Wolstenholme's congruence says that $\binom{2p-1{p-1\equiv1\pmod{p^3$ for all primes $p\ge5$. Kimball and Webb established an analogue of the congruence of Wolstenholme using Fibonomial coefficients. This note answers the question: `Is there a common generalization to the Wolstenholme and the Kimball and Webb congruences?'. Tinted by a positive answer, valid for all fundamental Lucas sequences, we go up the ladder. We give a broad generalization of several congruences such as Ljunggren et al's $\binom{kp{\ell p\equiv\binom{k\ell\pmod{p^3$, ($p\ge5$), or McIntosh's: $\binom{2p-1{p-1\equiv1-p^2\sum_{0<t<p\frac1{t^2\pmod{p^5$, ($p\ge7$), replacing ordinary binomials by generalized binomial coefficients $\binom{*{*_U$, where $U=U(P,Q)$ is an arbitrary fundamental Lucas sequence. That is, a sequence which satisfies $U_0=0$, $U_1=1$ and $U_{t+2=PU_{t+1-QU_t$, for all $t\ge0$.
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\title{The congruence of Wolstenholme and generalized binomial coefficients related to Lucas sequences} 

\author{
{\sc Christian ~Ballot} \\
{D\'epartement de Math\'ematiques et M\'ecanique}\\
{Universit\'e de Caen} \\
{F14032 Caen Cedex, France} \\
{christian.ballot@unicaen.fr}\\}

\date{}

\maketitle
\begin{abstract} Wolstenholme's congruence says that  
$\binom{2p-1}{p-1}\equiv1\pmod{p^3}$ for all primes $p\ge5$.  
Kimball and Webb established an analogue of the congruence of Wolstenholme using Fibonomial 
coefficients. This note answers the question: `Is there a common generalization  
to the Wolstenholme and the Kimball and Webb congruences?'.  
Tinted by a positive answer, valid for all fundamental Lucas sequences, we go up the ladder. 
We give a broad generalization of several congruences such as Ljunggren et al's 
$\binom{kp}{\ell p}\equiv\binom{k}\ell\pmod{p^3}$,  
($p\ge5$), or McIntosh's: $\binom{2p-1}{p-1}\equiv1-p^2\sum_{0<t<p}\frac1{t^2}\pmod{p^5}$,  
($p\ge7$), replacing ordinary binomials by generalized binomial coefficients $\binom{*}{*}_U$,  
where $U=U(P,Q)$ is an arbitrary fundamental Lucas sequence. That is, a sequence which satisfies 
$U_0=0$, $U_1=1$ and $U_{t+2}=PU_{t+1}-QU_t$, for all $t\ge0$.
\end{abstract}

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

 In 1862 Joseph 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}

 Charles 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 in the last 150 years. 
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{Glai1} p. 21, \cite{Glai2} p. 33). 

 Lemma 3 of the paper \cite{KW1}, which we rewrite as a theorem below, is an analogue of (\ref{eq:W+}). 

\begin{theorem} \label{thm:KW} Let $p$ be a prime at least $7$ whose rank of appearance $\rho$ in the Fibonacci 
sequence is equal to $p-\e_p$, 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} 

 If $A=(a_n)_{n\ge0}$ is a sequence of complex numbers where $a_0=0$ and all $a_n\not=0$ for $n>0$, 
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},& \text{ if }m\ge n\ge1;\\
1,& \text{ if }n=0;\\
0,& \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 
$m>n>0$; the other criterion is expressed in terms of ranks of appearance of prime powers in $A$. The equivalence 
of these two criteria was essentially rediscovered in \cite{KnWi}. 
When $A$ is the Fibonacci sequence 
these binomial coefficients are called {\it Fibonomials} and many papers have studied their properties. 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 Lucanomial} coefficients in the sequel. Ordinary binomials are Lucanomial 
coefficients with parameters $(P,Q)=(2,1)$, whereas the Fibonomials correspond to $(P,Q)=(1,-1)$.       

 
 Therefore it makes sense to look for a simple congruence for the general Lucanomial     
\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 
$U$, that is, the least positive integer $t$ such that $p\mid U_t$. It is known to exist 
for all primes $p$ not dividing $Q$ and to divide $p-\e_p$, where $\e_p$ is the Legendre 
character $(D\;|\;p)$ and $D$ is $P^2-4Q$. 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 $1$ is at least conform 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 
Lucanomial 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 Lucanomial 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 $p>5$ 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} 

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

\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 30 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{Stan} (see solution of exercice 1.14 p. 165).  

