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\begin{center} \vskip 1cm{\LARGE\bf 
Probabilistic Proof for the Generalization \\
\vskip .1in of Some Well-Known Binomial Identities }
\vskip 1cm 
\large 
Lucio Barabesi\\
Department of Economics and Statistics\\
University of Siena\\
Piazza San Francesco 7 \\
53100 Siena\\
Italy\\ 
\href{mailto:barabesi@unisi.it}{\tt barabesi@unisi.it}\\
\end{center}

\vskip .2 in

\begin{abstract} We provide probabilistic proofs for the
generalized version of some celebrated identities involving
binomial coefficients. The results are based on the
properties of the Beta distribution.\end{abstract}

\section{Introduction} \label{sec1} The probabilistic proof of
classic mathematical formulas has received considerable
attention for decades. For example, Chin \cite{Chin} proved the Wallis formula and Holst \cite{Holst} proved some Euler identities related to the zeta function by this approach. Spivey \cite{Spivey} devoted an entire chapter of his
recent monograph to the probabilistic proof of binomial
identities. In this setting, the present note focuses on generalizing some well-known binomial identities such as
\begin{equation} \label{eq1}
\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k}=4^n ,
\end{equation} 
or its alternating version
\begin{equation}\label{eq2}
\sum_{k=0}^n(-1)^k\binom{2k}{k}\binom{2n-2k}{n-k}= 
\begin{cases}
2^n\binom{n}{\frac{n}{2}},&amp;\text{if $n$ is even;}\\
0,&amp;\text{if $n$ is odd,}
\end{cases}
\end{equation}
where $n\in\mathbb{N}_0$. In a probabilistic approach, Chang and Xu \cite{Chang} and Pathak \cite{Pathak} proved identities \eqref{eq1} and \eqref{eq2}, respectively, on the basis of some properties of the Normal distribution. Vellaisamy and Zeleke \cite{Vellaisamy} also provided proofs of \eqref{eq1} and \eqref{eq2} using identities involving the Gamma and Beta functions. In the same vein, we aim to introduce some general binomial identities by means of certain features of the Beta distribution.

\section{Main results} \label{sec2} Let $Z$ be a random variable (r.v.)\ distributed according to the Beta law. The corresponding probability density function is
\begin{equation*}
f_Z(z)=\frac{1}{B(a,b)}\,z^{a-1}(1-z)^{b-1}\boldsymbol{1}_{[0,1]}(z) ,
\end{equation*}
where $a,b&gt;0$ are the shape parameters, while $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ is the usual Euler Beta function. In the following, the Beta law
with parameters $a$ and $b$ is denoted by $\mathcal{B}(a,b)$. In addition, if $r,s&gt;0$, then 
\begin{equation}\label{eq3}
\mathbf{E}[Z^r(1-Z)^s]=\frac{B(r+a,s+b)}{B(a,b)}.
\end{equation}
For further details on the Beta distribution see, e.g., Johnson et al.\ \cite{Johnson}.

Let us consider the random vector $(X_1,X_2)$. For $n\in\mathbb{N}_0$, a unifying formula for obtaining many binomial identities is
\begin{equation}\label{eq4}
\sum_{k=0}^n\binom{n}{k}\mathbf{E}[X_1^kX_2^{n-k}]=
\mathbf{E}[(X_1+X_2)^n],
\end{equation}
which can be easily assessed on the basis of the Binomial theorem. By a suitable choice of $(X_1,X_2)$, remarkable
formulas may generally be obtained from \eqref{eq4} (see, e.g.,
Vignat and Moll \cite{Vignat}). The results described here provide relevant applications
of \eqref{eq4}, concerning r.v.'s with the Beta distribution. The first result presents
a further (and simpler) probabilistic proof of an identity proposed
by Vellaisamy and Zeleke \cite{Vellaisamy}.
\begin{proposition}\label{p1}If $a,b&gt;0$, we have
\begin{equation}\label{eq5}
\sum_{k=0}^n\binom{n}{k}B(k+a,n-k+b)=B(a,b).
\end{equation}
\end{proposition}
\begin{proof}Let us assume that $X_1=Z$ and $X_2=1-Z$ in expression \eqref{eq4}, where the r.v.\ $Z$ is distributed according to the Beta law $\mathcal{B}(a,b)$. By means of
\eqref{eq3}, we have
\begin{equation*}
\mathbf{E}[X_1^kX_2^{n-k}]=\mathbf{E}[Z^k(1-Z)^{n-k}]=\frac{B(k+a,
n-k+b)}{B(a,b)}.
\end{equation*}
Since the r.v.\ $(X_1+X_2)$ is degenerate in such a way that $P(X_1+X_2=1)=1$, we also have $\mathbf{E}[(X_1+X_2)^n]=1$. Thus, identity \eqref{eq5} follows from \eqref{eq4} on substitution.
\end{proof}

