Proof of a conjecture related to divisibility properties of binomial coefficients

0.08 MB·

by  Quan-Hui Yang

Proof of a conjecture related to divisibility properties
of binomial coefficients∗
4 Quan-Hui Yang†
1
0
2
School of Mathematics and Statistics,

n
a
J
Nanjing University of Information Science and Technology,
8

]
T Nanjing 210044, P. R. China
N
.
h
t
a
Abstract
m
[
Let a,b and n be positive integers with a > b. In this note, we

2
v prove that
8
2bn 2an an
0
(2bn+1)(2bn+3) 3(a−b)(3a−b) .
1 (cid:18)bn(cid:19)(cid:12) (cid:18)an(cid:19)(cid:18)bn(cid:19)
1 (cid:12)
(cid:12)
. This confirms a recent conjectur(cid:12)e of Amdeberhan and Moll.
1
0
2010 Mathematics Subject Classifications: 11B65, 05A10
4
1
Keywords: binomial coefficients, p-adic order, divisibility proper-
:
v
i ties
X
r
a
1 Introduction
Let Z denote the set of all integers. In 2009, Bober [1] determined all cases
such that
(a n)!···(a n)!
1 k
∈ Z,
(b n)!···(b n)!
1 k+1
∗This work was supported by the National Natural Science Foundation of China,
Grant No. 11371195.
†Email: [email protected].
1
where a 6= b for all s,t, a = b and gcd(a ,...,a ,b ,...,b ) = 1.
s t s t 1 k 1 k+1
P P
Recently, Z.-W. Sun [12, 13] studied divisibility properties of binomial
coefficients and obtained some interesting results. For example,
2n 6n 3n
2(2n+1) ,
(cid:18)n(cid:19)(cid:12)(cid:18)3n(cid:19)(cid:18)n(cid:19)
(cid:12)
(cid:12)
(cid:12)
3n 15n 5n−1
(10n+1) .
(cid:18)n(cid:19)(cid:12)(cid:18)5n(cid:19)(cid:18)n−1(cid:19)
(cid:12)
(cid:12)
Later, GuoandKrattenthaler(see[5,(cid:12)7])gotsomesimilardivisibilityresults.
For related results, one can refer to [2]-[4] and [8]-[11].
Let
6n 3n 15n 5n−1
S = 3n n and t = 5n n−1 .
n 2(2(cid:0)n+(cid:1)(cid:0)1)(cid:1)2n n (1(cid:0)0n(cid:1)+(cid:0)1) 3(cid:1)n
n n
(cid:0) (cid:1) (cid:0) (cid:1)
In [6], Guo proved the following Sun’s conjectures.
Theorem A. (See [12, Conjecture 3(i)].) Let n be a positive integer. Then
3S ≡ 0 (mod 2n+3).
n
Theorem B. (See [13, Conjecture 1.3].) Let n be a positive integer. Then
21t ≡ 0 (mod 10n+3).
n
Some other results are also proved.
T. Amdeberhan and V. H. Moll proposed the following conjecture (See
Guo [6, Conjecture 7.1]).
Conjecture 1. Let a,b and n be positive integers with a > b. Then
2bn 2an an
(2bn+1)(2bn+3) 3(a−b)(3a−b) .
(cid:18)bn(cid:19)(cid:12) (cid:18)an(cid:19)(cid:18)bn(cid:19)
(cid:12)
(cid:12)
In this note, we give the proof o(cid:12)f this conjecture.
Theorem 1. Conjecture 1 is true.
Remark 1. Let a = 3 and b = 1. Then Theorem A follows from Theorem
1 immediately.
2
2 Proofs
For an integer n and a prime p, we write pkkn if pk|n and pk+1 ∤ n. We use
ν (n) to denote such integer k. It is well known that
p
