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A190304
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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=csch(1), t=sech(1).
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3
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1, 4, 6, 9, 12, 14, 16, 19, 21, 24, 27, 29, 32, 34, 36, 39, 42, 44, 47, 49, 51, 54, 56, 59, 62, 64, 66, 69, 71, 74, 77, 79, 82, 84, 86, 89, 91, 94, 97, 99, 101, 104, 106, 109, 112, 114, 116, 119, 121, 124, 127, 129, 132, 133, 136, 139, 141, 144, 147, 149, 151, 154, 156, 159, 162, 164, 167, 169, 171, 174, 177, 179, 182, 183, 186, 189
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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+[ns/r]+[nt/r],
b(n) = n+[nr/s]+[nt/s],
c(n) = n+[nr/t]+[ns/t], where []=floor.
Taking r=1, s=csch(1), t=sech(1) gives
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LINKS
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FORMULA
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A190304: a(n)=n+[n*csch(1)]+[n*sech(1)].
A190305: b(n)=n+[n*sinh(1)]+[n*tanh(1)].
A190306: c(n)=n+[n*cosh(1)]+[n*coth(1)].
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MATHEMATICA
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r=1; s=Csch[1]; t=Sech[1];
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}] (*A190304*)
Table[b[n], {n, 1, 120}] (*A190305*)
Table[c[n], {n, 1, 120}] (*A190306*)
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PROG
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(PARI) for(n=1, 30, print1(n + floor(n/sinh(1)) + floor(n/cosh(1)), ", ")) \\ G. C. Greubel, Dec 29 2017
(Magma) [n + Floor(n/Sinh(1)) + Floor(n/Cosh(1)): n in [1..30]]; // G. C. Greubel, Dec 29 2017
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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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