A258212 -id:A258212 - cp's OEIS frontend

A258244 Number of steps from n to 0, where allowable steps are x -> [x/e] if x = is in A022843 (the Beatty sequence for e) and x -> [e*x] otherwise, where [ ] = floor.

0, 2, 1, 3, 5, 3, 5, 7, 2, 4, 4, 6, 8, 6, 8, 10, 4, 6, 8, 6, 8, 8, 10, 12, 3, 5, 7, 5, 7, 5, 7, 9, 7, 9, 11, 9, 11, 13, 7, 9, 9, 11, 13, 11, 13, 15, 5, 7, 7, 9, 11, 9, 11, 13, 7, 9, 11, 9, 11, 9, 11, 13, 11, 13, 15, 13, 15, 4, 6, 8, 6, 8, 10, 8, 10, 12, 6, 8

Offset: 0

Clark Kimberling, Jun 08 2015

a(n) = number of edges from 0 to n in the tree at A258243.

			20->54->19->6->16->5->1->2->0, so that a(20) = 8.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A022843, A258243, A258212.

r = E; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258236 Number of steps from n to 0, where allowable steps are x -> [x/r] if x is in upper Wythoff sequence (A001950) and x -> [r*x] otherwise, where [ ] = floor and r = (3+sqrt(5))/2.

Original entry on oeis.org

0, 2, 1, 3, 5, 3, 5, 2, 4, 6, 4, 6, 8, 6, 8, 4, 6, 8, 6, 8, 3, 5, 7, 5, 7, 9, 7, 9, 5, 7, 9, 7, 9, 11, 9, 11, 7, 9, 11, 9, 11, 5, 7, 9, 7, 9, 11, 9, 11, 7, 9, 11, 9, 11, 4, 6, 8, 6, 8, 10, 8, 10, 6, 8, 10, 8, 10, 12, 10, 12, 8, 10, 12, 10, 12, 6, 8, 10, 8

Offset: 0

Clark Kimberling, Jun 05 2015

a(n) = number of edges from 0 to n in the tree at A258235.

			29->75->28->10->3->7->2->0, so that a(29) = 7.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A258235, A258212, A001950.

r = GoldenRatio^2; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258238 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A001951 (the Beatty sequence for sqrt(2)) and x -> [r*x] otherwise, where [ ] = floor and r = sqrt(2).

Original entry on oeis.org

0, 1, 2, 4, 3, 5, 7, 4, 6, 8, 10, 5, 7, 9, 9, 11, 6, 8, 8, 10, 12, 10, 12, 14, 7, 9, 9, 11, 11, 13, 15, 11, 13, 15, 17, 8, 10, 12, 10, 12, 14, 12, 14, 16, 18, 12, 14, 16, 16, 18, 9, 11, 11, 13, 15, 11, 13, 15, 17, 13, 15, 17, 17, 19, 21, 13, 15, 17, 19, 17

Offset: 0

Clark Kimberling, Jun 05 2015

a(n) = number of edges from 0 to n in the tree at A258237.

			36->25->17->24->16->11->7->4->2->1->0, so that a(36) = 10.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A258237, A258212, A001951.

r = Sqrt[2]; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258240 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A001952 (the Beatty sequence for 2 + sqrt(2)) and x -> [r*x] otherwise, where [ ] = floor and r = 2 + sqrt(2).

Original entry on oeis.org

0, 2, 4, 1, 3, 5, 3, 5, 7, 9, 5, 7, 9, 2, 4, 6, 8, 4, 6, 8, 6, 8, 10, 4, 6, 8, 10, 6, 8, 10, 8, 10, 12, 14, 10, 12, 14, 6, 8, 10, 8, 10, 12, 14, 10, 12, 14, 3, 5, 7, 9, 5, 7, 9, 7, 9, 11, 13, 9, 11, 13, 5, 7, 9, 7, 9, 11, 13, 9, 11, 13, 7, 9, 11, 13, 9, 11

Offset: 0

Clark Kimberling, Jun 05 2015

a(n) = number of edges from 0 to n in the tree at A258239.

			8->27->7->23->6->1->3->0, so that a(8) = 7.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A258239, A258212, A001952.

r = 2+Sqrt[2]; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258242 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A022838 (the Beatty sequence for sqrt(3)) and x -> [r*x] otherwise, where [ ] = floor and r = sqrt(3).

Original entry on oeis.org

0, 1, 3, 2, 4, 4, 3, 5, 5, 7, 5, 7, 4, 6, 8, 6, 8, 8, 10, 6, 8, 10, 5, 7, 7, 9, 11, 7, 9, 9, 11, 9, 11, 13, 7, 9, 9, 11, 11, 6, 8, 8, 10, 8, 10, 10, 12, 14, 8, 10, 10, 10, 12, 12, 14, 10, 12, 12, 14, 16, 8, 10, 10, 12, 10, 12, 14, 12, 14, 7, 9, 9, 9, 11, 11

Offset: 0

Clark Kimberling, Jun 07 2015

a(n) = number of edges from 0 to n in the tree at A258241.