The application of the Legendre duplication
formula for the Gamma function yields 
\begin{equation}\label{eq6}
B\left(k+\frac{1}{2},l+\frac{1}{2}\right)=\frac{\Gamma(k+\frac{1}{2})
\Gamma(l+\frac{1}{2})}{\Gamma(k+l+1)}=\frac{\pi\binom{2k}{k}
\binom{2l}{l}}{4^{k+l}\binom{k+l}{k}},
\end{equation} 
for $k,l\in\mathbb{N}_0$. Hence, Proposition \ref{p1} provides the
following corollary, which introduces a generalization of identity
\eqref{eq1}. 
\begin{corollary}\label{c2}If $r,s\in\mathbb{N}_0$, we have
\begin{equation}\label{eq7}
\sum_{k=0}^n\binom{n}{k}\frac{\binom{2k+2r}{k+r}\binom{2n-2k+2s}{n-k+s}}{
\binom{n+r+s}{k+r}}=4^n\,\frac{\binom{2r}{r}\binom{2s}{s}}{\binom{r+s}{r}}. 
\end{equation} 
\end{corollary} 
\begin{proof}By assuming
$a=r+\frac{1}{2}$ and $b=s+\frac{1}{2}$ in expression
\eqref{eq5}, we have identity \eqref{eq7} by means of \eqref{eq6}. 
\end{proof} 

\begin{remark}Identity \eqref{eq1} is obtained from \eqref{eq7} for $r=s=0$. In addition, for $r=n$ and $s=0$, we also obtain the interesting identity
\begin{equation*}
\sum_{k=0}^n\binom{n+k}{n}\binom{2n+2k}{n+k}\binom{2n-2k}{n-k}=4^n
\binom{2n}{n}^2.
\end{equation*} Furthermore, after some algebra, identity \eqref{eq7} may be written as
\begin{equation*}
\sum_{k=0}^n\binom{k+r}{r}\binom{2k+2r}{k+r}\binom{n-k+s}{s}
\binom{2n-2k+2s}{n-k+s}=4^n\,\binom{2r}{r}\binom{2s}{s}\binom{n+r+s}{n}.
\end{equation*} 
\end{remark}

