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Jun 20, 2014 at 15:56 vote accept Stanley Yao Xiao
Jun 20, 2014 at 15:45 answer added Emil Jeřábek timeline score: 5
Jun 20, 2014 at 15:18 comment added Stanley Yao Xiao I phrased thr question in terms of counting, which is hopefully more clear.
Jun 20, 2014 at 15:16 history edited Stanley Yao Xiao CC BY-SA 3.0
added 47 characters in body; edited title
Jun 20, 2014 at 15:05 comment added Robert Israel Can you clarify? Do you mean you are taking the product of a fixed number $n$ of primes randomly chosen from all the primes less than $Y$? Obviously if at least two of those primes are greater than $Y/k$ the product will be greater than $Y^2/k^2$...
Jun 20, 2014 at 14:38 comment added Emil Jeřábek Sorry, a factor of $X$ is missing in the last but one expression (but it is swamped in the final $o(1)$, so it doesn’t matter).
Jun 20, 2014 at 14:28 comment added Emil Jeřábek If I understand the question correctly, it is $2^{-\pi(Y)}$ times the number of square-free $Y$-smooth numbers below $X$. If $X=CY$ for fixed $C$, a $6/\pi^2$ fraction of numbers below $X$ are square-free, and all but an $O(1/\log X)$ fraction are $Y$-smooth, so the result is ~ $2^{-\pi(Y)}6/\pi^2=2^{-(1+o(1))Y/\log Y}$.
Jun 20, 2014 at 13:57 history edited Stanley Yao Xiao CC BY-SA 3.0
added 158 characters in body
Jun 20, 2014 at 13:43 history asked Stanley Yao Xiao CC BY-SA 3.0