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\begin{center}
\vskip 1cm{\LARGE\bf 
The Congruence of Wolstenholme for 
\vskip .02in
Generalized Binomial Coefficients
\vskip .14in
Related to Lucas Sequences
}
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\large
Christian Ballot \\
D\'epartement de Math\'ematiques et M\'ecanique\\
Universit\'e de Caen \\
F14032 Caen Cedex \\
France \\
\href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr} \\
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\begin{abstract} 
In recent years much research has been carried out on extending 
Wolstenholme classical congruence modulo the cube of a prime to higher prime powers. 
Here we show that this work can be done in much broader generality 
by replacing ordinary binomials by Lucasnomials, which are generalized binomial 
coefficients related to fundamental Lucas sequences. The paper builds on earlier work of 
Kimball and Webb in relation to the Fibonacci sequence and on recent work of the author related 
to congruences involving sums of quotients of Lucas sequences. The paper offers what may be 
a surprising line of development for very classical congruences.   
\end{abstract}

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

 In 1862 Wolstenholme \cite{Wo} established a now well-known congruence for binomial 
coefficients, namely 
 
 \begin{theorem}\label{thm:W} Let $p$ be a prime number $\ge5$. Then 
\begin{equation}\label{eq:W}
\binom{2p-1}{p-1}\equiv1\pmod {p^3}.
\end{equation}
\end{theorem}

 Babbage \cite{Bab}, in 1819, had actually shown that congruence (\ref{eq:W}) held modulo $p^2$ for 
all primes $p$ greater than $2$. There is a survey paper \cite{Mes} on the   
numerous generalizations of Theorem \ref{thm:W} discovered since the paper of Wolstenholme 
appeared in 1862. This survey also contains many other related results. 

 We focus first our attention on the sligthly more general congruence  
\begin{equation}\label{eq:W+}
\binom{(k+1)p-1}{p-1}\equiv1\pmod {p^3},
\end{equation} which holds for all primes $p\ge5$ and all nonnegative integers $k$. 
According to the survey \cite{Mes}, congruence (\ref{eq:W+}) was proved in 1900 by Glaisher 
\cite[p.\ 21]{Glai1}, \cite[p.\ 33]{Glai2}. 

 If $A=(A_n)_{n\ge0}$ is a sequence of complex numbers where $A_0=0$ and all $A_n\not=0$ for $n&gt;0$, 
then one defines, for $m$ and $n$ nonnegative integers, the {\it generalized} binomial coefficient 
\begin{equation}\label{eq:def}
\binom{m}{n}_A=\begin{cases}\frac{A_mA_{m-1}\dots A_{m-n+1}}{A_nA_{n-1}\dots A_1},&amp; \text{ if }m\ge n\ge1;\\
1,&amp; \text{ if }n=0;\\
0,&amp; \text{ otherwise.}\end{cases} 
\end{equation} The well-written paper \cite{Gou} contains a number of early references about these coefficients 
and investigated several of their general properties. We point out another early reference \cite{Wa},  
not often quoted, in which Ward gives two equivalent criteria that imply the integrality of the generalized 
coefficients $\binom{m}{n}_A$ of a sequence of integers $A$. One of them is that  
$A$ be a {\it strong divisibility sequence}, i.e., one for which $A_{\gcd(m,n)}=\gcd (A_m,A_n)$ for all 
$m&gt;n&gt;0$; the other criterion is expressed in terms of ranks of appearance of prime powers in $A$. 
The {\it rank of appearance} $\rho=\rho_A(x)$ of a nonzero integer $x$ with respect to $A$ is, 
if it exists, 
the least positive integer $t$ such $x$ divides the integer $A_t$.     
The equivalence of the two criteria of Ward was essentially rediscovered in \cite{KnWi}. 
When $A$ is the Fibonacci sequence $F=(F_n)_{n\ge0}$, defined by $F_0=0$, $F_1=1$ and 
$F_{n+2}=F_{n+1}+F_n$ for all $n\ge0$,  
these binomial coefficients are called {\it Fibonomials}. Many papers have studied their properties. 


\medskip

 Kimball and Webb \cite[Lemma 3]{KW1} proved an analogue of the Wolstenholme-Glaisher congruence (\ref{eq:W+}), 
which we rewrite as a theorem below. 

\begin{theorem} \label{thm:KW} Let $p$ be a prime at least $7$ whose rank of appearance $\rho$ in the Fibonacci 
sequence is of the form $p-\e_p$, where $\e_p$ is $\pm1$. Then for all integers $k\ge0$ 
\begin{equation}\label{eq:Fib}
\binom{(k+1)\rho-1}{\rho-1}_F\equiv\e_p^k\pmod {p^3},
\end{equation}
where the symbol $\binom{*}{*}_F$ stands for the Fibonomial coefficient.
\end{theorem} 



Some 
papers have considered the generalized binomial coefficients when $A$ is a fundamental Lucas sequence,  
that is, a sequence $U=U(P,Q)$ satisfying 
\begin{equation}\label{eq:second}
U_0=0,\; U_1=1\;\text{ and }U_{n+2}=PU_{n+1}-QU_n, \;\text{ for all }n\ge0,
\end{equation}
where $(P,Q)$ is a pair of integers, $Q$ nonzero. We will refer to these generalized 
binomials as {\it Lucasnomial} coefficients, or {\it Lucasnomials}, in the sequel. Ordinary 
binomials are Lucasnomial coefficients with parameters $(P,Q)=(2,1)$, whereas the Fibonomials 
correspond to $(P,Q)=(1,-1)$. Thus it makes sense to look for a simple congruence for the 
general Lucasnomial    
\begin{equation}
\binom{(k+1)\rho-1}{\rho-1}_U\pmod {p^3},
\end{equation}
valid for an arbitrary Lucas sequence $U$, that would encompass both the congruence (\ref{eq:W+}) and 
Theorem \ref{thm:KW}. 

 Here $\rho$ represents the rank of appearance of the prime $p$ in 
$U$. It is known to exist 
for all primes $p$ not dividing $Q$ and, for $p$ odd, to divide $p-\e_p$, where $\e_p$ is the Legendre 
character $(D\;|\;p)$ and $D$ is $P^2-4Q$. To obtain a congruence modulo $p^3$ it is necessary to 
require, as in Theorem \ref{thm:KW}, 
that the rank $\rho$ be maximal, i.e., be equal to $p-\e_p$. Note that the rank of any 
prime $p$ is maximal and equal to $p$ for $U_n=n$ ($D=0$, $\e_p=0$). However, the case $\e_p=0$  
only occurs for $p=5$ for the Fibonacci sequence $F=U(1,-1)$, a case that Theorem \ref{thm:KW} 
does not address. 
A calculation for $p=5$ yields  
\begin{equation}\label{eq:case0}
\binom{2\rho-1}{\rho-1}_F=\binom{9}{4}_F\equiv1\pmod {125}. 
\end{equation}  This residue of $1$ at least conforms to what one gets in (\ref{eq:W+}), but does not 
match the expression $\e_p^k$ of Theorem \ref{thm:KW} which would yield $0$. 

\medskip

 Thus, one needs to generalize the results of the paper \cite{KW1} from Fibonomial coefficients to 
Lucasnomial coefficients and include the case $\e_p=0$ in the analysis. However, some of the results 
leading to Theorem \ref{thm:KW} in \cite{KW1} seem, at first sight, to depend on idiosyncracies 
of the Fibonacci sequence. Thus, a few numerical calculations helped us believe in the  
existence of a generalization and were useful in guiding us to it.   
 

\begin{theorem} \label{thm:N} Let $U=U(P,Q)$ be a fundamental Lucas sequence with parameters $P$ and 
$Q$. Let $p\ge5$, $p\nmid Q$, be a prime whose rank of appearance $\rho$ in $U$   
is equal to $p-\e_p$, where $\e_p$ is the Legendre character $(D\;|\;p)$, $D=P^2-4Q$. 
Then for all integers $k\ge0$ 
\begin{equation}\label{eq:Luc}
\binom{(k+1)\rho-1}{\rho-1}_U\equiv(-1)^{k\e_p}Q^{k\rho(\rho-1)/2}\pmod {p^3},
\end{equation}
where the symbol $\binom{*}{*}_U$ stands for the Lucasnomial coefficient.
\end{theorem}

\begin{remark}\label{rem:pow} Theorem \ref{thm:N} implies that for all $k\ge0$ 
$$
\binom{(k+1)\rho-1}{\rho-1}_U\equiv\binom{2\rho-1}{\rho-1}_U^k\pmod{p^3}.
$$
\end{remark}

\begin{remark}\label{rem:kw} Congruence (\ref{eq:W+}), Theorem \ref{thm:KW} and, as readily checked, 
congruence (\ref{eq:case0}) are implied by 
Theorem \ref{thm:N}. Indeed, the sequence $A_n=n$ is $U_n(2,1)$, for which $Q=1$ and $\e_p=0$ 
for all primes.  
To see that Theorem \ref{thm:KW} is a corollary of Theorem \ref{thm:N}, it suffices 
to check that 
$$
\e_p=-(-1)^{\rho(\rho-1)/2},
$$ for every odd prime $p&gt;5$ of maximal rank in 
the Fibonacci sequence $U(1,-1)$. All primes of rank $p\pm1$ in the Fibonacci sequence 
must be congruent to $3\pmod 4$, since by Euler's criterion for Lucas sequences (\ref{eq:criterion}) 
we need 
to have $(-1\;|\;p)=-1$. If $\e_p=1$, that is, if $\rho=p-1$, then $\rho(\rho-1)\equiv2\pmod4$ 
so that $-(-1)^{\rho(\rho-1)/2}=+1=\e_p$. If $\e_p=-1$, that is, $\rho=p+1$, 
then $\rho(\rho-1)\equiv0\pmod4$ so $-(-1)^{\rho(\rho-1)/2}=-1=\e_p$. 
\end{remark} 

