Proof Since $ 1+\sum _{j=1}^k d_j = \sigma (n)-n$ , by switching the sign of $d_j$ for a given index $j \in \{ 1, 2, \ldots , k \}$ , we see that: The existence of an expansion of the form $ n=1+\sum _{j=1}^k c_jd_j $ for $c_j\in \{-1,1\}$ is equivalent to the existence of a (possibly empty) subset $T\subseteq \{1,\ldots ,k\}$ such that $ \sigma (n) - 2n = 2 \sum _{j \in T} d_j$ , as desired.

Proof Let y be a real number such that $y \geq 3$ . We proceed to define

(2.5) $$ \begin{align} \mathcal{R}_{y} = \{ m \geq 1 : (m, p_{\pi(y)}\# ) = 1 \}. \end{align} $$

We rewrite the right-hand side of (2.5) as a union of residue classes modulo $p_{\pi (y)}\#$ , and we find that it has a natural density equal to

(2.6) $$ \begin{align} \rho_{y} = \prod_{p \leq y} \left( 1 - \frac{1}{p} \right) = \frac{\varphi(p_{\pi(y)}\#) }{p_{\pi(y)}\# }. \end{align} $$

It is well known that the counting function for integers not exceeding $x> 0$ and coprime to a fixed modulus $q \in \mathbb {Z}$ is $\frac {\varphi (q)}{q} x + O(1)$ , with the error term $O(1)$ being uniform in x for every fixed q. Consequently, the relation

(2.7) $$ \begin{align} \left| \{ m \leq x : m \in \mathcal{R}_{y} \} \right| = \rho_{y} x + O_y(1) \end{align} $$

holds, with the error term in (2.7) being uniform in x (but dependent on y). Since $y \geq 3$ , we have that $2 \mid p_{\pi (y)}\#$ and $3 \mid p_{\pi (y)}\#$ . Consequently, the implication

(2.8) $$ \begin{align} m \in \mathcal{R}_{y} \Longrightarrow (m, 6) = 1 \end{align} $$

holds. Now, we define

(2.9) $$ \begin{align} G_{y}(m) = \sum_{ \substack{p \mid m \\ p> y} } \frac{1}{p-1}. \end{align} $$

If $m \in \mathcal {R}_{y}$ , then each prime $p \leq y$ does not divide m. So, if $m \in \mathcal {R}_{y}$ , then, since each prime divisor of m is strictly greater than y, the right-hand side of (2.9) may be rewritten so that $G_y(m) = \sum _{p \mid m} \frac {1}{p-1}$ .

Now, by writing a positive integer m so that $m = \prod _{p^b \Vert m} p^b$ , this together with the relation in (2.3) give us that

$$ \begin{align*} \frac{\sigma(m)}{m} & = \prod_{p^a \Vert m} \left( 1 + \frac{1}{p} + \cdots + \frac{1}{p^a} \right) \\ & \leq \prod_{p \mid m} \left( 1 - \frac{1}{p} \right)^{-1}. \end{align*} $$

If $u \in (0, 1)$ , then $-\log (1-u) \leq \frac {u}{1-u}$ . This together with the relation $\log \frac {\sigma (m)}{m} \leq \sum _{p \mid m} - \log \left ( 1 - \frac {1}{p} \right )$ give us that

(2.10) $$ \begin{align} \log \frac{\sigma(m)}{m} \leq \sum_{p \mid m} \frac{1}{p-1}. \end{align} $$

So, from (2.10) along with how the right-hand side of (2.9) reduces for $m \in \mathcal {R}_{y}$ , we find that

(2.11) $$ \begin{align} m \in \mathcal{R}_{y} \Longrightarrow \log \frac{\sigma(m)}{m} \leq G_y(m). \end{align} $$

By defining

(2.12) $$ \begin{align} S_{y}(x) = \sum_{\substack{m \leq x \\ m \in \mathcal{R}_{y} }} G_{y}(m), \end{align} $$

we proceed to rewrite the right-hand side of (2.12) so that

(2.13) $$ \begin{align} S_{y}(x) & = \sum_{\substack{m \leq x \\ m \in \mathcal{R}_{y} }} \sum_{ \substack{p \mid m \\ p> y} } \frac{1}{p-1} \nonumber \\ & = \sum_{p \in (y, x]} \frac{1}{p-1} \left| \{ m \leq x : m \in \mathcal{R}_{y}, p \mid m \} \right|. \end{align} $$

