A283233 2*A000201.

2, 6, 8, 12, 16, 18, 22, 24, 28, 32, 34, 38, 42, 44, 48, 50, 54, 58, 60, 64, 66, 70, 74, 76, 80, 84, 86, 90, 92, 96, 100, 102, 106, 110, 112, 116, 118, 122, 126, 128, 132, 134, 138, 142, 144, 148, 152, 154, 158, 160, 164, 168, 170, 174, 176, 180, 184, 186

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

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=(-1+sqrt(5))/2, t=(1+sqrt(5))/2 gives a=A283233, b=A283234, c=A005843.

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Formula

a(n) = 2*floor(n*r), where r = (1+sqrt(5))/2.

Mathematica

r = 1; s = (-1 + 5^(1/2))/2; t = (1 + 5^(1/2))/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}]  (* A283233 *)
Table[b[n], {n, 1, 120}]  (* A283234 *)
Table[c[n], {n, 1, 120}]  (* A005408 *)

Prog

(Python)
from math import isqrt
def A283233(n): return (n+isqrt(5*n**2))&-2 # Chai Wah Wu, Aug 10 2022

Crossrefs

Cf. A000201, A283234, A005843 (sequential union of A283233 and A283234), A005408.

Sequence in context: A057656 A247066 A084724 * A189400 A285342 A111051

Adjacent sequences: A283230 A283231 A283232 * A283234 A283235 A283236

Keyword

nonn,easy

Author

Clark Kimberling, Mar 03 2017

Status

approved