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\begin{center}
\vskip 1cm{\LARGE\bf Average of the Fibonacci Numbers
}
\vskip 1cm
\large
Amirali Fatehizadeh\\
Department of Mathematical Sciences \\
Sharif University of Technology\\
Tehran \\
Iran\\
\href{mailto: amirali.fatehizadeh@gmail.com}{\tt amirali.fatehizadeh@gmail.com} \\
\ \\ 
Daniel Yaqubi\\ 
Department of Computer Science\\
University of Torbat-e Jam\\
Torbat-e Jam \\
Iran\\ 
\href{mailto:daniel_yaqubi@yahoo.es}{\tt daniel\_yaqubi@yahoo.es} 
\end{center}

\vskip .2 in

\begin{abstract}
The arithmetic mean of the first $n$ Fibonacci numbers is not an integer for all $n$. However, for some values of $n$, it is. In this paper we consider the sequence of integers $n$ for which the average of the first $n$ Fibonacci numbers is an integer. We prove some interesting properties and present two related conjectures.
\end{abstract}


\section{Introduction}
The Fibonacci sequence $(F_n)_{n\geq 0}$ is defined by  $F_0=0$, $F_1=1$,  
and $F_{n}=F_{n-1}+F_{n-2}$ for $n\geq 2$; it is sequence \seqnum{A000045} in 
the {\it On-Line Encyclopedia of Integer Sequences} (OEIS) \cite{OEIS}.
Fibonacci numbers have been extensively studied \cite{Knu, Kos}. Numerous fascinating properties are known. For instance, the Fibonacci numbers
have a close relation to binomial coefficients:
\[F_{n+1}=\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n-i}{i}.\]

The average of the first $n$ terms of the Fibonacci sequence
is not always an integer.  For instance, for $n=3$ we have $
(1+1+2)/3= 4/3$, but  for $n=1, 2, 24, 48, \ldots$,
the quantities $\bigg(\frac{1}{n}\sum_{i=1}^n F_i\bigg)_{n\geq 1}$
are integers.

In this paper, we explore the following question:
Which terms of the sequence
\begin{eqnarray}
A_F(n) = \bigg(\frac{1}{n}\sum_{i=1}^n F_i\bigg)_{n\geq 1}
\end{eqnarray}
are integers?


We give a characterization of the values of $n$ for which
$A_F(n)$ is an integer. In other words,
we give an implicit necessary and sufficient condition in Theorem \ref{TN1} and explicit sufficient
conditions in the proof of Theorem \ref{Thm2} and in Theorems \ref{Thm3} and \ref{Thmn} on $n$ to make the first $n$ Fibonacci
numbers divisible by $n$.
 Moreover, we present a construction of finding infinitely many $n$ that satisfy the given conditions. 
 Further, we show that there are infinitely many $n$ for which $6$ is a divisor of the sum of the first $n$ Fibonacci numbers. 

 Finally, we show that $A_F(p)$ is not an integer if $p$ is an odd prime number. 
  
 Our work is based on the results in \cite{Kos, Lag, OEIS, Rob, Wal}. 


\section{Preliminaries}\label{Sec2}



First, we recall some definitions and important theorems  \cite{Apo, Ful, Som}. 
A fundamental identity that we use in this paper is \cite[Theorem 5.1]{Kos}

\begin{align}\label{Id:Fib_sum}
	\sum_{i=1}^n F_i=F_{n+2}-1.
\end{align}


The Lucas numbers $(L_n)_{n\geq 0}$, are defined by the same recurrence relation as the Fibonacci numbers with different initial values (see \seqnum{A000032}). 

\[L_0=2, \quad L_1=1,\quad  L_{n}=L_{n-1}+L_{n-2},\quad\textit{for}\quad n\geq 2.\]

The following relations between Fibonacci numbers and Lucas numbers can be found in \cite{Kos}:
\begin{align}
\label{Id:Fib_Luc2}
  F_{4k+1}-1&=F_{2k}L_{2k+1},\\
\label{Id:Fib_Luc3}
F_{4k+2}-1&=F_{2k}L_{2k+2},\\
\label{Id:Fib_Luc4}
F_{4k+3}-1&=F_{2k+2}L_{2k+1},\\
\label{Id:Fib_Luc5}
F_{2k}&=F_{k}L_{2k}.
\end{align}


An integer $a$ is called a \textit{quadratic residue modulo $p$} (with $p>2$) if $p\nmid a$ and there exists an integer $b$ such that  $a\equiv b^2$ (mod $p$). Otherwise, it is called a \emph{non-quadratic residue modulo $p$}.
 