Now, fix $y \geq 3$ . We rewrite the estimate in (2.7) so that

(2.14) $$ \begin{align} \left| \{ u \leq X : (u, p_{\pi(y)}\# ) = 1 \} \right| = \rho_y X + O_{y}(1), \end{align} $$

uniformly for real $X> 0$ , with the understanding that the error term in (2.14) depends only on y and not on X. Now, fix a prime $p> y$ . By writing $m = p u$ , we obtain that

$$ \begin{align*} \left| \left\{ u \leq \frac{x}{p} : (u, p_{\pi(y)}\# ) = 1 \right\} \right| & = \left| \left\{ u \leq \frac{x}{p} : u \in \mathcal{R}_{y} \right\} \right| \\ & = \left| \left\{ m \leq {x} : p \mid m, \, \frac{m}{p} \in \mathcal{R}_{y} \right\} \right| \\ & = \left| \left\{ m \leq {x} : p \mid m, \, \left( \frac{m}{p}, p_{\pi(y)}\# \right) = 1 \right\} \right| \\ & = \left| \left\{ m \leq {x} : p \mid m, \, \left( m, p_{\pi(y)}\# \right) = 1 \right\} \right| \\ & = \left| \left\{ m \leq {x} : p \mid m, \, m \in \mathcal{R}_{y} \right\} \right|, \end{align*} $$

with the understanding that u and m range over positive integers, and where, for the second-to-last equality above, we have exploited the property that $p \nmid p_{\pi (y)}\#$ , that is, since p does not divide the product of primes not exceeding $y < p$ . So, by setting $X = \frac {x}{p}$ in (2.14), this and the chain of five equalities above give us that

(2.15) $$ \begin{align} \left| \{ m \leq x : m \in \mathcal{R}_{y}, p \mid m \} \right| = \rho_y \frac{x}{p} + O_{y}(1), \end{align} $$

and we again emphasize that the error term in (2.15) depends only on y.

Now, from (2.13) and (2.15) together, we find that

(2.16) $$ \begin{align} S_{y}(x) = \rho_y x \sum_{p \in (y, x]} \frac{1}{ p (p-1)} + \left( \sum_{p \in (y, x]} \frac{1}{p-1} \right) O_{y}(1). \end{align} $$

From (2.16), we write

(2.17) $$ \begin{align} \frac{S_{y}(x)}{x} = \rho_y \sum_{p \in (y, x]} \frac{1}{ p (p-1)} + O_{y}\left( \frac{\log \log x}{x} \right). \end{align} $$

The convergence of $\sum _{p} \frac {1}{p(p-1)}$ together with the relation in (2.17) gives us that

(2.18) $$ \begin{align} \lim_{x \to \infty} \frac{S_{y}(x)}{x} = \rho_y \sum_{p> y} \frac{1}{p(p-1)}. \end{align} $$

Observe that $ \sum _{p> y} \frac {1}{p(p-1)} \to 0$ as $y \to \infty $ . So, for arbitrary $\varepsilon> 0$ , we may choose $y = y_{\varepsilon } \geq 3$ so that $ \sum _{p> y} \frac {1}{p(p-1)} < \log (1 + \varepsilon )$ .

