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%I #14 Jun 09 2021 02:34:09
%S 1,2,6,20,74,284,1116,4424,17622,70340,281076,1123736,4493828,
%T 17973080,71887896,287542736,1150153322,4600578044,18402241836,
%U 73608826664,294435025580,1177739540168,4710957036936,18843825900272,75375299107260
%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), b/2^m + 1/2^(2m)] 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]. 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.
%C 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
%H Yutaka Nishiyama, <a href="https://ijpam.eu/contents/2010-59-3/10/10.pdf">Pattern Matching Probabilities and Paradoxes as a New Variation on Penney's Coin Game</a>, International Journal of Pure and Applied Mathematics, Volume 59 No. 3 2010, 357-366.
%F a(n) = 4*a(n-1) - A003000(n) = 2*A105284(n-1).
%F a(2*n+1) = 6*a(2*n) - 8*a(2*n-1).
%F a(4*n) = 6*a(4*n-1) - 8*a(4*n-2) - a(n).
%F a(4*n+2) = 6*a(4*n+1) - 8*a(4*n) - 2*a(n).
%e 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.
%Y Cf. A003000, A045690, A105284.
%K nonn
%O 0,2
%A _Henry Bottomley_, May 19 2005
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