20161201, 19:12  #45 
Nov 2016
4731_{8} Posts 
S = conjectured smallest base b such that k is a Sierpinski number.
k S cover set/algebraic factors 1 8 1*8^n+1 = (1*2^n+1) * (1*4^n1*2^n+1) period=1 2 201446503145165177 (?) {3, 5, 17, 257, 641, 65537, 6700417} period=64 3 none 4 14 {3, 5} period=2 5 140324348 {3, 13, 17, 313, 11489} period=16 6 34 {5, 7} period=2 7 none 8 8 8*8^n+1 = (2*2^n+1) * (4*4^n2*2^n+1) period=1 9 177744 {5, 17, 41, 193} period=8 10 32 {3, 11} period=2 11 14 {3, 5} period=2 12 142 {11, 13} period=2 13 20 {3, 7} period=2 14 38 {3, 13} period=2 15 none 16 38 {3, 5, 17} period=4 17 278 {3, 5, 29} period=4 18 322 {17, 19} period=2 19 14 {3, 5} period=2 20 56 {3, 19} period=2 21 54 {5, 11} period=2 22 68 {3, 23} period=2 23 32 {3, 11} period=2 24 114 {5, 23} period=2 R = conjectured smallest base b such that k is a Riesel number. k R cover set/algebraic factors 1 none 2 none 3 none 4 9 4*9^n1 = (2*3^n1) * (2*3^n+1) period=1 5 none 6 24 {5} for even n, 6*24^n1 = (12*24^((n1)/2)1) * (12*24^((n1)/2)+1) for odd n, period=2 7 ? {3, 5, 17, 1201, 169553} period=16 8 20 {3, 7} period=2 9 4 9*4^n1 = (3*2^n1) * (3*2^n+1) period=1 10 32 {3, 11} period=2 11 14 {3, 5} period=2 12 142 {11, 13} period=2 13 20 {3, 7} period=2 14 8 {3, 5, 13} period=4 15 ? {7, 17, 113, 1489} period=8 16 9 16*9^n1 = (4*3^n1) * (4*3^n+1) period=1 17 none 18 50 {17} for even n, 18*50^n1 = (30*50^((n1)/2)1) * (30*50^((n1)/2)+1) for odd n, period=2 19 14 {3, 5} period=2 20 56 {3, 19} period=2 21 54 {5, 11} period=2 22 68 {3, 23} period=2 23 32 {3, 11} period=2 24 114 {5, 23} period=2 Last fiddled with by sweety439 on 20161201 at 19:16 
20161201, 19:26  #46 
Nov 2016
2,521 Posts 
The term 201446503145165177 for Sierpinski problem with k=2 is the smallest known nontrivial base b (i.e. gcd(b1, 2+1)=1) such that 2*b^n+1 is composite for all n>=1, it is not known whether there is a smaller such base b. Thus, the list lists "201446503145165177 (?)", with a question mark.
There is a pdf file https://www.rosehulman.edu/mathjour...4/v9n24pd.pdf for the research of the reverseSierpinski problem, but not the reverseRiesel problem, can someone find the smallest nontrivial base b such that 7*b^n1 (or 15*b^n1) is composite for all n>=1? (I have found a possible cover set, like the cover set of 5*b^n+1 and 9*b^n+1, they are taken by the odd prime factors of the generated Fermat number with these bases, but the smallest nontrivial base b such that 5*b^n+1 (or 9*b^n+1) is composite for all n>=1 is given by the OEIS sequence https://oeis.org/A263500) I want to find the smallest base b with this cover set. Last fiddled with by sweety439 on 20161201 at 19:31 
20161202, 00:59  #47  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{3}×719 Posts 
Quote:
For k=7 a Riesel b is 33559116638. I do not claim that this is minimal. Combine the following statements to form a covering set and input into your favourite chinese remainder theorem calculator. Code:
