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\begin{center}
\vskip 1cm{\LARGE\bf 
When the Large Divisors of a Natural \\
\vskip .1in
Number Are in Arithmetic Progression
}
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\large
H\`ung Vi\d{\^e}t Chu\\
Department of Mathematics\\
University of Illinois at Urbana-Champaign \\
Champaign, IL 61820\\
USA \\
\href{mailto:hungchu2@illinois.edu}{\tt hungchu2@illinois.edu} \\
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\begin{abstract}
Iannucci considered the positive divisors of a natural number $n$ that
do not exceed the square root of $n$ and found all numbers whose such
divisors are in arithmetic progression. Continuing the work, we define
\textit{large divisors} to be divisors at least $\sqrt{n}$ and find
all numbers whose large divisors are in arithmetic progression. The
asymptotic formula for the count of these numbers not larger than $x$
is observed to be $\frac{x\log\log x}{\log x}$.
\end{abstract}


\section{Introduction}
For a natural number $n$, let $L_n$ denote the set of positive divisors of $n$ that are at least $\sqrt{n}$ and strictly smaller than $n$; that is,
$$L_n\ :=\ \{d\,:\, d|n, \sqrt{n}\le d<n\}.$$
Also, define $$L'_n\ :=\ \{d\,:\, d|n, \sqrt{n}\le d\le n\}.$$ We call $L'_n$ the set of \textit{large divisors} of $n$. Clearly, we have $|L'_n| = |L_n| + 1$. In this paper, we will determine the set of all natural numbers $n$ such that either $L_n$ or $L'_n$ forms an arithmetic progression. Since $L_n\subset L_n'$, if $L_n'$ forms an arithmetic progression, then so does $L_n$. Hence, we will first focus our attention on $L_n$ and find all $n$ such that 
$$L_n\ =\ \{d, d+ a, d+2a,\ldots, d+(k-1)a\}$$
for some natural numbers $d, a$, and $k$. Note that $L_n$ can be empty and in that case, $L_n$ vacuously forms an arithmetic progression. Let $|L_n| = k\ge 0$.

Our work is a companion to a paper of Iannucci \cite{Ian}, who defined \textit{small divisors} of $n$ to be divisors not exceeding $\sqrt{n}$ and found all natural numbers whose small divisors are in arithmetic progression. For previous work on divisors in or not in arithmetic progression, see \cite{BFL, Var} and on small divisors, see \cite{BTK, Ian2}.

As usual, we have the divisor-counting function
$$\tau(n) \ :=\ \sum_{d|n}1.$$ Since $\tau(n)$ is multiplicative, for the $k$ distinct primes $p_1<p_2<\cdots<p_k$ and natural numbers $a_1, a_2, \ldots, a_k$, we have
\begin{align}\label{multau}\tau(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k})\ =\ (a_1+1)(a_2+1)\cdots (a_k+1).\end{align}
If $n = bc$ and $b\le c$, then $b\le \sqrt{n}\le c$; hence
\begin{align}\label{rela}\tau(n) \ =\ \begin{cases} 2|L'_n|, & \text{if $n$ is not a square;}\\ 2|L'_n|-1,  & \text{if $n$ is a square} \end{cases}\ =\ \begin{cases}2|L_n|+2, & \mbox{if $n$ is not a square;}\\ 2|L_n|+1,  & \text{if $n$ is a square}.\end{cases}\end{align}
\begin{theorem}\label{main}
Let $n$ be a natural number. If numbers in $L_n$ are in arithmetic progression, then one of the following holds:
\begin{itemize}
    \item [(i)] $n = 1$, and hence $L_n = \emptyset$.
    \item [(ii)] $n = p$ for some prime $p$, and hence $L_n = \emptyset$.
    \item [(iii)] $n = p^2$ for some prime $p$, and hence $L_n = \{p\}$.
    \item [(iv)] $n = p^3$ for some prime $p$, and hence $L_n = \{p^2\}$.
    \item [(v)] $n = pq$ for some primes $p<q$, and hence $L_n = \{q\}$.
    \item [(vi)] $n = p^4$ for some prime $p$, and hence $L_n = \{p^2, p^3\}$.
    \item [(vii)] $n = p^5$ for some prime $p$, and hence $L_n = \{p^3, p^4\}$.
    \item [(viii)] $n = p^2q$ for some primes $p<q$, and hence $L_n = \{p^2, pq\}$ or $L_n = \{q, pq\}$.
    \item [(ix)] $n = pq^2$ for some primes $p<q$, and hence $L_n = \{pq, q^2\}$.
    \item [(x)] $n = pqr$ for some primes $p<q<r$, $pq<r$ and $p=\frac{1}{2}(q+1)$, and hence $L_n = \{r, rp, rq\}$.
    \item [(xi)] $n = p^3q$ for some primes $p>q$ and $q = \frac{1}{2}(p+1)$, and hence $L_n = \{p^2, p^2q, p^3\}$.
\end{itemize}
\end{theorem}

