Telescoping method and congruences for double sums

0.2 MB·

by  Yan-Ping Mu

Preprint, arXiv:1601.03952
TELESCOPING METHOD AND CONGRUENCES FOR
DOUBLE SUMS
YAN-PING MU AND ZHI-WEI SUN
6
1
0
2 Abstract. In recent years,Z.-W. Sun proposedseveralconjectures on

n congruencesforfinitesumswithtermsinvolvingcombinatorialsequences
a such as central trinomial coefficients, Domb numbers and Franel num-
J
bers. These sums are double summations of hypergeometric terms. By

5 the telescoping method, we transform such a double summation into a
2
singlesum. Withthisnewapproach,weconfirmseveralopenconjectures

of Sun.
]
T
N
.
h
at 1. Introduction
m
[
For a sequence of integers a ,a ,a ,..., it is interesting to see whether
0 1 2

2
the arithmetic mean
2v a0 +a1 +···+an−1
5 n
9 is integral for any n ∈ Z+ = {1,2,3,...}. It is easy to see that
3
0 n−1 n−1
1 1
1. (2k+1)(−1)k = (−1)n−1 ∈ Z and (2k +1) = 1 ∈ Z.
n n2
0
k=0 k=0
6 X X
For the sequence of Ap´ery numbers
1
:
v k 2 2 k 2 2
k k +l k +l 2l
i A := = (k ∈ N = {0,1,2,...}),
X k
l l 2l l
r Xl=0 (cid:18) (cid:19) (cid:18) (cid:19) Xl=0 (cid:18) (cid:19) (cid:18) (cid:19)
a Z.-W. Sun [17] proved the congruence n−1(2k +1)A ≡ 0 (mod n), and
k=0 k
conjectured that n−1(2k+1)(−1)kA ≡ 0 (mod n), which was confirmed
k=0 kP
by V.J. Guo and J. Zeng [5]. For the sequence of Franel numbers
P
k 3
k
f := (k = 0,1,2,...)
k
l
l=0 (cid:18) (cid:19)
X
Keywordsandphrases. Telescopingmethod,centraltrinomialcoefficient,congruence,
Domb number, Franel number.
2010 Mathematics Subject Classification. Primary 11B75; Secondary 05A10, 11A07,
33F10.
The second author is the corresponding author. Both authors are supported by
the National Natural Science Foundation of China (grants 11471244 and 11571162,
respectively).
1
2 YAN-PINGMU ANDZHI-WEISUN
(for which J. Franel [2] found a recurrence), Z.-W. Sun [19] conjectured that
n−1
(3k +2)(−1)kf ≡ 0 (mod 2n2),
k
k=0
X
which was confirmed by Guo [3]. However, some such conjectural congru-
ences still remain open. For example, Sun [20, Conjecture 5.1] conjectured
that
n−1
(5k +4)D ≡ 0 (mod 4n),
k
k=0
X
where D ,D ,D ,... are the Domb numbers defined by
0 1 2
k 2
k 2l 2(k −l)
D := (k ∈ N).
k
l l k −l
l=0 (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
X
For various combinatorial interpretations of the Domb numbers, see L.B.
Richmond and J. Shallit [12], and N.J.A Sloane [13].
In this paper we study some conjectures of Sun [21, 23] on congruences
for sums of the form
n−1 k
S = F(k,l) (n ∈ Z+), (1.1)
n
k=0 l=0
XX
where F(k,l) is a bivariate hypergeometric term of k and l. To confirm
such conjectures, we try to search for two hypergeometric terms G (k,l)
1
and G (k,l) such that
2
F(k,l) = ∆ (G (k,l))+∆ (G (k,l)), (1.2)
k 1 l 2
where
G (k,l) = R (k,l)F(k,l) and G (k,l) = R (k,l)F(k,l)
1 1 2 2
with R (k,l) and R (k,l) being rational functions, and
1 2
∆ (G (k,l)) = G (k+1,l)−G (k,l)and∆ (G (k,l)) = G (k,l+1)−G (k,l).
