NLP

    Natural Language Processing
PoS

    Part of Speech
PLM

    Pretrained Language Model

# PatternRank: Leveraging Pretrained Language Models and Part of Speech for
Unsupervised Keyphrase Extraction

Tim Schopf, Simon Klimek, and Florian Matthes
Department of Computer Science, Technical University of Munich,
Boltzmannstrasse 3, Garching, Germany
{tim.schopf, simon.klimek<EMAIL_ADDRESS>

###### Abstract

Keyphrase extraction is the process of automatically selecting a small set of
most relevant phrases from a given text. Supervised keyphrase extraction
approaches need large amounts of labeled training data and perform poorly
outside the domain of the training data [Bennani-Smires et al., 2018]. In this
paper, we present PatternRank, which leverages pretrained language models and
part-of-speech for unsupervised keyphrase extraction from single documents.
Our experiments show PatternRank achieves higher precision, recall and
$\text{F}_{1}\text{-scores}$ than previous state-of-the-art approaches. In
addition, we present the
KeyphraseVectorizers111https://github.com/TimSchopf/KeyphraseVectorizers
package, which allows easy modification of part-of-speech patterns for
candidate keyphrase selection, and hence adaptation of our approach to any
domain.

##  1 INTRODUCTION

To quickly get an overview of the content of a text, we can use keyphrases
that concisely reflect its semantic context. Keyphrases describe the most
essential aspect of a text. Unlike simple keywords, keyphrases do not consist
solely of single words, but of several compound words. Therefore, keyphrases
provide more information about the content of a text compared to simple
keywords. Supervised keyphrase extraction approaches usually achieve higher
accuracy than unsupervised ones [Kim et al., 2012, Caragea et al., 2014, Meng
et al., 2017]. However, supervised approaches require manually labeled
training data, which often causes subjectivity issues as well as significant
investment of time and money [Papagiannopoulou and Tsoumakas, 2019]. In
contrast, unsupervised keyphrase extraction approaches do not have these
issues and are moreover mostly domain-independent.

Keyphrases and their vector representations are very versatile and can be used
in a variety of different Natural Language Processing (NLP) downstream tasks
[Braun et al., 2021, Schopf et al., 2022]. For example, they can be used as
features or input for document clustering and classification [Hulth and
Megyesi, 2006, Schopf et al., 2021], they can support extractive summarization
[Zhang et al., 2004], or they can be used for query expansion [Song et al.,
2006]. Keyphrase extraction is particularly relevant for the scholarly domain
as it helps to recommend articles, highlight missing citations to authors,
identify potential reviewers for submissions, analyze research trends over
time, and can be used in many different search scenarios [Augenstein et al.,
2017].

In this paper, we present PatternRank, an unsupervised approach for keyphrase
extraction based on Pretrained Language Models and Part of Speech (PoS). Since
keyphrase extraction is especially important for the scholarly domain, we
evaluate PatternRank on a specific dataset from this area. Our approach does
not rely on labeled data and therefore can be easily adapted to a variety of
different domains. Moreover, PatternRank does not require the input document
to be part of a larger corpus, allowing the keyphrase extraction to be applied
to individual short texts such as publication abstracts. Figure 1 illustrates
the general keyphrase extraction approach of PatternRank.

Figure 1: PatternRank approach for unsupervised keyphrase extraction. A single
text document is used as input for an initial filtering step where candidate
keyphrases are selected which match a defined PoS pattern. Subsequently, the
candidate keyphrases are ranked by a PLM based on their semantic similarity to
the input text document. Finally, the top-$N$ keyphrases are extracted as a
concise reflection of the input text document.

##  2 RELATED WORK

Most popular unsupervised keyphrase extraction approaches can be characterized
as either statistics-based, graph-based, or embedding-based methods, while Tf-
Idf is a common baseline used for evaluation [Papagiannopoulou and Tsoumakas,
2019].

YAKE uses a set of different statistical metrics including word casing, word
position, word frequency, and more to extract keyphrases from text [Campos et
al., 2020]. TextRank uses PoS filters to extract noun phrase candidates that
are added to a graph as nodes, while adding an edge between nodes if the words
co-occur within a defined window [Mihalcea and Tarau, 2004]. Finally, PageRank
[Page et al., 1999] is applied to extract keyphrases. SingleRank expands the
TextRank approach by adding weights to edges based on word co-occurrences [Wan
and Xiao, 2008]. RAKE generates a word co-occurrence graph and assigns scores
based on word frequency, word degree, or the ratio of degree and frequency for
keyphrase extraction [Rose et al., 2010]. Furthermore, Knowledge Graphs can be
used to incorporate semantics for keyphrase extraction [Shi et al., 2017].
EmbedRank leverages Doc2Vec [Le and Mikolov, 2014] and Sent2Vec [Pagliardini
et al., 2018] sentence embeddings to rank candidate keyphrases for extraction
[Bennani-Smires et al., 2018]. More recently, a PLM-based approach was
introduced that uses BERT [Devlin et al., 2019] for self-labeling of
keyphrases and subsequent use of the generated labels in an LSTM classifier
[Sharma and Li, 2019].

##  3 KEYPHRASE EXTRACTION APPROACH

Figure 1 illustrates the general keyphrase extraction process of our
PatternRank approach. The input consists of a single text document which is
being word tokenized. The word tokens are then tagged with PoS tags. Tokens
whose tags match a previously defined PoS pattern are selected as candidate
keyphrases. Then, the candidate keyphrases are fed into a PLM to rank them
based on their similarity to the input text document. The PLM embeds the
entire text document as well as all candidate keywords as semantic vector
representations. Subsequently, the cosine similarities between the document
representation and the candidate keyphrase representations are computed and
the candidate keyphrases are ranked in descending order based on the computed
similarity scores. Finally, the top-$N$ ranked keyphrases, which are most
representative of the input document, are extracted.

###  3.1 Candidate Selection with Part of Speech

In previous work, simple noun phrases consisting of zero or more adjectives
followed by one or more nouns were used for keyphrase extraction [Mihalcea and
Tarau, 2004, Wan and Xiao, 2008, Bennani-Smires et al., 2018]. However, we
define a more complex PoS pattern to extract candidate keyphrases from the
input text document. In our approach, the tags of the word tokens have to
match the following PoS pattern in order for the tokens to be considered as
candidate keyphrases:

$\displaystyle\Big{(}\big{(}\\{.*\\}\\{HYPH\\}\\{.*\\}\big{)}\\{NOUN\\}*\Big{)}\Big{|}$
(1)
$\displaystyle\Big{(}\big{(}\\{VBG\\}|\\{VBN\\}\big{)}?\\{ADJ\\}*\\{NOUN\\}+\Big{)}$

The PoS pattern quantifiers correspond to the regular expression syntax.
Therefore, we can translate the PoS pattern as arbitrary parts-of-speech
separated by a hyphen, followed by zero or more nouns OR zero or one verb
(gerund or present or past participle), followed by zero or more adjectives,
followed by one or more nouns.

###  3.2 Candidate Ranking with Pretrained Language Models

Earlier work used graphs [Mihalcea and Tarau, 2004, Wan and Xiao, 2008] or
paragraph and sentence embeddings [Bennani-Smires et al., 2018] to rank
candidate keyphrases. However, we leverage PLMs based on current transformer
architectures to rank the candidate keyphrases that have recently demonstrated
promising results [Grootendorst, 2020]. Therefore, we follow the general
EmbedRank [Bennani-Smires et al., 2018] approach for ranking, but use PLMs
instead of Doc2Vec [Le and Mikolov, 2014] and Sent2Vec [Pagliardini et al.,
2018] to create semantic vector representations of the entire text document as
well as all candidate keyphrases. In our experiments, we use SBERT [Reimers
and Gurevych, 2019] PLMs since they have been shown to produce state of the
art text representations for semantic similarity tasks. Using these semantic
vector representations, we rank the candidate keyphrases based on their cosine
similarity to the input text document.

