

A166310


Wythoff Triangle, T.


2



1, 2, 3, 4, 6, 8, 5, 7, 9, 11, 10, 12, 14, 16, 21, 13, 15, 17, 19, 24, 29, 18, 20, 22, 27, 32, 37, 42, 23, 25, 30, 35, 40, 45, 50, 55, 26, 28, 33, 38, 43, 48, 53, 58, 63, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 47, 52, 57, 62, 67, 72
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OFFSET

1,2


COMMENTS

(1) Every positive integer occurs exactly once, so that
this is a permutation of the natural numbers.
(2) Obtained from the preliminary Wyhoff triangle
(A166309) by arranging each row in increasing order.
(3) The difference between consecutive row terms is a
(4) Is the difference between consecutive column terms a
Fibonacci number?


REFERENCES

C. Kimberling, "The Wythoff triangle and unique representations of positive integers," Proceedings of the Fourteenth International Conference on Fibonacci Numbers and Their Applications," Aportaciones Matematicas Invertigacion 20 (2011) 155169.


LINKS



FORMULA

For a=1,2,3,... and b=0,1,...,a1, let P(a,b) be the
number of the row of the Wythoff array (A035513) that
precurses to (a,b). Then for each a, arrange the numbers P
(a,b) in increasing order.


EXAMPLE

The first nine rows of T:
1
2....3
4....6...8
5....7...9..11
10..12..14..16..21
13..15..17..19..24..29
18..20..22..27..32..37..42
23..25..30..35..40..45..50..55
26..28..33..38..43..48..53..58..63
Row 5 of the preliminary Wythoff triangle is
16,21,10,12,14, so that row 5 of the Wythoff triangle is
10,12,14,16,21. These are the row numbers of the Wythoff
array W (A035513) which precurse to pairs (5,b) for
b=0,1,2,3,4, not respectively. Example of precursion: row
16 of W is 40,65,105,...; then 6540=25, 4025=15,
2515=10, 1510=5, 105=5, 55=0, 50=5, so that the
initial pair (5,0) is reached in seven precursive steps.


MATHEMATICA

f[n_]:=f[n]=Fibonacci[n]; w[n_, k_] := f[k + 1] Floor[n GoldenRatio] + (n  1) f[k]; a[n_, k_] := w[n, Module[{z = 0}, ((While[w[#1, z] <= w[#1, z + 1], z]; z  1) &)[n] + k]]; z = 100; t = Table[a[n, k], {n, 1, z}, {k, 1, 2}] (* nth pair: 1st 2 terms of row n of leftjustified Wythoff array, A165357 *)
u = Table[t[[n]][[1]], {n, 1, z}]
v = Table[Flatten[Position[u, n]], {n, 1, z/5}]
TableForm[Table[Flatten[Position[u, n]], {n, 1, z/5}]] (* A166310 triangle, Clark Kimberling, Aug 01 2013 *)


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AUTHOR



STATUS

approved



