Timeline for Erdős-Szekeres for first differences
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| S Jun 20, 2015 at 8:47 | history | edited | Marco Golla | CC BY-SA 3.0 |
fix bad encoding in Wiki link
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| S Jun 20, 2015 at 8:47 | history | suggested | András Salamon | CC BY-SA 3.0 |
fix bad encoding in Wiki link
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| Jun 20, 2015 at 8:45 | review | Suggested edits | |||
| S Jun 20, 2015 at 8:47 | |||||
| Mar 16, 2012 at 13:14 | vote | accept | Seva | ||
| Mar 14, 2012 at 19:00 | comment | added | Sergey Norin | @Boris: I have changed the penultimate sentence slightly after your first comment. (The original version was unnecessarily strong and false.) As $\alpha_i(r'+1)>z$ and $\beta_i(s'+1)< z$, using the monotonicity of the $\alpha$ and $\beta$ sequences we see that to the right of the first $r'+1$ alphas there are at most $s'$ betas. | |
| Mar 14, 2012 at 14:17 | comment | added | Boris Bukh | @Sergey: I still do not follow. At the beginning of the "Otherwise,..." sentence we know that $r'\leq r-2$ and $s'\leq s-2$. The definition of $s'$ and $r'$ gives $\alpha_i(r')\leq z$, $\alpha_i(r'+1)>z$ and $\beta_i(s')\geq z$, $\beta_i(s'+1)<z$. This means that to the left $z$ there are precisely $r'$ alphas, and to the right there are $s'$ betas. This is different from what the penultimate sentence says. | |
| Mar 14, 2012 at 13:04 | history | edited | Sergey Norin | CC BY-SA 3.0 |
added 7 characters in body
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| Mar 14, 2012 at 12:58 | comment | added | Sergey Norin | @Boris: If $r'=r-1$ then $\alpha_i(r'+1)$ is not in the sequence. | |
| Mar 14, 2012 at 10:28 | comment | added | Boris Bukh | I do not follow the conclusion of the proof: how do you use that $r'\neq r-1$ and $s'\neq s-1$ to deduce the penultimate sentence? | |
| Mar 13, 2012 at 22:27 | history | edited | Sergey Norin | CC BY-SA 3.0 |
added 57 characters in body
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| Mar 13, 2012 at 22:04 | history | answered | Sergey Norin | CC BY-SA 3.0 |