Timeline for Asymptotic estimate for a random model of primes
Current License: CC BY-SA 3.0
6 events
when toggle format | what | by | license | comment | |
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Mar 28, 2016 at 23:57 | history | edited | user45947 | CC BY-SA 3.0 |
Fixed typo.
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Mar 28, 2016 at 23:55 | comment | added | user45947 | The correct expression should of course be $E[\pi_{rm}] = \sum_{n\leq x} W(\sqrt{n})$. I'll edit the original post accordingly. I assume only that $P(a \equiv b \pmod{p}) = 1/p$ for all $0\leq b < p$ and $p\leq \sqrt{x}$, or equivalently, that $P(a \equiv b \pmod{P(\sqrt{x}))} = 1/P(\sqrt{x})$ for all $0\leq b < P(\sqrt{x})$. | |
Mar 21, 2016 at 2:23 | comment | added | reuns | when you write $E[\pi_{rm}(x)] = \sum_{n \le x} W(\sqrt{x})$ you are assuming that $E[a] = 0$ and $P(a \equiv b \pmod p) = 1/p$ for every $b$ and prime $p \le x$ ? | |
Mar 9, 2016 at 19:08 | history | edited | user45947 | CC BY-SA 3.0 |
added 80 characters in body
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Mar 9, 2016 at 11:48 | history | asked | user45947 | CC BY-SA 3.0 |