A356180 a(n) = A022838(A001951(n)).

1, 3, 6, 8, 12, 13, 15, 19, 20, 24, 25, 27, 31, 32, 36, 38, 41, 43, 45, 48, 50, 53, 55, 57, 60, 62, 65, 67, 71, 72, 74, 77, 79, 83, 84, 86, 90, 91, 95, 96, 98, 102, 103, 107, 109, 112, 114, 116, 119, 121, 124, 126, 128, 131, 133, 136, 138, 142, 143, 145, 148

Offset: 1

Author

Clark Kimberling, Aug 24 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 A356088 to A356091.

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 A356180, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*sqrt(2)) and v(n) = floor(n*sqrt(3)), so r = sqrt(2), s = sqrt(3), r' = 2 + sqrt(2), s' = (3 + sqrt(3))/2.

Examples

			(1)  v o u = (1, 3, 6, 8, 12, 13, 15, 19, 20, 24, 25, 27, 31, 32, ...) = A356180
(2)  v' o u = (2, 4, 9, 11, 16, 18, 21, 26, 28, 33, 35, 37, 42, 44, ...) = A356181
(3)  v o u' = (5, 10, 17, 22, 29, 34, 39, 46, 51, 58, 64, 69, 76, ...) = A356182
(4)  v' o u' = (7, 14, 23, 30, 40, 47, 54, 63, 70, 80, 87, 94, 104, ...) = A356183
		

Crossrefs

Cf. A001951, A001952, A022838, A054406, A346308 (intersections), A356088 (reverse composites), A356181, A356182, A356183.

Programs

Mathematica

z = 800; zz = 100;
u = Table[Floor[n*Sqrt[2]], {n, 1, z}];  (* A001951 *)
u1 = Complement[Range[Max[u]], u];       (* A001952 *)
v = Table[Floor[n*Sqrt[3]], {n, 1, z}];  (* A022838 *)
v1 = Complement[Range[Max[v]], v];  (* A054406 *)
Table[v[[u[[n]]]], {n, 1, zz}]      (* A356180 *)
Table[v1[[u[[n]]]], {n, 1, zz}]     (* A356181 *)
Table[v[[u1[[n]]]], {n, 1, zz}]     (* A356182 *)
Table[v1[[u1[[n]]]], {n, 1, zz}]    (* A356183 *)

Python

from math import isqrt
def A356180(n): return isqrt(3*isqrt(n**2<<1)**2) # Chai Wah Wu, Sep 06 2022