A105284 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.

1, 3, 10, 37, 142, 558, 2212, 8811, 35170, 140538, 561868, 2246914, 8986540, 35943948, 143771368, 575076661, 2300289022, 9201120918, 36804413332, 147217512790, 588869770084, 2355478518468, 9421912950136, 37687649553630

Offset

0,2

Comments

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).

Links

Table of n, a(n) for n=0..23.

Formula

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).

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

Example

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.

Crossrefs

Sequence in context: A195350 A289810 A149044 * A211975 A357399 A219260

Adjacent sequences: A105281 A105282 A105283 * A105285 A105286 A105287

Keyword

nonn

Author

Henry Bottomley, Apr 25 2005

Status

approved