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\begin{center}
\vskip 1cm{\LARGE\bf 
More Congruences for Central Binomial Sums\\
\vskip .1in
with Fibonacci and Lucas Numbers 
}
\vskip 1cm
\large
Roberto Tauraso\\
Dipartimento di Matematica \\
Universit\`a di Roma Tor Vergata \\
00133 Roma\\
Italy \\
\href{mailto: tauraso@mat.uniroma2.it}{\tt tauraso@mat.uniroma2.it}
\end{center}

\vskip .2 in


\begin{abstract}
We mainly determine $\sum_{k=1}^{p-1}\binom{2k}{k}h_kx^k$ modulo a prime $p$ with $h_k=\sum_{j=1}^k\frac{1}{2j-1}$. We also provide some applications of this polynomial congruence for some special values of $x$ which involve the Fibonacci and Lucas numbers.
\end{abstract}

\section{Introduction}
In \cite{Le85}, Lehmer presented a long list of interesting series involving the central binomial coefficient $\binom{2k}{k}$ such as 
\begin{equation} \label{2kkkseries}
\sum_{k=1}^\infty\binom{2k}{k}\frac{x^k}{k} = -2 \ln\left(\frac{1+\sqrt{1-4x}}{2}\right).
\end{equation}
Since then, using many different approaches and methods, several other authors 
have  actively investigated this kind of sums. Among them we would like to mention \cite{CZ09,Bo12,Ch16}. In particular, Chen in \cite[Theorem 8]{Ch16} obtained
\begin{equation} \label{2kkhseries}
\sum_{k=1}^\infty\binom{2k}{k}h_kx^k = - \frac{\ln(1-4x)}{2\sqrt{1-4x}}
\end{equation}
with $h_k=\sum_{j=1}^k\frac{1}{2j-1}$.
In this work, as we did in \cite{MT13,Ta16,MT18}, we are going to consider the sums of the first $p-1$ terms of the above series, with a prime $p$, and evaluate them modulo $p$. 

The paper is organized as follows. The next section is devoted to the proofs of two polynomial congruences which are the finite analogues of the infinite series \eqref{2kkkseries} and  \eqref{2kkhseries}. In the third section, we specialize those polynomial congruences to some particular values $(u_n)_{n\geq 1}$ and $(v_n)_{n\geq 1}$ related to Fibonacci numbers $F_n$ (\seqnum{A000045}) and Lucas numbers $L_n$ (\seqnum{A000032}):
\begin{align*}
&amp;u_n=\frac{(-1)^n}{L_n^2}=\frac{(-1)^n}{L_{2n}+2(-1)^n}\to
-1,\frac{1}{9},-\frac{1}{16},\frac{1}{49},-\frac{1}{121},\frac{1}{324},-\frac{1}{841},\dots\\
&amp;v_n=\frac{(-1)^{n-1}}{5F_n^2}=\frac{(-1)^{n-1}}{L_{2n}-2(-1)^n}\to
\frac{1}{5},-\frac{1}{5},\frac{1}{20},-\frac{1}{45},\frac{1}{125},-\frac{1}{320},\frac{1}{845},\dots
\end{align*}
Notice that, up to sign, $u_n$ and $v_n$ are the reciprocals of \seqnum{A001254} and \seqnum{A099921} in the {\it On-Line Encyclopedia
of Integer Sequences} (OEIS).

For instance, after extending a result by Williams \cite{Wi82}, 
we show that for any integer $n\geq 1$, and for any prime $p&gt;2$ not dividing $5F_nL_n$,
\begin{align*}
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}\frac{u_n^k}{k}\equiv 2q_p(L_n)-\frac{5nF_n}{L_n}f_p\pmod{p},\\
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}h_{k}v_n^k\equiv
-\leg{5}{p}\left(
\frac{1}{2}q_p\left(\frac{L_n^2}{5F_n^2}\right)+(-1)^n\frac{2n}{F_{2n}}f_p
\right)\pmod{p}.
\end{align*}
Here $\legs{5}{p}$ stands for a Legendre symbol, and let
$$q_p(x)=\frac{x^{p-1}-1}{p},\quad\quad f_p=\frac{F_{p-\leg{5}{p}}}{p}$$
denote the {\it Fermat quotient} and the {\it Fibonacci quotient}, respectively. 

