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\begin{center}
\vskip 1cm{\LARGE\bf
Another Lucasnomial Generalization
\vskip .2in
of Wolstenholme's Congruence
}
\vskip 1cm
\large
Christian Ballot \\
D\'epartement de Math\'ematiques et Informatique\\
Universit\'e de Caen-Normandie \\
F14032 Caen Cedex \\
France \\
\href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr} \\
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\begin{abstract} If $p\ge5$ is a prime, then Wolstenholme's congruence
stipulates that $\binom{2p-1}{p-1}\equiv1\pmod{p^3}$. New
generalizations of this congruence to Lucasnomials
$\pmod{U_p^2V_p/V_1}$ are given, where $U$ and $V$
are a pair of Lucas sequences.
\end{abstract}
\section{Introduction}
\label{sec:1}
The binomial coefficient congruence
\begin{equation}\label{eq:W}
\binom{2p-1}{p-1}\equiv1\pmod {p^3},
\end{equation} valid for all primes $p\ge5$, was established by Wolstenholme
\cite{Wo} in 1862.
Glaisher (\cite[p.\ 21]{Glai1}, \cite[p.\ 33]{Glai2}) later gave the slightly
more general congruence
\begin{equation}\label{eq:W+}
\binom{(k+1)p-1}{p-1}\equiv1\pmod {p^3},
\end{equation} for all nonnegative
integers $k$ and all primes $p\ge5$.
An interest in finding an analogue, or a generalization of the
Wolstenholme or the Glaisher
congruences for Lucasnomials appears in various papers \cite{An, Ba1, KW1,
KW2, KW3, Shi}.
If $A=(a_n)_{n\ge0}$ is a sequence of integers, $a_n\not=0$ for
$n>0$, then generalized binomial
coefficients, $\binom{m}{n}_A$, with respect to $A$ are defined for
$m\ge n\ge0$ to be
$$
\binom{m}{n}_A=\frac{a_ma_{m-1}\dots a_{m-n+1}}{a_na_{n-1}\dots a_1},
$$ if $m\ge n\ge1$, and $1$ if $n=0$.
Lucasnomials $\binom{m}{n}_U$ are generalized binomial coefficients defined with
respect to a fundamental Lucas sequence $U$. They turn out always to
be integers. Given two nonzero
integers $P$ and $Q$, the {\it fundamental Lucas sequence} $U=U(P,Q)$ is the
second-order linear recurring sequence that satisfies the recursion
\begin{equation}\label{eq:rec}
U_{n+2}=PU_{n+1}-QU_n,
\end{equation} for all integers $n$, and has initial conditions
$U_0=0$ and $U_1=1$.
If $U_n\not=0$ for all $n\ge1$, then $U$ is said to be {\it
nondegenerate}. A necessary and sufficient condition for $U$ to be
nondegenerate is that $U_{12}\not=0$. Lucas sequences $U$ are {\it
divisibility} sequences, i.e., they satisfy
$$
m\mid n\implies U_m\mid U_n,
$$ for all $n\ge m\ge1$. If $\gcd(P,Q)=1$, then $U(P,Q)$ is called
{\it regular}. If $U$ is regular, then it satisfies
$$
\gcd(U_m,U_n)=|U_{\gcd(m,n)}|,
$$ for all nonnegative $m$ and $n$, not both zero. A sequence
with this property is a {\it strong divisibility} sequence.
A prime $p$ is {\it regular} with respect to $U(P,Q)$ if
$p\nmid\gcd(P,Q)$. An integer $m$ is said to be {\it regular} if all
its prime factors are regular. A {\it special} prime is one that
divides $\gcd(P,Q)$. By extension an integer $m$ is said to be {\it
special} if all its prime factors are special.
If $m\ge2$ is an integer, then the {\it rank}, $\rho=\rho(m)$, of $m$ is the least
$t\ge2$ for which $m\mid U_t$. It is guaranteed to exist if
$\gcd(m,Q)=1$. If $\gcd(m,Q)=1$, then the rank $\rho$ satisfies
$$
m\mid U_n\iff \rho\mid n.
$$ If $p\nmid Q$ is an odd prime, then $\rho(p)$ is a divisor
of $p-(D\,|\,p)$, where $D=P^2-4Q$ and $(D\,|\,p)$ is the Legendre
character of $D$ with respect to $p$. The rank of $p$ is {\it maximal}
if $\rho(p)=p-(D\,|\,p)$. To every Lucas sequence $U(P,Q)$,
there is an {\it associate}, or {\it companion} Lucas sequence $V$
which satisfies the same recursion (\ref{eq:rec}), but has initial
values $V_0=2$, $V_1=P$. If $D\not=0$ and $\a$ and $\b$ denote the
zeros of $x^2-Px+Q$, then for all $n\ge0$
\begin{equation}\label{eq:Bin}
U_n=\frac{\a^n-\b^n}{\a-\b}\quad\text{ and }\quad V_n=\a^n+\b^n.
\end{equation}
We won't say much more about Lucas sequences,
but refer interested readers to the original Lucas memoir \cite{Lu1}, and to
Chapter 4 of the book \cite{Wi}. The reason why it is not
unreasonable, with luck, to expect properties of ordinary binomial coefficients to extend to
Lucasnomials is that binomial coefficients are special
Lucasnomials. Indeed, the binomial coefficients are the Lucasnomials
attached to the fundamental Lucas sequence $U_n(2,1)=n$.
