Goldbach’s problem

数学1+1

Goldbach’s problem                    Su XiaoguangAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：[code]<SPAN style="FONT-FAMILY: Arial; COLOR: #333333; FONT-SIZE: 12pt; mso-font-kerning: 0pt; mso-ansi-language: EN" lang=EN></SPAN>[/code]A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1DeducedD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
- P2 ]5 F8 ]& Y. J3 N* D" a Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com

数学1+1

                  Goldbach’s problem (pdf)

                       Su Xiaoguang
2 k) E- E& T6 o$ u- K6 i8 o     

Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：

A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.

C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1

Deduced

D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge

1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}

; n  t4 m: f2 S

Key words: Germany，Goldbach，even number, Odd number ，prime number,

MR (2000) theme classification: 11 P32

Email:suxiaoguong@foxmail. com

数学1+1

                  Goldbach’s problem
- b6 H9 u' `; t* q& s9 {                    Su Xiaoguang
8 `# D( t0 ~3 s- v5 e6 ~4 M: `Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：
8 C8 |- o& O$ q3 c& k0 A" H( ^2 [
A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.
C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1
: P. o- \9 W, [' A8 cDeduced
. l% P& B3 `+ [, R# BD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
( O. c  c+ j8 @. @& D# R
- W7 }% k! k0 ~$ U/ g  {Key words: Germany，Goldbach，even number, Odd number ，prime number,
! d/ l, e+ V1 x" w$ IMR (2000) theme classification: 11 P32
/ ]7 V# p$ b4 [Email:suxiaoguong@foxmail. com
6 Y4 G2 H- z; ~3 t% V  G§ 1 Introduction
1 O9 X' ]) `8 C; ~          In 1742, the German mathematician Christian Goldbach (1690-1764)， Put forward two speculated about the relationship between positive integers and prime number，using analytical language expressed as:
2 n; J# u# Y- N4 G+ m% o2 h（A）For even number N
( @+ |) l- Q& m
% ~* K  X! T8 }9 S, A7 IN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0
; B% n' O1 v- e5 r+ ]& p
* i- g. R- X" ]  ~  H% ](B)  For odd number N
& Z3 J# F8 T8 }$ [/ [: l
# N8 v: G) L# K, @N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0
, _# W7 Z; U1 Y! c* e4 H8 o" r
This is the famous GOldbach conjecture。If the proposition (A) true, then the proposition (B) True。So, as long as we prove Proposition (A), Launched immediately conjecture (B) is correct
0 L, i% O5 E' Y) @$ }$ H% m( o+ i         
§2 Correlation set constructor
A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
+ G/ V& n: l6 I- Z& c$ I5 @ A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}
% G0 ]$ Q( J. ]5 `& Q( ]! G$ ` \cdots
A=\bigcup_{i=0}^{\infty }A_{i}\Rightarrow B(N)=\sum_{n_{1}+n_{2}=N.0\leq n _{1},n_{2}\leq N}1,B(x,N)=\sum_{N\leq x,B(N)\neq 0}1      （1）
p_{0}=2,p_{1}=3,p_{2}=5,\cdots C_{0}=\left \{ p_{0} +p_{0},p_{0}+p_{1},p_{0}+p_{2},\cdots \right \} C_{1}=\left \{ p_{1} +p_{0},p_{1}+p_{1},p_{1}+p_{2},\cdots \right \} C_{2}=\left \{ p_{2} +p_{0},p_{2}+p_{1},p_{2}+p_{2},\cdots \right \} C=\bigcup_{i=0}^{\infty }C_{i}\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N.p_{1},p_{2}\geq 3}1\wedge M(x)=\sum_{N\leq x,D(N)\neq 0}  （2）      
  §3    Ready  Theorem
Theorem 1
M_{i}=(x_{1}^{(i)},x_{2}^{(i)},\cdots ,x_{i}^{(i)},\cdots ),Is a countable set\Rightarrow M=\bigcup_{i=1}^{N}M_{i},Is a countable set
$ a% U( |( N' ~  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
4 B* ?' T% u8 F. m% ]" v7 O# HM_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots
- z0 H6 r; A5 E1 w# _M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
( y0 F5 }' J4 p$ |, X0 D$ P\cdots
9 k# v. [" X, x% O+ n; t. H\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable
     Theorem 2 (Prime number theorem)

