Acta Arith. 156(2012), no. 2, 123–141.
ON SUMS INVOLVING PRODUCTS OF
arXiv:1012.3141v6 [math.NT] 28 Apr 2014
THREE BINOMIAL COEFFICIENTS
Zhi-Wei Sun
Department of Mathematics, Nanjing University
Nanjing 210093, People’s Republic of China
[email protected]
http://math.nju.edu.cn/∼zwsun
Abstract. In this paper we mainly employ the Zeilberger algorithm to
study congruences for sums of terms involving products of three binomial
coefficients. Let p > 3 be a prime. We prove that
2k2 2k k
k+d
64k
p−1
X
k=0
(mod p2 )
≡0
for all d ∈ {0, . . . , p − 1} with d ≡ (p + 1)/2 (mod 2). If p ≡ 1 (mod 4)
and p = x2 + y 2 with x ≡ 1 (mod 4) and y ≡ 0 (mod 2), then we show
p−1
X
k=0
2k2 2k k
k+1
(−8)k
2
2
≡ 2p−2x (mod p ) and
2k 2k 2
k
k+1
(−8)k
p−1
X
k=0
≡ −2p (mod p2 )
by means of determining x mod p2 via
(p−1)/2
(−1)
(p−1)/4
x≡
X
k=0
k + 1 2k2
≡
8k
k
(p−1)/2
X
k=0
2k + 1 2k2
(−16)k k
(mod p2 ).
We also solve the remaining open cases of Rodriguez-Villegas’ conjectural
congruences on
p−1
X
k=0
2k2 3k
k
k
,
108k
p−1
X
k=0
2k2 4k
k
2k
,
256k
p−1
X
k=0
2k 3k 6k
k
k
3k
123k
modulo p2 .
2010 Mathematics Subject Classification. Primary 11B65; Secondary 05A10, 11A07,
11E25.
Keywords. Central binomial coefficients, super congruences, Zeilberger’s algorithm,
Schröder numbers, binary quadratic forms.
Supported by the National Natural Science Foundation (grant 11171140) of China
and the PAPD of Jiangsu Higher Education Institutions.
1
2
ZHI-WEI SUN
1. Introduction
Let p be an odd prime. It is known that (see, e.g., S. Ahlgren [A], L.
van Hamme [vH] and T. Ishikawa [I])
(p−1)/2
≡
X
k=0
2
3
−1/2
(−1)
k
k
4x − 2p (mod p2 ) if p = x2 + y 2 (2 ∤ x & 2 | y),
0 (mod p2 )
if p ≡ 3 (mod 4).
Clearly,
2k
−1/2
k
=
k
(−4)k
for all k ∈ N = {0, 1, 2, . . . },
and
2k
k
=
(2k)!
≡0
(k!)2
(mod p) for any k =
p+1
, . . . , p − 1.
2
k
Pp−1
2
After his determination of k=0 2k
k /m mod p (where m ∈ Z and m 6≡
0 (mod p)) in [Su1], the author [Su2, Su3] posed some conjectures on
Pp−1 2k3 k
/m mod p2 with m ∈ {1, −8, 16, −64, 256, −512, 4096}; for
k=0 k
example, the author [Su2] conjectured that
3
p−1 X
2k
≡
4x2 − 2p (mod p2 ) if ( p7 ) = 1 & p = x2 + 7y 2 (x, y ∈ Z),
if ( p7 ) = −1, i.e., p ≡ 3, 5, 6 (mod 7),
(1.1)
p
where (−) denotes the Legendre symbol. (It is known that if ( 7 ) = 1 then
p = x2 + 7y 2 for some x, y ∈ Z; see, e.g., [C, p. 31].) Quite recently Z.-H.
Sun [S2] made a certain progress on those conjectures; in particular, he
proved (1.1) in the case ( 7p ) = −1 and confirmed the author’s conjecture
3
Pp−1
on k=0 2k
/(−8)k mod p2 .
k
Let p = 2n+1 be an odd prime. It is easy to see that for any k = 0, . . . , n
we have
Qk
k 2 Y
2k
p2
n+k
j=1 (−(2j − 1) )
k
=
1−
≡
(mod p2 ).
2
k
2k
4k (2k)!
(2j
−
1)
(−16)
j=1
k=0
k
0 (mod p2 )
(1.2)
Based on this observation Z.-H. Sun [S2] studied the polynomial
fn (x) =
2
n X
2k
n+k
k=0
2k
k
xk
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
3
and found the key identity
fn (x(x + 1)) = Dn (x)2
(1.3)
in his approach to (1.1), where
n n X
2k k X n n + k k
n+k
x .
x =
Dn (x) :=
k
k
k
2k
k=0
k=0
Note that the numbers Dn = Dn (1) (n ∈ N) are the so-called central
Delannoy numbers and Pn (x) := Dn ((x−1)/2) is the Legendre polynomial
of degree n.
Recall that the Catalan numbers are the integers defined by
2n
2n
2n
1
(n ∈ N)
−
=
Cn =
n+1
n
n+1 n
while the Schröder numbers are given by
n n X
X
1
n n+k
n+k
Ck =
Sn :=
.
k
k
2k
k+1
k=0
k=0
We define the Schröder polynomial of degree n by
Sn (x) :=
n X
n+k
k=0
2k
Ck xk .
(1.4)
For basic information about Dn and Sn , the reader may consult [CHV],
[Sl], [St, pp. 178 and 185], and [Su4].
