In this paper, we derive and prove, by means of Binomial theorem and Faulhaber's formula, the following identity between $m$-order polynomials in \(T\)
\(\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k=T^{2m+1}, \ \ell=T\in\mathbb{N}.\)
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\hypersetup{
pdftitle={Another Power Identity involving Binomial Theorem and Faulhaber's formula},
pdfsubject={Mathematics, Number Theory, Combinatorics},
pdfauthor={Petro Kolosov},
pdfkeywords={Faulhaber's formula, Faulhaber's theorem, Binomial Theorem, Binomial coefficient, Binomial distribution,
Binomial identities, Power Sums, Finite differences}
}
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%--------Meta Data: Fill in your info------
%Го?споди, Иису?се Христе?, Бо?же наш, поми?луй нас
\author[Petro Kolosov]{Petro Kolosov}
\email{kolosovp94@gmail.com}
\keywords{Faulhaber's formula, Faulhaber's theorem, Binomial Theorem, Binomial coefficient, Binomial distribution,
Binomial identities, Power Sums, Finite differences}
\urladdr{https://kolosovpetro.github.io}
\subjclass[2010]{11C08 (primary), 41A10 (secondary)}
\title{Another Power Identity involving Binomial Theorem and Faulhaber's formula}
\begin{document}
\begin{abstract}
In this paper, we derive and prove, by means of Binomial theorem and Faulhaber's formula, the following identity
between $m$-order polynomials in $T$
$$\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k=T^{2m+1}, \ \ell=T\in\mathbb{N}.$$
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction and Main results}\label{intro}
The following identity holds in $m$-order polynomials $P_m(\ell,T)$
\begin{equation}\label{f1}
P_m(\ell,T) := \sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k
\Bigg|_{\ell=T\in\mathbb{N}}=T^{2m+1},
\end{equation}
where $\ell\in\mathbb{N}_1,m\in\mathbb{N}_1$ and $T\in\mathbb{R}$ and $U_m(\ell,k), \ A_{m,j}$ are coefficients dependent on $\ell$
and $m$, respectively.
The term $(-1)^{m-k}$ appears in r.h.s of the expression (\hyperref[f1]{1.1})
by means of sign-changing inside $\sum_{j=0}^m A_{m,j}k^j(T-k)^j$ by
Binomial expansion of $(T-k)^j$.
We start our discussion concerning the identity (\hyperref[f1]{1.1}) from the derivation of the partial case of the polynomial $P_m(\ell,T)$
for $m=1$, by means of Faulhaber's identity for $n^3$. The terms of the of the polynomial
$P_1(T,T)=\sum_{k=1}^{T}\sum_{j=0}^1 A_{1,j}k^j(T-k)^j=T^3$ over $k$ produce the $T$-th
row of the sequence \href{https://oeis.org/A287326}{A287326}, starting from $k=1$, \cite{3}.
To derive the partial case of the polynomial $P_m(\ell,T)$
for $m=1$, let's consider the Faulhaber's identities \cite{1} for odd powers $n^{2m+1}, \ m\in\mathbb{N}_0$
\begin{equation}\label{knuth1}
\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}
The coefficients $T(m,k)$ in these formula are related to what Riordan \cite{22} has called
central factorial numbers of the second kind. In his notation,
\begin{equation*}
x^m=\sum\limits_{1\leq k \leq m}T(m,k)x^{[k]}, \ x^{[k]}=x(x+\tfrac{k}{2}-1)(x+\tfrac{k}{2}-2)\cdots(x+\tfrac{k}{2}+1)
\end{equation*}
The coefficients $T(2m,2k)$ are always integers, because the $x^{[k+2]}=x^{[k]}(x^2/k^4)$ implies recurrence
\begin{equation*}
T(2m+2,2k)=k^2T(2m,2k)+T(2m,2k-2).
\end{equation*}
Central factorial numbers of the second kind can be calculated as follows:
\begin{equation}\label{knuth2}
(2k-1)!T(2n,2k)=\frac{1}{r}\sum_{j=0}^{r}(-1)^j\binom{2r}{j}(r-j)^{2n},
\end{equation}
where $r=n-k+1$. The formula (\hyperref[knuth2]{1.3}) was derived by Peter Luschny in \cite{27}.
The Faulhaber's formula and polynomial (\hyperref[f1]{1.1}) could be connected the following way
\begin{equation*}
\begin{split}
n^{2m+1}
&=\sum_{k=1}^{m+1}(2k-1)!T(2m+2,2k)\binom{n+k-1}{2k-1}\\
&=\sum_{k=1}^{n}\sum_{j=0}^m A_{m,j}k^j(n-k)^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(n,k)\cdot n^k\\
n^{2m-1}
&=\sum_{k=1}^{m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1}\\
&=\sum_{k=1}^{n}\sum_{j=0}^{m-1} A_{m-1,j}k^j(n-k)^j=\sum_{k=0}^{m-1}(-1)^{m-1-k}U_{m-1}(n,k)\cdot n^k.
\end{split}
\end{equation*}
The forward finite difference of odd power could be reached introducing the $r\in\mathbb{N}_1$ to the lower index of binomial coefficient of (\hyperref[knuth1]{1.2}) the following way
\begin{equation*}
\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}&= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-1}
\end{cases}
\end{equation*}
By the dentity $\Delta f(x)=\nabla f(x+1), \ h=1$, backward differences could be reached as well,
\begin{equation*}
\begin{cases}
r=1: \ \nabla (n+1)^1 = \binom{n}{1-r}&=\binom{n}{0}\\
r=1: \ \nabla (n+1)^3 = 6\binom{n+1}{3-r}+\binom{n}{1-r}&=6\binom{n+1}{2}+\binom{n}{0}\\
r=1: \ \nabla (n+1)^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: \ \nabla (n+1)^{2m-1} = \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-r}&= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-1}
\end{cases}
\end{equation*}
Continuing similarly, for every $r\geq 1$ we get $\Delta^r n^{2m-1}$ and $\nabla^r n^{2m-1}$. To derive the polynomial $P_m(T,T)=T^{2m+1}, \ T\in \mathbb{N}_1$, for $m=1$, recall the Faulhaber's identity $\Delta n^{3}=\nabla (n+1)^3 =6\binom{n+1}{2}+\binom{n}{0}$, thus, the perfect
cube in $T$ is:
\begin{equation}\label{f6}
\begin{split}
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}\\
=&\sum\limits_{k=1}^{T}\nabla (T+1)^3(k)=\sum\limits_{k=1}^{T}6\binom{k+1}{2}+\binom{k}{0}.
\end{split}
\end{equation}
Let's rewrite the expression (\hyperref[f6]{1.4}), taking to the attention the following identity $\tbinom{k+2}{2}=1+2+\cdots+(k+2)$, thus
\begin{equation}\label{f7}
\begin{split}
T^3
&=(1+6\cdot0)+(1+6\cdot0+6\cdot1)+\cdots+(1+6\cdot0+\cdots+6\cdot(T-1))\\
&=(1+6\cdot1)+(1+6\cdot1+6\cdot2)+\cdots+(1+6\cdot1+\cdots+6\cdot T)
\end{split}
\end{equation}
Factorising the expression (\hyperref[f7]{1.5}) we get
\begin{equation}\label{f8}
\begin{split}
T^3
&=T+(T-0)\cdot6\cdot0+(T-1)\cdot6\cdot1+\cdots+(T-(T-1))\cdot6\cdot(T-1)\\
&=T+(T-1)\cdot6\cdot1+(T-2)\cdot6\cdot2+\cdots+(T-T)\cdot6\cdot T
\end{split}
\end{equation}
Let's apply a compact sigma notation on the expression (\hyperref[f8]{1.6}), we have
\begin{equation}\label{cubbe}
\begin{split}
T^3
&=T+\sum\limits_{k=0}^{T-1}6k(T-k) = \sum\limits_{k=0}^{T-1}6k(T-k)+1\\
&=T+\sum\limits_{k=1}^{T}6k(T-k) = \sum\limits_{k=1}^{T}6k(T-k)+1=P_1(T,T),
\end{split}
\end{equation}
see (\hyperref[f1]{1.1}) for $P_1(T,T)$.
