20200327, 09:35  #23  
Quasi Admin Thing
May 2005
921_{10} Posts 
Quote:


20200327, 09:48  #24  
May 2007
Kansas; USA
10011110111110_{2} Posts 
Quote:
Because it is the lowest k that has a known covering set of numeric factors for R1024. A factor of 5 covers evenn and a factor of 41 covers oddn. Smaller k's k==(1 mod 3, 1 mod 11, & 1 mod 31) have a single trivial factor that eliminates them or others (k=9 & 36) need algebraic factors to eliminate them from consideration. k=29 does not have a known covering set and is not eliminated by one of the conditions above so we believe it will eventually have a prime. But like many other bases that is just a conjecture at this point. :) Last fiddled with by gd_barnes on 20200327 at 09:51 

20200327, 14:48  #25 
Nov 2016
2^{2}×3×179 Posts 
An example is Sierp base 311, all k's that between a multiple of 3 and a multiple of 13 have the covering set {3,13}, the first such k's are 14, 25, 53, 64, 92, 103, 131, 142, ..., however all such k's < 142 are also trivial (i.e. gcd(k+1,3111) is not 1), thus the conjectured k for Sierp base 311 is 142. (this conjecture is proven, all k's except 10, 46, and 76 have an easy prime)

20200327, 14:54  #26 
Romulan Interpreter
Jun 2011
Thailand
21BB_{16} Posts 
Then why did you eliminate 9 and 36?
(the tree I am barking at, is the fact that 81 is a square) 
20200327, 15:14  #27  
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}·7^{3} Posts 
Quote:
to get a riesel/sierpinski cover you have to use d_i divisors in each remainder class in the covering set, you can't use algebraic factors, you can't use even in those cases where you could use, like in your example for k=81 we could cover the n==0 mod 2 case, because 81*1024^n1 has an algebraic factor. For this reason we can rule out say k=9, it is sure that this sequence contains no prime, but it has no covering set. The proof could be hard/impossible(?), but just see the prime factorization for n=54: http://factordb.com/index.php?id=1100000000033081963 . ps. you could also cover the n==0 mod 1 so every integer for k=81 with an algebraic factor. Last fiddled with by R. Gerbicz on 20200327 at 15:17 

20200327, 15:43  #28 
Nov 2016
100001100100_{2} Posts 

20200327, 15:45  #29  
Nov 2016
2^{2}·3·179 Posts 
Quote:


20200327, 15:51  #30  
Nov 2016
2^{2}·3·179 Posts 
Quote:
Thus these k*2^n+1 cannot have a covering set: 1*2^n+1 (for n=0, the value is 2) 3*2^n+1 (for n=0, the value is 4) 7*2^n+1 (for n=0, the value is 8) 15*2^n+1 (for n=0, the value is 16) 31*2^n+1 (for n=0, the values is 32) 1*2^n1 (the dual form 2^n1, for n=1, the value is 1) 3*2^n1 (for n=0, the value is 2, also the dual form 2^n3, for n=2, the value is 1) 5*2^n1 (for n=0, the value is 4) 7*2^n1 (the dual form 2^n7, for n=3, the value is 1) 9*2^n1 (for n=0, the value is 8) 15*2^n1 (the dual form 2^n15, for n=4, the value is 1) 17*2^n1 (for n=0, the value is 16) 31*2^n1 (the dual form 2^n31, for n=5, the value is 1) 33*2^n1 (for n=0, the value is 32) etc. However, there is no proof that 5*2^n+1, 9*2^n+1, 11*2^n+1, 13*2^n+1, 17*2^n+1, ... have no covering set (i.e. the sequence of the smallest prime factor of k*2^n+1 is unbounded above). Last fiddled with by sweety439 on 20200327 at 15:54 

20200327, 19:59  #31 
May 2007
Kansas; USA
27BE_{16} Posts 
Put more simply for R1024, for k=81 all n's have a numeric covering set of factors [5, 41]. For k=9 and 36 that is not the case. They require algebraic factors for the even n to eliminate them.

20200327, 20:31  #32  
"Robert Gerbicz"
Oct 2005
Hungary
2534_{8} Posts 
Quote:
n==1 mod 2 is divisible by 5 n==2 mod 6 is divisible by 7 n==4 mod 6 is divisible by 13 Ofcourse you can go further so where you need the algebraic factor for less than 1/6 part of the cases. But you will need in at least one remainder class the algebraic factor (without a proof). 

20200328, 03:21  #33 
Romulan Interpreter
Jun 2011
Thailand
5×11×157 Posts 
Ok. Thanks all three (yep, including sweety, in spite of the fact that most of his/her posts are clutter, he/she gave the example with 125 )
I have learned something new today. I was thinking that stuff with algebraic factors are excluded completely (because they are always composite, so Mr. Riesel respective Sierpinsky, would have nothing to "prove" in this case. It is clear that, if you have algebraic factorization, then k*b^n1 is always composite. Nothing to prove). The "covering set" should be applied to what is left after getting rid of the sequences with algebraic factors. It seems I was wrong (which I still refuse to accept, but that is not your problem ) Last fiddled with by LaurV on 20200328 at 03:21 
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