A190082 n + [n*s/r] + [n*t/r]; r=1, s=sin(2*Pi/5), t=csc(2*Pi/5).

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 60, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 117, 120, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 177, 180, 183, 185, 188, 191, 194, 197, 200, 203, 206, 209, 212, 215, 218

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(2*Pi/5), t=csc(2*Pi/5) gives

a=A190082, b=A190083, c=A190084.

Links

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

Formula

A190082: a(n) = n + [n*sin(2*Pi/5)] + [n*csc(2*Pi/5)].

A190083: b(n) = n + [n*csc(2*Pi/5)] + [n*(csc(2*Pi/5))^2].

A190084: c(n) = n + [n*sin(2*Pi/5)] + [n*(sin(2*Pi/5))^2].

Mathematica

r=1; s=Sin[2*Pi/5]; t=Csc[2*Pi/5];
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}]  (* A190082 *)
Table[b[n], {n, 1, 120}]  (* A190083 *)
Table[c[n], {n, 1, 120}]  (* A190084 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sin(2*Pi/5)) + floor(n/sin(2*Pi/5)), ", ")) \\ G. C. Greubel, Mar 04 2018
(Magma) R:= RealField(); [n + Floor(n*Sin(2*Pi(R)/5)) + Floor(n/Sin(2*Pi(R)/5)): n in [1..100]]; // G. C. Greubel, Mar 04 2018

Crossrefs

Cf. A190083, A190084.

Sequence in context: A189386 A292661 A016789 * A165334 A189512 A190361

Adjacent sequences: A190079 A190080 A190081 * A190083 A190084 A190085

Keyword

nonn

Author

Clark Kimberling, May 04 2011

Status

approved