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\begin{center}
\vskip 1cm{\LARGE\bf Power Sums of Binomial Coefficients}
\vskip 1cm
\large
Askar Dzhumadil'daev and Damir Yeliussizov\\
Kazakh-British Technical University\\
59 Tole bi St\\
050000 Almaty \\
Kazakhstan\\
\href{mailto:dzhuma@hotmail.com}{\tt dzhuma@hotmail.com} \\
\href{mailto:yeldamir@gmail.com}{\tt yeldamir@gmail.com}
\end{center}

\vskip .2 in

\begin{abstract}
We establish an analog of Faulhaber's theorem for a power sum of binomial coefficients. 
We study reciprocal power sums of binomial coefficients and Faulhaber coefficients for a power sum of triangular numbers. 
\end{abstract}

\section{Introduction}\label{intro}

While the sum of powers of consecutive nonnegative integers was studied by many mathematicians from ancient times, two names should be especially mentioned: Jacob Bernoulli (1654-1705) and Johann Faulhaber (1580-1635). It is well known that
\begin{align*}
&amp;\sum_{i=1}^ni =\frac{n^2}{2}+\frac{n}{2}=\frac{n(n+1)}{2},\\
&amp;\sum_{i=1}^ni^2 =\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\frac{n(n+1)(2n+1)}{6},\\
&amp;\sum_{i=1}^n i^3 =\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}=\left(\frac{n(n+1)}{2}\right)^2.
\end{align*}
In general, 
$$\sum_{i=1}^{N-1} i^m = \frac{1}{m + 1} \sum_{i=0}^m \binom{m+1}{i}B_i N^{m+1-i},$$
where $B_i$ are Bernoulli numbers.

Faulhaber \cite{faulhaber, knuth} noticed that odd power sums  
can be represented as a polynomial in $t=n(n+1)/2.$ For example,
$$\sum_{i=1}^{n}i^3=t^2,$$
$$\sum_{i=1}^{n}i^5=\frac{4t^3-t^2}{3},$$
$$\sum_{i=1}^{n}i^7=\frac{12t^4-8t^3+2t^2}{6}.$$
He computed these sums up to degree $17.$ The first proof of Faulhaber's theorem was given by Jacobi  \cite{jacobi}.
The general formula for odd power sums can be written as
$$\sum_{i=1}^n i^{2p+1}=\frac{1}{2^{2p+2}(2p+2)}\sum_{i=0}^p \binom{2p+2}{2i}(2-2^{2i})B_{2i}
((8t+1)^{p+1-i}-1).$$
Faulhaber knew that odd power sums are divisible by $t^2$ and 
even power sums can be expressed in terms of odd power sums. If 
$$\sum_{i=1}^n i^{2p+1}=c_1 t^2+c_2 t^3+\cdots +c_p t^{p+1},$$ then 
$$\sum_{i=1}^n i^{2p}=\frac{2n+1}{2(2p+1)}(2c_1 t+3c_2 t^2+\cdots+(p+1)c_p t^p).$$

In our paper we consider the power sum of binomial coefficients 
$$f_{k,m}(N)=\sum_{i=0}^{N-1} \binom{i+k-1}{k}^m,$$
and for $k=1$ we obtain the usual power sum $$f_{1,m}(N)=\sum_{i=1}^{N-1}i^m.$$ 
We establish an analog of Faulhaber's theorem for any positive integer $k.$ Namely, we show that: 
\begin{itemize}
\item $f_{k,m}(N)$ is a polynomial in $N$ and it can therefore be considered as a polynomial $f_{k,m}(x)$ in any variable $x$;
\item $f_{k,m}(x)$ can be expressed as:

\qquad $\binom{x+k-1}{k+1}^2 Q_{k,m}((2x+k-2)^2),$ if $m, k$ are odd and $m &gt; 1$; 

\qquad $\binom{x+k-1}{k+1}(2x+k-2) Q_{k,m}((2x+k-2)^2),$ if $k$ is odd and $m$ is even; 

\qquad $\binom{x+k-1}{k+1} Q_{k,m}((2x+k-2)^2),$ otherwise; 

for some polynomials $Q_{k,m}(x)$ with rational coefficients.
\end{itemize}

For example, 
\begin{align*}
f_{3,3}(N) &amp;= \sum _{i=0}^{N-1} \binom{i+2}{3}^3 = \binom{N+2}{4}^2 \frac{(2N+1)^2 - 10}{15},\\
f_{3,2}(N) &amp;= \sum _{i=0}^{N-1} \binom{i+2}{3}^2 = \binom{N+2}{4} (2N+1) \frac{5(2N+1)^2 - 41}{420},\\
f_{2,2}(N) &amp;= \sum _{i=0}^{N-1} \binom{i+1}{2}^2 = \binom{N+1}{3} \frac{3 N^2 - 2}{10}.
\end{align*}

By Faulhaber's theorem, any odd power sum can be expressed as a combination of powers of triangular numbers  
$$\sum_{i=0}^N i^{2m-1} =\frac{1}{2m}\sum_{i=0}^{m-1} F_i(m) (N(N+1))^{m-i},$$
and any even power sum can be expressed as
$$
\sum_{i=0}^N i^{2m} = (N - \frac{1}{2}) \sum_{i = 0}^m \widetilde F_i(m) (N(N-1))^{m-i},
$$
where $\widetilde F_i(m) = \frac{m+1-i}{(2m+1)(m+1)} F_i(m+1).$
\footnote{$F_i(m)$ is referred to the sequences \seqnum{A093556}, \seqnum{A093557} and $\widetilde F_i(m)$ is referred to the sequences \seqnum{A093558}, \seqnum{A093559} in \cite{oeis}.}

Knuth \cite{knuth} showed that the coefficients $F_i(m)$ have many interesting properties.
Our generalization of Faulhaber's theorem tends to consider the inverse problem: expressing the power sum of triangular numbers $f_{2,m}(N)$ in terms of powers of $N$. 
We show that this expression can be presented as a combination of odd powers of $N,$ 
$$f_{2,m}(N)=\sum_{i=1}^{N-1} \left(\frac{i(i+1)}{2}\right)^m= \frac{1}{2^m}\sum_{i=0}^m \wlm_i{(m)} N^{2m-2i+1}. $$
We find the following duality relations between coefficients  $\wlm^{(m)}_i$ and $F_i(m), \widetilde F_i(m)$:
$$\wlm_i{(m+i)}=(-1)^{i-1}\frac{m+i}{m(2m+1)} F_i(-m),$$
$$\wlm_i{(m+i-1)}  = (-1)^i \widetilde F_i(-m).$$
Properties of $\wlm_i{(m)}$ which similar to the properties of  $F_i(m)$ are also established.
 
We study integer divisibility properties of $f_{k,m}(N)$ for integer $N.$ 
We consider an analog of the Riemann zeta function 
$$
\zeta(s) = \sum_{i=1}^{\infty} \frac{1}{i^s}
$$
for binomial coefficients. Let 
$$\zeta_{k}(m)=\sum_{i=1}^{\infty} \binom{i+k-1}{k}^{-m}.$$
We prove  that (for positive integers $k,m$ and $l = \lceil m/2\rceil$)
$$
\zeta_{k}(m) \in 
\begin{cases}
\mathbb{Q} + \mathbb{Q} \zeta(2) +  \dots + \mathbb{Q} \zeta(2 l),&amp;\text{if $km$ is even;}\\
\mathbb{Q} + \mathbb{Q} \zeta(3)  + \dots + \mathbb{Q} \zeta(2 l - 1),&amp;\text{if $km$ is odd and $km&gt;1$.} 
\end{cases}
$$ 

In case of $k=2$ we obtain the formula for $\zeta_{2}(m)$, 
\begin{align*}
 \frac{\zeta_{2}(m)}{2^m}
 	&amp;= (-1)^{m-1} \binom{2m-1}{m} + (-1)^m 2 \sum_{i=1}^{\lfloor m/2 \rfloor} \binom{2m-2i-1}{m-1}  \zeta(2i).
\end{align*}

The structure of our paper is as follows.  
In section \ref{pr-bin}, we consider the sum of products of binomial coefficients, which has combinatorial origin and is used in establishing polynomial properties of  $f_{k,m}(x).$ 
In section \ref{faul}, an analog of Faulhaber's theorem for the powers of binomial coefficients is proved.
In section \ref{int}, we study integer properties for $f_{k,m}(x)$ and for $f_{k,-1}$.
In section \ref{recip}, the properties of infinite sum $\zeta_{k}(m)$ are derived.
In section \ref{triang}, we focus on the partial case $k = 2$ and express the power sum of triangular numbers $f_{2,m}(N)$ as a sum of  powers of $N$.

\section{Sum of products of binomial coefficients} \label{pr-bin}

\subsection{A generalized Worpitzky identity}
Let 
\begin{equation}
\label{def1}
 f_{k_1, \ldots, k_m}(N) = \sum_{i=0}^{N-1} \binom{i+k_1-1}{k_1} \cdots \binom{i+k_m-1}{k_m},
\end{equation}
where $k_1 \le \dots \le k_m$. 

To study $f_{k_1,\ldots,k_m},$ we need a generalization of Worpitzky identity for multisets. 
To formulate this identity let us introduce Eulerian numbers for multisets. 