∞
n
(1) ν (n!) = ,
p (cid:22)pi(cid:23)
Xi=1
where ⌊x⌋ denotes the greatest integer not exceeding x.
Before the proof of Theorem 1, we give the following lemma.
Lemma 1. Let x and y be two real numbers. Then ⌊2x⌋ + ⌊y⌋ ≥ ⌊x⌋ +
⌊x−y⌋+⌊2y⌋.
Proof. Noting that 2x+y = x+(x−y)+2y, we only need to prove that
{2x}+{y} ≤ {x}+{x−y}+{2y}, where {z} denotes the fractional part
of z. We can prove it by comparing {x} and {y} with 1/2. We leave the
proof to the reader.
Proof of Theorem 1. Let
2an an 2bn (2an)!(bn)!
T(a,b,n) := = .
(cid:18)an(cid:19)(cid:18)bn(cid:19)(cid:30)(cid:18)bn(cid:19) (an)!(an−bn)!(2bn)!
By (1), for any prime p, we have
∞
2an bn an an−bn 2bn
ν (T(a,b,n)) = + − − − .
p (cid:18)(cid:22) pi (cid:23) (cid:22)pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23)(cid:19)
Xi=1
By Lemma 1, it follows that each term of ν (T(a,b,n)) is nonnegative.
p
Hence ν (T(a,b,n)) ≥ 0. Therefore, T(a,b,n) ∈ Z.
p
Since gcd(2bn+1,2bn+3) = 1, it suffices to prove that
2bn+1|3(a−b)(3a−b)T(a,b,n)
and
2bn+3|3(a−b)(3a−b)T(a,b,n).
3
Now we first prove the latter part. Suppose that pαk2bn+3 with α ≥ 1.
Then we shall prove
(2) pα|3(a−b)(3a−b)T(a,b,n).
Let pβka−b and pγk3a−b with β ≥ 0 and γ ≥ 0. Write τ = max{β,γ}.
If α ≤ τ, then (2) clearly holds. Now we assume α > τ.
Suppose that p ≥ 5. Next we prove
2an bn an an−bn 2bn
+ − − − = 1
(cid:22) pi (cid:23) (cid:22)pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23)
for i = τ + 1,τ + 2,...,α. Noting that p|2bn + 3 and p ≥ 5, we have
gcd(p,n) = 1. Otherwise, we have p|3, a contradiction.
By pαk2bn+3, it follows that 2bn ≡ pα−3 (mod pα) and bn ≡ (pα−3)/2
(mod pα).
Take an integer i ∈ {τ +1,τ +2,...,α}. Then 2bn ≡ pi −3 (mod pi)
and bn ≡ (pi −3)/2 (mod pi). Now we divide into several cases according
to the value of an (mod pi).
Case 1. an ≡ t (mod pi) with 0 ≤ t < (pi − 3)/2. It follows that
2an ≡ 2t (mod pi) and 0 ≤ 2t < pi −3. We also have
an−bn ≡ t−(pi −3)/2+pi (mod pi),
where 0 ≤ t−(pi −3)/2+pi < pi. Hence
2an bn an an−bn 2bn
+ − − −
(cid:22) pi (cid:23) (cid:22)pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23)
2an−2t bn−(pi −3)/2 an−t
= + −
pi pi pi
an−bn−(t−(pi −3)/2+pi) 2bn−(pi −3)
− −
(cid:18) pi (cid:19) pi
= 1.
Case 2. an ≡ (pi − 3)/2 (mod pi). Clearly, an − bn ≡ 0 (mod pi).
Since gcd(p,n) = 1, we have pi|a−b. However, pβka−b and β ≤ τ < i, a
contradiction.
4
Case 3. an ≡ (pi −1)/2 (mod pi). It follows that
3(pi −1) (pi −3)
3an−bn ≡ − ≡ 0 (mod pi).
2 2
By gcd(p,n) = 1, we have pi|3a−b. This contradicts pγk3a−b and γ < i.
Case 4. an ≡ t (mod pi) with (pi +1)/2 ≤ t < pi. It follows that
2an ≡ 2t−pi (mod pi), 0 ≤ 2t−pi < pi.
We also have
an−bn ≡ t−(pi −3)/2 (mod pi), 0 ≤ t−(pi −3)/2 < pi.
Hence