k 1 1 1 l 2 2 2
Though the denominator of the rational function R (k,l) or R (k,l) might
1 2
be zero for some 0 6 l 6 k 6 n−1, the functions G (k,l) and G (k,l) we
1 2
obtain are essentially well defined for 0 6 l 6 k 6 n−1 and hence (1.2) can
be verified directly. Once we have G (k,l) and G (k,l) in hand, the sum S
1 2 n
can be transformed to a single sum
n−1 n−1
S = G (n,l)−G (l,l) + G (k,k +1)−G (k,0) . (1.3)
n 1 1 2 2
l=0 k=0
X(cid:0) (cid:1) X(cid:0) (cid:1)
TELESCOPING METHOD AND CONGRUENCES FOR DOUBLE SUMS 3
We can use the Maple package DoubleSum given by Chen-Hou-Mu [1] or
the Mathematica package HolonomicFunctions given by C. Koutschan [8]
to compute G (k,l) and G (k,l). Usually, the two hypergeometric terms
1 2
G (k,l) and G (k,l) obtained directly by these two packages are not opti-
1 2
mal. We will remove some factors from the denominators of the resulting
R and R , and use the package DoubleSum or the package MultiSum [9] to
1 2
compute a suitable pair (G (k,l),G (k,l)). Once we get the simplification
1 2
(1.3) for S , it would be be convenient to deduce Sun’s conjectural congru-
n
ences for S . Using this powerful method, we confirm several sophisticated
n
open conjectures of Sun.
The second author [18, 19] established several supercongruences on sums
of Franel numbers. Our first theorem confirms a conjecture of him involving
Franel numbers.
Theorem 1.1 ([23, Conjecture 4.1(i)]). For any integer n > 1, we have
n−1
(9k2 +5k)(−1)kf ≡ 0 (mod n2(n−1)). (1.4)
k
k=0
X
Moreover, for any odd prime p we have
p−1
(9k2 +5k)(−1)kf ≡ 3p2(p−1)−16p3q (2) (mod p4), (1.5)
k p
k=0
X
where q (2) denotes the Fermat quotient (2p−1 −1)/p.
p
In [23] Sun introduced the polynomials
n 2
n 2k
g (x) := xk (n = 0,1,2,...)
n
k k
k=0(cid:18) (cid:19) (cid:18) (cid:19)
X
and obtained some related congruences. Such polynomials are interesting
since it is closely related to the Ap´ery polynomials
n 2 2 n 2 2
n n+k n+k 2k
A (x) := xk = xk (n ∈ N);
n
k k 2k k
k=0(cid:18) (cid:19) (cid:18) (cid:19) k=0(cid:18) (cid:19) (cid:18) (cid:19)
X X
in fact, Sun [23, (2.8)] showed that
n
n n+k
A (x) = (−1)n−kg (x).
n k
k k
k=0(cid:18) (cid:19)(cid:18) (cid:19)
X
4 YAN-PINGMU ANDZHI-WEISUN
Theorem 1.2 ([23, Conjecture 4.1(ii)]). For any n ∈ Z+, the number
n−1
1
(8k2 +12k +5)g (−1)
k
n2
k=0
X
is an odd integer. Moreover, for any prime p we have
p−1
(8k2 +12k +5)g (−1) ≡ 3p2 (mod p3). (1.6)
k
k=0
X
Remark. Guo, Mao and Pan [4] called g (x) (n ∈ N) Sun polynomials, and
n
they used the Zeilberger algorithm (cf. [10, pp.101-119]) to prove that
n−1
(8k2 +12k +5)g (−1) ≡ 0 (mod n) for all n ∈ Z+,
k
k=0
X
which is the first progress on [23, Conjecture 4.1(ii)].
Theorem 1.3 ([23, Conjecture 4.3]). For k = 0,1,2,... define
k 3
k
F := (−8)l.
k
l
l=0 (cid:18) (cid:19)
X
Then, for any positive integer n, the number
n−1
1
(6k +5)(−1)kF
k
n
k=0
X
is an odd integer.
Forn ∈ N, thecentral trinomial coefficientT denotesthecoefficient ofxn
n
intheexpansion of(x2+x+1)n. T hasmany combinatorial interpretations,
n
forexample, T countsthenumberofpermutationsofnsymbols, each−1,0,
n
or 1, which sum to 0. For b,c ∈ Z, Sun [21] defined T (b,c) as the coefficient
n
of xn in the expansion of (x2 +bx+c)n, and called those T (b,c) (n ∈ N)
n
generalized central trinomial coefficients.
