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\begin{document}

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\begin{center}
\vskip 1cm{\LARGE\bf 
The $p$-Adic Valuation of Lucasnomials 
\vskip .2in
When $p$ is a Special Prime
}
\vskip 1cm
\large
Christian Ballot \\
Normandie Universit\'e \\
UNICAEN, CNRS\\
Laboratoire de Math\'ematiques Nicolas Oresme\\
14000 Caen \\
France \\
\href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr} \\
\end{center}

\vskip .2 in

\begin{abstract} 
To determine the $p$-adic valuation of a Lucasnomial $\binom{m}{n}_U$,
i.e., of a generalized binomial coefficient with respect to a fundamental
Lucas sequence $U=U(P,Q)$, there is an adequate Kummer rule if $p$ is
a regular prime. No such rule exists if $p$ is a special prime, i.e.,
if $p$ divides $\gcd(P,Q)$.  We provide a complete description of the
$p$-adic valuation of Lucasnomials when $p$ is special, with some
numerical examples. Applications to the integrality of generalized
Lucasnomial Fuss-Catalan numbers, Lucasnomial ballot and Lucasnomial
Lobb numbers are also given.
\end{abstract}

\section{Introduction}
\label{sec:1}

A fundamental Lucas sequence $U=U(P,Q)=(U_n)_{n\ge0}$, where $P$ and $Q$
are nonzero integral parameters, is a second-order linear
recurring sequence with initial conditions $U_0=0$ and $U_1=1$,
which satisfies
$$
U_{n+2}=PU_{n+1}-QU_n,
$$ for all $n\ge0$. The fundamental Lucas sequence together with its
companion sequence,  $V(P,Q)$, which won't intervene in this paper, form
a pair of sequences which derive their name from the famous memoir of
Lucas of 1878 \cite{Lu1}, and have a long history of studies and
applications. Later, Lucas also dedicated a chapter of his number theory book \cite{Lu2}
to these sequences. Readers may consult \cite[Chap.\ 4]{Wi} for a good introduction to their properties. The Lucas sequence
$U$ is a {\it divisible} sequence. That is, if $m\mid n$, then $U_m\mid U_n$. If no term $U_t$, $t\ge1$, is zero, then $U$ is said to be {\it nondegenerate}.  

 For $U$ nondegenerate and $m$ and $n$ two positive integers, one defines the Lucasnomial, or Lucasnomial coefficient, $\binom{m+n}{n}_U$ as 
$$
\frac{U_{m+n}U_{m+n-1}\cdots U_1}{(U_mU_{m-1}\cdots U_1)(U_nU_{n-1}\cdots
  U_1)}=\frac{U_{m+n}\cdots U_{m+1}}{U_n\cdots U_1}.
$$ Lucasnomials have been proved to be integers in various ways, algebraic, arithmetic and combinatorial. One of the quickest ways to show their integrality is by the Lucasnomial identity 
$$
\binom{r}{s}_U=U_{s+1}\binom{r-1}{s}_U-QU_{r-s-1}\binom{r-1}{s-1}_U,
$$ followed by an induction, which is the way many papers have used (e.g., \cite[eq.\ (11)]{Ba3}, \cite[Lemma 1, eq.\ (6)]{HS}). 
If $U=U(2,1)$, then $U_n=n$ for all $n\ge0$, so that $\binom{m+n}{n}_U$ is the binomial coefficient
$\binom{m+n}{n}$. 

 Given a prime $p$, the $p$-adic
 valuation of an integer $x$ is the largest exponent $t\ge0$ such that
 $p^t$ divides $x$. It is denoted by $\nu_p(x)$. For a binomial 
 coefficient or a Lucasnomial, we write $\nu_p\binom{*}{*}_U$ instead
 of  
$$
\nu_p\bigg(\binom{*}{*}_U\bigg),
$$ to alleviate notation. 

By the well-known Kummer rule \cite{Ku}, the $p$-adic valuation of the
binomial coefficient $\binom{m+n}{n}$ is equal to the number of carries
that come up when you add $m$ and $n$ in base $p$. For instance, in base
$3$, $16=(121)_3$ and $5=(012)_3$. Exactly two carries occur when adding
$16$ and $5$ in base $3$. Thus $\nu_3\binom{21}{5}=2$.

The $p$-adic valuation of a rational number
$x/y$ is $\nu_p(x)-\nu_p(y)$. A prime $p$ is said to be {\it special} in
relation to the Lucas sequence $U(P,Q)$ whenever $p\mid\gcd(P,Q)$. A prime is said to be {\it regular} if it does not divide $Q$. Primes that divide $Q$, but not $P$, do not divide any $U_n$, $n\ge1$. 
 
For regular primes $p$ with respect to a Lucas sequence $U$, there is
a Kummer rule to help determine the $p$-adic valuation of Lucasnomials
$\binom{m+n}{n}_U$ (\cite[Section 4]{Ba1}, \cite{KW}). The rank, $\rho$,
of $p$ is the least positive integer $t$ such that $p\mid U_t$. If
$p$ is regular, then the rank of $p$ exists and is either $p$, or a
divisor of $p\pm1$. The valuation of $\binom{m+n}{n}_U$ is the number of
carries that occur to the left of the radix point when adding $m/\rho$
and $n/\rho$ in base $p$, plus $\nu_p(U_\rho)$ if a carry occurs across
the radix point. However, if $p=2$,  a carry occurring inbetween the
first two places left of the radix point, everything else being equal,
bears a  weight of $\nu_2(P^2-3Q)$ instead of $1$.

