摘要
设任一偶数2n,是否存在着一个仅依赖于2n的函数f(2n)?它能表示偶数表为两个素数之和的素数解的组数。本文首先把素数定理引入奇数列(一维空间),然后拓展到二维空间。在一维空间,素数定理-素数的分布函数π(x)~xlogx(x∞),从素数定理得到:P(N)~1logx及P(G)~2logx。P(G)作为数据处理的工具,用它解决了命题P2n(1,1)。在二维空间:素数的联合分布密度P(Px,Py)~1logxlogy,由它积分得到了分布函数π(x,y),利用π(x,y)可以估计圆内素点(Px,Py)的个数,并且解决了命题,P2n(1,1)2。P2n(1,1)和P2n(1,1)2的结果是用不同的方法建立的不同的数学模型,但是它们主阶的数值规律是一致的。这个问题本文得到解决。
Suppose any even integer 2n
. Is there or not a function f(2n) to be on
ly dependent upon 2n?It can express a number of group of prim
e solutions on representation of even integer as a sum of two primes.
In the paper,first,the prime theorem is led into odd sequence integer,i.e. one
dimensional space Second,it is generalized into twodimensional space.
In the one-dimensional space,the prime theoremprime distributed fun
ction is π(
x)~xlogx(x∞).P(
N)~1logxin N={1,2,3,4...,x}and
P(G)~2logxin G={1,3,5,...,x} are from the
prime theorem.P(G)is regraded as a data handling tool to solve
P2n(1,1). In the twodimensional space,independence of the prime distrib
ution
,the prime edge densities determine a prime joint density:P(Px,Py)
~1logxlogy.It is int
egrated to get a distributed fun
ction of the primes:π(x,y).A number of prime spots (
Px,Py) in a circle can be estimated by π(x,y),an
d P2n(1,1)2 has a reault here.
The two principal steps of the results of P2n(1,1) and
P2n(1,1)2 are the two mathematical models set up by d
ifferent methods. But the laws of their value are unanimous.
The problem is solved. This is a direct answer to the Goldbach's con
jecture.
出处
《安徽广播电视大学学报》
1999年第2期75-81,共7页
Journal of Anhui Radio & TV University
关键词
哥德巴赫猜想
素数定理
第二素数概率
偶数的素数解
哥德巴赫第一空间
哥德巴赫线
哥德巴赫猜想点
:Goldbach's conjecture
Prime theorem
second prime pro
bability
prime solution to even integer
Goldbach's first space
Goldbach line
Go
ldbach's conjecture spots.