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Re: Sieve needed for k*b1^m*b2^n+1

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  • Mike Oakes
    ... [snip] ... Really? That s false for k=0 mod 3; and for k=1, b1=3; and for... -Mike Oakes
    Message 1 of 2 , Jul 23, 2009
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      --- In primenumbers@yahoogroups.com, "Bryan" <beyastard@...> wrote:
      >
      > I've been trying to find a type of prime that has not been explored and
      > yet easily
      > factored for p+1 or p-1 and came up with the following:
      >
      > k*b1^m*b2^n+1 where b1 and b2 are prime bases > 2, bases differing and k
      > is even
      >
      > for example:
      >
      > 6904*7^987*11^654+1
      > 8020*13^731*31^257+1
      > 6460*257^112*523^388+1
      > 7702*5^2386*7^1257+1
      >
      [snip]
      > One thing I have noticed is for every k=2i+1, i>=0, the composite is
      > divisble by 3.

      Really?
      That's false for k=0 mod 3; and for k=1, b1=3; and for...

      -Mike Oakes
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