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

%I A191904 #22 May 14 2022 19:57:33
%S A191904 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,
%T A191904 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,
%U A191904 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
%N A191904 Square array read by antidiagonals up: T(n,k) = 1-k if k divides n, else 1.
%C A191904 Transposed variant of A177121. Array variant of A176079.
%H A191904 Mats Granvik, <a href="/A191904/a191904.txt">Mathematica program for recurrences.</a>
%F A191904 Conjecture: Sum_{n>=1} T(n,k)/n = log(k).
%F A191904 From _Mats Granvik_, Apr 24 2022: (Start)
%F A191904 Sum recurrence:
%F A191904 T(n, 1) = [n >= 1]*0;
%F A191904 T(n, k) = [n < k]*1;
%F A191904 T(n, k) = [n >= k](Sum_{i=1..k-1} T(n - i, k - 1) - Sum_{i=1..k-1} T(n - i, k)).
%F A191904 Product recurrence:
%F A191904 T(n, 1) = [n >= 1]*0;
%F A191904 T(n, k) = [n < k]*1;
%F A191904 T(n, k) = [n >= k](Product_{i=1..k-1} T(n - i, k - 1) - Product_{i=1..k-1} T(n - i, k)).
%F A191904 (End)
%e A191904 Table begins:
%e A191904 0..1..1..1..1..1..1..1..1...
%e A191904 0.-1..1..1..1..1..1..1..1...
%e A191904 0..1.-2..1..1..1..1..1..1...
%e A191904 0.-1..1.-3..1..1..1..1..1...
%e A191904 0..1..1..1.-4..1..1..1..1...
%e A191904 0.-1.-2..1..1.-5..1..1..1...
%e A191904 0..1..1..1..1..1.-6..1..1...
%e A191904 0.-1..1.-3..1..1..1.-7..1...
%e A191904 0..1.-2..1..1..1..1..1.-8...
%t A191904 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}]]
%Y A191904 Cf. A177121, A127093, A175992, A191907, A156734.
%Y A191904 Cf. A176079.
%K A191904 sign,easy,tabl
%O A191904 1,13
%A A191904 _Mats Granvik_, Jun 19 2011