Re: [PrimeNumbers] gen Cullen/Woodalls
- Thomas wrote:
> What do you think, does there exist a base b such that n*b^n-1 orNo base b has a finite covering set of factors.
> 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
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 prime.
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
Large bases are less likely to produce primes, because of the rate
the series grows by, but it is impossible to say there is no prime in
the series or there are infinite number of primes, for a given base.
For base 2 there are thought to be infinite but negligible number of
primes for Cullen/woodall.
We can kind of define a weight for each k, similar to nash weights.
This might help decide what bases are good and what not.
Since all factors of n*b^n+-1 are also factors of b^a-1, we can
search from a=1 to 256 to see how many candidates remain for n=100K
Have you done some searching. What base have you got stuck on? May be
some people from this group can help you find the prime for that
An intresting question, I myself was wondering about some time back.
- 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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