In order to demonstrate the similarities of the above formulas with the corresponding  series evaluations, notice that, by \eqref{2kkkseries} and \eqref{2kkhseries}, we obtain 
\begin{align*}
&amp;\sum_{k=1}^\infty\binom{2k}{k}\frac{u_n^k}{k}=
2\ln(L_n)-2n\ln\left(\frac{1+\sqrt{5}}{2}\right),\\
&amp;\sum_{k=1}^\infty\binom{2k}{k}h_{k}v_n^k = -\frac{\sqrt{5}F_n}{2L_n}\ln\left(\frac{L_n^2}{5F_n^2}\right).
\end{align*}


\section{Polynomial congruences}

The next congruence is equivalent to \cite[(32)]{MT18}. Here we give a new and self-contained proof.
\begin{theorem}
Let $p&gt;2$ be a prime. Then
\begin{equation}\label{mod2kkkx}\sum_{k=1}^{p-1} \binom{2k}{k}\frac{x^k}{k}\equiv
2\frac{1-\alpha^p-(1-\alpha)^p}{p}\pmod{p}
\end{equation}
where $x=\alpha(1-\alpha)$.
\end{theorem}
\begin{proof} Let $\beta=1-\alpha$. Then, for $1\leq k\leq p-1$, it follows
that
$$\binom{p}{k}=\frac{p(p-1)\cdots(p-(k-1))}{k!}\equiv\frac{p(-1)^{k-1}}{k}\pmod{p^2},$$
and we find
\begin{align*} 
(1-\alpha z)^p(1-\beta z)^p
&amp;=\sum_{k=0}^p\binom{p}{k}(-\alpha z)^k\sum_{k=0}^p\binom{p}{k}(-\beta z)^k\\
&amp;\equiv \left(1-\alpha^pz^p-p\sum_{k=1}^{p-1}\frac{\alpha^k z^k}{k}\right)
\left(1-\beta^pz^p-p\sum_{k=1}^{p-1}\frac{\beta^k z^k}{k}\right)\\
&amp;\equiv 1-(\alpha^p+\beta^p)z^p+(\alpha\beta)^pz^{2p}\\
&amp;\quad-p(1-\alpha^pz^p)\sum_{k=1}^{p-1}\frac{\beta^k z^k}{k}-p(1-\beta^pz^p)\sum_{k=1}^{p-1}\frac{\alpha^k z^k}{k}
\pmod{p^2}.
\end{align*}
Hence, we have
\begin{equation}\label{oa}
[z^{p}](1-\alpha z)^p(1-\beta z)^p \equiv -\alpha^p-\beta^p\pmod{p^2}.
\end{equation}
On the other hand,
\begin{align*} 
[z^{p}](1-\alpha z)^p(1-\beta z)^p &amp;=[z^{p}](1-z+xz^2)^p\\
&amp;=[z^{p}]\sum_{k=0}^p\binom{p}{k}x^kz^{2k}(1-z)^{p-k}\\
&amp;=[z^{p}]\sum_{k=0}^p\binom{p}{k}x^kz^{2k}\sum_{j=0}^{p-k}\binom{p-k}{j}(-z)^j\\
&amp;=\sum_{k=0}^{\frac{p-1}{2}}\binom{p}{k}\binom{p-k}{p-2k}(-1)^{p-2k}x^k\\
&amp;=-\sum_{k=0}^{\frac{p-1}{2}}\binom{p}{k}\binom{p-k}{k}x^k.
\end{align*}
Therefore, we get
\begin{align}\label{ob}
[z^{p}](1-\alpha z)^p(1-\beta z)^p &amp;\equiv
-1+p\sum_{k=1}^{\frac{p-1}{2}}(-1)^k\binom{p-k}{k}\frac{x^k}{k}\nonumber\\
&amp;\equiv-1+\frac{p}{2}\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k}\pmod{p^2}
\end{align}
where we used the fact that, for $1\leq k\leq \frac{p-1}{2}$,
$$\binom{p-k}{k}=\frac{(p-k)\dots(p-(2k-1))}{k!}
\equiv \frac{(-1)^k}{2}\binom{2k}{k} \pmod{p},$$
and $p$ divides $\binom{2k}{k}$ for $\frac{p-1}{2}&lt;k&lt;p$.
Combining \eqref{oa} with \eqref{ob} yields the desired result.
\end{proof}