##  4 EXPERIMENTS

In this section, we compare four different approaches for unsupervised
keyphrase extraction in the scholarly domain.

###  4.1 Data

In our experiments, we use the Inspec dataset [Hulth, 2003], which consists of
2,000 English computer science abstracts collected from scientific journal
articles between 1998 and 2002. Each abstract has assigned two different types
of keyphrases. First, controlled and manually assigned keyphrases that appear
in the thesaurus of the Inspec dataset but do not necessarily have to appear
in the abstract. Second, uncontrolled keyphrases that are freely assigned by
professional indexers and are not restricted to either the thesaurus or the
abstract. In our experiments, we consider the union of both types of
keyphrases as the ground truth.

###  4.2 Evaluation

For evaluation, we compare the performances of four different keyphrase
extraction approaches.

YAKE is a fast and lightweight approach for unsupervised keyphrase extraction
from single documents based on statistical features [Campos et al., 2020].

SingleRank applies a ranking algorithm to word co-occurrence graphs for
unsupervised keyphrase extraction from single documents [Wan and Xiao, 2008].

KeyBERT uses, similar to PatternRank, a PLM to rank candidate keyphrases
[Grootendorst, 2020]. However, KeyBERT uses simple word n-grams as candidate
keyphrases rather than word tokens that match a certain PoS pattern, as in our
PatternRank approach. For the KeyBERT experiments, we use the all-mpnet-
base-v2111https://huggingface.co/sentence-transformers/all-mpnet-base-v2 SBERT
model for candidate keyphrase ranking and an n-gram range of $[1,3]$ for
candidate keyphrase selection. This means that n-grams consisting of 1, 2 or 3
words are selected as candidate keyphrases.

PatternRank To select candidate keyphrases, we developed the
KeyphraseVectorizers222https://github.com/TimSchopf/KeyphraseVectorizers
package, which allows custom PoS patterns to be defined and returns matching
candidate keyphrases. We evaluate two different versions of the PatternRank
approach. $\text{PatternRank}_{NP}$ selects simple noun phrases as candidate
keyphrases and $\text{PatternRank}_{PoS}$ selects word tokens whose PoS tags
match the pattern defined in section 3.1. In both cases, the all-mpnet-base-v2
SBERT model is used for candidate keyphrase ranking.

We evaluate the models based on exact match, partial match, and the average of
exact and partial match. For each approach, we report Precision@N, Recall@N,
and $\text{F}_{1}\text{@N}$ scores, using the top-N extracted keyphrases
respectively. The gold keyphrases always remain the entire set of all manually
assigned keyphrases, regardless of N. Additionally, we lowercase the gold
keyphrases as well as the extracted keyphrases and remove duplicates. We
follow the approach of Rousseau and Vazirgiannis (2015) and calculate
Precision@N, Recall@N, and $\text{F}_{1}\text{@N}$ scores per document and
then use the macro-average at the collection level for evaluation. The exact
match approach yields true positives only for extracted keyphrases that have
an exact string match to one of the gold keyphrases. However, this evaluation
approach penalizes keyphrase extraction methods which predict keyphrases that
are syntactically different from the gold keyphrases but semantically similar
[Rousseau and Vazirgiannis, 2015, Wang et al., 2015]. The partial match
approach converts gold keyphrases as well as extracted keyphrases to unigrams
and yields true positives if the extracted unigram keyphrases have a string
match to one of the unigram gold keyphrases [Rousseau and Vazirgiannis, 2015].
The drawback of the partial match evaluation approach, however, is that it
rewards methods which predict keyphrases that occur in the unigram gold
keyphrases but are not appropriate for the corresponding document
[Papagiannopoulou and Tsoumakas, 2019]. For empirical comparison of keyphrase
extraction approaches, we therefore also report the average of the exact and
partial matching results.

| Method |  | @5 |  |  | @10 |  |  | @20 |
---|---|---|---|---|---|---|---|---|---|---
| P | R | $\text{F}_{1}$ | P | R | $\text{F}_{1}$ | P | R | $\text{F}_{1}$
Exact Match | YAKE | 26.16 | 11.71 | 15.37 | 20.88 | 18.45 | 18.50 | 16.45 | 27.78 | 19.65
SingleRank | 38.11 | 16.55 | 21.97 | 33.29 | 27.27 | 28.55 | 27.24 | 38.84 | 30.80
KeyBERT | 12.97 | 6.08 | 7.82 | 11.42 | 10.53 | 10.30 | 9.75 | 17.14 | 11.76
$\text{PatternRank}_{NP}$ | 41.15 | 18.09 | 23.92 | 34.60 | 28.33 | 29.66 | 25.88 | 36.69 | 29.19
$\text{PatternRank}_{PoS}$ | 41.76 | 18.44 | 24.35 | 36.10 | 29.63 | 30.99 | 27.80 | 39.42 | 31.37
Partial Match | YAKE | 77.45 | 19.49 | 29.91 | 68.20 | 33.46 | 42.67 | 59.69 | 45.58 | 48.69
SingleRank | 75.54 | 19.36 | 29.56 | 68.63 | 33.98 | 43.24 | 58.82 | 53.68 | 53.68
KeyBERT | 77.48 | 20.06 | 30.55 | 65.78 | 32.90 | 41.67 | 57.11 | 45.37 | 48.34
$\text{PatternRank}_{NP}$ | 83.64 | 21.93 | 33.29 | 75.27 | 37.62 | 47.69 | 62.78 | 56.69 | 57.03
$\text{PatternRank}_{PoS}$ | 82.49 | 21.61 | 32.79 | 74.79 | 37.50 | 47.48 | 63.21 | 57.66 | 57.71
Avg. Match | YAKE | 51.81 | 15.60 | 22.64 | 44.54 | 25.96 | 30.59 | 38.07 | 36.68 | 34.13
SingleRank | 56.83 | 17.96 | 25.77 | 50.96 | 30.63 | 35.90 | 43.03 | 46.26 | 42.24
KeyBERT | 45.23 | 13.07 | 19.19 | 38.60 | 21.72 | 25.99 | 33.43 | 31.23 | 30.05
$\text{PatternRank}_{NP}$ | 62.40 | 20.01 | 28.61 | 54.94 | 32.98 | 38.68 | 44.33 | 46.69 | 43.11
$\text{PatternRank}_{PoS}$ | 62.13 | 20.03 | 28.57 | 55.45 | 33.57 | 39.24 | 45.51 | 48.54 | 44.54
Table 1: Evaluation of our approach against state of the art using the Inspec
dataset. Precision (P), Recall (R), and $\text{F}_{1}$-score ($\text{F}_{1}$)
for N = 5, 10, and 20 are reported. The evaluation results are based on exact
match, partial match, and the average of exact and partial match. Two
variations of PatternRank are presented: $\text{PatternRank}_{NP}$ selects
simple noun phrases as candidate keyphrases and $\text{PatternRank}_{PoS}$
selects word tokens whose PoS tags match the pattern defined in section 3.1.
In both cases, a SBERT PLM is used for candidate keyphrase ranking.

