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标题: Goldbach’s problem [打印本页]

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作者: 数学1+1    时间: 2013-12-6 12:27
标题: Goldbach’s problem
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)}
2 S9 @  Q- Y, ^4 H Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com

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作者: 数学1+1    时间: 2013-12-6 13:21

                  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)}

9 U0 v% j1 c/ t4 Y# n4 e

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

MR (2000) theme classification: 11 P32

Email:suxiaoguong@foxmail. com

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作者: 数学1+1    时间: 2013-12-9 10:43
                  Goldbach’s problem
4 E# e1 d( U# ]                    Su Xiaoguang
Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：
( Z2 j2 w) d+ h+ S5 ^
- ?. x+ @; \( Y1 w: pA= \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
% c1 [+ o! r+ y+ P% `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
  B3 B" L! Q1 J7 H! t! v9 h' SEmail:suxiaoguong@foxmail. com
/ _& C8 H0 v! G1 T§ 1 Introduction
1 Q! i  c: a& d          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:
. z* n: t2 J/ F- a6 s' _' h（A）For even number N
4 b& l0 j# ^% e  R8 G
N\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0

; |" F9 ?: D% b. g. L(B)  For odd number N
( ?) G3 p$ G" e4 c/ O
) A, ?8 ^; D2 O6 UN\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0
: |1 d) e1 T0 j$ w: k
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
         
, W5 D' }9 @& T: h  n% t§2 Correlation set constructor
A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
" _4 g- i2 T  g9 X- U+ k; B 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    Ready  Theorem
1 H+ R+ z* p+ z6 M8 L% s/ w1 X8 P0 ?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
) V0 i' y" ~8 q9 B* F7 a; n  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
0 t: Y& l: C, Z5 `3 N! U  F\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
6 T& ]# i' v9 L2 K0 ]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
6 k0 z. k" b1 g* l7 m( D\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 y+ `' U8 Q/ W     Theorem 2 (Prime number theorem)
6 N: [+ l7 d- Z% J, R) Q9 E; z
\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
2 [& H1 O! Z2 _" d     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,
\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
0 n% x# Z% _- ySimilarly, according to Theorem 1, (2), C countable
Suppose
$ V* V: |% i* F- m      M_{1}(x)=minM(x)
according to （2）, Then we have
- V- M. k0 e5 x0 N/ I$ d( N' j( k. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
Theorem 4  For even number x
6 v: B7 a5 c+ n5 b2 t$ `9 ~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）
Proof: According to (2)，Then we have
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}}
' a/ n0 D9 C+ ^! I0 Y     Suppose
      M_{2}(x)=maxM(x)
' e: A: g5 d8 V; `\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( ^+ [9 J* G& v& `3 x=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 end
Theorem 5  For evem number N
1 s1 t. H8 ?* i6 S  T2 k; }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)
# K2 E5 k' x- C$ x5 @  y2 y3 N' ULet   c_{1}=min(\alpha ,\beta ),
9 f4 T3 `( x/ A- _6 zAccording to Theorem 3,Then we have
( u" ~, w% p* pD_{1}(N)=M_{1}(N)-M_{1}(N-2)
$ o$ ?, ]9 w, H1 n" JClear
D(N)\geq D_{1}(N)
+ v5 M% ]/ F; l5 U1 h\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)
$ c' K  g8 i" l* p\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
7 f% P" N1 S6 u' s' ~) }: KN\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
$ ^% F4 J2 k! J0 a3 e" GD(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
0 _( K4 Y1 j. _3 t7 t# W Theorem 6  For evem number N
# h' r2 }% l# y5 l3 F/ W  G+ AN> 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
D_{2}(N)=M_{2}(N)-M_{2}(N-2)
1 ^, L, L; A* H9 x: g- e5 _) l\because D(N)\leq D_{2}(N)
4 [9 H' g% S+ ZAccording to (5), Then we have
' o8 c0 g; ]: {7 jD(N)\leq
) C* u( M5 \, T; R& b5.0176\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
9 s2 @$ ]' ~$ b" P9 J) \% f- B; Q- ^N\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
; w  v/ s; R7 b6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
1 i6 `: ?6 Q% u" _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)}
% N4 g5 j& S+ J( x1 N- E6 c* d; V\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
+ Z+ I3 G1 O2 v( Y9 q5 `\because 2n+1=2(n-1)+3
According to Theorem 7,  Then we have
N\geq 9\Rightarrow T(N)\geq 1
; R; H# ^9 k6 w0 F& L2 C0 t0 }! c

! w7 R) Z1 e5 B  K4 ]( N    References
; H  ]. t  N+ d3 H[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
% S+ i1 g5 O* Q! F  m, e( N[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.
/ f9 b) H) s8 P3 x# L0 Q[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.

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作者: 数学1+1    时间: 2013-12-9 11:45
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。

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作者: 数学1+1    时间: 2013-12-9 13:04

                  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
; V, V/ u7 ~% U) I+ R      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
! O7 }' d  k6 F 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)
类似地，根据定理1, (2), C可数  
4 b+ I& [0 U$ N, {3 ?) T) o/ A& P4 p

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
/ _' o* P; R/ W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
& [. q" T# c& }# W 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
- N' L) u+ }( F9 ~' K       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
; X8 k2 ], v+ G8 i/ E- r4 A       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
) [6 H3 n. z+ O+ l, H       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
& m4 c& o- q5 L+ I      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
4 ?+ g( E7 z1 K      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
) K; Y2 p9 N- e1 D$ T$ z7 e( U      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.

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作者: 数学1+1    时间: 2013-12-9 13:56
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.

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作者: 数学1+1    时间: 2013-12-13 15:32
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)}
" t# `1 i: F) y

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作者: 数学1+1    时间: 2013-12-14 11:37
若N＞800000，
( E. v% j' N/ F+ @* X0 t则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×
' ~% \5 H; S- d* E/ ]N／{log[(N-2)／2]log(N-2)}
( N' N* s7 a, @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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作者: 1300611016    时间: 2013-12-28 18:21
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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作者: 1300611016    时间: 2013-12-28 18:31

) a* k" o( ?/ o* B, m# }1 G8 J
+ y6 u* q& o1 c: P9 C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
! Z( J% D! M5 M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。

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作者: 数学1+1    时间: 2013-12-29 21:30
1300611016:
      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。

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