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\begin{center}
\vskip 1cm{\LARGE\bf 
Lucasnomial Fuss-Catalan Numbers and \\
\vskip .1in
Related Divisibility Questions 
}
\vskip 1cm
\large
Christian Ballot\\
D\'epartement de Math\'ematiques et Informatique\\
Universit\'e de Caen-Normandie \\
France \\
\href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr} \\
\end{center}

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\begin{abstract} 
For all integers $n\ge1$ we define the generalized Lucasnomial
Fuss-Catalan numbers
$$C_{U,a,r}(n):=\frac{U_r}{U_{(a-1)n+r}}\binom{an+r-1}{n}_U,$$ and
prove their integrality.  Here $U$ is a fundamental Lucas sequence,
$a\ge2$ and $r\ge1$ are integers, and $\binom{*}{*}_U$ denotes a
Lucasnomial coefficient. If $U=I$, where $I_n=n$, then the
$C_{I,a,r}(n)$ are the usual generalized Fuss-Catalan numbers. With the
assumption that $U$ is regular, we show that $U_{(a-1)n+k}$ divides
$\binom{an}{n}_U$ for all $n\ge1$ but a set of asymptotic density $0$
if $k\ge1$, but only for a small set if $k\le0$. This small set is
finite when $U\not=I$ and at most of upper asymptotic density $1-\log
2$ when $U=I$.  We also determine
all triples $(U,a,k)$, where $k\ge2$, for which the
exceptional set of density $0$ is actually finite, and in fact empty.
\end{abstract}

\section{Introduction}

 The Catalan numbers $C_n$, which may be defined algebraically by the formula $\frac1{n+1}\binom{2n}{n}$, 
appear in all kinds of mathematical contexts and have numerous combinatorial interpretations. One may   
get seriously acquainted with them by consulting the recent book of Stanley \cite{St}. 
They admit several generalizations. Two of them are relevant to this paper.
First,
the generalized Fuss-Catalan numbers, defined for all integers $n\ge1$ as
\begin{equation}\label{eq:FC}
\frac{r}{(a-1)n+r}\binom{an+r-1}{n}, 
\end{equation} where $a\ge2$ and $r\ge1$ are fixed integers. 
Fuss-Catalan numbers, which correspond to $r=1$, are shown \cite[Exercise A14, pp.\ 108]{St} to have 
combinatorial interpretations that extend some of the interpretations 
for ordinary Catalan numbers. We note that Fuss-Catalan numbers are sometimes plainly called generalized Catalan 
numbers (e.g., \cite{Su}). The second generalization of interest to this paper are   
the Lucasnomial Catalan numbers 
\begin{equation}\label{eq:LC}
\frac{1}{U_{n+1}}\binom{2n}{n}_U,
\end{equation} where $U=(U_n)$ is a fundamental 
Lucas sequence and $\binom{2n}{n}_U$ is the generalized central binomial coefficient with respect to $U$. 
Both generalizations are known to yield integers only; see, e.g., \cite{Ba4, Ed}. In Section 2, the integrality of more 
general numbers is proved, namely the   
{\it generalized Lucasnomial Fuss-Catalan} numbers  
\begin{equation}\label{eq:3}
C_{U,a,r}(n):=\frac{U_r}{U_{(a-1)n+r}}\binom{an+r-1}{n}_U, 
\end{equation} where $U$ is a regular fundamental Lucas sequence, and $a\ge2$ and $r\ge1$ are given integers. The numbers 
\begin{equation}\label{eq:sfc}
C_{U,a,1}(n)=\frac{1}{U_{(a-1)n+1}}\binom{an}{n}_U,
\end{equation} where $r=1$ are simply referred to as {\it Lucasnomial Fuss-Catalan} numbers. 
Thus, ordinary Catalan and Fibonomial Catalan numbers which correspond, 
respectively, to $(U,a,r)=(I,2,1)$ and $(U,a,r)=(F,2,1)$ are particular instances of Lucasnomial  
Fuss-Catalan numbers. Throughout the paper, the letter $I$ denotes the identity sequence, i.e., $I_n=n$ for all 
$n\ge0$, while $F$ denotes the Fibonacci sequence defined by $F_{n+2}=F_{n+1}+F_n$, for all $n\ge0$, $F_0=0$ and 
$F_1=1$. 


 In Section 2, the integrality of various Lucasnomial generalizations of classical numbers is established. Indeed,  
Theorem \ref{thm:1} unconditionally establishes  
the integrality of all Lucasnomial Fuss-Catalan numbers, Theorem \ref{thm:2} proves 
the integrality of all generalized Lucasnomial Fuss-Catalan numbers with the restriction that $U$ be regular, Theorem 
\ref{thm:Lobb} deduces from Theorem \ref{thm:2} the integrality of the {\it generalized Lucasnomial Lobb} numbers 
$L_{m,s}^{U,a}$, defined by 
\begin{equation}\label{eq:Lobb}
L_{m,s}^{U,a}:=\frac{U_{as+1}}{U_{(a-1)m+s+1}}\binom{am}{(a-1)m+s}_U,
\end{equation} for $a\ge1$, $m&gt;s\ge0$, and, given the author's patronym we could hardly fail to show 
Theorem \ref{thm:ballot}, which establishes the integrality of all {\it Lucasnomial ballot} numbers $B_U(a,b)$, 
defined by 
\begin{equation}\label{eq:ballot}
B_U(a,b):=\frac{U_{a-b}}{U_{a+b}}\binom{a+b}{a}_U,
\end{equation} for all $a&gt;b\ge0$ integers.
 
 However, Section 2 apart, the rest of the paper is devoted to a variant of Lucasnomial Fuss-Catalan numbers, 
namely the numbers 
\begin{equation}\label{eq:k}
\frac1{U_{(a-1)n+k}}\binom{an}{n}_U,
\end{equation} where $k$ is a fixed integer. The investigation we launched into 
persues the research work conducted in two recent papers, one of Pomerance \cite{Po} 
and then one of the author \cite{Ba4}.   

 Given a triple $(U,a,k)$, we consider the set $D_{U,a,k}$ of all positive integers $n$ for which 
$U_{(a-1)n+k}$ divides the Lucasnomial $\binom{an}{n}_U$. When $n$ is in $D_{U,a,k}$, then the 
numbers in (\ref{eq:k}) are integers. When $k=1$, they are Lucasnomial Fuss-Catalan numbers and are 
integers for all $n\ge1$. Thus, $D_{U,a,1}=\mathbb N$. The main question the paper addresses 
is how does replacing $1$ by $k$ affects 
the sets $D_{U,a,k}$? How far astray do we get from the Catalan phenomenon? 

  In his clear and attractive paper \cite{Po}, Pomerance studied 
how often the middle binomial coefficients $\binom{2n}{n}$ are divisible by $n+k$, when $k$ is a fixed 
arbitrary integer. The enquiry brought out two chief phenomena: the singularity 
of the Catalan numbers and a drastic difference in behavior between the case $k\ge1$ and the case $k\le0$.  
Indeed, if $k\not=1$, then $\overline D_{I,2,k}$, the complementary set of $D_{I,2,k}$ in the positive 
integers, is infinite. That is, there are infinitely many $n$ for which $n+k$ does not divide $\binom{2n}{n}$. 
Secondly, if $k\ge1$, then $D_{I,2,k}$ has asymptotic density one, whereas if $k\le0$, although $D_{I,2,k}$ 
is infinite, its upper asymptotic density, whose value is still unknown 
--- see Section 3 --- is small and definitely less than $1/3$. 

 Before describing the content of the second paper \cite{Ba4}, we recall some notions. 

 A fundamental Lucas sequence $U=U(P,Q)$ is a binary linear recurrent sequence defined by the initial values $U_0=0$, $U_1=1$ 
and the recurrence 
\begin{equation}\label{eq:rec}
U_{n+2}=PU_{n+1}-QU_n,
\end{equation} for all integers $n\ge0$, where $P$ and $Q$ are nonzero integers. The discriminant $\Delta$ of 
$U(P,Q)$ is $P^2-4Q$, i.e., it is the discriminant of the characteristic polynomial associated with the recursion. 
 
 In stating our results we often use the notation $U(P,Q)$, with parameters $P$ and $Q$, to designate 
a Lucas sequence later simply referred to as $U$ and with terms $U_n$.
Note that 
all terms of a sequence $U$ are integers. With its initial conditions $U$ turns out to be a {\it divisibility}, or a {\it 
divisible} sequence, i.e., one that satisfies, for all positive integers $m$ and $n$, the property 
\begin{equation}\label{eq:div}
m\mid n\implies U_m\mid U_n.
\end{equation}
 
 A fundamental Lucas sequence is {\it nondegenerate} if $U_n\not=0$ for all $n\ge1$. 
Nondegeneracy occurs whenever the ratio of the zeros of $x^2-Px+Q$ is not a root of unity 
of order $\ge2$. Alternatively, the condition $U_{12}\not=0$ is necessary and sufficient to  
ensure the nondegeneracy of $U$ --- see, e.g., \cite[Section 2]{Ba1}. We sometimes use 
the abbreviation `NFLS', or `NFL-sequence', for a nondegenerate fundamental Lucas sequence. 

 A {\it regular} Lucas sequence $U(P,Q)$ is a NFLS such that $\gcd(P,Q)=1$.\footnote{The nondegeneracy condition simplifies 
to $U_3\not=0$, i.e., to not having $P^2=Q=1$ when $\gcd(P,Q)=1$.} 
This property is well-known 
to confer a stronger form of divisibility to the sequence $U$, i.e., for all nonnegative integers $m$ and $n$
\begin{equation}\label{eq:reg}
\gcd(U_m,U_n)=|U_{\gcd(m,n)}|.
\end{equation} Conversely, property (\ref{eq:reg}) is easily seen to imply $\gcd(P,Q)=1$. 

 A {\it Lucasnomial}, or a Lucasnomial coefficient, is a generalized binomial coefficient with 
respect to a nondegenerate fundamental Lucas sequence $U$. 
That is, for $m\ge n\ge1$, the Lucasnomial $\binom{m}{n}_U$ is 
defined as 
$$
\binom{m}{n}_U:=\frac{U_mU_{m-1}\cdots U_{m-n+1}}{U_nU_{n-1}\cdots U_1},
$$ and as $1$, if $m\ge0$ and $n=0$, and as $0$ otherwise. Thus, for instance,
$$
\binom{6}{3}_F=\frac{F_6\cdot F_5\cdot F_4}{F_3\cdot F_2\cdot F_1}=\frac{8\cdot5\cdot3}{2\cdot1\cdot1}=60.
$$ 
  
 The important Lucasnomial identity  
\begin{equation}\label{eq:2}
\binom{m}{n}_U=U_{n+1}\binom{m-1}{n}_U-QU_{m-n-1}\binom{m-1}{n-1}_U,\text{ for }m\ge n\ge1, 
\end{equation} easily follows from the Lucas identity \cite[Section 5]{Ba1}
$$
U_m=U_{n+1}U_{m-n}-QU_nU_{m-n-1}. 
$$ The integrality of all Lucasnomials may be seen inductively from (\ref{eq:2}). 


 There are two particular Lucas sequences $U(P,Q)$ for which the corresponding Lucasnomials have 
their own name. On the one hand, of course, 
the ordinary binomial coefficients $\binom{m}{n}=\binom{m}{n}_I$, for which $I=U(2,1)$ and $I_n=n$ for all $n\ge0$. 
On the other hand, we have the Fibonomials, $\binom{m}{n}_F$, derived from the much-studied Fibonacci 
sequence $F$ which is equal to $U(1,-1)$. Indeed, Fibonomials appeared earlier than general 
Lucasnomials in the mathematical literature.  


 In the sequel, we usually assume $P&gt;0$. Indeed, given $P$ and $Q$,  
$$
U(-P,Q)(n)=(-1)^{n-1}U(P,Q)(n), \text{ for all }n\ge0,
$$ so that 
Lucasnomials with respect to $U(-P,Q)$ are, up to sign, identical to corresponding Lucasnomials 
with respect to $U(P,Q)$. Therefore, the two sets $D_{U(\pm P,Q),a,k}$ are identical. 


 Lucasnomial Catalan numbers with respect to a NFLS $U$, defined in (\ref{eq:LC}), deserve their appellation  
as they are integral for all $n\ge 0$ \cite{Ba4, Ek, Go2, Sa2}. Gould \cite{Go1, Go2} might have been the first person to 
coin the term `Fibonomial Catalan number' for $U=F$ and to prove their integrality. To date and to our knowledge there is no 
combinatorial interpretation of Lucasnomial Catalan numbers for $U\not=I$, even though combinatorial interpretations of 
Lucasnomials themselves exist and have received attention  \cite{Be1, Be2, Sa1, Sa2}.   


 Because $I$ is a particular NFLS and because the Catalan phenomenon persists when we consider Lucasnomial 
Catalan numbers, it was natural to find out whether
\begin{enumerate}
\item The sets $D_{U,2,k}$ all missed infinitely many 
integers as soon as $k\not=1$, 

\item
The cleavage observed between the cases $k\ge1$ and $k\le0$ 
for $U=I$ remained true generally. 
\end{enumerate}

The answers \cite{Ba4} were essentially affirmative at least for all regular Lucas 
sequences if not for one exception with respect to question 1, namely the sequence $U(1,2)$. For this sequence,  
$D_{U,2,2}$ turns out to be the whole set of natural numbers. The behavioral dichotomy between $k\ge1$ and $k\le0$ is 
even sharper than for $U=I$. That is, if $\D\not=0$, then the sets $D_{U,2,k}$ are finite when $k\le0$, but  
still have asymptotic density $1$ when $k\ge1$. (The only two zero-discriminant regular Lucas sequences are $U(\pm2,1)$ 
and the only one with $P&gt;0$ is $U=I=U(2,1)$.)  


 This paper sets about discovering whether, for all regular Lucas sequences $U$ and all integers $k$, the divisor sets 
$D_{U,a,k}$ display the same features for $a&gt;2$ as for $a=2$. Section 3 is a study of the case $k\le0$: Theorem \ref{thm:3} 
shows the finiteness of all $D_{U,a,k}$ when $U$ is not $I$, while, in Theorem \ref{thm:I}, the sets $D_{I,a,k}$ 
are once more proven to be of asymptotic density less than $1/3$. That all sets $D_{I,a,k}$, $k\le0$, are infinite, 
is conjectured, but only proven, given an $a\ge3$, for infinitely many values of $k$ in Theorem \ref{thm:I2}. 
The section ends by discussing the current knowledge 
about the size of $D_{I,2,0}$, the set of integers $n\ge1$ that divide the middle binomial coefficient $\binom{2n}{n}$. 
It may seem like a simple case, yet with much food for thought leftover.  

 Section 4 establishes a number of lemmas and examines some examples that help to understand the direction we took   
and the methods we used, while 
speeding up the search carried out in Section 5. Indeed, Section 5 is devoted to the search for triples 
$(U,a,k)$, $k\ge2$, called {\it Catalan-like} triples, for which $D_{U,a,k}$ is the entire set of all natural numbers. We found 
exactly four new Catalan-like triples. Except for those five Catalan-like triples --- i.e., including the triple $\big(U(1,2),2,2\big)$ 
found in \cite{Ba4} --- the sets $D_{U,a,k}$, $k\ge2$,  all 
miss infinitely many positive integers. Theorem \ref{thm:G} summarizes the various partial results of Section 5. 

