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\begin{center}
\vskip 1cm{\LARGE\bf A Perfect Number Generalization\\
\vskip .12in
and Some Euclid-Euler Type Results
}
\vskip 1cm
\large
Tyler Ross\footnote{This project was supported by the Natural Science Foundation of China (Grant No.\ 12071421). }\\
School of Mathematical Sciences\\
Zhejiang University\\
Hangzhou, Zhejiang, 310058\\
China\\
\href{mailto:tylerxross@gmail.com}{\tt tylerxross@gmail.com} \\
\end{center}

\vskip .2 in
\begin{abstract}
In this paper, we introduce a new generalization of the perfect numbers, called $\mathcal{S}$-perfect numbers. Briefly stated, an $\mathcal{S}$-perfect number is an integer equal to a weighted sum of its proper divisors, where the weights are drawn from some fixed set $\mathcal{S}$ of integers. After a short exposition of the definitions and some basic results, we present our preliminary investigations into the $\mathcal{S}$-perfect numbers for various special sets $\mathcal{S}$ of small cardinality. In particular, we show that there are infinitely many $\{0, m\}$-perfect numbers and $\{-1,m\}$-perfect numbers for every $m \geq 1$. We also provide a characterization of the $\{-1,m\}$-perfect numbers of the form $2^kp$ ($k \geq 1$, $p$ an odd prime), as well as a characterization of all even $\{-1,1\}$-perfect numbers.
\end{abstract}

\section{Introduction} \label{sec1}

A positive integer $n &gt; 1$ is called a \textit{perfect number} if it is equal to the sum of its proper divisors; symbolically, if
\begin{equation}\nonumber
n=\sum_{\substack{1 \leq d &lt; n \\ d \mid n}}d. 
\end{equation}

It has been known since Euclid that any number of the form $n=2^{p-1}(2^p-1)$ where both $p$ and $2^p-1$ are prime is perfect. Centuries later, Euler proved the converse: if any $n &gt; 1$ is an even perfect number, then it is of the form $n=2^{p-1}(2^p-1)$, with both $p$ and $2^p-1$ prime. On the other hand, it is not known if there exist infinitely many perfect numbers, or if there exists even a single odd perfect number. To date, fifty-one even perfect numbers have been found, many by computer search, the largest of which corresponds to the prime number $p = 82589933$ (see Hassler \cite{hass} for a brief review of the early history, or Cai \cite{cai} for a more thorough treatment and a survey of the current state of research). The literature has been broadened by the introduction of various generalizations, to which this paper adds another, encompassing many of those previously put forward.

Let $\mathcal{S} \subset \mathbb{Z}$ be any collection of integers, and let $n \in \mathbb{Z}$ with $|n| &gt; 1$. Then we call $n$ an \textit{$\mathcal{S}$-perfect number of the first kind} if there exist integers $\lambda_1, \ldots, \lambda_k \in \mathcal{S}$ such that
\begin{equation}\nonumber
    1+\sum_{j=1}^{k}{\lambda_{j}d_{j}}=n  ,
\end{equation}
where $1 = d_0 &lt; d_1 &lt; \cdots &lt; d_k &lt; d_{k+1}=|n|$ are the positive divisors of $n$. We call $n$ an \textit{$\mathcal{S}$-perfect number of the second kind} if there exist integers $\lambda_0, \ldots, \lambda_k \in \mathcal{S}$ such that
\begin{equation}\nonumber
    \lambda_{0}+\sum_{j=1}^{k}{\lambda_{j}d_{j}}=n  .
\end{equation} 
If $n$ is an $\mathcal{S}$-perfect number, we refer to the sum $n=1+\sum\limits_{j=1}^{k}{\lambda_{j}d_{j}}$ (respectively $n=\lambda_0+\sum\limits_{j=1}^{k}{\lambda_{j}d_{j}}$) as an \textit{$\mathcal{S}$-presentation} of $n$, or simply a \textit{presentation} of $n$ when $\mathcal{S}$ is fixed.  

