I want to prove that the matrix $A$ of order $n$ with $(a_{ij})=\frac{2^{i+j}-2^{|i-j|}}{3}$ admits a Cholesky factorization. $A$ is trivially symmetric, but when trying to show it is positive definite I have had some issues. First, we prove $det(A)>0$ by induction:
Case $n=1$:
$A = \left(\frac{2^{2}-2^{0}}{3}\right)$, so $det(A)=1>0$.
We assume $det(A)>0$ for all matrices of order $n$.
Then, for order $n+1$ we develop the determinant from the first column, assuming $n$ is even: $$det(A) = (-1)^{1+1}a_{11}|A_{11}|+(-1)^{2+1}a_{21}|A_{21}|+\dots+(-1)^{n+1+1}a_{n+1,1}|A_{n+1,1}|$$ If we develop the $a_{ij}$ terms we have that: $$det(A)=\sum_{i=0}^n (-2)^i|A_{i+1,1}|$$
Now, we have to prove that $2^0|A_{11}|-2|A_{21}|+2^2|A_{31}|-\dots+2^n|A_{n+1,1}|>0$. But at this point I guess it is not enough that the $|A_{ij}|>0$ by induction hypothesis, because the sum is alternating. I have tried more things, such as applying the Cauchy-Schwarz inequality to the expression:
$$ \left( \begin{array}{} 2^0 \\ 2^2 \\ \vdots \\ 2^n \end{array} \right) \left( \begin{array}{} |A_{11}|, |A_{31}|, \dots, |A_{n+1,1}| \end{array} \right) > 2\left( \begin{array}{} 2^0 \\ 2^2 \\ \vdots \\ 2^{n-2} \end{array} \right)\left( \begin{array}{} |A_{21}|, |A_{41}|, \dots, |A_{n1}| \end{array} \right) $$
But I end up with a more difficult to prove inequality, because I can't find the proper relations between the cofactors of $A$.
Use multilinearity of the determinant. Let $A_n$ denote the $n \times n$ matrix which you defined. Subtract $2^{i-1}$ times the first row from the $i$-th row of $A_{n+1}$ for each $i \ge 2$. The result is $$\begin{bmatrix} 1 & \ast \\ 0 & A_n \end{bmatrix}.$$ Thus $\det(A_{n+1}) = \det(A_n) = 1$ by induction.