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A126999
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Zero-one fractional-part array for the golden ratio; a rectangular array T by antidiagonals.
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4
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1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1
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OFFSET
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1,1
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COMMENTS
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(Row 1) = (Column 1) = A005614 (infinite Fibonacci word).
See A187950 for connections to left-shifted sums of the infinite Fibonacci word.
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LINKS
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FORMULA
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T(n,k) = {nr} + {kr} - {nr+kr}, where r=(1+sqrt(5))/2 and { } denotes fractional part;, i.e., {x} = x - floor(x).
T(n,k) = [nr] + [kr] - [nr+kr], where []=floor.
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EXAMPLE
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Northwest corner:
1 0 1 1 0 1 0 1 1
0 0 1 0 0 0 0 1 0
1 1 1 1 0 1 1 1 1
1 0 1 0 0 1 0 1 1
0 0 0 0 0 0 0 1 0
1 0 1 1 0 1 1 1 1
T(3,3)=1 because 2{3x}-{6x}=1.
The antidiagonals form a triangle with these first six rows:
1
0 0
1 0 1
1 1 1 1
0 0 1 0 0
1 0 1 1 0 1
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MATHEMATICA
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r:=(1+Sqrt[5])/2;
T[k_, n_]:=Floor[n*r+k*r]-Floor[n*r]-Floor[k*r]
TableForm[Table[T[n, k], {k, 1, 10}, {n, 1, 10}]]
Table[T[n-k+1, k], {n, 1, 10}, {k, 1, n}]//Flatten (* modified by G. C. Greubel, May 30 2019 *)
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PROG
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(PARI) r=(1+sqrt(5))/2;
T(n, k) = ((n+k)*r)\1 - (n*r)\1 - (k*r)\1;
for(n=1, 10, for(k=1, n, print1(T(n-k+1, k), ", "))) \\ G. C. Greubel, May 30 2019
(Magma) r:=(1+Sqrt(5))/2; [[Floor((n+1)*r)-Floor((n-k+1)*r)-Floor(k*r): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019
(Sage)
def T(n, k): return floor((n+k)*golden_ratio) - floor(n*golden_ratio) - floor(k*golden_ratio)
[[T(n-k+1, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, May 30 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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