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Goldbach conjecture solution?
The proof process of Goldbach 1+ 1 conjecture;

N positive odd numbers from 1 to 2N- 1 are made into a positive odd spectrum 2NG, which contains only three computable numbers: 1 and 1.

The iP first odd set iPc constructed by vP with suboptimal less than √2N is k term, 3. There is no subprime mortgage greater than √2N.

Set wP to 1 From this, we can get the number spectrum of two 2NG parallel lines merged into n pairs, and the sum of the two numbers is equal to 2N, and the result is only

There are three kinds of computable number pairs: 1, edge number pair 1 ~ 2n- 1 and 2n- 1 are always 2 pairs, and 2, iPc number pair iPc2 is k.

Item, 3, wP+wP=2N is 1 item for wP2. If the distribution ratio is used to study the properties of N-2 logarithm after removing the edge number pairs, then

The distribution ratio of iPc2 is 1Pc2L, 2Pc2L, ... …kPc2L, and the same module is the joint fractional series1∨ 2/IP * ` i-1` ∏1p ∈ 3 (.

Through mathematical induction, it is proved that the (series) of K continued fraction series and the distribution ratio wP2L with WP2 are "1" each other.

1-k∑ 1P∈3: 1∨2/iP * ` I- 1 `∏ 1P∈3( 1- 1∨2/vP)= k∏ 1P∈3:( 1- 1∨

Where KP < √ 2N, iP can be divided by 2N and taken as 1∨2/iP= 1/iP, and vice versa. Displays the right side of the (1) equal sign.

If wp2l > 1/KP, the Goldbach conjecture of even number 1+ 1 is proved theoretically.

If the contents of number pairs in the number spectrum wP2 are represented by G(2N), then from G(6)= 1, G(8)=2, G( 10)≈ 1, G( 12)≈2, …

To 2N→∞ can be estimated.

g(2N)≈(N-2)×k∏ 1P∈3( 1- 1∨2/vP)_(2).

According to (2), the total number of logarithms starting from 6, the real trace number of G(2N) and the theoretical calculation number contained in even numbers can all be programmed by computer.

Do the following collection, calculation, comparison and verification table ↓

2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N 2N。

The ``````` contained before the ````` contained in `````.

Calculate G(2N)≈ with the original number spectrum ```````````````````````````````````````````````````.

Value `````````````````` Total number of instances ``````` (n-2) × k ∏ i ∈1p (1-2 ∨1)

``````````````` Logarithm' expression ``````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

06 ```1````````` None ````````1````````````````1× (1-0) ````` =

08 ` ` 2 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` `×( 1-0)` ` ` ` = = 2

10 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` `

12``4`````3````3`````2`````2≈`````4×( 1- 1/3)≈2

14 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` `

16 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` `( 1-2/3)= = 2。

18``7`````3````3`````4`````4≈`````7×( 1- 1/3)≈4

20 ``````````````````````````` none ````````````` 2 ≈ 8× (1-2/3) ≈ 2.

22 ````` 9````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

24`` 10````3````3`````6`````6≈```` 10×( 1- 1/3)≈6

26 `````11`````````````````````````````````````````````````````````````````````````````````````````````` ```````````1× (/kloc-

28 `````12 ```````````` None `````````12/3 (1-2/5).

30`` 13```3`5``3`5````6`````6≈```` 13×( 1- 1/3)( 1- 1/5)≈6

32 ```14 ```````````` None `````````14× (1-2/3) (1-2/5).

34 ````15 ```````````` None `````````15× (1-2/3) (1-2/5).

36`` 16```3`5```3`````6`````6≈```` 16×( 1- 1/3)( 1-2/5)≈6

38 ```17 ```````````` None `````````17× (1-2/3) (1-2/5).

40`` 18```3`5```5`````6`````4≈```` 18×( 1-2/3)( 1- 1/5)≈4

42`` 19```3`5```3`````8`````6≈```` 19×( 1- 1/3)( 1-2/5)≈6

44 ````` 20 ```````````` None ```````````` 4 ≈````` 20× (1-2/3) (1-2/5) = = 4.

46 ```` 21``````````` None ````````````` 21× (1-2/3) (1-2/5).

48``22```3`5```3````` 10````8≈````22×( 1- 1/3)( 1-2/5)≈8

50``23``3`5`7``5`````8`````4≈````23×( 1-2/3)( 1- 1/5)( 1-2/7)≈4

52 """"""""""""""""""""""""""""""""""""""""""""""( 1-2/3)( 1-2/5)( 1-2)

54'' 25'' 3' 5' 7'' 3'''' 10''' 8 ≈'' 25× ( 1- 1/3) ( 1-2/5)

56``26``3`5`7``7`````6`````4≈````26×( 1-2/3)( 1-2/5)( 1- 1/7)≈4

……

Among them, when the real trace number of G(2N) from 2N=6 is compared with the theoretical calculation, there are positive and negative errors between the two curves.

Mandarin ducks and kisses interweave, and both advance infinitely at the pace of yangko, during which no counterexample appears, according to the theoretical expression of (1) and the calculation of (2).

Proof, Goldbach even number 1+ 1 conjecture is proved.