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Array by antidiagonals: R(i,j)=number of the row of the Wythoff array which includes row(i+j)-row(i).
1

%I #13 Feb 12 2014 03:17:12

%S 1,1,1,1,1,1,2,1,3,1,1,3,1,1,1,3,1,2,1,1,1,2,4,1,3,2,1,1,1,2,3,1,3,1,

%T 1,1,4,1,8,3,1,2,1,3,1,3,6,1,2,4,1,3,1,1,1,2,3,4,1,2,3,1,2,1,1,1,5,2,

%U 11,4,1,8

%N Array by antidiagonals: R(i,j)=number of the row of the Wythoff array which includes row(i+j)-row(i).

%C The rows of the Wythoff array are essentially the positive Fibonacci sequences. If i>=1 and j>=1, then row(i+j)-row(i) is a positive Fibonacci sequence and therefore a tail of a row of the Wythoff array.

%H Clark Kimberling, <a href="/A186007/b186007.txt">Antidiagonals n=1..60, flattened</a>

%e Northwest corner:

%e 1....1....1....2....1....3....2....1....4....3

%e 1....1....1....3....1....4....2....1....6....3

%e 1....3....1....2....1....3....8....1....4....11

%e 1....1....1....3....1....3....2....1....4....3

%e 1....1....2....3....1....4....2....1....6....3

%e Let W be the Wythoff array (A035513).

%e row 8 of W: 19,31,50,81,...

%e row 2 of W: 4,7,11,18,...

%e (row 8)-(row 2): 15,24,39,63,... a tail of row 4,

%e so that R(2,6)=4.

%t w[{n_, k_}] := w[{n, k}] = Fibonacci[k + 1] Floor[n GoldenRatio] + (n - 1) Fibonacci[k];

%t f = Map[w[{Plus @@ #, {1, 2}}] - w[{#[[1]], {1, 2}}] &, Flatten[Table[{k, z - k + 1}, {z, 15}, {k, z}], 1]];

%t Module[{n, z}, Table[n = 1; While[(z = 1; While[First[f[[k]]] >= w[{n, z}], z++]); f[[k]] != {w[{n, z - 1}], w[{n, z}]}, n++]; n, {k, 1, Length[f]}]] (* _Peter J. C. Moses_, Apr 13 2013 *)

%Y Cf. A035513, A185735 (addition table for Wythoff rows).

%K nonn,tabl

%O 1,7

%A _Clark Kimberling_, Feb 10 2011

%E Corrections and additions by _Clark Kimberling_, Apr 13 2013