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\begin{center}
\vskip 1cm{\LARGE\bf Simplicial $d$-Polytopic Numbers \\
\vskip.1in
Defined on Lucas Sequences}
\vskip 1cm
\large
Ronald Orozco L\'opez\\
Departamento de Matem\'aticas \\
Universidad de los Andes\\
Bogot\'a, 111711\\
Colombia \\
\href{mailto:rj.orozco@uniandes.edu.co}{\tt rj.orozco@uniandes.edu.co} \\
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\begin{abstract}
This paper introduces the simplicial $d$-polytopic numbers defined on Lucas sequences. We establish basic identities and find $q$-identities. Furthermore, we find generating functions for the simplicial $d$-Lucas-polytopic numbers and for the squares of the Lucas-triangular numbers. Finally, we compute sums of reciprocals of Lucas sequences and Lucas-triangular numbers. We introduce an analogue of the zeta function defined on Lucas sequences. 
\end{abstract}


\section{Introduction}
There is growing research on analogue sequences of numbers defined in terms of Lucas sequences, including Catalan numbers \cite{bennet,ekhad}, Bernoulli and Euler polynomials \cite{kus}, Eulerian numbers \cite{ruiz}, among others. In this paper, we define simplicial $d$-polytopic numbers on Lucas sequences. The advantage of doing so is that we obtain, for free, analogs of Fibonacci, Pell, Jacobsthal, and Mersenne sequences, among others.

The simplicial polytopic numbers \cite{deza} are a family of sequences of figurate numbers corresponding to the $d$-dimensional simplex for each dimension $d$, where $d$ is a non-negative integer. For $d$ ranging from $1$ to $5$, we have the following simplicial polytopic numbers, respectively: non-negative numbers $\N$, triangular numbers $T_n$, tetrahedral numbers $\mathrm{Te}_n$, pentachoron numbers $P_n$ and hexateron numbers $H_n$. A list of the above sets of numbers is as follows:
\begin{align*}
    \N&=(0,1,2,3,4,5,6,7,8,9,\ldots),\\
    {T}&=(0,1,3,6,10,15,21,28,36,45,55,66,\ldots),\\
    \mathrm{Te}&=(0,1,4,10,20,35,56,84,120,165,\ldots),\\
    {P}&=(0,1,5,15,35,70,126,210,330,495,715,\ldots),\\
    {H}&=(0,1,6,21,56,126,252,462,792,1287,\ldots).
\end{align*}
The $n^{th}$ simplicial $d$-polytopic numbers $P_{n}^{d}$ are given by the formula
\begin{equation*}
    P_{n}^{d}=\binom{n+d-1}{d}=\frac{n^{(d)}}{d!},
\end{equation*}
where $x^{(d)}=x(x+1)(x+2)\cdots(x+d-1)$ is the rising factorial.
The generating function of the simplicial $d$-polytopic numbers is
\begin{equation*}
    \sum_{n=1}^{\infty}\binom{n+d-1}{d}x^{n}=\frac{x}{(1-x)^{d+1}}.
\end{equation*}
In this paper, the $n$-th simplicial $d$-Lucas-polytopic number is defined by
    \begin{equation*}
        \fibonomial{n+d-1}{d}_{s,t}=\frac{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}\cdots\brk[c]{n+d-1}_{s,t}}{\brk[c]{d}_{s,t}!},
    \end{equation*}
where $\{n\}_{s,t}$ is the Lucas analogue of the positive integer $n$. We determine basic identities for simplicial Lucas polytopic numbers, especially for Lucas-triangular and Lucas-tetrahedral numbers. These sequences are part of the 
On-Line Encyclopedia of Integer Sequences \cite{sloane}. Some known $q$-identities are found, \cite{sch,war}. We establish generating functions for the simplicial $d$-Lucas-polytopic numbers $\fibonomial{n+d-1}{d}_{s,t}$ and for the sequence $\fibonomial{n+1}{2}_{s,t}^2$. Finally, we introduce the Lucas-zeta function $\zeta_{s,t}(z)$ and find some values for $\zeta_{s,t}(1)$. In addition, we calculate reciprocal sums of Lucas sequences and Lucas-triangular numbers.


