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\begin{center}
\vskip 1cm{\LARGE\bf Symmetric Identities on Modified\\
\vskip .1in
Degenerate Bernoulli Polynomials
}
\vskip 1cm
\large
Qi-Peng Su and Hao Pan\\
School of Applied Mathematics\\
Nanjing University of Finance and Economics\\ 
Nanjing, Jiangsu 210023\\
China\\
\href{mailto:3093420060@qq.com}{\tt 3093420060@qq.com} \\
\href{mailto:haopan79@zoho.com}{\tt haopan79@zoho.com} \\

\end{center}

\vskip .2 in
\begin{abstract}
The modified degenerate Bernoulli polynomial $\fB_{n,\lambda}(x)$ introduced by Dolgy et al.\ is given by
$$
\sum_{n=0}^\infty\frac{\fB_{n,\lambda}(x)}{n!}t^n=\frac{t(1+\lambda)^{\frac{xt}\lambda}}{(1+\lambda)^{\frac t\lambda}-1}.
$$
In this paper, we prove some symmetric identities on modified degenerate Bernoulli polynomials, which generalize the result of
Fu, Pan, and Zhang.
\end{abstract}















\section{Introduction}

The Bernoulli polynomial $B_n(x)$ $(n\in \mathbb{N})$ is given by
$$
\sum_{n=0}^\infty\frac{B_n(x)}{n!}t^n=\frac{te^{xt}}{e^t-1}.
$$
In particular, $B_n:=B_n(0)$ is called the $n$-th Bernoulli number.
A curious identity of Miki \cite{Mi78} says that
\begin{equation}
\label{miki}
\sum_{k=2}^{n-2}\frac{B_kB_{n-k}}{k(n-k)}=\sum_{k=2}^{n-2}\binom{n}{k}\frac{B_kB_{n-k}}{k(n-k)}+2H_n\cdot\frac{B_n}{n}
\end{equation}
for $n\geqslant 4$, where the $n$-th harmonic number
$
H_n:=\sum_{i=1}^n{1}/{i}.
$
Different proofs of
\eqref{miki} were given by Shirantani and Yokoyama
\cite{SY82}, Gessel \cite{Ge05}, Dunne and
Schubert \cite{DS13}. Furthermore, for $n\geqslant 4$, Dunne and Schubert also proved a similar identity
\begin{equation}
\label{matiyasevich}
(n+2)\sum_{k=2}^{n-2}B_kB_{n-k}=2\sum_{k=2}^{n-2}\binom{n+2}{k}B_kB_{n-k}+n(n+1)B_n,
\end{equation}
which was conjectured by
Matiyasevich \cite{Ma97}.

In \cite{PS06}, Pan and Sun used the
difference-differential method to establish the polynomial extensions of the Miki identity and the Matiyasevich-Dunne-Schubert identity. Moreover, Sun and Pan \cite{SP06} proved the following general symmetric identity involving Bernoulli polynomials:
\begin{equation}
\label{sunpan} r\left[\begin{matrix}s&amp;t\\
x&amp;y\end{matrix}\right]_n+ s\left[\begin{matrix}t&amp;r\\
y&amp;z\end{matrix}\right]_n+ t\left[\begin{matrix}r&amp;s\\
z&amp;x\end{matrix}\right]_n=0,
\end{equation}
where $r+s+t=n$, $x+y+z=1$ and
$$
\left[\begin{matrix}s&amp;t\\
x&amp;y\end{matrix}\right]_n=\sum_{k=0}^n(-1)^k\binom{s}{k}\binom{t}{n-k}B_{n-k}(x)B_k(y).
$$
Subsequently, Fu, Pan, and Zhang \cite{FPZ09} gave a generalization of
\eqref{sunpan} involving sums of products of more Bernoulli
polynomials as follows:
\begin{align}
\label{FuPanZhang}
&amp;r_{m+1}\sum_{\substack{k_1,\ldots,k_m\geq 0\\
k_1+\cdots+k_m=n}}\prod_{j=1}^m\binom{r_j}{k_j}B_{k_j}(x_j)\notag\\
&amp;=-\sum_{i=1}^mr_i\sum_{\substack{k_1,\ldots,k_m\geq 0\\
k_1+\cdots+k_m=n}}\binom{r_{m+1}}{k_i}B_{k_i}(1-x_i)\prod_{\substack{1\leq
j\leq m\\j\not=i}}\binom{r_j}{k_j}B_{k_j}(x_j-x_i+\1_{j&gt;i}),
\end{align}
where $r_{m+1}=n-r_1-\cdots-r_m$ and $\1_{j&gt;i}=1$ or $0$ according
to whether $j&gt;i$ or not.

