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Distribution of square free positive integers

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  • Kermit Rose
    Distribution of square free positive integers Let N be the generic positive integer. Number of positive primes
    Message 1 of 2 , Jul 6, 2006
      Distribution of square free positive integers

      Let N be the generic positive integer.

      Number of positive primes < N


      = N/ln(N)

      Number of positive integers = product of two primes
      that are < N

      is

      (1/2) [(sqrt(N))/ln(sqrt(N)) ]^2

      = (1/2) N/[ (1/2) ln(N) ]^2

      = 2 N / [ ln(N) ]^2


      Number of positive integers = product of three primes
      that are < N

      is

      (1/6) [ N ]/ [ (1/3) ln(N) ]^3

      = (3^3/3!) N/ [ ln(N) ]^3


      Number of positive integers = product of k primes
      that are < N

      = (k^k/k!) N / [ln(N) ] ^k



      So,

      is it true that

      limit as N --> infinity

      of

      summation of k from 1 to infinity

      (-1)^k (k^k/k!) N / [ln(N) ]^k

      is equal to zero?


      Kermit < kermit@... >
    • Chris Caldwell
      ... I am not sure that the question, as written above, makes sense. k must be substantially smaller than N for this type of estimation to have any value, as N
      Message 2 of 2 , Jul 6, 2006
        Kermit Rose asks:
        > Number of positive integers = product of k primes that are < N
        > = (k^k/k!) N / [ln(N) ] ^k
        >
        > So, is it true that limit as N --> infinity of
        > summation of k from 1 to infinity
        >
        > (-1)^k (k^k/k!) N / [ln(N) ]^k
        >
        > is equal to zero?

        I am not sure that the question, as written above, makes sense. k must be substantially smaller than N for this type of estimation to have any value, as N must be at least the product of k primes; so N >> p_1 * p*2 * ... * p_k. This makes N >> e^p_k or very roughly N >> e^(k ln k) = k^k. So perhaps one might as about the limit as k --> infinity of

        (k^k/k!) k^k / [ln(k^k) ]^k

        Using a very rough Sterling appx k! is about (k/e)^k, so this latter limit is zero. Maybe use N = (a*k)^k to make some sense of this nonsensical limit.

        But if you just want the average number of prime factors of an integer, just ask that.

        CC
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