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Re: Ultimate Sierpinski

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  • 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 1 of 5 , Oct 14, 2008
      --- 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 2 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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