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

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  • Kevin Acres
    Hello David, ... In the end I didn t need the wiggle room. Only to modify my code to run sensibly with negative b, where, as I think you know, it is not as
    Message 1 of 143 , Jan 5, 2011
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      Hello David,


      At 10:29 PM 5/01/2011, djbroadhurst wrote:
      >--- In primenumbers@yahoogroups.com, Kevin Acres
      ><research@...> wrote:
      >
      > > Something like [18*x-448552937, [15788701, 15]] probably violates
      > > your search criteria.
      >
      >Negative primes not allowed, sorry. So this case

      In the end I didn't need the wiggle room. Only to modify my code to
      run sensibly with negative b, where, as I think you know, it is not
      as efficient as running with positive b.

      So, after ~30 15/16 chains I can now post:

      [18, -143835631, [44253593, 16]]

      [1, 44253593, Mat([44253593, 1])]
      [2, 652729043, Mat([652729043, 1])]
      [3, 11605287143, Mat([11605287143, 1])]
      [4, 208751332943, Mat([208751332943, 1])]
      [5, 3757380157343, Mat([3757380157343, 1])]
      [6, 67632698996543, Mat([67632698996543, 1])]
      [7, 1217388438102143, Mat([1217388438102143, 1])]
      [8, 21912991742002943, Mat([21912991742002943, 1])]
      [9, 394433851212217343, Mat([394433851212217343, 1])]
      [10, 7099809321676076543, Mat([7099809321676076543, 1])]
      [11, 127796567790025542143, Mat([127796567790025542143, 1])]
      [12, 2300338220220315922943, Mat([2300338220220315922943, 1])]
      [13, 41406087963965542777343, Mat([41406087963965542777343, 1])]
      [14, 745309583351379626156543, Mat([745309583351379626156543, 1])]
      [15, 13415572500324833126982143, Mat([13415572500324833126982143, 1])]
      [16, 241480305005846996141842943, Mat([241480305005846996141842943, 1])]
      [17, 4346645490105245930409337343, [17, 1; 29, 1; 2092457, 1;
      4213575300261521443, 1]]


      This was found just 11 hours into a search with negative b.


      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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