In \cite{Ba1}, one finds two distinct generalizations of
(\ref{eq:W+}) for Lucasnomials. The first, which
appeared in a weaker form in \cite[Lemma 6]{Shi},
received a fully detailed proof and is stated below as a theorem. It is
actually a concatenation of Theorems 3 and 7 in \cite{Ba1}.
\begin{theorem} \label{thm:1} Let $U=U(P,Q)$ be a fundamental Lucas sequence with parameters $P$ and
$Q$. If a prime $p\ge3$, $p\nmid Q$, has rank $\rho$ in $U(P,Q)$,
then the congruence
\begin{equation}\label{eq:LW}
\binom{(k+1)\rho-1}{\rho-1}_U\equiv(-1)^{k(\rho-1)}Q^{k\rho(\rho-1)/2}\pmod {p^\nu},
\end{equation} holds for all integers $k\ge0$ with
$$
\nu=2+[\,p\ge5\,]\cdot[\,\rho\text{ is maximal }].
$$
\end{theorem}
In the statement of the theorem we made use of the Iverson symbol
$[-]$, where $[\cal P]$ is $1$, if $\cal P$ is a true statement, and
$[\cal P]$ is $0$ otherwise. That is,
$$
\nu=\begin{cases}3,&\text{ if } \rho \text{ is maximal and }p>3;\\
2,&\text{ otherwise.}
\end{cases}
$$
In stating Theorem \ref{thm:1}, one would expect the Lucas
sequence $U$ to be nondegenerate. However, with the convention that
two zero-terms, one in the numerator, the other in the denominator of
a Lucasnomial, cancel out as $1$, the theorem holds even in the
degenerate case \cite{Ba1}.
The second generalization, \cite[Thm.\ 9]{Ba1}, went as follows:
\begin{theorem} \label{thm:2} Suppose $U(P,Q)$ is a nondegenerate
regular fundamental Lucas sequence and $p\ge3$ is a prime. Then
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^2},
$$ for all $k\ge0$.
\end{theorem}
When $|U_p|>1$, $p$ is the rank of $U_p$. Thus, as mentioned in \cite{Ba1},
if $|U_p|$ is prime, then Theorem \ref{thm:2}
follows from Theorem \ref{thm:1}. Indeed, the hypotheses of Theorem
\ref{thm:2} entail that if $|U_p|$ is prime, then $U_p\nmid Q$.
For if $U_p\mid Q$, then, by (\ref{eq:rec}), $U_p\equiv P^{p-1}\pmod{U_p}$, implying that
$U_p\mid P$. This would contradict the regularity of $U$. For
$U(1,-1)$, i.e., for the Fibonacci sequence, the congruence
in Theorem \ref{thm:2} follows from the statement of a problem posed by Ohtsuka \cite{Oh}.
Mention was made in \cite{Ba1} that the published solution to the
Ohtsuka problem \cite{Ba0} can be turned into a general proof of
Theorem \ref{thm:2}. This note provides a
proof of Theorem \ref{thm:3}, a more general theorem and a stronger congruence
than Theorem \ref{thm:2}, which, when $U$ is the Fibonacci
sequence, reduces to the initial problem \cite{Oh} posed by Ohtsuka.
We point out that the congruence holds irrespective of the regularity
of $U(P,Q)$.
\begin{theorem} \label{thm:3} Suppose $U(P,Q)$ is a nondegenerate
fundamental Lucas sequence and $p\ge3$ is a prime. Then the congruence
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^2V_p/P},
$$ holds for all $k\ge0$, where $V(P,Q)$ is the companion Lucas sequence
associated with $U$.
\end{theorem}
\begin{remark}\label{rem:1} Theorem \ref{thm:3} does not hold with the
modulus $U_p^2V_p$. For instance, with $P=5$, $Q=1$ and $p=5$, we
find that
$\binom{2p-1}{p-1}_U\equiv153\,318\,506\not\equiv1\pmod{U_p^2V_p}$. Here, $U_5^2V_5=766\,592\,525$, whereas
$U_5^2V_5/P=153\,318\,505$.
\end{remark}
Theorem \ref{thm:3} implies a stronger version of Theorem \ref{thm:2}
where $U(P,Q)$ need not be regular, which is worth pointing out and
stating.
\begin{theorem} \label{thm:2'} Suppose $U(P,Q)$ is a nondegenerate
fundamental Lucas sequence and $p\ge3$ is a prime. Then for all $k\ge0$
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^2}.
$$
\end{theorem}
Section \ref{sec:2} gives a proof of Theorem \ref{thm:3}. We proceed
roughly as follows. Put $M_p=U_pU_{2p}/P=U_p^2V_p/P$. Let $R_p$ be
the largest regular integer factor of $M_p$, i.e., the largest factor
of $M_p$ prime to $\gcd(P,Q)$. Then $M_p=R_pS_p$, where $S_p$ is the
largest special factor of $M_p$. To prove the congruence of Theorem \ref{thm:3} modulo
$R_p$, we will use a generalization of the
non-published proof of Ohtsuka, who took the kind initiative to send
it to the author in February 2015. Actually, for the prime $p=3$,
this approach immediately gives the congruence modulo $M_3$. (Two
alternatives would have been either to use a generalization of the
published solution \cite{Ba0} to Ohtsuka's problem \cite{Oh}, or to use Theorem
\ref{thm:1}. The first alternative is longer. The second can only
yield the congruence modulo $\gcd(U_p^2,R_p)$.) To prove
the congruence of Theorem \ref{thm:3} modulo $S_p$, given a
special prime $q$ it will suffice to show that both integers
$\binom{(k+1)p-1}{p-1}_U$ and $Q^{kp(p-1)/2}$ have a $q$-adic
valuation at least as high as that of $M_p$. We are able to prove
the congruence modulo $S_p$ in all cases because a full description
\cite{Ba3} of the $q$-adic
valuation of the terms of Lucas sequences exists. (Throughout the paper, if $m$
is an integer, $\nu_q(m)$ denotes its $q$-adic
valuation.) Actually, by Remark \ref{rem:2} of Section \ref{sec:4},
proving the regular case only would have been sufficient to imply
Theorem \ref{thm:3} in its full generality.
Section \ref{sec:3} gives some further results: In
Proposition \ref{prop:1}, we find a few instances of pairs
$(U(P,Q),p)$ when Theorem \ref{thm:3} holds with respect to the
modulus $U_p^3V_p/P$.
