A246355 Rectangular array: T(n,k) is the position in the infinite Fibonacci word s = A003849 at which the block s(2)..s(n+1) occurs for the k-th time.

2, 5, 2, 7, 5, 2, 10, 7, 7, 2, 13, 10, 10, 7, 2, 15, 13, 15, 10, 7, 2, 18, 15, 20, 15, 10, 10, 2, 20, 18, 23, 20, 15, 15, 10, 2, 23, 20, 28, 23, 20, 23, 15, 10, 2, 26, 23, 31, 28, 23, 31, 23, 15, 10, 2, 28, 26, 36, 31, 28, 36, 31, 23, 15, 10, 2, 31, 28, 41

Offset: 1

Author

Clark Kimberling, Aug 24 2014

Keywords

• nonn
• easy
• tabl

Comments

Assuming that every row of T is infinite, each row contains the next row as a proper subsequence. Row 1 of A246354 and row 1 of A246355 partition the positive integers.

Examples

			The upper Wythoff sequence, A001950 gives the positions of 1 in A003849, which begins thus:  0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1.  For n = 1, the block s(2)..s(2) is simply 1, which occurs at positions 2,5,7,10,13,... as in row 1 of T.  For n = 5, the block s(2)..s(6) is 1,0,0,1,0 which occurs at positions 2,7,10,15,20,23, ...
The first 6 rows follow:
2 .. 5 .. 7 ... 10 .. 13 .. 15 .. 18 ...
2 .. 5 .. 7 ... 10 .. 13 .. 15 .. 18 ...
2 .. 7 .. 10 .. 15 .. 20 .. 23 .. 28 ...
2 .. 7 .. 10 .. 15 .. 20 .. 23 .. 28 ...
2 .. 7 .. 10 .. 15 .. 20 .. 23 .. 28 ...
2 .. 10 . 15 .. 23 .. 31 .. 36 .. 44 ...
		

Crossrefs

Cf. A003849, A246354.

Programs

Mathematica

z = 1000; s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, 12]]; Flatten[Position[s, 1]]; b[m_, n_] := b[m, n] = Take[s, {m, n}]; z1 = 500; z2 = 12; t[k_] := t[k] = Take[Select[Range[1, z1], b[#, # + k] == b[2, 2 + k] &], z2]; Column[Table[t[k], {k, 0, z2}]](* A246355, array *)
w[n_, k_] := t[n][[k + 1]]; Table[w[n - k, k], {n, 0, z2 - 1}, {k, n, 0, -1}] // Flatten (* A246355, sequence *)

Formula

First 2 rows: A001950 (upper Wythoff numbers);

next 3 rows: A035336 (Wythoff BA numbers);

next 5 rows: A134861 (Wythoff BAA numbers);

next 8 rows: (Wythoff BAAA numbers).

(The patterns continue; in particular the number of identical consecutive rows is always a Fibonacci number, as in A000045.)