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.

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, 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

Offset

1,2

Comments

An aglet is a tag or sheath on the end of a lace to facilitate its passing through eyelet holes.

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.

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).

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.

Fieggen's site has pictures of aglets (and hints on repairing them). - N. J. A. Sloane May 16 2005

Links

Table of n, a(n) for n=1..50.

Ian Fieggen, Aglet repair

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, x))) for(n=3, 50, print1(a(n), ", "))

Crossrefs

See A106744 for another version.

Sequence in context: A358845 A121347 A303589 * A190228 A286667 A083120

Adjacent sequences: A106826 A106827 A106828 * A106830 A106831 A106832

Keyword

nonn

Author

Gerald McGarvey, May 22 2005

Status

approved