<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}
\usepackage{mathrsfs}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}}
\newcommand{\di}{\displaystyle}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf 
Another Proof of Zagier's Matrix Conjecture
}
\vskip 1cm
\large
Yawen Ma and Lee-Peng Teo\\
Department of Mathematics\\ 
Xiamen University Malaysia\\ 
Jalan Sunsuria, Bandar Sunsuria\\ 
43900, Sepang, Selangor\\ 
Malaysia \\
\href{mailto:m1102439194@gmail.com}{\tt m1102439194@gmail.com}\\
\href{mailto:lpteo@xmu.edu.my}{\tt lpteo@xmu.edu.my}
\end{center}

\vskip .2 in




\begin{abstract}
We prove a conjecture of Zagier about the inverse of a $(K-1)\times (K-1)$
matrix $A=A_{K}$ using elementary methods. This formula  allows one to
express the product of single zeta values $\zeta(2r)\zeta(2K+1-2r)$ for
$1\leq r\leq K-1$ in terms of the double zeta values $\zeta(2r,
2K+1-2r)$ for $1\leq r\leq K-1$ and $\zeta(2K+1)$.
\end{abstract}


\begin{center}
 \emph{Dedicated to Professor Don Zagier on the Occasion of His 70th Birthday.}\end{center}

\section{Introduction}
 This paper addresses a conjecture of Zagier  \cite{Zagier2012}.
For positive integers $k_1, \ldots, k_n$, with $k_n\geq 2$, define the multiple zeta value $\zeta(k_1, k_2, \ldots, k_n)$  by
\begin{align}\label{eq1_a}
\zeta\left(k_1, k_2, \ldots, k_n\right)&amp;=\sum_{1\leq m_1&lt; \cdots&lt; m_n }\frac{1}{m_1^{k_1}\cdots m_n^{k_n}}.
\end{align}Here $k=k_1+k_2+\cdots+k_n$ is called the 
{\it weight\/} of this multiple zeta value.

When $n=1$, we have the classical Riemann zeta value
$$\zeta(k)=\sum_{m=1}^{\infty}\frac{1}{m^k}.$$
When $n=2$,  the double sum
\begin{align}\label{eq2_a}
\zeta(k_1,k_2)=\sum_{m=2}^{\infty}\frac{1}{m^{k_2}}\sum_{j=1}^{m-1}\frac{1}{j^{k_1} }
\end{align}
was considered by Euler.

Let $H(0)=1$ and define
$$H(n)=\zeta(\underbrace{2, 2, \ldots, 2}_n)\quad\text{for $n\geq 1$}.$$ It is well known that for $n\geq 0$,
 $$H(n)=\frac{\pi^{2n}}{(2n+1)!}.$$
 When $a$ and $b$ are nonnegative integers, define
 $$H(a, b)=\zeta(\underbrace{2,\ldots,2}_a,3, \underbrace{2,\ldots,2}_b).$$
Zagier \cite{Zagier2012}  derived the formula
\begin{align}\label{eq30}
H(a, b)=2\sum_{r=1}^{K}(-1)^r\left( \binom{2r} {2a+2 } -\left(1-\frac{1}{2^{2r}}\right){2r\choose  2b+1  }\right)H(K-r)\zeta(2r+1),
\end{align}
where $K=a+b+1$.
Here for a real number  $n$ and a nonnegative integer $k$, we define the generalized binomial coefficient $\di {n\choose k}$ by
\begin{displaymath}
{n\choose k}=\begin{cases}\di\frac{n(n-1)\cdots (n-k+1)}{k! },\quad &amp; \text{if  $k\geq 1$};\\
\quad 1,\quad &amp; \text{if   $k=0$}.\end{cases}
\end{displaymath}
In particular, if $n$ is an integer and $n&lt;k$, then $\di{n\choose k}=0$.

 The formula  \eqref{eq30} expresses   $H(a, b)$ as a rational linear combination of the  $H(K-r)\zeta(2r+1)$. It can be used to prove that the following two sets
 \begin{align*}
 \mathscr{B}_1&amp;=\left\{H(a, K-1-a)\,|\, 0\leq a\leq K-1\right\}, \\
 \mathscr{B}_2&amp;=\left\{H(K-r)\zeta(2r+1)\,|\, 1\leq r\leq K\right\},
 \end{align*} span the same vector space over $\mathbb{Q}$.