Note that the congruence of Glaisher (\ref{eq:W+}) gives
$$
\binom{kp}{p}=\frac{kp}{p}\binom{kp-1}{p-1}\equiv\binom{k}{1}\pmod{p^3}.
$$ It was generalized further \cite{Bru} as follows: if $p\ge5$ is prime and
$\ell$ and $k$ are nonnegative
integers, then
\begin{equation}\label{eq:Bru}
\binom{kp}{\ell p}\equiv\binom{k}{\ell}\pmod{p^3}.
\end{equation}
Kimball and Webb \cite{KW4} gave an analogue of (\ref{eq:Bru}) for
Fibonomials, i.e., Lucasnomials with respect to the Fibonacci
sequence $U(1,-1)$, but modulo $p^2$. However, a Lucasnomial
generalization of (\ref{eq:Bru}) along the line of
Theorem \ref{thm:1} modulo $p^3$ exists \cite[Thm.\ 13]{Ba1}. We give another,
in line with Theorem \ref{thm:3}, in Theorem
\ref{thm:4}.
Section \ref{sec:3} contains yet another proof of Theorem \ref{thm:3}
for the case $p=3$. This proof generalizes the proof given in the
published solution \cite{Ba0} of Ohtsuka's problem \cite{Oh}.
The referee made some numerical experiments that suggested that a
polynomial version of Theorem \ref{thm:3} might hold in the ring
$\Z[P,Q]$ and that this polynomial version might also hold for $p=2$
whenever $4$ divides $k$. We added a new section to the paper
to address the referee's questions. Section \ref{sec:4} contains
Theorems \ref{thm:5} and \ref{thm:6} which prove the
referee's observations to be exact.
\section{Proof of Theorem \ref{thm:3}}
\label{sec:2}
We begin with a generalized Cassini identity. This identity is proved
\cite{Joh}
in a long and indirect manner using matrices and, probably, in other
places as well. We give a very short and direct proof which uses
the formulas (\ref{eq:Bin}) in the next lemma.
\begin{lemma} \label{lem:10} Suppose $U(P,Q)$ is a fundamental Lucas
sequence with nonzero discriminant $D=P^2-4Q$. Then, for all
$r\ge0$, we have the identity
$$
U_aU_b-U_cU_d=Q^r(U_{a-r}U_{b-r}-U_{c-r}U_{d-r}),
$$ provided $a+b=c+d$.
\end{lemma}
\begin{proof} The quantity $U_tU_{n-t}-Q^rU_{t-r}U_{n-t-r}$ is independent
of $t$. Indeed, if $\a$ and $\b$ are the distinct zeros of $x^2-Px+Q$,
then using $Q=\a\b$ we obtain
\begin{align*}
D(U_tU_{n-t}&-Q^rU_{t-r}U_{n-t-r})=(\a^t-\b^t)(\a^{n-t}-\b^{n-t})-Q^r(\a^{t-r}-\b^{t-r})(\a^{n-t-r}-\b^{n-t-r})\\
&=(\a^n+\b^n-\a^t\b^{n-t}-\b^t\a^{n-t})-Q^r(\a^{n-2r}+\b^{n-2r})+Q^r(\a^{t-r}\b^{n-t-r}+\b^{t-r}\a^{n-t-r})\\
&=V_n-Q^rV_{n-2r}.
\end{align*}
Therefore, for all integers $t$ and $s$, we see that
$$
U_tU_{n-t}-Q^rU_{t-r}U_{n-t-r}=U_sU_{n-s}-Q^rU_{s-r}U_{n-s-r},
$$ which yields the identity on putting $a=t$, $b=n-t$, $c=s$ and
$d=n-s$.
\end{proof}
\begin{lemma}\label{lem:11} For all odd primes $p$ and all integers
$k\ge1$, we have
$$
\binom{(k+1)p-1}{p-1}_U=\prod_{i=1}^{(p-1)/2}\bigg(\frac{U_{kp}U_{(k+1)p}}{U_iU_{p-i}}+Q^{kp}\bigg).
$$
\end{lemma}
\begin{proof} By Lemma \ref{lem:10}, we see that
$$
U_{(k+1)p-i}U_{kp+i}-U_{(k+1)p}U_{kp}=Q^{kp}(U_{p-i}U_i-U_pU_0).
$$ Thus,
$$
U_{(k+1)p-i}U_{kp+i}=U_{(k+1)p}U_{kp}+Q^{kp}U_{p-i}U_i.
$$ Therefore,
\begin{align*}
\binom{(k+1)p-1}{p-1}_U&=\prod_{i=1}^{p-1}\frac{U_{kp+i}}{U_i}=\prod_{i=1}^{(p-1)/2}
\frac{U_{(k+1)p-i}U_{kp+i}}{U_{p-i}U_i}\\
&=\prod_{i=1}^{(p-1)/2}\frac{U_{(k+1)p}U_{kp}+Q^{kp}U_{p-i}U_i}{U_{p-i}U_i}\\
&=\prod_{i=1}^{(p-1)/2}\bigg(\frac{U_{kp}U_{(k+1)p}}{U_iU_{p-i}}+Q^{kp}\bigg).
\end{align*}
\end{proof}
\begin{lemma}\label{lem:12} Theorem \ref{thm:3} holds for $p=3$.
\end{lemma}
\begin{proof} By Lemma \ref{lem:11}, we obtain
$$
\binom{(k+1)p-1}{p-1}_U=U_{kp}U_{(k+1)p}/P+Q^{kp}\equiv
Q^{kp}=Q^{kp(p-1)/2}\pmod{M_p},
$$ for $p=3$ since $U_2=P$ and $U_pU_{2p}$ divides $U_{kp}U_{(k+1)p}$, where
$M_p=U_pU_{2p}/P$.
\end{proof}
\begin{lemma}\label{lem:13} Suppose $q\nmid Q$ is a prime.
Then, for all $n\ge0$, $q\nmid\gcd(U_{n+1},U_n)$.
