G.f. for Matrix Inverse of Triangle with Known G.F.

[Paul D. Hanna]

Emeric Deutsch submitted a triangle enumerating Schroder paths:
http://www.research.att.com/projects/OEIS?Anum=A101275
with G.f.:  A(x,y) = 2/((2-y)*(1-x)+y*sqrt(1-6*x+x^2)) 
 
The matrix inverse of this triangle is:
http://www.research.att.com/projects/OEIS?Anum=A102051
with G.f.: B(x,y) = 2/(1+y+(1-y)*sqrt(1+4*x-4*x^2)) .
 
We can transform these o.g.f.s into each other by a change of variables: 
 
A(1-1/x,1-y)/x  =  B(x,y) ,
 
B(1/(1-x),1-y)/(1-x)  =  A(x,y) .
  
 
Now the $100,000 question arises: 
 
for a known g.f. of an invertible triangle T, what is the transform of
variables x,y, 
that converts the g.f. of T into the g.f.of the matrix inverse of T? 
 
Although such a transform would depend on the g.f. involved, 
perhaps there is a general formula? 
(at least for certain types of g.f.s). 
  
Pascal's triangle is the simplest example with G.f.: 1/(1-x-xy) 
so that  x' = -x, y' = -y, transforms the g.f. into the g.f. of the
matrix inverse. 
  
Perhaps the next easiest to find/study would be quadratic o.g.f. cases
like above.
 
Can anyone think of other nice examples of g.f.s of triangles 
that have simple known g.f.s for the matrix inverse? 
 
 From even a small collection of such pairs of g.f.s, 
perhaps a pattern could be found ... 
 
Thanks,
        Paul
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20050203/515639eb/attachment.htm>