 The proof that all $D_{U,a,k}$ are of asymptotic density $1$ when $k\ge1$ is established in Section 6 and 
finalized in Theorem \ref{thm:4}. One of the findings that comes out of the proof of Theorem \ref{thm:4} is that for 
each triple $(U,a,k)$, $U$ regular, $a\ge2$ and $k\ge1$, there is a minimal integer $m\ge1$ such that 
\begin{equation}\label{eq:min}
\frac{m}{U_{(a-1)n+k}}\binom{an}{n}_U\quad\text{ is integral for all }n\ge1.
\end{equation} We may say that the triple $(U,a,k)$ {\it belongs} to the integer $m$. Thus, amongst all regular $U$, 
the triples $(U,a,1)$, $a\ge2$, 
and the five Catalan-like triples are {\it all} the triples that belong to $1$. The triple $(F,3,2)$ 
is seen to belong to $3$ as an outcome of the proof of Corollary \ref{cor:1} of Section 4. This observation raises 
numerous questions such as 

\medskip

\noindent Question 1. What are all triples $(U,a,k)$, $U$ regular, that belong to $3$? 

\medskip

\noindent Question 2. Are there triples that belong to $m$ for all $m\ge1$? If not, what is the set of $m$ not represented 
by any triple? 

\medskip

\noindent Question 3. Can one find general criteria that describe the triples that belong to a given $m$? 

\medskip

One might consider a further more general research problem which we
tentatively state below.

\begin{problem}\label{prob:1} Can one obtain even broader theorems using the {\it generalized}  
Lucasnomial Fuss-Catalan numbers defined in (\ref{eq:3})? Perhaps, the following problem, or a variant of it, would just 
do: Given a quadruple $(U,a,r,k)$, $a\ge2$, $r\ge1$, 
$k$ an integer, one could study the sets $D_{U,a,r,k}$ of integers $n\ge1$ such that 
\begin{equation}\label{eq:ext}
\frac{U_r}{U_{(a-1)n+k}}\binom{an+r-1}{n}_U,
\end{equation} is an integer. What density would these sets obey? Would one find a corresponding cleavage between 
the cases $k&lt;r$ and $k\ge r$? Could one find all Catalan-like quadruples $(U,a,r,k)$ for which the numbers in 
(\ref{eq:ext}) are always integers? 
\end{problem}


 Some explicit Kummer rule was devised \cite{Ed} for another whole class of generalized binomial coefficients, namely 
those generalized with respect to multiplicative arithmetic functions with range in the positive integers. Thus, one may 
consider a study comparable to ours in this context. Let us state the problem more precisely. 

\begin{problem}\label{prob:2} If $(u_n)_{n\ge1}$ is a sequence of positive integers such that 
\begin{eqnarray*}
u_{mn}&amp;=&amp;u_mu_n,\quad\text{ if }\gcd(m,n)=1, \qquad\text{(multiplicative)}\\
u_m&amp;\mid &amp; u_n,\quad\quad\;\text{ if }m\mid n, \quad\qquad\qquad\text{(divisible)}
\end{eqnarray*}
then the generalized Fuss-Catalan numbers with respect to $u$ 
$$\frac{u_r}{u_{(a-1)n+r}}\binom{an+r-1}{n}_u$$ are shown to 
be integers \cite{Ed}. Studying, say when $r=1$, the sets $D_{u,a,k}$ of integers $n\ge1$ for which $u_{(a-1)n+k}$ 
divides $\binom{an}{n}_u$, would be an interesting project. 
In particular and for instance, let $u_n=\varphi(n)$, where $\varphi$ is the Euler totient function, 
a sequence both multiplicative and divisible. Then for $k\not=1$, what can be said of the sets 
$D_{u,2,k}$ of integers $n\ge1$ such that 
$$
\frac{1}{u_{n+k}}\binom{2n}{n}_u\quad\text{ is an integer?}
$$  
\end{problem}
(Some authors \cite{Ed2, LS} have studied generalized binomial coefficients with respect to the Euler function $\varphi$.)


 The paper is mostly very elementary, but draws on three main sources for its proofs. 
One source is the classification \cite{Ab, Bi} of 
all $k$-defective regular Lucas sequences. A prime $p$ is said to be a {\it primitive prime divisor} (p.p.d.) of the $n$th term 
of a Lucas sequence $U$ if and only if $p\mid U_n$, but $p\nmid U_m$, $1&lt;m&lt;n$. Given $k\ge2$, a regular Lucas sequence $U(P,Q)$ 
of discriminant $\D\not=0$ is said to be $k$-{\it defective} if and only if $U_k$ does not have primitive prime divisors  
not dividing $\D$. For instance, $F$ is $6$-defective as $F_6=8$ and $F_3=2$. The prime $5$ is a p.p.d.\ of $F_5=5$, but 
$F$ is $5$-defective as the discriminant $\D_F=5$. The primitive divisor theorem \cite[Theorem 1.4]{Bi} asserts that for regular  
nonzero-discriminant Lucas sequences $U$, $U_n$ has a primitive prime divisor not dividing $\D$, for all $n&gt;30$. So 
there are no $k$-defective regular Lucas sequences $U$, $U\not=I$, for $k&gt;30$. Moreover, tables of all $k$-defective 
regular $U$ 
were made \cite{Bi} for all values of $k$, $1&lt;k&lt;31$. Some errors remained and were later corrected \cite{Ab}. 
The tables \cite{Ab, Bi} list Lucas sequences $U$ in terms of $P$ and $\D$. 
For convenience, and so the readers can better follow some of the arguments in Section 5, we added a table, Table A, in 
an appendix in Section 7.  Table A gathers together the various tables \cite[Tables 1 and 3, pp.\ 78--79]{Bi}, \cite[p.\ 312]{Ab}, 
but indexes Lucas sequences with the parameters $P$ and $Q$ instead. We note here that our arguments use primitive divisors so 
that primes dividing $\D$ may occasionally help to conclude a particular argument case. 

 Note that in Section 5, Proposition \ref{prop:3},  
we could not show the sets $D_{I,a,k}$, for $k\ge2$, missed infinitely many integers using the primitive divisor 
theorem or the tables of $k$-defective regular Lucas sequences, and we had to come up with specific arguments.   


 Another extensive source of proofs throughout the paper are the Kummer rules for determining 
the $p$-adic valuation of binomial coefficients or Lucasnomials when $p$ is a prime. Kummer's rule \cite{Ku} gives 
the $p$-adic valuation of the binomial coefficient $\binom{m+n}{n}$ as the number of carries in the addition of 
$m$ and $n$ performed in base $p$. Kummer-like theorems were obtained \cite{KW}  
for generalized binomial coefficients with respect to regular sequences of positive integers, i.e., 
sequences satisfying property (\ref{eq:reg}), and an explicit 
theorem for Fibonomial coefficients \cite[Theorem 2]{KW}. An explicit theorem for Lucasnomials with respect 
to a generic NFLS $U(P,Q)$ is given in \cite[Theorem 4.2]{Ba2} and repeated below in Proposition \ref{prop:K}.  
 
 As for the two preceding papers \cite{Ba4, Po}, a key point is that $n\in D_{U,a,k}$ if and only if for each prime $p$, 
the $p$-adic valuation of $\binom{an}{n}_U$ is at least that of $U_{(a-1)n+k}$. 


 The $p$-adic valuation of the terms $U_n$ (or $U_{(a-1)n+k}$) of a Lucas sequence has been known to obey certain rules 
since at least the time of Lucas \cite[Section XIII]{Lu}, but is being reproved every now and then in various guises (one of the 
latest appeared in a recent issue of the Fibonacci Quarterly \cite{San2}). 
The theory of Lucas 
sequences is the third main source of our proofs with which we assume some familiarity 
from the readers. Chapter 4 of the book \cite{Wi} can serve as a valuable introduction. Still we recall 
some basic facts for completeness' sake. The rank $\rho_U(p)$, or $\rho$, of a prime $p$ in 
a Lucas sequence $U(P,Q)$ is the smallest positive index $t$ such that $p$ divides $U_t$. It is guaranteed to exist 
when $p\nmid Q$. 

 By \cite[Eq.~(4.4) and Theorem 4.1, Section 4]{Ba2}, we next state a proposition that gives     
the $p$-adic valuation of all terms of a NFL-sequence $U(P,Q)$ for all primes $p\nmid Q$. The content 
of Proposition \ref{prop:AR} is what classically constitutes the Lucas laws of appearance and repetition 
for primes and prime powers in Lucas sequences. 

\begin{prop}\label{prop:AR} Let $U(P,Q)$ be a nondegenerate fundamental Lucas sequence and $p\nmid Q$ be 
a prime of rank $\rho$ in $U$. Then, for all positive integers $m$ and $n$, we have  
\begin{align*}
\rho&amp;\text{ divides } p-(\D\,|\,p),\quad\text{ if }p\text{ is odd},\\
&amp;p\mid U_n\;\text{ if and only if }\; \rho\mid n,\\
\nu_p(U_{m\rho})&amp;=\nu_p(m)+\nu_p(U_\rho)+\delta\cdot[2\mid m]:=x+\nu+\delta_x,
\end{align*} where $\D=P^2-4Q$, $(*\,|\,*)$ is the Legendre symbol, $x=\nu_p(m)$, $\nu=\nu_p(U_\rho)$,   
$$
\delta=\nu_2\big((P^2-3Q)/2\big)\cdot[p=2]\cdot[2\nmid PQ], 
$$ and $\d_x=\d\cdot[x&gt;0]$.
\end{prop}

 The notation $x$, $\nu$, $\delta$ and $\delta_x$ is used consistently with the utilization of Proposition \ref{prop:AR} 
throughout the paper. In Proposition \ref{prop:AR}, we used the Iverson symbol $[-]$, defined by: 
$$
[\cal P]=\begin{cases}1,&amp; \text{ if } \cal P \text{ is a true statement;}\\
0,&amp; \text{ if not.} 
\end{cases}
$$ 
 Here is the explicit Kummer rule obtained in \cite[Section 4]{Ba1}. 

\begin{prop}\label{prop:K} {\rm (Kummer's rule for Lucasnomials)}
Let $U(P,Q)$ be a nondegenerate Lucas sequence and $p\nmid Q$ be 
a prime of rank $\rho$ in $U$. Let $m$ and $n$ be two positive integers. 
Then the $p$-adic valuation of the Lucasnomial $\binom{m+n}{n}_U$ is equal to the number of carries 
that occur to the left of the radix point when $m/\rho$ and $n/\rho$ are added in base-$p$ notation, 
plus $\nu_p(U_\rho)$ if a carry occurs across the radix point, plus $\delta$ 
if a carry occurs from the first to the second digit to the left of the radix point, where $\d$ was 
defined in Proposition \ref{prop:AR}.   
\end{prop}

 This proposition suggests we distinguish carries across or to the left of the radix point from the other  
carries. Thus, as in \cite{Ba2, Ba4}, when adding $m/\rho+n/\rho$ in base $p$, we call a carry {\it relevant} 
when it occurs across or to the left of the radix point. 

 Here is an illustrative example of the use of Proposition \ref{prop:K} with $U(1,-5)$ in order to determine 
the $2$-adic valuation of $\binom{20}{5}_U$. We see that $U_3=6$ so that $\rho(2)=3$ and $\nu=1$. Also, 
$P^2-3Q=16$ so that $\d=3$. 
Now, using some self-evident base-$2$ writing, we find that 
\begin{eqnarray*}
\frac{15}{\rho}&amp;=&amp;5\;=\;(101)_2,\\
\frac{5}{\rho}&amp;=&amp;1+\frac23\;=\;(1.\overline{10}\strut^\infty)_2. 
\end{eqnarray*} Hence, there is only one relevant carry from first to second digit left of the radix point in the base-$2$ 
sum of $15/\rho$ and $5/\rho$. 
Thus, by our Kummerian rule, $\nu_2\binom{20}{5}_U=1+0+\d=4$. Using Proposition \ref{prop:AR} one can directly check that  
$$
\nu_2\binom{20}{5}_U=\nu_2\bigg(\frac{U_{20}\cdot U_{19}\cdot U_{18}\cdot U_{17}\cdot U_{16}}
{U_5\cdot U_4\cdot U_3\cdot U_2\cdot U_1}\bigg)=\nu_2\bigg(\frac{U_{18}}{U_3}\bigg). 
$$ However, $\nu_2(U_{18})=\nu_2(U_{3\cdot2^1\cdot\rho})=1+\nu_2(U_3)+\d$. 



 We won't always mention the Propositions \ref{prop:AR} and \ref{prop:K} when using them. 



 The letter $p$ invariably denotes a prime number except in Lemma \ref{lem:5}. 
Although we usually write the $p$-adic valuation of an integer $m$, i.e., the highest exponent $e\ge0$ of $p$ 
such that $p^e$ divides $m$, as $\nu_p(m)$, we omit the parentheses when $m$ is a Lucasnomial coefficient $\binom{\ell}{k}_U$ 
and write instead $\nu_p\binom{\ell}{k}_U$. We also occasionally use the alternative notation $p^e||m$. 



 Given a subset $S$ of the natural numbers, we write $S(z)$ for the elements of $S$ not exceeding $z$ and 
$\#S(z)$ for the cardinality of $S(z)$. We say $S$ has asymptotic density $d$ to mean that 
the ratios $\#S(z)/z$ tend to $d$ as $z$ tends to infinity. The sentence    
``almost all positive integers" means all but a set of asymptotic density $0$. The upper asymptotic density 
of $S$ is the upper limit of the ratios $\#S(z)/z$ as $z$ tends to infinity.    