 On the other hand, Euler found that all double zeta values of odd weight can be expressed as rational linear combinations of products of Riemann zeta values. The following explicit formula was derived in \cite{Zagier2012}.  For $K\geq 2$ and  $1\leq r\leq K-1$, we have
\begin{align}\label{eq40}
\zeta(2r, 2K+1-2r)&amp;=-\frac{1}{2}\zeta(2K+1) +\sum_{s=1}^{K-1}A_{r,s}\zeta(2s)\zeta(2K+1-2s),
\end{align}where
\begin{equation}\label{eq11}A_{r,s}={ 2K-2s\choose 2r-1 }+{2K-2s\choose 2K-2r }.\end{equation}
Zagier   \cite{Zagier2012} used an elementary argument to show that the $(K-1)\times (K-1)$ matrix 
$$A=A_K=(A_{r,s})$$   has nonzero determinant, and thus it is invertible.  Using the fact that both $\zeta (2s)$ and $H(2s)$  are rational multiples of $\pi^{2s}$, this shows that
the set $$\mathscr{B}_3=\left\{ \zeta(2r, 2K+1-2r)\,|\, 1\leq r\leq K-1\right\}\cup\left\{\zeta(2K+1)\right\}$$ spans the same  $\mathbb{Q}$-vector space as the set $\mathscr{B}_2$.

Zagier  \cite{Zagier2012} formulated three conjectures about the matrix $A_K$. The first one is about its determinant, the second is about its $LU$-decomposition, and the third is about its inverse. The main objective of this  paper is to prove the third conjecture, which states a pair of conjectural formulas for $A^{-1}$.

Let $P$ and $Q$ be the $(K-1)\times (K-1)$ matrices with entries
\begin{equation}\label{eq2}\begin{split}
P_{s,r}&amp;=\frac{2}{2s-1}\sum_{n=0}^{2K-2s}{2r-1\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n,\\
Q_{s,r}&amp;=-\frac{2}{2s-1}\sum_{n=0}^{2K-2s}{ 2K-2r\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n.
\end{split}\end{equation}Here $B_n$ are the Bernoulli numbers.

 Zagier conjectured that both $P$ and $Q$ are the inverse of $A$. This implies that $P$ and $Q$ must be the same matrix. This conjecture was proved by  Ma    \cite{Ma2016} using generating functions. In this work, we use a different approach, which is interesting in its own right.

 In Section \ref{equalmatrix}, we prove that $P=Q$. In Section \ref{conjecture}, we prove that they indeed give the inverse of $A$.




\section{The equality of the two conjectural formulas}
\label{equalmatrix}
Recall that the Bernoulli numbers $B_n$ are defined by the generating function \cite[p.\ 12]{Andrews}:
$$\frac{t}{e^t-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}t^n.$$  One can compute $B_n$ recursively by $B_0=1$,  and
$$B_n=-n!\sum_{k=0}^{n-1}\frac{B_k}{k! (n-k+1)!},\quad\text{for}\; n\geq 1.$$It is an easy consequence of the definitions  that $B_1=-1/2$, and for any odd integer $n&gt;1$ we have $B_n=0$.

Let $K$ be an integer greater than or equal to 2, and let $P=\left(P_{s,r}\right)$ and $Q=\left(Q_{s,r}\right)$ be the $(K-1)\times (K-1)$ matrices with entries $P_{s,r}$ and $Q_{s,r}$ given by \eqref{eq2}.
Our goal is to prove that $P_{s,r}=Q_{s,r}$ for all $1\leq r,s\leq K-1$, using a generating function technique that is quite different from that used by Ma \cite{Ma2016}.  In Zagier's paper \cite{Zagier2012}, the summations of $n$ in \eqref{eq2} are taken to be until the term $n=2K-2s+1$. Since $s\leq K-1$, we find that $2K-2s+1$ is an odd integer greater  than 2, and so $B_{2K-2s+1}=0$. Thus the summations  can be taken to be until $n=2K-2s$ only.