On the other hand, in \cite{DKKS16}, Dolgy, Kim, Kwon, and Seo introduced the modified degenerate Bernoulli polynomial, which is different from Carlitz's degenerate Bernoulli polynomial \cite{Ca79}.
Define the modified degenerate Bernoulli polynomial $\fB_{n,\lambda}(x)$ by
$$
\sum_{n=0}^\infty\frac{\fB_{n,\lambda}(x)}{n!}t^n=\frac{t(1+\lambda)^{\frac{xt}\lambda}}{(1+\lambda)^{\frac t\lambda}-1},
$$
where $\lambda\in\C$.
Moreover, set $\fB_{n,\lambda}(x)=0$ if $n&lt;0$.
Since $(1+\lambda)^{\frac 1\lambda}$ tends to $e$ as $\lambda\to 0$,
clearly
$$
\lim_{\lambda\to 0}\fB_{n,\lambda}(x)=B_n(x).
$$
Various results related to the modified degenerate Bernoulli polynomials have been discussed in \cite{DKS16,LK17,PKK17,SKK19}.

In this short note, we extend \eqref{FuPanZhang} to the modified degenerate Bernoulli polynomials.
\begin{theorem}\label{main} Suppose that $m\geq 2$ and $n\geq 1$ are integers. Let $r_1,\ldots,r_m,\lambda$ be arbitrary complex numbers and $r_{0}=n-r_1-\cdots-r_m$. We have
\begin{align}\label{maine}
&amp;r_{0}\sum_{\substack{k_1,\ldots,k_m \geq 0\\ k_1+\cdots+k_m =n}}\prod_{j=1}^{m}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j)\nonumber\\
&amp;=-\sum_{i=1}^{m}r_i\sum_{\substack{k_1,\ldots,k_m \geq 0\\ k_1+\cdots+k_m =n}}\binom{r_{0}}{k_i}\fB_{k_i,\lambda}(1-x_{i})\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j-x_i+\1_{j&gt;i}).\nonumber\\
\end{align}
\end{theorem}
When $m=2$, we have the following extension of \eqref{sunpan}.
\begin{corollary}
\label{Corm2} For $s,t,\lambda\in\C$,
let
$$
\fD_{s,t,\lambda}(x,y)=\sum_{k=0}^n(-1)^k\binom{s}{k}\binom{t}{n-k}\fB_{n-k,\lambda}(x)\fB_{k,\lambda}(y).
$$
Then for each $n\geq 1$,
\begin{equation}\label{sunpanD}
r\cdot\fD_{s,t,\lambda}(x,y)+s\cdot\fD_{t,r,\lambda}(y,z)+t\cdot\fD_{r,s,\lambda}(z,x)=0,
\end{equation}
where $r,s,t,x,y,z\in\C$ satisfy that $r+s+t=n$ and $x+y+z=1$. In particular, we have the extensions of \eqref{miki} and \eqref{matiyasevich} as follows:
\begin{equation}\label{mikiD}
\sum_{k=1}^{n-1}\frac{\fB_{k,\lambda}}{k}\cdot\fB_{n-k,\lambda}-\sum_{k=1}^{n-1}\binom{n}{k}\frac{\fB_{k,\lambda}}{k}\cdot\fB_{n-k,\lambda}
=\frac{\lambda}{\ln(1+\lambda)}\cdot H_n\fB_{n,\lambda}+\frac{n}{2}\cdot\fB_{n-1,\lambda},
\end{equation}
and
\begin{equation}\label{matiyasevichD}
(n+2)\sum_{k=2}^{n-2}\fB_{k,\lambda}\fB_{n-k,\lambda}=2\sum_{k=2}^{n-2}\binom{n+2}{k}\fB_{k,\lambda}\fB_{n-k,\lambda}+\frac{\lambda}{\ln(1+\lambda)}\cdot n(n+1)\fB_{n,\lambda},
\end{equation}
where $\fB_{k,\lambda}=\fB_{k,\lambda}(0)$ and $n\geq4$.
\end{corollary}

