Let be a power function $f_{r,n}(s)$ defined as follows \begin{equation}\label{f1} f_{r,n}(s)= \begin{cases} s^r, \ &0\leq s\leq n,\\ 0, \ &\mathrm{otherwise}. \end{cases} \end{equation}
The following indentity involving convolution of $f_{r,n}(s)$ holds
$$(n+1)^{2m+1}=\sum_{r=0}^{m}A_{m,r}(f_{r,n}*f_{r,n})[n]$$
Since the $f_{r,n}(s)$ defined only on the interval $[0,n]$ of real set, we can conclude the following identity \begin{equation}\label{der1} (f_{r,n}*f_{r,n})[n]\equiv\sum_{m=0}^{n}m^r(n-m)^r \end{equation} Definition of $A_{m,j}$ coefficients says that \begin{equation} (n+1)^{2m+1}=\sum_{r=0}^{m}A_{m,r}(f_{r,n}*f_{r,n})[n]\equiv\sum_{r=0}^{m}A_{m,r}\sum_{k=0}^{n}k^r(n-k)^r \end{equation} Expanding $\sum_{k=0}^{n}k^r(n-k)^r$ and applying Faulhaber's formula, we get \begin{equation}\label{proof2} \begin{split} \sum_{k=0}^{n}k^r(n-k)^r &=\sum_{k=0}^{n} k^r \sum_{j} (-1)^j\binom{r}{j} n^{r-j}k^{j}=\sum_{j} (-1)^j\binom{r}{j} n^{r-j}\left(\sum_{k=0}^{n}k^{r+j}\right)\\ &=\sum_{j} \binom{r}{j} n^{r-j}\frac{(-1)^j}{r+j+1}\left[\sum_{s}\binom{r+j+1}{s}B_{s}n^{r+j+1-s}\right]\\ &=\sum_{j,s}\binom{r}{j}\frac{(-1)^j}{r+j+1}\binom{r+j+1}{s}B_{s}n^{2r+1-s}, \end{split} \end{equation} where $B_s$ are Bernoulli numbers. Now, we notice that \begin{equation}\label{proof3} \sum_{j} \binom{r}{j}\frac{(-1)^j}{r+j+1}\binom{r+j+1}{s} =\begin{cases} \frac{1}{(2r+1)\binom{2r}r}, & \text{if } s=0;\\ \frac{(-1)^r}{s}\binom{r}{2r-s+1}, & \text{if } s>0. \end{cases} \end{equation} In particular, the last sum is zero for $0<s\leq r$. Therefore, expression (2.3) takes the form \begin{equation} \sum_{k=0}^{n}k^r(n-k)^r=\frac{1}{(2r+1)\binom{2r}r}n^{2r+1}+\sum_{s>0}\frac{(-1)^s}{s}\binom{r}{2r-s+1}B_{s}n^{2r+1-s} \end{equation} Using the definition of $A_{m,j}$, we obtain the following identity for polynomials in $n$ \begin{equation} (\star)\sum_{r}\frac{1}{(2r+1)\binom{2r}r}n^{2r+1}A_{m,r}+\sum_{r,s>0}\frac{(-1)^s}{s}\binom{r}{2r-s+1}B_{s}n^{2r+1-s}A_{m,r}\equiv (n+1)^{2m+1} \end{equation}
Question: How to evaluate the coefficients $A_{m,r}$ from last step $(\star)$ ?
This problem is closely related to the question https://mathoverflow.net/questions/297900/, but still solution is unclear for me