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 $(\star)$ ?
Best Answer
EDIT 2018-04-16: Formulae are corrected.
I'm not sure about connection with $\beta_{mv}$, but we can obtain a recurrence formula for $A_{j,m}$ as follows.
First let us fix the unused values $A_{j,m}=0$ for $j<0$ or $j>m$, so we won't need to care about the summation range for $j$.
Expanding $(n-k)^j$ and using Faulhaber's formula, we get $$\sum_{k=0}^{n-1} (n-k)^j k^j = \sum_{k=0}^{n-1} \sum_i \binom{j}i n^{j-i} (-1)^i k^{i+j}$$ $$=\sum_{i} \binom{j}{i} n^{j-i} \frac{(-1)^i}{i+j+1} \left[ \sum_t \binom{i+j+1}t B_t n^{i+j+1-t} - B_{i+j+1}\right]$$ $$=\sum_{i,t} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t B_t n^{2j+1-t} - \sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} B_{i+j+1} n^{j-i}$$ where $B_t$ are Bernoulli numbers.
Now, we notice that $$\sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t =\begin{cases} \frac{1}{(2j+1)\binom{2j}j}, & \text{if } t=0;\\ \frac{(-1)^j}{t}\binom{j}{2j-t+1}, & \text{if } t>0. \end{cases} $$ In particular, the last sum is zero for $0<t\leq j$.
Hence, introducing $\ell=2j+1-t$ and $\ell=j-i$, respectively, we get $$\sum_{k=0}^{n-1} (n-k)^j k^j = \frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + \sum_{\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell} - \sum_{\ell} \binom{j}{\ell} \frac{(-1)^{j-\ell}}{2j+1-\ell} B_{2j+1-\ell}n^{\ell}$$ $$=\frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + 2\sum_{\text{odd }\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell}.$$
Using the definition of $A_{j,m}$, we obtain the following identity for polynomials in $n$: $$(\star)\qquad\sum_{j} A_{j,m} \frac{1}{(2j+1)\binom{2j}j}n^{2j+1} + 2 \sum_{j,\text{ odd }\ell} A_{j,m} \binom{j}{\ell} \frac{(-1)^j}{2j+1-\ell} B_{2j+1-\ell}n^{\ell} \equiv n^{2m+1}.$$
Taking the coefficient of $n^{2m+1}$ in $(\star)$, we get $A_{m,m} = (2m+1) \binom{2m}{m},$ and taking the coefficient of $x^{2d+1}$ for an integer $d$ in the range $m/2 \leq d < m$, we get $A_{d,m} = 0$.
Taking the coefficient of $n^{2d+1}$ in $(\star)$ for $m/4 \leq d < m/2$, we get $$A_{d,m} \frac{1}{(2d+1)\binom{2d}{d}} + 2 (2m+1) \binom{2m}{m} \binom{m}{2d+1} \frac{(-1)^m}{2m-2d} B_{2m-2d} = 0,$$ i.e. $$A_{d,m} = (-1)^{m-1} \frac{(2m+1)!}{d!d!m!(m-2d-1)!}\frac{1}{m-d} B_{2m-2d}.$$
Continue similarly, we can express $A_{d,m}$ for each integer $d$ in the range $m/2^{s+1}\leq d< m/2^s$ (iterating consecutively $s=1,2,\dots$) via previously determined values of $A_{j,m}$ as follows: $$A_{d,m} = (2d+1)\binom{2d}{d} \sum_{j\geq 2d+1} A_{j,m} \binom{j}{2d+1} \frac{(-1)^{j-1}}{j-d} B_{2j-2d}.$$ The same formula holds also for $d=0$.