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A107284
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), b/2^m + 1/2^(2m)] have been removed, with b and m positive integers, b < 2^m and m <= n.
0
1, 2, 6, 20, 74, 284, 1116, 4424, 17622, 70340, 281076, 1123736, 4493828, 17973080, 71887896, 287542736, 1150153322, 4600578044, 18402241836, 73608826664, 294435025580, 1177739540168, 4710957036936, 18843825900272, 75375299107260
OFFSET
0,2
COMMENTS
Removing all such intervals (without an upper limit on n) leaves a nowhere dense subset of [0,1]. It is of positive measure, namely 0.2677868402178891123766714035843..., the limit of a(n)/4^n. This is the same as the limit of A003000(n)/2^n and of A045690(n)/2^n and half the limit of A105284(n)/4^n.
It can be shown that this sequence also counts the pairs of binary sequences with Conway number 0. These Conway numbers arise in the analysis of Penney's game and measure to what degree two sequences overlap; see the Nishiyama paper in the links for further details. - Reed Phillips, Jun 09 2020
LINKS
Yutaka Nishiyama, Pattern Matching Probabilities and Paradoxes as a New Variation on Penney's Coin Game, International Journal of Pure and Applied Mathematics, Volume 59 No. 3 2010, 357-366.
FORMULA
a(n) = 4*a(n-1) - A003000(n) = 2*A105284(n-1).
a(2*n+1) = 6*a(2*n) - 8*a(2*n-1).
a(4*n) = 6*a(4*n-1) - 8*a(4*n-2) - a(n).
a(4*n+2) = 6*a(4*n+1) - 8*a(4*n) - 2*a(n).
EXAMPLE
At the start the interval [0,1] has measure 1 = 1/1. The first step removes the interval [1/4,3/4], leaving a subset with measure of 1/2 = 2/4. The second step in addition removes the intervals [3/16,1/4) and (3/4,13/16], leaving a subset with measure of 3/8 = 6/16. The third step in addition removes the intervals [7/64,9/64] and [55/64,57/64], leaving a subset with measure of 5/16 = 20/64.
CROSSREFS
KEYWORD
nonn
AUTHOR
Henry Bottomley, May 19 2005
STATUS
approved