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Re: consecutive p-smooth integers

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  • djbroadhurst
    ... My current record is s = 4.1067361598, achieved with x = 20375617 N = (x^2-3^4)*(x^2-2^10)/55440 = 3108993777544846030873003 David
    Message 1 of 42 , Aug 6, 2011
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      --- In primenumbers@yahoogroups.com, "djbroadhurst"
      <d.broadhurst@...> wrote:

      > Definition: A sequence of consecutive positive
      > integers, beginning with N, has strength
      > s = log(N)/log(p), where p is the largest prime
      > dividing any integer in the sequence.

      > Puzzle: Find a sequence of 5 consecutive integers with
      > strength s > 3.80.

      > I imagine that Jens might soon beat strength s = 4.009147298.

      My current record is s = 4.1067361598, achieved with
      x = 20375617
      N = (x^2-3^4)*(x^2-2^10)/55440 = 3108993777544846030873003

      David
    • Mark
      ... Too hard! It was difficult getting just over 2 for the first time with log(287946949)/log(15823). My only consolation is that the above is also good for 15
      Message 42 of 42 , Nov 8, 2011
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        --- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
        >
        > >> http://www.primefan.ru/stuff/math/maxs.xls
        > >> http://www.primefan.ru/stuff/math/maxs_plots.gif
        > >
        > > Thanks, Andrey. The gradients are fanning out better now:
        >
        > The files were updated again.
        >
        > Puzzle: find a chain of 13 consecutive p-smooth integers, starting at N,
        > with log(N)/log(p) greater than
        >
        > log(8559986129664)/log(58393) = 2.71328
        >
        > Best regards,
        >
        > Andrey
        >

        Too hard!

        It was difficult getting just over 2 for the first time with
        log(287946949)/log(15823).

        My only consolation is that the above is also good for 15 consecutive 15823-smooth integers.
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