%I #9 Mar 01 2023 14:28:45
%S 2,6,9,10,13,15,16,19,22,24,25,28,29,32,36,37,40,42,43,46,47,51,53,55,
%T 56,60,62,64,67,69,70,73,76,78,79,82,83,87,89,91,92,96,99,100,103,104,
%U 106,109,110,114,117,118,121,122,124,127,130,131,133,136,137
%N a(n) = 1 + A026430(A026430(n)).
%C 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:
%C (1) v o u, defined by (v o u)(n) = v(u(n));
%C (2) v' o u;
%C (3) v o u';
%C (4) v' o u.
%C Every positive integer is in exactly one of the four sequences. Their limiting densities are 4/9, 2/9, 2/9, 1/9, respectively.
%e (1) v o u = (2, 6, 9, 10, 13, 15, 16, 19, 22, 24, 25, 28, 29, 32, 36, ...) = A360136
%e (2) v' o u = (1, 5, 12, 14, 21, 23, 26, 33, 39, 41, 44, 50, 54, 59, 65, ...) = A360137
%e (3) v o u' = (4, 7, 11, 17, 20, 27, 31, 34, 38, 45, 49, 52, 58, 61, 66, ...) = A360138
%e (4) v' o u' = (3, 8, 18, 30, 35, 48, 57, 63, 72, 84, 93, 98, 111, 116, ...) = A360139
%t z = 2000; zz = 100;
%t u = Accumulate[1 + ThueMorse /@ Range[0, 600]]; (* A026430 *)
%t u1 = Complement[Range[Max[u]], u]; (* A356133 *)
%t v = u + 1; (* A285954 *)
%t v1 = Complement[Range[Max[v]], v]; (* A285953 *)
%t Table[v[[u[[n]]]], {n, 1, zz}] (* A360136 *)
%t Table[v1[[u[[n]]]], {n, 1, zz}] (* A360137 *)
%t Table[v[[u1[[n]]]], {n, 1, zz}] (* A360138 *)
%t Table[v1[[u1[[n]]]], {n, 1, zz}] (* A360139 *)
%o (Python)
%o def A360136(n): return 1+(m:=n+(n-1>>1)+(n-1&1|(n.bit_count()&1^1)))+(m-1>>1)+(m-1&1|(m.bit_count()&1^1)) # _Chai Wah Wu_, Mar 01 2023
%Y Cf. A026530, A359352, A285953, A359277 (intersections instead of results of composition), A359352-A360135, A360137-A360139.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Feb 03 2023