To prove Theorem \ref{main}, we first find all forms of $n$ when $|L_n| = k\le 3$ by case analysis, then show that $k$ cannot be larger than $3$. To find all $n$ such that $L'_n$ forms an arithmetic progression, we need only to check the 11 forms in Theorem \ref{main}. It is straightforward to prove the following corollary, so we omit the proof. 
\begin{corollary}\label{largeN}
Let $n$ be a natural number. If numbers in $L_n'$ are in arithmetic progression, then one of the following holds:
\begin{itemize}
    \item [(i)] $n = 1$, and hence $L'_n = \{1\}$.
    \item [(ii)] $n = p$, and hence $L'_n = \{p\}$.
    \item [(iii)] $n = p^2$ for some prime $p$, and hence $L'_n = \{p, p^2\}$.
    \item [(iv)] $n = p^3$ for some prime $p$, and hence $L'_n = \{p^2, p^3\}$.
    \item [(v)] $n = pq$ for some primes $p<q$, and hence $L'_n=\{q, pq\}$.
\end{itemize}
\end{corollary}



\section{Small cases of $|L_n|$}
Assuming $L_n$ is in arithmetic progression, we fully characterize $n$ when $|L_n|\le 3$ and prove that $|L_n|\neq 4$. 

\begin{lemma}\label{allforms}
If $L_n$ forms an arithmetic progression and $k\le 3$, then one of the items in Theorem \ref{main} is true. 
\end{lemma}

\begin{proof} We consider four cases corresponding to each $0\le k\le 3$. 

\bigskip

\noindent Case 1: If $k = 0$, then by \eqref{rela}, we have $\tau(n)\in \{1,2\}$. If $\tau(n) = 1$, then $n = 1$. If $\tau(n) = 2$, then $n = p$ for some prime $p$. Hence, $L_n = \emptyset$. This corresponds to items $(i)$ and $(ii)$ of the theorem.  

\bigskip

\noindent Case 2: If $k = 1$, then by \eqref{rela}, we have $\tau(n)\in\{3,4\}$. 

If $\tau(n) = 3$, then by \eqref{multau}, we have $n = p^2$ for some prime $p$, and hence $L_n = \{p\}$. This corresponds to item (iii) of the theorem. 

If $\tau(n) = 4$, then by \eqref{multau}, we have $n = p^3$ for some prime $p$ or $n = pq$ for some primes $p<q$. For the former, we get $L_n = \{p^2\}$ and for the latter, we get $L_n = \{q\}$, corresponding to items (iv) and (v) of the theorem. 

\bigskip

\noindent Case 3: If $k = 2$, then by \eqref{rela}, we have $\tau(n)\in \{5, 6\}$. 

If $\tau(n) = 5$, then by \eqref{multau}, we have $n = p^4$ for some prime $p$, and hence $L_n = \{p^2, p^3\}$. This corresponds to item (vi). 

If $\tau(n) = 6$, then by \eqref{multau}, we have $n = p^5$ for some prime $p$ or $n = p^2q$ or $pq^2$ for some primes $p<q$.

\begin{itemize}
    \item [] If $n = p^5$, then $L_n = \{p^3, p^4\}$.
    \item [] If $n = p^2q$ for some primes $p<q<p^2$, then $L_n = \{p^2, pq\}$. If $n = p^2q$ for some primes $p^2<q$, we get $L_n = \{q, pq\}$.
    \item [] If $n = pq^2$ for some primes $p<q$, then $L_n = \{pq. q^2\}$.
\end{itemize}
These correspond to items (vii), (viii), (ix). 

\bigskip

\noindent Case 4: If $k = 3$, then by \eqref{rela}, we have $\tau(n)\in \{7,8\}$.

If $\tau(n) = 7$, then by $\eqref{multau}$, we get $n = p^6$ for some prime $p$. Then $L_n = \{p^3, p^4, p^5\}$, which is impossible since $p^5-p^4\neq p^4-p^3$. 

