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A189926
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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/5), t=cos(Pi/5), where [] denotes the floor function.
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7
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1, 4, 6, 9, 11, 13, 16, 18, 21, 23, 25, 28, 30, 33, 35, 37, 39, 42, 45, 47, 49, 51, 54, 57, 59, 62, 63, 66, 69, 71, 74, 75, 78, 80, 83, 86, 87, 90, 92, 95, 98, 99, 102, 104, 107, 110, 112, 114, 116, 119, 121, 124, 126, 128, 131, 133, 136, 138, 140, 143, 145, 148, 150, 152, 155, 157, 160, 162, 164, 167, 169, 172, 174, 176, 179, 181, 184
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OFFSET
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1,2
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COMMENTS
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This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
a(n) = n + [n*s/r] + [n*t/r],
b(n) = n + [n*r/s] + [n*t/s],
c(n) = n + [n*r/t] + [n*s/t], where []=floor.
sin(Pi/5)=sqrt((5-sqrt(5))/8); cos(Pi/5)=(1+sqrt(5))/4.
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LINKS
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FORMULA
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A189926: a(n) = n + [n*sin(Pi/5)] + [n*cos(Pi/5)].
A189927: b(n) = n + [n*csc(Pi/5)] + [n*cot(Pi/5)].
A189928: c(n) = n + [n*sec(Pi/5)] + [n*tan(Pi/5)].
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MATHEMATICA
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r=1; s=Sin[Pi/5]; t=Cos[Pi/5];
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}] (*A189926*)
Table[b[n], {n, 1, 120}] (*A189927*)
Table[c[n], {n, 1, 120}] (*A189928*)
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PROG
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(PARI) for(n=1, 100, print1(n + floor(n*sin(Pi/5)) + floor(n*cos(Pi/5)), ", ")) \\ G. C. Greubel, Jan 13 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/5)) + Floor(n*Cos(Pi(C)/5)): n in [1..100]]; // G. C. Greubel, Jan 13 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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