The results of our evaluation are shown in Table 1. We can see that our
PatternRank approach outperforms all other approaches across all benchmarks.
In general, both approaches $\text{PatternRank}_{NP}$ and
$\text{PatternRank}_{PoS}$ perform fairly similarly, whereas
$\text{PatternRank}_{PoS}$ produces slightly better results in most cases. In
the exact match evaluation, $\text{PatternRank}_{PoS}$ consistently achieves
the best results of all approaches. Furthermore, $\text{PatternRank}_{PoS}$
also yields best results in the average mach evaluation for N = 10 and 20. In
the partial match evaluation, the $\text{PatternRank}_{NP}$ approach
marginally outperforms the $\text{PatternRank}_{PoS}$ approach and yields best
results for N = 5 and 10. However, as we mentioned earlier the partial match
evaluation approach, may wrongly reward methods which extract keyphrases that
occur in the unigram gold keyphrases but are not appropriate for the
corresponding document. Since the $\text{PatternRank}_{PoS}$ approach
outperforms the $\text{PatternRank}_{NP}$ approach in the more important exact
match and average match evaluations, we argue that selecting candidate
keyphrases based on the PoS pattern defined in Section 3.1 instead of simple
noun phrases helps to extract keyphrases predominantly occurring in the
scholarly domain. In contrast, skipping the PoS pattern-based candidate
keyphrase selection step results in a significant performance decline. KeyBERT
uses the same PLM to rank the candidate keyphrases as PatternRank, but uses
simple n-grams for candidate keyphrase selection instead of PoS patterns or
noun phrases. As a result, the KeyBERT approach consistently performs worst
among all approaches. As expected, YAKE was the fastest keyphrase extraction
approach because it is a lightweight method based on statistical features.
However, the extracted keyphrases are not very accurate and in comparison to
PatternRank, YAKE significantly performs worse in all evaluations. SingleRank
is the only approach that achieves competitive results compared to
PatternRank. Nevertheless, it consistently performs a few percentage points
worse than PatternRank across all evaluations. We therefore conclude that our
PatternRank achieves state-of-the-art keyphrase extraction results, especially
in the scholarly domain.

##  5 CONCLUSION

We presented the PatternRank approach which leverages PLMs and PoS for
unsupervised keyphrase extraction. We evaluated our approach against three
different keyphrase extraction methods: one statistics-based approach, one
graph-based approach and one PLM-based approach. The results show that the
PatternRank approach performs best in terms of precision, recall and
$\text{F}_{1}\text{-score}$ across all evaluations. Furthermore, we evaluated
two different PatternRank versions. $\text{PatternRank}_{NP}$ selects simple
noun phrases as candidate keyphrases and $\text{PatternRank}_{PoS}$ selects
word tokens whose PoS tags match the pattern defined in Section 3.1. While
$\text{PatternRank}_{PoS}$ produced better results in the majority of cases,
$\text{PatternRank}_{NP}$ still performed very well in all benchmarks. We
therefore conclude that the $\text{PatternRank}_{PoS}$ approach works
particularly well in the evaluated scholarly domain. Furthermore, since the
use of noun phrases as candidate keyphrases is a more general and domain-
independent approach, we propose using $\text{PatternRank}_{NP}$ as a simple
but effective keyphrase extraction method for arbitrary domains. Future work
may investigate how the PLM and PoS pattern used in this approach can be
adapted to different domains or languages.

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  *[NLP]: Natural Language Processing
  *[PoS]: Part of Speech
  *[PLM]: Pretrained Language Model
  *[PLMs]: Pretrained Language Model


# The sum of the reciprocals of the prime divisors of an odd perfect or odd
primitive non-deficient number

Joshua Zelinsky111Hopkins School, New Haven, CT, USA<EMAIL_ADDRESS>
###### Abstract

Write $T(n)$ as the sum of the reciprocals of the primes which divide $n$.
Write $H(n)=\prod_{p\mid n}p/(p-1)$ where the product is over the prime
divisors of $n$. We prove new bounds for $T(n)$ and $H(n)$ in terms of the
smallest prime factor of $n$, under the assumption that $n$ is an odd perfect
number. Some of the results also apply under the weaker assumption that $n$ is
odd and primitive non-deficient.

##  1 Introduction

Let $\sigma(n)$ be the sum of the positive divisors of $n$. For example,
$\sigma(4)=1+2+4=7$. A number is said to be perfect if $\sigma(n)=2n$. For
example, $\sigma(6)=1+2+3+6=12=2(6)$.

Perfect numbers form sequence A000396 in the OEIS. An old unsolved problem is
whether there is any odd integer $n$ which is perfect.

Given a positive integer $n$, we will write $T(n)$ as the sum of the
reciprocals of the primes which divide $n$. For example, $T(12)=1/2+1/3=5/6$.
A closely related function is

$H(n)=\prod_{p\mid n}\frac{p}{p-1}$

where the product is over the prime divisors of $n$. The main focus of this
paper is to prove new bounds for $T(n)$ and $H(n)$ in terms of the smallest
prime factor of $n$, under the assumption that $n$ is an odd perfect number.
Some of the results also apply to all odd non-deficient numbers.

The function $T(n)$ is closely related to the arithmetic derivative of $n$.
The arithmetic derivative is a function $D(n)$ defined by analogy to the usual
derivative from calculus. The function $D(n)$ is the unique function on the
natural numbers with the properties that $D(p)=1$ for any prime $p$ and for
any integers $m$ and $n$, one has $D(mn)=D(m)n+mD(n)$. When $n$ is squarefree,
$nT(n)=D(n)$ is the arithmetic derivative of $n$. The function $nT(n)$ is
always an integer and is sequence A069359 in the OEIS.

We will write $h(n)=\sigma(n)/n$. Note that we also have that

$h(n)=\sum_{d\mid n}\frac{1}{d}.$

In the literature, $h(n)$ often referred as the abundancy index of $n$. A
number is said to be abundant when $h(n)>2$, and said to be deficient when
$h(n)<2$. For many purposes, it is useful to group perfect and abundant
numbers together as non-deficient numbers. It is not hard to show that

$h(ab)\geq h(a),$ (1)

with equality if and only if $b=1$. Thus, if $n$ is perfect or abundant, then
$mn$ is abundant for any $m>1$.

Inequality (1) motivates the following definition. A primitive non-deficient
number is a number $n$ which is non-deficient and which has no proper divisor
$d$ of $n$ where $d$ is non-deficient. These numbers form sequence A006039 in
the OEIS.

All perfect numbers are primitive non-deficient, and thus theorems about odd
primitive non-deficient numbers are also theorems about odd perfect numbers.
Importantly, while we do not know if there are any odd perfect numbers, there
are many odd primitive non-deficient numbers, with $945$ being the smallest
example.

Inequality (1) also leads to the observation that no perfect number can be
divisible by any other perfect or abundant number. This observation is
frequently used to restrict what the prime divisors of an odd perfect number
can look like. For example, it is not hard to show that if $n$ is an odd
perfect number then $105\nmid n$, and similarly that $(3)(5)(11)(13)\nmid n$.

One of the basic properties of $h(n)$ is that $h(n)\leq H(n)$ with equality if
and only if $n=1$. In fact, $H(n)$ is also the best possible function which
bounds $h(n)$ from above and which only keeps track of which primes divide $n$
but not their multiplicity. In particular,

$\lim_{k\rightarrow\infty}h(n^{k})=H(n).$

When $n$ is perfect, one has $h(n)=2$, and thus $H(n)>2$. When $n$ is perfect,
$H(n)$ is a natural function to estimate.

Prior work on estimating $T(n)$ and $H(n)$ when $n$ is an odd perfect goes
back to a 1958 paper of Peri Sastri [15] who showed that if $n$ is an odd
perfect number, then

$\frac{1}{2}<T(n)<2\log\frac{\pi}{2}.$

Subsequently, tighter bounds were proven in a series of papers by Cohen,
Hagis, and Suryanarayana [4, 5, 18, 19], which included the result that
$T(n)<\log 2$.

We summarize their results in the Proposition below.

######  Proposition 1.

(Cohen, Hagis, and Suryanarayana) If $n$ is an odd perfect number, then $T(n)$
is bounded as given in the table below, broken down by the greatest common
divisor of $n$ and $15$ being as listed.

$\gcd(15,n)$ |  Lower bound for $T(n)$ |  Upper bound for $T(n)$
---|---|---
$1$ | $0.667450$ | $0.693148$
$3$ | $0.603831$ | $0.657304$
$5$ | $0.647387$ | $0.678036$
$15$ | $0.596063$ | $0.673770$

Note that the $0.693148$ in the case where $\gcd(15,n)=1$ is due to the
aforementioned bound that $T(n)<\log 2$.

