For an integer $n$, let $T_n$ be the $(n+1) \times (n+1)$ Toeplitz matrix built on the $2n$th row of Pascal's triangle, i.e., its $(i,j)$ entry equals $\binom{2n}{n+i-j}$. For example, $$ T_2 =\begin{pmatrix} 6 & 4 & 1 \\ 4 & 6 & 4 \\ 1 & 4 & 6 \end{pmatrix}. $$
Numerical experiments indicate that $T_n$ is always positive definite. How to prove this claim?
Since $$ 2^{2n} \cos^{2n} t = \big(e^{it} + e^{-it}\big)^{2n} = \sum_{k=-n}^{n}{2n \choose n+k} e^{2tki}, $$ we have $$ {2n \choose n+k} = \frac{2^{2n-1}}{\pi} \int_{-\pi}^\pi e^{-2tki}\cos^{2n}t\, dt. $$ Then, for any $z_0,\dots,z_n \in \mathbb C$, $$ \sum_{k,l=0}^n {2n \choose n+k-l} z_k \overline{z_l} = \frac{2^{2n-1}}{\pi} \int_{-\pi}^\pi \sum_{k,l=0}^n e^{-2t(k-l)i}z_k \overline{z_l} \cos^{2n}t\, dt \\ = \frac{2^{2n-1}}{\pi} \int_{-\pi}^\pi\left|\sum_{k=0}^n e^{-2tki}z_k \right| ^2 \cos^{2n}t\, dt\ge 0. $$ This proves that $T_n$ is positive definite.