Although, the expression (\hyperref[cubbe]{1.7}) is derived, essentially, from identity in finite differences of cubes, such as $T^3=\sum\nolimits_{k=0}^{T-1}\Delta T^3(k)$ and $T^3=\sum\nolimits_{k=1}^{T}\nabla (T+1)^3(k)$, the finite differences $\Delta T^3(k)\neq6k(T-k)+1$ and $\nabla (T+1)^3(k)\neq6k(T-k)+1$.
Over the manuscript we review the $P_m(\ell, T)$ in sense of backward difference $\nabla(T+1)^{2m+1}$, i.e we use the summation limits over $k$ as $1\leq k\leq \ell$ in (\hyperref[f1]{1.1}).
Therefore, we have reached the generating function of \href{https://oeis.org/A287326}{A287326} in r.h.s of (\hyperref[cubbe]{1.7}).
The corresponding coefficients $A_{m,j}$ in the definition $\sum_{j=0}^1A_{1,j}k^j(T-k)^j=1+6k(T-k)$ of generating
function are: $A_{1,0}=1, \ A_{1,1}=6$. Note that by definition the polynomials $\sum_{j=0}^mA_{m,j}k^j(T-k)^j$
should be displayed starting from the $A_{m,0}k^0(T-k)^0$, we will denote these polynomials starting from the $m$-th term,
as the order of summation doesn't change it's result. Let's construct the triangle \href{https://oeis.org/A287326}{A287326},
every $(\ell=T)$-th row sum starting from $k=1$ gives the terms of polynomial
$P_1(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^1 A_{1,j}k^j(T-k)^j=\sum_{k=1}^{\ell}6k(T-k)+1$, starting from $k=1$. We review the triangle \href{https://oeis.org/A287326}{A287326} in context of the partial case of polynomial $P_m(\ell, T), \ m=1, \ \ell=T\in\mathbb{N}_0$
\begin{table}[H]
\centering
\begin{tabular}{lrrrrrrrrrrr}
$\ell=T=0$& & & & & & 1 & & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=1$& & & & & 1 & & 1 & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=2$& & & & 1 & & 7 & & 1 & & & \\\noalign{\smallskip\smallskip}
$\ell=T=3$& & & 1 & & 13 & & 13 & & 1 & & \\\noalign{\smallskip\smallskip}
$\ell=T=4$& & 1 & & 19 & & 25 & & 19 & & 1 & \\\noalign{\smallskip\smallskip}
$\ell=T=5$&1 & & 25 & & 37 & & 37 & & 25 & & 1 \\\noalign{\smallskip\smallskip}
\end{tabular}
\caption{Triangle generated by the polynomial $\sum_{j=0}^1A_{1,j}k^j(T-k)^j=6k(T-k)+1, \ 0\leq k\leq \ell=T\in\mathbb{N}_0$,
sequence \href{https://oeis.org/A287326}{A287326} in OEIS.
Summation of the $(\ell=T)$-th row terms from $k=1$ gives $P_1(T,T)=T^3$, see (\hyperref[f1]{1.1}).} \label{fig_1}
\end{table}
The sum of the $(\ell=T)$-th row terms of \hyperref[fig_1]{table 1} starting from $k=1$ generates the partial case of (\hyperref[f1]{1.1})
for $m=1$ and $\ell=T\in\mathbb{N}$, that is $P_1(T,T)=T^3$. Binomial distribution of the row terms of \hyperref[fig_1]{table 1} can be easily proven
by reviewing of its generating function $\sum_{j=0}^1 A_{1,j}k^j(T-k)^j=6k(T-k)+1=6kT-6k^2+1$, which is parabolic for every
given variable $T$, and, therefore, is symmetrical over $\tfrac{T}{2}$.
Let's show a few properties of the generating function of $P_1(T,T)$, define the term $D_1(T,k)$
\begin{equation}\label{f5}
D_1(T,k):=\sum\limits_{1\leq j\leq0}A_{1,j}k^j(T-k)^j=6k(T-k)+1
\end{equation}
The following properties hold in terms of $D_1(T,k)$
\begin{enumerate}
\item Finite difference property, $\Delta(n^3)=D_1\left(\frac{n^2+n+2}{2}, 1\right)$
\item Binomial transform property, $D_1(n+1,k)=2D_1(n,k)-D_1(n-1,k)$
\item Symmetry, $D_1(n,k)=D_1(n, n-k), \ 0\leq k\leq n$
\end{enumerate}
Below we show initial ten polynomials $P_m(\ell,T)=\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k$, generated by the partial case of (\hyperref[f1]{1.1}) for $m=1$ over $\ell, \ 0\leq\ell\leq 10$:
\begin{equation}\label{u_1}
\begin{split}
P_1(\ell, T) &= \sum_{k=1}^{\ell}\sum_{j=0}^1 A_{1,j}k^j(T-k)^j =
\sum_{k=1}^{\ell}6k(T-k)+1=\sum_{k=0}^{1}(-1)^{1-k}U_1(\ell,k)\cdot T^k \\
&=\begin{cases}
\ell=1 : -5 + 6 T &= -U_1(1,0)\cdot T^0+U_1(1,1)\cdot T^1=P_1(1,T)\\
\ell=2 : -28 + 18 T &= -U_1(2,0)\cdot T^0+U_1(2,1)\cdot T^1=P_1(2,T)\\
\ell=3 : -81 + 36 T &= -U_1(3,0)\cdot T^0+U_1(3,1)\cdot T^1=P_1(3,T)\\
\ell=4 : -176 + 60 T &= -U_1(4,0)\cdot T^0+U_1(4,1)\cdot T^1=P_1(4,T)\\
\ell=5 : -325 + 90 T &= -U_1(5,0)\cdot T^0+U_1(5,1)\cdot T^1=P_1(5,T)\\
\ell=6 : -540 + 126 T &= -U_1(6,0)\cdot T^0+U_1(6,1)\cdot T^1=P_1(6,T)\\
\ell=7 : -833 + 168 T &= -U_1(7,0)\cdot T^0+U_1(7,1)\cdot T^1=P_1(7,T)\\
\ell=8 : -1216 + 216 T &= -U_1(8,0)\cdot T^0+U_1(8,1)\cdot T^1=P_1(8,T)\\
\ell=9 : -1701 + 270 T &= -U_1(9,0)\cdot T^0+U_1(9,1)\cdot T^1=P_1(9,T)\\
\ell=10: -2300 + 330 T &= -U_1(10,0)\cdot T^0+U_1(10,1)\cdot T^1=P_1(10,T)
\end{cases}\\
&=T^3 \ \mathrm{as} \ T\rightarrow \ell.
\end{split}
\end{equation}
The coefficients $U_1(\ell,k), \ 0\leq k\leq 1$ in (\hyperref[u_1]{1.9}) are terms of the sequence \href{https://oeis.org/A320047}{A320047}, \cite{28}.
One would ask, "Why to show the identity (\hyperref[f1]{1.1}) in terms of $(\ell=T)$-th row of \hyperref[fig_1]{table 1} we used the condition
$\ell=T\in\mathbb{N}_0$, but to show the identity (\hyperref[f1]{1.1}) in terms of (\hyperref[u_1]{1.9}) we apply the condition $T\to\ell$,
and $\ell\in\mathbb{N}$?"
- We used the condition $T\to\ell$, and $\ell\in\mathbb{N}$ to show the identity (\hyperref[f1]{1.1}) in terms of (\hyperref[u_1]{1.9}) as,
by definition $T\in\mathbb{R}$
in (\hyperref[f1]{1.1}) and the variable $\ell\in\mathbb{N}$, since the coefficient $U_m(\ell,k)$ is dependent on $\ell$, and,
therefore, $\ell$ is
always $\mathbb{N}_1$, so if we speak about approximation of monomial $T^{2m+1}$ by polynomial $\sum_{k=0}^{1}(-1)^{1-k}U_1(\ell,k)\cdot T^k$
in some particular point $\ell\in\mathbb{N}$, we say that $T\to\ell$.
