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

t[k] = Take[Select[Range[1, z1], b[#, # + k] == b[1, 1 + k] &], z2]

Cf. A004713, A246341.

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allocated Rectangular array: T(n,k) is the position in s = A004713 at which the block s(1)..s(n) occurs for Clark Kimberlingthe kth time.

1, 3, 1, 4, 4, 1, 6, 6, 4, 1, 8, 8, 6, 42, 1, 14, 14, 42, 66, 66, 1, 17, 20, 66, 72, 78, 136, 1, 18, 24, 72, 78, 89, 195, 195, 1, 19, 28, 78, 89, 136, 370, 437, 775, 1, 20, 37, 89, 96, 144, 437, 504, 782, 775, 1, 23, 42, 96, 113, 195, 504, 676, 1170, 782

1,2

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.

A004713 gives the positions of 1 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. The block consisting of the first 1 terms is 1, and it occurs at positions 1,3,4,6,8,14,..., so that this is row 1 of A246340. The block consisting of the first 5 terms is 1,0,1,1,0, and it begins at the positions shown in row 5.

The first 6 rows follow:

1 .. 3 ... 4 ... 6 ... 8 ... 14 ...

1 .. 4 ... 6 ... 8 ... 14 .. 20 ...

1 .. 4 ... 6 ... 42 .. 66 .. 72 ...

1 .. 42 .. 66 .. 72 .. 78 .. 89 ...

1 .. 66 . 78 .. 89 .. 136 . 144 ..

1 .. 136 . 195 . 370 . 437 . 504 ..

z = 150000; s = RealDigits[1/Sqrt[2], 2, z][[1]]; Flatten[Position[s, 1]];

b[m_, n_] := b[m, n] = Take[s, {m, n}]; z1 = 100000; 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}]](* A246340, array *)

w[n_, k_] := t[n][[k + 1]]; tt = Table[w[n - k, k], {n, 0, z2 - 1}, {k, n, 0, -1}] // Flatten (* A246340, sequence *)

Cf. A004713,A246341.

allocated

nonn,easy,tabl

Clark Kimberling, Aug 24 2014

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allocated for Clark Kimberling

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