A189367 - OEIS

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A189367

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

1, 4, 7, 9, 12, 15, 17, 19, 22, 25, 27, 30, 33, 35, 37, 40, 43, 45, 48, 51, 53, 56, 58, 60, 63, 66, 69, 71, 74, 76, 78, 81, 84, 87, 89, 92, 95, 96, 99, 102, 104, 107, 110, 113, 114, 117, 120, 122, 125, 128, 131, 133, 135, 138, 140, 143, 146, 149, 151, 153, 156, 158, 161, 164, 166, 169, 172, 174, 176, 179, 182, 184, 187, 190, 192, 194, 197, 200, 202, 205, 208, 210, 212, 215

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

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=sqrt(2), t=sqrt(3) gives f=A189367, g=A189368, h=A189369.

LINKS

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

MATHEMATICA

r = 2; s = Sqrt[2]; t = 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}] (* A189367 *)