The proofs of Theorem \ref{main} and Corollary \ref{Corm2} will be given in the next section.


\section{Proofs of Theorem \ref{main} and Corollary \ref{Corm2}}

First, we need the following lemma.
\begin{lemma}  Suppose that $m\geq 2$ and $s_1,\ldots,s_m $ be non-negative integers. Then
\begin{align}
&amp;\sum_{i=1}^{m}s_i\fB_{s_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\fB_{s_j,\lambda}(x_j)\nonumber\\
&amp;=\sum_{i=1}^{m}s_i\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =s_1+\cdots+s_m}}\fB_{k_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{s_j}{k_j}\fB_{k_j,\lambda}(x_j-x_i+\1_{j&gt;i}).
\end{align}
\end{lemma}
\begin{proof}
Clearly
\begin{align}\label{eq:1}
&amp;(t_1+\cdots+t_m)\prod_{j=1}^{m}\frac{t_j(\lambda+1)^{\frac{x_jt_j}\lambda}}{(\lambda+1)^{\frac{t_j}\lambda}-1}\nonumber\\
&amp;=\frac{(t_1+\cdots+t_m)}{(\lambda+1)^{\frac{t_1+\cdots+t_m}{\lambda}}-1}
\bigg(\sum_{i=1}^{m}((\lambda+1)^{\frac{t_i}\lambda}-1)(\lambda+1)^{\sum_{i&lt;j\leq m}\frac{t_j}\lambda}\bigg)\prod_{j=1}^{m}\frac{t_j(\lambda+1)^{\frac{x_jt_j}\lambda}}{(\lambda+1)^{\frac{t_j}\lambda}-1}\nonumber\\
&amp;=\sum_{i=1}^{m}\frac{(t_1+\cdots+t_m)}{(\lambda+1)^{\frac{t_1+\cdots+t_m}{\lambda}}-1} \cdot(\lambda+1)^{\sum_{i&lt;j\leq m}\frac{t_j}\lambda}\cdot t_i(\lambda+1)^{\frac{x_it_i}\lambda}\prod_{\substack{j=1\\j\neq i}}^{m}\frac{t_j(\lambda+1)^{\frac{x_jt_j}\lambda}}{(\lambda+1)^{\frac{t_j}\lambda}-1}\nonumber\\
&amp;=\sum_{i=1}^{m}\frac{t_i(t_1+\cdots+t_m)(\lambda+1)^{\frac{x_i(t_1+\cdots+t_m)}{\lambda}}}{(\lambda+1)^{\frac{t_1+\cdots+t_m}{\lambda}}-1}
\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{t_j(\lambda+1)^{\frac{(x_j-x_i+\1_{j&gt;i})t_j}{\lambda}}}{(\lambda+1)^{\frac{t_j}\lambda}-1}.
\end{align}
Let $[t_1^{s_1}\cdots t_m^{s_m}]f(t_1,\ldots,t_m)$ denote the coefficient of $t_1^{s_1}\cdots t_m^{s_m}$ in the formal power series $f(t_1,\ldots,t_m)$.
Then
\begin{align}\label{eq:2}
&amp;[t_1^{s_1}\cdots t_m^{s_m}](t_1+\cdots+t_m)\prod_{j=1}^{m}\frac{t_j(\lambda+1)^{\frac{x_jt_j}\lambda}}{(\lambda+1)^{\frac{t_j}\lambda}-1}\nonumber\\
&amp;=[t_1^{s_1}\cdots t_m^{s_m}](t_1+\cdots+t_m)