The binomials, listed in (\hyperref[u_1]{1.9}) are linear approximations of monomial $T^{3}$ in some neighborhood of the
point $\ell\in\mathbb{N}$ and $T\rightarrow \ell$. The following figure graphically shows the
binomials $\sum_{k=0}^{1}(-1)^{1-k}U_1(\ell,k)\cdot T^k, \ \ell=1,2,3$ among the cubic curve in $T$
\begin{figure}[H]
\centering
\includegraphics[width=400px, keepaspectratio]{figure_n_3.eps}
\captionof{figure}{Linear approximations of the monomial $T^{3}$ by binomials $P_1(1,T), \ P_1(2,T), \ P_1(3,T)$ in some neighborhood
of the point $\ell=1,2,3$ with $T\to\ell$, see (\hyperref[u_1]{1.11}).}
\end{figure}
Now, we have derived and discussed the pattern \href{https://oeis.org/A287326}{A287326}, generated by the $\sum_{j=0}^1A_{1,j}k^j(T-k)^j=6k(T-k)+1, \ 0\leq k\leq \ell=T\in\mathbb{N}_0$. Let's find analogs of the pattern \href{https://oeis.org/A287326}{A287326} for $m>1$. To find such analogs, let's determine the coefficients $A_{2,0}, \ A_{2,1}, \ A_{2,2}$ in the polynomial $\sum_{j=0}^2 A_{2,j}k^j(T-k)^j$, such that $P_2(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=T^5$, as $\ell=T\in\mathbb{N}$, by the (1.2). Firstly, to determine the coefficients coefficients $A_{2,0}, \ A_{2,1}, \ A_{2,2}$, let's rewrite the polynomial $P_2(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j,$ in extended view as follows
\begin{equation}\label{alb_3_1}
\begin{split}
&\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j\\
=A_{2,2}&\sum\limits_{1\leq k \leq \ell}k^2(T-k)^2+A_{2,1}\sum\limits_{1\leq k \leq \ell}k^1(T-k)^1+A_{2,0}\sum\limits_{1\leq k \leq \ell}k^0(T-k)^0\\
=A_{2,2}&\sum\limits_{1\leq k \leq \ell} k^2(T^2-2Tk+k^2)+A_{2,1}\sum\limits_{1\leq k \leq \ell} kT-k^2 + A_{2,0}\sum\limits_{1\leq k \leq \ell}k^0(T-k)^0\\
=A_{2,2}&\sum\limits_{1\leq k \leq \ell} k^2T^2-2Tk^3+k^4+A_{2,1}\sum\limits_{1\leq k \leq \ell} kT-k^2 + A_{2,0}\sum\limits_{1\leq k \leq \ell}1 \\
=A_{2,2}&T^2\left(\sum\limits_{1\leq k \leq \ell}k^2\right)-2A_{2,2}T\left(\sum\limits_{1\leq k \leq \ell}k^3\right)+A_{2,2}\left(\sum\limits_{1\leq k \leq \ell}k^4\right)+A_{2,1}T\left(\sum\limits_{1\leq k \leq \ell} k\right) \\
-A_{2,1}&\left(\sum\limits_{1\leq k \leq \ell}k^2\right)+A_{2,0}\left(\sum\limits_{1\leq k \leq \ell}1\right)=T^5, \ \mathrm{as} \ \ell=T\in\mathbb{N}.
\end{split}
\end{equation}
Above derivation could be generalised in terms of Binomial coefficients and Faulhaber's sum for every $m\in\mathbb{N}_0$ as follows
\begin{equation*}
\sum_{s=0}^{m}\sum_{j=0}^{m-s}(-1)^j A_{m,m-s} \binom{m-s}{j}\left(\sum_{k=1}^{\ell}k^{m+s}\right)T^{m+s}
\end{equation*}
Note that the part $A_{2,0}\sum\nolimits_{1\leq k \leq \ell}k^0(T-k)^0$ of (\hyperref[alb_3_1]{1.10}) gives an indeterminate form as $k=T$ since the $A_{2,0}(T-T)^0k^0$ contains the term $(T-T)^0=0^0$. Some textbooks leave the quantity $0^0$ undefined, because the functions $x^0$ and $0^x$ have different limiting values when $x$ decreases to $0$. For our purposes we will use the convention:
$$\forall x: \ x^0 = 1,$$
as it is a common agreement, see \cite{16}.
Above we have derived an expression containing sums of powers of successive natural exponents, where the powers are $\{1,2,3,4\}$.
These formulae contain so-called Bernoulli numbers, \cite{7}.
By Faulhaber's formula, the sums of successive powers in $k$ are following
\begin{equation}\label{alb_3_1_1}
\begin{split}
&\sum\limits_{1\leq k \leq \ell} k = \frac{\ell^2+\ell}{2}, \\
&\sum\limits_{1\leq k \leq \ell} k^2 = \frac{2\ell^3+3\ell^2+\ell}{6}, \\
&\sum\limits_{1\leq k \leq \ell} k^3 = \frac{\ell^4+2\ell^3+\ell^2}{4}, \\
&\sum\limits_{1\leq k \leq \ell} k^4 = \frac{6\ell^5+15\ell^4+10\ell^3-\ell}{30}.
\end{split}
\end{equation}
Next, we substitute the identities (\hyperref[alb_3_1_1]{1.11}), into (\hyperref[alb_3_1]{1.10}), respectively
\begin{equation}\label{alb_3_1_2}
\begin{split}
A_{2,2}T^2\left(\frac{2\ell^3+3\ell^2+\ell}{6}\right)
-&2A_{2,2}T\left(\frac{\ell^4+2\ell^3+\ell^2}{4}\right)+A_{2,2}\left(\frac{6\ell^5+15\ell^4+10\ell^3-\ell}{30}\right)\\
+&A_{2,1}T\left(\frac{\ell^2+\ell}{2}\right)-A_{2,1}\left(\frac{2\ell^3+3\ell^2+\ell}{6}\right)+A_{2,0}\ell.
\end{split}
\end{equation}
Factorising the expression (\hyperref[alb_3_1_2]{1.12}) and collecting the terms under common divisor with set $\ell=T\in\mathbb{N}$, we get
\begin{equation}\label{alb_3}
\frac{A_{2,2}T^5-A_{2,2}T+30A_{2,0}}{30}+A_{2,1}\frac{T^3-T}{6}=T^5.
\end{equation}
In order to satisfy (\hyperref[alb_3]{1.13}) for each $\ell=T\in\mathbb{N}$, coefficients $A_{2,0}, \ A_{2,1}, \ A_{2,2}$ should be a solutions of following system of equations
\begin{equation}\label{alb_5}
\begin{cases}
\frac{1}{30}A_{2,2} &=1, \\
A_{2,1} &=1, \\
30A_{2,0}-A_{2,2}&=0.
\end{cases}
\end{equation}
The solutions to the system (\hyperref[alb_5]{1.14}) are following: $A_{2,2}=30, \ A_{2,1}=0, \ A_{2,0}=1$. Hereby, polynomial $\sum_{j=0}^2 A_{2,j}k^j(T-k)^j$ takes the form
\begin{equation}\label{alb_6}
\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=30k^2(T-k)^2+1.