Let $p$ be an odd prime number. The \textit{Legendre symbol} is a function of $a$ and $p$ defined as  
\begin{displaymath}
\Leg{a}{p}=
\begin{cases}
+1, & \text{if $a$ is a quadratic residue modulo $p$ and $a\not\equiv\modd{0} {p}$}, \\
-1, & \text{if $a$ is a non-quadratic residue modulo $p$},\\
0, & \text{if $a\equiv\modd{0} {p}$}.
\end{cases}
\end{displaymath}
We note that for a prime number $p$ the Legendre symbol, $\Leg{5}{p}$, is equal to
\begin{displaymath}
\Leg{5}{p}=
\begin{cases}
+1, &\text{if $p\equiv \modd{\pm 1} {5}$}, \\
0, &\text{if $p\equiv \modd{0} {5}$} ,  \\
-1,&\text{if $p\equiv \modd{\pm 2}  {5}$}.
\end{cases}
\end{displaymath}

Consider the sequence of the Fibonacci numbers modulo $8$:
\[0,1,1,2,3,5,0,5,5,2,7,1,0,1,1,2,3,5,\ldots\] 
We observe that the reduced sequence is periodic. 

Lagrange \cite{Lag} proved that this property is true in general, i.e., that the Fibonacci sequence is periodic modulo $m$ for any positive integers $m>1$. 

\begin{definition}\label{Def1}
	For a given positive integer $m$, we call the least integer such that $(F_n,F_{n+1}) \equiv (0,1)$ (mod $m$) the \emph{(Pisano) period of the Fibonacci sequence modulo $m$} and denote it by $\pi(m)$.
\end{definition}

The sequence $\pi(n)$ is sequence \seqnum{A001175} in the OEIS \cite{OEIS}.
 
We recall as a lemma the fixed point theorem of Fulton and Morris \cite{Ful}.
\begin{lemma}[Fixed Point Theorem \cite{Ful}]\label{lemma_fixed point}
Let $m$ be a positive integer greater than $1$. Then $\pi(m) = m$ 
if and only if $m=(24)5^{\lambda-1}$ for some $\lambda>0$.
\end{lemma}


For instance, with $m=8$ we have $\pi(8)=12$ and $\alpha(8)=6$. The $12$ terms in the period form two sets of $6$ terms. The terms of the second half are $5$ times the corresponding terms in the first half (mod $8$). For the Lucas sequence $F_n=U_n(P,Q)$; Robinson \cite{Rob}, we have $t\equiv F_{\alpha(m)-1}(-Q)$ (mod $m$) is the multiplier between consecutive parts of length $\alpha(m)$ of the period. If the (mod $m$) order of $t$ is $r$ then $\pi(m)=r \alpha(m)$. Here $F_n=U_n(1,-1), (P,Q)=(1,-1), \alpha(8)=6, t=5, r=2;$ thus $\pi(8)=2\cdot 6=12$; Robinson \cite{Rob}.
The $12$ terms in the period form two sets of $6$ terms. The terms of the second half are $5$ times the corresponding terms in the first half (modulo $8$). The next definition is 

\begin{definition}\label{Def2}
	For a given positive integer, we call the least integer such that $(F_n,F_{n+1}) \equiv \sigma(0,1)$ (mod $m$) for some positive integer $\sigma$ the \emph{restricted period of the Fibonacci sequence modulo $m$} and denote it by $\alpha(m)$.
\end{definition}
Robinson \cite{Rob} proved the following theorems. 
\begin{theorem}\label{Rob1}
	\leavevmode
	\begin{itemize}
		\item [(i)] {$m\mid F_n$ if and only if $\alpha(m)\mid n$, and}
		\item [(ii)] {$m\mid F_n$ and  $m\mid F_{n+1}-1$ if and only if $\pi(m)\mid n$.}
	\end{itemize}
\end{theorem}

