A273262 Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th antidiagonal of the 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, 21, 1, 5, 7, 41, 1, 3, 7, 15, 31, 1, 33, 1, 3, 4, 13, 6, 59, 1, 37, 1, 3, 7, 3, 31, 21, 1, 5, 13, 53, 1, 3, 28, 29, 1, 45, 1, 3, 4, 5, 11, 4, 36, 39, 1, 9, 61, 1, 3, 34, 33, 1, 5, 19, 65

Offset: 1

Author

Omar E. Pol, May 20 2016

Keywords

• sign
• 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 contains negative terms.

First differs from A274532 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, 21;
1, 5, 7, 41;
1, 3, 7, 15, 31;
1, 33;
1, 3, 4, 13, 6, 59;
1, 37;
1, 3, 7, 3, 31, 21;
1, 5, 13, 53;
1, 3, 28, 29;
1, 45;
1, 3, 4, 5, 11, 4, 36, 39;
1, 9, 61;
1, 3, 34, 33;
1, 5, 19, 65;
...
For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the difference triangle of the divisors is
1, 2, 3, 6, 9, 18;
1, 1, 3, 3, 9;
0, 2, 0, 6;
2, -2, 6;
-4, 8;
12;
The antidiagonal sums give [1, 3, 4, 13, 6, 59] which is also the 18th row of the irregular triangle.
		

Crossrefs

Row lengths give A000005. Column 1 is A000012. Right border gives A161700. Row sums give A273103.

Cf. A000225, A161700, A187202, A272210, A273102, A273135, A273261, A273263, A274532.

Programs

Mathematica

Table[Map[Total, Table[#[[m - k + 1, k]], {m, Length@ #}, {k, m}], {1}] &@ NestWhileList[Differences, Divisors@ n, Length@ # > 1 &], {n, 27}] (* Michael De Vlieger, Jun 26 2016 *)

PARI

row(n) = {my(d = divisors(n)); my(nd = #d); my(m = matrix(#d, #d)); for (j=1, nd, m[1,j] = d[j];); for (i=2, nd, for (j=1, nd - i +1, m[i,j] = m[i-1,j+1] - m[i-1,j];);); vector(nd, i, sum(k=0, i-1, m[i-k, k+1]));}
tabf(nn) = for (n=1, nn, print(row(n)););
lista(nn) = for (n=1, nn, v = row(n); for (j=1, #v, print1(v[j], ", "));); \\ Michel Marcus, Jun 25 2016