\begin{remark}\label{rem:p2} Although throughout the paper we assume rank maximality,   
not all is lost when a prime does not have maximal rank. As will be easily inferred from 
the proof of Theorem \ref{thm:N}, we have for general rank a weaker congruence albeit valid for all $p\ge3$  
which we state below. 
\end{remark}

\begin{theorem} \label{thm:Ng} If a prime $p\ge3$, $p\nmid Q$, has rank $\rho$ in $U(P,Q)$,     
then for all integers $k\ge0$ 
$$
\binom{(k+1)\rho-1}{\rho-1}_U\equiv(-1)^{k(\rho-1)}Q^{k\rho(\rho-1)/2}\pmod {p^2}. 
$$
\end{theorem}

\begin{remark}\label{rem:Up} Various other Lucasnomial generalizations of the Glaisher congruence 
(\ref{eq:W+}), besides 
Theorems \ref{thm:N} and \ref{thm:Ng},  
are possible. For one instance, retaining $p$ instead of $\rho$ in the terms of the congruence 
but replacing $p$ in the modulus by $U_p$, we can prove the theorem. 
\end{remark}

\begin{theorem} \label{thm:Nm} Suppose $U(P,Q)$ is a fundamental Lucas sequence with  
$P$ and $Q$ coprime and $U_2U_3U_4U_6$ nonzero. Let $p\ge5$ be a prime. Then 
for all $k\ge0$ we have the congruence 
$$
\binom{(k+1)p-1}{p-1}_U\equiv Q^{kp(p-1)/2}\pmod {U_p^2}. 
$$ 
\end{theorem}

 When $U_p$ is a prime then $p$ is the rank of $U_p$ so that, in that case, Theorem \ref{thm:Nm} 
follows from Theorem \ref{thm:Ng}. For $U(1,-1)$, the Fibonacci sequence, the congruence 
in Theorem \ref{thm:Nm} may be deduced from the statement of a problem posed by Ohtsuka \cite{Oh}. 
Generalizing the published solution to this problem \cite{Ba0}, one can derive Theorem \ref{thm:Nm}. 

\medskip 


 Section \ref{sec:2} of the paper is devoted to some relevant additional remarks  
on Lucas sequences, some useful lemmas and to proofs of Theorem \ref{thm:N} and 
Theorem \ref{thm:Ng}. 

\medskip

 For all primes $p\ge5$ and all nonnegative integers $k$ and $\ell$, we have the congruence 
\begin{equation}\label{eq:Viggo}
\binom{kp}{\ell p}\equiv\binom{k}{\ell}\pmod{p^3}.
\end{equation} This congruence supersedes congruence (\ref{eq:W+}) and was first proved in a 
collective paper \cite{Bru} which 
appeared in 1952. It was reproved by Bailey some forty years later in the paper \cite{Bai}, where 
the case $(k,\ell)=(2,1)$, which is equivalent to Wolstenholme's congruence (\ref{eq:W}), 
is proved first before an induction on $k$ yielded congruence (\ref{eq:W+}) and 
another proof by induction gave (\ref{eq:Viggo}). Interestingly another simple 
argument, combinatorial, reduces the proof of (\ref{eq:Viggo}) to that of the 
case $(k,\ell)=(2,1)$ in the book \cite[solution of exercise 1.14, p.\ 165]{Stan}.  

 Similarly in \cite{KW1}, Theorem \ref{thm:KW} is used by the authors to 
produce an analogue of (\ref{eq:Viggo}) for the Fibonacci sequence $U=F$. 
That is, in our notation, for primes $p\ge7$ of rank $\rho=p-\epsilon_p$, 
where $\epsilon_p=\pm1$, their result \cite[p.\ 296]{KW1} states that
\begin{equation}\label{eq:KW2}
\binom{k\rho}{\ell\rho}_F\equiv\epsilon_p^{(k-\ell)\ell}\binom{k}{\ell}_{F'}\pmod{p^3},
\end{equation} where $F'_t=F_{\rho t}$ for all $t\ge0$, $k$, $\ell$ are integers satisfying $k\ge\ell\ge1$. 
Section \ref{sec:3} states and proves a congruence, Theorem \ref{thm:LjWe}, for Lucasnomials 
$\binom{k\rho}{\ell\rho}_U$ 
$\pmod{p^3}$ that subsumes the congruences (\ref{eq:Viggo}) and (\ref{eq:KW2}).  
Here again the proof of this more general result is easily derived from Theorem \ref{thm:N}. 
We raise in passing the question of the existence of a combinatorial argument that 
would reduce Theorem \ref{thm:LjWe} to the case $(k,\ell)=(2,1)$. Lucasnomial coefficients 
received two distinct combinatorial interpretations in the paper \cite{SaSa}, both of which 
were explained another time in \cite{BeRe}. Also a $q$-analogue 
of (\ref{eq:Viggo}) that uses $q$-binomial coefficients was established in 
the paper \cite{Stra}.    

\medskip

 In the fourth section, we selected three congruences for binomials $\binom{2p-1}{p-1}\pmod{p^5}$, namely 
(\ref{eq:i}), (\ref{eq:ii}) and  (\ref{eq:iii}),    
and establish for each a generalization to Lucasnomial coefficients $\binom{2\rho-1}{\rho-1}_U\pmod{p^5}$ for 
primes $p\ge7$ of maximal rank $\rho$ in $U$. Not to lengthen an already long introduction we only 
state the example of congruence (\ref{eq:iii}), i.e.,  
$$
\binom{2p-1}{p-1}\equiv1-p^2\sum_{0&lt;t&lt;p}\frac1{t^2}\pmod{p^5},
$$ which generalizes into 
$$
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\rho-1}Q^{\frac{\rho(\rho-1)}2}\bigg(1-4\frac{U_\rho^2}{V_\rho^2}
\sum_{0&lt;t&lt;\rho}\frac{Q^t}{U_t^2}\bigg)\pmod{p^5},
$$ where $U(P,Q)$ is a fundamental Lucas sequence and $V$ its companion Lucas sequence. The $V$  
sequence is defined by $V_0=2$, $V_1=P$ and the same recursion, $V_{n+2}=PV_{n+1}-QV_n$ for 
all $n\ge0$, as the $U$ sequence. Thus, the companion Lucas sequence of $U_n(2,1)=n$ is 
$V_n(2,1)=2$, for all $n\ge0$. At the basis of these results are congruences modulo a
prime $p$ or modulo $p^2$ for generalized harmonic sums $\sum_{t=1}^{\rho-1}(V_t/U_t)^\nu$ 
which are compiled in Lemma \ref{lem:psquare}. We recall that Theorem \ref{thm:W} 
is intimately linked with the congruence $\sum_{t=1}^{p-1}1/t\equiv0\pmod{p^2}$ for primes 
$p\ge5$.    

\medskip

Note that the condition that $p$ be of maximal rank in $U$ may be viewed as a quadratic 
analogue of Artin's conjecture which gives a positive density (equal to a positive 
rational number times Artin's constant) for the set of primes $p$ for which a given 
$a$ is a primitive root $\pmod p$, when $a$ is a non-square integer and $|a|\ge2$.  
Hooley \cite{Hoo} proved Artin's conjecture conditionally to some generalized Riemann hypotheses. 
So did Roskam \cite{Ros,Ros1} for the set of primes $p$ for which a fundamental unit 
of a quadratic field has maximal order modulo $(p)$. Thus, given $U(P,Q)$, $Q$ not a square, 
our theorems presumably should also concern sets of primes of positive densities.       

\medskip

 In recent years congruences for ordinary binomials $\binom{2p-1}{p-1}\pmod{p^e}$ 
have been established for larger and larger exponents $e$ \cite[pp.\ 4--6]{Mes}. 
No doubt there must be higher corresponding congruences for Lucasnomials. In fact, we end 
Section \ref{sec:4} with such a congruence, expressed in Theorem \ref{thm:Tau}, when $e=6$. 
Generalizations of (\ref{eq:i}) are stated in Theorems \ref{thm:pcinq1} and \ref{thm:pcinq4}, 
those of (\ref{eq:ii}) and the above congruence (\ref{eq:iii}) appear in Theorems  
\ref{thm:pcinq2} and \ref{thm:pcinq3} respectively. We added an appendix as a short fifth 
section where the integrality of all Lucasnomial coefficients $\binom{m}{n}_U$ is asserted 
for all $U$ Lucas sequences, including degenerate cases, provided we make a reasonable amendment 
to the definition (\ref{eq:def}).  

\medskip

 Familiarity with Lucas sequences is assumed throughout the paper, but the reader may want 
to consult the introduction of \cite{Ba3} and the references it mentions. Chapter 4 of the 
book \cite{Wi} is a useful introduction to these sequences. 