 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{KW1}, p. 296) states that
\begin{equation}\label{eq:KW2}
\binom{kp}{\ell p}_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 Lucanomials 
$\binom{k\rho}{\ell\rho}_U$ 
$\pmod{p^3}$ 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 $(k,\ell)=(2,1)$. Note that 
Lucanomial coefficients were given a combinatorial interpretation in \cite{Ben}. Also a $q$-analogue 
of (\ref{eq:Viggo}) that uses $q$-binomial coefficients was established in 
the paper \cite{Stra}.    

\medskip

 In a 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 Lucanomial 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.,  
$$
\binom{2p-1}{p-1}\equiv1-p^2\sum_{0<t<p}\frac1{t^2}\pmod{p^5},
$$ which generalizes into 
$$
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg[1-4\frac{U_\rho^2}{V_\rho^2}
\sum_{0<t<\rho}\frac{Q^t}{U_t^2}\bigg]\pmod{p^5},
$$ where $U(P,Q)$ is a fundamental Lucas sequence and $V(P,Q)$ is its companion sequence.   

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}, \cite{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 $\binom{2p-1}{p-1}\pmod{p^l}$ 
have been established for larger and larger values of $l$ (see \cite{Mes}, p. 4-6). 
No doubt there must be higher corresponding congruences for Lucanomials. In fact, we end the 
paper with such a congruence modulo ${p^6}$. 
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 Lucanomial coefficients $\binom{m}{n}_U$ is asserted 
for all $U$ Lucas sequences. 

\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

 Lucanomial 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 generalized to all strong divisibility 
sequences of positive integers \cite{KnWi}. That includes, in particular, all Lucas sequences $U(P,Q)$ 
with positive terms when $P$ and $Q$ are coprime. 

 Also a generalization 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 $p$, 
was achieved in terms of Lucanomials $\binom{mp+r}{np+s}_U$, under the 
hypothesis that $U(P,Q)$ is a Lucas sequence with $\gcd(P,Q)=1$, $P\not=0$ and 
$P^2Q\not=1$ (see \cite{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 developped 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 
Lucanomials $\binom{m}{n}_U$ are well defined as rational numbers. Although 
the hypotheses of Theorems \ref{thm:N}, \ref{thm:LjWe} or of the theorems of Section \ref{sec:4} 
if applied to a prime $p\ge11$ prevent the corresponding Lucanomials from having zero terms, 
this is not necessarily the case if $p=5$ or $p=7$.  
With $\gcd(P,Q)>1$, 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 
$p^a$ ($a\ge1$), $p\nmid2Q$, and for all $x\ge2$, we have  
\begin{equation}\label{eq:convex}
\#\;\{t\in[x],\;p^a\mid A_t\}\;\ge\;\#\;\{t\in[y],\;p^a\mid A_t\}+\#\;\{t\in[x-y],\;p^a\mid A_t\},
\end{equation} for all $y\in[x-1]$. 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, is 
well defined $p$-adically for all primes $p\nmid2Q$. In fact we can show it is always a rational integer.
\footnote{See our short Appendix}      

\medskip

 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} & = & U_sV_t+U_tV_s,\label{eq:1}\\
2V_{s+t} & = & V_sV_t+DU_sU_t,\label{eq:2}\\
V_t^2-DU_t^2 & = & 4Q^t,\label{eq:3}\\
U_{2t} & = & U_tV_t,\label{eq:4}\\
V_{2t} & = & V_t^2-2Q^t,\label{eq:5}\\
2Q^tU_{s-t} & = & 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 $2DQ$ (see \cite{Wi}, pp. 84--85). 

 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<t<\rho}\frac{V_t^\nu}{U_t^\nu} 
\text{ and }\Sigma_{1,1}:=\sum_{0<s<t<\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{less $\nu=0$ 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}, &\text{ if }\nu\text{ is odd };\\
0\pmod p, &\text{ if }\epsilon_p=-1\text{ or }0;\\
-2D^{\nu/2}\pmod p, & \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 rank $\rho$, then 
\begin{equation}\label{eq:+}
\Sigma_\nu\equiv0\pmod p, \quad\text{ when }\nu \text{ is odd.}
\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 Theorem 4.1 of \cite{Ba2}, or of Theorems 3 and 12 of \cite{Ba3}.) 


