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A141104 Lower Even Swappage of Upper Wythoff Sequence. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

1. lim (1/n)*A141104(n) = 1 + tau

2. Let S(n)=(1/2)*A141104(n). Is the complement of S equal to A004976?

REFERENCES

Vincent Russo and Loren Schwiebert, Beatty sequences, Fibonacci numbers and the golden ratio, http://www-personal.umich.edu/~vprusso/Fibonacci.pdf

LINKS

Table of n, a(n) for n=1..62.

FORMULA

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.

EXAMPLE

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,...).

CROSSREFS

Cf. A000201, A001950, A141105, A141106, A141107, A004976.

Sequence in context: A103799 A175817 A256773 * A047410 A255056 A164875

Adjacent sequences:  A141101 A141102 A141103 * A141105 A141106 A141107

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jun 02 2008, Aug 27 2008

STATUS

approved

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Last modified August 18 16:11 EDT 2017. Contains 290727 sequences.