Throughout this paper, we limit our investigation to positive $\mathcal{S}$-perfect numbers of the first kind unless otherwise indicated. We prove that there are infinitely many $\{0, m\}$-perfect numbers and infinitely many $\{-1,m\}$-perfect numbers for every $m \geq 1$. We also provide a characterization of the $\{-1,m\}$-perfect numbers of the form $2^kp$ $(k \geq 1\text{, } p \text{ an odd prime})$, as well as a characterization of all even $\{-1,1\}$-perfect numbers. The symbols $\tau$ and $\sigma$ indicate the familiar arithmetic functions
\nonumber \begin{align} 
    \tau(n) &amp;= \sum_{d \mid n}1, \\
    \sigma(n) &amp;= \sum_{d\mid n}d. 
\end{align}
\begin{example}The $\mathcal{S}$-perfect numbers generalize the perfect numbers ($\mathcal{S}=\{1\}$), as well several other previously defined generalizations of the perfect numbers. The $\{1,0\}$-perfect numbers (of the second kind) are the semiperfect numbers. For $k  \geq 1$, the $\{k\}$-perfect numbers of the first kind were introduced by Minoli and Bear \cite{min2} as $k$-hyperperfect numbers, and subsequently also studied by te Riele \cite{teri} and McCranie \cite{mccr}. The $\{-k\}$-perfect numbers are integers $n &lt; 0$ satisfying 
\begin{equation}\nonumber
    \sigma(|n|)=\frac{(k+1)(|n|+1)}{k}  .
\end{equation}
When $k = 1$ , these are the numbers with abundance $2$ (OEIS sequence \seqnum{A088831}; see Guy \cite[\S B2]{guy} for a discussion of related topics). Bege and Fogarasi \cite{bege} have investigated the case $k = 2$.
\end{example}
\begin{example}
We list here the first few $\mathcal{S}$-perfect numbers for various small $\mathcal{S}$.
\begin{itemize}
    \item $\mathcal{S}=\{1\}$: $6, 28, 496, 8128, 33550336, \ldots$ (perfect numbers, OEIS sequence \seqnum{A000396}).
    \item $\mathcal{S}=\{1,0\}$ (second kind): $6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 66 \ldots$ (semiperfect numbers, OEIS sequence \seqnum{A005835}); the smallest semiperfect number that is not a $\{1,0\}$-perfect number of the first kind is $66$.
    \item $\mathcal{S}=\{2\}$: $21, 2133, 19521, 176661, \ldots$ ($2$-hyperperfect numbers, OEIS sequence \seqnum{A007593}).
    \item $\mathcal{S}=\{3\}$: $325, \ldots$ ($3$-hyperperfect numbers); it is not known whether or not there are any more, but see McCranie \cite{mccr} for some conjectures and numerical data, including all hyperperfect numbers less than $10^{11}$.
    \item $\mathcal{S}=\{0,2\}$: $21, 63, 147, 171, 189, 225 \ldots$ (see Section \ref{sec2}).
    \item $\mathcal{S}=\{-1,2\}$: $21, 28, 52, 84, 112, 156, 189, 208, 228, \ldots$ (see Section \ref{sec2}). 
    \item $\mathcal{S}=\{-1,1\}$: $6, 12, 24, 28, 30, 40, 42, 48, 54, 56, 60, 66, 70, 78, 80, \ldots$ (see Section \ref{sec3}).
    \item $\mathcal{S}=\{1,2\}$: $6, 10, 21, 28, 44, 45, 50, 52, 99, 105, 117, 135, 136, \ldots$.
    \item $\mathcal{S}=\{1,3\}$: $6, 14, 15, 28, 44, 76, 110, 135, 152, 182, 184, 190, 231, \ldots$.
    \item $\mathcal{S}=\{2,3\}$: $21, 175, 325, 333, \ldots$.
\end{itemize}
\end{example}