 We begin with the following lemma.

\begin{lemma}
Let $s$ be a positive integer. Define the function $f_s(t)$ by
$$f_s(t)=\frac{t^{2s-1}}{e^t-1}.$$ 
If $m$ is a positive integer, we have the following relation that relates the derivatives of $f_s$ up to  order $m$:
\begin{equation}\label{eq1}\begin{split}
e^tf_s^{(m)}(t)&amp;=\sum_{p=0}^{m} (-1)^{m-p}{m\choose p} f_s^{(p)}(t)\\
&amp;+\sum_{p=0}^{\min(m,2s-1)}(-1)^{m-p} { m\choose p} \frac{(2s-1)!}{(2s-1-p)!}t^{2s-1-p}.
\end{split}\end{equation}
\end{lemma}
\begin{proof}
By the definition of $f_s(t)$, we have
\begin{align*}
f_s(t)=e^{-t}f_s(t)+t^{2s-1}e^{-t}.
\end{align*}Differentiate both sides $m$ times and apply the Leibnitz rule.
This gives
\begin{align*}
f_s^{(m)}(t)&amp;=\sum_{p=0}^m {m \choose p} f_s^{(p)}(t)\frac{d^{m-p}}{dt^{m-p}}e^{-t}+\sum_{p=0}^{m} {m \choose p} \frac{d^p}{dt^p}t^{2s-1}\frac{d^{m-p}}{dt^{m-p}}e^{-t}.\end{align*}Since
$$\frac{d^p}{dt^{p}}t^{2s-1}=0\hspace{1cm}\text{if}\; p&gt;2s-1,$$ we find that
\begin{align*}
 f_s^{(m)}(t)&amp;=\sum_{p=0}^{m} (-1)^{m-p} {m \choose p} e^{-t}f_s^{(p)}(t)+
\sum_{p=0}^{\min(m,2s-1)}(-1)^{m-p} {m \choose p}e^{-t} \frac{(2s-1)!}{(2s-1-p)!}t^{2s-1-p}.
\end{align*}Multiplying both sides by $e^t$ gives
\eqref{eq1}.
\end{proof}

Now we can prove the main theorem in this section.
\begin{theorem}
 If $K,r,s$ are integers with $K\geq2$ is an integer and $r, s&lt;K$, then
 \begin{equation}\label{eq5}
 \begin{split} &amp;\frac{(2s-2)!}{(2K-2r)!} \sum_{n=0}^{2K-2s}{ 2K-2r\choose 2K-2s-n+1}{ n+2s-2\choose n}B_n\\
 &amp;=-\frac{(2s-2)!}{(2K-2r)!} \sum_{n=0}^{2K-2s}{ 2r-1\choose 2K-2s-n+1}{ n+2s-2\choose n}B_n.\end{split}\end{equation} In particular, this implies that the matrices $P$ and $Q$ defined by \eqref{eq2} are equal.
\end{theorem}
\begin{proof}The left hand side of \eqref{eq5} can be rewritten as
\begin{equation}\label{eq6}\begin{split}&amp;
\sum_{n=\max(0, 2r-2s+1)}^{2K-2s}\frac{1}{(2K-2s-n+1)!(2s-2r+n-1)!} \frac{(n+2s-2)!}{n!}B_n,\end{split}
\end{equation}and the right hand side of \eqref{eq5} can be rewritten as
\begin{equation}\label{eq4}\begin{split}&amp;
-\sum_{n=\max(0, 2K-2s-2r+2)}^{2K-2s}\frac{(2r-1)!}{(2K-2s-n+1)!(2r+2s-2K+n-2)!} \frac{(n+2s-2)!}{(2K-2r)!}\frac{B_n}{n!}.\end{split}
\end{equation}
The equality \eqref{eq5} is proved by taking $m=2r-1$ in the equation \eqref{eq1} and  comparing the coefficients of $t^{2K-2r}$ on both sides. Namely, we want to compare the coefficients of $t^{2K-2r}$ on both sides of the equation
\begin{equation}\label{eq3}\begin{split}
e^tf_s^{(2r-1)}(t)&amp;=-\sum_{p=0}^{2r-1} (-1)^{p}{ 2r-1\choose p } f_s^{(p)}(t)\\&amp;-\sum_{p=0}^{\min(2r-1,2s-1)}(-1)^{p} { 2r-1\choose p } \frac{(2s-1)!}{(2s-1-p)!}t^{2s-1-p}.
\end{split}\end{equation}
Notice that
$$f_s(t)=t^{2s-2} \cdot \frac{t}{e^t-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}t^{n+2s-2}.$$
Therefore, we get
\begin{align*}
f_{s}^{(p)}(t)=\sum_{n=\max(0, p-2s+2)}^{\infty}B_n\frac{(n+2s-2)!}{(n+2s-2-p)!n!}t^{n+2s-2-p}.
\end{align*}First we consider the coefficient of $t^{2K-2r}$ in the left hand side of \eqref{eq3}, namely, the coefficient of $t^{2K-2r}$ in
\begin{align}\label{eq8}
e^tf_s^{(2r-1)}(t)=\sum_{k=0}^{\infty}\frac{t^k}{k!}\sum_{n=\max(0, 2r-2s+1)}^{\infty}B_n\frac{(n+2s-2)!}{(n+2s-2r-1)!n!}t^{n+2s-2r-1}.
\end{align}It is given by the expression \eqref{eq6}.

