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Re: its fanny

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  • David Broadhurst
    ... Dear Rachid Here is my attempt to present your observation with clarity and due attention to known results. Let M(n) = 2^n - 1 and MM(n) = 2^(2^n - 1) - 1.
    Message 1 of 2 , Jun 29, 2009
      --- In primenumbers@yahoogroups.com,
      sta staf <sta_staf@...> wrote:

      > A simple idea can be the source of an empire

      Dear Rachid

      Here is my attempt to present your observation
      with clarity and due attention to known results.

      Let M(n) = 2^n - 1 and MM(n) = 2^(2^n - 1) - 1.

      1) We know that MM(n) is prime n = 2, 3, 5, 7.

      2) We know that MM(n) is composite for n = 13, 17, 19, 31.

      3) We know nothing about the primality of MM(n) for any n > 31
      such that M(n) is prime, despite considerable effort on MM(61):
      http://anthony.d.forbes.googlepages.com/mm61prog.htm

      3) Let T(n) = (2^n - 2 - n)/n. Then, by construction,
      none of T(2) = 0, T(3) = 1 , T(5) = 5, T(7) = 17 is composite.

      4) It takes less than a second to show that T(n) is composite
      for n = 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203,
      2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209,
      44497, 86243, 110503, 132049, 216091, 756839.

      6) Rachid Baih elevates these simple observations to a
      conjecture, namely that if M(n) is prime, then MM(n)
      is composite if and only T(n) is composite.

      7) This conjecture is hard to falsify, since it is
      heuristically probable that each of MM(n) and T(n)
      is composite for all n > 7 such that M(n) is prime.

      David
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