\section{Preliminaries}

The Lucas sequences \cite{Sa1} on the parameters $s,t$ are defined by
\begin{equation*}
    \brk[c]{n+2}_{s,t}=s\{n+1\}_{s,t}+t\{n\}_{s,t}
\end{equation*}
with initial values $\brk[c]{0}_{s,t}=0$ and $\brk[c]{1}_{s,t}=1$, where $s\neq0$ and $t\neq0$. Below are some important specializations of Lucas sequences.
\begin{enumerate}
    \item If $s=2,t=-1$, then $\brk[c]{n}_{2,-1}=n$ are the positive integers.
    \item If $s=1,t=1$, then $\brk[c]{n}_{1,1}=F_n$ are the Fibonacci numbers
    \begin{equation*}
        F_{n}=(0,1,1,2,3,5,8,13,21,34,55,\ldots).
    \end{equation*}
    \item If $s=2,t=1$, then $\brk[c]{n}_{2,1}=P_n$, where $P_n$ are the Pell numbers
    \begin{equation*}
        P_n=(0,1,2,5,12,29,70,169,408\ldots).  
    \end{equation*}
    \item If $s=1,t=2$, then $\brk[c]{n}_{1,2}=J_n$, where $J_n$ are the Jacobsthal numbers
    \begin{equation*}
        J_n=(0,1,1,3,5,11,21,43,85,171,\ldots).
    \end{equation*}
    \item If $s=3,t=-2$, then $\brk[c]{n}_{3,-2}=M_n$, where $M_n=2^n-1$ are the Mersenne numbers
    \begin{equation*}
        M_n=(0,1,3,7,15,31,63,127,255,\ldots).
    \end{equation*}
    \item If $s=p+q,t=-pq$, then $\brk[c]{n}_{p+q,-pq}=\brk[s]{n}_{p,q}$, where $\brk[s]{n}_{p,q}$ are the $(p,q)$-numbers
    \begin{equation*}
        \brk[s]{n}_{p,q}=(0,1,\brk[s]{2}_{p,q},[3]_{p,q},[4]_{p,q},[5]_{p,q},[6]_{p,q},[7]_{p,q},[8]_{p,q}\ldots).
    \end{equation*}
    If $p=1$, we obtain the $q$-numbers $\brk[s]{n}_{q}=1+q+q^2+q^3+\cdots+q^{n-1}$.
    \item If $s=2t,t=-1$, then $\brk[c]{n}_{2t,-1}=U_{n-1}(t)$, where $U_n(t)$ are the Chebyshev polynomials of the second kind, with $U_{-1}(t)=0$.
\end{enumerate}
The Lucas constant is the ratio toward which adjacent terms in a Lucas sequence tend. This is the only positive zero of $x^{2}-sx-t=0$. We let $\varphi_{s,t}$ denote this constant, where
\begin{equation*}
    \varphi_{s,t}=\frac{s+\sqrt{s^{2}+4t}}{2}
\end{equation*}
and
\begin{equation*}
    \varphi_{s,t}^{\prime}=s-\varphi_{s,t}=-\frac{t}{\varphi_{s,t}}=\frac{s-\sqrt{s^{2}+4t}}{2}
\end{equation*}
denote the reciprocal of $\varphi_{s,t}$.
Some specializations of the constants $\varphi_{s,t}$ and $\varphi_{s,t}^\prime$ are
\begin{enumerate}
    \item If $s=2$ and $t=-1$, then $\varphi_{2,-1}=1$ and $\varphi_{2,-1}^\prime=1$.
    \item If $s=1$ and $t=1$, then $\varphi_{1,1}=\varphi=\frac{1+\sqrt{5}}{2}$ and $\varphi_{1,1}^\prime=\varphi^\prime=\frac{1-\sqrt{5}}{2}$.
    \item If $s=2$ and $t=1$, then $\varphi_{2,1}=1+\sqrt{2}$ and $\varphi_{2,1}^\prime=1-\sqrt{2}$.
    \item If $s=1$ and $t=2$, then $\varphi_{1,2}=2$ and $\varphi_{1,2}^\prime=-1$.
    \item If $s=3$ and $t=-2$, then $\varphi_{3,-2}=2$ and $\varphi_{3,-2}^\prime=1$.
    \item If $s=p+q$ and $t=-pq$, then $\varphi_{p+q,-pq}=p$ and $\varphi_{p+q,-pq}^\prime=q$.
    \item If $s=2t$ and $t=-1$, then $\varphi_{2t,-1}=\frac{t+\sqrt{t^2-1}}{2}$ and $\varphi_{2t,-1}^\prime=\frac{t-\sqrt{t^2-1}}{2}$.
\end{enumerate}
The Lucasnomial coefficients are defined by
\begin{equation*}
    \fibonomial{n}{k}_{s,t}=\frac{\brk[c]{n}_{s,t}!}{\brk[c]{k}_{s,t}!\brk[c]{n-k}_{s,t}!},
\end{equation*}
where $\brk[c]{n}_{s,t}!=\brk[c]{1}_{s,t}\brk[c]{2}_{s,t}\cdots\brk[c]{n}_{s,t}$. The Lucasnomial coefficients satisfy the following Pascal recurrence relationships. For $1\leq k\leq n-1$ we have
\begin{align}
    \fibonomial{n+1}{k}_{s,t}&=\varphi_{s,t}^{k}\fibonomial{n}{k}_{s,t}+\varphi_{s,t}^{\prime(n+1-k)}\fibonomial{n}{k-1}_{s,t},\label{eqn_pascal1}\\
    &=\varphi_{s,t}^{\prime(k)}\fibonomial{n}{k}_{s,t}+\varphi_{s,t}^{n+1-k}\fibonomial{n}{k-1}_{s,t}.\label{eqn_pascal2}
\end{align}
A proof of the above identities was provided by Corcino \cite{corcino}.

Set $s,t\in\R$, with $s\neq0$ and $t\neq0$. If $s^2+4t\neq0$, we define the Lucas-derivative $D_{s,t}$ of the function $f(x)$ as
\begin{displaymath}
(D_{s,t}f)(x)=
\begin{cases}
\frac{f(\varphi_{s,t}x)-f(\varphi_{s,t}^{\prime}x)}{(\varphi_{s,t}-\varphi_{s,t}^{\prime})x},&\text{ if }x\neq0;\\
f^{\prime}(0),&\text{ if }x=0,
\end{cases}
\end{displaymath}
provided $f(x)$ is differentiable at $x=0$. If $s^2+4t=0$, with $t<0$, we define the Lucas-derivative of the function $f(x)$ as
\begin{equation*}
    (D_{\pm2i\sqrt{t},t}f)(x)=f^{\prime}(\pm i\sqrt{t}x).
\end{equation*}
The Lucas-derivative $D_{s,t}$ fulfills the following properties.
\begin{itemize}
\item Linearity:
\begin{align*}
            D_{s,t}(\alpha f+\beta g)&=\alpha D_{s,t}f+\beta D_{s,t}g.
        \end{align*}
\item 
Product rules:
        \begin{equation*}
            D_{s,t}(f(x)g(x))=f(\varphi_{s,t}x)D_{s,t}g(x)+g(\varphi_{s,t}^\prime x)D_{s,t}f(x),
        \end{equation*}
and
\begin{equation*}
             D_{s,t}(f(x)g(x))=f(\varphi_{s,t}^{\prime}x)D_{s,t}g(x)+g(\varphi_{s,t}x)D_{s,t}f(x).
         \end{equation*}
\item 
Quotient rules:
         \begin{equation*}
             D_{s,t}\biggl(\frac{f(x)}{g(x)}\biggr)=\frac{g(\varphi_{s,t}x)D_{s,t}f(x)-f(\varphi_{s,t}x)D_{s,t}g(x)}{g(\varphi_{s,t}x)g(\varphi_{s,t}^{\prime}x)},
         \end{equation*}
and         
         \begin{equation*}
             D_{s,t}\biggl(\frac{f(x)}{g(x)}\biggr)=\frac{g(\varphi_{s,t}^{\prime}x)D_{s,t}f(x)-f(\varphi_{s,t}^{\prime}x)D_{s,t}g(x)}{g(\varphi_{s,t}x)g(\varphi_{s,t}^{\prime}x)}.
         \end{equation*}    
\end{itemize}
        
Define the $n$-th Lucas-derivative of the function $f(x)$ recursively as
\begin{equation*}
    D_{s,t}^nf(x)=D_{s,t}(D_{s,t}^{n}f(x)).
\end{equation*}


