

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
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OFFSET

1,11


LINKS

Lars Blomberg, Table of n, a(n) for n = 1..10000
Lars Blomberg, C# program for calculating the bfile


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 bfile from Lars Blomberg, Feb 12 2013


STATUS

approved