Once the preliminary congruence \eqref{mod2kkkx} is obtained, we are in the position to state the main theorem of this section.
\begin{theorem}
If $p&gt;2$ is a prime then
\begin{equation}\label{mod2kkhx}
\sum_{k=1}^{p-1} \binom{2k}{k}h_k x^k
\equiv\frac{1}{2}(1-4x)^{\frac{p-1}{2}}\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^{k}}{k}
-\sum_{k=1}^{p-1}\binom{2k}{k}\frac{y^{k}}{k}\right) \pmod{p}
\end{equation}
where $y=-\frac{x}{1-4x}$.
\end{theorem}
\begin{proof} For $1\leq k\leq \frac{p-1}{2}$,
$$\binom{\frac{p-1}{2}}{k}=\frac{(p-1)(p-3)\dots(p-(2k-1))}{2^kk!}\equiv
\frac{(-1)^k(2k-1)!!}{2^kk!}=\binom{2k}{k}\frac{1}{(-4)^k}\pmod{p}$$
which implies that
\begin{equation}\label{one}
\sum_{k=0}^{p-1}\binom{2k}{k}x^k\equiv \sum_{k=0}^{\frac{p-1}{2}}\binom{\frac{p-1}{2}}{k}(-4x)^k
\equiv(1-4x)^{\frac{p-1}{2}}\pmod{p}.
\end{equation}
Furthermore, by \cite[(14)]{Ta16}, we have that
\begin{equation}\label{two}
\left(\sum_{k=0}^{p-1}\binom{2k}{k}x^{k}\right)\cdot
\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^{k}}{k}\right)
\equiv 2\sum_{k=1}^{p-1}\binom{2k}{k}(H_{2k}-H_k)x^k\pmod{p}
\end{equation}
where $H_k=\sum_{j=1}^k\frac{1}{j}$ is the $k$-th {\it harmonic number}.

In view of the known identity $\sum_{k=1}^n \binom{n}{k}\frac{(-1)^{k}}{k}=-H_{n}$, it follows that
\begin{align}\nonumber
\sum_{k=1}^{p-1}\binom{2k}{k}H_k x^k
&amp;\equiv \sum_{k=1}^{\frac{p-1}{2}}\binom{\frac{p-1}{2}}{k} H_k(-4x)^k\\
\nonumber
&amp;= -\sum_{k=1}^{\frac{p-1}{2}}\binom{\frac{p-1}{2}}{k}(-4x)^k
\sum_{j=1}^{k}\binom{k}{j}\frac{(-1)^j}{j}\\
\nonumber
&amp;= -\sum_{j=1}^{\frac{p-1}{2}}\frac{(-1)^j}{j} 
\sum_{k=j}^{\frac{p-1}{2}}\binom{\frac{p-1}{2}}{k}\binom{k}{j}(-4x)^k\\
\nonumber
&amp;= -\sum_{j=0}^{\frac{p-1}{2}}\frac{(-1)^j}{j} \binom{\frac{p-1}{2}}{j}
(-4x)^{j}(1-4x)^{\frac{p-1}{2}-j}\\
\label{three}
&amp;\equiv -(1-4x)^{\frac{p-1}{2}}\sum_{j=1}^{p-1}
\binom{2j}{j}\frac{y^j}{j}\pmod{p}.
\end{align}
Finally, given that $h_k=H_{2k}-\frac{1}{2}H_k$ and combining \eqref{one}, \eqref{two}, and \eqref{three}, we obtain
\begin{align*}
\sum_{k=1}^{p-1} \binom{2k}{k}h_k x^k
&amp;=\sum_{k=1}^{p-1} \binom{2k}{k}(H_{2k}-H_k) x^k+\frac{1}{2}\sum_{k=1}^{p-1} \binom{2k}{k}H_k x^k\\
&amp;\equiv \frac{1}{2}(1-4x)^{\frac{p-1}{2}}\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^{k}}{k}
-\sum_{k=1}^{p-1}\binom{2k}{k}\frac{y^{k}}{k}\right) 
\pmod{p}.
\end{align*}
\end{proof}

If $x=\frac{1}{2}$ then $y=-\frac{x}{1-4x}=\frac{1}{2}$ 
and congruence \eqref{mod2kkhx} immediately implies the following corollary.
\begin{corollary}
For any prime $p&gt;2$, 
$$\sum_{k=1}^{p-1} \binom{2k}{k}\frac{h_k}{2^k}\equiv 0 \pmod{p}.$$
\end{corollary}

