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Jul 16, 2016 at 16:25 vote accept CommunityBot
Jul 16, 2016 at 15:27 comment added Fedor Petrov Though the question is not well formulated, it is, after all, quite reasonable.
Jul 16, 2016 at 15:25 history edited user6671 CC BY-SA 3.0
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Jul 16, 2016 at 15:16 answer added Fedor Petrov timeline score: 9
Jul 16, 2016 at 15:15 history edited user6671 CC BY-SA 3.0
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Jul 16, 2016 at 15:01 history edited user6671 CC BY-SA 3.0
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Jul 16, 2016 at 15:00 comment added Fedor Petrov but I think that $a_2=0$
Jul 16, 2016 at 14:56 comment added user6671 I edited the question to show the computation of $n=3$.
Jul 16, 2016 at 14:55 history edited user6671 CC BY-SA 3.0
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Jul 16, 2016 at 14:49 comment added Willie Wong My previous comment lists $b_n$ if they are proper subsets. If they are not then you add 2. They don't seem to agree with your $a_n$. (In particular, how is $a_2 = 2$ when $a_1 = 0$? If you count only proper subsets, $b_1 = 0$ and $b_2 = 2$. But if you count all subsets, $b_1 = 2$ and $b_2 = 4$.) Can you do a few examples to show how you computed "manually" those numbers?
Jul 16, 2016 at 14:42 history edited user6671 CC BY-SA 3.0
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Jul 16, 2016 at 14:42 comment added Willie Wong Do you require proper subsets? Also, your sequence seems wrong. For $n = 3$ you have $\{1\}, \{2\}, \{1,3\}, \{2,3\}$. For $n = 4$ I get at least 10: $\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}$ and their complements. For $n = 5$ you should have $$ \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\} $$ and their complements. So it seems you inserted an extra $2$ in your sequence?
Jul 16, 2016 at 14:18 comment added user6671 Yes, thats what I mean.
Jul 16, 2016 at 14:17 comment added Fedor Petrov You mean that this inequality holds for all sequences $0<x_1<\dots <x_n$?
Jul 16, 2016 at 13:51 history asked user6671 CC BY-SA 3.0