\end{equation}
Let's construct the triangle, every $(\ell=T)$-th row sum starting from $k=1$ gives the terms of polynomial $P_2(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=\sum_{k=1}^{\ell}30k^2(T-k)^2+1$, over $k$ in range $1\leq k\leq \ell=T\in\mathbb{N}_0$. We review the triangle in context of the partial case of polynomial $P_m(\ell, T), \ m=2, \ \ell=T\in\mathbb{N}_0$
\begin{table}[H]
\begin{tabular}{rccccccccccc}
$\ell=T=0$& & & & & & 1 & & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=1$& & & & & 1 & & 1 & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=2$& & & & 1 & & 31 & & 1 & & & \\\noalign{\smallskip\smallskip}
$\ell=T=3$& & & 1 & & 121 & & 121 & & 1 & & \\\noalign{\smallskip\smallskip}
$\ell=T=4$& & 1 & & 271 & & 481 & & 271 & & 1 & \\\noalign{\smallskip\smallskip}
$\ell=T=5$&1 & & 481 & & 1081 & & 1081 & & 481 & & 1 \\\noalign{\smallskip\smallskip}
\end{tabular}
\caption{Triangle generated by the polynomial $\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=30k^2(T-k)^2+1, \ 0\leq k \leq \ell=T\in\mathbb{N}_0$,
sequence \href{https://oeis.org/A300656}{A300656} in OEIS, \cite{8}.
Summation of the $(\ell=T)$-th row terms from $k=1$ gives $P_2(T,T)=T^5$, see (\hyperref[f1]{1.1}).} \label{fig_2}
\end{table}
By the identity (\hyperref[f1]{1.1}) in polynomials $P_m(\ell,T)$, the sum of the $(T=\ell)$-th row terms of the
\hyperref[fig_2]{table 2} over $k$ from 1 to $\ell=T\in\mathbb{N}$ gives $P_2(T,T)=T^5$.
Below we show initial ten polynomials, generated by the partial case of (\hyperref[f1]{1.1}) for $m=2$ in $\ell=1,2,...,10$:
\begin{small}
\begin{equation}\label{u_2}
\begin{split}
&P_2(\ell, T)=\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=\sum\limits_{k=1}^{\ell}30k^2(T-k)^2+1=
\sum_{k=0}^{2}(-1)^{2-k}U_2(\ell,k)\cdot T^k \\
&=\begin{cases}
\ell=1: 31 - 60 T + 30 T^2 &=U_2(1,0)\cdot T^0-U_2(1,1)\cdot T^1+U_2(1,2)\cdot T^2=P_2(1,T) \\
\ell=2: 512 - 540 T + 150 T^2 &=U_2(2,0)\cdot T^0-U_2(2,1)\cdot T^1+U_2(2,2)\cdot T^2=P_2(2,T)\\
\ell=3: 2943 - 2160 T + 420 T^2 &=U_2(3,0)\cdot T^0-U_2(3,1)\cdot T^1+U_2(3,2)\cdot T^2=P_2(3,T)\\
\ell=4: 10624 - 6000 T + 900 T^2 &=U_2(4,0)\cdot T^0-U_2(4,1)\cdot T^1+U_2(4,2)\cdot T^2=P_2(4,T)\\
\ell=5: 29375 - 13500 T + 1650 T^2 &=U_2(5,0)\cdot T^0-U_2(5,1)\cdot T^1+U_2(5,2)\cdot T^2=P_2(5,T)\\
\ell=6: 68256 - 26460 T + 2730 T^2 &=U_2(6,0)\cdot T^0-U_2(6,1)\cdot T^1+U_2(6,2)\cdot T^2=P_2(6,T)\\
\ell=7: 140287 - 47040 T + 4200 T^2 &=U_2(7,0)\cdot T^0-U_2(7,1)\cdot T^1+U_2(7,2)\cdot T^2=P_2(7,T)\\
\ell=8: 263168 - 77760 T + 6120 T^2 &=U_2(8,0)\cdot T^0-U_2(8,1)\cdot T^1+U_2(8,2)\cdot T^2=P_2(8,T)\\
\ell=9: 459999 - 121500 T + 8550 T^2 &=U_2(9,0)\cdot T^0-U_2(9,1)\cdot T^1+U_2(9,2)\cdot T^2=P_2(9,T)\\
\ell=10: 760000 - 181500 T + 11550 T^2 &=U_2(10,0)\cdot T^0-U_2(10,1)\cdot T^1+U_2(10,2)\cdot T^2=P_2(10,T)\\
\end{cases}\\
&=T^5 \ \mathrm{as} \ T\to\ell.
\end{split}
\end{equation}
\end{small}
The coefficients $U_2(\ell,k), \ 0\leq k\leq 2$ in (\hyperref[u_2]{1.16}) are terms of the sequence \href{https://oeis.org/A316349}{A316349}, \cite{17}.
Above polynomials are approximations of monomial $T^{5}$ in some neighborhood of the point $T\to \ell$, the following graph shows corresponding polynomials $\sum_{k=1}^{\ell}\sum_{j=0}^2 A_{2,j}k^j(T-k)^j=\sum_{k=0}^{2}(-1)^{2-k}U_2(\ell,k)\cdot T^k, \ \ell=1,2,..$ and the $T^{5}$ curve
\begin{figure}[H]
\centering
\includegraphics[width=400px, keepaspectratio]{figure_n_5.eps}
\captionof{figure}{Approximations of monomial $T^{5}$ by polynomials $P_2(1,T), \ P_2(2,T), \ P_2(3,T)$ in some neighborhood
of the point $\ell=1,2,3$, with $T\to\ell$ see (\hyperref[u_2]{1.16}).}
\end{figure}
Similarly, finding the coefficients $A_{3,0}, \ A_{3,1}, \ A_{3,2}, \ A_{3,3}$ in $P_3(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^3 A_{3,j}k^j(T-k)^j$,
we get $A_{3,0}=1, \ A_{3,1}=0, \ A_{3,2}=-14, \ A_{3,3}=140$, therefore, for each $\ell=T\in\mathbb{N}$:
\begin{equation*}
T^7=\sum\limits_{1\leq k \leq \ell} 140k^3(T-k)^3-14k^2(T-k)^2+1.
\end{equation*}
Below we show a few initial rows of the triangle, every $(\ell=T)$-th row sum starting from $k=1$ gives the terms of polynomial $P_3(\ell,T)=\sum_{k=1}^{\ell}\sum_{j=0}^3 A_{3,j}k^j(T-k)^j=\sum_{k=1}^{\ell}140k^3(T-k)^3-14k^2(T-k)^2+1$, over $k$ in range $1\leq k\leq \ell=T\in\mathbb{N}_0$. We review the triangle in context of the partial case of polynomial $P_m(\ell, T), \ m=3, \ \ell=T\in\mathbb{N}_0$
\begin{table}[H]
\begin{tabular}{rccccccccccc}
$\ell=T=0$& & & & & & 1 & & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=1$& & & & & 1 & & 1 & & & & \\\noalign{\smallskip\smallskip}
$\ell=T=2$& & & & 1 & & 127 & & 1 & & & \\\noalign{\smallskip\smallskip}
$\ell=T=3$& & & 1 & & 1093 & & 1093 & & 1 & & \\\noalign{\smallskip\smallskip}
$\ell=T=4$& & 1 & & 3793 & & 8905 & & 3793 & & 1 & \\\noalign{\smallskip\smallskip}
$\ell=T=5$&1 & & 8905 & & 30157 & & 30157 & & 8905 & & 1 \\\noalign{\smallskip\smallskip}
\end{tabular}
\caption{Triangle generated by the polynomial $\sum_{j=0}^3 A_{3,j}k^j(T-k)^j=140k^3(T-k)^3-14k^2(T-k)^2+1, \ 0\leq k \leq \ell=T\in\mathbb{N}_0$,
sequence \href{https://oeis.org/A300785}{A300785} in OEIS, \cite{9}.
Summation of the $(\ell=T)$-th row terms from $k=1$ gives $P_3(T,T)=T^7$, see (\hyperref[f1]{1.1}).} \label{fig_3}
\end{table}
As identity (\hyperref[f1]{1.1}) holds, the sum of the $(T=\ell)$-th row terms of \hyperref[fig_3]{table 3} over $k$ from 1 to $T=\ell$ equals to $T^{2m+1}, \ m=3$.
