3奇素数和的问题
哥德巴赫的任一个大于2的偶数都可以写成2个素数之和的猜想提出至今已过去282年,没人给出真正的证明;
但任一个大于等于5的整数都可以写成3个素数之和的猜想据说已经被证明。
不知为何熊一兵先生现在又关心起3奇素数和的问题,
3素数和,3奇素数和的统计数据在OEIS网站有不少网页,其中
A007963给出10000个2n+1型奇数可以写成3奇素数和的(无序写法)个数,
"
Number of (unordered) ways of writing 2n+1 as a sum of 3 odd primes."
0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 6, 8, 7, 9, 10, 10, 10, 11, 12, 12, 14, 16, 14, 16, 16, 16, 18, 20, 20, 20, 21, 21, 21, 27, 24, 25, 28, 27, 28, 33, 29, 32, 35, 34, 30, 37, 36, 34, 42, 38, 36, 46, 42, 42, 50, 46, 47, 53, 50, 45, 56, 54, 46, 62, 53, 48, 64, 59, 55, 68, 61, 59, 68
熊一兵先生所要的奇数9,11,99,101,999,1001,9999,10001的写法数全在其中。
9 4 1
11 5 1
99 49 30
101 50 37
999 499 769
1001 500 1094
9999 4999 28055
10001 5000 42615
另A087916给出10000个2n+1型奇数可以写成3奇素数和的(有序写法)个数,
Number of ordered ways to write 2n+1 as a sum of 3 odd primes.
0, 0, 0, 0, 1, 3, 6, 7, 9, 12, 16, 18, 21, 27, 30, 30, 34, 36, 42, 46, 48, 48, 51, 63, 60, 64, 81, 75, 76, 87, 87, 90, 102, 105, 97, 117, 114, 105, 144, 129, 126, 159, 141, 145, 177, 162, 160, 195, 186, 153, 207, 201, 171, 237, 210, 187, 255, 234, 222, 279
A054860给出10000个2n+1型奇数可以写成3素数和的(无序写法)个数,
Number of ways of writing 2n+1 as p + q + r where p, q, r are primes with p <= q <= r.
0, 0, 0, 1, 2, 2, 2, 3, 4, 3, 5, 5, 5, 7, 7, 6, 9, 8, 9, 10, 11, 10, 12, 13, 12, 15, 16, 14, 17, 16, 16, 19, 21, 20, 20, 22, 21, 22, 28, 24, 25, 29, 27, 29, 33, 29, 33, 35, 34, 30, 38, 36, 35, 43, 38, 37, 47, 42, 43, 50, 46, 47, 53, 50, 45, 57, 54, 47, 62, 53, 49, 65, 59, 55, 68
7 = 2 + 2 + 3 so a(3) = 1;
9 = 2 + 2 + 5 = 3 + 3 + 3 so a(4) = 2;
11 = 2 + 2 + 7 = 3 + 3 + 5 so a(5) = 2.
|