

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
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OFFSET

1,1


COMMENTS

2. Let S(n)=(1/2)*A141105(n). Is the complement of S equal to A035487?
Both 2 and 3 are true. These can be verified with the Walnut theoremprover, using a synchronized Fibonacci automaton for the sequence. This automaton accepts the Fibonacci (Zeckendorf) representation of n and x in parallel if and only if x = A141105(n).  Jeffrey Shallit, Jan 27 2024


LINKS



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}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



