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Ultimate Sierpinski

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  • Robert
    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, 2008
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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.
    • 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 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.

        --
        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 3 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 4 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 5 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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