The following proposition shows that, for most integers $n &gt; 1$, it is easy to find a set $\mathcal{S} \subset \mathbb{Z}$ such that $n$ is $\mathcal{S}$-perfect. For this reason we focus our discussion mainly on determining the $\mathcal{S}$-perfect numbers and related properties for fixed $\mathcal{S}$. This is somewhat at odds with the literature on $k$-hyperperfect numbers, in which the term hyperperfect number is used generically to refer to any integer that is $k$-hyperperfect for some $k \geq 1$.
\begin{proposition}
If $n \in \mathbb{Z}$ $(|n| &gt; 1)$ has at least two prime factors, then there exists a finite set $\mathcal{S} \subset \mathbb{Z}$ with $\#\mathcal{S} \leq \tau(n)-2$ such that $n$ is $\mathcal{S}$-perfect. If $n \in \mathbb{Z}$ is a prime power, then $n$ is not $\mathcal{S}$-perfect for any $\mathcal{S} \subset \mathbb{Z}$.
\end{proposition}
\begin{proof}
If $n$ has at least two prime factors, with positive divisors 
\begin{equation} \nonumber
1=d_0 &lt; d_1 &lt; \cdots &lt; d_k &lt; d_{k+1} = |n|,    
\end{equation}
then $\gcd(d_1, \ldots, d_k)=1$. It follows that the linear diophantine equation
\begin{equation} \nonumber
    \sum_{j=1}^{k}d_{j}x_{j}=n-1
\end{equation}
has solutions. The second claim is obvious.
\end{proof}

For $\mathcal{S} \subset \mathbb{Z}$, we denote the set of $\mathcal{S}$-perfect numbers by $\mathcal{P}(\mathcal{S})$, omitting curly brackets when $\mathcal{S}$ is given by enumeration of its elements. We have the following easy inclusions.

\begin{proposition}
If $(\mathcal{S}_\alpha)_{\alpha \in \mathcal{A}}$ is any family of subsets $\mathcal{S}_{\alpha} \subset \mathbb{Z}$, then
\nonumber \begin{align}
    &amp;\bigcup_{\alpha \in \mathcal{A}}\mathcal{P}(\mathcal{S}_\alpha) \subset \mathcal{P}(\bigcup_{\alpha \in \mathcal{A}}\mathcal{S}_\alpha), \\
    &amp;\mathcal{P}(\bigcap_{\alpha \in \mathcal{A}}\mathcal{S}_\alpha) \subset \bigcap_{\alpha \in \mathcal{A}}\mathcal{P}(\mathcal{S}_\alpha) .
\end{align}
\end{proposition}
\begin{proof}
Follows immediately from the definitions.
\end{proof}
\section{Some special cases} \label{sec2}

