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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 $(c_n)(c_n)^T(H_n)^{-1}$.

yanjunk
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  • First thought is to explore proof by induction. – A rural reader Oct 15 '22 at 02:04
  • The vector $c_n$ can be represented as $c_n=\tilde{H}_ne_1,$ where $\tilde{H}_n$ is the Hankel matrix starting from $1$ end $e_1$ is the first vector of standard basis. It seems a relation between $H_n$ and $\tilde{H}_n$ might be helpful. – Ryszard Szwarc Oct 15 '22 at 09:54

1 Answers1

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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, $v^tH_n^{-1}v=1-\frac{1}{n+1}=\frac{n}{n+1}.$

Daniel
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