In combinatorics, Zeilberger’s algorithm developed in [Z] (see also Chapter 6 of [PWZ, pp. 101-119]) is an algorithm which finds a polynomial recurrence for a terminating hypergeometric sum. For example, if we use
Mathematica 7 and input Zb[Binomial[n,k]∧ 3,{k,0,n},n,2], then we
3
Pn
obtain the following second-order recurrence for S(n) = k=0 nk :
−8(n + 1)2 S(n) − (7n2 + 21n + 16)S(n + 1) + (n + 2)2 S(n + 2) = 0.
Via the Schröder polynomials and the Zeilberger algorithm, we obtain
the following result.
Theorem 1.1. Let p be an odd prime.
(i) We have
p−1 2k2 2k X
k
k+d
≡0
k
64
k=0
(mod p2 )
(1.5)
4
ZHI-WEI SUN
for all d ∈ {0, 1, . . . , p − 1} with d ≡ (p + 1)/2 (mod 2).
(ii) If p ≡ 3 (mod 4), then
p−1
X
k=0
2k 2 2k
k
k+1
64k
≡ (2p + 2 − 2
p−1
2
(p − 1)/2
)
(p + 1)/4
(mod p2 )
(1.6)
Now we state our second theorem the first part of which plays a key
role in our proof of the second part.
Theorem 1.2. Let p ≡ 1 (mod 4) be a prime and write p = x2 + y 2 with
x ≡ 1 (mod 4) and y ≡ 0 (mod 2).
(i) We can determine x mod p2 in the following way:
(p−1)/2
(−1)
X k + 1 2k 2 (p−1)/2
X 2k + 1 2k 2
x≡
≡
k
8k
(−16)k k
(p−1)/4
k=0
(mod p2 ).
k=0
(1.7)
Also,
(p−1)/2
X
k=0
2
p−1
X
k 2k
Ck
p
(p−1)/4
k
2x
−
≡
−2
≡
(−1)
8k
8k
x
2k
k
(mod p2 ),
k=0
(p−1)/2
X
X k 2k 2
Ck
k
S(p−1)/2 ≡
≡ −8
(−16)k
(−16)k
k=0
k=0
p
≡(−1)(p−1)/4 2 2x −
(mod p2 ),
x
(p−1)/2 2 2k2
X k
3p
(p−1)/4
k
(mod p2 ),
x−
≡ (−1)
8k
4x
(p−1)/2
(1.8)
2k
k
(1.9)
(1.10)
k=0
and
(p−1)/2
X
k=0
2
k 2 2k
p
k
≡ (−1)(p+3)/4
k
(−16)
16x
(mod p2 ).
(1.11)
(ii) We have
p−1
X
k=0
2k 2 2k
k
k+1
(−8)k
and
p−1
X
k=0
≡ 2p − 2x2
2k 2
k+1
(−8)k
2k
k
≡ −2p
(mod p2 )
(mod p2 ).
(1.12)
(1.13)
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
5
Remark 1.1. Let p be an odd prime. We conjecture that
2
p−1
X
k + 1 2k
k=0
(
≡
8k
(p−1)/2
+
k
2( p2 )x
X
k=0
3
2
2k + 1 2k
(−16)k k
(mod p ) if p = x2 + y 2 (4 | x − 1 & 2 | y),
0 (mod p2 )
if p ≡ 3 (mod 4).
Motivated by his study of Gaussian hypergeometric series and CalabiYau manifolds, in 2003 Rodriguez-Villegas [RV] raised some conjectures
on congruences. In particular, he conjectured that for any prime p > 3 we
have
p−1 2k2 3k
p−1 2k2 4k
X
X
2
k
k
k
2k
≡ b(p) (mod p ),
≡ c(p) (mod p2 ),
k
k
108
256
k=0
k=0
(1.14)
and
p−1 2k 3k 6k
p
X
k
k
3k
a(p) (mod p2 ),
(1.15)
≡
123k
3
k=0
where
∞
X
n
a(n)q = q
n=1
∞
X
n=1
∞
X
c(n)q n = q
(1 − q 4n )6 = η(4z)6 ,
n=1
b(n)q n = q
n=1
∞
Y
∞
Y
∞
Y
(1 − q 6n )3 (1 − q 2n )3 = η 3 (6z)η 3 (2z),
n=1
(1 − q n )2 (1 − q 2n )(1 − q 4n )(1 − q 8n )2 = η 2 (8z)η(4z)η(2z)η 2 (z),
n=1
and the Dedekind η-function is given by
η(z) = q
1/24
∞
Y
(1 − q n ) (Im(z) > 0 and q = e2πiz ).
n=1
In 1892 F. Klein and R. Fricke [KF] proved that (see also [SB])
2
4x − 2p if p ≡ 1 (mod 4) and p = x2 + y 2 (2 ∤ x),
a(p) =
0
if p ≡ 3 (mod 4).
By [SB] we also have
2
4x − 2p if p ≡ 1 (mod 3) and p = x2 + 3y 2 with x, y ∈ Z,
b(p) =
0
if p ≡ 2 (mod 3),
6
ZHI-WEI SUN
and
c(p) =
(
4x2 − 2p
2
2
if ( −2
p ) = 1 and p = x + 2y with x, y ∈ Z,
0
if ( −2
p ) = −1, i.e., p ≡ 5, 7 (mod 8).
