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Re: consecutive p-smooth integers

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  • djbroadhurst
    ... No: I misread the gradients. Please scrap the above message. I am, however, surprised that your gradients seem to increase in an unusual manner: k=5:
    Message 1 of 42 , Aug 9, 2011
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

      > > Some related data is collected there:
      > > http://www.primefan.ru/stuff/math/maxs.xls
      >
      > Your straight-line fits of log(N) against sqrt(p),
      > for k=6 (i.e. 7 consecutive integers) and k=7
      > (i.e. 8) were at very small p. If you were to
      > extrapolate them to larger p, you would be led to
      > the absurdity that 8 consecutive p-smooth integers
      > are easier to find than 7, not so :-?

      No: I misread the gradients. Please scrap the above message.
      I am, however, surprised that your gradients
      seem to increase in an unusual manner:

      k=5: gradient = 0.6042
      k=6: gradient = 0.3917
      k=7: gradient = 0.3398

      I guess that may just be because your primes are so small?

      David (guilty of careless error in previous message: sorry!)
    • 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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