\end{lemma}
\begin{proof} If not, there must exist a minimal integer $m\ge1$ such
that $q$ divides $U_m$ and $U_{m+1}$. Since $q\nmid\gcd(U_1,U_2)$,
it must be that
$m\ge2$. But as $QU_{m-1}=PU_m-U_{m+1}$ and $q\nmid Q$, we see that $q\mid
U_{m-1}$. Thus, $q\mid\gcd(U_{m-1},U_m)$, which contradicts the
minimality of $m$.
\end{proof}
\begin{lemma}\label{lem:14} Let $q$ be a regular prime with respect to
$U(P,Q)$. Then, for all $m\ge n>0$, $\gcd(U_m,U_n)$ and
$U_{\gcd(m,n)}$ share the same $q$-adic valuation.
\end{lemma}
\begin{proof} If $q\mid Q$, then $q\nmid U_n$ for any $n>0$. Thus, the
result holds in this case. Suppose $q\nmid Q$. Certainly, because $U$ is a
divisibility sequence, the $q$-adic valuation of $U_{\gcd(m,n)}$ is
less than or equal to the $q$-adic valuation of
$\gcd(U_m,U_n)$. Assume $q^\ell\mid \gcd(U_m,U_n)$. Then by the
Lucas identity
$$
U_m=U_{n+1}U_{m-n}-QU_nU_{m-n-1},
$$ we see that $q^\ell\mid U_{n+1}U_{m-n}$. Since $q\nmid Q$ we know
by Lemma \ref{lem:13} that $q$ does not divide $U_{n+1}$. Thus, $q^\ell\mid
U_{m-n}$. Therefore, $q^\ell\mid U_r$, where $r$ is the first
remainder in the Euclidean division of $m$ by $n$. Reiterating the reasoning with $n$ and
$r$ in place of $m$ and $n$ and, further, with any two successive remainders in the
Euclidean division algorithm of $m$ by $n$ we find that $q^\ell$ divides
$U_{\gcd(m,n)}$.
\end{proof}
\begin{theorem}\label{thm:R_p} The congruence of Theorem \ref{thm:3}
holds modulo $R_p$, where $R_p$ is the regular part of
$M_p=U_pU_{2p}/P$.
\end{theorem}
\begin{proof} Both $U_{kp}U_{(k+1)p}$
and $U_iU_{p-i}$ are divisible by $U_2=P$ so that
$$
\frac{U_{kp}U_{(k+1)p}}{U_iU_{p-i}}=\frac{U_{kp}U_{(k+1)p}/P}{U_iU_{p-i}/P}.
$$ By Lemma \ref{lem:11}, the theorem will hold if we show that for all
$i$, $1\le i\le(p-1)/2$, $U_iU_{p-i}/P$ and $R_p$ are coprime
integers. Let $q$ be a prime factor of $R_p$. Since $i$ and $p-i$ are
coprime, Lemma \ref{lem:14} tells us that
$q\nmid\gcd(U_i,U_{p-i})$. Thus, with the notation $m\sim_qn$ meaning
that $\nu_q(m)=\nu_q(n)$, we obtain
\begin{align*}
\gcd(U_iU_{p-i},U_pU_{2p})&\sim_q\gcd(U_i,U_pU_{2p})\cdot\gcd(U_{p-i},U_pU_{2p})\\
&\sim_q\gcd(U_i,U_{2p})\cdot\gcd(U_{p-i},U_{2p})\\
&\sim_qU_1U_2=P
\end{align*} Therefore, $\gcd(U_iU_{p-i}/P,M_p)\sim_q1$.
\end{proof}
We now consider the congruence of Theorem \ref{thm:3} modulo special
primes. Hence, throughout the remainder of this section $q$
designates a special prime of $U(P,Q)$ with $P=q^aP'$ and $Q=q^bQ'$,
$a\ge1$, $b\ge1$ and $q\nmid P'Q'$.
The $q$-adic valuation of the terms $U_n$, ($n\ge1$), is simplest in the
case $b>2a$. In this case, we have \cite[Thm.\ 1.2]{Ba3} for all $n\ge1$
\begin{equation}\label{eq:b>2a}
\nu_q(U_n)=(n-1)a.
\end{equation}
\begin{lemma}\label{lem:15} Suppose $U(P,Q)$ is a
fundamental Lucas sequence and $p\ge5$ is a prime. If $q$ a special prime
with $b>2a$, then for all $k\ge0$
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {q^{\nu_q(M_p)}},
$$ where $M_p=U_pU_{2p}/P$.
\end{lemma}
\begin{proof} By (\ref{eq:b>2a}),
$\nu_q(M_p)=(p-1)a+(2p-1)a-a=3(p-1)a$. Thus, to prove the lemma it
suffices to see that both $\binom{(k+1)p-1}{p-1}_U$ and
$Q^{kp(p-1)/2}$ have a $q$-adic valuation of at least
$3(p-1)a$. Indeed, we have for all $k\ge1$ and $p\ge5$
\begin{equation}\label{eq:Q}
\nu_q(Q^{kp(p-1)/2})=bkp(p-1)/2>ap(p-1)>3(p-1)a.
\end{equation} Also as
$\binom{(k+1)p-1}{p-1}_U=\prod_{i=1}^{p-1}\frac{U_{kp+i}}{U_i}$, we
find using (\ref{eq:b>2a}) that
\begin{align*}
\nu_q\bigg(\binom{(k+1)p-1}{p-1}_U\bigg)&=\sum_{i=1}^{p-1}\big((kp+i-1)a-(i-1)a\big)\\
&=\sum_{i=1}^{p-1}kpa=k(p-1)pa>3(p-1)a.
\end{align*}
\end{proof}
We now address the case $b=2a$.
\begin{lemma}\label{lem:16} Suppose $U(P,Q)$ is a
fundamental Lucas sequence and $p\ge5$ is a prime. If $q$ is a special prime
with $b=2a$, then for all $k\ge0$
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {q^{\nu_q(M_p)}},
$$ where $M_p=U_pU_{2p}/P$.