 
\section{Integrality of Lucasnomial Fuss-Catalan numbers and other Lucasnomial extensions of classical numbers} 



 The forthcoming lemma leads to an unconditional algebraic proof of the integrality of  
Lucasnomial Fuss-Catalan numbers. As for the integrality of generalized Lucasnomial Fuss-Catalan numbers, 
we provide an arithmetic proof, alas conditional to the regularity of the Lucas sequence.  

\begin{lemma}\label{lem:1} Let $U$ be a nondegenerate fundamental Lucas sequence and $a\ge2$, $r\ge1$ 
be integers. 
Then the generalized Lucasnomial Fuss-Catalan numbers satisfy, for all $n\ge1$, the identity
\begin{equation}\label{eq:5}
C_{U,a,r}(n)=U_r\cdot\binom{an+r-2}{n-1}_U-Q\frac{U_rU_{(a-1)n+r-1}}{U_n}\cdot\binom{an+r-2}{n-2}_U.
\end{equation}
\end{lemma}
\begin{proof} Replacing $m$ by $an+r-1$ and $n$ by $(a-1)n+r-1$ in (\ref{eq:2}), we obtain 
$$
\binom{an+r-1}{(a-1)n+r-1}_U=U_{(a-1)n+r}\binom{an+r-2}{(a-1)n+r-1}_U-
Q U_{n-1}\binom{an+r-2}{(a-1)n+r-2}_U.
$$ That is, using the symmetry of the Lucasnomial triangle, 
$$
\binom{an+r-1}{n}_U=U_{(a-1)n+r}\binom{an+r-2}{n-1}_U-
Q U_{n-1}\binom{an+r-2}{n}_U.
$$
Multiplying through by $U_r/U_{(a-1)n+r}$, we find that 
$$
C_{U,a,r}(n)=U_r\binom{an+r-2}{n-1}_U-Q\frac{U_rU_{n-1}}{U_{(a-1)n+r}}\binom{an+r-2}{n}_U. 
$$ But 
\begin{eqnarray*}
\frac{U_{n-1}}{U_{(a-1)n+r}}\binom{an+r-2}{n}_U&amp;=&amp;\frac{U_{n-1}}{U_{(a-1)n+r}}\cdot\frac{U_{an+r-2}\cdots U_{(a-1)n+r-1}}{U_n\cdots U_1}\\
&amp;=&amp;\frac{U_{(a-1)n+r-1}}{U_n}\cdot\frac{U_{an+r-2}\cdots U_{(a-1)n+r+1}}{U_{n-2}\cdots U_1}\\
&amp;=&amp;\frac{U_{(a-1)n+r-1}}{U_n}\cdot\binom{an+r-2}{n-2}_U, 
\end{eqnarray*} which yields equation (\ref{eq:5}). 
\end{proof} 

\begin{theorem}\label{thm:1} Let $U$ be a nondegenerate fundamental Lucas sequence and $a\ge2$ be an integer. Then 
the Lucasnomial Fuss-Catalan numbers  
$$
C_{U,a,1}(n)=\frac{1}{U_{(a-1)n+1}}\binom{an}{n}_U \quad\text{ are integers for all }n\ge1. 
$$
\end{theorem} 
\begin{proof} As $U_n$ divides $U_{(a-1)n}$, setting $r=1$ in identity (\ref{eq:5}) we readily see that 
$C_{U,a,1}(n)$ is an integer. 
\end{proof}

 We will use the next small lemma a few times. 

\begin{lemma}\label{lem:notQ} Let $U(P,Q)$ be a fundamental Lucas sequence  
and $p$ be a prime. If $p\nmid\gcd(P,Q)$, then either $p\nmid Q$, or $p$ does not divide any term $U_n$, $n\ge1$, i.e., $p$ 
has no rank. 
\end{lemma}
\begin{proof} Suppose $p\mid Q$. Then $p\nmid P$. Using (\ref{eq:rec}) inductively, one finds  
$U_n\equiv P^{n-1}\pmod p$ for all $n\ge1$. Therefore, $p\nmid U_n$. 
\end{proof}


 Here is a straightforward observation about generalized Lucasnomial Fuss-Catalan numbers. 

\begin{remark} Ordinary Catalan numbers $C_{I,2,1}(n)=\frac1{n+1}\binom{2n}{n}$ have the well-known 
alternative representations 
$$
\frac{1}{2n+1}\binom{2n+1}{n}\quad\text{ and }\quad\frac{1}{n}\binom{2n}{n-1},
$$ which carry over to generalized Lucasnomial Fuss-Catalan numbers. That is, 
\begin{equation}\label{eq:4}
C_{U,a,r}(n)=\frac{U_r}{U_{an+r}}\binom{an+r}{n}_U=\frac{U_r}{U_n}\binom{an+r-1}{n-1}_U.
\end{equation}
\end{remark}

  Using the last form of $C_{U,a,r}(n)$ in (\ref{eq:4}), we produce 
an arithmetic proof of the integrality of $C_{U,a,r}(n)$. However, it assumes the regularity of $U$. 


\begin{theorem}\label{thm:2} Let $U(P,Q)$ be a regular fundamental Lucas sequence. Let $r\ge1$ and $a\ge2$ be integers. Then 
the generalized Lucasnomial Fuss-Catalan numbers
$$
\frac{U_r}{U_{(a-1)n+r}}\binom{an+r-1}{n}_U \quad\text{ are integral for all }n\ge1. 
$$
\end{theorem}
\begin{proof} By (\ref{eq:4}), 
$$
C_{U,a,r}(n)=\frac{U_r}{U_n}\binom{an+r-1}{n-1}_U.
$$ 
We claim that for all primes $p$ the $p$-adic valuation of $U_r\binom{an+r-1}{n-1}_U$ is 
at least that of $U_n$. If $p\mid Q$, then, by Lemma \ref{lem:notQ}, $p\nmid U_n$ and our claim holds.  
Now suppose $p\nmid Q$ and $p$ divides $U_n$. Then, 
by Proposition \ref{prop:AR}, 
$n=\l\rho p^x$ for some $x\ge0$ and some $\l$ 
prime to $p$, where $\rho$ is the 
rank of $p$, and $\nu_p(U_n)=x+\nu+\d_x$. Let us divide $r$ by $\rho$ and write 
$r=q\rho+r_1$ with $0\le q$ and $0\le r_1&lt;\rho$. Then 
\begin{eqnarray*} 
\frac{n-1}{\rho}&amp;=&amp;\l p^x-\frac{1}{\rho}=(\l p^x-1)+\frac{\rho-1}{\rho},\\
\frac{(a-1)n+r}{\rho}&amp;=&amp;\l(a-1)p^x+q+\frac{r_1}{\rho}. 
\end{eqnarray*} If the sum of the fractional parts 
of $(n-1)/\rho$ and $((a-1)n+r)/\rho$ is at least $1$, then a carry occurs across the radix point which 
ensures, independently of the value of $q$, a minimum of $x$ carries left of the radix point in 
the base-$p$ addition of $(n-1)/\rho$ and $((a-1)n+r)/\rho$. Indeed, the first $x$  
base-$p$ digits of $(n-1)/\rho$ left of the radix point are all $p-1$. Therefore, by Kummer's rule for Lucasnomials,   
$\nu_p\binom{an+r-1}{n-1}_U\ge x+\nu+\d_x=\nu_p(U_n)$. 

 If the fractional parts add up to less than $1$, 
i.e., if $(\rho-1)+r_1&lt;\rho$, then $r_1=0$ and $r=q\rho$. 
Let $p^i$, $i\ge0$, be the exact $p$-power dividing $q$. Then $p^{i+\nu+\d_i}||U_r$. 
If $i\ge x$ then $p^{x+\nu+\d_x}$ divides $U_r$ and our claim holds. If $i&lt;x$, then 
the base-$p$ addition of $p^x-1$ to $q$ produces $x-i$ carries. Note that, using the Iverson symbol, $\d_x=\d_i+\d_x\cdot[i=0]$.  
Hence, 
$$
\nu_p(U_r)+\nu_p\binom{an+r-1}{n-1}_U\ge (i+\nu+\d_i)+(x-i+\d\cdot[i=0]\cdot[x&gt;0])=x+\nu+\d_x=\nu_p(U_n).
$$ Thus, $C_{U,a,r}(n)$ is always an integer. 
\end{proof}

\begin{remark}\label{rem:0} We could have stated a stronger theorem by dropping the regularity assumption on 
$U(P,Q)$ and instead stating that for all primes $p\nmid\gcd(P,Q)$, the $p$-adic valuation of the 
numbers $C_{U,a,r}(n)$ is nonnegative, for all $n\ge1$. 
\end{remark} 

 We do not resist giving another related theorem with a different simple proof technique 
that partially impliesÂ&nbsp; Theorem \ref{thm:1} and Theorem \ref{thm:2}. 

\begin{theorem}\label{thm:2bis} Let $U(P,Q)$ be a regular fundamental Lucas sequence. Let $r\ge1$ and $m\ge n\ge1$ be integers 
with $\gcd(m+r,n)=1$. Then 
the generalized Lucasnomial Fuss-Catalan numbers
$$
T_{m,n,r}:=\frac{U_r}{U_{m+r}}\binom{m+n+r-1}{n}_U \quad\text{ are integers.} 
$$  
\end{theorem}
\begin{proof} We see that $U_{m+r}T_{m,n,r}$ is an integer and that 
$$
U_nT_{m,n,r}=U_r\binom{m+n+r-1}{n-1}_U, 
$$ is also integral. Since $U$ is regular and $\gcd(m+r,n)=1$, we find that $\gcd(U_{m+r},U_n)=1$ and immediately infer 
that $T_{m,n,r}$ is the greatest common divisor of the two integers $U_{m+r}T_{m,n,r}$ and $U_nT_{m,n,r}$. Hence, $T_{m,n,r}$ 
is an integer. 
\end{proof}

 Putting $m=(a-1)n$ in $T_{m,n,r}$, we see that Theorem \ref{thm:2} holds albeit with the restriction that 
$\gcd(r,n)=1$. If $r=1$, then we obtain Theorem \ref{thm:1} though only for $U$ regular. 



 Lobb numbers $L_{m,s}=L_{m,s}^2$ and generalized Lobb numbers $L_{m,s}^a$ are defined by the expression
$$
L_{m,s}^a:=\frac{as+1}{(a-1)m+s+1}\binom{am}{(a-1)m+s};
$$ see \cite[Eq.~(11)]{Ch}.   

 The integrality of generalized Lucasnomial Lobb numbers (\ref{eq:Lobb}) which are a natural 
Lucasnomial extension of the   
generalized Lobb numbers is the object of the next theorem. 

\begin{theorem}\label{thm:Lobb} If $U$ is a regular Lucas sequence, $a\ge1$, $m&gt;s\ge0$ are integers, then 
the generalized Lucasnomial Lobb numbers
$$ 
L_{m,s}^{U,a}=\frac{U_{as+1}}{U_{(a-1)m+s+1}}\binom{am}{(a-1)m+s}_U
$$ are integers. 
\end{theorem}
\begin{proof} Putting $r=as+1$ and $n=m-s$ in the expression (\ref{eq:3}) for $C_{U,a,r}(n)$ yields $L_{m,s}^{U,a}$, 
so that, by Theorem \ref{thm:2}, the numbers $L_{m,s}^{U,a}$ are integral. Indeed, 
\begin{eqnarray*}
C_{U,a,r}(n)&amp;=&amp;\frac{U_{as+1}}{U_{(a-1)(m-s)+as+1}}\cdot\binom{a(m-s)+(as+1)-1}{m-s}_U\\
&amp;=&amp;\frac{U_{as+1}}{U_{(a-1)m+s+1}}\cdot\binom{am}{m-s}_U\\
&amp;=&amp;\frac{U_{as+1}}{U_{(a-1)m+s+1}}\cdot\binom{am}{(a-1)m+s}_U=L_{m,s}^{U,a}. 
\end{eqnarray*}
\end{proof}

 Similarly, the natural Lucasnomial extension of ballot numbers is shown to always yield integers. 
Ordinary ballot numbers are defined by $B(m,n)=\frac{m-n}{m+n}\binom{m+n}{n}$; see (14) in \cite{Ch}. 
 
\begin{theorem}\label{thm:ballot} If $U$ is a regular Lucas sequence, $m&gt;n\ge0$ are integers, then 
the Lucasnomial ballot numbers
$$
B_U(m,n)=\frac{U_{m-n}}{U_{m+n}}\binom{m+n}{n}_U
$$ are integers. 
\end{theorem}
\begin{proof} If $m$ and $n$ have different parities, then replacing $m$ by $(m+n-1)/2$ and $s$ by $(m-n-1)/2$ in 
$L_{m,s}^{U,2}$ precisely yields $B_U(m,n)$. 

 However, to obtain a proof for all $m&gt;n\ge0$, we first notice that 
\begin{equation}\label{eq:ballot2}
B_U(m,n)=\frac{U_{m-n}}{U_{m+n}}\binom{m+n}{n}_U=\frac{U_{m-n}}{U_n}\binom{m+n-1}{n-1}_U. 
\end{equation} So, if $B_U(m,n)$ is a nonintegral rational number, there must exist a prime number $p$ with respect to 
which $B_U(m,n)$ has negative $p$-adic valuation. In particular, $p$ divides 
both $U_{m+n}$ and $U_n$. By Lemma \ref{lem:notQ}, $p\nmid Q$. Thus, $m=\mu p^y\rho$ and $n=\eta p^z\rho$ with 
$p\nmid\mu\eta$, where $\rho$ is the rank of $p$. If $y\not=z$, then $\nu_p(m-n)=\nu_p(m+n)$ and so 
$\nu_p(U_{m-n})=\nu_p(U_{m+n})$, which would contradict the negativity of $\nu_p(B_U(m,n))$. 
Therefore, $y=z$ and $\nu_p(U_n)=z+\nu+\d_z\le\nu_p(U_{m-n})$, because $m-n=(\mu-\eta)p^z\rho$. 
But, by the last expression in (\ref{eq:ballot2}), this contradicts the negativity of the $p$-adic valuation of $B_U(m,n)$. 
\end{proof}


\section{The smallness of $D_{U,a,k}$ when $k\le0$}


 We provide two main theorems to assess the smallness of $D_{U,a,k}$ when $k\le0$ and $U(P,Q)$ is regular.  
The first theorem treats sequences $U$ with $P^2\not=4Q$, while the second 
treats the zero-discriminant case.  

 Our first theorem extends \cite[Theorem 5.1]{Ba4}, which treated the case $a=2$, to all values of $a\ge2$.     
 

\begin{theorem}\label{thm:3} Suppose $U(P,Q)$ is a regular Lucas sequence with $P^2-4Q\not=0$.  
Assume $a\ge2$ and $k\ge0$ are fixed integers. Then there are at most finitely many integers $n\ge1$ such that 
$U_{(a-1)n-k}$ divides $\binom{an}{n}_U$. 
\end{theorem}
\begin{proof} Since the case $a=2$ 
corresponds to \cite[Theorem 5.1]{Ba4}, we may assume $a\ge3$. Put 
$$
M:=\text{ max }\bigg\{\frac{2k}{a-2}, \frac{30+k}{a-1}\bigg\}.
$$ Suppose $n&gt;M$. Since $n$ is larger than $(30+k)/(a-1)$, the primitive divisor theorem 
\cite[Theorem 1.4, p.\ 80]{Bi} 
ensures that $U_{(a-1)n-k}$ has a primitive prime divisor, say $p$. That is, $\rho=\rho_U(p)=(a-1)n-k$. 
Thus, we find that   
\begin{eqnarray*}
\frac{n}{\rho}&amp;=&amp;0+\frac{n}{(a-1)n-k},\\
\frac{(a-1)n}{\rho}&amp;=&amp;\frac{(a-1)n}{(a-1)n-k}=1+\frac{k}{(a-1)n-k}. 
\end{eqnarray*} Observe that $n&gt;2k/(a-2)$ implies that $(a-1)n-k&gt;0$ and that $(n+k)/((a-1)n-k)&lt;1$. Thus, we see 
that $n/((a-1)n-k)$ and $k/((a-1)n-k)$ are the fractional parts of, respectively, $n/\rho$ 
and $(a-1)n/\rho$, and they add up to less than $1$.   
Hence, in the base-$p$ addition of $n/\rho$ and $(a-1)n/\rho$, 
there is no relevant carry as $0+1&lt;p$. By Kummer's rule for Lucasnomials, 
$p$ does not divide $\binom{an}{n}_U$. 
Therefore, for all $n&gt;M$, we find that $U_{(a-1)n-k}$ does not divide $\binom{an}{n}_U$. 
\end{proof}

 We now look at the case of regular Lucas sequences with null discriminant, which essentially corresponds 
to $U_n=I_n$ and ordinary binomial coefficients. Theorem \ref{thm:I} below extends \cite[Theorem 3]{Po} 
that covered the case $a=2$. 

\begin{theorem}\label{thm:I} Suppose $U(P,Q)$ is a regular Lucas sequence with $P^2-4Q=0$, i.e., 
$U_n=n$ for all $n\ge1$, or $U_n=(-1)^{n-1}n$ for all $n\ge1$.  
Assume $a\ge2$ and $k\ge0$ are fixed integers. Then the upper asymptotic density of the set of integers $n\ge1$ 
such that $U_{(a-1)n-k}$ divides $\binom{an}{n}_U$ is at most $1-\log 2$. 
\end{theorem}
\begin{proof} It suffices to consider the case $U=I$. The proof we 
give is an adaptation of the proof of \cite[Theorem 3]{Po}. Moreover, 
this adapted proof yields the same upper bound of $1-\log2$ for the 
upper asymptotic density of $D_{I,a,k}$, $k\le0$, as the one obtained when $a=2$.
We begin by observing that if $(a-1)n-k$ has a prime factor $p&gt;\sqrt{2(a-1)n}$ and $p&gt;ak$, then 
$\nu_p((a-1)n-k)&gt;\nu_p\binom{an}{n}$ so that $n\not\in D_{I,a,-k}$. Indeed, say $(a-1)n-k=cp$. Then 
$$
c\le\frac{(a-1)n}{p}&lt;\frac{(a-1)n}{\sqrt{2(a-1)n}}=\frac{\sqrt{2(a-1)n}}{2}&lt;\frac{p}{2}.
$$ Hence, $(a-1)n=cp+k$ with $c&lt;p/2$ and $k&lt;p/a&lt;p$. Dividing $c$ by $a-1$, we write $c=q(a-1)+r$ with $0\le r\le a-2$. 
Thus, $n=qp+(rp+k)/(a-1)$. Noting that $q\le c&lt;p/2$ and that 
$$
\frac{rp+k}{a-1}+k=\frac{rp+ak}{a-1}&lt;\frac{(r+1)p}{a-1}\le p,
$$ we find that no carry occurs in the base-$p$ addition of $(a-1)n$ and $n$. 
By the rule of Kummer, $p\nmid\binom{an}{n}$. 

