RE: [PrimeNumbers] Distribution of square free positive integers
- Kermit Rose asks:
> Number of positive integers = product of k primes that are < NI 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!) 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?
(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.