\end{lemma}
\begin{proof} If $b=2a$, then, as seen in the proof of \cite[Thm.\ 2.2]{Ba3}, we
have for all $n\ge0$, $U_n=q^{(n-1)a}U'_n$, where $U'$ is the Lucas
sequence $U(P',Q')$. Thus,
$$
\nu_q(M_p)=3(p-1)a+\nu_q(M'_p),
$$ where $M'_p=U'_pU'_{2p}/P'$. Since
$q$ is regular with respect to $U'$, we find by Theorem \ref{thm:R_p}
that
\begin{equation}\label{eq:11}
\binom{(k+1)p-1}{p-1}_{U'}\equiv
(Q')^{kp(p-1)/2}\pmod{q^{\nu_q(M'_p)}}.
\end{equation} Since $U_n=q^{(n-1)a}U'_n$, we see that
$$
\binom{(k+1)p-1}{p-1}_U=q^{kp(p-1)a}\binom{(k+1)p-1}{p-1}_{U'}.
$$ But we also find that
$$
Q^{kp(p-1)/2}=q^{kp(p-1)a}\cdot(Q')^{kp(p-1)/2}.
$$ Thus multiplying the congruence (\ref{eq:11}) through by $q^{kp(p-1)a}$
we obtain
$$
\binom{(k+1)p-1}{p-1}_U\equiv
Q^{kp(p-1)/2}\pmod{q^{\nu_q(M'_p)+kp(p-1)a}}.
$$ Since $kp(p-1)a>3(p-1)a$, we may degrade the modulus in the
previous congruence and prove our lemma.
\end{proof}
If $b<2a$, then by \cite[Thm.\ 1.2]{Ba3}
\begin{equation}\label{eq:12}
\nu_q(U_{2n+1})=bn,
\end{equation} while
\begin{equation}\label{eq:13}
\nu_q(U_{2n})=bn+(a-b)+\nu_q(n)+c,
\end{equation} where $c$ is a nonzero constant only if $q=2$ or $3$, $2a=b+1$ and
$q\mid n$.
\begin{lemma}\label{lem:17} Suppose $U(P,Q)$ is a
fundamental Lucas sequence and $p\ge5$ is a prime. If $q$ a special prime
with $b<2a$, then for all $k\ge0$
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {q^{\nu_q(M_p)}},
$$ where $M_p=U_pU_{2p}/P$.
\end{lemma}
\begin{proof} By equations (\ref{eq:12}) and (\ref{eq:13}), we
calculate that
$$
\nu_q(M_p)=3b(p-1)/2+\nu_q(p).
$$ Indeed, by (\ref{eq:13}), for $c$ to be nonzero, we need $q$ to
divide $p$, i.e., $q=p$. As $p\ge5$, $q$ is not equal to $2$ or $3$.
Thus, $c=0$.
Again we verify below that $\nu_q(Q^{kp(p-1)/2})$
exceeds $\nu_q(M_p)$ for all $k\ge1$. For
$$
\nu_q(Q^{kp(p-1)/2})=kbp(p-1)/2\ge5b(p-1)/2>3b(p-1)/2+\nu_q(p).
$$ Now, assuming $k$ is {\bf even}, we write
$$
\binom{(k+1)p-1}{p-1}_U=\prod_{i=1}^{p-1}\frac{U_{kp+i}}{U_i}=\prod_{i=1}^{(p-1)/2}
\frac{U_{kp+2i}}{U_{2i}}\cdot\frac{U_{kp+2i-1}}{U_{2i-1}}.
$$
In evaluating the $q$-adic valuation of $\binom{(k+1)p-1}{p-1}_U$, we
make two observations. First, there are exactly $(p-1)/2$ even-indexed
terms in both the numerator and the denominator of the above product.
So the contribution of the quantities $(a-b)$ from equation
(\ref{eq:13}) cancel out.
The terms involving a nonzero quantity $c$ require $q$ to divide their
index. We claim there are at least as many such indices among the
`$kp+2i$' as among the `$2i$' so
their total contribution to the $q$-adic valuation of
$\binom{(k+1)p-1}{p-1}_U$ is nonnegative. If $q=2$ this is clearly
true. If $q=3$, then $3$ divides an integer in the interval
$[1,(p-1)/2]$ exactly $\lfloor(p-1)/6\rfloor$ times. It divides an
integer in $[1+\frac{kp}2,\frac{p-1}2+\frac{kp}2]$ exactly
$\lfloor(kp+p-1)/6\rfloor-\lfloor(kp)/6\rfloor$ times. But as
for any two real numbers $x$ and $y$, $\lfloor x+y\rfloor\ge\lfloor
x\rfloor+\lfloor y\rfloor$, we see that
$$
\bigg\lfloor\frac{kp+p-1}{6}\bigg\rfloor\ge\bigg\lfloor\frac{p-1}{6}\bigg\rfloor+
\bigg\lfloor\frac{kp}{6}\bigg\rfloor.
$$ The total contribution of the quantities $\nu_q(n)$ which appear in
(\ref{eq:13}) for even-indexed terms is given by
$$
\nu_q\bigg(\prod_{i=1}^{(p-1)/2}\frac{kp/2+i}{i}\bigg)=\nu_q\bigg(\binom{kp/2+(p-1)/2}{(p-1)/2}\bigg)\ge0,
$$ since binomial coefficients are integers.
Thus, using again equations (\ref{eq:12}) and (\ref{eq:13}), we
deduce that the $q$-adic valuation of $\binom{(k+1)p-1}{p-1}_U$ is at
least
$$
b\sum_{i=1}^{(p-1)/2}\big((kp/2+i)-i\big)+b\sum_{i=1}^{(p-1)/2}\big((kp/2+i-1)-(i-1)\big)=
kbp(p-1)/2>\nu_q(M_p),
$$ proving our claim. The case $k$ {\bf odd} can be treated similarly
obtaining again the lower bound $kbp(p-1)/2$ for
$\nu_q\big(\binom{(k+1)p-1}{p-1}_U\big)$.
