20161205, 16:57  #56 
Nov 2016
2^{2}·3·5·47 Posts 
Besides, do you miss any primes (that were found recently)? At beginning, you missed two primes: 4*737^269302+1 and 4*72^11198491. I don't know whether you miss other primes.
Last fiddled with by sweety439 on 20161205 at 17:20 
20161205, 18:56  #57  
May 2007
Kansas; USA
2^{3}·3·431 Posts 
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There should be no primes missing and the bases remaining lists should be correct because I have been updating them with primes found by the CRUS project. As you found the only problems were with search depth and k's remaining. The latter of those is mostly irrelevant for this particular effort but I show them as a curiosity. I will correct the lists in my records so that they are correct the next time that they are posted. Thanks for checking everything. Last fiddled with by gd_barnes on 20161205 at 19:00 

20161206, 12:10  #58  
Nov 2016
2^{2}×3×5×47 Posts 
Quote:
(1) b%3=2, b%5=3, b%1201=1194, b%17=10, b%169553=7 (2) b%3=2, b%5=3, b%1201=1194, b%17=7, b%169553=169546 (3) b%3=2, b%5=3, b%1201=7, b%17=11, b%169553=152746 (4) b%3=2, b%5=3, b%1201=7, b%17=6, b%169553=16807 (5) b%3=2, b%5=2, b%1201=858, b%17=3, b%169553=169210 (6) b%3=2, b%5=2, b%1201=858, b%17=14, b%169553=343 (7) b%3=2, b%5=2, b%1201=343, b%17=5, b%169553=145331 (8) b%3=2, b%5=2, b%1201=343, b%17=12, b%169553=24222 Which situation can give the smallest b, by the chinese remainder theorem? (note that the b should not congruent to 1 mod 2 or 1 mod 3, or k=7 become a trivial k) You said that you do not claim that this is minimal. Last fiddled with by sweety439 on 20161206 at 12:24 

20161206, 12:22  #59 
Nov 2016
2^{2}×3×5×47 Posts 
Like my research for Riesel k=15, there are 4 situations:
(1)b%7=6, b%113=98, b%17=2, b%1489=15 The smallest such b is 12196414. (2)b%7=6, b%113=98, b%17=15, b%1489=200 The smallest such b is 11047091. (3)b%7=6, b%113=15, b%17=8, b%1489=397 The smallest such b is 1099279. (4)b%7=6, b%113=15, b%17=9, b%1489=1092 The smallest such b is 8241218. The smallest number of which is 1099279. However, gcd(151, 10992791) is not 1. Thus, 15 is a trivial k of R1099279 and not a Riesel number to base 1099279. Besides, the second such k of (3) is 1099279+7*113*17*1489=21121862, which is much large. Thus, the (conjectured) smallest b such that 15 is a Riesel number to base b is 8241218 (I have checked that gcd(151, 82412181)=1) Last fiddled with by sweety439 on 20161206 at 12:25 
20161206, 16:01  #60  
"Nuri, the dragon :P"
Jul 2016
Good old Germany
2×13×31 Posts 
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I´m not that sort of math pro, but this is somethink like NumberFields@Home is doing. (It sound like that...) I can give you contact to the project admin, Eric. He knows a lot about the chinese reminder theorem an a lot of other math stuff. NF@H was created to prove his Doctorwork Just let me know if you want his private Mail. There is the project description: https://numberfields.asu.edu/NumberF...scription.html And his dissertation: https://numberfields.asu.edu/NumberF...ssertation.pdf . 

20161206, 21:16  #61 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·29·101 Posts 
I am not aware of any method that is better than running the CRT on all of these combinations.
However, it might not be necessary. It may be possible to find the smallest b with a given covering set easily once you know a possible k. This can be done using similar methods to what you did in post #49 for k=15. This ignores the fact that there could be a different covering set with a larger period that produces a smaller b. 
20170107, 20:44  #62 
May 2007
Kansas; USA
2^{3}×3×431 Posts 
I have now searched k=11 and 12 for all bases <= 1030. Therefore all k=2 thru 12 for all bases <= 1030 have been completed. All k=2 thru 7 have been searched to n=25K for all bases and k=8 thru k=12 have been searched to n=5K for all bases.
Attached are all primes for n<=5K found by my effort, n>5K found by CRUS, and bases remaining for each k. There have been some updates for k=2 thru 10 so all of k=2 thru 12 are included. Below are all exclusions including bases with trivial factors, algebraic factors, and covering sets for k=11 and 12. Exclusions for k<=10 were previously posted. Code:
Riesel k=11: b==(1 mod 2) has a factor of 2 b==(1 mod 5) has a factor of 5 b==(14 mod 15) has a covering set of [3, 5] Riesel k=12: b==(1 mod 11) has a factor of 11 b==(142 mod 143) has a covering set of [11, 13] base 307 has a covering set of [5, 11, 29] base 901 has a covering set of [7, 11, 13, 19] Sierp k=11: b==(1 mod 2) has a factor of 2 b==(1 mod 3) has a factor of 3 b==(14 mod 15) has a covering set of [3, 5] Sierp k=12: b==(1 mod 13) has a factor of 13 b==(142 mod 143) has a covering set of [11, 13] bases 562, 828, and 900 have a covering set of [7, 13, 19] base 563 has a covering set of [5, 7, 13, 19, 29] base 597 has a covering set of [5, 13, 29] bases 296 and 901 have a covering set of [7, 11, 13, 19] base 12 is a GFN with no known prime Admin edit: Moved all attachments to 1st post of this thread. Last fiddled with by gd_barnes on 20200830 at 11:33 Reason: remove attachments 
20170117, 02:15  #63 
Romulan Interpreter
Jun 2011
Thailand
24A3_{16} Posts 
Hey Gary, if you are still interested in this, I got a bit hooked with 2*b^n1 for small bases (b<=2048), and found (in your area of interest, b<=1030) that 2*303^401741 is prime.

20170120, 10:14  #64 
May 2007
Kansas; USA
2^{3}·3·431 Posts 
Great thanks! I keep my files updated with all primes found for these from everyone. I will note that one.

20170120, 16:13  #65 
Nov 2016
2^{2}×3×5×47 Posts 
Some additional exclusions for k<=12 and b<=2048:
Riesel k=6: all bases b=6*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 24, 54, 294, 384, 864, 1014, 1734, 1944) Riesel k=11: all bases b=11*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 44, 99, 539, 704, 1584, 1859) Last fiddled with by sweety439 on 20170120 at 16:44 
20170120, 21:40  #66  
May 2007
Kansas; USA
10100001101000_{2} Posts 
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