In this section we investigate the $\{0,m\}$-perfect numbers and $\{-1,m\}$-perfect numbers for arbitrary $m \geq 1$. The former are dispatched quite easily via the following lemma.
\begin{lemma}
If $n \in \mathcal{P}(0, m)$ for some $m \geq 1$, then also $(m+1)n \in \mathcal{P}(0, m)$.
\end{lemma}
\begin{proof}
If
\begin{equation}\nonumber
    n=1+\sum_{j=1}^{k}{\lambda_{j}d_{j}}
\end{equation}
is a $\{0, m\}$-presentation of $n$, then
\begin{equation}\nonumber
    (m+1)n = 1+\sum_{j=1}^{k}{\lambda_{j}d_{j}} + mn
\end{equation}
is a $\{0,m\}$-presentation of $(m+1)n$.
\end{proof}
Therefore it suffices to exhibit a single $n \in \mathcal{P}(0,m)$ to generate infinitely many $\{0,m\}$-perfect numbers, which gives the following theorem.
\begin{theorem}
There exist infinitely many $\{0,m\}$-perfect numbers for all $m \geq 1$.
\end{theorem}
\begin{proof}
Note that
\begin{equation} \nonumber
    (m+1)(m^2+m+1)=1+m(m+1)+m(m^2+m+1)
\end{equation}
is $\{0, m\}$-perfect for any $m \geq 1$.
\end{proof}
The $\{-1,m\}$-perfect numbers are more interesting. We focus on the $\{-1,m\}$-perfect numbers of the form $n=2^kp$, where $p$ is an odd prime. The following lemma and corollary will be useful for proving that there are infinitely many such numbers, and characterizing their occurrences among numbers of the same form. We make frequent use of the $2$-adic valuation $\nu_{2}(n) = \max(k \geq 0: 2^{k} \text{ divides } n)$.
\begin{lemma} \label{lem2}
Let $0 \leq s \leq t$, $m \geq 1$. Then the numbers of the form $n=\sum\limits_{j=s}^t\lambda_j\cdot2^j$ with $\lambda_s, \ldots, \lambda_t \in \{-1,m\}$ are precisely the numbers
\begin{equation}\nonumber
    n \equiv \modd{-2^s(2^{t-s+1}-1)} {2^s(m+1)}
\end{equation}
in the interval
\begin{equation} \nonumber
    -2^s(2^{t-s+1}-1) \leq n \leq 2^sm(2^{t-s+1}-1).
\end{equation}
\end{lemma}
\begin{proof}
It is easy to see that
\begin{equation} \nonumber
    \sum_{j=s}^t \lambda_j \cdot 2^j \equiv \modd{-\sum_{j=s}^t 2^j = -2^s(2^{t-s+1}-1)} {2^s(m+1)}
\end{equation}
for any $\lambda_s, \ldots, \lambda_t \in \{-1,m\}$. Therefore the different choices of $\lambda_s, \ldots, \lambda_t \in \{-1,m\}$ give $2^{t-s+1}$ different numbers
\begin{equation}\nonumber
    n \equiv \modd{-2^s(2^{t-s+1}-1)} {2^s(m+1)}
\end{equation}
in the interval
\begin{equation} \nonumber
    -2^s(2^{t-s+1}-1) \leq n \leq \sum_{j=s}^t 2^jm = 2^sm(2^{t-s+1}-1).
\end{equation}
Since there are exactly $2^{t-s+1}$ such numbers, we are done.
\end{proof}
\begin{corollary} \label{cor1}
Fix $m \geq 1$ and set $\beta = \nu_{2}(m+1)$. If $n=\sum\limits_{j=s}^t \lambda_j \cdot 2^j$ for some $0 \leq s \leq t$ and some $\lambda_s, \ldots, \lambda_t \in \{-1,m\}$, with $t \geq s+\beta-1$, then also $n=\sum\limits_{j=s}^{t+\alpha} \Lambda_j \cdot 2^j$ for some $\Lambda_s, \ldots, \Lambda_{t+\alpha} \in \{-1,m\}$ whenever $2^\alpha \equiv \modd{1} {(m+1)/2^\beta}$.
\end{corollary}
\begin{proof}
If $2^\alpha \equiv \modd{1} {(m+1)/2^\beta}$, then
\begin{equation}\nonumber
    2^{t+\alpha+1} \equiv \modd{2^{t+1}} {2^{t+1-\beta}(m+1)}.
\end{equation}
If moreover $t+1-\beta \geq s$, we also have 
\begin{equation}\nonumber
    2^{t+\alpha+1} \equiv \modd{2^{t+1}} {2^s(m+1)},
\end{equation}
so
\begin{equation}\nonumber
-2^s(2^{t-s+1}-1) \equiv \modd{-2^s(2^{t+\alpha-s+1}-1)} {2^s(m+1)},
\end{equation}
as required by the conditions in Lemma \ref{lem2}.
\end{proof}
\begin{theorem} \label{thrm1}
Fix $m \geq 1$, and set $\beta = \nu_{2}(m+1)$. If both $2^k p$, $2^{k+\alpha}p \in \mathcal{P}(-1,m)$ for some odd prime $p$ and some integers $k, \alpha \geq 1$, then 
\begin{equation}\nonumber
2^\alpha \equiv \modd{1} {(m+1)/2^\beta}.
\end{equation}
Conversely, if $2^k p \in \mathcal{P}(-1,m)$ for some odd prime $p$ and some integer $k \geq \beta$, then also $2^{k+\alpha} p \in \mathcal{P}(-1,m)$ whenever $2^\alpha \equiv \modd{1} {(m+1)/2^\beta}$.
\end{theorem}
\begin{proof}
Suppose first that both $2^kp,$ and $2^{k+\alpha}p \in \mathcal{P}(-1,m)$, with presentations
\begin{align}
    \label{eqn1} \tag{1} 2^kp  &amp;=  1+\sum_{j=1}^{k}\lambda^{(1)}_{j} \cdot 2^j + \sum_{j=0}^{k-1}\lambda^{(2)}_{j} \cdot 2^j p ,  \\
    \label{eqn2} \tag{2} 2^{k+\alpha}p  &amp;= 1+\sum_{j=1}^{k+\alpha}\Lambda^{(1)}_{j} \cdot 2^j + \sum_{j=0}^{k+\alpha-1}\Lambda^{(2)}_{j} \cdot 2^j p
\end{align}
respectively. Note that every $\lambda^{(i)}_j, \Lambda^{(i)}_j \equiv \modd{-1} {m+1}$; we reduce the first equation to find \begin{equation} \nonumber
    2^kp  \equiv \modd{1-\sum_{j=1}^{k}2^j - \sum_{j=0}^{k-1} 2^j p} {m+1},
\end{equation}
or $(2^{k+1}-1)(p+1) \equiv \modd{2} {m+1}$, from which it follows easily that $p+1$ must be a unit modulo $(m+1)/2^\beta$. 

