Re: [PrimeNumbers] gen Cullen/Woodalls
- Indeed,the non-existence of a finite covering set does
not prove infinite number of primes.Some time ago i
had constructed a "possible"-infinite covering set for
the Sierpinski numbers which used the algebraic
factoring of 4*X^4+1.
--- Jens Kruse Andersen <jens.k.a@...> wrote:
> Thomas wrote:__________________________________
> > What do you think, does there exist a base b such
> that n*b^n-1 or
> > n*b^n+1 never is prime?
> > I have no answer to that question, but maybe it is
> easy for some of
> > you to create such a base b, so that n*b^n +1/-1
> always have small
> > factors.
> No base b has a finite covering set of factors.
> If M is a finite set then let A be the product of
> the members.
> If n=k*A for any k, then n*b^n+/-1 clearly never has
> a factor in M.
> This does not necessarily mean there always is a
> There could be an algebraic factor I have
> overlooked. I know little about that.
> If there is not then heuristics seem to support
> infinitely many primes for all
> Jens Kruse Andersen
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