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\begin{document}
\title{Linear equations involving iterates of $\sigma(N)$}
\date{}
\author{Tomohiro Yamada}
\maketitle

\begin{abstract}
We study integers $N$ satisfying the equation $\sigma(\sigma(N))=A\sigma(N)+BN$.
\end{abstract}

\section{Introduction}\label{intro}
We denote by $\sigma(N)$ the sum of divisors of $N$.
$N$ is said to be perfect if $\sigma(N)=2N$ and multiperfect
if $\sigma(N)=kN$ for some integer $k$.  It is not known whether or not
an odd perfect/multiperfect number exists.  There are many known conditions
which must be satisfied by such a number.  But these results are far from
answering whether or not an odd perfect/multiperfect number exists.

Suryanarayana \cite{Sur} called $N$ to be superperfect if $\sigma(\sigma(N))=2N$
and Pomerance \cite{Pom2} called $N$ to be super multiply perfect if $\sigma(\sigma(N))=kN$
for some integer $k$.
More generally, Cohen and te Riele \cite{CtR} called $N$ to be $(m, k)$-perfect
if $\sigma^{(m)}(N)=kN$, where $\sigma^{(m)}(N)$ denotes $\sigma(\sigma^{(m-1)}(N))$
with $\sigma^{(0)}(N)=N$.

We introduce another analogous notion of perfect/multiperfect numbers.
We call $N$ to be $(n; a_0: \ldots: a_n)$-perfect if $\sum_{i=0}^{n}a_{n-i}\sigma^{(i)}(N)=0$
and $(n; a_1, \ldots, a_n)$-perfect if $N$ is $(n; -1: a_1: \ldots: a_n)$-perfect.
$N$ is $(2; A, B)$-perfect if $\sigma(\sigma(N))=A\sigma(N)+BN$.

We begin by noting that almost all integers are $(2; A, B)$-perfect for some integers $A, B$.
This fact follows from Katai and Subbarao \cite[Theorem 1]{KS}.  They proved that
for any fixed $m\geq 1$, we have $(N, \sigma^{(m)}(N))=\prod_{p, p^a\mid\mid N} p^a$,
where $p$ runs over all primes below $x_2^m$, for all integers $N\leq x$ with $o(x)$ exceptions
(Here we use the notation $x_0=x, x_{i+1}=\max\{1, \log x_i\}$ for all $i\geq 0$, where
$\log x$ is the natural logarithm of $x$, introduced in \cite{KW}).
Hence, for almost all integers $N$, $(N, \sigma(N))$ divides $\sigma^{(2)}(N)$,
which implies that there exist some integers $A, B$ such that $\sigma(\sigma(N))=A\sigma(N)+BN$.


On the other hand, we can show that for any {\it fixed} positive integers $A, B$,
the set of $(2; A, B)$-perfect numbers has density zero.
\begin{thm}\label{th11}
Let $A, B, C$ be integers not all zero and satisfying $AC\leq 0$.
Then the number of $(2; A: B: C)$-perfect numbers below $x$ is
at most $x\exp(-(2^{-1/3}+o(1))(\log x)^{1/3}(\log\log x)^{2/3})$.
\end{thm}

Our argument is similar to the argument of Pomerance \cite{Pom1} to study the distribution
of integers $n$ satisfying $\sigma(n)\equiv a\pmod{n}$.  His argument rests on the fact that
almost all integers $n$ can be written as $mp$, where $p\nmid m$ is prime, large and uniquely
determined by $m$ except some special cases.
Our argument adopts a factorization of $p+1$ into $lq$ with $q$ large and $(l, q)=1$,
to show that $q$ is uniquely determined by $m$ and $l$ under some condition.