Let ${\bf m}=1^{k_1}\cdots m^{k_m}$ be a multiset where $l$ repeats $k_l$ times, $l=1,\ldots,m.$ 
Let $S_{{\bf m}}$ be the set of permutations of ${\bf m}$ and set $K=k_1+\cdots+k_m.$ 

For a permutation $\sigma=\sigma(1)\ldots \sigma(K)\in S_{{\bf m}},$ let $i$ be a {\it descent} index if $i=K$ or $\sigma(i)&gt;\sigma(i+1), i&lt;K.$ A descent number $\des(\sigma)$ is defined as a
 number of  descent indices of $\sigma$ and Eulerian number $a_{{\bf m},p}$ is defined as a number of permutations with $p$ descents,
$$a_{{\bf m},p}=|\{\sigma\in S_{\bf m} | des(\sigma)=p\}|.$$
For ${\bf m}=1^1 2^1 \cdots m^1,$ we obtain the usual Eulerian numbers $a_{m,p}$ (\seqnum{A008292} in \cite{oeis}) and the well-known Worpitzky identity
$$x^m =  \sum_{p &gt; 0} \binom{x + m - p}{m} a_{m,p}.$$ 

%{\bf Example.} 
\begin{example}
Let ${\bf m}=1^22^2.$ Then 
$$S_{1^2 2^2}=\{1122,1212,2112,2121,2211,1221\},$$
and 
$$\des(1122)=1, \des(1212)=2, \des(2112)=2, \des(2121)=3,$$
$$\des(2211)=2, \des(1221)=2.$$
Therefore,
$$a_{1^22^2,1}=1,\; a_{1^22^2,2}=4, \; a_{1^22^2,3}=1, \; 
\mbox{and } a_{1^22^2,i}=0, \mbox{ if $i\ne1,2,3$}.$$
\end{example}

\begin{theorem} [\cite{DzhWorp}] \label{Worpitzky} For any nonnegative integers $k_1,\ldots,k_m,$ 
$$
\prod_{i=1}^m\binom{x+k_i-1}{k_i}=\sum_{p&gt; 0} \binom{x+K -p}{K} a_{{\bf m},p},
$$
where $a_{{\bf m},p}$  are Eulerian numbers of permutations of the multiset ${\bf m}= 1^{k_1}\cdots m^{k_m}.$ 
\end{theorem}

%{\bf Example.}  
\begin{example}
If ${\bf m}=1^22^2,$ then  
$$\binom{x+1}{2}^2=\binom{x+3}{4}+4\binom{x+2}{4}+\binom{x+1}{4}. $$
\end{example}

For a more detailed overview and other properties of Eulerian numbers on multisets and generalized Worpitzky identity, see \cite{DzhWorp}.

\subsection{Sum of products of binomial coefficients as a polynomial}
\begin{theorem} \label{ttt1} Let  $0\le k_1\le  \dots \le k_m.$ Then the sum {\rm  (\ref{def1})} induces a polynomial
$f_{k_1, \dots, k_m}(x)$ of degree $K+1$ with rational coefficients.  
As a polynomial with rational coefficients, the polynomial $f_{k_1,\ldots,k_m}(x)$ is divisible  by $\binom{x+k_m-1}{k_m+1}$.
\end{theorem}

\begin{proof}
By Theorem \ref{Worpitzky},
$$
f_{k_1,\ldots,k_m}(N)=
\sum_{p&gt; 0} \sum_{i=0}^{N-1}\binom{i+K -p}{K} a_{{\bf m},p}=\sum_{p&gt; 0} \binom{N+K -p}{K+1} a_{{\bf m},p},$$
for any positive integer $N.$ Therefore, we can substitute any variable $x$ in $N$ and see that 
$$f_{k_1,\ldots,k_m}(x)\in \mathbb{Q}[x], \qquad \deg\, f_{k_1,\ldots,k_m}(x)=K+1.$$

If $k_m=0,$ then $k_1=\cdots=k_m=0,$ and 
$$f_{0,\ldots,0}(N)=N-1.$$
Therefore, $f_{0,\ldots,0}(x)=x-1.$ 
So, in this case divisibility of $f_{k_1,\ldots,k_m}(x)$ by $\binom{x+k_m-1}{k_m+1}$ is evident.

Suppose now that $k_m&gt;0.$ Let us consider a difference  polynomial 
$$
\Delta f(x) = f_{k_1, \dots, k_m}(x+1) - f_{k_1, \dots, k_m}(x).
$$
By  (\ref{def1}),
$$
\Delta f(x) = \binom{x+k_1-1}{k_1} \cdots \binom{x+k_m-1}{k_m}.
$$
Therefore, $\Delta f(x)$ has $k_m$ zeros: $0,-1,\dots, -(k_m-1)$. Hence,
\begin{align*}
f_{k_1, \dots, k_m}(1) - f_{k_1, \dots, k_m}(0) &amp;= \Delta f(0) = 0,\\
f_{k_1, \dots, k_m}(0) - f_{k_1, \dots, k_m}(-1) &amp;= \Delta f(-1) = 0,\\
\dots\\
f_{k_1, \dots, k_m}(-(k_m-2)) - f_{k_1, \dots, k_m}(-(k_m-1)) &amp;= \Delta f(-(k_m-1)) = 0.
\end{align*}
If $k_m&gt;0,$ then $\binom{i+k_m-1}{k_m}=0$ for $i=-k_m+1.$ Therefore, 
$$f_{k_1,\ldots,k_m}(-k_m+1)=0.$$
We thus obtain a polynomial $f_{k_1, \dots, k_m}(x)$ with 
$k_m+1$ zeros 
$$1,0,-1,\dots, -(k_m-1).$$ This means that  $f_{k_1, \dots, k_m}(x)$ is divisible by $\binom{x+k_m-1}{k_m+1}$.
\end{proof}

Set $f_{k,m}(x)=f_{k,k,\ldots,k}(x).$ In other words, $f_{k,m}(x)$ is a polynomial defined by the following relations 
\begin{equation}
 f_{k,m}(N) =
 \sum_{i=0}^{N-1} \binom{i+k-1}{k}^m.
 \label{def}
\end{equation}

 \begin{corollary} The polynomial $f_{k,m}(x)$ has the following properties
 \begin{itemize}
 \item  $f_{k,m}(x)\in \mathbb{Q}[x]$,
 \item $\deg\,f_{k,m}(x)=km+1,$
 \item $f_{k,m}(x)$ is divisible by $\binom{x+k-1}{k+1}.$
 \end{itemize}
 \end{corollary}
A more detailed version of this result is given in the next section.

\section{Faulhaber's theorem for powers of binomial coefficients} \label{faul}

\subsection{Formulation of the main result}
We know that 
$f_{1,m}(N) = \sum_{i=1}^{N-1} i^m$ is 
a polynomial in  $N$ of degree $m+1$. 
By Faulhaber's theorem \cite{faulhaber, knuth}, the polynomial $f_{1,m}(x)$ is 
divisible by the polynomial $f_{1,1}(x) = x(x-1)/2$.
For odd $m,$ the polynomial $f_{1,m}(x)$  is divisible by $f_{1,1}(x)^2$ and
the quotient is a polynomial in  $f_{1,1}(x).$
For even $m,$ the polynomial  $f_{1,m}(x)$ can be presented  
as product of $f_{1,1}(x)(2x-1)$ and  
a polynomial in $f_{1,1}(x)$.

The following theorem is an analog of Faulhaber's theorem
for the sum of powers of binomial coefficients.

\begin{theorem}
\label{main1}
There exist polynomials $Q_{k,m}(x) \in \mathbb{Q}[x],$ such that
\begin{equation*}
 f_{k,m}(x) = \begin{cases}
   \binom{x+k-1}{k+1}^2 Q_{k,m}((2x+k-2)^2), &amp; \text{if $m, k$ are odd, $m &gt; 1$;} \\
   \\
   \binom{x+k-1}{k+1}(2x+k-2) Q_{k,m}((2x+k-2)^2), &amp; \text{if $k$ is odd, $m$ is even;}\\
   \\
   \binom{x+k-1}{k+1} Q_{k,m}((2x+k-2)^2), &amp; \text{otherwise.}\\
 \end{cases}
\end{equation*}
\end{theorem}

Note that $f_{k,1}(x) = \binom{x+k-1}{k+1}$ 
and our theorem states that the polynomial $f_{k,m}(x)$ 
is divisible by $f_{k,1}(x)$. Moreover, if $m$ and $k$ are odd $(m &gt; 1),$ 
then $f_{k,m}(x)$ is divisible by $f_{k,1}(x)^2$.

Let us show Theorem \ref{main1} for some small values of $k.$ 

If $k=1,$ then there are polynomials
$Q_{1,m}(x) \in \mathbb{Q}[x],$ such that the polynomial
$f_{1,m}(N) = \sum_{i=1}^{N-1} i^m$
can be expressed as:
$$
 f_{1,m}(x) = 
 \begin{cases}
   \binom{x}{2}^2 Q_{1,m}((2x-1)^2), &amp; \text{if $m$ is odd and $m&gt;1$;}\\
   \\
   \binom{x}{2}(2x-1) Q_{1,m}((2x-1)^2), &amp; \text{if $m$ is even.}\\
 \end{cases}
$$
Note that the polynomial $Q_{1,m}((2x-1)^2) = Q_{1,m}(8\frac{x(x-1)}{2} + 1)$ can be written as
a polynomial in $\binom{x}{2} = \frac{x(x-1)}{2}$.  Hence, Faulhaber's theorem is a particular case of Theorem \ref{main1}.

If $k = 2,$ then for any $m &gt; 0$,
there exists a polynomial $Q_{2,m}(x) \in \mathbb{Q}[x],$ such that
$$
 f_{2,m}(x) = \binom{x+1}{3} Q_{2,m}(x^2).
$$
Since 
$$\binom{x+1}{3}=\frac{x(x^2-1)}{6},$$
this implies that $f_{2,m}(x)$ is an odd  polynomial.

If $k = 3,$ then $f_{3,3}(x) = \binom{x+2}{4}^2 Q_{3,3}(x)$, where
$$
  Q_{3,3}(x) = \frac{1}{15}(4x^2 + 4x - 9).
$$
Then for each positive integer $m,$ $(m &gt; 1),$ 
there exists a polynomial 
$Q_{3,m}(x) \in \mathbb{Q}[x],$ such that
$$
  f_{3,m}(x) = \binom{x+2}{4}^2 Q_{3,m}(Q_{3,3}(x)), \text{ if $m$ is odd, and}
$$
$$
  f_{3,m}(x) = \binom{x+2}{4} (2x+1) Q_{3,m}(Q_{3,3}(x)), \text{ if $m$ is even.}
$$

\subsection{Reflective functions}
The proof of Theorem \ref{main1} is based on the notion of reflective functions introduced by Knuth \cite{knuth}.
The function $f(x)$ is called \textit{$r$-reflective} if 
for all $x,$ we have
$$f(x) = f(-x-r);$$
and $f(x)$ is called \textit{anti-$r$-reflective} if 
for all $x,$ we have
$$f(x) = -f(-x-r).$$
In other words, reflective functions are even or odd functions shifted by $r/2$.