Via an advanced approach involving the p-adic Gamma function and Gauss
and Jacobi sums (see K. Ono [O, Chapter 11] for an introduction to this
method), E. Mortenson [M] managed to provide a partial solution of (1.14)
and (1.15), with the following congruences still open:
p−1
X
k=0
2k 2 3k
k
k
108k
2k 2 4k
k
2k
256k
p−1
X
k=0
p−1
X
≡ b(p) = 0 (mod p2 )
2k
k
k=0
3k 6k
k
3k
123k
if p ≡ 5 (mod 6),
(1.16)
≡ c(p) (mod p2 ) if p ≡ 3 (mod 4),
(1.17)
≡ −a(p) (mod p2 ) if p ≡ 5 (mod 6).
(1.18)
Concerning (1.16)-(1.18), Mortenson’s approach [M] only allowed him to
show that for each of them the squares of both sides of the congruence are
congruent modulo p2 .
Our following theorem confirms (1.16)-(1.18) and hence completes the
proof of (1.14) and (1.15). So far, all conjectures of Rodriguez-Villegas
[RV] involving at most three products of binomial coefficients have been
proved!
Theorem 1.3. Let p > 3 be a prime.
(i) Given d ∈ {0, . . . , p − 1}, we have
p−1
X
2k
k+d
k=0
p−1
X
2k
k+d
k=0
p−1
X
k=0
2k
k+d
2k
k
108k
3k
k
2k
k
256k
4k
2k
3k
k
123k
6k
3k
≡ 0 (mod p2 )
if d ≡
≡ 0 (mod p )
if d ≡
≡ 0 (mod p2 )
if d ≡
2
1 + ( p3 )
(mod 2), (1.19)
2
1 + ( −2
p )
2
1 + ( −1
)
p
2
(mod 2), (1.20)
(mod 2). (1.21)
(ii) If p ≡ 3 (mod 8) and p = x2 + 2y 2 with x, y ∈ Z, then
p−1
X
k=0
2k 2 4k
k
2k
256k
≡ 4x2 − 2p
(mod p2 ).
(1.22)
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
7
(iii) If p ≡ 5 (mod 12) and p = x2 + y 2 with 2 ∤ x and 2 | y, then
p−1
X
2k
k
k=0
3k 6k
k
3k
123k
≡ 2p − 4x2
(mod p2 ).
(1.23)
In the cased = 1, Theorem
3k 1.3(i) yields the following new result. (Note
2k
3k
2k
that k k+1 = 2 k+1 k for any k ∈ N.)
Corollary 1.1. Let p > 3 be a prime. Then
p−1
X
2k 2 3k
k
k+1
108k
k=0
p−1 4k 2k 2k X 2k k k+1
256k
k=0
p−1
X
k=0
6k
3k
3k
2k
k
k+1
123k
≡ 0 (mod p2 )
if p ≡ 1 (mod 3),
(1.24)
≡ 0 (mod p2 )
if p ≡ 1, 3 (mod 8),
(1.25)
if p ≡ 1 (mod 4).
(1.26)
≡ 0 (mod p2 )
We will prove Theorems 1.1-1.3 in Sections 2-4 respectively.
2. Proof of Theorem 1.1
Lemma 2.1. For any positive integer n we have
n X
2k
n + k 2k
xk−1 (x + 1)k+1 = n(n + 1)Sn (x)2 .
k+1
k
2k
(2.1)
k=1
Proof. Observe that
2n
n n X
X
X
n+l
n+k
l
k
Cl x =
am (n)xm ,
Ck x
Sn (x) =
2l
2k
m=0
2
l=0
k=0
where
am (n) :=
m X
n+k
k=0
2k
n+m−k
Cm−k .
Ck
2m − 2k
Also, the coefficient of xm on the left-hand side of (2.1) coincides with
m+1
X
k+1
2k
2k
bm (n) :=
k+1 m+1−k
k
k=1
m X
k+2
n + k + 1 2k + 2 2k + 2
.
=
m−k
k
k+1
2k + 2
k=0
n+k
2k
8
ZHI-WEI SUN
Thus, for the validity of (2.1) it suffices to show that bm (n) = n(n+1)am (n)
for all m = 0, 1, . . . . Obviously, a0 (n) = 1 and b0 (n) = n(n + 1). Also,
a1 (n) = n(n + 1) and b1 (n) = n2 (n + 1)2 . By the Zeilberger algorithm via
Mathematica 7 we find that both um = am (n) and um = bm (n) satisfy
the following recursion:
(m + 2)(m + 3)(m + 4)um+2
=2(2mn2 + 5n2 + 2mn + 5n − m3 − 6m2 − 11m − 6)um+1
− (m + 1)(m − 2n)(m + 2n + 2)um .
So bm (n) = n(n + 1)am (n) for all m ∈ N. This proves (2.1). 2 2k k
Pp−1
2
Proof of Theorem 1.1. We first determine k=0 2k
k
k+1 /64 mod p
via Lemma 2.1, which actually led the author to the study of (1.5).
Recall the following combinatorial identity (cf. [Su2, (4.3)]):
n X
(−1)(n−1)/2 C(n−1)/2 /2n if 2 ∤ n,
Ck
n+k
=
(2.2)
(−2)k
2k
0
if
2
|
n.
k=0
Set n = (p − 1)/2. Applying (2.1) with x = −1/2 we get
2
n X
1
2k
2k
1
n+k
.
= n(n + 1)Sn −
k + 1 (−2)k−1 2k+1
k
2k
2
k=1
Thus, with the helps from (1.2) and (2.2), we have
p−1
X
k=0
2k 2 2k
k
k+1
64k
n X
1
2k
n + k 2k
≡
k + 1 (−4)k
k
2k
k=1
2
1
= − n(n + 1)Sn −
2
2
0 (mod p )
if p ≡ 1 (mod 4)
≡
2
C(n−1)/2
/22n+2 (mod p2 ) if p ≡ 3 (mod 4).