If $\tau(n) = 8$, then by $\eqref{multau}$, we  get $n=pqr$ for some distinct primes $p,q,r$ or $p^3q$ for some distinct primes $p, q$.

\begin{itemize}
    \item [] If $n = pqr$, we may assume that $p<q<r$. Two subcases are either $r>pq$ or $r<pq$.
    \begin{itemize}
        \item [] $r> pq$: We have $L_n = \{r, pr, qr\}$ and so, $qr - pr = pr-r$, which implies that $p = \frac{1}{2}(q+1)$. This is item (x).
        \item [] $r<pq$: We have $L_n = \{pq, pr, qr\}$ and so, $qr - pr = pr - pq$, which implies that $p = \frac{qr}{2r-q}$. So, either $(2r-q)| q$ or $(2r-q)|r$. However, both are impossible since $2r-q>r>q$. 
    \end{itemize}
    \item [] If $n = p^3q$, two subcases are either $p<q$ or $p>q$.
    \begin{itemize}
        \item [] $p<q$: If $p<q<p^3$, then $L_n = \{p^3, pq, p^2q\}$. Either $p^2q - pq = pq - p^3$ or $p^2q-p^3 = p^3 - pq$. It is easy to see that both cases are impossible. If $q>p^3$, then $L_n = \{q, pq, p^2q\}$. Since $p^2q - pq = pq - q$, we get $p^2=2p-1$, which implies that $p = 1$, a contradiction. 
        \item [] $p>q$: $L_n = \{p^2,  p^2q, p^3\}$. So, $p^3 - p^2q = p^2q-p^2$. Then $q = \frac{1}{2}(p+1)$. This is item (xi).
    \end{itemize}
\end{itemize}
\end{proof}



\begin{lemma}\label{not4}
Our set $L_n$ cannot have exactly $4$ elements.
\end{lemma}
\begin{proof}
We prove this by contradiction. Suppose that $|L_n| = 4$. By \eqref{rela}, we have $\tau(n) \in \{9,10\}$.
\begin{itemize}
    \item[] If $\tau(n) =9$, then \eqref{multau} implies that $n = p^8$ for some prime $p$ or $n = p^2q^2$ for some primes $p<q$.
    \begin{itemize}
        \item [] If $n = p^8$, then $L_n = \{p^4, p^5, p^6, p^7\}$, which cannot form an arithmetic progression.
        \item [] If $n = p^2q^2$ for $p<q$, then $L_n = \{pq, p^2q, q^2, pq^2\}$. So, $pq^2 + pq = p^2q + q^2$, which implies that $p=q$, a contradiction.
        \end{itemize}
    \item[] If $\tau(n) = 10$, either $n = p^{9}$ for some prime $p$ or $n = pq^4$ for distinct primes $p,q$.
    \begin{itemize}
        \item [] If $n = p^9$, then $L_n = \{p^5, p^6, p^7, p^8\}$, which cannot form an arithmetic progression.
        \item [] If $n = pq^4$, we have four subcases.
        \begin{itemize}
            \item [] $p<q$: $L_n = \{pq^2, q^3, pq^3, q^4\}$, so $pq^2 + q^4 = q^3 + pq^3$, which implies that $p=q$, a contradiction.
            \item [] $q<p<q^2$: $L_n = \{q^3, pq^2, q^4, pq^3\}$, so $q^3 + pq^3 = pq^2+q^4$, which implies that $p =q$, a contradiction. 
            \item [] $q^2<p<q^4$: $L_n = \{pq, q^4, pq^2, pq^3\}$. Either $pq^3+pq=pq^2+q^4$ or $pq^3+q^4=pq+pq^2$. The former gives $q=1$, while the latter gives $p = -\frac{q^3}{q^2-q-1}$. Both pose a contradiction. 
            \item [] $q^4<p$: $L_n = \{p, pq, pq^2, pq^3\}$, so $p+pq^3 = pq+pq^2$, which implies that $q=1$, a contradiction. 
        \end{itemize}
    \end{itemize}
\end{itemize}
Therefore, $|L_n| \neq 4$.
\end{proof}




\section{Proof of Theorem \ref{main}}
By Lemmas \ref{allforms} and \ref{not4}, to prove Theorem \ref{main}, it suffices to prove that $|L_n|\le 4$. 
\begin{proof}[Proof of Theorem \ref{main}]
We prove this by contradiction. Suppose that $k = |L_n|\ge 5$. Recall that
$$L_n \ =\ \{d, d + a, d+2a, \ldots, d+(k-1)a\}$$
for some natural numbers $d$ and $a$. Let $\gcd(d,a) = \ell$. Write $d = \ell k_1$ and $a = \ell k_2$. Clearly, $\gcd(k_1, k_2) = 1$, so there exist integers $s, t$ such that $sk_1 + tk_2 = 1$. 

