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%I #5 Jun 13 2017 23:39:45
%S 1,0,1,0,6,1,0,48,12,1,0,605,186,18,1,0,11196,3892,414,24,1,0,280440,
%T 106089,12021,732,30,1,0,8981460,3620379,429345,27152,1140,36,1,0,
%U 353283128,149740555,18386361,1196910,51445,1638,42,1
%N Triangle, read by rows, given by the product R^-1*P^3 using triangular matrices P=A113370, R=A113389.
%C Complementary to A114152, which gives R^3*P^-1.
%e Triangle R^-1*P^3 begins:
%e 1;
%e 0,1;
%e 0,6,1;
%e 0,48,12,1;
%e 0,605,186,18,1;
%e 0,11196,3892,414,24,1;
%e 0,280440,106089,12021,732,30,1; ...
%e Compare to R^2 (A113392):
%e 1;
%e 6,1;
%e 48,12,1;
%e 605,186,18,1;
%e 11196,3892,414,24,1;
%e 280440,106089,12021,732,30,1; ...
%e Thus R^-1*P^3 equals R^2 shift right one column.
%o (PARI) T(n,k)=local(P,Q,R,W);P=Mat(1);for(m=2,n+1,W=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,W[i,j]=1,if(j==1, W[i,1]=1,W[i,j]=(P^(3*j-2))[i-j+1,1]));));P=W); Q=matrix(#P,#P,r,c,if(r>=c,(P^(3*c-1))[r-c+1,1])); R=matrix(#P,#P,r,c,if(r>=c,(P^(3*c))[r-c+1,1])); (R^-1*P^3)[n+1,k+1]
%Y Cf. A113392 (R^2), A113370 (P), A113381 (Q), A113389 (R); A114150 (R^2*Q^-1=Q^3*P^-2), A114151 (R^-2*Q^3=Q^-1*P^2), A114152 (R^3*P^-1), A114154 (R^3*Q^-2), A114155 (Q^-2*P^3); A114156 (P^-1), A114158 (Q^-1), A114159 (R^-1).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 15 2005
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