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order of 2 in Z_p

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  • LEGalup
    Greetings all, let o_p(2) = order of 2 in F_p. in other words, o_p(2) = card{ 2^0 (mod p), 2^1 (mod p), ... , 2^p (mod p) } we know that it is an open
    Message 1 of 2 , Nov 15, 2009
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      Greetings all,
      let o_p(2) = order of 2 in F_p.
      in other words, o_p(2) = card{ 2^0 (mod p), 2^1 (mod p), ... , 2^p (mod
      p) }
      we know that it is an open conjecture that o_p(2) = p-1 for p a
      prime.but does anyone actually know what is:
      limsup o_p(2)/p = ?where p goes over all primes?
      thanks,

      lou



      [Non-text portions of this message have been removed]
    • djbroadhurst
      ... Counterexample: o_137(2) = (137 - 1)/2 = 68 o_p(2) is provably a divisor of p-1 for prime p. By the Artin conjecture, it is a proper divisor of p-1 for
      Message 2 of 2 , Nov 15, 2009
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        --- In primenumbers@yahoogroups.com,
        "LEGalup" <legalup@...> wrote:

        > let o_p(2) = order of 2 in F_p.
        ...
        > o_p(2) = p-1 for p prime

        Counterexample: o_137(2) = (137 - 1)/2 = 68

        o_p(2) is provably a divisor of p-1 for prime p.
        By the Artin conjecture, it is a proper divisor
        of p-1 for about
        62.604418638079771194527194565348358488837 percent
        of the primes. Here are the first few:

        forprime(p=3,137,if(znorder(Mod(2,p))<p-1,print1(p" ")))
        7 17 23 31 41 43 47 71 73 79 89 97 103 109 113 127 137

        David
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