《哥德巴赫猜想的验证》其中的要点

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《哥德巴赫猜想的验证》青岛 王新宇用中文推荐其中的要点：
    Goldbach conjecture verification
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Introduction News Results Top 50 Contributors References Links Contact  
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Introduction
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.
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.
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)
                         n                         p-1
N2(n) = 2 C       ----------------     PRODUCT     --- ,
           twin   (log n)(log n-2)   p odd prime   p-2
                                     divisor of n
where
                       p(p-2)
C     =    PRODUCT     -------  = 0.66016181584686957392...
twin    p odd prime   (p-1)^2
is the twin primes constant. In [3], Crandall and Pomerance suggest replacing the factor
       n
----------------
(log n)(log n-2)
appearing in the formula of N2(n) by the asymptotically equivalent factor
        n-2         dx
INTEGRAL      --------------- .
        2     log(x) log(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.
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.
News
February 1, 2005: 2·10^17 reached.
March 20, 2005: 10^17 double checked.
December 26, 2005: 3·10^17 reached.
J...........
November 30, 2011: discovery of the minimal Goldbach partition 2795935116574469638=9629+P19.
January 16, 2012: discovery of the minimal Goldbach partition 3325581707333960528=9781+P19.
February 6, 2012: 34·10^17 reached.
February 13, 2012: 35·10^17 reached.
April 4, 2012: 4·10^18, our desired verification limit, reached.
June 18, 2012: 2·10^17 double checked.
September 7, 2012: 3·10^17 double checked.
Computational results
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 [22k, 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 2·10^18 in November 2010, and in April 2012 have finally reached 4·10^18.
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).
0000 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 0013 0014 0015 0016 0017 0018 0019
... (same state) ...
..........
... (same state) ...
3980 3981 3982 3983 3984 3985 3986 3987 3988 3989 3990 3991 3992 3993 3994 3995 3996 3997 3998 3999
We have tested all consecutive even numbers up to 4·10^18, and, so far, double-checked our results up to 3·10^17 (in all, 480205 intervals of 10^12 were double-checked). About 772.3 single-core CPU years were used to do all this.
As far as we are aware, the previous record of computation was 4·10^14 [5]. As expected, no counter-example of the conjecture was found. In this table [24k, compressed with gzip] we present all values of S(p) we were able to compute, as well as counts of the number of times each (small) prime was used in a minimal Goldbach partition. The record-holders, i.e., numbers larger than all previous ones of the same kind, are clearly marked in the table, which extends the tables 1 and 2 of [4] and table 1 of [5]. The following figure presents a graph with the available values of S(p).
The values of S(p) are bounded, for our empirical data, by the functions
                         0.4                                     0.4
                  0.4   p                                 0.4   p
S_min(p) = 0.06  p     e        and    S_max(p) = 11.05  p     e     .
The two kinds of record-holders marked in the table mentioned above correspond to the values of S(p) closer to one of the two bounds. For large p the values of S(p) appear to be slowly approaching the upper bound; hence, the asymptotic growth rate of S(p) probably has a different functional form. In [6] it was stated, based on probabilistic considerations, that p should not grow faster than log^2 S(p) log log S(p). Our data is not enough to resolve this conjecture; the black curve corresponds to the line 1.527 log^2 S(p) log log S(p) (according to the conjecture, all data points should be above the line). It was found that for all our data p can be reasonably well approximated by 0.33(log S(p) log log S(p))^2.
Let D(x;p) be the relative frequency of occurrence of the prime p in the minimal Goldbach partition of the even numbers not larger than x. The following figure presents a graph of this function, computed for our current verification limit of the Goldbach conjecture.

Besides the expected near exponential decay of D(x;p), it is interesting to observe that there exists a distinct difference of behavior in the values of this function when p is a multiple of three plus one (white dots) and when it is not (yellow dots).
