A189364 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(2), t=(1+sqrt(5))/2.

3, 7, 11, 15, 20, 23, 27, 31, 35, 40, 43, 47, 52, 55, 60, 63, 68, 72, 75, 80, 83, 88, 92, 95, 100, 104, 108, 112, 116, 120, 124, 128, 132, 137, 140, 144, 148, 152, 157, 160, 164, 168, 172, 177, 180, 185, 189, 192, 197, 200, 205, 209, 212, 217, 220, 225, 229, 233, 237, 241, 245, 249, 253, 257, 261, 265, 269, 274, 277, 281, 285, 289, 294, 297, 302, 305, 309, 314, 317, 322, 326, 329, 334

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

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

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

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

Taking r=1, s=sqrt(2), t=(1+sqrt(5))/2 gives f=A189364, g=A189365, h=A189366.

Links

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

Mathematica

r = 1; s = Sqrt[2]; t = (1 + Sqrt[5])/2;
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t]
Table[f[n], {n, 1, 120}]  (* A189364 *)
Table[g[n], {n, 1, 120}]  (* A189365 *)
Table[h[n], {n, 1, 120}]  (* A189366 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sqrt(2)) + floor(n*(1+sqrt(5))/2), ", ")) \\ G. C. Greubel, Apr 20 2018
(Magma) [n + Floor(n*Sqrt(2)) + Floor(n*(1+Sqrt(5))/2): n in [1..100]]; // G. C. Greubel, Apr 20 2018

Crossrefs

Cf. A189365, A189366.

Sequence in context: A220520 A330165 A228436 * A022797 A190884 A310211

Adjacent sequences: A189361 A189362 A189363 * A189365 A189366 A189367

Keyword

nonn

Author

Clark Kimberling, Apr 20 2011

Status

approved