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 A161895 Write down binary n as a string of 0's and 1's. Consider the runs of 1's (bounded by 0's or by the edge of the string) alternating with the runs of 0's (bounded by 1's or by the edge of the string) in the string. Then, a(n) = the number of positive binary integers that contain the same lengths of runs of 1's as of the runs of 1's in binary n, and contain the same lengths of runs of 0's as of the runs of 0's in binary n. 1
 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 4, 2, 2, 2, 1, 3, 4, 3, 2, 2, 1, 2, 4, 1, 4, 3, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 4, 2, 2, 1, 3, 6, 2, 6, 4, 2, 2, 2, 3, 6 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,11 LINKS Lars Blomberg, Table of n, a(n) for n = 1..10000 Lars Blomberg, C# program for calculating the b-file EXAMPLE 77 in binary is 1001101. There is a run of two 0's, and is a run of one 0. There is a run of two 1's, and are two runs of one 1 each. There are six binary integers (including 1001101 itself) that contain the same lengths of runs of 1's and the same lengths of runs of 0's. (These are: 1001011, 1001101, 1010011, 1011001, 1100101, and 1101001.) So a(77) = 6. CROSSREFS Cf. A161819, A161820, A161821, A161822. Sequence in context: A092400 A269974 A269975 * A048138 A165022 A030338 Adjacent sequences:  A161892 A161893 A161894 * A161896 A161897 A161898 KEYWORD nonn,base AUTHOR Leroy Quet, Jun 21 2009 EXTENSIONS a(39)-a(83) and b-file from Lars Blomberg, Feb 12 2013 STATUS approved

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