\end{proof}
Thus gathering together Lemmas \ref{lem:15}, \ref{lem:16} and
\ref{lem:17}, we have shown the following theorem.
\begin{theorem} \label{thm:S_p} Let $U=U(P,Q)$ be a nondegenerate fundamental
Lucas sequence, $p\ge5$ a prime, $k\ge0$ an
integer. Then
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod{S_p},
$$ where $S_p$ is the largest special factor of $M_p=U_pU_{2p}/P$.
\end{theorem}
Since $M_p=R_pS_p$ and $\gcd(R_p,S_p)=1$, putting together
Lemma \ref{lem:12} for the case $p=3$ and Theorems \ref{thm:R_p} and
\ref{thm:S_p}, we have a proof of Theorem \ref{thm:3}.
\section{Further complementary results}
\label{sec:3}
By Theorem \ref{thm:1}, the condition `$|U_p|$ is prime and has maximal rank in $U$' is
a sufficient condition for the congruence in Theorem \ref{thm:2} to hold
modulo $U_p^3$. The next proposition describes all the cases when
this rare condition is met. We recall first that a regular $U$ is called
{\it $n$-defective} if all primes of rank $n$ divide $D$. We know
that if $n>30$ and $U$ is regular, then $U$ is never $n$-defective
\cite{Bi}. Moreover, all cases of defectiveness were described in
several tables \cite{Ab, Bi} using the parameters $P$ and $D$. A single
table \cite[p. 33]{Ba2} describes all cases of defectiveness using
the parameters $P$ and $Q$.
\begin{proposition} \label{prop:1} Suppose $p\ge5$ is prime and $|U_p|$ is a
prime of maximal rank in $U$, where $U=U(P,Q)$, $P>0$, is a
fundamental Lucas sequence other than $I=U(2,1)$. Then, either
$$
p=5\quad\text{ and }\quad (P,Q)\in\{(1,-1),(1,4),(2,11)\},
$$ or
$$
p=7\quad\text{ and }\quad (P,Q)=(1,2).
$$ Thus, in these four cases, we find that
\begin{equation}\label{eq:oui}
\binom{(k+1)p-1}{p-1}_U\equiv Q^{\frac{kp(p-1)}2}\pmod{U_p^3V_p/P},
\end{equation} for all $k\ge0$.
\end{proposition}
\begin{proof} Note that the hypotheses imply that $U(P,Q)$ is
regular. For if a prime $q$ divides $\gcd(P,Q)$, then, by \cite[Thm.\
1.1]{Ba3}, $U_p$ is at least divisible by $q^{(p-1)/2}$. As $p\ge5$,
this would contradict the primality of $|U_p|$. Because $|U_p|$
divides $U_n$ at $n=p$ the rank of $|U_p|$ must divide $p$ and thus
be equal to $p$. If this rank is
maximal, then $\rho(|U_p|)=p=|U_p|\pm1$, or $|U_p|$. As $p\ge5$, there are no
two primes $|U_p|$ and $p$ one apart from each other. Hence,
$U_p=\pm p$. Not surprisingly this Diophantine equation has few
solutions since $U_n$ grows exponentially. But the condition
$U_p=\pm p$ implies that $p\mid D$ and that $U$ is $p$-defective.
Therefore, $p\le30$ and we need only check those $U$ in \cite[Table
A, p. 33]{Ba2}
that are $p$-defective with $5\le p\le29$. Actually, there are only seven $U$
that are $5$-defective, two that are $7$-defective and one which
is $13$-defective.
Suppose $p=5$. The seven sequences correspond to
$(P,Q)=(1,-1),\;(1,2),\; (1,3),\; (1,4),\;$ $(2,11),\; (12,55)$ and
$(12,377)$. The discriminant $D=P^2-4Q$ is divisible by $5$ only for
$(P,Q)=(1,-1),\;(1,4)$ and $(2,11)$. Since $U_5=P^4-3P^2Q+Q^2$, it is easy to check that
$U_5$ is $\pm5$ in these three cases. If $p=7$, then $U(1,2)$ and
$U(1,5)$ are $7$-defective and their respective seventh terms are
$7$ and $1$. The only $13$-defective sequence is $U(1,2)$ with
$D=-7$ not divisible by $13$. Since $p=U_p$, $p\nmid Q$. By the
identity $V_p^2-DU_p^2=4Q^p$, we see that $\gcd(U_p,V_p)=1$. Thus,
the congruence (\ref{eq:oui}) holds.
\end{proof}
Thus, for instance, if $U=U(1,2)$, $p=7$ and $k=1$, then
$U_p=7$, $V_p=-13$ and
$$
\binom{2p-1}{p-1}_U-Q^{\frac{p(p-1)}{2}}=-9\cdot11\cdot17\cdot23-2^{21}=-7^3\cdot13\cdot499\equiv0\pmod{U_p^3V_p}.
$$ However, again, the congruence modulo $U_p^3$ holds as a
consequence of Theorem \ref{thm:1}.