Note that the 
\begin{itemize}
\item sum of two (anti)-$r$-reflective functions is (anti)-$r$-reflective; 
\item product of two $r$-reflective functions is $r$-reflective; 
\item product of anti-$r$-reflective and $r$-reflective is anti-$r$-reflective function and 
\item product of two anti-$r$-reflective functions is $r$-reflective.
\end{itemize}

\begin{lemma}
\label{kl1}
Let $\nabla f(x) = f(x) - f(x-1)$. 
Suppose that $f(0) = f(-r) = 0$ and the function $f$ is defined on the set of integers. 
Then the following is true:
\begin{itemize}
\item if the function $\nabla f$ is $(r-1)$-reflective, then $f$ is anti-$r$-reflective and
\item if the function $\nabla f$ is anti-$(r-1)$-reflective, then $f$ is $r$-reflective.
\end{itemize}
\end{lemma}

\begin{proof}
Suppose that $\nabla f$ is $(r-1)$-reflective. Then we have
$$f(N) - f(0) = \sum_{i=1}^{N} \nabla f(i) = \sum_{i=1}^{N} \nabla f(-i-r+1) = f(-r) - f(-N-r),$$
which gives $f(N) = -f(-N-r)$ and this implies that $f$ is anti-$r$-reflective.

Now if $\nabla f$ is anti-$(r-1)$-reflective, then
$$f(N) - f(0) = \sum_{i=1}^{N} \nabla f(i) = -\sum_{i=1}^{N} \nabla f(-i-r+1) = -f(-r) + f(-N-r).$$
So, $f(N) = f(-N-r)$ and $f$ is $r$-reflective.
\end{proof}

\begin{lemma}[\cite{knuth}, Lemma 4]
\label{knuth4}
A polynomial $f(x)$ is $r$-reflective if and only if it can be presented 
as a polynomial in  $x(x+r)$ (or $(2x+r)^2$); it is anti-$r$-reflective if and only
if it can be presented as $2x+r$ times a polynomial in  $x(x+r)$ (or $(2x+r)^2$).
\end{lemma}

\begin{lemma}\label{11} The polynomial 
 $\binom{x+k-1}{k}$ is $(k-1)$-reflective
if $k$ is even and anti-$(k-1)$-reflective if $k$ is odd.
\end{lemma}
 
\begin{proof}
The proof follows from the identity $\binom{x+k-1}{k} = (-1)^k \binom{-x}{k}$.
\end{proof}

By Theorem \ref{ttt1}, there exist polynomials $g_{k,m}(x) \in \mathbb{Q}[x],$ 
such that 
$$
f_{k,m}(x) = \binom{x+k-1}{k+1} g_{k,m}(x).
$$

\begin{lemma}\label{fg-refl}
Reflective properties of functions $f_{k,m}$ and $g_{k,m}$ are 
\begin{itemize}
\item $f_{k,m}(x)$ is $(k-2)$-reflective if $km+1$ is even
and anti-$(k-2)$-reflective if $km+1$ is odd.
\item $g_{k,m}(x)$ is $(k-2)$-reflective if $(m-1)k$ is even
and anti-$(k-2)$-reflective if $(m-1)k$ is odd.
\end{itemize}
\end{lemma}

\begin{proof}
Let $\nabla f = f(x)-f(x-1)$.
Since 
$\nabla f_{k,m}(x)= \binom{x+k-2}{k}^m,$
we see that $\nabla f_{k,m}(x)$ is $(k-3)$-reflective if $km$ is even
and anti-$(k-3)$-reflective, otherwise.  By Theorem \ref{ttt1}, 
$f(0) = f(-(k-2)) = 0$. Thus, by Lemma \ref{kl1}, $f_{k,m}(x)$ is anti-$(k-2)$-reflective 
if $km$ is even and $(k-2)$-reflective, otherwise.

Note that $g_{k,m} = f_{k,m}/\binom{x+k-1}{k+1}.$
Therefore by  Lemma \ref{11},
$g_{k,m}(x)$ is $(k-2)$-reflective if $km - k$ is even
and anti-$(k-2)$-reflective, otherwise. 
\end{proof}

\begin{lemma}
Let $k$ be an odd number.
Then $f_{k,m}(x)$ is divisible by 
$\binom{x+k-1}{k+1} (2x+k-2)$ if $m$ is even
and is divisible by $\binom{x+k-1}{k+1}^2$ if $m$ is odd and $m &gt; 1$.
\label{divk}
\end{lemma}

\begin{proof} Suppose that  $m$ is even.
By Lemma \ref{fg-refl}, the function $g_{k,m}$ is anti-$(k-2)$-reflective.
The anti-$(k-2)$-reflectivity condition for  $x = \frac{2-k}{2}$ gives us
$$g_{k,m}(-\frac{2-k}{2} - k + 2) = g_{k,m}(\frac{2-k}{2}) = -g_{k,m}(\frac{2-k}{2}).$$
Hence, $g_{k,m}(\frac{2-k}{2}) = 0$ and $g(x)$ is divisible by $(2x + k - 2)$.

Now consider the case $m$ is odd $(m &gt; 1)$.  
By Lemma \ref{fg-refl}, the function $g_{k,m}$ is $(k-2)$-reflective. 
We have 
\begin{align*}
 f_{k,m}(x+1) - f_{k,m}(x) &amp;= 
 \binom{x+k}{k+1} g_{k,m}(x+1) - \binom{x+k-1}{k+1} g_{k,m}(x) %\\
 = \binom{x+k-1}{k}^m.
\end{align*}
Hence,
\begin{equation*}
(x+k)g_{k,m}(x+1) - (x-1)g_{k,m}(x) = (k+1)\binom{x+k-1}{k}^{m-1}.
\end{equation*}
Therefore, for $i = 0, -1, \dots, -(k-1),$ we obtain 
$$
 (k+i)g_{k,m}(i+1) - (i-1)g_{k,m}(i) = 0.
$$
In other words,
\begin{align*}
k g_{k,m}(1) &amp;= -g_{k,m}(0),\\
(k-1) g_{k,m}(0) &amp;= -2g_{k,m}(-1),\\
(k-2) g_{k,m}(-1) &amp;= -3 g_{k,m}(-2),\\
&amp;\vdots\\
g_{k,m}(-(k-2)) &amp;= -k g_{k,m}(-(k-1)).
\end{align*}
Hence,
$$g_{k,m}(1) = (-1)^k g_{k,m}(-(k-1)) = -g_{k,m}(-(k-1)).$$
Since $g_{k,m}(x)$ is $(k-2)$-reflective, the reflectivity condition for $x = 1$ gives us 
$$g_{k,m}(1) = g_{k,m}(-(k-1)).$$
So, $g_{k,m}(1) = 0$ and 
$$g_{k,m}(-(k-1)) = \cdots =g_{k,m}(-1) = g_{k,m}(0) = 
g_{k,m}(1) =0.$$
Thus, $g_{k,m}(x)$ is divisible by $\binom{x+k-1}{k+1}$ and
$f_{k,m}(x)$ is divisible by $\binom{x+k-1}{k+1}^2$.
\end{proof}

\subsection{Proof of Theorem~\ref{main1}.}
Let $m $ and $k$ be two odd positive integers and $m &gt; 1$. By 
Lemma \ref{divk}, there exist polynomials $R_{k,m}(x)$
such that $g_{k,m}(x) = \binom{x+k-1}{k+1}R_{k,m}(x)$.
The function $g_{k,m}(x)$ is $(k-2)$-reflective and therefore, 
$R_{k,m}(x)$ (according to parity of $k$) is also $(k-2)$-reflective.
So, by Lemma \ref{knuth4}, there exist polynomials $Q_{k,m}(x) \in \mathbb{Q}[x]$
such that $R_{k,m}(x) = Q_{k,m}((2x+k-2)^2)$. In this case 
$$
 f_{k,m}(x) = \binom{x+k-1}{k+1}^2 Q_{k,m}((2x+k-2)^2).
$$

Now assume that $k$ is odd and $m$ is even. 
Then the function $g_{k,m}$ is anti-$(k-2)$-reflective.
By Lemma \ref{knuth4}, there exist polynomials $Q_{k,m}(x) \in \mathbb{Q}[x],$
such that $g_{k,m}(x) = (2x+k-2)Q_{k,m}((2x+k-2)^2)$.
Therefore,
$$
 f_{k,m}(x) = \binom{x+k-1}{k+1}(2x+k-2) Q_{k,m}((2x+k-2)^2).
$$
In all other  cases
$g_{k,m}$ is $(k-2)$-reflective.
By Lemma \ref{knuth4}, there exist polynomials $Q_{k,m}(x) \in \mathbb{Q}[x]$
such that $g_{k,m}(x) = Q_{k,m}((2x+k-2)^2)$. We have
$$
 f_{k,m}(x) = \binom{x+k-1}{k+1}Q_{k,m}((2x+k-2)^2).
$$
$\square$

\section{Integer divisibility for \texorpdfstring{$f_{k,m}(x)$}{Lg}} \label{int}

\subsection{Formulation of the main result}
By Theorem \ref{main1}, 
the polynomial $f_{k,m}(x)$ is divisible by $\binom{x+k-1}{k+1}^2$  if  $m$ and $k$ are odd $(m &gt; 1)$. 
In particular, $f_{k,m}(x)$ is divisible by $x^2$. 
By Theorem \ref{main1}, $f_{k,m}(x)$ is
divisible by $x$ for any $k,m$. 
Here divisibility refers to the divisibility of  polynomials with rational coefficients.
Now in this section, we study divisibility properties of $f_{k,m}(x)$ for integer $x.$

The divisibility properties of Theorem \ref{main1} do not hold for integers. 
Let us give counter-examples for all three cases: 
\begin{align*}
 \frac{f_{3,5}(7)}{7^2} &amp;= \frac{1}{7^2}\sum_{i=1}^{6} \binom{i+2}{3}^5 =  \frac{86650668}{7} \not\in \mathbb{Z},\\
 \frac{f_{3,2}(5)}{5} &amp;= \frac{1}{5}\sum_{i=1}^{4} \binom{i+2}{3}^2 =  \frac{517}{5} \not\in \mathbb{Z},\\
 \frac{f_{4,4}(11)}{11} &amp;= \frac{1}{11}\sum_{i=1}^{10} \binom{i+3}{4}^4 =  \frac{335469880502}{11} \not\in \mathbb{Z}.
\end{align*}
Since $f_{k,m}(x)\in \mathbb{Q}[x],$  for any given $k$ and $m$, there is a sufficiently large prime $p,$ such that 
$f_{k,m}(p)$ is divisible by $p$ and by $p^2$ if $m,k$ are odd and $m &gt; 1$. The following theorem is a more detailed version of these statements. 

