# Nested Subsets of Superabundant Numbers

Agustín Besteiro

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Submitted on 14 Oct 2025

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# Nested Subsets of Superabundant Numbers

Agustín Besteiro¹,²

¹Centro de Matemática Aplicada, Instituto de Tecnologías Emergentes, Universidad Nacional de San Martín, Buenos Aires, Argentina. (abesteiro@unsam.edu.ar)

²Universidad Abierta Interamericana, Buenos Aires, Argentina.

October 14, 2025

## Abstract

We introduce a step-by-step selection of integers depending on a denominator placed between the logarithmic mean and a logarithmic term. At the critical choice (the logarithmic-mean boundary) the procedure recovers exactly the extremely abundant numbers. For larger denominators, it yields nested infinite supersets within the superabundant numbers. For subcritical choices (strictly below a fixed fraction of the logarithmic mean), it stops after a finite number of steps. Under a convex interpolation, the resulting families are infinite, decrease along the parameter, and their intersection is the extremely abundant numbers.

## 1 Introduction

Let $\sigma(n) = \sum_{d|n} d$ and define

$$F(n) := \frac{\sigma(n)}{n}, \quad f(n) := \frac{\sigma(n)}{n \log \log n}.$$

An integer $n \ge 1$ is superabundant (SA) if $F(n) > F(m)$ for all $m < n$ (see [AE44, AF09]). Let $n_0 = 10080$ and call $n \ge n_0$ extremely abundant (XA) if $f(n) > f(m)$ for all $n_0 \le m < n$; in other words, XA are the record-holders of $f$ from $n_0$ onward (see, e.g., [NY14, Naz13]). It is well known that $\limsup_{n \to \infty} f(n) = e^\gamma$ (Grönwall [Gro13]), and that the Riemann Hypothesis is equivalent to Robin's inequality $\sigma(n) < e^{\gamma n \log \log n}$ for $n > 5040$ (see [Rob84, RNR97, Bro17]). These facts place $f(n)$ at the core of the interface between multiplicative structure and analytic bounds: the SA condition captures global records of $F$, while XA captures records of the normalized quantity $f$ that naturally appears in Grönwall's theorem and Robin's criterion. Beyond SA and XA, Nazardonyavi introduced two auxiliary families, denoted $X'$ and $X''$, in their study of the fragility of RH (see [NY14]). These two families sit strictly between SA and XA:

$$SA \supseteq X' \supseteq X'' \supseteq XA.$$

In particular, $X'$ and $X''$ are infinite sets.

In this note we introduce a simple one-parameter selection rule that produces nested subsets of SA by comparing the growth of $F$ against an explicit monotone threshold depending on a denominator $D(A, X)$, with $A = \log a$ and $X = \log n$. The rule interpolates between the logarithmic mean $L_m(A, X)$ and $X$. At the critical choice $D = L_m$ the selection reduces to comparing $F(n)/\log \log n$ against $F(a)/\log \log a$, so the procedure recovers exactly the record-holders of $f$ and therefore enumerates XA. For any monotone denominator with $L_m \le D \le X$, each output of the rule is a new record of $F$ and the resulting set lies inside SA while still containing XA. When $D$ runs along the convex interpolation $D_\theta = (1-\theta)L_m + \theta X$ with $\theta \in (0, 1]$,
the associated families form a decreasing nested chain as $\theta \to 0$, every family in the chain is infinite, and their intersection is precisely XA. If, on the other hand, D is pushed strictly below $L_m$ by a fixed factor, the rule cannot keep producing new records indefinitely and the construction terminates after finitely many steps; we refer to this finite-termination phenomenon as subcritical shutdown.

The proofs rely on elementary ingredients. A calculus identity for the logarithmic mean shows that the step threshold is strictly increasing in $n$, which guarantees that every selected value is a record of $F$. To prove infinitude of the supercritical families we only need that the series of reciprocals of the primes diverges. To prove subcritical shutdown we translate the acceptance condition into a multiplicative gain for $f(n)$ and compare the cumulative effect along the sequence with Grönwall's limsup; if the process did not stop, one would contradict the $e^\gamma$ barrier.