Now, define

(2.19) $$ \begin{align} \mathcal{B}_{y, \varepsilon} = \{ m \in \mathcal{R}_{y} : G_{y}(m) \geq \log(1 + \varepsilon) \}. \end{align} $$

An application of the formulation of Markov’s inequality in (2.4) then gives us that

(2.20) $$ \begin{align} \log(1 + \varepsilon) \, \left| \{ m \leq x : m \in \mathcal{B}_{y, \varepsilon} \} \right| & \leq \sum_{\substack{ m \leq x \\ m \in \mathcal{R}_{y}}} G_{y}(m) \nonumber \\ & = S_{y}(x), \end{align} $$

recalling (2.12). From (2.18) along with (2.20), we find that

$$ \begin{align*}\overline{d}(\mathcal{B}_{y, \varepsilon}) \leq \frac{\rho_y}{\log(1 + \varepsilon)} \sum_{p> y} \frac{1}{p(p-1)}.\end{align*} $$

Recalling the density of $\mathcal {R}_{y}$ shown in (2.7), we find that $ \underline {d}\big ( \mathcal {R}_{y} \setminus \mathcal {B}_{y, \varepsilon } \big ) \geq \rho _y - \overline {d}(\mathcal {B}_{y, \varepsilon })$ . Consequently, we obtain the relations

(2.21) $$ \begin{align} \underline{d}\big( \mathcal{R}_{y} \setminus \mathcal{B}_{y, \varepsilon} \big) \geq \rho_y \left( 1 - \frac{1}{\log(1 + \varepsilon)} \sum_{p> y} \frac{1}{p(p-1)} \right) > 0. \end{align} $$

Now, if $ m \in \mathcal {R}_{y} \setminus \mathcal {B}_{y, \varepsilon }$ , then a combined application of the implication in (2.11) together with the complement of the right-hand side of (2.19) gives us that $\log \frac {\sigma (m)}{m} \leq G_y(m) < \log (1 + \varepsilon )$ , that is, so that

(2.22) $$ \begin{align} m \in \mathcal{R}_{y} \setminus \mathcal{B}_{y, \varepsilon} \Longrightarrow \frac{\sigma(m)}{m} < 1 + \varepsilon. \end{align} $$

So, a combined application of the lower bound in (2.21) together with the implications in both (2.8) and (2.22) gives us that $\mathcal {R}_{y} \setminus \mathcal {B}_{y, \varepsilon }$ is a set of positive integers m that is of positive lower density such that both $ (m,6)=1$ and $ \frac {\sigma (m)}{m}<1+\varepsilon $ hold.

Proof We henceforth assume that (2.24) holds. Since m and $6$ are coprime, we find that $(18, m) = 1$ . Since $\sigma $ is multiplicative, we see that $\sigma (18m) = 39 \sigma (m)$ , so that (2.23) then gives us that $\sigma (18m) = 39(m + s(m))$ . Since $m> 0$ and $s(m) \geq 0$ , we find that $\sigma (18m) - 2(18m) = 3 m + 39 s(m)> 0$ . It follows that $18m$ is strictly abundant, in particular $18 m \in \mathcal {A}$ , and it remains only to be shown that also $18 m \not \in \mathcal {P}$ .

By way of contradiction, suppose that $18 m \in \mathcal {P}$ . So, by Lemma 2.1, the integer

(2.25) $$ \begin{align} H = \frac{\sigma(18m) - 2(18m)}{2} \end{align} $$

can be written as a sum of distinct divisors of $18m$ apart from $1$ and $18m$ . We then rewrite (2.25) so that

$$ \begin{align*}H = \frac{3m + 39 s(m)}{2}.\end{align*} $$

With the assumption that (2.24) holds, we find that $39 s(m) < \frac {m}{2}$ . So, again with the assumption of (2.24) holding, we find that

(2.26) $$ \begin{align} \frac{3m}{2} \leq H = \frac{3m + 39 s(m)}{2} < \frac{3m + \frac{m}{2}}{2} = \frac{7m}{4} < 2m. \end{align} $$

Now, let $\mathscr {S}$ be a sum of proper (distinct) nontrivial divisors of $18 m$ . Since $(18, m) = 1$ , each divisor of $18m$ is of the form $ed$ for $e \mid 18$ and $d \mid m$ . We consider the two cases given by $d = m$ and $d < m$ . For the $d = m$ case, any such divisors appearing in the expansion of $\mathscr {S}$ as a sum of divisors of $18 m$ provide a contribution of $A m $ to $\mathscr {S}$ , for some integer $A \geq 0$ . We then write $\mathscr {S} = A m + B$ , where

$$ \begin{align*}0 \leq B \leq \sum_{e \mid 18} \sum_{\substack{d \mid m \\ d < m}} e d.\end{align*} $$