\begin{theorem}
\label{p1}
Let $N,k,m$ be positive integers. Let $M$ be an odd positive integer 
such that  $\gcd(M, k!) = 1$ and $M-k+1 \le N \le M+1$.
Then
$$
 f_{k,m}(N) = \sum_{i=0}^{N-1} \binom{i+k-1}{k}^m \equiv 0 \pmod{M}
$$
in the following cases
\begin{itemize}
\item $m, k$ are odd numbers, or
\item  $\gcd(M, (km + 1)!) = 1$ ($k,m$ may be even or odd).
\end{itemize}
If  $m, k$ are odd numbers and $m &gt; 1$, then 
$$
 f_{k,m}(N) = \sum_{i=0}^{N-1} \binom{i+k-1}{k}^m \equiv 0 \pmod{M^{2}}
$$
in the following cases 
\begin{itemize}
\item  $m$ is divisible by $M$;
\item  $\gcd(M, (km + 1)!) = 1$.
\end{itemize}
\end{theorem}

\begin{remark}
Theorem \ref{p1} do not present all conditions for divisibility of $f_{k,m}(N)$ by $M$ and $M^2$. For example, 
if $M = 13, k = 4, m = 14, N = 13,$ then
$$
f_{4, 14}(13) = \sum_{i = 0}^{12} \binom{i+3}{4}^{14 }= 13 \times 6075433069762635003999567344079450237278856,
$$
but $4, 14$ are even and $\gcd(13, 57!) \ne 1$.
Similarly, if $M = 7, k = 3, m = 9, N = 7,$ then
$$
f_{3,9}(7) = \sum_{i = 0}^{6} \binom{i+2}{3}^9 = 7^2 \times 112153022185284,
$$
but $9$ is not divisible by $7$ and $\gcd(7, 28!) \ne 1$.
\end{remark}

\subsection{Proof}
To prove Theorem \ref{p1} we need some preliminary facts.

\begin{lemma}
\label{pk}
Let $M$ be an odd positive integer number such that $\gcd(M, k!) = 1$. Then for odd numbers 
$k,m$ and for all integers $i$ such that $1 \le i \le M-k,$ the following relation holds
\begin{align*}
 \binom{M-i+k-1}{k}^m &amp;\equiv 
 -\binom{i+k-1}{k}^m %\\
 + M m \binom{i+k-1}{k}^m \sum_{j=0}^{k-1} \frac{1}{i+j} \pmod{M^{2}}.
\end{align*}
\end{lemma}

\begin{proof}
 Let us consider an expression $\binom{M-i+k-1}{k}$ as a polynomial in $M,$

 \begin{align*}
  \binom{M-i+k-1}{k} &amp;= \frac{(M-i+k-1)\cdots(M-i)}{k!} \\
  &amp;= a_k M^{k} + \dots + a_1 M + a_0 \\
  &amp;= \frac{M^{k}}{k!} + \dots + M (-1)^{k-1} \binom{i+k-1}{k} \sum_{j=0}^{k-1} \frac{1}{i+j} %\\
  + (-1)^{k}\binom{i+k-1}{k}.
 \end{align*}
 Note that for $0 \le j \le k-1,$ all numbers $i + j$ are relatively prime with $M$.
 Therefore,
 \begin{align*}
  \binom{M-i+k-1}{k} 
  &amp;\equiv 
  (-1)^{k}\binom{i+k-1}{k} %\\
  + (-1)^{k-1} M \binom{i+k-1}{k} \sum_{j=0}^{k-1} \frac{1}{i+j} \pmod{M^{2}}.
 \end{align*}
 Hence,
 \begin{align*}
  \binom{M-i+k-1}{k}^m \equiv \left( -\binom{i+k-1}{k} + M \binom{i+k-1}{k} \sum_{j=0}^{k-1} \frac{1}{i+j}\right)^m %\\
  \pmod{M^{2}}
 \end{align*}
On expanding the right side of this congruence, we obtain the result. 
\end{proof}

\begin{lemma}
\label{fz}
$(km+1)!f_{k,m}(x) \in \mathbb{Z}[x]$.
\end{lemma}

\begin{proof} By Theorem \ref{Worpitzky}, there are some integers $a_j$ for which 
we can present  $f_{k,m}(x)$
in the form $\sum_{j &gt; 0} a_j \binom{x+km-j}{km+1}$.
Note that  $(km+1)!\binom{x+km-j}{km+1} \in \mathbb{Z}[x].$
\end{proof}
                         
\begin{proof}[Proof of Theorem \ref{p1}]%{\bf Proof of Theorem \ref{p1}.}
For  $0 \le i \le M-k$ with odd $k,$ by  Lemma \ref{pk}, 
$$
 \binom{M-i+k-1}{k}^m + \binom{i+k-1}{k}^m \equiv 
 0 \pmod{M}.
$$
Therefore, 
\begin{align*}
 \sum_{i=0}^{N-1} \binom{i+k-1}{k}^m &amp; \equiv \sum_{i=1}^{M-k} \binom{i+k-1}{k}^m \\
 &amp;= \frac{1}{2}\sum_{i=1}^{M-k}\left(\binom{M-i+k-1}{k}^m + \binom{i+k-1}{k}^m\right) %\\
 \equiv 0 \pmod{M}.
\end{align*}

Now suppose that $m$ is divisible by $N$. Hence, by Lemma \ref{pk}, we have
\begin{align*}
 \binom{M-i+k-1}{k}^m &amp; \equiv 
 -\binom{i+k-1}{k}^m + M m \binom{i+k-1}{k}^m \sum_{j=0}^{k-1} \frac{1}{i+j} \\ 
 &amp;\equiv -\binom{i+k-1}{k}^m \pmod{M^{2}}.
\end{align*}
Therefore,
\begin{align*}
 \sum_{i=0}^{N-1} \binom{i+k-1}{k}^m &amp; \equiv \sum_{i=1}^{M-k} \binom{i+k-1}{k}^m \\
 &amp;= \frac{1}{2}\sum_{i=1}^{M-k}\left(\binom{M-i+k-1}{k}^m + \binom{i+k-1}{k}^m\right) %\\
 \equiv 0 \pmod{M^{2}}.
\end{align*}

Since $\gcd(M, (km+1)!) = ~1,$ by  Lemma \ref{fz} and 
Theorem \ref{main1}, $(km+1)! f_{k,m}(x) \in \mathbb{Z}[x]$ 
 is divisible
 by $x(x+1)\cdots(x+k-1)$ and
by $(x(x+1)\cdots(x+k-1))^2$ if $m, k$ are odd numbers and $m &gt; 1$.
Here divisibility refers to the divisibility of polynomials with integer coefficients. 
\end{proof}

%{\bf Remark.} 
\begin{remark}
In Theorem \ref{p1} we can change $M$ to a power of some prime 
number $p$. 
The property $\gcd(M, (km+1)!) = 1$ can be changed by an 
inequality $p &gt; km + 1$.
\end{remark}

\subsection{Integer divisibility in case \texorpdfstring{$k=2$}{Lg}}

\begin{theorem}
\label{tr-p}
Assume that $p$ is an odd prime number and $1 \le m \le p-1$. 
Then 
$$
  f_{2,m}(p) = \sum_{i=1}^{p-1} \binom{i+1}{2}^m \equiv \begin{cases}
     -2 \pmod{p}, &amp; \text{if $m = p-1$;}\\
     -\frac{1}{2^m} \binom{m}{p-1-m} \pmod{p}, &amp; \text{otherwise.}
  \end{cases}
$$
\end{theorem}

\begin{proof}
The following fact is known:
  $$
    \sum_{i=1}^{p-1} i^t \equiv \begin{cases}
       0 \pmod{p}, &amp; \text{ if $t$ is not divisible by $p-1$;}\\
       -1 \pmod{p}, &amp; \text{ otherwise.}
    \end{cases}
  $$
  Therefore, 
  \begin{align*}
     f_{2,m}(p) &amp;= \sum_{i=1}^{p-1} \binom{i+1}{2}^m 
     = \frac{1}{2^m} \sum_{i=1}^{p-1} \sum_{j=0}^{m} \binom{m}{j} i^{m + j}%\\
     = \frac{1}{2^m} \sum_{j=0}^{m} \binom{m}{j} \sum_{i=1}^{p-1} i^{m + j}
  \end{align*}
 If $m = p-1,$ then  
 $m+j$ is divisible by $p-1$ only in two cases: $j = 0, m.$ Hence,
  $$
   f_{2,p-1}(p) \equiv - \frac{1}{2^{p-1}} \left(\binom{p-1}{0} + \binom{p-1}{p-1} \right) \equiv -2 \pmod{p}.
  $$
 If $1\le m&lt;p-1,$  there is only one integer $j \in [0,m]$ such that $m + j$ is divisible by $p-1.$ Namely, $j = p - 1 - m$.  In this case,
  $$
   f_{2,m}(p) \equiv - \frac{1}{2^{m}} \binom{m}{p-1-m} \pmod{p}.
  $$
\end{proof}

\begin{remark}
By Theorem \ref{tr-p}, if $p - 1 - m &gt; m,$ then $f_{2,m}(p)\equiv 0 \pmod{p}.$
This fact is compatible with Theorem \ref{p1}, because $p &gt; 2m + 1$.
\end{remark}

\subsection{The case \texorpdfstring{$f_{k,-1}(N)$}{Lg}}

\begin{theorem}
\label{q-2}
Let $N,k$ be positive integer numbers and $M$ be a positive integer such that $\gcd(M, k!) = 1$.
Then, the rational number $q$ defined as
$$
 q = f_{(k,-1)}(N) = \sum_{i=1}^{N-1} \frac{1}{\binom{i+k-1}{k}},
$$
is divisible by $M$ (its denominator is relatively prime with $M$)
in the following cases:

(i) $N \equiv 1 \pmod{M}$;