Top 50
The following table presents the 50 largest p (least primes of a Goldbach partition) found so far, with S(p)<4·10^18, either by contributors to this project or by other discoverers (those have an * before the discoverer name). Repeated values of p were excluded from this list.
  Rank     p     S(p)     Discoverer   
  1     9781     3325 58170 73339 60528     Silvio Pardi   
  2     9629     2795 93511 65744 69638     Silvio Pardi   
  3     9341     906 03057 95622 79642     John Fettig & Nahil Sobh   
  4     9203     1348 11357 94295 47486     Siegfried "Zig" Herzog   
  5     9161     887 12380 30778 37868     Siegfried "Zig" Herzog   
  6     9091     3164 06916 06618 44912     Silvio Pardi   
  7     9001     3893 00922 74334 20582     Tomás Oliveira e Silva   
  8     8971     2588 35699 18831 39892     Tomás Oliveira e Silva   
  9     8951     914 47723 42519 16254     John Fettig & Nahil Sobh   
  10     8941     555 27435 15567 50822     Siegfried "Zig" Herzog   
..............
  40     8539     941 90839 15563 21548     Siegfried "Zig" Herzog   
  41     8527     1295 15748 05397 26954     Siegfried "Zig" Herzog   
  42     8521     1176 80059 43587 37918     Siegfried "Zig" Herzog   
  43     8513     3431 34334 19613 51016     Silvio Pardi   
  44     8501     255 32912 66885 55994     Siegfried "Zig" Herzog   
  45     8467     2576 43841 05051 31868     Tomás Oliveira e Silva   
  46     8461     2565 58105 68187 05782     Tomás Oliveira e Silva   
  47     8447     2764 13588 98014 80338     Silvio Pardi   
  48     8443     121 00502 23040 07026     Tomás Oliveira e Silva   
  49     8431     2290 65390 58512 00938     Tomás Oliveira e Silva   
  50     8419     1730 12004 22605 42848     Tomás Oliveira e Silva   
Contributors
The following table presents some details about the contribution of all (past and present) individuals or organizations which donated computing power to this project.
  Name     Location     Number of  
  tasks     Number of first (known) occurrences   
  Minimal Goldbach partitions     Prime gaps   
  Tomás Oliveira e Silva     All     1999424     325 (0)     70 (0)   
  DETUA     1450498     191 (0)     51 (0)   
  Home     400468     11 (0)     12 (0)   
  Kraken     109000     4 (0)     1 (0)   
..........
  All     All     4480205     460 (0)     139 (0)   
This work was partially supported by the National Center for Supercomputing Applications and utilized the NCSA Xeon cluster.
This research was partially supported by an allocation of advanced computing resources supported by the National Science Foundation. Part of the computations were performed on Kraken (a Cray XT5) at the National Institute for Computational Sciences
References
[1] Richard K. Guy, Unsolved problems in number theory, third edition, Springer-Verlag, 2004.
[2] G. H. Hardy and J. E. Littlewood, Some problems of `partitio numerorum';; III: on the expression of a number as a sum of primes, Acta Mathematica, vol. 44, pp. 1-70, 1922.  
.......
Additional links
Our prime gaps page.
MathSoft';s Hardy-Littlewood Constants page.
Cracking Goldbach';s Conjecture (European Grid Infrastructure case studies)
--------------------------------------------------------------------------------
Tomás Oliveira e Silva
Departamento de Eletrónica, Telecomunica??es e Informática
Universidade de Aveiro
..........