Similarly, bounds on $H(n)$ were proved by Suryanarayana [18]. These bounds
can also broken down into the same major cases.

######  Proposition 2.

(Suryanarayana) If $n$ is an odd perfect number, then $H(n)$ is bounded
according to the greatest common divisor of $n$ and 15 as follows.

$\gcd(15,n)$ | $H(n)<$
---|---
$1$ | $2.014754$
$3$ | $2.096234$
$5$ | $2.031002$
$15$ | $2.165439$

The upper bounds in the table in Proposition 2 are rounded up values of
numbers all of which rational multiples of $\zeta(3)$ where $\zeta(s)$ is the
Riemann zeta function.

Throughout the rest of this paper, unless stated otherwise, we will assume
when we have a number $n$ that $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{k}^{a_{k}}$ and the $p_{i}$ are primes where $p_{1}<\cdots<p_{k}$.

When $n$ is non-deficient, we will write $S(n)$ to be what we call the surplus
of $n$, defined by $S(n)=H(n)-2$. The surplus played a major role in related
prior work by the author[23]. Using this notation, Lemma 6 from that paper
becomes:

######  Lemma 3.

(Zelinsky) Let $n$ be a primitive non-deficient number with
$n=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}$ with $p_{1}<\ldots<p_{k}$. Let
$R=\mathrm{rad}(n)=p_{1}p_{2}\cdots p_{k}$. Assume further that

$S(n)\geq 2+\alpha$

for some positive $\alpha<1$ Then there exists an $i$, $1\leq i\leq k$ such
that

$p_{i}^{a_{i}+1}<\max\\{\frac{2\left(k+2+p_{1}\right)}{\alpha},k(k+1)\\}.$

The main results of this paper are the following two propositions.

######  Proposition 4.

Assume $n$ is a primitive non-deficient number with prime factorization given
by $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ with primes $p_{1}$,
$p_{2}$, $\ldots p_{k}$ and $p_{1}<\ldots<p_{k}$, and surplus $S=S(n)$. Then
$T(n)\geq\log 2-\frac{25}{64p_{1}}+\frac{S}{2}-\frac{S^{2}}{4}$.

We also have the following upper bound for $T(n)$.

######  Proposition 5.

Assume $n$ a primitive non-deficient number. with
$n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$, and $p_{1}<\cdots<p_{k}$.
Assume that the $a_{i}$ are even for all $i$ except some $t$ where $a_{t}$ is
odd. Assume further that $\frac{p_{1}+2}{2}\geq p_{1}\geq 11$. Then

$T(n)<\log 2-\frac{11}{50p_{1}^{2}}.$

Proposition 5 will give a bound for $T(n)$ for $n$ when $n$ is odd perfect
numbers since any odd perfect number $n$ will satisfy the hypotheses of the
proposition. We will also prove slightly stronger but more complicated bounds.

If $n$ is an odd perfect number, then other types of bounds relating $p_{1}$
to $n$ also exist. The most basic and classic result in this regard is due to
Servais [17] who proved the bound for odd perfect numbers, but the same
argument applies to any non-deficient number. Servais proved:

######  Proposition 6.

(Servais) Assume that $n$ is a non-deficient number with
$n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ where the $p_{i}$ are
primes with $p_{1}<\cdots<p_{k}$. Then $p_{1}\leq k$.

###### Proof.

Assume that $n$ is non-deficient. Thus,

$2\leq h(n)<H(n)=\prod_{i=1}^{k}\frac{p_{i}}{p_{i}-1}$

Since $\frac{x}{x-1}$ is a decreasing function for positive $x$ and the $i$th
prime after $p_{1}$ is at least $p_{1}+i$, we must have

$2<\left(\frac{p_{1}}{p_{1}-1}\right)\left(\frac{p_{1}+1}{p_{1}}\right)\left(\frac{p_{1}+2}{p_{1}+1}\right)\cdots\left(\frac{p_{1}+k-1}{p+k-2}\right)=\frac{p+k-1}{p-1}.$

Thus

$2p_{1}-2<k-1,$

and so $k>p_{1}-1$, and so $k\geq p_{1}$. ∎

Grün [9] observed that one can tighten this bound by taking into account that
every prime other than 2 is odd. This yields a tighter but still linear bound.

Norton [14] used the prime number theorem to give a tighter bound for $k$ in
terms of $p_{1}$ as well as bounds on the size of $n$ in terms of $p_{1}$.
These were slightly tightened by the author [20]. The bounds of Norton and the
author are asymptotically best possible when restricted to non-deficient
numbers, but there is likely much room for improvement if one is restricted to
odd perfect numbers.

Similar bounds relating $k$ to $p_{i}$ for small $i$ also exist. Kishore [10]
proved bounds similar to Grün relating small factor primes to $k$. Kishore’s
bounds are linear bounds on $k$ in terms of the size of the first few smallest
prime factors. The author [21] then adopted Norton’s argument for small prime
factors in a way that again is essentially asymptotically best possible for
non-deficient numbers.

In a different direction, bounds relating $n$ and its large prime factors also
exist. Acquaah and Konyagin [1] showed that if $n$ is an odd perfect number
then

$p_{k}^{3}<3n.$ (2)

The proof for the bound of Acquaah and Konyagin carefully uses that $n$ is
perfect, and therefore the proof does not apply to all primitive non-deficient
numbers. One might wonder if Acquaah and Konyagin’s bound can be extended to
apply to all primitive non-deficient numbers. The answer is no. Consider for
example $n=9765$ which is an odd primitive non-deficient number. The largest
prime factor of $9765$ is 31, and $(3(9765))^{\frac{1}{3}}=30.826\cdots$. It
is not hard to show that if $n$ is primitive non-deficient number then one
must have $p_{k}^{2}<2n$. This weaker bound is essentially best possible.
However, empirically, the vast majority of odd primitive deficient numbers
satisfy Inequality (2). Acquaah and Konyagin’s bound is notable also for being
strong enough to rule out some “spoof” perfect numbers, especially that of
Descartes’s example.

Descartes noted that $D=198585576189$ looks almost like an odd perfect number.
In particular, one may factor it as $D=3^{2}7^{2}11^{2}13^{2}22021$. One has
then that

$\sigma(D)=(3^{2}+3+1)(7^{2}+7+1)(11^{2}+11+1)(13^{2}+13+1)(22021+1)=2D$

where we ignore that 22021 is in fact not a prime number. Examples similar to
Descartes’s example were extensively studied in [3] which greatly generalized
this notion as follows. Given an integer $n$, define a _factorization_ of $n$
to be an expression of the form

$n=\prod_{i=1}^{k}x_{i}^{b_{i}},$

where the $x_{i}$ are integers and the $b_{i}$ are positive integers. Notice
that a factorization can also be thought of as a multiset of ordered pairs of
the form $(x_{i},b_{i})$.

Define a function $\tilde{\sigma}$ on the multiset of such ordered pairs as
follows.

$\tilde{\sigma}\Big{(}\\{(x_{i},b_{i}):1\leq i\leq
k\\}\Big{)}=\prod_{i=1}^{k}\left(\sum_{j=0}^{b_{i}}x_{i}^{j}\right).$

A factorization as above is _spoof perfect_ if
$\tilde{\sigma}(\prod_{i=1}^{k}x_{i}^{b_{i}})=2\prod_{i=1}^{k}x_{i}^{b_{i}}$.
If $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ with the $p_{i}$
distinct primes, then

$\tilde{\sigma}({(p_{1},a_{1}),\cdots(p_{k},a_{k})}=\sigma(n).$

Thus $\tilde{\sigma}$ provides a generalization of the classical $\sigma$
function.

By the above remark, any actual perfect number $n$ gives rise to a spoof
perfect factorization.

