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Irregular triangle (or "upper Wythoff tree", or Beatty tree for r = (golden ratio)^2), T, of all nonnegative integers, each exactly once, as determined from the upper Wythoff sequence as described in Comments.
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%I #5 Jun 07 2015 18:02:42

%S 0,2,1,7,3,5,20,10,15,8,54,4,6,23,28,41,21,143,9,11,13,16,18,57,62,75,

%T 109,55,376,31,36,44,49,22,24,26,29,42,146,151,164,198,287,144,986,12,

%U 14,17,19,65,70,78,83,96,112,117,130,56,58,60,63,76,110

%N Irregular triangle (or "upper Wythoff tree", or Beatty tree for r = (golden ratio)^2), T, of all nonnegative integers, each exactly once, as determined from the upper Wythoff sequence as described in Comments.

%C Suppose that r is an irrational number > 1, and let s = r/(r-1), so that the sequences u and v defined by u(n) = floor(r*n) and v(n) = floor(s*n), for n >=1 are the Beatty sequences of r and s, and u and v partition the positive integers.

%C The tree T has root 0 with an edge to 2, and all other edges are determined as follows: if x is in u(v), then there is an edge from x to floor(r + r*x) and an edge from x to ceiling(x/r); otherwise there is an edge from x to floor(r + r*x). (Thus, the only branchpoints are the numbers in u(v).)

%C Another way to form T is by "backtracking" to the root 0. Let b(x) = floor[x/r] if x is in (u(n)), and b(x) = floor[r*x] if x is in (v(n)). Starting at any vertex x, repeated applications of b eventually reach 0. The number of steps to reach 0 is the number of the generation of T that contains x. (See Example for x = 29).

%C See A258212 for a guide to Beatty trees for various choices of r.

%e Rows (or generations, or levels) of T:

%e 0

%e 2

%e 1 7

%e 3 5 20

%e 10 15 8 54

%e 4 6 23 28 41 21 143

%e 9 11 13 16 18 57 62 75 109 55 376

%e Generations 0 to 9 of the tree are drawn by the Mathematica program. In T, the path from 0 to 29 is (0,2,7,3,10,28,75,29). The path obtained by backtracking (i.e., successive applications of the mapping b in Comments) is (29,75,28,10,3,7,2,0).

%t r = (3 + Sqrt[5])/2; k = 1000; w = Map[Floor[r #] &, Range[k]];

%t f[x_] := f[x] = If[MemberQ[w, x], Floor[x/r], Floor[r*x]];

%t b := NestWhileList[f, #, ! # == 0 &] &;

%t bs = Map[Reverse, Table[b[n], {n, 0, k}]];

%t generations = Table[DeleteDuplicates[Map[#[[n]] &, Select[bs, Length[#] > n - 1 &]]], {n, 10}]

%t paths = Sort[Map[Reverse[b[#]] &, Last[generations]]]

%t graph = DeleteDuplicates[Flatten[Map[Thread[Most[#] -> Rest[#]] &, paths]]]

%t TreePlot[graph, Top, 0, VertexLabeling -> True, ImageSize -> 1400]

%t Map[DeleteDuplicates, Transpose[paths]] (* _Peter J. C. Moses_, May 21 2015 *)

%Y Cf. A000201, A001950, A258236 (path-length from 0 to n).

%K nonn,tabf,easy

%O 1,2

%A _Clark Kimberling_, Jun 05 2015