Review the main result of [mathoverflow.net/questions/297900](https://mathoverflow.net/questions/297900/coefficients-in-the-sum-sum-k-0n-1-sum-j-0ma-j-mn-kjkj-n2m1), 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](https://oeis.org/A302971) and [A304042](https://oeis.org/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](https://mathoverflow.net/questions/297900/coefficients-in-the-sum-sum-k-0n-1-sum-j-0ma-j-mn-kjkj-n2m1). 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](https://mathoverflow.net/questions/297900/coefficients-in-the-sum-sum-k-0n-1-sum-j-0ma-j-mn-kjkj-n2m1),
\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](https://oeis.org/A028896) and [A275709](https://oeis.org/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](https://kolosovpetro.github.io/mathematica_codes/Um(n,k)_coefficients2.txt). 
The [PDF-analog](https://kolosovpetro.github.io/pdf/OEIS_Um(n,k)_coefficients.pdf) of this question with extended data of $U\_m(n,k)$ coefficients up to $T=40$ is available at this [link](https://kolosovpetro.github.io/pdf/OEIS_Um(n,k)_coefficients.pdf).



Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$
First off, as I explained in the comments, the identity (1.3) should contain $U\_m(T,k)$ rather than $U\_m(n,k)$ (the latter does not make any sense), and so the correct identity (1.3) (for polynomials in $n$) states:
$$(1.3)\quad\sum\_{k=1}^T\sum\_{j=0}^m A\_{m,j}k^j(n-k)^j\equiv \sum\limits\_{0\leq k \leq m}(-1)^{m-k}U\_m(T,k)\cdot n^k.$$
Also, to avoid confusion, it is important to notice that while coefficients $A\_{m,j}$ come from the [previous question](https://mathoverflow.net/q/297900), there their indices are interchanged (i.e., $A\_{j,m}$ instead of $A\_{m,j}$).


Said that, it is not hard to express the polynomials $U\_m(T,k)$ in terms of $A\_{m,j}$ and Bernoulli numbers. To do so, let's expand the binomial $(n-k)^j$ in the l.h.s. of (1.3) and change of the order of summation:
\begin{split}
\sum\_{k=1}^T\sum\_{j=0}^m A\_{m,j}k^j(n-k)^j &= \sum\_{k=1}^T\sum\_{j=0}^m A\_{m,j}k^j\sum\_{t=0}^j\binom{j}{t}n^t(-1)^{t-t}k^{j-t}\\
&=\sum\_{t=0}^m n^t \sum\_{k=1}^T\sum\_{j=t}^m \binom{j}{t}A\_{m,j}k^{2j-t}(-1)^{j-t}.
\end{split}
Now, taking the coefficient of $n^t$ in (1.3) gives:
$$U\_m(T,t) = (-1)^m \sum\_{k=1}^T\sum\_{j=t}^m \binom{j}{t}A\_{m,j}k^{2j-t}(-1)^j.$$


From this formula it may be not immediately clear why $U\_m(T,t)$ represent polynomials in $T$. However, this can be seen if we change the summation order again and use Faulhaber's formula to obtain:
$$U\_m(T,t) = (-1)^m \sum\_{j=t}^m \binom{j}{t}A\_{m,j} \frac{(-1)^j}{2j-t+1}\sum\_{\ell=0}^{2j-t} \binom{2j-t+1}{\ell}B\_{\ell}T^{2j-t+1-\ell}.$$
Introducing $k=2j-t+1-\ell$, we further get the formula:
$$U\_m(T,t) = (-1)^m \sum\_{k=1}^{2m-t+1} T^k \sum\_{j=t}^m \binom{j}{t}A\_{m,j} \frac{(-1)^j}{2j-t+1}\binom{2j-t+1}{k}B\_{2j-t+1-k},$$
which allows easily compute the coefficient of $T^k$ in $U\_m(T,t)$ for each $k$.


**N.B.** In the above formulae, we assume that $B\_1=+\frac{1}{2}$.