Theorem 1.4 ([21, Conjecture 5.2]). Let b,c ∈ Z. For any n ∈ Z+, we
have
n−1
(8ck+4c+b)T (b,c2)2(b−2c)2(n−1−k) ≡ 0 (mod n). (1.7)
k
k=0
X
If p is an odd prime not dividing b(b−2c), then
p−1 T (b,c2)2 b2 −4c2
(8ck +4c+b) k ≡ p(b+2c) (mod p2), (1.8)
(b−2c)2k p
k=0 (cid:18) (cid:19)
X
·
where ( ) denotes the Legendre symbol.
p
TELESCOPING METHOD AND CONGRUENCES FOR DOUBLE SUMS 5
In the case b = c = 1, Theorem 1.4 yields the following consequence.
Corollary 1.5 ([21, Conjecture 1.1(i)]). For any positive integer n, we have
n−1
(8k +5)T2 ≡ 0 (mod n), (1.9)
k
k=0
X
If p is a prime, then
p−1
p
(8k +5)T2 ≡ 3p (mod p2). (1.10)
k 3
Xk=0 (cid:16) (cid:17)
Our next theorem confirms a conjecture of Sun involving the Domb num-
bers.
Theorem 1.6 ([15, Conjecture 5.15(ii)]). For any integer n > 1, the num-
ber
n−1
1
a = (3k2 +k)D 16n−1−k
n 2n3(n−1) k
k=0
X
is an integer. Moreover, a is odd if and only if n is a power of two. Also,
n
for any prime p > 3 we have the supercongruence
p−1 3k2 +k
D ≡ −4p4q (2) (mod p5). (1.11)
16k k p
k=0
X
Our last theorem is about some conjectures of the second author posed
in a preprint form arXiv:1407.0967v6 of [23].
Theorem 1.7. (i) For any integer n > 1, we have
n−1
(12k4 +25k3 +21k2 +6k)(−1)kf ≡ 0 (mod 4n2(n−1)). (1.12)
k
k=0
X
If p is an odd prime, then
p−1
(12k4 +25k3 +21k2 +6k)(−1)kf ≡ −4p3 (mod p4). (1.13)
k
k=0
X
(ii) For any n ∈ Z+ we have
n−1
(12k3 +34k2 +30k+9)g ≡ 0 (mod 3n3), (1.14)
k
k=0
X
6 YAN-PINGMU ANDZHI-WEISUN
where g := g (1) = k k 2 2l . If p is an odd prime, then
k k l=0 l l
p−1 P (cid:0) (cid:1) (cid:0) (cid:1) 3p2 p
(12k3 +34k2 +30k+9)g ≡ 1+3 (mod p4). (1.15)
k
2 3
Xk=0 (cid:16) (cid:16) (cid:17)(cid:17)
(iii) For any n ∈ Z+ we have
n−1
(18k5 +45k4 +46k3 +24k2 +7k +1)(−1)kA ≡ 0 (mod n4). (1.16)
k
k=0
X
Also, for any prime p > 3 we have
p−1
(18k5 +45k4 +46k3 +24k2 +7k +1)(−1)kA ≡ −2p4 +3p5 (mod p7).
k
k=0
X
(1.17)
Remark. The second author even conjectured that
p−1
(18k5 +45k4 +46k3 +24k2 +7k +1)(−1)kA
k
(1.18)
k=0
X
12
≡−2p4 +3p5 +(6p−8)p5Hp−1 − p9Bp−5 (mod p10)
5
foranyprimep > 5,whereH denotesthen-thharmonicnumber 1/k,
n 0<k6n
and B ,B ,B ,... are the Bernoulli numbers.
0 1 2 P
2. Proof of Theorem 1.1
In this section, we illustrate in detail how to prove Theorem 1.1 by the
telescoping method.
Proof of Theorem 1.1. By an identity of V. Strehl [14], for any k ∈ N we
have
k 2
k 2l
f = . (2.1)
k
l k
l=0 (cid:18) (cid:19) (cid:18) (cid:19)
X
Therefore, for any integer n > 1, the left hand side of the congruence (1.4)
can be written as
n−1 k
F(k,l),
k=0 l=0
XX
where
2
k 2l
F(k,l) = (9k2 +5k)(−1)k .
l k
(cid:18) (cid:19) (cid:18) (cid:19)
TELESCOPING METHOD AND CONGRUENCES FOR DOUBLE SUMS 7
By the command GetDen(F, k, l) of the package DoubleSum, we obtain
the estimated denominators of R (k,l) and R (k,l):
1 2
d (k,l) = k(9k +5)(l+1)(−2l +k −1)(−2l−2+k),
1
d (k,l) = k(9k +5)(k −l+1)2.
