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%I #9 Jan 12 2018 12:33:11
%S 1,7,1,315,26,1,45682,2600,63,1,15646589,675194,11655,124,1,
%T 10567689552,366349152,4861458,37944,215,1,12503979423607,
%U 361884843866,3882676581,23641468,100835,342,1,23841011541867520,591934698991168,5318920238688
%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)^3)^(n-k)/(n-k)! for n >= k >= 1.
%C Define triangular matrix P by P(n,k) = (-k^3-3k^2-3k)^(n-k)/(n-k)!, then M = P*D*P^-1 = A103237 satisfies: M^3 + 3M^2 + 3M = 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 A082172 as a triangular matrix. The first column is A082160 (quasi-acyclic automata with 3 inputs).
%F For n > k >= 1: 0 = Sum_{m=k..n} C(n-k, m-k)*(1-(m+1)^3)^(n-m)*T(m, k). For n > k >= 1: 0 = Sum_{j=k..n} C(n-k, j-k)*(1-(k+1)^3)^(j-k)*T(n, j).
%e Rows of unreduced fractions T(n,k)/(n-k)! begin:
%e [1/0! ],
%e [7/1!, 1/0! ],
%e [315/2!, 26/1!, 1/0! ],
%e [45682/3!, 2600/2!, 63/1!, 1/0! ],
%e [15646589/4!, 675194/3!, 11655/2!, 124/1!, 1/0! ],
%e [10567689552/5!, 366349152/4!, 4861458/3!, 37944/2!, 215/1!, 1/0! ], ...
%e forming the inverse of matrix P where P(n,k) = A103247(n,k)/(n-k)!:
%e [1/0! ],
%e [ -7/1!, 1/0! ],
%e [49/2!, -26/1!, 1/0! ],
%e [ -343/3!, 676/2!, -63/1!, 1/0! ],
%e [2401/4!, -17576/3!, 3969/2!, -124/1!, 1/0! ],
%e [ -16807/5!, 456976/4!, -250047/3!, 15376/2!, -215/1!, 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)^3)^(r-c)/(r-c)!))); return(if(n<k|k<1,0,(P^-1)[n,k]*(n-k)!))}
%Y Cf. A103248, A103237, A082172, A082160.
%K nonn,tabl,frac
%O 1,2
%A _Paul D. Hanna_, Feb 02 2005
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