A190053 a(n) = n + [n*s/r] + [n*t/r]; r=2, s=sin(Pi/3), t=csc(Pi/3).

1, 3, 5, 7, 9, 11, 14, 15, 17, 19, 21, 23, 25, 28, 29, 31, 33, 35, 37, 39, 42, 43, 45, 47, 49, 52, 53, 56, 57, 59, 61, 63, 66, 67, 70, 71, 74, 75, 77, 80, 81, 84, 85, 88, 89, 91, 94, 95, 98, 99, 102, 104, 105, 108, 109, 112, 113, 116, 118, 119, 122, 123, 126, 127, 130, 132, 134, 136, 137, 140, 141, 144, 146, 148, 150, 151, 154, 156, 158, 160

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 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=2, s=sin(Pi/3), t=csc(Pi/3) gives a=A190053, b=A190054, c=A190055.

Links

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

Formula

A190053: a(n) = n + [(n/2)*sin(Pi/3)] + [(n/2)*csc(Pi/3)].

A190054: b(n) = n + [2n*csc(Pi/3)] + [n*(csc(Pi/3))^2].

A190055: c(n) = n + [2n*sin(Pi/3)] + [n*(sin(Pi/3))^2].

Mathematica

r=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}]  (* A190053 *)
Table[b[n], {n, 1, 120}]  (* A190054 *)
Table[c[n], {n, 1, 120}]  (* A190055 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sin(Pi/3)/2) + floor(n/(2*sin(Pi/3))), ", ")) \\ G. C. Greubel, Jan 10 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/3)/2) + Floor(n/(2*Sin(Pi(C)/3))): n in [1..100]]; // G. C. Greubel, Jan 10 2018

Crossrefs

Cf. A190054, A190055.

Sequence in context: A024323 A118820 A117521 * A327209 A285590 A195179

Adjacent sequences: A190050 A190051 A190052 * A190054 A190055 A190056

Keyword

nonn

Author

Clark Kimberling, May 04 2011

Status

approved