A258212 - OEIS

A258212

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

%I #4 Jun 07 2015 18:02:07

%S 0,1,3,2,6,4,11,8,7,19,5,12,14,32,9,21,24,20,53,16,13,15,33,35,40,87,

%T 10,22,25,27,55,58,66,54,142,17,37,42,45,34,36,41,88,90,95,108,231,29,

%U 23,26,28,56,59,61,67,69,74,144,147,155,176,143,375,18,38

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

%C Let r = (1+ sqrt(5))/2 = golden ratio. Let u(n) = floor[n*r] and v(n) = floor[n*r^2], so that u = (u(n)) = A000201 = lower Wythoff sequence and v = (v(n)) = A001950 = upper Wythoff sequence; it is well known that u and v partition the positive integers. The tree T has root 0 with an edge to 1, 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, and b(x) = floor[r*x] if x is in v. 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 = 35).

%C In the procedure just described, r can be any irrational number > 1. Beatty trees and backtracking sequences for selected r are indicated here:

%C r Beatty tree for r backtracking sequence, (b(n))

%C (1+sqrt(5))/2 A258212 A258215

%C (3+sqrt(5))/2 A258235 A258236

%C sqrt(2) A258237 A258238

%C 2+sqrt(2) A258239 A258240

%C sqrt(3) A258241 A258242

%C e A258243 A258244

%C Pi A258245 A258246

%C sqrt(8) A258247 A258248

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

%e 0

%e 1

%e 3

%e 6 2

%e 11 4

%e 19 7 8

%e 32 12 14 5

%e 53 20 21 24 9

%e 87 33 35 13 40 15 16

%e Generations 0 to 10 of the tree are drawn by the Mathematica program. In T, the path from 0 to 35 is (0,1,3,6,11,7,12,21,35). The path obtained by backtracking (i.e., successive applications of the mapping b in Comments) is (35,21,12,7,11,6,3,1,0).

%t r = GoldenRatio; 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, 11}]

%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 -> 700]

%t Map[DeleteDuplicates, Transpose[paths]] (*The numbers in each level of the tree*)

%t (* _Peter J. C. Moses_, May 21 2015 *)

%Y Cf. A000201, A001950, A258212 (path-length from 0 to n).

%K nonn,tabf,easy

%O 1,3

%A _Clark Kimberling_, Jun 05 2015