【原】Every even number not less than 6 is the sum of tw

Every even number not less than 6 is the sum of two odd primes

Cui Kun

266200,Jimo, Qingdao, Shandong, China  E-mail: [e-mail]

Abstract: ：Every even number not less than 6 is the sum of two odd primes,

abbreviated as (1 + 1),expressed in，after in-depth study, it is found that:

【1】Truth formula equation: =+2π- ，

【2】Odd composite logarithm density theorem： =

【3】The inference Q = 3 + q1 + q2 of the theorem of three primes gives a general proof of

    ≥ 1

【4】）Is an increasing function, inference:)≥

 and extended to  ≥  ≥1,

【5】According to the double sieve method and prime theorem,

It can be further deduced that: r2() ≥ [ ] ≥1

key words:

Prime number theorem; Three prime number theorem; Increasing function

1: Re agree that 1 is an odd prime number, and the truth formula equation is:

  =C+2π-  

Number of odd sums in even :

There are  different odd numbers in total,there are  different odd sums:

There are four odd sum classifications related to:

[1](odd prime, odd prime), abbreviated as 1 + 1, so that there are

[2] (odd composite number, odd composite number), abbreviation: C + C,

so that there are  

[3](odd prime, odd composite), abbreviation: 1 + C,so that there are  

[4](odd composite number, odd prime number), abbreviation: C + 1,

so that there are W

According to its symmetry:  =W

Let  have (π - 1) + 1 = π different odd primes, then:

 +C+W+M=  .......................〈1〉

M= π- ................................〈2〉

M=W.......................................〈3〉

With the above formulas (1), (2) and (3), it is obtained:

 =C+2π-  

Where  and C are natural numbers, and π is a non-zero natural number,

Even number ≥ 6

For example:30

π(30)=10,namely：1，3，5，7，11，13，17，19，23，29

C(30)=3,namely：（9，21），（15，15），（21，9）

M(30)=2,namely：（3，27）,（5，25）

W(30)=2,namely：（27，3），（25，5）

r2(30)=8,namely:（1，29），（7，23），（11，19），（13，17），

（17，13），（19，11），（23，7），（29，1）

r2(30)=C(30)+2π(30)- = 3+2*10-15=8

2.Lemma: =

prove：   + -

  + -

 =0，≤)

 =0

  + -

0= + 0 -

 =

3. ≥1

prove：

According to the theorem of three prime numbers thoroughly proved by Peruvian mathematician Harold heovgert:

Each odd number greater than or equal to 9 is the sum of three odd primes, and each odd primes can be reused.

It is expressed by the following formula:

Q is each odd number ≥ 9, odd prime number: q1 ≥ 3, q2 ≥ 3, q3 ≥ 3,

Then Q = q1 + q2 + q3

According to the law of combination of addition and exchange, let q1 ≥ q2 ≥ q3 ≥ 3, then:

Q+3=q1+q2+q3+3

Q+3-q3=3+q1+q2

Obviously, when there is and only q3 = 3, Q = 3 + q1 + q2, which is the inference of the theorem of three primes

Therefore, every odd number Q greater than or equal to 9 is the sum of 3 + two odd primes.

According to the inference Q = 3 + q1 + q2, each even number not less than 6 = Q-3 = q1 + q2

Therefore, "every even number not less than 6 is the sum of two odd primes",

That is, there is always ≥1

For example, take any large odd number: 309, please prove that 306 is the sum of two odd primes.

Proof: according to the theorem of three prime numbers, we have: 309 = q1 + q2 + q3

According to the law of additive commutative Association, there must be a problem: three prime numbers:

 q1 ≥ q2 ≥ q3 ≥ 3

Then:

309+3=3+q1+q2+q3

309+3-q3=3+q1+q2

Obviously, when q3 = 3, 309 = 3 + q1 + q2

Then:306=q1+q2

) is an increasing function

Proof:

)=C)+2π) -

= 2C()+4π() - 2 ()

When x+1,then there：

C()+2π()-()≡ *[C()+2π()-()]

When ，

【1】According to lemma：

 = ..................<1>

 = .....................<2>

<1>/<2>：

 = 1

C() ~C()……...（a）

【2】According to the prime theorem：

  =1........<3>

  =1...........<4>

<3>/<4>：

  =1.