For the case $P_3(\ell,T)$, the following polynomials $\sum_{k=1}^{\ell}\sum_{j=0}^3 A_{3,j}k^j(T-k)^j=\sum_{k=0}^{3}(-1)^{3-k}U_3(\ell,k)\cdot T^k, \ \mathrm{for} \ \ell=1,2,...,10$ can be generated
\begin{small}
\begin{equation}\label{u_3}
\begin{split}
P_3(\ell,T)
&=\sum_{k=1}^{\ell}\sum_{j=0}^3 A_{3,j}k^j(T-k)^j=\sum\limits_{1\leq k \leq \ell}140k^3(T-k)^3-14k^2(T-k)^2+1=\sum_{k=0}^{3}(-1)^{3-k}U_3(\ell,k)\cdot T^k\\
&=\begin{cases}
\ell=1 :& -125 + 406 T - 420 T^2 + 140 T^3=P_3(1,T)\\
\ell=2 :& -9028 + 13818 T - 7140 T^2 + 1260 T^3=P_3(2,T)\\
\ell=3 :& -110961 + 115836 T - 41160 T^2 + 5040 T^3=P_3(3,T)\\
\ell=4 :& -684176 + 545860 T - 148680 T^2 + 14000 T^3=P_3(4,T)\\
\ell=5 :& -2871325 + 1858290 T - 411180 T^2 + 31500 T^3=P_3(5,T)\\
\ell=6 :& -9402660 + 5124126 T - 955500 T^2 + 61740 T^3=P_3(6,T)\\
\ell=7 :& -25872833 + 12182968 T - 1963920 T^2 + 109760 T^3=P_3(7,T)\\
\ell=8 :& -62572096 + 25945416 T - 3684240 T^2 + 181440 T^3=P_3(8,T)\\
\ell=9 :& -136972701 + 50745870 T - 6439860 T^2 + 283500 T^3=P_3(9,T)\\
\ell=10 :& -276971300 + 92745730 T - 10639860 T^2 + 423500 T^3=P_3(10,T)
\end{cases} \\
&=T^7 \ \mathrm{as} \ T\to\ell.
\end{split}
\end{equation}
\end{small}
The coefficients $U_3(T,k), \ 0\leq k\leq 3$ in (\hyperref[u_3]{1.17}) are terms of the sequence \href{https://oeis.org/A316387}{A316387}, \cite{18}.
Above polynomials are approximations of monomial $T^{7}$ in some neighborhood of the point $T\rightarrow \ell$. The following graph shows corresponding polynomials $\sum_{k=0}^{3}(-1)^{3-k}U_3(\ell,k)\cdot T^k, \ T=1,2,3$ and the $T^{7}$ curve
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{figure_n_7.eps}
\captionof{figure}{Approximations of monomial $T^{7}$ by polynomials $P_3(1,T), \ P_3(2,T), \ P_3(3,T)$ in some neighborhood of the point
$\ell=1,2,3$, with $T\to\ell$, see (\hyperref[u_3]{1.17}).}
\end{figure}
Now, let generalise the the identity $P_m(T,T)=T^{2m+1}, \ T\in\mathbb{N}$ for every integer $m\geq0$, as we have already discussed the cases $m=1,2,3$. Firstly, let's find a recurrence formula for coefficients $A_{m,j}, \ 0\leq j \leq m$, see definition (\hyperref[f1]{1.1}). To reach a recurrent formula of $A_{m,j}, \ m\geq0$, first let's fix the unused values of $A_{m,j}=0, \ \mathrm{for} \ j<0 \ \mathrm{or} \ j>m$, so we don't need to consider the summation range for $j$. Let's rewrite the polynomial $P_m(\ell,T)$ again
\begin{equation}\label{gen_5_1}
P_m(\ell,T) \ = \ \sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j \ \equiv \ \sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k,
\end{equation}
Since the symmetry holds
\begin{equation*}
D_m(T,k):=\sum_{j=0}^m A_{m,j}k^j(T-k)^j, \ D_m(T,k)=D_m(T-k,k)
\end{equation*}
We can rewrite (\hyperref[gen_5_1]{1.18}) as follows
\begin{equation}\label{gen_5_2}
\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j=\sum_{k=0}^{\ell-1}\sum_{j=0}^m A_{m,j}k^j(T-k)^j
\end{equation}
By expanding $(T-k)^j$ in r.h.s of (\hyperref[gen_5_2]{1.19}) and using Faulhaber's formula \cite{11}, the result is
\begin{equation}\label{gen_5}
\begin{split}
&\sum_{k=0}^{\ell-1} (T-k)^j k^j = \sum_{k=0}^{\ell-1} \sum_{i} \binom{j}i T^{j-i} (-1)^i k^{i+j}\\
&=\sum_{i} \binom{j}{i} T^{j-i} \frac{(-1)^i}{i+j+1} \left[ \sum_{t} \binom{i+j+1}t B_t \ell^{i+j+1-t} - B_{i+j+1}\right] \\
&=\underbrace{\sum_{i,t}\binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t B_tT^{j-i}\ell^{i+j+1-t}}_{(\star)}
- \underbrace{\sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} B_{i+j+1} T^{j-i}}_{(\diamond)}
\end{split}
\end{equation}
where $B_t$ are Bernoulli numbers \cite{14}. Now, we notice that
\begin{equation}\label{gen_6}
\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}
\end{equation}
In particular, the last sum is zero for $0<t\leq j$. Now, by substituting to the $(\star)$ part of
(\hyperref[gen_5]{1.20}) the result of (\hyperref[gen_6]{1.21}), we have
\begin{equation}\label{gen_6_1_2}
\begin{split}
\sum_{i,t}\binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}{t} B_tT^{j-i}\ell^{i+j+1-t}
&= \frac{1}{(2j+1)\binom{2j}j}T^{j}\ell^{j+1}\\
&+\sum_{t>0} \frac{(-1)^j}{t}\binom{j}{2j-t+1} B_t T^{j-i}\ell^{i+j+1-t}
\end{split}
\end{equation}
By means of (\hyperref[gen_6_1_2]{1.22}), expression (\hyperref[gen_5]{1.20}) takes the form
\begin{equation}\label{gen_6_1}
\begin{split}
\sum_{k=0}^{\ell-1} (T-k)^j k^j
&=\underbrace{\frac{1}{(2j+1)\binom{2j}j}T^{j}\ell^{j+1}+\sum_{t>0}\frac{(-1)^j}{t}\binom{j}{2j-t+1} B_t T^{j-i}\ell^{i+j+1-t}}_{(\star)} \\
&- \underbrace{\sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} B_{i+j+1} T^{j-i}}_{(\diamond)}
\end{split}
\end{equation}
We have to remember that if the sum over some variable $i$ contains $\binom{j}{i}$, then instead of limiting its summation range to $i\in[0,j]$, we can let $i\in[-\infty,+\infty]$ since $\binom{j}{i} = 0$ for $i$ outside the range $i\in[0,j]$ (i.e., when $i<0$ or $i>j$). It's much easier to review such sum as sum from $-\infty$ to $+\infty$ (unless specified otherwise), where only a finite number of terms are nonzero, this fact is also discussed in \cite{12}. Therefore, we haven't shown detailed bounds of summation in above derivation.
Now, we keep our attention on (\hyperref[gen_6_1]{1.23}).
To combine or cancel identical terms across the two sums in (\hyperref[gen_6_1]{1.23}) more easily, let introduce $\kappa=2j+1-t$ to $(\star)$ and $\kappa=j-i$ to $(\diamond)$, respectively
\begin{equation}\label{gen_7}
\begin{split}
&\sum_{k=0}^{\ell-1} (T-k)^j k^j= \frac{1}{(2j+1)\binom{2j}j} T^{j}\ell^{j+1}
+\sum_{\kappa} \frac{(-1)^j}{2j+1-\kappa}\binom{j}{\kappa}B_{2j+1-\kappa}T^{j-i}\ell^{\kappa+i-j} \\
&-\sum_{\kappa} \binom{j}{\kappa} \frac{(-1)^{j-\kappa}}{2j+1-\kappa} B_{2j+1-\kappa}T^{\kappa}
\end{split}
\end{equation}
Let be $T=\ell$ in (\hyperref[gen_7]{1.24}), thus
\begin{equation}\label{gen_7_1}
\begin{split}
&\sum_{k=0}^{T-1} (T-k)^j k^j= \frac{1}{(2j+1)\binom{2j}j} T^{2j+1}
+\sum_{\kappa} \frac{(-1)^j}{2j+1-\kappa}\binom{j}{\kappa}B_{2j+1-\kappa}T^{\kappa} \\
&-\sum_{\kappa} \binom{j}{\kappa} \frac{(-1)^{j-\kappa}}{2j+1-\kappa} B_{2j+1-\kappa}T^{\kappa}\\
&=\frac{1}{(2j+1)\binom{2j}j} T^{2j+1} + 2\sum_{\text{odd }\kappa}
\frac{(-1)^j}{2j+1-\kappa}\binom{j}{\kappa}B_{2j+1-\kappa}T^{\kappa}.
