Timeline for Strange (or stupid) arithmetic derivation
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2017 at 19:27 | comment | added | Sebastien Palcoux | Richard J. Mathar put this sequence in OEIS on Jan 04 2012: oeis.org/A203639 | |
| Aug 20, 2014 at 16:27 | answer | added | A. Bellmunt | timeline score: 5 | |
| Feb 1, 2014 at 23:15 | comment | added | Roland Bacher | A similar question is mathoverflow.net/questions/25974/calculus-on-rationals. | |
| Feb 1, 2014 at 15:19 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
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| Jan 21, 2014 at 15:27 | history | edited | user9072 |
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| Jan 13, 2014 at 2:52 | answer | added | Daniel Soltész | timeline score: 5 | |
| Oct 19, 2013 at 19:00 | vote | accept | Daniel Soltész | ||
| Oct 19, 2013 at 18:59 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
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| Sep 28, 2013 at 18:50 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
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| Sep 22, 2013 at 14:08 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
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| Sep 22, 2013 at 11:54 | comment | added | Lev Borisov | Why wouldn't you make it distributive over products of primes: seems a bit more natural? | |
| Sep 22, 2013 at 11:29 | comment | added | Martin Brandenburg | Note that this differs from the usual arithmetic derivative en.wikipedia.org/wiki/Arithmetic_derivative. | |
| Sep 22, 2013 at 5:29 | answer | added | user6976 | timeline score: 22 | |
| Sep 21, 2013 at 18:27 | answer | added | Vidit Nanda | timeline score: 3 | |
| Sep 21, 2013 at 16:34 | comment | added | Felipe Voloch | I suspect $p^{p^p}$ and so on will need several steps before hitting a cycle so the answer to the second question should be no. For example $3^{27}$ takes $10$ steps before hitting a fixed point. | |
| Sep 21, 2013 at 15:47 | history | asked | Daniel Soltész | CC BY-SA 3.0 |