E-mail address: tos@ua.pt
Home page: http://www.ieeta.pt/~tos/goldbach.html  
      青岛 王新宇 用中文推荐其中的要点：
题目是：哥德巴赫猜想的验证（外国用计算机研究哥德巴赫猜想的成果）。
目录分：主体解简介，文献，首尾解成果， 前50个首尾解，贡献者，工具书类，链接，联系人。
主体解简介：
哥德巴赫猜想是[1对称1]的对称素数问题，现代说法：每一个大于2的偶数都可以表示为“两个素数的和”。设n是大于2的偶数，若n = p + q，p与q都是素数，p≤q，则是符合哥德巴赫和式的n。命r(n)表示n的符合哥德巴赫和式的个数。公式解没包含首和尾两个平方根数的素数的解，称为主体解，简约为解。
最小的哥德巴赫和式的素数就是给定偶数平方根数内的素数，是补充到主体解，使解精准的关键。
命“偶数两个边缘的两个平方根数内的素数如果满足哥德巴赫和式”为首尾解。首尾解和中的小素数称为
最小的哥德巴赫素数“P”，对应“最小的哥德巴赫素数”的偶数称为“S(P)”,图表示“对应关系”。
前50个最小的哥德巴赫(注：都是小于给定偶数平方根数的素数且满足哥德巴赫和式的素数的数值，)
2011年11月30日。发现的最小的哥德巴赫279593511657446963 = 9629 +p19(表示书写19位数的素数)。
2012年1月16日。 发现的最小的哥德巴赫332558170733396052 = 9781 +p19(表示书写19位数的素数)。
“9629 ,9781”都是小于给定偶数平方根数的素数且满足哥德巴赫和式的素数的数值，即：P素数。
哥德巴赫和式包含都熟悉的哈代公式的(主体)解,也包含“与小于给定偶数平方根数的素数对称的素数”解，本验证提供了“计算机成果”，很宝贵，值得爱求“准确解”的人学习和深入。
待续：
             qdxinyu
          2012.12.24
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     青岛 王新宇 用中文推荐英文《哥德巴赫猜想的验证》其中的要点：
题目是：哥德巴赫猜想的验证（外国用计算机研究哥德巴赫猜想的成果）。
目录分：主体解简介，文献，首尾解成果， 前50个首尾解，贡献者，工具书类，链接，联系人。
主体解简介：
哥德巴赫猜想是[1对称1]的对称素数问题，现代说法：每一个大于2的偶数都可以表示为“两个素数的和”。设n是大于2的偶数，若n = p + q，p与q都是素数，p≤q，则是符合哥德巴赫和式的n。命r(n)表示n的符合哥德巴赫和式的个数,加数交换左右算一个解(和的个数与双素数组相等),命R(n)为表示n的符合哥德巴赫和式的个数,两个素数的顺序重要,加数交换左右算两个解(和的个数与和中素数种类相等)。当{n/2}不是素数时,R(n)=2r(n)，当{n/2}是素数时,R(n)= 2r(N)-1。Goldbach猜想：n为大于2的每一个偶数，r(n)> 0 或R(n) > 0。
哈代和李特维而德猜想，当n充分大时，R(N)趋近于（几个分数式的乘积）
`````````````n``````````````P-1
R(n)=2*C*----------------∏-----
..........ln(n)[ln(n)-2]....P-2
∏是连乘积运算(级数)符号,下标表示P是整除n的素数,C的下标twin表示是孪生素数常数。
(注：该公式解没包含首和尾两个平方根数的素数的解,称为主体解,很多学者简约为全解)，
(注：原文“N(n)”或许是R(n)笔误,公式比国内常用R(n)=2C{n/ln^2(n)}∏[(p-1)/(p-2)]准确了些),
`````P(P-2)
C =∏----- ------- = 0.66016181584686957392.....
.....(P-1)^2  
``````````````````````````dx      
下限2,上限(n-2),积分--------------
....................ln(n)[ln(n)-2]
数值的证据支持这一猜想渐近公式很强。10^10的公式解，偏离实际不超过40150，2^40偏离实际不会超过401900。
通过增加平方根内小的素数的哥德巴赫和式继续研究r(n)。研究首尾平方根数内素数的哥德巴赫和式。
文献
2005年2月1日。达到2·10^17。2006年6月5日。达到4·10^17。2007年4月25日。达到10 ^18。.....