\section{$d$-Lucas-polytopic numbers}

\subsection{Definition and basic properties}
\begin{definition}
The $n$-th simplicial $d$-Lucas-polytopic number is defined by
    \begin{equation*}
        \fibonomial{n+d-1}{d}_{s,t}=\frac{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}\cdots\brk[c]{n+d-1}_{s,t}}{\brk[c]{d}_{s,t}!}.
    \end{equation*}
The Lucas analogues of the triangular numbers, tetrahedral numbers, pentachoron numbers, and hexateron numbers are
\begin{align*}
    {T}_{s,t}&=\Bigg\{{T}_{n}(s,t)=\fibonomial{n+1}{2}_{s,t}:n\geq0\Bigg\},\\
    \mathrm{Te}_{s,t}&=\Bigg\{\mathrm{Te}_{n}(s,t)=\fibonomial{n+2}{3}_{s,t}:n\geq0\Bigg\},\\
    {P}_{s,t}&=\Bigg\{{P}_{n}(s,t)=\fibonomial{n+3}{4}_{s,t}:n\geq0\Bigg\},\\
    {H}_{s,t}&=\Bigg\{{H}_{n}(s,t)=\fibonomial{n+4}{5}_{s,t}:n\geq0\Bigg\}.
\end{align*}
\end{definition}
From the Pascal recurrence in Eqs.\ (\ref{eqn_pascal1}) and (\ref{eqn_pascal2})
we have
\begin{align}
    \fibonomial{n+d}{d}_{s,t}&=\varphi_{s,t}^{d}\fibonomial{n+d-1}{d}_{s,t}+\varphi_{s,t}^{\prime n}\fibonomial{n+d-1}{d-1}_{s,t},\label{eqn_gspn_recu1}\\
    &=\varphi_{s,t}^{\prime d}\fibonomial{n+d-1}{d}_{s,t}+\varphi_{s,t}^{n}\fibonomial{n-d-1}{d-1}_{s,t}\label{eqn_gspn_recu2}.
\end{align}
It is a well-known fact that the sum of the first $n$ terms of a sequence of $d$-polytopic numbers is the $n$-th term of a sequence of $(d+1)$-polytopic numbers, i.e,
\begin{equation*}
    \sum_{k=1}^{n}P_{n}^d=P_{n}^{d+1}.
\end{equation*}
We then obtain the Lucas analogue of the above formula.
\begin{theorem}\label{theo_reduc}
For all $n\geq1$, the sum of simplicial $d$-Lucas-polytopic numbers is
    \begin{align*}
        \fibonomial{n+d}{d+1}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{(d+1)(n-k)}\varphi_{s,t}^{\prime(k-1)}\fibonomial{k+d-1}{d}_{s,t},\\
        \fibonomial{n+d}{d+1}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{\prime(d+1)(n-k)}\varphi_{s,t}^{k-1}\fibonomial{k+d-1}{d}_{s,t}.
    \end{align*}
\end{theorem}
\begin{proof}
The proof is by induction on $n$. When $n=1$, then $\fibonomial{d+1}{d+1}_{s,t}=\fibonomial{d}{d}_{s,t}$. For $n=2$, it follows that
\begin{align*}
    \fibonomial{d+2}{d+1}_{s,t}&=\varphi_{s,t}^{d+1}\fibonomial{d}{d}_{s,t}+\varphi_{s,t}^\prime\fibonomial{d+1}{d}_{s,t}.
\end{align*}
Suppose the statement is true for $n$ and we prove the statement for $n+1$. We have
\begin{align*}
    \fibonomial{n+d+1}{d+1}_{s,t}&=\varphi_{s,t}^{d+1}\fibonomial{n+d}{d+1}_{s,t}+\varphi_{s,t}^{\prime n}\fibonomial{n+d}{d}_{s,t}\\
    &=\varphi_{s,t}^{d+1}\sum_{k=1}^{n}\varphi_{s,t}^{(d+1)(n-k)}\varphi_{s,t}^{\prime(k-1)}\fibonomial{k+d-1}{d}_{s,t}+\varphi_{s,t}^{\prime n}\fibonomial{n+d}{d}_{s,t}\\
    &=\sum_{k=1}^{n}\varphi_{s,t}^{(d+1)(n+1-k)}\varphi_{s,t}^{\prime(k-1)}\fibonomial{k+d-1}{d}_{s,t}+\varphi_{s,t}^{\prime n}\fibonomial{n+d}{d}_{s,t}\\
    &=\sum_{k=1}^{n+1}\varphi_{s,t}^{(d+1)(n+1-k)}\varphi_{s,t}^{\prime(k-1)}\fibonomial{k+d-1}{d}_{s,t}.
\end{align*}
The proof is reached.
\end{proof}
From Theorem \ref{theo_reduc} we obtain the following known result about $q$-binomial coefficients:
\begin{equation*}
    \qbinom{n+d}{d+1}_{q}=\sum_{k=1}^{n}q^{k-1}\qbinom{k+d-1}{d}_{q}=\sum_{k=1}^{n}q^{(d+1)(n-k)}\qbinom{k+d-1}{d}_{q}.
\end{equation*}


\section{Some specializations}

\subsection{Lucas-triangular numbers}

Some specializations of Lucas-triangular numbers are
\begin{align*}
    \fibonomial{n+1}{2}_{1,1}&=F_{n}F_{n+1}=(0,1,2,6,15,40,104,273,\ldots),\\
\end{align*}
\begin{align*}
    \fibonomial{n+1}{2}_{2,1}&=\frac{1}{2}P_{n}P_{n+1}=(0,1,5,30,174,1015,5915,\ldots),\\
    \fibonomial{n+1}{2}_{1,2}&=J_{n}J_{n+1}=(0,1,2,6,15,55,231,903,3655,\ldots),\\
    \fibonomial{n+1}{2}_{3,-2}&=\frac{1}{3}(2^n-1)(2^{n+1}-1)\\
    &=(0,1,7,35,155,651,2667,10795,43435,174251,\ldots).
\end{align*}
The $\fibonomial{n+1}{2}_{1,1}$-numbers are known as golden rectangle numbers, (\seqnum{A001654} in \cite{sloane}). The $\fibonomial{n+1}{2}_{2,1}$-numbers may be called Pell triangles, (\seqnum{A084158} in \cite{sloane}). The $\fibonomial{n+1}{2}_{1,2}$-numbers are known as Jacobsthal oblong numbers, (\seqnum{A084175} in \cite{sloane}). The $\fibonomial{n+1}{2}_{3,-2}$-numbers are the Gaussian binomial coefficients $\qbinom{n}{2}_{q}$ for $q=2$, (\seqnum{A006095} in \cite{sloane}).

From Eqs.\ (\ref{eqn_gspn_recu1}) and (\ref{eqn_gspn_recu2}), and Theorem \ref{theo_reduc}, we obtain
\begin{align*}
    \fibonomial{n+2}{2}_{s,t}&=\varphi_{s,t}^2\fibonomial{n+1}{2}+\varphi_{s,t}^{\prime n}\brk[c]{n+1}_{s,t},\\
    \fibonomial{n+2}{2}_{s,t}&=\varphi_{s,t}^{\prime2}\fibonomial{n+1}{2}_{s,t}+\varphi_{s,t}^{n}\brk[c]{n+1}_{s,t},\\
    \fibonomial{n+1}{2}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{2(n-k)}\varphi_{s,t}^{\prime(k-1)}\brk[c]{k}_{s,t},\\
    \fibonomial{n+1}{2}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{\prime2(n-k)}\varphi_{s,t}^{k-1}\brk[c]{k}_{s,t}.
\end{align*}
From the above identities, we obtain the identity of Warnaar \cite{war}
\begin{equation*}
    \qbinom{n+1}{2}_{q}=\sum_{k=1}^{n}\frac{1-q^k}{1-q}q^{2(n-k)}=\sum_{k=1}^{n}q^{k-1}\frac{1-q^k}{1-q}.
\end{equation*}

