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

1,2

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.

Links

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

Formula

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

Example

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).

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 *)

Crossrefs

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

Sequence in context: A293977 A282649 A102454 * A085790 A257339 A210882

Adjacent sequences: A232637 A232638 A232639 * A232641 A232642 A232643

Keyword

nonn,easy

Author

Clark Kimberling, Nov 28 2013

Status

approved