 The case $\nu$ even can be treated with the very same arguments used in the last part of the proof 
of Theorem 4, p. 5, of \cite{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{Ba1}.   
\end{proof}    

\begin{lemma} \label{lem:Sig} Let $U=U(P,Q)$ be a fundamental Lucas sequence. If $p\nmid 6Q$ is a prime    
of maximal rank $\rho$ in $U$, then 
$$
\Sigma_{1,1}\equiv\begin{cases} 0\pmod p,& \text{ if }\e_p=0
\text{ or }-1;\\
D\pmod p,& \text{ if }\e_p=1.\end{cases}
$$
\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 $U=U(P,Q)$ be a fundamental Lucas sequence. If $p\nmid Q$ is an odd prime of even 
rank $\rho$ in $U$ and $k\ge1$ is an odd integer, then 
$$
\frac{V_{k\rho}}{2}\equiv-Q^{k\rho/2}\pmod {p^2}.
$$
\end{lemma}
\begin{proof} Since $p$ divides $U_{k\rho}$, but not $U_{k\rho/2}$, we find by 
(\ref{eq:4}) that $p$ divides $V_{k\rho/2}$. Therefore, from 
(\ref{eq:5}) with $t=k\rho/2$, we deduce that $V_{k\rho}\equiv-2Q^{k\rho/2}\pmod {p^2}$. 
\end{proof} 

\begin{lemma} \label{lem:V2} Let $V=V(P,Q)$ be a companion Lucas sequence. Let $m$ be an integer $\ge2$. 
Suppose $V_t\equiv\pm2Q^{t/2}\pmod m$. Then 
$$
V_{2t}\equiv2Q^t\pmod m.
$$
\end{lemma}
\begin{proof} We have $V_{2t}=V_t^2-2Q^t\equiv2Q^t\pmod m$.
\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*}
2^{\rho-1}\prod_{t=1}^{\rho-1}U_{k\rho+t}&=&\prod_{t=1}^{\rho-1}(V_{k\rho} U_t+U_{k\rho} V_t)\\
&\equiv&(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}\\
&\equiv&(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},
\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. 
If $\rho$ is $p+1$ and $k$ is odd, then by Lemma \ref{lem:Vro} there is an integer 
$\lambda$ such that $\frac{V_{k\rho}}{2}=-Q^{k\rho/2}+\lambda p^2$. 
Raising members of the previous equation to the $p$th power gives $(V_{k\rho}/2)^p\equiv-Q^{k\rho p/2}
\pmod{p^3}$. But $-1=(-1)^{-k}$ so the theorem follows in that case. 

If $\rho$ is $p+1$ and $k=2^a\ell$, where $\ell$ is odd and $a\ge1$, then, by Lemma \ref{lem:Vro}, 
we have $V_{\ell\rho}\equiv-2Q^{\ell\rho/2} \pmod{p^2}$. Applying $a$ times Lemma \ref{lem:V2} 
we see that $V_{k\rho}\equiv2Q^{k\rho/2}\pmod{p^2}$. As we did in the case $k$ odd, we raise both 
sides of the congruence to the $p$th power to obtain $(V_{k\rho}/2)^p\equiv Q^{k\rho(\rho-1)/2}=
(-1)^{k\e_p}Q^{k\rho(\rho-1)/2}\pmod{p^3}$ and 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{equation}\label{eq:long}
\binom{(k+1)\rho-1}{\rho-1}_U\equiv\bigg(\frac{V_{k\rho}}{2}\bigg)^p\bigg[2\bigg(\frac2{V_{k\rho}}\bigg)^2
-Q^{k\rho}\bigg(\frac2{V_{k\rho}}\bigg)^4\bigg]\pmod{p^3}.
\end{equation}  
By Lemma \ref{lem:Vro}, we have $V_{k\rho}/2\equiv-Q^{k\rho/2}\pmod {p^2}$ in case $k$ is odd. Using Lemma 
\ref{lem:V2}, as for the case $\rho=p+1$, we get that $V_{k\rho}/2\equiv Q^{k\rho/2}\pmod {p^2}$ if 
$k$ is even. Thus, generally, $V_{k\rho}/2\equiv(-1)^kQ^{k\rho/2}\pmod {p^2}$.  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, with $\alpha_{p,k}:=$ the bracket factor of the righthand side of (\ref{eq:long}), we find that modulo $p^3$
\begin{eqnarray*}
\alpha_{p,k}&\equiv&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\\
&\equiv&(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)\\
&=&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 
finding a global and more natural approach at least for Theorems \ref{thm:pcinq2} and \ref{thm:pcinq3}.  

