A191904 Square array read by antidiagonals up: T(n,k) = 1-k if k divides n, else 1.

0, 0, 1, 0, -1, 1, 0, 1, 1, 1, 0, -1, -2, 1, 1, 0, 1, 1, 1, 1, 1, 0, -1, 1, -3, 1, 1, 1, 0, 1, -2, 1, 1, 1, 1, 1, 0, -1, 1, 1, -4, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, -1, -2, -3, 1, -5, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, -1, 1, 1, 1, 1, -6, 1, 1, 1, 1, 1, 1, 0, 1, -2, 1, -4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, -1, 1, -3, 1, 1, 1, -7, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Mats Granvik, Jun 19 2011

Keywords

• sign
• easy
• tabl

Comments

Transposed variant of A177121. Array variant of A176079.

Examples

			Table begins:
0..1..1..1..1..1..1..1..1...
0.-1..1..1..1..1..1..1..1...
0..1.-2..1..1..1..1..1..1...
0.-1..1.-3..1..1..1..1..1...
0..1..1..1.-4..1..1..1..1...
0.-1.-2..1..1.-5..1..1..1...
0..1..1..1..1..1.-6..1..1...
0.-1..1.-3..1..1..1.-7..1...
0..1.-2..1..1..1..1..1.-8...
		

Links

• Mats Granvik, Mathematica program for recurrences.

Crossrefs

Cf. A177121, A127093, A175992, A191907, A156734.

Cf. A176079.

Programs

Mathematica

nn = 30; t[n_, k_] := t[n, k] = If[Mod[n, k] == 0, -(k - 1), 1]; MatrixForm[Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]

Formula

Conjecture: Sum_{n>=1} T(n,k)/n = log(k).

From Mats Granvik, Apr 24 2022: (Start)

Sum recurrence:

T(n, 1) = [n >= 1]*0;

T(n, k) = [n < k]*1;

T(n, k) = [n >= k](Sum_{i=1..k-1} T(n - i, k - 1) - Sum_{i=1..k-1} T(n - i, k)).

Product recurrence:

T(n, 1) = [n >= 1]*0;

T(n, k) = [n < k]*1;

T(n, k) = [n >= k](Product_{i=1..k-1} T(n - i, k - 1) - Product_{i=1..k-1} T(n - i, k)).

(End)