![]() |
![]() |
#1 |
Jan 2017
38 Posts |
![]()
It is well known that if q | 2^p - 1, then q = 2px + 1 and q = 8r +(-) 1.
when I do some research on Fp (Fibonacci prime), I guess : 1: if q | Fp, then q = 2px + (-1)^x 2: F(2n+1) must have a prime factor q, q = 2(2n+1)x + (-1)^x 3: ... At this moment , I try to factor C261 ( F1565 = F5* F313 * C261), which has 261 digital. If guess is right, then C261 must have a prime factor = 2*1565*x + (-1)^x. The GIMPS factoring code creates a modified sieve of Eratosthenes with each bit representing a potential 2kp+1 factor. The sieve then eliminates any potential factors that are divisible by prime numbers below 40,000 or so. Also, bits representing potential factors of 3 or 5 mod 8 are cleared. This process eliminates roughly 95% of potential factors. Since I don't know assemble language, I could not understand such code. I tried to code it by myself, but the code efficiency is very low. I have no other good choice but ask for help : Could someone provide a efficient sieve ? Thanks a lot! |
![]() |
![]() |
![]() |
#3 |
Jun 2003
484510 Posts |
![]()
Trial factoring is the wrong tool for numbers this size. You should use ECM.
Last fiddled with by axn on 2017-01-05 at 04:35 Reason: speeling |
![]() |
![]() |
![]() |
#5 |
Jan 2017
3 Posts |
![]()
Thanks for your advice.
|
![]() |
![]() |
![]() |
Thread Tools | |
![]() |
||||
Thread | Thread Starter | Forum | Replies | Last Post |
Prime numbers test primality - with proof written in invisible ink | Godzilla | Miscellaneous Math | 40 | 2018-10-17 00:11 |
Looking for British written American history on Kindle | jasong | Homework Help | 3 | 2015-01-30 21:36 |
Math of LLR(written by Jean Penne) | TTn | 15k Search | 3 | 2006-01-06 00:08 |
Help w/ python. | a216vcti | Programming | 7 | 2005-10-30 00:37 |
Available Ranges suitable for P4s | garo | Lone Mersenne Hunters | 1 | 2003-09-10 21:10 |