\section{Lucanomials $\binom{k\rho}{\ell\rho}_U\pmod{p^3}$}
\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    
equal to $p-\e_p$, $\epsilon_p$ being $0$ or $\pm1$. 
Then, for all nonnegative integers $k$ and $\ell$, we have  
\begin{equation}\label{eq:Lju}
\binom{k\rho}{\ell\rho}_U\equiv\binom{k}{\ell}_{U'}(-1)^{\ell(k-\ell)\e_p}Q^{\ell(k-\ell)\rho(\rho-1)/2}\pmod {p^3},
\end{equation}
where $U'$ is the sequence $U_\rho\times U(V_\rho,Q^\rho)$. 
\end{theorem} 
\begin{proof} We only need a proof in case $k>\ell\ge1$. With convention (\ref{eq:conv}) we may write 
\begin{eqnarray*} \binom{k\rho}{\ell\rho}_U & = & \frac{U_{k\rho}U_{k\rho-1}\cdots U_{(k-\ell)\rho+1}}
{U_{\ell\rho}U_{\ell\rho-1}\cdots U_1}\\
& = & \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}}\\
& = & \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}}\\
& = &  \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}\\
&\equiv &\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} })\\
&=&\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, 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{Lucanomials $\binom{2\rho-1}{\rho-1}_U\pmod{p^5}$} 
\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}&\equiv&1+p\sum_{0<t<p}\frac1{t}+p^2\sum_{0<s<t<p}\frac{1}{st}\pmod{p^5}\label{eq:i}\\
                 &\equiv&1+2p\sum_{0<t<p}\frac1{t}\pmod{p^5}\label{eq:ii}\\
                 &\equiv&1-p^2\sum_{0<t<p}\frac1{t^2}\pmod{p^5}\label{eq:iii}.
\end{eqnarray} 

 We will find congruences for the Lucanomial 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 Lucanomials 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{McI}, p. 385. 

\medskip

 To complete the notation introduced in (\ref{eq:nota}) we define the symbols 
$\Sigma_{1,\nu}$ ($\nu=2$ or $3$), $\Sigma_{2,2}$, $\Sigma_{1,1,1}$, $\Sigma_{1,1,2}$ and $\Sigma_{1,1,1,1}$,  
respectively, as the sums 
$$
\sum_{s,\,t}\frac{V_sV_t^\nu}{U_sU_t^\nu},\;\sum_{s<t}\frac{V_s^2V_t^2}{U_s^2U_t^2},\;
\sum_{r<s<t}\frac{V_rV_sV_t}{U_rU_sU_t},\;\sum_{\substack{r<s,\\t\in(0,\rho)}}\frac{V_rV_sV_t^2}{U_rU_sU_t^2},\;
\sum_{q<r<s<t}\frac{V_qV_rV_sV_t}{U_qU_rU_sU_t},
$$  where in each sum $q$, $r$, $s$ and $t$ are distinct integers in the interval $(0,\rho)$ and 
$\rho$ is the rank of a prime $p$.  

