A190079 n + [n*s/r] + [n*t/r]; r=1, s=cos(Pi/5), t=sec(Pi/5).

2, 5, 8, 11, 15, 17, 20, 23, 27, 30, 32, 35, 39, 42, 45, 47, 51, 54, 57, 60, 62, 66, 69, 72, 75, 79, 81, 84, 87, 91, 94, 96, 99, 103, 106, 109, 111, 114, 118, 121, 124, 126, 130, 133, 136, 139, 143, 145, 148, 151, 155, 158, 160, 163, 166, 170, 173, 175, 178, 182, 185, 188, 190, 194, 197, 200, 203, 207, 209, 212, 215, 218, 222

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=cos(Pi/5), t=sec(Pi/5) gives

a=A190079, b=A190080, c=A190081.

Links

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

Formula

A190079: a(n) = n + [n*cos(Pi/5)] + [n*sec(Pi/5)].

A190080: b(n) = n + [n*sec(Pi/5)] + [n*(sec(Pi/5))^2].

A190081: c(n) = n + [n*cos(Pi/5)] + [n*(cos(Pi/5))^2].

Mathematica

r=1; s=Cos[Pi/5]; t=Sec[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}]  (*A190079*)
Table[b[n], {n, 1, 120}]  (*A190080*)
Table[c[n], {n, 1, 120}]  (*A190081*)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*cos(Pi/5)) + floor(n/cos(Pi/5)), ", ")) \\ G. C. Greubel, Feb 15 2018
(Magma) R:= RealField(); [n + Floor(n*Cos(Pi(R)/5)) + Floor(n/Cos(Pi(R)/5)): n in [1..100]]; // G. C. Greubel, Feb 15 2018

Crossrefs

Cf. A190080, A190081.

Sequence in context: A353985 A118518 A329604 * A184872 A163249 A295624

Adjacent sequences: A190076 A190077 A190078 * A190080 A190081 A190082

Keyword

nonn

Author

Clark Kimberling, May 04 2011

Status

approved