The distribution of twin prime numbers is discussed. The research method of corresponding prime number distribution is proposed. The distribution of prime numbers corresponding to integers and composite numbers is discussed. Through the corresponding prime distribution rate of integers and composite numbers, it is found that the corresponding prime distribution rate of composite numbers approaches the corresponding prime distribution rate of integers. The distribution principle of corresponding prime number of composite number is proved. The twin prime distribution theorem is obtained. The number of twin prime numbers is thus obtained. It provides a practical way to study the conjecture of twin prime numbers.
In the long process of human understanding the world, there are many problems that attract many scholars and confuse many wise men.
Mathematicians found that the interval between two adjacent prime numbers is 2. This is a prime number, which is called twin prime number [
Example 1
Twin prime numbers
11 and 13,
59 and 61,
That is, 11 and 11 + 2,
59 and 59 + 2,
In this way, there are infinite primes, which are called twin prime conjecture.
The twin prime distribution problem is the corresponding prime distribution.
Set the integer k, let’s look at the sequence,
k , 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ⋯ k , k + 2 , 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ⋯ k + 2 ,
In k + 2, integers
2, 3, 4, 5, ...
are called the corresponding number of k.
In k + 2, prime numbers
2, 3, 5, 7, ...
are called the corresponding prime numbers of k. That is, the prime number corresponding to an integer.
In k, the composite numbers
4, 6, 8, 9, ...,
correspond to the prime numbers
11, 17, ...,
which are called the corresponding prime numbers of the composite numbers.
The key to the distribution of twin prime numbers is to study the corresponding prime numbers of integers and composite numbers.
Before studying the twin prime distribution problem, let’s look at the prime distribution problem.
Chebyshev function is the focus of mathematicians’ research on prime distribution. Let’s look at the mathematician’s method.
In 1852, the Russian mathematician Chebyshev proposed the function [
θ ( x ) = ∑ p ≤ x log p ~ x , ( x → ∞ )
Get
θ ( x ) = log 2 + log 3 + log 5 + log 7 + log 11 + log 13 + ⋯ + log p
Set x > y, for large y, log p ≤ log x , Obtained by Chebyshev function
( π ( x ) − π ( y ) ) log y < θ ( x ) − θ ( y ) < ( π ( x ) − π ( y ) ) log x
Obviously θ ( x ) − θ ( y ) < θ ( x ) , can get
( π ( x ) − π ( y ) ) log y < θ ( x ) < π ( x ) log x
Set y = xλ, we get [
( π ( x ) − π ( x λ ) ) log x λ < θ ( x ) < π ( x ) log x
Obviously estimate π(xλ) < xλ, get
λ ( π ( x ) − x λ ) log x < θ ( x ) < π ( x ) log x (1.1)
By (1.1) get
λ π ( x ) − x λ x log x < θ ( x ) x < π ( x ) log x x
Can get
λ ( π ( x ) x − x λ x ) log x < θ ( x ) x < π ( x ) log x x
λ < 1, can confirm
x λ x → 0
By π(x) > xλ, we can get
λ π ( x ) log x x < θ ( x ) x < π ( x ) log x x
λ ~1, can get:
lim x → ∞ π ( x ) log x x = lim x → ∞ θ ( x ) x
By Chebyshev function
θ ( x ) ~ x , ( x → ∞ )
The prime theorem is proved
π ( x ) ~ x log x , ( x → ∞ ) (1.2)
Example 2
x = 1018, calculated by (1.2)
π ( 10 18 ) ≈ 24127471216847323
π ( 10 18 ) = 24739954287740860
The method of proof is to construct unequal forms, carry out unequal transformation, delete
x λ x
by limit, and then obtain the prime theorem.
However, some scholars can’t understand it. We can improve the traditional proof method.
We improve the proof method of (1.1).
Might as well set
λ = 1 − 2 log log x log x
Obviously, x tends to infinity, so we can confirm λ → 1 , and we get [
x λ = x ( log x ) 2 (2.1)
By (1.1) and (2.1) can get
λ ( π ( x ) − x ( log x ) 2 ) log x < θ ( x ) < π ( x ) log x
Now, prove π(x) > xλ, You can delete it xλ/x, however, It’s complicated.
We use another method to prove the prime theorem.
