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

\vskip .2 in


\begin{abstract}
We present several polynomial congruences about sums with central
binomial coefficients and harmonic numbers. In the final section we
collect some new congruences involving Fibonacci and Lucas numbers.
\end{abstract}

\section{Introduction}

Recently, the following identity was proposed by Knuth in the problem section of the {\it American Mathematical Monthly} \cite{Kn}:
\begin{equation}\label{Ka}
\left(\sum_{k=1}^{\infty}\binom{2k}{k}\frac{x^k}{k}\right)^2=4\sum_{k=1}^{\infty}\binom{2k}{k}(H_{2k-1}-H_k)\frac{x^k}{k},
\end{equation}
where $H_n=\sum_{k=1}^n 1/k$ is the $n$-th harmonic number.
Playing around with this formula, we realized that there is a corresponding polynomial congruence, namely, for all prime numbers $p$, 
\begin{equation}\label{Kd}
\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k}\right)^2
\equiv 4\sum_{k=1}^{p-1}\binom{2k}{k}(H_{2k-1}-H_k)\frac{x^k}{k} \pmod{p}.
\end{equation}
By using this congruence together with some previous results given in \cite{MT13,MT16}, we find that for all prime numbers $p&gt;3$,
\begin{equation}\label{Kg}
\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}x^k}{k}
\equiv (2x-\alpha)^p\pounds_2(-\beta/\alpha)
+2\alpha^p\pounds_2(\beta/\alpha)\pmod{p}
\end{equation}
where $\LL_2(x)=\sum_{k=1}^{p-1} \frac{x^k}{k^2}$ is the {\em finite dilogarithm} and
\begin{equation*}
\aa=\frac{1}{2}\left(1+\sqrt{1-4x}\right)\qquad\mbox{and}\quad 
\bb=\frac{1}{2}\left(1-\sqrt{1-4x}\right).
\end{equation*}
These kind of congruences have been actively investigated and many interesting 
formulas have been discovered (see the references in \cite{MT13,MT16}).
For example, by letting $x=1$ in \eqref{Kg}, we recover the congruence 
\begin{equation}\label{Kc}
\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}}{k}\equiv 
\frac{7}{12}\leg{p}{3}B_{p-2}\left(1/3\right)\pmod{p} 
\end{equation}
which appeared in \cite{MS}, where $\leg{x}{y}$ denotes the Legendre symbol,
and $B_n(x)$ is the $n$-th Bernoulli polynomial. 
Moreover, we show several congruences involving Fibonacci numbers $F_n$ and Lucas numbers $L_n$. 
Two of them are as follows: for all prime numbers $p&gt;5$,
\begin{align}
\label{Ke}
&amp;\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{2k}F_{3k}\equiv 
\frac{13}{10}\leg{p}{5}q_L^2\pmod{p},\\
\label{Kf}
&amp;\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{2k}L_{3k}\equiv 
\frac{5}{2}\,q_L^2\pmod{p},
\end{align}
where $q_L=(L_p-1)/p$ is the so-called {\em Lucas quotient}.

The paper is organized into four sections. The next section  is devoted to a brief introduction to the finite polylogarithm. In Section 3 we present the proofs of the main theorems about the polynomial congruences and in the final section we establish various congruences involving Fibonacci numbers.

\section{The finite polylogarithm}

The {\sl classical polylogarithm} function is defined for complex 
$|z|&lt;1$ and all positive integers $d$ by the power series
\[\Li_d(z)=\sum_{k=1}^{\infty} \frac{z^k}{k^d}.\]
It is well known that the polylogarithm can be extended analytically to a wider range of $z$ and it satisfies several remarkable identities such as the two reflection properties,
\[\Li_2(z)+\Li_2(1/z)=-\frac{\pi^2}{6}-\frac{\ln^2(-z)}{2}\quad\mbox{and}\quad
\Li_2(z)+\Li_2(1-z)=\frac{\pi^2}{6}-\ln(z)\ln(1-z).\]
These identities allow the explicit evaluation of the polylogarithm at some special values, such as
\begin{equation*}
\Li_2(1) =\zeta(2)=\frac{\pi^2}{6},\quad
\Li_2(1/2)= \frac{\pi^2}{12}-\frac{\ln^2(2)}{2},\quad
\Li_2(\phi_-)=-\frac{\pi^2}{15}+\frac{\ln^2(\phi_+)}{2}.
\end{equation*}
where $\phi_{\pm}=(1\pm\sqrt{5})/2$.