\sum_{k_j=0}^\infty\frac{\fB_{s_j,\lambda}(x)}{s_j!}t_j^{s_j}\nonumber\\
&amp;=\sum_{i=1}^{m}\frac{\fB_{s_i-1,\lambda}(x_i,\lambda)}{(s_i-1)!}\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{\fB_{s_j,\lambda}(x_j)}{s_j!}.
\end{align}
On the other hand, we have
\begin{align}\label{eq:3}
&amp;[t_1^{s_1}\cdots t_m^{s_m}]\sum_{i=1}^{m}\frac{t_i(t_1+\cdots+t_m)(\lambda+1)^{\frac{x_i(t_1+\cdots+t_m)}{\lambda}}}{(\lambda+1)^{\frac{t_1+\cdots+t_m}{\lambda}}-1}
\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{t_j(\lambda+1)^{\frac{t_j(x_j-x_i+\1_{j&gt;i})}{\lambda}}}{(\lambda+1)^{\frac{t_j}\lambda}-1}\nonumber\\
\end{align}
\begin{align}
&amp;=[t_1^{s_1}\cdots t_m^{s_m}]\sum_{i=1}^{m}t_i{\sum_{k_1,\ldots,k_m\geq0}^{\infty}}\frac{\fB_{k_1+\cdots+k_m,\lambda}(x_i)}{k_1!k_2!\cdots k_m!}t_1^{k_1}\cdots t_m^{k_m}
\prod_{\substack{1\leq j\leq m\\j\neq i}}\sum_{s_j=0}^{\infty}\frac{\fB_{s_j,\lambda}(x_j-x_i+\1_{j&gt;i})}{s_j!}t_j^{s_j}\nonumber\\
&amp;=[t_1^{s_1}\cdots t_m^{s_m}]\sum_{i=1}^{m}t_i\sum_{k_1,k_2,\ldots,k_m\geq0}^{\infty}\frac{\fB_{k_1+\cdots+k_m,\lambda}(x_i)}{k_1!k_2!\cdots k_m!}t_1^{k_1}t_2^{k_2}\cdots t_m^{k_m}\cdot\nonumber\\
&amp;\cdot\prod_{\substack{1\leq j\leq m\\j\neq i}}\sum_{s_j-k_j=0}^{\infty}\frac{\fB_{s_j-k_j,\lambda}(x_j-x_i+\1_{j&gt;i})}{(s_j-k_j)!}t_j^{s_j-k_j}\nonumber\\
&amp;=\sum_{i=1}^{m}\sum_{k_1,\ldots,k_{i-1},k_{i+1},\ldots,k_m \geq 0}\frac{\fB_{k_1+\cdots+k_{i-1}+s_i-1+k_{i+1}+\cdots+k_m,\lambda}(x_i)}{k_1!\cdots k_{i-1}!(s_i-1)!k_{i+1}!\cdots k_m!}\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{\fB_{s_j-k_j,\lambda}(x_j-x_i+\1_{j&gt;i})}{(s_j-k_j)!}.
\end{align}
Combining \eqref{eq:1}, \eqref{eq:2}, and \eqref{eq:3}, we obtain that
\begin{align*}\label{eq:4}
&amp;\sum_{i=1}^{m}s_i\fB_{s_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\fB_{s_j,\lambda}(x_j)=s_1!s_2!\cdots s_m!\sum_{i=1}^{m}\frac{\fB_{i-1,\lambda}(x_i)}{(s_i-1)!}\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{\fB_{s_j,\lambda}(x_j)}{s_j!}
\nonumber\\
&amp;=s_1!\cdots s_m!\sum_{i=1}^{m}\sum_{k_1,\ldots,k_{i-1},k_{i+1},\ldots,k_m \geq 0}\frac{\fB_{k_1+\cdots+k_{i-1}+s_i-1+k_{i+1}+\cdots+k_m}(x_i,\lambda_1\lambda_2\cdots\lambda_m)}{k_1!k_2!\cdots k_{i-1}!(s_i-1)!k_{i+1}!\cdots k_m!}\cdot\nonumber\\
