Distribution of square free positive integers

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• 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@... >
• ... 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
> 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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