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Re: [PrimeNumbers] Re: consecutive p-smooth integers

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  • Robert Gerbicz
    2011/8/6 djbroadhurst ... N=28998978021284765 is good with strength~3.92795. ? N=28998978021284765;for(i=0,4,print(factor(N+i))) [5,
    Message 1 of 42 , Aug 6, 2011
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      2011/8/6 djbroadhurst <d.broadhurst@...>

      > **
      >
      >
      >
      >
      > --- In primenumbers@yahoogroups.com,
      > "Jens Kruse Andersen" <jens.k.a@...> wrote:
      >
      > > There are some computational results for 3 and 4 integers in
      > > http://www.mersenneforum.org/showthread.php?t=5647
      >
      > Here is a puzzle about sequences of 5 consecutive integers.
      >
      > 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.
      >
      > Example: The sequence of 5 consecutive integers
      > beginning with N = 612130418113087340 has strength
      > s = log(N)/log(48437) =~ 3.7964, as shown here:
      >
      > for(k=0,4,print(factor(612130418113087340+k)[,1]~));
      >
      > [2, 5, 17, 263, 347, 929, 1237, 17167]
      > [3, 13, 31, 71, 157, 757, 1367, 4877]
      > [2, 7, 37, 1481, 2069, 14591, 26431]
      > [19, 23, 467, 3257, 19013, 48437]
      > [2, 3, 43, 479, 1109, 13007, 42923]
      >
      > Puzzle: Find a sequence of 5 consecutive integers with
      > strength s > 3.80.
      >
      > Comment: A solution may be found in less than 200 seconds.
      >
      > David Broadhurst
      >
      >
      >

      N=28998978021284765 is good with strength~3.92795.

      ? N=28998978021284765;for(i=0,4,print(factor(N+i)))
      [5, 1; 13, 1; 2663, 1; 4261, 1; 4919, 1; 7993, 1]
      [2, 1; 3, 1; 11, 1; 59, 1; 313, 1; 337, 1; 4547, 1; 15527, 1]
      [47, 1; 241, 2; 863, 1; 1129, 1; 10903, 1]
      [2, 5; 7, 1; 1549, 1; 1613, 1; 6871, 1; 7541, 1]
      [3, 6; 19, 2; 137, 2; 2423, 2]

      The code in PARI-GP (note that it would be much faster to use sieve), it ran
      about 150 minutes: (prints the record strength also for these type of
      numbers)

      T(n)=f=factor(n);return(log(n)/log(f[matsize(f)[1],1]))
      r=0;n=3;while(1,n++;if(T(n)>r/2&&T(n-1)>r/2&&T(n+1)>r/2&&\
      T(n-2)>r/2&&T(n+2)>r/2&&T(n*n-2)>r&&T(n*n-3)>r,\
      r=100.0;for(i=0,4,r=min(r,T(n*n-i)));print(r" "n*n-4)))

      In this way N=n*n-4, so N;N+3 and N+4 has got an algebraic factorization,
      increasing our chances to find a large strength. (I've used a slightly
      different strength definition.)


      [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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