20161125, 07:26  #34  
May 2007
Kansas; USA
2^{3}×3×431 Posts 
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I found all of my files of primes where n<=5000 for these efforts. They are not sorted quite like you want but you should be able to get them into your desired format. They will be attached to the next post. Last fiddled with by gd_barnes on 20161125 at 08:03 

20161125, 07:55  #35 
May 2007
Kansas; USA
2^{3}×3×431 Posts 
Attached are all primes for k=2 thru k=7 for bases<=1030 where n<=5000. These are sorted by nvalue. Combining these primes with all previously posted primes for n>5000 in post #27 should give all known primes for these k and base ranges.
Last fiddled with by gd_barnes on 20161125 at 07:56 
20161128, 12:14  #36 
Nov 2016
101100000100_{2} Posts 
There is another problem, the reversed Sierpinski/Riesel problem.
For fixed k, find the smallest base b such that all numbers of the form k*b^n+1 (k*b^n1) are composite. If k is of the form 2^n1 (2^n+1), except k=1 (k=9), it is conjectured for every nontrivial base b, there is a prime of the form k*b^n+1 (k*b^n1). However, for all other k's, there is a base b such that all numbers of the form k*b^n+1 (k*b^n1) are composite. For the Sierpinski (k*b^n+1) cases: (for k up to 24, only tested bases b<=1030) S = conjectured smallest base b such that k is a Sierpinski number. k S remaining bases b with no known primes 1 8 proven 2 201446503145165177 (?) {218, 236, 365, 383, 461, 512, 542, 647, 773, 801, 836, 878, 908, 914, 917, 947, 1004, ...} 3 none {718, 912, ...} 4 14 proven 5 140324348 {308, 326, 512, 824, ...} 6 34 proven 7 none {1004, ...} 8 8 proven 9 177744 {244, 592, 724, 884, 974, 1004, ...} 10 32 proven 11 14 proven 12 142 {12} 13 20 proven 14 38 proven 15 none {398, 650, 734, 874, 876, 1014, ...} 16 38 {32} 17 278 {68, 218} 18 322 {18, 74, 227, 239, 293} 19 14 proven 20 56 proven 21 54 proven 22 68 {22} 23 32 proven 24 114 {79} Last fiddled with by sweety439 on 20161128 at 12:28 
20161128, 17:51  #37 
Nov 2016
2^{2}·3·5·47 Posts 
This is I found primes for Sierpinski k=10 with base b<=1030. I didn't search very far, so there are many bases remaining.
I take Sierpinski k=10 because of https://oeis.org/A088622 https://oeis.org/A088782 and https://oeis.org/A088783. The smallest prime obtained as the concatenation of a power of b followed by a 1, is in fact the smallest prime of the form 10*b^n+1. Thus, this is the reverse Sierpinski problem with k=10. A088783 lists the bases for which 10 is a Sierpinski number or a trivial k (gcd(10+1, b1) is not 1), but only up to 166, since when this sequence was created, 10 is still a remaining k for S173. Recently, a prime 10*173^264234+1 was found, next term of this sequence should be 177. However, the correct value of the term after 177 is still unknown, since 10 is a remaining k for S185 (at n=1M). Last fiddled with by sweety439 on 20161128 at 17:53 
20161128, 20:43  #38  
May 2007
Kansas; USA
2^{3}·3·431 Posts 
Quote:
1. k=1 is not applicable because all odd b have a trivial factor of 2 and all even b are GFNs. 2. For k=2, b=512 is removed since 2*512^n+1=2^(9n+1)+1. 3. For k=12, b=12 is removed since 12*12^n+1=12^(n+1)+1; k=12 is proven. 4. For k=16, b=32 is removed since 16*32^n+1=2^(5n+4)+1; k=16 is proven. 5. For k=18, b=18 is removed since 18*18^n+1=18^(n+1)+1. 6. For k=22, b=22 is removed since 22*12^n+1=22^(n+1)+1; k=22 is proven. One question: How did you determine the conjectures for k=2, k=5, and k=9? Last fiddled with by gd_barnes on 20161129 at 03:47 

20161129, 05:28  #39  
May 2007
Kansas; USA
2^{3}×3×431 Posts 
Quote:
10*460^751+1 10*708^17562+1 10*830^436+1 10*927^4752+1 10*954^1506+1 10*1012^426+1 

20161129, 12:41  #40 
Nov 2016
2820_{10} Posts 
There is an OEIS sequence for the reverseSierpinski problem https://oeis.org/A263500.
However, no OEIS sequence for the reverseRiesel problem, this is my research of this problem: R = conjectured smallest base b such that k is a Riesel number. k R remaining bases b with no known primes 1 none (for all base b>2, 1 is a trivial k, and for base b=2, there is a prime with n=2: 1*2^21) 2 none 3 none 4 9 proven 5 none 6 24 proven 7 ? (I found no information of this term, can someone find it?) 8 20 proven 9 4 proven 10 32 proven 11 14 proven 12 142 {65, 98} 13 20 proven 14 8 proven 15 ? (I found no information of this term, can someone find it?) 16 9 proven 17 none 18 50 proven 19 14 proven 20 56 proven 21 54 proven 22 68 {38, 62} 23 32 proven 24 114 {64} There are also OEIS sequence for Sierpinski/Riesel problem: Sierpinski problem: https://oeis.org/A123159 Riesel problem: https://oeis.org/A273987 Last fiddled with by sweety439 on 20161129 at 13:41 
20161129, 12:55  #41 
Nov 2016
2^{2}×3×5×47 Posts 
Thanks for some k=10 prime.
About one month ago, I take my effort to find k=10 prime for b=269, 278, 282, 284, 356. There are some bases b<=1030 remain for k=10: 185 (1M) 338 (100K) 417 (400K) 432 ? 449 ? 537 ? 614 ? 668 ? 671 ? 726 ? 728 ? 743 (200K) 744 (100K) 773 (200K) 786 ? 827 ? 863 ? 869 ? 885 ? 929 ? 935 (200K) 959 ? 977 (100K) 986 ? 1000 (GFN, searched up to (2^251)/31) 1004 ? These bases are either not started or have a conjectured k<10. Last fiddled with by sweety439 on 20161129 at 12:58 
20161129, 18:46  #42 
Nov 2016
2^{2}×3×5×47 Posts 
For the remaining bases for the reverseRiesel problem for k=2, 3, 5, 7, 15 and 17:
k R remaining bases b with no known primes 2 none {303, 522, 578, 581, 992, 1019, ...} 3 none {588, 972, ...} 5 none {338, 998, ...} 7 ? {308, 392, 398, 518, 548, 638, 662, 848, 878, ...} 15 ? {454, 552, 734, 856, ...} 17 none {98, 556, 650, 662, 734, ...} 
20161129, 18:54  #43  
May 2007
Kansas; USA
2^{3}·3·431 Posts 
Quote:
Just my two cents. Last fiddled with by gd_barnes on 20161129 at 19:10 

20161129, 18:57  #44 
Nov 2016
2^{2}·3·5·47 Posts 
According to https://web.archive.org/web/20160507...t/riesel2.html, 3*2^181231 is prime, that is, 24*64^30201 is prime. Thus, base 64 can be removed for the reverseRiesel problem with k=24, and the reverseRiesel problem with k=24 is proven. (This is the smallest prime of the form 24*64^n1 with n>1, since 18123 is the smallest number greater than 3 and = (3 mod 6) in the k=3 list.
Last fiddled with by sweety439 on 20161129 at 18:59 