A189524 a(n) = n + [n*s/r] + [n*t/r]; r=2, s=1+sqrt(2), t=1+sqrt(3).

3, 6, 10, 13, 17, 21, 24, 27, 31, 35, 39, 42, 45, 49, 53, 56, 60, 63, 66, 71, 74, 78, 81, 84, 89, 92, 95, 99, 103, 106, 110, 113, 117, 121, 124, 128, 131, 134, 139, 142, 146, 149, 152, 157, 160, 163, 167, 170, 174, 178, 181, 185, 188, 192, 196, 199, 202, 207, 210, 213, 217, 220, 225, 228, 231, 235, 238, 242, 246, 249, 252, 256, 260, 264, 267, 270, 274, 278, 281, 285, 288, 292, 296

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=2, s=1+sqrt(2), t=1+sqrt(3) gives f=A189524, g=A189525, h=A189526.

Links

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

Mathematica

r=2; s=1+2^(1/2); t=1+3^(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}]  (* A189524 *)
Table[g[n], {n, 1, 120}]  (* A189525 *)
Table[h[n], {n, 1, 120}]  (* A189526 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*(sqrt(2)+1)/2) + floor(n*(1+sqrt(3))/2), ", ")) \\ G. C. Greubel, Apr 20 2018
(Magma) [n + Floor(n*(Sqrt(2)+1)/2) + Floor(n*(1+Sqrt(3))/2): n in [1..100]]; // G. C. Greubel, Apr 20 2018

Crossrefs

Cf. A189525, A189526.

Sequence in context: A310055 A356086 A310056 * A288205 A049880 A276219

Adjacent sequences: A189521 A189522 A189523 * A189525 A189526 A189527

Keyword

nonn

Author

Clark Kimberling, Apr 23 2011

Status

approved