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A246341

Rectangular array: T(n,k) is the position in s = A004713 at which the block s(2)..s(n+1) occurs for the k-th time.

2, 5, 2, 7, 5, 2, 9, 7, 16, 2, 10, 13, 22, 22, 2, 11, 16, 26, 26, 87, 2, 12, 22, 30, 49, 94, 196, 2, 13, 26, 39, 67, 137, 438, 776, 2, 15, 30, 43, 79, 196, 505, 783, 776, 2, 16, 39, 49, 87, 345, 512, 1171, 783, 783, 2, 21, 43, 67, 90, 371, 677, 1184, 1171

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OFFSET

1,1

COMMENTS

Assuming that every row of T is infinite, each row contains the next row as a proper subsequence. Row 1 of A246340 and row 1 of A246341 partition the positive integers.

LINKS

Table of n, a(n) for n=1..63.

EXAMPLE

A246339 gives the positions of 0 in the base 2 expansion of 1/sqrt(2), which begins thus: 1,0,1,1,0,1,0,1,0,0,0,0,0,1,0,0,1,1,1,1. For n = 1, the block s(2)..s(2) is simply 0, which occurs at positions 1,3,4,6,8,... as in row 1 of T. For n = 5, the block s(2)..s(6) is 0,1,1,0,1, which occurs at positions 2,87,94,137, ...

The first 6 rows:

2 .. 5 ... 7 ... 9 ... 10 .. 11 .. 12 .. 13 ...

2 .. 5 ... 7 ... 13 .. 16 .. 22 .. 26 .. 30 ...

2 .. 16 .. 22 .. 26 .. 30 .. 39 .. 43 .. 49 ...

2 .. 22 .. 26 .. 49 .. 67 .. 79 .. 87 .. 90 ...

2 .. 87 .. 94 .. 137 . 196 . 345 . 371 . 438 ...

2 .. 196 .438 . 505 . 512 . 677 . 776 . 783 ...

MATHEMATICA

z = 200000; s = RealDigits[1/Sqrt[2], 2, z][[1]]; Flatten[Position[s, 0]];

b[m_, n_] := b[m, n] = Take[s, {m, n}]; z1 = 150000; 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}]] (* A246341, array *)