|
%I #22 Mar 26 2022 14:29:56
%S 1,0,1,-1,1,1,-1,0,1,1,-2,0,1,1,1,-1,-1,0,1,1,1,-2,-1,0,1,1,1,1,-2,-1,
%T 0,0,1,1,1,1,-2,-1,-1,0,1,1,1,1,1,-1,-2,-1,0,0,1,1,1,1,1
%N Triangle read by rows: A000012 * A054525 regarded as infinite lower triangular matrices.
%C Row sums = 1.
%C Left border = A002321, the Mertens function.
%C A134541 * [1,2,3,...] = A002088: (1, 2, 4, 6, 10, 12, 18, 22, ...).
%F Recurrence: T(n, k) = If n >= k then If k = 1 then 1 - Sum_{i=1..n-1} T(n, k + i)/(i + 1)^(s - 1) else T(floor(n/k) else 1)) else 0). - _Mats Granvik_, Apr 17 2016
%e First few rows of the triangle:
%e 1;
%e 0, 1;
%e -1, 1, 1;
%e -1, 0, 1, 1;
%e -2, 0, 1, 1, 1;
%e -1, -1, 0, 1, 1, 1;
%e -2, -1, 0, 1, 1, 1, 1;
%e -2, -1, 0, 0, 1, 1, 1, 1;
%e -2, -1, -1, 0, 1, 1, 1, 1, 1;
%e -1, -2, -1, 0, 0, 1, 1, 1, 1, 1;
%e ...
%t Clear[t, s, n, k, z, x]; z = 1; nn = 10; t[n_, k_] := t[n, k] = If[n >= k, If[k == 1, 1 - Sum[t[n, k + i]/(i + 1)^(s - 1), {i, 1, n - 1}], t[Floor[n/k], 1]], 0]; Flatten[Table[Table[Limit[t[n, k], s -> z], {k, 1, n}], {n, 1, nn}]] (* _Mats Granvik_, Jul 22 2012 *) (* updated _Mats Granvik_, Apr 10 2016 *)
%Y Cf. A054525, A002321, A002088.
%Y Matrix inverse of A176702. - _Mats Granvik_, Apr 24 2010
%K tabl,sign,more
%O 1,11
%A _Gary W. Adamson_, Oct 31 2007
|