 We now fix a sufficiently large $z&gt;0$ and estimate the number of $n$ in $(ak^2,z]$ that have a prime factor 
$p&gt;\sqrt{2(a-1)z}$. The lower bound $ak^2$ for $n$ is sufficient to imply $p&gt;ak$. In the interval $(ak^2,z]$, there 
are $z/p+O(1)$ multiples of a prime $p&gt;\sqrt{2(a-1)z}$.  
As $p&gt;\sqrt{z}\ge\sqrt{n}$, no integer $n$ may have two such prime factors. Hence, we find that 
$$
\#{\overline D_{I,a,-k}}(z)\ge\sum_{\sqrt{2(a-1)z}&lt;p\le z}\frac{z}{p}+O(\pi(z)),
$$ which using Mertens' estimate $\sum_{p\le z}\frac1p=\log\log z+M+o(1)$, where $M$ is a small constant, leads to 
$$
\#{\overline D_{I,a,-k}}(z)\ge z\big(\log\log z-\big(\log\frac12+\log\log(2(a-1)z)\big)\big)+o(z)\ge z\log2+o(z).
$$ This readily implies the upper density of $D_{I,a,-k}$ is bounded above by $1-\log 2$.
\end{proof}

 We conjecture that the set $D_{I,a,-k}$ is nevertheless always infinite. Given $a\ge2$, we only prove the infinitude 
of $D_{I,a,-k}$ for infinitely many values of $k$ all multiples of $a-1$. For this, we extend an idea found in 
\cite[Theorems 2.2 and 3.2]{Ul} and also in \cite{Po}, where the infinitely many integers $n$ of the form $pq+k$, 
where $p$ and $q$ are both primes, 
$p&gt;k$ and $3p/2&lt;q&lt;2p$, are shown to belong to $D_{I,2,-k}$.  

\begin{theorem}\label{thm:I2} Let $a\ge2$ and $m\ge0$ be integers with $\gcd(a-1,m+1)=1$. Put $k=m(a-1)$. 
Then the set $D_{I,a,-k}$ is infinite. Furthermore, there is a positive 
constant $c_a$ such that for all large enough $z$
$$
\#D_{I,a,-k}(z)\ge\frac{c_az}{\log^2 z}.
$$
\end{theorem}
\begin{proof} For $a\ge2$ and $k=m(a-1)$, we consider integers $n$ of the form $pq+m$, where $p$ and $q$ are primes, $p&gt;k$, and 
\begin{equation}\label{eq:+}
q=p+d\quad\text{ with }\quad\frac{p}{a}&lt;d&lt;\frac{p}{a-1}. 
\end{equation} Therefore, $(a-1)n-k=(a-1)pq$. Since $n=qp+m=p^2+dp+m$ and $(a-1)n=(a-1)p^2+(a-1)dp+k$, we see 
that the base-$p$ addition of $n$ and $(a-1)n$ produces at least one carry. Similarly, $n=pq+m$ and 
$(a-1)n=(a-1)pq+k=(a-2)q^2+d_qq+k$ with $q-p&lt;d_q&lt;q$. Indeed, $(a-1)p=(a-1)(q-d)&gt;(a-1)(q-p/(a-1))=(a-2)q+(q-p)$ 
and $(a-1)p&lt;(a-1)q&lt;(a-2)q+q$. Thus, the base-$q$ addition of $n$ and $(a-1)n$ produces at least one carry. 
Hence, $pq$ divides $\binom{an}{n}$. From the lemmas of Section 6, one has that $a-1$ divides $\binom{an}{n}$ 
for almost all $n$. However, here, the integers $n$ we consider have a special form. 

 So, in addition to $p$ and $q$ satisfying (\ref{eq:+}), we further 
 impose the condition that $p\equiv m+1\pmod{(a-1)^2}$ 
and $q\equiv-1\pmod{(a-1)^2}$. Suppose $r$ is a prime factor of $a-1$ and $a-1=r^\a\l$, where $r$ and $\l$ 
are coprime. Then $p\equiv m+1\pmod{r^{2\a}}$ and $q\equiv-1\pmod{r^{2\a}}$. Thus, $pq+m\equiv-1\pmod{r^{2\a}}$, 
which says that the $r$-ary expansion of $n$ terminates with $2\a$ digits all $r-1$.  
Furthermore, the base-$r$ expansion of $(a-1)n=r^\a\cdot\l(pq+m)$ ends with a nonzero digit followed 
by $\a$ zero digits. Therefore, the base-$r$ addition of $n$ and $(a-1)n$ generates a minimum of $\a$ carries. 
By the Kummer rule, $r^{\a}$ divides $\binom{an}{n}$. As it is true of all prime factors $r$ of $a-1$, 
$a-1$ divides $\binom{an}{n}$. 
We conclude that for all such integers $n$, $(a-1)pq=(a-1)n-k$ divides $\binom{an}{n}$. 

Therefore,  for all large enough real numbers $z\ge1$, we find that
\begin{equation*}
\#D_{I,a,-k}(z)\ge\sum_{k&lt;p\le\sqrt{z}/2}S_p\ge S_z:=\sum_{\sqrt{z}/3&lt;p\le\sqrt{z}/2}S_p,
\end{equation*} where the sum 
$$
S_p:=\sum_{\frac{a+1}ap&lt;q&lt;\frac{a}{a-1}p}1,\;\text{ is taken over all primes }q\equiv-1\pmod{(a-1)^2}, 
$$ and the primes $p$ satisfy $p\equiv m+1\pmod{(a-1)^2}$.  

 Using the prime number theorem for primes in arithmetic progressions and the fact that $a/(a-1)$ is larger than $(a+1)/a$, 
each inner sum $S_p$ in $S_z$ is seen to be at least $c_1\sqrt{z}/\log z$, for some positive constant $c_1$. Indeed, if $\pi(x;\;a,\;b)$ 
denotes the number of primes $p\le x$ with $p\equiv a\pmod{b}$, then, as $z\to\infty$, 
\begin{eqnarray*}
S_p&amp;=&amp;\pi\bigg(\frac{a}{a-1}p;\;-1,\;(a-1)^2\bigg)-\pi\bigg(\frac{a+1}{a}p;\;-1,\;(a-1)^2\bigg)\\
&amp;=&amp;\frac1{\varphi\big((a-1)^2\big)}\frac{p}{\log p}\bigg(\frac{a}{a-1}\big(1+o(1)\big)-\frac{a+1}{a}\big(1+o(1)\big)\bigg)\\
&amp;\sim&amp; c\frac{p}{\log p}\ge\frac23c\frac{\sqrt{z}}{\log z}\big(1+o(1)\big), 
\end{eqnarray*} where $c=\big(\varphi((a-1)^2)a(a-1)\big)^{-1}$, $\varphi$ is the Euler totient function and where,  
in the last inequality, we lavishly used $p&gt;\sqrt{z}/3$ and $\log p\le\log(\sqrt{z}/2)$. 

 Thus, as $z$ tends to infinity, $S_z$ is at least equivalent to 
$$
\frac23c\frac{\sqrt{z}}{\log z}\sum_{\frac{\sqrt{z}}3&lt;p\le\frac{\sqrt{z}}2}1,
$$ where the sum is over primes $p\equiv m+1\pmod{(a-1)^2}$. 
Now, using again the prime number theorem for primes in arithmetic progressions, we see that, as $z$ tends to infinity,  
\begin{eqnarray*}
\sum_{\frac{\sqrt{z}}3&lt;p\le\frac{\sqrt{z}}2}1&amp;=&amp;\pi\bigg(\frac{\sqrt{z}}{2};\;m+1,\;(a-1)^2\bigg)-\pi\bigg
(\frac{\sqrt{z}}{3};\;m+1,\;(a-1)^2\bigg)\\
&amp;=&amp;\frac1{\varphi\big((a-1)^2\big)}\bigg(\frac{\sqrt{z}}{\log z}\big(1+o(1)\big)-\frac23\frac{\sqrt{z}}{\log z}\big(1+o(1)\big)\bigg)\\
&amp;=&amp;\frac1{3\varphi\big((a-1)^2\big)}\frac{\sqrt{z}}{\log z}\big(1+o(1)\big).
\end{eqnarray*}
Hence, we obtain that, as $z\to\infty$, $\#D_{I,a,-k}(z)$ is at least asymptotically equivalent to $c_az/(\log z)^2$, 
where 
$$
c_a=\frac2{9a(a-1)\big(\varphi(a^2-2a+1)\big)^2}.
$$ 
\end{proof} 

 To conclude the section, the bound of $1-\log2$ found in Theorem \ref{thm:I} being less than $1/3$ amply 
demonstrates that the cleavage observed for ordinary binomial coefficients by Pomerance \cite{Po} between the cases $k\le0$ and 
$k\ge1$ persists for all $a\ge2$ and, by Theorem \ref{thm:3}, for all regular Lucas sequences $U$. However, 
a smaller upper asymptotic density for $D_{I,2,0}$ was sought after by Sanna \cite{San}, who successfully 
brought it down from $1-\log 2$ to $1-\log 2-0.05551$. Recall that, as a consequence of \cite[Theorem 4]{Po}, 
the lower and upper densities of $D_{I,2,k}$ are identical for all $k\le0$, so we may as well assume $k=0$. 
Actually, sequence \seqnum{A014847} 
in the OEIS \cite{Sl} is an enumeration of the set $D_{I,2,0}$, and, in 2002, Cloitre observed on numerical evidence 
that it seemed that the quotient of the $n$th term of this sequence over $n$ tended to a limit between 9 and 10. 
If that were true this would indicate a density between $1/10$ and $1/9$. According to the data in \cite[Table 5]{Ul}, 
$\#D_{I,2,0}(2^{26})=8{,}225{,}813$ which yields a quotient $\#D_{I,2,0}(x)/x$ of about $0.1226$ for $x=2^{26}$. 
The existence of a positive lower asymptotic density 
for $D_{I,2,0}$ was conjectured by Pomerance \cite[bottom of page 7]{Po}. At the West Coast Number Theory Conference 
of 2016, Pomerance asked whether this set has 
a positive lower density and whether it has a density, and, on that occasion, St\u anic\u a 
\cite[Problem 016:04]{My} conjectured 
that, for $z\ge3700$, we actually would have   
$$
\frac{z}{(\log\log z)^3}\le\# D_{I,2,0}(z)\le\frac{z}{(\log\log z)^2},
$$ implying, in particular, a zero density. Note that these bounds are quite a notch higher than the $cz/\log^2z$ lower 
bound mentioned in \cite[Section 6]{Po} or proven in Theorem \ref{thm:I2}.  

\section{Preliminaries to the study of the case $k\ge1$}

 Once a Lucas sequence $U$ and the integers $a\ge2$ and $k\ge1$ have been fixed, we define, for every prime $p$, the set 
\begin{equation}\label{eq:ap}
A_p=A_p(U,a,k):=\bigg\{n\ge1\;:\;\nu_p(U_{(a-1)n+k})&gt;\nu_p\binom{an}{n}_U\bigg\}.
\end{equation} Note that 
\begin{equation}\label{eq:ap'}
\overline D_{U,a,k}=\bigcup_pA_p,
\end{equation} where the union is over all primes. We proceed to show that $A_p$ is empty except, possibly, for the 
finitely many primes $p$ of rank less than $ak$. 

\begin{lemma}\label{lem:2} Suppose $U(P,Q)$ is a nondegenerate fundamental Lucas sequence, 
while $a\ge2$ and $k\ge1$ are fixed 
integers. Assume $p\nmid Q$ is a prime of rank $\ge ak$. Then, $A_p$ is empty. That is, for all $n\ge1$, 
$$
\nu_p\binom{an}{n}_U\ge\nu_p\big(U_{(a-1)n+k}\big). 
$$
\end{lemma} 
\begin{proof} Note that 
$$
\frac1{U_{(a-1)n+k}}\binom{an}{n}_U=\frac{\prod_{j=1}^{k-1}U_{(a-1)n+j}}{\prod_{i=0}^{k-1}U_{n-i}}
\binom{an}{n-k}_U.
$$ Thus, if our claim is wrong, then $p$ must divide some factor, say $U_{n-e}$, $0\le e\le k-1$, in the product 
$\prod_{i=0}^{k-1}U_{n-i}$. Thus, $\rho$ divides $n-e$, where $\rho$ is the rank of $p$ in $U$. 
But $p$ must also divide $U_{(a-1)n+k}$, that is, $\rho$ divides $(a-1)n+k$. As $n\equiv e\pmod \rho$, we see 
that $\rho$ divides $(a-1)e+k$. However, $0&lt;(a-1)e+k&lt;ak$ and $\rho\ge ak$. Hence, we have a contradiction. 
\end{proof} 



 We begin by putting to use Lemma \ref{lem:2} to obtain a complete description of a particular 
set $D_{U,a,k}$ with $a\ge3$ and $k\ge2$. An example with $a=2$ had already been explicitly 
computed in \cite[Proposition 4.2]{Ba4}.   

\begin{cor}\label{cor:1} The set $\overline D_{F,3,2}$ of integers $n\ge1$ such that $F_{2n+2}$ does not divide 
$\binom{3n}{n}_F$ is precisely the set
$$
\{2\cdot3^x-1;\;x\ge0\}.
$$
\end{cor}
\begin{proof} By Lemma \ref{lem:2}, 
$\overline D_{F,3,2}$ is the union of the $A_p$ over all primes $p$ of rank less than $6$. Only the primes 
$2$, $3$ and $5$ have rank less than $6$ in the Fibonacci sequence. Let $p$ denote one of the primes 
$2$, $3$ or $5$, and $\rho\ge3$ denote its rank. For an integer $n\ge1$ to  
be in $A_p$, we need $p$ to divide $F_{2n+2}$. Thus, $2n+2$ is of the form $\l\rho p^x$ with 
$x\ge0$ and $\l\ge1$ two integers, where we assume $\l$ prime to $p$. Hence, we find that    
\begin{align}\begin{split}
\frac{2n}{\rho}&amp;=(\l-1)p^x+p^x-1+\frac{\rho-2}{\rho},\\
\frac{n}{\rho}&amp;=\bigg\lf\frac{n}{\rho}\bigg\rf+\bigg\{\frac{n}{\rho}\bigg\}.
\end{split}\label{eq:8} 
\end{align}
Because $\rho&gt;2$, we see that $(\rho-2)/\rho$ is the fractional part of $2n/\rho$. 
Moreover, $\rho\mid 2n+2$ implies $\rho\nmid n$. Thus, 
$\big\{\frac{n}{\rho}\big\}\ge\frac1{\rho}$. 

