A189457 a(n) = n+[ns/r]+[nt/r]; r=2, s=sqrt(2), t=1+sqrt(2).

2, 5, 8, 10, 14, 17, 19, 22, 25, 29, 31, 34, 37, 39, 43, 46, 49, 51, 54, 58, 60, 63, 66, 68, 72, 75, 78, 80, 84, 87, 89, 92, 95, 99, 101, 104, 107, 109, 113, 116, 118, 121, 124, 128, 130, 133, 136, 138, 142, 145, 148, 150, 153, 157, 159, 162, 165, 169, 171, 174, 177, 179, 183, 186, 188, 191, 194, 198, 200, 203, 206, 208, 212, 215, 218, 220, 223, 227, 229, 232, 235, 237, 241, 244

Offset: 1

Author

Clark Kimberling, Apr 22 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.

With r=2, s=sqrt(2), t=1+sqrt(2), a=A189457, b=A189458, c=A186222 (conjectured).

Links

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

Crossrefs

Cf. A189458, A186222.

Programs

Magma

[Floor(n + Floor(n*Sqrt(2)/2) + Floor(n*(1+Sqrt(2))/2)): n in [1..120]]; // G. C. Greubel, Aug 19 2018

Mathematica

r=2; s=2^(1/2); t=1+2^(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}]  (*A189457*)
Table[b[n], {n, 1, 120}]  (*A189458*)
Table[c[n], {n, 1, 120}]  (*A186222, conjectured*)

PARI

vector(120, n, floor(n+floor(n*sqrt(2)/2)+floor(n*(1+sqrt(2))/2))) \\ G. C. Greubel, Aug 19 2018