Proof of Sun's conjecture on the divisibility of certain binomial sums

0.07 MB·English

by  Victor J. W. Guo

Proof of Sun’s conjecture on the divisibility of
certain binomial sums
Victor J. W. Guo
Department of Mathematics, East China Normal University,
Shanghai 200062,People’s Republic of China
[email protected], http://math.ecnu.edu.cn/~jwguo
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2 Abstract. In this paper, we prove the following result conjectured by Z.-W. Sun:

n
a n
J 3n 6k 3k 6(n k) 3(n k)
(2n 1) (cid:12) − −
21 − (cid:18)n(cid:19)(cid:12)(cid:12)Xk=0(cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n−k)(cid:19)(cid:18) n−k (cid:19)
(cid:12)
by showing that the left-hand side(cid:12)divides each summand on the right-hand side.
]
T
N
Keywords. congruences, binomial coefficients, super Catalan numbers, Stirling’s formula
.
h
t
a
1 Introduction
m
[

In [2], Z.-W. Sun proved some new series for 1/π as well as related congruences on sums
1
v of binomial coefficients, such as
7
7 ∞ n
n 6k 3k 6(n k) 3(n k) 1
8
− − = ,
4 864n (cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19) π
. Xn=0 Xk=0 − −
1
0
and, for any prime p > 3,
3
1
: p−1 n
v n 6k 3k 6(n k) 3(n k)
2
i − − 0 (mod p ).
X 864n (cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19) ≡
Xn=0 Xk=0 − −
r
a
Sun [2] also proposed many interesting related conjectures, one of which is
Conjecture 1.1 [2, Conjecture 4.2]For n = 0,1,2,... define
n
1 6k 3k 6(n k) 3(n k)
s := − − .
n (2n 1) 3n (cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19)
− n Xk=0 − −
(cid:0) (cid:1)
Then s Z for all n. Also,
n
∈
lim √n s = 64. (1.1)
n
n→∞
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Sun himself has proved that sn 0 (mod 8) for n 1 and sp−1 (p 1)/6 (mod p)
≡ ≥ ≡ ⌊ − ⌋
for any prime p. In this paper, we shall prove that Conjecture 1.1 is true by establishing
the following two theorems.
Theorem 1.2 For 0 k n, there holds
≤ ≤
1 6k 3k 6(n k) 3(n k)
− − Z.
(2n 1) 3n (cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19) ∈
− n − −
(cid:0) (cid:1)
Note that, in [3,4], Sun proved many similar results on the divisibility of binomial coeffi-
cients.
Theorem 1.3 For n 2 and 0 k < n/2, there holds
≥ ≤
6k 3k 6(n k) 3(n k)
− −
(cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19)
− −
6k +6 3k +3 6(n k 1) 3(n k 1)
− − − − ,
≥ (cid:18)3k +3(cid:19)(cid:18)k +1(cid:19)(cid:18)3(n k 1)(cid:19)(cid:18) n k 1 (cid:19)
− − − −
and so
2 6n n+1 6n
s . (1.2)
n
2n 1(cid:18)3n(cid:19) ≤ ≤ 2n 1(cid:18)3n(cid:19)
− −
It is easy to see that (1.1) follows from (1.2) and Stirling’s formula
n n
n! √2πn .
∼ e
(cid:16) (cid:17)
2 Proof of Theorem 1.2
We need the following two lemmas.
Lemma 2.1 Let m,n 1 and 0 k n. Then
≥ ≤ ≤
mn 2mk mk 2m(n k) m(n k)
− − .
(cid:18) n (cid:19)(cid:12)(cid:18)mk (cid:19)(cid:18) k (cid:19)(cid:18)m(n k)(cid:19)(cid:18) n k (cid:19)
(cid:12) − −
(cid:12)
Proof. Observe that (cid:12)
2mk mk 2m(n k) m(n k) mn
− −
(cid:18)mk (cid:19)(cid:18) k (cid:19)(cid:18)m(n k)(cid:19)(cid:18) n k (cid:19)(cid:30)(cid:18) n (cid:19)
− −
(2mk)!(2mn 2mk)! (m 1)n n
= − − . (2.1)
(mk)!(mn mk)!(mn)!(cid:18)(m 1)k(cid:19)(cid:18)k(cid:19)
− −
The proof then follows from the fact that numbers of the form
(2a)!(2b)!
,
a!b!(a+b)!
called the super Catalan numbers, are integers (see [1,5]).
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Lemma 2.2 Let 0 k n be integers. Then
≤ ≤
(6k)!(6n 6k)!(2n)!
(2n 1) − .
− (cid:12)(3k)!(3n 3k)!(3n)!(2k)!(2n 2k)!
(cid:12) − −
(cid:12)
Or equivalently, (cid:12)
(6k)!(6n 6k)!(2n)!(2n 2)!
− − Z.
(3k)!(3n 3k)!(3n)!(2k)!(2n 2k)!(2n 1)! ∈
− − −
Corollary 2.3 Let n 1. Then (2n 1) divides 6n .
≥ − 3n
(cid:0) (cid:1)
The order in which a prime p enters n! can be written as
∞
n
ord n! = , (2.2)
p (cid:22)pi(cid:23)
Xi=1
where x denotes the greatest integer less than or equal to x. In order to prove Lemma
⌊ ⌋
2.2, we first establish the following result.
Lemma 2.4 Let m 2 and 0 k n be integers. Then
