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Re: Prime chains x-->Ax+B

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  • djbroadhurst
    ... Those terms are also coprime to 3, so you might, with equal arbitrariness, have set C = (2/1)^2*(3/2)^2 = 9, as I did on first seeing your exercise . The
    Message 1 of 143 , Dec 1, 2010
      --- In primenumbers@yahoogroups.com,
      "mikeoakes2" <mikeoakes2@...> wrote:

      > As you know, my "4" came from the fact that the 3rd and 4th
      > terms of the sequence are known to be odd

      Those terms are also coprime to 3, so you might, with
      equal arbitrariness, have set C = (2/1)^2*(3/2)^2 = 9,
      as I did on first seeing your "exercise".

      The asymptotic constant seems to lie between 4 and 9,
      but is hard (for me) to estimate accurately.

      David (over-exercised :-)
    • 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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