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)}
0 y& K* b7 j# \9 o% l4 ]: Z Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com
3 e- }6 d4 X6 {3 r

数学1+1

1300611016:
0 n+ M% Z- K% x+ q; L      你如果能多读几本数论方面的著作，你就能理解哥德巴赫猜想，理解哥德巴赫猜想中D(N)表示什么？也就不感觉烦了。

1300611016

* [  v0 h' h- |5 C7 z$ w5 Q5 e
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
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。

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

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)]×
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。

数学1+1

D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
: t! S/ X! e8 e) Z, `" D$ P. l1.83150(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)}
. t( r% ?: {& X6 S" K

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

                  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
+ ?5 f' y) y0 J& ?      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
+ [0 F  H% M: H$ o 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
3 M9 z% ?; \6 [( o) C+ 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  

设      M_{1}(x)=minM(x)

根据（2），那么我们有.
9 K1 O: U; V# l8 f$ p 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
( V5 p% |" m+ t! i1 j  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
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
( E# j: p8 N3 y: u: 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
3 c; V  v7 C  I: S       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- n; ^" K' b% J# \' h. A* F       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
      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
. l! Y" E! X  m. b* Y      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

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
0 h+ ]: k* v, F3 mAbstract: 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.
  M0 V6 a0 @  v! `6 u0 d; H* qC=\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
0 X: @( D1 V, h$ _Deduced
D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
7 r% W0 q1 ~) r! j/ r# D7 Q1.8432(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)}

0 l, B3 U  D- ?3 r. C) aKey words: Germany，Goldbach，even number, Odd number ，prime number,
MR (2000) theme classification: 11 P32
Email:suxiaoguong@foxmail. com
! T: K. |. U! c8 {# u0 @§ 1 Introduction
          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:
+ w: s6 F+ `2 u: e  p（A）For even number N
4 }3 M" F3 X9 v. a$ J; }
0 C# O. I: J" I0 ZN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0

# m7 J! R6 H3 G4 m(B)  For odd number N
# s  E* d* Y/ l. [9 M
N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0
: d5 n% U8 B" z# Y6 p
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
         
$ W! f( N! E  e) N0 c§2 Correlation set constructor
/ O0 M/ w# I4 @A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
$ R2 E2 |$ \. E1 ~- @ 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    Ready  Theorem
Theorem 1
6 O, t9 L) \$ v7 ]# |+ c& TM_{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}
# p& ~# X" I7 o' F: o\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
) D2 W3 _; u6 ~; X0 X, fM_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots
! t+ g" D7 `! {7 UM_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
: D$ u; `7 e0 |+ a9 D\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
1 a4 N. ^: E* u' J) n3 \) _     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)  
/ D& `% ~# T4 k  V& Q7 n, x9 ]9 I$ j  \because A_{i},A_{j} Countable,
\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
! L, \5 o2 K% M/ [( F2 o) gSimilarly, according to Theorem 1, (2), C countable
9 D, f1 T4 k! O& ?/ t9 v' o, } Suppose
3 f' }( D" B' W! b  F4 l6 W9 Y( a      M_{1}(x)=minM(x)
according to （2）, Then we have
) x- E' j3 I2 I2 `. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
Theorem 4  For even number x
6 ]4 L9 M0 Z1 _- Y/ [$ jx>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）
: T& w1 P! \) Y1 TProof: According to (2)，Then we have
" B2 ?! v  X. v+ A4 ~& T, O. OM(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}}
6 \' Y. S, @  T. [& N+ S* H     Suppose
      M_{2}(x)=maxM(x)
  s: g5 e$ f; E1 \\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
& m# {5 j. h! S! x5 ^=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
1 k* Q5 Z/ L: z9 T; S8 Z§4 Goldbach's problem end
" W, O3 J9 q, ?; eTheorem 5  For evem number N
N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
: b4 p: ?' i" T; r, s0 Z     Proof: According to Theorem 2
/ K; z* p9 {2 e6 R" O4 ]N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)
Let   c_{1}=min(\alpha ,\beta ),
According to Theorem 3,Then we have
D_{1}(N)=M_{1}(N)-M_{1}(N-2)
Clear
  q1 s+ G; t9 a' i  _/ e& S0 p; pD(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)
3 f9 t! j) |+ Z0 n0 z\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
& [$ C+ y% l. D, M8 u9 k% T+ iN\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
0 u" F; R2 ?5 S9 i6 H$ W( m9 @  AD(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
' Y: T* v: n# w' t. ~' f$ c5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
$ x% i4 L5 P0 [$ _# _Proof : According to (4)
) _- l2 g5 n# U. F* |Let  c_{2}=max(\alpha ,\beta )
/ r# P5 c( ?# p: j' eAccording to Theorem 4,Then we have
D_{2}(N)=M_{2}(N)-M_{2}(N-2)
9 @& Y% g( ^6 O, O9 d( M! \\because D(N)\leq D_{2}(N)
According 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)}
/ m  S! M4 U7 g1 U) KTheorem 7 (Goldbach Theorem)  
For evem number N
5 j0 i. v3 C7 R+ e" o  JN\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
Proof : According to Shen Mok Kong verification
6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
) G7 U, G, s$ k( JAccording 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
! m# H# N8 Y9 T% uLemma 1 For odd number N
N\geq 9\Rightarrow
0 I& g( ^2 {4 K7 Y0 D7 zT(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
\because 2n+1=2(n-1)+3
According to Theorem 7,  Then we have
N\geq 9\Rightarrow T(N)\geq 1

/ b4 ?/ @* j/ n0 F4 D1 u! o
    References
& f* q2 ]  y4 T[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
6 }7 M4 O  p+ G7 y" b[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.
4 ?- X4 L3 T$ h- ]6 Y' O1 |
1 ]+ x+ p7 o/ O/ t/ X