bounds on π⁢(n)

bounds on

π

⁢

(

n

)

This article shows that

ln

⁡

2

⁢

(

n

ln

⁡

n

)

-

1

≤

π

⁢

(

n

)

≤

8

⁢

ln

⁡

2

⁢

(

n

ln

⁡

n

)

and thus that the function

π

⁢

(

n

)

/

n

ln

⁡

n

is bounded above and below.

Definition 1

If

n

is an

integer

, write

n

=

∏

p

i

r

i

where the

p

i

are distinct primes. Then

o

⁢

r

⁢

d

p

⁢

(

n

)

=

{

r

i

if

⁢

p

=

p

i

0

𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Lemma 1

o

⁢

r

⁢

d

p

⁢

(

n

!

)

=

∑

i

=

1

∞

⌊

n

p

i

⌋

Proof.

Note that the number of

multiples

of

p

≤

n

is simply

⌊

n

p

⌋

. But that does not result in

o

⁢

r

⁢

d

p

⁢

(

n

!

)

since some of those

⌊

n

p

⌋

numbers may have additional factors of

p

. So divide each by

p

; then clearly

o

⁢

r

⁢

d

p

⁢

(

n

!

)

=

⌊

n

p

⌋

+

o

⁢

r

⁢

d

p

⁢

(

⌊

n

p

⌋

!

)

The result follows inductively. Note that the sum is actually finite.

Lemma 2

If

(

m

+

n

n

)

=

∏

q

i

,

q

i

=

p

i

r

i

, all

p

i

distinct, then

∀

i

,

q

i

≤

m

+

n

.

Proof.

Consider

q

1

=

p

c

. Choose

a

such that

p

a

≤

m

+

n

<

p

a

+

1

; it suffices to show

c

≤

a

.

c

=

o

⁢

r

⁢

d

p

⁢

(

(

m

+

n

n

)

)

=

o

⁢

r

⁢

d

p

⁢

(

(

m

+

n

)

!

)

-

o

⁢

r

⁢

d

p

⁢

(

m

!

)

-

o

⁢

r

⁢

d

p

⁢

(

n

!

)

=

∑

1

∞

(

⌊

m

+

n

p

i

⌋

-

⌊

m

p

i

⌋

-

⌊

n

p

i

⌋

)

Now, in general, if

⌊

x

⌋

=

r

,

⌊

y

⌋

=

s

, then

⌊

x

+

y

⌋

=

r

+

s

or

r

+

s

+

1

. So the sum above has at most

a

terms (since

p

a

+

1

>

m

+

n

), each of which is either

0

or

1

, and thus

c

≤

a

.

Theorem 1

(Chebyshev)

If

n

≥

2

π

⁢

(

n

)

≥

ln

⁡

2

⁢

(

n

ln

⁡

n

)

-

1

Proof.

Choose

r

such that

0

<

r

<

n

, and write

(

n

r

)

=

∏

p

i

λ

i

. Each

p

i

λ

i

≤

n

by the preceding lemma, and there are at most

π

⁢

(

n

)

terms on the right-hand side. Thus

n

π

⁢

(

n

)

≥

(

n

r

)

⁢

∀

r

Summing

from

r

=

1

to

r

=

n

-

1

, we get

(

n

-

1

)

⁢

n

π

⁢

(

n

)

≥

∑

r

=

1

n

-

1

(

n

r

)

=

2

n

-

2

But

n

π

⁢

(

n

)

≥

2

since

n

≥

2

, so

n

π

⁢

(

n

)

+

1

≥

2

n

so

(

π

⁢

(

n

)

+

1

)

⁢

ln

⁡

n

≥

n

⁢

ln

⁡

2

and thus

π

⁢

(

n

)

≥

ln

⁡

2

⁢

(

n

ln

⁡

n

)

-

1

Theorem 2

(Chebyshev)

π

⁢

(

n

)

≤

8

⁢

ln

⁡

2

⁢

(

n

ln

⁡

n

)

Proof.

Note that if

p

is prime,

n

≤

p

≤

2

⁢

n

, then

p

divides

(

2

⁢

n

n

)

since

p

occurs in the

numerator

but not in the denominator. Thus

∏

n

≤

p

≤

2

⁢

n

p

divides

(

2

⁢

n

n

)

as well, and thus

∏

n

≤

p

≤

2

⁢

n

p

≤

(

2

⁢

n

n

)

≤

2

2

⁢

n

and therefore

n

π

⁢

(

2

⁢

n

)

-

π

⁢

(

n

)

≤

2

2

⁢

n

Taking logs, we get

π

⁢

(

2

⁢

n

)

-

π

⁢

(

n

)

≤

2

⁢

n

⁢

ln

⁡

2

ln

⁡

n

=

2

⁢

ln

⁡

2

⁢

n

ln

⁡

n

Let

f

⁢

(

n

)

=

π

⁢

(

n

)

⁢

ln

⁡

n

. Then

f

⁢

(

2

⁢

n

)

-

f

⁢

(

n

)

=

π

⁢

(

2

⁢

n

)

⁢

ln

⁡

2

⁢

n

-

π

⁢

(

n

)

⁢

ln

⁡

n

=

(

π

⁢

(

2

⁢

n

)

-

π

⁢

(

n

)

)

⁢

ln

⁡

n

+

π

⁢

(

2

⁢

n

)

⁢

(

ln

⁡

2

⁢

n

-

ln

⁡

n

)

=

(

π

⁢

(

2

⁢

n

)

-

π

⁢

(

n

)

)

⁢

ln

⁡

n

+

π

⁢

(

2

⁢

n

)

⁢

ln

⁡

2

≤

2

⁢

n

⁢

ln

⁡

2

+

2

⁢

n

⁢

ln

⁡

2

=

4

⁢

n

⁢

ln

⁡

2

Then

f

⁢

(

2

t

)

=

(

f

⁢

(

2

t

)

-

f

⁢

(

2

t

-

1

)

)

+

(

f

⁢

(

2

t

-

1

)

-

f

⁢

(

2

t

-

2

)

)

+

…

+

(

f

⁢

(

2

)

-

f

⁢

(

1

)

)

≤

4

⁢

ln

⁡

2

⁢

(

2

t

-

1

+

2

t

-

2

+

…

+

2

0

)

≤

4

⁢

ln

⁡

2

⋅

2

t

Note that

f

is

monotonically increasing

, so if we choose

t

with

2

t

-

1

≤

n

<

2

t

, then

f

⁢

(

n

)

≤

f

⁢

(

2

t

)

≤

4

⁢

ln

⁡

2

⋅

2

t

≤

4

⁢

ln

⁡

2

⋅

2

⁢

n

=

8

⁢

n

⁢

ln

⁡

2

Title

bounds on

π

⁢

(

n

)

Canonical name

BoundsOnpin

Date of creation

2013-03-22 16:23:51

Last modified on

2013-03-22 16:23:51

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11A41

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