A189469 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1+sqrt(2), t=1+sqrt(3).

5, 11, 18, 23, 30, 36, 42, 48, 54, 61, 67, 72, 79, 85, 91, 97, 104, 110, 115, 122, 128, 135, 140, 146, 153, 159, 165, 171, 178, 183, 189, 196, 202, 208, 214, 220, 227, 232, 239, 245, 251, 257, 263, 270, 275, 282, 288, 294, 300, 306, 313, 319, 324, 331, 337, 343, 349, 356, 362, 367, 374, 380, 387, 392, 398, 405, 411, 417, 423, 429, 435, 441, 448, 454, 460, 466, 472, 479, 484, 491, 497, 503

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(3) gives f=A189469, g=A189470, h=A189471.

Links

Ivan Panchenko, Table of n, a(n) for n = 1..10000

Mathematica

r=1; s=1+Sqrt[2]; t=1+Sqrt[3];
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}]  (* A189469 *)
Table[g[n], {n, 1, 120}]  (* A189470 *)
Table[h[n], {n, 1, 120}]  (* A189471 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*(1+sqrt(2))) + floor(n*(1+sqrt(3))), ", ")) \\ G. C. Greubel, Apr 13 2018
(Magma) [n + Floor(n*(1+Sqrt(2))) + Floor(n*(1+Sqrt(3))): n in [1..100]]; // G. C. Greubel, Apr 13 2018

Crossrefs

Cf. A189470, A189471.

Sequence in context: A314265 A189520 A189406 * A314266 A314267 A314268

Adjacent sequences: A189466 A189467 A189468 * A189470 A189471 A189472

Keyword

nonn

Author

Clark Kimberling, Apr 22 2011

Status

approved