Therefore (1.5) with d = 1 holds if p ≡ 1 (mod 4). In the case p ≡ 3
(mod 4), clearly
2 2
(p−1)/2
2
C(n−1)/2
(p+1)/4 p−1
=
22n+2
4 × 2p−1
2
1
(p − 1)/2
≡
(1 − 2p)(1 + p qp (2)) (p + 1)/4
2
(p − 1)/2
(mod p2 )
≡(1 + 2p − p qp (2))
(p + 1)/4
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
9
where qp (2) = (2p−1 − 1)/p, and hence (1.6) holds.
For d = 0, 1, 2, . . . set
p−1 2k2 2k 2k 2 2k
X
X
k
k+d
k
k+d
=
.
ud =
64k
64k
k=0
d6k<p
By the Zeilberger algorithm we find the recursion
2
2p − 2
2p
(2p − 1)2 (d + 1)
2
2
.
(2d + 1) ud − (2d + 3) ud+2 =
p−1
p+d+1
64p−1 p
Note that
2p − 2
= pCp−1 ≡ 0
p−1
(mod p).
If 0 6 d < p − 2, then
2p
2p
2p − 1
=
≡ 0 (mod p)
p+d+1
p+d+1 p+d
and hence
(2d + 1)2 ud ≡ (2d + 3)2 ud+2
(mod p2 ).
For d ∈ {0, . . . , p − 3} with d ≡ (p + 1)/2 (mod 2), clearly p 6= 2d + 1 < 2p
and hence
ud+2 ≡ 0 (mod p2 ) =⇒ ud ≡ 0
(mod p2 ).
If d ∈ {p − 1, p − 2} and d ≡ (p + 1)/2 (mod 2), then d > (p + 1)/2 and
hence ud ≡ 0 (mod p2 ). So (1.5) holds for all d ∈ {0, . . . , p − 1} with
d ≡ (p + 1)/2 (mod 2).
So far we have completed the proof of Theorem 1.1. 3. Proof of Theorem 1.2
Lemma 3.1. For any n ∈ N we have
2
2 3 n n X
X
2(n − k)
2k
k
2k
n−k
.
(−16)
=
n−k
k
n−k
k
k=0
(3.1)
k=0
Proof. For n = 0, 1, both sides of (3.1) take the values 1 and 8 respectively. Let un denote the left-hand side of (3.1) or the right-hand side of
(3.1). Applying the Zeilberger algorithm via Mathematica 7, we obtain
the recursion
(n + 2)3 un+2 = 8(2n + 3)(2n2 + 6n + 5)un+1 − 256(n + 1)3 un (n ∈ N).
So, by induction (3.1) holds for all n = 0, 1, 2, . . . .
10
ZHI-WEI SUN
Lemma 3.2. Let p be an odd prime. Then
2
2 p−1
n X
2(n − k)
n + 1 X 2k
n−k
k
8n
n=0
k=0
2
2 p−1
n X
2n + 1 X 2k
2(n − k)
≡
n−k
k
(−16)n
n=0
k=0
−1
≡p
(mod p3 ).
p
Proof. In view of Lemma 3.1, we have
2
2 p−1
n X
2(n − k)
n + 1 X 2k
n−k
k
8n
n=0
k=0
3 p−1
n X
n + 1 X 2k
k
=
(−16)n−k
n
n
−
k
k
8
n=0
k=0
p−1 2k3 p−1−k
X
X
k (−16)j
k
(k
+
j
+
1)
=
j
8k
8j
j=0
k=0
(p−1)/2
≡
X
k=0
(p−1)/2
=
X
k=0
≡
p−1
X
k=0
2k 3 k
(k
8k
2k 3
k
8k
k k X
X
k−1
k
j−1
j
(−2)
(−2) − 2k
+ 1)
j
−
1
j
j=1
j=0
(k + 1)(−1)k − 2k(−1)k−1
3
3k + 1 2k
(mod p3 ).
(−8)k k
In [Su3] the author conjectured that
3
p−1
X
−1
3k + 1 2k
≡p
+ p3 Ep−3
k
k
(−8)
p
(mod p4 )
k=0
provided p > 3, where E0 , E1 , E2 , . . . are Euler numbers given by
n X
n
En−k = 0 (n = 1, 2, 3, . . . ).
E0 = 1 and
k
k=0
2|k
The last congruence is still open but [GZ] confirmed that
3
p−1
X
−1
3k + 1 2k
≡p
(mod p3 ).
(−8)k k
p
k=0
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
So we have
2
2 p−1
n X
n + 1 X 2k
−1
2(n − k)
≡p
n−k
k
8n
p
n=0
(mod p3 ).
k=0
Similarly,
2
2 p−1
n X
2(n − k)
2n + 1 X 2k
n−k
k
(−16)n
n=0
k=0
3 p−1
n X
2n + 1 X 2k
k
(−16)n−k
=
n
(−16)
n
−
k
k
n=0
k=0
(p−1)/2
k k 2k 3 X
X
X
k−1
k
k
≡
+ 2k
(2k + 1)
j−1
(−16)k
j
j=0
k=0
j=1
3
p−1
X
3k + 1 2k
−1
≡
≡p
(mod p3 ).
k
(−8)
k
p
k=0
This concludes the proof. Lemma 3.3. Let p be an odd prime. Then
(p−1)/2
2k 2
k
8k
(p−1)/2
2k
k
Ck
2
2
2
(mod p3 ),
+
≡2p
k
8
p
k=0
k=0
2
(p−1)/2
(p−1)/2 2k
2k
X k
X
−1
2
k
k Ck
8
(mod p3 ),
+
≡2p
(−16)k
(−16)k
p
k=0
k=0
(p−1)/2
2k 2
X
2
2
2
k
(2k + 4k + 1) k ≡p
(mod p3 ),
8
p
k=0
(p−1)/2
2k 2
X
−1
2
2
k
(mod p3 ).