Let $$M \ =\ \lcm{(d, d + a, d+2a, \ldots, d+(k-1)a)};$$
that is, $M$ denotes the least common multiple of all numbers in $L_n$. Write 
\begin{align*}M &\ =\ \lcm{(\ell k_1, \ell k_1+ \ell k_2, \ell k_1 + 2\ell k_2, \ldots, \ell k_1 + (k-1)\ell k_2)}\\
&\ =\ \ell\cdot \lcm{(k_1, k_1+k_2, k_1+2k_2, \ldots, k_1+(k-1)k_2)}.
\end{align*}
We claim that $\gcd(k_1+(k-2)k_2, k_1+(k-1)k_2) = 1$. Indeed, let \begin{align*}x &\ =\ k_1+(k-1)k_2 \\y &\ =\ k_1+(k-2)k_2.\end{align*} We have \begin{align*}k_2 &\ =\ x-y\\ k_1&\ =\ x - (k-1)k_2 \ =\ x - (k-1)(x-y).\end{align*} Because $sk_1+tk_2 = 1$, we have
$$s(x-(k-1)(x-y))+t(x-y) \ =\ 1.$$ So, $$(t+s-s(k-1))x+((k-1)s-t)y\ =\ 1,$$ which implies that $\gcd(x,y) = 1$.
Hence, $$N \ =\ \ell(k_1+(k-2)k_2)(k_1+(k-1)k_2) \ =\ \ell\cdot\lcm{(k_1+(k-2)k_2,k_1+(k-1)k_2)} \mbox{ divides } M.$$
Because $N$ divides $M$ and $M$ divides $n$, we know that $N$ divides $n$. Clearly, $N>\ell(k_1+(k-1)k_2) = d+(k-1)a \ge \sqrt{n}$. Because $N\notin L_n$, we get $N=n$. So, $\ell(k_1+(k-3)k_2)$ divides $N$. Hence, 
$$k_1+(k-3)k_2\mbox{ divides }(k_1+(k-2)k_2)(k_1+(k-1)k_2).$$
Using the same argument as above, we know that $\gcd(k_1+(k-3)k_2,k_1+(k-2)k_2) = 1$. So, 
$$k_1+(k-3)k_2\mbox{ divides }k_1+(k-1)k_2.$$
Write $k_1+(k-1)k_2 = u(k_1+(k-3)k_2)$ for some integer $u\ge 2$. Simplifying the equation, we get
$$\frac{3u-1}{u-1} \ =\ \frac{k_1+kk_2}{k_2}\ =\ \frac{k_1}{k_2} + k \ > \ 5.$$
So, $u<2$. This contradicts that $u\ge 2$. Therefore, it must be that $|L_n|<5$, as desired. 
\end{proof}
\begin{remark}
We can estimate how often a natural number $n$ not larger than $x>0$ has its large divisors form an arithmetic progression. Let $f(x)$ be the function counting such numbers not larger than $x$.

The number of $n$ not larger than $x$ that is either of form $p, p^2$, or $p^3$ for a prime $p$ is asymptotic to
$$\sum_{i=1}^3 \pi(x^{1/i})\ \sim\ \sum_{i=1}^3 \frac{ix^{1/i}}{\log x}.$$
By a result of Landau \cite[\textsection 56]{Lan}, the number of $n\le x$ of the form $pq$ for primes $p<q$ is asymptotic to $$\frac{x\log\log x}{\log x}.$$ Combined with Corollary \ref{largeN}, we know that 
$$f(x)\ \sim\ \frac{x\log\log x}{\log x},$$
which is similar to the asymptotic formula for the case of small divisors \cite{Ian}.
\end{remark} 



\begin{thebibliography}{99}
\bibitem{BFL}
W. Banks, J. Friedlander, and F. Luca, Integers without divisors from a fixed arithmetic progression, \textit{Forum Math.} \textbf{20} (2008), 1005--1037. 

\bibitem{BTK}
D. Bruijn, N. Tengbergen, and D. Kruyswijk, On the set of divisors of a number,  \textit{Nieuw Arch. Wiskd.} \textbf{23} (1951), 191--193.

\bibitem{Ian}
D. E. Iannucci, When the small divisors of a natural number are in arithmetic progression, \textit{Integers} \textbf{18} (2018).

\bibitem{Ian2}
D. E. Iannucci, On sums of the small divisors of a natural number, preprint, 2018. Available at
\url{https://arxiv.org/abs/1910.11835}.

\bibitem{Lan}
E. Landau, \textit{Handbuch der Lehre von der Vorteilung der Primzahlen}, B. G. Teubner, 1909.

\bibitem{Var}
P. Varbanek and P. Zarzycki, Divisors of the Gaussian integers in an arithmetic progression, \textit{J. Number Theory} \textbf{33} (1989), 152--169.
\end{thebibliography}
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B25.

\noindent \emph{Keywords:} divisor, arithmetic progression.

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\vspace*{+.1in}
\noindent
Received January 29 2020;
revised version received May 10 2020; May 12 2020. 
Published in {\it Journal of Integer Sequences},  June 9 2020.

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