TITLE: How to solve this determinant: $a_{ij}=|i-j|+1$?
QUESTION [5 upvotes]: I have to solve determinant of the following form:
$$a_{ij}=|i-j|+1$$
It looks like this: 
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
2 & 1 & 2 & 3 & \cdots & n-1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & n-3 & \cdots & 1
\end{pmatrix}
$$
It looks something like Toeplitz matrix, but I haven't found any method of solving it. I would appreciate also a kind of hint that would help.
EDIT: OEIS gives a formula for absolute value: $(n+1)\cdot2^{n-2}$
http://oeis.org/A001792
Thanks in advance!

REPLY [2 votes]: What do you get when you substract (i+1)th row - ith row?
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & -1
\end{pmatrix}
$$
What do you get when you add nth column to all other columns?
$$\begin{pmatrix}
n+1 & n+2 & n+3 & n+4 & \cdots & n \\
0 & -2 & -2 & -2 & \cdots & -1 \\
0 & 0 & -2 & -2 & \cdots & -1 \\
0 & 0 & 0 & -2 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & -1
\end{pmatrix}$$
What is determinant of the last matrix?