Timeline for Number of products of distinct primes lying in an interval
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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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
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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 |
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Jun 20, 2014 at 13:43 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |