A232640 Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 2*x + 1 are in S, and duplicates are deleted as they occur.

1, 2, 3, 5, 4, 7, 6, 11, 9, 8, 15, 13, 12, 23, 10, 19, 17, 16, 31, 14, 27, 25, 24, 47, 21, 20, 39, 18, 35, 33, 32, 63, 29, 28, 55, 26, 51, 49, 48, 95, 22, 43, 41, 40, 79, 37, 36, 71, 34, 67, 65, 64, 127, 30, 59, 57, 56, 111, 53, 52, 103, 50, 99, 97, 96, 191

Offset: 1

Author

Clark Kimberling, Nov 28 2013

Keywords

• nonn
• easy

Comments

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 2*x + 1 are in S. Then S is the set of 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) = (5,4,7), etc. Concatenating these gives A232640, a permutation of the positive integers. The number of numbers in g(n) is F(n), where F = A000045, the Fibonacci numbers. It is helpful to show the results as a tree with the terms of S as nodes, an edge from x to x + 1 if x + 1 has not already occurred, and an edge from x to 2*x + 1 if 2*x + 1 has not already occurred.

Examples

			Each x begets x + 1 and 2*x + 1, but if either has already occurred it is deleted. Thus, 1 begets 2 and 3; then 2 begets only 5, and 3 begets (4,7), so that g(3) = (5,4,7).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000045, A003754, A135533, A232559, A232639, A232641.

Programs

Mathematica

z = 14; g[1] = {1}; g[2] = {2}; g[n_] := Riffle[g[n - 1] + 1, 2 g[n - 1] + 1]; j[2] = Join[g[1], g[2]]; j[n_] := Join[j[n - 1], g[n]]; g1[n_] := DeleteDuplicates[DeleteCases[g[n], Alternatives @@ j[n - 1]]]; g1[1] = g[1]; g1[2] = g[2]; t = Flatten[Table[g1[n], {n, 1, z}]]  (* this sequence *)
Table[Length[g1[n]], {n, 1, z}]  (* A000045 *)
Flatten[Table[Position[t, n], {n, 1, 200}]]  (* A232641 *)

Formula

Conjecture: a(n) = A135533(A003754(n+1)) for n > 0. - Mikhail Kurkov, Feb 26 2023