In this context, Descartes number $D$ gives rise to a spoof factorization
given by the ordered pairs $\\{(3,2),(7,2),(11,2),(13,2),(22021,1)\\}$. The
number $D$ with this factorization is particularly notable for being the only
known example where all the $x_{i}$ are positive and odd. However, many other
examples exist where one is allowed to have negative values for $x_{i}$.

The existence of Descartes’s number, along with similar examples, place heavy
restrictions on the effectiveness of different sorts of arguments about odd
perfect numbers; any restrictions which apply to spoofs cannot show that no
odd perfect numbers cannot exist without also showing that spoofs do not
exist. Many results about odd perfect numbers do not avoid this issue.
However, Acquaah and Konyagin’s bound, which carefully uses that the largest
prime factor is actually a prime, manages to be tighter than the best possible
bound that applies to spoofs. In particular,
$(3(198585576189))^{\frac{1}{3}}=8414.396\ldots<22021$.

Bounds similar to that of Inequality (2) have been proven for the the second
and third largest prime factors of an odd perfect number $n$. In particular,
the author [22] proved that

$p_{k-1}<(2N)^{1/5},$ (3)

and that

$p_{k}p_{k-1}<6^{1/4}N^{1/2}.$ (4)

Note that the bound in Equation (3), unlike Inequality (2) is not so strong as
to avoid the Descartes example. In particular,
$(2(198585576189))^{\frac{1}{5}}=208.831\cdots>13.$

Bibby, Vyncke, and the author [2] proved a similar bound for $p_{k-3}$ as well
as a bound on $p_{k}p_{k-1}p_{k-2}$. That paper also includes a weak bound of
a similar sort on $p_{k-i}$ for any $i$.

A major role in most work on this topic is played by Euler’s theorem on odd
perfect numbers.

######  Lemma 7.

(Euler) If $n$ is an odd perfect number, then $n=q^{e}m^{2}$ where $q$ is
prime, $\gcd(q,m)=1$, and where $q\equiv e\equiv 1\pmod{4}.$

When we have an odd perfect number $n$ in the above form, we will refer to $q$
as the special prime. Note that Euler’s theorem is in modern terms an easy
exercise in modular arithmetic, but its ease reflects that it is a very weak
statement. The same restriction applies just as well to any odd $n$ where one
only assumes that $\sigma(n)\equiv 2\pmod{4}$.

##  2 Proofs of the main bounds

Before we prove our main results, we need a few technical lemmata.

######  Lemma 8.

If $0\leq x\leq\frac{1}{11}$, then

$1+x+x^{2}\geq e^{x+\frac{2x^{2}}{5}},$

with equality if and only if $x=0$.

###### Proof.

We note that this is true for $x=0$, and so we may assume that $x\neq 0$.
Proving this Lemma is the same as showing that in the desired range that

$e^{x+\frac{2x^{2}}{5}}-(1+x+x^{2})\leq 0.$

Expanding this quantity using the Taylor series for $e^{x}$ and we obtain

$\begin{split}&e^{x+\frac{x^{2}}{3}}-(1+x+x^{2})=\\\
&\frac{-1}{10}x^{2}+\frac{2}{5}x^{3}+\frac{2}{25}x^{4}+\frac{(x+\frac{2x^{2}}{5})^{3}}{6}+\frac{(x+\frac{2x^{2}}{5)^{4}}}{24}+\frac{(x+\frac{2x^{2}}{5})^{5}}{120}+\cdots.\end{split}$
(5)

Thus, the result can be proved if we prove that if $x\leq\frac{1}{11}$, then

$\frac{2}{5}x+\frac{2}{25}x^{2}+\frac{(x+\frac{2x^{2}}{5})^{3}}{6x^{2}}+\frac{(x+\frac{2x^{2}}{5})^{4}}{24x^{2}}+\frac{(x+\frac{2x^{2}}{5})^{5}}{120x^{2}}+\cdots<\frac{1}{10}..$
(6)

We will break up the right-hand side of Equation (6) and estimate its first
three terms, and then estimate the remaining terms.

If $x\leq\frac{1}{11}$, then

$\frac{2}{5}x+\frac{2}{25}x^{2}+\frac{(x+\frac{2x^{2}}{5})^{3}}{6x^{2}}=\frac{2}{5}x+\frac{2}{25}x^{2}+\frac{(1+\frac{2x}{5})^{2}(x+\frac{2x^{2}}{5})}{6}<0.053891.$

We need to also estimate

$\frac{(x+\frac{2x^{2}}{5})^{4}}{24x^{2}}+\frac{(x+\frac{2x^{2}}{5})^{5}}{120x^{2}}+\cdots.$

We have:

$\begin{split}\frac{(x+\frac{2x^{2}}{5})^{4}}{24x^{2}}+\frac{(x+\frac{2x^{2}}{5})^{5}}{120x^{2}}+\cdots&\leq\frac{(1+\frac{2x}{5})^{2}(x+\frac{2x^{2}}{5})^{2}}{24}\sum_{i=0}^{\infty}\left(\frac{(x+\frac{2x^{2}}{5})}{5}\right)^{i}\\\
&\leq\frac{1}{20}\sum_{i=0}^{\infty}\left(\frac{(\frac{1}{11}+\frac{2(\frac{1}{11})^{2}}{5})}{5}\right)^{i}\\\
&\leq 0.0459663.\end{split}$ (7)

Now, note that $0.053891+0.0459663=0.0998573<\frac{1}{10}$ which proves the
result. ∎

Note that Lemma 8 is nearly identical to a similar Lemma in Cohen [4] but the
conclusion is tighter, though at the cost of being true for a smaller range of
values of $x$.

######  Lemma 9.

Let $q$ be a prime with $q\geq 3$ . Let $M(q)=\sum_{p\geq q}\frac{1}{p^{2}}$
where $p$ ranges over all primes at least $q$. Then $M(q)<\frac{1}{q\log
q}\left(1+\frac{2\log\log q}{\log q}\right).$

###### Proof.

Let $q$ be a prime which is at least 19. Assume that $M(q)=\sum_{p\geq
q}\frac{1}{p^{2}}$. From Lemma 12 and Lemma 13, we have

$\begin{split}M(q)\leq\sum_{n\geq\pi(q)}\frac{1}{(n\log
n)^{2}}&\leq\int_{\frac{q}{\log q}}^{\infty}\frac{1}{(x\log x)^{2}}dx\\\
&<\int_{\frac{q}{\log q}}^{\infty}\frac{2+\log x}{(x\log
x)^{3}}dx=\frac{1}{\frac{q}{\log q}\log\frac{q}{\log q}}.\end{split}$ (8)

It is now straightforward to verify that

$\frac{1}{\frac{q}{\log q}\log\frac{q}{\log q}}<\frac{1}{q\log
q}\left(1+\frac{2\log\log q}{\log q}\right),$

and so the result is proven for $q\geq 19$. Using the inequality for $q\geq
19$, one can verify directly that the same inequality holds for $q=3$, $5$,
$7,$ $11$, and $13$. ∎

Note that Lemma 9 is asymptotically best possible. However, we suspect that if
$q$ is sufficiently large that one in fact has $M(q)\leq\frac{1}{q\log q}$.

######  Lemma 10.

Let $q$ be an odd prime. Set $E=\prod_{p\geq q}\frac{p^{2}}{p^{2}-1}$. Then

$\log E\leq\frac{1}{q\log q}\left(1+\frac{2\log\log q}{\log
q}\right)+\frac{4}{q^{3}}.$

###### Proof.

We have

$\log E=\sum_{p\geq q}\log\left(1+\frac{1}{p^{2}-1}\right)\leq\sum_{p\geq
q}\frac{1}{p^{2}-1}=\sum_{p\geq q}\frac{1}{p^{2}}+\frac{1}{p^{2}(p^{2}-1)}.$

We now note that $\sum_{p\geq q}\frac{1}{p^{2}(p^{2}-1)}\leq\frac{4}{q^{3}}$,
and combine that with Lemma 9. ∎

######  Lemma 11.

If $x$ is an integer and $x\geq 2$, then
$\log(1+\frac{1}{x-1})\leq\frac{1}{x}+\frac{25}{32}{x^{2}}$.