There are a few additional results for the $p$-adic valuation of
Lucasnomials of the form $\binom{p^b}{p^a}_U$ \cite[Thm.\ 7.1]{Ba1} as
well as for various Fibonomial coefficients \cite{PhPo}. Fibonomials are
the Lucasnomials that correspond to the Fibonacci sequence $F=U(1,-1)$.

There is no Kummer rule for special
primes. Our main theorem, Theorem \ref{thm:1}, provides formulas
to express in a concise and useful manner this valuation. Theorem
\ref{thm:1} is based on the complete description \cite{Ba4} of the
$p$-adic valuation of the terms of a Lucas sequence when $p$ is a special
 prime. If $p$ is special, then we write $P=p^aP'$ and $Q=p^bQ'$, where
 $a$ and $b$ are positive integers and $p\nmid P'Q'$. Note that the prime
 $p$ is a regular prime with respect to the Lucas sequence $U'=U(P',Q')$,
 with a rank $\rho'\ge3$ since $U'_2=P'$. Define the condition $\cP_0$
 as \begin{equation}\label{eq:-1}
\cP_0: \quad 2\le p\le3 \quad \text{ and } \quad 2a=b+1.  \end{equation}
If $p$ is special, we will often refer to the following four equalities,
which, gathered together, form the content of \cite[Thm 1.2]{Ba4}.
 \begin{equation}\label{eq:0} {\bf If }\;b\ge2a,\; {\bf then }\quad
 \nu_p(U_k)=\begin{cases}(k-1)a, &amp;\text{ if }b&gt;2a;\\
               (k-1)a+\nu_p(U'_k),&amp;\text{ if }b=2a,
              \end{cases} \end{equation}
 while, {\bf if} $b&lt;2a$, {\bf then} \begin{align}\begin{split}
\nu_p(U_{2k+1})&amp;=bk,\\
\nu_p(U_{2k})&amp;=bk+(a-b)+\nu_p(k)+\nu_p(P'^2-Q')\cdot[\cP_0]\cdot[p\mid
k], \end{split}\label{eq:0+} \end{align} where $\cP_0$ is defined in
(\ref{eq:-1}) and $[-]$ denotes the Iverson symbol.

 We recall that, given a condition $\cP$, the Iverson function evaluated
 at $\cP$ is defined as
$$ [\cP]=\begin{cases}1,&amp;\text{ if }\cP \text{ is true};\\ 0,&amp;\text{
if }\cP \text{ is false}.  \end{cases} $$ We are now ready to state our
theorem.  \begin{theorem} \label{thm:1} Let $U=U(P,Q)$ be a nondegenerate
fundamental
  Lucas sequence, $p$ be a special prime where we set $P=p^aP'$,
  $Q=p^bQ'$, $a$ and
 $b$ positive integers with $p\nmid P'Q'$. Suppose $m$ and $n$ are two
 positive integers. Let $r_m = (m \bmod 2p)$ and $r_n = (n \bmod 2p)$,
 i.e., $r_m$ and
$r_n$ are the respective remainders of the euclidean division of $m$
and $n$ by $2p$. Let $\cP_1$ and $\cP_2$ be the following conditions:
\begin{itemize} \item $\cP_1$: \quad $m$ and $n$ are odd; \item
$\cP_2$: \quad $2\le p\le3$ and $2a=b+1$ and $r_m+r_n\ge2p$.
\end{itemize} Then $$ \nu_p\binom{m+n}{n}_U=\begin{cases}
amn,&amp;\text{ if }b&gt;2a;\\ 
amn+\nu_p\binom{m+n}{n}_{U'},&amp;\text{ if }b=2a;\\
b\lfloor\frac{mn}{2}\rfloor+\nu_p\binom{\lf\frac{m+n}{2}\rf}{\lf\frac{n}{2}\rf}+
\big(a+\nu_p(\frac{m+1}2)\big)\cdot[\cP_1] +c\cdot[\cP_2],&amp;\text{
if }b&lt;2a, \end{cases} $$ 
where $U'$ is the Lucas sequence $U(P',Q')$,
$c=\nu_p(P'^2-Q')$ and $[-]$ denotes the Iverson symbol.
\end{theorem}

\begin{remark} The expression of the valuation of $\binom{m+n}{n}_U$
  when $b&lt;2a$ reduces much if at least one of $m$ or $n$ is even and 
$p\ge5$ or $2a&gt;b+1$. In those cases, this valuation is 
\begin{equation}\label{eq:1}
b\frac{mn}{2}+\nu_p\binom{\lf\frac{m+n}{2}\rf}{\lf\frac{n}{2}\rf}.
\end{equation}
\end{remark}

