%I #8 Dec 15 2014 05:17:46
%S 1,1,2,3,4,3,16,20,9,4,127,156,63,16,5,1363,1664,648,144,25,6,18628,
%T 22684,8703,1840,275,36,7,311250,378572,144243,29824,4200,468,49,8,
%U 6173791,7504640,2849400,582640,79775,8316,735,64,9,142190703,172785512
%N Triangular matrix, read by rows, that satisfies: T(n,k) = [T^2](n-1,k) when n>k>=0, with T(n,n) = (n+1).
%C Column 0 forms A082161. Column 1 forms A102087. Row sums form A102088.
%F T(n, 0) = A082161(n) for n>0, with T(0, 0) = 1.
%F G.f. for column k: T(k, k) = k+1 = Sum_{n>=0} T(n+k, k)*x^n*prod_{j=1, n+1} (1-(j+k)*x).
%e Rows of T begin:
%e [1],
%e [1,2],
%e [3,4,3],
%e [16,20,9,4],
%e [127,156,63,16,5],
%e [1363,1664,648,144,25,6],
%e [18628,22684,8703,1840,275,36,7],
%e [311250,378572,144243,29824,4200,468,49,8],
%e [6173791,7504640,2849400,582640,79775,8316,735,64,9],...
%e Matrix square T^2 equals T excluding the main diagonal:
%e [1],
%e [3,4],
%e [16,20,9],
%e [127,156,63,16],
%e [1363,1664,648,144,25],...
%e G.f. for column 0: 1 = (1-x) + 1*x*(1-x)(1-2x) + 3*x^2*(1-x)(1-2x)(1-3x) + ... + T(n,0)*x^n*(1-x)(1-2x)(1-3x)*..*(1-(n+1)*x) + ...
%e G.f. for column 1: 2 = 2(1-2x) + 4*x*(1-2x)(1-3x) + 20*x^2*(1-2x)(1-3x)(1-4x) + ... + T(n+1,1)*x^n*(1-2x)(1-3x)(1-4x)*..*(1-(n+2)*x) + ...
%e G.f. for column 2: 3 = 3(1-3x) + 9*x*(1-3x)(1-4x) + 63*x^2*(1-3x)(1-4x)(1-5x) + ... + T(n+2,2)*x^n*(1-3x)(1-4x)(1-5x)*..*(1-(n+3)*x) + ...
%p {T(n,k)=local(A=matrix(1,1),B);A[1,1]=1; for(m=2,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=j,if(j==1,B[i,j]=(A^2)[i-1,1], B[i,j]=(A^2)[i-1,j]));));A=B);return(A[n+1,k+1])}
%t T[n_, n_] := n+1; T[n_, k_] /; k>n = 0; T[n_, k_] /; k == n-1 := n^2; T[n_, k_] := T[n, k] = Coefficient[1-Sum[T[i, k]*x^i*Product[1-(j+k)*x, {j, 1, i-k+1}], {i, k, n-1}], x, n]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 15 2014, after PARI script *)
%o (PARI) {T(n,k)=if(n<k,0,if(n==k,k+1, polcoeff(1-sum(i=k,n-1,T(i,k)*x^i*prod(j=1,i-k+1,1-(j+k)*x+x*O(x^n))),n)))}
%Y Cf. A082161, A102087, A102088.
%Y Cf. A102316.
%K nonn,tabl
%O 0,3
%A _Paul D. Hanna_, Dec 29 2004