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A299762
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Irregular triangle T(n,k) read by rows in which row n lists the positive integers whose sum of divisors is n, or 0 if no such integer exists.
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7
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1, 0, 2, 3, 0, 5, 4, 7, 0, 0, 0, 6, 11, 9, 13, 8, 0, 0, 10, 17, 0, 19, 0, 0, 0, 14, 15, 23, 0, 0, 0, 12, 0, 29, 16, 25, 21, 31, 0, 0, 0, 22, 0, 37, 18, 27, 0, 20, 26, 41, 0, 43, 0, 0, 0, 33, 35, 47, 0, 0, 0, 0, 0, 34, 53, 0, 28, 39, 49, 0, 0, 24, 38, 59, 0, 61, 32, 0, 0, 0, 0, 67, 0, 0, 0, 30, 46, 51, 55, 71, 0, 73
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OFFSET
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1,3
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COMMENTS
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Essentially the same as the triangle described in the example section of A085790, but with 0's added in empty rows.
Are the records the same as A008578?
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LINKS
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FORMULA
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sigma(T(n,k)) = n, if T(n,k) >= 1.
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EXAMPLE
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First 24 rows of triangle T(n,k):
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. n / k: 1 2 3 ...
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| 1| 1;
| 2| 0;
| 3| 2;
| 4| 3;
| 5| 0;
| 6| 5;
| 7| 4;
| 8| 7;
| 9| 0;
|10| 0;
|11| 0;
|12| 6, 11;
|13| 9;
|14| 13;
|15| 8;
|16| 0;
|17| 0;
|18| 10, 17;
|19| 0;
|20| 19;
|21| 0;
|22| 0;
|23| 0;
|24| 14, 15, 23;
...
For n = 23 there are no positive integers whose sum of divisors is 23, so T(23, 1) = 0, which is the only element in the 23rd row of the triangle.
For n = 24 there are three positive integers whose sum of divisors is 24; they are 14, 15 and 23, since sigma(14) = 1 + 2 + 7 + 14 = 24, sigma(15) = 1 + 3 + 5 + 15 = 24 and sigma(23) = 1 + 23 = 24, so the 24th row of the triangle is [14, 15, 23].
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MATHEMATICA
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With[{nn = 74}, ReplacePart[ConstantArray[{0}, nn], PositionIndex@ Array[DivisorSigma[1, #] &, nn]]] // Flatten (* Michael De Vlieger, Mar 16 2018 *)
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CROSSREFS
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Row n has A054973(n) positive integers.
Positive terms in the first column give A002192.
Indices of the rows that contain a zero give A007369.
Indices of the rows that contain positive terms give A002191.
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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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