Search: id:A106829 Results 1-1 of 1 results found. %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, Aglet repair %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 Search completed in 0.001 seconds