~)…………(b)

【3】

 ≡ 1

 = 1

  = 1..............（c)

Substitute equations (a) and (b) into equation (c),

And order =,

The numerator and denominator on the left of the equation are the same as divided by -

 - =1-

 = 1

~*

>>1，≥，inference：≥

Therefore:)=C)+2π) -  is an increasing function

And inference:  ≥ extended to ≥ [] ≥1

Since even numbers can be divided into square even numbers and non square even numbers:

【1】≥, even ≥ 6

【2】≥ [],even number ≥ 6

5. According to the double sieve method and prime theorem,

It is further deduced that: r2() ≥ [  ] ≥1

Proof:

For conjugate reciprocal sequence A, B:

A:｛1,3,5,7,9,……,（-1）｝

B:｛（-1）,……,9,7,5,3,1｝

Steps of double screen method:

Establish the following reciprocal sequence:

The first term is 1, the last term is -1, and the tolerance is 2

The first term is -1 and the last term is 1,

Sequence B of equal differences with tolerance - 2

Obviously,= A + B

The odd prime set {Pr} is obtained according to the Ehrlich sieve method:

｛1，3，5，…，pr｝,pr<

In order to obtain the (1 + 1) tabular method number of even , carry out step-by-step operation according to the double screen method:

Step 1: use 3 pairs of sieves to obtain the true residual ratio m1

Step 2: use 5 pairs of sieves to obtain the true residual ratio m2

Step 3:The remaining reciprocal sequence is sieved with 7 pairs to obtain the true residual ratio m3

…

And so on to:

Step r: Use pr double screen to obtain the true residual ratio mr after the remaining reciprocal sequence

In this way, the double screen method (1 + 1) for even number n is completed. According to the multiplication principle, there are:

r2()=（/2）*m1*m2*m3*…*mr

r2()=(/2)∏mr

For example：70

[√70]=8，{Pr}={1,3,5,7},

3|/70，m1=13/35

5|70, m2=10/13

7|70, m3=10/10

According to the truth formula：

r2(70)

=(70/2)*m1*m2*m3

=35*13/35*10/13*10/10

=10

r2(70)=10

The first step of the double screen method: first screen the sequence of A. according to the prime number theorem, there are at least [  ] odd primes in A,

That is, there are at least [  ] odd primes in the conjugate reciprocal sequence AB at this time

Step 2: filter the B-sequence, and the sieve is the same   

Then, according to the multiplication principle, it is deduced that there are at least :

r2() ≥ [  ] ≥1,odd prime number in the conjugate sequence AB

For example: 70

A	1	3	5	7	9	11	13	15	17	19	21	23	25	27	29	31	33	35	37	39	41	43	45	47	49	51	53	55	57	59	61	63	65	67	69
B	69	67	65	63	61	59	57	55	53	51	49	47	45	43	41	39	37	35	33	31	29	27	25	23	21	19	17	15	13	11	9	7	5	3	1

Step 1: filter the sequence of a first. There are at least  [  ]  = [  ] = 16

odd primes in A, and π (70) = 19,

That is, there are at least [  ]  = [  ] = 16

odd primes in the conjugate reciprocal sequence AB at this time.

Step 2: screen the b sequence, and the sieve is the same    , so it is deduced that there are at least:

r2 (70) ≥ [70 / (ln70) ^ 2] = 3 ,odd primes,r2(70)=10

A	1	3	5	7	9	11	13	15	17	19	21	23	25	27	29	31	33	35	37	39	41	43	45	47	49	51	53	55	57	59	61	63	65	67	69
B	69	67	65	63	61	59	57	55	53	51	49	47	45	43	41	39	37	35	33	31	29	27	25	23	21	19	17	15	13	11	9	7	5	3	1

Therefore,  r2( ) ≥[  ] ≥1

 odd prime number, that is,

each even number n not less than 6 is the sum of two odd prime numbers.

6. Conclusion:

【1】=C+2π-

【2】 =

【3】Inference of three prime theorem：Q=3+q1+q2

【4】)=C)+2π) -  Is an increasing function

~*，inference：≥

【5】inference：≥ ,further expand into≥ [] ≥1

【6】 r2( )≥ [  ] ≥1

7. Reference

[1] Major Arcs for Goldbach's Theorem. Arxiv [Reference date 2013-12-18]

[2] Minor arcs for Goldbach's problem．Arxiv [Reference date 2013-12-18]