 Section \ref{sec:2} contains a proof of Theorem \ref{thm:1}. In Section \ref{sec:3} we present three examples of calculations of a Lucasnomial valuation
 using the main theorem. In doing so we use the classical rule of
 Kummer to evaluate the $p$-adic valuation of some binomial coefficients, and
 its generalizations to Lucasnomials \cite{Ba1, KW}, which we recalled
 earlier. We are pleased to apply Theorem \ref{thm:1} to complete 
 the arithmetic proof of the integrality of generalized Lucasnomial
 Fuss-Catalan numbers given in \cite{Ba3}. 
 This is done in Theorem \ref{thm:2} of Section \ref{sec:4}, together
 with an introduction to these numbers. Immediate corollaries are proved which
 complete earlier proofs of the integrality of the generalized
 Lucasnomial Lobb numbers and of the Lucasnomial ballot numbers.    

\section{Proof of Theorem \ref{thm:1}}
\label{sec:2}
 
There are three cases to examine. 

\bigskip

\noindent {\bf Case 1: $b&gt;2a$.} In this case, (\ref{eq:0}) says that, for all $k\ge1$, 
$\nu_p(U_k)=(k-1)a$. Thus $\nu_p(U_\ell U_{\ell-1}\cdots
 U_1)=\sum_{i=1}^{\ell-1}ia=a\frac{(\ell-1)\ell}2$. Hence, 
\begin{equation}\label{eq:2}
\nu_p\binom{m+n}{n}_U=\frac{a}2((m+n-1)(m+n)-(n-1)n-(m-1)m)=amn.
\end{equation}

\bigskip

\noindent {\bf Case 2: $b=2a$.} By (\ref{eq:0}), and more precisely by
\cite[Proof of Thm 2.2]{Ba4}, we have $U_k=p^{(k-1)a}U_k'$,  where
$U'=U(P',Q')$. Thus, using (\ref{eq:2}), we find that
\begin{equation}\label{eq:2'}
\binom{m+n}{n}_U=p^{amn}\binom{m+n}{n}_{U'},
\end{equation} and the result we seeked follows. 

\bigskip

\noindent {\bf Case 3: $b&lt;2a$.} By (\ref{eq:0+}) and writing
$c=\nu_p(P'^2-Q')$, we find that
\begin{align*}
\nu_p(U_\ell\cdots
U_1)&amp;=\sum_{i=1}^{\lf\frac{\ell-1}2\rf}\nu_p(U_{2i+1})+
\sum_{i=1}^{\lf\frac{\ell}2\rf}\nu_p(U_{2i})\\
&amp;=b\sum_{i=1}^{\lf\frac{\ell-1}2\rf}i+b\sum_{i=1}^{\lf\frac{\ell}2\rf}i+(a-b)\big\lfloor\frac{\ell}2\big\rfloor+\nu_p(\big\lfloor\frac{\ell}2\big\rfloor!)+\bigg\lfloor\frac{\lf\frac{\ell}2\rf}{p}\bigg\rfloor\cdot
c\cdot[\cP_0],
\end{align*} which by combining the two sums on the RHS of the previous
equation and using the fact \cite[pp.\ 71--72]{GKP} that
$\big\lfloor\frac{\lf\frac{\ell}2\rf}{p}\big\rfloor=\big\lfloor\frac{\ell}{2p}\big\rfloor$, yields 
\begin{equation}\label{eq:5}
\nu_p(U_\ell\cdots U_1)=b\bigg\lfloor\frac{\ell-1}2\bigg\rfloor\bigg\lfloor\frac{\ell+1}2\bigg\rfloor+\frac{b\ell}2\cdot[2\mid\ell]
+(a-b)\bigg\lfloor\frac{\ell}2\bigg\rfloor+\nu_p(\big\lfloor\frac{\ell}2\big\rfloor!)+\bigg\lfloor\frac{\ell}{2p}\bigg\rfloor\cdot
c\cdot[\cP_0].
\end{equation}
 Note that $\nu_p\binom{m+n}{n}_U=\nu_p(U_{m+n}\cdots U_1)-\nu_p(U_m\cdots
U_1)-\nu_p(U_n\cdots U_1)$. Guided by equation (\ref{eq:5}), we first
take care of the first two terms of the RHS of (\ref{eq:5}). An
elementary calculation gives  
\begin{align*}
\left\lfloor\frac{m+n-1}2\right\rfloor&amp;\left\lfloor\frac{m+n+1}2\right\rfloor
-
\left\lfloor\frac{m-1}2\right\rfloor\left\lfloor\frac{m+1}2\right\rfloor
-\left\lfloor\frac{n-1}2\right\rfloor\left\lfloor\frac{n+1}2\right\rfloor\\
&amp;+\left(\frac{m+n}2\cdot[2\mid m+n]-\frac{m}2\cdot[2\mid
m]-\frac{n}2\cdot[2\mid n]\right)\\
&amp;=\begin{cases}\frac{mn+1}2,&amp;\text{ if }m\text{ and }n\text{ are both
    odd};\\
\frac{mn}2,&amp;\text{ otherwise.}
\end{cases}
\end{align*} 
 Since
\begin{equation}\label{eq:6}
\left\lfloor\frac{m+n}2\right\rfloor-\left\lfloor\frac{m}2\right\rfloor-
\left\lfloor\frac{n}2\right\rfloor=\begin{cases}
1,&amp;\text{ if }m\text{ and }n\text{ are both
    odd};\\
0,&amp;\text{ otherwise,}
\end{cases}
\end{equation} 
the contribution of the third terms of the RHS of (\ref{eq:5}) to 
$\nu_p\binom{m+n}{n}_U$ is nil unless $m$ and $n$ are both odd when it
is $a-b$. Hence, so far, the contribution of the first three terms 
of the RHS of (\ref{eq:5}) is 
$$
a\cdot[\cP_1]+b\begin{cases}\frac{mn-1}2,&amp;\text{ if }m\text{ and }n\text{ are 
    odd};\\
\frac{mn}2,&amp;\text{ otherwise;}
\end{cases}\;=\;a\cdot[\cP_1]+b\cdot\left\lfloor\frac{mn}2\right\rfloor.
$$ The contribution of the fourth terms
$\nu_p(\big\lfloor\frac{\ell}2\big\rfloor!)$ to
$\nu_p\binom{m+n}{n}_U$ is 
\begin{equation}\label{eq:6+}
C_4:=\nu_p(\big\lfloor\frac{m+n}2\big\rfloor!)-\nu_p(\big\lfloor\frac{m}2\big\rfloor!)-\nu_p(\big\lfloor\frac{n}2\big\rfloor!). 
\end{equation}
By (\ref{eq:6}), if $m$ and $n$ are not each odd,
then the contribution $C_4$ is exactly equal to
$$
\nu_p\binom{\left\lfloor\frac{m+n}2\right\rfloor}{\left\lfloor\frac{n}2
\right\rfloor}.
$$ If $m$ and $n$ are odd, then, by (\ref{eq:6}),
$$
\left\lfloor\frac{m+n}2\right\rfloor-\left\lfloor\frac{n}2\right\rfloor=
\left\lfloor\frac{m}2\right\rfloor+1=\frac{m+1}2. 
$$ In particular,
$\frac{m+1}2!=\big\lfloor\frac{m}2\big\rfloor!\cdot\frac{m+1}2$.
Hence, in this case, the contribution, $C_4$, of the fourth terms in (\ref{eq:6+}) is 
\begin{align*}
 &amp;\nu_p(\big\lfloor\frac{m+n}2\big\rfloor!)-
\nu_p(\big\lfloor\frac{n}2\big\rfloor!)- 
\nu_p(\frac{m+1}2!)+\nu_p(\frac{m+1}2)\\
&amp;=\nu_p\binom{\big\lfloor\frac{m+n}2\big\rfloor}{\big\lfloor\frac{n}2\big\rfloor}+\nu_p\bigg(\frac{m+1}2\bigg).
\end{align*}

