Timeline for Integer points avoiding three on a line, four on a circle
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2018 at 15:11 | comment | added | Manfred Weis | Maybe also of interest: Achim Flammenkamp's solutions to the no-3-in-a-line problem | |
| Dec 26, 2018 at 15:04 | history | edited | Manfred Weis |
added the flags "pythagorean-triples" and "prime-number" because of the relation to these
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| Dec 26, 2018 at 15:00 | answer | added | Manfred Weis | timeline score: 2 | |
| Dec 25, 2018 at 14:39 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
added 134 characters in body
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| Dec 25, 2018 at 14:14 | comment | added | Joseph O'Rourke | @AaronMeyerowitz: Nice reasoning! Maybe $g_4(4) = 5$... | |
| Dec 25, 2018 at 5:50 | comment | added | Aaron Meyerowitz | In your 6 point diagram, any point on the central line is equidistant from the two on the left and also equidistant from the two on the right. Some point is equidistant from all 4 of them. | |
| Dec 24, 2018 at 22:07 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
added 84 characters in body
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| Dec 24, 2018 at 21:49 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
added 249 characters in body
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| Dec 24, 2018 at 15:30 | comment | added | Joseph O'Rourke | @BorisBukh: Thanks for the correction. | |
| Dec 24, 2018 at 15:30 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
deleted 6 characters in body
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| Dec 24, 2018 at 15:25 | comment | added | Boris Bukh | You should adjust the number $d+2$ upward, as a curve of degree $d$ has $\binom{d+1}{2}-1$ degrees of freedom. If you make this adjustment, then, I believe, that the answer should be $o(n)$ for $d\geq d_0$, but that is certainly wide open. | |
| Dec 24, 2018 at 14:06 | history | asked | Joseph O'Rourke | CC BY-SA 4.0 |