The {\sl finite polylogarithm} function is the partial sum of the above series over the range $0&lt;k&lt;p$ where $p$ is a prime
\[\LL_d(x)=\sum_{k=1}^{p-1} \frac{x^k}{k^d}.\]
It satisfies some nice properties that resemble the ones satisfied by the classical polylogarithm. 

Here we restrict our attention to $\LL_2(x)$ (see \cite{MT13} for more details): for all prime numbers $p&gt;3$,
\begin{align*}
\LL_2(x) &amp;\equiv x^p\LL_2(1/x) \pmod{p},\\
\LL_1(1-x) &amp;\equiv -Q_p(x)-p\LL_2(x) \pmod{p^2},\\
\LL_2(x) &amp;\equiv \LL_2(1-x)+x^p\LL_2(1-1/x) \pmod{p},\\
x^p\LL_2(x)+(1-x)^p\LL_2(1-x) &amp;\equiv \frac{1}{2}\,Q_p^2(x) \pmod{p}.
\end{align*}
where
$$Q_p(x)=xq_p(x)+(1-x)q_p(1-x),
\quad\mbox{with}\quad q_p(x)=\frac{x^{p-1}-1}{p}.$$
Several congruences for special values of $\LL_2(x)$ are known:
$$\LL_2(1) \equiv \LL_2(-1) \equiv 0, \quad
\LL_2(2) \equiv  2\LL_2(1/2)\equiv  -q_p^2(2) \pmod{p}.$$
Moreover
\begin{equation*}
\begin{split}
&amp;\LL_2\bigl((1{\pm}i)/2\bigr)\equiv
-\frac{q_p^2(2)}{8}\,+\frac{1}{4}\left(\leg{-1}{p}\pm i\right)E_{p-3} \pmod{p},\\
&amp;\LL_2(\omega_6^{\pm 1}) \equiv \frac{1}{8}\biggl(\leg{p}{3}\pm i\frac{\sqrt{3}}{3}\biggr)B_{p-2}(1/3),\pmod{p}
\end{split}
\end{equation*}
where $\omega_6=(1\pm i\sqrt{3})/2$ and $E_n$ is $n$-th Euler number.
Finally, for all prime numbers $p&gt;5$ we have
\begin{equation*}
\begin{split}
\LL_2(\phi_{\pm})&amp;\equiv
\mp \frac{\sqrt{5}}{10}\leg{p}{5}q_L^2\pmod{p},\\
\LL_2(\phi^2_{\pm})&amp;\equiv -\frac{1}{2}\biggl(1\pm \frac{\sqrt{5}}{5}\leg{p}{5}\biggr)q_L^2 \pmod{p},\\
\LL_2(-\phi_{\pm})&amp;\equiv -\frac{1}{4}\biggl(1\pm \frac{\sqrt{5}}{5}\leg{p}{5}\biggr)q_L^2 \pmod{p}.
\end{split}
\end{equation*}
Notice that the Lucas quotient satisfies (see \cite{Wi}),
$$q_L=Q(\phi_{\pm})\equiv\frac{5\,F_{p-\leg{p}{5}}}{2p} \pmod{p}.$$



\section{Polynomial congruences for central binomial sums}

In \cite{MT13,MT16}, we studied various sum involving the central binomial coefficients. 
In particular, it has been shown that for all prime numbers $p&gt;3$,
\begin{align}
\label{Na}
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}x^k\equiv \sum_{k=1}^{p-1}\binom{\frac{p-1}{2}}{k}(-4x)^k
\equiv(1-4x)^{(p-1)/2}\pmod{p},\\
\label{Nb}
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k}\equiv 
\LL_1(\aa)+\LL_1(\bb)\pmod{p},\\
\label{Nc}
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k^2}\equiv 
2\LL_2(\aa)+2\LL_2(\bb)\pmod{p},\\
&amp;\label{Nd}\sum_{k=1}^{p-1} \binom{2k}{k}H_k^{(2)} x^{k}
\equiv \frac{2(\LL_2(\bb)-\LL_2(\aa))}{\sqrt{1-4x}}
\pmod{p}.
\end{align}
where $H_n^{(2)}=\sum_{k=1}^{n}1/k^2$.