\begin{theorem}\label{Rob2}
	If $p$ is a prime, then
	\begin{itemize}
		\item [(i)] {$\alpha(p)\mid (p-\Leg{5}{p})$,}
		\item [(ii)] {if $p\equiv\modd{\pm 1} {5}$, then $\pi(p) \mid (p-1)$, and}
		\item[(iii)] {if $p\equiv \modd{\pm 2} {5}$, then $\pi(p)\mid 2(p+1)$.}
	\end{itemize}
\end{theorem} 

The exponent of the multiplier of the Fibonacci sequence modulo $p$,
$t\equiv\modd{F_{\alpha(p)-1}} {p}$ is $\frac{\pi(p)}{\alpha(p)}$ and can only take the values $1$, $2$ and $4$.  
  
For a positive integer $n$ and a prime $p$, \emph{the $p$-adic valuation of $n$}, $\nu_p(n)$, is the exponent of the highest power of $p$ that divides $n$. 

 
\textit{Legendre's classical formula} for the $p$-adic valuation of the factorials  is well known:
\[\nu_p(n!)=\sum_{i=1}^{\infty}\left\lfloor \frac{n}{p^i}\right\rfloor.\] 
We recall Lengyel's lemma \cite{Leg} about the $p$-adic evaluation of Fibonacci numbers in cases $p=2,3$ and $5$.
\begin{lemma}[\cite{Leg}, Lemmas 1 and 2]
  	\label{Leg1}
  	For all $n\geq 0$, we have $\nu_5(F_n)=\nu_5(n)$. On the other hand, 
  	\begin{displaymath}
  		\nu_2(F_n)=
  		\begin{cases}
  			0, &\text{if $n\equiv \modd{1,2} {3}$}; \\
  			1, &\text{if $n\equiv \modd{3} {6}$};  \\
  			1, &\text{if $n\equiv \modd{6} {12}$};\\
  			\nu_2(n)+2, &\text{if $n\equiv \modd{0} {12}$},
  		\end{cases}
  	\end{displaymath}
  	and
  	\begin{displaymath}
  		\nu_3(F_n)=
  		\begin{cases}
  			0, & \text{if $n\not\equiv \modd{0} {4}$}; \\
  			\nu_3(n)+1, &  \text{if $n\equiv \modd{0} {4}$}.
  		\end{cases}
  	\end{displaymath}
  \end{lemma}

\section{Main results}\label{Sec3}
We focus now on the average of the first $n$ Fibonacci numbers, $A_F(n)$.

\begin{theorem}\label{TN1}
Let $n$ be a positive integer. Then $n\mid\sum_{i=1}^n F_i$ if and only if
$F_{n+2}\equiv \modd{1} {n}$.
\begin{proof}
We have
\begin{eqnarray*}
n\mid\sum_{i=1}^n F_i&\Longleftrightarrow & \sum_{i=1}^n F_i \equiv  \modd{0} {n}\\
&\Longleftrightarrow &F_{n+2}-1\equiv \modd{0} {n}\\
&\Longleftrightarrow&F_{n+2}\equiv \modd{1} {n}.
\end{eqnarray*}
\end{proof}
\end{theorem}
\begin{lemma}\label{Lemma}
Let $n$ be a positive integer. If $\pi(n)=n$ or $\pi(n)=n+1$, then
\[ n\mid\sum_{i=1}^n F_i.\] 
\end{lemma}
\begin{proof}
According to the definition of the period of a Fibonacci sequence,
the congruence
$F_{k}\equiv F_{k+\pi(m)}$ (mod $m$) holds for any integer $k$.  In particular, for $k=1$ and $k=2$, we have
\begin{eqnarray*}
	F_{1}\equiv F_{1+\pi(n)}\text{(mod $n$)}\quad \mbox{and} \quad F_{2}\equiv F_{2+\pi(n)}\text{ (mod $n$)}.
\end{eqnarray*}
This implies that if  $\pi(n) = n+1$, then $1 = F_{1}\equiv F_{n+2}$ (mod $n$) and if $\pi(n) = n$, then $1=F_{2}\equiv F_{n+2}$ (mod $n$). 
By Theorem \ref{TN1} and Identity (\ref{Id:Fib_sum}) the statement follows.
\end{proof}
\begin{theorem}\label{Thm2}
There are infinitely many even numbers $n$ such that
\[n \mid\sum_{i=1}^n F_i.\]
\end{theorem}
\begin{proof}
Lemma \ref{lemma_fixed point} states that  
for $n=(24)5^{k-1}$ and $k\geq 1$,
we have $\pi(n)=n$. Hence, Lemma \ref{Lemma} implies the theorem.
\end{proof}

