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%I #20 Feb 15 2022 11:10:52
%S 1,-2,1,-3,0,1,0,-2,0,1,-5,0,0,0,1,6,-3,-2,0,0,1,-7,0,0,0,0,0,1,0,0,0,
%T -2,0,0,0,1,0,0,-3,0,0,0,0,0,1,10,-5,0,0,-2,0,0,0,0,1
%N Inverse triangle of A126988.
%C Row sums give A023900.
%C Left column is A055615.
%C A127139 * [1, 2, 3, ...] = [1, 0, 0, 0, ...].
%C A127139 * [1, 0, 0, 0, ...] = A055615.
%C A127140 is the square of A127139.
%F Inverse triangle of A126988.
%e First few rows of the triangle:
%e 1;
%e -2, 1;
%e -3, 0, 1;
%e 0, -2, 0, 1;
%e -5, 0, 0, 0, 1;
%e 6, -3, -2, 0, 0, 1;
%e -7, 0, 0, 0, 0, 0, 1;
%e ...
%t nn = 10; s = 0; t[1, 1] = 1; t[n_, k_] := t[n, k] = If[k == 1, -Sum[t[n, k + i]/(i + 1)^(s - 1), {i, 1, n - 1}], If[Mod[n, k] == 0, t[n/k, 1], 0], 0]; Flatten[Table[Table[t[n, k], {k, 1, n}], {n, 1, nn}]] (* _Mats Granvik_, Mar 12 2016 *)
%Y Cf. A126988, A055615, A023900, A127140.
%K tabl,sign
%O 1,2
%A _Gary W. Adamson_, Jan 06 2007