20150108, 11:52  #45 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
5·23·41 Posts 

20150217, 02:03  #46 
Sep 2006
The Netherlands
677 Posts 
Good Early Morning! Fearing my hack in tool producing what to test for 69 * 2^n  1 has bug
3M  4M had roughly 53892 exponents to test and range am starting now at a few cores 4M  5M has roughly 53990 exponents to test Sounds weird to me such "huge" range has more exponents to test. http://www.mersenneforum.org/showthr...t=18255&page=5 shows both ranges sieved to 400P. Note i might have used tad older abcd file for 3m4m range to do this quick compare, whereas in reality i had upgraded in between testing the abcd file, so i suddenly had less to test then. Yet i'm bit confused why this difference is there, anyone? 
20150217, 06:27  #47 
"Curtis"
Feb 2005
Riverside, CA
1000011100101_{2} Posts 
Both ranges span 1 million, both sieved to the same level, and the number of tests differs by 100. What is it you find strange? Can you rephrase your question?

20150217, 09:27  #48  
Feb 2003
3·5·127 Posts 
Quote:
And (just for comparison) for the range 5M6M there are 53731 exponents. As VBCurtis already mentioned: The difference is quite small and is just the typical fluctuation in the distribution of surviving exponents after sieving. 

20150217, 10:30  #49  
May 2009
Russia, Moscow
3^{2}×5^{2}×11 Posts 
Quote:
Code:
51323 1M/t17_b2_k5.npg 48098 2M/t17_b2_k5.npg 45757 3M/t17_b2_k5.npg 34694 4M/t17_b2_k5.npg 34752 5M/t17_b2_k5.npg 34774 6M/t17_b2_k5.npg Code:
45919 1M/t17_b2_k33.npg 46129 2M/t17_b2_k33.npg 46271 3M/t17_b2_k33.npg 44182 4M/t17_b2_k33.npg 43896 5M/t17_b2_k33.npg 43877 6M/t17_b2_k33.npg 

20150217, 10:45  #50 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
5·23·41 Posts 
First there was some kind of sieve gap and second they forgot to take out the algebraic factors from k=5.
More in this thread: http://www.mersenneforum.org/showthread.php?t=19170 
20150217, 11:38  #51  
Feb 2003
3·5·127 Posts 
Quote:
The number of factors for a given sieve range (from p1 to p2) can be estimated by N1*(1log(p1)/log(p2)), where N1 is the number of candidates at sieve level p1. Then the number of candidates surviving the sieve up to p2 should be roughly N2 = N1*log(p1)/log(p2). By taking N1=46000 at p1=100P (=100*10^15) we get N2=44427 at p2=400P. This makes the counts given above for k=33 (and all the other k's except k=5) quite plausible for me... Last fiddled with by Thomas11 on 20150217 at 11:58 

20150217, 14:51  #52 
Sep 2006
The Netherlands
677 Posts 
Thanks for the explanations and most interesting estimation formula!
At the risk of being wrong, i tend to remember when i trial factored Wagstaff ( (2^n + 1) / 3 ) that odds dramatic low that at a reasonable large domain, less exponents would be left than in a similar domain, given the same sieve depth. Yet quite possible that with the much tougher sieving that's needed for Riesel, that sieve depths, though impossible to do at home that deep, still aren't deep enough for statistical logics to become reality. 
20150221, 12:53  #53 
Sep 2006
The Netherlands
677 Posts 
Current status.
Everything tested once up to: 69*2^36143051 is not prime. LLR Res64: 09A4B35ED567A38D Time : 12454.741 sec. 
20150404, 19:25  #54 
Sep 2006
The Netherlands
1010100101_{2} Posts 
Everything tested once up to 3.79M
Odds ticking away there still is a new gem < 4M 
20150405, 06:07  #55 
Nov 2003
2×1,811 Posts 
Hi diep,
Congrats on your dedication and a great run from n=2.5M, approaching 4M soon. No worries about no new primes. The more composites, the more reasons to rejoice since every new test has higher and higher probability to produce a new prime! :) 