≡p
(8k + 4k + 1)
(−16)k
p
X k
k=0
X
11
12
ZHI-WEI SUN
Proof. By induction, for every n = 0, 1, 2, . . . we have
2k2
2
1
(2n + 1)2 2n
k
,
=
k+1
8k
(n + 1)8n n
k=0
2k2
2
n
X
(2n + 1)2
2n
1
k
,
=
8k +
k
n
n
k + 1 (−16)
(n + 1)(−16)
k=0
2
n
2k 2
X
(2n + 1)2 2n
2
k
(2k + 4k + 1) k =
,
n
8
8n
k=0
2
n
2k 2
X
(2n + 1)2 2n
2
k
(8k + 4k + 1)
.
=
(−16)k
(−16)n
n
n X
2k +
k=0
Applying these identities with n = (p−1)/2 we immediately get the desired
congruences. Let p ≡ 1 (mod 4) be a prime and write p = x2 +y 2 with x ≡ 1 (mod 4)
and y ≡ 0 (mod 2). In 1828 Gauss showed the congruence (p−1)/2
≡ 2x
(p−1)/4
(mod p). In 1986, S. Chowla, B. Dwork and R. J. Evans [CDE] used Gauss
and Jacobi sums to prove that
p 2p−1 + 1 (p − 1)/2
2x −
≡
2
2x
(p − 1)/4
(mod p2 ),
(3.2)
which was first conjectured by F. Beukers. (See also [BEW, Chapter 9]
and [HW] for further related results.) In 2009, the author (see [Su2])
conjectured that
(p−1)/2
X
k=0
2k 2
k
8k
(p−1)/2
≡
X
k=0
2k 2
k
(−16)k
≡ (−1)
(p−1)/4
p 2x −
2x
(mod p2 ),
(3.3)
and this was confirmed by Z.-H. Sun [S1] via (3.2) and the Legendre polynomials.
Proof of Theorem 1.2(i). By (1.2),
(p−1)/2
S(p−1)/2 ≡
X
k=0
2k
k
Ck
(−16)k
(mod p2 ).
In view of this and Lemma 3.3 and (3.3), it suffices to show (1.7).
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
2k
k
As p |
13
for all k = (p + 1)/2, . . . , p − 1, we have
2
2 p−1
n X
n + 1 X 2k
2(n − k)
8n
n−k
k
n=0
k=0
p−1 2k2 p−1
X
X n + 1 2(n − k)2
k
=
n−k
8k
8n−k
k=0
n=k
p−1 2k2 p−1−k
X k + j + 1 2j 2
X
k
=
j
8k
8j
j=0
k=0
(p−1)/2
X
≡
k=0
(p−1)/2
=2
2k 2 (p−1)/2
X
k
8k
j=0
X
k=0
k=0
Similarly,
2
(k + 1) + (j + 1) − 1 2j
8j
j
2 p−1 2k2 2
2k 2 (p−1)/2
X (j + 1) 2j
X
j
k
k
−
(mod p2 ).
k
j
k
8
8
8
j=0
2
2 p−1
n X
2n + 1 X 2k
2(n − k)
n−k
k
(−16)n
n=0
k=0
2 p−1 2k2 2
(p−1)/2
2k 2 (p−1)/2
X (2j + 1) 2j
X
X
j
k
k
−
(mod p2 ).
≡2
k
j
k
(−16)
(−16)
(−16)
j=0
k=0
k=0
Combining these with Lemma 3.2 and (3.3), we immediately obtain (1.7). Lemma 3.4. Let p ≡ 1 (mod 4) be a prime. Write p = x2 + y 2 with
x ≡ 1 (mod 4) and y ≡ 0 (mod 2). Then
p (mod p2 ).
(3.4)
D(p−1)/2 ≡ (−1)(p−1)/4 2x −
2x
Proof. By (1.2),
(p−1)/2
D(p−1)/2 ≡
X
k=0
2k 2
k
(−16)k
(mod p2 ).
So (3.4) follows from (3.3). Remark 3.1. If p is a prime with p ≡ 3 (mod 4), then n = (p − 1)/2 is odd
and hence
2
n
n
2k 2
X
X
k k
k −1/2
Dn ≡
(−1)
=
(−1)
k
16k
k=0
k=0
2
2 X
n
n
X
n−k n
k n
= 0 (mod p).
=
(−1)
≡
(−1)
k
k
k=0
k=0
14
ZHI-WEI SUN
The following result was conjectured by the author [Su2] and confirmed
by Z.-H. Sun [S2].
Lemma 3.5. Let p be an odd prime. Then
p−1
X
k=0
2k 3
k
(−8)k
≡
4x2 − 2p (mod p2 )
0 (mod p2 )
if 4 | p − 1 & p = x2 + y 2 (2 ∤ x),
if p ≡ 3 (mod 4).