 It remains to evaluate the contribution of the fifth terms in
 (\ref{eq:5}). It is 
$$
\bigg(\bigg\lfloor\frac{m+n}{2p}\bigg\rfloor-\bigg\lfloor\frac{m}{2p}\bigg\rfloor
-\bigg\lfloor\frac{n}{2p}\bigg\rfloor\bigg)\cdot
c\cdot[\cP_0].
$$ Since for two real numbers $x$ and $y$, we have
$$
\lf x+y\rf-\lf x\rf-\lf y\rf=\begin{cases}0,&amp;\text{ if
  }\{x\}+\{y\}&lt;1;\\
1,&amp;\text{ otherwise,}
\end{cases}
$$ where $\{-\}$ denotes the fractional part, we see that this
contribution is nonzero provided the sum of the fractional parts of
$\frac{m}{2p}$ and of $\frac{n}{2p}$ is at
least $1$, i.e., if $r_m+r_n\ge2p$. We conclude by observing that 
$[\cP_2]=[\cP_0]\cdot[r_m+r_n\ge2p]$, where $\cP_2$ was defined 
in Theorem \ref{thm:1}. \qed


\section{Examples}
\label{sec:3}

 We compute the $p$-adic valuation of a few Lucasnomials via Theorem 
\ref{thm:1}. 

\begin{example}\label{ex:1} Suppose $U=U(33,30)$ and $p=3$. Then
  $a=1$, $b=1$, $P'=11$, $Q'=10$ and $P'^2-Q'=111$. Note that $b&lt;2a$. We want to
  compute, say,  
  $\nu_3\binom{36}{13}_U$. Thus $m=23$ and $n=13$. Hence, following
  Theorem \ref{thm:1}, we obtain
$$
\nu_3\binom{36}{13}_U=1\cdot\bigg\lfloor\frac{23\times13}{2}\bigg\rfloor+\nu_3\binom{\frac{36}2}{\lf\frac{13}2\rf}+\bigg(1+\nu_3\big(\frac{23+1}2\big)\bigg)\cdot1+\nu_3(111)\cdot[\cP_2].
$$ Since $m=3\cdot6+5$ and $n=2\cdot6+1$, the sum
$r_m+r_n=5+1=2p$. Also, $2\le p\le3$ and $2a=2=b+1$. Hence,
$[\cP_2]=1$. It follows that
$$
\nu_3\binom{36}{13}_U=149+\nu_3\binom{12+6}{6}+(1+1)+1.
$$ But $12=9+3=(110)_3$ and $6=2\cdot3=(020)_3$ so there is a unique
carry in the base-3 addition of 12 and 6. Thus, by Kummer's rule
\cite{Ku}, we find that $\nu_3\binom{18}{6}=1$. Hence, 
$$
\nu_3\binom{36}{13}_U=153.
$$
\end{example}

