A161895 - OEIS

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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, 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: A269974 A269975 A308069 * 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