The term on the right hand side of \eqref{eq3} can be written as $T_1+T_2$, where
\begin{align*}
T_1&amp;=-\sum_{p=0}^{2r-1} (-1)^{p} { 2r-1\choose p } f_s^{(p)}(t)\\
&amp;=-\sum_{p=0}^{2r-1}(-1)^p\frac{(2r-1)!}{p!(2r-1-p)!}\sum_{n=\max(0,  p-2s+2)}^{\infty}B_n\frac{(n+2s-2)!}{(n+2s-2-p)!n!}t^{n+2s-2-p}\\
&amp;=-\sum_{n=0}^{\infty}\sum_{p=0}^{\min(2r-1,n+2s-2)}(-1)^p\frac{(2r-1)!}{p!(2r-1-p)!}B_n\frac{(n+2s-2)!}{(n+2s-2-p)!n!}t^{n+2s-2-p},\\
T_2&amp;=-\sum_{p=0}^{\min(2r-1,2s-1)}(-1)^{p} \frac{(2r-1)!}{p!(2r-1-p)!}\frac{(2s-1)!}{(2s-1-p)!}t^{2s-1-p}.
\end{align*}
The term $T_2$ contains a term in $t^{2K-2r}$ if and only if $2s-1\geq 2K-2r$, or equivalently, $r+s\geq K+1$. In this case the coefficient of $t^{2K-2r}$ in $T_2$ is
\begin{align}\label{eq9}
\frac{(2r-1)!}{(2r+2s-2K-1)!(2K-2s)!}\frac{(2s-1)!}{(2K-2r)!}.
\end{align}For the term $T_1$, the coefficient of $t^{2K-2r}$ is
\begin{equation}\label{eq7}\begin{split}
&amp;-\sum_{n=\max(0, 2K-2s-2r+2)}^{2K-2s}(-1)^n\frac{(2r-1)!}{(2K-2s-n+1)!(2r+2s-2K+n-2)!} \frac{(n+2s-2)!}{(2K-2r)!}\frac{B_n}{n!}.
\end{split}\end{equation}
When $r+s\leq K$,  we find that $2K-2s-2r+2\geq 2$. Hence, the sum over
$n$ in \eqref{eq7} does not contain the $n=1$ term. Since $B_n=0$ when
$n &gt; 2$ is an odd integer, we find that \eqref{eq7} is equal to
 \eqref{eq4}. Since there is no contribution from $T_2$ to the term $t^{2K-2r}$ when $r+s\leq K$, this proves that when $r+s\leq K$, the coefficient of $t^{2K-2r}$ in the right hand side of
 \eqref{eq3} is \eqref{eq4}.

 When $r+s\geq K+1$, the  $n=1$ term is present in \eqref{eq7}. Using the fact that $\di B_1=-\frac{1}{2}$, we find that this term is given by
 \begin{align*}
-\frac{1}{2}\frac{(2r-1)!}{(2r+2s-2K-1)!(2K-2s)!}\frac{(2s-1)!}{(2K-2r)!},
\end{align*}which is $-1/2$ of the term \eqref{eq9}. Summing the coefficients of $t^{2K-2r}$ from $T_1$ and $T_2$, we find that the sum is equal to \eqref{eq4}. Therefore, when $r+s\geq K+1$, the coefficient of $t^{2K-2r}$ in the right hand side of
 \eqref{eq3} is \eqref{eq4}.

