A228853 Nodes of tree generated as follows: (1,2) is an edge, and if (x,y) is an edge, then (y,y+x) and (y,2y+x) are edges.

1, 2, 3, 5, 7, 8, 11, 12, 13, 17, 18, 19, 21, 26, 27, 29, 30, 31, 34, 41, 43, 44, 45, 46, 47, 49, 50, 55, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 79, 80, 81, 89, 97, 99, 100, 101, 104, 105, 106, 108, 109, 111, 112, 115, 116, 117, 119, 121, 123, 128

Offset: 1

Author

Clark Kimberling, Sep 05 2013

Keywords

• nonn
• easy
• nice

Comments

As a tree, infinitely many branches are essentially linearly recurrent sequences. The extreme cases, (1,2) -> (2,3) -> (3,5) -> ... and (1,2) -> (2,5) -> (5,12) -> ..., contribute A000045 (Fibonacci numbers) and A000129 (Pell numbers) to A228853.

Suppose that (u,v) and (v,w) are consecutive edges. The continued fraction of w/v is obtained from the continued fraction of v/u by prefixing 1 if w = v + u, or 2 if w = 2v + u. Consequently, if each edge is labeled with 1 or 2 in the obvious way, then the continued fraction of w/v is the sequence of 1s and 2s, in reverse order, from the node 2 to the node w, with 2 attached at the end. (See Example, Part 2.)

Is A228853 essentially A141832? (If so, the answer to the question in Comments at A141832 is that A141832 is infinite.)

Yes; the initial node (1,2) adds a single 2 to the end of the fraction, and subsequent edges prepend 1's and 2's. - Charlie Neder, Oct 21 2018

Examples

			Part 1: Taking the first generation of edges of the tree to be G(1) = {(1,2)}, the edge (1,2) grows G(2) = {(2,3), (2,5)}, which grows G(3) = {(3,5), (3,8), (5,7), (5,12)}, ... Expelling duplicate nodes and sorting leave {1, 2, 3, 5, 7, 8, ...}.
Part 2: The branch 2, 3, 8, 11, 19, 30, 49, 128, 305 has edge-labels 1, 2, 1, 1, 1, 1, 2, 2, so that 305/128 = [2, 2, 1, 1, 1, 1, 2, 1, 2].
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A141832, A228854, A228855, A228856, A000045, A000129.

Programs

Mathematica

f[x_, y_] := {{y, x + y}, {y, x + 2 y}}; x = 1; y = 2; t = {{x, y}}; u = Table[t = Flatten[Map[Apply[f, #] &, t], 1], {12}]; v = Flatten[u]; w = Flatten[Prepend[Table[v[[2 k]], {k, 1, Length[v]/2}], {x, y}]]; Sort[Union[w]]