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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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\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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\vspace*{+.1in}
\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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