A logical approach to validate the Goldbach conjecture: Paper 1/3

M. Khare, K. Chitta
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Abstract

This paper is a first of the series of three papers which provide a general proof to validate the Goldbach conjecture. This conjecture states that every even number can be expressed as a summation of two prime numbers. At the onset, concept of successive-addition of‐digits‐of‐an‐integer‐number (SADN) and its properties in terms of basic algebraic functions like addition, multiplication and subtraction are discussed. SADN classifies odd numbers into 3 sequences—the S1, S3 and S5 sequences—which comprise of odd numbers having SADN 7 or 4 or 1; SADN 3 or 9 or 6 and SADN 5 or 2 or 8 respectively. The S1 and S5 sequences are of interest in the analysis. Furthermore, composites on the S1 sequence are derived as products of intra-sequence elements of the S1 and S5 sequences while composites on the S5 sequence are derived as products of inter-sequence elements of the S1 and S5 sequences. SADN also shows why such combinations for even numbers of SADN (1, 4, 7) will be found on the S5 sequence while those for even numbers of SADN (2, 5, 8) will lie on the S1 sequence and both the sequences have a role to play in identifying the prime number combinations for even numbers with SADN (3, 6, 9). Thereafter, analysis moves to calculating the total number of combinations for a given even number that would include combinations in the nature of two composites (c1 + c2); prime & composite (p + c) and two primes (p1 + p2). Identifying the total number of c1 + c2 and p + c combinations yield the number of p1 + p2 combinations. The logic employed in present discussion shows that at least one such p1 + p2 combination exists for the even numbers having SADN digit within 1 to 9. This encompasses all even numbers and hence generalizes this method for all even numbers.
验证哥德巴赫猜想的逻辑方法:论文 1/3
本文是三篇系列论文中的第一篇,为验证哥德巴赫猜想提供了一般性证明。该猜想指出,每个偶数都可以表示为两个质数的求和。本文首先讨论了整数数位连续加法(SADN)的概念及其在加法、乘法和减法等基本代数函数方面的性质。SADN 将奇数分为 3 个序列--S1、S3 和 S5 序列--分别由 SADN 为 7 或 4 或 1、SADN 为 3 或 9 或 6 和 SADN 为 5 或 2 或 8 的奇数组成。S1 和 S5 序列是本次分析的重点。此外,S1 序列上的合成物是 S1 和 S5 序列内元素的产物,而 S5 序列上的合成物则是 S1 和 S5 序列间元素的产物。SADN 还说明了为什么 SADN(1、4、7)的偶数组合会出现在 S5 序列上,而 SADN(2、5、8)的偶数组合会出现在 S1 序列上,这两个序列在确定 SADN(3、6、9)的偶数质数组合方面都有作用。之后,分析将转向计算给定偶数的组合总数,其中包括两个合成数(c1 + c2)、质数与合成数(p + c)以及两个质数(p1 + p2)的组合。找出 c1 + c2 和 p + c 的组合总数,就得出了 p1 + p2 的组合数。本讨论采用的逻辑表明,对于 SADN 位数在 1 到 9 之间的偶数,至少存在一个这样的 p1 + p2 组合。这涵盖了所有偶数,从而将此方法推广到所有偶数。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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