A356217 a(n) = A022839(A000201(n)).

2, 6, 8, 13, 17, 20, 24, 26, 31, 35, 38, 42, 46, 49, 53, 55, 60, 64, 67, 71, 73, 78, 82, 84, 89, 93, 96, 100, 102, 107, 111, 114, 118, 122, 125, 129, 131, 136, 140, 143, 147, 149, 154, 158, 160, 165, 169, 172, 176, 178, 183, 187, 190, 194, 196, 201, 205, 207

Offset: 1

Author

Clark Kimberling, Oct 02 2022

Keywords

• nonn
• easy

Comments

This is the first of four sequences that partition the positive integers. Suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their (increasing) complements, and consider these four sequences:

(1) v o u, defined by (v o u)(n) = v(u(n));

(2) v' o u;

(3) v o u';

(4) v' o u'.

Every positive integer is in exactly one of the four sequences. For the reverse composites, u o v, u o v', u' o v, u' o v', see A356104 to A356107.

Assume that if w is any of the sequences u, v, u', v', then lim_{n->oo} w(n)/n exists and defines the (limiting) density of w. For w = u,v,u',v', denote the densities by r,s,r',s'. Then the densities of sequences (1)-(4) exist, and

1/(r*r') + 1/(r*s') + 1/(s*s') + 1/(s*r') = 1.

For A356217 u, v, u', v', are the Beatty sequences given by u(n) = floor(n*(1+sqrt(5))/2) and v(n) = floor(n*sqrt(5)), so that r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.

Examples

			(1)  v o u = (2, 6, 8, 13, 17, 20, 24, 26, 31, 35, 38, 42, ...) = A356217
(2)  v' o u = (1, 5, 7, 10, 14, 16, 19, 21, 25, 28, 30, 34, ...) = A356218
(3)  v o u' = (4, 11, 15, 22, 29, 33, 40, 44, 51, 58, 62, 76, ...) = A190509
(4)  v' o u' = (3, 9, 12, 18, 23, 27, 32, 36, 41, 47, 50, 56, ...) = A356220
		

Crossrefs

Cf. u = A000201, u' = A001950, v = A022839, v' = A108598, A351415 (intersections), A356104 (reverse composites), A356218, A190509, A356220.

Programs

Mathematica

z = 1000;
u = Table[Floor[n*(1 + Sqrt[5])/2], {n, 1, z}];  (* A000201 *)
u1 = Complement[Range[Max[u]], u];  (* A001950 *)
v = Table[Floor[n*Sqrt[5]], {n, 1, z}];  (* A022839 *)
v1 = Complement[Range[Max[v]], v];  (* A108598 *)
Table[v[[u[[n]]]], {n, 1, z/4}]   (* A356217 *)
Table[v1[[u[[n]]]], {n, 1, z/4}]  (* A356218 *)
Table[v[[u1[[n]]]], {n, 1, z/4}]  (* A190509 *)
Table[v1[[u1[[n]]]], {n, 1, z/4}] (* A356220 *)

Python

from math import isqrt
def A356217(n): return isqrt(5*(n+isqrt(5*n**2)>>1)**2) # Chai Wah Wu, Oct 14 2022