A360403 a(n) = A360393(A026430(n)).

1, 4, 9, 13, 19, 22, 24, 31, 36, 40, 42, 49, 51, 58, 64, 66, 73, 76, 78, 85, 87, 94, 99, 103, 106, 112, 117, 121, 126, 129, 133, 139, 144, 148, 150, 157, 159, 166, 171, 175, 178, 184, 189, 193, 199, 202, 204, 210, 213, 220, 225, 229, 235, 238, 240, 246, 253

Offset: 1

Author

Clark Kimberling, Mar 11 2023

Keywords

• nonn
• easy

Comments

This is the second 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. Their limiting densities are 4/9, 2/9, 2/9, 1/9, respectively (and likewise for A360394-A360401).

Examples

			(1)  v o u = (3, 7, 10, 11, 14, 16, 17, 20, 23, 25, 26, 29, 30, 33, 37, ...) = A360402
(2)  v' o u = (1, 4, 9, 13, 19, 22, 24, 31, 36, 40, 42, 49, 51, 58, 64, ...) = A360403
(3)  v o u' = (5, 8, 12, 18, 21, 28, 32, 35, 39, 46, 50, 53, 59, 62, 67, ...) = A360404
(4)  v' o u' = (2, 6, 15, 27, 34, 45, 55, 60, 69, 81, 91, 96, 108, 114, ...) = A360405
		

Links

• Winston de Greef, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A026530, A360392, A360393, A360394-A3546352 (intersections instead of results of compositions), A360398-A360401 (results of reversed compositions), A360402, A360404, A360405.

Programs

Mathematica

z = 2000; zz = 100;
u = Accumulate[1 + ThueMorse /@ Range[0, 600]]; (* A026430 *)
u1 = Complement[Range[Max[u]], u];  (* A356133 *)
v = u + 2;  (* A360392 *)
v1 = Complement[Range[Max[v]], v];  (* A360393 *)
Table[v[[u[[n]]]], {n, 1, zz}]    (* A360402 *)
Table[v1[[u[[n]]]], {n, 1, zz}]   (* A360403 *)
Table[v[[u1[[n]]]], {n, 1, zz}]   (* A360404 *)
Table[v1[[u1[[n]]]], {n, 1, zz}]  (* A360405 *)

Python

def A360393(n):
    if n < 3: return [0, 1, 2][n]
    return 3*n - 5 - (n-3).bit_count() % 2
def A026430(n): return n+(n-1>>1)+(n-1&1|(n.bit_count()&1^1))
def A360403(n): return A360393(A026430(n)) # Winston de Greef, Mar 24 2023