			16->27->15->8->4->6->3->1->0, so that a(16) = 8.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A022838, A258241, A258212.

r = Sqrt[3]; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258248 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A022842 (the Beatty sequence for sqrt(8)) and x -> [r*x] otherwise, where r = sqrt(8) and [ ] = floor.

Original entry on oeis.org

0, 2, 1, 3, 5, 3, 5, 7, 2, 4, 6, 4, 6, 8, 6, 8, 4, 6, 8, 6, 8, 10, 8, 10, 12, 3, 5, 7, 5, 7, 9, 7, 9, 5, 7, 9, 7, 9, 11, 9, 11, 13, 7, 9, 11, 9, 11, 13, 5, 7, 7, 9, 11, 9, 11, 13, 7, 9, 11, 9, 11, 13, 11, 13, 15, 9, 11, 11, 13, 15, 13, 15, 17, 4, 6, 8, 6, 8

Offset: 0

Clark Kimberling, Jun 08 2015

a(n) = number of edges from 0 to n in the tree at A258247.

			13->36->12->33->11->3->8->2->0, so that a(12) = 8.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A022842, A258247, A258212.

r = Sqrt[8]; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258215 Number of steps from n to 0, where allowable steps are x -> [x/r] if x is in lower Wythoff sequence (A000201) and x -> [r*x] otherwise, where [ ] = floor and r = (1+sqrt(5))/2 = golden ratio.

Original entry on oeis.org

0, 1, 3, 2, 4, 6, 3, 5, 5, 7, 9, 4, 6, 8, 6, 8, 8, 10, 12, 5, 7, 7, 9, 11, 7, 9, 11, 9, 11, 11, 13, 15, 6, 8, 10, 8, 10, 10, 12, 14, 8, 10, 10, 12, 14, 10, 12, 14, 12, 14, 14, 16, 18, 7, 9, 9, 11, 13, 9, 11, 13, 11, 13, 13, 15, 17, 9, 11, 13, 11, 13, 13, 15

Offset: 0

Clark Kimberling, Jun 05 2015

a(n) = number of edges from 0 to n in the tree at A258212.

			35->21->12->7->11->6->3->1->0, so that a(35) = 8.

Clark Kimberling, Table of n, a(n) for n = 0..10000

Cf. A258212, A000201.

r = GoldenRatio; w = Table[Floor[r*n], {n, 1, 1000}];
f[x_] := If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
g[x_] := Drop[FixedPointList[f, x], -1];
Table[-1+ Length[g[n]], {n, 0, 100}]

A258235 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.

Original entry on oeis.org

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, 109, 55, 376, 31, 36, 44, 49, 22, 24, 26, 29, 42, 146, 151, 164, 198, 287, 144, 986, 12, 14, 17, 19, 65, 70, 78, 83, 96, 112, 117, 130, 56, 58, 60, 63, 76, 110

Offset: 1

Clark Kimberling, Jun 05 2015

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

			Rows (or generations, or levels) of T:
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  109  55  376
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).

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

r = (3 + Sqrt[5])/2; k = 1000; w = Map[Floor[r #] &, Range[k]];
f[x_] := f[x] = If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
b := NestWhileList[f, #, ! # == 0 &] &;
bs = Map[Reverse, Table[b[n], {n, 0, k}]];
generations = Table[DeleteDuplicates[Map[#[[n]] &, Select[bs, Length[#] > n - 1 &]]], {n, 10}]
paths = Sort[Map[Reverse[b[#]] &, Last[generations]]]
graph = DeleteDuplicates[Flatten[Map[Thread[Most[#] -> Rest[#]] &, paths]]]
TreePlot[graph, Top, 0, VertexLabeling -> True, ImageSize -> 1400]
Map[DeleteDuplicates, Transpose[paths]] (* Peter J. C. Moses, May 21 2015 *)

A258237 Irregular triangle (Beatty tree for r = sqrt(2)), T, of all nonnegative integers, each exactly once, as described in Comments.

Original entry on oeis.org

0, 1, 2, 4, 3, 7, 5, 11, 8, 16, 6, 12, 24, 9, 18, 17, 35, 14, 13, 25, 26, 50, 10, 19, 21, 38, 36, 72, 15, 28, 27, 31, 52, 51, 55, 103, 20, 22, 37, 39, 41, 45, 73, 74, 79, 147, 32, 29, 56, 53, 59, 65, 106, 104, 113, 209, 23, 42, 40, 46, 76, 75, 80, 84, 93

Offset: 1

Clark Kimberling, Jun 05 2015

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.
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).)
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 = 17).
See A258212 for a guide to Beatty trees for various choices of r.

