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%I #15 Jan 05 2019 05:16:26
%S 1,3,1,39,8,1,1206,176,15,1,69189,7784,495,24,1,6416568,585408,29430,
%T 1104,35,1,881032059,67481928,2791125,84600,2135,48,1,168514815360,
%U 11111547520,389244600,9841728,204470,3744,63,1,42934911510249
%N Unreduced numerators of the elements T(n,k)/(n-k)!, read by rows, of the triangular matrix P^-1, which is the inverse of the matrix defined by P(n,k) = (1-(k+1)^2)^(n-k)/(n-k)! for n >= k >= 1.
%C Define a triangular matrix P where P(n,k) = (-k^2-2*k)^(n-k)/(n-k)!; then M = P*D*P^-1 = A103236 satisfies M^2 + 2*M = SHIFTUP(M) where D is the diagonal matrix consisting of {1,2,3,...}. The operation SHIFTUP(M) shifts each column of M up 1 row. Essentially equal to square array A082171 as a triangular matrix. The first column is A082163 (enumerates acyclic automata with 2 inputs).
%F For n > k >= 1: 0 = Sum_{m=k..n} C(n-k, m-k)*(1-(m+1)^2)^(n-m)*T(m, k).
%F For n > k >= 1: 0 = Sum_{j=k..n} C(n-k, j-k)*(1-(k+1)^2)^(j-k)*T(n, j).
%e Rows of unreduced fractions T(n,k)/(n-k)! begin:
%e [1/0!],
%e [3/1!, 1/0!],
%e [39/2!, 8/1!, 1/0!],
%e [1206/3!, 176/2!, 15/1!, 1/0!],
%e [69189/4!, 7784/3!, 495/2!, 24/1!, 1/0!],
%e [6416568/5!, 585408/4!, 29430/3!, 1104/2!, 35/1!, 1/0!], ...
%e forming the inverse of matrix P where P(n,k) = A103247(n,k)/(n-k)!:
%e [1/0!],
%e [ -3/1!, 1/0!],
%e [9/2!, -8/1!, 1/0!],
%e [ -27/3!, 64/2!, -15/1!, 1/0!],
%e [81/4!, -512/3!, 225/2!, -24/1!, 1/0!],
%e [ -243/5!, 4096/4!, -3375/3!, 576/2!, -35/1!, 1/0!], ...
%o (PARI) {T(n,k)=local(P);if(n>=k&k>=1, P=matrix(n,n,r,c,if(r>=c,(1-(c+1)^2)^(r-c)/(r-c)!))); return(if(n<k|k<1,0,(P^-1)[n,k]*(n-k)!))}
%Y Cf. A103247, A103236, A082171, A082163.
%K nonn,tabl,frac
%O 1,2
%A _Paul D. Hanna_, Feb 02 2005
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