\vskip .2in
The following proposition provides the alternating version of identity \eqref{eq5} for $a=b$. Vellaisamy and Zeleke \cite{Vellaisamy} proposed a different probabilistic proof of this proposition based on the properties of the Gamma distribution, though if the present proof is simpler.
\begin{proposition}\label{p4}If $a&gt;0$, we have
\begin{equation}\label{eq8}
\sum_{k=0}^n(-1)^k\binom{n}{k}B(k+a,n-k+a)=2^{1-2a}B\left(\frac{n+1}{2},a
\right),
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd. 
\end{proposition}
\begin{proof}Let us assume that $X_1=-Z$ and
$X_2=1-Z$ in expression \eqref{eq4}, where the r.v.\ $Z$ is distributed according to the Beta law
$\mathcal{B}(a,a)$. Using expression $\eqref{eq3}$, we have
\begin{equation*}
\mathbf{E}[X_1^kX_2^{n-k}]=(-1)^k\mathbf{E}[Z^k(1-Z)^{n-k}]=(-1)^k\,
\frac{B(k+a,n-k+a)}{B(a,a)}.
\end{equation*}
In addition, the r.v.\ $Y=(1-2Z)^2$ is distributed according to the Beta
law $\mathcal{B}(\frac{1}{2},a)$. For $n$ even we have
\begin{equation*}
\mathbf{E}[(X_1+X_2)^n]=\mathbf{E}[(1-2Z)^n]=\mathbf{E}[Y^{
\frac{n}{2}}]=\frac{B(\frac{n+1}{2},a)}{B(\frac{1}{2},a)}=\frac{B(
\frac{n+1}{2},a)}{2^{2a-1}B(a,a)},
\end{equation*}
where the last equality follows from the Legendre duplication formula for the Gamma function. Finally, since the distribution of $Z$ is symmetric with respect
to $\mathbf{E}[Z]=\frac{1}{2}$, we have
$\mathbf{E}[(1-2Z)^n]=0$ for $n$ odd. Thus, \eqref{eq8} follows
from \eqref{eq4}.
\end{proof}
Proposition \ref{p4} gives rise to the following corollary, where a
generalization of identity \eqref{eq2} is obtained.
\begin{corollary}\label{c5}If $r\in\mathbb{N}_0$, we have
\begin{equation}\label{eq9}
\sum_{k=0}^n(-1)^k\,\binom{n}{k}\frac{\binom{2k+2r}{k+r}
\binom{2n-2k+2r}{n-k+r}}{\binom{n+2r}{k+r}}=2^n\,\frac{\binom{n}{\frac{n}{2}}
\binom{2r}{r}}{\binom{\frac{n}{2}+r}{r}},
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd.
\end{corollary}
\begin{proof}Assuming $a=r+\frac{1}{2}$ in expression \eqref{eq8} and
using \eqref{eq6}, we obtain \eqref{eq9}.
\end{proof}

\begin{remark}Identity \eqref{eq2} is promptly obtained from \eqref{eq9} for
$r=0$. After some algebra, for $n$ even identity
\eqref{eq9} may also be written as follows
\begin{equation*}
\begin{aligned}[t]
\sum_{k=0}^n(-1)^k\,&amp;\binom{k+r}{r}\binom{2k+2r}{k+r}\binom{n-k+r}{r}
\binom{2n-2k+2r}{n-k+r}\\
&amp;=2^n\,\binom{2r}{r}\binom{n+2r}{\frac{n}{2}+2r}\binom{\frac{n}{2}+2r}{\frac{n}{2}+r}. 
\end{aligned} 
\end{equation*}
\end{remark}

\vskip .2in
We recall that the bivariate Selberg integral is given by
\begin{equation*} 
\begin{aligned}[t]
S_2(\alpha,\beta,\gamma)&amp;=\int_0^1\int_0^1t_1^{\alpha-1}(1-t_1)^{
\beta-1}t_2^{\alpha-1}(1-t_2)^{\beta-1}|t_1-t_2|^{2\gamma}\,dt_1dt_2\\
&amp;=\frac{\Gamma\left(\alpha\right)\Gamma(\beta)\Gamma(\alpha+\gamma)\Gamma(
\beta+\gamma)\Gamma(1+2\gamma)}{\Gamma(\alpha+\beta+\gamma)\Gamma(\alpha+
\beta+2\gamma)\Gamma(1+\gamma)},
\end{aligned} 
\end{equation*}
for $\Re(\alpha)&gt;0$, $\Re(\beta)&gt;0$ and
$\Re(\gamma)&gt;-\min(\frac{1}{2},\Re(\alpha),\Re(\beta))$ (see,
e.g., Forrester and Warnaar \cite{Forrester}). By means of the Legendre
duplication formula, we also have
\begin{equation}\label{eq10}
S_2(\alpha,\beta,\gamma)=2^{2\gamma}\,\frac{B(\alpha,\beta+\gamma)B(\alpha\,+\gamma,\beta)B(\alpha+\beta+\gamma,\gamma)}{B(\gamma,\frac{1}{2})},
\end{equation}
and
\begin{equation}\label{eq11}
S_2(\alpha,\alpha,\gamma)=2^{1-2\alpha}\,B\left(\alpha\,,\gamma+\frac{1}{2}\right)B(\alpha,\alpha+\gamma).
\end{equation}
Moreover, if $Z_1$ and $Z_2$ are independent r.v.'s distributed
according to the Beta law $\mathcal{B}(a,b)$, for $r$ even it is
apparent that
\begin{equation}\label{eq12}
\mathbf{E}[(Z_1-Z_2)^r]=\frac{S_2(a,b,\frac{r}{2})}{B(a,b)^2}.
\end{equation}