\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
0 a8 m! A9 {; D* k6 g     Theorem 3  For even number x
x>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ]
Proof: According to Theorem 1, (1)  
! Z9 d( G' j- [% ?5 j8 w0 a5 q  \because A_{i},A_{j} Countable,
. [& o) i) t6 m- z+ A  K0 T2 f \therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
3 Z; @7 Y/ u9 K7 w) fSimilarly, according to Theorem 1, (2), C countable
Suppose
      M_{1}(x)=minM(x)
according to （2）, Then we have
. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
5 X+ e9 @, _+ j- X; _ Theorem 4  For even number x
4 G# ~' u6 W: u; z1 F" ox>800000\wedge M_{2}(x)=maxM(x)\Rightarrow M_{2}(x)=4\pi (\frac{x}{2})\pi (x)-2\pi ^ {2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)                        （3）
Proof: According to (2)，Then we have
) k& l" N1 i5 |! @. y5 M$ ~& c- |, PM(x)=\sum_{N\leq x,D(N)\neq 0}1< \sum_{3\leq p_{1},p_{2}\leq \frac{x}{2}}1+\sum_{3\leq p_{1}\leq \frac{x}{2},\frac{x}{2}< p_{2}< x}+\sum_{\frac{x}{2}<p _{1}< x,3\leq p_{2}\leq \frac{x}{2}}
. U8 z$ ]3 t& p: |) G/ _     Suppose
7 {- U, l& Z/ a; p      M_{2}(x)=maxM(x)
/ U& p7 g$ o1 w! {9 k( _9 x+ S\therefore M_{2}(x)
=\frac{1}{2}\cdot 2\pi (x)\left [ 2\pi (\frac{x}{2})-1 \right ]-\frac{1}{2}\left [ 2\pi (\frac{x}{2})-\pi (x) \right ]\left [ 2\pi (\frac{x}{2})-\pi (x) +1\right ]\cdot 2
=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
* p# x: W( c+ u8 y. S§4 Goldbach's problem end
$ J0 t  u. Q# g  xTheorem 5  For evem number N
N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
     Proof: According to Theorem 2
N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)
) l7 Z$ S% d/ k" |' V5 HLet   c_{1}=min(\alpha ,\beta ),
& L* b) F# @: O% N' dAccording to Theorem 3,Then we have
! l  J5 b2 ~% S: |, n; W+ ID_{1}(N)=M_{1}(N)-M_{1}(N-2)
0 n5 D1 N) \: H: X; Y* RClear
0 p" }' q9 e4 L1 B; J. \D(N)\geq D_{1}(N)
) P* N0 M. c7 y* e1 M- L9 M4 Z\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt
+ D) q: @" q0 Z: a. {\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)
+ f5 K1 Y% q0 |# J$ L- X\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
Theorem 6  For evem number N
N> 800000\Rightarrow D(N)\leq
$ z, P7 y5 [: a5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
Proof : According to (4)
; L6 G6 ]8 W4 ?# d) V  T! o' HLet  c_{2}=max(\alpha ,\beta )
( Q9 C; W3 E0 ^; XAccording to Theorem 4,Then we have
/ e4 D( [; _8 N% J# a* A" N& }D_{2}(N)=M_{2}(N)-M_{2}(N-2)
4 w" z% I6 T" H8 d* P\because D(N)\leq D_{2}(N)
/ y4 a. k' n$ I- D! m, a' K6 tAccording to (5), Then we have
D(N)\leq
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
Theorem 7 (Goldbach Theorem)  
For evem number N
! h* @9 p* k/ x0 `N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
4 P" O. z3 v- Z) m8 S% |  XProof : According to Shen Mok Kong verification
6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
According to Theorem 5, Theorem 6, Then we have
N> 800000\Rightarrow 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
\therefore N\geq 6\Rightarrow D(N)\geq 1
+ V* }' f3 ~; i* X' ?Lemma 1 For odd number N
! E6 |+ Y: G9 X8 I8 r- zN\geq 9\Rightarrow
3 |$ R4 Q5 q* |. Y+ j& t, o" }T(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 1
Proof et  n\geq 4
. f+ [+ A8 c9 ?1 ]\because 2n+1=2(n-1)+3
According to Theorem 7,  Then we have
N\geq 9\Rightarrow T(N)\geq 1


9 D2 n" V+ n; S* _    References
; `% u2 }0 V# e3 i% ]2 P5 H+ s[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.
[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.
0 ?- ]( p6 H- p  K( ?0 d

数学1+1

阅读本帖需具备阅读LATEX文件的知识，作者有一word文件上传，有兴趣的读者可下载阅读。

数学1+1

                  Goldbach’s problem

                    Su Xiaoguang

摘要：哥德巴赫问题是解析数论的一个重要问题。作者研究了

A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.

C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1

Deduced

D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge

1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}

Key words: Germany，Goldbach，even number, Odd number ，prime number,

MR (2000) theme classification: 11 P32

Email:suxiaoguong@foxmail. com

§ 1  引言
9 L* X) M! D% n4 N      1742年，德国数学家Christian Goldbach提出了关于正整数和素数之间关系的两个推测，用分析的语言表述为:

（A）对于偶数N

N\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0

(B)  对于奇数N

N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0

        这就是著名的哥德巴赫猜想，如果命题（A）真，那么命题（B）真，所以，只要我们证明命题（A），立即推出猜想（B）是正确的

         

