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
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.
Programs
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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)
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