If b = 2 mod 3 then p  7*b^n1 if n = 0 mod 2 If b = 2 mod 5 then p  7*b^n1 if n = 3 mod 4 If b = 3 mod 5 then p  7*b^n1 if n = 1 mod 4 If b = 3 mod 17 then p  7*b^n1 if n = 5 mod 16 If b = 5 mod 17 then p  7*b^n1 if n = 1 mod 16 If b = 6 mod 17 then p  7*b^n1 if n = 11 mod 16 If b = 7 mod 17 then p  7*b^n1 if n = 15 mod 16 If b = 10 mod 17 then p  7*b^n1 if n = 7 mod 16 If b = 11 mod 17 then p  7*b^n1 if n = 3 mod 16 If b = 12 mod 17 then p  7*b^n1 if n = 9 mod 16 If b = 14 mod 17 then p  7*b^n1 if n = 13 mod 16 If b = 7 mod 1201 then p  7*b^n1 if n = 7 mod 8 If b = 343 mod 1201 then p  7*b^n1 if n = 5 mod 8 If b = 858 mod 1201 then p  7*b^n1 if n = 1 mod 8 If b = 1194 mod 1201 then p  7*b^n1 if n = 3 mod 8 If b = 7 mod 169553 then p  7*b^n1 if n = 15 mod 16 If b = 343 mod 169553 then p  7*b^n1 if n = 5 mod 16 If b = 16807 mod 169553 then p  7*b^n1 if n = 3 mod 16 If b = 24222 mod 169553 then p  7*b^n1 if n = 1 mod 16 If b = 145331 mod 169553 then p  7*b^n1 if n = 9 mod 16 If b = 152746 mod 169553 then p  7*b^n1 if n = 11 mod 16 If b = 169210 mod 169553 then p  7*b^n1 if n = 13 mod 16 If b = 169546 mod 169553 then p  7*b^n1 if n = 7 mod 16 

20161202, 15:28  #48 
Nov 2016
2521_{10} Posts 
There is no base b such that k is a Sierpinski number if and only if k is of the form 2^n1 for positive integer n but k != 1. Similarly, there is no base b such that k is a Sierpinski number if and only if k is of the form 2^n1 for positive integer n but k != 9 (or k=1, since k=1 is a trivial k for all the Riesel bases b except 2, but R2 has an easy prime for k=1: 1*2^21). According to the Catalan conjecture (which was proven), 1 and 9 are the only two perfect powers next to a power of 2. If this k is of these forms, then k+1 (or k1) is a power of 2 and the only prime factor of it is 2, since the n=0 case would cause the number to be a power of 2. Thus, 2 must be in the cover set, so both of k and b must be odd and both of b1 and k+1 are divisible by 2, and this k is a trivial k. The unique possible situation is: there is algebraic factors of k*b^n+1. Thus, there exists an n such that k*b^n is a perfect power, and since the n=0 case would cause the number to be a power of 2, n=0 must be in the algebraic n's. Thus, n must be a perfect power, due to the Catalan conjecture, for the Sierpinski case, n must be 1, and for the Riesel case, n must be 9.
Similarly, if there exists an n such that all of the prime factor of b^n+k are prime factors of b or prime factors of k, then the sequence of the smallest prime factor of k*b^n+1 is unbounded above, since b^n+k is the dual of k*b^n+1. Besides, if there exists an n such that all of the prime factor of b^nk are prime factors of b or prime factors of k, then the sequence of the smallest prime factor of k*b^n1 is unbounded above, since b^nk is the dual of k*b^n1. If k*b^n+1 (or k*b^n1) is composite for all n>=1 and with a cover set, then (b^n+k)/gcd(b^n, k) (or (b^nk)/gcd(b^n, k)) is also composite for all n>=1 and with the same cover set. We can find a dual prime to prove the dual Sierpinski/Riesel conjecture base b, or the mixed Sierpinski/Riesel conjecture base b. In fact, we want to find the smallest positive integer k such that gcd(k+1, b1)=1 (+ for Sierpinski,  for Riesel) but the sequence of the smallest prime factor of k*b^n+1 (+ for Sierpinski,  for Riesel) is bounded above, it is conjectured that these k's are the same as the CRUS k's (the smallest k such a full cover set). However, it is not proven, e.g. is the sequence of the smallest prime factor of 5*2^n+1 unbounded above? Even this case is not proven. Last fiddled with by sweety439 on 20161202 at 15:47 
20161202, 15:46  #49 
Nov 2016
2,521 Posts 
33559116638 may be the smallest base b with this cover set: {3, 5, 17, 1201, 169553} for 7*b^n1. Like this number, 140324348, it may be the smallest b with this cover set: {13, 17, 313, 11489} for 5*b^n+1.