\end{split}
\end{equation}
Now, using the definition of $A_{m,j}$,
\begin{equation*}
\sum_{k=0}^{\ell-1}\sum_{j} A_{m,j}k^j(T-k)^j=T^{2m+1}, \ \mathrm{as} \ \ell=T\in\mathbb{N},
\end{equation*}
From (\hyperref[gen_7_1]{1.25}) we can obtain the following identity for polynomials in $T$
\begin{equation}\label{gen_8}
\begin{split}
&\sum_{j}A_{m,j}\frac{1}{(2j+1)\binom{2j}j}T^{2j+1}
+ 2\sum_{j, \text{ odd }\kappa} A_{m,j} \binom{j}{\kappa} \frac{(-1)^j}{2j+1-\kappa} B_{2j+1-\kappa}T^{\kappa}\\
&\equiv n^{2m+1}, \ \mathrm{as} \ T\to\ell.
\end{split}
\end{equation}
Taking the coefficient of $n^{2j+1}$ in the above expression, 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 $A_{m,d}=0$.
Taking the coefficient of $n^{2d+1}$ in (\hyperref[gen_8]{1.26}) for $m/4 \leq d < m/2$ , we get
\begin{equation*}\label{gen_10}
A_{m,d} \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,
\end{equation*}
i.e
\begin{equation*}\label{gen_11}
A_{m,d} = (-1)^{m-1} \frac{(2m+1)!}{d!d!m!(m-2d-1)!}\frac{1}{m-d} B_{2m-2d}.
\end{equation*}
Continuing similarly, we can express $A_{m,j}$ for each integer $j$ in the range $m/2^{s+1}\leq j< m/2^s$ (iterating consecutively $s=1,2,...$) via the previously determined values of $A_{m,d}, \ d<j$ as follows
\begin{equation*}\label{gen_12}
A_{m,j} = (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}.
\end{equation*}
The same formula holds also for $m=0$. Note that the $m$ in above sum must satisfy $m\geq2j+1$ to return a nonzero term $A_{m,j}$.
\begin{defn}
We define here a generating function of sequence of coefficients $A_{m,j}$ as follows
\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*}
\end{defn}
Five initial rows of the triangle generated by $A_{m,j}, \ j\geq0, \ 0\leq j \leq m$ are
\begin{table}[H]
\begin{tabular}{rccccccccccc}
$m=0$& & & & & & 1 & & & & & \\\noalign{\smallskip\smallskip}
$m=1$& & & & & 1 & & 6 & & & & \\\noalign{\smallskip\smallskip}
$m=2$& & & & 1 & & 0 & & 30 & & & \\\noalign{\smallskip\smallskip}
$m=3$& & & 1 & & -14 & & 0 & & 140 & & \\\noalign{\smallskip\smallskip}
$m=4$& & 1 & & -120 & & 0 & & 0 & & 630 & \\\noalign{\smallskip\smallskip}
$m=5$&1 & & -1386 & & 660 & & 0 & & 0 & & 2772 \\\noalign{\smallskip\smallskip}
\end{tabular}
\caption{Triangle generated by $A_{m,j}, \ j\geq0, \ 0\leq j \leq m$, see \hyperref[gen_13]{definition (1.27)}.} \label{fig_4}
\end{table}
Note that starting from row $m\geq11$ the terms of the \hyperref[fig_4]{table 4} consist of fractional numbers, for example, $A_{11,1}=800361655623.6$. One can find a complete list of the numerators and denominators of $A_{m,j}$ in OEIS under the identifiers \href{https://oeis.org/A302971}{A302971} and \href{https://oeis.org/A304042}{A304042}, respectively, see \cite{17},\cite{18}. Note that
\begin{equation*}
\sum_{j}A_{m,j} \ = \ 2^{2m+1}-1.
\end{equation*}
As we have found a recurrence for $A_{m,j}$ coefficients, let's back to the identity (\hyperref[f1]{1.1})
\begin{equation*}
\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j \ \equiv \ \sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k
\end{equation*}
to find a recurrence for coefficients $U_m(\ell,k)$ for every $m\geq1$. To understand the nature of the coefficients $U_m(T,k)$,
express the polynomials $U_m(\ell,k)$ in terms of $A_{m,j}$ and Bernoulli numbers. To do so, let's expand the binomial
$(T-k)^j$ in the l.h.s. of (\hyperref[f1]{1.1}) and change of the order of summation:
\begin{equation}\label{sum_2}
\begin{split}
\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j &= \sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j\sum_{t=0}^j\binom{j}{t}T^t(-1)^{j-t}k^{j-t}\\
&=\sum_{t=0}^m T^t \sum_{k=1}^{\ell}\sum_{j=t}^m \binom{j}{t}A_{m,j}k^{2j-t}(-1)^{j-t}.
\end{split}
\end{equation}
Now, taking the coefficient of $T^t$ in (\hyperref[sum_2]{1.28}) gives:
\begin{equation*}
U_m(\ell,t) = (-1)^m \sum_{k=1}^{\ell}\sum_{j=t}^m \binom{j}{t}A_{m,j}k^{2j-t}(-1)^j.
\end{equation*}
From this formula it may be not immediately clear why as $T\to\ell$ the $U_m(\ell,t)$ represent polynomials in $T$. However, this can be seen if we change the summation order again and use Faulhaber's formula to obtain:
\begin{equation}\label{maxal1}
U_m(T,t) = (-1)^m \sum_{j=t}^m \binom{j}{t}A_{m,j} \frac{(-1)^j}{2j-t+1}\sum_{l=0}^{2j-t} \binom{2j-t+1}{l}B_{l}T^{2j-t+1-l}.
\end{equation}
Introducing $\kappa=2j-t+1-l$, to (\hyperref[maxal1]{1.29}) we further get the formula:
\begin{equation*}
U_m(T,t) = (-1)^m \sum_{k=1}^{2j-t+1} T^k \sum_{j=t}^m \binom{j}{t}A_{m,j} \frac{(-1)^j}{2j-t+1}\binom{2j-t+1}{\kappa}B_{2j-t+1-\kappa},
\end{equation*}
which allows easily compute the coefficient of $T^k$ in $U_m(T,t)$ for each $\kappa$. In above formulae we assume that $B_1=+\frac12$.
\subsection{Error of approximation}
Generally, the monomial $T^{2m+1}, \ \ell,m\in\mathbb{N}, \ T\in\mathbb{R}$ could be approximated by (\hyperref[f1]{1.1}) as follows
\begin{equation*}
\lim_{T\to\ell}\left[\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j\right] = \lim_{T\to\ell}\left[\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k\right]=T^{2m+1}.
