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\begin{center}
\vskip 1cm{\LARGE\bf
Sums of Powers of Integers and the Sequence \\
\vskip .1in
A304330}
\vskip 1cm
\large
Jos\'{e} L. Cereceda\\
Collado Villalba, 28400 (Madrid)\\
Spain\\
\href{mailto:jl.cereceda@movistar.es}{\tt jl.cereceda@movistar.es} \\
\end{center}

\vskip .2in

\begin{abstract}
For integers $k \geq 1$, let $S_k(n)$ denote the sum of the $k$th powers of the first $n$ positive integers. In this paper, we derive a new formula expressing $2^{2k}$ times $S_{2k}(n)$ as a sum of $k$ terms involving the numbers in the $k$th row of the integer sequence \seqnum{A304330}, which is closely related to the central factorial numbers with even indices of the second kind. Furthermore, we provide an alternative proof of Knuth's formula for $S_{2k}(n)$ and show that it can equally be expressed in terms of \seqnum{A304330}. Moreover, we obtain corresponding formulas for $2^{2k-1}S_{2k-1}(n)$ and determine the Faulhaber form of both $S_{2k}(n)$ and $S_{2k+1}(n)$ in terms of \seqnum{A304330} and the Legendre-Stirling numbers of the first kind.
\end{abstract}


\section{Introduction and main results}

For integers $k \geq 1$, consider the sum of the $k$th powers of the first $n$ positive integers $S_k(n) = 1^k + 2^k + \cdots + n^k$. Recently, the author \cite[Theorem 2]{cere} has shown that $S_{2k}(n)$ can be expressed in the form
\begin{equation}\label{int1}
S_{2k}(n) = \frac{1}{2} \sum \frac{(2k)!}{b_1! b_2! \cdots b_k!} \prod_{r=1}^{k} \biggl( \frac{1}{4^r (2r)!}
\biggl )^{b_r} 2^m m! \binom{2n+m+1}{2m+1},
\end{equation}
where the summation takes place over all $k$-tuples of nonnegative integers $(b_1,b_2,\ldots,b_k)$ satisfying the constraint $b_1 + 2b_2 + \cdots + k b_k = k$, and where $m = b_1 + b_2 +\cdots +b_k$. Since $m$ runs over all the integers in the set $\{1,2,\ldots,k \}$, the above formula implies that $S_{2k}(n)$ can be expressed in the polynomial form
\begin{equation*}
S_{2k}(n) = \sum_{m=1}^k p_{k,m} \binom{2n+m+1}{2m+1},
\end{equation*}
for certain (nonzero) rational coefficients $p_{k,1}, p_{k,2}, \ldots, p_{k,k}$ which are independent of $n$. Indeed, as can be checked for $k =1,2,3,4$, formula \eqref{int1} yields
\begin{align*}
S_2(n) &amp; = \frac{1}{4} \binom{2n+2}{3}, \\
S_4(n) &amp; = \frac{1}{16} \binom{2n+2}{3} + \frac{3}{4} \binom{2n+3}{5}, \\
S_6(n) &amp; = \frac{1}{64} \binom{2n+2}{3} + \frac{15}{16} \binom{2n+3}{5} + \frac{45}{8} \binom{2n+4}{7}, \\
S_8(n) &amp; = \frac{1}{256} \binom{2n+2}{3} + \frac{63}{64} \binom{2n+3}{5} + \frac{315}{16} \binom{2n+4}{7}
+ \frac{315}{4} \binom{2n+5}{9}.
\end{align*}
If we now multiply these equations by the corresponding factor $2^{2k}$, $k =1,2,3,4$, we are left with the following all-integer formulas:
\begin{align*}
2^2 S_2(n) &amp; = \binom{2n+2}{3}, \\
2^4 S_4(n) &amp; = \binom{2n+2}{3} + 12 \binom{2n+3}{5}, \\
2^6 S_6(n) &amp; = \binom{2n+2}{3} + 60 \binom{2n+3}{5} + 360 \binom{2n+4}{7}, \\
2^8 S_8(n) &amp; = \binom{2n+2}{3} + 252 \binom{2n+3}{5} + 5040 \binom{2n+4}{7} + 20160 \binom{2n+5}{9}.
\end{align*}

At this point, it is pertinent to bring up the sequence \seqnum{A304330} in the On-Line Encyclopedia of Integer Sequences (OEIS) \cite{OEIS}, whose general term is given by
\begin{equation}\label{exprkm}
R(k,m) = \sum_{j=0}^m (-1)^j \binom{2m}{j} (m-j)^{2k}, \quad 0 \leq m \leq k.
\end{equation}
Table \ref{tab:1} displays the first few rows of the numerical triangle for the sequence \seqnum{A304330}. As we will see now, there is an intimate connection between this sequence and the sequence of central factorial numbers with even indices of the second kind $T(2k,2m)$, listed as \seqnum{A036969} in the OEIS. Following the notation introduced by Gelineau and Zeng \cite{gelineau}, in this paper we will write $U(k,m)$ to refer to $T(2k,2m)$. As shown by Butzer et al.\ \cite[Proposition 2.4 (xiii)]{butzer}, the numbers $U(k,m)$ admit the explicit formula
\begin{equation*}
U(k,m) = \frac{2}{(2m)!} \sum_{j=0}^m (-1)^{m+j} \binom{2m}{m-j} j^{2k},
\end{equation*}
or, equivalently,
\begin{equation*}
U(k,m) = \frac{2}{(2m)!} \sum_{j=0}^m (-1)^j \binom{2m}{j} (m-j)^{2k},
\end{equation*}
where it is assumed that $0 \leq m \leq k$. Thus, $R(k,m)$ and $U(k,m)$ are related by
\begin{equation}\label{int2}
R(k,m) = \frac{(2m)!}{2} U(k,m).
\end{equation}

\begin{table}[ttt]
\centering
\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline
$k \backslash m$ &amp; $m=0$ &amp; $m=1$ &amp; $m=2$ &amp; $m=3$ &amp; $m=4$ &amp; $m=5$ &amp; $m=6$ \\
\hline\hline
$k=0$ &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$k=1$ &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$k=2$ &amp; 0 &amp; 1 &amp; 12 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$k=3$ &amp; 0 &amp; 1 &amp; 60 &amp; 360 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$k=4$ &amp; 0 &amp; 1 &amp; 252 &amp; 5040 &amp; 20160 &amp; 0 &amp; 0  \\ \hline
$k=5$ &amp; 0 &amp; 1 &amp; 1020 &amp; 52920 &amp; 604800 &amp; 1814400 &amp; 0  \\ \hline
$k=6$ &amp; 0 &amp; 1 &amp; 4092 &amp; 506880 &amp; 12640320 &amp; 99792000 &amp; 239500800  \\ \hline
\end{tabular}
\caption{Triangular array of the numbers $R(k,m)$ up to $k=6$.}\label{tab:1}
\end{table}


