A190085 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(1/2), t=cos(1/2).

1, 3, 6, 8, 11, 13, 16, 18, 20, 22, 25, 27, 30, 32, 35, 37, 39, 41, 44, 46, 49, 51, 54, 56, 57, 60, 62, 65, 67, 70, 72, 75, 76, 79, 81, 84, 86, 89, 91, 94, 95, 98, 100, 103, 105, 108, 110, 113, 115, 116, 119, 121, 124, 126, 129, 131, 134, 135, 138, 140, 143, 145, 148, 150, 153, 154, 157, 159, 162, 164, 167, 169, 171, 173, 175, 178, 180

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 f(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define g(n) and h(n) as the ranks of n/s and n/t. It is easy to prove that

f(n) = n + [n*s/r] + [n*t/r],

g(n) = n + [n*r/s] + [n*t/s],

h(n) = n + [n*r/t] + [n*s/t], where []=floor.

Taking r=1, s=sin(1/2), t=cos(1/2) gives f=A190085, g=A190086, h=A190087.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

A190085: f(n) = n + [n*sin(1/2)] + [n*cos(1/2)], this sequence.

A190086: g(n) = n + [n*csc(1/2)] + [n*cot(1/2)].

A190087: h(n) = n + [n*sec(1/2)] + [n*tan(1/2)].

Mathematica

r=1; s=Sin[1/2]; t=Cos[1/2];
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}]  (* A190085 *)
Table[g[n], {n, 1, 120}]  (* A190086 *)
Table[h[n], {n, 1, 120}]  (* A190087 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sin(1/2)) + floor(n*cos(1/2)), ", ")) \\ G. C. Greubel, Mar 04 2018
(Magma) [n + Floor(n*Sin(1/2)) + Floor(n*Cos(1/2)): n in [1..100]]; // G. C. Greubel, Mar 04 2018

Crossrefs

Cf. A190086, A190087.

Sequence in context: A189679 A286666 A190229 * A360926 A303590 A358844

Adjacent sequences: A190082 A190083 A190084 * A190086 A190087 A190088

Keyword

nonn

Author

Clark Kimberling, May 04 2011

Status

approved