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Re: [PrimeNumbers] Ultimate Sierpinski

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  • Jens Kruse Andersen
    ... 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 1 of 5 , Oct 14, 2008
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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
    • Chris Caldwell
      ... 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 2 of 5 , Oct 14, 2008
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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>)
      • Robert
        ... 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 3 of 5 , Oct 14, 2008
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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.
        • Robert
          ... 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 4 of 5 , Oct 16, 2008
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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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