Thus, we have shown that the coefficient of $t^{2K-2r}$ in the left hand side of \eqref{eq3} is \eqref{eq6}, and the coefficient of $t^{2K-2r}$ in the right hand side of \eqref{eq3} is \eqref{eq4}. This completes the proof of the theorem.


\end{proof}
As we mentioned before, this theorem was proved by Ma \cite{Ma2016}  using a different method, with the help of Carlitz's Bernoulli number identity \cite{Carlitz1971, Prodinger2014}. Our proof uses the generating function of the Bernoulli numbers directly.

\begin{remark} Carlitz's identity says that for any nonnegative integers $m$ and $n$, we have
\begin{equation}\label{eq10} (-1)^m\sum_{k=0}^m{ m\choose k } B_{n+k}=(-1)^n\sum_{k=0}^n{ n\choose k } B_{m+k}.\end{equation}
Prodinger \cite{Prodinger2014} gave a short  proof using an exponential generating function of two variables.  Here we show that  this identity can be derived directly from \eqref{eq1} by setting $s=1$. Namely, we consider the generating function  of the Bernoulli numbers
$$f(t)=\frac{t}{e^t-1}.$$
Since \eqref{eq10} is symmetric in $m$ and $n$, it is sufficient to consider the  case $0\leq m&lt; n$. In this case, $n\geq 1$.
Equation \eqref{eq1} says that
$$e^tf^{(m)}(t)=(-1)^m t+(-1)^{m-1}m+\sum_{k=0}^{m}(-1)^{m-k}{ m\choose k }f^{(k)}(t).$$
This gives
\begin{align*}
\sum_{l=0}^{\infty}\frac{t^l}{l!}\sum_{k=m}^{\infty}\frac{B_k}{(k-m)!}t^{k-m}=(-1)^mt+(-1)^{m-1}m+\sum_{k=0}^m (-1)^{m-k}{m\choose k }\sum_{l=k}^{\infty}\frac{B_l}{(l-k)!}t^{l-k}.
\end{align*}Compare the coefficients of $t^n$ on both sides, we have
\begin{align*}
\sum_{k=m}^{m+n}\frac{1}{(k-m)!(m+n-k)!}B_k&amp;=(-1)^m\delta_{n,1}+\sum_{k=0}^{m}(-1)^{m-k}{ m\choose k } \frac{B_{n+k}}{n!}.
\end{align*}If $n=1$, the last sum contains the term $(-1)^m B_1$, which can be combined with $(-1)^m\delta_{n,1}$ to yield
$(-1)^{m-1}B_1$. Since $B_k=0$ when $k$ is odd and greater than 2, we find that
 \begin{align*}
\sum_{k=m}^{m+n}\frac{n!}{(k-m)!(m+n-k)!}B_k&amp;= \sum_{k=0}^{m}(-1)^{m-n}{ m\choose k } B_{n+k}.
\end{align*}Multiplying $(-1)^n$ on both sides and shifting the summation variables on the left hand side, we obtain
\begin{align*}
(-1)^n\sum_{k=0}^{n}{ n\choose k}B_{m+k}&amp;= (-1)^m\sum_{k=0}^{m} { m\choose k } B_{n+k},
\end{align*}
which is Carlitz's identity.

\end{remark}

\section{The proof of the conjecture}\label{conjecture}
In this section, we prove that the inverse of the matrix $A=A_K$ is the matrix $P$.
\begin{theorem}[Zagier's conjecture]  \quad
Let $K&gt;1$ be an integer, and let $A=A_K$ be the $(K-1)\times (K-1)$ matrix defined by \begin{equation}\label{eq12}A_{r,s}={ 2K-2s\choose 2r-1 }+{ 2K-2s\choose 2K-2r }.\end{equation} The inverse of $A$ is the matrix $P$ defined by  one of the following two formulas that are equal.
\begin{equation}\label{eq13}\begin{split}
P_{s,r}&amp;=\frac{2}{2s-1}\sum_{n=0}^{2K-2s}{ 2r-1\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n,\\
&amp;=-\frac{2}{2s-1}\sum_{n=0}^{2K-2s}{ 2K-2r\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n.
\end{split}\end{equation}
\end{theorem}
\begin{proof}
Define the $(K-1)\times (K-1)$ matrices $B$ and $C$ by
\begin{align*}
B_{r,s}={ 2K-2s\choose 2r-1 },\hspace{1cm}C_{r,s}={2K-2s\choose 2K-2r},
\end{align*}so that $A=B+C$. Notice that $B_{r,s}=0$ if $r+s&gt; K$, and $C_{r,s}=0$ if $r&lt;s$.