We can see that the integer coefficients that appear in the previous expressions of $2^{2k} S_{2k}(n)$, $k =1,2,3,4$, are precisely the entries in the $k$th row of the triangle in Table~\ref{tab:1}. We could then conjecture that this rule effectively applies for successive values of $k$ and, eventually, for every $k$. The following theorem (which is proven in Section \ref{sec:2}) confirms that this is indeed the case.
\begin{theorem}\label{th:1}
For integers $k \geq 1$, we have
\begin{equation}\label{th1}
2^{2k} S_{2k}(n) = \sum_{m=1}^k R(k,m) \binom{2n+m+1}{2m+1}.
\end{equation}
\end{theorem}


On the other hand, in his excellent paper on Johann Faulhaber and power sums, Knuth \cite[p.\ 285]{knuth} derived the following alternative, explicit formulas for $S_{2k}(n)$, $k=1,\ldots,6$:
\begin{align*}
S_2(n) &amp; = T_1(n),  \\
S_4(n) &amp; = T_1(n) + 12 T_2(n), \\
S_6(n) &amp; = T_1(n) + 60 T_2(n) + 360 T_3(n), \\
S_8(n) &amp; = T_1(n) + 252 T_2(n) + 5040 T_3(n) + 20160 T_4(n), \\
S_{10}(n) &amp; =  T_1(n) + 1020 T_2(n) + 52920 T_3(n) + 604800 T_4(n) + 1814400 T_5(n), \\
S_{12}(n) &amp; =  T_1(n) + 4092 T_2(n) + 506880 T_3(n) + 12640320 T_4(n) + 99792000 T_5(n) \\
&amp; \qquad \qquad \qquad \qquad \quad \,\, + 239500800 T_6(n),
\end{align*}
where
\begin{equation*}
T_m(n) = \frac{2n+1}{2m+1} \binom{n+m}{2m}.
\end{equation*}

Knuth also found \cite[pp.\ 285--286]{knuth} the following relationship between the power sums $S_{2k}(n)$ and $S_{2k-1}(n)$.
\begin{proposition}[Knuth, 1993]\label{pr:1}
The formula
\begin{equation*}
\frac{S_{2k}(n)}{2n+1} = a_1 \binom{n+1}{2} + a_2 \binom{n+2}{4} + \cdots + a_k \binom{n+k}{2k},
\end{equation*}
holds if and only if
\begin{equation*}
S_{2k-1}(n) = \frac{3}{1}a_1 \binom{n+1}{2} + \frac{5}{2}a_2 \binom{n+2}{4} + \cdots +
\frac{2k+1}{k} a_k \binom{n+k}{2k}.
\end{equation*}
\end{proposition}

Since the coefficients $a_1, a_2, \ldots, a_k$ for $S_{2k-1}(n)$ are known (see the formula for $S_{2k-1}(n)$ at the bottom of p.\ 284 of Knuth's paper \cite{knuth}), Proposition \ref{pr:1} enables one to fully determine $S_{2k}(n)$ for arbitrary $k$. The explicit expressions of the above formulas for $S_{2k}(n)$ and $S_{2k-1}(n)$ are then given as follows.
\begin{proposition}[Knuth, 1993]\label{pr:2}
For integers $k\geq 1$, we have
\begin{align}
S_{2k}(n) &amp; = \sum_{m=1}^k R(k,m) \, \frac{2n+1}{2m+1} \binom{n+m}{2m}, \label{pr21} \\[-2mm]
\intertext{and}
S_{2k-1}(n) &amp; = \sum_{m=1}^k  R(k,m) \, \frac{1}{m} \binom{n+m}{2m}. \label{pr22}
\end{align}
\end{proposition}

The rest of the paper is organized as follows. In Section \ref{sec:2}, we firstly prove the above formulas \eqref{th1}, \eqref{pr21}, and \eqref{pr22}. As in the case of \eqref{th1}, our proof of \eqref{pr21} and \eqref{pr22} relies on the respective binomial identity as well as on a pair of properties of the central factorial numbers with even indices. As a by-product, we obtain formulas for the power sums $T_{2k}(n) = \sum_{i=1}^n (2i -1)^{2k}$ and $\Omega_{2k} = \sum_{i=1}^n (-1)^{n-i} i^{2k}$. Furthermore, we derive a couple of formulas for the product $2^{2k-1} S_{2k-1}(n)$ involving the numbers $R(k,m)$ (Theorem \ref{th:9} and equation \eqref{odd2}, respectively). In Section~\ref{sec:3}, we deal with the so-called Faulhaber form of the power sums $S_{2k}(n)$ and $S_{2k+1}(n)$ and provide an alternative representation of the Faulhaber coefficients in terms on $R(k,m)$ and the Legendre-Stirling numbers of the first kind (Proposition \ref{prop:11}). We conclude the paper in Section \ref{sec:4} with some additional remarks.

Next we summarize the main results found in this paper, highlighting the close relationship between the integer sequence \seqnum{A304330} (with general term given by \eqref{exprkm}) and the three varieties of power sums considered in this paper, namely, $S_k(n) = 1^k + 2^k + \cdots + n^k$, $T_k(n) = 1^k + 3^k + \cdots + (2n-1)^k$, and $\Omega_k(n) = n^k - (n-1)^k + \cdots + (-1)^{n-1} 1^k$. (Note that the formulas for $S_{2k}(n)$ and $S_{2k-1}(n)$ were previously obtained in Knuth's paper \cite{knuth}.)

For integers $k \geq 1$, we have
\begin{align*}
2^{2k} S_{2k}(n) &amp; = \sum_{m=1}^k R(k,m) \binom{2n+m+1}{2m+1},  \\
S_{2k}(n) &amp; = \sum_{m=1}^k R(k,m) \, \frac{2n+1}{2m+1} \binom{n+m}{2m}, \\
T_{2k}(n) &amp; = \sum_{m=1}^k R(k,m) \binom{2n+m}{2m+1}, \\
\Omega_{2k}(n) &amp; = \sum_{m=1}^k R(k,m) \binom{n+m}{2m}, \\
S_{2k-1}(n) &amp; = \sum_{m=1}^k R(k,m) \, \frac{1}{m} \binom{n+m}{2m}, \\[-2mm]
\intertext{and}
2^{2k-1} S_{2k-1}(n) &amp; = \sum_{m=1}^k  Q_{k,m}(n) \binom{n+m}{2m-1},
\end{align*}
where
\begin{displaymath}
Q_{k,m}(n) = \begin{cases}
  n^{2k-1}, &amp; \text{for $m=1$;} \\
  2 \, \displaystyle{\sum_{j=m}^k \binom{2k-1}{2j-2} R(j-1,m-1) n^{2k-2j+1}}, &amp; \text{for $m \geq 2$.}
\end{cases}
\end{displaymath}