\medskip

 Lucasnomial coefficients have already been the object of generalizations of 
classical arithmetic properties of ordinary binomial coefficients. 
Kummer's theorem giving the exact power of a prime $p$ 
in the binomial coefficient $\binom{m+n}{n}$ as the number of carries in the addition of $m$ 
and $n$ in radix $p$ was extended to generalized binomials $\binom{m+n}{n}_A$, when $A$ is a strong divisibility 
sequence of positive integers \cite{KnWi}. That includes, in particular, all Lucas sequences $U(P,Q)$ 
with positive terms when $P$ and $Q$ are coprime. A further Kummer rule pertaining to generalized 
binomials $\binom{*}{*}_U$, valid for an arbitrary nondegenerate fundamental $U$ Lucas sequence, 
appears in the preprint \cite{Ba}. 

 Also various generalizations of the celebrated theorem of Lucas:  
$$
\binom{mp+r}{np+s}\equiv\binom{m}{n}\binom{r}{s}\pmod p,
$$ where $r$ and $s$ are nonnegative integers less than the prime $p$, 
were achieved in terms of Lucasnomials $\binom{mp+r}{np+s}_U$, often 
under restrictive hypotheses on the Lucas sequence $U(P,Q)$ \cite{We,Ho,HuSu}. 

 In fact both the theorems of Kummer and of Lucas had been generalized in an earlier paper 
\cite{Fr} but with respect to $q$-binomial coefficients.      
 
\section{Preliminaries and a proof of Theorem \ref{thm:N}}
\label{sec:2}

 Lucas theory is often developed with the two hypotheses that  $U(P,Q)$ is  
nondegenerate and $\gcd(P,Q)$ is $1$. 
The Lucas sequence $U(P,Q)$ is called {\it degenerate} whenever the ratio of the 
zeros $\alpha$ and $\beta$ of $x^2-Px+Q$ is a root of unity. We do not make any of these assumptions 
here. If $U$ is degenerate then we must have $U_2U_3U_4U_6=0$. Indeed, if $\alpha\not=\beta$ 
then $U_t=\frac{\alpha^t-\beta^t}{\alpha-\beta}$ and the ratio $\alpha/\beta$, lying in the 
quadratic field $\mathbb Q(\sqrt{D})$, must be a second, third, fourth or sixth root of unity. 
Thus, some terms of the sequence $U$ will be $0$, but rather than discard those Lucas sequences 
from our analysis, we make a small amendment 
to the definition (\ref{eq:def}) to ensure that the corresponding 
Lucasnomials $\binom{m}{n}_U$ are well-defined as rational numbers. Although 
the hypotheses of Theorems \ref{thm:N} and \ref{thm:LjWe} or of the theorems of Section \ref{sec:4} 
if applied to a prime $p\ge11$ prevent the corresponding Lucasnomials from having zero terms, 
this is not necessarily the case if $p=5$ or $p=7$.  
With $\gcd(P,Q)&gt;1$, the 
Lucas sequence $A=U(P,Q)$ is no longer a strong divisibility sequence. Nevertheless $A$, or $\lambda A$, 
$\lambda$ an integer, satisfies some `convexity' property. Namely for all prime powers 
$p^a$ ($a\ge1$), $p\nmid Q$, and for all $x$ and $y\ge1$, we have  
\begin{equation}\label{eq:convex}
\#\;\{t\in[x+y],\;p^a\mid A_t\}\;\ge\;\#\;\{t\in[x],\;p^a\mid A_t\}+\#\;\{t\in[y],\;p^a\mid A_t\}. 
\end{equation} Here, if $z$ is an integer $\ge1$, $[z]$ denotes the 
set of natural numbers $1,2,\hdots,z$. This property holds because for such prime powers $p^a$, 
we have $p^a\mid U_t$ iff $\rho(p^a)\mid t$, where $\rho(p^a)$ is the rank of appearance 
of $p^a$ in $U$, and because $\lfloor x+y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor$ for 
all real numbers $x$ and $y$. 

\medskip 

 The convention we adopt for the generalized binomials $\binom{m}{n}_A$ of definition (\ref{eq:def}) 
is that if there are zero terms in the product $\prod_{i=1}^n\frac{A_{m+1-i}}{A_i}$ then 
\begin{equation}\label{eq:conv}
\text{a }0\text{ in the numerator and a }0\text{ in the denominator 
{\bf cancel out} as a }1.
\end{equation} 

 With convention (\ref{eq:conv}), property (\ref{eq:convex}) satisfied by $A=\lambda U$, for all Lucas sequences $U$, 
guarantees that the generalized binomial $\binom{m}{n}_A$ is a well-defined rational number. 
Indeed this property implies that the number of $0$ terms in the numerator of  
$\prod_{i=1}^n\frac{A_{m+1-i}}{A_i}$ is at least that of its denominator. It also implies that $\binom{m}{n}_A$, $m$ 
and $n$ nonnegative integers, has nonnegative $p$-adic valuation for all primes $p\nmid Q$. 
In fact we can show it is always a rational integer.
\footnote{See our short Appendix.}      

\medskip

 As already mentioned, to each fundamental Lucas sequence $U(P,Q)$ we associate a {\it companion} Lucas sequence $V=V(P,Q)$ 
which obeys recursion (\ref{eq:second}), but has initial values $V_0=2$ and $V_1=P$. 
The following identities are all classical ones and are all valid no matter what the value  
of $\gcd(P,Q)$ is. We will use them throughout the paper.   

\begin{eqnarray}
2U_{s+t} &amp; = &amp; U_sV_t+U_tV_s,\label{eq:1}\\
2V_{s+t} &amp; = &amp; V_sV_t+DU_sU_t,\label{eq:2}\\
V_t^2-DU_t^2 &amp; = &amp; 4Q^t,\label{eq:3}\\
U_{2t} &amp; = &amp; U_tV_t,\label{eq:4}\\
V_{2t} &amp; = &amp; V_t^2-2Q^t,\label{eq:5}\\
2Q^tU_{s-t} &amp; = &amp; U_sV_t-U_tV_s.\label{eq:6}
\end{eqnarray}

 We referred to Euler's criterion for Lucas sequences in our introduction. The criterion  
states that 
\begin{equation}\label{eq:criterion}p\mid U_{(p-\epsilon_p)/2} \;\text{ iff }\;
Q \text{ is a square modulo }p,
\end{equation} where $U(P,Q)$ is a fundamental Lucas sequence and $p$ is a prime that 
does not divide $2DQ$ \cite[pp. 84--85]{Wi}. 

 Note that our theorems and the lemmas of Section \ref{sec:4} all deal with primes $p\ge5$ 
of maximal rank. In their statements, we sometimes  
omit to mention the condition $p\nmid Q$, because that condition is necessary. 
Indeed, if $p\mid Q$, then, by (\ref{eq:second}), $U_t\equiv P^{t-1}\pmod p$. Thus, $p$ has 
no rank, because if $p$ divided $P$, then $\rho(p)$ would be equal to $2$, as $U_2=P$, a 
contradiction.   

\medskip

 Given a prime $p$ of rank $\rho$ and a nonnegative integer $\nu$, we write 
\begin{equation}\label{eq:nota}
\Sigma_\nu:=\sum_{0&lt;t&lt;\rho}\frac{V_t^\nu}{U_t^\nu} 
\text{ and }\Sigma_{1,1}:=\sum_{0&lt;s&lt;t&lt;\rho}\frac{V_sV_t}{U_sU_t}.
\end{equation}

 The proof of Theorem \ref{thm:N} we are about to write uses a few lemmas which we  
state first. 

\begin{lemma} \label{lem:psquare} Let $(U,V)$ be a pair of Lucas 
sequences with parameters $P$ and $Q$. Let $\nu$ be a nonnegative integer. 
If $p\nmid Q$ is a prime at least $\nu+3$ of maximal rank 
$\rho$, i.e., of rank $p-\e_p$, where $\e_p=0$ or $\pm1$, then 
\footnote{unless $\nu=0$ and $\epsilon_p=0$, when $\Sigma_0\equiv-1\pmod p$.} 
\begin{equation}\label{eq:lot}
\Sigma_\nu\equiv\begin{cases}0\pmod {p^2}, &amp;\text{ if }\nu\text{ is odd};\\
0\pmod p, &amp;\text{ if }\epsilon_p=-1\text{ or }0;\\
-2D^{\nu/2}\pmod p, &amp; \text{ if }\nu\text{ is even and }\epsilon_p=1.\end{cases}
\end{equation} Moreover, if $p$ is an odd prime not dividing $Q$ of any rank $\rho$, then 
\begin{equation}\label{eq:+}
\Sigma_\nu\equiv0\pmod p, \quad\text{ for all odd }\nu. 
\end{equation}  
\end{lemma} 
\begin{proof} The case $\nu$ odd of (\ref{eq:lot}) is Theorem 3 of \cite{Ba1}. 
(The case $\nu=1$ first appeared, nearly complete, as the main theorem of the paper 
\cite{KW2}, but also, nearly, as a corollary of the main theorem of \cite{Pan}, and as a particular 
case of \cite[Thm.\ 4.1]{Ba2}, or of \cite[Thms.\ 3 and 12]{Ba3}.) 


 The case $\nu$ even can be treated with the very same arguments used in the last part of the proof 
of Theorem 4 of \cite[p.\ 5]{Ba1}. (The basic facts, noted first in \cite{KW2}, are that, by (\ref{eq:6}), 
all $V_t/U_t$ are distinct $\pmod p$ for $t\in(0,\rho)$ and no $V_t/U_t$ is $\pm\sqrt{D}\pmod p$ 
by (\ref{eq:3}); also $p\mid\sum_{t=1}^pt^e$ if $p-1\nmid e$). The condition 
$p\ge\nu+3$ is a sufficient condition which guarantees that $p-1\nmid\nu$ for $\nu\ge2$ even. 

