p1p2p3p4p5p6p7是公差大于零的素数等差数列，求p7的最小值

nyy

p1、p2、p3、p4、p5、p6、p7是公差大于零的素数等差数列，求p7的最小值
在网络上看到的，别人用笔算的。

nyy

1. Clear["Global`*"];(*Clear all variables*)
   
2. aaa=Table[Prime[k],{k,2,1000}];(*素数表,从第2个到第1000个*)
   
3. Do[ lst={0,1,2,3,4,5,6}*d;(*公差*)
   
4.     If[Union@PrimeQ[p1+lst]=={True},(*必须都是素数*)
   
5.         Print[{p1,d,p1+lst}](*输出求解结果:第一个素数、公差、所有素数*)
   
6.     ],
   
7. {p1,aaa},(*第一个素数必然来自上面的素数表*)
   
8. {d,2,1000,2}](*公差只可能是偶数*)
   

复制代码

我只会穷举法：假设第一个素数在前1000个素数里面，再假设公差小于1000，然后穷举，找出
求解结果中最小的，那就是得到的结果。

{7,150,{7,157,307,457,607,757,907}}
{47,210,{47,257,467,677,887,1097,1307}}
{179,210,{179,389,599,809,1019,1229,1439}}
{193,420,{193,613,1033,1453,1873,2293,2713}}
{199,210,{199,409,619,829,1039,1249,1459}}
{409,210,{409,619,829,1039,1249,1459,1669}}
{619,210,{619,829,1039,1249,1459,1669,1879}}
{829,210,{829,1039,1249,1459,1669,1879,2089}}
{881,210,{881,1091,1301,1511,1721,1931,2141}}
{1061,840,{1061,1901,2741,3581,4421,5261,6101}}
{1091,210,{1091,1301,1511,1721,1931,2141,2351}}
{1453,210,{1453,1663,1873,2083,2293,2503,2713}}
{1619,420,{1619,2039,2459,2879,3299,3719,4139}}
{1637,630,{1637,2267,2897,3527,4157,4787,5417}}
{1753,840,{1753,2593,3433,4273,5113,5953,6793}}
{2239,420,{2239,2659,3079,3499,3919,4339,4759}}
{2267,630,{2267,2897,3527,4157,4787,5417,6047}}
{2659,420,{2659,3079,3499,3919,4339,4759,5179}}
{3499,210,{3499,3709,3919,4129,4339,4549,4759}}
{3623,840,{3623,4463,5303,6143,6983,7823,8663}}
{3709,210,{3709,3919,4129,4339,4549,4759,4969}}
{3919,210,{3919,4129,4339,4549,4759,4969,5179}}
{4259,420,{4259,4679,5099,5519,5939,6359,6779}}
{4493,840,{4493,5333,6173,7013,7853,8693,9533}}
{5569,630,{5569,6199,6829,7459,8089,8719,9349}}
{5651,840,{5651,6491,7331,8171,9011,9851,10691}}
{5849,420,{5849,6269,6689,7109,7529,7949,8369}}
{6043,840,{6043,6883,7723,8563,9403,10243,11083}}
{6529,420,{6529,6949,7369,7789,8209,8629,9049}}
{6883,840,{6883,7723,8563,9403,10243,11083,11923}}
{7723,840,{7723,8563,9403,10243,11083,11923,12763}}

上面的第一个就是求解结果！

nyy

nyy 发表于 2022-12-1 09:48
我只会穷举法：假设第一个素数在前1000个素数里面，再假设公差小于1000，然后穷举，找出
求解结果中最小 ...

从上面的求解结果可以“归纳”出：公差是30的倍数

nyy

nyy 发表于 2022-12-1 09:48
我只会穷举法：假设第一个素数在前1000个素数里面，再假设公差小于1000，然后穷举，找出
求解结果中最小 ...

这不是一个多简洁的代码，但是确实能在相对短的时间内给出结果！

nyy

谁能证明p1的最小值是7？不过只是猜测，我猜测p1的最小值是7，没找到以3或者5开头的7个素数等差数列！

northwolves

5+6k*d,5+6k(d+1),5+6k(d+2),5+6k(d+3),5+6k(d+4),5+6k(d+5) 里至少有一个是5的倍数

northwolves

3，3+d,3+2d,3+3d...