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A114155 Triangle, read by rows, given by the product Q^-2*P^3 using triangular matrices P=A113370, Q=A113381.

%I A114155 #5 Jun 13 2017 23:40:29
%S A114155 1,-1,1,3,2,1,6,6,5,1,-8,37,45,8,1,-501,429,635,120,11,1,-13623,7629,
%T A114155 12815,2556,231,14,1,-409953,185776,343815,71548,6556,378,17,1,
%U A114155 -14544683,5817106,11651427,2508528,233706,13391,561,20,1
%N A114155 Triangle, read by rows, given by the product Q^-2*P^3 using triangular matrices P=A113370, Q=A113381.
%C A114155 Complementary to A114154, which gives R^3*Q^-2. Column 0 equals column 0 of P^-1 (A114157).
%e A114155 Triangle Q^-2*P^3 begins:
%e A114155 1;
%e A114155 -1,1;
%e A114155 3,2,1;
%e A114155 6,6,5,1;
%e A114155 -8,37,45,8,1;
%e A114155 -501,429,635,120,11,1;
%e A114155 -13623,7629,12815,2556,231,14,1;
%e A114155 -409953,185776,343815,71548,6556,378,17,1; ...
%e A114155 Compare to Q (A113381):
%e A114155 1;
%e A114155 2,1;
%e A114155 6,5,1;
%e A114155 37,45,8,1;
%e A114155 429,635,120,11,1;
%e A114155 7629,12815,2556,231,14,1;...
%e A114155 Thus Q^-2*P^3 shift left one column equals Q.
%o A114155 (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])); (Q^-2*P^3)[n+1,k+1]
%Y A114155 Cf. A114157 (column 0), 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), A114153 (R^-1*P^3), A114154 (R^3*Q^-2); A114156 (P^-1), A114158 (Q^-1), A114159 (R^-1).
%K A114155 sign,tabl
%O A114155 0,4
%A A114155 _Paul D. Hanna_, Nov 15 2005