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%I #15 Feb 06 2018 11:45:21
%S 1,1,0,1,-1,0,1,-2,-2,0,1,-3,-3,-1,0,1,-4,-3,2,-1,0,1,-5,-2,8,4,5,0,1,
%T -6,0,16,9,16,1,0,1,-7,3,25,9,18,-3,13,0,1,-8,7,34,0,4,-35,6,4,0,1,-9,
%U 12,42,-21,-26,-90,-33,-31,0,0,1,-10,18,48,-56,-66,-145,-56,-66,-72,2,0
%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*x^j)^k.
%F G.f. of column k: Product_{j>=1} (1 - j*x^j)^k.
%e G.f. of column k: A_k(x) = 1 - k*x + (1/2)*k*(k - 5)*x^2 - (1/6)*k*(k^2 - 15*k + 20)*x^3 + (1/24)*k*(k^3 - 30*k^2 + 155*k - 150)*x^4 - (1/120)*k*(k^4 - 50*k^3 + 575*k^2 - 1750*k + 624)*x^5 + ...
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, -1, -2, -3, -4, -5, ...
%e 0, -2, -3, -3, -2, 0, ...
%e 0, -1, 2, 8, 16, 25, ...
%e 0, -1, 4, 9, 9, 0, ...
%e 0, 5, 16, 18, 4, -26, ...
%t Table[Function[k, SeriesCoefficient[Product[(1 - i x^i)^k, {i, 1, n}], {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten
%o (PARI) first(n, k) = my(res = matrix(n, k)); for(u=1, k, my(col = Vec(prod(j=1, n, (1 - j*x^j)^(u-1)) + O(x^n))); for(v=1, n, res[v, u] = col[v])); res \\ _Iain Fox_, Dec 28 2017
%Y Columns k=0..32 give A000007, A022661, A022662, A022663, A022664, A022665, A022666, A022667, A022668, A022669, A022670, A022671, A022672, A022673, A022674, A022675, A022676, A022677, A022678, A022679, A022680, A022681, A022682, A022683, A022684, A022685, A022686, A022687, A022688, A022689, A022690, A022691, A022692.
%Y Main diagonal gives A297324.
%Y Antidiagonal sums give A299209.
%Y Cf. A266964, A297321, A297325, A297328.
%K sign,tabl
%O 0,8
%A _Ilya Gutkovskiy_, Dec 28 2017
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