\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 & = & 3\Sigma_{1,2}+6\Sigma_{1,1,1},\\
\Sigma_1\cdot\Sigma_{1,1} & = & \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 $p>5$, 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 
$$
\binom{2\rho-1}{\rho-1}_U\equiv
\bigg(\frac{V_\rho}{2}\bigg)^{\rho-1}\bigg[1+\frac{U_\rho}{V_\rho}\sum_{0<t<\rho}
\frac{V_t}{U_t}+\frac{U_\rho^2}{V_\rho^2}\sum_{0<s<t<\rho}\frac{V_sV_t}{U_sU_t}+R\bigg]\pmod{p^5},
$$
$$\text{ where }\;R=\frac{\epsilon_p(1+\epsilon_p)}2\frac{D^2U_\rho^4}{V_\rho^4}=
\begin{cases}0,\;\,\quad\qquad\text{ if }\e_p=0\text{ or }-1;\\
D^2U_\rho^4/V_\rho^4,\text{ if }\e_p=1.
\end{cases}
$$
\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.  
It is particularly appealing because it only contains two terms, no more than (\ref{eq:ii}), 
and is valid regardless of the values 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
$$
\binom{2\rho-1}{\rho-1}_U\equiv
(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg[1+2\frac{U_\rho}{V_\rho}\sum_{0<t<\rho}\frac{V_t}{U_t}
\bigg]\pmod{p^5}.
$$ 
\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 & = & \prod2Q^tU_{\rho-t}=\prod(U_\rho V_t-V_\rho U_t)\\
& = & (-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 
\begin{equation}\label{eq:red}
(-1)^{\rho-1}Q^{\rho(\rho-1)/2}\equiv\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}\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}
&\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)\\
&\equiv & 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 
$$
\sum_{0<t<\rho}\frac{4Q^t}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\Sigma_{\nu+2}-D\Sigma_\nu\equiv
\begin{cases} 0\pmod{p^2},\;\text{ if }\nu\text{ is odd};\\
0\pmod p,\;\;\text{ if }\nu\text{ is even.}
\end{cases}
$$
\end{lemma}
\begin{proof} We have 
$$
\sum_{0<t<\rho}\frac{4Q^t}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\sum_{0<t<\rho}
\frac{(V_t^2-DU_t^2)}{U_t^2}\frac{V_t^\nu}{U_t^\nu}=\Sigma_{\nu+2}-D\Sigma_\nu. 
$$
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 $\rho$
$$
-2\Sigma_1\equiv\frac{U_\rho}{V_\rho}\sum_{0<t<\rho}\frac{4Q^t}{U_t^2}
\pmod {p^4}.
$$
\end{lemma}
\begin{proof} All sums are over an index $t$ running from $1$ to $\rho-1$. 
\begin{eqnarray*}
-2\Sigma_1&=&-\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}),\\
&=&-2U_\rho\sum\frac{2Q^t}{U_t(U_\rho V_t-U_tV_\rho)}, \text{ using }(\ref{eq:6}),\\
&=&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]}\\
&\equiv&\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*}
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
$$
\binom{2\rho-1}{\rho-1}_U\equiv
(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg[1-4\frac{U_\rho^2}{V_\rho^2}\sum_{0<t<\rho}\frac{Q^t}{U_t^2}
\bigg]\pmod{p^5}.
$$ 
\end{theorem}
\begin{proof} In the congruence for the Lucanomial $\binom{2\rho-1}{\rho-1}_U$ 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 
$$
\frac{U_\rho}{V_\rho}\Sigma_1\equiv-\frac12\frac{U_\rho^2}{V_\rho^2}\sum_{0<t<\rho}\frac{4Q^t}{U_t^2}\pmod {p^5}.
$$ By Lemma \ref{lem:c},  
$$
\sum_{0<t<\rho}\frac{4Q^t}{U_t^2}=\Sigma_2-D(\rho-1).
$$ 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
$$
(-1)^{\e_p}Q^{\frac{\rho(\rho-1)}2}\bigg[1+\frac{U_\rho}{V_\rho}\sum_{0<t<\rho}\frac{V_t}{U_t}+
\frac{U_\rho^2}{V_\rho^2}\sum_{0<s<t<\rho}\frac{V_sV_t}{U_sU_t}-\frac12D
\frac{U_\rho^2}{V_\rho^2}(\rho-1)\bigg]\pmod{p^5}.
$$
\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   
$$
\binom{2p-1}{p-1}\equiv1+2p\sum_{0<t<p}\frac1t+\frac{2p^3}3\sum_{0<t<p}\frac1{t^3}\pmod{p^6},
$$ 
for all primes $p\ge7$, and also generalizes our Theorem \ref{thm:pcinq2}. 

\begin{theorem} 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 
$$
\binom{2\rho-1}{\rho-1}_U\equiv(-1)^{\rho-1}Q^{\frac{\rho(\rho-1)}2}\bigg[1+2\frac{U_\rho}
{V_\rho}\sum_{0<t<p}\frac{V_t}{U_t}+
\frac23\frac{U_\rho^3}{V_\rho^3}\sum_{0<t<p}\frac{V_t^3}{U_t^3}\bigg]\pmod{p^6}.
$$
\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 $p^2$). 
First we extend the definitions made before Lemma \ref{lem:sigmas} to define analogously 
the sums $\Sigma_{1,4}$, $\Sigma_{1,1,3}$, $\Sigma_{2,3}$, $\Sigma_{1,2,2}$ and $\Sigma_{1,1,1,2}$.  
The expressions $\Sigma_1\cdot\Sigma_4-\Sigma_5$, $\Sigma_3\cdot\Sigma_{1,1}$, $\Sigma_1\cdot\Sigma_{1,3}$ 
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 $p^2$,  the two expressions $\Sigma_1\cdot\Sigma_{1,1,1,1}$ 
and $\Sigma_1^5-\Sigma_5$ are linear combinations of $\Sigma_{1,1,1,1,1}$ and $\Sigma_{1,1,1,2}$.  
Because these two expressions are each $0\pmod{p^2}$ we deduce 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} & \equiv & -\Sigma_3,\\
\Sigma_{1,2}+3\Sigma_{1,1,1} & \equiv &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 $\epsilon_p=1$ so that 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} & \equiv & -\Sigma_3,\\
\Sigma_{1,2}+3\Sigma_{1,1,1} & \equiv & 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 Lucanomials}

