the infinite lower triangular matrix with nonzero entries on the main diagonal, where $h_{i,j}=[x^i](g(x)(f(x)^j)$ for $i\geq j$, namely, $h_{i,j}$ equals the coefficient of $x^i$ in the expansion of the series $f(x)(g(x)^j)$. If an infinite lower triangular matrix $H$ can be constructed in this way from two generating functions $g(x)$ and $f(x)$, then it is called a [*Riordan array*]{} and is denoted by $H=(g(x),f(x))=(g,f)$.

Suppose we multiply the matrix $H=(g,f)$ by a column vector $(a_0,a_1,\cdots)^T$ and get a column vector $(b_0,b_1,\cdots)^T$. Let $A(x)$ and $B(x)$ be the generating functions for the sequences $(a_0,a_1,\cdots)$ and $(b_0,b_1,\cdots)$ respectively. Then the method of Riordan arrays asserts that $$B(x)=g(x)A(f(x)).$$

For the matrix identity (\[eq2.2\]), let $g(x)$ be the generating function for Catalan numbers $(1, 2, 5, 14, \ldots)$: $$g(x) = {1 -2x-
\sqrt{1-4x} \over 2x^2}.$$ Let $f(x)=xg(x)$. From the recurrence relation (\[r2.2\]) one may derive that the generating function for the sequence in the $j$-th $(j\geq 1)$ column in the matrix in (\[eq2.2\]) equals $g(xg)^{j-1}$. Let $H$ be the Riordan array $(g, xg)$. Since the generating function of $(1,2,3,4\cdots)^T$ equals $A(x)=\frac{1}{(1-x)^2}$, it follows that $B(x)=g(x)A(xg(x))=\frac{1}{1-4x}$ equals the generating function for the right hand side of (\[eq2.2\]). Thus we obtain the identity (\[eq2.2\]).

Let us consider 