&amp;\cdot\prod_{\substack{1\leq j\leq m\\j\neq i}}\frac{\fB_{s_j-k_j}(x_j-x_i+\textbf{1}_{j&gt;i},\lambda_j)}{(s_j-k_j)!}\nonumber\\
&amp;=\sum_{i=1}^{m}s_i\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =s_1+\cdots+s_m}}\fB_{k_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{s_j}{k_j}\fB_{k_j,\lambda}(x_j-x_i+\textbf{1}_{j&gt;i}).
\end{align*}
Thus we get the desired result.
\end{proof}
Now we are ready to complete the proof of Theorem \ref{main}.
\begin{proof}[Proof of Theorem \ref{main}]
Clearly
\begin{align*}
&amp;(n-r_1-\cdots-r_m)\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n}}\prod_{j=1}^{m}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j)\nonumber\\
&amp;=\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n}}\sum_{i=1}^{m}(k_i-r_i)\binom{r_i}{k_i}\fB_{k_i,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j)\nonumber\\
\end{align*}
\begin{align*}
&amp;=-\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n}}\sum_{i=1}^{m}(k_i+1)\binom{r_i}{k_i+1}\fB_{k_i,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j)\nonumber\\
&amp;=-\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\sum_{i=1}^{m}k_i\binom{r_i}{k_i}\fB_{k_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j).\nonumber
\end{align*}
By Lemma 2.1, we have
\begin{align*}
&amp;\sum_{\substack{k_1,k_2,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\sum_{i=1}^{m}k_i\binom{r_i}{k_i}\fB_{k_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\ j\neq i}}\binom{r_j}{k_j}\fB_{k_j,\lambda}(x_j)\\
&amp;=\sum_{\substack{k_1,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\prod_{j=1}^m\binom{r_j}{k_j}\cdot\bigg(\sum_{i=1}^{m}k_i\fB_{k_i-1,\lambda}(x_i)\prod_{\substack{1\leq j\leq m\\ j\neq i}}\fB_{k_j,\lambda}(x_j)\bigg)\nonumber\\
&amp;=\sum_{\substack{k_1,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\prod_{j=1}^m\binom{r_j}{k_j}\cdot\bigg(\sum_{i=1}^{m}k_i\sum_{\substack{l_1,\ldots,l_m \geq 0\\l_1+\cdots+l_m =n}}\fB_{l_i,\lambda}(x_{i})\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{k_j}{l_j}\fB_{l_j,\lambda}(x_j-x_i+\1_{j&gt;i})\bigg)\nonumber\\