(ii) $N \equiv 1-k \pmod{M}$ and $k$ is odd.
\end{theorem}

To prove this result we need one 

\begin{lemma}
\label{harm-bin}
Suppose that $k &gt; 1$. Then 
$$
 f_{k,-1}(N) = \sum_{i=1}^{N-1} \frac{1}{\binom{i+k-1}{k}} 
 = \frac{k}{k-1}\left(1 - \frac{1}{\binom{N + k - 2}{k-1}} \right).
$$
\end{lemma}

\begin{proof} Let $\phi(x) = \frac{1}{\binom{x + k - 1}{k-1}}$. 
Then it is easy to verify that 
$$
 \frac{1}{\binom{i + k - 1}{k}} = \frac{k}{k-1} \left(\frac{1}{\binom{i + k - 2}{k-1}} - \frac{1}{\binom{i + k - 1}{k-1}} \right) = -\frac{k}{k-1} \nabla \phi(i).
$$
Therefore, 
\begin{align*}
 f_{(k,-1)}(N) &amp;= \sum_{i=1}^{N-1} \frac{1}{\binom{i+k-1}{k}}
 = -\frac{k}{k-1} \sum_{i=1}^{N-1} \nabla \phi(i) \\
 &amp;= \frac{k}{k-1} (\phi(0) - \phi(N-1)) 
 = \frac{k}{k-1}\left(1 - \frac{1}{\binom{N + k - 2}{k-1}} \right).
\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{q-2}]%{\bf Proof of Theorem \ref{q-2}.} 
By Lemma \ref{harm-bin},
$$
 f_{k,-1}(N) = \frac{k!}{k-1} \left( \frac{1}{(k-1)!} - \frac{1}{(N+k-2)\cdots N}\right).
$$
Notice that for both cases (i) and (ii), the numbers $N, \dots, N+k-2$ are relatively prime with $M$.
So, if $N \equiv 1 \pmod{M},$ then 
$$(N+k-2)\cdots N \equiv (k-1)! \pmod{M}.$$ 
If $N \equiv 1-k \pmod{M}$ and $k$ is odd, then 
$$(N+k-2)\cdots N \equiv (-1)^{k-1}(k-1)! \equiv (k-1)! \pmod{M}.$$ 
%$\square$
\end{proof}

\section{Power sum of reciprocals of binomial coefficients}\label{recip}

\subsection{Formulation of the main result}
In this section, we consider the case of power sums of binomial coefficients  with negative powers, 
$$
 \zeta_{k}(m) = \sum_{i = 1}^{\infty} \binom{i+k-1}{k}^{-m}.
$$
For $k = 1$ we have 
$$\zeta_{1}(m) = \zeta(m) = \sum_{i = 1}^{\infty} \frac{1}{i^m},$$ 
where $\zeta(m)$ is a Riemann zeta function.

In particular, for $m = 1,$ by Lemma \ref{harm-bin}, one can obtain the exact value
$$
 \zeta_{k}(1) = \sum_{i = 1}^{\infty} \binom{i+k-1}{k}^{-1} = \frac{k}{k-1}.
$$

It is known that for any positive integer $m,$ 
\begin{equation}
\label{zetaeven}
  \zeta(2m) = (-1)^{m + 1} \frac{(2\pi)^{2m}}{2(2m)!}B_{2m},
\end{equation}
where $B_{2m}$ is a Bernoulli number.

We prove similar results for binomial coefficients.
Some examples that follow from our results:  
\begin{align*}
 \zeta_{2}(2) &amp;= \sum_{i = 1}^{\infty} \binom{i+1}{2}^{-2} = \frac{4}{3}\pi^2 - 12,\\
 \zeta_{2}(3) &amp;= \sum_{i = 1}^{\infty} \binom{i+1}{2}^{-3} = -8\pi^2 + 80,\\
 \zeta_{3}(2) &amp;= \sum_{i = 1}^{\infty} \binom{i+2}{3}^{-2} = 9\pi^2 - \frac{351}{4}.
\end{align*}
Similarly,
\begin{align*}
 \zeta_{3}(3) &amp;= \sum_{i = 1}^{\infty} \binom{i+2}{3}^{-3} = \frac{783}{4}-162 \zeta(3), \\
 \zeta_{5}(3) &amp;= \sum_{i = 1}^{\infty} \binom{i+4}{5}^{-3} = -\frac{1298125}{96}+11250 \zeta(3), \\
 \zeta_{3}(5) &amp;= \sum_{i = 1}^{\infty} \binom{i+2}{3}^{-5} = \frac{576639}{16}-\frac{47385}{2} \zeta(3) - 7290 \zeta(5).
\end{align*}

Below we show that these relations are based on reflectivity properties of binomial coefficients, and 
that $\zeta_{k}(m)$ can be expressed as a linear combination
of certain values of the Riemann zeta function. 

\begin{theorem}\label{a1a1}
For any positive integers $k$ and $m$, there exist rationals
$\lambda_0, \lambda_1, \dots, \lambda_{\lceil m/2 \rceil},$ 
such that
$$
 \zeta_{k}(m) = \begin{cases}
   \lambda_0 + \displaystyle\sum_{i=1}^{\lceil m/2 \rceil} \lambda_i \zeta(2i), &amp; \text{ if $km$ is even;}\\
   \lambda_0 + \displaystyle\sum_{i=1}^{\lceil m/2 \rceil} \lambda_i \zeta(2i-1), &amp; \text{ if $km$ is odd.}
 \end{cases}
$$
\end{theorem}

\subsection{Proof of Theorem \ref{a1a1}}
 Let $F(x) = \binom{x+k-1}{k}^{-m}$. Then $\zeta_{k}(m) = \sum_{i=1}^{\infty} F(i)$. 
 Since the polynomial $F(x)$ has $k$ zeros $x=0,1,\ldots,k-1,$ 
 \begin{align}
 \label{part}
  F(x) = \frac{1}{\binom{x+k-1}{k}^{m}} &amp;= \sum_{j=0}^{k-1} \sum_{i=1}^m \frac{a_{i,j}}{(x+j)^i}%\\
  =\frac{1}{2} \sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}}{(x+j)^i} + \frac{a_{i,k-1-j}}{(x+k-1-j)^i} \right) %\nonumber
 \end{align}
 for some rational numbers $a_{i,j}$ ($1 \le i \le m, 0 \le j \le k-1$).

By Lemma \ref{11},  $1/F(x)$ is $(k-1)$-reflective  if $km$ is even, 
 and  $1/F(x)$ is anti-$(k-1)$-reflective, if $km$ is odd. 
 In other words,  $F(x) = e F(-x-k+1)$ for almost all $x$ (except zeros of denominator), 
 where $e = \pm 1.$ 
 Thus, by (\ref{part}) and based on the reflectivity property mentioned above,
 \begin{align*}
  &amp;\sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}}{(x+j)^i} + \frac{a_{i,k-1-j}}{(x+k-1-j)^i} \right) %\\
  =
  e &amp;\sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}}{(-x-k+1+j)^i} + \frac{a_{i,k-1-j}}{(-x-j)^i} \right)
 \end{align*}
 or
 $$
  \sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}-e(-1)^i a_{i,k-1-j}}{(x+j)^i} + \frac{a_{i,k-1-j} - e(-1)^i a_{i,j}}{(x+k-1-j)^i} \right) 
  = 0.
 $$
 Therefore, 
 $$
 a_{i,j}-e(-1)^i a_{i,k-1-j} = 0
 $$
 for all $i,j$ ($1 \le i \le m, 0 \le j \le k-1$).
 Hence,  equation (\ref{part}) can be rewritten as
 \begin{equation*}
  F(x) = \frac{1}{\binom{x+k-1}{k}^{m}} = 
  \frac{1}{2} \sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}}{(x+j)^i} + e(-1)^i \frac{a_{i,j}}{(x+k-1-j)^i} \right).
 \end{equation*}
 Finally, 
 \begin{align*}
   \zeta_{k}(m) &amp;= \sum_{x=1}^{\infty} F(x) \\
   &amp;= \sum_{x=1}^{\infty} \frac{1}{2}\sum_{j=0}^{k-1} \sum_{i=1}^{m} 
  \left(\frac{a_{i,j}}{(x+j)^i} + e(-1)^i \frac{a_{i,j}}{(x+k-1-j)^i} \right)\\
  &amp;= \frac{1}{2} \sum_{j=0}^{k-1} \sum_{i=1}^{m} a_{i,j} \left(\zeta(i)-\sum_{l=1}^{j}\frac{1}{l^i} + e(-1)^i\zeta(i) - e(-1)^i \sum_{l=1}^{k-1-j}\frac{1}{l^i}\right)\\
  &amp;= c_0 + \sum_{i=1}^{m} c_i (\zeta(i) + e(-1)^i \zeta(i)),
 \end{align*}
 for some rational constants $c_i$ ($0 \le i \le m$). 
 Note that value $(\zeta(i) + e(-1)^i \zeta(i))$ vanishes if $e = 1$ and $i$ is odd,
 or if $e = -1$ and $i$ is even. 
$\square$

Presentations of infinite series as a linear combination of odd (even) values of zeta functions 
play an important role in studying the irrationality problems of zeta functions. See,
 for example, the proof of Rivoal \cite{rivoal} that the sequence  $\zeta(3), \zeta(5), \dots$
 contains infinitely many irrational values, or
result of Zudilin \cite{zudilin} that at least one of four numbers $\zeta(5), \zeta(7), \zeta(9), \zeta(11)$ is irrational.

Different kinds of sums of inverses of binomial coefficients were studied in many works, e.g., the sums of other types by Sofo \cite{sofo}, Sprugnoli \cite{sprugnoli}, Sury, Wang and Zhao \cite{sury}, Yang and Zhao \cite{yang}. 

