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)}
; N8 O/ [1 U- `1 i 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
     

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)}

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
4 W, i3 ]+ U! U4 A                    Su Xiaoguang
5 s8 d! @, `/ d' B1 K+ TAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：

$ Z% R, F; b6 P! ^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.
8 b/ U0 N& N, G( K7 z& WC=\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
' D" Q# h2 L* e+ b' p$ vDeduced
) o+ K4 N$ V- S/ L, g# OD(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)}

5 [- ]& d3 k2 T6 s6 g% x. yKey words: Germany，Goldbach，even number, Odd number ，prime number,
MR (2000) theme classification: 11 P32
Email:suxiaoguong@foxmail. com
§ 1 Introduction
' F" j+ |/ q. M: t: {* F! I! s          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:
! F1 A- @: M! H: g* K2 }（A）For even number N

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

% L- ~* `0 H- e7 T6 ]- B9 J' a(B)  For odd number N
* U: `2 r8 T/ q* _; j7 {8 m. Q% z
! T: F2 |8 Q) o# s2 v& \" H: MN\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0

/ u9 N" |) G( D2 _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
9 V4 K! k4 F+ O( \% |/ }* ?1 \: X         
- i# G: p, ~' O. I$ D9 P6 e4 Q§2 Correlation set constructor
% Q# w1 U1 a/ sA_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
7 J5 {# H  Y+ D3 D A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}
1 b9 G* }) j) \: T7 G8 ?: _ \cdots
; ~% p- b, L( T4 q1 t4 zA=\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）
$ D' @1 A( V2 [, u2 z, R7 cp_{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）      
1 e2 w" o0 N' t( Q+ V! {) X" o  §3    Ready  Theorem
Theorem 1
/ ~: i. d# T0 f0 ^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
+ r' Y1 z8 W9 d3 t5 s% t: E' m, p  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
; l+ k; `$ j1 k9 g2 h\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
+ ]* e9 C; O& P( K& `M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots
! K! J9 H, i& j& t9 g; d$ ]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
     Theorem 2 (Prime number theorem)

\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
     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)  
- A$ o& }! d7 K8 Z( [. B  \because A_{i},A_{j} Countable,
: K: ~+ ?9 p! _* E: L \therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
* c  D% n9 X$ T: G& Y/ LSimilarly, according to Theorem 1, (2), C countable
1 ?$ ]- w! @) [ Suppose
8 b& |2 o$ v$ ~8 I, Z" u6 K      M_{1}(x)=minM(x)
- J+ S* H" m+ j2 }; vaccording to （2）, Then we have
. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
Theorem 4  For even number 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）
3 B" V. Q* _4 I8 rProof: According to (2)，Then we have
, W: C# l7 b% ^* n) O! k1 }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}}
     Suppose
      M_{2}(x)=maxM(x)
; m. Q  i* h0 _/ D\therefore M_{2}(x)
) S) s/ ]1 Q1 a7 Y) ^& z" A=\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)
: A: r/ Y. {7 @§4 Goldbach's problem end
  G4 M! u- n% r! hTheorem 5  For evem number N
3 l, H3 H  Y  d) G) QN> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
2 E+ ~" |5 d8 X4 @) j& P* c" g     Proof: According to Theorem 2
N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)
Let   c_{1}=min(\alpha ,\beta ),
5 E; s  V" G4 \+ HAccording to Theorem 3,Then we have
- b& e  G8 ^8 B0 y/ A# DD_{1}(N)=M_{1}(N)-M_{1}(N-2)
Clear
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
( H/ k2 M2 v) f, K' I\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)
3 W" e$ P/ @. d* y3 X5 a, oN\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
! H/ g7 S2 ?5 w# V Theorem 6  For evem number N
& O( B) L( G  [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)}
Proof : According to (4)
Let  c_{2}=max(\alpha ,\beta )
According to Theorem 4,Then we have
% E; y, V+ y: t' X( kD_{2}(N)=M_{2}(N)-M_{2}(N-2)
  M0 a' M0 u) `* v\because D(N)\leq D_{2}(N)
According to (5), Then we have
4 v4 l* f: \9 v; m! R, {D(N)\leq
0 w3 ~: ?5 Z- \9 Y3 x" y1 U5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
Theorem 7 (Goldbach Theorem)  
* L+ T3 U2 [& z0 Z9 ]6 R& UFor evem number N
N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
  t. w8 u3 |0 I# i7 h6 OProof : According to Shen Mok Kong verification
. F: P' |+ b, |0 C( J6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
- d. h; V3 O- K3 E3 o. oAccording 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
Lemma 1 For odd number N
0 T. n3 \- L$ X/ G/ _N\geq 9\Rightarrow
6 i" e& [8 q) i/ rT(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
; m/ X# f. }: L2 |6 Z; m\because 2n+1=2(n-1)+3
According to Theorem 7,  Then we have
N\geq 9\Rightarrow T(N)\geq 1

1 M" O1 e, g  D$ E& }6 I# |
' r! `+ `& C( b4 r  }5 t9 t    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.

0 `; \( E* S1 ?1 z3 ~

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
" w5 X+ Z# P& V0 m      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
" ~: S: L$ v# T1 g$ i9 a3 I) N* l 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
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)
4 S) B# z4 z0 r0 k6 S- z2 ^类似地，根据定理1, (2), C可数  

设      M_{1}(x)=minM(x)

根据（2），那么我们有.
3 M: J7 U3 s# N; U 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
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
) T1 U/ r3 ~4 }) o 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
% T% a6 _  s" K. o6 C* }      D_{1}(N)=M_{1}(N)-M_{1}(N-2)

显然
# Y  ?0 t5 p! {$ o       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
  y; b# @: b; n9 a% ~9 c5 V; z       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
/ s# B4 ^6 U9 z; a2 z9 I" D* v+ a       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
& ]1 {: Y5 w0 |" {6 ~2 l+ J0 \3 o. [      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
      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
1.83150(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)}

数学1+1

若N＞800000，
8 {; A( J  r% ^5 r9 q0 |则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×
N／{log[(N-2)／2]log(N-2)}
这就是哥德巴赫公式，有兴趣的读者不妨检测一下。

1300611016

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

1300611016

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
5 `) n! \: N# @& y3 a+ h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。