Subtracting \eqref{eqn1} from \eqref{eqn2} and reducing again modulo $m+1$ gives
\begin{equation} \nonumber
    2^kp(2^\alpha - 1) \equiv \modd{-\sum_{j=k+1}^{k+\alpha}2^j - \sum_{j=k}^{k+\alpha-1} 2^j p} {m+1},
\end{equation}
or 
\begin{equation}\nonumber
    2^{k+1}(p+1)(2^{\alpha}-1) \equiv \modd{0} {m+1},
\end{equation}
so that also
\begin{equation}\nonumber
    2^{k+1}(p+1)(2^{\alpha}-1) \equiv \modd{0} {(m+1)/2^\beta}.
\end{equation}
Since both $2^{k+1}$ and $p+1$ are units modulo $(m+1)/2^\beta$, we conclude that
\begin{equation}\nonumber
    2^{\alpha}-1 \equiv \modd{0} {(m+1)/2^\beta}.
\end{equation}

Conversely, suppose $k \geq \beta$ and $2^k p \in \mathcal{P}(-1,m)$ for some odd prime $p$, with presentation
given by \eqref{eqn1}. Fix any $\alpha \geq 1$ such that $2^\alpha \equiv \modd{1} {(m+1)/2^\beta}$. We have 
\begin{equation} \tag{3} \label{eqn3}
    2^{k+\alpha}p  =  1+\sum_{j=1}^{k}\lambda^{(1)}_{j} \cdot 2^j + \sum_{j=0}^{k-1}\lambda^{(2)}_{j} \cdot 2^j p + \sum_{j=k}^{k+\alpha-1}2^jp.
\end{equation}
Since $k \geq \beta$, we can use Corollary \ref{cor1} to find $\Lambda^{(1)}_{1}, \ldots, \Lambda^{(1)}_{k+\alpha}$ such that 
\begin{equation} \nonumber
    \sum_{j=1}^{k+\alpha}\Lambda^{(1)}_j \cdot 2^j =\sum_{j=1}^{k}\lambda^{(1)}_j \cdot 2^j.
\end{equation}
As for the remaining sum in \eqref{eqn3}, set
\begin{equation}\nonumber
    A=\sum_{j=0}^{k-1}\lambda^{(2)}_{j} \cdot 2^j  + \sum_{j=k}^{k+\alpha-1}2^j.
\end{equation}
Reducing modulo $m+1$,
\begin{equation} \nonumber
    A \equiv \modd{2^{k+\alpha}-2^{k+1}+1 \equiv -(2^{k+\alpha}-1)}  {m+1},
\end{equation}
where we have made use of the hypotheses $2^\alpha \equiv \modd{1} {(m+1)/2^\beta}$ and $k \geq \beta$ to substitute $2^{k+\alpha} \equiv \modd{2^k} {m+1}$. Therefore $A$ satisfies the conditions of Lemma \ref{lem2} (with $s=0$, $t=k+\alpha-1$), so we can find $\Lambda^{(2)}_{0}, \ldots, \Lambda^{(2)}_{k+\alpha-1}$ such that $A = \sum\limits_{j=0}^{k+\alpha-1}\Lambda^{(2)}_j$. 

