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A191439
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Dispersion of ([n*sqrt(2)+n+1/2]), where [ ]=floor, by antidiagonals.
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2
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1, 2, 3, 5, 7, 4, 12, 17, 10, 6, 29, 41, 24, 14, 8, 70, 99, 58, 34, 19, 9, 169, 239, 140, 82, 46, 22, 11, 408, 577, 338, 198, 111, 53, 27, 13, 985, 1393, 816, 478, 268, 128, 65, 31, 15, 2378, 3363, 1970, 1154, 647, 309, 157, 75, 36, 16, 5741, 8119, 4756
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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....12...29
3....7....17...41...99
4....10...24...58...140
6....14...34...82...198
8....19...46...111..268
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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[2];
f[n_] := Floor[n*x+n+1/2] (* 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}]] (* A191439 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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