			Rows (or generations, or levels) of T:
0
1
2
4
3   7
5   1
8   16
6   12  24
9   18  17  35
14  13  25  26  50
10  19  21  38  36  72
15  28  27  31  52  51  55  103
Generations 0 to 10 of the tree are drawn by the Mathematica program.  In T, the path from 0 to 36 is (0,1,2,4,7,11,16,24,17,25,36).  The path obtained by backtracking (i.e., successive applications of the mapping b in Comments) is (36,25,17,24,16,11,7,4,2,1,0).

Cf. A001951, A001952, A258238 (path-length from 0 to n), A258212

r = Sqrt[2]; k = 1000; w = Map[Floor[r #] &, Range[k]];
f[x_] := f[x] = If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
b := NestWhileList[f, #, ! # == 0 &] &;
bs = Map[Reverse, Table[b[n], {n, 0, k}]];
generations = Table[DeleteDuplicates[Map[#[[n]] &, Select[bs, Length[#] > n - 1 &]]], {n, 11}]
paths = Sort[Map[Reverse[b[#]] &, Last[generations]]]
graph = DeleteDuplicates[Flatten[Map[Thread[Most[#] -> Rest[#]] &, paths]]]
TreePlot[graph, Top, 0, VertexLabeling -> True, ImageSize -> 700]
Map[DeleteDuplicates, Transpose[paths]] (* Peter J. C. Moses,May 21 2015 *)

A258239 Irregular triangle (Beatty tree for r = 2 + sqrt(2)), T, of all nonnegative integers, each exactly once, as determined in Comments.

Original entry on oeis.org

0, 3, 1, 13, 6, 4, 47, 2, 17, 23, 14, 163, 5, 7, 10, 51, 61, 81, 48, 559, 27, 37, 15, 18, 20, 24, 167, 177, 211, 279, 164, 1911, 8, 11, 54, 64, 71, 85, 95, 129, 49, 52, 62, 82, 563, 573, 607, 723, 955, 560, 6527, 30, 40, 16, 19, 21, 25, 28, 38, 170, 180, 187

Offset: 1

Clark Kimberling, Jun 05 2015

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.
The tree T has root 0 with an edge to 3, 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).)
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 = 8).
See A258212 for a guide to Beatty trees for various choices of r.

			Rows (or generations, or levels) of T:
0
3
1   13
6   4   47
2   23  17  14   163
10  7   81  5    61   51   48   559
37  27  24  279  20   18   211  15   177   167  164   1911
Generations 0 to 7 of the tree are drawn by the Mathematica program.  In T, the path from 0 to 8 is (0,3,1,6,23,7,27,8).  The path obtained by backtracking (i.e., successive applications of the mapping b in Comments) is (8,27,7,23,6,1,3,0).

Cf. A001951, A001952, A258240 (path-length, 0 to n), A258212

r = 2+Sqrt[2]; k = 1000; w = Map[Floor[r #] &, Range[k]];
f[x_] := f[x] = If[MemberQ[w, x], Floor[x/r], Floor[r*x]];
b := NestWhileList[f, #, ! # == 0 &] &;
bs = Map[Reverse, Table[b[n], {n, 0, k}]];
generations = Table[DeleteDuplicates[Map[#[[n]] &, Select[bs, Length[#] > n - 1 &]]], {n, 8}]
paths = Sort[Map[Reverse[b[#]] &, Last[generations]]]
graph = DeleteDuplicates[Flatten[Map[Thread[Most[#] -> Rest[#]] &, paths]]]
TreePlot[graph, Top, 0, VertexLabeling -> True, ImageSize -> 700]
Map[DeleteDuplicates, Transpose[paths]] (* Peter J. C. Moses,May 21 2015 *)

cp's OEIS Frontend

A258244 Number of steps from n to 0, where allowable steps are x -> [x/e] if x = is in A022843 (the Beatty sequence for e) and x -> [e*x] otherwise, where [ ] = floor.

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A258236 Number of steps from n to 0, where allowable steps are x -> [x/r] if x is in upper Wythoff sequence (A001950) and x -> [r*x] otherwise, where [ ] = floor and r = (3+sqrt(5))/2.

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A258238 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A001951 (the Beatty sequence for sqrt(2)) and x -> [r*x] otherwise, where [ ] = floor and r = sqrt(2).

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Mathematica

A258240 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A001952 (the Beatty sequence for 2 + sqrt(2)) and x -> [r*x] otherwise, where [ ] = floor and r = 2 + sqrt(2).

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Mathematica

A258242 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A022838 (the Beatty sequence for sqrt(3)) and x -> [r*x] otherwise, where [ ] = floor and r = sqrt(3).

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Mathematica

A258248 Number of steps from n to 0, where allowable steps are x -> [x/r] if x = is in A022842 (the Beatty sequence for sqrt(8)) and x -> [r*x] otherwise, where r = sqrt(8) and [ ] = floor.

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Mathematica

A258215 Number of steps from n to 0, where allowable steps are x -> [x/r] if x is in lower Wythoff sequence (A000201) and x -> [r*x] otherwise, where [ ] = floor and r = (1+sqrt(5))/2 = golden ratio.

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Mathematica

A258235 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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Mathematica