Our strategy of the proof is to  show that the matrix $PA=PB+PC$ is indeed the identity matrix, by showing that if $s$ and $s'$ are positive integers less than $K$, then
\begin{align}\label{eq17}
(PB)_{s,s'}+(PC)_{s,s'}=\begin{cases} 1,\quad &amp; \text{if $s=s'$};\\0,\quad &amp; \text{if $s\neq s'$}.\end{cases}
\end{align}We use the following two elementary identities of 
binomial coefficients.
If $n\geq 1$, then
\begin{equation}\label{eq14}\begin{split}
\sum_{k=0}^{\lfloor n/2\rfloor}{ n\choose 2k }&amp;=2^{n-1},\\
\sum_{k=0}^{\lfloor (n-1)/2\rfloor}{ n\choose 2k+1 }&amp;=2^{n-1}.
\end{split}\end{equation}First we compute $(PB)_{s,s'}$ using the first formula in \eqref{eq13} for $P_{s,r}$. The cases where $s\leq s'$ and $s&gt;s'$ are considered separately.

If $s\leq s'$, then for $r\leq K-s'$, we have $r\leq K-s$. Hence, $2K-2s-2r+2\geq 2$. Therefore,
\begin{align*}
(PB)_{s,s'} &amp;=\frac{2}{2s-1}\sum_{r=1}^{K-s'}{ 2K-2s'\choose 2r-1 }\sum_{n=  2K-2s-2r+2}^{2K-2s}{ 2r-1\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n\\
&amp;=\frac{2}{2s-1}\sum_{n=2s'-2s+2}^{2K-2s}\sum_{r=K-s-\frac{n}{2}+1}^{K-s'}{2K-2s'\choose 2r-1 }{ 2r-1\choose 2K-2s-n+1 } {n+2s-2\choose n }B_n.
\end{align*}
Notice that the summation over $n$ only contains terms with $n$ even, since $2s'-2s+2\geq 2$.

 Using the binomial identity
 \begin{displaymath}
 {a\choose b}{b\choose c}={a\choose a-c}{a-c \choose b-c},
 \end{displaymath}we have
\begin{align*}
{ 2K-2s'\choose 2r-1 }{ 2r-1\choose 2K-2s-n+1 } &amp;= { 2K-2s'\choose 2s-2s'+n-1 }{ 2s-2s'+n-1\choose 2r+2s-2K+n-2}.
\end{align*}Therefore,
\begin{align*}
(PB)_{s,s'}&amp;=\frac{2}{2s-1}\sum_{n=2s'-2s+2}^{2K-2s}   { 2K-2s'\choose 2s-2s'+n-1 }{ n+2s-2\choose n }B_n  \\&amp;\hspace{3cm} \times\sum_{r=K-s-\frac{n}{2}+1}^{K-s'}{2s-2s'+n-1\choose 2r+2s-2K+n-2 }.
\end{align*} For $n\geq 2s'-2s+2$, we have $2s-2s'+n-1\geq 1$. The first formula  in \eqref{eq14} implies that
\begin{equation}\label{eq22}\begin{split}
\sum_{r=K-s-\frac{n}{2}+1}^{K-s'}{ 2s-2s'+n-1\choose 2r+2s-2K+n-2 }
&amp;=\sum_{r=0}^{s-s'+n/2-1}{ 2s-2s'+n-1\choose 2r }\\&amp;=2^{2s-2s'+n-2}.\end{split}\end{equation}
This shows that when $s\leq s'$, we have
\begin{align}\label{eq25}
(PB)_{s,s'}&amp;=\frac{2}{2s-1}\sum_{n=2s'-2s+2}^{2K-2s}  { 2K-2s'\choose 2s-2s'+n-1 }{ n+2s-2\choose n}2^{2s-2s'+n-2}B_n.
\end{align}
When $s&gt;s'$, we get
\begin{align*}
(PB)_{s,s'}&amp;=\frac{2}{2s-1} \sum_{n=0}^{2K-2s}\sum_{K-s-\frac{n}{2}+1\leq r\leq K-s'}{ 2K-2s'\choose 2r-1 }{ 2r-1\choose 2K-2s-n+1 } {n+2s-2\choose n }B_n.
\end{align*}Splitting out the   $n=1$ term, we have
\begin{align*}
(PB)_{s,s'}&amp;=\frac{2}{2s-1} \sum_{\substack{0\leq n\leq 2K-2s\\
n\,\text{is even}}}{ 2K-2s'\choose 2s-2s'+n-1 }{ n+2s-2\choose n }B_n\\
&amp;\hspace{4cm}\times\sum_{r=K-s-\frac{n}{2}+1}^{K-s'} { 2s-2s'+n-1\choose 2r+2s-2K+n-2 } \\
 &amp;\quad +\frac{2}{2s-1} { 2K-2s'\choose 2s-2s' }{2s-1\choose 1 }B_1\sum_{r=K-s +1}^{K-s'} { 2s-2s'\choose 2r+2s-2K-1 }.\end{align*}
Notice that the second formula in \eqref{eq14} gives
$$\sum_{r=K-s +1}^{K-s'}  {2s-2s'\choose 2r+2s-2K-1 } =\sum_{r=0}^{s-s'-1} {2s-2s'\choose 2r+1 } =2^{2s-2s'-1}.$$Together with \eqref{eq22}, we 
find that when $s&gt;s'$, we have
\begin{align*}
(PB)_{s,s'}&amp;=\frac{2}{2s-1} \sum_{n=0}^{2K-2s}  { 2K-2s'\choose 2s-2s'+n-1 }  { n+2s-2\choose n }2^{2s-2s'+n-2}B_n.
\end{align*}

