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\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 $n>0$,
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 {\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 $p>5$ 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_{01$, 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} & = & 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$ \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\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 $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}.
$$
\end{example}
\section{Lucasnomials $\binom{2\rho-1}{\rho-1}_U\pmod{p^5}$}
\label{sec:4}
The congruence of Wolstenholme has been studied to prime powers higher than the third. In particular,
we have, for all primes $p\ge7$,
\begin{eqnarray}
\binom{2p-1}{p-1}&\equiv&1+p\sum_{05$, which yields the values of $\Sigma_2$ and $\Sigma_4\pmod p$, we deduce that
$$
\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_{00$, 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 $m>n\ge1$ by observing that
\begin{eqnarray*}
U_{n+1}\binom{m-1}{n}_U-QU_{m-n-1}\binom{m-1}{n-1}_U & = &\\
\big(U_{n+1}\frac{U_{m-n}}{U_n}-QU_{m-n-1}\big)\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
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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\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.
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\noindent
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