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A189932
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a(n) = n + [n*r/s] + [n*t/s]; r=1, s=(sin(Pi/5))^2, t=(cos(Pi/5))^2, where [] denotes the floor function.
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4
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4, 10, 16, 22, 28, 34, 40, 46, 52, 56, 62, 68, 74, 80, 86, 92, 98, 104, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 166, 172, 178, 184, 190, 196, 202, 208, 214, 218, 224, 230, 236, 242, 248, 254, 260, 266, 272, 276, 282, 288, 294, 300, 306, 312, 318, 324, 328, 334, 340, 346, 352, 358, 364, 370, 376, 382, 386, 392, 398, 404
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OFFSET
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1,1
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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.
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LINKS
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FORMULA
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A005408: a(n) = n + [n*(sin(Pi/5))^2] + [n*(cos(Pi/5))^2] = 2*n - 1.
A189932: b(n) = n + [n*(csc(Pi/5))^2] + [n*(cot(Pi/5))^2].
A189933: c(n) = n + [n*(sec(Pi/5))^2] + [n*(tan(Pi/5))^2]. (End)
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MATHEMATICA
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r=1; s=Sin[Pi/5]^2; t=Cos[Pi/5]^2;
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}] (*A005408*)
Table[b[n], {n, 1, 120}] (*A189932*)
Table[c[n], {n, 1, 120}] (*A189933*)
Table[b[n]/2, {n, 1, 120}] (*A189934*)
Table[c[n]/2, {n, 1, 120}] (*A189935*)
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PROG
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(PARI) for(n=1, 100, print1(n + floor(n/(sin(Pi/5))^2) + floor(n/(tan(Pi/5))^2), ", ")) \\ G. C. Greubel, Jan 13 2018
(Magma) C<i> := ComplexField(); [n + Floor(n/(Sin(Pi(C)/5))^2) + Floor(n/(Tan(Pi(C)/5))^2): 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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