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A246340
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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 k-th time.
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2
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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 ..
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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