

A106744


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.


1



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
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OFFSET

1,3


COMMENTS

Assistance in extending sequence given by Gerald McGarvey
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


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 pickup and p=2/3 that they won't, so the mean no. of pickups is (1/3)*2 + (2/3)*1 = 4/3, for which the ceiling is 2.


PROG

(PARI) choose(n, k)=gamma(n+1)/(gamma(nk+1)*gamma(k+1)) a(n)=ceil(solve(x=n/2, n, 2^(2*(nx))*choose(n, 2*(nx))(1/2)*choose(2*n, 2*x))) for(n=3, 50, print1(a(n), ", ")) (McGarvey)


CROSSREFS

See A106829 for another version.
Sequence in context: A228297 A303788 A319288 * A072932 A029915 A108141
Adjacent sequences: A106741 A106742 A106743 * A106745 A106746 A106747


KEYWORD

nonn


AUTHOR

Neil Fernandez, May 16 2005


EXTENSIONS

More terms from Gerald McGarvey, May 17 2005


STATUS

approved



