A189386 a(n) = n+[ns/r]+[nt/r]; r=1, s=sqrt(2), t=1/sqrt(3), []=floor.

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 34, 38, 41, 44, 47, 50, 53, 55, 59, 62, 65, 68, 70, 74, 77, 80, 83, 86, 89, 91, 95, 98, 101, 104, 106, 110, 112, 116, 119, 121, 125, 127, 131, 133, 137, 140, 142, 146, 148, 152, 155, 157, 161, 163, 167, 169, 173, 176, 178, 182, 184, 188, 190, 193, 197, 199, 203, 205, 208, 211, 214, 218, 220, 224, 226, 229, 233, 235, 239, 241, 244, 247

Offset: 1

Author

Clark Kimberling, Apr 21 2011

Keywords

• nonn

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

a(n)=n+[ns/r]+[nt/r],

b(n)=n+[nr/s]+[nt/s],

c(n)=n+[nr/t]+[ns/t], where []=floor.

Taking r=1, s=sqrt(2), t=1/sqrt(3) gives

a=A189386, b=A189387, c=A189388.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A189387, A189388, A189361, A189383, A189395.

Programs

Mathematica

r=1; s=2^(1/2); t=3^(-1/2);
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
Table[a[n], {n, 1, 120}]  (*A189386*)
Table[b[n], {n, 1, 120}]  (*A189387*)
Table[c[n], {n, 1, 120}]  (*A189388*)