--- In

primenumbers@yahoogroups.com, "Robert" <robert_smith44@...> wrote:

>

> --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"

> <jens.k.a@> wrote:

> >

> > Robert wrote:

> > > Does a base b exist such that k*b^n+/1 is composite for all k and n,

> > > i.e. all k are Sierpinski/Riesel, or provide trivial results for

> all n.

> >

> > By Dirichlet's theorem, for any value of b^n there are infinitely

many k

> > such that k*b^n+1 is prime, and infinitely many k such that

k*b^n-1 is

> > prime.

> >

> > --

> > Jens Kruse Andersen

Yet I still find this counterintuitive.

Take a base b=p[m]#+1, where p[m]is the mth prime and is quite large,

# is the factorial.

The smallest k that can give rise to a prime is k=p[m-1], and other k

that can give rise to a prime are those where any of p[1] to p[m] are

not factors of k-1, i.e. k=primes greater than p[m], although there

are also a large (but I think finite) number of Sierpinski numbers and

(possibly) total or partial factorisations which eliminate some of

these k. Some of the k remaining will possibly provide zero primes for

increasing n.

If m is large enough, then the chance that there are infinite primes

in k*(p[m])^n+1, k and n integer, seems implausible to me.

The only way I can rationalise this is by saying to myself that there

are an infinite number of primes that are greater than the p[m], and

that there are only a finite number of these that are Sierpinski

numbers or provide factorisations or partial factorisations.

Therefore, there are still infinite k, then as long as there is a

greater than zero chance that the power series k.b^n+1 will provide at

least one prime, then there will be an infinite number of primes.