\begin{theorem}\label{eqn_tri_sq}
For all $n\in\N$,
\begin{enumerate}
    \item The Lucas analogue of the identity $T_{n}+T_{n-1}=n^2$ is
    \begin{equation*}
        \fibonomial{n+1}{2}_{s,t}=t\fibonomial{n}{2}_{s,t}+\brk[c]{n}^2.
    \end{equation*}
    \item  Another Lucas analogue of the identity $T_{n}+T_{n-1}=n^2$ is
    \begin{equation*}
        \varphi_{s,t}^{\prime(n-1)}\fibonomial{n+1}{2}_{s,t}+\varphi_{s,t}^{n+1}\fibonomial{n}{2}_{s,t}=\frac{\brk[a]{n}_{s,t}}{\brk[c]{2}_{s,t}}\brk[c]{n}_{s,t}^2,
    \end{equation*}
    where $\brk[a]{n}_{s,t}=\varphi_{s,t}^n+\varphi_{s,t}^{\prime n}$.
    \item The Lucas analogue of the alternating sum squares $T_{n}=\sum_{k=1}^{n}(-1)^{n-k}k^2$ is
    \begin{equation*}
        \fibonomial{n+1}{2}_{s,t}=\sum_{k=1}^{n}t^{n-k}\brk[c]{k}_{s,t}^2.
    \end{equation*}
\end{enumerate}
\end{theorem}
\begin{proof}
We have
    \begin{align*}
        \fibonomial{n+1}{2}_{s,t}&=\frac{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}{\brk[c]{2}_{s,t}}\\
        &=\frac{\brk[c]{n}_{s,t}(s\brk[c]{n}_{s,t}+t\brk[c]{n-1}_{s,t})}{\brk[c]{2}_{s,t}}\\
        &=s\frac{\brk[c]{n}_{s,t}^2}{s}+t\frac{\brk[c]{n}_{s,t}\brk[c]{n-1}_{s,t}}{\brk[c]{2}_{s,t}}\\
        &=\brk[c]{n}_{s,t}^2+t\fibonomial{n}{2}_{s,t}.
    \end{align*}
Statement 2 is proved as follows:
\begin{align*}
    \varphi_{s,t}^{n+1}\fibonomial{n}{2}_{s,t}+\varphi_{s,t}^{\prime(n-1)}\fibonomial{n+1}{2}_{s,t}&=\varphi_{s,t}^{n+1}\frac{\brk[c]{n}_{s,t}\brk[c]{n-1}_{s,t}}{\brk[c]{2}_{s,t}}+\varphi_{s,t}^{\prime(n-1)}\frac{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}{\brk[c]{2}_{s,t}}\\
    &=\frac{\brk[c]{n}_{s,t}}{\brk[c]{2}_{s,t}}(\varphi_{s,t}^{n+1}\brk[c]{n-1}_{s,t}+\varphi_{s,t}^{\prime(n-1)})\brk[c]{n+1}_{s,t})\\
    &=\frac{\brk[c]{n}_{s,t}}{\brk[c]{2}_{s,t}}\brk[c]{2n}_{s,t}\\
    &=\frac{\brk[a]{n}_{s,t}}{\brk[c]{2}_{s,t}}\brk[c]{n}_{s,t}^2.
\end{align*}
By iterating statement (1), we obtain statement (3).
\end{proof}
If we set $s=1+q$, and $t=-q$ in the previous theorem, then
\begin{align}
   \qbinom{n+2}{2}_{q}&=-q\qbinom{n+1}{2}_{q}+\biggl(\frac{1-q^{n+1}}{1-q}\biggr)^2\label{eqn_qbinom}
\end{align}
and
\begin{equation*}
   \qbinom{n}{2}_{q}+q^{n-1}\qbinom{n+1}{2}_{q}=\frac{1+q^n}{1+q}\brk[s]{n}_{q}^2. 
\end{equation*}
By iterating Eq.~(\ref{eqn_qbinom}), we obtain a result due to Schlosser \cite{sch}:
\begin{equation*}
    \qbinom{n+1}{2}_{q}=\sum_{k=1}^{n}(-q)^{n-k}\biggl(\frac{1-q^k}{1-q}\biggr)^2.
\end{equation*}


\begin{theorem}
For all $n\in\N$,
\begin{equation}\label{eqn_tri_cube}
    \fibonomial{n+2}{2}_{s,t}^2-t^2\fibonomial{n+1}{2}_{s,t}^2=\biggl(\frac{\brk[c]{n+2}_{s,t}+t\brk[c]{n}_{s,t}}{s}\biggr)\brk[c]{n+1}_{s,t}^3.
\end{equation}
\end{theorem}
\begin{proof}
From Theorem \ref{eqn_tri_sq}, we have that
\begin{align*}
    \fibonomial{n+2}{2}_{s,t}^2&=\biggl(t\fibonomial{n+1}{2}_{s,t}+\brk[c]{n+1}_{s,t}^2\biggr)^2\\
    &=t^2\fibonomial{n+1}{2}_{s,t}^2+2t\fibonomial{n+1}{2}_{s,t}^2\brk[c]{n+1}_{s,t}^2+\brk[c]{n+1}_{s,t}^4\\
    &=t^2\fibonomial{n+1}{2}_{s,t}^2+2t\frac{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}^3}{\brk[c]{2}_{s,t}}+\brk[c]{n+1}_{s,t}^4\\
    &=t^2\fibonomial{n+1}{2}_{s,t}^2+\biggl(\frac{2t}{s}\brk[c]{n}_{s,t}+\brk[c]{n+1}_{s,t}\biggr)\brk[c]{n+1}_{s,t}^3\\
    &=t^2\fibonomial{n+1}{2}_{s,t}^2+\biggl(\frac{t\brk[c]{n}_{s,t}+\brk[c]{n+2}_{s,t}}{s}\biggr)\brk[c]{n+1}_{s,t}^3.
\end{align*}
The remainder of the proof follows straightforwardly.
\end{proof}
Eq.~(\ref{eqn_tri_cube}) is the Lucas analogue of the identity
\begin{equation*}
    {T}_{n+1}^2-{T}_{n}^2=n^3.
\end{equation*}
From Warnaar \cite{war} we have
\begin{equation*}
    \qbinom{n+2}{2}_{q}^2-q^2\qbinom{n+1}{2}_{q}^2=\biggl(\frac{1-q^{2(n+1)}}{1-q^2}\biggr)\biggl(\frac{1-q^{n+1}}{1-q}\biggr)^2.
\end{equation*}