\begin{theorem}\label{Thm3}
Let $\alpha$ be a non-negative integer. Then
\begin{equation}\label{equation:1}
3\cdot2^{\alpha+3} \mid \sum_{i=1}^{3\cdot2^{\alpha+3}}F_i.
\end{equation}
\end{theorem}
\begin{proof}
 According to the Fibonacci identity (\ref{Id:Fib_sum}), we have 
\begin{align*}
    \sum_{i=1}^{3\cdot2^{\alpha+3}}F_i=F_{3\cdot2^{\alpha+3}+2}-1.
\end{align*} It is sufficient to show that $3\cdot2^{\alpha+3}\mid F_{3\cdot2^{\alpha+3}+2}-1$.


By Identities (\ref{Id:Fib_Luc3}) and (\ref{Id:Fib_Luc5}) we have
\begin{align*}
F_{3\cdot2^{\alpha+3}+2}-1&=F_{3\cdot2^{\alpha+2}}L_{3\cdot2^{\alpha+2}+2}\\
&=F_{3\cdot2^{\alpha+1}}L_{3\cdot2^{\alpha+1}}L_{3\cdot2^{\alpha+2}+2}\\
&=F_{3\cdot2^{\alpha}}L_{3\cdot2^{\alpha}}L_{3\cdot2^{\alpha+1}}L_{3\cdot2^{\alpha+2}+2}\\
&\quad \vdots\\
&=F_3L_3L_6\cdots L_{3\cdot2^{\alpha}}L_{3\cdot2^{\alpha+1}}L_{3\cdot2^{\alpha+2}+2}.
\end{align*}
But $F_3=2,~L_3=4$ and each of $L_6,\ldots, L_{3\cdot2^{\alpha+1}}$ are even numbers.  Since $L_{3\cdot2^{\alpha+2}+2}$ is divisible  by $3$, (\ref{equation:1}) holds.
\end{proof}
Next we present a generalization of the previous theorem. 
\begin{theorem}\label{Thmn}
Let $\alpha,\beta$ and $\gamma$ be positive integers. For $n=2^{\alpha+3}\cdot3^{\beta+1}\cdot5^{\gamma}$ it holds
\begin{equation}
18\mid\frac{1}{n}\sum_{i=1}^n F_i 
\end{equation}
\end{theorem}
\begin{proof}
Consider $n=2^{\alpha+3}\cdot3^{\beta+1}\cdot5^{\gamma}$, then $n\equiv 0$ (mod $4$). Using Identity (\ref{Id:Fib_sum}) and  (\ref{Id:Fib_Luc3})
we have
\begin{eqnarray}\label{Eq9}
\sum_{i=1}^n F_i=F_{2^{\alpha+3}\cdot3^{\beta+1}\cdot5^{\gamma}+2}-1=
F_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}}
L_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}+2}
\end{eqnarray}
By Lengyel's lemma \ref{Leg1} we can write
\begin{align*}
\nu_2(F_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}})&=\alpha+4,\\
\nu_3(F_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}})&=\beta+2,\\
\nu_5(F_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}})&=\gamma.
\end{align*}
In (\ref{Eq9}) the Lucas factor $L_{2^{\alpha+2}\cdot3^{\beta+1}\cdot5^{\gamma}+2}$ is also divisible by $3$.
This implies
\begin{equation}
18\mid\frac{1}{n}\sum_{i=1}^n F_i,
\end{equation}
which concludes the proof.
\end{proof}