 The additional congruence (\ref{eq:+}) for $\nu$ odd, but without the restrictions that $\rho$ 
be maximal and $p\ge\nu+3$, 
is a consequence of the congruence $\pmod{p^2}$ on the sixth line of the proof of Theorem 4 of \cite[p.\ 4]{Ba1}.   
\end{proof}    

\begin{lemma} \label{lem:Sig} Let $(U,V)$ be a pair of Lucas sequences with parameters $P$ and $Q$. 
If $p\nmid 6Q$ is a prime of maximal rank $\rho$ in $U$, then 
$$
\Sigma_{1,1}\equiv\begin{cases} 0\pmod p,&amp; \text{ if }\e_p=0
\text{ or }-1;\\
D\pmod p,&amp; \text{ if }\e_p=1.\end{cases}
$$
\end{lemma}
\begin{proof} We have $\Sigma_1^2=\Sigma_2+2\Sigma_{1,1}$ so that $\Sigma_{1,1}\equiv-\frac12\Sigma_2
\pmod p$, since, by Lemma \ref{lem:psquare}, $p^4$ divides $\Sigma_1^2$ and $\Sigma_2$ is either $0$ 
or $-2D\pmod p$. 
\end{proof}

\begin{lemma} \label{lem:Vro} Let $V=V(P,Q)$ be a companion Lucas sequence.  
Let $p\nmid Q$ be an odd prime of rank $\rho$ and $t\ge0$ an integer. Then modulo $p^2$ we have 
$$
V_{t\rho}\equiv\begin{cases}2Q^{t\rho/2}, &amp;\text{ if }t\text{ is even};\\
-2Q^{t\rho/2}, &amp;\text{ if }t\text{ is odd and }\rho\text{ is even}.\end{cases} 
$$
\end{lemma}
\begin{proof} Assume $t$ is even. By (\ref{eq:3}), $V_{t\rho/2}^2\equiv4Q^{t\rho/2}\pmod{p^2}$ 
since $p$ divides $U_{t\rho/2}$. However, by (\ref{eq:5}), $V_{t\rho}=V_{t\rho/2}^2-2Q^{t\rho/2}
\equiv2Q^{t\rho/2}\pmod{p^2}$.  

 Suppose $t$ odd and $\rho$ even. Since $p$ divides $U_{t\rho}$, but not $U_{t\rho/2}$, we see by 
(\ref{eq:4}) that $p$ divides $V_{t\rho/2}$. So by  
(\ref{eq:5}) we find that $V_{t\rho}\equiv-2Q^{t\rho/2}\pmod {p^2}$. 
\end{proof} 


 We are now ready for a proof of Theorem \ref{thm:N}. 

\begin{proof} We have 
$$
\binom{(k+1)\rho-1}{\rho-1}_U=\frac{\prod_{t=1}^{\rho-1}U_{k\rho+t}}{\prod_{t=1}^{\rho-1}U_t}.
$$
By the addition formula (\ref{eq:1}), we find that 
\begin{eqnarray}\label{eq:alph}
2^{\rho-1}\prod_{t=1}^{\rho-1}U_{k\rho+t}&amp;=&amp;\prod_{t=1}^{\rho-1}(V_{k\rho} U_t+U_{k\rho} V_t)\nonumber\\
&amp;\equiv&amp;(V_{k\rho}^{\rho-1}+V_{k\rho}^{\rho-2}U_{k\rho}\Sigma_1+V_{k\rho}^{\rho-3}U_{k\rho}^2
\Sigma_{1,1})\times\prod_{t=1}^{\rho-1}U_t\label{eq:dev}\\
&amp;\equiv&amp;(V_{k\rho}^{\rho-1}+V_{k\rho}^{\rho-3}U_{k\rho}^2
\Sigma_{1,1})\times\prod_{t=1}^{\rho-1}U_t\pmod {p^3},\nonumber
\end{eqnarray} since $p$ divides $U_{k\rho}$ and, by Lemma \ref{lem:Sig}, $\Sigma_1$ is $0\pmod {p^2}$. 

 We first examine the cases $\rho$ is $p+1$ and $\rho$ is $p$. In those cases 
$U_{k\rho}^2\Sigma_{1,1}$ is $0\pmod{p^3}$ by Lemma \ref{lem:Sig}. 
Hence, 
$$
\binom{(k+1)\rho-1}{\rho-1}_U\equiv\bigg(\frac{V_{k\rho}}{2}\bigg)^{\rho-1}\pmod{p^3}. 
$$ 
If $\rho$ is $p$, then, by (\ref{eq:3}) and the fact that 
$p^3\mid DU_{k\rho}^2$, we see that $V_{k\rho}^2\equiv4Q^{k\rho}\pmod{p^3}$. 
Therefore, 
$$ 
\binom{(k+1)\rho-1}{\rho-1}_U\equiv (Q^{k\rho})^{(\rho-1)/2}\pmod{p^3},
$$ yielding the result in that case. 
Suppose $\rho$ is $p+1$. By Lemma \ref{lem:Vro} there is an integer, or a half-integer,  
$\lambda$ such that $\frac{V_{k\rho}}{2}=(-1)^kQ^{k\rho/2}+\lambda p^2$. 
Raising both members of the previous equation to the $p$th power gives $(V_{k\rho}/2)^p\equiv(-1)^kQ^{k\rho p/2}
\pmod{p^3}$. But $(-1)^k=(-1)^{-k}=(-1)^{k\e_p}$ so the theorem follows. 

Suppose now $\e_p$ is $1$, that is, $\rho$ is $p-1$. By Lemma \ref{lem:Sig}, $\Sigma_{1,1}\equiv D\pmod p$ so 
that $U_{k\rho}^2\Sigma_{1,1}\equiv DU_{k\rho}^2\pmod{p^3}$. But, by (\ref{eq:3}), $DU_{k\rho}^2=
V_{k\rho}^2-4Q^{k\rho}$. Therefore, we have 
$$
2^{\rho-1}\binom{(k+1)\rho-1}{\rho-1}_U\equiv2V_{k\rho}^{\rho-1}-4Q^{k\rho}V_{k\rho}^{\rho-3}\pmod{p^3}.
$$
This gives 
\begin{eqnarray*}\label{eq:long}
\binom{(k+1)\rho-1}{\rho-1}_U &amp; \equiv &amp; \bigg(\frac{V_{k\rho}}{2}\bigg)^p\cdot\alpha_{p,k}\pmod{p^3},\\
\text{with }\qquad\qquad\qquad \alpha_{p,k} &amp; := &amp; 2\bigg(\frac2{V_{k\rho}}\bigg)^2
-Q^{k\rho}\bigg(\frac2{V_{k\rho}}\bigg)^4.
\end{eqnarray*}  
By Lemma \ref{lem:Vro}, $V_{k\rho}/2\equiv(-1)^kQ^{k\rho/2}\pmod {p^2}$.  Raising the previous congruence 
to the $p$th power yields $(V_{k\rho}/2)^p\equiv(-1)^kQ^{kp\rho/2}\pmod{p^3}$, while inverting 
it yields the existence of an integer $\mu$ such that $2/V_{k\rho}\equiv(-1)^kQ^{-k\rho/2}+\mu p^2
\pmod{p^3}$. Thus, we find that, modulo $p^3$, 
\begin{eqnarray*}
\alpha_{p,k}&amp;\equiv&amp;2\big((-1)^kQ^{-k\rho/2}+\mu p^2\big)^2
-Q^{k\rho}\big((-1)^kQ^{-k\rho/2}+\mu p^2\big)^4\\
&amp;\equiv&amp;(2Q^{-k\rho}+(-1)^k4Q^{-k\rho/2}\mu p^2)-Q^{k\rho}(Q^{-2k\rho}+(-1)^k4Q^{-3k\rho/2}\mu p^2)\\
&amp;=&amp;Q^{-k\rho}.
\end{eqnarray*}
Thus, we end up with 
$$
\binom{(k+1)\rho-1}{\rho-1}_U\equiv(-1)^kQ^{kp\rho/2}Q^{-k\rho}=(-1)^{k\e_p}Q^{k\rho(p-2)/2}\pmod{p^3},
$$
which yields the theorem.
\end{proof}

 The above proof is the first that came to us. It proceeds case by case according to whether the 
value of the rank of $p$ is $p+1$, $p$ or $p-1$ and, thus, appears somewhat miraculous. Although we 
initially wrote case by case proofs for the higher congruences of Section \ref{sec:4}, we ended up 
presenting a global and thus less seemingly miraculous approach at least for Theorems \ref{thm:pcinq2} 
and \ref{thm:pcinq3}. 

\medskip

 We now prove Theorem \ref{thm:Ng} using the elements of the proof of Theorem \ref{thm:N}. 

