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Search: id:A136494
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| A136494 |
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Number of permutation symmetries in the binary expansion of n. |
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+0
1
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| 1, 1, 1, 2, 2, 2, 2, 6, 6, 4, 4, 6, 4, 6, 6, 24, 24, 12, 12, 12, 12, 12, 12, 24, 12, 12, 12, 24, 12, 24, 24, 120
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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One can find the number of 0s in 1s in a string recursively by finding the number of 0s and 1s in disjoint substrings. Then just follow the formula.
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FORMULA
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a(n) = A093659! * A139329!
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EXAMPLE
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a(14) = 6 because there are 3! permutation symmetries of 1s * the 0! permutation symmetries of 0s.
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CROSSREFS
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Cf. A139329, A093659.
Sequence in context: A151704 A110023 A116863 this_sequence A048764 A038714 A139554
Adjacent sequences: A136491 A136492 A136493 this_sequence A136495 A136496 A136497
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KEYWORD
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base,nonn
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AUTHOR
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Max Sills (maxwell.sills(AT)case.edu), Apr 13 2008
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