We note that this result does not seem to apply to the case $AC>0$.
If $p, 2p-1$ are both prime, then $n=2p-1$ satisfies $2\sigma(\sigma(n))-3\sigma(n)+6n=0$ and therefore
$n$ is $(2; 2: -3: 6)$-perfect.  More generally, we can easily confirm the following result.

\begin{thm}\label{th12}
Let $R$ be an arbitrary positive integer.  If $p$ and $N=Rp-1$ are both prime, then
$N$ is $(2; R: -(R+1)\sigma(R): R\sigma(R))$-perfect.
\end{thm}
\begin{cor}\label{cor11}
If $R$ is a $k$-multiperfect number and $p, N=Rp-1$ are both prime, then
$N$ is $(2; k(R+1): -kR)$-perfect.
\end{cor}

A well-known conjecture states that, for any even integer $R$, the number of primes $p<x$ with $Rp-1$
also prime is asymptotically equal to $cx/(\log x)^2$ for some constant $c>0$ depending on $R$,
contrary to the above estimate $O(x\exp(-(2^{-1/3}+o(1))(\log x)^{1/3}\\(\log\log x)^{2/3}))$.


\section{Notations and Preliminary Lemmas}\label{lemmas}
We denote by $P(n), p(n)$ the largest and smallest prime factor of $n$ respectively.
For the positive real number $x$, let us denote $x_0=x, x_{i+1}=\max\{1, \log x_i\}$
as mentioned in the previous section.  We denote by $c$ some positive constant
not necessarily same at every occurence.  Furthermore, we denote by $x, y, z$
real numbers and we put $u=\log x/\log y$ and $v=\log y/\log z$.

\begin{lem}\label{lm21}
Denote by $\Psi(x, y)$ the number of integers $n\leq x$ divisible by no prime $>y$.
If $y>x_1^2$, then we have $\Psi(x, y)<x\exp(-(1+o(1))u\log u)$ as $x, u$ tend to infinity.
\end{lem}

\begin{proof}
This follows from a well-known theorem of de Bruijn \cite{Bru}.  For details
on the distribution of integers free from large prime factors, we refer the readers
to \cite[Chapter III. 5]{Ten}, where a simple proof of the lemma is also given.
\end{proof}

\begin{lem}\label{lm22}
Let
\begin{equation}
s(x, k)=\sum_{p\leq x, p\equiv -1\pmod{k}}\frac{1}{p}.
\end{equation}
Uniformly in $k$ and $x\geq e^2$, we have
\begin{equation}
s(x, k)\ll\frac{x_2}{\varphi(k)}.
\end{equation}
\end{lem}

\begin{proof}
This inequality can be immediately obtained using partial summation and the Brun-Titchmarsh inequality.
The complete proof is given in \cite[Lemma 2]{KW}.
\end{proof}

\begin{lem}\label{lm23}
Let
\begin{equation*}
S(x)=\{n\mid n\leq x, p^a\mid n\mbox{ for some }p, a\mbox{ with }p^a>y, a\geq 2\}.
\end{equation*}
Then we have $\#S(x)\ll xy^{-1/2}$.
\end{lem}

\begin{proof}
Let $\Pi(t)$ be the number of perfect powers below $t$.
It is clear that $\Pi(t)<t^{1/2}+t^{1/3}+\ldots+t^{1/k}<t^{1/2}+ct^{1/3}\log t\ll t^{1/2}$,
where $k=\floor{(\log x)/(\log 2)}$.

Let us denote by $\gamma_p$ the smallest integer $\gamma$ for which $p^{\gamma}>y$ and $\gamma>1$.
Clearly we have $\#S(x)\leq x\sum_{p\leq x}p^{-\gamma_p}$.  Since $p^{\gamma_p}>y$, we have by partial summation
\begin{equation}
\sum_{p\leq x}\frac{1}{p^{\gamma_p}}\leq\frac{\Pi(x)}{x}-\frac{\Pi(y)}{y}+\int_y^x\frac{\Pi(t)}{t^2}dt\ll y^{-1/2}.
\end{equation}
This prove the lemma.
\end{proof}

