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Search: id:A163755
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| A163755 |
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a(0)=1. For n>=1, write n in binary. Let b(n,m) be the length of the mth run of 0's or 1's, reading right to left. Then a(n) = product{m=1 to M} p(m)^b(n,m), where p(m) is the mth prime, and M is the number of runs of 0's and 1's in binary n. |
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+0
1
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| 1, 2, 6, 4, 12, 30, 18, 8, 24, 90, 210, 60, 36, 150, 54, 16, 48
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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This sequence is a permutation of the terms of sequence A055932.
Clarification: By "run" of 0's or 1's in binary n, it is meant a group either entirely of 0's, and bounded by 1's or the edge of the binary number interpreted as a string, or entirely of 1's, and bounded by 0's or the edge of the string. In other words, the runs of 0's alternate with the runs of 1's.
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LINKS
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Andrew Weimholt, Table of n, a(n) for n = 0..1000
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EXAMPLE
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13 in binary is 1101. So, reading right to left, there is a run of one 1, followed by a run of one 0, followed by a run of two 1's. So, the lengths of the runs are 1,1,2. Therefore a(13) = p(1)^1 * p(2)^1 * p(3)^2 = 2^1 * 3^1 * 5^2 = 150.
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CROSSREFS
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Cf. A055932
Sequence in context: A127730 A118416 A046204 this_sequence A100851 A064462 A111807
Adjacent sequences: A163752 A163753 A163754 this_sequence A163756 A163757 A163758
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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), Aug 03 2009
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