\medskip

{\it Proof of Theorem \ref{thm:Ng}.} Since, in (\ref{eq:alph}), $U_{k\rho}$ is $0\pmod p$ 
and, by (\ref{eq:+}), $\Sigma_1$ is also $0\pmod p$, we 
find that for all primes $p\ge3$ 
$$
\binom{(k+1)\rho-1}{\rho-1}_U\equiv\bigg(\frac{V_{k\rho}}{2}\bigg)^{\rho-1}\pmod{p^2}.
$$

By (\ref{eq:3}), $(V_{k\rho}/2)^2\equiv4Q^{k\rho}/4=Q^{k\rho}\pmod{p^2}$. 
Thus, if $\rho$ is odd, then $(V_{k\rho}/2)^{\rho-1}\equiv \big(Q^{k\rho}\big)^{(\rho-1)/2}=
(-1)^{k(\rho-1)}Q^{k\rho(\rho-1)/2}\pmod {p^2}$. If $k$ is even, 
then, by Lemma \ref{lem:Vro}, $V_{k\rho}/2\equiv Q^{k\rho/2}\pmod{p^2}$ and 
the result holds by raising the congruence to the power $\rho-1$. If $k$ is odd and $\rho$ 
even, then, again by Lemma \ref{lem:Vro}, $V_{k\rho}/2\equiv-Q^{k\rho/2}=(-1)^kQ^{k\rho/2}
\pmod {p^2}$, 
which raised to the power $\rho-1$ yields the theorem. \qed




 

\section{Lucasnomials $\binom{k\rho}{\ell\rho}_U\pmod{p^3}$}
\label{sec:3}
 

Here is our common generalization of the Ljunggren et al. congruence (\ref{eq:Viggo}) 
and Kimball and Webb's theorem (\ref{eq:KW2}). 

\begin{theorem}\label{thm:LjWe} Let $U,\,V$ be a pair of Lucas sequences with parameters $P$ and 
$Q$. Let $p\ge5$, $p\nmid Q$, be a prime whose rank of appearance $\rho$ is    
$p\pm1$ or $p$. 
Then, for all nonnegative integers $k$ and $\ell$, we have  
\begin{equation}\label{eq:Lju}
\binom{k\rho}{\ell\rho}_U\equiv\big((-1)^{\rho-1}Q^{\rho(\rho-1)/2}\big)^{\ell(k-\ell)}
\binom{k}{\ell}_{U'}\pmod {p^3},
\end{equation}
where $U'$ is the sequence $U_\rho\times U(V_\rho,Q^\rho)$. 
\end{theorem} 
\begin{proof} Note that if $\rho$ is maximal, then the factor $(-1)^{\e_p}$ of Theorem \ref{thm:N} 
may be replaced by $(-1)^{\rho-1}$. 
We only need a proof in case $k&gt;\ell\ge1$. With convention (\ref{eq:conv}) we may write 
\begin{eqnarray*} \binom{k\rho}{\ell\rho}_U &amp; = &amp; \frac{U_{k\rho}U_{k\rho-1}\cdots U_{(k-\ell)\rho+1}}
{U_{\ell\rho}U_{\ell\rho-1}\cdots U_1}\\
&amp; = &amp; \frac{U_{k\rho}U_{(k-1)\rho}\cdots U_{(k-\ell+1)\rho}}{U_{\ell\rho} U_{(\ell-1)\rho}\cdots 
U_\rho}\cdot\frac{\prod_{i=k-\ell}^{k-1}
\prod_{t=1}^{\rho-1}U_{i\rho+t}}{\prod_{i=0}^{\ell-1}\prod_{t=1}^{\rho-1}U_{i\rho+t}}\\
&amp; = &amp; \binom{k}{\ell}_{U'}\cdot\frac{\prod_{i=k-\ell}^{k-1}\prod_{t=1}^{\rho-1}U_{i\rho+t}}
{\big(\prod_{t=1}^{\rho-1}U_t\big)^\ell}\cdot\frac{\big(\prod_{t=1}^{\rho-1}U_t\big)^\ell}
{\prod_{i=0}^{\ell-1}\prod_{t=1}^{\rho-1}U_{i\rho+t}}\\
&amp; = &amp;  \binom{k}{\ell}_{U'}\cdot\prod_{i=k-\ell}^{k-1}\binom{(i+1)\rho-1}{\rho-1}_U\cdot\bigg(
\prod_{i=0}^{\ell-1}\binom{(i+1)\rho-1}{\rho-1}_U\bigg)^{-1}\\
&amp;\equiv &amp;\binom{k}{\ell}_{U'}\cdot\binom{2\rho-1}{\rho-1}_U^{\sum_{i=k-\ell}^{k-1}i-\sum_{i=0}^{\ell-1}i}
\qquad(\text{ by Remark \ref{rem:pow} })\\
&amp;=&amp;\binom{k}{\ell}_{U'}\cdot\binom{2\rho-1}{\rho-1}_U^{\ell(k-\ell)}\pmod{p^3},
\end{eqnarray*}
yielding, by Theorem \ref{thm:N}, the theorem. 
\end{proof}

\begin{remark} If $p\ge3$, $p\nmid Q$, is a prime and no assumption is made about its rank, 
then congruence (\ref{eq:Lju}) holds modulo $p^2$. This is established by following the proof 
of Theorem \ref{thm:LjWe} and using Theorem \ref{thm:Ng}.
\end{remark}

\begin{remark} If, in Theorem \ref{thm:LjWe}, $U_\rho\not=0$ then we might as well set $U'$ equal to 
$U(V_\rho,Q^\rho)$. 
\end{remark}

\begin{remark} If $U=U(2,1)$, then $U_t=t$ and $U'_t=pt$, or $U'_t=t$ by the above remark. 
Thus the theorem implies that 
$$
\binom{kp}{\ell p}\equiv\binom k\ell_{U'}=\binom k\ell\pmod{p^3},
$$ which is the classical congruence (\ref{eq:Viggo}) of Ljunggren et alii. For $U=U(1,-1)$     
and $\epsilon_p=\pm1$ we saw in Remark \ref{rem:kw} that $\epsilon_p=-(-1)^{\rho(\rho-1)/2}
=-Q^{\rho(\rho-1)/2}$ so that Theorem \ref{thm:LjWe} implies (\ref{eq:KW2}).
\end{remark}
 
 Since we took care of including all cases of Lucas sequences in our theorems, we provide 
an example of an application of Theorem \ref{thm:LjWe} to a degenerate Lucas sequence. 

\begin{example} Consider $U(2,2)$.  Its first terms are 
$$0,\;1,\;2,\;2,\;0,-4,-8,-8,\;0,16,32,32,\;0,\dots$$ 
So Theorem \ref{thm:LjWe} applies to $p=5$ since its rank is maximal and equal to $4$. Choose, 
say $k=3$ and $\ell=2$. By our extended definition of (\ref{eq:def}), we have $\binom{3}{2}_{U'}=1$ 
and $(-1)^{\ell(k-\ell)\e_p}Q^{\ell(k-\ell)\rho(\rho-1)/2}=2^{12}$. Computing 
$\binom{12}{8}_U$ we may verify the congruence modulo $125$, which in that case is an equality, since 
$$
\binom{12}{8}_U=\frac{U_{11}\cdot U_{10}\cdot U_9}{U_3\cdot U_2\cdot U_1}=
\frac{16\cdot32\cdot32}{2\cdot2\cdot1}=2^{12}.
$$
\end{example}


\section{Lucasnomials $\binom{2\rho-1}{\rho-1}_U\pmod{p^5}$} 
\label{sec:4}


 The congruence of Wolstenholme has been studied to prime powers higher than the third. In particular, 
we have, for all primes $p\ge7$,
\begin{eqnarray}
\binom{2p-1}{p-1}&amp;\equiv&amp;1+p\sum_{0&lt;t&lt;p}\frac1{t}+p^2\sum_{0&lt;s&lt;t&lt;p}\frac{1}{st}\pmod{p^5}\label{eq:i}\\
                 &amp;\equiv&amp;1+2p\sum_{0&lt;t&lt;p}\frac1{t}\pmod{p^5}\label{eq:ii}\\
                 &amp;\equiv&amp;1-p^2\sum_{0&lt;t&lt;p}\frac1{t^2}\pmod{p^5}\label{eq:iii}.
\end{eqnarray} 

 We will find congruences for the Lucasnomial coefficients $\binom{2\rho-1}{\rho-1}_U$, valid for a general 
fundamental Lucas sequence $U$, modulo the fifth power of a prime 
of maximal rank $\rho$, which generalize the three congruences above. Expanding 
the binomial $\binom{2p-1}{p-1}$, as was done more generally for Lucasnomials in the proof 
of Theorem \ref{thm:N}, one falls naturally on the congruence (\ref{eq:i}). This expansion 
appears, for instance, in the proof of Proposition 1 in \cite{Mes1}.  
Congruence (\ref{eq:ii}) is a special case of 
Theorem 3 of the paper \cite{Zh} and was known to hold for primes $p\ge5$ modulo $p^4$ much 
earlier, while congruence (\ref{eq:iii}) appears in \cite[p.\ 385]{McI}. 

\medskip

 Given a prime $p$ of rank $\rho$ we make a formal definition to complete 
the notation introduced in (\ref{eq:nota}). 

\begin{definition} If $\nu\ge1$ is an integer, then
$$
\Sigma_{\nu,\,\cdots,\nu\;(k\text{ times})}:=\sum\bigg(\frac{V_{t_1}}{U_{t_1}}\bigg)^{\nu}\cdots\bigg(\frac{V_{t_k}}{U_{t_k}}\bigg)^{\nu}, 
$$ the sum being over all $(t_1,\dots,t_k)$ in $(0,\rho)^k$, $t_1&lt;t_2&lt;\cdots&lt;t_k$. 