\begin{example}\label{ex:2} Consider the Lucas sequence
  $U=U(4,-2)$ and compute the $2$-adic valuation of $\binom{22}{7}_U$. 
  We find that $a=2$ and $b=1$ so that $b&lt;2a$. Here $m=15$ and $n=7$. 
  Also $P'^2-Q'=2$. 
  Thus, by Theorem \ref{thm:1},
$$
\nu_2\binom{22}{7}_U=\bigg\lfloor\frac{7\times15}2\bigg\rfloor+\nu_2\binom{11}3+\bigg(2+\nu_2\big(\frac{15+1}2\big)\bigg)\cdot1+\nu_2(2)\cdot[\cP_2].
$$ By Kummer's rule \cite{Ku}, we have $\nu_2\binom{11}3=0$ because there are
no carries in the base-2 addition of $8=(1000)_2$ and
$3=(11)_2$. Also, $[\cP_2]=0$ because, of the three sub-conditions that
make up $\cP_2$, one is false, namely $2a=4\not=b+1=2$. Hence, 
$$
\nu_2\binom{22}{7}_U=52+0+(2+3)+0=57.
$$
\end{example}

\begin{example}\label{ex:3} Let $U=U(20,25)$ and let us evaluate the
  $5$-adic valuation of the Lucasnomial $\binom{25}{11}_U$. Here
  $m=14$, $n=11$, $p=5$, $a=1$, $b=2$ and $U'=U(4,1)$. Since $b=2a$, we must have 
$$
\nu_5\binom{25}{11}_U=amn+\nu_5\binom{25}{11}_{U'}=14\times11+\nu_5\binom{25}{11}_{U'}.
$$ The initial values of $U'$ are $0$, $1$, $4$ and $15$. Thus,
$\rho'$, the rank of $5$ in $U'$, is $3$. To compute
$\nu_5\binom{25}{11}_{U'}$, as explained in the introduction, 
we tally carries in the base-5 addition of
$m/\rho'$ and $n/\rho'$ \cite[Thm.\ 4.2]{Ba1}. Now 
\begin{eqnarray*}
\frac{14}{\rho'}&amp;=&amp;4+\frac23=(4)_5+\frac23,\\
\frac{11}{\rho'}&amp;=&amp;3+\frac23=(3)_5+\frac23,
\end{eqnarray*} so there are two carries, one across the radix point 
which accounts for $\nu_5(U_{\rho'})=\nu_5(15)=1$ in the valuation of
  $\nu_5\binom{25}{11}_{U'}$ and another since $1+4+3&gt;5$, which
  accounts for $1$. Hence, 
$$
\nu_5\binom{25}{11}_U=154+2=156.
$$
\end{example}

\section{Applications}
\label{sec:4}

 Given two natural numbers $r\ge2$ and $s\ge1$, one defines for all
 integers $t\ge1$ the generalized Fuss-Catalan numbers as
\begin{equation}\label{eq:7}
C_{r,s}(t)=\frac{s}{(r-1)t+s}\binom{rt+s-1}{t}. 
\end{equation} When $s=1$, we find the Fuss-Catalan numbers, and when in addition
$r=2$, we end up with the well-known Catalan numbers
$$
C(t)=\frac1{t+1}\binom{2t}{t},
$$ with their many occurrences and combinatorial interpretations
\cite{St}. The generalized Fuss-Catalan numbers possess at least one
combinatorial interpretation counting Raney sequences \cite[pp.\
359--363]{GKP}. So, in particular, they are integers. 

 Given a nondegenerate Lucas sequence $U=U(P,Q)$, the generalized Lucasnomial 
Fuss-Catalan numbers, $C_{U,r,s}(t)$, are defined in an
analogous manner \cite{Ba3}, i.e., 
\begin{equation}\label{eq:8}
C_{U,r,s}(t)=\frac{U_s}{U_{(r-1)t+s}}\binom{rt+s-1}{t}_U. 
\end{equation} In fact, this is more than a definition by analogy, since for the
Lucas sequence $U=U(2,1)$ we fall back on the generalized Fuss-Catalan
numbers defined in (\ref{eq:7}). Lucasnomial Fuss-Catalan numbers \cite{Ba3} correspond to the case $s=1$, while if $s=1$ and $r=2$, then we find numbers
considered much earlier, the Lucasnomial Catalan numbers
\cite{Ba2, Go1, Go2, SaSa}. Only very recently has a combinatorial
interpretation of the Lucasnomial Fuss-Catalan numbers
$$
C_{U,r,1}(t)=\frac1{U_{(r-1)t+1}}\binom{rt}{t}_U,
$$ been discovered \cite{Ben1, Ben2} for all $r\ge2$. 
 The Lucasnomial Fuss-Catalan numbers were shown to be always integers
 by an algebraic argument \cite[Thm.\ 6]{Ba3}. This proof did not seem
 to extend to generalized Lucasnomial Fuss-Catalan numbers. But,
 we were nearly able to prove their integrality via an arithmetic
 argument. That is, for all regular primes $p$, i.e., primes $p$ not
 dividing $\gcd(P,Q)$, the $p$-adic valuation of $C_{U,r,s}(t)$ is
 nonnegative \cite[Thm.\ 9 and Rmk.\ 10, p.\ 11]{Ba3}. Consequently, if
 $U(P,Q)$ is regular, i.e., if $\gcd(P,Q)=1$, then the generalized Lucasnomial Fuss-Catalan
 numbers $C_{U,r,s}(t)$ are already known to be integral. With the help of Theorem
 \ref{thm:1}, we are able to prove their integrality in all cases, by
 showing that if $p\mid\gcd(P,Q)$, then $\nu_p(C_{U,r,s}(t))\ge0$. 