 If $\big\{\frac{n}{\rho}\big\}\ge\frac2{\rho}$, 
then there is a carry across the radix point and, due to the $x$ consecutive $p-1$ digits in the $p$-ary expansion 
of $p^x-1$, this carry over the radix point guarantees at least $x$ 
further carries left of that point in the base-$p$ addition 
of $2n/\rho$ to $n/\rho$. Thus, by the Kummer rule for Lucasnomials, the $p$-adic valuation of $\binom{3n}{n}_F$ 
is at least $x+1+\d_x$, which is $\nu_p(F_{2n+2})$. We conclude that if $\big\{\frac{n}{\rho}\big\}\ge\frac2{\rho}$, 
then $n\not\in A_p$. 

Hence, $\big\{\frac{n}{\rho}\big\}=\frac1\rho$, i.e., an $n\in A_p$ must be of the form $q\rho+1$. 
In particular, $2n+2=2q\rho+4$ so that $\rho\mid 4$, and, as $F_2=1$, $\rho=4$. But $F_4=3$. 
Therefore, $A_2$ and $A_5$ are empty and 
we now assume $p=3$. Since $n=4q+1$, equations (\ref{eq:8}) become 
\begin{align}\begin{split}
\frac{2n}{\rho}&amp;=2q+\frac{1}{2},\\
\frac{n}{\rho}&amp;=q+\frac{1}{4}.
\end{split}\label{eq:9} 
\end{align}
Since $2n+2=\l\cdot 4\cdot3^x=4(2q+1)$, we see that $2q+1=(3j+i)3^x$, where we set $\l=3j+i$ with $i=1$ or $2$.  
Therefore, 
\begin{align}\begin{split}
2q&amp;=(3j+i-1)\cdot3^x+(22\cdots 2)_3,\quad(x\;2\text{'s})\\
q&amp;=\frac{3j+i-1}2\cdot3^x+(11\cdots 1)_3,\quad(x\;1\text{'s})
\end{split}\label{eq:10} 
\end{align} where $(dd\cdots d)_3$ with $x$ $d$'s stands for $d(3^{x-1}+3^{x-2}+\cdots+1)$. 
If the integer $\ell=(3j+i-1)/2\not=0$,  
then we obtain  from (\ref{eq:10}) that the base-$3$ addition of $2q$ and $q$ produces at least $\nu_3(F_{2n+2})=x+1$ 
carries. Indeed, as easily seen, and as shown in \cite[Lemma 2.2]{Ba2}, adding $\ell\ge1$ and $(p-1)\ell$ in 
base $p$, $p$ a prime, yields at least one carry. However, if $3j+i-1=0$, the base-$3$ addition ``\,$2q+q$\," only produces $x$ 
carries so that $n$ belongs to $A_3$. But then $2q+1=3^x$ and $n=2\cdot3^x-1$. This proves the corollary.    
\end{proof}

\begin{remark}\label{rem:2} The proof of Corollary \ref{cor:1} shows that $\frac3{F_{2n+2}}\binom{3n}{n}_F$ is an integer 
for all $n\ge1$. We also note that $D_{F,3,2}$ has asymptotic density one 
in the positive integers, but that it misses infinitely many integers. Do these facts hold in general?
\end{remark}

 We will provide answers in due course, but we begin with the latter fact, which is that $\overline D_{F,3,2}$ 
is infinite. 


  
The only regular Lucas sequences $U(P,Q)$ with $P&gt;0$ for which there exists a $k\ge2$ 
such that $\overline D_{U,2,k}$ is finite --- it is in fact empty --- corresponds to $(P,Q)=(1,2)$ 
\cite[Theorem 3.5]{Ba4}. We want to find out whether, when $a\ge3$, such examples occur. 
We say a triple $(U,a,k)$, with $k\ge2$, is a {\it Catalan-like} triple if and only if, for all natural 
numbers $n$,   
$$
\frac1{U_{(a-1)n+k}}\binom{an}{n}_U \quad\text{ is an integer.}
$$ 

 We proceed with a lemma which states conditions sufficient to guarantee the infinitude of $A_p(U,a,k)$. 
It gives a minimal infinite subset of $A_p$ when $p$ satisfies some rank condition. Actually, the set 
$A_3$ in Corollary \ref{cor:1} is equal to that minimal subset.    

\begin{lemma}\label{lem:3} Let $U(P,Q)$ be a nondegenerate fundamental Lucas sequence and $a\ge2$,  
$k\ge2$ be integers. Assume there exists a prime $p\nmid Q$ of rank $\rho$ in $U$, where 
$$
\rho=k+\ell(a-1),
$$ for some $\ell$, $1\le\ell\le k-1$.  Then $A_p(U,a,k)$ is infinite and contains all integers $n$ 
of the form 
$$\frac{\rho p^x-k}{a-1}, \text{ for all }x \text{ divisible by }\varphi(a-1),
$$ where $\varphi$ denotes the Euler totient function.
\end{lemma} 
\begin{proof} As $\rho\ge2+(a-1)=a+1$ and $p\ge\rho-1$, we have $p\ge a$. Hence, $p\nmid a-1$. 
Thus, the condition $\varphi(a-1)$ divides $x$ guarantees that $p^x-1$ is divisible by $a-1$. 
Also, because $\rho\equiv k\pmod{a-1}$, for each $x\ge0$ divisible by $\varphi(a-1)$, there is a unique 
$n=n_x\ge1$ such that 
$$
(a-1)n+k=\rho p^x.
$$ Therefore, for these integers $n$, we find that
\begin{align}\begin{split}
\frac{(a-1)n}{\rho}&amp;=p^x-1+\frac{\rho-k}{\rho},\\
\frac{n}{\rho}&amp;=\frac{p^x-1}{a-1}+\frac{\ell}{\rho},
\end{split}\label{eq:11} 
\end{align} where we see that $(\rho-k)/\rho$ and $\ell/\rho$ are the fractional parts of, respectively, 
$(a-1)n/\rho$ and $n/\rho$. The sum of these two fractional parts is less than $1$ since $\rho-(k-\ell)&lt;\rho$. 
By the Kummer rule for Lucasnomials, we see that $\nu_p\binom{an}{n}_U\le x+\d_x$. (In fact, it is $x+\d_x$ since,  
if $x&gt;0$, the integer $(p^x-1)/(a-1)$ is prime to $p$ and the base-$p$ addition of $(a-1)n/\rho$ and 
$n/\rho$ produces exactly $x$ carries left of the radix point.) But, $\nu_p\big(U_{(a-1)n+k}\big)=x+\nu+\d_x
&gt; x+\d_x$.  
\end{proof} 

 We add some power to Lemma \ref{lem:3} by 
observing that if $U_k$ possesses a primitive prime divisor $p$ prime to $Q$, then the equations in 
(\ref{eq:11}) remain valid. 
Thus, if $p\nmid a-1$, then, again, for all integers $n$ of the form $k(p^x-1)/(a-1)$, $x$ a multiple of 
$\varphi(a-1)$, $U_{(a-1)n+k}$ 
does not divide $\binom{an}{n}_U$. However, the condition $p\nmid a-1$ is no longer necessarily true. 
We state this observation as an additional lemma. 

\begin{lemma}\label{lem:4} Let $U(P,Q)$ be a nondegenerate fundamental Lucas sequence and $a\ge2$,  
$k\ge2$ be integers. Assume there exists a prime $p\nmid Q(a-1)$ of rank $k$.  
Then $A_p(U,a,k)$ is an infinite set.
\end{lemma}

 One can see that the hypotheses of Lemma \ref{lem:3} are the weakest when $k=2$. Indeed, for $k=2$, 
we need a prime of rank $a+1$ to assert that $A_p$ is infinite. Some lemmas 
will help reach a 
conclusion whenever there are no primes of rank $a+1$. Inspired by the proof of Corollary \ref{cor:1}, 
we establish a first supplementary lemma for the case $k=2$. 

\begin{lemma}\label{lem:N} Let $U(P,Q)$ be a nondegenerate fundamental Lucas sequence and $a\ge2$ be 
an integer. If $p\nmid Q$ is a prime of rank $\rho$, with $\rho&gt;2$ and $\rho\nmid a+1$, then $A_p(U,a,2)$ is empty. 
\end{lemma}
\begin{proof} The argument is close to that of Corollary \ref{cor:1} so we abbreviate it.  
If $n\in A_p$, then there is an $x\ge0$ and a $\l\ge1$ not divisible by $p$ such that $(a-1)n+2=\l\rho p^x$. 
Hence, 
$$
\frac{(a-1)n}{\rho}=(\l-1)p^x+p^x-1+\frac{\rho-2}\rho.
$$ In the base-$p$ addition of $n/\rho$ to $(a-1)n/\rho$ a carry across the radix point generates 
at least $x$ carries left of that point. As a consequence $\nu_p\binom{an}{n}_U\ge\nu_p\big(U_{(a-1)n+2}\big)$ 
and $n\not\in A_p$. Since $n\in A_p$, either $\rho\mid n$ or $n=q\rho+1$ for some integer $q$. As 
$\rho\mid(a-1)n+2$, if $\rho\mid n$, then $\rho=2$. If $n=q\rho+1$, then 
$$
(a-1)n+2=(a-1)q\rho+(a+1)=\l\rho p^x,
$$ so that $\rho\mid a+1$. 
\end{proof} 

 This, together with Lemma \ref{lem:notQ}, yields the immediate corollary. 

\begin{cor}\label{cor:2} With the hypotheses of Lemma \ref{lem:N} and $k=2$, if $P=1$ and $U_{a+1}=\pm1$, then 
$\overline D_{U,a,2}$ is empty. On the other hand, if $|U_{a+1}|&gt;1$ and $\rho(p)=a+1$, then, 
by Lemma \ref{lem:3}, $\overline D_{U,a,2}$ is infinite. If $a$ and $P$ are even and $Q$ is odd, then $\overline D_{U,a,2}$ 
is infinite by Lemma \ref{lem:4}. 
\end{cor}

 Here is a theorem, which under fairly broad hypotheses, tells us that $A_p$ cannot be a finite nonempty 
set. 

\begin{theorem}\label{thm:D} Let $U(P,Q)$ be a NFL-sequence and $a\ge2$, $k\ge1$ be integers. 
Let $p\nmid Q(a-1)$ be a prime of rank at least $k$. Then $A_p$ is either empty or infinite. 
\end{theorem}
\begin{proof} Denoting as usual the rank of $p$ by $\rho$, suppose $A_p$ is not empty. Then there is an 
integer $n_0\ge1$ in $A_p$ and integers $\l\ge1$, $p\nmid\l$, $x_0\ge0$ such that 
\begin{equation}\label{eq:star}
(a-1)n_0+k=\l\rho p^{x_0}. 
\end{equation} In particular, $\nu_p\big(U_{(a-1)n_0+k}\big)=x_0+\nu+\d_{x_0}$.   

 As $n_0\in A_p$ and $(a-1)n_0/\rho=\l p^{x_0}-1+(\rho-k)/\rho$, it must be, by the Kummer rule for Lucasnomials, 
that 
$$
\bigg\{\frac{n_0}{\rho}\bigg\}+\frac{\rho-k}{\rho}&lt;1.
$$ We are about to show that, with the same integer $\l$ which appears in (\ref{eq:star}), there are 
infinitely many integers $n\ge1$ such that $(a-1)n+k=\l\rho p^x$, for some $x\ge0$, which satisfy 
$\big\{\frac{n}{\rho}\big\}=\big\{\frac{n_0}{\rho}\big\}$. By the above analysis for $n_0$, this implies 
all such $n$ are in $A_p$. Now $(a-1)n+k=\l\rho p^x$ if and only if 
\begin{equation}\label{eq:star+}
(a-1)(n-n_0)=\l\rho(p^x-p^{x_0})
\end{equation} As $p\nmid a-1$, $p^x\equiv p^{x_0}\pmod{a-1}$ occurs for all $x$ satisfying $x\equiv x_0\pmod h$, 
where $h$ is the multiplicative order of $p\pmod{a-1}$. For each $s\ge0$, put $x_s=x_0+sh$ and define $t_s$ by  
$p^{x_s}=p^{x_0}+t_s(a-1)$. Note that $(t_s)_{s\ge0}$ is an increasing sequence of integers.  
Putting $p^{x_s}-p^{x_0}=t_s(a-1)$ into $(\ref{eq:star+})$ we find that $n=n_s=n_0+t_s\l\rho$ 
satisfies $(\ref{eq:star+})$ with the fractional part of $n/\rho$ equal to that of $n_0/\rho$.  
\end{proof}

 Theorem \ref{thm:D} will come in handy in the next section, but mostly in the guise of the next corollary. 

\begin{cor}\label{cor:S} Suppose $U(P,Q)$ is regular, $a\ge2$ and $k=2$. Let $p\nmid a-1$ be a prime. 
If $p\mid U_{a+1}$, then $A_p(U,a,2)$ is infinite. 
\end{cor}
\begin{proof} By (\ref{eq:reg}), if $p\mid U_{a+1}$, then $p\nmid U_a$. Hence, $\nu_p(U_{(a-1)+2})&gt;\nu_p\binom{a}{1}_U$ and 
$1\in A_p(U,a,2)$. By Lemma \ref{lem:notQ}, $p\nmid Q$. So by Theorem \ref{thm:D}, $A_p$ is infinite. 
\end{proof}    
   

\section{Chasing all Catalan-like triples $(U,a,k)$, $k\ge2$,\\ 
and a proof that $\overline D_{U,a,k}$ is otherwise infinite}

 We first study the case $k\ge3$. 


\begin{prop}\label{prop:1} Let $a\ge3$ and $k\ge3$. Then for all nonzero-discriminant regular Lucas 
sequences $U(P,Q)$, there are infinitely many integers $n\ge1$ for which $U_{(a-1)n+k}$ does not 
divide $\binom{an}{n}_U$. 
\end{prop}
\begin{proof} Suppose first $k\ge4$. To have a chance at finding a counterexample to our proposition, we need, 
by Lemma \ref{lem:3}, to find regular sequences $U$ that are $n$-defective for at least three indices $n$ 
in arithmetic progression, namely at least at $k+a-1$, $k+2(a-1)$ and $k+3(a-1)$ knowing that $k+a-1\ge6$.  
Inspecting Table A of Section 7, we discover only two such instances, namely the sequence $U(1,2)$ which 
is $n$-defective at $6$, $12$ and $18$ and also at $n=8$, $13$ and $18$. In the case of $6$, $12$ and $18$, the common difference  
is $6$. So $a-1=6$ and $k+a-1=6$. This yields $k=0$, a contradiction. In the second case, as $U(1,2)$ is not $23$-defective, 
there is no counter-example if $k\ge5$. So assume $k=4$. Solving $k+a-1=8$ 
for $a-1$ yields $a-1=4$. However, as the three 
indices $n=8$, $13$ and $18$ are $5$ apart, we would have needed $a-1=5$. 

