Hankel matrix of Catalan numbers Recall that the $n$-th Catalan number $C_n=\frac{1}{n+1}{2n\choose n}$ counts the number of
paths connecting $(0, 0)$ to $(n, n)$ that travel along the grid of integer lattice points of
$R^2$ where each path moves up or right in one-unit steps and no path extends above
the line $y = x$.
In linear algebra, a Hankel matrix of Catalan numbers is defined as following:
$$H_n^t=(C_{i+j+t})_{0\leq i,j\leq n-1}=
\begin{bmatrix}
    c_{t} & c_{t+1} & c_{t+2} & \dots  & c_{t+n-1} \\
    c_{t+1} & c_{t+2} & c_{t+3} & \dots  & c_{t+n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    c_{t+n-1} & c_{t+n} & c_{t+n+1} & \dots  & c_{t+2n-2}
\end{bmatrix}
$$
How can I calculate the Hankel determinant of Catalan numbers for $t=1$?
Is it possible obtain the Hankel determinant of Catalan numbers for $t>1$?
 A: Following Angina Seng's suggestion, let $G$ be the following locally finite, directed acyclic graph:

Let $a_1, \ldots, a_n, b_1, \ldots, b_n$ be as labeled, and let the weight of each edge be $1$. Then $M$ as defined in the Wikipedia article for the Lindström-Gessel-Viennot Lemma is exactly $H_n := H^0_n$, since $M_{i,j}$ is exactly the number of paths connecting $a_i$ to $b_j$, a "staircase" of height $i + j - 2$.
The LGV Lemma is that the determinant of $M$ is given by the following formula:
$$\sum_{P: A \rightarrow B} \text{sign}(\sigma (P))\prod_{P_i \in P} \omega(P_i)$$
where the sum is taken over all families of non-intersecting paths $P = \{P_i\}_{i = 1}^n$ taking the vertices $\{a_1,\ldots,a_n\}$ to the vertices $\{b_1,\ldots,b_n\}$ in some order, $\sigma(P)$ denotes the sign of the permutation of $\{1,\ldots,n\}$ taking $i$ to the index of the end vertex of the path starting at $a_i$, and $\omega(P_i)$ denotes the product of the weights of the edges in the path $P_i$.
It's clear that the only possible family of non-intersecting paths is the one taking each $a_k$ to $b_k$, so the sum has only one term, for which $\sigma(P)$ is the identity, so $\text{sign}(\sigma(P)) = 1$. Furthermore, the weight of every edge was defined to be $1$, so the product of $\omega(P_i)$ is also simply $1$.
So $\det(H^0_n) = \det(M) = 1$ for all $n$. By shifting all the $a_i$ down one diagonal, the same argument applies verbatim, so $\det(H^1_n)$ is also $1$ for all $n$. When $t > 1$ the argument starts to fail because there will exist more than one family of non-intersecting paths connecting $\{a_1,\ldots,a_n\}$ to $\{b_1,\ldots,b_n\}$. (But $\sigma(P)$ will always be the identity, for whatever that's worth.)
