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%I #7 Mar 31 2012 23:00:51
%S 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,
%T 23,24,25,26,27,28,29,30,31,32,33,33,34,35,36,37,38,39,40,41,42,43,44,
%U 45
%N 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.
%C Assistance in extending sequence given by _Gerald McGarvey_
%C Fieggen's site has pictures of aglets (and hints on repairing them). - _N. J. A. Sloane_ May 16 2005
%H Ian Fieggen, <a href="http://www.fieggen.com/shoelace/agletrepair.htm">Aglet repair</a>
%e 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.
%o (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)
%Y See A106829 for another version.
%K nonn
%O 1,3
%A _Neil Fernandez_, May 16 2005
%E More terms from _Gerald McGarvey_, May 17 2005
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