Determinant of a Toeplitz integer sequence I have a conjecture that
$$\det \mathrm{Toep}(0,\dots,N)= (-1)^N N2^{N-1},$$
where
$\mathrm{Toep}(0,\dots,N)$ is the symmetric Toeplitz matrix with first column is equal to $(0,\dots,N)$.
Does anyone have an idea about how to prove that?
 A: One possible way is by using Schur's complement in the first $N$ rows and then induction. $N = 2$ is easily verifiable. Now let $A_N$ be your matrix, given $N$. Then $A_{N-1}$ is precisely the $N \times N$ sub matrix of $A_N$ formed by the first $N$ rows and $N$ columns . Schur's complement tells us that:
$$\det A_N = \det A_{N-1} \det(- v A_{N-1}^{-1} v^t) = (-1) v A_{N-1}^{-1} v^t \det A_{N-1} $$ where $v = (N,N-1,\ldots,1)$ is the last row of $A_N$, excluding the zero. Now to calculate $A_{N-1}^{-1} v^t$ let us do a trick. If $A_{N-1}^{-1} v^t = x^t$, then $x^t$ is the solution to the linear system $A_{N-1} x^t = v^t$. Since the first and last elements of each row of $A_{N-1}$ add to $N-1$, it is not hard to see that $$x = \left(\frac{1}{N-1},0,0,\ldots,0,\frac{N}{N-1}\right)$$ does the job, therefore $v A_{N-1}^{-1} v^t = 2N/N-1$. Supposing that your conjecture is valid until $N-1$ then
$$\det A_N =  (-1) v A_{N-1}^{-1} v^t \det A_{N-1} = (-1) \frac{2N}{N-1} (-1)^{N-1} (N-1) 2^{N-2},$$
which is the desired formula.