It is worth pointing out that, based on a numerical experiment, the above congruence seems to have the following conjectural $q$-analogue: for any odd integer $n$,
\begin{equation}\label{conj}
\sum_{k=1}^{n-1}\qbin{2k}{k}\frac{q^k}{\prod_{j=1}^k (1+q^j)}\sum_{j=1}^k \frac{q^{2j-1}}{[2j-1]_q}\equiv (-1)^{\frac{n-1}{2}}\frac{(n-1)(q-1)}{2}q^{\frac{n^2-1}{4}}\pmod{\Phi_n(q)}
\end{equation}
where $[n]_q=(1-q^n)/(1-q)$, for $0\leq k\leq n$,
$$\qbin{n}{k}=\frac{(q;q)_{n}}{(q;q)_{k}(q;q)_{n-k}}$$
is the $q$-binomial coefficient with $(z;q)_n=(1-z)(1-zq)\cdots(1-zq^{n-1})$, and 
$\Phi_n(q)$ denotes the $n$-th cyclotomic polynomial.

\section{Congruences involving Fibonacci and Lucas numbers}

We first establish the next two congruences modulo $p^2$,
which extend \cite[(2.9) and (2.10)]{Wi82}. 
\begin{lemma}
Let $\eps=\legs{5}{p}$ and let $p$ be a prime different from $2$ and $5$. Then for any integer $n\geq 0$,
\begin{align}\label{Lnp}
&amp;L_{np}\equiv
L_n+\frac{5n}{2}F_nF_{p-\eps} \pmod{p^2},\\
\label{Fnp}
&amp;F_{np}\equiv
\eps\left(F_n+\frac{n}{2}L_nF_{p-\eps}\right) \pmod{p^2}.
\end{align}
\end{lemma}
\begin{proof} We show both congruences by induction with respect to $n$. 
They are are clearly true for $n=0$. 
For $n=1$, by \cite[(2.9)]{Wi82},  we have
$L_{p-\eps}\equiv 2\eps\pmod{p^2}$.
Therefore
$$ 
L_p=\frac{1}{2}\left(L_{p-\eps}L_{\eps}+5F_{p-\eps}F_{\eps}\right)
\equiv \eps L_{\eps}+\frac{5}{2}F_{p-\eps}F_{\eps}
=L_1+\frac{5}{2}F_1F_{p-\eps} \pmod{p^2}
$$
and 
$$F_{p}=\frac{1}{2}\left(L_{p-\eps}F_{\eps}+F_{p-\eps}L_{\eps}\right)
\equiv \eps F_{\eps}+\frac{\eps}{2}L_{1}F_{p-\eps}
\equiv \eps\left(F_1+\frac{1}{2}L_1F_{p-\eps}\right)  \pmod{p^2}
$$
where we applied the identities
$$2L_{n+m}=L_nL_m+5F_nF_m, \qquad 2F_{n+m}=L_nF_m+F_nL_m,$$
and $\eps L_{\eps}=1=L_1$, $F_{\eps}=1=F_1$.

Inductive step for \eqref{Lnp}: for $n\geq 3$, by the multiple-angle recurrence for $L_{np}$, and because $p$ divides  $F_{p-\eps}$, we get
\begin{align*} 
L_{np}&amp;=L_pL_{(n-1)p}-(-1)^pL_{(n-2)p}\\
&amp;\equiv \left(1+\frac{5}{2}F_{p-\eps}\right)
\left(L_{n-1}+\frac{5(n-1)}{2}F_{n-1}F_{p-\eps}\right)
+\left(L_{n-2}+\frac{5(n-2)}{2}F_{n-2}F_{p-\eps}\right) \\
&amp;\equiv (L_{n-1}+L_{n-2})+\frac{5}{2}F_{p-\eps}\left(L_{n-1}+(n-1)F_{n-1}+(n-2)F_{n-2}\right) \\
&amp;\equiv L_{n}+\frac{5}{2}F_{p-\eps}\left(F_{n}+(n-1)F_{n-1}+(n-1)F_{n-2}\right) \\
&amp;\equiv L_{n}+\frac{5n}{2}F_nF_{p-\eps}
\pmod{p^2}.
\end{align*}
A similar argument shows the inductive step for \eqref{Fnp}:
\begin{align*} 
F_{np}&amp;=L_pF{(n-1)p}-(-1)^pF_{(n-2)p}\\
&amp;\equiv \left(1+\frac{5}{2}F_{p-\eps}\right)\left(F_{n-1}+\frac{(n-1)}{2}L_{n-1}F_{p-\eps}\right)+\left(F_{n-2}+\frac{(n-2)}{2}L_{n-2}F_{p-\eps}\right) \\
&amp;\equiv (F_{n-1}+F_{n-2})+
\frac{5}{2}F_{p-\eps}\left(F_{n-1}+(n-1)L_{n-1}+(n-2)L_{n-2}\right) \\
&amp;\equiv F_{n}+\frac{5}{2}F_{p-\eps}\left(L_{n}+(n-1)L_{n-1}+(n-1)L_{n-2}\right) \\
&amp;\equiv F_{n}+\frac{5n}{2}L_nF_{p-\eps}
\pmod{p^2}.
\end{align*}
\end{proof}


