In a computation of Pontryagin-numbers of certain manifolds (see the appendix of https://arxiv.org/pdf/2109.10306.pdf for more context) we came across the following elementary problem:

Consider the matrix $A^m = (A^m_{ij})_{i=6\dots m,\ j=0\dots\lfloor\frac{m}{2}\rfloor}$ with $$A^m_{ij}:=\sum_{n=0}^{\lfloor{\frac i2}\rfloor} 2^{m-n}\ i^{n}\ \binom{j}{n}\binom{m-2j}{i-(j+n)}$$ Question: For for $m\ge13$, does this matrix have full rank $= \lfloor\frac{m}{2}\rfloor+1$?

The matrix $A^m$ looks intimidating, but for $j>i$, we have $\binom{m-2j}{i-(j+n)}=0$ and hence $A^m_{ij}=0$. Therefore, $A^m$ has the following form, where the asterisks represent non-zero entries. $$A = \begin{pmatrix} * &\dots & * & 0 & \dots & 0\\ & & & \ddots & \ddots & \vdots\\ \vdots & & & & * & 0\\ & & & & & * \\ \vdots & & & & & \vdots \\ * & & \dots & \dots & & * \end{pmatrix}$$

In the first row, there are $7$ non-zero entries, so the rank of $A$ is at least $\lfloor\frac{m}{2}\rfloor-5$.

Computer calculations have shown, that $A^m$ has actually full rank for $m\le500$. Even more, the square matrix given by the rows $i=6\dots\lfloor\frac{m}{2}\rfloor+6$ is invertible.