## Ultimate Sierpinski

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• 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.
Message 1 of 5 , Oct 14 9:57 AM
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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. --
Message 2 of 5 , Oct 14 10:24 AM
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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
• ... n. ... Yes! Ignore my odd answer and go with this one Robert. (Unless you want k=5 and b=1, this works for all n )
Message 3 of 5 , Oct 14 11:07 AM
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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.

Yes! Ignore my odd answer and go with this one Robert. (Unless you
want k=5 and b=1, this works for all n <grin>)
• ... all n. ... Thank you Jens and Chris. Dirichlet is still a surprising result to me even after all the time I have spent looking at k*b^n+/-1. It has caught
Message 4 of 5 , Oct 14 11:12 PM
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--- 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
>

Thank you Jens and Chris. Dirichlet is still a surprising result to me
even after all the time I have spent looking at k*b^n+/-1. It has
caught me out more than once.
• ... 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 5 of 5 , Oct 16 1:14 AM
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--- 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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