Primes in The Interval (2n, 3n)

Int. J. Contemp. Math. Sci., Vol. 1, 2006, no.
13, 617 - 621

Primes in the Interval [2n, 3n]
M. El Bachraoui
School of Science and Engineering
Al Akhawayn University in Ifrane
P.O.Box 2096, Ifrane 53000, Morocco
m.elbachraoui@aui.ma
Abstract. Is it true that for all integer n > 1 and k n there exists a
prime number in the interval [kn, (k + 1)n]? The case k = 1 is the Bertrands
postulate which was proved for the rst time by P. L. Chebyshev in 1850, and
simplied later by P. Erd os in 1932, see [2]. The present paper deals with the
case k = 2. A positive answer to the problem for any k n implies a positive
answer to the old problem whether there is always a prime in the interval
[n
2
, n
2
+ n], see [1, p. 11].
Keywords: prime numbers
Mathematics Subject Classication: 51-01
1. the result
Throughout the paper ln(x) is the logarithm with base e of x and (x) is
the number of prime numbers not greater than x. We let n run through the
natural numbers and p through the primes.
Lemma 1.1. The following inequalities hold:
1. If n is even then
3n
2
n
<
6.75
n
.
2. If n is even such that n > 152 then
3n
2
n
>
6.5
n
.
3. If n is odd and n > 7 then
3n+1
2
n
<
6.75
n1
.
618 M. El Bachraoui
4. If n > 945 then
(
6.5
27
)
n
> (3n)
3n
2
.
Proof. (1,2) By induction on n. We have
3
2
< 6.75 and
3154
2
154
>
6.5
154
.
Assume now that the two inequalities hold for
3n
2n
. Then
3n + 3
2n + 2
3n
2n
(3n + 1)(3n + 2)(3n + 3)

(n + 1)(2n + 1)(2n + 2)
=
3n
2n
3(3n + 1)(3n + 2)
(2n + 1)(2n + 2)
=
3n
2n
27n
2
+ 27n + 6
4n
2
+ 6n + 2
.
It now suces to note that for all n
27n
2
+ 27n + 6
4n
2
+ 6n + 2
< 6.75
and for all n > 12
6.5 <
27n
2
+ 27n + 6
4n
2
+ 6n + 2
.
(3) By induction on n. We have
14
9
< (6.75)
4
. Assume now that the result is
true for
3n+2
2n+1
. Then
3n + 5
2n + 3
3n + 2
2n + 1
3(3n + 4)(3n + 5)
2(n + 2)(2n + 3)
< (6.75)
n
6.75
= (6.75)
n+1
.
(4) Note that the following three inequalities are equivalent:
(
6.5
27
)
n
> (3n)
3n
2
nln
6.5
27
>
3n
2
ln 3n
2
3
ln
6.5
27
>
ln 3n
n
.
Then the result follows since the function
ln3x
x
is decreasing and
2
3
ln
6.5
27
>
ln(3 946)
946
.
Primes in the Interval [2n, 3n] 619
Lemma 1.2. 1. If n is even then
n
2
<p
3n
4
p
n<p
3n
2
p <
3n
2
n
.
2. If n is odd then
n+1
2
<p
3n
4
p
n<p
3n+1
2
p <
3n+1
2
n
.
Proof. (1) We have
3n
2
n
=
(n + 1)
3n
2
n
2
!
. (1.1)
Then clearly
n<p
3n
2
p divides
3n
2
n
. Furthermore, if
n
2
< p
3n
4
then 2p
occurs in the numerator of (1.1) but p does not occur in the denominator.
Then after simplication of 2p with an even number from the denominator
we get the prime factor p in
3n
2
n
. Thus
n
2
<p
3n
4
p divides
3n
2
n
too and the

required inequality follows.
(2) Similar to the rst part.
To prove Bertrands postulate, P. Erd os needed to check the result for n =
2, . . . , 113, refer to [3, p. 173]. Our theorem requires a separate check for
n = 2, . . . , 945 but we omitted to list them for reasons of space.
Theorem 1.3. For any positive integer n > 1 there is a prime number between
2n and 3n.
Proof. It can be checked (using Mathematica for instance) that for n = 2, . . . , 945
there is always a prime between 2n and 3n. Now let n > 945. As
3n
2n
=
(2n + 1)(2n + 2) 3n
1 2 n
, (1.2)
the product of primes between 2n and 3n, if there are any, divides
3n
2n
. Fol-
lowing the notation used in [3], we let
T
1
=
3n
p
(p)
, T
2
=
3n<p2n
p
(p)
, T
3
=
2n+1p3n
p,
such that
3n
2n
= T
1
T
2
T
3
. (1.3)
The prime decomposition of
3n
2n
implies that the powers in T