\begin{lem}\label{lm24}
  If $y>z>2x_1^2$ and $v>x_2$, then the number of integers $\leq x$
divisible by some prime $p\geq y$ with $P(p+1)<z$ is
at most $x\exp(-(1+o(1))v\log v)$.
\end{lem}

\begin{proof}
  The number of such integers at most
\begin{equation}\label{eq21}
\sum_{y\leq p\leq x, P(p+1)<z}\frac{x}{p}\leq x\sum_{y\leq m\leq x, P(m+1)<z}\frac{1}{m},
\end{equation}
where $p$ and $m$ respectively run over primes and integers satisfying the described conditions.
By partial summation, we find that the last sum is at most
\begin{equation}
\frac{\Psi(y, z)}{y}+\int_y^{2x}\frac{\Psi(t, z)dt}{t^2}.
\end{equation}

Since we have $\Psi(t, z)/t<\exp(-(1+o(1))v\log v)$ uniformly for $t\in [y, 2x]$ by Lemma \ref{lm21},
the last sum in (\ref{eq21}) can be bounded from above by
\begin{equation}
\exp(-(1+o(1))v\log v) \left(1+\int_y^{2x}\frac{dt}{t}\right).
\end{equation}
This integral is $O(x_1)=O(\exp v)$ since, by assumption, $x_2<v$.  Thus we obtain
\begin{equation}
x\sum_{y<m\leq x, P(m+1)<z}\frac{1}{m}=x\exp(-(1+o(1))v\log v).
\end{equation}
This completes the proof.
\end{proof}


\section{Proof of Theorem \ref{th11}}
Let $y, z$ be real numbers and put $u=\log x/\log y$ and $v=\log y/\log z$.
We choose $y, z$ later so that $x>y>z>x_1^2, v>x_2$ and $u, v, y, z$ tend to infinity as $x$ does so.

Let 
\begin{equation}
S_1=\{n\mid n\leq x, P(n)\leq y\}
\end{equation}
and
\begin{equation}
S_2=\{n\mid n\leq x, p^a\mid n\mbox{ for some }p, a\mbox{ with }p^a\geq z, a\geq 2\}.
\end{equation}

We immediately obtain $\#S_1=O(x\exp(-(1+o(1))u\log u))$ by Lemma \ref{lm21}
and $\#S_2=O(x/z^{1/2})$ by Lemma \ref{lm23}.

Denote by $S_3$ the set of integers $n\leq x$ not in $S_1\cup S_2$
which can be written in the form $mp$, where $p$ is a prime $\geq y$,
$m$ is an integer not divisible by $p$ and $(\sigma(m), p+1)$ is divisible by
some prime $q\geq z$.

Let $n$ be an integer in $S_3$ and write $n=mp$ in the above way.
Then $q\mid\sigma(r^a)$ for some prime $r$ dividing $m$
and some integer $a$ with $r^a\mid\mid m$.  Since $q\geq z$, we have $a=1$ by the assumption $n\not\in S_2$.
Hence, $m$ is divisible by some prime $r$ congruent to $-1\pmod{q}$.

Since $m\leq x/p$, the number of integers $n$ satisfying $n=mp, q\mid (\sigma(m), p+1)$
for some $q\geq z$ is at most
\begin{equation}
\sum_{q\geq z}\sum_{p\equiv -1\pmod{q}}\sum_{r\equiv -1\pmod{q}}\frac{x}{pr}\leq\sum_{q\geq z}\frac{cxx_2^2}{q^2}\leq\frac{cxx_2^2}{z},
\end{equation}
by Lemma \ref{lm22}.  Hence, we have $\#S_3=O(xx_2^2/z)$.