2
We remove the factor k(−2l−2+k) from d (k,l), and the factor (k−l+1)2
1
from d (k,l), and then use the command
2
SolveR(F, n, k, l, (9k+5)(l+1)(-2l+k-1), k(9k+5), 0, "s")
to obtain the rational functions
(k −l)2(3k +4l−3)
R (k,l) =
1
(9k +5)(l+1)(−2l+k −1)
and
(−2l +k)(3k −l −2)
R (k,l) = .
2
k(9k +5)
Therefore
F(k,l) = ∆ (G (k,l))+∆ (G (k,l)),
k 1 l 2
where
1 2l k −1 l
G (k,l) = (−1)k−1k2(3k +4l−3) ,
1
l+1 l l k −1−l
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
2
k 2l
G (k,l) = (−1)k−1(2l−k)(3k −l−2) .
2
l k
(cid:18) (cid:19) (cid:18) (cid:19)
Noting that
G (l,l) = G (k,k +1) = G (k,0) = 0,
1 2 2
we get
n−1
S = G (n,l)
n 1
l=0
X
n−1
1 2l n−1 l
=(−1)n−1n2 (3n+4l−3)
l+1 l l n−1−l
l=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X
n−1
1 2l n−1 l
=(−1)n−13n2(n−1)
l+1 l l n−1−l
l=0 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X
n−1
1 2l n−2 l
+(−1)n−14n2(n−1) .
l+1 l l−1 n−1−l
l=1 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X
Since C = 2l /(l +1) is the l-th Catalan number which is an integer, we
l l
derive immediately that S is divisible by n2(n−1).
(cid:0) (cid:1) n
8 YAN-PINGMU ANDZHI-WEISUN
Now let p be an odd prime. By the above,
p−1
3p−3+4l p−1 2l l
S = p2 . (2.2)
p
l +1 l l p−1−l
l=(p−1)/2 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X
For l = (p+1)/2,...,p−2, we clearly have p | 2(l+1) and hence
l+1
(cid:0) (cid:1)
2l l l +1 2l+1 −l +(p−1−l)−1
= (−1)l
l p−1−l 2l+1 l p−1−l
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
(−1)l(l+1) 2(l +1) 2(p−1−l)−p
=
2(2l+1) l+1 p−1−l
(cid:18) (cid:19)(cid:18) (cid:19)
(−1)l(l+1) 2(l +1) 2(p−(l +1))
≡
2(2l+1) l+1 p−(l+1)
(cid:18) (cid:19)(cid:18) (cid:19)
(−1)lp
≡ (mod p2)
2l+1
with the help of [15, Lemma 2.1]. Therefore
S 3p−3+4l p−1 (−1)lp
p
≡
p2 l+1 l 2l+1
(p−1)/2<l<p−1 (cid:18) (cid:19)
X
2
3p−3+4(p−1)/2 p−1 3p−3+4(p−1) 2(p−1)
+ +
(p+1)/2 (p−1)/2 p p−1
(cid:18) (cid:19) (cid:18) (cid:19)
4l−3 10(p−1)
≡p + (1−pH(p−1)/2)2
(l+1)(2l+1) p+1
(p−1)/2<l<p−1
X
7(p−1) 2p−1
+ (mod p2)
2p−1 p−1
(cid:18) (cid:19)
by noting that
(p−1)/2
p−1 p
(−1)(p−1)/2 = 1− ≡ 1−pH(p−1)/2 (mod p2).
(p−1)/2 j
(cid:18) (cid:19) j=1 (cid:18) (cid:19)
Y
Clearly,
(p−1)/2
1 1
Hp−1 = + ≡ 0 (mod p)
k p−k
k=1 (cid:18) (cid:19)
X
and
p−1
2p−1 p
= 1+ ≡ 1+pHp−1 ≡ 1 (mod p2).
p−1 k
(cid:18) (cid:19) kY=1(cid:16) (cid:17)
TELESCOPING METHOD AND CONGRUENCES FOR DOUBLE SUMS 9
Thus, we deduce that
S 7 10
p
≡p −
p2 l +1 2l+1
(p−1)/2<l<p−1(cid:18) (cid:19)
X
−10(p−1)2 1−2pH(p−1)/2 −7(p−1)(2p+1)
(p−3)/2
(cid:0) 2 (cid:1) 1
≡7p Hp−1 −H(p−1)/2 − −10p
p+1 p+2j
(cid:18) (cid:19) j=1
X
+10(2p−1)(1−2pH(p−1)/2)+7(p+1)
H(p−1)/2 1
≡−7pH(p−1)/2 −14p−10p −
2 p−1
(cid:18) (cid:19)
+10(2p−1+2pH(p−1)/2)+7(p+1)
=8pH(p−1)/2 −24p+27p−3 (mod p2).