Furthermore, $S_{2k}(n)$ and $S_{2k+1}(n)$ can be expressed in the Faulhaber form $S_{2k}(n) = S_2(n) \sum_{r=1}^k b_{k,r} \big( S_1(n) \big)^{r-1}$ and $S_{2k+1}(n) = \big( S_1(n) \big)^2 \sum_{r=1}^k c_{k,r} \big( S_1(n) \big)^{r-1}$, with coefficients $b_{k,r}$ and $c_{k,r}$ given by
\begin{align*}
b_{k,r} &amp; = \sum_{m=r}^k  \frac{3 \cdot 2^r}{(2m+1)!} \, R(k,m) Ps_{m}^{(r)}, \\[-3mm]
\intertext{and}
c_{k,r} &amp; = \sum_{m=r}^k  \frac{2^{r+1}}{(2m+2)! (m+1)} \, R(k+1,m+1) Ps_{m+1}^{(r+1)},
\end{align*}
where $Ps_{m}^{(r)}$ are the Legendre-Stirling numbers of the first kind.

In addition, $\Omega_{2k}(n)$ can be expressed in the form $\Omega_{2k}(n) = \sum_{r=1}^k d_{k,r} \big( S_1(n) \big)^{r}$, with coefficients $d_{k,r}$ given by
\begin{equation*}
d_{k,r} = \sum_{m=r}^k  \frac{2^{r}}{(2m)!} \, R(k,m) Ps_{m}^{(r)}.
\end{equation*}



\section{Sums of powers of integers involving the sequence \seqnum{A304330}}\label{sec:2}

In this section we successively prove the formulas for $2^{2k} S_{2k}(n)$, $S_{2k}(n)$, and $S_{2k-1}(n)$ stated in Theorem \ref{th:1} and Proposition \ref{pr:2} above. Then we obtain formulas for $T_{2k}(n)$ and $\Omega_{2k}(n)$, and derive a formula for $2^{2k-1}S_{2k-1}(n)$ as given in Theorem \ref{th:9}. Lastly, we present an alternative formula for $2^{2k-1}S_{2k-1}(n)$ in equation \eqref{odd2}.


\subsection{Proof of formula \eqref{th1}}

To prove \eqref{th1}, we need the following lemma.

\begin{lemma}\label{lm:3}
For integers $n \geq 1$, we have
\begin{equation*}
\sum_{i=1}^n (2i) \binom{2i+k-1}{2k-1} = k \binom{2n+k+1}{2k+1},
\end{equation*}
where $k$ is an arbitrary, fixed positive integer.
\end{lemma}

For later use we put the above binomial identity in the equivalent form
\begin{equation}\label{lm31}
\sum_{i=1}^n 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) = \frac{(2n-k+1)(2n-k+2) \cdots (2n+k+1)}{2k+1},
\end{equation}
and refer the reader to the Appendix for a proof of \eqref{lm31}.

For the proof of \eqref{th1}, we make use of two well-known properties of the (signed) central factorial numbers with even indices of the first kind $t(2k,2m)$ (OEIS \seqnum{A204579}). In what follows we denote these numbers by $u(k,m)$, in accordance with the notation of Gelineau and Zeng \cite{gelineau}. The first property is the definition of $u(k,m)$ through the generating function
\begin{equation}\label{th11}
\sum_{m=1}^k u(k,m) x^m = x (x -1^2) (x -2^2) \cdots (x - (k-1)^2 ), \quad k \geq 1,
\end{equation}
(see, e.g., \cite[Equation (4.15)]{coffey} and Equation (11) (with $r=0$) of \cite{shiha}). The second property of interest is the central factorial inversion, which is a direct consequence of the orthogonality relations $\sum_{m=i}^k u(k,m) U(m,i) = \sum_{m=i}^k U(k,m) u(m,i) = \delta_{k,i}$ ($\delta_{k,i}$ being the Kronecker delta). This property is stated in the following lemma.

\begin{lemma}[{{\cite[Proposition 2.3]{butzer}}}]\label{lm:4}
If $(a_k )_{k \geq 1}$ and $(b_k )_{k \geq 1}$ are two sequences of real numbers (where we tacitly assume that $a_0 = b_0 =0$), there holds the inversion formula
\begin{equation*}
a_k = \sum_{m=1}^k u(k,m) b_m  \,\, \Longleftrightarrow \,\, b_k = \sum_{m=1}^k U(k,m) a_m, \quad k \geq 1.
\end{equation*}
\end{lemma}

Therefore, setting $x =(2i)^2$ in \eqref{th11} and multiplying through by $2$ gives
\begin{align*}
\sum_{m=1}^k u(k,m) 2^{2m+1} i^{2m} &amp; = 8 i^2 ((2i)^2 -1^2) ((2i)^2 -2^2 ) \cdots
((2i)^2 - (k-1)^2 )   \\
&amp; = 4i (2i-k+1) (2i-k+2) \cdots (2i+k-1).
\end{align*}
Furthermore, summing from $i = 1$ to $n$ on both sides of the last equation and employing \eqref{lm31} in the right-hand side, we get
\begin{equation*}
\sum_{m=1}^k u(k,m) 2^{2m+1} S_{2m}(n) = (2k)! \binom{2n+k+1}{2k+1}.
\end{equation*}
This allows us to apply Lemma \ref{lm:4} with $a_k = (2k)! \binom{2n+k+1}{2k+1}$ and $b_m = 2^{2m+1} S_{2m}(n)$ to finally obtain the following formula for $2^{2k} S_{2k}(n)$:
\begin{equation*}
2^{2k} S_{2k}(n) = \frac{1}{2} \sum_{m=1}^k U(k,m) (2m)! \binom{2n+m+1}{2m+1},
\end{equation*}
which, in view of \eqref{int2}, is just the formula \eqref{th1}.

