A299762 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.

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

Offset: 1

Author

Omar E. Pol, Mar 12 2018

Keywords

• nonn
• tabf

Comments

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?

Examples

			First 24 rows of triangle T(n,k):
-----------------------
. n / k:  1   2   3 ...
-----------------------
| 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].
		

Crossrefs

Row sums give A258913.

Column 1 gives A051444.

Right border gives A057637.

Positive terms give A085790.

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.

Cf. A000203, A007368, A007609, A008578.

Programs

Mathematica

With[{nn = 74}, ReplacePart[ConstantArray[{0}, nn], PositionIndex@ Array[DivisorSigma[1, #] &, nn]]] // Flatten (* Michael De Vlieger, Mar 16 2018 *)

Formula

sigma(T(n,k)) = n, if T(n,k) >= 1.