§2相关集的构造
' H; l9 G3 ]7 s A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}

A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}

A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}

\cdots

A=\bigcup_{i=0}^{\infty }A_{i}\Rightarrow B(N)=\sum_{n_{1}+n_{2}=N.0\leq n _{1},n_{2}\leq N}1,B(x,N)=\sum_{N\leq x,B(N)\neq 0}1      （1）

p_{0}=2,p_{1}=3,p_{2}=5,\cdots C_{0}=\left \{ p_{0} +p_{0},p_{0}+p_{1},p_{0}+p_{2},\cdots \right \} C_{1}=\left \{ p_{1} +p_{0},p_{1}+p_{1},p_{1}+p_{2},\cdots \right \} C_{2}=\left \{ p_{2} +p_{0},p_{2}+p_{1},p_{2}+p_{2},\cdots \right \} C=\bigcup_{i=0}^{\infty }C_{i}\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N.p_{1},p_{2}\geq 3}1\wedge M(x)=\sum_{N\leq x,D(N)\neq 0}  （2）      

  §3    预备定理
  E" o  p3 A/ V# Q* P, U1 L/ {定理 1

M_{i}=(x_{1}^{(i)},x_{2}^{(i)},\cdots ,x_{i}^{(i)},\cdots ),Is a countable set\Rightarrow M=\bigcup_{i=1}^{N}M_{i},Is a countable set

  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}

\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots

M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots

M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots

\cdots

\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable

     定理2 (素数定理)

\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}

      定理3  对于偶数x

x>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ]

证明 根据定理1, (1)  

  \because A_{i},A_{j} Countable,

\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
类似地，根据定理1, (2), C可数  
* _+ D- c% V# i! B

设      M_{1}(x)=minM(x)

根据（2），那么我们有.
M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]

定理4  对于偶数x

x>800000\wedge M_{2}(x)=maxM(x)\Rightarrow M_{2}(x)=4\pi (\frac{x}{2})\pi (x)-2\pi ^ {2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)                        （3）

证明: 根据（2），那么我们有
  M(x)=\sum_{N\leq x,D(N)\neq 0}1< \sum_{3\leq p_{1},p_{2}\leq \frac{x}{2}}1+\sum_{3\leq p_{1}\leq \frac{x}{2},\frac{x}{2}< p_{2}< x}+\sum_{\frac{x}{2}<p _{1}< x,3\leq p_{2}\leq \frac{x}{2}}

     设   M_{2}(x)=maxM(x)

\therefore M_{2}(x)

=\frac{1}{2}\cdot 2\pi (x)\left [ 2\pi (\frac{x}{2})-1 \right ]-\frac{1}{2}\left [ 2\pi (\frac{x}{2})-\pi (x) \right ]\left [ 2\pi (\frac{x}{2})-\pi (x) +1\right ]\cdot 2

=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)

§4 Goldbach's problem 终结
( I. ~$ a0 i1 m8 a/ Q) U0 g定理 5  对于偶数N

N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}

    证明: 根据定理2
N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)

让  c_{1}=min(\alpha ,\beta ),

根据定理3，然后我们有
      D_{1}(N)=M_{1}(N)-M_{1}(N-2)

显然
% V2 O, r: S) L+ ^       D(N)\geq D_{1}(N)

\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt

\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)

\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)

N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow

D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}

定理6  对于偶数N

N> 800000\Rightarrow D(N)\leq

5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}

证明: 根据(4)

让  c_{2}=max(\alpha ,\beta )

根据定理4，然后我们有
$ T, o5 q+ C! I$ Q. n- l2 k; I       D_{2}(N)=M_{2}(N)-M_{2}(N-2)

\because D(N)\leq D_{2}(N)

根据(5)，那么我们有
       D(N)\leq

5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}

定理7 (Goldbach Theorem)  

对于偶数N

N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1

证明: 根Shen Mok Kong 的验证
, s2 H* [4 g) C' o5 P# Y      6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1

根据定理5, 定理 6, 然后我们有
      N> 800000\Rightarrow 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}

\therefore N\geq 6\Rightarrow D(N)\geq 1

引理1 对于奇数N

N\geq 9\Rightarrow

T(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 1

证明: 让 n\geq 4

\because 2n+1=2(n-1)+3

根据定理7，然后我们有
# l' N; j  e, \4 h6 Z      N\geq 9\Rightarrow T(N)\geq 1

    References

[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.

[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.

[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.

数学1+1

我国数学家华罗庚，闵嗣鹤均对M(x)的下界做过研究，潘承洞，潘承彪对D(N)的上界做过研究，他们留下了遗憾，也留下了经验.

数学1+1

D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
# `0 P6 [% _1 ]# K0 O; Y1.83150(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 4.36166\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
: p. Q* \' ?2 s

数学1+1

若N＞800000，
则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×
; q; G1 R! R0 ^# C6 n1 rN／{log[(N-2)／2]log(N-2)}
2 \5 A: s& _: t3 r这就是哥德巴赫公式，有兴趣的读者不妨检测一下。

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楼主的帖子怎么样？赶紧试试这里的快速回复给楼主点评论

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太烦，可以用一个简明的形式，如·同偶质数对·形式展开详细见http://www.madio.net/thread-202136-1-1.html
% u6 L; E9 `0 H" P; E一般的用简明浅显的形式表述更容易推广，如能用初等数学表述这一问题，可以尝试一下。但不妨碍专业研究。