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

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  • Kevin Acres
    Hello David, ... I won t jump into this one just yet. I m turning over about 56000 8/16 chains per hour currently for Puzzle 2. My hope is that I ll get one
    Message 1 of 143 , Dec 1, 2010
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      Hello David,

      At 02:18 PM 2/12/2010, djbroadhurst wrote:
      >--- 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

      I won't jump into this one just yet. I'm turning over about 56000
      8/16 chains per hour currently for Puzzle 2. My hope is that I'll
      get one 16/16 per 8^8 8/16 and find another five 15/16 chains on the way.

      I had a quick check though, but my current script doesn't lend itself
      to pluperfect chains let alone plupluperfect.


      Best Regards,

      Kevin.
    • 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
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        --- 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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