 So we now assume $k=3$. Using Table A, we now search for sequences that are both $(a+2)$ 
and $(2a+1)$-defective. For $a=3$, again the sequence $U(1,2)$ is both $5$ and $7$-defective. But 
$U_7$ has the primitive prime divisor $7$, which was discarded from Table A because 
$7$ divides the discriminant $P^2-4Q$. 
But that $p$ divides $\Delta$ does not invalidate Lemma \ref{lem:3}. Examining all values of $a$, $4\le a\le14$, we only 
find one candidate sequence when $a=6$, i.e., again $U(1,2)$, which is $8$ and $13$-defective as seen earlier. 
To show that $\overline D_{U,6,3}$, with $U=U(1,2)$, is infinite, it suffices to prove that the set $A_3(U,6,3)$, 
defined in (\ref{eq:ap}), is infinite. Note that $3\nmid a-1$. Moreover, $U_6=\binom{6}{1}_U=5$ is not divisible by 
$U_{(a-1)\cdot1+3}=U_8=-3$. That is, $1\in A_3$. By Theorem \ref{thm:D}, $A_3$ is infinite. 
\end{proof}

  We now study the case $k=2$. 

\begin{theorem}\label{thm:E} The four triples $\big(U(1,2),a,2\big)$, for $a=4$ and $a=12$, $\big(U(1,3),4,2\big)$ 
and $\big(U(1,5),6,2\big)$ are all four Catalan-like triples. That is, 
$$
\frac1{U_{(a-1)n+k}}\binom{an}{n}_U,
$$ is integral for all natural numbers $n$. 
\end{theorem}
\begin{proof} The first few terms of the most frequently defective Lucas sequence $U(1,2)$ are
$$
0,\,1,\,1,\,-1,\,-3,\,-1,\,5,\,7,\,-3,\,-17,\,-11,\,23,\,45,\,-1,\,-91,\,-89,\,93,\,271 \text{ and }85. 
$$ The first terms up to $U_7$ are 
$$\begin{cases} 
0,\,1,\,1,\,-2,\,-5,\,1,\,16,\,13,&amp; \text{ for }U(1,3);\\
0,\,1,\,1,\,-4,\,-9,\,11,\,56,\,1,&amp; \text{ for }U(1,5). 
\end{cases}$$ For $U(1,2)$ and $U(1,3)$, we see that $U_5=\pm1$. Therefore, by Corollary \ref{cor:2}, $\overline D_{U,4,2}$ 
is empty. For $U(1,5)$, $U_7=1$ so $\overline D_{U,6,2}$ is empty by the same Corollary \ref{cor:2}. Similarly, for $U(1,2)$, 
we find that $U_{13}=-1$ so $\overline D_{U,12,2}$ is empty. 
\end{proof} 

\begin{prop}\label{prop:2} Let $U(P,Q)$ be a nonzero-discriminant regular Lucas 
sequence, $a\ge3$ be an integer and $k=2$. Then, except for the four exceptions of Theorem \ref{thm:E}, 
there exist infinitely many integers $n\ge1$ for which $U_{(a-1)n+k}$ does not 
divide $\binom{an}{n}_U$. 
\end{prop}
\begin{proof} We display the terms $U_2$, $U_3$, $U_4$ and $U_6$ of $U(P,Q)$ as polynomials in $P$ and $Q$:
$$
U_2=P\;,U_3=P^2-Q,\;U_4=P(P^2-2Q),\;U_6=P(P^2-Q)(P^2-3Q).
$$
We proceed case by case depending on the value of $a$. By Lemma \ref{lem:3}, the only values of $a$ to consider 
are those for which there is some $U$ with $U_{a+1}$ free of primitive prime divisors.  

\medskip

\noindent{\bf Case $a=3$.} If $U_4$ has a primitive prime divisor, then, as $a+1=4$, we obtain that $\overline D_{U,3,2}$ 
is infinite as a consequence of Lemma \ref{lem:3}. (Note that the hypothesis $p\nmid Q$ holds by Lemma \ref{lem:notQ}.)
If $U_4$ has no p.p.d., then $P^2-2Q=\pm1$. So $P$ is odd $\ge1$. 
If $P=1$, then $Q=1$, so that $P^2-Q=U_3=0$, which contradicts the nondegeneracy of $U$. Hence, $P$ is odd $&gt;1$ 
and $U_2=U_k$ admits a p.p.d.\ $p\ge3$. As $p\nmid a-1=2$, we reach the result sought by applying Lemma \ref{lem:4}. 
 


\medskip

\noindent{\bf Case $a=4$.} Since $a+1=5$ we need only consider the seven sequences listed as $5$-defective in 
Table A for, by Lemma \ref{lem:3}, they are the only ones for which $\overline D_{U,4,2}$ 
could be finite. But of these seven sequences three have a discriminant divisible by $5$. Thus, $5$ is a 
primitive divisor of $U_{a+1}$ and we may discard them by Lemma \ref{lem:3}. Two others are $U(1,2)$ and 
$U(1,3)$ for which, as seen in Theorem \ref{thm:E}, $\overline D_{U,4,2}$ is empty. The remaining two are 
$U(12,55)$ and $U(12,377)$ for which $p=2$ has rank $2$ with $2\nmid a-1$ so that Lemma \ref{lem:4} ensures 
our claim holds.   



\medskip

\noindent{\bf Case $a=5$.} This might be the most intricate case because there are four infinite families of regular Lucas 
sequences that are $(a+1)$-defective, i.e., $6$-defective, according to Table A. However, 
for two of these four infinite families $3\mid P$. Hence, $p=3$ has rank $k$ and $p\nmid a-1$ so that, by 
Lemma \ref{lem:4}, $\overline D_{U,5,2}$ is infinite. The two remaining families are {\bf 1.} 
$(P,Q)=\big(P,(P^2-1)/3\big)$ for all $P&gt;3$, $3\nmid P$, and {\bf 2.} $(P,Q)=\big(P,(P^2-(-2)^i)/3\big)$, 
where $i\ge1$, $P\equiv\pm1\pmod 6$, $(P,i)\not=(1,1)$. In both families, if $P$ has an odd prime divisor $p$, 
then, as $p\nmid a-1$, $\overline D_{U,5,2}$ is infinite by Lemma \ref{lem:4}. If not, then for the family {\bf 1.} 
$P=2^\a$ for some $\a\ge2$. Thus, $Q=(P^2-1)/3$ is odd. Moreover,  
$P^2-3Q=1$ and $4^\a-3Q=1$ implies that $Q\equiv1\pmod 4$. Thus, $U_3=P^2-Q=(2P^2+1)/3$ is $\ge11$ and 
$\equiv3\pmod 4$. Hence, there exists a prime $p\ge3$, $p\equiv3\pmod 4$ of rank $3$. Thus, we obtain an infinite 
sequence of integers $(n_x)_{x\ge1}$ defined by $4n_x+2=2\cdot3\cdot p^x$. Putting $n=n_x$ we see that $4n/3=2p^x-1+1/3$. 
Noting that $2p^x-1=4q+1$ for some integer $q$, we find that $n/3=q+1/4+1/12=q+1/3$. Therefore, the base-$p$ 
addition of $(a-1)n/\rho$ and $n/\rho$ generates exactly $x$ carries left of the radix point and none across 
that point. Therefore, $n_x\in A_p$ for all $x\ge1$ and $\overline D_{U,5,2}$ is infinite. Now for the second 
family {\bf 2.}, if $U_2$ does not have an odd p.p.d., then $P=1$. Thus, as $(P,i)\not=(1,1)$, we have $i\ge2$ and,  
putting $j=i-2$, we find that $U_3=1-Q=\big(2+(-2)^{j+2}\big)/3$, $j\ge0$, is divisible by $2$. We 
prove that $A_2$ is infinite. For each $\l\ge1$ odd, $\nu_2(U_{6\l})=j+3$. Indeed, $\nu_2(U_{6\l})=\nu_2(U_6)$.  
But $U_6=(1-Q)(1-3Q)$ and $1-Q=\big(2+(-2)^{j+2})/3$ has $2$-adic valuation $1$, whereas $1-3Q=(-2)^{j+2}$.     

 For each $t\ge1$ define the odd integer $\l=(2\cdot8^t+5)/7$, which may be written in terms of the base-$2$ 
representation of $\l-1$ as
$$
\l=(\l-1)+1=(\overline{010}\strut^t)_2+1,
$$ where $\overline{010}\strut^t=010\,010\cdots010$ (the string $010$ being repeated $t$ times). 
To each such $\l$, corresponds a unique integer $n\ge1$ such that $4n+2=6\l$. Since $4n/3=2\l-1+1/3=
(\overline{100}^t)_2+1+1/3$ and, thus, $n/3=(\overline{001}^t)_2+1/4+1/12=(\overline{001}^t)_2+1/3$, 
we see that the base-$2$ addition of $(a-1)n/\rho$ and $n/\rho$ creates a single carry from position $1$ to 
position $2$ left of the radix point. 
Therefore, $\nu_2\binom{5n}{n}_U=1+\d=j+2$, whereas $\nu_2(U_{6\l})=j+3$. Thus, $A_2$ is infinite. 



 The only values of $a$ left for consideration are: 6, 7, 9, 11, 12, 17 and 29. For the other values, there 
are no $(a+1)$-defective sequences $U$. 



\medskip

\noindent{\bf Case $a=6$.} The only $7$-defective sequences according to Table A, Section 7, are $(P,Q)=(1,2)$ 
and $(1,5)$. However, $7$ is a primitive divisor of $U_7=U_{a+1}$ for the first sequence since $7\mid\Delta$. 
So $A_7$ is infinite by Lemma \ref{lem:3}. The second sequence $U$ has parameters $(P,Q)=(1,5)$ and 
$(U,6,2)$ was identified as a Catalan-like triple in Theorem \ref{thm:E}. 



\medskip

\noindent{\bf Case $a=7$.} There are only two $8$-defective regular Lucas sequences $U(2,7)$ and $U(1,2)$. For 
$(P,Q)=(2,7)$, $U_2=2$ and $U_8=-40$ so that, by Lemma \ref{lem:N}, we need only consider $A_2$ and $A_5$. 
Since $5\nmid a-1$, but $5\mid U_{a+1}$, $A_5$ is infinite by Corollary \ref{cor:S}. (In fact, $A_2$ is 
infinite as well. Indeed, for each even $x\ge2$, there is an $n\ge1$ which 
solves the equation $6n+2=2\cdot2^x$. Since for all such $x$ and $n$ we find that $6n/2=2^x-1$  
is integral, the base-$2$ addition of $(a-1)n/\rho$ and $n/\rho$ does not produce 
a carry across the radix point so that $\nu_2\binom{7n}{n}_U\le x&lt;1+x=\nu_2\big(U_{6n+2}\big)$.) 
For $(P,Q)=(1,2)$, we show that $A_3$ contains $\{1+4\cdot3^t,\;t\ge0\}$. Indeed, $\rho(3)=4$ and, for 
$n=1+4\cdot3^t$, we see that  
$6n/4=2\cdot3^{t+1}+1+1/2$ while $n/4=3^t+1/4$. Their base-$3$ addition raises no carry at all. So $3\nmid\binom{7n}{n}_U$. 
However, $4$ divides $6n+2$ implies that $\nu_3\big(U_{6n+2})\ge1$. 



\medskip

\noindent{\bf Case $a=9$.} As usual if $U_{a+1}$ has a p.p.d.\ then $\overline D_{U,a,2}$ is infinite by Lemma \ref{lem:3}. 
So we look at Table A for the $10$-defective sequences. There are three: $(P,Q)=(2,3)$, $(5,7)$ and $(5,18)$. For the 
last two $P=5$ so the prime $5$ has rank $2$ and does not divide $a-1$. Hence, by Lemma \ref{lem:4}, $A_5$ is infinite. 
For $U(2,3)$, $U_5=-11$. So $11\mid U_{a+1}$, but $11\nmid a-1$. Hence, $A_{11}$ is infinite by Corollary \ref{cor:S}. 



\medskip

\noindent{\bf Case $a=11$.} There are six $12$-defective sequences corresponding to $(P,Q)=(1,-1)$, $(1,2)$, $(1,3)$, $(1,4)$, 
$(1,5)$ and $(2,15)$. For $(1,-1)$, $(1,2)$, $(1,4)$ and $(1,5)$, $3\nmid Q$. So, as $\rho(3)$ is either $2$, $3$ 
or $4$, we see that $3\mid U_{12}$. As $3\nmid a-1$, we conclude that $A_3$ is infinite by Corollary \ref{cor:S}.  
For $U(1,3)$, $U_3=-2$ and $U_6=16$. Although $2\mid a-1$, we claim that $A_2$ contains 
$$\bigg\{\frac{3\cdot2^{4t+1}-1}5\;;\;t\ge0\bigg\}.$$ Indeed, if $n=(3\cdot2^{4t+1}-1)/5$, then $10n+2=3\cdot2^{4t+2}$. 
Thus, $\nu_2(U_{10n+2})=(4t+2)+1+\d=4t+5$. But $10n/3=2^{4t+2}-1+1/3$, while $n/3=(2^{4t+2}-1)/10+1/30\equiv3/10+1/30=1/3
\pmod 1$. Therefore, by the Kummer rule for Lucasnomials, $\nu_2\binom{11n}{n}_U\le(4t+2)+\d=4t+4&lt;\nu_2(U_{10n+2})$.  
For $U(2,15)$, $U_3=-11$. Hence, $11\mid U_{12}$ and $11\nmid a-1$, so that $A_{11}$ is infinite by Corollary \ref{cor:S}. 



\medskip

\noindent{\bf Case $a=12$.} The only $13$-defective sequence is $U(1,2)$. We saw in Theorem \ref{thm:E} that $\big(U(1,2),12,2\big)$ 
is a Catalan-like triple. 



\medskip

\noindent{\bf Cases $a=17$ and $a=29$.} The sequence $U(1,2)$ is the only $18$-defective and the only $30$-defective regular Lucas 
sequence. As $U_6=5$, we see that $5\mid U_{18}$ and $5\mid U_{30}$. Since $5\nmid 17-1$ and $5\nmid 29-1$, we conclude,  
with the help of Corollary \ref{cor:S}, that $A_5(U,17,2)$ and $A_5(U,29,2)$ are both infinite sets. 
\end{proof}

 For the sake of completeness, we now address the case of zero-discriminant regular Lucas sequences $U$, i.e., of 
$U_n=I_n$ or $U_n=(-1)^{n-1}I_n$. Clearly it suffices to study the case of $U=I$. 

\begin{prop}\label{prop:3} The sets $\overline D(I,a,k)$ are infinite for all $a\ge2$ and all $k\ge2$. 
\end{prop}
\begin{proof} Let $a$ and $k$ be integers exceeding $1$. Let $g=\gcd(a-1,k)$. The argument is split 
into three complementary cases. 