As a result we find the following representation of $R(k,m)$, which follows straightforwardly by comparing the formulas \eqref{int1} and \eqref{th1}.
\begin{corollary}
For positive integers $k$ and $m$, with $1 \leq m \leq k$, we have
\begin{equation}\label{col5}
R(k,m) = 2^{2k+m-1} m! \sum \frac{(2k)!}{b_1! b_2! \cdots b_k!}
\prod_{r=1}^{k} \biggl( \frac{1}{4^r (2r)!} \biggl)^{b_r},
\end{equation}
where the summation extends over all $k$-tuples of nonnegative integers $(b_1,b_2,\ldots,b_k)$ fulfilling the conditions $b_1 + 2b_2 + \cdots + k b_k = k$ and $b_1 + b_2 +\cdots +b_k = m$.
\end{corollary}

Using the above representation, we get the following diagonals $R(k,k-s)$ for $s =0,1,2,3,4$:
\begin{align*}
R(k,k) &amp; = \frac{1}{2} (2k)!,  \quad k \geq 1, \\
R(k,k-1) &amp; = \frac{k-1}{24} (2k)!,  \quad k \geq 2, \\
R(k,k-2) &amp; = \frac{(k-2)(5k-11)}{2880} (2k)!,  \quad k \geq 3, \\
R(k,k-3) &amp; = \frac{(k-3)(35k^2 -231k +382)}{725760} (2k)!,  \quad k \geq 4, \\
R(k,k-4) &amp; = \frac{(k-4)(175k^3 -2310k^2 + 10181k -14982)}{174182400} (2k)!, \quad k \geq 5.
\end{align*}

From the above particular cases, we can guess the general pattern
\begin{equation*}
R(k,k-s) = (2k)! (k-s) P_s(k), \,\,\, \text{for} \,\,\, k \geq s+1 \,\,\, \text{and} \,\,\, s\geq 1,
\end{equation*}
where $P_s(k)$ is a polynomial in $k$ of degree $s-1$ with coefficients having alternating signs.


\subsection{Proof of formulas \eqref{pr21} and \eqref{pr22}}

Next we prove \eqref{pr21}. For this, we need the following lemma.

\begin{lemma}\label{lm:6}
For integers $n \geq 1$, we have
\begin{equation*}
\sum_{i=1}^n i \binom{i+k-1}{2k-1} = \frac{k(2n+1)}{2k+1} \binom{n+k}{2k},
\end{equation*}
where $k$ is an arbitrary, fixed positive integer.
\end{lemma}

Likewise, for later use we put the above binomial identity in the equivalent form
\begin{equation}\label{lm51}
\sum_{i=1}^n 2i (i-k+1)(i-k+2)\cdots (i+k-1) = \frac{(2n+1)(n-k+1)(n-k+2) \cdots (n+k)}{2k+1},
\end{equation}
and refer the reader to the Appendix for a proof of \eqref{lm51}.

Now, setting $x =i^2$ in \eqref{th11} and summing from $i = 1$ to $n$ on both sides gives
\begin{equation*}
\sum_{m=1}^k u(k,m) S_{2m}(n) = \frac{1}{2} \sum_{i=1}^n 2i (i-k+1) (i-k+2) \cdots (i+k-1).
\end{equation*}
By virtue of \eqref{lm51}, this can be expressed as
\begin{equation*}
\sum_{m=1}^k u(k,m) S_{2m}(n) = \frac{(2k)!}{2} \, \frac{2n+1}{2k+1} \binom{n+k}{2k}.
\end{equation*}
Thus, taking $a_k = \frac{(2k)!}{2} \, \frac{2n+1}{2k+1} \binom{n+k}{2k}$ and $b_m = S_{2m}(n)$ in Lemma \ref{lm:4}, we obtain the following formula for $S_{2k}(n)$:
\begin{equation*}
S_{2k}(n) = \sum_{m=1}^k U(k,m) \frac{(2m)!}{2} \, \frac{2n+1}{2m+1} \binom{n+m}{2m},
\end{equation*}
which, by relation \eqref{int2}, is the same as formula \eqref{pr21}.

On the other hand, to show \eqref{pr22}, set $x =i^2$ in \eqref{th11} and divide both sides by $i$. This gives
\begin{equation*}
\sum_{m=1}^k u(k,m)  i^{2m-1} = (i-k+1) (i-k+2) \cdots (i+k-1).
\end{equation*}
Thus, summing over $i$ on both sides of the above equation and making use of the well-known binomial identity $\sum_{i=1}^n \binom{i+k-1}{2k-1} = \binom{n+k}{2k}$ (see \cite[Identity 58]{spivey}), we  get
\begin{equation*}
\sum_{m=1}^k u(k,m) S_{2m-1}(n) = (2k-1)! \binom{n+k}{2k}.
\end{equation*}
Finally, by inverting the equation above, we recover the known formula \cite[p.\ 284]{knuth}
\begin{equation*}
S_{2k-1}(n) = \sum_{m=1}^k (2m-1)! U(k,m) \binom{n+m}{2m},
\end{equation*}
which can be put in the form \eqref{pr22} upon using \eqref{int2}.

\begin{remark}
Formula \eqref{pr21} can be derived directly from \eqref{pr22} and the well-known recursive formula (see, e.g., \cite[Proposition 1]{ziel})
\begin{equation*}
S_k(n) = (n+1) S_{k-1}(n) - \sum_{i=1}^n S_{k-1}(i).
\end{equation*}
Indeed, putting $2k$ instead of $k$ in the above equation and substituting $S_{2k-1}(n)$ and $S_{2k-1}(i)$ from \eqref{pr22} gives
\begin{align*}
S_{2k}(n) &amp; = (n+1)\sum_{m=1}^k  R(k,m) \, \frac{1}{m} \binom{n+m}{2m}
- \sum_{m=1}^k  R(k,m) \, \frac{1}{m} \sum_{i=1}^n \binom{i+m}{2m} \\
&amp; = \sum_{m=1}^k  R(k,m) \, \frac{1}{m} \biggl( (n+1) \binom{n+m}{2m}
- \frac{n+m+1}{2m+1} \binom{n+m}{2m} \biggl) \\
&amp; = \sum_{m=1}^k  R(k,m) \, \frac{2n+1}{2m+1} \binom{n+m}{2m}.
\end{align*}
\end{remark}


\subsection{Formulas for $T_{2k}(n)$ and $\Omega_{2k}(n)$}

For integers $k \geq 1$, the sum of the $k$th powers of the first $n$ odd integers $T_k(n) = 1^k + 3^k + \cdots + (2n-1)^k$ can be expressed as $T_k(n) = S_k(2n) - 2^k S_k(n)$. Consequently, for the case of even powers, from \eqref{th1} and \eqref{pr21} we have
\begin{equation*}
T_{2k}(n) = \sum_{m=1}^k R(k,m) \, \frac{4n+1}{2m+1} \binom{2n+m}{2m}
- \sum_{m=1}^k R(k,m) \binom{2n+m+1}{2m+1}.
\end{equation*}
Furthermore, since
\begin{equation*}
\frac{4n+1}{2m+1} \binom{2n+m}{2m} - \frac{2n+m+1}{2m+1} \binom{2n+m}{2m} = \frac{2n-m}{2m+1}
\binom{2n+m}{2m} = \binom{2n+m}{2m+1},
\end{equation*}
this becomes
\begin{equation}\label{oddi}
T_{2k}(n) = \sum_{m=1}^k R(k,m) \binom{2n+m}{2m+1}.
\end{equation}

\begin{remark}
Merca \cite[Equation (3.1)]{merca} proved that $T_{2k}(n) = \frac{2^{2k}}{2k+1} B_{2k+1}\big( n +\frac{1}{2} \big)$, where $B_k(n)$ is the $k$th Bernoulli polynomial evaluated at $n$. Thus, letting $n \to n - \frac{1}{2}$ in Merca's result, and using \eqref{oddi}, gives the identity
\begin{equation*}
B_{2k+1}(n) = \frac{2k+1}{2^{2k}} \sum_{m=1}^k R(k,m) \binom{2n+m-1}{2m+1}, \quad k \geq 1.
\end{equation*}
Note that this identity can also be obtained directly from \eqref{th1} and the well-known relationship $S_{2k}(n) = \frac{1}{2k+1} ( B_{2k+1}(n+1) - B_{2k+1})$, once one realizes that $B_{2k+1} =0$ for all $k \geq 1$.
\end{remark}

