A189380 a(n) = n + floor(n*s/r) + floor(n*t/r); r=1, s=-1+sqrt(2), t=1+sqrt(2).

3, 6, 11, 14, 19, 22, 25, 30, 33, 38, 41, 44, 49, 52, 57, 60, 65, 68, 71, 76, 79, 84, 87, 90, 95, 98, 103, 106, 111, 114, 117, 122, 125, 130, 133, 136, 141, 144, 149, 152, 155, 160, 163, 168, 171, 176, 179, 182, 187, 190, 195, 198, 201, 206, 209, 214, 217, 222, 225, 228, 233, 236, 241, 244, 247, 252, 255, 260, 263, 266, 271, 274, 279, 282, 287, 290, 293, 298, 301, 306, 309, 312, 317

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=-1+sqrt(2), t=1+sqrt(2) gives f=A189380, g=A189381, h=A189382.

Links

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

Formula

a(n) = 5*floor(n * (sqrt(2) - 1)) + 3*floor(n * (2 - sqrt(2))) + 3. - Miko Labalan, Dec 04 2016

Mathematica

r=1; s=-1+2^(1/2); t=1+2^(1/2);
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}]  (* A189380 *)
Table[g[n], {n, 1, 120}]  (* A189381 *)
Table[h[n], {n, 1, 120}]  (* A189382 *)

Prog

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

Crossrefs

Cf. A189381, A189382.

Sequence in context: A182669 A026368 A246976 * A047398 A047924 A267519

Adjacent sequences: A189377 A189378 A189379 * A189381 A189382 A189383

Keyword

nonn

Author

Clark Kimberling, Apr 20 2011

Status

approved