For the k=15 case, I want to find the smallest base with this cover set: {7, 17, 113, 1489}, this is my research: If b = 6 mod 7 then 7  15*b^n1 if n = 0 mod 2 If b = 98 mod 113 then 113  15*b^n1 if n = 1 mod 4 If b = 15 mod 113 then 113  15*b^n1 if n = 3 mod 4 If b = 8 mod 17 then 17  15*b^n1 if n = 1 mod 8 If b = 2 mod 17 then 17  15*b^n1 if n = 3 mod 8 If b = 9 mod 17 then 17  15*b^n1 if n = 5 mod 8 If b = 15 mod 17 then 17  15*b^n1 if n = 7 mod 8 If b = 1092 mod 1489 then 1489  15*b^n1 if n = 1 mod 8 If b = 200 mod 1489 then 1489  15*b^n1 if n = 3 mod 8 If b = 397 mod 1489 then 1489  15*b^n1 if n = 5 mod 8 If b = 15 mod 1489 then 1489  15*b^n1 if n = 7 mod 8 I found the smallest base is 1099279, but it is also a trivial base (i.e. gcd(151, 10992791) is not 1). Thus, it is the next smallest base, 8241218. Last fiddled with by sweety439 on 20161202 at 15:47 
20161202, 15:49  #50 
Nov 2016
2521_{10} Posts 
Thanks! I completed my lists for k up to 24:
S = conjectured smallest base b such that k is a Sierpinski number. k S cover set/algebraic factors 1 8 1*8^n+1 = (1*2^n+1) * (1*4^n1*2^n+1) period=1 2 201446503145165177 (?) {3, 5, 17, 257, 641, 65537, 6700417} period=64 3 none 4 14 {3, 5} period=2 5 140324348 {3, 13, 17, 313, 11489} period=16 6 34 {5, 7} period=2 7 none 8 8 8*8^n+1 = (2*2^n+1) * (4*4^n2*2^n+1) period=1 9 177744 {5, 17, 41, 193} period=8 10 32 {3, 11} period=2 11 14 {3, 5} period=2 12 142 {11, 13} period=2 13 20 {3, 7} period=2 14 38 {3, 13} period=2 15 none 16 38 {3, 5, 17} period=4 17 278 {3, 5, 29} period=4 18 322 {17, 19} period=2 19 14 {3, 5} period=2 20 56 {3, 19} period=2 21 54 {5, 11} period=2 22 68 {3, 23} period=2 23 32 {3, 11} period=2 24 114 {5, 23} period=2 R = conjectured smallest base b such that k is a Riesel number. k R cover set/algebraic factors 1 none 2 none 3 none 4 9 4*9^n1 = (2*3^n1) * (2*3^n+1) period=1 5 none 6 24 {5} for even n, 6*24^n1 = (12*24^((n1)/2)1) * (12*24^((n1)/2)+1) for odd n, period=2 7 33559116638 {3, 5, 17, 1201, 169553} period=16 8 20 {3, 7} period=2 9 4 9*4^n1 = (3*2^n1) * (3*2^n+1) period=1 10 32 {3, 11} period=2 11 14 {3, 5} period=2 12 142 {11, 13} period=2 13 20 {3, 7} period=2 14 8 {3, 5, 13} period=4 15 8241218 {7, 17, 113, 1489} period=8 16 9 16*9^n1 = (4*3^n1) * (4*3^n+1) period=1 17 none 18 50 {17} for even n, 18*50^n1 = (30*50^((n1)/2)1) * (30*50^((n1)/2)+1) for odd n, period=2 19 14 {3, 5} period=2 20 56 {3, 19} period=2 21 54 {5, 11} period=2 22 68 {3, 23} period=2 23 32 {3, 11} period=2 24 114 {5, 23} period=2 Note: The Sierpinski case the base b does not exist if and only if k is of the form 2^n1 and k != 1. The Riesel case the base b does not exist if and only if k is of the form 2^n+1 and k != 9, or k = 1. Last fiddled with by sweety439 on 20161202 at 15:50 
20161202, 16:02  #51 
Nov 2016
2,521 Posts 
It is only conjectured these are the smallest Sierpinski/Riesel base for these k's, like the original Sierpinski/Riesel problems, these are also only conjectured (e.g. 6694 is the smallest base 22 Sierpinski number, and 4461 is the smallest base 22 Riesel number, both of them are not proven and with 1 k remain). However, it is very hard (as I think) to proven the Sierpinski k=2, k=5, and k=9 case and the Riesel k=7 and k=15 case. Besides, there is no base b such that k=3, 7, 15, ... is a Sierpinski number, and there is no base b such that k=2, 3, 5, 17, ... is a Riesel number (of course for Riesel k=1 case, but k=1 is a trivial k for all base b>=3 and there is a prime for base b=2: 1*2^21. Thus, 1 is not a Riesel number to any base b). However, it is also only conjectured, it is unknown that for every nontrivial base b, there is a prime of the form 3*b^n+1, 7*b^n+1, 15*b^n+1, 2*b^n1, 3*b^n1, 5*b^n1, 17*b^n1, ..., I think since there are infinitely many bases b need to be tested, unlike the 2*b^n+1, 5*b^n+1, 9*b^n+1, 7*b^n1, 15*b^n1, ... cases (these cases there are only a finite number of base b need to be tested), the problem may be still unsolved at 2200!