\medskip

\noindent{\bf Case 1.} Suppose $g&gt;1$. Let $p$ be a prime factor of $g$ and put $n=p^x$ for some $x\ge0$. 
Then the $p$-ary expansion of $(a-1)n$ ends with $x+1$ zeros, while $n$ has a single digit $1$ at position $x$. 
Thus, there is no carry in the base-$p$ addition of $(a-1)n$ and $n$. By Kummer's rule, $p$ does not 
divide $\binom{an}{n}$. But $p$ divides $(a-1)n+k$. So $(a-1)n+k$ does not divide $\binom{an}{n}$. 
Therefore, for all $x\ge0$, $p^x$ belongs to $\overline D(I,a,k)$. 



\medskip

\noindent{\bf Case 2.} Assume $g=1$ and $k\nmid a$. Then there is a prime $p$ such that $p^\k||k$, $p^u||a$ 
for some $\k&gt;u$. We write $a=p^ua'$. Put $n=p^x$ for some $x&gt;\k$. Then 
\begin{eqnarray*}
(a-1)n&amp;=&amp;an-n=a'p^{x+u}-p^x=(a'-1)p^{x+u}+(p-1)p^{x+u-1}+\cdots+(p-1)p^x\\
n&amp;=&amp;p^x. 
\end{eqnarray*} Their base-$p$ addition creates exactly $u$ carries since the carry into position $x+u$ 
does not propagate any further to the left. Indeed, $p\nmid a'\implies a'-1\not\equiv p-1\pmod p$. 
By the rule of Kummer, $p^u||\binom{an}{n}$. However, $p^\k$ divides $(a-1)n+k$. Therefore, all $p^x$, 
$x&gt;\k$, are in $\overline D(I,a,k)$.




\medskip


\noindent{\bf Case 3.} Assume $a=k\ell$ for some $\ell\ge1$. Thus, $g=1$. 
Let $p$ be a prime factor of $k$ and write $k=p^\k k'$, $p\nmid k'$. 
Let $h$ be the multiplicative order of $p$ modulo $a-1$. Let $t&gt;0$ be an integer large enough so that $p^x&gt;k$, 
where $x=\k+ht$. Then $k'p^x-k=k(p^{ht}-1)$ is a multiple of $a-1$. Define $n=n_t$ as $(a-1)n=k'p^x-k$. 
The $p$-ary expansion of $k'p^x-k$ has the form 
$$
(k'-1)p^x+d_{x-1}p^{x-1}+\cdots+d_\k p^\k,
$$ where 
the $d_i$'s are $p$-ary digits. If $d_x$ is the $(x+1)$-st $p$-ary digit of $k'p^x-k$, i.e., if $(k'-1)p^x
\equiv d_xp^x\pmod{p^{x+1}}$, then $d_x\not=p-1$, because $d_x=p-1$  
would imply that $p\mid k'$. Moreover, $(a-1)n&lt;k' p^x=kp^{x-\k}\le kp^{x-1}$ implies that 
$$
n&lt;\frac{k}{a-1}p^{x-1}\le\frac{k}{k-1}p^{x-1}\le2p^{x-1}\le p^x.
$$ The $p$-ary expansion of $n$ has at most $x$ significant digits. Thus, the base-$p$ addition of 
$n$ and $(a-1)n$ generates at the most $x-\k$ carries since the least significant $\k$ digits of 
$(a-1)n$ are all zero. Thus, 
$$
\nu_p\binom{an}{n}\le x-\k&lt;x=\nu_p\big((a-1)n+k\big).
$$ Hence, $n_t\in A_p$, for all $t$ large enough.
\end{proof}

 Gathering together Propositions \ref{prop:1}, \ref{prop:2} and \ref{prop:3}, the theorems of 
Sections 2 and 3 and \cite[Theorem 3.5]{Ba4}, 
we state a general theorem. 

\begin{theorem}\label{thm:G} Suppose $U(P,Q)$ is a regular Lucas sequence and $a\ge2$ and $k$ are integers. 
Then there are infinitely many natural numbers $n$ such that 
$$
U_{(a-1)n+k}\text{ does not divide }\binom{an}{n}_U,
$$ unless either $k=1$ or, $k=2$, $P=\pm1$ and $(Q,a)$ is one of the five ordered pairs 
$(2,2)$, $(2,4)$, $(2,12)$, $(3,4)$ or $(5,6)$, in which cases 
$$
U_{(a-1)n+k}\text{ divides }\binom{an}{n}_U,\quad\text{ for all }n\ge1.
$$ \end{theorem} 

 We derive a couple of corollaries, one for the sequence $U(1,2)$ which stands out conspicuously, as 
had already been noted in \cite[Corollary 3.6 and the subsequent remark]{Ba4}, and one for the 
Fibonacci sequence. 

\begin{cor}\label{cor:H} For the two Lucas sequences $U(\pm1,2)=(U_n)_{n\ge0}$, we see that for all $n\ge1$ 
$$
U_{n+1}U_{n+2}\mid\binom{2n}{n}_U,\; U_{3n+1}U_{3n+2}\mid\binom{3n}{n}_U\text{ and }U_{11n+1}
U_{11n+2}\mid\binom{12n}{n}_U. 
$$
\end{cor}

 The next corollary is an extension of \cite[Proposition 3.2]{Ba4}. It would also hold if we replaced $F$ by $I$. 
Indeed, Fibonomial Fuss-Catalan numbers bear the same singular status as Fuss-Catalan numbers do. 
 
\begin{cor}\label{cor:F} Let $a\ge2$ and $k$ be integers. Let $F$ denote the Fibonacci sequence. 
Then there are infinitely many $n\ge1$ for which the numbers 
$$
\frac{1}{F_{(a-1)n+k}}\binom{an}{n}_F
$$ are not integers, unless $k=1$ when they are integers for all $a\ge2$ and $n\ge1$. 
\end{cor}

\section{The density of $D_{U,a,k}$ when $k\ge1$}

 Our objective is to show that $D_{U,a,k}$ has asymptotic density $1$ in the set of positive 
integers whenever $k\ge1$. Here is an indicative roadmap of the route we intend to follow. 
As in \cite[Section 4]{Ba4}, where we proved the case $a=2$, we proceed 
by showing that the complementary set of $D_{U,a,k}$ in the positive integers, $\overline D_{U,a,k}$, 
has asymptotic density $0$. By Lemma \ref{lem:2}, there are only at most finitely many primes $p$ 
for which $A_p$ is not empty. Thus, by (\ref{eq:ap'}), we only need to prove that $A_p$ is of density zero  
for each prime $p$. We will establish an estimate for the number of words of length $\ell$ over an 
alphabet of $p$ letters which miss a given block of $b$ consecutive identical letters, as $\ell$ tends to 
infinity. This result is then converted into an upper bound for the number of integers $n\le z$ 
for which the base-$p$ addition of 
$n/\rho$ and $(a-1)n/\rho$ generates less than a given number of carries.   
This upper bound turns out to be $o(z)$, as $z$ tends to infinity. Integers $n$ in $A_p$ may be 
viewed as generating few carries in the addition of $n/\rho$ and $(a-1)n/\rho$, at least fewer than 
the $p$-adic valuation of $U_{(a-1)n+k}$. We split $A_p$ into the union of all $A_p^x$, 
where $A_p^x$ is the subset of $A_p$ of integers $n$ for which $x$ is the exact exponent of $p$ in the equation  
$(a-1)n+k=\l\rho p^x$, $p\nmid\l$. For $x$ smaller than a bound $u$, integers in $A_p$ are easily 
seen to generate less than $a_p$ carries in the base-$p$ addition of $n/\rho$ and $(a-1)n/\rho$, 
where $a_p$ is a fixed bound that depends on $p$ and $u$. 
So there are no more than $o(z)$ such integers as $z$ tends to infinity. For $x&gt;u$, we will see that the 
form of $(a-1)n/\rho$ generally induces a minimum of $x-u$ carries right of position $x$ and left of the radix 
point. Thus, no matter how large $x$ we only need a bounded fixed number of carries left of position $x$ 
to guarantee $n$ is not $A_p$. This gives the desired conclusion.



 It is well-known \cite[Theorem 143, p.\ 120]{HW} that almost all natural numbers, when expressed in any scale, 
contain every possible sequence of digits. But we need to quantify somewhat this `almost all' statement. 
The editor-in-chief mentioned the work of Guibas and Odlyzko (\cite{GO} and their further papers) whose general 
results may well imply our lemma. 

\begin{lemma}\label{lem:5} Let $p\ge2$ and $b\ge1$ be integers. Let $x_n$ be the number of strings 
of $p$-ary digits of length $n$ which do not contain a block of $b$ consecutive digits all equal to $p-1$. Then, 
as $n$ tends to infinity, 
$$
x_n\sim c\cdot\o^n p^n,
$$ for some $\o\in(0,1)$ and some positive constant $c$. 
\end{lemma}
\begin{proof} The sequence $(x_n)$ forms a linear recurrent sequence 
with annihilating polynomial 
\begin{equation}\label{eq:12}
P(X)=X^b-(p-1)X^{b-1}-(p-1)X^{b-2}-\cdots-(p-1).
\end{equation}  To see that $(x_n)$ is linear recurrent, we only consider 
the case $b=3$ as an example, but the reasoning may be carried out in all generality. 
We say that a string is {\it admissible} if it does not contain a block of three consecutive digits 
equal to $p-1$. Define $y_n$ as the number of length-$n$ admissible strings that end with 
two consecutive $p-1$ digits, $z_n$ as the number of length-$n$ admissible strings ending 
with only one $p-1$ digit and $t_n$ for those ending with a digit at most $p-2$. Then we readily 
see that for $n\ge2$, $z_n=t_{n-1}$ and $t_n=(p-1)x_{n-1}$. Consequently we see that for $n\ge3$
\begin{eqnarray*}
x_{n+1}&amp;=&amp;(p-1)y_n+p(z_n+t_n)\\
&amp;=&amp;(p-1)(y_n+z_n+t_n)+(t_n+t_{n-1})\\
&amp;=&amp;(p-1)x_n+(p-1)x_{n-1}+(p-1)x_{n-2}.
\end{eqnarray*}
Noting that $x_i=p^i$ for $0\le i\le b-1$ and $x_b=p^b-1$, we see that the polynomial $P$ in (\ref{eq:12})
must be the characteristic polynomial of $(x_n)$, i.e., its minimal annihilating polynomial. 
Indeed suppose to the contrary that there is an $1\le s&lt;b$ and constants $a_i$, $1\le i\le s$, 
such that $x_{n+s}=\sum_{i=1}^sa_ix_{n+s-i}$ for all $n\ge0$. Then 
$$
x_b=\sum_{i=1}^sa_ix_{b-i}=p\sum_{i=1}^sa_ix_{b-i-1}=px_{b-1}=p^b,
$$ contradicting the fact that $x_b=p^b-1$. 

Using for instance \cite[Lemma 3]{Ba3}, we know the polynomial $P$ in (\ref{eq:12}) has a unique simple 
dominant real zero $\a&gt;1$. Because $P$ is the characteristic polynomial 
of $(x_n)$, the closed-form expression of $x_n$ must contain a nonzero term in $\a^n$. Therefore, 
$x_n\sim c\a^n$, as $n$ tends to infinity, where $c$ is some positive constant that depends only on $p$ and 
$b$. Note that $P(x)$ is increasing for $x&gt;p$. Indeed, the derivative $P'$ is positive, since for $x&gt;p$
\begin{eqnarray*}
P'(x)&amp;=&amp;bx^{b-1}-(p-1)\big((b-1)x^{b-2}+(b-2)x^{b-3}+\cdots+1\big)\\
&amp;&gt;&amp;bx^{b-1}-(p-1)(b-1)(x^{b-2}+x^{b-3}+\cdots+1)\\
&amp;=&amp;bx^{b-1}-(p-1)(b-1)\frac{x^{b-1}-1}{x-1}\\
&amp;&gt;&amp;bx^{b-1}-(b-1)(x^{b-1}-1)=x^{b-1}+b-1&gt;0.
\end{eqnarray*}
Since $P(p)=1&gt;0$, the polynomial $P$ has no zero larger than $p$. Thus, $\a&lt;p$. Our claim holds with $\o=\a/p$. 
\end{proof}

\begin{lemma}\label{lem:6} Let $p$ be a prime, $\rho\ge1$, $a\ge2$ and $a_p\ge1$ be integers. Then  
the set $C_{&lt;a_p}$ of integers $n$ such that the base-$p$ addition of $n/\rho$ and $(a-1)n/\rho$ produces less than 
$a_p$ relevant carries satisfies 
$$
\#C_{&lt;a_p}(z)\ll z\cdot\o^D,\text{ as } z\to\infty, 
$$ for some positive $\o&lt;1$ and $D=1+\lf\log_p z\rf$. 
In particular, $\#C_{&lt;a_p}(z)$ is $o(z)$, as $z$ tends to infinity. 
\end{lemma}
\begin{proof} Let $z\ge1$ be a large real number. Thus, if $n\le z$, then the number of $p$-ary digits of $n/\rho$ is at most $D$.  
Let $b\ge a_p+\th$, where the integer $\th\ge1$ satisfies $p^{\th}&gt;a-1$, but $p^{\th-1}\le a-1$.  
If $n$ belongs to $C_{&lt;a_p}$, then the $p$-adic expansion of $\lf n/\rho\rf$ cannot contain a block of $b$ consecutive 
$p-1$ digits. 
Indeed, otherwise there exist $i,l\ge0$ such that 
\begin{eqnarray*}
\lf n/\rho\rf&amp;=&amp;lp^{i+b}+(p-1)(p^{i+b-1}+p^{i+b-2}+\cdots+p^i)+m\\
&amp;=&amp;lp^{i+b}+(p^{i+b}-p^i)+m=(l+1)p^{i+b}-m',
\end{eqnarray*} with $m&lt;p^i$ and $1\le m'=p^i-m\le p^i$.  
Therefore there are integers $c_1\ge1$ and $0\le c_2\le a-2$ and an $\e\in[0,1)$ such that 
$$(a-1)n/\rho=c_1 p^{i+b}-m'(a-1)+c_2+\e$$ with $m'(a-1)&lt;p^{i+\th}$. Hence, $1\le m'(a-1)-c_2&lt;p^{i+\th}$
so that the base-$p$ digits of $(a-1)n/\rho$ at all positions 
$i+b-1$, $i+b-2$, down to position $i+\th$, are all positive. Therefore, the addition of $n/\rho$ and 
$(a-1)n/\rho$ would produce at least $b-\th\ge a_p$ carries. Note that there are $\rho$ values of $n$ 
for which $\lf n/\rho\rf$ is the same positive integer $m$. Thus, $\#C_{&lt;a_p}(z)$ is at most equal to 
$\rho$ times the number of $p$-ary strings of length $D$ not containing $a_p+\th$ consecutive $p-1$ digits. By 
Lemma \ref{lem:5}, there exist $\o&lt;1$ and a positive $c$ such that
$$
\#C_{&lt;a_p}(z)\le \rho c(1+o(1)) p^D\o^D\ll z\o^D.
$$ 
\end{proof}

\begin{remark}\label{rem:1} The number $\#C_{&lt;a_p}^{[x,D]}(z)$ of integers $n\le z$ for which less than 
$a_p$ carries occur between  
positions $x$ and $D$ in the base-$p$ addition of $n/\rho$ and $(a-1)n/\rho$ is seen from the proof of Lemma 
\ref{lem:6} to be $\ll\o^{D-x+1}z$. Indeed, the presence of $a_p+\th$ consecutive $p-1$ digits at positions $i$, 
$i+1$, \ldots, $i+b-1$, with $i\ge x$, generates at least $a_p$ carries left of position $x$. 
By Lemma \ref{lem:5}, $\#C_{&lt;a_p}^{[x,D]}(z)\ll \rho\cdot c\o^{D-x+1}p^{D-x+1}\cdot p^x\ll\o^{D-x+1}z$, where 
the constant implied by the Vinogradov symbol ``$\ll$" does not depend on $x$.   
\end{remark}

\begin{lemma}\label{lem:7} Let $U(P,Q)$ be a regular Lucas sequence and $a\ge2$, $k\ge1$ be integers. 
Given a prime $p$ we let $A_p$ denote the set of natural numbers $n$ 
for which the $p$-adic valuation of $U_{(a-1)n+k}$ is larger than that of $\binom{an}{n}_U$. 
Then the asymptotic density of $A_p$ is zero. 
\end{lemma}
\begin{proof} If $n$ belongs to $A_p$, then there is a unique $x\ge0$ 
such that $(a-1)n+k=\l\rho p^x$, with $p\nmid\l$ and $\rho$ the rank of $p$. 
We define $A_p^x$ as the subset of $A_p$ of integers $n$ 
which correspond to that $x$. Also, for an integral constant $c\ge0$, we use the notation $A_p^{&lt;c}$ and 
$A_p^{\ge c}$ to denote the union of the $A_p^x$ over, respectively, all $x&lt;c$, or all $x\ge c$. 