In \cite[Proposition 5]{Bo}, Boyadzhiev used the following Euler-type series transformation formula to handle series with central binomial coefficients: if $a_n=\sum_{k=0}^{n}\binom{n}{k}(-1)^k b_k$ then in a neighborhood of $x=0$,
$$\sum_{k=0}^{\infty}\binom{2k}{k} a_k x^k= 
\frac{1}{\sqrt{1-4x}}\sum_{j=0}^{\infty}\binom{2j}{j}b_j \left(\frac{-x}{1-4x}\right)^j.$$
It turns out that something similar holds for finite sum congruences:
\begin{align}
\nonumber
\sum_{k=0}^{p-1}\binom{2k}{k} a_k x^k
&amp;\equiv \sum_{k=0}^{(p-1)/2}\binom{\frac{p-1}{2}}{k} a_k (-4x)^k
= \sum_{k=0}^{(p-1)/2}\binom{\frac{p-1}{2}}{k}(-4x)^k
\sum_{j=0}^{k}\binom{k}{j}(-1)^jb_j\\
\nonumber
&amp;= \sum_{j=0}^{(p-1)/2}(-1)^jb_j 
\sum_{k=j}^{(p-1)/2}\binom{\frac{p-1}{2}}{k}\binom{k}{j}(-4x)^k\\
\nonumber
&amp;= \sum_{j=0}^{(p-1)/2}(-1)^jb_j \binom{\frac{p-1}{2}}{j}
(-4x)^{j}(1-4x)^{\frac{p-1}{2}-j}\\
\label{Ba}
&amp;\equiv (1-4x)^{\frac{p-1}{2}}\sum_{j=0}^{p-1}
\binom{2j}{j}b_j \left(\frac{-x}{1-4x}\right)^j\pmod{p}.
\end{align}
In the next theorem we apply the above transformation.

\begin{theorem}
For all prime numbers $p&gt;3$,
\begin{align}
&amp;\label{Ne}
\sum_{k=1}^{p-1}\binom{2k}{k} H_k x^k
\equiv -2(1-4x)^{\frac{p-1}{2}}\LL_1\left(-\frac{\bb}{\sqrt{1-4x}}\right)
\pmod{p},\\
&amp;\label{Nf}
\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_k x^k}{k} 
\equiv 2(1-4x)^{\frac{p}{2}}\left(
\LL_2\left(\frac{\aa}{\sqrt{1-4x}}\right)-\LL_2\left(-\frac{\bb}{\sqrt{1-4x}}\right)\right)
\pmod{p}.
\end{align}
\end{theorem}
\begin{proof} It is easy to verify by induction that
$$\sum_{k=1}^n (-1)^{k}{n\choose k}H_{k}(1)=-\frac{1}{n}
\quad\mbox{and}\quad
\sum_{k=1}^n (-1)^{k}{n\choose k}H_{k}(2)=-\frac{H_n}{n}.
$$
Moreover
$$\aa\left(\frac{-x}{1-4x}\right)=\frac{\aa}{\sqrt{1-4x}}
\quad\mbox{and}\quad \bb\left(\frac{-x}{1-4x}\right)=-\frac{\bb}{\sqrt{1-4x}}.
$$
Hence, by \eqref{Ba} and \eqref{Nb},
\begin{align*}
\sum_{k=1}^{p-1}\binom{2k}{k} H_k x^k
&amp;\equiv -(1-4x)^{\frac{p-1}{2}}\sum_{j=1}^{p-1}
\binom{2k}{k}\frac{1}{k} \left(\frac{-x}{1-4x}\right)^k\\
&amp;\equiv -(1-4x)^{\frac{p-1}{2}}\left(
\LL_1\left(\frac{\aa}{\sqrt{1-4x}}\right)+\LL_1\left(-\frac{\bb}{\sqrt{1-4x}}\right)\right)\\
&amp;\equiv -(1-4x)^{\frac{p-1}{2}}\left(
\LL_1\left(1-\frac{\aa}{\sqrt{1-4x}}\right)+\LL_1\left(-\frac{\bb}{\sqrt{1-4x}}\right)\right)\\
&amp;\equiv -2(1-4x)^{\frac{p-1}{2}}\LL_1\left(-\frac{\bb}{\sqrt{1-4x}}\right)
\pmod{p},
\end{align*}
where we also used $\LL_1(x)\equiv \LL_1(1-x)$. Thus the proof of \eqref{Ne} is complete. 

