Product of Inverse Hankel Matrix Consider $H_n$, the $n\times n$ Hankel matrix of the Catalan numbers starting from $2$:
$$H_n = \begin{bmatrix}
2 & 5 & 14 & 42 & 132\\
5 & 14 & 42 & 132 & 429\\
14 & 42 & 132 & 429 & 1430 & \cdots\\
42 & 132 & 429 & 1430 & 4862\\
132 & 429 & 1430 & 4862 & 16796\\
&&\vdots\end{bmatrix}$$
It is known that $\text{det}(H_n) = n + 1$. (see Hankel Matrix)
Consider the column vector,
$$c_n = \begin{bmatrix}1 \\ 2 \\ 5 \\ 14 \\ \vdots \end{bmatrix}$$
that contains the first $n$ Catalan numbers.
I have found a pattern that I have checked up to $n=240$, that $$(c_n)^T(H_n)^{-1}(c_n) = \frac{n}{n+1}$$
Is there any method I can take to prove this, or is there a counterexample? Note also that this product is the only non-zero eigenvalue of $(c_n)(c_n)^T(H_n)^{-1}$.

In the provided link, it says that
if $S_{n+1}=\begin{pmatrix} 1 & v^t\\
v & H_n  \\
 \end{pmatrix}$, where $v^t=(1,2,5,\ldots)$, then $\det(S_{n+1})=1$.
Notice that $$\begin{pmatrix} 1 & -v^tH_n^{-1}\\
0 & Id  \\
 \end{pmatrix}\begin{pmatrix} 1 & v^t\\
v & H_n  \\
 \end{pmatrix}\begin{pmatrix} 1 & 0^t\\
-H_n^{-1}v & Id  \\
 \end{pmatrix}=\begin{pmatrix} 1-v^tH_n^{-1}v & 0^t\\
0 & H_n  \\
 \end{pmatrix}.$$
Now, $\det\begin{pmatrix} 1 & -v^tH_n^{-1}\\
0 & Id  \\
 \end{pmatrix}=\det\begin{pmatrix} 1 & 0^t\\
-H_n^{-1}v & Id  \\
 \end{pmatrix}=1$, which implies
$$1=\det\begin{pmatrix} 1 & v^t\\
v & H_n  \\
 \end{pmatrix}=(1-v^tH_n^{-1}v).\det(H_n)=(1-v^tH_n^{-1}v).(n+1)$$
Therefore, $v^tH_n^{-1}v=1-\frac{1}{n+1}=\frac{n}{n+1}.$
