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A191543
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Dispersion of (floor(8n/3)), by antidiagonals.
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1
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1, 2, 3, 5, 8, 4, 13, 21, 10, 6, 34, 56, 26, 16, 7, 90, 149, 69, 42, 18, 9, 240, 397, 184, 112, 48, 24, 11, 640, 1058, 490, 298, 128, 64, 29, 12, 1706, 2821, 1306, 794, 341, 170, 77, 32, 14, 4549, 7522, 3482, 2117, 909, 453, 205, 85, 37, 15, 12130, 20058
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OFFSET
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1,2
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COMMENTS
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Background discussion: Suppose that s is an increasing sequence of positive integers, that the complement t of s is infinite, and that t(1)=1. The dispersion of s is the array D whose n-th row is (t(n), s(t(n)), s(s(t(n)), s(s(s(t(n)))), ...). Every positive integer occurs exactly once in D, so that, as a sequence, D is a permutation of the positive integers. The sequence u given by u(n)=(number of the row of D that contains n) is a fractal sequence. Examples:
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LINKS
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EXAMPLE
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Northwest corner:
1...2....5....13...34
3...8....21...56...149
4...10...26...69...184
6...16...42...112...298
7...18...48...128..341
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MATHEMATICA
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(* Program generates the dispersion array T of the complement of increasing sequence f[n] *)
r=40; r1=12; c=40; c1=12; f[n_] :=Floor[8n/3]] (* complement of column 1 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]]
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191543 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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