On the other hand, for integers $k \geq 1$, define the alternating sum of the $k$th powers of the first $n$ integers as
\begin{equation*}
\Omega_k(n) = n^k - (n-1)^k + \cdots + (-1)^{n-1} 1^k.
\end{equation*}
It is easily seen that $\Omega_k(2n) = S_k(2n) - 2T_k(n)$ and $\Omega_k(2n-1) = 2T_k(n) - S_k(2n-1)$. Hence, for the case of even powers, from \eqref{pr21} and \eqref{oddi} we find that $\Omega_{2k}(2n) = \sum_{m=1}^k R(k,m) \binom{2n+m}{2m}$ and $\Omega_{2k}(2n-1) = \sum_{m=1}^k R(k,m) \binom{2n+m-1}{2m}$. This means that
\begin{equation}\label{alter}
\Omega_{2k}(n) = \sum_{m=1}^k R(k,m) \binom{n+m}{2m},
\end{equation}
holds for all $k,n \geq 1$. In particular, $\Omega_2(n) = \binom{n+1}{2}$. See equation \eqref{vien1} below for an alternative representation of $\Omega_{2k}(n)$ as a polynomial in $S_1(n)$ of degree $k$.


\subsection{Corresponding formulas for $2^{2k-1}S_{2k-1}(n)$}

An explicit formula for $2^{2k-1} S_{2k-1}(n)$ involving the numbers $R(k,m)$ is given by the following theorem.
\begin{theorem}\label{th:9}
For integers $k \geq 1$, we have
\begin{equation}\label{th9}
2^{2k-1} S_{2k-1}(n) = \sum_{m=1}^k  Q_{k,m}(n) \binom{n+m}{2m-1},
\end{equation}
where $Q_{k,m}(n)$ is an odd polynomial in $n$ of degree $2k-2m+1$ given by
\begin{displaymath}
Q_{k,m}(n) = \begin{cases}
  n^{2k-1}, &amp; \text{for $m=1$;} \\
  2 \, \displaystyle{\sum_{j=m}^k \binom{2k-1}{2j-2} R(j-1,m-1) n^{2k-2j+1}}, &amp; \text{for $m \geq 2$.}
\end{cases}
\end{displaymath}
\end{theorem}
\begin{proof}
Formula \eqref{th9} follows readily by combining the representation of $R(k,m)$ in \eqref{col5} and the following formula for $2^{2k-1} S_{2k-1}(n)$ given in \cite[Theorem 3]{cere}, namely
\begin{multline*}
2^{2k-1} S_{2k-1}(n) = (n+1) n^{2k-1} + \sum_{j=1}^{k-1} 4^j n^{2k-2j-1} \binom{2k-1}{2j}
\sum \frac{(2j)!}{b_1! b_2! \cdots b_j!}  \\
\times \, \prod_{r=1}^{j} \biggl( \frac{1}{4^r (2r)!} \biggl)^{b_r} 2^m m! \binom{n+m+1}{2m+1},
\end{multline*}
where, for each $j =1,2,\ldots,k-1$, the rightmost summation is taken over all \mbox{$j$-tuples} of nonnegative integers $(b_1,b_2,\ldots,b_j)$ satisfying the constraint $b_1 + 2b_2 + \cdots + j b_j = j$, and where $m = b_1 + b_2 +\cdots +b_j$.
\end{proof}

It is to be noted that formula \eqref{th9} above actually involves a double summation, which makes it more complicated than its counterpart \eqref{th1}. We believe, however, that the structure of such a formula is interesting in its own right. In this respect, it is worth to emphasize the striking odd-parity property of the polynomials $Q_{k,m}(n)$. As an example, for $k=5$, we have
\begin{align*}
Q_{5,1}(n) &amp; = n^9, \,\,\, Q_{5,2}(n) = 72n^7 + 252n^5 +168n^3 + 18n, \\[1mm]
Q_{5,3}(n) &amp; = 3024 n^5 + 10080n^3 +4536n, \,\,\, Q_{5,4}(n)= 60480 n^3 + 90720n, \,\,\, Q_{5,5}(n)= 362880 n,
\end{align*}
from which it follows that
\begin{align*}
2^9 S_9(n) &amp; = n^9 \binom{n+1}{1} + (72n^7 + 252n^5 +168n^3 + 18n) \binom{n+2}{3} \\
&amp; + (3024 n^5 + 10080n^3 +4536n) \binom{n+3}{5} \\
&amp; + (60480 n^3 + 90720n) \binom{n+4}{7} + 362880 n \binom{n+5}{9}.
\end{align*}

On the other hand, by setting $x =(2i)^2$ in \eqref{th11}, dividing through by $2i$, summing over $n$, and inverting the resulting equation, one arrives at the following formula for $2^{2k-1} S_{2k-1}(n)$:
\begin{equation}\label{odd2}
2^{2k-1} S_{2k-1}(n) = \sum_{m=1}^k  R(k,m) \, \frac{F_m(n)}{m},
\end{equation}
where
\begin{equation*}
F_m(n) = \sum_{i=1}^n \binom{2i+m-1}{2m-1},
\end{equation*}
is a polynomial in $n$ of degree $2m$. The first few of these polynomials are $F_1(n) = 2\binom{n+1}{2}$ and
\begin{align*}
F_2(n) &amp; = \frac{1}{3} (2n^2 +2n -1) \binom{n+1}{2}, \\
F_3(n) &amp; = \frac{2}{15} (8n^2 +8n -3) \binom{n+2}{4}, \\
F_4(n) &amp; = \frac{1}{105} (8n^4 +16n^3 -24n^2 -32n +9) \binom{n+2}{4}, \\
F_5(n) &amp; = \frac{2}{315} (16n^4 +32n^3 -46n^2 -62n +15) \binom{n+3}{6}, \\
F_6(n) &amp; = \frac{1}{10395} (32n^6 +96n^5 -304n^4 -768n^3 +677n^2 +1077n -225) \binom{n+3}{6}, \\
F_7(n) &amp; = \frac{2}{135135} (256n^6 +768n^5 -2400n^4 -6080n^3 +5168n^2 +8336n -1575) \binom{n+4}{8},
\end{align*}
from which it can be inferred that $F_m(n)$ factorizes as
\begin{equation*}
F_m(n) = G_m(n) \binom{n + \lfloor \frac{m+1}{2} \rfloor}{2 \lfloor \frac{m+1}{2} \rfloor}, \quad m \geq 1,
\end{equation*}
where $\lfloor \cdot \rfloor$ denotes the floor function and $G_m(n)$ is a polynomial in $n$ of degree $2m - 2 \lfloor \frac{m+1}{2} \rfloor$. Unfortunately, there appears to be no recognizable closed-form expression for $G_m(n)$.

