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Re: Lucas super-pseudoprime puzzle

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  • djbroadhurst
    ... Yes. So far I have run the faster test up to n = 9*10^9. Confirmation of the claim that the next solution has n 10^10 should not take much longer. I did
    Message 1 of 33 , May 24, 2010
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      --- In primenumbers@yahoogroups.com,
      "mikeoakes2" <mikeoakes2@...> wrote:

      >> n is a solution if and only if it is an odd
      >> square-free composite integer such that for each prime p|n
      >> n = +/- 1 mod p-1 ... [1]
      >> n = +/- 1 mod p+1 ... [2]

      > It might be interesting to program this test as it might be
      > significantly faster for large n, not so?

      Yes. So far I have run the faster test up to n = 9*10^9.
      Confirmation of the claim that the next solution has
      n > 10^10 should not take much longer.

      I did not see Jacobs, Rayes and Trevisan remark
      on the fact that n cannot be divisible 3, so
      please pat yourself on the back, Mike.

      David
    • 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
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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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