A190006 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/3), t=csc(Pi/3).

2, 5, 8, 11, 14, 17, 21, 23, 26, 29, 32, 35, 39, 42, 44, 47, 50, 53, 56, 60, 63, 66, 68, 71, 74, 78, 81, 84, 87, 89, 92, 95, 99, 102, 105, 108, 111, 113, 117, 120, 123, 126, 129, 132, 134, 138, 141, 144, 147, 150, 153, 157, 159, 162, 165, 168, 171, 174, 178, 180, 183, 186, 189, 192, 196, 199, 202, 204, 207, 210, 213, 217, 220

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 + [n*s/r] + [n*t/r],

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

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

Taking r=1, s=sin(Pi/3), t=csc(Pi/3) gives a=A190006, b=A190007, c=A190008.

Links

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

Formula

A190006: a(n) = n + [n*sin(Pi/3)] + [n*csc(Pi/3].

A190007: b(n) = n + [n*csc(Pi/3)] + [n*(csc(Pi/3))^2].

A190008: c(n) = n + [n*sin(Pi/3)] + [n*(sin(Pi/3))^2].

Mathematica

r=1; s=Sin[Pi/3]; t=Csc[Pi/3];
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}]  (* A190006 *)
Table[b[n], {n, 1, 120}]  (* A190007 *)
Table[c[n], {n, 1, 120}]  (* A190008 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sin(Pi/3)) + floor(n/sin(Pi/3)), ", ")) \\ G. C. Greubel, Jan 11 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/3)) + Floor(n/Sin(Pi(C)/3)): n in [1..100]]; // G. C. Greubel, Jan 11 2018

Crossrefs

Cf. A180007, A190008.

Sequence in context: A184905 A279773 A184910 * A135677 A355135 A192585

Adjacent sequences: A190003 A190004 A190005 * A190007 A190008 A190009

Keyword

nonn

Author

Clark Kimberling, May 03 2011

Status

approved