

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



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



EXTENSIONS



STATUS

approved



