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A191445
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Dispersion of ([(n+1)*sqrt(3)]), where [ ]=floor, by antidiagonals.
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1
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1, 3, 2, 6, 5, 4, 12, 10, 8, 7, 22, 19, 15, 13, 9, 39, 34, 27, 24, 17, 11, 69, 60, 48, 43, 31, 20, 14, 121, 105, 84, 76, 55, 36, 25, 16, 211, 183, 147, 133, 96, 64, 45, 29, 18, 367, 318, 256, 232, 168, 112, 79, 51, 32, 21, 637, 552, 445, 403, 292, 195, 138
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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...3...6...12..22
2...5...10..19..34
4...8...15..27..48
7...13..24..43..76
9...17..31..55..96
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MATHEMATICA
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(* Program generates the dispersion array T of increasing sequence f[n] *)
r=40; r1=12; c=40; c1=12; x = Sqr[3];
f[n_] := Floor[n*x+x] (* 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}]] (* A191445 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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