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A106744
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Given n shoelaces, each with two aglets; sequence gives number of aglet pairs that must be picked up to guarantee that the probability that no shoelace is left behind is > 1/2.
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1
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1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Assistance in extending sequence given by Gerald McGarvey (Gerald.McGarvey(AT)comcast.net)
Fieggen's site has pictures of aglets (and hints on repairing them). - N. J. A. Sloane (njas(AT)research.att.com) May 16 2005
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LINKS
| Ian Fieggen, Aglet repair
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EXAMPLE
| a(2)=2 because given 2 shoelaces, p=1/3 that the first two aglets picked up will be on a single shoelace, requiring another pick-up and p=2/3 that they won't, so the mean no. of pick-ups is (1/3)*2 + (2/3)*1 = 4/3, for which the ceiling is 2.
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PROG
| (PARI) choose(n, k)=gamma(n+1)/(gamma(n-k+1)*gamma(k+1)) a(n)=ceil(solve(x=n/2, n, 2^(2*(n-x))*choose(n, 2*(n-x))-(1/2)*choose(2*n, 2*x))) for(n=3, 50, print1(a(n), ", ")) (McGarvey)
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CROSSREFS
| See A106829 for another version.
Sequence in context: A003005 A006163 A053757 * A072932 A029915 A108141
Adjacent sequences: A106741 A106742 A106743 * A106745 A106746 A106747
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KEYWORD
| nonn
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AUTHOR
| N Fernandez (primeness(AT)borve.org), May 16 2005
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EXTENSIONS
| More terms from Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), May 17 2005
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