As regards \eqref{Nf}, Eqns.~\eqref{Ba} and \eqref{Nd} imply 
\begin{align*}
\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_k x^k}{k} 
&amp;\equiv -(1-4x)^{\frac{p-1}{2}}\sum_{j=1}^{p-1}
\binom{2k}{k}H_k^{(2)}\left(\frac{-x}{1-4x}\right)^k\\
&amp;\equiv 2(1-4x)^{\frac{p}{2}}\left(
\LL_2\left(\frac{\aa}{\sqrt{1-4x}}\right)-\LL_2\left(-\frac{\bb}{\sqrt{1-4x}}\right)\right)
\pmod{p}.
\end{align*}
\end{proof}
In the next theorem, we establish \eqref{Kd}, the analogous congruence for the series \eqref{Ka}.
\begin{theorem}
For all prime numbers $p&gt;3$,
\begin{align}
&amp;\label{Ng}
\left(\sum_{k=1}^{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-1}-H_k)x^k\pmod{p},\\
&amp;\label{Nh}\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k}\right)^2\equiv 4\sum_{k=1}^{p-1}\binom{2k}{k}(H_{2k-1}-H_k)\frac{x^k}{k}\pmod{p}.
\end{align}
\end{theorem}
\begin{proof}
Since $p$ divides $\binom{2k}{k}$ for $(p-1)/2&lt;k&lt;p$, it follows that
\begin{align*}
\left(\sum_{k=1}^{p-1}\binom{2k}{k}x^{k}\right)\cdot
\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^{k}}{k}\right)
&amp;\equiv\sum_{n=1}^{p-1}x^n
\sum_{k=1}^{n-1}\left(
\frac{1}{k}\binom{2k}{k}\binom{2(n-k)}{n-k}\right) \pmod{p}.
\end{align*}
In a similar way,
\begin{align*}
\left(\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^{k}}{k}\right)^2
&amp;\equiv\sum_{n=1}^{p-1}x^n
\sum_{k=1}^{n-1}\left(
\frac{1}{k(n-k)}\binom{2k}{k}\binom{2(n-k)}{n-k}\right)\\
&amp;\equiv\sum_{n=1}^{p-1}\frac{x^n}{n}
\sum_{k=1}^{n-1}\left(
\left(\frac{1}{k}+\frac{1}{n-k}\right)\binom{2k}{k}\binom{2(n-k)}{n-k}\right)\\
&amp;\equiv 2\sum_{n=1}^{p-1}\frac{x^n}{n}
\sum_{k=1}^{n-1}\left(
\frac{1}{k}\binom{2k}{k}\binom{2(n-k)}{n-k}\right)\pmod{p}.
\end{align*}
Therefore, it suffices to show by induction that
$$\sum_{k=1}^{n-1}F(n,k)=2(H_{2n-1}-H_n)\quad\mbox{where}\quad
F(n,k)=\frac{1}{k}\binom{2k}{k}\binom{2(n-k)}{n-k}\binom{2n}{n}^{-1}.$$
It holds for $n=1$, and it is straightforward to verify that
$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)\;\;\mbox{with}\;\;
G(n,k)=-\frac{k^2(2n-2k+1)F(n,k)}{(n+1)(2n+1)(n+1-k)}.
$$
Hence, by the inductive assumption,
\begin{align*}
\sum_{k=1}^{n}F(n+1,k)
&amp;=\sum_{k=1}^{n}F(n,k)+\sum_{k=1}^{n}(G(n,k+1)-G(n,k))\\
&amp;=2(H_{2n-1}-H_n)+F(n,n)+G(n,n+1)-G(n,1)\\
&amp;=2(H_{2n-1}-H_n)+\frac{1}{n}+0+\frac{(2n-1)F(n,1)}{(n+1)(2n+1)n}\\
&amp;=2(H_{2n-1}-H_n)+\frac{1}{n}+\frac{1}{(n+1)(2n+1)}=2(H_{2n+1}-H_{n+1}).
\end{align*}
\end{proof}