As
(p−1)/2 (p−1)/2 (p−1)/2
p p p−1 p
− ≡ = = 2p−1−1 (mod p2),
2k 2k 2k −1 2k
k=1 k=1 (cid:18) (cid:19) k=1 (cid:18) (cid:19)
X X X
we have pH(p−1)/2 ≡ −2(2p−1 −1) (mod p2), i.e.,
H(p−1)/2 ≡ −2qp(2) (mod p). (2.3)
So, we finally obtain the congruence
S
p ≡ 8p(−2q (2))+3p−3 (mod p2)
p2 p
which gives (1.5). This concludes our proof of Theorem 1.1.
3. Proofs of Theorems 1.2 and 1.3
In this section, we prove Theorems 1.2 and 1.3. The method is similar to
the one used in Section 2. We will give related R (k,l) and R (k,l) directly.
1 2
Proof of Theorem 1.2. Let
2
k 2l
F(k,l) = (8k2 +12k+5) (−1)l
l l
(cid:18) (cid:19) (cid:18) (cid:19)
be the summand and
n−1 k
S = F(k,l) with n ∈ Z+.
n
k=0 l=0
XX
10 YAN-PINGMU ANDZHI-WEISUN
We find that
(k −l)2(2k +5l+3) l2
R (k,l) = and R (k,l) = − .
1 (8k2 +12k +5)(l+1) 2 8k2 +12k +5
Thus we derive that
n−1 n−1 2
n−1 1 2l
S = R (n,l)F(n,l) = n2 (2n+5l+3)(−1)l .
n 1
l l+1 l
l=0 l=0 (cid:18) (cid:19) (cid:18) (cid:19)
X X
Noting that C = 2l /(l+1) ∈ Z, we have
l l
(cid:0) (cid:1)n−1 n−1 2 2l
S ≡ n2 (−1)l ≡ n2 (mod 2n2)
n
l l
l=0 (cid:18) (cid:19) (cid:18) (cid:19)
X
since 2l = 2 2l−1 for l = 1,2,3,.... Therefore, S /n2 is an odd integer.
l l−1 n
It (cid:0)is s(cid:1)traig(cid:0)htfor(cid:1)ward to check that S = −20 ≡ 3×22 (mod 23). So (1.6)
2
holds for p = 2. Now let p be an odd prime. Since
p−1
≡ (−1)l (mod p) for all l = 0,1,...,p−1,
l
(cid:18) (cid:19)
we have
S p−1 (−1)l 2l p−1 2l p−1
p ≡ (5l+3) = 5 (−1)l −2 (−1)lC (mod p).
p2 l+1 l l l
l=0 (cid:18) (cid:19) l=0 (cid:18) (cid:19) l=0
X X X
Clearly,
p−1 p−1
2l −1/2
(−1)l = 4l
l l
l=0 (cid:18) (cid:19) l=0 (cid:18) (cid:19)
X X
p−1
(p−1)/2 5 p
≡ 4l = 5(p−1)/2 ≡ = (mod p)
l p 5
Xl=0 (cid:18) (cid:19) (cid:18) (cid:19) (cid:16) (cid:17)
and
p−1
5 5 5 p
(−1)lC ≡ − 1− = −1 (mod p)
l
2 p 2 5
Xl=1 (cid:18) (cid:18) (cid:19)(cid:19) (cid:16)(cid:16) (cid:17) (cid:17)
by [16, Lemma 2.1]. We thus derive that
S ≡ 3p2 (mod p3),
p
completing the proof of Theorem 1.2.
Proof of Theorem 1.3. For n ∈ Z+ let
n−1 n−1 k
S = (6k +5)(−1)kF = F(k,l)
n k
k=0 k=0 l=0
X XX