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A191542
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Dispersion of (2*floor(n*sqrt(3))), by antidiagonals.
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1
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1, 2, 3, 6, 10, 4, 20, 34, 12, 5, 68, 116, 40, 16, 7, 234, 400, 138, 54, 24, 8, 810, 1384, 478, 186, 82, 26, 9, 2804, 4794, 1654, 644, 284, 90, 30, 11, 9712, 16606, 5728, 2230, 982, 310, 102, 38, 13, 33642, 57524, 19842, 7724, 3400, 1072, 352, 130, 44, 14
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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....6....20...68
3...10...34...116..400
4...12...40...138..478
5...16...54...186..644
7...24...82...284..982
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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_] :=2*Floor[n*Sqrt[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}]] (* A191542 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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