%I
%S 1,2,3,6,4,9,7,18,5,12,10,27,8,21,19,54,15,13,36,11,30,28,81,24,22,63,
%T 20,57,55,162,16,45,14,39,37,108,33,31,90,29,84,82,243,25,72,23,66,64,
%U 189,60,58,171,56,165,163,486,17,48,46,135,42,40,117,38
%N Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 3*x are in S, and duplicates are deleted as they occur.
%D Let S be the set of numbers defined by these rules: 1 is in S, and if x is in S, then x + 1 and 3*x are in S. Then S is the set of all positive integers, which arise in generations. Deleting duplicates as they occur, the generations are given by g(1) = (1), g(2) = (2,3), g(3) = (6,4,9), g(4) = (7,18,5,8,10,27), etc. Concatenating these gives A232561, a permutation of the positive integers. The number of numbers in g(n) is A001590(n), the nth tribonacci number. It is helpful to show the results as a tree with the terms of S as nodes and edges from x to x + 1 if x + 1 has not already occurred, and an edge from x to 3*x if 3*x has not already occurred.
%H Clark Kimberling, <a href="/A232561/b232561.txt">Table of n, a(n) for n = 1..1000</a>
%e Each x begets x + 1 and 3*x, but if either has already occurred it is deleted. Thus, 1 begets 2 and 3; in the next generation, 2 begets only 6, and 3 begets 4 and 9.
%t z = 8; g[1] = {1}; g[2] = {2, 3};
%t g[n_] := Riffle[g[n  1] + 1, 3 g[n  1]]
%t j[2] = Join[g[1], g[2]]; j[n_] := Join[j[n  1], g[n]];
%t g1[n_] := DeleteDuplicates[DeleteCases[g[n], Alternatives @@ j[n  1]]];
%t g1[1] = g[1]; g1[2] = g[2];
%t t = Flatten[Table[g1[n], {n, 1, z}]] (* A232561 *)
%t Table[Length[g1[n]], {n, 1, z}] (* A001590 *)
%t t1 = Flatten[Table[Position[t, n], {n, 1, 200}]] (* A232562 *)
%Y Cf. A232560, A232562, A001590.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Nov 26 2013
