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a(n) = A026430(1 + A360392(n)).
9

%I #4 Mar 04 2023 15:26:25

%S 5,8,10,12,15,16,18,21,24,26,27,30,31,35,37,39,42,44,45,48,50,52,55,

%T 57,59,61,65,66,69,70,72,75,78,80,81,84,86,88,91,93,95,98,100,102,105,

%U 107,108,111,113,116,118,120,123,125,126,129,132,134,135,138

%N a(n) = A026430(1 + A360392(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) u o v, defined by (u o v)(n) = u(v(n));

%C (2) u o v';

%C (3) u' o v;

%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 (and likewise for A360394-A360397 and A360402-A360405).

%e (1) u o v = (5, 8, 10, 12, 15, 16, 18, 21, 24, 26, 27, 30, 31, 35, 37, 39, ...) = A360398

%e (2) u o v' = (1, 3, 6, 9, 14, 19, 23, 28, 33, 36, 41, 46, 51, 54, 60, 63, 68, ...) = A360399

%e (3) u' o v = (7, 13, 20, 22, 29, 32, 34, 40, 47, 49, 53, 58, 62, 67, 74, 76, ...) = A360400

%e (4) u' o v' = (2, 4, 11, 17, 25, 38, 43, 56, 64, 71, 79, 92, 101, 106, 119, ...) = A360401

%t z = 2000;

%t u = Accumulate[1 + ThueMorse /@ Range[0, 600]]; (* A026430 *)

%t u1 = Complement[Range[Max[u]], u]; (* A356133 *)

%t v = u + 2; (* A360392 *)

%t v1 = Complement[Range[Max[v]], v]; (* A360393 *)

%t zz = 100;

%t Table[u[[v[[n]]]], {n, 1, zz}] (* A360398 *)

%t Table[u[[v1[[n]]]], {n, 1, zz}] (* A360399 *)

%t Table[u1[[v[[n]]]], {n, 1, zz}] (* A360400 *)

%t Table[u1[[v1[[n]]]], {n, 1, zz}] (* A360401 *)

%Y Cf. A026530, A356133, A360392, A360393, A360399, A286355, A286356, A360394 (intersections instead of results of composition), A360402-A360405.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Feb 10 2023