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A114150 Triangle, read by rows, given by the product R^2*Q^-1 = Q^3*P^-2 using triangular matrices P=A113370, Q=A113381, R=A113389.

%I A114150 #5 Jun 13 2017 23:39:52
%S A114150 1,4,1,28,7,1,326,91,10,1,5702,1722,190,13,1,136724,43764,4945,325,16,
%T A114150 1,4226334,1415799,163705,10751,496,19,1,161385532,56096733,6617605,
%U A114150 437723,19896,703,22,1
%N A114150 Triangle, read by rows, given by the product R^2*Q^-1 = Q^3*P^-2 using triangular matrices P=A113370, Q=A113381, R=A113389.
%C A114150 Complementary to A114151, which gives R^-2*Q^3 = Q^-1*P^2.
%e A114150 Triangle R^2*Q^-1 = Q^3*P^-2 begins:
%e A114150 1;
%e A114150 4,1;
%e A114150 28,7,1;
%e A114150 326,91,10,1;
%e A114150 5702,1722,190,13,1;
%e A114150 136724,43764,4945,325,16,1;
%e A114150 4226334,1415799,163705,10751,496,19,1; ...
%e A114150 Compare to P (A113370):
%e A114150 1;
%e A114150 1,1;
%e A114150 1,4,1;
%e A114150 1,28,7,1;
%e A114150 1,326,91,10,1;
%e A114150 1,5702,1722,190,13,1; ...
%e A114150 Thus R^2*Q^-1 = Q^3*P^-2 equals P shift left one column.
%o A114150 (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^2*Q^-1)[n+1,k+1]
%Y A114150 Cf. A113370 (P), A113381 (Q), A113389 (R); A114151 (R^-2*Q^3=Q^-1*P^2), A114152 (R^3*P^-1), A114153 (R^-1*P^3), A114154 (R^3*Q^-2), A114155 (Q^-2*P^3); A114156 (P^-1), A114158 (Q^-1), A114159 (R^-1).
%K A114150 nonn,tabl
%O A114150 0,2
%A A114150 _Paul D. Hanna_, Nov 15 2005