\begin{proposition}\label{p7}If $a&gt;0$, we have
\begin{equation}\label{eq13}
\sum_{k=0}^n(-1)^k\binom{n}{k}B(k+a,a)B(n-k+a,a)=2^{1-2a}B\left(\frac{n+1}{2},
a\right)B\left(\frac{n}{2}+a,a\right),
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd.
\end{proposition} 
\begin{proof}Let us assume that $X_1=-Z_1$ and
$X_2=Z_2$ in expression \eqref{eq4}, where $Z_1$ and $Z_2$ are independent r.v.'s distributed according to the Beta law $\mathcal{B}(a,a)$. From \eqref{eq3} it follows that
\begin{equation*}
\mathbf{E}[X_1^kX_2^{n-k}]=(-1)^k\mathbf{E}[Z_1^k]
\mathbf{E}[Z_2^{n-k}]=(-1)^k\,\frac{B(k+a,a)B(n-k+a,a)}{B(a,a)^2}. 
\end{equation*}
In addition, from \eqref{eq11} and \eqref{eq12}, for $n$ even we have
\begin{equation*}
\mathbf{E}[(X_1+X_2)^n]=\mathbf{E}[(Z_1-Z_2)^n]=\frac{S_2(a,a,
\frac{n}{2})}{B(a,a)^2}=2^{1-2a}\,\frac{B(\frac{n+1}{2},a)B(\frac{n}{2}+a,a)}{B(a,a)^2}.
\end{equation*}
Since $Z_1$ and $Z_2$ are independent r.v.'s
distributed according to the same law, the distribution of the r.v.\
$(Z_1-Z_2)$ is symmetric with respect to zero and we have $\mathbf{E}[(Z_1-Z_2)^n]=0$ for $n$ odd. Expression
\eqref{eq13} therefore follows from \eqref{eq4} on substitution.
\end{proof}

\begin{corollary}\label{c8}If $r\in\mathbb{N}_0$, we have
\begin{equation}\label{eq14}
\sum_{k=0}^n(-1)^k\,\binom{n}{k}\frac{\binom{2k+2r}{k+r}
\binom{2n-2k+2r}{n-k+r}}{\binom{k+2r}{r}\binom{n-k+2r}{r}}=\frac{\binom{n}{
\frac{n}{2}}\binom{n+2r}{\frac{n}{2}+r}}{\binom{\frac{n}{2}+r}{r}\binom{\frac{n}{2}+2r}{r}},
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd.
\end{corollary}
\begin{proof}If we assume $a=r+\frac{1}{2}$ in expression \eqref{eq13}, the result follows from \eqref{eq6}.
\end{proof}

\begin{remark} By assuming that $r=0$ in identity \eqref{eq14}, we have 
\begin{equation*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=
\begin{cases} 
\binom{n}{\frac{n}{2}}^2,&amp;\text{if $n$ is even;}\\
0,&amp;\text{if $n$ is odd,} 
\end{cases}
\end{equation*}
an identity considered by Petkov\v{s}ek et al. \cite[Example 3.6.2]{Petko}, and Mik\a'\i c \cite{Mikic}, among others, although its probabilistic proof does not seem available in the literature. Moreover, identity \eqref{eq14} may also be written
\begin{equation*} 
\begin{aligned}[t]
\sum_{k=0}^n(-1)^k\,&amp;\binom{k+r}{r}\binom{2k+2r}{k+r}\binom{n-k+r}{r}
\binom{2n-2k+2r}{n-k+r}\binom{n+4r}{k+2r}\\
&amp;=\binom{\frac{n}{2}+r}{r}\binom{\frac{n}{2}+2r}{r}\binom{n+2r}{
\frac{n}{2}+r}\binom{n+4r}{\frac{n}{2}+2r}.
\end{aligned}
\end{equation*}
\end{remark}

