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Eratosthenes and a mean four factors

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  • gulland68
    When you take your Sieve of Eratosthenes and consider values of n well beyond the square of the highest prime that you have selected to give a filter, does
    Message 1 of 1 , Apr 17, 2007
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      When you take your Sieve of Eratosthenes and consider values of n well
      beyond the square of the highest prime that you have selected to give
      a filter, does anybody know what sort value of n it would be when,
      among all those places - I guess you could call them component
      coincidences - where more than one filter's component falls, the mean
      number of filter components will be four?(That's obviously counting
      all powers of p as being represented by one sole sieve component i.e.
      there's no new filter made especially for squares of 3, for cubes of
      3, for squares of 5 etc.)

      Cheers,

      Tom
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