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 $p>\sqrt{2(a-1)n}$ and $p>ak$, then $\nu_p((a-1)n-k)>\nu_p\binom{an}{n}$ so that $n\not\in D_{I,a,-k}$. Indeed, say $(a-1)n-k=cp$. Then $$ c\le\frac{(a-1)n}{p}<\frac{(a-1)n}{\sqrt{2(a-1)n}}=\frac{\sqrt{2(a-1)n}}{2}<\frac{p}{2}. $$ Hence, $(a-1)n=cp+k$ with $c

0$ and estimate the number of $n$ in $(ak^2,z]$ that have a prime factor $p>\sqrt{2(a-1)z}$. The lower bound $ak^2$ for $n$ is sufficient to imply $p>ak$. In the interval $(ak^2,z]$, there are $z/p+O(1)$ multiples of a prime $p>\sqrt{2(a-1)z}$. As $p>\sqrt{z}\ge\sqrt{n}$, 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}

k$ and $3p/2 \sqrt{z}/3$ and $\log p\le\log(\sqrt{z}/2)$.
Thus, as $z$ tends to infinity, $S_z$ is at least equivalent to
$$
\frac23c\frac{\sqrt{z}}{\log z}\sum_{\frac{\sqrt{z}}3 \nu_p\binom{an}{n}_U\bigg\}.
\end{equation} Note that
\begin{equation}\label{eq:ap'}
\overline D_{U,a,k}=\bigcup_pA_p,
\end{equation} where the union is over all primes. We proceed to show that $A_p$ is empty except, possibly, for the
finitely many primes $p$ of rank less than $ak$.
\begin{lemma}\label{lem:2} Suppose $U(P,Q)$ is a nondegenerate fundamental Lucas sequence,
while $a\ge2$ and $k\ge1$ are fixed
integers. Assume $p\nmid Q$ is a prime of rank $\ge ak$. Then, $A_p$ is empty. That is, for all $n\ge1$,
$$
\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, $0<(a-1)e+k a-1$, but $p^{\th-1}\le a-1$.
If $n$ belongs to $C_{k$, and
\begin{equation}\label{eq:+}
q=p+d\quad\text{ with }\quad\frac{p}{a}

**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 $x>p$. Indeed, the derivative $P'$ is positive, since for $x>p$
\begin{eqnarray*}
P'(x)&=&bx^{b-1}-(p-1)\big((b-1)x^{b-2}+(b-2)x^{b-3}+\cdots+1\big)\\
&>&bx^{b-1}-(p-1)(b-1)(x^{b-2}+x^{b-3}+\cdots+1)\\
&=&bx^{b-1}-(p-1)(b-1)\frac{x^{b-1}-1}{x-1}\\
&>&bx^{b-1}-(b-1)(x^{b-1}-1)=x^{b-1}+b-1>0.
\end{eqnarray*}
Since $P(p)=1>0$, the polynomial $P$ has no zero larger than $p$. Thus, $\a**__ D/2$. 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 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
$n$ & $(P,\,Q)$,\quad $P>0$,\quad$(P,\,Q)\not=(2,\,1)$ \\
\hline
& \\
\hline
$2$ & $(1,\,Q),\;Q\not=-3;\qquad(2^i,\,2j+1),\;i\ge1$ \\
\hline
$3$ & $(P,\,P^2\pm1);\qquad(P,\,P^2\pm3^i),\;3\nmid P$ \\
\hline
$4$ & $\big(P,\,(P^2\pm1)/2\big),\;P\text{ odd};\qquad(2i,\,2i^2\pm1)$ \\
\hline
$5$ & $(1,\,-1)\quad(1,\,2)\quad(1,\,3)\quad(1,\,4)\quad(2,\,11)\quad(12,\,55)\quad(12,\,377)$ \\
\hline
& $\big(P,\,(P^2-1)/3\big),\;3\nmid P,\;P\ge4;\qquad\big(P,\,(P^2\pm3)/3\big),\;3\mid P;$ \\
$6$ & $\qquad\big(P,\,(P^2-(-2)^i)/3\big),\;P\equiv\pm1\pmod 6,\;i\ge1,\; (P,i)\not=(1,1);\qquad$ \\
& $\big(P,\,(P^2\pm3\cdot2^i)/3\big),\;P\equiv3\pmod 6,\;i\ge1$ \\
\hline
$7$ & $(1,\,2)\quad(1,\,5)$ \\
\hline
$8$ & $(1,\,2)\quad(2,\,7)$ \\
\hline
$10$ & $(2,\,3)\quad(5,\,7)\quad(5,\,18)$ \\
\hline
$12$ & $(1,\,-1)\quad(1,\,2)\quad(1,\,3)\quad(1,\,4)\quad(1,\,5)\quad(2,\,15)$ \\
\hline
$13$ & $(1,\,2)$ \\
\hline
$18$ & $(1,\,2)$ \\
\hline
$30$ & $(1,\,2)$ \\
\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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