设D(n)表示方程n=p1+p2的解数,其中p1,p2为奇素数,若D(n)>0,则我们称n为偶数G o ldbach数.主要目的是利用初等和解析方法从两个不同的角度来研究偶数G o ldbach数的均值性质,并给出了两个相同的渐近公式,从而为进一步证明偶数G o ldb...设D(n)表示方程n=p1+p2的解数,其中p1,p2为奇素数,若D(n)>0,则我们称n为偶数G o ldbach数.主要目的是利用初等和解析方法从两个不同的角度来研究偶数G o ldbach数的均值性质,并给出了两个相同的渐近公式,从而为进一步证明偶数G o ldbach猜想的正确性提供了有力的证据.展开更多
In the paper: the representation of large ev en integer as a sum of two primes is proved to be right independently by each of W-progression ∑(∞)(X n=1D)(n+1)(n-1)!of the discovery and the prime theorem. I t is induc...In the paper: the representation of large ev en integer as a sum of two primes is proved to be right independently by each of W-progression ∑(∞)(X n=1D)(n+1)(n-1)!of the discovery and the prime theorem. I t is induced as two following problems which are solved for getting results of ration: Is there a function of f(2n) to be only depend ent upon 2n or not? And it can express a number of group of prime solutions on r epresentatio n of even integer as a sum of two primes. In one－ dimensional space, the prime t heorem is led into odd sequence integer to find P(G)～2 log n is regarded as a data handling tool for setting a mathematical model of ran dom sampling, get:P2n(1，1)n＞2 2n-P 2=P 1=f(2n)～(2nlogn/2log2nlog2n(2n→∞). The prime theorem π(x) is gene ralized to the two-dimensional space: π(x,y). A mathematical model of average values is set up by π(x,y), get: P2n(1，1)2 (X n＞2 2n=P 1+P 2)=f(2n)2～(2n log22n SX) (2n→∞). But for expressing a number of group of prime solutions of even integer，the laws of values of principal steps of the two different functions f(2n) and f(2n) 2 are unanimous. Thus, the proof of different ways lead to the same result and determines a forceful declaration: Goldbach’s conjecture is proved to be a right theorem.展开更多
<正> 设自然数 n 的标准分解式为n=2~αP_1~α 1 P_2~α 2…p_k~α k,(1)其中 k,α为非负整数,p_i(1≤i≤k)为互异奇素数,α_i(1≤i≤k)为正整数.文[1]获得表 n 为连续自然数的和之方法数为f(n)={(α_1+1)(α_2+1)…(α_k+1),k≥1...<正> 设自然数 n 的标准分解式为n=2~αP_1~α 1 P_2~α 2…p_k~α k,(1)其中 k,α为非负整数,p_i(1≤i≤k)为互异奇素数,α_i(1≤i≤k)为正整数.文[1]获得表 n 为连续自然数的和之方法数为f(n)={(α_1+1)(α_2+1)…(α_k+1),k≥1,1,k=0. (2)(上面的表达式略有改进,原文忽略了 k=0的情形.)本文推广此问题,得到如下结果:定理:设 n 的标准分解式如(1)式。展开更多
文摘设D(n)表示方程n=p1+p2的解数,其中p1,p2为奇素数,若D(n)>0,则我们称n为偶数G o ldbach数.主要目的是利用初等和解析方法从两个不同的角度来研究偶数G o ldbach数的均值性质,并给出了两个相同的渐近公式,从而为进一步证明偶数G o ldbach猜想的正确性提供了有力的证据.
文摘In the paper: the representation of large ev en integer as a sum of two primes is proved to be right independently by each of W-progression ∑(∞)(X n=1D)(n+1)(n-1)!of the discovery and the prime theorem. I t is induced as two following problems which are solved for getting results of ration: Is there a function of f(2n) to be only depend ent upon 2n or not? And it can express a number of group of prime solutions on r epresentatio n of even integer as a sum of two primes. In one－ dimensional space, the prime t heorem is led into odd sequence integer to find P(G)～2 log n is regarded as a data handling tool for setting a mathematical model of ran dom sampling, get:P2n(1，1)n＞2 2n-P 2=P 1=f(2n)～(2nlogn/2log2nlog2n(2n→∞). The prime theorem π(x) is gene ralized to the two-dimensional space: π(x,y). A mathematical model of average values is set up by π(x,y), get: P2n(1，1)2 (X n＞2 2n=P 1+P 2)=f(2n)2～(2n log22n SX) (2n→∞). But for expressing a number of group of prime solutions of even integer，the laws of values of principal steps of the two different functions f(2n) and f(2n) 2 are unanimous. Thus, the proof of different ways lead to the same result and determines a forceful declaration: Goldbach’s conjecture is proved to be a right theorem.
文摘<正> 设自然数 n 的标准分解式为n=2~αP_1~α 1 P_2~α 2…p_k~α k,(1)其中 k,α为非负整数,p_i(1≤i≤k)为互异奇素数,α_i(1≤i≤k)为正整数.文[1]获得表 n 为连续自然数的和之方法数为f(n)={(α_1+1)(α_2+1)…(α_k+1),k≥1,1,k=0. (2)(上面的表达式略有改进,原文忽略了 k=0的情形.)本文推广此问题,得到如下结果:定理:设 n 的标准分解式如(1)式。