 The question of the integrality of Lucanomials has appeared in various places, but 
we want to formally prove that with convention (\ref{eq:conv}) they are integral  
in full generality.  

\begin{prop} Let $U=(U_n)$ be a Lucas sequence with parameters $P$ and $Q$. With 
the adoption of convention (\ref{eq:conv}) the Lucanomial coefficients 
$\binom{m}{n}_U$ are rational integers for all nonnegative integers $m$ and $n$. 
\end{prop}
\begin{proof} If all $U_n$, $n>0$, are nonzero then the frequently used induction argument 
(see \cite{HuSu}, Lemma 1; or \cite{Ben}) based on the general Lucas identity 
$U_{n+1}U_{m-n}-QU_nU_{m-n-1}=U_m$ works fine. 
(The induction is on $m$. So one proves the integrality of the Lucanomial $\binom{m}{n}_U$ 
for $m>n\ge1$ by observing that
\begin{eqnarray*}
U_{n+1}\binom{m-1}{n}_U-QU_{m-n-1}\binom{m-1}{n-1}_U & = &\\
\bigg(U_{n+1}\frac{U_{m-n}}{U_n}-QU_{m-n-1}\bigg)\cdot\binom{m-1}{n-1}_U & = &\\
\frac{U_m}{U_n}\cdot\binom{m-1}{n-1}_U=\binom{m}{n}_U&,&
\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 
Lucanomial $\binom{m}{n}_U$ 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 Lucanomials 
follows readily.  
\end{proof}
   


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-----------------------------------------------------------------------------------

2010 {\it Mathematics Subject Classification}: 11A07, 11B65, 11B39.

{\it Keywords}: generalized binomial coefficients, Wolstenholme's congruence, Lucas sequences, rank of appearance.    


