哥德巴赫猜想验证

yangchuanju

Goldbach conjecture verification
https://sweet.ua.pt/tos/goldbach.html

Goldbach conjecture verification
哥德巴赫猜想验证

Introduction引言
The Goldbach conjecture is one of the oldest unsolved problems in number theory [1, problem C1]. In its modern form, it states that every even number larger than two can be expressed as a sum of two prime numbers.
哥德巴赫猜想是数论中最古老的未解问题之一[1，问题C1]。在其现代形式中，它表示每一个大于2的偶数都可以表示为两个素数的和。

Let n be an even number larger than two, and let n=p+q, with p and q prime numbers, p<=q, be a Goldbach partition of n. Let r(n) be the number of Goldbach partitions of n. The number of ways of writing n as a sum of two prime numbers, when the order of the two primes is important, is thus R(n)=2r(n) when n/2 is not a prime and is R(n)=2r(n)-1 when n/2 is a prime. The Goldbach conjecture states that r(n)>0, or, equivalently, that R(n)>0, for every even n larger than two.
设n是大于2的偶数，设n=p+q，有p和q个质数，p<=q，是n的哥德巴赫分块，设r(n)是n的哥德巴赫分块数，当两个质数的顺序重要时，把n写成两个质数之和的方式数，因而当n/2不是质数时是R(n)=2r(n)，当n/2是质数时是R(n)=2r(n)-1。哥德巴赫猜想指出，对于每一个大于2的偶数n，r(n)>0，或者等价地，R(n)>0。

In their famous memoir [2, conjecture A], Hardy and Littlewood conjectured that when n tends to infinity, R(n) tends asymptotically to (i.e., the ratio of the two functions tends to one)
在他们著名的回忆录[2，猜想A]中，哈代和利特伍德推测当n趋于无穷大时，R(n)趋于渐近(即两个函数的比值趋于1)

                         n                         p-1
N2(n) = 2 C       ----------------     PRODUCT     --- ,
           twin   (log n)(log n-2)   p odd prime   p-2
                                     divisor of n

改写为——R2(n)=2*C*N/[ln(N)*ln(N-2)]*∏(p-1)/(p-2)
改写式与R2(n)=2*C*N[ln(N)^2*∏(p-1)/(p-2)略有差别！

yangchuanju

哈李研究了哥猜素数对数关系式，并总结出渐进对数式，称之是哈李对数式有什么不可？

cuikun-186

杨老师你好，鲁思顺的9602是怎幺算出来的？

yangchuanju

cuikun-186 发表于 2026-9-17 10:27
杨老师你好，鲁思顺的9602是怎幺算出来的？

鲁思顺哥猜数计算式乱的很，论文式是一套大打折扣的“加强式”，约相当于真实哥猜数的20%多一点；
后鲁思顺认为哥猜已被他证明，认为论文中的加强系数有些过头，同时也想算一算给定偶数的哥猜数到底有多少，于是来了较弱的加强系数4/9，由此算得9602的哥猜数不少于79；
后9602的单计真实哥猜数77被证实，此时鲁思顺慌忙将加强系数改为3/7，于是9602的哥猜数变成不少于76啦！

再往后，伟大的69169被鲁思顺发现，再次将计算式改为与论文一致的加强（打折）计算式，只是将素数连乘积转换成合数连乘积，算得大于69169的偶数的哥猜数不少于100！

wangyangke

小学语文都没有过关，还那么轻薄无知；不知羞耻！

yangchuanju

（接1楼）
where
                       p(p-2)
C     =    PRODUCT     -------  = 0.66016181584686957392...
twin    p odd prime   (p-1)^2
【改写式】C=∏p*(p-2)/(p-1)^2=0.66016181584686957392...
is the twin primes constant. In [3], Crandall and Pomerance suggest replacing the factor
其中C=……是孪生素数常数。在[3]中，克兰德尔和波默兰斯建议替换该因子

       n
----------------
(log n)(log n-2)
【改写式】n/[ln(n)*ln(n-2)]
appearing in the formula of N2(n) by the asymptotically equivalent factor
出现在N2(n)公式中的渐近等价因子

        n-2         dx
INTEGRAL      --------------- .
        2     log(x) log(n-x)
【附注】INTEGRAL——积分，下限2，上限n-2，被积函数1/[ln(x)*ln(n-x)]

The numerical evidence supporting this conjectured asymptotic formula is very strong. Up to 10^10, the Crandall-Pomerance formula does not deviate from R(n) by more than 40150, and up to 2^40 it does not deviate from R(n) by more than 401900.
支持这个推测的渐近公式的数值证据非常有力。直到10^10，克兰德尔-博美伦斯公式与R(n)的偏差不超过40150，直到2^40，它与R(n)的偏差不超过401900。