If $1\le\nu_1&lt;\nu_2$ are two integers, then $\Sigma_{\nu_1,\cdots,\nu_1,\;\nu_2,\cdots,\nu_2\;}$, where 
$\nu_u$, $u=1$ or $2$, is respectively repeated $k_u$ times, is defined as  
$$
\sum\bigg(\frac{V_{t_1}}{U_{t_1}}\bigg)^{\nu_1}\cdots
\bigg(\frac{V_{t_{k_1}}}{U_{t_{k_1}}}\bigg)^{\nu_1}\bigg(\frac{V_{s_1}}{U_{s_1}}\bigg)^{\nu_2}\cdots
\bigg(\frac{V_{s_{k_2}}}{U_{s_{k_2}}}\bigg)^{\nu_2},
$$ the sum being over all $(t_1,t_2,\dots,t_{k_1})$ and $(s_1,s_2,\dots,s_{k_2})$ such that 
$t_i\not=s_j$, for all $i$ and $j$, and $0&lt;t_1&lt;t_2&lt;\cdots&lt;t_{k_1}&lt;\rho$, $0&lt;s_1&lt;s_2&lt;\cdots&lt;s_{k_2}&lt;\rho$. The notation can be extended to more than two distinct $\nu$ exponents, but we won't need such 
sums in this paper.   

 Thus, for instance, the sums $\Sigma_{1,3}$, $\Sigma_{2,2}$, $\Sigma_{1,1,2}$ and $\Sigma_{1,1,1,1}$ 
are respectively  
$$
\sum_{s,\,t}\frac{V_sV_t^3}{U_sU_t^3},\;\sum_{s&lt;t}\frac{V_s^2V_t^2}{U_s^2U_t^2},\;
\;\sum_{\substack{r&lt;s,\\t\in(0,\rho)}}\frac{V_rV_sV_t^2}{U_rU_sU_t^2},\;
\sum_{q&lt;r&lt;s&lt;t}\frac{V_qV_rV_sV_t}{U_qU_rU_sU_t},
$$  where in each sum $q$, $r$, $s$ and $t$ are distinct integers in the interval $(0,\rho)$.
\end{definition}

\begin{lemma}\label{lem:sigmas} We have for all primes $p\ge7$ of maximal ranks
$$
\Sigma_{1,1,1}\equiv0\pmod{p^2} \text{ and } \Sigma_{1,1,1,1}\equiv\begin{cases}
0\pmod p,\quad\text{ if }\e_p=0 \text{ or }-1;\\
D^2\pmod p,\quad\text{ if }\e_p=1.
\end{cases}
$$
\end{lemma}
\begin{proof} We have the linear system 
\begin{eqnarray*} 
\Sigma_1^3-\Sigma_3 &amp; = &amp; 3\Sigma_{1,2}+6\Sigma_{1,1,1},\\
\Sigma_1\cdot\Sigma_{1,1} &amp; = &amp; \Sigma_{1,2}+3\Sigma_{1,1,1}. 
\end{eqnarray*}
Because $p^2$ divides both $\Sigma_1$ and $\Sigma_3$, $\Sigma_1^3-\Sigma_3$ and $\Sigma_1\cdot\Sigma_{1,1}$ 
are each $0\pmod {p^2}$. Since the determinant of the system is prime to $p$, 
$\Sigma_{1,2}$ and $\Sigma_{1,1,1}$ are both $0\pmod {p^2}$. 

 From Lemma \ref{lem:psquare} with $p&gt;5$, which yields the values of $\Sigma_2$ and $\Sigma_4\pmod p$, we deduce that   
$$
\Sigma_{2,2}=\frac12\big(\Sigma_2^2-\Sigma_4\big)\equiv\begin{cases}0\pmod p,
\text { if }\e_p=0\text{ or }-1;\\
3D^2\pmod p,\text { if }\e_p=1.
\end{cases}
$$ 

 Now $\Sigma_{1,3}=\Sigma_1\cdot\Sigma_3-\Sigma_4\implies\Sigma_{1,3}\equiv-\Sigma_4\pmod p$. Moreover, 
$2\Sigma_{1,1,2}+2\Sigma_{2,2}+\Sigma_{1,3}=\Sigma_{1,2}\cdot\Sigma_1\equiv0\pmod p$.

 Thus, $\Sigma_{1,1,2}$ is $0\pmod p$, if $\e_p$ is $0$ or $-1$, and $\Sigma_{1,1,2}$ is 
$-4D^2\pmod p$, if $\e_p$ is $1$.   

Therefore, as $6\Sigma_{1,1,1,1}=\Sigma_{1,1}^2-\Sigma_{2,2}-2\Sigma_{1,1,2}$, we obtain, 
using Lemma \ref{lem:Sig}, the 
desired congruences for $\Sigma_{1,1,1,1}$.  
\end{proof}

 Our first theorem is a generalization of congruence (\ref{eq:i}). 

\begin{theorem}\label{thm:pcinq1} Let $(U,V)$ be a pair of Lucas sequence with parameters $P$ and $Q$. 
Let $p$ be a prime at least $7$ of maximal rank $\rho$ equal to $p-\e_p$. Then 
$$
\binom{2\rho-1}{\rho-1}_U\equiv
\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\bigg(1+\frac{U_\rho}{V_\rho}\sum_{0&lt;t&lt;\rho}
\frac{V_t}{U_t}+\frac{U_\rho^2}{V_\rho^2}\sum_{0&lt;s&lt;t&lt;\rho}\frac{V_sV_t}{U_sU_t}+R\bigg)\pmod{p^5},
$$
$$\text{ where }\;R=\frac{\epsilon_p(1+\epsilon_p)}2\frac{D^2U_\rho^4}{V_\rho^4}=
\begin{cases}0,\;\,\quad\qquad\text{ if }\e_p=0\text{ or }-1;\\
D^2U_\rho^4/V_\rho^4,\text{ if }\e_p=1.
\end{cases}
$$
\end{theorem} 
\begin{proof} Expanding the product $2^{\rho-1}\prod_{t=1}^{\rho-1}U_{\rho+t}=
\prod_{t=1}^{\rho-1}(V_\rho U_t+U_\rho V_t)$  
as we did early in the proof of Theorem \ref{thm:N}, but up to the fourth power of $U_\rho$, 
yields that $2^{\rho-1}\binom{2\rho-1}{\rho-1}_U$ is congruent to  
$$
V_\rho^{\rho-1}+V_\rho^{\rho-2}U_\rho\Sigma_1+
V_\rho^{\rho-3}U_\rho^2\Sigma_{1,1}+V_\rho^{\rho-4}U_\rho^3\Sigma_{1,1,1}+
V_\rho^{\rho-5}U_\rho^4\Sigma_{1,1,1,1}\pmod {p^5}.
$$ Applying the congruences obtained in Lemma \ref{lem:sigmas} to the last two terms 
of the above sum yields the theorem.
\end{proof}



 We now prove a congruence formula that generalizes (\ref{eq:ii}), but 
also generalizes Theorem \ref{thm:N} when $k=1$. The method of proof  
brings out the factor $(-1)^{\epsilon_p} Q^{\rho(\rho-1)/2}$ naturally, albeit  
in the equivalent form $(-1)^{\rho-1} Q^{\rho(\rho-1)/2}$.  
It is particularly appealing because it only contains two terms, no more than (\ref{eq:ii}), 
and is valid regardless of the value of the maximal rank $\rho$.    

\begin{theorem}\label{thm:pcinq2} Let $(U,V)$ be a pair of Lucas sequence with parameters $P$ and $Q$. 
Let $p$ be a prime at least $7$ of maximal rank $\rho$ equal to $p-\e_p$. Then
$$
\binom{2\rho-1}{\rho-1}_U\equiv
(-1)^{\rho-1}Q^{\frac{\rho(\rho-1)}2}\bigg(1+2\frac{U_\rho}{V_\rho}\sum_{0&lt;t&lt;\rho}\frac{V_t}{U_t}
\bigg)\pmod{p^5}.
$$ 
\end{theorem}
\begin{proof} All unmarked sums and products are for $t$ running from $1$ to $\rho-1$. Note that 
$\prod U_t=\prod U_{\rho-t}$. Thus by (\ref{eq:6}) we may write 
\begin{eqnarray*}
2^{\rho-1}Q^{\sum t}\prod U_t &amp; = &amp; \prod2Q^tU_{\rho-t}=\prod(U_\rho V_t-V_\rho U_t)\\
&amp; = &amp; (-V_\rho)^{\rho-1}\prod\bigg(1-\frac{U_\rho}{V_\rho}\frac{V_t}{U_t}\bigg)\prod U_t.
\end{eqnarray*}
Therefore 
$$
(-1)^{\rho-1}Q^{\rho(\rho-1)/2}=\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}
\prod\bigg(1-\frac{U_\rho}{V_\rho}\frac{V_t}{U_t}\bigg),
$$ so that $(-1)^{\rho-1}Q^{\rho(\rho-1)/2}$ is congruent to 
\begin{equation}\label{eq:red}
\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}
\bigg(1-\frac{U_\rho}{V_\rho}\Sigma_1+
\frac{U_\rho^2}{V_\rho^2}\Sigma_{1,1}-\frac{U_\rho^3}{V_\rho^3}\Sigma_{1,1,1}+
\frac{U_\rho^4}{V_\rho^4}\Sigma_{1,1,1,1}\bigg)\pmod {p^5}.
\end{equation}
Note that from (\ref{eq:red}) we recover the congruence
\begin{equation}\label{eq:blanc}
(-1)^{\rho-1}Q^{\rho(\rho-1)/2}\equiv\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\pmod {p^2}.
\end{equation} 
Subtracting the expansion in (\ref{eq:red}) from that of $\binom{2\rho-1}{\rho-1}_U$ obtained in the proof 
of Theorem \ref{thm:pcinq1}, we find that 
\begin{eqnarray*}
\binom{2\rho-1}{\rho-1}_U-(-1)^{\rho-1}Q^{\rho(\rho-1)/2}
&amp;\equiv&amp;\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\bigg(2\frac{U_\rho}{V_\rho}\Sigma_1+
2\frac{U_\rho^3}{V_\rho^3}\Sigma_{1,1,1}\bigg)\\
&amp;\equiv &amp; 2\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\frac{U_\rho}{V_\rho}\Sigma_1\pmod {p^5},
\end{eqnarray*} since $\Sigma_{1,1,1}$ is $0\pmod {p^2}$ by Lemma \ref{lem:sigmas}.
In the above congruence as $\frac{U_\rho}{V_\rho}\Sigma_1$ is $0\pmod {p^3}$ we may,  
by (\ref{eq:blanc}), replace $\big(\frac{V_\rho}{2}\big)^{\rho-1}$ by  
$(-1)^{\rho-1}Q^{\rho(\rho-1)/2}$ and deduce our theorem.  
\end{proof}