Now we are ready to show that our main result \eqref{Kg} and 
the congruence corresponding to the series \cite[Theorem 6]{Ch}: for $|x|&lt;1/4$,
\begin{equation}\label{Kb}
\sum_{k=1}^{\infty}\binom{2k}{k}H_{2k}x^k=
\frac{1}{\sqrt{1-4x}}\left(
\ln\left(\frac{1+\sqrt{1-4x}}{2}\right)-2\ln(\sqrt{1-4x})\right).
\end{equation}

\begin{theorem}
For all prime numbers $p&gt;3$,
\begin{align}
&amp;\label{Ni}
\sum_{k=1}^{p-1}\binom{2k}{k}H_{2k}x^k
\equiv (1-4x)^{(p-1)/2}\left(\pounds_1(\beta)
-2\pounds_1\left(-\frac{\beta}{\sqrt{1-4x}}\right)\right)\pmod{p}\\
&amp;\label{Nl}
\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}x^k}{k}
\equiv (2x-\alpha)^p\pounds_2(-\beta/\alpha)
+2\alpha^p\pounds_2(\beta/\alpha)\pmod{p}.
\end{align}
\end{theorem}
\begin{proof} As regards \eqref{Ni}, since 
$H_{2k}=\frac{1}{2k}+(H_{2k-1}-H_{k})+H_k$,
it follows immediately that, 
$$\sum_{k=1}^{p-1}\binom{2k}{k}H_{2k}x^k
=\frac{1}{2}\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k}
+\sum_{k=1}^{p-1}\binom{2k}{k}(H_{2k-1}-H_{k})x^k
+\sum_{k=1}^{p-1}\binom{2k}{k}H_kx^k
$$
and we apply \eqref{Na}, \eqref{Ng}, and \eqref{Ne}. In a similar way, for \eqref{Nl},
$$\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}x^k}{k}
=\frac{1}{2}\sum_{k=1}^{p-1}\binom{2k}{k}\frac{x^k}{k^2}
+\sum_{k=1}^{p-1}\binom{2k}{k}(H_{2k-1}-H_{k})\frac{x^k}{k}
+\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_k x^k}{k}
$$
and then we use \eqref{Nc}, \eqref{Nh}, and \eqref{Nf}.
\end{proof}


As a remark, we point out that although the series \eqref{Kb} does not converge for $x=1/4$,
by letting $f(x)$ be the left-hand side of \eqref{Kb} then
$$\sum_{k=1}^{\infty}\binom{2k}{k}\frac{H_{2k}}{4^k k}=
\int_0^{\frac{1}{4}}\frac{f(x)}{x}\,dx=\frac{5\pi^2}{12}.$$
On the other hand, it can be verified that the congruence \eqref{Nl} holds
even for $x=1/4$, and for all prime numbers $p&gt;3$,
$$\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}}{4^k k}\equiv \pounds_2(1)\equiv 0\pmod{p}.$$