(3.5)
Remark 3.2. Fix an odd prime p = 2n + 1. By (1.2) and (1.3) we have
p−1
X
k=0
2k 3
k
(−8)k
≡
2
n X
2k
n+k
k=0
2k
k
2k = Dn2
(mod p2 ).
Hence (3.5) follows from Lemma 3.4 and Remark 3.1.
Lemma 3.6. For any positive integer n we have
2
n X
Sn (x)
2k + 1 k
2k
n+k
x (x + 1)k+1 =
(Dn−1 (x) + Dn+1 (x)).
2
(k + 1)
2
k
2k
k=0
(3.6)
Proof. Note that
Sn (x)(Dn−1 (x) + Dn+1 (x)) =
2n+1
X
cm (n)xm
m=0
where
m X
n+1+m−k
n−1+m−k
2m − 2k
n+k
+
Ck
cm (n) =
2m − 2k
2m − 2k
m−k
2k
k=0
m X
2m − 2k (m + n − k)2 − n(2m − 2k − 1)
n+m−k
n+k
Ck
=2
.
m−k
2m − 2k
2k
(m + n − k)(n − m + k + 1)
k=0
By the Zeilberger algorithm we find that um = cm (n)/2 satisfies the recursion
(m + 2)(m + 3)2 (m2 + 5m + 6 + 4n(n + 1))um+2 + 2P (m, n)um+1
=(m + 2)((2n + 1)2 − m2 )(m2 + 7m + 12 + 4n(n + 1))um
(3.7)
where P (m, n) denotes the polynomial
m5 + 11m4 + 45m3 + 83m2 + 64m + 12 + 20n4 − 40n3 − 58n2 − 38n
− 25mn + m2 n + 2m3 n − 33mn2 + m2 n2 + 2m3 n2 − 16mn3 − 8mn4 .
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
15
Clearly the coefficient of xm on the left-hand side of (3.6) coincides with
2 m X
k + 1 2k + 1
2k
n+k
dm (n) =
.
m − k (k + 1)2
k
2k
k=0
By the Zeilberger algorithm um = dm (n) also satisfies the recursion (3.7).
Thus we have dm (n) = cm (n) by induction on m. So (3.6) holds. Proof of Theorem 1.2(ii). Write p = 2n + 1. By (2.1),
n X
n(n + 1) 2
2k
2k
n+k
Sn .
2k =
2
k+1
k
2k
k=0
Thus, by (1.2) and (1.9) we have
p−1
X
k=0
2k 2 2k
k
k+1
(−8)k
n X
2k
2k
n+k
2k
≡
k+1
k
2k
k=0
2
≡
p −1
4(4x2 − 4p) (mod p2 )
8
and hence (1.12) holds.
Now we consider (1.13). Observe that
2k
k+1
2
2
2k + 1
2k
= 1−
2
k
(k + 1)
for k = 0, 1, 2, . . . ,
and
2
2
p
2(p − 1)
2(p − 1)
2p − 1 2p − 2
=
≡ −p
(p − 1) + 1
p−2
p−1
2p − 1 p − 1
(mod p2 ).
Thus we have
p−1
X
k=0
2k 2
k+1
(−8)k
2k
k
≡ −p +
n
X
k=0
2k 3
k
(−8)k
3
n
X
(2k + 1) 2k
k
−
(k + 1)2 (−8)k
(mod p2 ). (3.8)
k=0
By (1.2) and (3.6) with x = 1,
3
2
n n
X
X
(2k + 1) 2k
n + k 2k (2k + 1)2k
k
≡
k
(k + 1)2 (−8)k
2k
(k + 1)2
k=0
k=0
=
Sn
(Dn−1 + Dn+1 ) (mod p2 ).
4
16
ZHI-WEI SUN
It is known (cf. [Sl] and [St, p. 191]) that
(n + 1)Dn+1 = 3(2n + 1)Dn − nDn−1
and
Dn+1 − 3Dn = 2nSn .
Thus
n(Dn−1 + Dn+1 ) =3(2n + 1)Dn − Dn+1
=3(2n + 1)Dn − (3Dn + 2nSn ) = 2n(3Dn − Sn )
and hence
3
n
X
(2k + 1) 2k
Sn
k
≡
(3Dn − Sn ) (mod p2 ).
2
k
(k + 1) (−8)
2
k=0
With the help of (1.9) and (3.4), we have
2p
p p
Sn
− 4x −
3 2x −
(3Dn − Sn ) ≡ 2x −
2
x
2x
x
and hence
3
n
X
(2k + 1) 2k
k
≡ 4x2 − p
2
k
(k + 1) (−8)
(mod p2 )
(mod p2 ).
k=0
Combining this with (3.5) and (3.8), we immediately obtain (1.13). 4. Proof of Theorem 1.3
Lemma 4.1. Let p be an odd prime. Then, for any p-adic integer x 6≡
0, −1 (mod p) we have
3 k
p−1 X
−x
2k
k=0
k
64
≡
x+1
p
X
p−1 k=0
2k
k
2 k
x
4k
(mod p).
2k
64(x + 1)2
(4.1)
Proof. Taking n = (p − 1)/2 in the following identity of MacMahon (see,
e.g., [G, (6.7)])
n 3
X
n
k=0
k
n X
2k n − k k
n+k
x (1 + x)n−2k
x =
k
k
2k
k
k=0
and noting (1.2) and the basic facts
2k
−1/2
n
= k k
≡
k
k
(−4)
(mod p)
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
and
4k
−1/2 − k
n−k
2k
=
≡
(−4)k
k
k
17
(mod p),
we immediately get (4.1). Proof of Theorem 1.3. (i) For d = 0, 1, 2, . . . , we define
f (d) =
p−1
X
k=0
2k
k+d
2k
k
108k
and
h(d) =
3k
k
p−1
X
k=0
,
g(d) =
p−1
X
2k
k+d
k=0
2k
k+d
3k
k
3k
12
6k
3k
2k
k
256k
4k
2k
,
.
