mersenneforum.org  

Go Back   mersenneforum.org > Other Stuff > Open Projects

Reply
 
Thread Tools
Old 2010-11-21, 01:14   #12
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

22×1,607 Posts
Default

Quote:
Originally Posted by R. Gerbicz View Post
suppose that res=p#(n)%p (one mulmod) then:
p#(n)/p(k)+1==0 mod p if and only if p(k)==p-res mod p
p#(n)/p(k)-1==0 mod p if and only if p(k)==res mod p
Not quite true:

p(4347)# % 1006729441 = (41579# % 1006729441) = 31729

41579# % 31729 != 0

because 1006729441 % 31729 = 0. In fact 1006729441 = 31729 ^ 2.

I have to add a check to make sure res % p(k) > 0 because I see other cases where res is a multiple of p(k).
rogue is offline   Reply With Quote
Old 2010-11-21, 02:10   #13
axn
 
axn's Avatar
 
Jun 2003

514510 Posts
Default

Quote:
Originally Posted by rogue View Post
Not quite true:

p(4347)# % 1006729441 = (41579# % 1006729441) = 31729

41579# % 31729 != 0

because 1006729441 % 31729 = 0. In fact 1006729441 = 31729 ^ 2.

I have to add a check to make sure res % p(k) > 0 because I see other cases where res is a multiple of p(k).
First off, I'm assuming that where you said "41579# % 31729 != 0", you meant "= 0". I think it might be easier if you don't sieve with semiprimes having factors < p(max).
axn is offline   Reply With Quote
Old 2010-11-21, 04:06   #14
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

22×1,607 Posts
Default

Quote:
Originally Posted by axn View Post
First off, I'm assuming that where you said "41579# % 31729 != 0", you meant "= 0".
Oops. You are correct.
rogue is offline   Reply With Quote
Old 2010-11-21, 13:25   #15
Mini-Geek
Account Deleted
 
Mini-Geek's Avatar
 
"Tim Sorbera"
Aug 2006
San Antonio, TX USA

102548 Posts
Default

3530-3549 complete, and all k's had a prime. Here are the primes:
Code:
p(3530)#*p(46)+1
p(3531)#*p(664)-1
p(3532)#*p(44)-1
p(3533)#*p(207)+1
p(3534)#*p(1708)+1
p(3535)#*p(1)+1
p(3536)#*p(717)-1
p(3537)#*p(1077)+1
p(3538)#*p(154)+1
p(3539)#*p(1894)-1
p(3540)#*p(67)-1
p(3541)#*p(383)+1
p(3542)#*p(3186)+1
p(3543)#*p(816)+1
p(3544)#*p(384)-1
p(3545)#*p(1148)-1
p(3546)#*p(376)+1
p(3547)#*p(1516)-1
p(3548)#*p(860)-1
p(3549)#*p(1964)+1
Edit: all proven prime

Last fiddled with by Mini-Geek on 2010-11-21 at 13:49
Mini-Geek is offline   Reply With Quote
Old 2010-11-21, 14:07   #16
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

22·1,607 Posts
Default

After about 10 hours of sieving, the new sieve is already 100x deeper than the old one (1.7e10 vs 1.7e8). The old sieve took about 24 days to get as far as it did. There are 5.75 million candidates left and it is still removing candidates at about 7 per second. There is a long way to go with sieving. The optimal removal rate for the range is about 4 or 5 per minute. I don't know how long it will take to get to that rate.

The only mistake in my sieve is that I could have written to an ABC file and used the number_primes option with PFGW, but didn't. I'm still writing to the k.in and j.in files used by the PFGW script.

I'll provide files for the divide side if anyone wants to work on it, but I suggest that everyone holds off until I've sieved deeper.

On an ancillary note, although we are trying to disprove the conjecture, should we also try to show that there is a prime of each form for each p(k)? If there is, then that would be a new conjecture. If not (and I suspect not), then that also implies that this conjecture is wrong.
rogue is offline   Reply With Quote
Old 2010-11-21, 19:48   #17
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

22·1,607 Posts
Default

Quote:
Originally Posted by rogue View Post
On an ancillary note, although we are trying to disprove the conjecture, should we also try to show that there is a prime of each form for each p(k)? If there is, then that would be a new conjecture. If not (and I suspect not), then that also implies that this conjecture is wrong.
p(11)#/p(j)-1 is not prime for any j <= 11.
rogue is offline   Reply With Quote
Old 2010-11-22, 17:26   #18
Harvey563
 
Harvey563's Avatar
 
Apr 2004

2738 Posts
Default

p(14)*p(j)+1 is not prime for any j <= 14.
Harvey563 is offline   Reply With Quote
Old 2010-11-22, 17:33   #19
Harvey563
 
Harvey563's Avatar
 
Apr 2004

11×17 Posts
Default

p(19)#/p(j)+1 is not prime for any j <= 19
Harvey563 is offline   Reply With Quote
Old 2010-11-22, 17:58   #20
Harvey563
 
Harvey563's Avatar
 
Apr 2004

11×17 Posts
Default

p(29)#*p(j)-1 is not prime for any j <=29

Last fiddled with by Harvey563 on 2010-11-22 at 18:03
Harvey563 is offline   Reply With Quote
Old 2010-11-25, 00:48   #21
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

22×1,607 Posts
Default An update

The current sieve is at 1.3e11 with a removal rate of just under 1 per second (about 50 per minute). The removal rate is about 10x faster than what it should be before it's done. There are currently just under 5.3e6 candidates left.

I don't know how much longer sieving will take, but I don't expect a huge percentage of tests to be removed either. I'm going to take 3900 <= k < 4000.
rogue is offline   Reply With Quote
Old 2010-11-29, 22:02   #22
rogue
 
rogue's Avatar
 
"Mark"
Apr 2003
Between here and the

642810 Posts
Default

Close, but no cigar. I found a k for which there was no j on the divide side that yielded a PRP, but found on on the multiply side after going through 50% of the candidates on that side.
rogue is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
What is the problem here? didgogns Msieve 1 2016-11-15 03:31
problem I have science_man_88 Miscellaneous Math 2 2010-10-10 16:36
Intel Atom revisited hj47 Hardware 15 2010-07-08 20:19
51 problem Neves Miscellaneous Math 5 2004-02-10 22:59
51 problem Neves Puzzles 15 2004-02-05 23:11

All times are UTC. The time now is 17:16.


Sun Oct 17 17:16:41 UTC 2021 up 86 days, 11:45, 0 users, load averages: 2.23, 2.17, 1.99

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.