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