Timeline for Prove: If $P_n$ is $n$-$th$ prime number then $P_{n+m} \ge P_n+P_m$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2018 at 1:45 | vote | accept | Đào Thanh Oai | ||
| S Jun 28, 2018 at 1:36 | history | suggested | David G. Stork | CC BY-SA 4.0 |
Improved English and layout
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| Jun 28, 2018 at 1:00 | review | Suggested edits | |||
| S Jun 28, 2018 at 1:36 | |||||
| Jun 27, 2018 at 22:18 | history | edited | GH from MO |
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| Jun 27, 2018 at 17:12 | comment | added | Đào Thanh Oai | My conjecture equivalent to $\pi(x+y) \le \pi(x)+\pi(y)+1$ | |
| Jun 27, 2018 at 15:53 | comment | added | Đào Thanh Oai | There is a good answer, if we can convert my conjecture with form $\pi(x+y) \le \pi(a)+\pi(b)$ where $a, b=f(x,y)$ after that we can compare with k-tuplet conjecture. | |
| Jun 27, 2018 at 7:45 | comment | added | Đào Thanh Oai | @GregMartin I am sorry, May You see answer below and my comment below? | |
| Jun 27, 2018 at 3:54 | history | edited | GH from MO |
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| Jun 27, 2018 at 3:54 | answer | added | GH from MO | timeline score: 13 | |
| Jun 27, 2018 at 2:50 | comment | added | Đào Thanh Oai | I am sorry, how these are equivalent, may you show me? | |
| Jun 27, 2018 at 2:47 | comment | added | Greg Martin | I didn't claim that $\pi(x)=P_x$; I claimed that your inequality is equivalent to my inequality. | |
| Jun 27, 2018 at 2:42 | comment | added | Đào Thanh Oai | $\pi(x)$ is not equivalent to $P_x$ | |
| Jun 27, 2018 at 2:41 | comment | added | Greg Martin | This is equivalent to the well-known conjecture $\pi(x+y) \le \pi(x)+\pi(y)$, which is now widely believed to be false because it contradicts the prime $k$-tuples conjecture. | |
| Jun 27, 2018 at 2:38 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
added 25 characters in body
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| Jun 27, 2018 at 2:24 | history | asked | Đào Thanh Oai | CC BY-SA 4.0 |