For comparison, we write down the expression of $2^9 S_9(n)$ that is obtained from \eqref{odd2}
\begin{align*}
2^9 S_9(n) &amp; = (340n^2 +340n -168) \binom{n+1}{2} \\
&amp; + (11520 n^4 + 23040n^3 -15744n^2 -27264n +5904) \binom{n+2}{4} \\
&amp; + (36864 n^4 + 73728n^3 -105984n^2 -142848n +34560)\binom{n+3}{6}.
\end{align*}


\section{Faulhaber form of $S_{2k}(n)$ and $S_{2k+1}(n)$}\label{sec:3}

Edwards' seminal work \cite{edw} triggered an intensive research on the so-called Faulhaber polynomials, named after the German mathematician Johann Faulhaber (1580--1635); see, e.g., \cite{bazso}, \cite[Section 3]{beardon}, \cite{chen,coffey}, \cite[Section 12]{gessel}, and \cite{kellner,knuth,krishna,ziel}). As is well-known, the power sums $S_{2k}(n)$ and $S_{2k+1}(n)$ (with $k \geq 1$) can be expressed in the Faulhaber form as
\begin{align}
S_{2k}(n) &amp; = S_2(n)  \big[ b_{k,1} + b_{k,2} S_1(n) + b_{k,3} \big( S_1(n) \big)^2 + \cdots + b_{k,k} \big( S_1(n)
\big)^{k-1} \big],  \label{fau1}  \\[-2mm]
\intertext{and}
S_{2k+1}(n) &amp; =  \big ( S_1(n) \big)^2  \big[ c_{k,1} + c_{k,2} S_1(n) + c_{k,3} \big( S_1(n) \big)^2 + \cdots + c_{k,k}
\big( S_1(n) \big)^{k-1} \big], \label{fau2}
\end{align}
where the Faulhaber coefficients $b_{k,r}$ and $c_{k,r}$, $r =1,2,\ldots,k$, are (nonzero) rational numbers satisfying the relation
\begin{equation}\label{fau3}
b_{k,r} = \frac{3r+3}{4k+2} \, c_{k,r}.
\end{equation}

\begin{remark}
An explicit formula for $c_{k,r}$ in terms of the Bernoulli numbers was first obtained by Gessel and Viennot \cite[Equation (12.10)]{gessel}. In our notation, this last formula can be rewritten in the form (cf.\ \cite[Equation (3.5)]{cere2} and \cite[Equation (5-3)]{kellner})
\begin{equation*}
c_{k,r} = (-1)^{r-1} \frac{2^{r+1}}{r+1} \sum_{m=0}^{\lfloor \frac{r-1}{2} \rfloor} \binom{2r-1-2m}{r}
\binom{2k+1}{2m+1} B_{2k-2m}.
\end{equation*}
\end{remark}

In the next proposition we show how Knuth's formulas \eqref{pr21} and \eqref{pr22} for, respectively, $S_{2k}(n)$ and $S_{2k+1}(n)$, can be converted into the corresponding Faulhaber form by exploiting the relationship between $\binom{n+m}{2m}$ and the Legendre-Stirling numbers of the first kind.
\begin{proposition}\label{prop:11}
For integers $k \geq 1$, the power sum polynomials $S_{2k}(n)$ and $S_{2k+1}(n)$ can be expressed in the Faulhaber form \eqref{fau1} and \eqref{fau2}, respectively, with coefficients $b_{k,r}$ and $c_{k,r}$ given by
\begin{align}
b_{k,r} &amp; = \sum_{m=r}^k  \frac{3 \cdot 2^r}{(2m+1)!} \, R(k,m) Ps_{m}^{(r)}, \label{coeb} \\
c_{k,r} &amp; = \sum_{m=r}^k  \frac{2^{r+1}}{(2m+2)! (m+1)} \, R(k+1,m+1) Ps_{m+1}^{(r+1)}, \label{coec}
\end{align}
for $r =1,2, \ldots,k$, and where $Ps_{m}^{(r)}$ are the Legendre-Stirling numbers of the first kind.
\end{proposition}
\begin{proof}
Let us first recall that, according to \cite[Theorem 5.5]{andrews}, the (signed) Legendre-Stirling numbers of the first kind $Ps_{m}^{(r)}$ (OEIS \seqnum{A129467}) satisfy the horizontal generating function (for integers $r$ and $m$ fulfilling $1\leq r \leq m$)
\begin{equation*}
\langle x \rangle_m = \sum_{r=1}^m Ps_m^{(r)} x^r,
\end{equation*}
where $\langle x \rangle_m$ is the generalized falling factorial defined by
\begin{equation*}
\langle x \rangle_m = \prod_{j=0}^{m-1} (x - j(j+1)).
\end{equation*}
(For the sake of illustration, the first values of $Ps_{m}^{(r)}$ are listed in Table \ref{tab:2}.) Now, invoking the binomial expansion (see \cite[Equation (5.1)]{coffey})
\begin{equation}\label{fau4}
\binom{n+m}{2m} = \frac{1}{(2m)!} \sum_{r=1}^m Ps_m^{(r)} \big( n(n+1) \big)^r,
\end{equation}
and combining \eqref{fau4} with \eqref{pr21}, we obtain
\begin{align*}
S_{2k}(n) &amp; = (2n+1) \sum_{m=1}^k \sum_{r=1}^m \frac{2^r}{(2m+1)!} \, R(k,m) Ps_{m}^{(r)} \big( S_1(n) \big)^r  \\
&amp; = S_2(n) \sum_{r=1}^k  \Biggl( \sum_{m=r}^k  \frac{3 \cdot 2^r}{(2m+1)!} \, R(k,m) Ps_{m}^{(r)} \Biggl) \big( S_1(n) \big)^{r-1},
\end{align*}
which is of the form \eqref{fau1} with the coefficients $b_{k,r}$ given by \eqref{coeb}.