Next we show that $A_F(p)$ is not an integer for an odd prime number $p$ . 

\begin{theorem}\label{10}
Let $p$ be an odd prime number. Then
\begin{equation}
p\nmid \sum_{i=1}^p F_i.
\end{equation}
\end{theorem}
\begin{proof}
Assume that $p$ is an odd prime number such that
\[p\mid \sum_{i=1}^p F_i.\]
We investigate the cases $p=4k+1$ and $p=4k+3$ separately.
We can assume that $p >5$.
 \begin{itemize}
\item[] \textbf{Case I.} 
Suppose that $p=4k+1$. According to Identities (\ref{Id:Fib_sum}) and (\ref{Id:Fib_Luc4}),
 we have
 
\begin{equation}
p\mid \sum_{i=1}^p F_i=F_{4k+3}-1=F_{2k+2}L_{2k+1}.
\end{equation}
Hence, $p\mid F_{2k+2}$ or $p\mid L_{2k+1}$. Suppose that $p\mid F_{2k+2}$. According to the first case of Theorem \ref{Rob2}, we have 
 \[\alpha(p)\mid p-\Leg{5}{p}=4k+1-\Leg{5}{4k+1}.\]

The first case of Theorem \ref{Rob1} leads to
 $\alpha(p)\mid 2k+2$. Then 
\[\alpha(p)\mid 4k+4-\left(4k+1-\Leg{5}{4k+1}\right).\]
This implies
\[\alpha(p)\mid 3+\Leg{5}{4k+1}.\]
This is impossible, since $\Leg{5}{p}=\pm 1$ .

Now suppose that  $p\mid L_{2k+1}$. From Identity (\ref{Id:Fib_Luc5}), we have $p\mid F_{4k+2}$. So, by Theorems \ref{Rob1} and \ref{Rob2}, we have 
\[\alpha(p)\mid p-\Leg{5}{p}=4k+1-\Leg{5}{4k+1},\]
and
$\alpha(p)\mid 4k+2$. This implies
\[\alpha(p)\mid 4k+2-\left(4k+1-\Leg{5}{4k+1}\right)= 1+\Leg{5}{4k+1}.\]
We have again a contradiction, since  $\left(\frac{5}{p}\right)=\pm 1$.
\item[] \textbf{Case II.} 
Now assume that  $p=4k+3$. We follow the argument given in
Case I. By using Identities (\ref{Id:Fib_sum}) and
 (\ref{Id:Fib_Luc2}), we can write
\begin{equation}
p\mid \sum_{i=1}^p F_i=F_{4(k+1)+1}-1=F_{2(k+1)}L_{2(k+1)+1}.
\end{equation}
The prime number $p=4k+3$ must divide $F_{2(k+1)}$ or $L_{2(k+1)+1}$. Suppose first that $p\mid F_{2(k+1)}$.  Similarly as in the first case, by Theorems \ref{Rob1} and \ref{Rob2}, we have
\[\alpha(p)\mid p-\Leg{5}{p}=4k+3-\Leg{5}{4k+3},\]
and
$\alpha(p)\mid 2(k+1)=2k+2$. This implies
\[\alpha(p)\mid 4k+4-\left(4k+3-\Leg{5}{4k+3}\right)= 1+\Leg{5}{4k+3}.\]
This is impossible, since $\Leg{5}{p}=\pm 1$.

Now let $p\mid L_{2(k+1)+1}$.  From the identity (\ref{Id:Fib_Luc5}), we have $p\mid F_{4k+6}$. According to Theorems \ref{Rob1} and \ref{Rob2}, we can write  
\[\alpha(p)\mid p-\Leg{5}{p}=4k+3-\Leg{5}{4k+3},\]
and
$\alpha(p)\mid 4k+6$. Then
\[\alpha(p)\mid 4k+6-\left(4k+3-\Leg{5}{4k+3}\right)= 3+\Leg{5}{4k+3}.\]
This is impossible, since $\Leg{5}{p}=\pm 1$.
\end{itemize}
The proof of the result follows immediately from the two cases above.

