实用哥猜数表
270年前哥德巴赫猜想任意大于等于4的偶数都可以表示成两个素数之和,270年来尚未被证明。
其间数学界就哥猜进行了多方面研究和探讨,其中求算各种偶数可拆分成多少个素数对只是证明哥猜的一个窗口,即求算各种偶数的哥猜数。
在OEIS网站中有相当多网页涉及哥猜数,给出许多实用的哥猜数表,下面简要介绍一些,共网友参考。
网页A045917和A002375给出前2万个连续偶数的无序哥猜数,两网页的区别是:A045917中的数据是两素数之和,A002375中的数据是两奇素数之和;由于偶素数只有一个2,故两网页的差别仅仅是前者偶数4的哥猜数是1,后者偶数4 的哥猜数是0;其余偶数完全相同。本表仅给出前部部分数据,其余数据请读者查阅原网页表格。
A045917 From Goldbach problem: number of decompositions of 2n into unordered sums of two primes.
0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 4, 4, 2, 3, 4, 3, 4, 5, 4, 3, 5, 3, 4, 6, 3, 5, 6, 2, 5, 6, 5, 5, 7, 4, 5, 8, 5, 4, 9, 4, 5, 7, 3, 6, 8, 5, 6, 8, 6, 7, 10, 6, 6, 12, 4, 5, 10, 3, 7, 9, 6, 5, 8, 7, 8, 11, 6, 5, 12, 4, 8, 11, 5, 8, 10, 5, 6, 13, 9, 6, 11, 7, 7, 14, 6, 8, 13, 5, 8, 11, 7, 9……
A002375 From Goldbach conjecture: number of decompositions of 2n into an unordered sum of two odd primes.
0, 0, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 4, 4, 2, 3, 4, 3, 4, 5, 4, 3, 5, 3, 4, 6, 3, 5, 6, 2, 5, 6, 5, 5, 7, 4, 5, 8, 5, 4, 9, 4, 5, 7, 3, 6, 8, 5, 6, 8, 6, 7, 10, 6, 6, 12, 4, 5, 10, 3, 7, 9, 6, 5, 8, 7, 8, 11, 6, 5, 12, 4, 8, 11, 5, 8, 10, 5, 6, 13, 9, 6, 11, 7, 7, 14, 6, 8, 13, 5, 8, 11, 7, 9 ……
网页A020481和 A020482分别给出前2万个连续偶数第一个分拆数的最小素数和最大素数。
A020481 Least p with p, q both prime, p+q = 2n.
2, 3, 3, 3, 5, 3, 3, 5, 3, 3, 5, 3, 5, 7, 3, 3, 5, 7, 3, 5, 3, 3, 5, 3, 5, 7, 3, 5, 7, 3, 3, 5, 7, 3, 5, 3, 3, 5, 7, 3, 5, 3, 5, 7, 3, 5, 7, 19, 3, 5, 3, 3, 5, 3, 3, 5, 3, 5, 7, 13, 11, 13, 19, 3, 5, 3, 5, 7, 3, 3, 5, 7, 11, 11, 3, 3, 5, 7, 3, 5, 7, 3, 5, 3, 5, 7, 3, 5, 7, 3, 3, 5, 7, 11, 11, 3, 3, 5, 3, 3, 5, 7 ……
A020482 Greatest p with p, q both prime, p+q = 2n.
2, 3, 5, 7, 7, 11, 13, 13, 17, 19, 19, 23, 23, 23, 29, 31, 31, 31, 37, 37, 41, 43, 43, 47, 47, 47, 53, 53, 53, 59, 61, 61, 61, 67, 67, 71, 73, 73, 73, 79, 79, 83, 83, 83, 89, 89, 89, 79, 97, 97, 101, 103, 103, 107, 109, 109, 113, 113, 113, 109, 113, 113, 109, 127, 127, 131, 131……
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