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A228853
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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.
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9
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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
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OFFSET
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1,2
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COMMENTS
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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.)
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
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LINKS
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EXAMPLE
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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].
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MATHEMATICA
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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]]
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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