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Re: Lucas super-pseudoprime puzzle

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  • mikeoakes2
    ... Proof: V(1,1,3*m)=w^(3*m)+w_bar^(3*m)=(-1)^m+(-1)^m=-2 for odd m, where w=(1+sqrt(-3))/2 and w_bar=(1-sqrt(-3))/2 QED Mike
    Message 1 of 33 , May 22, 2010
      --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
      >
      > After a few mins with pfgw, the following seems to be true:
      > If n = 0 mod 3, then (V(1,1,n)-1)<> 0 mod n
      >
      > So, your n cannot have 3 as a factor (?)

      Proof:
      V(1,1,3*m)=w^(3*m)+w_bar^(3*m)=(-1)^m+(-1)^m=-2 for odd m,
      where
      w=(1+sqrt(-3))/2
      and
      w_bar=(1-sqrt(-3))/2
      QED

      Mike
    • mikeoakes2
      ... You did very much what I did. I nearly fell off the chair when I averaged everything out and saw 0.57... :-) ... Not very. Would you buy an appeal to
      Message 33 of 33 , May 27, 2010
        --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
        >
        > I tried 1/n^c:
        >
        > v=[1237.1, 328.7, 105.4, 28.01, 6.22 , 1.510, 0.439, 0.0939];
        > print(vector(8,k,-log(v[k]/10^6)/(k+4)/log(10)));
        >
        > [0.5815, 0.5805, 0.5682, 0.5691, 0.5785, 0.5821, 0.5780, 0.5856]
        >
        > and then A/n^c, using the first datum to remove A:
        >
        > print(vector(7,k,-log(v[k+1]/v[1])/k/log(10)));
        >
        > [0.5756, 0.5348, 0.5484, 0.5747, 0.5827, 0.5750, 0.5885]
        >
        > In both cases c =~ Euler looks rather convincing,
        > given the statistics. Well spotted, Sir!

        You did very much what I did.
        I nearly fell off the chair when I averaged everything out and saw 0.57... :-)

        > How strongly are you committed to A = 1, for the average,
        > given the variability of the overall factor with a?

        Not very.
        Would you buy an appeal to Occam's razor, mon vieux?

        Mike
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