Denote by $S_4$ the set of integers $n\leq x$ divisible by some prime $p\geq y$ with $P(p+1)<z$ or $q^2\mid(p+1)$ for some $q\geq z$.
By Lemma \ref{lm24}, the number of integers $n\leq x$ divisible by some prime $p\geq y$
with $P(p+1)<z$ is $O(x\exp(-(1+o(1))v\log v))$.
The number of integers $n\leq x$ divisible by some prime $p$ with $q^2\mid(p+1)$ for some $q\geq z$ is at most
\begin{equation}
x\sum_{q\geq z}\sum_{p\equiv 1\pmod{q^2}}\frac{1}{p}\leq cxx_2\sum_{q\geq z}\frac{1}{q^2}\leq \frac{cxx_2}{zx_1^{1/3}}\ll\frac{x}{z},
\end{equation}
by the assumption that $z>x_1^2$.

Combining these estimates yield $\#S_4=O(x(1/z+\exp(-(1+o(1))v\log v)))$.

We may assume that at least one of $A, C$ is nonzero since the equation
$A\sigma(\sigma(N))+B\sigma(N)+CN=0$ does not hold if exactly one of $A, B, C$ is nonzero.
Now let us denote by $T$ the set of $(2; A: B: C)$-perfect numbers $n\leq x$ belonging to none of $S_i (i=1, 2, 3, 4)$.
We assume that $n\in T$.  
Since $n\not\in S_1\cup S_2$, we have $P(n)>y$ and $P(n)^2\nmid n$.  Thus $n$ can be expressed as $n=mp$, $p>y$
and $p\nmid m$.
Now it follows from $n\not\in S_3$ that $(\sigma(m), p+1)$ has no prime factor $\geq z$.
Let $T_m$ denote the set of such integers.  Moreover, we write $p+1=N_1N_2$ in the way $P(N_1)<z\leq p(N_2)$ and
divide each of $T_m$ into sets $T_{m, N_1}$ according to the value of $N_1$.  Since $n\not\in S_4$, we have $p+1$ is divisible by
some prime $Q\geq z$ exactly once.  By the definition of $N_1, N_2$, we have $N_2=N_3Q$ and $T_{m, N_1}$
can be covered by sets $T_{m, N_1, N_3}$ according to the value of $N_3$.

We shall show each $T_{m, N_1, N_3}$ consists at most one element.
We have $\sigma(n)=\sigma(m)(p+1)$ and $\sigma(\sigma(n))=\sigma(M_1N_1)\sigma(M_2)\sigma(N_2)$,
where $M_1, M_2$ are uniquely determined by $\sigma(m)=M_1M_2$, $P(M_1)<z\leq p(M_2)$.
Furthermore, noting that $Q$ does not divide $(p+1)/Q$ by the assumption that $n\not\in S_4$, we obtain
$\sigma(N_2)=\sigma(N_3)(Q+1)$.
Since $A\sigma(\sigma(n))+B\sigma(m)(p+1)+Cmp=0$, we have
\begin{equation}
A\sigma(M_1N_1)\sigma(M_2)\sigma(N_3)(Q+1)+(B\sigma(m)+Cm)N_1N_3Q-Cm=0.
\end{equation}
Denoting
\begin{equation}
C_1=A\sigma(M_1N_1)\sigma(M_2)\sigma(N_3), C_2=(B\sigma(m)+Cm)N_1N_3, C_3=Cm,
\end{equation}
this equation can be written
\begin{equation}
(C_1+C_2)Q=(C_3-C_1).
\end{equation}
Since $AC\leq 0$, we have $C_1\neq C_3$ and therefore $Q$ can be uniquely determined as
\begin{equation}
Q=\frac{C_3-C_1}{C_1+C_2}.
\end{equation}
This is the crucial point where we use the assumption $AC\leq 0$.
The uniqueness of $Q$ yields that $\#T_{m, N_1, N_3}\leq 1$ for any $m, N_1, N_3$.