Last fiddled with by sweety439 on 20161202 at 16:03 
20161202, 19:46  #52 
Nov 2016
2,521 Posts 
These are the Sierpinski k=12 primes up to b=1030.

20161205, 05:39  #53 
May 2007
Kansas; USA
10100000000110_{2} Posts 
I have now searched all k=2 thru k=10 for bases <= 1030. As previously stated all k=2 thru 7 have been searched to n=25K for all bases. The new effort for k=8 thru k=10 has been searched to n=5K for all bases.
Also done with the effort were: 1. Research on the CRUS pages and prime files for k=8 thru 10 to add primes for n>5K. 2. Put primes found for n>5K into separate files for all bases. 3. Check all exclusions, that is bases with trivial factors, algebraic factors, and covering sets, previously posted by Sweety for k=8 thru k=10. All were correct with the exception of k=8 on both sides where I found some additional covering sets. I previously posted all exclusions for k=2 thru k=7. Shown below are exclusions for k=8 thru 10: Code:
Riesel k=8: b==(1 mod 7) has a factor of 7 b==(20 mod 21) has a covering set of [3, 7] *b==(83, 307 mod 455) has a covering set of [5, 7, 13] (This includes bases 83, 307, 538, 762, 993.) b=m^3 proven composite by full algebraic factors Riesel k=9: b==(1 mod 2) has a factor of 2 b==(4 mod 5): odd n has a factor of 5; even n has algebraic factors b=m^2 proven composite by full algebraic factors Riesel k=10: b==(1 mod 3) has a factor of 3 b==(32 mod 33) has a covering set of [3, 11] Sierp k=8: b==(1 mod 3) has a factor of 3 b==(20 mod 21) has a covering set of [3, 7] *b==(47 or 83 mod 195) has a covering set of [3, 5, 13] (This includes bases 47, 83, 242, 278, 437, 473, 632, 668, 827, 863, 1022.) *base 467 has a covering set of [3, 5, 7, 19, 37] *base 722 has a covering set of [3, 5, 13, 73, 109] b=m^3 proven composite by full algebraic factors base 128 is a GFN with no possible prime Sierp k=9: b==(1 mod 2) has a factor of 2 b==(1 mod 5) has a factor of 5 Sierp k=10: b==(1 mod 11) has a factor of 11 b==(32 mod 33) has a covering set of [3, 11] base 1000 is a GFN with no known prime Attached are the following: 1. Primes for all n<=5000. 2. Primes for all n>5000. 3. Remaining bases and their search limits for each k. Sometime in the next month I will search all bases for k=11 and k=12 to n=5K. I also hope to eventually search all k=8 thru 12 to n=25K like I did for k=2 thru 7 many years ago. Doing so requires a lot of personal effort because nearly all k/base combos must be sieved separately. Last fiddled with by gd_barnes on 20161205 at 06:32 
20161205, 12:51  #54 
Nov 2016
4731_{8} Posts 
Thanks very much!

20161205, 14:03  #55 
Nov 2016
4731_{8} Posts 
The files are not right: 10*611^n1 (R611, k=10) is already tested to n=200K, not just 5K.
Besides, why the "number of remaining k's" column of 7*1004^n+1 lists 2k, but that of 10*1004^n+1 lists 1k? S1004 has conjecture only k=4, and with only k=2 (for k<4) remaining, but k=7 and k=10 are also remaining with k's > conjecture k. Last fiddled with by sweety439 on 20161205 at 14:03 