Next we compute $(PC)_{s,s'}$. Using the second expression in \eqref{eq13} for $P_{s,r}$, we have
\begin{equation}\label{eq2022_1}\begin{split}
(PC)_{s,s'}
&amp;=-\frac{2}{2s-1}\sum_{r=s'}^{K-1}{ 2K-2s'\choose 2K-2r } \sum_{n=\max(0,2r-2s+1)}^{2K-2s}{2K-2r\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n.
\end{split}\end{equation} Using the binomial identity
 \begin{displaymath}
 {a\choose b}{b\choose c}={a\choose a-c}{a-c \choose a-b},
 \end{displaymath}we have
\begin{align*}
{ 2K-2s'\choose 2K-2r }{ 2K-2r\choose 2K-2s-n+1 } ={ 2K-2s'\choose 2s-2s'+n-1 }{ 2s-2s'+n-1\choose 2r-2s' }.
\end{align*}
Now we discuss the cases $s&lt;s'$, $s=s'$ and $s&gt;s'$ separately.

When $s=s'$, we have
\begin{align*}
(PC)_{s,s}&amp;=-\frac{2}{2s-1}\sum_{r=s}^{K-1}{ 2K-2s\choose 2K-2r } \sum_{n= 2r-2s+1 }^{2K-2s}{ 2K-2r\choose 2K-2s-n+1 }{ n+2s-2\choose n }B_n.
\end{align*}In this case, we have an $n=1$ term when $r=s$. When $r&gt;s$, the summation over $n\geq 2r-2s+1$ can be replaced by the summation over $n\geq 2r-2s+2$.  The term with $r=s$ and $n=1$ contributes the term 1. Therefore,
\begin{align*}
(PC)_{s,s}&amp;= 1-\frac{2}{2s-1}\sum_{r=s}^{K-1}{ 2K-2s\choose 2K-2r } \sum_{n=2r-2s+2}^{2K-2s}{ 2K-2r\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n\\
&amp;=1-\frac{2}{2s-1}\sum_{n=2}^{2K-2s}\sum_{r=s}^{s+n/2-1}{2K-2s\choose n-1 }{ n-1\choose 2r-2s } { n+2s-2\choose n }B_n\\
&amp;=1-\frac{2}{2s-1}\sum_{n=2}^{2K-2s}{2K-2s\choose n-1 } { n+2s-2\choose n }2^{n-2}B_n.
\end{align*}The last equality follows from the first equation in \eqref{eq14}. Comparing to the   $s=s'$ case in \eqref{eq25}   shows that
$$(PB+PC)_{s,s}=1.$$

