Determinant of a Toeplitz matrix How can I calculate the determinant of the following Toeplitz matrix?
\begin{bmatrix}
1&2&3&4&5&6&7&8&9&10\\
2&1&2&3&4&5&6&7&8&9 \\
3&2&1&2&3&4&5&6&7&8 \\
4&3&2&1&2&3&4&5&6&7 \\
5&4&3&2&1&2&3&4&5&6 \\
6&5&4&3&2&1&2&3&4&5 \\
7&6&5&4&3&2&1&2&3&4 \\
8&7&6&5&4&3&2&1&2&3 \\
9&8&7&6&5&4&3&2&1&2 \\
10&9&8&7&6&5&4&3&2&1 \\
\end{bmatrix}

We define the following $n \times n$ (symmetric) Toeplitz matrix
$${\rm A}_n := \begin{bmatrix}
          1      & 2      & 3      & \dots  & n-1    & n     \\
          2      & 1      & 2      & \dots  & n-2    & n-1   \\
          3      & 2      & 1      & \dots  & n-3    & n-2   \\
          \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
          n-1    & n-2    & n-3    & \dots  & 1      & 2     \\
          n      & n-1    & n-2    & \dots  & 2      & 1     \\
         \end{bmatrix}$$
Hence,
$${\rm A}_{n+1} = \begin{bmatrix} {\rm A}_n & {\rm c}_n\\ {\rm c}_n^\top & 1\end{bmatrix}$$
where ${\rm c}_n = {\rm A}_n {\rm e}_n + {\Bbb 1}_n$. Computing the determinant,
$$\det \left( {\rm A}_{n+1} \right) = \det \begin{bmatrix} {\rm A}_n & {\rm c}_n\\ {\rm c}_n^\top & 1\end{bmatrix} = \left( 1 - {\rm c}_n^\top {\rm A}_n^{-1} {\rm c}_n \right) \det \left( {\rm A}_n \right)$$
where
$$\begin{aligned} {\rm c}_n^\top {\rm A}_n^{-1} {\rm c}_n &= \left( {\rm A}_n {\rm e}_n + {\Bbb 1}_n \right)^\top {\rm A}_n^{-1} \left( {\rm A}_n {\rm e}_n + {\Bbb 1}_n \right)\\ &= \underbrace{{\rm e}_n^\top {\rm A}_n {\rm e}_n}_{= 1} + \underbrace{{\rm e}_n^\top {\Bbb 1}_n}_{= 1} + \underbrace{{\Bbb 1}_n^\top {\rm e}_n}_{= 1} + \underbrace{{\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n}_{= \frac{2}{n+1}} = 3 + \frac{2}{n+1}\end{aligned}$$
and, thus,
$$\boxed{ \quad \det \left( {\rm A}_{n+1} \right) = -2 \left( \frac{n+2}{n+1} \right) \det \left( {\rm A}_n \right) \quad }$$
and, since $\det \left( {\rm A}_1 \right) = 1$, after some work, we eventually conclude that
$$\color{blue}{\boxed{ \quad \det \left( {\rm A}_n \right) = (-1)^{n-1} \left( n + 1 \right) 2 ^{n-2} \quad }}$$
which is related to integer sequence A001792, as pointed out by Somos.

Addendum
How to show the following?
$${\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n = \frac{2}{n+1}$$
Note that the $n$-th column of matrix ${\rm A}_n$ is the reversal of its first column. Hence,
$${\rm A}_n \left( {\rm e}_1 + {\rm e}_n \right) = (n+1) {\Bbb 1}_n$$
Left-multiplying both sides by ${\Bbb 1}_n^\top {\rm A}_n^{-1}$,
$$\underbrace{{\Bbb 1}_n^\top {\rm A}_n^{-1} {\rm A}_n \left( {\rm e}_1 + {\rm e}_n \right)}_{= {\Bbb 1}_n^\top \left( {\rm e}_1 + {\rm e}_n \right) = 2} = (n+1) {\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n$$
and, thus,
$${\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n = \frac{2}{n+1}$$

SymPy code
>>> from sympy import *
>>> [ Matrix(n, n, lambda i,j: 1 + abs(i-j)).det() for n in range(1,11) ]
[1, -3, 8, -20, 48, -112, 256, -576, 1280, -2816]