\begin{lemma}\label{lem:c}  Suppose $\nu$ is a nonnegative integer. 
Let $p\ge\nu+5$ be a prime of maximal rank, say $\rho$. Then 
$$
\sum_{0&lt;t&lt;\rho}\frac{4Q^t}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\Sigma_{\nu+2}-D\Sigma_\nu\equiv
\begin{cases} 0\pmod{p^2},\;\text{ if }\nu\text{ is odd};\\
0\pmod p,\;\;\text{ if }\nu\text{ is even.}
\end{cases}
$$
\end{lemma}
\begin{proof} We have 
$$
\sum_{0&lt;t&lt;\rho}\frac{4Q^t}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\sum_{0&lt;t&lt;\rho}
\frac{(V_t^2-DU_t^2)}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\Sigma_{\nu+2}-D\Sigma_\nu. 
$$
If $\nu$ is odd, then, $p\ge\nu+5$ implies, by Lemma \ref{lem:psquare}, 
that both $\Sigma_\nu$ and $\Sigma_{\nu+2}$ are $0\pmod{p^2}$. If $\nu$ is even, 
then both $\Sigma_{\nu+2}$ and $D\Sigma_\nu$ are $0\pmod p$, when $\rho$ is $p$ or $p+1$,  
by Lemma \ref{lem:psquare}. If $\rho$ is $p-1$, then by the same lemma 
$\Sigma_{\nu+2}-D\Sigma_\nu\equiv-2D^{\frac{\nu+2}2}-D(-2D^{\nu/2})\equiv0\pmod p$. 
\end{proof}  


\begin{lemma}\label{lem:s1} We have for all primes $p\ge7$ of maximal rank $\rho$
$$
-2\Sigma_1\equiv\frac{U_\rho}{V_\rho}\sum_{0&lt;t&lt;\rho}\frac{4Q^t}{U_t^2}
\pmod {p^4}.
$$
\end{lemma}
\begin{proof} All sums are over an index $t$ running from $1$ to $\rho-1$. 
\begin{eqnarray*}
-2\Sigma_1&amp;=&amp;-\sum\bigg(\frac{V_t}{U_t}+\frac{V_{\rho-t}}{U_{\rho-t}}\bigg)=
-2U_\rho\sum\frac1{U_tU_{\rho-t}}, \text{ by }(\ref{eq:1}),\\
&amp;=&amp;-2U_\rho\sum\frac{2Q^t}{U_t(U_\rho V_t-U_tV_\rho)}, \text{ using }(\ref{eq:6}),\\
&amp;=&amp;2\frac{U_\rho}{V_\rho}\sum \frac{2Q^t}{U_t^2\big(1-\frac{V_t}{U_t}
\frac{U_\rho}{V_\rho}\big)}\\
&amp;\equiv&amp;\frac{U_\rho}{V_\rho}\sum\frac{4Q^t}{U_t^2}\bigg(1+\frac{V_t}{U_t}\frac{U_\rho}{V_\rho}
+\frac{V_t^2}{U_t^2}\frac{U_\rho^2}{V_\rho^2}\bigg)\pmod{p^4},
\end{eqnarray*}
yielding the lemma because, by Lemma \ref{lem:c}, $U_\rho^{\nu+1}\sum\frac{4Q^t}{U_t^2}
\frac{V_t^\nu}{U_t^\nu}$ is $0\pmod{p^4}$, for $\nu=1$ and $\nu=2$, if $p\ge7$. 
\end{proof} 

 From Theorem \ref{thm:pcinq2}, it is not difficult to reach a third theorem that 
generalizes (\ref{eq:iii}).


\begin{theorem}\label{thm:pcinq3} Let $(U,V)$ be a pair of Lucas sequence with parameters $P$ and $Q$. 
Let $p$ be a prime at least $7$ of maximal rank $\rho$ equal to $p-\e_p$. Then
$$
\binom{2\rho-1}{\rho-1}_U\equiv
(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg(1-4\frac{U_\rho^2}{V_\rho^2}\sum_{0&lt;t&lt;\rho}\frac{Q^t}{U_t^2}
\bigg)\pmod{p^5}.
$$ 
\end{theorem}
\begin{proof} In the congruence for the Lucasnomial $\binom{2\rho-1}{\rho-1}_U$ of Theorem \ref{thm:pcinq2}  
we may replace $2\frac{U_\rho}{V_\rho}\Sigma_1$ by $-\frac{U_\rho^2}{V_\rho^2}\sum\frac{4Q^t}{U_t^2}$ 
since by Lemma \ref{lem:s1} the two expressions are congruent modulo ${p^5}$. 
\end{proof}

\begin{remark} In stating Theorem \ref{thm:pcinq3} we chose the expression 
$-4\frac{U_\rho^2}{V_\rho^2}\sum\frac{Q^t}{U_t^2}$ 
rather than $-\frac{U_\rho^2}{V_\rho^2}\Sigma_2+\frac{U_\rho^2}{V_\rho^2}(\rho-1)D$ 
because it contains only one term; that term is $0\pmod{p^3}$ and it reduces to 
$-p^2\sum\frac1{t^2}$ for $U=U(2,1)$.  
\end{remark}

\begin{lemma}\label{lem:go} We have for all primes $p\ge7$ of maximal rank $\rho$
$$
\frac{U_\rho}{V_\rho}\Sigma_1\equiv\frac{U_\rho^2}{V_\rho^2}\Sigma_{1,1}-\frac12
\frac{U_\rho^2}{V_\rho^2}(\rho-1)D\pmod {p^5}.
$$
\end{lemma}
\begin{proof} By Lemma \ref{lem:s1}, we see that 
$$
\frac{U_\rho}{V_\rho}\Sigma_1\equiv-\frac12\frac{U_\rho^2}{V_\rho^2}\sum_{0&lt;t&lt;\rho}\frac{4Q^t}{U_t^2}\pmod {p^5}.
$$ By Lemma \ref{lem:c},  
$$
\sum_{0&lt;t&lt;\rho}\frac{4Q^t}{U_t^2}=\Sigma_2-D(\rho-1).
$$ Thus, as $\Sigma_2=\Sigma_1^2-2\Sigma_{1,1}\equiv-2\Sigma_{1,1}\pmod {p^4}$, 
the lemma follows. 
\end{proof}


 By using Lemma \ref{lem:go} and Theorem \ref{thm:pcinq2} we obtain another generalization of (\ref{eq:i}) 
slightly different from that given in Theorem \ref{thm:pcinq1}, which we now state. 

\begin{theorem}\label{thm:pcinq4} Let $(U,V)$ be a pair of Lucas sequences with parameters $P$ and $Q$. 
Let $p$ be a prime at least $7$ of maximal rank $\rho$ equal to $p-\e_p$. Then
$\binom{2\rho-1}{\rho-1}_U$ is congruent to
$$
(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg(1+\frac{U_\rho}{V_\rho}\sum_{0&lt;t&lt;\rho}\frac{V_t}{U_t}+
\frac{U_\rho^2}{V_\rho^2}\sum_{0&lt;s&lt;t&lt;\rho}\frac{V_sV_t}{U_sU_t}-\frac12D
\frac{U_\rho^2}{V_\rho^2}(\rho-1)\bigg)\pmod{p^5}.
$$
\end{theorem}

 We end the paper with a congruence for $\binom{2\rho-1}{\rho-1}_U$ modulo ${p^6}$. It generalizes 
Theorem 2.4 of \cite{Tau} which says that   
$$
\binom{2p-1}{p-1}\equiv1+2p\sum_{0&lt;t&lt;p}\frac1t+\frac{2p^3}3\sum_{0&lt;t&lt;p}\frac1{t^3}\pmod{p^6},
$$ 
for all primes $p\ge7$, and also generalizes our Theorem \ref{thm:pcinq2}. 

\begin{theorem}\label{thm:Tau} Let $(U,V)$ be a pair of Lucas sequences with parameters $P$ and $Q$. 
Let $p$ be a prime at least $7$ of maximal rank $\rho$. Then 
$$
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\rho-1}Q^{\frac{\rho(\rho-1)}2}\bigg(1+2\frac{U_\rho}
{V_\rho}\sum_{0&lt;t&lt;p}\frac{V_t}{U_t}+
\frac23\frac{U_\rho^3}{V_\rho^3}\sum_{0&lt;t&lt;p}\frac{V_t^3}{U_t^3}\bigg)\pmod{p^6}.
$$
\end{theorem} 
\begin{proof} We proceed as in Lemma \ref{lem:sigmas} to show that $\Sigma_{1,1,1,1,1}\equiv0\pmod{p}$ 
(in fact $0$ modulo $p^2$). By Lemma \ref{lem:psquare}, the expressions 
$\Sigma_1\cdot\Sigma_4-\Sigma_5$, $\Sigma_{1,1}\cdot\Sigma_3$, 
$\Sigma_1\cdot\Sigma_{1,3}$ 
and $\Sigma_1\cdot\Sigma_{2,2}$ are all $0\pmod{p^2}$, so we deduce, successively, that the  
sums $\Sigma_{1,4}$, $\Sigma_{1,1,3}$, $\Sigma_{2,3}$ and $\Sigma_{1,2,2}$ are each $0\pmod{p^2}$. 
Therefore, modulo $p^2$, we obtain the linear system  
\begin{eqnarray*} 
\Sigma_1\cdot\Sigma_{1,1,1,1} &amp; \equiv &amp; 5\,\Sigma_{1,1,1,1,1}+\Sigma_{1,1,1,2},\\
\Sigma_1^5-\Sigma_5 &amp; \equiv &amp;120\,\Sigma_{1,1,1,1,1}+60\,\Sigma_{1,1,1,2}. 
\end{eqnarray*}
As its determinant, $2^2\cdot3^2\cdot5$, is prime to $p$, and its left members are 
each $0\pmod{p^2}$, we find that $\Sigma_{1,1,1,1,1}\equiv0\pmod{p^2}$. 


