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A190431
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a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,3,1) and []=floor.
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5
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2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 3, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 3, 2, 1, 3, 1, 0, 2, 1, 3, 2, 1, 2, 1, 0, 2, 1, 3, 2, 0, 2, 1, 3, 2, 1, 3, 1, 0
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OFFSET
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1,1
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COMMENTS
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Write a(n) = [(b*n+c)*r] - b*[n*r] - [c*r]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
Examples:
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LINKS
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FORMULA
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a(n) = floor((3*n+1)*(1+sqrt(5))/2) - 3*floor(n*(1+sqrt(5))/2) - 1. - G. C. Greubel, Apr 06 2018
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MATHEMATICA
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r = GoldenRatio; b = 3; c = 1;
f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
t = Table[f[n], {n, 1, 320}] (* A190431 *)
Flatten[Position[t, 0]] (* A190432 *)
Flatten[Position[t, 1]] (* A190433 *)
Flatten[Position[t, 2]] (* A190434 *)
Flatten[Position[t, 3]] (* A190435 *)
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PROG
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(PARI) for(n=1, 100, print1(floor((3*n+1)*(1+sqrt(5))/2) - 3*floor(n*(1+sqrt(5))/2) - 1, ", ")) \\ G. C. Greubel, Apr 06 2018
(Magma) [Floor((3*n+1)*(1+Sqrt(5))/2) - 3*Floor(n*(1+Sqrt(5))/2) - 1: n in [1..100]]; // G. C. Greubel, Apr 06 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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