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Re: Prime chains x-->Ax+B [puzzle 2]

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  • djbroadhurst
    ... Some way back, you reported using a wheel of size 510510 = 17# and you reported the factorization ... In such a case,
    Message 1 of 143 , Jan 4, 2011
      --- In primenumbers@yahoogroups.com,
      Kevin Acres <research@...> wrote:

      > A review of logs of 14+ chains gives me 17 15/16 amongst 226 14/16.
      > Some of these are outside of my preferred x < 51561510 search area,
      > where the yield of 15/16 becomes somewhat less.

      Some way back, you reported using a wheel of size 510510 = 17#
      and you reported the factorization
      > [16, 3052362829787719304602274593,
      > [43, 1; 80048363, 1; 886778685131718377, 1]]

      In such a case,

      prod(p=2,17,1-isprime(p)/p)*log(3052362829787719304602274593) =~ 11.4

      estimates the mean number of 15's that you might need for a 16.

      By my reckoning you now have about 23 runs of 15.
      The a priori probability of a 16th prime was about
      1 - exp(-23/11.4) = 87%
      so it appears that you have been unlucky.

      David
    • djbroadhurst
      ... Suppose that we want to start with a square and get a square. Then we must solve the Diophantine equation y^2 = a*x^2 + b For any pair (a,b), Dario will
      Message 143 of 143 , Jan 7, 2011
        --- In primenumbers@yahoogroups.com,
        Kevin Acres <research@...> wrote:

        > x=a*x+b either is a square or has a square as a major factor.

        Suppose that we want to start with a square and get a square.
        Then we must solve the Diophantine equation
        y^2 = a*x^2 + b

        For any pair (a,b), Dario will tell us all the solutions:
        http://www.alpertron.com.ar/QUAD.HTM

        David
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