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Re: Prime chains x-->Ax+B

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  • mikeoakes2
    ... That s a really neat script, and (after formatting it properly:-) I believe it. But you were clearly running 64-bit pari, since ... gives only 9-digit
    Message 1 of 143 , Dec 4, 2010
      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > > asking that
      > > N = 3^t+4
      > > M = (N-2)*(N-3)+3
      > > be simultaneously prime. The Poisson mean for
      > > such cases with t > x >> 1 may be estimated as
      > > C/(2*log(3)^2)*sum(t>x,1/t^2) =~ C/(2*log(3)^2)/x
      > > where C is a constant of order unity.
      >
      > Sieving to depth 10^8, with the aid of "znorder" and
      > "polrootsmod" in Pari-GP, I estimated
      > C ~ 6.5, but the convergence is not good:
      >
      > {mp=nextprime(10^8);default(primelimit,mp);
      > default(realprecision,5);n=1;g=1.;forprime(p=5,mp,
      > if(p>10^n,print([g*log(p)^2*exp(2*Euler),p]);n++);
      > a=znorder(Mod(3,p));b=znorder(Mod(-4,p));c=if(a%b,0,1);
      > f=lift(polrootsmod((x+1)*(x+2)+3,p));for(k=1,#f,r=f[k];
      > if(r,b=znorder(Mod(r,p));c+=if(a%b,0,1)));g*=1-c/a);}
      >
      > [7.6000, 11]
      > [7.4962, 101]
      > [7.4031, 1009]
      > [7.5940, 10007]
      > [7.2959, 100003]
      > [7.0154, 1000003]
      > [6.7590, 10000019]
      > [6.5402, 100000007]
      >
      > David (puzzled by this slow convergence)

      That's a really neat script, and (after formatting it properly:-) I believe it.
      But you were clearly running 64-bit pari, since
      > default(realprecision,5);
      gives only 9-digit precision in 32-bit pari, and that's not accurate enough.
      To make the code portable, you should have written something like
      \p 18
      default(format,"g0.5");

      I have gone up to nextprime(4*10^9) and have 2 new points for you:-
      [6.35205959664247940, 1000000007]
      [6.25064108108829438, 4000000007]

      Convergence still slow.

      Mike
    • 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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