Timeline for Weighted maximal number of disjoint singly-generated ideals in the divisibility poset for $\{1,2,\ldots,n\}$
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 30, 2015 at 0:30 | comment | added | kodlu | @FedorPetrov:Thanks, clarified the question and corrected the terminology, I am looking at not chains but ideals generated by single elements. Thus, if I chose $i$, I have to choose all the multiples of $i$ in $\{1,\ldots,n\}$ as part of my ideal $C_i.$ | |
| Nov 30, 2015 at 0:29 | history | edited | kodlu | CC BY-SA 3.0 |
corrected terminology
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| Nov 28, 2015 at 9:28 | comment | added | Fedor Petrov | Is $m$ fixed? If not, why is not the answer $H_n$ for chains $C_i=\{i\}$? | |
| Nov 27, 2015 at 23:18 | comment | added | kodlu | @Jan-ChristophSchlage-Puchta:Thanks, please see edit. | |
| Nov 27, 2015 at 23:17 | comment | added | kodlu | @IlyaBogdanov:Thanks for your comment please see edit. | |
| Nov 27, 2015 at 23:17 | history | edited | kodlu | CC BY-SA 3.0 |
corrected error and extended description
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| Nov 27, 2015 at 12:02 | comment | added | Jan-Christoph Schlage-Puchta | Is there a condition which ensures that the chains have a certain length? Why can't you take $C_x=\{x\}$ and obtain costs $\sim\log n$? | |
| Nov 27, 2015 at 6:54 | comment | added | Ilya Bogdanov | The set up is not clear for me, sorry. Are $y_1,y_2,\dots$ fixed numbers? If yes --- are they chosen so that $C_x\cap C_{x'}=\varnothing$, or it is our constraint to choose $x_1,x_2,\dots$ so that the resulting sets are disjoint? | |
| Nov 26, 2015 at 23:39 | history | asked | kodlu | CC BY-SA 3.0 |