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%I #15 Mar 07 2020 11:42:51
%S 1,0,1,0,0,1,-1,0,0,1,0,-4,0,0,1,0,0,-10,0,0,1,19,0,0,-20,0,0,1,0,133,
%T 0,0,-35,0,0,1,0,0,532,0,0,-56,0,0,1,-1513,0,0,1596,0,0,-84,0,0,1,0,
%U -15130,0,0,3990,0,0,-120,0,0,1,0,0,-83215,0
%N Triangle read by rows, e.g.f. exp(x*(z+1/2))/((exp(3*x/2) + 2*cos(sqrt(3)*x/2))/3).
%F Matrix inverse is A215061.
%F T(n,k) = A215064(n,k) - A215062(n,k) + [n==k].
%F |T(3*n,0)| = A002115(n).
%e [0] [1]
%e [1] [0, 1]
%e [2] [0, 0, 1]
%e [3] [-1, 0, 0, 1]
%e [4] [0, -4, 0, 0, 1]
%e [5] [0, 0, -10, 0, 0, 1]
%e [6] [19, 0, 0, -20, 0, 0, 1]
%e [7] [0, 133, 0, 0, -35, 0, 0, 1]
%e [8] [0, 0, 532, 0, 0, -56, 0, 0, 1]
%e [9] [-1513, 0, 0, 1596, 0, 0, -84, 0, 0, 1]
%o (Sage)
%o def triangle(f, dim):
%o var('x,z')
%o s = f.series(x, dim+2)
%o P = [factorial(i)*s.coefficient(x,i) for i in range(dim)]
%o for k in range(dim): print([k], [P[k].coefficient(z,i) for i in (0..k)])
%o def A215060_triangle(dim) :
%o var('x, z')
%o f = exp(x*(z+1/2))/((exp(3*x/2)+2*cos(sqrt(3)*x/2))/3)
%o return triangle(f, dim)
%o A215060_triangle(12)
%Y Cf. A215061, A215062, A215063, A215064, A215065.
%K sign,tabl
%O 0,12
%A _Peter Luschny_, Aug 01 2012
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