&amp;=\sum_{\substack{k_1,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\sum_{i=1}^{m}r_i\binom{r_i-1}{k_i-1}\sum_{\substack{l_1,\ldots,l_m \geq 0\\l_1+\cdots+l_m =n}}\fB_{l_i,\lambda}(x_{i})\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{l_j}\binom{r_j-l_j}{k_j-l_j}\fB_{l_j,\lambda}(x_j-x_i+\1_{j&gt;i})\nonumber\\
&amp;=\sum_{i=1}^{m}r_i\sum_{\substack{l_1,\ldots,l_m \geq 0\\l_1+\cdots+l_m =n}}\fB_{l_i,\lambda}(x_{i})\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j}{l_j}\fB_{l_j,\lambda}(x_j-x_i+\textbf{1}_{j&gt;i})\cdot\nonumber\\
&amp;\cdot\bigg(\sum_{\substack{k_1,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\binom{r_i-1}{k_i-1}\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j-l_j}{k_j-l_j}\bigg).\nonumber
\end{align*}
According to the Chu-Vandermonde identity,
\begin{align*}
&amp;\sum_{\substack{k_1,\ldots,k_m \geq 0\\k_1+\cdots+k_m =n+1}}\binom{r_i-1}{k_i-1}\prod_{\substack{1\leq j\leq m\\j\neq i}}\binom{r_j-l_j}{k_j-l_j}\\
&amp;=\binom{r_1+\cdots+r_m-1-l_1-\cdots-l_{i-1}-l_{i+1}-\cdots-l_m}{k_1+\cdots+k_m-1-l_1-\cdots-l_{i-1}-l_{i+1}-\cdots-l_m}\\
&amp;=\binom{r_1+\cdots+r_m-1-n+l_i}{l_i}=(-1)^{l_i}\binom{n-r_1-\cdots-r_m}{l_i}.
\end{align*}
Finally, since
$$
\frac{t(1+\lambda)^{\frac{(1-x)t}\lambda}}{(1+\lambda)^{\frac t\lambda}-1}=
\frac{(-t)(1+\lambda)^{\frac{x(-t)}\lambda}}{(1+\lambda)^{\frac {-t}\lambda}-1},
$$
we have
\begin{equation}\label{B1x}
\fB_{l,\lambda}(1-x)=(-1)^{l}\fB_{l,\lambda}(x)
\end{equation}
for each $l\geq 0$.
All are done.
\end{proof}
\begin{proof}[Proof of Corollary \ref{Corm2}]
Using \eqref{B1x} and applying to Theorem \ref{main} with $m=2$, $r_1=s$, $r_2=t$, $x_1=1-y$ and $x_2=x$, we have
\begin{align*}
&amp;r\cdot\fD_{s,t,\lambda}(x,y)=r\sum_{k=0}^n\binom{s}{k}\binom{t}{n-k}\fB_{k,\lambda}(1-y)\fB_{n-k,\lambda}(x)\nonumber\\
&amp;=-s\sum_{k=0}^n\binom{r}{k}\binom{t}{n-k}\fB_{k,\lambda}(y)\fB_{n-k,\lambda}(y+x)\\
&amp;-t\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}\fB_{k,\lambda}(1-x)\fB_{n-k,\lambda}(1-x-y)\\
&amp;=-s\cdot\fD_{t,r,\lambda}(y,1-x-y)-t\cdot\fD_{r,s,\lambda}(1-x-y,x).
\end{align*}
Thus, Eq.~\eqref{sunpanD} is concluded.