\begin{theorem}
For all $n\in\N$,
\begin{equation}\label{eqn_cub_sq}
    \sum_{k=1}^{n}t^{2(n-k)}\biggl(\frac{\brk[c]{k+1}_{s,t}+t\brk[c]{k-1}_{s,t}}{s}\biggr)\brk[c]{k}_{s,t}^3=\fibonomial{n+1}{2}_{s,t}^2.
\end{equation}
\end{theorem}
\begin{proof}
\begin{align*}
    \fibonomial{n+1}{2}_{s,t}^2&=\biggl(\fibonomial{n+1}{2}_{s,t}^2-t^2\fibonomial{n}{2}_{s,t}^2\biggr)+\biggl(t^2\fibonomial{n}{2}_{s,t}^2-t^4\fibonomial{n-1}{2}_{s,t}^2\biggr)\\
    &\hspace{2cm}+\cdots+\biggl(t^{2n-2}\fibonomial{2}{2}_{s,t}^2-t^{2n}\fibonomial{1}{2}_{s,t}^2\biggr)\\
\end{align*}
\begin{align*}
    &=\biggl(\frac{\brk[c]{n+1}_{s,t}+t\brk[c]{n-1}_{s,t}}{s}\biggr)\brk[c]{n}_{s,t}^3+t^2\biggl(\frac{\brk[c]{n}_{s,t}+t\brk[c]{n-2}_{s,t}}{s}\biggr)\brk[c]{n-1}_{s,t}^3+\\
    &\cdots+t^{2n-2}\biggl(\frac{\brk[c]{2}_{s,t}}{s}\biggr)\brk[c]{1}_{s,t}^3\\
    &=\sum_{k=1}^{n}t^{2(n-k)}\biggl(\frac{\brk[c]{k+1}_{s,t}+t\brk[c]{k-1}_{s,t}}{s}\biggr)\brk[c]{k}_{s,t}^3.
\end{align*}    
\end{proof}
Eq.~(\ref{eqn_cub_sq}) can be written as
\begin{equation}\label{eqn_sq_trian}
    \sum_{k=1}^{n}t^{2(n-k)}\biggl(\frac{\varphi_{s,t}^{2k}-\varphi_{s,t}^{\prime2k}}{\varphi_{s,t}^2-\varphi_{s,t}^{\prime2}}\biggr)\biggl(\frac{\varphi_{s,t}^k-\varphi_{s,t}^{\prime k}}{\varphi_{s,t}-\varphi_{s,t}^\prime}\biggr)^2=\sum_{k=1}^{n}t^{2(n-k)}[k]_{\varphi^2,\varphi^{\prime2}}[k]_{\varphi,\varphi^\prime}^2=\fibonomial{n+1}{2}_{s,t}^2.
\end{equation}
Eq.~(\ref{eqn_cub_sq}) is the Lucas analogue of the identity
\begin{equation*}
    \sum_{k=1}^{n}k^3=\biggl(\sum_{k=1}^{n}k\biggr)^2.
\end{equation*}

\begin{corollary}
For all $n\in\N$, we have
\begin{enumerate}
    \item The Fibonacci analogue of the sum of cubes:
    \begin{equation*}
        \sum_{k=1}^{n}(F_{k+1}+F_{k-1})F_{k}^3=F_{n}^2F_{n+1}^2.
    \end{equation*}
    \item The Pell analogue of the sum of cubes:
    \begin{equation*}
        \sum_{k=1}^{n}(P_{k+1}+P_{k-1})P_{k}^3=\frac{1}{2}P_{n}^2P_{n+1}^2.
    \end{equation*}
    \item The Jacobsthal analogue of the sum of cubes:
    \begin{equation*}
        \sum_{k=1}^{n}4^{n-k}(J_{k+1}+2J_{k-1})J_{k}^3=J_{n}^2J_{n+1}^2.
    \end{equation*}
    \item The Mersenne analogue of the sum of cubes:
    \begin{equation*}
        \sum_{k=1}^{n}4^{n-k}(2^{k}+1)(2^k-1)^3=\frac{1}{3}(2^n-1)^2(2^{n+1}-1)^2.
    \end{equation*}
    \item The $q$-identity of Warnaar \cite{war}:
    \begin{equation*}
    \sum_{k=1}^{n}q^{2(n-k)}\biggl(\frac{1-q^{2k}}{1-q^2}\biggr)\biggl(\frac{1-q^k}{1-q}\biggr)^2=\qbinom{n+1}{2}_{q}^2.
\end{equation*}
\end{enumerate}
\end{corollary}

\subsection{Lucas-tetrahedral numbers}
Some specializations of Lucas-tetrahedral numbers are
\begin{align*}
    \fibonomial{n+2}{3}_{1,1}&=\frac{1}{2}F_{n}F_{n+1}F_{n+2}\\
    &=(0,1,3,15,60,260,1092,4641,19635,\ldots),\\
    \fibonomial{n+2}{3}_{2,1}&=\frac{1}{10}P_{n}P_{n+1}P_{n+2}\\
    &=(1, 12, 174, 2436, 34307, 482664,\ldots),\\
    \fibonomial{n+2}{3}_{1,2}&=\frac{1}{3}J_{n}J_{n+1}J_{n+2}\\
    &=(0,1,5,55,385,3311,25585,208335,\ldots),\\
    \fibonomial{n+2}{3}_{3,-2}&=\frac{1}{21}(2^n-1)(2^{n+1}-1)(2^{n+2}-1)\\
    &=(0,1, 15, 155, 1395, 11811, 97155,\ldots).
\end{align*}
The sequences $\fibonomial{n+2}{3}_{1,1}$, $\fibonomial{n+2}{3}_{2,1}$, and $\fibonomial{n+2}{3}_{3,-2}$ correspond to \seqnum{A001655}, \seqnum{A099930}, and \seqnum{A006096}, respectively, in \cite{sloane}. The sequence $\fibonomial{n+2}{3}_{1,2}$ has not been previously investigated.
From Eqs.\ (\ref{eqn_gspn_recu1}) and (\ref{eqn_gspn_recu2}), and Theorem \ref{theo_reduc}, 
\begin{align*}
    \fibonomial{n+3}{3}_{s,t}&=\varphi_{s,t}^3\fibonomial{n+2}{3}_{s,t}+\varphi_{s,t}^{\prime n}\fibonomial{n+2}{2}_{s,t},\\
    \fibonomial{n+3}{3}_{s,t}&=\varphi_{s,t}^{\prime3}\fibonomial{n+2}{3}_{s,t}+\varphi_{s,t}^{n}\fibonomial{n+2}{2}_{s,t},\\
    \fibonomial{n+2}{3}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{3(n-k)}\varphi_{s,t}^{\prime(k-1)}\fibonomial{k+1}{2}_{s,t},\\
    \fibonomial{n+3}{3}_{s,t}&=\sum_{k=1}^{n}\varphi_{s,t}^{\prime3(n-k)}\varphi_{s,t}^{k-1}\fibonomial{k+1}{2}_{s,t}.
\end{align*}