Thus we obtain a presentation
\begin{equation} \nonumber
    2^{k+\alpha}p = 1 + \sum_{j=1}^{k+\alpha}\Lambda^{(1)}_j \cdot 2^j + \sum_{j=0}^{k+\alpha-1}\Lambda^{(2)}_j \cdot 2^j p.
\end{equation}
\end{proof}
It follows that a single $\{-1,m\}$-perfect number of the form $2^kp$, with $k \geq \beta$ and $p$ an odd prime, is sufficient to generate infinitely many. The following theorem provides a construction.
\begin{theorem} \label{thrm2}
Fix $m \geq 1$, and set $\beta = \nu_{2}(m+1)$. Choose $\alpha &gt; \beta$ such that
\begin{equation}\nonumber
    2^\alpha \equiv \modd{1} {(m+1)/2^\beta}.
\end{equation}
If some
\begin{equation}\nonumber
    p \equiv \modd{2(2^{\alpha+1}-1)-1} {2(m+1)}
\end{equation}
is prime, then $2^kp \in \mathcal{P}(-1,m)$ for some $k \geq \alpha$.
\end{theorem}
\begin{proof} Set $N  = 2^{\alpha+1}-1$, and note that $\alpha &gt; \beta$ implies that $N^2 \equiv \modd{1} {2(m+1)}$. If
\begin{equation}\nonumber
    p \equiv \modd{2(2^{\alpha+1}-1)-1} {2(m+1)},
\end{equation}
then 
\begin{equation} \nonumber
    Np \equiv \modd{2N^2-N \equiv 2-N = 3-2^{\alpha+1}} {2(m+1)}.
\end{equation}
That is,
\begin{equation}\nonumber
    Np-1 \equiv \modd{-2(2^{\alpha}-1)} {2(m+1)}.
\end{equation}
Choose $k\geq \alpha$ with $2^k \equiv \modd{2^\alpha} {2(m+1)}$ such that $Np-1 \leq 2m(2^k-1)$. Then by Lemma \ref{lem2}, there are some $\lambda^{(1)}_{1}, \ldots, \lambda^{(1)}_{k} \in \{-1, m\}$ such that 
\begin{equation} \nonumber
    Np = 1 + \sum_{j=1}^{k}\lambda^{(1)}_{j} \cdot 2^j.
\end{equation}
We also have $2^k-N \equiv \modd{-(2^k-1)} {2(m+1)}$, so \begin{equation} \nonumber
    2^k-N = \sum_{j=0}^{k-1}\lambda^{(2)}_{j} \cdot 2^j.
\end{equation}
for some $\lambda^{(2)}_{1}, \ldots, \lambda^{(2)}_{k-1} \in \{-1, m\}$. Therefore
\begin{equation} \nonumber
    1 + \sum_{j=1}^{k}\lambda^{(1)}_{j} \cdot 2^j + \sum_{j=0}^{k-1}\lambda^{(2)}_{j} \cdot 2^jp = Np + (2^k-N)p = 2^kp
\end{equation}
is a presentation.
\end{proof}
\begin{corollary}
There exist infinitely many $\{-1,m\}$-perfect numbers for every $m \geq 1$.
\end{corollary}
\begin{proof}
With $\alpha$, $\beta$ as in Theorem \ref{thrm2}, we have
\begin{equation}\nonumber
    2(2^{\alpha+1}-1)-1 \equiv \modd{1} {(m+1)/2^\beta}.
\end{equation}
Then since $2(2^{\alpha+1}-1)-1$ is odd, it follows also that 
\begin{equation}\nonumber
    \gcd(2(2^{\alpha+1}-1)-1,2(m+1))=1,
\end{equation}
so there do in fact exist primes $p \equiv \modd{2(2^{\alpha+1}-1)-1} {2(m+1)}$.
\end{proof}
\section{The $\{-1, 1\}$-perfect numbers} \label{sec3}
In the previous section, we obtained a characterization of the $\{-1,m\}$-perfect numbers (for $m \geq 1$) of the form $2^kp$, where $p$ is an odd prime. When $m=1$, this can be extended to a characterization of all even $\{-1,1\}$-perfect numbers. The $\{-1,1\}$-perfect numbers have a certain aesthetic appeal owing to the formal similarity between the sum involved in a $\{-1,1\}$-presentation and the divisor sum involved in the definition of perfect numbers.

