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A189515
a(n) = n + [ns] + [nt]; s=Pi/2, t=2/Pi.
3
2, 6, 8, 12, 15, 18, 21, 25, 28, 31, 35, 37, 41, 43, 47, 51, 53, 57, 60, 63, 66, 70, 73, 76, 79, 82, 86, 88, 92, 96, 98, 102, 105, 108, 111, 114, 118, 121, 124, 127, 131, 133, 137, 141, 143, 147, 149, 153, 156, 159, 163, 166, 169, 172, 176, 178, 182, 185, 188, 192, 194, 198, 201, 204, 208, 211, 214, 217, 220, 223, 227, 230, 233, 237, 239, 243, 246, 249, 253, 255, 259, 262, 265
OFFSET
1,1
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=1, s=Pi/2, t=2/Pi gives a=A189515, b=A189516, c=A189517.
FORMULA
a(n) = n + A140758(n) + floor(2*n/Pi). - R. J. Mathar, Sep 30 2011
MATHEMATICA
r=1; s=Pi/2; t=2/Pi;
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}] (*A189515*)
Table[b[n], {n, 1, 120}] (*A189516*)
Table[c[n], {n, 1, 120}] (*A189517*)
CROSSREFS
Sequence in context: A328770 A138626 A178406 * A190344 A287000 A282358
KEYWORD
nonn
AUTHOR
Clark Kimberling, Apr 23 2011
STATUS
approved