 We divide $k$ by $\rho$, say $k=q\rho+r$ with $0\le r&lt;\rho$. Define 
$u$ as the smallest positive integer $t$ that satisfies $p^t&gt;q+1$. Then $A_p$ splits into the two  
subsets $A_{p}^{&lt;u}$ and $A_{p}^{\ge u}$. We begin with showing that $A_{p}^{&lt;u}$ has zero density. 
Define $a_p$ as $u+\nu+\d$. We show the inclusion $A_{p}^{&lt;u}\subset C_{&lt;a_p}$, where $C_{&lt;a_p}$ 
was defined in Lemma \ref{lem:6}. Suppose $n\in A_p^{&lt;u}$. Thus, there is an $x$, $0\le x&lt;u$, such that 
\begin{equation}\label{eq:sup}
\nu_p\binom{an}{n}_U&lt;\nu_p(U_{(a-1)n+k})\le x+\nu+\d&lt;a_p,
\end{equation} so that, by the Kummer rule for Lucasnomials, $n$ is in $C_{&lt;a_p}$. But, by Lemma \ref{lem:6}, 
$C_{&lt;a_p}$ has zero density. Hence, $A_{p}^{&lt;u}$ is a zero-density set as well.  



 We take note, for later use, that if we had defined $a_p$ as $u+v+\nu+\d$, where $v\ge0$ is some integer, 
then the same reasoning would have led to $\#A_{p}^{&lt;u+v}(z)=o(z)$. 


 
 We now turn our attention to the case when $x\ge u$. Since 
\begin{equation}\label{eq:18}
\frac{(a-1)n}{\rho}=(\l-1)p^x+p^x-q-1+\frac{\rho-r}{\rho}
\end{equation} 
and $p^x-q-1=(p^x-p^u)+(p^u-q-1)$, we see that $p^x-q-1$ has a string of $x-u$ consecutive $p-1$ digits at 
the positions $x-1,x-2,\ldots,u$. If the $u$-th digit of $n/\rho$, $d_u(n/\rho)$, is $\ge1$, then a minimum 
of $x-u$ carries 
occur in the addition of $(a-1)n/\rho$ and $n/\rho$, and they occur before reaching the position $x$. 
Therefore, with the notation of Lemma \ref{lem:6}, if $n$ belongs to $A_p^x(z)$,  
then there are no $a_p+\th$ consecutive $p-1$ digits in-between positions $x$ and $D$ in the base-$p$ 
expansion of $n/\rho$, where again $a_p=u+\nu+\d$. Thus, by Remark \ref{rem:1}, we obtain an upper bound 
for $\#A_p^x(z)$, namely 
$$
\#A_p^x(z)\ll z\o^{D-x+1}, 
$$ where $0&lt;\o&lt;1$ and the constant implied by the symbol $\ll$ depends only on $U$, $a$, $k$ and $p$. 
We now estimate the size of $A_p^{\ge u}$ by first considering 
its subset $B:=\cup A_p^x$ over 
all $x$'s in $[u,D/2]$. Thus, we obtain 
$$
\#B(z)\le\sum_{x=u}^{\lf D/2\rf}\#A_p^x(z)\ll z\sum_{x=u}^{\lf D/2\rf}\o^{D-x+1}
\le z\frac{\o^{\lc D/2\rc+1}}{1-\o},
$$ which is a $o(z)$-function, as $z$ tends to infinity. 

Let us now fix an integer $x&gt; D/2$. For all $n$ in $A_p^x$, there is a unique $\l$ satisfying $(a-1)n+k=\l\rho p^x$. 
Thus, we get an upper estimate for $\#A_p^x(z)$ by bounding above the corresponding number of $\l$'s. 
Indeed, 
$$
\l\le\frac{(a-1)n+k}{\rho p^x}\le\frac{(a-1)z+k}{\rho p^{D/2}}\ll z^{1/2},
$$ as $z$ tends to infinity. Hence,  
$$
\sum_{D/2&lt;x\le D}\#A_p^x(z)=O(D\sqrt{z})=O(z^{1/2}\log z)=o(z).
$$ Therefore, $\#A_p^{\ge u}(z)=o(z)$. 



 But what if $d_u(n/\rho)=0$? We will see that the above reasoning remains valid provided 
we make some adjustment. The euclidean division of $\l-1$ in (\ref{eq:18}) by $a-1$ gives $\l-1=Q(a-1)+R$  
with $0\le R\le a-2$. Thus, 
\begin{eqnarray*}
\frac{(a-1)n}{\rho}&amp;=&amp;Q(a-1)p^x+(R+1)p^x-q-1+\frac{\rho-r}{\rho},\\
\frac{n}{\rho}&amp;=&amp;Qp^x+S, 
\end{eqnarray*} where $S&lt;\big((a-1)p^x-q\big)/(a-1)\le p^x$. 
 Let $v$ and $w$ be the respective numbers of $p$-ary digits of $a-1$ and $R$. We have $w\le v$. 
Assume $x\ge u+v$. Again $p^x-q-1=(p^x-p^u)+(p^u-q-1)$ so that the base-$p$ expansion of 
$(R+1)p^x-q-1$ has the shape 
$$
\underbrace{*\cdots*}_{\text{$w$ digits}}\;\underbrace{(p-1)(p-1)\cdots(p-1)}_{\text{$x-u$ times}}\;
\underbrace{*\cdots*}_{\text{$u$ digits}}
$$ When one performs the division algorithm most of us learn in elementary school, although here in base $p$ 
rather than $10$, of $(R+1)p^x-q-1$ by 
$a-1$, the quotient has, to the left of the radix point, $v$ or $v-1$ digits less than the dividend.  
That is, $\lf S\rf$ must have $w+x-v$ or $w+x-v+1$ $p$-ary digits. Moreover, in that string of digits 
two nonzero digits are separated by at most $v$ zero digits so that the quotient has to have at least 
one nonzero digit in every string of $v+1$ digits. Because of the $x-u\ge v$ consecutive $p-1$ 
digits in $(R+1)p^x-q-1$, with a moment of thought one sees that the addition of $S$ to $(R+1)p^x-q-1$ 
has to generate a minimum of $x-u-v$ carries left of the radix point. 
Thus, we now have a deficit of at most $u+v+\nu+\d$ carries to fill in in the addition of $(a-1)n/\rho$ and $n/\rho$ 
left of position $x$ to ensure that $n$ is not in $A_p$. 
Hence, defining $a_p$ as $u+v+\nu+\d$, we can argue that $\#A_p^{&lt;u+v}(z)$ and $\#A_p^{\ge u+v}(z)$ 
are both $o(z)$, as $z$ tends to infinity, in the same respective ways we argued that $\#A_p^{&lt;u}(z)$ 
and $\#A_p^{\ge u}(z)$ were each $o(z)$.  
\end{proof}

\begin{theorem}\label{thm:4} Let $U(P,Q)$ be a regular Lucas sequence and $a\ge2$, $k\ge1$ be integers. 
Then for almost all integers $n\ge1$, we find that 
$$
\frac{1}{U_{(a-1)n+k}}\binom{an}{n}_U \text{is an integer.} 
$$ 
\end{theorem} 
\begin{proof} The set $\overline D_{U,a,k}$ is the union of the $A_p$'s over all primes $p$. 
However, by Lemma \ref{lem:2}, all but finitely many of them are empty. Since each of the finitely 
many nonempty ones have density zero by Lemma \ref{lem:7}, we see that $\overline D_{U,a,k}$ itself 
has zero density. 
\end{proof}

 Incidentally, the proof of Lemma \ref{lem:7} provides an answer to one of the three questions 
posed in Remark \ref{rem:2}. We state the answer as a theorem. 

\begin{theorem}\label{thm:5} Let $U(P,Q)$ be a regular Lucas sequence and $a\ge2$, $k\ge1$ be integers. 
Then there exists an integer $m=m(U,a,k)$, $m\ge1$, such that for all integers $n\ge1$
\begin{equation}\label{eq:19}
\frac{m}{U_{(a-1)n+k}}\binom{an}{n}_U \text{is an integer.} 
\end{equation} 
\end{theorem}
\begin{proof} By Lemma \ref{lem:2}, for all but finitely many primes $p$, $A_p$ is empty. But if $A_p$ 
is empty, then $\nu_p(U_{(a-1)n+k})\le\nu_p\binom{an}{n}_U$ for all $n\ge1$. Thus, the existence of an 
$m$, for which (\ref{eq:19}) holds for all $n$, depends on whether for the remaining finitely many primes $p$ the 
difference $\nu_p(U_{(a-1)n+k})-\nu_p\binom{an}{n}_U$ is bounded above. By the proof of Lemma \ref{lem:7} 
this is true. Indeed, with the notation of the lemma, given $A_p$, we saw (\ref{eq:sup}) that for $n\in A_p^{&lt;u+v}$, 
the $p$-adic valuation of $U_{(a-1)n+k}$ is bounded above by $a_p=u+v+\nu+\d$. If $n\in A_p^{\ge u+v}$, 
then we found that at least $x-u-v$ carries occurred in the base-$p$ addition of $n/\rho$ and $(a-1)n/\rho$ 
prior to reaching position $x$. Thus the difference $\nu_p(U_{(a-1)n+k})-\nu_p\binom{an}{n}_U$ is at most 
equal to $a_p$. if $\a_p\ge0$ is the least $e$ for which $\nu_p(U_{(a-1)n+k})-\nu_p\binom{an}{n}_U\le e$ 
for all $n\ge1$, then $m=\prod p^{\a_p}$, the product being over all primes of rank less than $ak$, 
is the least integer to satisfy (\ref{eq:19}).    
\end{proof}

\section{Appendix: The table of $n$-defective Lucas sequences}

 Here is a table of $n$-defective nonzero-discriminant regular Lucas sequences $U(P,Q)$, with $P&gt;0$, for all $n\ge2$. 
We did nothing more than concatenate in one table 
the data contained in the three tables \cite[Tables 1 and 3, pp.\ 78--79]{Bi}, \cite[p.\ 312]{Ab}, except 
for parametrizing the sequences in terms of $P$ and $Q$ rather than $P$ and $\D$. 
  
 Semi-columns are used to separate different parametric families of $n$-defective sequences, while commas are used 
to specify constraints on the parameters.  
Parameters $i$ and $j$ are integers. The case $n=6$ has three lines. To alleviate the table we did not repeat at various lines 
that whenever a parametrization yields the sequence $I=U(2,1)$, it should be discarded. 

\begin{center}
\begin{tabular} {|c|c|}
\hline
 $n$ &amp;  $(P,\,Q)$,\quad $P&gt;0$,\quad$(P,\,Q)\not=(2,\,1)$       \\
\hline
   &amp; \\
\hline
 $2$ &amp; $(1,\,Q),\;Q\not=-3;\qquad(2^i,\,2j+1),\;i\ge1$     \\
\hline
 $3$ &amp; $(P,\,P^2\pm1);\qquad(P,\,P^2\pm3^i),\;3\nmid P$     \\
\hline
 $4$ &amp; $\big(P,\,(P^2\pm1)/2\big),\;P\text{ odd};\qquad(2i,\,2i^2\pm1)$     \\
\hline
 $5$ &amp; $(1,\,-1)\quad(1,\,2)\quad(1,\,3)\quad(1,\,4)\quad(2,\,11)\quad(12,\,55)\quad(12,\,377)$     \\
\hline
     &amp; $\big(P,\,(P^2-1)/3\big),\;3\nmid P,\;P\ge4;\qquad\big(P,\,(P^2\pm3)/3\big),\;3\mid P;$     \\
 $6$ &amp; $\qquad\big(P,\,(P^2-(-2)^i)/3\big),\;P\equiv\pm1\pmod 6,\;i\ge1,\; (P,i)\not=(1,1);\qquad$ \\
     &amp; $\big(P,\,(P^2\pm3\cdot2^i)/3\big),\;P\equiv3\pmod 6,\;i\ge1$ \\
\hline
 $7$ &amp; $(1,\,2)\quad(1,\,5)$     \\
\hline
 $8$ &amp; $(1,\,2)\quad(2,\,7)$     \\
\hline
 $10$ &amp; $(2,\,3)\quad(5,\,7)\quad(5,\,18)$     \\
\hline
 $12$ &amp; $(1,\,-1)\quad(1,\,2)\quad(1,\,3)\quad(1,\,4)\quad(1,\,5)\quad(2,\,15)$     \\
\hline
 $13$ &amp; $(1,\,2)$     \\
\hline
 $18$ &amp; $(1,\,2)$     \\
\hline
 $30$ &amp; $(1,\,2)$     \\
\hline
\end{tabular}
\smallskip
\center{\rm{Table A.}}
\center{List of all $n$-defective regular Lucas sequences $U(P,Q)$, $n\ge2$}
\end{center}


\section{Acknowledgments}

 We thank the referee for his fast return on a long paper and the referee and the editor-in-chief 
for their various inputs. We appreciated the remarks of J.-P. B\'ezivin on his reading of the former papers 
\cite{Ba4, Po}; Theorem \ref{thm:2bis} is derived from a remark of his. On a short private visit to Nicosia, 
Cyprus, last December, P. Damianou and E. Ieronymou  
kindly helped secure a fruitful office in the mathematics department for me. 

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 11B65, 11B39, 05A10.

\noindent \emph{Keywords: }
generalized binomial coefficient, Lucasnomial, generalized
Fuss-Catalan number, Lobb number, ballot number, Lucas sequence, carry,
Kummer's rule, asymptotic density, integrality, divisibility.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A001764},
\seqnum{A003150},
\seqnum{A014847},
\seqnum{A107920}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  March 6 2018;
revised version received   June 6 2018.
Published in {\it Journal of Integer Sequences}, August 22 2018.

\bigskip
\hrule
\bigskip

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