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Re: consecutive p-smooth integers

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  • John
    Is the quest for series of numbers with what essentially is the property of having extraordinarily low factors a Critical Theoretical Question, of is it in the
    Message 1 of 42 , Aug 6, 2011
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      Is the quest for series of numbers with what essentially is the property of having extraordinarily low factors a Critical Theoretical Question, of is it in the nature of finding very interesting, dare I say "merely" very interesting, facts?


      --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:
      >
      > Andrey Kulsha wrote:
      > > The case of two consecutive integers was already studied
      > > (Sloane's A002072, A145605), but there are little information
      > > about larger chains. It there any interesting results known?
      >
      > There are some computational results for 3 and 4 integers in
      > http://www.mersenneforum.org/showthread.php?t=5647
      >
      > It resulted in http://oeis.org/A122464 about trios.
      > Fred Schneider found:
      > 1348770149848002 = 2 * 3 * 7 * 23 * 41 * 61^2 * 149 * 239 * 257
      > 1348770149848001 = 19^3 * 89 * 103 * 229 * 283 * 331
      > 1348770149848000 = 2^6 * 5^3 * 11 * 29 * 109 * 151 * 163 * 197
      >
      > --
      > Jens Kruse Andersen
      >
    • 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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