Re: Ultimate Sierpinski

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• ... many k ... k*b^n-1 is ... 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.
Message 1 of 5 , Oct 16, 2008
--- 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.
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