A105284 - OEIS

%I #21 Oct 04 2014 09:08:25

%S 1,3,10,37,142,558,2212,8811,35170,140538,561868,2246914,8986540,

%T 35943948,143771368,575076661,2300289022,9201120918,36804413332,

%U 147217512790,588869770084,2355478518468,9421912950136,37687649553630

%N a(n)/4^n is the measure of the subset of [0,1] remaining when all intervals of the form [b/2^m - 1/2^(2m+1), b/2^m + 1/2^(2m+1)] have been removed, with b and m positive integers, b<2^m and m<=n.

%C Removing all such intervals (without an upper limit on n) leaves a nowhere dense subset of [0,1]. However, since each step removes additional points of measure no more than 1/2^(n+1), this nowhere dense subset must make up at least half of [0,1] and so be of positive measure. In fact its measure is 0.535573680435778224753342807..., the limit of a(n)/A000302(n).

%F a(n) = 4*a(n-1)-A045690(n) for n>0. a(2n+2)=6*a(2n+1)-8*a(2n); a(4n+3)=6*a(4n+2)-8*a(4n+1)+a(n); a(4n+5)=6*a(4n+4)-8*a(4n+3)+2*a(n).

%F a(n) = A045690(2n+2). a(2n+1)=4*a(2n)-a(n); a(2n+2)=4*a(2n+1)-2*a(n). - _Mamuka Jibladze_, Sep 30 2014

%e At the start the interval [0,1] has measure 1=1/1. The first step removes the interval [3/8,5/8], leaving a subset with a measure of 3/4. The second step in addition removes the intervals [7/32,9/32] and [23/32,25/32], leaving a subset with a measure of 5/8=10/16. The third step in addition removes the intervals [15/128,17/128], [47/128,3/8), (5/8,81/128] and [111/128,113/128], leaving a subset with a measure of 37/64.

%K nonn

%O 0,2

%A _Henry Bottomley_, Apr 25 2005