Our goal is thus twofold. First, we place XA at a critical boundary inside a flexible selection framework that sits naturally between $L_m$ and $X$, clarifying why XA is the exact output at the logarithmic-mean choice and why any supercritical relaxation ($D \ge L_m$) still stays within SA. Second, we show that pushing below the boundary forces finite termination, which pinpoints the sharpness of the logarithmic-mean threshold. No assumption on the Riemann Hypothesis is required: all supercritical families are infinite unconditionally, and the critical family coincides with XA, whose infinitude is itself equivalent to RH via the classical results quoted above.

**Notation.** We write $A := \log a$, $X := \log n$, and $L := X - A = \log(n/a)$. The logarithmic mean of $A < X$ is

$$L_m(A, X) := \frac{X - A}{\log X - \log A} = \frac{L}{\log \log n - \log \log a}. \quad (1)$$

We will use $\gamma$ for the Euler-Mascheroni constant.

## Preliminary results

**Theorem 1.1** (Grönwall [Gro13]). We have $\limsup_{n \to \infty} \frac{\sigma(n)}{n \log \log n} = e^\gamma$.

**Theorem 1.2** (Robin's criterion [Rob84]). The Riemann Hypothesis is equivalent to $\sigma(n) < e^\gamma n \log \log n$ for all $n > 5040$.

**Lemma 1.3** (Mertens' product [Mer74]). As $y \to \infty$,

$$\lim_{y \to \infty} \frac{\prod_{p \le y} \left(1 - \frac{1}{p}\right)^{-1}}{e^{\gamma} \log y} = 1.$$

Consequently,

$$\lim_{y \to \infty} \frac{\prod_{p \le y} \left(1 + \frac{1}{p}\right)}{(e^{\gamma}/\zeta(2)) \log y} = 1.$$

**Remark 1.4** (On RH and XA). Combining Theorems 1.1 and 1.2 yields the dichotomy for XA: under RH there are infinitely many XA, while if RH fails there are only finitely many XA (see [NY14]).
## 2 Greedy families via admissible denominators

Recall the classical greedy algorithms from optimization—procedures that make a locally optimal choice at each step without lookahead—we adopt the following first-fit rule.

Fix a base $a \ge n_0$ and consider candidates $n > a$. Given a denominator $D = D(A, X)$, define the step threshold

$$ T_D(a, n) := 1 + \frac{L}{D(A, X) \log \log a} \quad (A = \log a, X = \log n, L = X - A). \quad (2) $$

Starting from $n_1 := n_0$, we generate inductively a sequence $(n_k)$ by the greedy rule

$$ n_{k+1} := \min \left\{ n > n_k : \frac{F(n)}{F(n_k)} > T_D(n_k, n) \right\}. \quad (3) $$

We denote by $X^{(D)}$ the set of values produced by (3).

**Definition 2.1 (Admissible denominators).** We call $D$ admissible if for each fixed base $a$ the map $n \mapsto T_D(a, n)$ is strictly increasing on $n > a$.

The examples we use are:

* $D(A, X) = X$ (this set is denoted $X'$);

* $D(A, X) = \frac{A+X}{2}$ (denoted $X''$);

* The convex family $D_{\theta}(A, X) = (1 - \theta)L_m(A, X) + \theta X$, $\theta \in [0, 1]$, which interpolates between $L_m$ and $X$.

All these choices are admissible; see Lemma 3.1.

