%I #12 Sep 27 2017 17:23:46
%S 1,1,0,1,1,0,1,2,2,0,1,3,8,5,0,1,4,18,40,15,0,1,5,32,135,240,52,0,1,6,
%T 50,320,1215,1664,203,0,1,7,72,625,3840,12636,12992,877,0,1,8,98,1080,
%U 9375,53248,147987,112256,4140,0,1,9,128,1715,19440,162500,831488,1917999,1059840,21147,0
%N Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. exp(exp(k*x)-1).
%F O.g.f. of column k: 1/(1 - k*x/(1 - k*x/(1 - k*x/(1 - 2*k*x/(1 - k*x/(1 - 3*k*x/(1 - k*x/(1 - 4*k*x/(1 - ...))))))))), a continued fraction.
%F E.g.f. of column k: exp(exp(k*x)-1).
%F A(n,k) = exp(-1)*k^n*Sum_{j>=0} j^n/j!.
%F A(n,k) = k^n * Bell(n). - _Alois P. Heinz_, Sep 26 2017
%e E.g.f. of column k: A_k(x) = 1 + k*x/1! + 2*k^2*x^2/2! + 5*k^3*x^3/3! + 15*k^4 x^4/4! + 52*k^5*x^5/5! + ...
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 2, 8, 18, 32, 50, ...
%e 0, 5, 40, 135, 320, 625, ...
%e 0, 15, 240, 1215, 3840, 9375, ...
%e 0, 52, 1664, 12636, 53248, 162500, ...
%p A:= (n, k)-> k^n * combinat[bell](n):
%p seq(seq(A(n, d-n), n=0..d), d=0..12); # _Alois P. Heinz_, Sep 26 2017
%t Table[Function[k, n! SeriesCoefficient[Exp[Exp[k x] - 1], {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
%t Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-((-1)^(i + 1) (i - 1) + i + 3) k x/4, 1, {i, 0, n}]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
%Y Columns k=0..3 give A000007, A000110, A055882, A247452.
%Y Rows n=0..2 give A000012, A001477, A001105.
%Y Main diagonal gives A292914.
%K nonn,tabl
%O 0,8
%A _Ilya Gutkovskiy_, Sep 26 2017