###### Proof.

One may verify directly that this is satisfied for $x=2$, $x=3$ or $x=4$. So
we may assume that $x\geq 5$. We have from the Taylor series for $\log(1+t)$,
that if $0\leq t<1$ that

$\log(1+t)\leq t-\frac{t^{2}}{2}+\frac{t^{3}}{3}.$ (9)

We can substitute $t=\frac{1}{x-1}$ into Inequality (9) to obtain:

$\log(1+\frac{1}{x-1})\leq\frac{1}{x-1}-\frac{1}{2(x-1)^{2}}+\frac{1}{3(x-1)^{3}}.$
(10)

We have then

$\frac{1}{x-1}-\frac{1}{2(x-1)^{2}}+\frac{1}{3(x-1)^{3}}=\frac{1}{x}+\frac{1}{x(x-1)}-\frac{1}{2(x-1)^{2}}+\frac{1}{3(x-1)^{3}}.$
(11)

Thus, the result can be proved if we show that

$\frac{1}{x(x-1)}-\frac{1}{2(x-1)^{2}}+\frac{1}{3(x-1)^{3}}\leq\frac{25}{32x^{2}}.$

We have that

$\frac{1}{x(x-1)}-\frac{1}{2(x-1)^{2}}+\frac{1}{3(x-1)^{3}}=\frac{3x^{2}-7x+6}{6x(x-1)^{3}}$

Since $x\geq\frac{3}{2}$, we have that

$\frac{3x^{2}-7x+6}{6x(x-1)^{3}}\leq\frac{3x^{2}-3x}{6x(x-1)^{3}}=\frac{1}{2(x-1)^{2}}.$

Since $x\geq 5$, $x-1\geq\frac{4}{5}x$, and so

$\frac{1}{2(x-1)^{2}}\leq\frac{25}{32x^{2}}.$

∎

######  Lemma 12.

If $p\geq 19$ is a prime number, then $\pi(p)\geq\frac{p}{\log p}+1$.

###### Proof.

We have from Rosser and Schoenfeld’s classic bounds on the prime number
theorem [16], that if $x$ is a real number and $x\geq 59$ then

$\pi(x)\geq\frac{x}{\log x}\left(1+\frac{1}{2\log x}\right).$

This implies that when $p$ is prime and $p\geq 59$, then
$\pi(p)\geq\frac{p}{\log p}+1$. We need to then only check that the inequality
is true for the primes which are between 19 and 53. ∎

We also need the following bound from Rosser and Schoenfeld [16].

######  Lemma 13.

If $n\geq 1$, then $P_{n}\geq n\log n$, where $P_{n}$ is the $n$th prime
number.

We now prove Proposition 4.

###### Proof.

Assume that $n$ is a non-deficient number with factorization as above. Then

$2+S=H(n)=\prod_{i=1}^{k}\frac{p_{i}}{p_{i}-1}.$ (12)

We take the logarithm of both sides of Equation (12) to obtain that

$\log(2+S)<\sum_{i=1}^{k}\log\frac{p_{i}}{p_{i}-1}=\sum_{i=1}^{k}\log\left(1+\frac{1}{p_{i}-1}\right)$
(13)

We then apply Lemma 11 to obtain that

$\begin{split}\log(2+S)&<\sum_{i=1}^{k}\frac{1}{p_{i}}+\frac{25}{32p_{i}^{2}}=\sum_{i=1}^{k}\frac{1}{p_{i}}+\frac{25}{32}\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}\\\
&\leq\sum_{i=1}^{k}\frac{1}{p_{i}}+\frac{25}{32}\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}<\sum_{i=1}^{k}\frac{1}{p_{i}}+\frac{25}{64p_{1}}.\end{split}$
(14)

In the last step of the above inequality, we are using that the primes are odd
to get an upper bound on $\sum_{i=1}^{\infty}\frac{1}{p_{i}^{2}}$.

Since $S<1$ we have $\log(2+S)\geq\log 2+\frac{S}{2}-\frac{S^{2}}{4}$, which
implies the desired result. ∎

Proposition 4 helps illustrate more explicitly why prior results cannot
progress beyond $\log 2$ as their general upper bound.

Similarly, we may use Lemma 9 to obtain:

######  Proposition 14.

Assume $n$ is a primitive non-deficient number with prime factorization given
by $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ with primes $p_{1}$,
$p_{2}$, $\cdots p_{k}$ and $p_{1}<\ldots<p_{k}$, and surplus $S=S(n)$. Then

$T(n)\geq\log 2-\frac{25}{32p_{1}\log p_{1}}\left(1+\frac{2\log\log
p_{1}}{\log p_{1}}\right)+\frac{S}{2}-\frac{S^{2}}{4}.$

We now prove Proposition 5.

###### Proof.

The proof given here is essentially a tightening of the proof given in Cohen
[4] to show that $T(n)<\log 2$. There are three changes. First, we use Lemma 8
rather than Cohen’s Lemma. Second, we keep more careful track of the second
order terms. Third, we extend the result to primitive non-deficient numbers of
a specific form which includes odd perfect numbers, rather than just prove the
result for odd perfect numbers.

Assume as given. Then we have:

$2=\frac{\sigma(n)}{n}=\prod_{1\leq i\leq
k}(1+\frac{1}{p_{i}}+\cdots+\frac{1}{p_{i}^{a_{i}}}\geq(1+\frac{1}{p_{t}})\prod_{\begin{subarray}{c}1\leq
i\leq k\\\ \i\neq t\end{subarray}}(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}).$
(15)

Since every $p_{i}$ is assumed to be at least 11, we may apply Lemma 8 to
Equation (15) to obtain that

$2\geq(1+\frac{1}{p_{t}})\prod_{\begin{subarray}{c}1\leq i\leq k\\\ i\neq
t\end{subarray}}e^{\frac{1}{p_{i}}+\frac{2}{5p_{i}^{2}}},$ (16)

and hence

$\log 2\geq\log\left(1+\frac{1}{p_{t}}\right)+\sum_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq t\end{subarray}}\frac{1}{p_{i}}+\frac{2}{5p_{i}^{2}}.$ (17)

Since $\log(1+x)>x-\frac{x^{2}}{2}$, we obtain from Equation (17) that

$\log
2>\sum_{i=1}^{k}\frac{1}{p_{i}}+\frac{2}{5}\sum_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq t\end{subarray}}\frac{1}{p_{i}^{2}}-\frac{1}{2p_{t}^{2}}.$
(18)

As we have $2p_{1}-1\leq p_{t}$, and $p_{1}\geq 3$, we have that
$\frac{5}{3}p_{1}\leq p_{1}$. Thus,
$\frac{9}{50}\frac{1}{p_{1}^{2}}\geq\frac{1}{2p_{t}^{2}}$, which combines with
Equation (18) to obtain the desired bound. ∎

We obtain as a corollary:

######  Corollary 15.

If $n$ is an odd perfect number with smallest prime factor $p_{1}\geq 11$,
then $T(n)<\log 2-\frac{11}{50p_{1}^{2}}.$

###### Proof.

To prove this we just need to show that any odd perfect number satisfies the
properties of Proposition 5. But this is essentially just Euler’s theorem for
odd perfect numbers, along with the fact that if $p_{t}$ is the special prime
of $n$, then

$p_{t}+1\mid\sigma(p_{t}^{a_{t}}),$

and hence $p_{t}+1\mid 2n$. Thus, $n$ must be divisible any odd prime factor
of $p_{t}+1$, and hence $p_{1}\leq\frac{p_{t}+1}{2}$. ∎

The main usage of Proposition 5 was in the case that our $n$ was an odd
perfect number, and those likely do not exist. Thus, it is worth asking if
there are any numbers which actually satisfy the hypotheses of Proposition 5.
In fact, such numbers do exist. An example is
$n=(3^{2})(5)(11^{2})(13^{2})(17^{2})(19^{2})(23^{2})$. We strongly suspect
that there are infinitely many such numbers, but do not see how to prove that.