### 2.1 Three benchmark cases: $X'$, $X''$, and $XA$

When $D = X$ we recover $X'$; when $D = \frac{A+X}{2}$ we recover $X''$. The critical case $D = L_m$ is special because of the identity (1):

$$ T_{L_m}(a, n) = 1 + \frac{L}{L_m \log \log a} = 1 + \frac{\log \log n - \log \log a}{\log \log a} = \frac{\log \log n}{\log \log a}. \quad (4) $$

Thus the greedy rule reads

$$ \frac{F(n)}{F(a)} > \frac{\log \log n}{\log \log a} \iff \frac{F(n)}{\log \log n} > \frac{F(a)}{\log \log a}, $$

i.e. the successor is the next record of $f(n) = F(n)/\log \log n$. Therefore, the $D = L_m$ process enumerates exactly the $XA$ numbers:

$$ X^{(L_m)} = XA. \quad (5) $$

## 3 Main results

We now analyze the structural properties of these sets. Proofs are given in Section 4.

**Lemma 3.1 (Monotonicity of the threshold).** For each fixed base $a$, the function $n \mapsto T_D(a, n)$ is strictly increasing in the following cases:

(a) $D(A, X) = X$;
$$ (b) \ D(A, X) = \frac{A + X}{2}; $$

$$ (c) \ D_{\theta}(A, X) = (1 - \theta)L_m(A, X) + \theta X \text{ with any } \theta \in [0, 1]. $$

In particular all three choices are admissible.

**Theorem 3.2.** Let $D$ be admissible. Then every element produced by the greedy rule (3) is a record of $F(n)$; hence $X^{(D)} \subseteq \text{SA}$. In particular, $X'$, $X'' \subseteq \text{SA}$.

**Theorem 3.3.** Let $D$ be such that for every $a$ and every $n > a$ satisfy $L_m \le D \le X$. Then $\text{XA} \subseteq X^{(D)}$. Moreover, for $D = L_m$ we have equality (5).

**Theorem 3.4** (Nested chains tending to XA). Fix a decreasing sequence $\theta_0 > \theta_1 > \theta_2 > \cdots \downarrow 0$ and set $D_j := D_{\theta_j} = (1 - \theta_j)L_m + \theta_j X$. Then the associated sets are nested

$$ X^{(\theta_{j+1})} \subseteq X^{(\theta_j)} \quad (j \ge 0), $$

all of them are infinite, and

$$ \bigcap_{j \ge 0} X^{(\theta_j)} = \text{XA}. $$

**Theorem 3.5** (Subcritical shutdown). Fix $\varepsilon \in (0, 1)$ and suppose $D \le (1 - \varepsilon)L_m$ pointwise. Then the greedy construction (3) terminates after finitely many steps.

**Remark 3.6.** The critical boundary $D = L_m$ is sharp: $D \ge L_m$ yields infinite nested families containing XA (Theorem 3.4), while $D \le (1 - \varepsilon)L_m$ forces finiteness (Theorem 3.5). At $D = L_m$ itself we recover precisely XA.

# 4 Proofs

## Proof of Lemma 3.1

For fixed $a$ the threshold is $T_D(a, n) = 1 + h(A, X)/\log \log a$ with $h(A, X) := L/D(A, X)$. Thus we must show $X \mapsto h(A, X)$ is increasing.

(a) $D(A, X) = X$. Then $h(A, X) = (X - A)/X = 1 - A/X$ is strictly increasing in $X$.

(b) $D(A, X) = \frac{A+X}{2}$. Then $h(A, X) = \frac{2(X - A)}{A + X}$ has derivative $\frac{2(A + X) - 2(X - A)}{(A + X)^2} = \frac{4A}{(A + X)^2} > 0$.

(c) $D_{\theta} = (1 - \theta)L_m + \theta X$ with $\theta \in [0, 1]$. Write $g(X) := L_m(A, X) = \frac{X - A}{\log X - \log A}$. Then

$$ h(A, X) = \frac{X - A}{(1 - \theta)g(X) + \theta X}. $$

Differentiating and simplifying,

$$ \frac{\partial h}{\partial X} = \frac{(1 - \theta)(g(X) - Lg'(X)) + \theta A}{((1 - \theta)g(X) + \theta X)^2}. $$

A direct computation shows $g(X) - Lg'(X) = \frac{L^2}{X(\log X - \log A)^2} > 0$, hence the numerator is positive (note also $A > 0$). Therefore $\partial h/\partial X > 0$ and $T_{D_{\theta}}(a, n)$ is strictly increasing in $n$. $\square$
### Proof of Theorem 3.2