We first refine slightly the relevant special case of Lemma \ref{lem2}.
\begin{lemma} \label{lem3}
If $n \in \mathbb{Z}$, then $n=1+\sum\limits_{j=1}^{k}\lambda_j \cdot 2^j$ for some $k \geq 1$ and some $\lambda_1, \ldots, \lambda_k \in \{-1,1\}$ if and only if $n \equiv \modd{3} {4}$.
\end{lemma}
\begin{proof}
Choose $k \geq 1$ such that $-2(2^k-1) \leq n-1 \leq 2(2^k-1)$; then Lemma \ref{lem2} applies ($m=1$, $s=1$, $t=k$).
\end{proof}
\begin{lemma} \label{lem4}
Let $n \in \mathbb{Z}$, and let $p$ be prime, with $p$ not dividing $n$. Then 
\begin{enumerate}[label=(\alph*)]
    \item \label{item1} if $n \in \mathcal{P}(-1,1)$ , then also  $np^k \in \mathcal{P}(-1,1)$ for all $k \geq 1$;
    \item \label{item2} if $np \in \mathcal{P}(-1,1)$, then also $np^{2k-1} \in \mathcal{P}(-1,1)$ for all $k \geq 1$.
\end{enumerate}
\end{lemma}
\begin{proof}
\ref{item1} If
\begin{equation}\nonumber
    n = 1+\sum_{j=1}^{k}{\lambda_{j}d_{j}}
\end{equation}
and
\begin{equation}\nonumber
    np^k = 1+\sum_{s=1}^{t}{\Lambda_{s}D_{s}},
\end{equation}
where $k \geq 0$, are presentations of $n$ and $np^k$ respectively, then
\begin{equation}\nonumber
    np^{k+1}=1+\sum_{s=1}^{t}{\Lambda_{s}D_{s}} - np^k +p^{k+1}(1+\sum_{j=1}^{k}{\lambda_{j}d_{j}})
\end{equation}
is a presentation of $np^{k+1}$. The proof of \ref{item2} is similar.
\end{proof}
\begin{lemma} \label{lem5}
If $n \in \mathcal{P}(-1,1)$, then also $2n \in \mathcal{P}(-1,1)$.
\end{lemma}
\begin{proof}
If $n$ is odd, this follows from Lemma \ref{lem4}. Suppose $n$ is even, and
\begin{equation}\nonumber
    n=1+\sum_{j=1}^{k}\lambda_{j}d_j
\end{equation}
is a presentation of $n$. Then
\begin{equation}\nonumber
    2n=1+\sum_{j=1}^{k}\lambda_{j}d_j+n.
\end{equation}
The proper divisors of $2n$ missing from this sum have the form $2d_j$ for some divisor $d_j$ of $n$ with $1&lt;d_j&lt;n$ (since $n$ is even). Replace all such $\lambda_{j}d_{j}$ in the sum with $-\lambda_{j}d_j+\lambda_{j}(2d_{j})$ to obtain a presentation of $2n$.
\end{proof}
\begin{theorem} \label{thrm3}
If $d \geq 1$ is odd and not a square, then $2^k d \in \mathcal{P}(-1,1)$ for all but finitely many $k \geq 1$. Conversely if $2^k d \in \mathcal{P}(-1,1)$ for some $k \geq 0, d \geq 1$, then $d$ is not square.
\end{theorem}
\begin{proof}
In light of Lemmas \ref{lem4} and \ref{lem5}, it suffices to show that, for every odd prime $p$, there exists some $k \geq 1$ (depending on $p$), such that $2^kp \in \mathcal{P}(-1,1)$. Choose (Lemma \ref{lem3}) $k \geq 1$ and $\lambda_1, \ldots, \lambda_k \in \{-1, 1\}$ such that
\begin{equation} \nonumber
    1+\sum_{j=1}^{k}\lambda_{j}2^{j} = 
    \begin{cases}
        p, &amp; \text{if $p \equiv 3 \pmod{4}$;}\\
        3p, &amp; \text{if $p \equiv 1 \pmod{4}$.}
    \end{cases}
\end{equation}
Then 
\begin{equation} \nonumber
    2^{k}p=1+\sum_{j=1}^{k}\lambda_{j}2^{j}+(-1)^{(p+1)/2}p+\sum_{j=1}^{k-1}2^{j}p
\end{equation}
is a presentation, as required.

