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A274532
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Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th antidiagonal of the absolute difference table of the divisors of n.
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3
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1, 1, 3, 1, 5, 1, 3, 7, 1, 9, 1, 3, 4, 13, 1, 13, 1, 3, 7, 15, 1, 5, 19, 1, 3, 10, 17, 1, 21, 1, 3, 4, 5, 11, 28, 1, 25, 1, 3, 16, 25, 1, 5, 7, 41, 1, 3, 7, 15, 31, 1, 33, 1, 3, 4, 13, 14, 47, 1, 37, 1, 3, 7, 7, 25, 39, 1, 5, 13, 53, 1, 3, 28, 41, 1, 45, 1, 3, 4, 5, 11, 12, 22, 61, 1, 9, 61, 1, 3, 34, 49, 1, 5, 19, 65
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OFFSET
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1,3
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COMMENTS
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If n is prime then row n contains only two terms: 1 and 2*n-1.
Row 2^k gives the first k+1 positive terms of A000225, k >= 0.
Note that this sequence is not the absolute values of A273262.
First differs from A273262 at a(41).
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LINKS
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EXAMPLE
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Triangle begins:
1;
1, 3;
1, 5;
1, 3, 7;
1, 9;
1, 3, 4, 13;
1, 13;
1, 3, 7, 15;
1, 5, 19;
1, 3, 10, 17;
1, 21;
1, 3, 4, 5, 11, 28;
1, 25;
1, 3, 16, 25;
1, 5, 7, 41;
1, 3, 7, 15, 31;
1, 33;
1, 3, 4, 13, 14, 47;
1, 37;
1, 3, 7, 7, 25, 39;
1, 5, 13, 53;
1, 3, 28, 41;
1, 45;
1, 3, 4, 5, 11, 12, 22, 61;
1, 9, 61;
1, 3, 34, 49;
1, 5, 19, 65;
...
For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the absolute difference triangle of the divisors is
1, 2, 3, 6, 9, 18;
1, 1, 3, 3, 9;
0, 2, 0, 6;
2, 2, 6;
0, 4;
4;
The antidiagonal sums give [1, 3, 4, 13, 14, 47] which is also the 18th row of the irregular triangle.
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MATHEMATICA
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Table[Map[Total, Table[#[[m - k + 1, k]], {m, Length@ #}, {k, m}], {1}] &@ NestWhileList[Abs@ Differences@ # &, Divisors@ n, Length@ # > 1 &], {n, 27}] // Flatten (* Michael De Vlieger, Jun 27 2016 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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