Fix $k$ and let $a = n_k$. By admissibility, $m \mapsto T_D(a, m)$ is strictly increasing on $m > a$. By minimality of $n_{k+1}$ in (3), for any $m$ with $a < m < n_{k+1}$ we have

$$ \frac{F(m)}{F(a)} \le T_D(a, m) < T_D(a, n_{k+1}) < \frac{F(n_{k+1})}{F(a)}. $$

If $m \le a$ then $F(m) \le F(a) < F(n_{k+1})$. Hence $F(n_{k+1}) > F(m)$ for all $m \le n_{k+1}$, i.e. $n_{k+1}$ is a record of $F$ and therefore lies in SA. Induction on $k$ yields $X^{(D)} \subseteq$ SA. $\square$

### Proof of Theorem 3.3

Let $x$ be XA and let $a$ be the last selected value strictly less than $x$ in the greedy process (any $a \ge n_0$ works). Since $x$ is a record of $f$, we have

$$ \frac{F(x)}{F(a)} > \frac{\log \log x}{\log \log a}. \qquad (6) $$

If $L_m \le D \le X$, then by (1)

$$ 1 + \frac{L}{D \log \log a} \le 1 + \frac{L}{L_m \log \log a} = \frac{\log \log x}{\log \log a}. $$

Combining with (6) gives $F(x)/F(a) > T_D(a, x)$. Let $n_1$ be the minimal $n > a$ with $F(n)/F(a) > T_D(a, n)$; then $n_1 \le x$ because $x$ already satisfies the inequality. If $n_1 < x$, reset the base to $a_1 := n_1$ and repeat the same step: $F(x)/F(a_1) > T_D(a_1, x)$, so the next minimal choice $n_2$ obeys $a_1 < n_2 \le x$. Iterating, we obtain a strictly increasing chain

$$ a < n_1 < n_2 < \dots \le x $$

which, being a chain of integers bounded above by $x$, must reach $x$ after finitely many steps. Therefore $x \in X^{(D)}$. For $D = L_m$, (4) shows the successor is exactly the next record of $f$, hence $X^{(L_m)} =$ XA. $\square$

### Proof of Theorem 3.4

Fix $a$ and $n > a$. If $\theta' < \theta$ then $D_{\theta'} \le D_{\theta}$, hence

$$ T_{D_{\theta'}}(a, n) = 1 + \frac{L}{D_{\theta'} \log \log a} \ge 1 + \frac{L}{D_{\theta} \log \log a} = T_{D_{\theta}}(a, n). $$

Thus the acceptance threshold increases as $\theta \to 0$, and minimality in (3) implies

$$ X^{(\theta_{j+1})} \subseteq X^{(\theta_j)}. $$

*Infinitude.* For $\theta \in (0, 1]$ we have $D_{\theta} \ge \theta X$, hence for base $a$ and all $n > a$,

$$ T_{D_{\theta}}(a, n) \le 1 + \frac{L}{\theta X \log \log a} \le 1 + \frac{1}{\theta \log \log a}. $$

By Grönwall, for any $c < e^{\gamma}$ there exist infinitely many integers $n$ with $f(n) = \sigma(n)/(n \log \log n) \ge c$. Choose an $n > a$ large enough such that

$$ \frac{c \log \log n}{f(a) \log \log a} > 1 + \frac{1}{\theta \log \log a}. $$
Then

$$ \frac{F(n)}{F(a)} = \frac{f(n) \log \log n}{f(a) \log \log a} \geq \frac{c \log \log n}{f(a) \log \log a} > 1 + \frac{1}{\theta \log \log a} \geq T_{D_{\theta}}(a, n). $$

Replacing the base by $a' := n$ and iterating yields an infinite sequence (note that $1/(\theta \log \log a')$ decreases as the base grows). Hence $X^{(\theta)}$ is infinite for every fixed $\theta > 0$.