Obviously xλ < 2xλ, by (1.1) can get
λ ( π ( x ) − 2 x λ ) log x < θ ( x ) < π ( x ) log x
We get
λ ( π ( x ) − 2 x λ ) log x < θ ( x ) and θ ( x ) < π ( x ) log x
Set positive number a and b, can get
π ( x ) < θ ( x ) λ log x + 2 x λ = a and π ( x ) > θ ( x ) log x = b
We can get
b < π ( x ) < a
Obviously
b < a + b 2 < a
We get the arithmetic mean of π ( x )
π ( x ) ≈ a + b 2
Can get
π ( x ) ≈ 1 2 ( θ ( x ) log x + θ ( x ) λ log x ) + x λ
x tends to infinity, λ → 1 , and (2.1) get
π ( x ) ~ 1 2 ( θ ( x ) log x + θ ( x ) log x ) + x ( log x ) 2
Can get
π ( x ) ~ θ ( x ) log x + x ( log x ) 2
The prime number theorem can be obtained from Chebyshev function
θ ( x ) ~ x , ( x → ∞ )
We get:
π ( x ) ~ x log x + x ( log x ) 2 , ( x → ∞ ) (2.2)
Example 3
x = 1018, calculated by (2.2)
π ( 10 18 ) ≈ 24709606084167120
π ( 10 18 ) = 24739954287740860
The prime theorem can also be obtained in another way.
We use a new method to study the prime theorem.
Set prime number p ≤ x,
μ ( x ) = 2 1 / x + 3 1 / x + 5 1 / x + 7 1 / x + 11 1 / x + 13 1 / x + ⋯ + p 1 / x (3.1)
Set e = 2.718…, by (3.1) can get
e 1 / x + e 1 / x + e 1 / x + e 1 / x + 11 1 / x + 13 1 / x + ⋯ + p 1 / x < 2 1 / x + 3 1 / x + 5 1 / x + 7 1 / x + 11 1 / x + 13 1 / x + ⋯ + p 1 / x
Can get [
π ( x ) e 1 / x < 2 1 / x + 3 1 / x + 5 1 / x + 7 1 / x + 11 1 / x + 13 1 / x + ⋯ + p 1 / x < π ( x ) p 1 / x
π ( x ) e 1 / x < π ( x ) x 1 / x
Can get
π ( x ) e 1 / x < π ( x ) x 1 / x < π ( x ) x 1 / x + x 1 / x
We get
π ( x ) e 1 / x x 1 / x < π ( x ) < π ( x ) x 1 / x + x 1 / x x 1 / x
Arithmetic mean of π ( x ) :
π ( x ) ≈ 1 2 π ( x ) e 1 / x + π ( x ) x 1 / x + x 1 / x x 1 / x
Can get
2 π ( x ) x 1 / x ≈ π ( x ) e 1 / x + π ( x ) x 1 / x + x 1 / x
Get
2 π ( x ) x 1 / x − π ( x ) x 1 / x ≈ π ( x ) e 1 / x + x 1 / x
We can get
π ( x ) x 1 / x ≈ π ( x ) e 1 / x + x 1 / x
Obviously prove
lim x → ∞ π ( x ) x 1 / x π ( x ) e 1 / x + x 1 / x = 1
From this, we can get a new prime theorem
π ( x ) x 1 / x ~ π ( x ) e 1 / x + x 1 / x , ( x → ∞ )
We can get:
π ( x ) x 1 / x − π ( x ) e 1 / x ~ x 1 / x , ( x → ∞ )
Can get
π ( x ) ~ x 1 / x x 1 / x − e 1 / x , ( x → ∞ )
That is
π ( x ) ~ 1 x 1 / x − e 1 / x , ( x → ∞ ) (3.2)
Example 4
x = 1018, calculated by (3.2)
π ( 10 18 ) ≈ 24723998785919976
π ( 10 18 ) = 24739954287740860
The problem of prime distribution was discussed earlier. We continue to discuss the distribution of twin prime numbers.
The twin prime distribution problem is the corresponding prime distribution.
Set Integer x = 16, let’s look at the sequence of k and k + 2
k , 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 , k + 2 , 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ,
k + 2 is called the corresponding number of k
In k + 2, prime
2, 3, 5, 7, 11, 13,
a total of 6. The table shows
π ( 16 ) = 6
and the proportion of prime numbers is:
π ( 16 ) 16 = 6 16
Generally [
ρ x = π ( x ) x (4.1)
Here (4.2) is called the prime distribution rate of integers.
