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Re: Another large pair of consecutive smooth numbers

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  • djbroadhurst
    ... Definition: For positive integer n let p be the largest prime divisor of n*(n+1) and S(n) = log(n)/log(p) be the figure of merit for the smoothness of the
    Message 1 of 2 , Apr 20, 2012
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      --- In primenumbers@yahoogroups.com,
      "WarrenS" <warren.wds@...> wrote:

      > max prime involved in either 2573827

      Definition: For positive integer n let p be the largest
      prime divisor of n*(n+1) and S(n) = log(n)/log(p) be
      the figure of merit for the smoothness of the
      consecutive integers n and n+1.

      Then Warren's 125-digit example had

      S(n) =~ log(7.5828465*10^124)/log(2573827) =~ 19.480

      Here is a 135-digit example with greater merit:

      {f(y)=
      (y^2-11^4)*(y^2-35^2)*(y^2-47^2)*
      (y^2-94^2)*(y^2-146^2)*(y^2-148^2)/
      67440294559676054016000 - 1;}

      n = f(12971885307194);

      {if(type(n)=="t_INT",F=factor(n*(n+1))[,1];
      p=F[#F];print("n has "#Str(n)" digits");
      print("max prime is "p);
      default(realprecision,5);
      print("S(n) =~ "log(n)/log(p)));}

      n has 135 digits
      max prime is 6244451
      S(n) =~ 19.797

      David
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