%I
%S 1,1,0,1,0,1,1,16,0,0,1,0,24,0,1,1,128,0,256,0,0,1,0,288,0,200,
%T 0,1,1,1024,0,24576,0,1296,0,0,1,0,3456,0,12000,0,784,0,1,1,
%U 8192,0,2621440,0,590976,0,4096,0,0
%N Antidiagonal expansion of rational polynomial with factors: p(x,n) = If[n == 0, 1/(1  x), x*ChebyshevU[n, x]/ChebyshevT[n + 1, x]].
%C Row sums are {1, 1, 2, 15, 22, 383, 486, 26895, 16238, 3224703, ...}.
%C The rational function here is associated with tan(n*arctan(x)).
%F p(x,n) = If[n == 0, 1/(1  x), x*ChebyshevU[n, x]/ChebyshevT[n + 1, x]];
%F a(n,m) = (n+1)^m*expansion(p(x,n));
%F t(n,m) = antidiagonal(t(n,m)).
%e {1},
%e {1, 0},
%e {1, 0, 1},
%e {1, 16, 0, 0},
%e {1, 0, 24, 0, 1},
%e {1, 128, 0, 256, 0, 0},
%e {1, 0, 288, 0, 200, 0, 1},
%e {1, 1024, 0, 24576, 0, 1296, 0, 0},
%e {1, 0, 3456, 0, 12000, 0, 784, 0, 1},
%e {1, 8192, 0, 2621440, 0, 590976, 0, 4096, 0, 0}
%t p[x_, n_] = If[n == 0, 1/(1  x), x*ChebyshevU[n, x]/ChebyshevT[n + 1, x]];
%t a = Table[Table[(n + 1)^(m + 1)*SeriesCoefficient[Series[p[x, n], {x, 0, 50}], m], { m, 0, 20}], {n, 0, 20}];
%t Table[Table[a[[m, n  m + 1]], {m, 1, n}], {n, 1, 10}];
%t Flatten[%]
%K sign,tabl,uned
%O 0,8
%A _Roger L. Bagula_, Feb 15 2010