\vskip .2in
A more general identity involving Beta functions is provided in
the following proposition. A suitable choice of
parameters in this formula may in turn produce new binomial identities,
as in the cases of Corollaries \ref{c2}, \ref{c5} and \ref{c8}.
\begin{proposition}\label{p10}If $a,b&gt;0$, we have
\begin{equation} \label{eq15}
\begin{aligned}[t]
\sum_{k=0}^n(-1)^k\,&amp;\binom{n}{k}B(k+a,n-k+b)B(n-k+a,k+b)\\
&amp;=2^n\,\frac{B(a,\frac{n}{2}+b)B(\frac{n}{2}+a,b)B(\frac{n}{2}+a+b,
\frac{n}{2})}{B(\frac{n}{2},\frac{1}{2})}, 
\end{aligned}
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd.
\end{proposition}
\begin{proof}Let us assume that $X_1=-Z_1(1-Z_2)$
and $X_2=(1-Z_1)Z_2$ in expression \eqref{eq4}, where $Z_1$ and $Z_2$ are independent
r.v.'s distributed according to the Beta law $\mathcal{B}(a,b)$. Using \eqref{eq3} it follows that
\begin{equation*} 
\begin{aligned}[t]
\mathbf{E}[X_1^kX_2^{n-k}]&amp;=(-1)^k\mathbf{E}[Z_1^k(1-Z_1)^{n-k}] \,
\mathbf{E}[Z_2^{n-k}(1-Z_2)^k]\\
&amp;=(-1)^k\,\frac{B(k+a,n-k+b)B(n-k+a,k+b)}{B(a,b)^2}.
\end{aligned} 
\end{equation*}
Using \eqref{eq10} and \eqref{eq12}, for $n$ even we have
\begin{equation*} 
\begin{aligned}[t]
\mathbf{E}[(X_1+X_2)^n]&amp;=\mathbf{E}[(Z_1-Z_2)^n]=\frac{S_2(a,b,
\frac{n}{2})}{B(a,b)^2}\\
&amp;=2^n\,\frac{B(a,\frac{n}{2}+b)B(\frac{n}{2}+a,b)B(\frac{n}{2}+a+b,
\frac{n}{2})}{B(a,b)^2B(\frac{n}{2},\frac{1}{2})}.
\end{aligned} 
\end{equation*}
In turn, the distribution of the r.v.\ $(Z_1-Z_2)$ is symmetric
with respect to zero and we have
$\mathbf{E}[(Z_1-Z_2)^n]=0$ for $n$ odd. Expression
\eqref{eq15} therefore follows from \eqref{eq4} on substitution.
\end{proof}

\begin{remark}Assuming $a=b$ in identity \eqref{eq15}, since we have 
\begin{equation*}
2^n\,\frac{B(a,\frac{n}{2}+a)^2B(\frac{n}{2}+2a,\frac{n}{2})}{B(\frac{n}{2},
\frac{1}{2})}=2^{1-2a}B\left(\frac{n+1}{2},a\right)B\left(\frac{n}{2}+a,a
\right),
\end{equation*}
on the basis of Propositions \ref{p7} and \ref{p10}, it also follows that 
\begin{equation*}
\sum_{k=0}^n(-1)^k\binom{n}{k}B(k+a,n-k+a)^2=\sum_{k=0}^n(-1)^k
\binom{n}{k}B(k+a,a)B(n-k+a,a).
\end{equation*}
\end{remark}

\begin{corollary}\label{c12}If $r,s\in\mathbb{N}_0$, we have
\begin{equation}\label{eq16}
\sum_{k=0}^n(-1)^k\,\binom{n}{k}\frac{\binom{2k+2r}{k+r}\binom{2k+2s}{k+s}
\binom{2n-2k+2r}{n-k+r}\binom{2n-2k+2s}{n-k+s}}{\binom{n+r+s}{k+r}
\binom{n+r+s}{k+s}}
=4^n\,\frac{\binom{n}{\frac{n}{2}}\binom{2r}{r}\binom{2s}{s}\binom{n+2r}{
\frac{n}{2}+r}\binom{n+2s}{\frac{n}{2}+s}}{\binom{\frac{n}{2}+r+s}{r}\binom{
\frac{n}{2}+r+s}{s}\binom{n+r+s}{\frac{n}{2}}},
\end{equation}
for $n$ even, while the sum vanishes for $n$ odd.
\end{corollary}
\begin{proof}Assuming $a=r+\frac{1}{2}$ and $b=s+\frac{1}{2}$
in Proposition \ref{p10} and applying \eqref{eq6}, the result follows.
\end{proof}