\begin{theorem} \label{thm:2} The generalized Lucasnomial Fuss-Catalan
 numbers 
$$
C_{U,r,s}(t)=\frac{U_s}{U_{(r-1)t+s}}\binom{rt+s-1}{t}_U
$$ are integral for all nondegenerate fundamental Lucas sequences $U=U(P,Q)$, all
integers $r\ge2$, $s\ge1$ and $t\ge1$.
\end{theorem}
\begin{proof} If $s=1$, then $C_{U,r,s}(t)$ is a Lucasnomial Fuss-Catalan
  number which, as mentioned prior to stating the theorem, is known to be an integer \cite[Thm.\ 6]{Ba3}. Thus, our proof 
assumes $s\ge2$. Furthermore, we readily see that  
\begin{equation}\label{eq:9}
C_{U,r,s}(t)=\frac{U_s}{U_t}\binom{rt+s-1}{t-1}_U.
\end{equation} Since $U_t=1$ for $t=1$, we also assume $t\ge2$. 

  As mentioned earlier, by \cite[Thm.\ 9]{Ba3}, it suffices to show the $p$-adic valuation of
  $C_{U,r,s}(t)$ is nonnegative if $p\mid\gcd(P,Q)$. We continue using
  the notation of Theorem \ref{thm:1} and, in particular, the
  exponents of $p$ in $P$ and $Q$ are respectively denoted by $a$ and
  $b$. The letter $c$ denotes the constant $\nu_p(P'^2-Q')$. Using (\ref{eq:9}), we see that 
\begin{equation}\label{eq:10}
\nu_p(C_{U,r,s}(t))=\nu_p(U_s)-\nu_p(U_t)+\nu_p\binom{rt+s-1}{t-1}_U.
\end{equation} We will use Theorem \ref{thm:1} to evaluate the
$p$-adic valuation of the above Lucasnomial
$\binom{rt+s-1}{t-1}_U$. Thus, in the notation of Theorem \ref{thm:1},
\begin{equation}\label{eq:11}
m=(r-1)t+s\qquad\text{ and }\qquad n=t-1.
\end{equation}

\noindent{\bf Case I: $b&gt;2a$}.
Then, by equation (\ref{eq:0})
and Theorem \ref{thm:1} 
$$
\nu_p(C_{U,r,s}(t))=(s-1)a-(t-1)a+a((r-1)t+s)(t-1):=aN(s,t).
$$ This valuation is clearly positive since $(r-1)t+s\ge2+2=4$. Hence,
Case 1 is proved. 

\bigskip

\noindent{\bf Case II: $b=2a$.}  Since $U_k=p^{a(k-1)}U'_k$, we readily see, by (\ref{eq:10}), that 
\begin{align*} 
\nu_p(C_{U,r,s}(t))&amp;=aN(s,t)+\nu_p(U'_s)-\nu_p(U'_t)+\nu_p\binom{rt+s-1}{t-1}_{U'}\\
&amp;=aN(s,t)+\nu_p(C_{U',r,s}(t)),
\end{align*} where $N(s,t)$ was shown to be positive in Case 1. Since $p$ is a regular prime with respect to $U'$, we
know that $\nu_p(C_{U',r,s}(t))\ge0$ by \cite[Rmk.\ 10]{Ba3}. 

\noindent{\bf Case III: $b&lt;2a$.} 
  By equations (\ref{eq:0+}), Theorem \ref{thm:1} and (\ref{eq:10}), we see 
  that the quantity 
\begin{equation}\label{eq:12}
M(s,t):=\bigg\lfloor\frac{s}2\bigg\rfloor-\bigg\lfloor\frac{t}2\bigg\rfloor+
\bigg\lfloor\frac{((r-1)t+s)(t-1)}2\bigg\rfloor
\end{equation} will appear in the calculation. So 
we begin with inequalities true in all parity cases for $s$ and $t$. 
\begin{align*}
M(s,t)&amp;\ge\frac{s-1}{2}-\frac{t}2+\frac{((r-1)t+s)(t-1)-1}2\\
&amp;\ge\frac{s-t-1}2+\frac{t(t-1)}2+\frac{s(t-1)}2-\frac12\\
&amp;\ge\frac{s-t-1}2+\frac{t}2+(t-1)-\frac12\\
&amp;=\frac{s-1}2+t-\frac32\ge1,
\end{align*} since $s\ge2$ and $t\ge2$. In fact, we also have  
\begin{equation}\label{eq:13}
M(s,t)\ge\frac{s-1}2+t-\frac32\ge\frac{t}2+\frac{t}2-1\ge\frac{t}2.
\end{equation}
 We break up the argument into four subcases depending on the parity of $s$ and $t$. 