Intersection equals XA. The inclusion $XA \subseteq X^{(\theta)}$ holds for every $\theta$ by Theorem 3.3; hence $XA \subseteq \bigcap_j X^{(\theta_j)}$. For the reverse inclusion, suppose $x \notin XA$ and let $a$ be the last record of $f$ before $x$ (so $f(x) \le f(a)$). By (4), $F(x)/F(a) \le \log \log x / \log \log a = T_{L_m}(a, x)$. Since $T_{D_{\theta}}(a, x) \to T_{L_m}(a, x)$ as $\theta \to 0$, there exists $j_0$ such that $F(x)/F(a) \le T_{D_{\theta_j}}(a, x)$ for all $j \ge j_0$. Then $x \notin X^{(\theta_j)}$ for $j \ge j_0$, which contradicts $x$ lying in the intersection. Thus $\bigcap_j X^{(\theta_j)} = XA$. $\square$

## Proof of Theorem 3.5

Let $(n_k)$ be generated by (3) with base $a \ge n_0$ and write $A_k := \log \log n_k$, $r_k := A_{k+1}/A_k \ge 1$. Under $D \le (1 - \varepsilon)L_m$ the acceptance condition reads

$$ \frac{F(n_{k+1})}{F(n_k)} > 1 + \frac{1}{1-\varepsilon} \left( \frac{A_{k+1}}{A_k} - 1 \right). \qquad (7) $$

Since $F(n) = f(n) \log \log n$, (7) is equivalent to

$$ \frac{f(n_{k+1})}{f(n_k)} > \frac{1}{r_k} + \frac{1}{1-\varepsilon} \left(1 - \frac{1}{r_k}\right) =: h(r_k), \quad r_k \ge 1. $$

Since

$$ h(r) = \frac{1}{r} + \frac{1}{1-\varepsilon} \left(1 - \frac{1}{r}\right) = 1 + \frac{\varepsilon}{1-\varepsilon} \cdot \frac{r-1}{r}, $$

we have, for $r \in [1, 2]$,

$$ h(r) \ge 1 + \frac{\varepsilon}{2(1-\varepsilon)}(r-1). $$

Therefore, for all sufficiently large $k$ (so that $r_k - 1 \le 1$),

$$ \log \frac{f(n_{k+1})}{f(n_k)} > \log\left(1 + \frac{\varepsilon}{2(1-\varepsilon)}(r_k - 1)\right) \ge \frac{\varepsilon}{4(1-\varepsilon)}(r_k - 1), $$

where the last step uses $\log(1 + u) \ge u/2$ for $0 \le u \le 1$. Hence

$$ \log f(n_{k+1}) - \log f(n_k) \ge \frac{\varepsilon}{4(1-\varepsilon)}(r_k - 1). $$

Assume, for contradiction, that the sequence $(n_k)$ is infinite. Since $(n_k)$ is strictly increasing then $A_k = \log \log n_k \to \infty$ and

$$ \sum_{k=K_0}^{K-1} \log r_k = \log \frac{A_K}{A_{K_0}} \xrightarrow{K \to \infty} \infty. $$

Since $\log r \le r - 1$ for $r \ge 1$, it follows that $\sum_{k \ge K_0} (r_k - 1) = \infty$. Summing the inequality

$$ \log f(n_{k+1}) - \log f(n_k) \ge c(r_k - 1) \quad (c > 0) $$

from $k = K_0$ to $K - 1$ gives

$$ \log f(n_K) - \log f(n_{K_0}) \ge c \sum_{k=K_0}^{K-1} (r_k - 1) \xrightarrow{K \to \infty} \infty, $$

hence $f(n_K) \to \infty$. This contradicts Grönwall's bound $\limsup_{n \to \infty} f(n) = e^{\gamma} < \infty$. Therefore the process cannot run indefinitely and terminates after finitely many steps. $\square$
# Acknowledgments

The author is supported by Universidad Abierta Interamericana, CONICET and Universidad Nacional de San Martín.

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