%I #21 Jul 21 2024 00:27:56
%S 1,1,1,15,8,1,1024,368,27,1,198581,53672,2727,64,1,85102056,18417792,
%T 710532,11904,125,1,68999174203,12448430408,386023509,4975936,38375,
%U 216,1,95264160938080,14734002979456,381535651512,3977848832,23945000
%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) = (-k^3)^(n-k)/(n-k)! for n >= k >= 1.
%C Define a triangular matrix P where P(n,k) = (-k^3)^(n-k)/(n-k)!, then M = P*D*P^-1 = A102098 satisfies M^3 = 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.
%C Essentially equal to square array A082170 as a triangular matrix. The first column is A082162 (enumerates acyclic automata with 3 inputs).
%F For n > k >= 1: 0 = Sum_{m=k..n} C(n-k, m-k)*(-m^3)^(n-m)*T(m, k).
%F For n > k >= 1: 0 = Sum_{j=k..n} C(n-k, j-k)*(-k^3)^(j-k)*T(n, j).
%e Rows of unreduced fractions T(n,k)/(n-k)! begin:
%e [1/0!],
%e [1/1!, 1/0!],
%e [15/2!, 8/1!, 1/0!],
%e [1024/3!, 368/2!, 27/1!, 1/0!],
%e [198581/4!, 53672/3!, 2727/2!, 64/1!, 1/0!],
%e [85102056/5!, 18417792/4!, 710532/3!, 11904/2!, 125/1!, 1/0!], ...
%e forming the inverse of matrix P where P(n,k) = A103246(n,k)/(n-k)!:
%e [1/0!],
%e [-1/1!, 1/0!],
%e [1/2!, -8/1!, 1/0!],
%e [-1/3!, 64/2!, -27/1!, 1/0!],
%e [1/4!, -512/3!, 729/2!, -64/1!, 1/0!], ...
%o (PARI) {T(n,k)=my(P);if(n>=k&k>=1, P=matrix(n,n,r,c,if(r>=c,(-c^3)^(r-c)/(r-c)!))); return(if(n<k||k<1,0,(P^-1)[n,k]*(n-k)!))}
%Y Cf. A103246, A102098, A082170, A082162.
%K nonn,tabl,frac
%O 1,4
%A _Paul D. Hanna_, Feb 02 2005