\bigskip
\noindent {\bf Subcase 1:} $s$ and $t$ are odd. Then we see by (\ref{eq:0+}) and Theorem
 \ref{thm:1} that 
$$
\nu_p(C_{U,r,s}(t))\ge b\cdot M(s,t)\ge b&gt;0.$$

\bigskip
\noindent{\bf Subcase 2:} $s$ is even and $t$ is odd. 
Again with the use of 
(\ref{eq:0+}) and Theorem \ref{thm:1}, we find that
$\nu_p(C_{U,r,s}(t))$ satisfies the lower bound 
$$
\nu_p(C_{U,r,s}(t))\ge b\cdot M(s,t)+(a-b)\ge a&gt;0.
$$

\bigskip
\noindent {\bf Subcase 3:} $s$ is odd and $t$ even. 
By (\ref{eq:0+}), Theorem \ref{thm:1}, and noting that $m=(r-1)t+s$ and $n=t-1$ are both odd, we obtain, using (\ref{eq:10}) again,   
\begin{align*}
\nu_p(C_{U,r,s}(t))&amp;\ge
bM(s,t)-(a-b)-\nu_p(t/2)-c\cdot[\cP_0]\cdot[p\mid t/2]\\
&amp;+\nu_p\binom{(m+n)/2}{(n-1)/2}+(a+\nu_p((m+1)/2))+c\cdot[\cP_0]
\cdot[r_m+r_n\ge2p]\\
&amp;=bM(s,t)+b+\nu_p\binom{(m+n)/2}{(n-1)/2}+\nu_p((m+1)/2)
-\nu_p((n+1)/2)\\
&amp;+c\cdot[\cP_0]\cdot\big([r_m+r_n\ge2p]-[p\mid t/2]\big)\\
&amp;\ge2b,
\end{align*} because $M(s,t)\ge1$ and 
$$
\nu_p\binom{(m+n)/2}{(n-1)/2}+\nu_p((m+1)/2)
-\nu_p((n+1)/2)\ge0,
$$ since this is the valuation of the binomial
coefficient
$$
\frac{(m+1)/2}{(n+1)/2}\binom{(m+n)/2}{(n-1)/2}=\binom{(m+n)/2}{(n+1)/2}.
$$ Also the quantity $H(s,t):=[\cP_0]\cdot\big([r_m+r_n\ge2p]-[p\mid
t/2]\big)$ is nonnegative. Indeed, if $H(s,t)&lt;0$, then $2\le p\le3$
and $p\mid t/2$. We are about to check these two conditions imply 
$r_m+r_n\ge2p$ so that $H(s,t)\ge0$, a contradiction. Assume first
$p=2$. Then $p\mid t/2$ implies $4\mid t$. But $n=t-1$ implies $r_n=3$
since $2p=4$. Since $m=(r-1)t+s$ and $s$ is odd,  $r_m=1$ or
$3$. Hence, $r_m+r_n\ge4=2p$. 

 Now suppose that $p=3$. Then $3\mid t/2$ implies $6\mid t$. Since $n=t-1$,
 the remainder, $r_n$, of the division of $n$ by $2p$, is $5$. But
 $m=(r-1)t+s$ so $m\equiv s\pmod6$. Now $s$ is odd so
 $r_m\not=0$. Hence, $r_m+r_n\ge6=2p$. 

\bigskip
\noindent{\bf Subcase 4:} $s$ and $t$ are both even. 
Thus $m$ is even and $n=t-1$ is odd. Using equations (\ref{eq:0+}) 
 and Theorem \ref{thm:1}, we see that 
\begin{align*}
\nu_p(C_{U,r,s}(t))&amp;\ge
bM(s,t)+\nu_p(s/2)-\nu_p(t/2)+\nu_p\binom{(m+n-1)/2}{(n-1)/2}\\
&amp;+c\cdot[\cP_0]\cdot\big([\,p\mid (s/2)\,]-[\,p\mid (t/2)\,]+
[r_m+r_n\ge2p]\big)\\
&amp;\ge bM(s,t)+\nu_p(s/2)-\nu_p(t/2)\ge M(s,t)-\nu_p(t/2).
\end{align*}
The second inequality above holds because the quantity 
$$
G(s,t):=[\cP_0]\cdot\big([p\mid s/2]-[p\mid t/2]+[r_m+r_n\ge2p]\big),
$$ is nonnegative. If not, $p=2$ or $3$, $p\mid t/2$ and $p\nmid
s/2$. But these conditions imply $r_m+r_n\ge2p$ so that $G(s,t)\ge0$. 
Indeed, if $p=2$, then $4\mid t$. Since $n=t-1$, $r_n=3$. Moreover,
$2\nmid s/2$ implies $s\equiv2\pmod4$. Now $m=(r-1)t+s$ so $r_m=2$. So
$r_m+r_n$ is 5 and exceeds $2p=4$. If $p=3$, then $6\mid t$ and
$r_n=5$. Moreover, $3\nmid s$ implies $r_m\not=0$. Again,
$r_m+r_n\ge2p=6$ as claimed. 

