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consecutive p-smooth integers

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  • Andrey Kulsha
    An integer is called p-smooth for a given prime p, if its prime factors don t exceed p. Sometimes a few consecutive integers are p-smooth (of course, we are
    Message 1 of 42 , Aug 1 4:18 AM
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      An integer is called p-smooth for a given prime p, if its prime factors don't exceed p. Sometimes a few consecutive integers are p-smooth (of course, we are interested in non-trivial cases, where the smallest integer in the chain exceeds the prime next to p): for example, 4374 and 4375 are 7-smooth, while 2430, 2431 and 2432 are 19-smooth. There are also some chains of larger length (all 41-smooth):
      212380, 212381, 212382
      1517, 1518, 1519, 1520, 1521
      285, 286, 287, 288, 289, 290
      Another six integers, from 3294850 to 3294855, are 239-smooth. Eight integers from 4895 to 4902 are 89-smooth, while 15 integers from 48503 ti 48517 are 379-smooth.

      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?

      Thanks,

      Andrey

      [Non-text portions of this message have been removed]
    • 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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