\begin{theorem}
For all $n\geq0$,
    \begin{equation}\label{eqn_iden_tetra}
        \fibonomial{n+3}{3}_{s,t}=st\fibonomial{n+2}{3}_{s,t}+\brk[c]{n+1}_{s,t}\fibonomial{n+2}{2}_{s,t}.
    \end{equation}
\end{theorem}
\begin{proof}
We have
    \begin{align*}
        \fibonomial{n+3}{3}_{s,t}&=\frac{\brk[c]{n+1}_{s,t}\brk[c]{n+2}_{s,t}\brk[c]{n+3}_{s,t}}{\brk[c]{3}_{s,t}!}\\
        &=\frac{\brk[c]{n+1}_{s,t}\brk[c]{n+2}_{s,t}}{\brk[c]{3}_{s,t}!}(s\brk[c]{n+2}_{s,t}+t\brk[c]{n+1}_{s,t})\\
        &=\frac{\brk[c]{n+1}_{s,t}\brk[c]{n+2}_{s,t}}{\brk[c]{3}_{s,t}!}((s^2+t)\brk[c]{n+1}_{s,t}+st\brk[c]{n}_{s,t})\\
        &=\brk[c]{n+1}_{s,t}\fibonomial{n+2}{2}_{s,t}+st\fibonomial{n+2}{3}_{s,t}.
    \end{align*}
\end{proof}
The above theorem is Prop.~2.2 given by Sagan and Savage \cite{bennet}. Eq.~(\ref{eqn_iden_tetra}) is the Lucas analogue of the identity
\begin{equation*}
    \mathrm{Te}_{n+1}+2\mathrm{Te}_{n}=(n+1){T}_{n+1}.
\end{equation*}
If we choose $s=1+q$ and $t=-q$ in Eq.~(\ref{eqn_iden_tetra}), we obtain 
\begin{equation*}
        \qbinom{n+3}{3}_{q}=-(1+q)q\qbinom{n+2}{3}_{q}+\frac{1-q^{n+1}}{1-q}\qbinom{n+2}{2}_{q}.
    \end{equation*}

\section{Generating functions}

\begin{theorem}\label{theo_nder_geo}
For all $n\geq1$,
    \begin{equation}\label{eqn_nder_geo}
        D_{\varphi,\varphi^\prime}^{n}\biggl(\frac{1}{1-x}\biggr)=\frac{\brk[c]{n}_{s,t}!}{(\varphi_{s,t}^nx;q)_{n+1}},
    \end{equation}
where $(a;q)_{n}=\prod_{k=0}^{n-1}(1-aq^k)$ is the $q$-shifted factorial.
\end{theorem}
\begin{proof}
Note that
\begin{equation*}
    D_{\varphi,\varphi^\prime}\biggl(\frac{1}{1-x}\biggr)=\frac{1}{(1-\varphi_{s,t}x)(1-\varphi_{s,t}^\prime x)}=\frac{\brk[c]{1}_{s,t}!}{(\varphi_{s,t}x;q)_{2}}.
\end{equation*}
Suppose that Eq.~(\ref{eqn_nder_geo}) is true for $n$ and let us prove by induction for $n+1$. As
\begin{equation*}
    D_{\varphi,\varphi^\prime}(\varphi_{s,t}^nx;q)_{n+1}=-\brk[c]{n+1}_{s,t}(\varphi_{s,t}^n\varphi_{s,t}^\prime x;q)_{n},
\end{equation*}
then
\begin{align*}
    D_{\varphi,\varphi^\prime}^{n+1}\biggl(\frac{1}{1-x}\biggr)&=D_{\varphi,\varphi^\prime}D_{\varphi,\varphi^\prime}^{n}\biggl(\frac{1}{1-x}\biggr)\\
    &=D_{\varphi,\varphi^\prime}\biggl(\frac{\brk[c]{n}_{s,t}!}{(\varphi_{s,t}^nx;q)_{n+1}}\biggr)\\
    &=\frac{-\brk[c]{n}_{s,t}!D_{\varphi,\varphi^\prime}(\varphi_{s,t}^nx;q)_{n+1}}{(\varphi_{s,t}^{n+1}x;q)_{n+1}(\varphi_{s,t}^n\varphi_{s,t}^{\prime}x;q)_{n+1}}\\
    &=\frac{\brk[c]{n+1}_{s,t}!(\varphi_{s,t}^{n}\varphi_{s,t}^{\prime}x;q)_{n}}{(\varphi_{s,t}^{n+1}x;q)_{n+1}(\varphi_{s,t}^{n}\varphi_{s,t}^{\prime}x;q)_{n+1}}\\
    &=\frac{\brk[c]{n+1}_{s,t}!}{(1-(\varphi_{s,t}q)^{n+1}x)(\varphi_{s,t}^{n+1}x;q)_{n+1}}\\
    &=\frac{\brk[c]{n+1}_{s,t}!}{(\varphi_{s,t}^{n+1}x;q)_{n+2}}.
\end{align*}
The proof is completed.
\end{proof}

\begin{theorem}
For all $d\in\N$,
    \begin{equation*}
        \sum_{n=1}^{\infty}\fibonomial{n+d-1}{d}_{s,t}x^n=\frac{x}{(\varphi_{s,t}^{d}x;q)_{d+1}}.
    \end{equation*}
\end{theorem}
\begin{proof}
From Theorem \ref{theo_nder_geo},
    \begin{align*}
        \frac{x}{(\varphi_{s,t}^{d}x;q)_{d+1}}&=\frac{x}{\brk[c]{d}_{s,t}!}D_{\varphi,\varphi^\prime}^d\biggl(\frac{1}{1-x}\biggr)\\
        &=\frac{x}{\brk[c]{d}_{s,t}!}D_{\varphi,\varphi^\prime}^{d}\biggl(\sum_{n=0}^{\infty}x^n\biggr)\\
        &=\frac{x}{\brk[c]{d}_{s,t}!}\sum_{n=d}^{\infty}\brk[c]{n}_{s,t}\brk[c]{n-1}_{s,t}\cdots\brk[c]{n-d+1}_{s,t}x^{n-d}\\
        &=\frac{x}{\brk[c]{d}_{s,t}!}\sum_{n=0}^{\infty}\frac{\brk[c]{d+n}_{s,t}!}{\brk[c]{n}_{s,t}!}x^n\\
        &=\sum_{n=1}^{\infty}\fibonomial{n+d-1}{d}_{s,t}x^{n}.
    \end{align*}
\end{proof}

