Quote:
Originally Posted by robert44444uk
Some notes from unpublished work by Prof Caldwell
Definition of a Sierpinski number
An integer k > 1 is a Sierpinski number base b if gcd(k+1,b1) = 1 and
k.bn+1 is composite for all n > 0.
• gcd(k+1,b1) = 1 avoids trivial covers (1covers).
• k > 1 avoids leading Generalized Fermat divisors. (May use Strong Sierpinski for GFN’s included, and may toss out further GFN’s in the weak case above.)
• n > 0 avoids removing k = p1 as a multiplier for all primes p and all bases b. (Shouldn’t b be involved in the choice of k?)

Thanks for the info. I understand everything but the last two here. On the 2nd and 3rd ones, can you give specific examples on what you are referring to. Pardon the ignorance. I can understand best by examples. I suspect the 2nd one is related to the issue of 22*22^n+1 and 484*22^n+1 but am not sure.
So by extension on the definition of a Sieprpinski number base b, we can draw the same conclusion on Riesel numbers that have 1covers by changing k+1 to k1. That is:
An integer k > 1 is a Riesel number base b if gcd(k1,b1) = 1 and
k.bn1 is composite for all n > 0.
• gcd(k1,b1) = 1 avoids trivial covers (1covers).
In my research and creation of web pages for all conjectures base 232, I had already observed this and will put it in 'sentence format' for ease of everyone else's reference (mainly my own)
:
A k cannot be a Riesel nor a Sierpinski k if all values of n have a single trival factor. And the elimination of these k's is directly related to the factorization of the base minus 1. Siting specific examples for both types, we have:
Riesel:
Base 2 (none, i.e. there is no factorization of 1)
Base 3 (k==1 mod 2 eliminated with a factor of 2, i.e. the factorization of 2 is simply 2).
Base 4 (k==1 mod 3 eliminated with a factor of 3, i.e. the factorization of 3 is simply 3).
Base 5 (k==1 mod 2 eliminated with a factor of 2, i.e. the factorization of 4 is 2*2)
Base 6 (k==1 mod 5 eliminated with a factor of 5).
Base 7 (k==1 mod 2 eliminated with a factor of 2 AND k==1 mod 3 eliminated with a factor of 3, i.e. the factorization of 6 is 2*3).
Sierpinski:
Bases 2 & 3; same as above
Base 4 (k==2 mod 3 eliminated with a factor of 3).
Base 5; same as above
Base 6; (k==4 mod 5 eliminated with a factor of 5).
Base 7 (k==1 mod 2 eliminated with a factor of 2 AND k==2 mod 3 eliminated with a factor of 3, i.e. the factorization of 6 is 2*3).
And continuing to larger Sierpinski bases we have base 30 where only k==28 mod 29 are gone with a factor of 29 and base 31 where k==1 mod 2 are gone with a factor of 2, k==2 mod 3 are gone with a factor of 3, and k==4 mod 5 are gone with a factor of 5.
Robert, I'm going to send you links to web pages in a PM that I have already created with a lot of information for bases 232 for this. I have come up with a most unusual proof for the base 12 Riesel conjecture and I think part of it may be related to the last item that you are referring to here. I will post the proof in the next post here for everyone to review.
You did a lot of coordination on these conjectures and if you could, I'd like to get your feedback on the pages before opening them up to everyone. I also have a reservation page to get the ball rolling again on this.
Thanks,
Gary