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Consider a triangle of integers
$a_{1,0}$
$a_{2,1}, \, a_{2,0}$
$a_{3,2}, \, a_{3,1}, \, a_{3,0}$
$\quad \vdots$
and its "transpose"
$a_{1,0}$
$a_{2,0}, \, a_{2,1}$
$a_{3,0}, \, a_{3,1}, \, a_{3,2}$
$\quad \vdots$
Given a bivariate generating function $f(x,y)$ with series $$ a_{1,0}*y + (a_{2,1}x + a_{2,0})y^2 + (a_{3,2}x^2 + a_{3,1}*x + a_{3,0})y^3 + \cdots $$ what is the generating function which gives the coefficients of the polynomials of $x$ as the transposed triangle, i.e. how can $g(x,y)$ be obtained with series $$ a_{1,0}y + (a_{2,1} + a_{2,0}x)y^2 + (a_{3,2} + a_{3,1}x + a_{3,0}x^2)y^3 + \cdots $$

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  • $\begingroup$ I wouldn't call that the transpose. $\endgroup$ Commented May 20, 2011 at 22:47

1 Answer 1

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$g(x, y) = \frac{1}{x} f \left( \frac{1}{x}, xy \right)$. First you should know that there is a more basic trick here where you reverse the coefficients of a polynomial by substituting $x \mapsto \frac{1}{x}$ and clearing denominators, and after that it's not hard to tweak everything else to work out.

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