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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
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 Wythoff triangle (A166309) by arranging each row in increasing order.
(3) The difference between consecutive row terms is a Fibonacci number (A000045).
(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) 155-169.
FORMULA
For a=1,2,3,... and b=0,1,...,a-1, 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 65-40=25, 40-25=15, 25-15=10, 15-10=5, 10-5=5, 5-5=0, 5-0=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}] (* n-th pair: 1st 2 terms of row n of left-justified Wythoff array, A165357 *)
u = Table[t[[n]][[1]], {n, 1, z}]
v = Table[Flatten[Position[u, n]], {n, 1, z/5}]
Flatten[v] (* A166310 sequence *)
TableForm[Table[Flatten[Position[u, n]], {n, 1, z/5}]] (* A166310 triangle, Clark Kimberling, Aug 01 2013 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Oct 11 2009
STATUS
approved