Let’s look at the prime number corresponding to the composite number.
Set the composite number c. for the convenience of discussion, temporarily set 0 as the composite number to see the sequence of composite numbers c and c + 2
c , 0 4 6 8 9 10 12 14 , c + 2 , 2 6 8 10 11 12 14 16 ,
In c, there are 8 composite numbers, and the table is F = 8.
In c + 2, there are 2 primes, 2, 11. The table is c (16) = 2.
The proportion of prime numbers is:
c ( 16 ) F = 2 8
Generally:
ρ c = c ( x ) F (4.2)
Here (4.2) is called the prime distribution rate of composite numbers.
Example 5
(4.1) and (4.2) Partial calculation:
A composite number is part of an integer. The greater the proportion of the composite number in the integer, the closer the distribution rate of the corresponding prime number of the composite number is to the distribution rate of the corresponding prime number of the integer.
Now let’s discuss the corresponding prime of prime.
Let the prime number p. for the convenience of discussion, temporarily set 1 as the prime number to see the sequence of prime numbers p and p + 2
p , 1 2 3 5 7 11 13 , p + 2 , 3 4 5 7 9 13 15 ,
In p + 2, there are four primes: 3, 5, 7, 13. The table is
L ( 16 ) = 4
Represents the number of primes in p + 2. Can get
L ( 16 ) = π ( 16 ) − c ( 16 ) = 4
According to statistics, if x is larger, then L(x) is more.
If we can prove that L(x) is infinite, we can prove the twin prime conjecture.
You can get from the front
π ( 16 ) = c ( 16 ) + L ( 16 ) = 2 + 4
Generally:
π ( x ) = c ( x ) + L ( x ) (5.1)
By (5.1) can get
c ( x ) = π ( x ) − L ( x )
Set number of set composite numbers
F = x − π ( x )
By (4.2) we can get
c ( x ) F = π ( x ) − L ( x ) x − π ( x )
π ( x ) − L ( x ) x − π ( x ) = π ( x ) ( 1 − L ( x ) π ( x ) ) x ( 1 − π ( x ) x ) = c ( x ) F
We get
π ( x ) x F c ( x ) = 1 − π ( x ) x 1 − L ( x ) π ( x )
Set x tends to infinity, so we can confirm [
π ( x ) x → 0 and L ( x ) π ( x ) → 0
We can get
lim x → ∞ π ( x ) x F c ( x ) = 1
Set λ, can get
π ( x ) x F c ( x ) = λ
Example 6
By λ Partial calculation:
we get:
c ( x ) = F π ( x ) λ x (5.2)
Here (5.2) is called the corresponding prime number principle of composite numbers.
We continue to discuss the distribution of twin prime numbers.
The number of twin prime numbers is obtained from (5.1)
L ( x ) = π ( x ) − c ( x ) (6.1)
By (5.2) and (6.1), we get [
L ( x ) = π ( x ) − F π ( x ) λ x (6.2)
Here (6.2) is called the twin prime distribution theorem [
Example 7
Calculated by part (6.2):
We confirm the number of twin prime numbers
From (6.2), we can get:
L ( x ) = π ( x ) ( 1 − F λ x ) = π ( x ) ( λ x − F λ x ) (7.1)
Can confirm
L ( x ) > L λ ( x ) = π ( x ) ( λ x − F ( λ + 1 ) x )
Example 8
By (7.1) Partial calculation:
x tends to infinity, λ → 1 , can get [
L ( x ) > π ( x ) ( x − F 2 x )
Set F = x − π ( x ) , we get
L ( x ) > π 2 ( x ) 2 x , ( x → ∞ )
Set F = x − π ( x ) , and π ( x ) > x / log x , we get:
L ( x ) > x 2 ( log x ) 2 , ( x → ∞ ) (7.2)
From (7.2), L(x) can be confirmed there is an infinite number.
The twin prime conjecture is correct.
We discuss the prime distribution
π ( x ) ~ 1 x 1 / x − e 1 / x , ( x → ∞ )
The twin prime conjecture is discussed
L ( x ) > x 2 ( log x ) 2 , ( x → ∞ )
The twin prime conjecture is correct.
The author declares no conflicts of interest regarding the publication of this paper.
Liu, D. (2022) Twin Prime Distribution Problem. Journal of Applied Mathematics and Physics, 10, 1352-1361. https://doi.org/10.4236/jamp.2022.104095