By the definition of $T_{m, N_1, N_3}$, each element of $T$ must belong to $T_{m, N_1, N_3}$
for at least one triple $(m, N_1, N_3)$.
Since $N_1N_3=(p+1)/Q\leq (x/m+1)/z\leq 2x/(mz)$, we have
\begin{equation}
\#T\leq\sum_{m\leq x/y}\sum_{N_1}\sum_{N_3}1\leq\sum_{m\leq x/y}\frac{2x}{mz}\leq\frac{cxx_1}{z}\leq\frac{cx}{z^{1/2}}.
\end{equation}

Now we conclude that the number of $(2; A, B)$-perfect numbers $\leq x$ is at most
$\#S_1+\#S_2+\#S_3+\#S_4+\#T$, which is
\begin{equation}
O(x(z^{-1/2}+\exp(-(1+o(1))u\log u)+\exp(-(1+o(1))v\log v)))
\end{equation}
by those estimates for $\#S_i$'s  and $\#T$ given above.

In order to search the optimal estimate, we put $\log z=c_1x_1^{1/3}x_2^{2/3}, \log y=c_2x_1^{2/3}x_2^{1/3}$
and we have $u\log u=(1+o(1))c_2^{-1}x_1^{1/3}x_2^{2/3}$ and $v\log v=(1+o(1))(c_2/c_1)x_1^{1/3}x_2^{2/3}$.
We see that $\max\{c_1/2, c_2/c_1, 1/c_2\}\geq 2^{-1/3}$ with the equality attained when
we choose $c_1=2^{2/3}, c_2=2^{1/3}$.  This choice gives the desired estimate.  This completes
the proof of Theorem \ref{th11}.


We remark that $(2; R: -(R+1)\sigma(R): R\sigma(R))$-perfect numbers given in Theorem \ref{th12}
correspond the case $m=N_1=1, C=A\sigma(R)=-(B+C)R$.


\section{Acknowledgments}
The author thanks Prof. Florian Luca for his advice on exceptional triples $(A, B, C)$.

\begin{thebibliography}{}
\bibitem{Bru}
N. G. de Bruijn, On the number of positive integers $\leq x$ and free of prime factors $>y$, {\em Nederl. Akad. Wetensch. Proc. Ser. A} \textbf{54} (1951), 50--60.
\bibitem{CtR}
G. L. Cohen and H. J. J. te Riele, Iterating the Sum-of-Divisors Function, {\em Experiment. Math.} \textbf{5} (1996), 91--100.
\bibitem{KS}
I. K\'atai and M. V. Subbarao, Some further remarks on the $\varphi$ and $\sigma$ functions, {\em Annales Univ. Sci. Budapest. Sect. Comput.} \textbf{26} (2006), 51--63.
\bibitem{KW}
I. K\'atai and M. Wijsmuller, On the iterates of the sum of unitary divisors, {\em Acta Math. Hungar.} \textbf{79}(1998), 149--167.
\bibitem{Pom1}
C. Pomerance, On the congruences $\sigma(n)\equiv a\pmod{n}$ and $n\equiv a\pmod{\varphi(n)}$, {\em Acta Arith.} \textbf{26} (1975), 265--272.
\bibitem{Pom2}
C. Pomerance, On multiply perfect numbers with a special property, {\em Pacific J. Math.} \textbf{57} (1975), 511--517.
\bibitem{Sur}
D. Suryanarayana, Super perfect numbers, {\em Elem. Math.} \textbf{24} (1969), 16--17.
\bibitem{Ten}
G. Tenenbaum, {\em Introduction to analytic and probabilistic number theory}, Cambridge University Press, 1995.
\end{thebibliography}{}
\vskip 12pt

{\small Tomohiro Yamada}\\
{\small Department of Mathematics\\Faculty of Science\\Kyoto University\\Kyoto, 606-8502\\Japan}\\
{\small e-mail: \protect\normalfont\ttfamily{tyamada@math.kyoto-u.ac.jp}}
\end{document}