Consider a sum $$\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j$$ which returns an odd power $n^{2m+1}$ of $n$, for $\ m=0,1,2,...$ given fixed $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$. The coefficients $A_{0,m}, \ A_{1,m},....$ are solutions of system of equations (refer to .txt-file with mathematica codes for $m=1,2,...12$). Coefficients $A_{0,m}, \ A_{1,m},....$ are arranged to the PDF-table. For example, $$\sum_{k=0}^{n-1}\sum_{j=0}^{2}A_{j,2}(n-k)^jk^j=\sum_{k=0}^{n-1}30k^2(n-k)^2+1=n^5$$ Our coefficients are closely related to to coefficients $\beta_{mv}$ from C. Jordan Calculus of Finite Differences, pp. 448 - 450.

**Question**: Could the coefficients $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$ be reached in a recurrent way using $\beta_{mv}$? This question also can be stated as: does there exist such a function $f(\beta_{mv})=A_{v,m}$?

**Additional Question** How exactly are our coefficients $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$ connected with $\beta_{mv}$?

**UPDATE 10-Mar-2019**

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}

Now we want to connect our identity with the discrete convolution of $f_{r,n}(s)$. 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)^s}{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}

It seems that for inteval $[0,n]$ result of $A_{m,r}$ shpuld be different...

**Question:** How to evaluate the coefficients $A_{m,r}$ from last step $(\star)$ ?