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2010-11-21, 01:14   #12
rogue

"Mark"
Apr 2003
Between here and the

22×1,607 Posts

Quote:
 Originally Posted by R. Gerbicz 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).

2010-11-21, 02:10   #13
axn

Jun 2003

514510 Posts

Quote:
 Originally Posted by rogue 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).

2010-11-21, 04:06   #14
rogue

"Mark"
Apr 2003
Between here and the

22×1,607 Posts

Quote:
 Originally Posted by axn First off, I'm assuming that where you said "41579# % 31729 != 0", you meant "= 0".
Oops. You are correct.

 2010-11-21, 13:25 #15 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 102548 Posts 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
 2010-11-21, 14:07 #16 rogue     "Mark" Apr 2003 Between here and the 22·1,607 Posts 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.
2010-11-21, 19:48   #17
rogue

"Mark"
Apr 2003
Between here and the

22·1,607 Posts

Quote:
 Originally Posted by rogue 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.

 2010-11-22, 17:26 #18 Harvey563     Apr 2004 2738 Posts p(14)*p(j)+1 is not prime for any j <= 14.
 2010-11-22, 17:33 #19 Harvey563     Apr 2004 11×17 Posts p(19)#/p(j)+1 is not prime for any j <= 19
 2010-11-22, 17:58 #20 Harvey563     Apr 2004 11×17 Posts p(29)#*p(j)-1 is not prime for any j <=29 Last fiddled with by Harvey563 on 2010-11-22 at 18:03
 2010-11-25, 00:48 #21 rogue     "Mark" Apr 2003 Between here and the 22×1,607 Posts 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.
 2010-11-29, 22:02 #22 rogue     "Mark" Apr 2003 Between here and the 642810 Posts 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.

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