An easy consequence of Proposition 15 and Proposition 14 is the following
upper bound on $S(n)$.

######  Corollary 16.

Assume $n$ is an odd perfect number with $p_{1}\geq 11$. Then
$S(n)<\frac{25}{8p_{1}\log p_{1}}$.

###### Proof.

Assume as given. From Proposition 15 and Proposition 14 we have

$\frac{25}{16p_{1}\log
p_{1}}\geq\frac{S(n)}{2}-\frac{S(n)^{2}}{4}+\frac{11}{50p_{1}^{2}}.$ (19)

From earlier mentioned results on $S(n)$, we have $S(n)<\frac{1}{5}$, and so
$\frac{S(n)^{2}}{4}\leq\frac{S(n)}{20}.$ Thus, we have from Equation (19) that

$\frac{25}{16p_{1}\log p_{1}}\geq\frac{9}{10}S(n),$

which implies that

$S(n)\leq\frac{125}{72p_{1}\log p_{1}}.$ (20)

Equation (20) implies that

$\frac{S(n)^{2}}{4}\leq\left(\frac{125}{72p_{1}\log
p_{1}}\right)^{2}<\frac{11}{50p_{1}^{2}}.$ (21)

The last step in the chain of inequalities in Equation (21) is using that
$p\geq 11$ and then estimating $\log 11$.

Equation (21) combines with Equation (19) to get that

$\frac{25}{16p_{1}\log p_{1}}\geq\frac{S(n)}{2},$

which implies the claimed bound. ∎

We can tighten this somewhat using the following method. The idea of the
following Lemma is implicitly used in Suryanarayana’s proof [18] but we are
making the general form explicit here.

######  Lemma 17.

Let $n$ be a number of the form $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{k}^{a_{k}}$ where $p_{1},p_{2},\cdots p_{k}$ are primes with $3\leq
p_{1}<\cdots<p_{k}$, and where all $a_{i}$ are even for $1\leq i\leq k$ except
for a single $a_{j}$ which is odd. Then

$\frac{H(n)}{h(n)}\leq\frac{p_{j}^{2}}{p_{j}^{2}-1}\prod_{1\leq i\leq k,i\neq
j}\frac{p^{3}}{p^{3}-1}.$

See Suryanarayana [18] for the proof.

We are now in a position where we can progress on our tighter estimate for
$H(n)$.

######  Lemma 18.

Let $n$ be a number of the form $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{k}^{a_{k}}$ where $p_{1},p_{2},\ldots p_{k}$ are primes with $3\leq
p_{1}<\ldots<p_{k}$, and where all $a_{i}$ are even for $1\leq i\leq k$ except
for a single $a_{j}$ which is odd. Then

$H(n)\leq h(n)\left(1+\frac{3}{4p_{1}^{2}}\right).$

###### Proof.

Assume as given. We have via logic similar to Lemma 17, that

$H(n)\leq
h(n)\left(\frac{p_{j}^{2}}{p_{j}^{2}-1}\right)\prod_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq j\end{subarray}}\frac{p^{3}}{p^{3}-1},$ (22)

which is equivalent to

$\log H(n)\leq\log
h(n)+\log\left(\frac{p_{j}^{2}}{p_{j}^{2}-1}\right)+\sum_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq j\end{subarray}}\log\frac{p^{3}}{p^{3}-1}.$ (23)

Since $x\geq\log(1+x)$, we have from Equation (23)

$\log H(n)\leq\log h(n)+\frac{1}{p_{j}^{2}-1}+\sum_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq j\end{subarray}}\log\frac{1}{p^{3}-1}.$ (24)

Since $x\geq\log(1+x)$, Equation (24) implies that

$\log H(n)\leq\log h(n)+\frac{1}{p_{j}^{2}}+\sum_{\begin{subarray}{c}1\leq
i\leq k\\\ i\neq j\end{subarray}}\frac{1}{p_{i}^{3}}.$ (25)

It is not hard to see that

$\frac{1}{p_{j}^{2}}+\sum_{1\leq i\leq k,i\neq
j}\frac{1}{p_{i}^{3}}\leq\frac{3}{4p_{1}^{2}},$

which combines with 25 to imply the desired result. The remainder of the proof
involves an estimate essentially identical in form to that made in the proof
of Proposition 5. ∎

As a corollary we obtain:

######  Corollary 19.

If $n$ is an odd perfect number, then

$H(n)\leq 2+\frac{9}{4p_{1}^{2}}.$

Note that Corollary 19 is better than the bounds from Suryanarayana when
$p\geq 13$. The analogous bound for Proposition 19 is satisfied by Descartes
number. However, this is likely not satisfied by all of the most general forms
of spoofs in [3]. In particular, since one of the forms of spoofing allowed
there is to have repeated copies of a prime factor be treated as distinct
primes. But the proof for Proposition 19 uses bounds on series involving the
the sums of primes and estimating those sums.

An argument similar to the proof of Corollary 19 can be used to show the
following:

######  Proposition 20.

There is a constant $C$ such that if $n$ is a primitive non-deficient number
with smallest prime factor $p_{1}$, then

$H(n)\leq 2+\frac{C}{p_{1}}.$

##  3 Related questions

The following seems like a basic question but as far as we are aware is not in
the literature. Let $X$ be the set of real numbers defined as follows: The set
$X$ consists of those real numbers $x$, such that there exist infinitely many
odd primitive non-deficient $n$ with $H(n)\leq x$. What is the supremum of
$X$?

A major difficulty in improving the bounds for an odd perfect number for both
$S(n)$ and $T(n)$ is the behavior of the special prime. In particular, there
is difficulty in the case where special prime is raised to just the first
power. There are a variety of modulo restrictions regarding when prime $q$ can
be a special prime. We mentioned earlier that one must have $(3)(5)(7)\nmid n$
for an odd perfect number $n$, and thus one cannot have any special prime $q$
where $q\equiv-1\pmod{105}$. But other similar restrictions are known.

We also have the following slightly more subtle folklore result which as far
as the author is aware is not explicitly in the literature.

######  Proposition 21.

If $q\equiv-1\pmod{165}$, then $q$ is not the special prime for any odd
perfect number $q$.

Proposition 21 has a curious nature. If one has not seen it before, one might
try to prove that no odd perfect number can be divisible by 165 from which the
result would then follow immediately. However, any attempt to prove that 165
cannot divide $n$ using current techniques fails. The key to proving
Proposition 21 is to instead prove the following, weaker statement.

######  Lemma 22.

Let $n$ be an odd perfect number. If $165\mid n$, then one must have
$5\mid\mid n$.

###### Proof.

Assume that $n$ is an odd perfect number with $165\mid n$. If $3^{2}\mid n$,
then $\sigma(3^{2})\mid n$, which would contradict the fact that one cannot
have $(3)(5)(11)(13)\mid n$. Thus, we must have $3^{4}\mid n$. Now, $5^{2}\mid
n$, then one must have $3^{4}5^{2}11^{2}\mid n$, and $3^{4}5^{2}11^{2}$ is
abundant. Thus we must have $5\mid\mid n$ which proves the desired result. ∎

Proposition 21 follows from Lemma 22 since an odd perfect number must have
exactly one special prime factor $q$. If $q$ is the special prime for $n$, and
$q\equiv-1\pmod{165}$, then $5$ is not the special prime but $165\mid n$. This
contradicts Lemma 22.

A natural question in this context is if a prime $q$ can be ruled out as being
the special prime where the contribution from $q$ itself is used in the method
above. A slightly weaker property is for there to exist a prime $q$ such that
$q\equiv 1\pmod{4}$, where $H(\frac{q+1}{2})<2$ and $H(q\frac{q+1}{2})>2$. We
will show that even this weaker property cannot occur. We will need a Lemma
first.