\end{equation*}
In this subsection we arrange the tables with values of corresponding polynomials $P_m(\ell, T)$ and monomials $T^{2m+1}$ to rate the quality of the approximation of $T^{2m+1}, \ m\geq 0$ by (1.1). We begin from the case $m=1$ and $T=3$, therefore, the approximating polynomials is $P_1{3, T}=-81+36T$. By solving the equation $T^3-36T+81=0$ we receive the following roots $T_1=3$, $T_2=3.90833$, it follows that our polynomial approximates the monomial $T^{2m+1}$ in neighborhood of $T_1$ and $T_2$, the following table contains the values of $P_1(3, T)$ and $T^{3}$ in neighborhood of the root $T_1=3$
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^3$ & $P_1(3, T)=-81+36T$ \\ \hline
2.95 & 25.67238 & 25.2 \\ \hline
2.96 & 25.93434 & 25.56 \\ \hline
2.97 & 26.19807 & 25.92 \\ \hline
2.98 & 26.46359 & 26.28 \\ \hline
2.99 & 26.73090 & 26.64 \\ \hline
$\underline{3.00}$ & $\underline{27.00000}$ & $\underline{27}$ \\ \hline
3.01 & 27.27090 & 27.36 \\ \hline
3.02 & 27.54361 & 27.72 \\ \hline
3.03 & 27.81813 & 28.08 \\ \hline
3.04 & 28.09446 & 28.44 \\ \hline
3.05 & 28.37263 & 28.8 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^3$ and $P_1(3, T)$ in neighborhood of $T_1=3$.} \label{tab:figure2}
\end{table}
Below we show graphically the intersection of of $F(T)=T^3$ and $P_1(3, T)$
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{n_3_by_polynomial_at_point_3.eps}%
\caption{Intersection of $F(T)=T^3$ and $P_1(3, T)$ in point $T_1=3$.}%
\end{figure}
As it is mentioned above the equation $T^3-36T+81=0$ has two roots $T_1=3$, $T_2=3.90833$, let's discuss the case $T_2=3.90833$, the following table represents the values of $P_1(3, T)$ and $T^{3}$ in neighborhood of the root $T_2=3.90833$
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^3$ & $P_1(3,T)=-81+36T$ \\ \hline
3.86000 & 57.51246 & 57.96 \\ \hline
3.87000 & 57.96060 & 58.32 \\ \hline
3.88000 & 58.41107 & 58.68 \\ \hline
3.89000 & 58.86387 & 59.04 \\ \hline
3.90000 & 59.31900 & 59.4 \\ \hline
$\underline{3.90833}$ & $\underline{59.69991}$ & $\underline{59.69988}$ \\ \hline
3.92000 & 60.23629 & 60.12 \\ \hline
3.93000 & 60.69846 & 60.48 \\ \hline
3.94000 & 61.16298 & 60.84 \\ \hline
3.95000 & 61.62988 & 61.2 \\ \hline
3.96000 & 62.09914 & 61.56 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^3$ and $P_1(3, T)$ in neighborhood of $T_1=3$.}
\end{table}
Below we show graphically the intersection of $F(T)=T^3$ and $P_1(3, T)$
\begin{figure}[H]
\centering
\includegraphics[width=460px, keepaspectratio]{n_3_by_polynomial_at_point_3_root_2.eps}%
\caption{Intersection of $F(T)=T^3$ and $P_1(3, T)$ in point $T_2=3.90833$.}%
\end{figure}
Next, let's review the case $m=2$ and $T=3$, the approximating polynomial is $P_2(3, T)=2943 - 2160 T + 420 T^2$. By solving the equation $T^5-(2943 - 2160 T + 420 T^2)=0$ we receive the following roots $T_1=3$, $T_2=3.40551$, $T_3=3.98704$, it follows that our polynomial approximates the monomial $T^{2m+1}$ in neighborhood of $T_1$, $T_2$ and $T_3$, the following tables contain the values of $P_2(3, T)$ and $T^{5}$ in neighborhood of the roots $T_1=3$, $T_2$ and $T_3$
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^5$ & $P_2(3, T)=2943 - 2160 T + 420 T^2$ \\ \hline
2.95 & 223.41384 & 226.05000 \\ \hline
2.96 & 227.22628 & 229.27200 \\ \hline
2.97 & 231.09058 & 232.57800 \\ \hline
2.98 & 235.00728 & 235.96800 \\ \hline
2.99 & 238.97691 & 239.44200 \\ \hline
$\underline{3.00}$ & $\underline{243.00000}$ & $\underline{243.00000}$ \\ \hline
3.01 & 247.07709 & 246.64200 \\ \hline
3.02 & 251.20872 & 250.36800 \\ \hline
3.03 & 255.39544 & 254.17800 \\ \hline
3.04 & 259.63780 & 258.07200 \\ \hline
3.05 & 263.93634 & 262.05000 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^5$ and $P_2(3, T)$ in neighborhood of $T_1=3$.}
\end{table}
Below we show graphically the intersection of $F(T)=T^5$ and $P_2(3, T)$
\begin{figure}[H]
\centering
\includegraphics[width= 450px, keepaspectratio]{n_5_by_polynomial_at_point_3_1.eps}%
\caption{Intersection of $F(T)=T^5$ and $P_2(3, T)$ in point $T_1=3$.}%
\end{figure}
Next, let's review the values of $F(T)=T^5$ and $P_2(3, T)$ in the neighborhood of the root $T_2=3.40551$ as follow table shows
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^5$ & $P_2(3, T)=2943 - 2160 T + 420 T^2$ \\ \hline
3.35000 & 421.91410 & 420.45000 \\ \hline
3.36000 & 428.24903 & 427.03200 \\ \hline
3.37000 & 434.65983 & 433.69800 \\ \hline
3.38000 & 441.14717 & 440.44800 \\ \hline
3.39000 & 447.71175 & 447.28200 \\ \hline
$\underline{3.40551}$ & $\underline{458.04780}$ & $\underline{458.04771}$ \\ \hline
3.41000 & 461.07534 & 461.20200 \\ \hline
3.42000 & 467.87574 & 468.28800 \\ \hline
3.43000 & 474.75615 & 475.45800 \\ \hline
3.44000 & 481.71727 & 482.71200 \\ \hline
3.45000 & 488.75980 & 490.05000 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^5$ and $P_2(3, T)$ in neighborhood of $T_2=3.40551$.}
\end{table}
Below we show graphically the intersection of $F(T)=T^5$ and $P_2(3, T)$ in point $T_2=3.40551$
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{n_5_by_polynomial_at_point_3_2.eps}
\caption{Intersection of $F(T)=T^5$ and $P_2(3, T)$ in point $T_2=3.90833$.}%
\end{figure}
Now, let's review the values of $F(T)=T^5$ and $P_2(3, T)$ in the neighborhood of the root $T_2=3.90833$ as follow table shows
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^5$ & $P_2(3,T)=2943-2160T+420 T^2$ \\ \hline
3.93000 & 937.48160 & 941.05800 \\ \hline
3.94000 & 949.46970 & 952.51200 \\ \hline
3.95000 & 961.58012 & 964.05000 \\ \hline
3.96000 & 973.81381 & 975.67200 \\ \hline
3.97000 & 986.17170 & 987.37800 \\ \hline
3.98000 & 998.65472 & 999.16800 \\ \hline
$\underline{3.98704}$ & $\underline{1007.51835}$ & $\underline{1007.51854}$ \\ \hline
4.00000 & 1024.00000 & 1023.00000 \\ \hline
4.01000 & 1036.86416 & 1035.04200 \\ \hline
4.02000 & 1049.85728 & 1047.16800 \\ \hline
4.03000 & 1062.98034 & 1059.37800 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^5$ and $P_2(3, T)$ in neighborhood of $T_3=3.98704$.}
\end{table}
Below we show graphically the intersection of $F(T)=T^5$ and $P_2(3, T)$ in point $T_3=3.98704$
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{n_5_by_polynomial_at_point_3_3.eps}
\caption{Intersection of $F(T)=T^5$ and $P_2(3, T)$ in point $T_3=3.98704$.}%
\end{figure}
Now, let's keep our attention to the case $m=3$ and $\ell=2$, the approximating polynomial is $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$. By solving the equation $T^7-(-9028 + 13818 T - 7140 T^2 + 1260 T^3)=0$ we receive the following roots $T_1=2$, $T_2=2.8605$, $T_3=2.98955$, it follows that our polynomial approximates the monomial $T^{7}$ in neighborhood of $T_1$, $T_2$ and $T_3$, the following tables contain the values of $P_2(3, T)$ and $T^{5}$ in neighborhood of the roots $T_1$, $T_2$ and $T_3$, note that we have arranged the roots $T_2$ and $T_3$ in single table
