −2qp (2) + pqp (2)2 + (−1)(p−1)/2 pEp−3
≡
≡ −2(−1)(p−1)/2 pEp−3 (mod p2 ). Hence n−1 k=0
2k2
H2k k ≡ −2(−1)(p−1)/2 Ep−3 (mod p). (2k + 1)16k
Therefore we complete the proof of (1.6). 2 3. Proof of Theorem 1.2 p−l 2 Proof of (1.7). For any prime p > 3, note that p−1 = Π2j l=1 l ≡ 1 − pH2j (mod p ) 2j for all 0 ≤ j ≤ (p − 3)/2. Then with (1.1) and Lemma 2.3 we have k 2j 2 p−1 p−1 k/2 2j Pk j = j 8k 16 j=0 k=0
k=0
(p−1)/2
=
j=0
2j 2 j
16j
2 p−1 (p−1)/2 4p−1
(p−1)/2 2j 2 p−1 k p j = 16j 2j + 1 2j j=0
k=2j
≡
+p
≡4
+p
j
(2j + 1)16j
j=0
2j 2
(p−3)/2 p−1
2j 2
(p−3)/2
j=0
j
(2j + 1)16j
(1 − pH2j )
(p−3)/2
−p
2
j=0
2j 2 j
H2j
(mod p3 ).
(2j + 1)16j
So by (1.3) we have p−1 Pk k=0
8k
≡ 1 + 2pqp (2) + p2 qp (2)2 + p(−2qp (2) − pqp (2)) 2j 2
(p−3)/2
− p2
j=0
j
H2j
(2j + 1)16j
(p−3)/2
≡ 1 − p2
j=0
2j 2 j
H2j
(2j + 1)16j
(mod p3 ).
Then with (1.6) we can easily get that p−1 Pk k=0
8k
≡ 1 + 2(−1)(p−1)/2 p2 Ep−3 (mod p3 ).
(1.7) in the case of p = 3 can be verified directly. So we complete the proof of (1.7). 2
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.8 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
8
Proof of (1.8). First we know that in [8] Z.-W. Sun proved that p−1
2k2 k
k=0
16k
≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ).
By taking m = n and x = 1/2 in [3, (1.1)] we can get that n n+k 1 = 2n . k 2k
k=0
With those two results and (1.2) we can obtain 2 p−1 p−1 p−1 n k Pn Sn 1 2k (−4)n−k = = n n n 16 8 8 k n − k n=0 n=0 n=0 k=0
n
p−1
1 = n (−2) n=0 =
p−1 k=0
2k2 k
2k k
k=0
p−1−k
(−4)k
n=0
p−1 k p−1 2k2 k 1 n−k k = n − k (−4)k (−4)k (−2)n
2
k=0
k n
(−2)n+k
=
p−1 k=0
n (p−1)/2 2k2 k k 1 k = + − 8k n=0 n 2 k=0
(p−1)/2
=
k=0
It is known that
2k2 k 1 k + 8k 2
2k2 k
≡
k
8k
n=0
k=(p+1)/2
k
k=(p+1)/2
k
n
(−2)n
2k2
p−1
2k2
p−1
n=k
k
p−1−k
8k
p−1−k
8k
n=0
p−1−k n=0
n k 1 − 2 n
n k 1 − . n 2
≡ 0 (mod p2 ) for each k = (p + 1)/2, . . . , p − 1, and
p−1−k n=0
2k2
n p−1−k −k + n − 1 1 n k 1 = − 2 2 n n n=0
p−1−k
n 1 p−1−k+n 1 = 2p−1−k ≡ k (mod p). 2 2 n
k=0
So we have p−1 k=(p+1)/2
Hence
2k2 k
8k
p−1−k n=0
k n
1 − 2
n ≡
p−1 k=(p+1)/2
2k2 k
16k
(mod p3 ).
JID:YJNTH
AID:5738 /FLA
[m1L; v1.215; Prn:12/05/2017; 16:29] P.9 (1-9)
G.-S. Mao / Journal of Number Theory ••• (••••) •••–•••
9
p−1 p−1 2k2 Pk k ≡ ≡ (−1)(p−1)/2 − p2 Ep−3 (mod p3 ). k 16k 16 k=0 k=0
Therefore we finish the proof of (1.8). 2 Acknowledgments The author would like to thank Prof. Z.-W. Sun and Prof. Victor J.-W. Guo for some helpful comments. References [1] V.J.W. Guo, Proof of a supercongruence conjectured by Z.-H. Sun, Integral Transforms Spec. Funct. 25 (2014) 1009–1015. [2] F. Jarvis, H.A. Verrill, Supercongruences for the Catalan–Larcombe–French numbers, Ramanujan J. 22 (2010) 171–186. [3] Tom H. Koornwinder, Michael J. Schlosser, On an identity by Chaundy and Bullard. I, Indag. Math. (N.S.) 19 (2) (2008) 239–261. [4] P. Larcombe, D. French, On the ‘other’ Catalan numbers: a historical formulation re-examined, Congr. Numer. 143 (2000) 33–64. [5] F. Morley, Note on the congruence 24n ≡ (−1)n (2n)!/(n!)2 , where 2n + 1 is a prime, Ann. Math. 9 (1895) 168–170. [6] R. Osburn, C. Schneider, Gaussian hypergeometric series and supercongruences, Math. Comp. 78 (2009) 275–292. [7] Z.-H. Sun, Congruences involving Bernoulli and Euler numbers, J. Number Theory 128 (2) (2008) 280–312. [8] Z.W. Sun, Super congruences and Euler numbers, Sci. China Math. 54 (12) (2011) 2509–2535. [9] Z.-H. Sun, Some conjectures on congruences, preprint, arXiv:1103.5384v5, 2013. [10] Z.W. Sun, Number theory: arithmetic in Shangri-La, in: S. Kanemitsu, H. Li, J. Liu (Eds.), Proc. 6th China–Japan Seminar, Shanghai, August 15–17, 2001, World Sci., Singapore, 2013, pp. 244–258. [11] Z.W. Sun, p-Adic congruences motivated by series, J. Number Theory 134 (1) (2014) 181–196. [12] Z.W. Sun, A new series for π 3 and related congruences, Internat. J. Math. 26 (8) (2015) 1550055 (23 pages). [13] D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: From Neolithic Scots to John McKay, in: CRM Proc. Lecture Notes, vol. 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349–366.