Note that Theorem \ref{thm:3} yields for all $k\ge1$
$$
\binom{kp}{p}_U=\binom{kp-1}{p-1}_U\cdot\frac{U_{kp}}{U_p}\equiv
Q^{(k-1)p(p-1)/2}\binom{k}{1}_{U'}\pmod{U_p^2V_p/P},
$$ where $U_n'=U_{np}$. This can be generalized to all Lucasnomials of
the type $\binom{kp}{\ell p}_U$.
\begin{theorem} \label{thm:4} Suppose $U=U(P,Q)$ is a fundamental
Lucas sequence, $p\ge3$ is a prime and $k\ge\ell\ge0$ are
integers. Then
\begin{equation}\label{eq:14}
\binom{kp}{\ell p}_U\equiv
Q^{(k-\ell)\ell\binom{p}{2}}\binom{k}{\ell}_{U'}\pmod {U_p^\nu V_p/P},
\end{equation} where $U_n'=U_{pn}$ and $\nu=2+[\,p\ge5\,]\cdot[\,U_p=\pm p\,].$
\end{theorem}
\begin{proof} It suffices to reproduce the proof of \cite[Thm.\
13]{Ba1} replacing $\rho$ by $p$ and the modulus $p^3$ by $U_p^\nu
V_p/P$. The key point in the proof of \cite[Thm.\ 13]{Ba1} was
\cite[Rmk.\ 4]{Ba1}, which has an equivalent here,
namely
$$
\binom{(k+1)p-1}{p-1}_U\equiv\binom{2p-1}{p-1}_U^k\pmod {U_p^\nu
V_p/P},
$$ by Theorem \ref{thm:3} and Proposition \ref{prop:1}.
\end{proof}
For the sake of curiosity we give another proof of Theorem
\ref{thm:3} for the case $p=3$, which generalizes the proof published
for $p=3$ in the Fibonacci case \cite[p.\ 191]{Ba0}.
\noindent{\it Another proof of Theorem \ref{thm:3} for $p=3$, i.e., of
Lemma \ref{lem:12}.}
Put
$$
M:=\frac{U_3^2V_3}P=U_3\cdot\frac{U_6}{P}=(P^2-Q)(P^4-4P^2Q+3Q^2)=P^6-5QP^4+7P^2Q^2-3Q^3.
$$ One may observe that
$a_k:=\binom{3(k+1)-1}{3-1}_U=U_{3k+2}U_{3k+1}/P$ is a recurrent
sequence. It is of the form
$(A\a^{3k}+B\b^{3k})\cdot(C\a^{3k}+D\b^{3k})$ for some constants
$A$, $B$, $C$ and $D$, where $\a$ and $\b$ are the zeros of
$x^2-Px+Q$. Thus, $(a_k)$ is
annihilated by the cubic polynomial
$$
C(x)=(x-\a^6)(x-\b^6)(x-Q^3)=x^3-(Q^3+V_6)x^2+(Q^3V_6+Q^6)x-Q^9.
$$ For
$k=-1,0$ and $1$, one can check that $a_k\equiv
Q^{3k}\pmod{M}$. For instance,
$$
a_1=U_4U_5/P=(P^2-2Q)(P^4-3P^2Q+Q^2)=M+Q^3\equiv Q^3\pmod{M}.
$$ Thus, that $a_k\equiv Q^{3k}\pmod{M}$, for all $k\ge0$, easily follows by induction on
noting that
$$C(x)=(x^3-Q^9)-(Q^3+V_6)(x^2-Q^3x).\;\qed$$
\section{A polynomial version of Theorem \ref{thm:3}}
\label{sec:4}
The referee said he made some quick numerical
experiments which indicated the congruence of Theorem
\ref{thm:3} may hold in the polynomial ring $\Z[P,Q]$ and asked
whether, if true, this statement is implied by Theorem \ref{thm:3}.
The two statements would then be clearly equivalent. We point out
that the foregoing second proof of Theorem \ref{thm:3}, for the case
$p=3$, at the end of Section \ref{sec:3}, proves the
$\Z[P,Q]$-statement is true when $p=3$. We had played with a
similar proof for the case $p=5$, but writing a proof along these
lines for general $p$ seemed cumbersome. However, we are able to
answer the referee's question in the positive
in the next theorem. Note that the nondegeneracy hypothesis is
no longer needed.
\begin{theorem} \label{thm:5} Let $p\ge3$ be a prime number,
$k\ge0$ an integer and $\{U,V\}$ a pair of Lucas sequences with
parameters $P$ and $Q$. Then the congruence
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^2V_p/P},
$$ holds in the ring $\Z[P,Q]$.
\end{theorem}
\begin{proof} As noted by Lucas \cite[p. 312-13]{Lu2}, $U_n$ and $V_n$
are homogeneous polynomials in $\Z[P,Q]$ of respective degrees $n-1$
and $n$ if one views the variable $Q$ as being of degree
$2$. Moreover, as easily checked by induction on $n$ using the
recursion (\ref{eq:rec}) each $U_n(P,Q)$, $n\ge2$, is, as a polynomial
in $P$, monic of leading term $P^{n-1}$, and each $V_n(P,Q)$,
$n\ge1$, is also monic in $P$ with leading term $P^n$. Thus the
modulus $M(P,Q)=M_Q(P):=U_p^2V_p/P$ is a monic polynomial in $P$ in
$\Z[Q][P]$. All Lucasnomials are also polynomials in $\Z[P,Q]$. This
is often shown by induction using the identity:
$$
\binom{m}{n}_U=U_{n+1}\binom{m-1}{n}_U-QU_{m-n-1}\binom{m-1}{n-1}_U.
$$
Thus, we find that $f(P,Q):=\binom{(k+1)p-1}{p-1}_U-Q^{kp(p-1)/2}$
is a polynomial in $\Z[P,Q]$. Let us put $f(P,Q)=f_Q(P)\in\Z[Q][P]$. Then
the euclidean division of $f_Q(P)$ by
$M_Q(P)$ yields two polynomials $q_Q(P)$ and $r_Q(P)$ in $\Z[Q][P]$
satisfying
\begin{equation}\label{eq:15}
f_Q(P)=q_Q(P)\cdot M_Q(P)+r_Q(P),
\end{equation} with the degree in $P$ of $r_Q(P)$ less than the degree
of $M_Q(P)$. Indeed, $M_Q(P)$ is a monic polynomial in $P$ so we know $q_Q(P)$
has polynomial coefficients in $\Z[Q]$. Therefore, $r_Q(P)$ is also in
$\Z[Q][P]$. Let us fix $Q$ to some nonzero value $y$ in
$\Z$. We may choose an integer value $x$ for $P$
large enough so that both $U_{12}(x,y)\not=0$ and
$|M_y(x)|>|r_y(x)|$. Thus, $U(x,y)$ is a nondegenerate fundamental
Lucas sequence. Therefore, by Theorem \ref{thm:3}, the integer
$M_y(x)$ divides the integer $f_y(x)$. It follows from (\ref{eq:15})
that $M_y(x)$ divides $r_y(x)$. Thus, $r_y(x)=0$ as an integer.