2an bn an an−bn 2bn
+ − − −
(cid:22) pi (cid:23) (cid:22)pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23) (cid:22) pi (cid:23)
2an−(2t−pi) bn−(pi −3)/2 an−t
= + −
pi pi pi
an−bn−(t−(pi −3)/2) 2bn−(pi −3)
− −
(cid:18) pi (cid:19) pi
= 1.
Therefore, ν (T(a,b,n)) ≥ α−τ, and then
p
ν (3(a−b)(3a−b)T(a,b,n)) ≥ α.
p
That is, (2) holds.
Now we deal with the case p = 3. If 9|n, then 3|2bn+3 and 9 ∤ 2bn+3.
It follows that α = 1, and then (2) clearly holds. If 9 ∤ n, then we can follow
the proof of the case p ≥ 5 above. In Case 2, by an − bn ≡ 0 (mod 3i),
we have 3i−1|a−b. In Case 3, we have 3i−1|3a−b. So, if i ≥ τ + 2, then
i−1 ≥ τ +1. It is a contradiction in both cases. Hence
2an bn an an−bn 2bn
+ − − − = 1
(cid:22) 3i (cid:23) (cid:22)3i (cid:23) 3i (cid:22) 3i (cid:23) (cid:22) 3i (cid:23)
j k
for i = τ + 2,τ + 3,...,α. It follows that ν (T(a,b,n)) ≥ α − τ − 1, and
3
then
ν (3(a−b)(3a−b)T(a,b,n)) ≥ α.
3
5
That is, (2) also holds. Hence, 2bn+3|3(a−b)(3a−b)T(a,b,n).
The proof of 2bn+1|3(a−b)(3a−b)T(a,b,n) is very similar. We omit
it here.
This completes the proof of Theorem 1.
References
[1] J.W. Bober, Factorial ratios, hypergeometric series, and a family of
step functions, J. Lond. Math. Soc. 79 (2009), 422-444.
[2] N.J. Calkin, Factors of sums of powers of binomial coefficients, Acta
Arith. 86 (1998), 17-26.
[3] H.Q. Cao and H. Pan, Factors of alternating binomial sums, Adv. in
Appl. Math. 45 (2010), 96-107.
[4] N. Fine, Binomial coefficients modulo a prime, Amer. Math. Monthly
54 (1947), 589-592.
[5] V.J.W. Guo, Proof of Suns conjecture on the divisibility of certain bi-
nomial sums, Electron. J. Combin. 20(4) (2013), #P20.
[6] V.J.W. Guo, Proof of two divisibility properties of binomial coefficients
conjectured by Z.-W. Sun, available at http://arxiv.org/abs/1312.7548,
2013.
[7] V.J.W. Guo and C. Krattenthaler, Some divisibility properties of bino-
mial and q-binomial coefficients, J. Number Theory 135 (2014), 167-
184.
[8] V.J.W. Guo, F. Jouhet and J. Zeng, Factors of alternating sums
of products of binomial and q-binomial coefficients, Acta Arith. 127
(2007), 17-31.
6
[9] V.J.W. Guo and J. Zeng, Factors of binomial sums from the Catalan
triangle, J. Number Theory 130 (2010), 172-186.
[10] V.J.W. Guo and J. Zeng, Factors of sums and alternating sums involv-
ing binomial coefficients and powers of integers, Int. Number Theory 7
(2011), 1959-1976.
[11] M. Razpet, On divisibility of binomial coefficients, Discrete Math. 135
(1994), 377-379.
[12] Z.-W. Sun, On divisibility of binomial coefficients, J. Austral. Math.
Soc. 93 (2012), 189-201.
[13] Z.-W. Sun, Products and sums divisible by central binomial coefficients,
Electron. J. Combin. 20(1) (2013), #P9.
7