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Jan 21, 2018 at 3:03 review Reopen votes
Jan 21, 2018 at 11:00
Jan 14, 2018 at 16:19 comment added anonim Apparently, this is (among the others), one of the fundamental properties, of the Conway's sequence; and, the following paper `On Conway's recursive sequence' by Kudo and Vakil gives a proof for the doubling inequality. So, in fact, this was really appropriate for MO; even though it was marked to be put on hold as off-topic...
Jan 14, 2018 at 16:01 comment added anonim Thanks everyone, who paid attention; I did not know the connection to Hofstadter-Conway sequence; otherwise I would have put it at the beginning.
Jan 14, 2018 at 15:44 comment added user6976 @FedorPetrov: It is not the $10000 Conway problem. See the reference in the comment by Robert Israel. The paper MR1083608 in the Monthly solved Convey's problem, and, I think, contains a solution of this problem as well. There must be a simpler solution though.
Jan 14, 2018 at 12:23 review Reopen votes
Jan 14, 2018 at 16:23
Jan 14, 2018 at 9:19 history closed user6976
Anthony Quas
YCor
David Handelman
Alexey Ustinov
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Jan 14, 2018 at 7:55 comment added user6976 @FanZheng: Yes, you are right. about subadditivity.
Jan 14, 2018 at 6:26 comment added Fedor Petrov I agree that the question is appropriate on MO. There were several interesting posts coming from the competitions problems. I suppose that if OP asked the same without mentioning Silk Road olympiad, but Conway and 10,000 dollars instead, this would be upvoted and popular and nobody would try to close it.
Jan 14, 2018 at 6:01 answer added Fan Zheng timeline score: 2
Jan 14, 2018 at 5:45 comment added Fan Zheng That gives some justification of the existence of this question on mathoverflow.
Jan 14, 2018 at 5:22 comment added Robert Israel See also OEIS sequence A004001.
Jan 14, 2018 at 5:16 comment added Gerry Myerson Wikipedia (en.wikipedia.org/wiki/Hofstadter_sequence) calls this the Hofstadter–Conway \$10,000 sequence.
Jan 14, 2018 at 5:01 comment added anonim Induction is well-defined here, I just have a sequence, -if you don't like this notation, then use, $a_n=a_{a_{n-1}} + a_{n-a_{n-1}}$. I already said that, this is an olympaid problem (see:beginning of my post.) For Mark Sapir, I know, this place is mostly for research level math, but there indeed is a section with elementary proofs; and I have seen olympiad problems being posted here. By the way, the sequence is not subadditive. Fan, thanks for the comment, nice observation.
Jan 14, 2018 at 4:46 comment added Fan Zheng Just some observation from the first 64 terms: for $k\ge1$ it seems that $a(2^k)=2^{k-1}$; thus the inequality is saturated infinitely often.
Jan 14, 2018 at 4:38 comment added Fan Zheng @MarkSapir Why does it follow that $a(n)$ is sub-additive? Maybe my olympiad skill has gone rust but I can see that immediately. Counterexample: $a(6)=4$, $a(2)=1$, $a(4)=2$.
Jan 14, 2018 at 4:05 comment added user6976 It is obviously well defined because $a(n+1)$ is either equal to $a(n)$ or to $a(n)+1$ for every $n$, so $a(n)< n$ for $n\ge 2$. But this site is not for olympiad problems.
Jan 14, 2018 at 4:02 comment added YCor Then this context should be included, if you want to convince people here that this does not belong on another site as it is at first sight. Also, when you define something as definition by induction, but it's not clear that this is well-defined, writing explicitly that it's indeed well-defined (if so) would show that you have thought about the problem, and not just copying the statement of an exercise as it regularly occurs here.
Jan 14, 2018 at 3:43 comment added anonim I don't think so. This was previously asked in artofproblemsolving forums; which has more people, that could tackle these type of problems, compared to SE; yet, no solution still, and in fact, seems really hard one. That is why I asked it here, under the tags; elementary-proofs; and, added that, this is a past olympiad problem. So I think this is the appropriate place, and, I would appreciate a help.
Jan 14, 2018 at 3:04 comment added YCor Your question will be better welcome at MathSE, where you should comment on the definition (is this induction meaningful? not a priori), explained what you've tried, etc.
Jan 14, 2018 at 2:55 review Close votes
Jan 14, 2018 at 9:19
Jan 14, 2018 at 2:37 history edited YCor
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Jan 14, 2018 at 2:22 history asked anonim CC BY-SA 3.0