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)}
Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com

数学1+1

1300611016:
7 ?# \2 O3 Y# ]! m" v3 d/ N      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

% h0 r) S8 Y' H) m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，
$ x; m, }7 i- ]0 |. k: X$ g则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×
' [* ^' y( l3 P6 zN／{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
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)}
8 o- ^* W' f" |: g6 N

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
. s) [: }9 M' E; n2 R5 d$ a      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
& p% b# k4 J  B/ K8 s) }9 Q# a 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
1 D7 d. O& H# V* q2 f5 V6 O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)
8 H3 `4 v4 O4 i( n2 W) J类似地，根据定理1, (2), C可数  
# @$ A% o, P3 @- Q* c1 l  H

设      M_{1}(x)=minM(x)

根据（2），那么我们有.
/ t1 z) H/ `8 d" {8 R0 _8 j 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
* s" D2 j+ I( F5 `7 ]3 b: n7 o  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
1 @. E) z6 O4 S# ?5 M9 M 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
5 G( x* u0 _5 I, Y       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
% S3 F) [" s. l: P; f9 W5 X       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
+ m1 u; i6 ^; p4 S      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
8 N1 L8 z+ V  n+ v" R      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
& F8 h# S! H$ M, rAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：
) y3 S, H- D0 Q$ e2 p8 W3 ^
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.
1 ]. `1 T  n- n% N  s' L- AC=\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
  b7 P# y; ~5 m' S1 k: LDeduced
2 A8 F  n' a5 e: H& vD(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
+ F/ c1 R) z2 N" d, h$ w% NEmail:suxiaoguong@foxmail. com
§ 1 Introduction
4 y5 p0 t' e  |, e* H$ k          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:
* }) h. [* F; r/ T8 Y（A）For even number N

: f) }7 W; C" w3 GN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0
- t' H5 w; ^5 K$ Y
(B)  For odd number N

: o# U' G& B$ |2 y9 pN\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0

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 z8 V' g) o1 r  F         
; f2 C0 |$ _' o§2 Correlation set constructor
6 I$ X2 o: V, h( l" XA_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
8 Z5 B" @- E# n$ _! j. J A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}
8 @) U, ^+ S9 k \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）      
& S' Y! A  j0 y- z  i  §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
  .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
5 g+ H; e/ [5 W# T3 a- u1 CM_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
+ _0 t8 m8 B: ^; H# p( g\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)

7 T& T) a3 ?2 c6 L$ J\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)  
  \because A_{i},A_{j} Countable,
! W1 o/ [% J: C, a" m \therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
Similarly, according to Theorem 1, (2), C countable
* X! z# J$ X9 A# j2 {5 Q) ]! b" I Suppose
" ~( ?& \* E8 W% F! H2 S+ K      M_{1}(x)=minM(x)
$ u6 R3 n! r2 Eaccording to （2）, Then we have
* [( J$ \2 G3 ?: \. 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）
+ r# ^# o* ~7 ?( CProof: According to (2)，Then we have
5 {1 g, P+ `) R" a( ]: Y5 Y  I! F4 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}}
     Suppose
" A8 r5 i& Y6 k# s, U0 g      M_{2}(x)=maxM(x)
( R+ b# c' p3 s2 \\therefore M_{2}(x)
+ x7 ?+ m1 v& V5 s' Z; S=\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
( t, |1 s1 [) [1 b7 e3 r) ]=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
* Z. L0 M- g, z' G8 ]§4 Goldbach's problem end
& k3 E; E9 D5 q( @+ ]0 q5 h5 rTheorem 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
( U0 V  b" G# Y+ y- XN> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)
Let   c_{1}=min(\alpha ,\beta ),
& h! h9 X5 u. O# fAccording to Theorem 3,Then we have
D_{1}(N)=M_{1}(N)-M_{1}(N-2)
Clear
4 M  ~. C' u1 Z( m5 p+ QD(N)\geq D_{1}(N)
7 I& m0 l! h2 C\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt
4 L, x' M2 e- C8 v\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)
/ }) ~0 V) ?- Y\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
+ f$ P6 x+ \3 |3 _" H7 z3 W) f; qD(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
% k0 M' ]/ D2 d) Q. F Theorem 6  For evem number N
N> 800000\Rightarrow D(N)\leq
) a+ R& \; j$ K/ B* n4 Y5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
4 r  Z' N) h  s$ G' R) `Proof : According to (4)
! i6 h/ r5 @* {, sLet  c_{2}=max(\alpha ,\beta )
According to Theorem 4,Then we have
D_{2}(N)=M_{2}(N)-M_{2}(N-2)
\because D(N)\leq D_{2}(N)
' H/ `8 i+ A) }. l( cAccording to (5), Then we have
- ]( l! L$ ?5 z8 _1 t  W6 H) x* GD(N)\leq
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
' ~* m6 f+ s$ A4 p1 k2 ^$ l: B0 WTheorem 7 (Goldbach Theorem)  
For evem number N
- F1 T7 b. B/ i$ Z1 u2 nN\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
1 |% O& L9 |# Z0 V- N6 C4 i8 ~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)}
1 H+ Q  o0 V7 p1 \\therefore N\geq 6\Rightarrow D(N)\geq 1
Lemma 1 For odd number 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
Proof et  n\geq 4
. V1 {  }7 v6 k\because 2n+1=2(n-1)+3
According to Theorem 7,  Then we have
6 a0 d, ^/ Q  x8 t! w1 GN\geq 9\Rightarrow T(N)\geq 1


+ M; M3 X6 i/ n7 b! L" `% d$ z2 v    References
" q" d$ z: g% t[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
0 E6 H* H& l# h! B/ c& q* S[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.
0 G/ w0 e& [9 p2 o5 g- g1 u  v7 {0 o( u[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.

5 L) e& K9 d# D* E, C