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A189796
n+[ns/r]+[nt/r]; r=2, s=(sin(1))^2, t=(cos(1))^2.
3
1, 2, 4, 5, 6, 8, 10, 11, 13, 14, 15, 17, 18, 20, 22, 23, 25, 26, 27, 29, 31, 32, 34, 35, 36, 38, 39, 41, 43, 44, 45, 47, 48, 50, 52, 53, 55, 56, 57, 59, 60, 62, 64, 65, 66, 68, 69, 71, 73, 74, 76, 77, 78, 80, 82, 83, 85, 86, 87, 89, 90, 92, 94, 95, 97, 98, 99, 101, 103, 104, 106, 107, 108, 110, 111, 113, 115, 116, 117, 119, 120, 122, 124, 125, 127
OFFSET
1,2
COMMENTS
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=2, s=(sin(1))^2, t=(cos(1))^2 gives
MATHEMATICA
r=2; s=Sin[1]^2; t=Cos[1]^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}] (*A189796*)
Table[b[n], {n, 1, 120}] (*A189797*)
Table[c[n], {n, 1, 120}] (*A189798*)
CROSSREFS
Sequence in context: A285143 A274436 A330330 * A284818 A192584 A337140
KEYWORD
nonn
AUTHOR
Clark Kimberling, Apr 27 2011
STATUS
approved