%I #5 Jun 13 2017 23:29:56
%S 1,4,22,212,3255,70777,2022897,72375484,3130502129,159476810183,
%T 9376968779265,626244735454991,46892450411406465,3894861818247549265,
%U 355651177699555693544,35432761283736539730108
%N Column 0 of triangle A113384.
%F A113384 equals the matrix square of A113381, which has the property: column k of A113381^2 = column 0 of A113381^(3*k+2) for k>=0.
%o (PARI) a(n)=local(A,B);A=Mat(1);for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(3*j-2))[i-j+1,1]));));A=B); (matrix(#A,#A,r,c,if(r>=c,(A^(3*c-1))[r-c+1,1]))^2)[n+1,1]
%Y Cf. A113381, A113384, A113386 (column 1).
%K nonn
%O 0,2
%A _Paul D. Hanna_, Nov 14 2005