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Re: [PrimeNumbers] Primaliy test for a series

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  • plesala
    Pardon for the clumsy statement I made. Of course the summation I give is easy to simplify. Actually my interest is summations that are slightly more
    Message 1 of 7 , Feb 22, 2007
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      Pardon for the clumsy statement I made. Of course the summation I give is easy to simplify. Actually my interest is summations that are slightly more complicated.

      Whenever I try to search for primes of the form
      6*n*(2^p - 1- n) + 2^p - 1, where n = 1, 2, 3, ... and 2^p - 1 is a Mersenne prime, I find that the factors are of the form

      2^(3*k) - 1 = (2^3 - 1)( (2^3)^(k - 1) + (2^3)^k - 2) + ...
      (2^3)^2 + (2^3) + 1),

      or alternatively

      2^(3*k) + 1 = (2^3 + 1)( (2^3)^(k - 1) - (2^3)^(k - 2) + ...
      + (2^3)^2 - (2^3) + 1), where k is odd.

      I wonder whether some of these factor summations cannot yield fairly large primes.

      Peter.
      ----- Original Message -----
      From: plesala
      To: primenumbers@yahoogroups.com
      Sent: Wednesday, February 21, 2007 1:54 PM
      Subject: [PrimeNumbers] Primaliy test for a series


      Hi to all,

      How does one test primality of a series like

      1 + 2 + 2^2 + 2^3 + ... + 2^(2*n - 1),

      for example, using primeform?

      Thank you.
      Peter.




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