Loading ...
Sorry, an error occurred while loading the content.

RE: [PrimeNumbers] Distribution of square free positive integers

Expand Messages
  • 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 1 of 2 , Jul 6, 2006
    • 0 Attachment
      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
    Your message has been successfully submitted and would be delivered to recipients shortly.