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Inverse triangle of A126988.
3

%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