2010年11月6日。达到 2·10 ^ 18。2011年11月30日。发现的最小的哥德巴赫279593511657446963 = 9629 +P19(19位数的素数)。2012年1月16日。发现的最小的哥德巴赫332558170733396052 = 9781 +P19(19位数的素数)。2012年4月4日。 达到4·10 ^ 18。
首尾解成果
一项计划，发现大于四的每个偶数的最小素数的哥德巴赫和式。在2012年四月终于达到了4·10 ^ 18。下表概述了这大规模计算。每一个细节的一个间隔10^15,其背景颜色；表明其计算状态(绿色)核对过，(黄色)尚未完全检查，正如预期的那样，没有反例的猜想被发现。
“哥德巴赫和式中的最小素数与偶数对应关系图的图象是有界的，我们的实证数据，由函数
S_最小(P)=0.06*(P^0.4)*[e^(P^0.4)] 和S_最大(p)=11.05*(P^0.4)*[e^(P^0.4)]。
2公式对应图象S(P)两界限。有人指出，黑线对应1.527*[ln^2(S(p))]*lnln(S(p))（猜想，所有的数据点应在线上）。结果发现，所有的数据P可以合理的近似0.33*[ln(S(p))*lnln(S(p)]^2。
4·10 ^ 18前50个首尾解
下表列出了50大p（哥德巴赫和式的最小素数）,S(P)临近 4·10 ^ 18，
排序;哥德巴赫和式的最小素数P;偶数S(P); 发现者
197813325 58170 73339 60528
296292795 93511 65744 69638
39341906 03057 95622 79642
最小的哥德巴赫和式的素数就是给定偶数平方根数内的素数，是补充到主体解，使解精准的关键。
命“偶数两个边缘的两个平方根数内的素数如果满足哥德巴赫和式”为首尾解。首尾解和中的小素数称为
最小的哥德巴赫素数“P”，对应“最小的哥德巴赫素数”的偶数称为“S(P)”,图表示“对应关系”。
前50个最小的哥德巴赫(注：都是小于给定偶数平方根数的素数且满足哥德巴赫和式的素数的数值，)
2011年11月30日。发现的最小的哥德巴赫279593511657446963 = 9629 +p19(表示书写19位数的素数)。
2012年1月16日。 发现的最小的哥德巴赫332558170733396052 = 9781 +p19(表示书写19位数的素数)。
“9629 ,9781”都是小于给定偶数平方根数的素数且满足哥德巴赫和式的素数的数值，即：P素数。
  英文R(n)公式比国内R(n)=2C{n/ln^2(n)}∏[(p-1)/(p-2)]准确了些,是因为素数定理的公式：
n/ln(n)《 n/[ln(n)-1],前者特别适合求解素数,孪生素数,哥德巴赫和式的下限。运算还简单。
后者因求解的素数个数准确些，用其分母的平方数"ln^2(n)-2ln(n)+1"求哥德巴赫和式自然准些,
适合求趋近解,(自然不利于求下限,)分母采用ln^2(n)-2ln(n)也可行。
    qdxinyu
   2012.12.24

任在深

注意！
     世界数学界根本不承认上述证明！

qdxy

   全世界的数学界都承认下述公式               
                        n                         p-1
N2(n) = 2 C       ----------------     PRODUCT     ---
          twin   (log n)(log n-2)   p odd prime   p-2
                     
       p-1     p(p-2)          n   
= 2 ∏ ---- ∏ ------- {-----------------}
       p-2    (p-1)^2   log^2(n)-2log(n)
       p-1           1          n   
= 2 ∏ ---- ∏{1- -------}  {------------}
       p-2        (p-1)^2    (log(n)-1)^2
       p-1           1          n   
≥2 ∏ ---- ∏{1- -------}  {--------}
       p-2        (p-1)^2    log^2(n)
      p-1     (1.32)n   
≥ ∏ ----   {--------}
      p-2     log^2(n)