A190056 a(n) = n + [n*s/r] + [n*t/r]; r=1/2, s=sin(Pi/3), t=csc(Pi/3).

4, 9, 14, 19, 24, 29, 35, 39, 44, 50, 55, 59, 65, 70, 74, 79, 85, 90, 94, 100, 105, 110, 115, 120, 125, 131, 135, 140, 145, 150, 155, 160, 166, 170, 175, 181, 186, 190, 196, 201, 206, 210, 216, 221, 225, 231, 236, 241, 246, 251, 256, 262, 266, 271, 277, 281, 286, 291, 297, 301, 306, 312, 317, 321, 327, 332, 337, 342, 347, 352, 356

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=1/2, s=sin(Pi/3), t=csc(Pi/3) gives a=A190056, b=A190057, c=A190058.

Links

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

Formula

A190056: a(n) = n + [2*n*sin(Pi/3)] + [2*n*csc(Pi/3)].

A190057: b(n) = n + [(n/2)*csc(Pi/3)] + [n*(csc(Pi/3))^2].

A190058: c(n) = n + [(n/2)*sin(Pi/3)] + [n*(sin(Pi/3))^2].

Mathematica

r=1/2; 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}]  (* A190056 *)
Table[b[n], {n, 1, 120}]  (* A190057 *)
Table[c[n], {n, 1, 120}]  (* A190058 *)

Prog

(PARI) for(n=1, 30, print1(n + floor(2*n*sin(Pi/3)) + floor(2*n/sin(Pi/3)), ", ")) \\ G. C. Greubel, Jan 10 2018
(Magma) C<i> := ComplexField(); [n + Floor(2*n*Sin(Pi(C)/3)) + Floor(2*n/Sin(Pi(C)/3)): n in [1..30]]; // G. C. Greubel, Jan 10 2018

Crossrefs

Cf. A190057, A190056, A190053.

Sequence in context: A313117 A016897 A189517 * A313118 A313119 A293807

Adjacent sequences: A190053 A190054 A190055 * A190057 A190058 A190059

Keyword

nonn

Author

Clark Kimberling, May 04 2011

Status

approved