\subsection{Sums of reciprocals of powers of triangular numbers}

Recall that $$\zeta_{2}(m)=\sum_{x=1}^{\infty} \frac{2^m}{(x(x+1))^m}.$$
In this subsection we give exact presentation of $\zeta_2(m)$ as a combination of binomial coefficients and Bernoulli numbers. Namely, we prove the following result. 

\begin{theorem} \label{qq}
\begin{align*}
 \sum_{x=1}^{\infty} \frac{1}{(x(x+1))^m} 
 	&amp;= (-1)^{m-1} \binom{2m-1}{m} %\\
	+ (-1)^m \sum_{i=1}^{\lfloor m/2 \rfloor} \binom{2m-2i-1}{m-1} (-1)^{i + 1} \frac{(2\pi)^{2i}}{(2i)!}B_{2i}.
\end{align*}
\end{theorem}

To prove this theorem we need the following 

\begin{lemma}
\label{frac2}
$$
 \frac{1}{\binom{x+1}{2}^m} = 
 2^m \sum_{i=0}^{m-1} \binom{m+i-1}{i} 
 	\left((-1)^{i} \frac{1}{x^{m-i}}+(-1)^{m}\frac{1}{(x+1)^{m-i}} \right).
$$                                                          
\end{lemma}

\begin{proof} We use induction on $m$. 
For $m = 1$ we have $$\frac{1}{\binom{x+1}{2}} = 2\left(\frac{1}{x} - \frac{1}{x+1}\right).$$
Assume that our statement holds for $m$ and let us prove it for $m + 1$.
In our proof we need the following formulas ($j$ is a positive integer):
\begin{align*}
 \frac{1}{x^j(x+1)} &amp;= \frac{1}{x^j} - \frac{1}{x^{j-1}} + \dots + (-1)^j \frac{1}{x+1} \text{ and}\\
 \frac{1}{x(x+1)^j} &amp;= \frac{1}{x} - \frac{1}{x+1} - \dots - \frac{1}{(x+1)^j},
\end{align*}
which follow from: 
\begin{align*}
 1 - (-x)^j &amp;= (1 + x) (1 + (-x) + \dots + (-x)^{j-1}) \text{ and}\\
 1 - (x + 1)^j &amp;= -x (1 + (x+1) +  \dots + (x+1)^{j-1}).
\end{align*}
We have
\begin{align*}
   &amp;\frac{1}{(x(x+1))^{m+1}}
   = \frac{1}{(x(x+1))^{m}} \left( \frac{1}{x} - \frac{1}{x+1} \right)\\
   &amp;= \sum_{i=0}^{m-1} \binom{m+i-1}{i} \left(\frac{(-1)^{i} }{x^{m-i}}+\frac{(-1)^{m}}{(x+1)^{m-i}} \right)
   \left( \frac{1}{x} - \frac{1}{x+1} \right)\\
   &amp;= \sum_{i=0}^{m} \binom{m+i}{i} \left(\frac{(-1)^{i} }{x^{m-i+1}}+\frac{(-1)^{m+1}}{(x+1)^{m-i+1}} \right)
\end{align*}
(since $\sum_{j=0}^{i} \binom{m + j - 1}{j} = \binom{m + i}{i}$ ).
\end{proof}

\begin{proof}[Proof of Theorem \ref{qq}]%{\bf Proof of Theorem \ref{qq}.} 
By Lemma \ref{frac2},
\begin{align*}
     \sum_{x=1}^{\infty} \frac{1}{(x(x+1))^m}
     &amp;= \sum_{x=1}^{\infty} \sum_{i=0}^{m-1} \binom{m+i-1}{i} 
 	   \left( \frac{(-1)^i}{x^{m-i}}+\frac{(-1)^{m}}{(x+1)^{m-i}} \right)\\
     &amp;= \sum_{i=0}^{m-1} \binom{m+i-1}{i} 
           \left( \sum_{x=1}^{\infty} \frac{(-1)^i}{x^{m-i}} + \sum_{x=1}^{\infty} \frac{ (-1)^m}{(x+1)^{m-i}}\right).\\
\end{align*}
Thus, the last expression can be written as
\begin{align*}
     &amp;\sum_{i=0}^{m-1} \binom{m+i-1}{i} 
           \left((-1)^i \zeta(m-i) + (-1)^m (\zeta(m-i) - 1) \right) \\
     &amp;= (-1)^{m-1} \sum_{i=0}^{m-1} \binom{m+i-1}{i}%\\
      + 
        \sum_{i=0, (m-i) \text{ even}}^{m-1}\binom{m+i-1}{i} (-1)^i  2\zeta(m-i)\\
     &amp;= (-1)^{m-1} \binom{2m-1}{m} + 
 	   (-1)^m 2 \sum_{i=1}^{\lfloor m/2 \rfloor} \binom{2m-2i-1}{m-1} \zeta(2i)\\
     &amp;= (-1)^{m-1} \binom{2m-1}{m}
     + 
 	   (-1)^m \sum_{i=1}^{\lfloor m/2 \rfloor} \binom{2m-2i-1}{m-1} (-1)^{i + 1} \frac{(2\pi)^{2i}}{(2i)!}B_{2i}.
\end{align*}
\end{proof}

\section{Faulhaber coefficients and coefficients of the polynomial \texorpdfstring{$f_{2,m}(x)$}{Lg}}\label{triang}

\subsection{Duality between \texorpdfstring{$F$- and $\wlm$-Faulhaber coefficients}{Lg}}
Recall that any number in the form $\binom{n}{2}=\frac{n(n-1)}{2}$ is called triangular (\seqnum{A000217} in \cite{oeis}). 
In this section we study analogs of Faulhaber coefficients for power sums of triangular numbers.
To stress similarities between Faulhaber coefficients $F_i(m)$ and $\wlm_i{(m)}$ (see definitions below), we use special notation for formulas like (\ref{A1}), (\ref{B1}), (\ref{A2}), (\ref{B2}), etc.  

We define Faulhaber coefficients \cite{ges, knuth} as  numbers $F_i(m),$ such that
\begin{equation*}\label{A1}
 \sum_{i=0}^{N} i^{2m-1} = \frac{1}{2m} \sum_{i=0}^{m-1} F_i(m) (N(N+1))^{m-i}. 
 \tag{A1}
\end{equation*}
These coefficients can be extended for all real numbers $x$ by $F_i(x)$.
In fact, $F_i(x)$ is a polynomial in  $x$. 
In Sloane's OEIS \cite{oeis}, $F_i(m)$ is referred to the sequences \seqnum{A093556}, \seqnum{A093557} and 
\begin{equation}\label{tildeA}
\widetilde F_i(m) = \frac{m+1-i}{(2m+1)(m+1)} F_i(m+1)
\end{equation}
 is referred to the sequences \seqnum{A093558}, \seqnum{A093559}.

Knuth  \cite{knuth}  has  established the following properties of these coefficients,

\begin{equation*}
F_0(x) = 1; \quad  \sum_{j=0}^k \binom{x - j}{2k+1 - 2j} F_j(x) = 0, \quad k &gt; 0. 
\label{A2}\tag{A2}
\end{equation*}

\begin{equation*}\label{A3}
F_k(x) = x(x-1)\cdots(x-k+1) \times q_k(x),
\tag{A3}
\end{equation*}
where $q_k(x)$ is 
\begin{itemize}
\item a polynomial of degree $k$ 
\item with leading coefficient equal to $(2 - 2^{2k})B_{2k}/(2k)!$; 
\item and $q_k(k+1) = 0$ if $k &gt; 0$.
\end{itemize}
Below we show that all zeros of this polynomial are real and distinct. 

A recurrence formula due to Jacobi yields:
\begin{equation*}\label{A4}
(2x - 2k) (2x - 2k + 1)F_k(x) + (x-k+1)(x-k)F_{k-1}(x) = 2x(2x-1)F_k(x-1).
\tag{A4}
\end{equation*}

Gessel and Viennot \cite{ges} obtained the following explicit formula:
\begin{equation*}\label{A5}
F_k(m) = (-1)^{m-k} \sum_{j} \binom{2m}{m-k-j} \binom{m-k+j}{j} \frac{m-k-j}{m-k+j} B_{m+k+j}, \tag{A5}
\end{equation*}
for $0 \le k &lt; m.$

In terms of determinants, this formula  can be written as 
\begin{equation*}\label{A6}
 %\label{a-det}
 F_k(m) = (-1)^{(m-k)} \frac{1}{(m-1)\cdots(m-k)}
 \det \left|\binom{m-k+i}{2i - 2j + 3} \right|_{i,j=1,\dots,k}.
 \tag{A6}
\end{equation*}
The last determinant has a combinatorial interpretation. 

\begin{theorem}[Gessel, Viennot, Theorem 14, \cite{ges}]
The number of sequences of positive integer numbers
$a_1, \dots, a_{3k}$ satisfying inequalities
$a_{3i-2} &lt; a_{3i-1} &lt; a_{3i}$, $a_{3i+1} \le a_{3i-1}$, $a_{3i+2} \le a_{3i}$
and $a_{3i} \le m-k+i,$ for all $i,$ is equal to
$$
 \det \left|\binom{m-k+i}{2i - 2j + 3} \right|_{i,j=1,\dots,k}.
$$
\end{theorem}

According to Theorem \ref{main1}, the polynomial $f_{2,m}(x)$ is odd.
Therefore we can  consider the coefficients
$\wlm_i{(m)}$ defined by 
\begin{equation*}\label{B1}
f_{2,m}(N) 2^m = \sum_{i=0}^{N-1} (i(i+1))^{m} = \sum_{i = 0}^{m} \wlm_i{(m)} N^{2m-2i+1}.
\tag{B1}
\end{equation*}

The following relations are $\wlm_i^{(m)}$ analogs of relations (\ref{A2})--(\ref{A6}),
\begin{equation*}\label{B2}
 \wlm_0{(m)} = 1; \quad \sum_{j=0}^{k} \binom{m + 2k - j}{2k+1 - 2j} (-1)^j \wlm_j{(m+j)} = 0.
 \tag{B2}
\end{equation*}

\begin{equation*}\label{B3}
 \wlm_i{(m)} = \frac{m(m-1)\cdots(m-i)}{2m-2i+1} \times h_i(m),
 \tag{B3}
\end{equation*}
where $h_i(m)$ is 
\begin{itemize}
\item a polynomial of degree $i$ 
\item with leading coefficient $(-1)^{i-1}(2-2^{2i})B_{2i}/(2i)!$; 
\item and $h_i(-1) = 0$ if $i &gt; 0$.
\end{itemize}
Proof of these properties will be presented in Theorem \ref{wh} below. 
In fact, Theorem \ref{lm-fh} tells us that $h_i(m) = q_i(i-m)$.

