A190361 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(5/4), t=sqrt(4/5).

2, 5, 8, 11, 14, 17, 20, 23, 27, 29, 32, 35, 38, 41, 44, 47, 51, 54, 56, 59, 62, 65, 68, 71, 74, 78, 81, 84, 86, 89, 92, 95, 98, 102, 105, 108, 111, 113, 116, 119, 122, 125, 129, 132, 135, 138, 141, 143, 146, 149, 153, 156, 159, 162, 165, 168, 170, 173, 176, 180, 183, 186, 189, 192, 195, 198, 200, 204, 207, 210, 213, 216

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(5/4), t=sqrt(4/5) gives

f=A190361, g=A190362, h=A190363.

Links

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

Formula

A190361: f(n) = n + [n*sqrt(5/4)] + [n*sqrt(4/5)].

A190362: g(n) = n + [n*sqrt(4/5)] + [4*n/5].

A190363: h(n) = 2*n + [n*sqrt(5/4)] + [n/4].

Mathematica

r=1; s=(5/4)^(1/2); t=1/s;
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}]  (* A190361 *)
Table[g[n], {n, 1, 120}]  (* A190362 *)
Table[h[n], {n, 1, 120}]  (* A190363 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sqrt(5/4)) + floor(n*sqrt(4/5)), ", ")) \\ G. C. Greubel, Apr 05 2018
(Magma) [n + Floor(n*Sqrt(5/4)) + Floor(n*Sqrt(4/5)): n in [1..100]]; // G. C. Greubel, Apr 05 2018

Crossrefs

Cf. A190362, A190363, A190347.

Sequence in context: A190082 A165334 A189512 * A184905 A279773 A184910

Adjacent sequences: A190358 A190359 A190360 * A190362 A190363 A190364

Keyword

nonn

Author

Clark Kimberling, May 09 2011

Status

approved