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A066879
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n such that there are as many 1's as 0's in the base 2 expansion of Floor(n/2).
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3
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4, 5, 18, 19, 20, 21, 24, 25, 70, 71, 74, 75, 76, 77, 82, 83, 84, 85, 88, 89, 98, 99, 100, 101, 104, 105, 112, 113, 270, 271, 278, 279, 282, 283, 284, 285, 294, 295, 298, 299, 300, 301, 306, 307, 308, 309, 312, 313, 326, 327, 330, 331, 332, 333, 338, 339, 340
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| n such that there are as many odd as even terms in the orbit f(n), f(f(n)), f(f(f(n))), ..., 1, where f(k) = Floor(k/2).
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FORMULA
| A037861(Floor(n/2)) = 0.
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EXAMPLE
| floor(18/2) = 9 = 1001 (base 2) has the same number of 1's as 0's. So 18 is a term of the sequence.
Also the orbit corresponding to 18 is 9, 4, 2, 1, which has an equal number (i.e. 2) of odd and even terms.
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CROSSREFS
| Complement is the union of 1 and A126388.
Sequence in context: A119997 A010361 A060289 * A134750 A051949 A026902
Adjacent sequences: A066876 A066877 A066878 * A066880 A066881 A066882
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KEYWORD
| easy,nonn
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AUTHOR
| Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Jan 21 2002
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EXTENSIONS
| Extended and edited by John W. Layman (layman(AT)math.vt.edu), Jan 30 2002
New definition by Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Jun 10 2011
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