Let us order the r(n) Goldbach partitions of n by increasing order of the smallest prime of the partition. More precisely, let us denote the two primes of the i-th Goldbach partition of n by p(n;i) and q(n;i), with p(n;i) <= q(n;i) and p(n;i) < p(n;i+1). In order to verify the Goldbach conjecture for a given n, it is sufficient to find one of its Goldbach partitions. Our strategy will be to find the minimal Goldbach partition n=p(n;1)+q(n;1), i.e., the one that uses the smallest possible prime number p(n)=p(n;1). As in [4], for every prime q we will denote by S(q) the least even number n such that p(n)=q.
让我们通过增加分区的最小素数的阶来对n的r(n)个哥德巴赫分区进行排序。更准确地说，让我们用p(n；I)和q(n；I)，带p(n；I)< = q(n；I)和p(n；I)< p(n；i+1)。为了验证给定n的哥德巴赫猜想，找到它的一个哥德巴赫分区就足够了。我们的策略是找到最小哥德巴赫分区n = p(n；1)+q(n；1)，即使用最小可能素数p(n)= p(n；1).如[4]所示，对于每个质数q，我们用S(q)表示最小偶数n，这样p(n)=q。

yangchuanju

Computational results计算结果
We have implemented a program that finds the minimal Goldbach partition of every even integer larger than four. In order to do this efficiently, the computation intensive parts of the program were written in assembly language (for the IA32 instruction set). A very efficient cache friendly implementation of the segmented sieve of Eratosthenes was used to generate the prime numbers (see our speed comparison chart [23KiB, PDF] between several Intel and AMD CPUs). For each interval of 10^12 integers, we record the number of times each (small) prime is used in a minimal Goldbach partition, as well as the even integer where it was first needed. Because it takes very little extra time, we also record information about the gaps between consecutive primes, viz., how many times each gap occurs, and its first occurrence. On a single core of a 3.3GHz core i3 processor, testing an interval of 10^12 integers near 10^18 takes close to 48 minutes. The execution time of the program grows very slowly, like log(N), where N is the last integer of the interval being tested, and it uses an amount of memory that is roughly given by 13 sqrt(N) / log(N). The program ran on the spare time of many computers, either under GNU/Linux or under Windows XP. We have reached 2&#8226;10^18 in November 2010, and in April 2012 have finally reached 4&#8226;10^18.
我们已经实现了一个程序，可以找到每个大于4的偶数整数的最小哥德巴赫划分。为了有效地做到这一点，程序的计算密集型部分是用汇编语言编写的(针对IA32指令集)。厄拉多塞分段筛的一个非常高效的高速缓存友好的实现被用来生成素数(参见我们的速度比较图[23kb，PDF]在几个英特尔和AMD处理器之间)。对于10^12整数的每个区间，我们记录每个(小的)素数在最小哥德巴赫分区中的使用次数，以及最先需要它的偶数整数。因为它只需要很少的额外时间，我们还记录了连续素数之间的间隔信息，即。，每个间隙出现多少次，以及它的第一次出现。在3.3GHz酷睿i3处理器的单核上，在10^18附近测试10^12整数的时间间隔需要将近48分钟。程序的执行时间增长非常缓慢，就像log(N)，其中N是被测试间隔的最后一个整数，它使用的内存量大致由13 sqrt(N) / log(N)给出。该程序在许多计算机的空闲时间运行，无论是在GNU/Linux下还是在Windows XP下。2010年11月我们到达了2个10^18，2012年4月我们终于到达了4个10^18.
The following table presents an overview of the current status of this massive computation. Each cell represents an interval of 10^15; its background color indicates its computational status (green for double-checked, yellow for single-checked, and red for not yet done or not yet fully checked), and its brightness indicates if counts of the primes in each of the 32 residue classes modulo 120 are available (bright) or not (not so bright) to perform an initial check of the correctness of the computation on each interval of 10^15 (prime counts for each interval of 10^12 are also available to perform correctness checks).
下表概述了这种大规模计算的当前状态。每个单元代表10^15的一个区间；它的背景颜色指示它的计算状态(绿色表示双重检查，黄色表示单次检查，红色表示尚未完成或尚未完全检查)，它的亮度指示模120的32个剩余类中的每一个的素数计数是否可用(亮)或不可用(不太亮)，以对10^15的每个区间执行计算正确性的初始检查(10^12的每个区间的素数计数也可用于执行正确性检查)。