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Re: consecutive p-smooth integers

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  • djbroadhurst
    ... May we see your data for length 5 with N between 20 and 30 digits? Here is mine :-) [N, s = log(N)/log(p)] [5787885182600784208790, 3.826403843] in 70
    Message 1 of 42 , Aug 7 3:00 AM
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      --- In primenumbers@yahoogroups.com,
      Andrey Kulsha <Andrey_601@...> 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.

      > it's little sense to use log(N)/log(p)

      May we see your data for length 5 with N between 20 and 30 digits?

      Here is mine :-)

      [N, s = log(N)/log(p)]
      [5787885182600784208790, 3.826403843] in 70 seconds
      [195110934522453734763104, 4.009147298] in 169 seconds
      [3108993777544846030873003, 4.106736160] in 229 seconds
      [877496832307054822313934323, 4.173526105] in 3033 seconds
      [21299560799614314335258375426, 4.194840344] in 8660 seconds
      [51196989520720340392524462575, 4.217628079] in 5324 seconds

      Best regards, as ever

      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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