 Now, from (\ref{eq:13}), we know $M(s,t)\ge t/2$. Say
$\nu_p(t/2)=u$. If $u=0$, then $\nu_p(C_{U,r,s}(t))\ge M(s,t)&gt;0$. If
$u\ge1$, then  
$$
t/2\ge p^u\ge2^u\ge(1+1)^u=\sum_{k=0}^u\binom{u}{k}\ge1+\binom{u}{1}=1+u.
$$ Hence, $\nu_p(C_{U,r,s}(t))\ge t/2-u\ge1$. 
\end{proof}  

\begin{remark} Actually, if $p$ is special, then all generalized
  Lucasnomial Fuss-Catalan numbers $C_{U,r,s}(t)$ have a positive $p$-adic
  valuation if $st&gt;1$. Indeed, the proof of Theorem \ref{thm:1} 
  shows that for $s\ge2$ and $t\ge2$, $\nu_p(C_{U,r,s}(t))&gt;0$. If
  $t=1$ and $s\ge2$, then, from (\ref{eq:9}), we see that 
  $\nu_p(C_{U,r,s}(t))\ge\nu_p(U_s)$, and $\nu_p(U_s)\ge\lf s/2\rf&gt;0$
  by (\ref{eq:0}) or (\ref{eq:0+}). If $s=1$ and $t\ge2$, then, by \cite[eq.\ (14)]{Ba3}, 
  \begin{equation}\label{eq:14}
  C_{U,r,s}(t)=\binom{rt-1}{t-1}_U-Q\frac{U_{(r-1)t}}{U_t}\binom{rt-1}{t-2}_U.
  \end{equation} Noting that Theorem \ref{thm:1} implies that, for $m$ and $n$  positive, $\binom{m+n}{n}_U$ has positive valuation, the first Lucasnomial on 
  the RHS of (\ref{eq:14}) has positive $p$-valuation. Moreover, $U$ being a divisible sequence $U_{(r-1)t}/U_t$ is an integer and $\nu_p(Q)&gt;0$. Hence, the second term on the RHS of (\ref{eq:14}) also has positive valuation. 
\end{remark} 

 Generalized Lucasnomial Lobb numbers and generalized Lucasnomial
 ballot numbers were shown to be integers under the hypothesis that 
$U(P,Q)$ be regular \cite{Ba3}. The restriction that $U$ be regular
can now be removed. Thus, we have the two corollaries. 

\begin{corollary}\label{cor:1} The generalized Lucasnomial Lobb 
 numbers 
$$
L_{m,s}^{U,a}=\frac{U_{as+1}}{U_{(a-1)m+s+1}}\binom{rm}{(r-1)m+s}_U
$$ are integral for all nondegenerate fundamental Lucas sequences $U=U(P,Q)$, all
integers $a\ge1$ and $m&gt;s\ge0$.
\end{corollary}
\begin{proof} As shown in the proof of \cite[Thm.\ 12]{Ba3}, each Lobb
  number is a generalized Lucasnomial Fuss-Catalan number.
\end{proof}

\begin{corollary}\label{cor:2} The generalized Lucasnomial ballot  
 numbers 
$$
B_U(s,t)=\frac{U_{s-t}}{U_{s+t}}\binom{s+t}{t}_U
$$ are integral for all nondegenerate fundamental Lucas sequences
$U=U(P,Q)$ and all integers $s&gt;t\ge0$.
\end{corollary}
\begin{proof} Setting $v=s-t$ we see that 
\begin{align*}
B_U(s,t)&amp;=\frac{U_{s-t}}{U_{s+t}}\binom{s+t}{t}_U=\frac{U_{s-t}}{U_{t}}\binom{s+t-1}{t-1}_U\\
&amp;=\frac{U_{v}}{U_{t}}\binom{2t+v-1}{t-1}_U,
\end{align*} which, by (\ref{eq:9}), is the generalized Lucasnomial
Fuss-Catalan number $C_{U,2,v}(t)$. Thus 
the result follows from Theorem \ref{thm:1}. 
\end{proof}

\section{Acknowledgments}
\label{sec:5} 

 We thank the referee and the editor-in-chief for their
 formal recommendations which hopefully will have helped make this calculatory paper more reader-friendly. 


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B65; Secondary 11B39.

\noindent \emph{Keywords: } generalized binomial coefficient,
$p$-adic valuation, Lucas sequence, generalized Catalan number, Kummer's rule.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 3 2020;
revised versions received  December 12 2020; December 31 2020.
Published in {\it Journal of Integer Sequences},
December 31 2020.

\bigskip
\hrule
\bigskip

\noindent
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