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^7$ & $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ \\ \hline
1.96 & 111.12007 & 113.47136 \\ \hline
1.97 & 115.14990 & 117.00398 \\ \hline
1.98 & 119.30436 & 120.59792 \\ \hline
1.99 & 123.58664 & 124.26074 \\ \hline
$\underline{2.00}$ & $\underline{128.00000}$ & $\underline{128}$ \\ \hline
2.01 & 132.54776 & 131.82326 \\ \hline
2.02 & 137.23333 & 135.73808 \\ \hline
2.03 & 142.06015 & 139.75202 \\ \hline
2.04 & 147.03177 & 143.87264 \\ \hline
2.05 & 152.15178 & 148.1075 \\ \hline
2.06 & 157.42385 & 152.46416 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^7$ and $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ in neighborhood of $T_1=2$.}
\end{table}
Below we arrange a plot of intersection of $F(T)=T^7$ and $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ in $T_1=2$
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{n_7_by_polynomial_at_point_2_1.eps}
\caption{Intersection of $F(T)=T^7$ and $P_2(2, T)$ in point $T_1=2$.}%
\end{figure}
Let's show the values of the $F(T)=T^7$ and polynomial $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ in neighborhood of two other roots,
$T_1=2.86046, \ T_2=2.98955$
\begin{table}[H]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$T$ & $F(T)=T^7$ & $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ \\ \hline
2.81000 & 1383.38875 & 1379.35766 \\ \hline
2.82000 & 1418.22050 & 1415.09168 \\ \hline
2.83000 & 1453.80129 & 1451.52962 \\ \hline
2.84000 & 1490.14449 & 1488.67904 \\ \hline
2.85000 & 1527.26367 & 1526.5475 \\ \hline
$\underline{2.86046}$ & $\underline{1566.94022}$ & $\underline{1566.940217}$ \\ \hline
2.87000 & 1603.88519 & 1604.47178 \\ \hline
2.88000 & 1643.41563 & 1644.54272 \\ \hline
2.89000 & 1683.77827 & 1685.36294 \\ \hline
2.90000 & 1724.98763 & 1726.94 \\ \hline
2.91000 & 1767.05848 & 1769.28146 \\ \hline
2.92000 & 1810.00577 & 1812.39488 \\ \hline
2.93000 & 1853.84467 & 1856.28782 \\ \hline
2.94000 & 1898.59054 & 1900.96784 \\ \hline
2.95000 & 1944.25897 & 1946.4425 \\ \hline
2.96000 & 1990.86576 & 1992.71936 \\ \hline
2.97000 & 2038.42692 & 2039.80598 \\ \hline
2.98000 & 2086.95867 & 2087.70992 \\ \hline
$\underline{2.98955}$ & $\underline{2134.25217}$ & $\underline{2134.252175}$ \\ \hline
3.00000 & 2187.00000 & 2186 \\ \hline
3.01000 & 2238.54314 & 2236.40126 \\ \hline
3.02000 & 2291.12403 & 2287.65008 \\ \hline
3.03000 & 2344.76002 & 2339.75402 \\ \hline
3.04000 & 2399.46868 & 2392.72064 \\ \hline
\end{tabular}
\caption{Table of values of $F(T)=T^7$ and $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ in neighborhood of $T_2=2.8605$, $T_3=2.98955$.}
\end{table}
Below we show graphically the intersection of $F(T)=T^7$ and $P_3(2, T)=-9028 + 13818 T - 7140 T^2 + 1260 T^3$ in point $T_3=3.98704$ in points $T_2=2.8605$, $T_3=2.98955$
\begin{figure}[H]
\centering
\includegraphics[width=450px, keepaspectratio]{n_7_by_polynomial_at_point_2_2.eps}
\caption{Intersection of $F(T)=T^5$ and $P_2(3, T)$ in points $T_2=2.8605$, $T_3=2.98955$.}%
\end{figure}
\section{Acknowledgements}
We would like to thank to Dr. Max Alekseyev (Department of Mathematics and Computational Biology,
George Washington University) for sufficient help in the derivation of $A_{m,j}$ coefficients,
Dr. Hansruedi Widmer for his useful comments concerning the derivation of coefficients $A_{m,j}$,
Dr. Ron Knott (Visiting Fellow, Dept. of Mathematics at University of Surrey) for his useful
suggestions on the writing of this article, and to Mr. Albert Tkaczyk his help in derivation
of the sequences \href{https://oeis.org/A300656}{A300656}, \href{https://oeis.org/A300785}{A300785}.
Also, we'd like to thank to OEIS editors Michel Marcus, Peter Luschny, Jon E. Schoenfield and others
for their patient, faithful volunteer work and for useful comments and suggestions during
the editing of sequences, concerned with this manuscript. We, also, greatly appreciate the
preliminary proofread and suggestions by
\href{http://www.cantab.net/users/bnr/}{www.cantab.net/users/bnr/}.
\section{Conclusion}
In this manuscript we have derived and discussed the polynomials $P_m(\ell,T)$,
such that converge to the monomial $T^{2m+1}, \ m\in\mathbb{N}_0$ as $T\to\ell, \ \ell\in\mathbb{N}$.
Polynomial $P_m(\ell,T)$ gives weak approximation of the monomial $T^{2m+1}$ mainly in neighborhood of the solution
of the equation $T^{2m+1}-P_m(\ell,T)=0$.
\section{Supplementary Files}
We provide the following supplementary files to support our study:
\begin{itemize}
\item \href{https://kolosovpetro.github.io/arxiv_1603_02468/identity_1_1.txt}
{\textsf{https://kolosovpetro.github.io/arxiv\_1603\_02468/identity\_1\_1.txt}} -
Mathematica program, implementation the l.h.s of identity (\hyperref[f1]{1.1}), that is $T^{2m+1}=\sum_{k=1}^{\ell}\sum_{j=0}^m A_{m,j}k^j(T-k)^j, \ \ell=T\in\mathbb{N}.$
\item \href{https://kolosovpetro.github.io/arxiv_1603_02468/identity_1_1_r_h_s.txt}
{\textsf{https://kolosovpetro.github.io/arxiv\_1603\_02468/identity\_1\_1\_r\_h\_s.txt}} -
Mathematica program, implementation the r.h.s of identity (\hyperref[f1]{1.1}), that is $T^{2m+1}=\sum_{k=0}^{m}(-1)^{m-k}U_m(\ell,k)\cdot T^k, \ \ell=T\in\mathbb{N}.$
\item \href{https://kolosovpetro.github.io/arxiv_1603_02468/u_coefficients_in_row.txt}
{\textsf{https://kolosovpetro.github.io/arxiv\_1603\_02468/u\_coefficients\_in\_row.txt}} -
Mathematica program, lists the coefficients $U_m(\ell,k)$ in (\hyperref[f1]{1.1}) for given $m$. (By default $m=2$)
\end{itemize}
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Petro, Kolosov, et al., Entry "Consider coefficients $U_m(n,k)$ defined by the identity $\sum_{k=1..l} \sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (t-k)^j = \sum_{k=0..m} (-1)^{(m-k)} * U_m(l,k) * t^k$ that holds for all positive integers $l,m,t$. This sequence gives 4-column table read by rows, where $n$-th row lists coefficients $U_3(n,k) \ for \ k = 0, 1, 2, 3$.", \href{https://oeis.org/A316387}{A316387}, 2018.
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\bibitem{28}
Petro Kolosov, "Consider coefficients $U(m,l,k)$ defined by the identity $\sum_{k=1..l} \sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = \sum_{k=0..m} (-1)^{(m-k)} * U(m,l,k) * T^k$ that holds for all positive integers $l,m,T$. This sequence gives 2-column table read by rows, where n-th row lists coefficients $U(1,n,k)$ for $k = 0, 1$ and $n \geq 1$. ", \href{https://oeis.org/A320047}{A320047}, 2018.
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