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A141104
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Lower Even Swappage of Upper Wythoff Sequence.
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4
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2, 4, 6, 10, 12, 14, 18, 20, 22, 26, 28, 30, 34, 36, 38, 40, 44, 46, 48, 52, 54, 56, 60, 62, 64, 68, 70, 72, 74, 78, 80, 82, 86, 88, 90, 94, 96, 98, 102, 104, 106, 108, 112, 114, 116, 120, 122, 124, 128, 130, 132, 136, 138, 140, 142, 146, 148, 150, 154, 156, 158, 162
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OFFSET
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1,1
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COMMENTS
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1. lim (1/n)*A141104(n) = 1 + tau
2. Let S(n)=(1/2)*A141104(n). Is the complement of S equal to A004976?
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REFERENCES
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Vincent Russo and Loren Schwiebert, Beatty sequences, Fibonacci numbers and the golden ratio, http://www-personal.umich.edu/~vprusso/Fibonacci.pdf
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LINKS
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Table of n, a(n) for n=1..62.
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FORMULA
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Let a = (1,3,4,6,8,9,11,12,...) = A000201 = lower Wythoff sequence; let b = (2,5,7,10,13,15,18,...) = A001950 = upper Wythoff sequence. For each odd b(n), let a(m) be the least number in a such that after swapping b(n) and a(m), the resulting new a and b are both increasing. A141104 is the sequence obtained by thus swapping all odds out of A001950.
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EXAMPLE
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Start with
a = (1,3,4,6,8,9,11,12,...) and b = (2,5,7,10,13,15,18,...).
After first swap,
a = (1,3,5,6,8,9,11,12,...) and b = (2,4,7,10,13,15,18,...).
After 2nd swap,
a = (1,3,5,7,8,9,11,12,...) and b = (2,4,6,10,13,15,18,...).
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CROSSREFS
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Cf. A000201, A001950, A141105, A141106, A141107, A004976.
Sequence in context: A103799 A175817 A256773 * A047410 A255056 A164875
Adjacent sequences: A141101 A141102 A141103 * A141105 A141106 A141107
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KEYWORD
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nonn
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AUTHOR
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Clark Kimberling, Jun 02 2008, Aug 27 2008
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STATUS
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approved
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