\begin{theorem} 
The generating function of the squares of the Lucas-triangular numbers is
    \begin{multline*}
        \sum_{n=1}^{\infty}\fibonomial{n+1}{2}_{s,t}^2x^n\\=\frac{x+(\brk[c]{4}\varphi_{s,t}-\varphi_{s,t}^4)x^2-\brk[c]{3}_{s,t}\varphi_{s,t}^3\varphi_{s,t}^{\prime3}x^3+(\brk[c]{3}_{s,t}\varphi_{s,t}^7\varphi_{s,t}^{\prime3}-\brk[c]{4}_{s,t}\varphi_{s,t}^6\varphi_{s,t}^{\prime3})x^4}{(1-t^2x)(1-\varphi_{s,t}^{\prime4}x)(\varphi_{s,t}^4x;q)_{4}}.
    \end{multline*}
\end{theorem}
\begin{proof}
From Eq.~(\ref{eqn_sq_trian}),
    \begin{align*}
        &\sum_{n=1}^{\infty}\fibonomial{n+1}{2}_{s,t}^2x^n\\
        &\hspace{1cm}=\sum_{n=1}^{\infty}\sum_{k=1}^{n}t^{2(n-k)}\brk[s]{k}_{\varphi^2,\varphi^{\prime2}}\brk[s]{k}_{\varphi,\varphi^\prime}^2x^n\\
        &\hspace{1cm}=\sum_{k=1}^{\infty}\brk[s]{k}_{\varphi^2,\varphi^{\prime2}}\brk[s]{k}_{\varphi,\varphi^\prime}^2x^k\sum_{n=k}^{\infty}t^{2(n-k)}x^n\\
        &\hspace{1cm}=\sum_{k=1}^{\infty}\brk[s]{k}_{\varphi^2,\varphi^{\prime2}}\brk[s]{k}_{\varphi,\varphi^\prime}^2x^k\sum_{n=0}^{\infty}(t^{2}x)^{n}\\
        &\hspace{1cm}=\frac{1}{1-t^2x}(xD_{\varphi,\varphi^\prime})^2(xD_{\varphi^2,\varphi^{\prime2}})\left\{\frac{x}{1-x}\right\}\\
        &\hspace{1cm}=\frac{x+(\brk[c]{4}_{s,t}\varphi_{s,t}-\varphi_{s,t}^4)x^2-\brk[c]{3}_{s,t}\varphi_{s,t}^3\varphi_{s,t}^{\prime3}x^3+(\brk[c]{3}_{s,t}\varphi_{s,t}^7\varphi_{s,t}^{\prime3}-\brk[c]{4}_{s,t}\varphi_{s,t}^6\varphi_{s,t}^{\prime3})x^4}{(1-t^2x)(1-\varphi_{s,t}^{\prime4}x)(\varphi_{s,t}^4x;q)_{4}}.
    \end{align*}
\end{proof}
The $q$-analogue of the previous theorem is
\begin{equation*}
        \sum_{n=1}^{\infty}\qbinom{n+1}{2}_{q}^{2}x^n=\frac{x+(\brk[s]{4}_{q}-1)x^2-\brk[s]{3}_{q}q^{3}x^3+(\brk[s]{3}_{q}q^{3}-\brk[s]{4}_{q}q^{3})x^4}{(1-q^2x)(1-q^{4}x)(x;q)_{4}}.
    \end{equation*}

\section{Sum of reciprocals}

The Lucas-zeta function, or $(s,t)$-zeta function, is the function defined by
\begin{equation*}
    \zeta_{s,t}(z)=\sum_{n=1}^{\infty}\frac{1}{\brk[c]{n}_{s,t}^z}=1+\frac{1}{s^z}+\frac{1}{(s^2+t)^z}+\frac{1}{(s^3+2st)^z}+\cdots.
\end{equation*}
Egami \cite{egami} and Navas \cite{navas} independently studied the zeta function $\zeta_{1,1}(z)$. Landau \cite{landau} studied the problem of evaluating $\zeta_{1,1}(1)$. The function $\zeta_{s,t}(z)$ is convergent when $z>0$ and when either $\varphi_{s,t}>1$ and $0<\vert q\vert<1$ or $\varphi_{s,t}^\prime>1$ and $\vert q\vert>1$. Take $z=\sigma+i\beta$. Then $\zeta_{s,t}(z)$ converges when $\Re(z)>0$. Some specializations of $\zeta_{s,t}(1)$ are
\begin{align*}
    \zeta_{1,1}(1)=\zeta_{F}(1)&=\sum_{n=1}^{\infty}\frac{1}{F_{n}}=1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+\cdots\doteq 3.359885666243\ldots,\\
    \zeta_{2,1}(1)=\zeta_{P}(1)&=\sum_{n=1}^{\infty}\frac{1}{P_{n}}=1+\frac{1}{2}+\frac{1}{5}+\frac{1}{12}+\frac{1}{29}+\cdots\doteq 1.81781609195402\ldots,\\
    \zeta_{1,2}(1)=\zeta_{J}(1)&=\sum_{n=1}^{\infty}\frac{1}{J_{n}}=1+1+\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+\frac{1}{21}+\cdots\doteq 2.67186147\ldots,\\
    \zeta_{3,-2}(1)=\zeta_{M}(1)&=\sum_{n=1}^{\infty}\frac{1}{M_{n}}=1+\frac{1}{3}+\frac{1}{7}+\frac{1}{15}+\frac{1}{31}+\cdots\doteq 1.57511520737327\ldots,\\
    \zeta_{2t,-1}(1)=\zeta_{U}(1)&=\sum_{n=1}^{\infty}\frac{1}{U_{n-1}(t)}=1+\frac{1}{2t}+\frac{1}{4t^2-1}+\frac{1}{8t^3-4t}+\frac{1}{16t^4-12t^2+1}+\cdots,
\end{align*}
with $t\neq0,\cos\frac{k\pi}{n+1}$, and $k=1,2,\ldots,n$.

\begin{theorem}
    \begin{equation*}
        \sum_{n=1}^{\infty}\frac{t^n}{\brk[c]{2n}_{s,t}}=\sqrt{s^2+4t}\biggl(L\bigl(\varphi_{s,t}^{\prime2}/t\bigr)-L\bigl(\varphi_{s,t}^{\prime4}/t^2\bigr)\biggr),
    \end{equation*}
where $L(q)=\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}$ is the Lambert function.
\end{theorem}
\begin{proof}
Take into account that
\begin{equation*}
    \frac{t^n}{\brk[c]{2n}_{s,t}}=\sqrt{s^2+4t}\biggl(\frac{(\varphi_{s,t}^{\prime2}/t)^n}{1-(\varphi_{s,t}^{\prime2}/t)^n}-\frac{(\varphi_{s,t}^{\prime4}/t^2)^n}{1-(\varphi_{s,t}^{\prime4}/t^2)^n}\biggr)
\end{equation*}
and sum for all $n\geq1$.
\end{proof}

\begin{theorem}
    \begin{equation*}
        \sum_{n=1}^{\infty}\frac{1}{\brk[c]{2n-1}_{s,1}}=-\frac{\sqrt{s^2+4}}{4}\theta_{2}(\varphi_{s,1}^{\prime2})^2,
    \end{equation*}
where $\theta_{2}(q)=\sum_{n=-\infty}^{\infty}q^{(n+1/2)^2}$.
\end{theorem}
\begin{proof}
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{\brk[c]{2n-1}_{s,1}}&=-\frac{\sqrt{s^2+4}}{2}\sum_{n=1}^{\infty}\frac{2(\varphi_{s,1}^{\prime2})^{n-1/2}}{1+(\varphi_{s,1}^{\prime2})^{2n-1}}\\
    &=-\frac{\sqrt{s^2+4}}{4}\sum_{a=-\infty}^{\infty}\frac{2(\varphi_{s,1}^{\prime2})^{a-1/2}}{1+(\varphi_{s,1}^{\prime2})^{2a-1}}\\
    &=-\frac{\sqrt{s^2+4}}{4}\theta_{2}(\varphi_{s,1}^{\prime2})^2.
\end{align*}    
\end{proof}



