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A141105 Upper Even Swappage of Upper Wythoff Sequence. 4
2, 6, 8, 10, 14, 16, 18, 20, 24, 26, 28, 32, 34, 36, 40, 42, 44, 48, 50, 52, 54, 58, 60, 62, 66, 68, 70, 74, 76, 78, 82, 84, 86, 90, 92, 94, 96, 100, 102, 104, 108, 110, 112, 116, 118, 120, 124, 126, 128, 130, 134, 136, 138, 142, 144, 146, 150, 152, 154, 158, 160, 162 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

1. lim (1/n)*A141105(n) = 1 + tau.

2. Let S(n)=(1/2)*A141105(n). Is the complement of S equal to A035487?

3. Is A141105 = 1+A141106?

LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000

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 greatest number in a such that after swapping b(n) and a(m), the resulting new a and b are both increasing. A141105 is the sequence obtained by thus swapping all odds out of A001950.

a(n) = round(n*tau^2), with tau = (1+sqrt(5))/2 (the golden ratio A001622).

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 = (1,3,4,5,8,9,11,12,...) and b = (2,6,7,10,13,15,18,...).

After 2nd swap,

a = (1,3,4,5,7,9,11,12,...) and b = (2,6,8,10,13,15,18,...).

MATHEMATICA

r = GoldenRatio^2; f[x_] := Floor[x];

g[x_] := If[Mod[f[x], 2] == 0, f[x], f[x] + 1];

t = Table[g[r*n], {n, 1, 100}]

(* Clark Kimberling, Nov 03 2012 *)

CROSSREFS

Cf. A000201, A001622, A001950, A141104, A141106, A141107, A004976.

Sequence in context: A258663 A166447 A075332 * A047395 A284794 A187692

Adjacent sequences:  A141102 A141103 A141104 * A141106 A141107 A141108

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jun 02 2008

STATUS

approved

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Last modified October 20 02:18 EDT 2019. Contains 328244 sequences. (Running on oeis4.)