A283234 2*A001950.

4, 10, 14, 20, 26, 30, 36, 40, 46, 52, 56, 62, 68, 72, 78, 82, 88, 94, 98, 104, 108, 114, 120, 124, 130, 136, 140, 146, 150, 156, 162, 166, 172, 178, 182, 188, 192, 198, 204, 208, 214, 218, 224, 230, 234, 240, 246, 250, 256, 260, 266, 272, 276, 282, 286, 292

Offset: 1

Author

Clark Kimberling, Mar 03 2017

Keywords

• nonn
• easy

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

Crossrefs

Cf. A000201, A283233, A005843.

Programs

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 *)

Python

from math import isqrt
def A283234(n): return ((n+isqrt(5*n**2))&-2)+(n<<1) # Chai Wah Wu, Aug 10 2022

Formula

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