\medskip

Since $\Sigma_{1,1,1,1,1}$ is 
$0\pmod p$, both the congruence for $\binom{2\rho-1}{\rho-1}_U$, derived from the proof 
of Theorem \ref{thm:pcinq1}, and congruence (\ref{eq:red}) remain valid when we raise  
the modulus from $p^5$ to $p^6$. 
Hence, 
\begin{equation}\label{eq:vent}
\binom{2\rho-1}{\rho-1}_U-(-1)^{\rho-1}Q^{\rho(\rho-1)/2}
\equiv\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\bigg(2\frac{U_\rho}{V_\rho}\Sigma_1+
2\frac{U_\rho^3}{V_\rho^3}\Sigma_{1,1,1}\bigg)\pmod {p^6}. 
\end{equation}
Suppose first that $\epsilon_p=-1$ or $\epsilon_p=0$. Then, as $\Sigma_{1,1}\equiv0\pmod p$, 
we find that (\ref{eq:blanc}) is valid modulo $p^3$. Thus, we may replace $(V_\rho/2)^{\rho-1}$ 
in (\ref{eq:vent}) by $(-1)^{\rho-1}Q^{\rho(\rho-1)/2}$ and obtain that 
\begin{equation}\label{eq:eau}
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\rho-1}Q^{\rho(\rho-1)/2}
\bigg(1+2\frac{U_\rho}{V_\rho}\Sigma_1+
2\frac{U_\rho^3}{V_\rho^3}\Sigma_{1,1,1}\bigg)\pmod {p^6}. 
\end{equation} Looking at the linear system at the start of the proof of Lemma \ref{lem:sigmas} modulo 
$p^3$ we find the system of congruences
\begin{eqnarray*} 
3\Sigma_{1,2}+6\Sigma_{1,1,1} &amp; \equiv &amp; -\Sigma_3,\\
\Sigma_{1,2}+3\Sigma_{1,1,1} &amp; \equiv &amp;0. 
\end{eqnarray*}
Solving for $\Sigma_{1,1,1}$, we see that $\Sigma_{1,1,1}\equiv\frac{\Sigma_3}{3}\pmod{p^3}$, 
which inserted in congruence (\ref{eq:eau}) yields the theorem. 

Suppose now $\epsilon_p=1$. By (\ref{eq:red}) and Lemma \ref{lem:Sig}, congruence (\ref{eq:blanc}), 
when the modulus is increased to $p^3$, becomes 
$$
(-1)^{\rho-1}Q^{\rho(\rho-1)/2}\equiv(V_\rho/2)^{\rho-1}(1+DU_\rho^2/V_\rho^2)
\pmod {p^3}.
$$
Thus we may replace $(V_\rho/2)^{\rho-1}$ in (\ref{eq:vent}) by $(-1)^{\rho-1}Q^{\rho(\rho-1)/2}
(1-DU_\rho^2/V_\rho^2)$, multiply out the resulting expression and remove the term 
in $U_\rho^5\Sigma_{1,1,1}$ which is $0\pmod {p^7}$ to find that 
$$
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\rho-1}Q^{\rho(\rho-1)/2}\bigg(1+2\frac{U_\rho}{V_\rho}\Sigma_1
+2\frac{U_\rho^3}{V_\rho^3}(\Sigma_{1,1,1}-D\Sigma_1)\bigg)\pmod {p^6}.
$$
Because $\Sigma_1$ is $0\pmod{p^2}$  and $\Sigma_{1,1}\equiv D\pmod{p}$, the linear system 
of Lemma \ref{lem:sigmas} taken modulo $p^3$ is 
\begin{eqnarray*} 
3\Sigma_{1,2}+6\Sigma_{1,1,1} &amp; \equiv &amp; -\Sigma_3,\\
\Sigma_{1,2}+3\Sigma_{1,1,1} &amp; \equiv &amp; D\Sigma_1. 
\end{eqnarray*}
Solving for $\Sigma_{1,1,1}$ yields $\Sigma_{1,1,1}\equiv D\Sigma_1+\Sigma_3/3$ and  
the theorem holds.  
\end{proof} 

\section{Appendix on the integrality of Lucasnomials}

 Lucas, with a nearly complete justification, indirectly asserted  
the integrality of Lucasnomials in his memoir \cite[p.\ 203]{Lu} by stating that the 
product of $n$ consecutive terms of a 
(nondegenerate) $U$ sequence is divisible by the product $U_1\dots U_n$. Various proofs 
have appeared, often with restrictions on the sequence $U$. 
In fact, they have been shown to be integral via a combinatorial argument \cite{Ben}. But 
with convention (\ref{eq:conv}) we want to prove their integrality in full generality.  

\begin{proposition} Let $U=(U_n)$ be a Lucas sequence with parameters $P$ and $Q$. With 
the adoption of convention (\ref{eq:conv}) the Lucasnomial coefficients 
$\binom{m}{n}_U$ are rational integers for all nonnegative integers $m$ and $n$. 
\end{proposition}

\begin{proof} If all $U_n$, $n&gt;0$, are nonzero then the frequently used induction argument 
\cite{Hog,Ho,HuSu} based on the general Lucas identity 
$U_{n+1}U_{m-n}-QU_nU_{m-n-1}=U_m$ works fine. We repeat the argument here. 
The induction is on $m$. So one proves the integrality of the Lucasnomial $\binom{m}{n}_U$ 
for $m&gt;n\ge1$ by observing that
\begin{eqnarray*}
U_{n+1}\binom{m-1}{n}_U-QU_{m-n-1}\binom{m-1}{n-1}_U &amp; = &amp;\\
\big(U_{n+1}\frac{U_{m-n}}{U_n}-QU_{m-n-1}\big)\cdot\binom{m-1}{n-1}_U &amp; = &amp;\\
\frac{U_m}{U_n}\cdot\binom{m-1}{n-1}_U=\binom{m}{n}_U&amp;,&amp;
\end{eqnarray*} 
completing the induction. 
If some term $U_n$, $n\ge1$, is $0$ then $U$ is degenerate and, as we saw early in 
Section \ref{sec:2}, $\rho(\infty)\in\{2,3,4,6\}$, where $\rho(\infty)$ is the least 
positive integer $t$ such that $U_t=0$. Note that we may always assume $m\ge2n$. Thus the 
Lucasnomial $\binom{m}{n}_U$ is the quotient of a product of $n$ consecutive $U$ terms 
of indices all larger than $n$ divided by $U_nU_{n-1}\cdots U_1$. 
If $\rho(\infty)=2$, i.e., $U_2=P=0$, then 
$U_{2k+1}=(-1)^kQ^k$ and $U_{2k}=0$, ($k\ge0$). Then $\binom{m}{n}_U$ is up to sign 
a positive power of $Q$. If $\rho(\infty)=3$, then, as $U_3=P^2-Q$, the first few terms of $U$ are 
$0,1,P,0,-P^3,-P^4,0,P^6,P^7,0,\cdots$. So $|U_t|=P^{t-1}$ if $3\nmid t$. 
If $\rho(\infty)=4$, then, as $U_4=P^3-2PQ$ and $P\not=0$, 
$P^2=2Q$ and we see that $|U_t|=2^{\lfloor t/2\rfloor}(P')^{t-1}$ if $4\nmid t$, 
where $P=2P'$. Omitting the $0$ terms when $4\mid t$, powers of $2$ and $P'$ 
in $U_t$ are nondecreasing functions of $t$. A similar result holds for $\rho(\infty)$ equal to 
$6$ when $P^2=3Q$ and, omitting terms divisible by $6$, powers of $3$ and of $P'$ in $U_t$ 
are nondecreasing functions of $t$, where in this case $P=3P'$. The integrality of the Lucasnomials 
follows readily.  
\end{proof}
   
\section{Acknowledgments} 

 We thank an anonymous referee for his clear report and few formal suggestions.   

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\bigskip
\hrule
\bigskip

\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.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 24 2014;
revised versions received  December 18 2014; February 25 2015; March 21 2015; 
March 26 2015.
Published in {\it Journal of Integer Sequences}, May 19 2015.

\bigskip
\hrule
\bigskip

\noindent
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