\end{document}

1409.8629.extracted_bib
{Bab C. Babbage, Demonstration of a theorem relating to prime numbers, {Edinburgh Philosophical J., {1 (1819), 46--49.
1409.8629.extracted_bib
{Bai D. F. Bailey, Two $p^3$ variations of Lucas' theorem, {J. Number Theory, {35 (1990), no. 2, 208--215.
1409.8629.extracted_bib
{Ba1 C. Ballot, On a congruence of Kimball and Webb involving Lucas sequences, {J. Integer Seq., {17 (2014), Article 14.1.3.
1409.8629.extracted_bib
{Ba2 C. Ballot, Lucas sequences with cyclotomic root field, {Dissertationes Math., {490 (2013), 92 pp.
1409.8629.extracted_bib
{Ba3 C. Ballot, A further generalization of a congruence of Wolstenholme, {J. Integer Seq., {15 (2012), Article 12.8.6.
1409.8629.extracted_bib
{Ben A. Benjamin and S. Plott, A combinatorial approach to Fibonomial coefficients, {Fibonacci Quart., {46/47 (2008/09), no. 1, 7--9.
1409.8629.extracted_bib
{Bru V. Brun, J. O. Stubban, J. E. Fjeldstad, R. Tambs Lyche, K. E. Aubert, W. Ljunggren, E. Jacobsthal. On the divisibility of the difference between two binomial coefficients. Den 11te Skandinaviske Matematikerkongress, Trondheim, 1949, pp. 42--54. {Johan Grundt Tanums Forlag, Oslo, 1952.
1409.8629.extracted_bib
{Fr R. D. Fray, Congruence properties of ordinary and q-binomial coefficients. {Duke Math. J., {34, (1967) 467--480.
1409.8629.extracted_bib
{Glai1 J. W. L. Glaisher, Congruences relating to the sums of products of the first $n$ numbers and to other sums of products, {Q. J. Math. {31 (1900), 1--35.
1409.8629.extracted_bib
{Glai2 J. W. L. Glaisher, On the residues of the sums of products of the first $p-1$ numbers, and their powers, to modulus $p^2$ or $p^3$, {Q. J. Math. {31 (1900), 321--353.
1409.8629.extracted_bib
{Gou H. W. Gould, The bracket function and Fonten\'e-Ward generalized binomial coefficients with application to Fibonomial coefficients, {Fibonacci Quart.\/, {7.1 (1969) 23--40, 55.
1409.8629.extracted_bib
{Hoo C. Hooley, On Artin's conjecture, {J. Reine Angew. Math. {225 (1967), 209--220.
1409.8629.extracted_bib
{HuSu H. Hu and Z-W Sun, An extension of Lucas' theorem, {Proc. Amer. Math. Soc. {129 (2001), no. 12, 3471--3478.
1409.8629.extracted_bib
{KW1 W. Kimball and W. Webb, A congruence for Fibonomial coefficients modulo $p^3$, {Fibonacci Quart.\/, {33 (1995) 290--297.
1409.8629.extracted_bib
{KW2 W. Kimball and W. Webb, Some generalizations of Wolstenholme's theorem, {Applications of Fibonacci Numbers {8 (Rochester, NY, 1998), Kluwer Acad. Publ., Dordrecht, (1999), 213--18.
1409.8629.extracted_bib
{KnWi D. Knuth and H. Wilf, The power of a prime that divides a generalized binomial coefficient, {J. Reine Angew. Math. {396 (1989), 212--219.
1409.8629.extracted_bib
{McI R. J. McIntosh, On the converse of Wolstenholme's Theorem, {Acta Arith.\/, {71 (1995), 381--389.
1409.8629.extracted_bib
{Mes R. Me\v{strovi\`c, Wolstenholme's theorem: Its generalizations and extensions in the last hundred and fifty years (1862-2012), preprint {\tt http://arxiv.org/abs/1111.3057v2[math.NT].
1409.8629.extracted_bib
{Mes1 R. Me\v{strovi\`c, Congruences for Wolstenholme primes, preprint {\tt http://arxiv.org/abs/1108.4178[math.NT].
1409.8629.extracted_bib
{Pan H. Pan, A generalization of Wolstenholme's harmonic series congruence, {Rocky Mountain J. Math.\/, {38 (2008), 1263--1269.
1409.8629.extracted_bib
{Stan R. Stanley, {Enumerative combinatorics, Volume 1. Second edition. Cambridge Studies in Advanced Mathematics, 49. Cambridge University Press, Cambridge, (2012).
1409.8629.extracted_bib
{Stra A. Straub, A q-analog of Ljunggren's binomial congruence, 23rd International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2011), 897--902, {Discrete Math. Theor. Comput. Sci. Proc., AO, Assoc. Discrete Math. Theor. Comput. Sci., Nancy, 2011
1409.8629.extracted_bib
{Ros H. Roskam, A quadratic analogue of Artin's conjecture on primitive roots. {J. Number Theory {81 (2000), no. 1, 93--109.
1409.8629.extracted_bib
{Ros1 H. Roskam, Erratum: ``A quadratic analogue of Artin's conjecture on primitive roots'' [J. Number Theory 81 (2000), no. 1, 93--109. {J. Number Theory {85 (2000), no. 1, 108.
1409.8629.extracted_bib
{Tau R. Tauraso, More congruences for central binomial coefficients. {J. Number Theory, {130 (2010), no. 12, 2639--2649.
1409.8629.extracted_bib
{Wa M. Ward, Note on divisibility sequences. {Bull. Amer. Math. Soc. {42 (1936), no. 12, 843--845.
1409.8629.extracted_bib
{Wi H. C. Williams, {\'Edouard Lucas and primality testing, Wiley, Canadian Math. Soc. Series of Monographs and Advanced Texts, (1998).
1409.8629.extracted_bib
{Wo J. Wolstenholme, On certain properties of prime numbers, {Q. J. Pure Appl. Math.\/, {5 (1862), 35--39.
1409.8629.extracted_bib
{Zh J. Zhao, Bernoulli numbers, Wolstenholme's theorem, and $p^5$ variations of Lucas' theorem. {J. Number Theory, {123 (2007), no. 1, 18--26.