\begin{table}[ttt]
\centering
\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline
$m \backslash r$ &amp; $r=0$ &amp; $r=1$ &amp; $r=2$ &amp; $r=3$ &amp; $r=4$ &amp; $r=5$ &amp; $r=6$ \\
\hline\hline
$m=0$ &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$m=1$ &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$m=2$ &amp; 0 &amp; $-2$ &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$m=3$ &amp; 0 &amp; 12 &amp; $-8$ &amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ \hline
$m=4$ &amp; 0 &amp; $-144$ &amp; 108 &amp; $-20$ &amp; 1 &amp; 0 &amp; 0  \\ \hline
$m=5$ &amp; 0 &amp; 2880 &amp; $-2304$ &amp; 508 &amp; $-40$ &amp; 1 &amp; 0  \\ \hline
$m=6$ &amp; 0 &amp; $-86400$ &amp; 72000 &amp; $-17544$ &amp; 1708 &amp; $-70$ &amp; 1 \\ \hline
\end{tabular}
\caption{The Legendre-Stirling numbers of the first kind $Ps_{m}^{(r)}$ up to $m=6$.}\label{tab:2}
\end{table}

Likewise, by using \eqref{fau4} in \eqref{pr22}, we obtain
\begin{align*}
S_{2k+1}(n) &amp; = \sum_{m=1}^{k+1} \sum_{r=1}^m \frac{2^r}{(2m)! m} \, R(k+1,m) Ps_{m}^{(r)} \big( S_1(n) \big)^r  \\
&amp; = \sum_{r=1}^{k+1} \Biggl( \sum_{m=r}^{k+1}  \frac{2^r}{(2m)! m} \, R(k+1,m) Ps_{m}^{(r)} \Biggl ) \big( S_1(n) \big)^{r}.
\end{align*}
The key point to observe here is that the term in $S_1(n)$ in the above expansion of $S_{2k+1}(n)$ does vanish because the derivative of $S_{2k+1}(n)$ with respect to $n$ (considering $n$ as a continuous variable) evaluated at $n=0$, namely, $S_{2k+1}^{\prime}(0) = (-1)^{2k+1} B_{2k+1}$, is equal to zero for all $k \geq 1$. This means that
\begin{equation}\label{fau5}
\sum_{m=1}^{k+1}  \frac{1}{(2m)! m} \, R(k+1,m) Ps_{m}^{(1)} =0,
\end{equation}
and then
\begin{align*}
S_{2k+1}(n)  &amp; = \sum_{r=2}^{k+1} \Biggl( \sum_{m=r}^{k+1} \frac{2^r}{(2m)! m} \, R(k+1,m)
Ps_{m}^{(r)} \Biggl)
\big( S_1(n) \big)^{r} \\
&amp; = \big( S_1(n) \big)^2 \sum_{r=1}^{k} \Biggl( \sum_{m=r}^{k}  \frac{2^{r+1}}{(2m+2)! (m+1)} \,
R(k+1,m+1) Ps_{m+1}^{(r+1)} \Biggl) \big( S_1(n) \big)^{r-1},
\end{align*}
which is of the form \eqref{fau2} with the coefficients $c_{k,r}$ given by \eqref{coec}.
\end{proof}

\begin{remark}
Thanks to relation \eqref{fau3}, the coefficients $c_{k,r}$ can also be expressed as
\begin{equation*}
c_{k,r} = \frac{2k+1}{r+1} \sum_{m=r}^k  \frac{2^{r+1}}{(2m+1)!} \, R(k,m) Ps_{m}^{(r)}.
\end{equation*}
\end{remark}

On the other hand, substituting the expression in \eqref{fau4} for $\binom{n+m}{2m}$ into \eqref{alter}, we obtain the following Faulhaber-like formula for the alternating sum of even powers $\Omega_{2k}(n)$:
\begin{equation}\label{vien1}
\Omega_{2k}(n) = \sum_{r=1}^k d_{k,r} \big( S_1(n) \big)^{r}, \quad k \geq 1,
\end{equation}
where the coefficients $d_{k,r}$ are given by
\begin{equation*}
d_{k,r} = \sum_{m=r}^k  \frac{2^{r}}{(2m)!} \, R(k,m) Ps_{m}^{(r)}.
\end{equation*}

\begin{remark}
Gessel and Viennot \cite[Theorem 28]{gessel} established the following formula for $\Omega_{2k}(n)$:
\begin{equation}\label{vien2}
\Omega_{2k}(n) = \frac{1}{2} \sum_{r=1}^k \mathfrak{s}(k,r) \big( n(n+1)\big)^r, \quad k \geq 1,
\end{equation}
where $\mathfrak{s}(n,r)$ denotes the so-called Sali\'{e} numbers (OEIS \seqnum{A065547}). By comparing \eqref{vien1} and \eqref{vien2}, and recalling \eqref{int2}, we get the following representation of the (signed) Sali\'{e} numbers in terms of $U(k,m)$ and $Ps_m^{(r)}$:
\begin{equation*}
\mathfrak{s}(k,r) = \sum_{m=r}^k U(k,m) Ps_m^{(r)}.
\end{equation*}
\end{remark}


\section{Concluding remarks}\label{sec:4}

We conclude this paper with the following remarks:

From \eqref{fau5} one can deduce the following horizontal recurrence relation for the numbers $R(k,m)$.
\begin{proposition}
For integers $k \geq 2$, we have
\begin{equation*}
\sum_{m=1}^k (-1)^m \frac{( (m-1)! )^2}{(2m)!} \, R(k,m) = 0.
\end{equation*}
\end{proposition}
\begin{proof}
This follows immediately by setting $k \to k-1$ in \eqref{fau5} and using the fact that $ Ps_{m}^{(1)} = (-1)^{m-1} m!(m-1)!$ \cite[Theorem 2]{guo}.
\end{proof}

Let us further note that the above recurrence relation can be written in terms of $U(k,m)$ as
\begin{equation*}
\sum_{m=1}^k (-1)^m ( (m-1)! )^2  U(k,m) = 0,  \quad k \geq 2.
\end{equation*}

Moreover, since $b_{k,1} = 6 B_{2k}$, it follows from \eqref{coeb} that
\begin{equation}\label{b2k}
B_{2k} = \sum_{m=1}^k (-1)^{m-1} \frac{m!(m-1)!}{(2m+1)!} \, R(k,m),  \quad k \geq 1,
\end{equation}
in accordance with the formula for $B_{2k}$ given by Coffey et al.\ \cite[p.\ 29]{coffey}. Hence, combining \eqref{b2k} with the classical Bernoulli's formula \cite[p.\ 48]{knoebel}, $S_k(n) = \frac{1}{k+1} n^{k+1} + \frac{1}{2}n^k + \frac{1}{k+1} \sum_{j=2}^{k} \binom{k+1}{j} B_j n^{k+1-j}$, we obtain that
\begin{equation*}
S_k(n) = \frac{n^{k+1}}{k+1} + \frac{1}{2}n^k - \frac{1}{k+1} \sum_{j=1}^{\lfloor \frac{k}{2} \rfloor}
\sum_{m=1}^j (-1)^m \frac{m!(m-1)!}{(2m+1)!} \binom{k+1}{2j} R(j,m) n^{k+1-2j},
\end{equation*}
which holds for all $k \geq 1$. Note that setting $n=1$ and $k = 2r+1$ in the above formula gives the identity
\begin{equation*}
\sum_{j=1}^r \sum_{m=1}^j (-1)^{m-1} \frac{m!(m-1)!}{(2m+1)!}
\binom{2r+2}{2j} R(j,m) = r, \quad r \geq 1.
\end{equation*}