Since there are arbitrarily many such integer values $x$
for $P$, i.e., more than the degree of $r_y(P)$, we deduce that
$r_y(P)=0$ as a polynomial. Thus, the polynomial coefficients in
$\Z[Q]$ of $r_Q(P)$ have $y$ as a zero. Since $y$ was arbitrary,
$r_Q(P)$ must be the zero polynomial in the ring $\Z[P,Q]$.
By equation (\ref{eq:15}) we conclude that
$M(P,Q)$ divides $f(P,Q)$ in $\Z[P,Q]$.
\end{proof}
\begin{remark}\label{rem:2} In the proof of Theorem \ref{thm:5} having
fixed a nonzero value $y$ for $Q$ we could have made the
additional requirement on the integer value $x$ for $P$ that it be
prime to $y$. Thus, it is enough to have Theorem \ref{thm:3} hold in
the regular case to imply Theorem \ref{thm:5}, which in turn implies
Theorem \ref{thm:3} in full generality.
\end{remark}
The referee's computations also seemed to indicate the polynomial version
in $\Z[P,Q]$ of Theorem \ref{thm:3} held for the case $p=2$
whenever $k$ is a multiple of $4$. Theorem \ref{thm:3} does not
consider the case $p=2$, as the Wolstenholme congruence
itself does not even hold modulo $p^2$, when $p=2$. However, we can
easily prove the referee's observation is true when $4\mid k$. It is actually
true for the higher modulus $U_p^3V_p/P$. Thus, we make this an
additional theorem.
\begin{theorem} \label{thm:6} Suppose $\{U,V\}$ is a pair of
Lucas sequences with parameters $P$ and $Q$, $k\ge0$ is an integer divisible by
$4$ and $p=2$. Then the congruence
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^3V_p/P},
$$ holds in the ring $\Z[P,Q]$.
\end{theorem}
\begin{proof} Since
$U_2^3V_2/P=P^2(P^2-2Q)$, the congruence to verify becomes for $p=2$
$$
U_{2k+1}\equiv Q^k\pmod{P^2(P^2-2Q)}.
$$ Putting $k=4n$, we proceed by induction on $n$. The congruence
holds true for
$n=0$ and $n=1$ as is easily checked. For $n=1$, we obtain
$U_9-Q^4=P^2(P^6-7P^4Q+15P^2Q^2-10Q^3)$ and, using
$P^2\equiv2Q\pmod{P^2-2Q}$, we see that the factor
$$
P^6-7P^4Q+15P^2Q^2-10Q^3\equiv(8-28+30-10)Q^3=0\pmod{P^2-2Q}.
$$ The linear recurrent sequence $(U_{8n+1})$ satisfies
\begin{equation}\label{eq:last}
U_{8(n+2)+1}=V_8U_{8(n+1)+1}-Q^8U_{8n+1},
\end{equation} for all $n\ge0$. Now
$$
V_8-2Q^4=P^2(P^6-8P^4Q+20P^2Q^2-16Q^3),
$$ and the factor $P^6-8P^4Q+20P^2Q^2-16Q^3$, using
$P^2\equiv2Q\pmod{P^2-2Q}$,
is seen to be congruent to $(8-32+40-16)Q^3=0$ modulo $P^2-2Q$. Thus,
assuming $U_{8n+1}\equiv Q^{4n}$ and $U_{8(n+1)+1}\equiv Q^{4n+4}$
modulo $P^2-2Q$, we obtain inductively by (\ref{eq:last}) that
$$
U_{8(n+2)+1}\equiv 2Q^4\cdot Q^{4n+4}-Q^8\cdot
Q^{4n}=Q^{4(n+2)}\pmod{P^2(P^2-2Q)},
$$ proving the claim.
\end{proof}
We remark that if $k=1$, $2$ and $3$, the $\Z[P,Q]$-congruence of
Theorem \ref{thm:6} does not hold even when degrading the modulus to
$U_2^2V_2/P$. However, to complete the picture we can prove, using the
line of proof of Theorem \ref{thm:6}, the following proposition.
\begin{proposition} \label{prop:2} Suppose $\{U,V\}$ is a pair of
Lucas sequences with parameters $P$ and $Q$ and $p=2$. Then the
congruence
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {M(P,Q)},
$$ holds in the ring $\Z[P,Q]$, where
$$
M(P,Q)=\begin{cases} P^2-2Q=U_2V_2/P,& \text{ if }k=4n+1;\\
P^2=U_2^2=U_2^3/P,& \text{ if }k=4n+2.
\end{cases}
$$
\end{proposition}
\section{Acknowledgments}
\label{sec:5}
As mentioned in the introduction, H. Ohtsuka took the pleasant
initiative of sending me the unpublished solution of his Fibonacci Quarterly
problem H-737 a few years ago. Part of the proof of Theorem
\ref{thm:3} we chose to write is a direct extension of his technique. Elif Tan kindly carried
out some of the numerical computations made to contend that Theorem \ref{thm:3}
held when $U(P,Q)$ is not a regular sequence. We thank the referee
for his interest and valuable questions which were addressed in Section \ref{sec:4}.
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51} (2013), 186.
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\end{thebibliography}
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A07; Secondary 11B65, 11B39.
\noindent \emph{Keywords: } generalized binomial coefficient,
Wolstenholme's congruence, Lucas sequence, rank of appearance.
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\noindent
Received August 9 2019;
revised version received December 19 2019; December 24 2019; March 11
2020.
Published in {\it Journal of Integer Sequences}, March 17 2020.
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