2
are less than 2,
see [3, p. 24] for the prime decomposition of
n
j
. Moreover, we claim that if a

prime p satises
3n
4
< p n then its power in T
2
is 0. Clearly, a prime p with
620 M. El Bachraoui
this condition appears in the denominator of (1.2) but 2p does not, and 3p
appears in the numerator of (1.2) but 4p does not. This way p cancels and the
claim follows. Furthermore, if
3n
2
< p 2n then its power in T
2
is 0 because
such a prime p is neither in the denominator nor in the numerator of (1.2) and
2p > 3n. Now by Lemma 1.2 and the fact that
px
p < 4
x
, refer to [3, p.
167], we have that:
If n is even then
T
2
<
3n<p
n
2
p
n
2
<p
3n
4
p
n<p
3n
2
p
< 4
n
2
3n
2
n
< 4
n
2
(6.75)
n
2
=
27
n
.
(1.4)
If n is odd then
T
2
<
3n<p
n+1
2
p
n+1
2
<p
3n
4
p
n<p
3n+1
2
p
< 4
n+1
2
3n+1
2
n
< 4
n+1
2
(6.75)
n1
2
= 4
27
n1
<
27
n
.
(1.5)
Thus by (1.4) and (1.5) we nd the following upper bound for T
2
:
T
2
<
27
n
. (1.6)
In addition, the prime decomposition of
3n
2n
yields the following upper bound

for T
1
:
T
1
< (3n)
(
3n)
. (1.7)
See [3, p. 24]. Then by virtue of Lemma 1.1(2), equality (1.3), and the
inequalities(1.6), and (1.7) we nd
(6.5)
n
< T
1
T
2
T
3
< (3n)
(
3n)
27
n
T
3
,
which implies that
T
3
>
6.5
27
n
1
(3n)
(
3n)
.
Primes in the Interval [2n, 3n] 621
But (
3n)
3n
2
. Then
T
3
>
6.5
27
n
1
(3n)
3n/2
> 1, (1.8)
where the second inequality follows by Lemma 1.1(4). Consequently, the prod-
uct T
3
of primes between 2n and 3n is greater than 1 and therefore the existence
of such numbers follows.
Corollary 1.4. For any positive integer n 2 there exists a prime number p
satisfying
n < p <
3(n + 1)
2
.
Proof. The result is clear for n = 2. For even n > 2 the result follows by
Theorem 1.3. Assume now that n = 2k + 1 for a positive integer k 1. Then
by Theorem 1.3 there is a prime p satisfying
2(k + 1) < p < 3(k + 1) =
3(n + 1)
2
,
and the result follows.
References
[1] T. M. Apostol. Introduction to Analytic Number Theory. Undergraduate Texts in Math-
ematics. Springer, 1998. xii+338 pp.
[2] P. Erdos. Beweis eines satzes von tschebyschef. Acta Litt. Univ. Sci., Szeged, Sect. Math.,
5:194198, 1932.
[3] P. Erdos and J. Suranyi. Topics in the theory of numbers. Undergraduate Texts in Math-
ematics. Springer Verlag, 2003. viii+287 pp.
Received: May 16, 2006

Primes in The Interval (2n, 3n)

Uploaded by

Copyright:

Available Formats

Primes in The Interval (2n, 3n)

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Primes in The Interval (2n, 3n)

Uploaded by

Copyright:

Available Formats

Int. J. Contemp. Math. Sci., Vol. 1, 2006, no.

13, 617 - 621

< 6.75 and

(3n + 1)(3n + 2)(3n + 3)

too and the

implies that the powers in T

. Moreover, we claim that if a

yields the following upper bound

You might also like