20101129, 23:05  #23 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010101110_{2} Posts 
This search is much like twin prime searches: you must first find a relatively rare occurrence, (in twin prime case a prime; in this case a multiply or divide with no primes) then be lucky enough to find another equally rare occurrence (assuming this conjecture is wrong and the twin prime conjecture is true) on the same number (in twin prime a prime two more, or less, than the prime you just found; in this case no primes on the other of multiply or divide side). Just like with a twin search, finding the first rare occurrence is just the first step. So unfortunately "Close, but no cigar" is an overlyoptimistic statement. If you had searched one full side and about 90% of the other side with no primes, then you'd probably expect under a 50% chance of finding a prime in the last 10%, and then you really would be close.
Last fiddled with by MiniGeek on 20101129 at 23:09 
20101129, 23:13  #24 
"Mark"
Apr 2003
Between here and the
2·5·647 Posts 
The current sieve depth is 2.9e11 with a removal rate of 1 every three seconds. There are 5.14e6 candidates remaining. With the current range I am testing I am testing about 20% of the candidates that remain after sieving and with the other range about 25%. With the entire range I am sieving, the average test takes about 25 seconds, so that implies that the sweet spot for sieving is a removal rate about of about 1 every 6 seconds.
I can now provide some estimates for the amount of time to do the entire range (3500<k<5000). When sieving completes, there will be about 5 million candidates. Remove the ranges that have been done or are reserved leaves about 4.5 million candidates. With an average test time of 25 seconds and having to test about 25% of the candidates in the range means that remaining part of this range should take about 325 days on a single core. When sieving is done I will post ranges of 100 k along with some scripts to use with PFGW. If enough people join the search, I will commence sieving from 5000<=k<10000 after the holidays. Eventually I will post the sieving code, which is portable to multiple platforms. I might even change PRPNet to support this project. There is no proof for the conjecture. I have no idea how one would prove it. I think the better bet is to show that the conjecture is false. 
20101201, 19:36  #25 
"Mark"
Apr 2003
Between here and the
2·5·647 Posts 
Here are the results from 39503599:
Code:
p(3950)#/p(503)1 p(3951)#/p(2354)1 p(3952)#/p(243)+1 p(3953)#/p(234)1 p(3954)#/p(2693)+1 p(3955)#/p(169)1 p(3956)#/p(1599)+1 p(3957)#/p(1302)+1 p(3958)#/p(476)+1 p(3959)#/p(549)+1 p(3960)#/p(66)+1 p(3961)#/p(353)1 p(3962)#/p(152)1 p(3963)#/p(177)+1 p(3964)#/p(1811)+1 p(3965)#/p(735)+1 p(3966)#/p(747)1 p(3967)#/p(136)1 p(3968)#/p(725)1 p(3969)#/p(1508)+1 p(3970)#/p(1375)1 p(3971)#/p(109)1 p(3972)#/p(571)+1 p(3973)#/p(433)+1 p(3974)#/p(1084)1 p(3975)#/p(217)+1 p(3976)#/p(96)+1 p(3977)#/p(1042)+1 p(3978)#/p(69)+1 p(3979)#/p(728)1 p(3980)#/p(925)+1 p(3981)#/p(248)1 p(3982)#/p(1690)1 p(3983)#/p(1470)+1 p(3984)#/p(455)1 p(3985)#/p(113)+1 p(3986)#/p(490)1 p(3987)#/p(361)+1 p(3988)#/p(2657)+1 p(3989)#/p(357)+1 p(3990)#/p(937)+1 p(3991)#/p(1065)1 p(3992)#/p(254)+1 p(3993)#/p(223)+1 p(3994)#/p(194)1 p(3995)#/p(525)+1 p(3996)#/p(1954)1 p(3997)#/p(335)+1 p(3998)#/p(570)+1 p(3999)#/p(368)+1 I'm taking 40004099. 
20101203, 01:38  #26 
"Mark"
Apr 2003
Between here and the
2×5×647 Posts 
Here is 39003949:
Code:
p(3900)#/p(118)+1 p(3901)#/p(3462)+1 p(3902)#/p(3021)+1 p(3903)#/p(139)1 p(3904)#/p(416)+1 p(3905)#/p(646)+1 p(3906)#/p(485)+1 p(3907)#/p(618)1 p(3908)#/p(1411)+1 p(3909)#/p(235)+1 p(3910)#/p(121)+1 p(3911)#/p(121)+1 p(3912)#/p(819)1 p(3913)#/p(183)+1 p(3914)#/p(2204)1 p(3915)#/p(706)+1 p(3916)#/p(306)+1 p(3917)#/p(290)1 p(3918)#/p(249)1 p(3919)#/p(99)1 p(3920)#/p(1937)1 p(3921)#/p(622)1 p(3922)#/p(1309)1 p(3923)#*p(1938)1 p(3924)#/p(272)1 p(3925)#/p(734)1 p(3926)#/p(382)+1 p(3927)#/p(1976)1 p(3928)#/p(1855)+1 p(3929)#/p(28)+1 p(3930)#/p(993)1 p(3931)#/p(182)1 p(3932)#/p(1026)+1 p(3933)#/p(449)1 p(3934)#/p(915)+1 p(3935)#/p(2787)1 p(3936)#/p(2008)+1 p(3937)#/p(391)+1 p(3938)#/p(9)+1 p(3939)#/p(304)+1 p(3940)#/p(565)+1 p(3941)#/p(1111)+1 p(3942)#/p(1235)+1 p(3943)#/p(2734)1 p(3944)#/p(1917)1 p(3945)#/p(536)+1 p(3946)#/p(806)+1 p(3947)#/p(61)+1 p(3948)#/p(495)+1 p(3949)#/p(41)1 
20101203, 14:29  #27 
"Mark"
Apr 2003
Between here and the
14506_{8} Posts 
I have posted a zip file here. It contains the following:
1) The source to p10sieve, the sieving program. I have not tested it on x86 (yet). I have tested it on PPC. I'll eventually get around to testing an x86 build. I don't expect any significant bugs in an x86 build as most of the logic (both x86 and PPC) was borrowed from psieve. This program does the divide side (both + and ) only. I haven't added the multiply side yet because it is waiting on a change to PFGW on how PFGW scripts handle input files. I want p10sieve to output a j.in file with a '/' or '*' in the first column, but PFGW won't allow for it. 2) 13 pairs of k.in and j.in files in groups of 100 distinct k values. 35003549 and 39004099 are excluded. Most (if not all) of these files can be done on a single core in less than a month. 3) The script to use with PFGW for this project. It presumes k.in and j.in are in the same folder as PFGW and the script. 4) A script to doublecheck. If you find a k for which no PRP is found, modify this script file and run through PFGW. It will test all multiply and divide numbers (both + and ) for the k to find a PRP. If none are found, then the conjecture is disproven. Note that you should use f with this script. 
20101203, 14:44  #28  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
2×5×7×61 Posts 
Does this mean you're done sieving?
Quote:
Quote:
Last fiddled with by MiniGeek on 20101203 at 14:54 

20101203, 15:49  #29  
"Mark"
Apr 2003
Between here and the
1946_{16} Posts 
Quote:
Yes, I am done sieving. I sieved to about 4e11. p10dc.sript (the doublecheck script) does both, but it would be easy to remove the tests you don't want to do. You are correct, one could use the poorly sieved file to do the multiply side if the divide side doesn't yield a prime. If you want to capture residues, that is up to you. I haven't been. 

20101203, 16:12  #30 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010101110_{2} Posts 
Hmm, I'm getting another corrupt download, and it's not ~double the size of the last one. Last one was 4.26 MB, this one is 5.16 MB and has an MD5 of D50207D6A5AF34642CE6931EA5FD7C91 (consistent over two downloads  one through my browser Chrome and the other through wget, to be sure the problem isn't with me).

20101203, 16:48  #31  
Jun 2003
5196_{10} Posts 
Quote:


20101203, 17:48  #32 
"Mark"
Apr 2003
Between here and the
2·5·647 Posts 
Apparently my web provider is having a problem with the file and it is too large to attach here. If someone wants to post the 10 MB file, let me know via PM. Give me your email and I'll send it to you.

20101203, 18:23  #33  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
2·5·7·61 Posts 
Quote:


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