%I A106829
%S A106829 1,2,4,5,7,9,10,12,14,15,17,19,21,22,24,26,28,30,31,33,35,37,39,41,42,
%T A106829 44,46,48,50,52,53,55,57,59,61,63,65,66,68,70,72,74,76,78,80,81,83,85,
               87,89
%N A106829 Given n shoelaces, each with two aglets; sequence gives number of aglets 
               that must be picked up to guarantee that the probability that no 
               shoelace is left behind is > 1/2.
%C A106829 An aglet is a tag or sheath on the end of a lace to facilitate its passing 
               through eyelet holes.
%C A106829 There are C(2n,k) ways of picking up k aglets out of 2*n aglets. All 
               of them are equally likely. Picking up k aglets leaves 2n - k aglets 
               not picked up. Each of these 2n - k aglets must be on a different 
               shoelace in order to get all the shoelaces. There are C(n,2n - k) 
               combinations of shoelaces with a not-picked aglet.
%C A106829 Each not-picked aglet can be on either end, so we multiply C(n,2n - k) 
               by 2^(2n - k) for a total of 2^(2n - k) * C(n,2n - k) ways to get 
               all the shoelaces. So we want So 2^(2n - k) * C(n,2n - k) > (1/2) 
               * C(2n,k).
%C A106829 The following PARI code can be used to calculate k, it solves for 50% 
               probability, pretending that the problem is continuous, then takes 
               the ceiling to get to the realistic k value required.
%C A106829 Fieggen's site has pictures of aglets (and hints on repairing them). 
               - N. J. A. Sloane (njas(AT)research.att.com) May 16 2005
%H A106829 Ian Fieggen, <a href="http://www.fieggen.com/shoelace/agletrepair.htm">
               Aglet repair</a>
%o A106829 (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,x))) for(n=3,50,print1(a(n),
               ","))
%Y A106829 See A106744 for another version.
%Y A106829 Sequence in context: A026351 A047212 A121347 this_sequence A083120 A001614 
               A050731
%Y A106829 Adjacent sequences: A106826 A106827 A106828 this_sequence A106830 A106831 
               A106832
%K A106829 nonn
%O A106829 1,2
%A A106829 Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), May 22 2005