A166310 - OEIS

%I #9 Nov 02 2014 12:18:35

%S 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,

%T 37,42,23,25,30,35,40,45,50,55,26,28,33,38,43,48,53,58,63,31,36,41,46,

%U 51,56,61,66,71,76,34,39,44,49,54,59,64,69,74,79,84,47,52,57,62,67,72

%N Wythoff Triangle, T.

%C (1) Every positive integer occurs exactly once, so that

%C     this is a permutation of the natural numbers.

%C (2) Obtained from the preliminary Wyhoff triangle

%C     (A166309) by arranging each row in increasing order.

%C (3) The difference between consecutive row terms is a

%C     Fibonacci number (A000045).

%C (4) Is the difference between consecutive column terms a

%C     Fibonacci number?

%D 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.

%F For a=1,2,3,... and b=0,1,...,a-1, let P(a,b) be the

%F number of the row of the Wythoff array (A035513) that

%F precurses to (a,b). Then for each a, arrange the numbers P

%F (a,b) in increasing order.

%e The first nine rows of T:

%e 1

%e 2....3

%e 4....6...8

%e 5....7...9..11

%e 10..12..14..16..21

%e 13..15..17..19..24..29

%e 18..20..22..27..32..37..42

%e 23..25..30..35..40..45..50..55

%e 26..28..33..38..43..48..53..58..63

%e Row 5 of the preliminary Wythoff triangle is

%e 16,21,10,12,14, so that row 5 of the Wythoff triangle is

%e 10,12,14,16,21. These are the row numbers of the Wythoff

%e array W (A035513) which precurse to pairs (5,b) for

%e b=0,1,2,3,4, not respectively. Example of precursion: row

%e 16 of W is 40,65,105,...; then 65-40=25, 40-25=15,

%e 25-15=10, 15-10=5, 10-5=5, 5-5=0, 5-0=5, so that the

%e initial pair (5,0) is reached in seven precursive steps.

%t 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 *)

%t u = Table[t[[n]][[1]], {n, 1, z}]

%t v = Table[Flatten[Position[u, n]], {n, 1, z/5}]

%t Flatten[v]  (* A166310 sequence *)

%t TableForm[Table[Flatten[Position[u, n]], {n, 1, z/5}]]  (* A166310 triangle, Clark Kimberling, Aug 01 2013 *)

%Y Cf. A035513, A165357, A166309.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Oct 11 2009