In the next two theorems we specialize the polynomial congruences \eqref{mod2kkkx} and \eqref{mod2kkhx} obtained in the previous section to the numbers $u_n$ and $v_n$.
\begin{theorem}
Let $n\geq 1$ and let $p&gt;2$ be a prime not dividing $5L_nF_n$. Then
\begin{align}\label{mod2kkku}
&amp;\sum_{k=1}^{p-1} \binom{2k}{k}\frac{u_n^k}{k}\equiv
-q_p(u_n)-\frac{5nF_n}{L_n}f_p \pmod{p}\\
\label{mod2kkkv}
&amp;\sum_{k=1}^{p-1} \binom{2k}{k}\frac{v_n^k}{k}\equiv
-q_p(v_n)-\frac{nL_n}{F_n}f_p\pmod{p}.
\end{align}
\end{theorem}
\begin{proof} Let $\varphi_{\pm}=(1\pm\sqrt{5})/2$. 
For $\alpha=\varphi_+^n/L_n$, we have that
$$\alpha(1-\alpha)=\frac{\varphi_+^n(L_n-\varphi_+^n)}{L_n^2}=\frac{\varphi_+^n(\varphi_-^n)}{L_n^2}=\frac{(-1)^n}{L_n^2}=u_n.$$
Let $x=u_n$ in \eqref{mod2kkkx}. Then we find
\begin{align*}\sum_{k=1}^{p-1} \binom{2k}{k}\frac{u_n^k}{k}&amp;\equiv
2\frac{1-\alpha^p-(1-\alpha)^p}{p}\\
&amp;=\frac{2}{pL_n^p}\left(L_n^p-\varphi_+^{np}-(L_n-\varphi_+^n)^p\right)\\
&amp;= \frac{2}{L_n^p}\cdot \frac{L_n^p-L_{np}}{p}\\
&amp;= \frac{2}{L_n^{p-1}}q_p(L_n)-\frac{2}{L_n^p}\cdot\frac{L_{np}-L_n}{p} \\
&amp;\equiv 2q_p(L_n)-\frac{2}{L_n}\cdot \frac{5nF_nF_{p-\eps}}{2p}\\
&amp;\equiv -q_p(u_n)-\frac{5nF_n}{L_n}f_p \pmod{p}
\end{align*}
where we applied Fermat's little theorem, \eqref{Lnp}, and $F_nL_n=F_{2n}$. Also recall that 
$q_p(x\cdot y)\equiv q_p (x) + q_p (y) \pmod{p}$.

For $\alpha=\varphi_+^n/(\sqrt{5}F_n)$, we have
$$\alpha(1-\alpha)=\frac{\varphi_+^n(\sqrt{5}F_n-\varphi_+^n)}{5F_n^2}=\frac{\varphi_+^n(-\varphi_-^n)}{5F_n^2}=\frac{(-1)^{n-1}}{5F_n^2}=v_n.$$
In the same manner, setting $x=v_n$ in \eqref{mod2kkkx}, we obtain
\begin{align*}\sum_{k=1}^{p-1} \binom{2k}{k}\frac{v_n^k}{k}&amp;\equiv
2\frac{1-\alpha^p-(1-\alpha)^p}{p}\\
&amp;=\frac{2}{p5^{p/2}F_n^p}\left(5^{p/2}F_n^p-\varphi_+^{np}-(\sqrt{5}F_n-\varphi_+^n)^p\right)\\
&amp;=\frac{2}{5^{p/2}F_n^p}\cdot \frac{5^{p/2}F_n^p-\sqrt{5}F_{np}}{p}\\
&amp;=\frac{2}{F_n^{p-1}}q_p(F_n)-\frac{2}{5^{\frac{p-1}{2}}F_n^p}\cdot \frac{F_{np}-5^{\frac{p-1}{2}}F_n}{p}\\
&amp;\equiv 2q_p(F_n)+q_p(5)-\frac{2\eps}{F_n}\cdot \frac{F_{np}-\eps F_n}{p}\\
&amp;\equiv -q_p(v_n)-\frac{nL_n}{F_n}f_p\pmod{p}, 
\end{align*}
where we used \eqref{Fnp}, and  
$$5^{\frac{p-1}{2}}\equiv \leg{5}{p}\left(1+\frac{p}{2}q_p(5)\right)\pmod{p^2},$$
which is a consequence of \cite[Lemma 4.1]{MT13}.
\end{proof}

