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A004759
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Binary expansion starts 111.
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7
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7, 14, 15, 28, 29, 30, 31, 56, 57, 58, 59, 60, 61, 62, 63, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| This is the minimal recursive sequence such that a(1)=7, A007814(a(n))= A007814(n) and A010060(a(n))=A010060(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 23 2009]
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FORMULA
| a(2n) = 2a(n), a(2n+1) = 2a(n) + 1 + 6[n==0].
a(n) = n + 6 * 2^floor(log2(n)) = A004758(n) + A053644(n).
a(n+1)=min{m>a(n): A007814(m)=A007814(n+1) and A010060(m)=A010060(n+1)}. a(2^k)-a(2^k-1)=A103204(k+2),k>=1. [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 23 2009]
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EXAMPLE
| 30 in binary is 11110, so 30 is in sequence.
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PROG
| (PARI) a(n)=n+6*2^floor(log(n)/log(2))
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CROSSREFS
| Cf. A004754 (10), A004755 (11), A004756 (100), A004757 (101), A004758 (110).
Cf. A004760, A053644, A062050, A076877.
A007814 A010060 A103204 A159559 A159560 A159615 A159619 A159629 A159698 [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 23 2009]
Sequence in context: A069137 A141164 A004781 * A062056 A173024 A115770
Adjacent sequences: A004756 A004757 A004758 * A004760 A004761 A004762
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Edited by Ralf Stephan (ralf(AT)ark.in-berlin.de), Oct 12 2003
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