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Search: id:A161895
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| A161895 |
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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. |
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+0
1
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| 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, 1, 1, 2, 4
(list; graph; listen)
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OFFSET
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1,11
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EXAMPLE
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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.
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CROSSREFS
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A161819, A161820, A161821, A161822
Sequence in context: A136044 A103414 A092400 this_sequence A048138 A165022 A030338
Adjacent sequences: A161892 A161893 A161894 this_sequence A161896 A161897 A161898
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KEYWORD
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base,more,nonn
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Jun 21 2009
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