Since $\sum _{e \mid 18} e = \sigma (18) = 39$ and since $\sum _{\substack {d \mid m \\ d < m}} d = \sigma (m) - m = s(m)$ , we find that $0 \leq B \leq 39 s(m)$ . So, the sum $\mathscr {S}$ is of the form $ \mathscr {S} = A m + B $ for a nonnegative integer A and for an integer $B \in [0, 39s(m)]$ . If $A \leq 1$ , then $\mathscr {S} \leq m + 39 s(m)$ , and from (2.24), we would then have that $\mathscr {S} < \frac {3m}{2}$ , so that $\mathscr {S} < H$ , recalling (2.26). Also, if $2 \leq A$ , then $2m \leq \mathscr {S}$ , but (2.26) gives us that $H < 2m$ , so that $H < \mathscr {S}$ . So, we have shown that no sum of proper, nontrivial divisors of $18 m$ can be equal to the right-hand side of (2.25), contradicting Lemma 2.1. Since we have shown that $18 m \not \in \mathcal {P}$ , we have completed our proof.

Proof By setting $\varepsilon = \frac {1}{78}$ in Lemma 2.2, we find that there exists a set $\mathcal {M}$ of positive integers such that

(2.27) $$ \begin{align} \underline{d}(\mathcal{M})> 0 \end{align} $$

and such that: for each member m in $\mathcal {M}$ , we have that $(m, 6) = 1$ and that $\frac {\sigma (m)}{m} < \frac {79}{78}$ . Again for a member m of $\mathcal {M}$ , we set $s(m) = \sigma (m) - m$ . We then find that $\frac {s(m)}{m} = \frac {\sigma (m)}{m} - 1 < \frac {1}{78}$ . So, if $m \in \mathcal {M}$ , then m satisfies all of the required conditions in Lemma 2.3. So, if $m \in \mathcal {M}$ , then $18 m$ is strictly abundant and $18 m \not \in \mathcal {P}$ . Recalling the definition of $\mathcal {A}$ in (2.2), we thus find: if $m \in \mathcal {M}$ , then $18m \in \mathcal {A} \setminus \mathcal {P}$ . By then writing $\mathcal {E}$ in place of the family $\{ 18 m : m \in \mathcal {M} \}$ , we obtain the containment

(2.28) $$ \begin{align} \mathcal{E} \subseteq \mathcal{A} \setminus \mathcal{P}. \end{align} $$

Now, observe that

(2.29) $$ \begin{align} \frac{ \left| \{ n \leq x : n \in \mathcal{E} \} \right| }{x} = \frac{1}{18} \frac{ \left| \{ m \leq \frac{x}{18} : m \in \mathcal{M} \} \right| }{x/18}. \end{align} $$

Taking lower limits with $x \to \infty $ , the relation in (2.29) then gives us that

(2.30) $$ \begin{align} \underline{d}(\mathcal{E}) = \frac{1}{18} \underline{d}(\mathcal{M}). \end{align} $$

So, from (2.27) and (2.30) together, we find that

(2.31) $$ \begin{align} \underline{d}(\mathcal{E})> 0. \end{align} $$

Now, recall that every $\{ -1, 1\}$ -perfect number is a nondeficient number, as observed by Ross [Reference Ross12] (and recall that this also follows from Lemma 2.1). So, we find that the containment $\mathcal {P} \subseteq \mathcal {A}$ holds. From this containment together with the containment in (2.28), we find: for all $x \geq 1$ , the families $\{ n \leq x : n \in \mathcal {P} \}$ and $\{ n \leq x : n \in \mathcal {E} \}$ are disjoint and are both contained in the family $\{ n \leq x : n \in \mathcal {A} \}$ . Consequently, we obtain the relation

(2.32) $$ \begin{align} \frac{\left| \{ n \leq x : n \in \mathcal{P} \} \right|}{x} + \frac{\left| \{ n \leq x : n \in \mathcal{E} \} \right|}{x} \leq \frac{\left| \{ n \leq x : n \in \mathcal{A} \} \right|}{x}. \end{align} $$

From (2.32), we find that

(2.33) $$ \begin{align} \overline{d}(\mathcal{P}) + \underline{d}(\mathcal{E}) \leq d(\mathcal{A}). \end{align} $$

From (2.31) and (2.33) together, we have, as a consequence, that $\overline {d}(\mathcal {P}) < d(\mathcal {A})$ , as desired.

