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Re: Prime chains x-->Ax+B [puzzle 11]

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  • djbroadhurst
    ... Puzzle 11: Find a plupluperfect chain of 12, i.e. a triplet (A, B, p[1]) such that A = 1 mod 11, B is coprime to 11, and the iteration p[n+1] = A*p[n] + B
    Message 1 of 143 , Dec 1, 2010
      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > Here's a plupluperfect chain of 8:
      >
      > p=3;for(k=1,9,print([k,factor(p)]);p=140820*p-422453)
      > [1, Mat([3, 1])]
      > [2, Mat([7, 1])]
      > [3, Mat([563287, 1])]
      > [4, Mat([79321652887, 1])]
      > [5, Mat([11170075159124887, 1])]
      > [6, Mat([1572969983907966164887, 1])]
      > [7, Mat([221505633133919795338964887, 1])]
      > [8, Mat([31192423257918585579633034964887, 1])]
      > [9, [7, 1; 627502434740013603046274854822137841, 1]] \\ Bounded!

      Puzzle 11: Find a plupluperfect chain of 12, i.e. a triplet
      (A, B, p[1]) such that A = 1 mod 11, B is coprime to 11,
      and the iteration p[n+1] = A*p[n] + B yields a chain of
      12 increasing primes, p[n], for n = 1 to 12.

      Comments: Clearly, p[2] = 11 and hence B = 11 - A*p[1], with
      p[1] = 2, 3, 5 or 7. So this boils down to 4 sub-problems of
      Carmody-scale difficulty (maybe easy for Phil; harder for mere
      mortals like Kevin, Mike, or me). I did not yet find a solution.

      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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