Review the main result of mathoverflow.net/questions/297900, that is the identity \begin{equation}\label{f1} n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j, \end{equation} where $A_{m,j}$ is from sequences A302971 and A304042. In this question we discuss the polynomial \begin{equation}\label{f2} \sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \ m\geq0 \ \mathrm{integer} \end{equation} That is generated by the identity \begin{equation}\label{f3} (1.3)\quad\sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \end{equation} where $T=1,2,3...$ and $m\geq0, \ m=\mathrm{const}$. The coefficient $A_{m,j}$ is generated by \begin{equation*}\label{gen_13} A_{m,j}:= \begin{cases} 0, & \mathrm{if } \ j<0 \ \mathrm{or } \ j>m \\ (2j+1)\binom{2j}{j} \sum_{d=2j+1}^{m} A_{m,d} \binom{d}{2j+1} \frac{(-1)^{d-1}}{d-j} B_{2d-2j}, & \mathrm{if } \ 0 \leq j < m \\ (2j+1)\binom{2j}{j}, & \mathrm{if } \ j=m \\ \end{cases} \end{equation*} Derivation of coefficients $A_{m,j}$ is discussed in mathoverflow.net/questions/297900. In particular, the right part of (1.3) returns odd power $2m+1$ of $T\in\mathbb{N}$ when $n=T$ \begin{equation*} T^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(T,k)\cdot T^k \end{equation*}
Detailed derivation of the polynomials, consisting the coefficient $U_m(n,k)$.
Consider the identity discussed in mathoverflow.net/questions/297900, \begin{equation} n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j, \end{equation} Let show a few examples of polynomials $\sum\nolimits_{j\geq0}A_{m,j}k^j(n-k)^j$ for $m=1,2,3$. We denote the part $\sum\nolimits_{j\geq0}A_{m,j}k^j(n-k)^j$ of the left part of equation (1.3) as \begin{equation}\label{f4} D_m(n,k)=\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j \end{equation} Therefore, for $m=1,2,3$ we have corresponding $D_m(n,k)$ as \begin{equation} (1.6)\quad\begin{cases} D_{1}(n,k)=1+6k(n-k), & \\ D_{2}(n,k)=1-0k(n-k)+30k^2(n-k)^2, & \\ D_{3}(n,k)=1-14k(n-k)+0k^2(n-k)^2+140k^3(n-k)^3, & \end{cases} \end{equation} The coefficients in $D_{t=1,2,3}(n,k)$ are the terms of corresponding row of triangle https://oeis.org/A302971. Now, we show an example of generation of polynomials from the right part of (1.3) for $m=1$,
Example 1.
Let be $m=1$, then we rewrite the left hand side of (1.3) as \begin{equation} (1.8)\quad\sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{1,j}k^j(n-k)^j \end{equation} Next, let substitute the polynomial $D_1(n,k)$ from (1.6) into equation (1.8) and let be $T=1,...,10$, then \begin{equation} \sum\limits_{1\leq k \leq T}1+6k(n-k)=\begin{cases} T=1 :& -5 + 6 n \\ T=2 :& -28 + 18 n \\ T=3 :& -81 + 36 n \\ T=4 :& -176 + 60 n \\ T=5 :& -325 + 90 n \\ T=6 :& -540 + 126 n \\ T=7 :& -833 + 168 n \\ T=8 :& -1216 + 216 n \\ T=9 :& -1701 + 270 n \\ T=10:& -2300 + 330 n \end{cases} \end{equation} Coefficients of above polynomials are terms of sequences A028896 and A275709. Let show the case for $m=2$ and $T=1,...,10$, again we recall the corresponding polynomial $D_2(n,k)$ from (1.6) and substitute it into left part of (1.3), \begin{equation} \sum\limits_{1\leq k \leq T}1-0k(n-k)+30k^2(n-k)^2=\begin{cases} T=1 :& 31 - 60 n + 30 n^2 \\ T=2 :& 512 - 540 n + 150 n^2 \\ T=3 :& 2943 - 2160 n + 420 n^2 \\ T=4 :& 10624 - 6000 n + 900 n^2 \\ T=5 :& 29375 - 13500 n + 1650 n^2 \\ T=6 :& 68256 - 26460 n + 2730 n^2 \\ T=7 :& 140287 - 47040 n + 4200 n^2 \\ T=8 :& 263168 - 77760 n + 6120 n^2 \\ T=9 :& 459999 - 121500 n + 8550 n^2 \\ T=10:& 760000 - 181500 n + 11550 n^2 \end{cases} \end{equation} Similarly, let show an example for $m=3$ and $T=1,...,10$, \begin{equation} \sum\limits_{1\leq k \leq T}1-14k(n-k)+0k^2(n-k)^2+140k^3(n-k)^3=\begin{cases} T=1 :& -125 + 406 n - 420 n^2 + 140 n^3\\ T=2 :& -9028 + 13818 n - 7140 n^2 + 1260 n^3\\ T=3 :& -110961 + 115836 n - 41160 n^2 + 5040 n^3\\ T=4 :& -684176 + 545860 n - 148680 n^2 + 14000 n^3\\ T=5 :& -2871325 + 1858290 n - 411180 n^2 + 31500 n^3\\ T=6 :& -9402660 + 5124126 n - 955500 n^2 + 61740 n^3\\ T=7 :& -25872833 + 12182968 n - 1963920 n^2 + 109760 n^3\\ T=8 :& -62572096 + 25945416 n - 3684240 n^2 + 181440 n^3\\ T=9 :& -136972701 + 50745870 n - 6439860 n^2 + 283500 n^3\\ T=10 :& -276971300 + 92745730 n - 10639860 n^2 + 423500 n^3 \end{cases} \end{equation} We can observe, that in every upper example, the resulting polynomial for every $m, T$ has the following form \begin{equation} \sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \end{equation} Therefore, the following question is stated
Question 1. Is there a recurrent that gives the coefficients $U_m(n,k)$ otherwise then by the identity \begin{equation}\label{f3_1} \sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \end{equation} i.e is there any function $F(n,m)$ such that $F(m,n)=U_m(n,k)$ but different from relation (1.3) ?
Above examples could be generated using Mathematica code Um(n,k)_coefficients2.txt. The PDF-analog of this question with extended data of $U_m(n,k)$ coefficients up to $T=40$ is available at this link.