Next we consider the case $s&lt;s'$. Since $r\geq s'$ in the summation \eqref{eq2022_1}, we find that  $2r-2s+1\geq 3$. Therefore,
\begin{align*}
(PC)_{s,s'}&amp;=-\frac{2}{2s-1}\sum_{r=s'}^{K-1}  \sum_{n=2r-2s+2}^{2K-2s}{ 2K-2s'\choose 2s-2s'+n-1 }{ 2s-2s'+n-1\choose 2r-2s' }  {n+2s-2\choose n }B_n\\
&amp;=-\frac{2}{2s-1}\sum_{n=2s'-2s+2}^{2K-2s}{ 2K-2s'\choose 2s-2s'+n-1 }{n+2s-2\choose n }B_n\sum_{r=s'}^{s+n/2-1}{ 2s-2s'+n-1\choose 2r-2s' } \\
&amp;=-\frac{2}{2s-1}\sum_{n=2s'-2s+2}^{2K-2s}{2K-2s'\choose 2s-2s'+n-1 }{n+2s-2\choose n }2^{2s-2s'+n-2}B_n\\
&amp;=-(PB)_{s,s'}.
\end{align*}Finally, we consider the case $s&gt;s'$.  In this case
 \begin{align*}
(PC)_{s,s'}&amp;=-\frac{2}{2s-1} \sum_{n=0}^{2K-2s}\sum_{s'\leq r\leq s+n/2-1} { 2K-2s'\choose 2K-2r } {2K-2r\choose 2K-2s-n+1 } { n+2s-2\choose n }B_n.
\end{align*}Splitting out the   $n=1$ term, we have
\begin{align*}
(PC)_{s,s'}
&amp;=-\frac{2}{2s-1} \sum_{\substack{0\leq n\leq 2K-2s\\n\,\text{is even}}} { 2K-2s'\choose 2s-2s'+n-1 }{ n+2s-2\choose n }B_n\sum_{r=s'}^{s+n/2-1}{ 2s-2s'+n-1\choose 2r-2s' } \\
&amp;  \quad -\frac{2}{2s-1} { 2K-2s'\choose 2s-2s' }{ 2s-1\choose 1 }B_1\sum_{r=s'}^s{ 2s-2s'\choose 2r-2s' }  
\end{align*}
\begin{align*}
&amp;=-\frac{2}{2s-1} \sum_{n=0}^{2K-2s} {2K-2s'\choose 2s-2s'+n-1 }{ n+2s-2\choose n }2^{2s-2s'+n-2}B_n\\
&amp;=-(PB)_{s,s'}.
\end{align*}This completes the proof of \eqref{eq17}, and so the assertion of the theorem is proved.
\end{proof}



 \section{Acknowledgment}
  This work is supported by the XMUM Research Fund XMUMRF/2018-C2/IMAT/0003. We would like to thank the anonymous referee and the editor for their suggestions and comments.

\begin{thebibliography}{99}


\bibitem{Carlitz1971} L. Carlitz,
\newblock Problem 795,
\newblock {\em Math. Mag.} {\bf 44} (1971), 107.


\bibitem{Andrews} G. E. Andrews, R. Askey, and R. Roy,
\newblock {\em Special Functions},
\newblock Cambridge University Press,  1999.

\bibitem{Ma2016} D. Ma,
\newblock Inverse of some matrix related to double zeta values of odd weight,
\newblock  {\em J. Number Theory} {\bf 166} (2016), 166--180.

\bibitem{Prodinger2014} H. Prodinger,
\newblock Carlitz's symmetric Bernoulli number identity,
\newblock  {\em J. Integer Sequences} {\bf 17} (2014), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL17/Prodinger/prod3.html}{Article 14.4.1}.

\bibitem{Zagier2012}  D. Zagier,
\newblock Evaluation of the multiple zeta values $\zeta(2,\ldots,2,3,2,\ldots,2)$,
\newblock {\em Ann. of Math.} {\bf 175} (2012), 977--1000.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B65; Secondary 11B68, 11M32.

\noindent \emph{Keywords:} 
double zeta value, Bernoulli number, generating function.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 1 2022;
revised versions received  March 2 2022; June 28 2022; June 29 2022.
Published in {\it Journal of Integer Sequences}, June 29 2022.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


</pre></body></html>