≥ ≤ ≤
6k 6n 6k 2n 2n 2
+ − + + −
(cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23) (cid:22) m (cid:23)
3k 3n 3k 3n 2k 2n 2k 2n 1
+ − + + + − + − , (2.3)
≥ (cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23) (cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22) m (cid:23)
unless k 1 (mod 3), m = 3, and n 2 (mod 3).
6≡ ≡
Proof. For any real numbers x and y, it is well known that
2x + 2y x + y + x+y ,
⌊ ⌋ ⌊ ⌋ ≥ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋
x+y x + y .
⌊ ⌋ ≥ ⌊ ⌋ ⌊ ⌋
It follows that
6k 6n 6k 2n 3k 3n 3k 3n 2k 2n 2k
+ − + + − + + + − .
(cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23) ≥ (cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23) (cid:22)m(cid:23) (cid:22) m (cid:23)
Suppose that (2.3) does not hold. Then we must have
6k 6n 6k 3k 3n 3k 3n
+ − = + − + , (2.4)
(cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23) (cid:22) m (cid:23) (cid:22)m(cid:23)
2n 2k 2n 2k
= + − , (2.5)
(cid:22)m(cid:23) (cid:22)m(cid:23) (cid:22) m (cid:23)
2n 2 2n 1
− < − , (2.6)
(cid:22) m (cid:23) (cid:22) m (cid:23)
3
and so, by (2.6), m 2n 1, then by (2.5), m 2k or m 2n 2k. If m 2k, then m k
| − | | − | |
(since m 2n 1 means that m is odd), the identity (2.4) implies that
| −
6n 3n
= 2 . (2.7)
(cid:22)m(cid:23) (cid:22)m(cid:23)
Since m 2n 1, the identity (2.7) implies that
| −
2n 1 3 n+1
− + = 2 . (2.8)
m (cid:22)m(cid:23) (cid:22) m (cid:23)
If m 4, then the left-hand side of (2.8) equals (2n 1)/m, while the right-hand side of
≥ −
(2.8) belongs to
2n+2 2n 2n 2
, , − ,... ,
(cid:26) m m m (cid:27)
a contradiction, and so m 3. Since m 2 is odd, we must have m = 3. Hence, n 2
≤ ≥ ≡
(mod 3) and k 0 (mod 3). Similarly, if m 2n 2k, then we deduce that m = 3, n 2
≡ | − ≡
(mod 3) and k 2 (mod 3). This proves the lemma.
≡
Proof of Lemma 2.2. For any prime p = 3, by (2.2) and (2.3), we have
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ord (6k)!+ord (6n 6k)!+ord (2n)!+ord (2n 2)!
p p p p
− −
ord (3k)!+ord (3n 3k)!+ord (3n)!+ord (2k)!+ord (2n 2k)!+ord (2n 1)!.
p p p p p p
≥ − − −
(2.9)
For p = 3, since ord (3j)! = j +ord j!, the inequality (2.9) reduces to
3 3
ord (2n)!+ord (2n 2)! ord k!+ord (n k)!+ord n!+ord (2n 1)!. (2.10)
3 3 3 3 3 3
− ≥ − −
Noticing that
(2n)!(2n 2)! 1 2n n 2n 2 2n n
− = = 4 − Z,
k!(n k)!n!(2n 1)! 2n 1(cid:18)n(cid:19)(cid:18)k(cid:19) (cid:18) (cid:18)n 1(cid:19)−(cid:18)n(cid:19)(cid:19)(cid:18)k(cid:19) ∈
− − − −
the inequality (2.10) holds. Namely, the inequality (2.9) is true for p = 3. This completes
the proof.
Proof of Theorem 1.2. By (2.1), one sees that
1 6k 3k 6(n k) 3(n k)
− −
(2n 1) 3n (cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19)
− n − −
1(cid:0) (cid:1) (6k)!(6n 6k)! 2n n
= − ,
2n 1(3k)!(3n 3k)!(3n)!(cid:18)2k(cid:19)(cid:18)k(cid:19)
− −
which is an integer divisible by n in view of Lemma 2.2.
k
(cid:0) (cid:1)
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3 Proof of Theorem 1.3
Let
6k 3k 6(n k) 3(n k)
A = − − .
n,k
(cid:18)3k(cid:19)(cid:18)k (cid:19)(cid:18)3(n k)(cid:19)(cid:18) n k (cid:19)
− −
Then, for n 2 and 0 k < n/2, we have
≥ ≤
A (36nk +31n 36k2 36k 5)(n 2k 1)
n,k
1 = − − − − − 0,
A − (6k +5)(6k +1)(n k)2 ≥
n,k+1
−
i.e., An,k An,k+1. Since An,k = An,n−k, for n 2, we have
≥ ≥
n
6n 3n 6n 3n
2A = 2 A (n+1)A . = (n+1)
n,0 n,k n,0
(cid:18)3n(cid:19)(cid:18)n(cid:19) ≤ ≤ (cid:18)3n(cid:19)(cid:18)n(cid:19)
Xk=0
In other words, the inequality (1.2) holds.
References
[1] I. Gessel, Super ballot numbers. J. Symb. Comput. 14 (1992), 179–194.
[2] Z.-W. Sun, Some new series for 1/π and related congruences, preprint, arXiv:1104.3856.
[3] Z.-W. Sun, Products and sums divisible by central binomial coefficients, Electron. J. Com-
bin. 20(1) (2013), #P9.
[4] Z.-W. Sun, On divisibility of binomial coefficients, J. Austral. Math. Soc., in press.
[5] S.O. Warnnar, A q-rious positivity, Aequat. Math. 81 (2011), 177–183.
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