This question is part of series of the questions, as follows:

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$,

Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$.

Consider the Faulhaber's identities (see D. Knuth, "Johann Faulhaber and Sums of Powers", Knuth, p. 9) for odd powers $n^{2m+1}, \ m\in\mathbb{N}_0$ \begin{equation*} (\diamond)\quad\quad\begin{cases} n^1 &= \binom{n}{1}\\ n^3 &= 6\binom{n+1}{3}+\binom{n}{1}\\ n^5 &= 120\binom{n+2}{5}+30\binom{n+1}{3}+\binom{n}{1}\\ &\vdots\\ n^{2m-1} &= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1} \end{cases} \end{equation*} Introduce the $r$ to the lower index of binomial coefficient of $(\diamond)$, then \begin{equation*} (\Box)\quad\begin{cases} r=1: \ \Delta n^1 &= \binom{n}{1-r}=\binom{n}{0}\\ r=1: \ \Delta n^3 &= 6\binom{n+1}{3-r}+\binom{n}{1-r}=6\binom{n+1}{2}+\binom{n}{0}\\ r=1: \ \Delta n^5 &= 120\binom{n+2}{5-r}+30\binom{n+1}{3-r}+\binom{n}{1-r}=120\binom{n+2}{4}+30\binom{n+1}{2}+\binom{n}{0}\\ &\vdots\\ r=1: \ \Delta n^{2m-1} &= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-r} \end{cases} \end{equation*} Continuing similarly, for every $r>0$ we get $\Delta^r n^{2m-1}$.

Therefore, by the Faulhaber's identity $\Delta n^{3}=6\binom{n+1}{2}+\binom{n}{0}$, the perfect cube in $T$ is: \begin{equation*} (1.1)\quad\quad T^3=\sum\limits_{k=0}^{T-1}\Delta(T^3)(k)=\sum\limits_{k=0}^{T-1}6\binom{k+1}{2}+\binom{k}{0}. \end{equation*} Let's rewrite the expression (1.1) in extended view, taking to attention the identity $\tbinom{k+1}{1}=1+2+\cdots+(k+1)$, thus \begin{equation*} (1.2)\quad\quad T^3=(1+6\cdot0)+(1+6\cdot0+6\cdot1)+\cdots+(1+6\cdot0+\cdots+6\cdot(T-1)) \end{equation*} Factorising expression (1.2), we get \begin{equation*} (1.3)\quad\quad T^3=T+(T-0)\cdot6\cdot0+(T-1)\cdot6\cdot1+\cdots+(T-(T-1))\cdot6\cdot(T-1) \end{equation*} Applying the sigma notation on the expression (1.3), we have \begin{equation*} (1.4)\quad\quad T^3=T+\sum\limits_{k=0}^{T-1}6k(T-k) = \sum\limits_{k=0}^{T-1}6k(T-k)+1 \end{equation*} Therefore, we have reached the generating function $6k(T-k)+1$ the case for $m=1$ in the question

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$.

The problem: Can we find similar to (1.3), (1.4) expressions for powers $2m+1, \ m>1$, only by means of binomial identities, extended form of the sum of finite difference (in sense of Faulhaber, (1.0), $(\Box)$), of corresponding power and its extended form ?

And, If we can, are they equal to $\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$ ?

We assume that pattern $n^{2m+1}=\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$ is unknown.

In other words, the problem asks to pass through the case $r=1$ in $(\Box)$: $$(\star)\quad \quad \Delta n^{2m+1}=\Sigma_{k=1}^{m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-r}$$ to (in sense of $(\star)$): $$n^{2m+1}=\sum_{k=0}^{n-1}\Delta n^{2m+1}(k)=\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$$.

Please, note that formulae notation in present question may be different in some places from notation in question Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$. More details on this question, as well as detailed derivations are available at this link.

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