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21509Re: Lucas super-pseudoprime puzzle

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  • mikeoakes2
    May 27, 2010
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      --- 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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