Proof Let a denote an odd abundant square. Let p be a prime such that $p> \sigma (a)$ . We then write n in place of $ap$ . Since $\sigma (a)> 2a$ (from the assumption that a is strictly abundant), we find that $p> \sigma (a) > 2a > a$ , so that $p \nmid a$ . Moreover, since $a \geq 1$ , we have that $p> 2$ , so that n is odd. Since $p \nmid a$ , the exponent of p in the prime factorization of $n = a p$ is $1$ ; in particular, n is not a square.

Now, since $(p, a) = 1$ and since $\sigma $ is multiplicative, we find that $\sigma (n) = \sigma (a) (p+1)$ . As a consequence, we have that

(3.1) $$ \begin{align} \sigma(n) - 2 n = (\sigma(a) - 2 a) p + \sigma(a). \end{align} $$

It follows that n is abundant.

Since a is an odd square, we write

(3.2) $$ \begin{align} a = \prod_{i=1}^{\omega(a)} p_{a_i}^{2c_i}, \end{align} $$

with $a_i> 1$ and $c_i \geq 1$ for all $i \in \{ 1, 2, \ldots , \omega (a) \}$ . From (2.3) and (3.2) together, we find that

(3.3) $$ \begin{align} \sigma(a) = \prod_{i=1}^{\omega(a)} \big( 1 + p_{a_i} + p_{a_i}^{2} + \cdots + p_{a_{i}}^{2 c_{i}} \big), \end{align} $$

with each factor on the right of (3.3) being odd, that is, so that $\sigma (a)$ is odd. So, the integer $\Delta := \sigma (a) - 2 a$ is odd. Also, with the assumption that a is abundant, we find that $\Delta> 0$ . So, there exists an integer $q \geq 0$ satisfying $\Delta = 2 q + 1$ . As a consequence of (3.1), we obtain that $ \sigma (n) - 2 n = \Delta p + \sigma (a)$ , and we define $ H(n) := \frac {\sigma (n) - 2 n}{2}. $ From this definition, we find that

$$ \begin{align*} H(n) & = \frac{\Delta p + \sigma(a)}{2} \\ & = q p + \frac{p + \sigma(a)}{2}. \end{align*} $$

Now, by way of contradiction, suppose that $n \in \mathcal {P}$ . So, by Lemma 2.1, the integer $H(n)$ can be written as a sum of proper (distinct) nontrivial divisors of n. Now, let $\mathscr {S}$ denote a sum of nontrivial and proper divisors of n. Since $n = ap$ and since a and p are relatively prime, each divisor of n is either a divisor d of a or is of the form $dp$ , again for a divisor d of a. This gives us that $\mathscr {S} = A p + B$ for some integer $A \geq 0$ and an integer B, which satisfies the bounds whereby $0 \leq B \leq \sigma (a)$ . From the condition $\sigma (a) < p$ , we find that

(3.4) $$ \begin{align} q p + \sigma(a) < q p + \frac{p + \sigma(a)}{2} = H(n). \end{align} $$

In a similar fashion, we have that

(3.5) $$ \begin{align} H(n) = q p + \frac{p + \sigma(a)}{2} < q p + p. \end{align} $$

If $A \leq q$ , then $Ap + B \leq q p + \sigma (a) < H(n)$ , from (3.4). If $A \geq q + 1$ , then $A p + B \geq (q+1) p> H(n)$ , from (3.5). So, no divisor sum of the form $\mathscr {S}$ can be equal to $H(n)$ , contradicting (via Lemma 2.1) our assumption that $n \in \mathcal {P}$ .

So, we have shown that n is odd, nonsquare, and abundant, with $n \not \in \mathcal {P}$ .