- Thomas wrote:

> What do you think, does there exist a base b such that n*b^n-1 or

No 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

> 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 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

b.

--

Jens Kruse Andersen - Thomas,

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

to 110K.

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

base.

An intresting question, I myself was wondering about some time back.

--Harsh Aggarwal - 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

> 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

> b.

>

> --

> Jens Kruse Andersen

>

>

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