A246354 Rectangular array: T(n,k) is the position in the infinite Fibonacci word s = A003849 at which the block s(1)..s(n) occurs for the k-th time.
1, 3, 1, 4, 4, 1, 6, 6, 4, 1, 8, 9, 6, 6, 1, 9, 12, 9, 9, 6, 1, 11, 14, 12, 14, 9, 6, 1, 12, 17, 14, 19, 14, 9, 9, 1, 14, 19, 17, 22, 19, 14, 14, 9, 1, 16, 22, 19, 27, 22, 19, 22, 14, 9, 1, 17, 25, 22, 30, 27, 22, 30, 22, 14, 9, 1, 19, 27, 25, 35, 30, 27, 35
Offset: 1
Examples
The lower Wythoff sequence, A000201 gives the positions of 0 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(1)..s(1) is simply 0, which occurs at positions 1,3,4,6,8,... as in row 1 of T. For n = 5, the block s(1)..s(5) is 0,1,0,0,1, which occurs at positions 1,6,9,14,19, ... The first 7 rows follow: 1 .. 3 .. 4 ... 6 ... 8 ... 9 ... 11 .. 12 ... 1 .. 4 .. 6 ... 9 ... 12 .. 14 .. 17 .. 19 ... 1 .. 4 .. 6 ... 9 ... 12 .. 14 .. 17 .. 19 ... 1 .. 6 .. 9 ... 14 .. 19 .. 22 .. 27 .. 30 ... 1 .. 6 .. 9 ... 14 .. 19 .. 22 .. 27 .. 30 ... 1 .. 6 .. 9 ... 14 .. 19 .. 22 .. 27 .. 30 ... 1 .. 9 .. 14 .. 22 .. 30 .. 35 .. 43 .. 48 ...
Programs
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Mathematica
z = 1000; s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, 12]]; Flatten[Position[s, 0]]; 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[1, 1 + k] &], z2]; Column[Table[t[k], {k, 0, z2}]] (* A246354, array *) w[n_, k_] := t[n][[k + 1]]; Table[w[n - k, k], {n, 0, z2 - 1}, {k, n, 0, -1}] // Flatten (* A246354, sequence *)
Formula
First row: A000201 (lower Wythoff numbers);
next 2 rows: A003622 (Wythoff AA numbers);
next 3 rows: A134859 (Wythoff AAA numbers);
next 5 rows: A151915 (Wythoff AAAA numbers).
(The patterns continue; in particular the number of identical consecutive rows is always a Fibonacci number, as in A000045.)
Comments