By the Zeilberger algorithm, we find the recursive relations:
(3d + 1)(3d + 2)f (d) − (3d + 4)(3d + 5)f (d + 2)
2p − 2 3p − 3
2p
(3p − 1)(3p − 2)(d + 1)
,
=
p−1
p−1
p+d+1
108p−1 p
(4.2)
(4d + 1)(4d + 3)g(d) − (4d + 5)(4d + 7)g(d + 2)
(4p − 1)(4p − 3)(d + 1)
2p − 2 4p − 4
2p
=
,
2p − 2
p−1
p+d+1
256p−1 p
(4.3)
(6d + 1)(6d + 5)h(d) − (6d + 7)(6d + 11)h(d + 2)
(6p − 1)(6p − 5)(d + 1)
3p − 3 6p − 6
2p
=
.
3p − 3
p−1
p+d+1
1728p−1 p
(4.4)
and
Recall that
2p−2
p−1
= pCp−1 ≡ 0 (mod p). Also,
3p − 2
3p − 3
≡ 0 (mod p),
=p
(3p − 2)
p
p−1
4p − 2
4p − 4
≡ 0 (mod p),
=p
(4p − 3)
2p
2p − 2
3p(3p − 1)(3p − 2) 6p − 3
6p − 6
=
(6p − 5)
≡ 0 (mod p).
3p − 3
3p
(6p − 3)(6p − 4)
If 0 6 d < p − 1, then
2p
2p
≡0
=
p−1−d
p+d+1
(mod p).
18
ZHI-WEI SUN
So, by (4.2)-(4.4), for any d ∈ {0, . . . , p − 1} we have
(3d + 1)(3d + 2)f (d) ≡(3d + 4)(3d + 5)f (d + 2) (mod p2 ),
(4.5)
2
(4d + 1)(4d + 3)g(d) ≡(4d + 5)(4d + 7)g(d + 2) (mod p ),
(4.6)
(6d + 1)(6d + 5)h(d) ≡(6d + 7)(6d + 11)h(d + 2) (mod p2 ).
(4.7)
Fix 0 6 d 6 p − 1. If d ≡ (1 + ( 3p ))/2 (mod 2), then it is easy to verify
that {3d + 1, 3d + 2} ∩ {p, 2p} = ∅, hence (3d + 1)(3d + 2) 6≡ 0 (mod p)
and thus by (4.5) we have
f (d + 2) ≡ 0 (mod p2 ) =⇒ f (d) ≡ 0 (mod p2 ).
))/2 (mod 2), then {4d + 1, 4d + 3} ∩ {p, 3p} = ∅, hence
If d ≡ (1 + ( −2
p
(4d + 1)(4d + 3) 6≡ 0 (mod p) and thus by (4.6) we have
g(d + 2) ≡ 0 (mod p2 ) =⇒ g(d) ≡ 0 (mod p2 ).
If d ≡ (1 + ( −1
p ))/2 (mod 2), then {6d + 1, 6d + 3} ∩ {p, 3p, 5p} = ∅, hence
(6d + 1)(6d + 3) 6≡ 0 (mod p) and thus (4.7) yields
h(d + 2) ≡ 0 (mod p2 ) =⇒ h(d) ≡ 0 (mod p2 ).
Since
f (p) = f (p + 1) = g(p) = g(p + 1) = h(p) = h(p + 1) = 0,
by the last paragraph, for every d = p + 1, p, . . . , 0 we have the desired
(1.19)-(1.21).
(ii) Assume that p ≡ 3 (mod 8) and p = x2 + 2y 2 with x, y ∈ Z. Since
4x2 6≡ 0 (mod p) and Mortenson [M] already proved that the squares of
both sides of (1.22) are congruent modulo p2 , (1.22) is reduced to its mod
p form. Applying (4.1) with x = 1 we get
p−1
X
k=0
2k 3
k
(−64)k
X
p−1
2
≡
p
k=0
2k 2 4k
k
2k
256k
(mod p).
By [A, Theorem 5(3)], we have
−1
p
X
n 2 n+k
n
(−1)k ≡ 4x2 − 2p (mod p),
k
k
k=0
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
19
where n = (p − 1)/2. For k = 0, . . . , n clearly
2 2 −(p + 1)/2
(p − 1)/2
n+k
n
k
(−1) =
k
k
k
k
3
3
2k
−1/2
= k k (mod p),
≡
k
(−64)
therefore
p−1
X
k=0
2k 3
k
(−64)k
≡
−1
p
(4x2 − 2p) (mod p)
and hence (1.22) follows.