A recurrence formula for $\wlm_i{(m)}$ is given by
\begin{equation*}\label{B4}
 2(m-i)(2m-2i+1)\wlm_i{(m)} = 2m(2m-1)\wlm_i{(m-1)} + m(m-1) \wlm_{i-1}{(m-2)}.
 \tag{B4}
\end{equation*}

Coefficients $\wlm_i{(m)}$ are closely related to $F_i(m).$

\begin{theorem}\label{lm-fh}
\begin{equation}
 \wlm_i{(m+i)} = (-1)^{i-1} \frac{m+i}{m(2m+1)} F_i(-m)
\end{equation}
\end{theorem}

Another formulation of Theorem \ref{lm-fh} is given by
$$ \wlm_i{(m)} = (-1)^{i-1} \frac{m}{ (m-i)(2m-2i+1)} F_i(i-m).$$

\begin{proof} From relations (\ref{A4}) and (\ref{B4}), we can see that the sequences 
$\wlm_i{(m+i)}$ and $F_i(-m)$ 
satisfy the same recurrence relations
and they have equal initial values
$F_0(x) = \wlm_0{(x)}= 1.$
\end{proof}

Knuth \cite{knuth} proved that 
$$
 \sum_{i=0}^{N-1} i^{2m} = (N - \frac{1}{2}) \sum_{i=0}^{m} \widetilde F_i(m) (N(N-1))^{m-i},
$$
with relation (\ref{tildeA}) between the coefficients $\widetilde F_i(m)$ and 
$F_i(m)$. 

\begin{corollary} \begin{equation}
 \wlm_i{(m+i-1)} = (-1)^i \widetilde F_i(-m).
\end{equation}
\end{corollary}

The following is a general formula for $\wlm_i{(m)}$ 
\begin{equation*}\label{B5}
 \wlm_i{(m)} 
 = \frac{1}{(2m-2i+1)} \sum_{k=0}^{2i} \binom{m}{2i-k} \binom{2m-2i+k}{k} B_k.
 \tag{B5}
\end{equation*}
Let us rewrite this formula in terms of determinants 
\begin{equation*}\label{B6}
 \wlm_k{(m)} = \frac{(-1)^{(k-m+1)} (k-1)!}{(2m-2k+1)(m-1)!}
  \det \left|
  \binom{m + i - 2j + 2}{2i - 2j + 3}
 \right|_{i,j = 1, \dots, k},
 \tag{B6}
\end{equation*}
which follows from (\ref{A6}) and Theorem \ref{lm-fh}. 

Using \cite[Theorem 16]{ges}, we obtain one more combinatorial interpretation for the last determinant.

\begin{theorem}
The number of sequences of positive integers $a_1, \dots, a_{3k}$ such that 
$a_{3i-2} \ge a_{3i-1} \ge a_{3i}$, $a_{3i+1} \ge a_{3i-1}$, $a_{3i+2} \ge a_{3i}$
and $a_{3i-2} \le m-i,$ for all $i,$ is equal to
$$
 \det \left|\binom{m+i-2j+2}{2i - 2j + 3} \right|_{i,j=1,\dots,k}.
$$
\end{theorem}

\begin{proof}  Let $\lambda = (\lambda_1, \dots, \lambda_k)$ and $\mu =(\mu_1, \dots, \mu_k)$
be two non-increasing sequences of nonnegative integers.
\textit{Plane partition of shape} $\lambda - \mu$ is a filling of 
the corresponding diagram with integers which are weakly
decreased in every row and column. Such a diagram can be  obtained from Ferrer diagram for $\lambda$ by removing
the diagram for $\mu$ (for details, see \cite{ges}) . 

By \cite[Theorem 16]{ges}, the number of plane partitions of shape
$\lambda - \mu$  
whose parts in row $i$ are at most $A_i$ and at least $B_i$,
where $A_i \ge A_{i+1} - 1$ and $B_i \ge B_{i+1} - 1,$ is equal to 
$$
 \det \left|\binom{A_j - B_i + \lambda_i - \mu_j} {\lambda_i - \mu_j + j - i} \right|_{i,j=1,\dots,k}.
$$
Taking  $\lambda_i = k + 3 - i$, $\mu_j = k - j$, $A_j = m-j$ and $B_i = 1,$ 
we obtain the result.
\end{proof}

\subsection{The polynomial \texorpdfstring{$f_{2,m}(x)$}{Lg}}

Set $\lambda_{m,i} = \wlm_i{(m)} / 2^m.$

\begin{theorem} \label{ffm2}
The polynomial $f_{2,m}(x)=\sum_{i=0}^{m}\lambda_{m,i}x^{2m-2i+1}$ 
has the following properties.

{\bf (I)}  It satisfies the equation
\begin{equation}\label{fm2}
f_{2,m}''(x) = m(2m-1)f_{2,m-1}(x) + \frac{m(m-1)}{4} f_{2,m-2}(x),\qquad m\ge 2,
\end{equation}
$$f_{0,2}(x)=x, \qquad  f_{2,1}(x)= \binom{x+1}{3}.$$

{\bf (II)} The following recurrence relation holds for coefficients $\lambda_{m,i},$  
\begin{equation}\label{lm-rec}
  \lambda_{m,i} = \frac{m(2m-1)}{2(m-i)(2m-2i+1)}\lambda_{m-1,i} + \frac{m(m-1)}{8(m-i)(2m-2i+1)} \lambda_{m-2,i-1},
\end{equation}
where  $0 &lt; i &lt; m.$

{\bf (III)} The following general formula holds
\begin{align}
\label{sum-l}                                       
 \lambda_{m,i} 
 = \frac{1}{2^m (2m-2i+1)} \sum_{k=0}^{2i} \binom{m}{2i-k} \binom{2m-2i+k}{k} B_k.
\end{align}
\end{theorem}

\begin{proof}

{\bf (I)} 
Since $\Delta f_{2,m} (x) = \binom{x+1}{2}^m,$ we have 
$$
 \Delta f_{2,m}'' (x) = m(2m-1) \binom{x+1}{2}^{m-1} + \frac{m(m-1)}{4}\binom{x+1}{2}^{m-2},
$$
and $$f_{2,m}''(0)=0.$$
Therefore, (\ref{fm2}) is true. 

{\bf (II)}  (\ref{lm-rec}) is a consequence of (\ref{fm2}).

{\bf (III)} We have 
\begin{align*}
  2^m f_{2,m}(N) &amp;= 
    \sum_{i=0}^{N-1} (i(i+1))^m = 
    \sum_{i=0}^{N-1} \sum_{j=0}^m \binom{m}{j} i^{m+j} =
    \sum_{j=0}^m \binom{m}{j} \sum_{i=0}^{N-1} i^{m+j} \\
    &amp;= \sum_{j=0}^{m}  \frac{1}{m+j+1}\binom{m}{j}\sum_{k=0}^{m+j} \binom{m+j+1}{k} B_k N^{m+j+1-k}\\
    &amp;= \sum_{k=1}^{2m+1} N^k \left( \sum_{j=0}^m \frac{1}{m+j+1} \binom{m}{j} \binom{m+j+1}{k} B_{m+j+1-k}\right).
\end{align*}
Therefore,
\begin{align*}
 \lambda_{m,i} &amp;= \frac{1}{2^m}\sum_{j \ge m - 2i}^{m} \frac{1}{m+j+1} \binom{m}{j} \binom{m+j+1}{2m-2i+1} B_{j+2i-m}\\
 &amp;= \frac{1}{2^m} \sum_{k=0}^{2i} \frac{1}{2m-2i+k+1} \binom{m}{m+k-2i} \binom{2m-2i+k+1}{k} B_k\\
 &amp;=\frac{1}{2^m (2m-2i+1)} \sum_{k=0}^{2i} \binom{m}{2i-k} \binom{2m-2i+k}{k} B_k.
\end{align*}
\end{proof}

\subsection{The polynomial part of \texorpdfstring{$\wlm_i{(m)}$}{Lg}}
Let us consider the function $\wh_i(m)$ defined as
 \begin{align*}
  \wh_i(m) &amp;= (-1)^i (i+1)(2m-2i+1) \frac{\wlm_i{(m)}}{\binom{m}{i}} %\\
  = (-1)^i(i+1) \sum_{k=0}^{2i} \frac{\binom{m}{2i-k} \binom{2m-2i+k}{k}}{\binom{m}{i}} B_k.
\end{align*} 
 In fact, $\wh_i(m) = h_i(m)/(i+1)!$, where $h_i(m)$ is defined in relation (\ref{B3}).

For example, 
$$
 \wh_0(m) = 1,$$
 $$\wh_1(m) = \frac{m+1}{3},$$
 $$\wh_2(m) = \frac{1}{60}(7m-6)(m+1),
$$
$$
 \wh_3(m) = \frac{(m+1)(31m^2-97m+60)}{630},
 $$
 $$
 \wh_4(m) = \frac{(m+1)(127m^3 - 833m^2 + 1606m - 840)}{5040}.
 $$

Analogous to Knuth's relation (\ref{A3}), polynomial parts of $\wlm_i{(m)}$ are interesting by themselves.
Below we study properties of $\wh_i(x).$ 

\begin{theorem} \label{wh} Let $i \ge 0.$  

{\bf (I)} The function $\wh_i(x)$ is a polynomial of degree $i$.

{\bf (II)} The following formula holds for $\wh_i(m)$
\begin{align}\label{hp}
  &amp;\wh_i(m) \\
  &amp;= (-1)^i (i+1)\left(\sum_{k=0}^{i} \frac{\binom{m-i}{i-k} \binom{2m-2i+k}{k}}{\binom{2i-k}{i}} B_k + \sum_{k=i+1}^{2i} \frac{\binom{i}{k-i} \binom{2m-2i+k}{k}}{\binom{m-2i+k}{k-i}} B_k \right) \nonumber\\
  &amp;=(-1)^i \frac{(i+1)!}{(2i)!}\left(\sum_{k=0}^{i} p_k(m) \binom{2i}{k} B_k + \sum_{k=i+1}^{2i} q_k(m)\binom{2i}{k} B_k \right)\nonumber.
\end{align}
Here, $p_k(m), q_k(m)$ are polynomials given by
$$
 p_k(m) = (m-i)\cdots(m-2i+k+1) (2m-2i+k)\cdots(2m-2i+1)
$$
and
\begin{align*}
 q_k(m) &amp;= \frac{(2m-2i+k)\cdots(2m-2i+1)}{(m-2i+k)\cdots(m-i+1)} \\
 &amp;= 2^{k-i} (2m-2i+1)(2m-2i+3)\cdots\\
 &amp;\times(2m-4i+2k+1)\dots(2m-2i+k).
\end{align*}
{\rm (}If $k = 2i,$ the last term $(2m-2i+k)$ in the product of $q_k(m)$ is cancelled{\rm )}.

