A190335 a(n) = n + [n*s/r] + [n*t/r]; r=2, s=sqrt(2), t=1/s.

1, 3, 6, 7, 9, 12, 13, 15, 18, 20, 21, 24, 26, 27, 30, 32, 35, 36, 38, 41, 42, 44, 47, 48, 50, 53, 55, 56, 59, 61, 62, 65, 67, 70, 71, 73, 76, 77, 79, 82, 83, 85, 88, 90, 91, 94, 96, 97, 100, 102, 105, 106, 108, 111, 112, 114, 117, 119, 120, 123, 125, 126, 129, 131, 132, 135, 137, 140, 141, 143, 146, 147, 149, 152, 154, 155

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=1/s gives

f=A190335, g=A190336, h=A190337.

Links

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

Formula

A190335: f(n) = n + [n*sqrt(2)] + [n/sqrt(8)].

A190336: g(n) = n + [n/sqrt(2)] + [n/2].

A190337: h(n) = 3*n + [n*sqrt(8)].

Mathematica

r=2; s=2^(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}]  (*A190335*)
Table[g[n], {n, 1, 120}]  (*A190336*)
Table[h[n], {n, 1, 120}]  (*A190337*)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sqrt(2)) + floor(n/sqrt(8)), ", ")) \\ G. C. Greubel, Apr 04 2018
(Magma) R:=RealField(); [n + Floor(n*Sqrt(2)) + Floor(n/Sqrt(8)): n in [1..100]]; // G. C. Greubel, Apr 04 2018

Crossrefs

Cf. A190336, A190337.

Sequence in context: A286686 A082847 A047242 * A083951 A284879 A208327

Adjacent sequences: A190332 A190333 A190334 * A190336 A190337 A190338

Keyword

nonn

Author

Clark Kimberling, May 08 2011

Status

approved