\section{Acknowledgments}

The author would like to thank the anonymous referee and the editor-in-chief for the careful reading of a first version of the paper and for many insightful comments and suggestions that greatly improved the quality of the paper. The author also wants to thank Dionisio F.\ Y\'{a}\~{n}ez (University of Valencia) for sending me a copy of the paper \cite{butzer}.


\appendix\section{Appendix}

In this section we prove the relations \eqref{lm31} and \eqref{lm51}, which are equivalent to the binomial identities appearing in Lemma \ref{lm:3} and Lemma \ref{lm:6}, respectively.

\subsection{Proof of relation \eqref{lm31}}

We prove \eqref{lm31} by induction on $n$. To this end, we distinguish the cases of even and odd $k$.

\begin{itemize}[leftmargin=*]

\item Even $k$. It is evident that both sides of \eqref{lm31} are identically zero when $n &lt; \frac{k}{2}$. Therefore, it is enough to show that
\begin{equation}\label{lm32}
\sum_{i= \frac{k}{2}}^n 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) = \frac{(2n-k+1)(2n-k+2) \cdots (2n+k+1)}{2k+1},
\end{equation}
for $n \geq \frac{k}{2}$. Relation \eqref{lm32} holds trivially for $n = \frac{k}{2}$. Assume the statement is true for $n = \frac{k}{2} + m$, where $m$ is an arbitrary, fixed nonnegative integer. We prove it for $n = \frac{k}{2} + m +1$
\begin{align*}
\sum_{i= \frac{k}{2}}^{\frac{k}{2} +m+1} &amp; 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) \\[-2mm]
 &amp; = \sum_{i= \frac{k}{2}}^{\frac{k}{2} +m} 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) \\
 &amp; \qquad + (2k +4m+4)(2m+3)(2m+4) \cdots (2m+2k+1) \\[2mm]
 &amp; = \frac{(2m+1)(2m+2) \cdots (2m+2k+1)}{2k+1} \\
 &amp; \qquad + (2k +4m+4)(2m+3)(2m+4) \cdots (2m+2k+1) \\[2mm]
 &amp; = \frac{(2m+3)(2m+4) \cdots (2m+2k+3)}{2k+1}.
\end{align*}
Thus \eqref{lm32} holds for all $n \geq \frac{k}{2}$ by induction.

\item Odd $k$. Likewise, for odd $k$ it is enough to show that
\begin{equation}\label{lm33}
\sum_{i=\frac{k+1}{2}}^n 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) = \frac{(2n-k+1)(2n-k+2) \cdots (2n+k+1)}{2k+1},
\end{equation}
for $n \geq \frac{k+1}{2}$. Clearly, relation \eqref{lm33} is satisfied for $n = \frac{k+1}{2}$. Assuming that \eqref{lm33} is true for $n = \frac{k+1}{2}+m$ (for any given nonnegative integer $m$), we prove it for $n = \frac{k+1}{2}+m +1$
\begin{align*}
\sum_{i=\frac{k+1}{2}}^{\frac{k+1}{2}+m+1} &amp; 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) \\[-2mm]
 &amp; = \sum_{i=\frac{k+1}{2}}^{\frac{k+1}{2}+m} 4i (2i-k+1)(2i-k+2)\cdots (2i+k-1) \\
 &amp; \qquad + (2k +4m+6)(2m+4)(2m+5) \cdots (2m+2k+2) \\[2mm]
 &amp; = \frac{(2m+2)(2m+3) \cdots (2m+2k+2)}{2k+1} \\
 &amp; \qquad + (2k +4m+6)(2m+4)(2m+5) \cdots (2m+2k+2) \\[2mm]
 &amp; = \frac{(2m+4)(2m+5) \cdots (2m+2k+4)}{2k+1},
\end{align*}
which means that \eqref{lm33} holds for all $n\geq \frac{k+1}{2}$ by induction. This completes the proof of \eqref{lm31}.
\end{itemize}


\subsection{Proof of relation \eqref{lm51}}

Similarly, we prove \eqref{lm51} by induction on $n$. Note that both sides of \eqref{lm51} are identically zero whenever $n &lt; k$. Therefore, it suffices to show that
\begin{equation}\label{lm52}
\sum_{i=k}^n 2i (i-k+1)(i-k+2)\cdots (i+k-1) = \frac{(2n+1)(n-k+1)(n-k+2) \cdots (n+k)}{2k+1},
\end{equation}
for $n \geq k$. Clearly, \eqref{lm52} is correct for $n =k$. Assume \eqref{lm52} holds for $n = k +m$, with $m$ being any given nonnegative integer. Then
\begin{align*}
\sum_{i=k}^{k +m+1} &amp; 2i (i-k+1)(i-k+2)\cdots (i+k-1) \\[-2mm]
 &amp; = \sum_{i=k}^{k +m} 2i (i-k+1)(i-k+2)\cdots (i+k-1) \\
 &amp; \qquad + (2k+2m+2)(m+2)(m+3) \cdots (m+2k) \\[2mm]
 &amp; = \frac{(2k+2m+1)(m+1)(m+2) \cdots (m+2k)}{2k+1} \\
 &amp; \qquad + (2k +2m+2)(m+2)(m+3) \cdots (m+2k) \\[2mm]
 &amp; = \frac{(2k+2m+3)(m+2)(m+3) \cdots (m+2k+1)}{2k+1}.
\end{align*}
Thus \eqref{lm52} holds for all $n \geq k$ by induction, and \eqref{lm51} is proven.


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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B57; Secondary 11B83.

\noindent \emph{Keywords: }
power sum, central factorial number with even indices of the first and second kind, Knuth's formula for the power sum, binomial identity, Faulhaber form of the power sum, Legendre-Stirling number of the first kind.

\bigskip
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\noindent (Concerned with sequences
\seqnum{A036969},
\seqnum{A065547},
\seqnum{A129467},
\seqnum{A204579}, and
\seqnum{A304330}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received April 24 2024;
revised versions received April 25 2024; December 23 2024;
January 1 2025; January 21 2025.
Published in {\it Journal of Integer Sequences},
January 22 2025.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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