{\bf (III)} The following  recurrence relation holds for $\wh_i(x)$ 
\begin{equation}
\label{rec-h}
 (2i+1-2x)\wh_i(x) = \frac{1}{2}(i+1)\wh_{i-1}(x-2) - (2x-1) \wh_i(x-1).
\end{equation}

{\bf (IV)} The initial value for $\wh_i(x)$ is given by:
\begin{equation}\label{h0}
 \wh_i(0) = (-1)^i (i+1) B_{2i}.
\end{equation}

{\bf (V)} If $x$ approaches $\infty,$ then 
\begin{equation}\label{hh0}
 \wh_i(x) \sim (-1)^i \frac{(i+1)!}{(2i)!} (2 - 2^{2i})B_{2i} x^i.
\end{equation}

{\bf (VI)} If $i&gt;0,$ then all zeros of the polynomial $\wh_i(x)$ are real. Moreover, it has  one negative zero $x_0=-1$ and if $i&gt;1,$ the other $i-1$ zeros $x_1, \dots, x_{i-1}$ 
are positive, distinct and satisfy the following inequalities 
$$
 0 &lt; x_1 &lt; 1, \; \; 2 &lt; x_2 &lt; 3,\;\;  3 &lt; x_3 &lt; 4, \;  \dots, \; i-1 &lt; x_{i-1} &lt; i.
$$
\end{theorem}

\begin{proof}
{\bf (I)} Note that the terms $\binom{m}{2i-k}$ and $\binom{2m-2i+k}{k}$ are polynomials in $m$ of degree $2i-k$
and $k$ respectively. Hence, $\binom{m}{2i-k}\binom{2m-2i+k}{k}$ is a polynomial of degree $2i$
and it vanishes at $m = 0, 1, \dots, i-1$ (if $i &gt; 0$). This  
means that it is divisible by $m(m-1)\cdots(m-i+1)$ or $\binom{m}{i}$. 
So, $\frac{\binom{m}{2i-k} \binom{2m-2i+k}{k}}{\binom{m}{i}}$ is a polynomial in $m$
of degree $i$.
Therefore, $\wh_i(m)$ is a polynomial in $m$ of degree $i$.

{\bf (II)} To prove (\ref{hp}), we have 
\begin{align*}
  &amp;\wh_i(m)
  = (-1)^i(i+1) \sum_{k=0}^{2i} \frac{\binom{m}{2i-k} \binom{2m-2i+k}{k}}{\binom{m}{i}} B_k\\
  &amp;= (-1)^i (i+1) \sum_{k=0}^{2i} \frac{i!(m-i)!}{(2i-k)!(m-2i+k)!} \binom{2m-2i+k}{k} B_k \nonumber\\
  &amp;= (-1)^i (i+1)\left(\sum_{k=0}^{i} \frac{\binom{m-i}{i-k} \binom{2m-2i+k}{k}}{\binom{2i-k}{i}} B_k + \sum_{k=i+1}^{2i} \frac{\binom{i}{k-i} \binom{2m-2i+k}{k}}{\binom{m-2i+k}{k-i}} B_k \right)\nonumber\\
  &amp;= (-1)^i (i+1) %\times \nonumber\\
  \sum_{k=0}^{i} \frac{(m-i)\cdots(m-2i+k+1) (2m-2i+k)\cdots(2m-2i+1)i!}{k!(2i-k)!} B_k \nonumber\\
  &amp;+ (-1)^i (i+1)\sum_{k=i+1}^{2i} \frac{i! (2m-2i+k)\cdots(2m-2i+1)}{k!(2i-k)!(m-2i+k)\cdots(m-i+1)} B_k \nonumber\\
  &amp;=(-1)^i \frac{(i+1)!}{(2i)!}\left(\sum_{k=0}^{i} p_k(m) \binom{2i}{k} B_k + \sum_{k=i+1}^{2i} q_k(m)\binom{2i}{k} B_k \right)\nonumber.
\end{align*}

{\bf (III)} We use (\ref{lm-rec}) for $x = m &gt; i$, where $m$ is a positive integer. We obtain  (\ref{rec-h}) for positive integers $x=m&gt;i.$
Since $\wh_i(x)$ is a polynomial, recurrence relation  (\ref{rec-h})  must hold for all real numbers $x$.

{\bf (IV)} A substitution $m = 0$ in  (\ref{hp}) gives us 
\begin{align*}
 \wh_i(0) &amp;= (-1)^i(i+1) \left(\sum_{k=0}^{i} \frac{\binom{-i}{i-k} \binom{-2i+k}{k}}{\binom{2i-k}{i}} B_k + \sum_{k=i+1}^{2i} \frac{\binom{i}{k-i} \binom{-2i+k}{k}}{\binom{-2i+k}{k-i}} B_k \right)\\
 &amp;= (-1)^i(i+1) \sum_{k=0}^{2i} \frac{1}{2} \binom{2i}{k} B_k = (-1)^i (i+1) B_{2i}.
\end{align*}
Thus, (\ref{h0}) is proved. 

{\bf (V)} Let us calculate the leading coefficient $A$ of $\wh_i(m)$.
By (\ref{hp}), we have
\begin{align*}
 A = (-1)^i \frac{(i+1)!}{(2i)!} \sum_{k=0}^{2i} 2^k \binom{2i}{k} B_k.
\end{align*}
Let $B_n(x)$ be the Bernoulli polynomial,
$$
 B_n(x)= \sum_{k=0}^{n}\binom{n}{k} B_k x^{n-k}.
$$
It is well known that $B_n(1/2) = (2^{1-n} - 1)B_n$.
Therefore, 
\begin{align*}
 A &amp;= (-1)^i \frac{(i+1)!}{(2i)!} 2^{2i}\sum_{k=0}^{2i} (1/2)^{2i-k} \binom{2i}{k} B_k\\
 &amp;= (-1)^i \frac{(i+1)!}{(2i)!} 2^{2i} B_{2i}(1/2) \\
 &amp;= (-1)^i \frac{(i+1)!}{(2i)!} 2^{2i} (2^{1-2i} - 1)B_{2i} = (-1)^i \frac{(i+1)!}{(2i)!} (2 - 2^{2i})B_{2i},
\end{align*}
which yields (\ref{hh0}).

{\bf (VI)} 
We have
\begin{align*}
 \wh_i(-1) &amp;= (-1)^i\frac{(i+1)}{\binom{-1}{i}} \sum_{k=0}^{2i} \binom{-1}{2i-k} \binom{-2-2i+k}{k} B_k\\
 &amp;= (i+1) \sum_{k=0}^{2i} \binom{2i+1}{k} B_k = 0.
\end{align*}

The part of the statement {\bf (VI)} that relates to  positive roots is evident  for $i = 2$. 
Suppose that $i \ge 3$.

Let $\sgn(x)=-1,$ if $x&lt;0;$ $\sgn(x)=0,$ if $x=0;$ and $\sgn(x)=1,$ if $x&gt;0.$ 

Putting $m = 1$ in (\ref{rec-h}), we have
$$
 (2i-1)\wh_i(1) = - \wh_i(0).
$$
Equation (\ref{h0}) gives us
\begin{equation}\label{1oct} 
\wh_i(0) = (-1)^i(i+1)B_{2i},
\end{equation}
\begin{equation}\label{2oct}
\wh_i(1) = (-1)^{i+1} \frac{i+1}{2i-1}B_{2i}.
\end{equation}
Therefore, 
$$\sgn(\wh(0)) = -\sgn(\wh(1)).$$ 
Hence,  there exists a real zero $x_1 \in (0,1)$ of $\wh_i(x)$.

Putting $m = 2$ in (\ref{rec-h}), by (\ref{1oct}) and (\ref{2oct}), we have 
$$
  \wh_i(2) = \frac{1}{2i-3} \left(\frac{1}{2}(-1)^{i-1}i B_{2i-2} + 3(-1)^i \frac{i+1}{2i-1}B_{2i} \right).
$$
By (\ref{zetaeven}), for any positive integer $i,$ 
$$\sgn(B_{2i}) = (-1)^{i+1}.$$
So, 
$$\sgn(\wh_i(2)) = \sgn(\wh_i(1)).$$
Therefore, relation (\ref{rec-h}) for $m = 3$  gives
$$
  \wh_i(3) = \frac{1}{2i-5}\left((-1)^i\frac{11}{2}\frac{i}{2i-3}B_{2i-2} + 3 (-1)^{i+1}\frac{i+1}{2i-1}B_{2i} \right).
$$
Hence, 
$$\sgn(\wh(3)) = -\sgn(\wh(2)).$$
Induction on $i$ and on $k$  shows that  for any $k \in [2,i]$ 
$$
 \sgn(\wh_i(k)) = (-1)^{i+k}\sgn(B_{2i}). 
$$
As we discussed above, this relation is also true for $k = 2,3$.

By inductive hypothesis,
$$
 \sgn(\wh_i(k-1)) = -\sgn(\wh_{i-1}(k-2)).
$$
Recurrence relation (\ref{rec-h}) yields
\begin{align*}
 \sgn(\wh_i(k)) &amp;= \sgn\left(\frac{1}{2i-5} \left( \frac{1}{2}\wh_{i-1}(k-2) - (2k-1)\wh_i(k-1)\right)\right)\\
 &amp;= \sgn(\wh_{i-1}(k-2) - \wh_i(k-1)) \\
 &amp;= \sgn((-1)^{i+k-3} B_{2i-2} - (-1)^{i+k-1} B_{2i})\\
 &amp;= (-1)^{i+k}\sgn(B_{2i}).
\end{align*}
It remains to note that
$$
 \sgn(\wh(2)) = -\sgn(\wh(3)) = \dots = (-1)^{i-1}\sgn(\wh(i-1)) = (-1)^{i}\sgn(\wh(i)).
$$
\end{proof}

\section{Acknowledgments} We are grateful to the anonymous referee for careful reading
the text and essential remarks.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B65; Secondary 11B68, 05A10.

\noindent \emph{Keywords: } 
binomial coefficient, Bernoulli number, Faulhaber's theorem.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000217},
\seqnum{A008292},
\seqnum{A093556}, 
\seqnum{A093557},
\seqnum{A093558},  and
\seqnum{A093559}.)


\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received  May 12 2012;
revised versions received   October 7 2012; January 4 2013.
Published in {\it Journal of Integer Sequences}, January 5 2013.

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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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