cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A074049 Tree generated by the Wythoff sequences: a permutation of the positive integers.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 8, 13, 6, 10, 11, 18, 12, 20, 21, 34, 9, 15, 16, 26, 17, 28, 29, 47, 19, 31, 32, 52, 33, 54, 55, 89, 14, 23, 24, 39, 25, 41, 42, 68, 27, 44, 45, 73, 46, 75, 76, 123, 30, 49, 50, 81, 51, 83, 84, 136, 53, 86, 87, 141, 88, 143, 144, 233, 22, 36, 37
Offset: 1

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Author

Clark Kimberling, Aug 14 2002

Keywords

Comments

Write t=tau=(1+sqrt(5))/2 and let S be generated by these rules: 1 is in S and if x is in S, then f(x) := [t*x] and g(x) := [(t+1)*x] are in S. Then S is the set of positive integers and the present permutation of S is obtained by arranging S in rows according to the order in which they are generated by f and g, starting with x=1.
The formula indicates the manner in which these numbers arise as a tree: 1 stems to 2, which branches to (3,5), and thereafter, each number branches to a pair:
3->(4,7) and 5->(8,13), etc.
The numbers >1 in the lower Wythoff sequence A000201 occupy the first place in each pair, and the numbers >2 in the upper Wythoff sequence A001950 occupy the second place. The pairs, together with (1,2) are the Wythoff pairs, much studied as the solutions of the Wythoff game. The Wythoff pairs also occur, juxtaposed, in the Wythoff array, A035513.

Examples

			First levels of the tree:
...................1
...................2
...........3.................5
.......4.......7........8........13
.....6..10...11..18....12..20...21..34
		

Crossrefs

Equals A048680(n-1) + 1.

Programs

  • Mathematica
    a = {1, 2}; row = {a[[-1]]}; r = GoldenRatio; s = r/(r - 1); Do[a = Join[a, row = Flatten[{Floor[#*{r, s}]} & /@ row]], {n, 5}]; a (* Ivan Neretin, Nov 09 2015 *)

Formula

Array T(n, k) by rows: T(0, 0)=1; T(1, 0)=2;
T(n, 2j) = floor(tau*T(n-1, j));
T(n, 2j+1) = floor((tau+1)*T(n-1, j))
for j=0,1,...,2^(n-1)-1, n>=2.

Extensions

Extended by Clark Kimberling, Dec 23 2010