A189515 a(n) = n + [ns] + [nt]; s=Pi/2, t=2/Pi.

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.

Links

Table of n, a(n) for n=1..83.

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

Cf. A189516, A189517.

Sequence in context: A328770 A138626 A178406 * A190344 A287000 A282358

Adjacent sequences: A189512 A189513 A189514 * A189516 A189517 A189518

Keyword

nonn

Author

Clark Kimberling, Apr 23 2011

Status

approved