\begin{theorem}
Let $n\geq 1$ and let $p&gt;2$ be a prime not dividing $5L_nF_n$. Then
\begin{equation}\label{hu}
\sum_{k=1}^{p-1} \binom{2k}{k}h_k u_n^k
\equiv \leg{5}{p}\left(
\frac{1}{2}q_p\left(\frac{v_n}{u_n}\right)+(-1)^n\frac{2n}{F_{2n}}f_p
\right) \pmod{p}
\end{equation}
and 
\begin{equation}\label{hv}
\sum_{k=1}^{p-1} \binom{2k}{k}h_k v_n^k\equiv 
-\sum_{k=1}^{p-1} \binom{2k}{k}h_k u_n^k\pmod{p}.
\end{equation}
\end{theorem}
\begin{proof} Setting $x=u_n$, the identity $L_{n}^2-5F_n^2 = 4(-1)^n$ yields
$$1-4x=\frac{L_n^2-4(-1)^n}{L_n^2}=\frac{5F_n^2}{L_n^2},$$
and therefore
$$y=\frac{-x}{1-4x}=-\frac{(-1)^n}{L_n^2}\cdot \frac{L_n^2}{5F_n^2}=\frac{(-1)^{n-1}}{5F_n^2}=v_n.$$
Moreover
$$(1-4x)^{\frac{p-1}{2}}=5^{\frac{p-1}{2}}\frac{F_n^{p-1}}{L_n^{p-1}}\equiv \leg{5}{p}\pmod{p}.$$
Thus, by \eqref{mod2kkhx}, \eqref{mod2kkku}, and \eqref{mod2kkkv}, we obtain \eqref{hu}. Indeed,
\begin{align*}\sum_{k=1}^{p-1} \binom{2k}{k}h_k u_n^k
&amp;\equiv\frac{1}{2}(1-4u_n)^{\frac{p-1}{2}}\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{u_n^{k}}{k}
-\sum_{k=1}^{p-1}\binom{2k}{k}\frac{v_n^{k}}{k}\right)\\
&amp;\equiv \frac{1}{2}\leg{5}{p}\left(
q_p(v_n)-q_p(u_n)+
\left(\frac{L_n}{F_n}-\frac{5F_n}{L_n}\right)nf_p\right) \\
&amp;\equiv \frac{1}{2}\leg{5}{p}\left(
q_p\left(\frac{v_n}{u_n}\right)+\frac{4(-1)^n}{F_{2n}}nf_p\right) \pmod{p}.
\end{align*}
It remains to show \eqref{hv}.
Since
$$1-4v_n=1-4y=1+\frac{4x}{1-4x}=(1-4x)^{-1}=(1-4u_n)^{-1},$$
it follows that
$$(1-4v_n)^{\frac{p-1}{2}}=(1-4u_n)^{-\frac{p-1}{2}}\equiv \leg{5}{p}^{-1}\equiv \leg{5}{p}\pmod{p}$$
and we are done because of \eqref{mod2kkhx}.
\end{proof}

\section{Acknowledgment}

The author wishes to thank the anonymous referee for the valuable comments and corrections. We are indebted to him/her for pointing out the conjectural congruence \eqref{conj}. We are also grateful to Prof.\ Hongwei Chen for his suggestions that have made this paper more readable.



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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11A07; Secondary 05A10, 11B39.

\noindent \emph{Keywords:} congruence, central binomial coefficient, harmonic number, Fibonacci number, Lucas number.

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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A001254}, and
\seqnum{A099921}.)

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\vspace*{+.1in}
\noindent
Received June 17 2023;
revised versions received June 18 2023; October 1 2023; October 2 2023.
Published in {\it Journal of Integer Sequences}, October 3 2023.

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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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