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Re: Ultimate Sierpinski

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  • 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 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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