Timeline for Are there infinitely long arithmetic progressions in every increasing sequence of positive integers with bounded gaps between consecutive terms?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2023 at 18:24 | answer | added | bof | timeline score: 2 | |
| Dec 6, 2023 at 14:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 8, 2023 at 14:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 11, 2023 at 0:20 | comment | added | Zach Teitler | Doesn't this simple idea work? Enumerate the positive increasing arithmetic sequences (all initial values and step sizes) and at the nth step, select a term of the nth sequence; the first term that's more than 1 greater than the previous selected term. So, no consecutive numbers and intersects every infinite arithmetic sequence. The complement has gaps of size 1 and fails to contain any infinite arithmetic sequence. | |
| Apr 10, 2023 at 14:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 11, 2022 at 13:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 11, 2022 at 12:10 | answer | added | Karl Fabian | timeline score: 1 | |
| Nov 10, 2022 at 12:27 | comment | added | Terry Tao | The obvious random construction will almost surely produce a counterexample. Baire category argument also works. | |
| Nov 10, 2022 at 9:49 | comment | added | Ilya Bogdanov | I think it suffices to take a Sturmian word and interpret its symbols as gaps of lengths 1 and 2... | |
| Nov 10, 2022 at 3:13 | comment | added | Kai Wang | @StanleyYaoXiao I mean the second, the infinite set {a+kd: k \in N}. The finite arithmetic progression version has been addressed in the linked question. | |
| Nov 9, 2022 at 22:01 | comment | added | Stanley Yao Xiao | Do you mean arbitrarily long arithmetic progressions or do you mean such a set must include all terms of the form $a + kd$ for some $a,d$ and $k \in \mathbb{N}$? | |
| Nov 9, 2022 at 21:33 | history | asked | Kai Wang | CC BY-SA 4.0 |