\end{proof}
\section{Concluding remarks}
We conclude this paper with two interesting conjectures concerning averages of Fibonacci numbers.
\begin{conjecture}\label{Con4}
There are infinitely many odd integers $n$ that divide the sum of the first $n$ Fibonacci numbers.
\end{conjecture}

\begin{conjecture}\label{Con3}
There are infinitely many pairs of positive integers $(n,n+1)$ such that 
\[n\mid \sum_{i=1}^n F_i\quad \text{and}\quad n+1\mid \sum_{i=1}^{n+1} F_i.\]
\end{conjecture}


Some values of the Conjectures \ref{Con4} and  \ref{Con3} are given as sequences \seqnum{A331976} and \seqnum{A331977}, respectively.
Currently $33$ such pairs have been found and given in \seqnum{A331977}.  
The first few pairs are $(1, 2)$, $(6479, 6480)$, $(11663, 11664)$, $(34943,34944)$, $(47519,47520)$, $(51983,51984)$.

Most of the results can be easily stated for the average of the first $n$ Lucas numbers with a similar proof (for example, Theorem \ref{Thmn}).  

We note that the study of the average of related number sequences, as generalized Fibonacci numbers, Pell-Lucas numbers might be lead to interesting results as well. 

\section{Acknowledgments}
The authors would like thank the anonymous referee for her/his careful reading and valuable suggestions. Special thanks go to Be\'{a}ta B\'{e}nyi for helpful remarks.

\begin{thebibliography}{99}

\bibitem{Apo} T.~M.~ Apostol,  \textit{Introduction to Analytic Number Theory}, Springer, 1998.

\bibitem{Bel} H.~Belbachir and S.~Celia,  The generalized bi-periodic Fibonacci sequence modulo $m$, \textit{J. Integer Sequences} \textbf{24} (2021), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL24/Salhi/salhi4.html}{Article 21.9.4}.


\bibitem{Ben} T.~A.~Benjamin and J.~Q.~Jennifer, \textit{Proofs that Really Count: the Art of Combinatorial proof}, American Mathematical Society,  2003.

\bibitem{Ful} J.~Fulton and W.~Morris,  On arithmetical functions related to the Fibonacci numbers, \textit{Acta Arith.} \textbf{16} (1969), 105--110.

\bibitem{Knu} D.~E.~Knuth, \textit{The Art of Computer Programming}, Addison--Wesley, 1968.

\bibitem{Kos} T.~Koshy, \textit{Fibonacci and Lucas Numbers with Applications}, John Wiley and Sons, 2001.

\bibitem{Lag} J.~L.~Lagrange, \textit{Oeuvres de Lagrange},
Gauthier-Villars, 1882.

\bibitem{Leg} T.~Lengyel, The order of Fibonacci and Lucas numbers, \textit{Fibonacci Quart.} \textbf{33} (1995), 234--239.

\bibitem{Rie} P.~Ribenboim, \textit{The Little Book of Big Primes}, Springer, 1991.

\bibitem{Rob} D.~W.~Robinson, The Fibonacci matrix modulo $m$, \textit{Fibonacci Quart.} \textbf{1} (1963), 29--36.

\bibitem{OEIS} N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2021. Available at \url{https://oeis.org}.

\bibitem{Som} L.~Somer and M.~Krizek, Fixed points and upper bounds for the rank of appearance in Lucas sequences, \textit{Fibonacci Quart.} \textbf{51} (2013), 291--306.

\bibitem{Wal} D.~D.~Wall Fibonacci series modulo $m$, \textit{Amer. Math. Monthly} \textbf{67} (1960), 525--532.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B50, 11A07 .

\noindent \emph{Keywords:} Fibonacci number, Lucas number,  Pisano period, rank of appearance, restricted period. 

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A001175},
\seqnum{A111035},
\seqnum{A124456},
\seqnum{A331870},
\seqnum{A331976}, and
\seqnum{A331977}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 30 2021;
revised version received  August 27 2021; November 2 2021; February 7 2022.
Published in {\it Journal of Integer Sequences}, February 7 2022.

\bigskip
\hrule
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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