Given the invertible matrix ${\bf A} \in {\Bbb R}^{n \times n}$, we have the linear system ${\bf A} {\bf x} = {\bf 1}_n$, whose (unique) solution is denoted by ${\bf x}_0 := {\bf A}^{-1} {\bf 1}_n$. We would like to compute its $\infty$-norm, $\| \, {\bf x}_0 \|$. Unless $\bf A$ has plenty of structure, it might be too ambitious to find $\| \, {\bf x}_0 \|$ in closed-form.
Let $\bf A$ be given by the entrywise product${\bf A} := {\bf P}_n \odot {\bf H}_n ({\boldsymbol{\alpha}})$, where ${\bf P}_n$ is a symmetric $n \times n$Pascal matrix and ${\bf H}_n : {\Bbb R}^{2n-1} \to {\Bbb R}^{n \times n}$ is a Hankel matrix-builder. Using the entrywise inverse, the linear system ${\bf A} {\bf x} = {\bf 1}_n$ can be rewritten as ${\bf H}_n ({\boldsymbol{\alpha}}) \, {\bf x} = \left( \, {\bf P}_n^{\odot -1}\right) {\bf 1}_n =: {\bf b}$ and solved via forward substitution or some variation of Levinson recursion.