# Recursive prime brother by Brillhart – Lehmer – Selfridge algorithm

Define:

p[k,i]=ABS[1+2*n[k,i]*p[k-1,1]*p[k-1,2]],n[k,1] is the integer with minimum ABS[n[k,1]] that makes p[k,1] a prime number, and n[k,2] is the integer with second minimum ABS[n[k,2]] that makes p[k,2] a prime number

The primality of p[k,i] can be proven using Brillhart – Lehmer – Selfridge algorithm recursively by using p[k-1,1] and p[k-1,2] as helper since n is a small integer, by reducing k to 1.

With this idea, taking

p[0,1]=1, p[0,2]=1

We got the n[i,j] ( columns : i; rows: k):

k |
i=1 |
i=2 |

1 | 1 | -2 |

2 | -1 | 1 |

3 | 1 | -2 |

4 | -3 | -5 |

5 | -11 | 19 |

6 | 51 | 94 |

7 | 7 | 33 |

8 | 147 | -165 |

9 | 15 | -29 |

10 | 5 | 339 |

11 | 412 | -1260 |

12 | 356 | 848 |

13 | 4809 | -5641 |

14 | -5215 | -5539 |

15 | 37695 | 41772 |

16 | 5343 | -6180 |

17 | -31463 | -36980 |

18 | 181802 | -292989 |

19 | 70660 | – |

The last three, p[19,1] makes top 5000 list.

The proof will be posted in the reply of this one.

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certificate at http://bitc.bme.emory.edu/~lzhou/prime_certs/pm2train_19.1.tar.gz