\section{Congruences with Fibonacci and Lucas numbers}

By looking at this table and by using the values of $\LL_1$ and $\LL_2$,
we can easily obtain the explicit values of the congruences established in the previous section.
\begin{center}
\begin{tabular}{|c|c|c|}
\hline $x$ &amp; $\aa$ &amp; $\bb$\\
\hline  
$1$ &amp; $\omega_6$ &amp; $\omega_6^{-1}$\\[3pt] 
$-1$ &amp; $\phi_+$ &amp; $\phi_-$\\[3pt] 
$-2$ &amp; $2$ &amp; $-1$\\[3pt] 
$1/2$ &amp; $(1+i)/2$ &amp; $(1-i)/2$\\[3pt] 
$1/3$ &amp; $(1+\omega_6)/3$ &amp; $(1+\omega_6^{-1})/3$\\[3pt] 
$1+i$ &amp; $1-i$ &amp;$i$ \\[3pt] 
$1-i$ &amp; $1+i$ &amp;$-i$ \\[3pt] 
$\pm i\sqrt{3}$ &amp; $1+\omega_6^{\mp 1}$ &amp;$-\omega_6^{\mp 1}$ \\[3pt] 
$-\phi^3_{-}$ &amp; $-\phi_-$ &amp; $\phi_-^2$\\[3pt] 
$-\phi^3_{+}$ &amp; $\phi_+^2$ &amp;$-\phi_+$ \\[3pt] 
\hline
\end{tabular}
\end{center}
For example, for all prime numbers $p&gt;3$,
by taking $x=1,1/2,1/3$ in \eqref{Nl}, we get
respectively \eqref{Kc}, and the next two congruences,
\begin{equation*}
\begin{split}
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}}{2^k k}\equiv 
\frac{3}{16}\leg{-1}{p}B_{p-2}\left(1/4\right)\pmod{p},\\
&amp;\sum_{k=1}^{p-1}\binom{2k}{k}\frac{H_{2k}}{3^k k}\equiv 
\frac{2}{9}\leg{p}{3}B_{p-2}\left(1/3\right)\pmod{p}.
\end{split}
\end{equation*}
To order to get the congruences with $F_n$ and $L_n$ we need 
consider the cases $x=-\phi^3_{\pm}$.  
If $x=-\phi^3_{-}$ then $2x-\alpha=-\phi^4_{-}$ and
\begin{align*}
\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{2k}\phi^{3k}_{-}
&amp;\equiv (-\phi^4_{-})^p\pounds_2(\phi_{-})
+2(-\phi_{-})^p\pounds_2(-\phi_{-})\\
&amp;\equiv \frac{1}{2}\left(-7+3\leg{p}{5}\sqrt{5}\right)\pounds_2(\phi_{-})
+\left(-1+\leg{p}{5}\sqrt{5}\right)\pounds_2(-\phi_{-})\\
&amp;\equiv \left(\frac{5}{4}-\frac{13}{20}\leg{p}{5}\sqrt{5}\right)q_L^2
\pmod{p}.
\end{align*}
where we used the fact that $2\phi_{\pm}^p\equiv 1\pm\leg{p}{5}\sqrt{5}$ (mod $p$).
In a similar way, we find that
\begin{align*}
\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{2k}\phi^{3k}_{\pm}
&amp;\equiv \left(\frac{5}{4}\pm\frac{13}{20}\leg{p}{5}\sqrt{5}\right)q_L^2
\pmod{p}, \\
\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{k}\phi^{3k}_{\pm}
&amp;\equiv \left(\frac{1}{2}\pm\frac{3}{10}\leg{p}{5}\sqrt{5}\right)q_L^2
\pmod{p}.
\end{align*}
Since 
$\sqrt{5}F_{3k}=\phi^{3k}_{+}-\phi^{3k}_{-}$ and $L_{3k}=\phi^{3k}_{+}+\phi^{3k}_{-}$,
it follows that for $p&gt;5$, \eqref{Ke} and \eqref{Kf} hold and also we find that
\begin{equation*}
\begin{split}
&amp;\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{k}F_{3k}\equiv 
\frac{3}{5}\leg{p}{5}q_L^2\pmod{p},\\
&amp;\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\binom{2k}{k}H_{k}L_{3k}\equiv 
q_L^2\pmod{p}.
\end{split}
\end{equation*}



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\bigskip
\hrule
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\noindent {\it 2010 Mathematics Subject Classification:} Primary 11A07; Secondary 05A10.
 
\bigskip

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


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000045},
\seqnum{A000984},
\seqnum{A001008},
\seqnum{A014445}, and
\seqnum{A014448}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 14 2016;
revised version received May 10 2016.
Published in {\it Journal of Integer Sequences}, June 2 2016.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

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