A274532 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.

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

Offset: 1

Author

Omar E. Pol, Jun 27 2016

Keywords

• nonn
• tabf

Comments

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).

Examples

			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.
		

Crossrefs

Row lengths give A000005. Column 1 is A000012. Row sums give A187215.

Cf. A000225, A187203, A272121, A273132, A273135, A273262, A274531.

Programs

Mathematica

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 *)