However, the proof of \eqref{mikiD} is a little more complicated.
Note that   $\binom{-1}{k}=(-1)^k$ and
$$
\frac{d}{ds}\binom{s}{k}\bigg|_{s=0}=\frac{(-1)^k}{k}.
$$
Taking the derivative with respect to $s$ on both sides of \eqref{sunpanD} and letting $s=0$ and $t=-1$, we obtain that
\begin{align}\label{sunpanDs0t1}
0&amp;=-(n+1)\sum_{k=1}^n(-1)^{n-k}\frac{\fB_{k,\lambda}(y)}{k}\cdot\fB_{n-k,\lambda}(x)
-(-1)^n\fB_{n,\lambda}(x)\fB_{0,\lambda}(y)\notag\\
&amp;+\sum_{k=0}^n\binom{n+1}{n-k}\fB_{n-k,\lambda}(y)\fB_{k,\lambda}(z)+(-1)^{n}\sum_{k=0}^{n-1}\binom{n+1}{k}\frac{\fB_{n-k,\lambda}(z)}{n-k}\cdot\fB_{k,\lambda}(x)\notag\\
&amp;+(-1)^n\bigg(1+\frac{n+1}{n}+\cdots+\frac{n+1}{2}\bigg)\fB_{0,\lambda}(z)\fB_{n,\lambda}(x).
\end{align}
Note that $\fB_{n,\lambda}(1)=(-1)^n\fB_{n,\lambda}$ by \eqref{B1x}.
Substituting $x=1$ and $y=z=0$ in \eqref{sunpanDs0t1}, we get
\begin{align}\label{sunpanD1}
0&amp;=-(n+1)\sum_{k=1}^n\frac{\fB_{k,\lambda}}{k}\cdot\fB_{n-k,\lambda}
-\fB_{n,\lambda}\fB_{0,\lambda}\notag\\
&amp;+\sum_{k=0}^n\binom{n+1}{n-k}\fB_{n-k,\lambda}\fB_{k,\lambda}+\sum_{k=0}^{n-1}(-1)^{n-k}\binom{n+1}{k}\frac{\fB_{n-k,\lambda}}{n-k}\cdot\fB_{k,\lambda}\notag\\
&amp;+\bigg(1+\frac{n+1}{n}+\cdots+\frac{n+1}{2}\bigg)\fB_{0,\lambda}(z)\fB_{n,\lambda}.
\end{align}
Since
$$t=\frac{t(1+\lambda)^{\frac{t}\lambda}-t}{(1+\lambda)^{\frac t\lambda}-1}=\sum_{k=0}^\infty\frac{\fB_{k,\lambda}(1)-\fB_{k,\lambda}}{k!}t^k,$$
we have
$\fB_{k,\lambda}(1)=\fB_{k,\lambda}$ for each $k\geq 2$. Thus by \eqref{B1x}, $\fB_{k,\lambda}=0$ if $k\geq3$ is odd, i.e.,
$(-1)^k\fB_{k,\lambda}=0$ for each $k\geq2$.
It follows from \eqref{sunpanD1} that
\begin{align}
&amp;(n+1)\sum_{k=1}^{n-1}\frac{\fB_{k,\lambda}}{k}\cdot\fB_{n-k,\lambda}+n(n+1)\fB_{1,\lambda}\fB_{n-1,\lambda}\nonumber\\
&amp;=(n+1)\sum_{k=1}^{n-1}\bigg{(}\binom{n}{k+1}+\frac{1}{n+1}\binom{n+1}{k}\bigg{)}\frac{\fB_{n-k,\lambda}}{n-k}\cdot\fB_{k,\lambda}+(n+1)H_n\fB_{n,\lambda}\fB_{0,\lambda}\nonumber\\
&amp;=(n+1)\sum_{k=1}^{n-1}\binom{n}{k}\frac{\fB_{k,\lambda}}{k}\cdot\fB_{n-k,\lambda}+(n+1)H_n\fB_{n,\lambda}\fB_{0,\lambda}.
\end{align}
Clearly, $$\fB_{0,\lambda}=\lim\limits_{t \to 0}\frac{t}{(1+\lambda)^{\frac t\lambda}-1}=\frac{\lambda}{\ln(1+\lambda)},\qquad\fB_{1,\lambda}=\lim\limits_{t \to 0}\frac{d}{dt}\bigg{(}\frac{t}{(1+\lambda)^{\frac t\lambda}-1}\bigg{)}=-\frac{1}{2}.$$
Thus \eqref{mikiD} is concluded.\\

Finally, substituting $r=n+2$, $s=t=-1$, $x=1$ and $y=z=0$  in \eqref{sunpanD}, we can easily get \eqref{matiyasevichD}.
\end{proof}

\section{Acknowledgment}

The authors thank the anonymous referee for very helpful suggestions.
The authors are supported by the National Natural Science Foundation of China 
(Grant No.~12071208).

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 05A10, 11B65.

\noindent \emph{Keywords: } modified degenerate Bernoulli polynomial,
symmetric identity, Miki's identity, Matiyasevich's identity.

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\noindent (Concerned with sequences
\seqnum{A027641} and \seqnum{A027642}.)

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\vspace*{+.1in}
\noindent
Received August 27 2023;
revised version received April 11 2024; April 24 2024.
Published in {\it Journal of Integer Sequences}, May 12 2024.

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