(iii) Finally we suppose p ≡ 5 (mod 12) and write p = x2 + y 2 with x
odd and y even. Once again it suffices to show the mod p form of (1.23)
in view of Mortenson’s work [M]. As Z.-H. Sun observed,
3k 6k
k − 5/6 2k
(p − 5)/6 + k 2k
k
3k
=
≡
(mod p)
k
2k
(−432)k
k
2k
for all k = 0, 1, 2, . . . . If p/6 < k < p/3 then p | 6k
; if p/3 < k < p/2
3k
3k
2k
then p | k ; if p/2 < k < p then p | k . Thus
2 k
(p−5)/6 p−1 2k 3k 6k
X
X
1
2k
(p − 5)/6 + k
k
k
3k
−
≡
k
2k
123k
4
k=0
k=0
2
1
(mod p) (by (1.3)),
=D2n −
2
where n = (p − 5)/12. Note that
1
1
2n
D2n −
=
n
n
2
(−4)
by [G, (3.133) and (3.135)], and
(p − 1)/2
n 2n
≡ 12(−432)
n
(p − 1)/4
(mod p)
by P. Morton [Mo]. Therefore
2
2
2
(p−1)/2 2
1
(p − 1)/2
1 2n
12
(p−1)/4
D2n −
(mod p).
≡
= n
≡
(p − 1)/4
n
2
16
126n+2
p
Thus, by applying Gauss’ congruence (p−1)/2
≡ 2x (mod p) (cf. [BEW,
(p−1)/4
(9.0.1)] or [HW]) we immediately get the mod p form of (1.23) from the
above.
The proof of Theorem 1.3 is now complete. Acknowledgment. The author would like to thank the referee for helpful
comments.
20
ZHI-WEI SUN
References
[A]
[BEW]
[CHV]
[CDE]
[C]
[G]
[GZ]
[HW]
[I]
[KF]
[M]
[Mo]
[O]
[PWZ]
[RV]
[Sl]
[St]
[SB]
[S1]
[S2]
[Su1]
[Su2]
[Su3]
[Su4]
S. Ahlgren, Gaussian hypergeometric series and combinatorial congruences,
in: Symbolic Computation, Number Theory, Special Functions, Physics and
Combinatorics (Gainesville, FI, 1999), pp. 1-12, Dev. Math., Vol. 4, Kluwer,
Dordrecht, 2001.
B. C. Berndt, R. J. Evans and K. S. Williams, Gauss and Jacobi Sums, John
Wiley & Sons, 1998.
J. S. Caughman, C. R. Haithcock and J. J. P. Veerman, A note on lattice chains
and Delannoy numbers, Discrete Math. 308 (2008), 2623–2628.
(p−1)/2
S. Chowla, B. Dwork and R. J. Evans, On the mod p2 determination of (p−1)/4
,
J. Number Theory 24 (1986), 188–196.
D. A. Cox, Primes of the Form x2 + ny 2 , John Wiley & Sons, 1989.
H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co.,
1972.
J. Guillera and W. Zudilin, “Divergent” Ramanujan-type supercongruences,
Proc. Amer. Math. Soc. 140 (2012), 765–777.
R. H. Hudson and K. S. Williams, Binomial coefficients and Jacobi sums, Trans.
Amer. Math. Soc. 281 (1984), 431–505.
T. Ishikawa, Super congruence for the Apéry numbers, Nagoya Math. J. 118
(1990), 195–202.
F. Klein and R. Fricke, Vorlesungen uber die Theorie der elliptischen Modulfunktionen, Teubner, Leipzig, 1892.
E. Mortenson, Supercongruences for truncated n+1Fn hypergeometric series
with applications to certain weight three newforms, Proc. Amer. Math. Soc.
133 (2005), 321–330.
P. Morton, Explicit identities for invariants of elliptic curves, J. Number Theory 120 (2006), 234–271.
K. Ono, Web of Modularity: Arithmetic of the Coefficients of Modular Forms
and q-series, Amer. Math. Soc., Providence, R.I., 2003.
M. Petkovšek, H. S. Wilf and D. Zeilberger, A = B, A K Peters, Wellesley,
1996.
F. Rodriguez-Villegas, Hypergeometric families of Calabi-Yau manifolds, in:
Calabi-Yau Varieties and Mirror Symmetry (Toronto, ON, 2001), pp. 223-231,
Fields Inst. Commun., 38, Amer. Math. Soc., Providence, RI, 2003.
N. J. A. Sloane, Sequences A001850 and A006318 in OEIS (On-Line Encyclopedia of Integer Sequences), http://oeis.org.
R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press,
Cambridge, 1999.
J. Stienstra and F. Beukers, On the Picard-Fuchs equation and the formal
Brauer group of certain elliptic K3-surfaces, Math. Ann. 271 (1985), 269–304.
Z.-H. Sun, Congruences concerning Legendre polynomials, Proc. Amer. Math.
Soc. 139 (2011), 1915–1929.
Z.-H. Sun, Congruences concerning Legendre polynomials (II), arXiv:1012.3898.
Z.-W. Sun, Binomial coefficients, Catalan numbers and Lucas quotients, Sci.
China Math. 53 (2010), 2473–2488.
Z.-W. Sun, On congruences related to central binomial coefficients, J. Number
Theory 131 (2011), 2219–2238.
Z.-W. Sun, Super congruences and Euler numbers, Sci. China Math. 54 (2011),
2509–2535.
Z.-W. Sun, On Delannoy numbers and Schroder numbers, J. Number Theory
131 (2011), 2387–2397.
SUMS INVOLVING PRODUCTS OF BINOMIAL COEFFICIENTS
[vH]
[Z]
21
L. van Hamme, Some conjectures concerning partial sums of generalized hypergeometric series, in: p-adic Functional Analysis (Nijmegen, 1996), pp. 223–236,
Lecture Notes in Pure and Appl. Math., Vol. 192, Dekker, 1997.
D. Zeilberger, A fast algorithm for proving terminating hypergeometric series
identities, Discrete Math. 80 (1990), 207–211.