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%I #21 Nov 10 2017 05:55:10
%S 1,1,-1,1,-1,-1,1,-1,-4,0,1,-1,-16,-23,0,1,-1,-64,-713,-229,1,1,-1,
%T -256,-19619,-64807,-2761,0,1,-1,-1024,-531185,-16757533,-9688425,
%U -42615,1,1,-1,-4096,-14347883,-4294435855,-30499541197,-2165979799,-758499,0
%N Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of Product_{j>=1} (1-j^(k*j)*x^j) in powers of x.
%H Seiichi Manyama, <a href="/A294653/b294653.txt">Antidiagonals n = 0..52, flattened</a>
%F A(0,k) = 1 and A(n,k) = -(1/n) * Sum_{j=1..n} (Sum_{d|j} d^(1+k*j)) * A(n-j,k) for n > 0.
%e Square array begins:
%e 1, 1, 1, 1, 1, ...
%e -1, -1, -1, -1, -1, ...
%e -1, -4, -16, -64, -256, ...
%e 0, -23, -713, -19619, -531185, ...
%e 0, -229, -64807, -16757533, -4294435855, ...
%t rows = 10;
%t col[k_] := col[k] = CoefficientList[Product[(1 - j^(k*j)*x^j), {j, 1, rows + 3}] + O[x]^(rows + 3), x];
%t A[n_, k_] := col[k][[n + 1]];
%t (* or: *)
%t A[0, _] = 1; A[n_, k_] := A[n, k] = -(1/n)*Sum[DivisorSum[j, #^(1 + k*j) &]*A[n - j, k], {j, 1, n}];
%t Table[A[n - k, k], {n, 0, rows - 1}, {k, n, 0, -1}] // Flatten (* _Jean-François Alcover_, Nov 10 2017 *)
%Y Columns k=0..1 give A010815, A292312.
%Y Rows n=0..2 give A000012, (-1)*A000012, (-1)*A000302.
%Y Cf. A283675, A294758.
%K sign,tabl
%O 0,9
%A _Seiichi Manyama_, Nov 06 2017
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