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A051127
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Table T(n,k) = k mod n read by antidiagonals (n >= 1, k >= 1).
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11
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0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 0, 0, 0, 2, 1, 0, 1, 1, 3, 2, 1, 0, 0, 2, 0, 3, 2, 1, 0, 1, 0, 1, 4, 3, 2, 1, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 1, 2, 3, 1, 5, 4, 3, 2, 1, 0, 0, 0, 0, 2, 0, 5, 4, 3, 2, 1, 0, 1, 1, 1, 3, 1, 6, 5, 4, 3, 2, 1, 0, 0, 2, 2, 4, 2, 0, 6, 5, 4, 3, 2, 1, 0, 1, 0, 3, 0, 3, 1, 7, 6, 5, 4, 3, 2, 1
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OFFSET
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1,9
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COMMENTS
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Note that the upper right half of this sequence when formatted as a square array is essentially the same as this whole sequence when formatted as an upper right triangle. Sums of antidiagonals are A004125. - Henry Bottomley, Jun 22 2001
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LINKS
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FORMULA
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As a linear array, the sequence is a(n) = A004736(n) mod A002260(n) or a(n) = ((t*t+3*t+4)/2-n) mod (n-(t*(t+1)/2)), where t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 17 2012
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EXAMPLE
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0 0 0 0 0 0 0 0 0 0 ...
1 0 1 0 1 0 1 0 1 0 ...
1 2 0 1 2 0 1 2 0 1 ...
1 2 3 0 1 2 3 0 1 2 ...
1 2 3 4 0 1 2 3 4 0 ...
1 2 3 4 5 0 1 2 3 4 ...
1 2 3 4 5 6 0 1 2 3 ...
1 2 3 4 5 6 7 0 1 2 ...
1 2 3 4 5 6 7 8 0 1 ...
1 2 3 4 5 6 7 8 9 0 ...
1 2 3 4 5 6 7 8 9 10 ...
1 2 3 4 5 6 7 8 9 10 ...
1 2 3 4 5 6 7 8 9 10 ...
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MATHEMATICA
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T[n_, m_] = Mod[n - m + 1, m + 1]; Table[Table[T[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%] (* Roger L. Bagula, Sep 04 2008 *)
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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