\begin{theorem}\label{theo_recip_tri}
    \begin{equation}\label{eqn_tri_reci}
        \sum_{n=1}^{\infty}\frac{1}{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}=\varphi_{s,t}-(1+t)\ln_{s,t}\biggl(1+\frac{\varphi_{s,t}}{t}\biggr),
    \end{equation}
where
\begin{equation*}
    \ln_{s,t}(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{\brk[c]{n}_{s,t}}
\end{equation*}
is a Lucas analogue of the logarithm function.
\end{theorem}
\begin{proof}
Suppose that $0<\vert q\vert<1$, and $q=\varphi_{s,t}^\prime/\varphi_{s,t}$, and set $a_{n}=1/\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}$. As 
\begin{equation*}
    \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow\infty}\frac{\brk[c]{n}_{s,t}}{\brk[c]{n+2}_{s,t}}=\lim_{n\rightarrow\infty}\frac{1-q^n}{\varphi_{s,t}^2-\varphi_{s,t}^{\prime2}q^n}=\frac{1}{\varphi_{s,t}^{2}},
\end{equation*}
then the series in the left-hand of Eq.~(\ref{eqn_tri_reci}) is convergent only if $\vert\varphi_{s,t}\vert>1$. If $\vert q\vert>1$, then the series in the left-hand of Eq.~(\ref{eqn_tri_reci}) is convergent only if $\vert\varphi_{s,t}^\prime\vert>1$.
By partial fractions
    \begin{align*}
        \frac{1}{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}=\frac{A}{\brk[c]{n}_{s,t}}+\frac{B}{\brk[c]{n+1}_{s,t}}=\frac{A}{\brk[c]{n}_{s,t}}+\frac{B}{\varphi_{s,t}\brk[c]{n}_{s,t}+\varphi_{s,t}^{\prime n}},
    \end{align*}
where $A=(-\varphi_{s,t}/t)^n$ and $B=(-\varphi_{s,t}/t)^{n+1}$. Then
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}&=\sum_{n=1}^{\infty}\biggl(\frac{(-\varphi_{s,t}/t)^{n}}{\brk[c]{n}_{s,t}}+\frac{t(-\varphi_{s,t}/t)^{n+1}}{\brk[c]{n+1}_{s,t}}\biggr)\\
    &=-\frac{\varphi_{s,t}}{t}+(1+t)\sum_{n=2}^{\infty}\frac{(-\varphi_{s,t}/t)^{n}}{\brk[c]{n}_{s,t}}\\
    &=-\frac{\varphi_{s,t}}{t}-(1+t)\biggl(\ln_{s,t}\biggl(1+\frac{\varphi_{s,t}}{t}\biggr)-\frac{\varphi_{s,t}}{t}\biggr)\\
    &=\varphi_{s,t}-(1+t)\ln_{s,t}\biggl(1+\frac{\varphi_{s,t}}{t}\biggr).
\end{align*}
\end{proof}
The function $\ln_{s,t}(1-x)$ is convergent for all $x\in(-\vert\varphi_{s,t}\vert,\vert\varphi_{s,t}\vert)$.

Some specializations of Theorem \ref{theo_recip_tri} are
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{F_{n}F_{n+1}}&=\frac{1+\sqrt{5}}{2}-2\ln_{F}\biggl(\frac{3+\sqrt{5}}{2}\biggr),\\
    \sum_{n=1}^{\infty}\frac{1}{P_{n}P_{n+1}}&=1+\sqrt{2}-2\ln_{P}(2+\sqrt{2}),\\
\end{align*}
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{J_{n}J_{n+1}}&=2-3\ln_{J}(2),\\
    \sum_{n=1}^{\infty}\frac{1}{M_{n}M_{n+1}}&=2+\ln_{M}(0),
\end{align*}
where $\ln_{F}\equiv\ln_{1,1}$, $\ln_{P}\equiv\ln_{2,1}$, $\ln_{J}\equiv\ln_{1,2}$, and $\ln_{M}\equiv\ln_{3,-2}$.
A very important remark is that $\ln_{M}(0)$ is finite.

\begin{corollary}\label{coro_sum-alt}
    \begin{equation*}
        \sum_{n=1}^{\infty}\frac{(-t)^{n}}{\brk[c]{n}_{s,t}\brk[c]{n+1}_{s,t}}=\frac{s+\sqrt{s^2+4t}}{2}.
    \end{equation*}
\end{corollary}
\begin{proof}
If $t=-1$ in the previous theorem, then
\begin{equation*}
    \sum_{n=1}^{\infty}\frac{1}{\brk[c]{n}_{s,-1}\brk[c]{n+1}_{s,-1}}=\varphi_{s,-1}.
\end{equation*}
Note that $\brk[c]{n}_{s,-1}=(i/\sqrt{t})^{n-1}\brk[c]{n}_{a,t}$, where $s=ia/\sqrt{t}$. Then
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{\brk[c]{n}_{s,-1}\brk[c]{n+1}_{s,-1}}&=
    \sum_{n=1}^{\infty}\frac{1}{(i/\sqrt{t})^{2n-1}\brk[c]{n}_{a,t}\brk[c]{n+1}_{a,t}}\\
    &=\frac{i}{\sqrt{t}}\sum_{n=1}^{\infty}\frac{(-t)^n}{\brk[c]{n}_{a,t}\brk[c]{n+1}_{a,t}}=\frac{i}{\sqrt{t}}\varphi_{a,t}.
\end{align*}
The proof is completed.
\end{proof}
Some specializations of Corollary \ref{coro_sum-alt} are
\begin{align*}
    \sum_{n=1}^{\infty}\frac{(-1)^{n}}{F_{n}F_{n+1}}&=\frac{1+\sqrt{5}}{2}.\\
    \sum_{n=1}^{\infty}\frac{(-1)^{n}}{P_{n}P_{n+1}}&=1+\sqrt{2}.\\
    \sum_{n=1}^{\infty}\frac{(-2)^{n}}{J_{n}J_{n+1}}&=2.\\
    \sum_{n=1}^{\infty}\frac{2^{n}}{M_{n}M_{n+1}}&=2.\\
    \sum_{n=1}^{\infty}\frac{1}{U_{n-1}(t)U_{n}(t)}&=t+\sqrt{t^2-1}.
\end{align*}





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\bigskip

\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 05A15, 11B65.

\noindent \emph{Keywords: } simplicial $d$-polytopic number, generalized Fibonacci polynomial, generalized triangular number, generalized tetrahedral number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001654},
\seqnum{A001655},
\seqnum{A006095},
\seqnum{A006096},
\seqnum{A084158},
\seqnum{A084175}, and
\seqnum{A099930}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 18 2025;
revised versions received  August 13 2025; September 20 2025.
Published in {\it Journal of Integer Sequences}, September 20 2025.

\bigskip
\hrule
\bigskip

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


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