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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
OFFSET
1,11
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
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