Conversely, suppose $n &gt; 1$ is $\{1,-1\}$-perfect with presentation
\begin{equation}\nonumber
    n=1+\sum_{j=1}^{k}\lambda_{j}d_{j}.
\end{equation}
Then
\begin{equation} \nonumber
    \sigma(n) = 2+\sum_{j=1}^{k}(1+\lambda_{j})d_{j}
\end{equation} is even, since every $1+\lambda_{j}=0 \text{ or } 2$. But it is well known that $\sigma(n)$ is even if and only if $n$ is not square or twice a square.
\end{proof}
We conclude with a few further remarks and conjectures concerning the $\{-1, 1\}$-perfect numbers. Recall that an abundant number (OEIS sequence \seqnum{A005101}) is a positive integer $n$ satisfying $\sigma(n) \geq 2n$. Evidently, every positive $\{-1,1\}$-perfect number is abundant, but not every abundant number is $\{-1,1\}$-perfect; the first few abundant numbers that are not also $\{-1,1\}$-perfect are $18, 20, 36, 72, \ldots$. We conjecture that almost every abundant number is $\{-1,1\}$-perfect
\begin{conjecture}
The positive $\{-1,1\}$-perfect numbers have a density equal to the density $\mathcal{A}$ of the abundant numbers, for which we have the bounds $0.2474 &lt; \mathcal{A} &lt; 0.2480$, obtained by Del\'eglise \cite{dele}.
\end{conjecture}
The smallest odd abundant number (OEIS sequence \seqnum{A005231}) is $945$, which is also $\{-1,1\}$-perfect, as, in fact, is every odd abundant number smaller than $10^4$. On the other hand, theoretical considerations show that not every odd abundant number is $\{-1,1\}$-perfect: if $n \in \mathbb{Z}$ is odd and abundant, then also $n^2$ is odd and abundant, since the set of abundant numbers is closed under multiplication;  but Theorem \ref{thrm3} shows that $n^2$ cannot be $\{-1,1\}$-perfect. So, for example, $945^2=893025$ is odd and abundant, but not $\{-1,1\}$-perfect. We propose the following conjecture.
\begin{conjecture}
Every nonsquare odd abundant number is $\{-1,1\}$-perfect.
\end{conjecture}
\section{Acknowledgments}\label{sec4}
The author would like to thank the anonymous referee and the editor for their helpful comments and suggestions.
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\bibitem{dele} M. Del\'eglise, Bounds for the density of abundant numbers, \textit{Exp. Math.} \textbf{7} (1998), 137--143.

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\href{https://cs.uwaterloo.ca/journals/JIS/VOL3/mccranie.html}{Article 00.1.3}.

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\bibitem{oeis}
{OEIS Foundation Inc.}, The {O}n-{L}ine {E}ncyclopedia of {I}nteger
  {S}equences, 2023.
\newblock Published electronically at \url{https://oeis.org}.

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\end{thebibliography}





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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11A67, 11Y55.

\noindent \emph{Keywords: } perfect number, semiperfect number, hyperperfect number, abundant number.

\bigskip
\hrule
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\noindent (Concerned with sequences 
\seqnum{A000396},
\seqnum{A005101},
\seqnum{A005231},
\seqnum{A005835},
\seqnum{A007593}, and
\seqnum{A088831}.)

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\hrule
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\vspace*{+.1in}
\noindent
Received March 7 2024;
revised versions received  March 8 2024; August 28 2024.
Published in {\it Journal of Integer Sequences}, September 8 2024.

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\noindent
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