\begin{remark}For $r=s=0$, expression \eqref{eq16} provides the following
nice binomial identity 
\begin{equation*}
\sum_{k=0}^n(-1)^k\,\frac{\binom{2k}{k}^2\binom{2n-2k}{n-k}^2}{\binom{n}{k}}=
\begin{cases}
4^n\binom{n}{\frac{n}{2}}^2,&amp;\text{if $n$ is even;}\\ 
0,&amp;\text{if $n$ is odd.} 
\end{cases}
\end{equation*}
\end{remark}

\begin{remark}We emphasize that the results given by Adell and
Lekuona \cite{Adell} and Chang and Xu \cite{Chang} can be 
obtained in an alternative way, using the properties of the Beta distribution and
the Dirichlet distribution (the well-known multivariate
generalization of the Beta distribution), in a manner similar to that followed in the present note.
\end{remark}

\section{\bf Acknowledgments} The author thanks the editor-in-chief and the referee for several helpful comments.

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\bibitem{Adell} J. A. Adell and A. Lekuona, Binomial identities and moments of random variables, {\it Amer. Math. Monthly} {\bf 125} (2018), 365--369. 

\bibitem{Chang} G. Chang and C. Xu, Generalization and probabilistic proof of a combinatorial identity, {\it Amer. Math. Monthly} {\bf 118} (2011), 175--177.

\bibitem{Chin} W. Chin, A probabilistic proof of a Wallis-type formula for the Gamma function, {\it Amer. Math. Monthly} {\bf 127} (2020), 75--79.

\bibitem{Forrester} P. J. Forrester and S. O. Warnaar, The importance of the Selberg integral, {\it  Bull. Amer. Math. Soc.} {\bf 45} (2008), 489--534.

\bibitem{Holst} L. Holst, Probabilistic proofs of Euler identities, {\it  J. Appl. Probab.} {\bf 45} (2008), 489--534.

\bibitem{Johnson} N. L. Johnson, S. Kotz, and N. Balakrishnan, \textit{Continuous Univariate Distributions}, Vol. 2, 2nd ed., Wiley, 1995.

\bibitem{Mikic} J. Mik\a'\i c, Two new identities involving the Catalan numbers and sign-reversing involutions, {\it J. Integer Sequences} {\bf 22} (2019),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL22/Mikic/mikic43.html}{Article
19.7.7}.

\bibitem{Pathak} A. K. Pathak, A simple probabilistic proof for the alternating convolution of the central binomial coefficients, {\it Amer. Statist.} {\bf 72} (2018), 287--288.

\bibitem{Petko} M. Petkov\v{s}ek, H. S. Wilf, and D. Zeilberger, \textit{A = B}, A. K. Peters/CRC Press, 1996.

\bibitem{Spivey} M. Z. Spivey, \textit{The Art of Proving Binomial Identities}, CRC Press, 2019.

\bibitem{Vignat} C. Vignat and V. H. Moll, A probabilistic approach to some binomial identities, {\it Elem. Math.} {\bf 70} (2015), 55--66.

\bibitem{Vellaisamy} P. Vellaisamy and A. Zeleke, Probabilistic proofs
of some Beta-function identities, {\it J. Integer Sequences} {\bf 22}
(2019), \href{https://cs.uwaterloo.ca/journals/JIS/VOL22/Vellaisamy/vellai5.html}{Article 19.6.6}.

\end{thebibliography}

\bigskip 
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\bigskip 
\noindent {\it 2020 Mathematics Subject Classification:} Primary 05A19; Secondary 60C05.

\noindent \emph{Keywords:} binomial identity, beta distribution, Selberg integral, probabilistic proof.

\bigskip
\hrule
\bigskip
	
\noindent 
(Concerned with sequences \seqnum{A098430} and \seqnum{A002894}.)
	
\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  October 13 2022;
revised versions received  October 14 2022; December 15 2022; January 13 2023.
Published in {\it Journal of Integer Sequences}, January 17 2023.

\bigskip
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\noindent
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