Timeline for Eulerian ordering of the integers modulo n
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2018 at 16:38 | comment | added | Sebastien Palcoux | The first bonus question is now posted in this sequel. | |
| Mar 29, 2018 at 22:50 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
Interesting data on the number of Eulerian orderings for n=6,10.
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| Mar 29, 2018 at 10:31 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
minor edit + explanation of some Sage functions
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| Mar 29, 2018 at 10:28 | vote | accept | Sebastien Palcoux | ||
| Mar 28, 2018 at 22:47 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
modification of the Sage program for handling partial Eulerian ordering and completion
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| Mar 28, 2018 at 6:50 | comment | added | user44191 | @GerhardPaseman It may be useful to think of this as a sequence of unit boxes in a larger box. The side lengths are each of the prime factors. The condition is then: is there a sequence such that for $j < i$, there's some box in an axis direction from the $i$th box, and that axis direction has to be one where the $i$th and $j$th boxes have different coordinates. My algorithm below then just "fills out" the boxes "lexicographically". | |
| Mar 28, 2018 at 0:19 | answer | added | user44191 | timeline score: 4 | |
| Mar 27, 2018 at 15:15 | comment | added | Sebastien Palcoux | @user44191: Yes! Your example is an Eulerian ordering (see the edited checking). I'm impressed by your comment. How did you get this idea? Do you have a proof? Please write it as an answer (even if you don't have a proof). | |
| Mar 27, 2018 at 15:13 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
typo edit + data update + more details about the number-theoretic reformulation + checking program
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| Mar 27, 2018 at 10:09 | comment | added | user44191 | I should add: given $i, j$, let $k$ be $i$, except with the largest digit (in the mixed-base) where $j$ differs from $i$ changed. | |
| Mar 27, 2018 at 9:59 | comment | added | user44191 | I think the following algorithm will generally produce an "Eulerian ordering": Choose an ordering of the prime factors of $n$, e.g. (5, 3, 2) for 30. For $0 \leq i < n$, write $i$ in mixed base (e.g. $23 = 1_2 1_3 3_5)$. Let $d_p(i)$ denote the digit corresponding to $i$. Then let $r_i = (\sum_{p \in P} d_p(i) \frac{n}{p}) (mod n)$. For example, for 30 (with $5, 3, 2$), this leads to the sequence $0, 6, 12, 18, 24, 10, 16, 22, 28, 4, 20, 26, 2, 8, 14,15, 21, 27, 3, 9, 25, 1, 7, 13, 19, 5, 11, 17, 23, 29$. Is this sequence Eulerian? | |
| Mar 27, 2018 at 8:41 | comment | added | Sebastien Palcoux | @GerhardPaseman: I wanted to hide the motivation because I believed this question interesting for itself in number theory... I edited a last paragraph explaining that the motivation comes from algebraic combinatorics, and justifying in great detail the name of Eulerian ordering. | |
| Mar 27, 2018 at 8:41 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
The motivation comes from algebraic combinatorics, inspired from "Shelling the coset poset".
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| Mar 27, 2018 at 1:05 | comment | added | Gerhard Paseman | I am finding it challenging to translate the condition into something I can mentally operate. Can you give some motivation for the name and for the need for their being a suitable index k? Gerhard "Usually Finds Coprimality Quite Intuitive" Paseman, 2018.03.26. | |
| Mar 27, 2018 at 0:36 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |