20170502, 00:34  #56 
May 2005
3·7^{2}·11 Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.61M
127*128^n1 tested till n=1.6M 
20170502, 02:16  #57 
"Curtis"
Feb 2005
Riverside, CA
4004_{10} Posts 
I am almost to 6M on 127*2^n1. Might you prepare a 6M7M file for me with the tests you've run removed? If that's inconvenient, I'll settle for a results file of your tests from 6M to 7M.

20170502, 05:24  #58 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
231B_{16} Posts 
Even before reaching 6M, by knowing that Borys already searched for these 127*2^(7*n)1 (reaching at least 7*n <= 11,200,000), you could simply remove all n :: 7n from your candidate list.

20170702, 23:05  #59 
May 2005
3×7^{2}×11 Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.63M
127*128^n1 tested till n=1.66M 
20171027, 08:40  #60 
May 2005
3×7^{2}×11 Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.66M
127*128^n1 tested till n=1.71M (sieving) 
20171202, 15:31  #61 
May 2005
3·7^{2}·11 Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.67M
127*128^n1 tested till n=1.72M 
20180101, 14:19  #62 
May 2005
1617_{10} Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.68M
127*128^n1 tested till n=1.73M 
20180101, 19:54  #63 
Nov 2016
6FE_{16} Posts 
Please see the project http://mersenneforum.org/showthread.php?t=21818&page=6 for the primes of the form (b+1)*b^n+1 and (probable) primes of the form b^n+(b+1). (b^n+(b+1) is the dual of (b+1)*b^n+1) I have searched these forms for all bases 2<=b<=1024.
Last fiddled with by sweety439 on 20180101 at 19:55 
20180202, 17:18  #64 
May 2005
651_{16} Posts 
status report
k=2 and k=4 @ base=3 tested till n=1.69M
127*128^n1 tested till n=1.74M 
20180412, 21:28  #65  
Nov 2016
11011111110_{2} Posts 
Quote:


20180415, 03:39  #66 
"Curtis"
Feb 2005
Riverside, CA
2^{2}·7·11·13 Posts 
I'm ready to search 78M, and may as well learn how to remove such candidates myself. How would I go about removing such n in a nonmanual fashion?