######  Lemma 23.

Suppose that $p_{1},p_{2},\ldots p_{m}$ are distinct odd primes, where

$\frac{p_{1}}{p_{1}-1}\cdots\frac{p_{m}}{p_{m}-1}\leq 2,$

and $q$ is an odd prime such that

$\left(\frac{p_{1}}{p_{1}-1}\cdots\frac{p_{m}}{p_{m}-1}\right)\frac{q}{q-1}>2.$

Then $q\leq 2(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1)-1$.

###### Proof.

Assume as given. Since $\frac{p_{1}}{p_{1}-1}\cdots\frac{p_{m}}{p_{m}-1}$
reduces to some fraction with an odd numerator, this product cannot be equal
to 2. Furthermore, the denominator of this fraction is at most
$(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1)$. Thus we have

$\frac{p_{1}}{p_{1}-1}\cdots\frac{p_{m}}{p_{m}-1}\leq
2-\frac{1}{(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1)}.$ (26)

By similar logic, we have

$\frac{p_{1}}{p_{1}-1}\cdots\frac{p_{m}}{p_{m}-1}\frac{q}{q-1}\geq
2+\frac{1}{(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1)(q-1)}.$ (27)

Combining Equation (27) and Equation (26), we obtain that

$\left(2-\frac{1}{(p_{1}-1)\cdots(p_{m}-1)}\right)\frac{q}{q-1}\geq
2+\frac{1}{(p_{1}-1)\cdots(p_{m}-1)(q-1)},$ (28)

which is equivalent to the desired inequality.

∎

Let ${\rm\,odd}(m)$ be the largest odd divisor of $m$. From Lemma 23 above we
have

######  Proposition 24.

There is no prime $q$ such that $H({\rm\,odd}(q+1))<2$, and
$H(q{\rm\,odd}(p+1))>2$.

Proposition 24 says that if one can make an abundancy argument to rule out
prime $q$ using $q$’s contribution, then one did not in fact need to use $q$
at all. That is, one can rule out $q$ being special based solely on using
abundancy considerations to show that there there is no odd perfect number
with $\sigma(\frac{p+1}{2}\mid n)$.

For higher powers the situation is slightly different. For example,
$H(\sigma(5^{5}))<2$, but $H(5\sigma(5^{5})>2$. In fact, one can use an
abundancy argument to show that one cannot have any odd perfect number $n$
with $5^{5}\mid\mid n$, even though a simple abundancy argument fails to be
sufficient to show that $\frac{\sigma(5^{5})}{2}\nmid n$. A variant of this
argument seems to have been first explicitly noted by Sorli and Cohen [6],
although it was implicitly used in prior work.

When the exponent is even, similar examples are more difficult to come by but
they do exist. Consider for example $7^{944}$. One has that
$H(\sigma(7^{944}))<2$ but $H(7\sigma(7^{944}))>2$. A little work shows that
one cannot have an odd perfect number $n$ where $7^{944}\mid\mid n$. The
exponent $944$ arises because $\sigma(7^{944})=\frac{7^{945}-1}{6}$, and since
$945=(3^{3})(5)(7)$ we get many small prime factors in $\sigma(7^{944})$.

The discussion in the previous three paragraphs motivates three questions:

######  Question 25.

Does there exist an odd prime $p$ such that $H(\sigma(p^{2}))<2$, but where
$H(p\sigma(p^{2}))>2$?

If Question 25 has an affirmative answer, then Lemma 23 is essentially best
possible. Note that we do not need to take the odd part in the above question
unlike in Lemma 23 since $p^{2}+p+1$ is always odd.)

######  Question 26.

Let $\Phi_{m}(x)$ be the $m$th cyclotomic polynomial. Is there any $m$ and odd
prime $p$ such that $H({\rm\,odd}(\Phi(p))<2$ and $H(p{\rm\,odd}(\Phi(p))>2$?

An affirmative answer to Question 25 would imply a positive answer to Question
26, since it would be the case when $m=2$.

######  Question 27.

What is the smallest $n$ such that $n=p^{a}$ where $p$ is an odd prime, $a$ is
even, and where $H(n)<2$ and $H(p\sigma(n))>2$?

There is a thematic connection between this question and prior work where
Cohen and Sorli [6] used essentially a similar abundancy argument to show that
$p^{a}$ is a component in an odd perfect number and $p$ is the special prime,
then $\sigma(\sigma(p^{a}))\leq 3p^{a}-1$. They also showed that if $p$ is not
special, then $\sigma(\sigma(p^{a}))\leq 2p^{a}-2$.

Dris [8] showed that if $n=m^{2}p_{j}^{a_{j}}$ and $p_{j}>m$, then $a_{j}=1$.
We use this opportunity to show an alternate proof of this result using
connected ideas to the above which also strengthens it slightly.

######  Proposition 28.

Assume that $n$ is an odd perfect number where $n=m^{2}q^{e}$, where $(q,m)=1$
and $q\equiv e\equiv 1\pmod{4}$ . Then either $e=1$, or

$q<2^{\frac{1}{5}}m^{\frac{2}{5}}.$

###### Proof.

Assume that $e>1$. Thus, $e\geq 5$. Since $\gcd(q,\sigma(q^{e}))=1$, We must
have $q^{5}\mid\sigma(m^{2})$, and so

$q^{5}\leq\sigma(m^{2})<2m^{2},$

which implies the desired inequality. ∎

We can improve the constant in Proposition 28 slightly more with only a small
amount of effort.

######  Proposition 29.

Assume that $n$ is an odd perfect number where $n=m^{2}q^{e}$ where $(q,m)=1$
and $q\equiv e\equiv 1\pmod{4}$. Then either $e=1$, or

$q<\left(\frac{2}{(3^{2})(5^{2})(11^{2})(13^{2})}\right)^{\frac{1}{5}}m^{\frac{2}{5}}.$

###### Proof.

Assume we have $e>1$. We now consider two cases, $e=5$ and $e\geq 9$. Assume
that $q=5$. It follows from results of Nielsen [13] that $m$ must have at
least $9$ distinct prime factors. This means that there have to be at least
four primes $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$ with exponents $a_{1}$,
$a_{2}$, $a_{3}$, $a_{4}$ such that for $i=1,2,3,4$ we have
$p_{i}^{a_{i}}\mid\mid m^{2}$, and $q\nmid\sigma(p_{i}^{a_{i}})$.

Thus we have

$q^{5}\mid\sigma\left(\frac{m^{2}}{p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}p_{4}^{a_{4}}}\right).$
(29)

Inequality (29) then implies that

$q^{5}<2\frac{m^{2}}{p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}p_{4}^{a_{4}}}.$
(30)

Since no odd perfect number can be divisible by 3, 5, and 7 we must have
$p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}p_{4}^{a_{4}}\geq(3^{2})(5^{2})(11^{2})(13^{2})$,
which together with Inequality (30) implies the inequality in question. If
$e\geq 9$, then we have by the same logic as the proof of the previous
proposition that $q^{9}<2m^{2}$, which is weaker than the desired inequality
unless $m<48289774$, and there is no odd perfect number with such a small $m$
value.

∎

The constant in the previous proposition can be improved further by a variety
of means, such as by using that no odd perfect number is divisible by $3,5,11$
and $13$. Similarly, one can use that if $p$ is a prime where $p^{a}\mid\mid
m^{2}$, and $(q,a\sigma(p^{a}))=1$, which us to further control $h(m^{2})$.
However, this involves arduous case checking followed by taking a $5$th root.
Thus, improvement in the constant ends up being small proportional to the
large amount of effort of such a calculation.

Note that in this context, the weakest bounds in our main inequalities occur
when $p_{j}$ is essentially as small as possible. In contrast, in many other
contexts, a small $p_{j}$ results in tighter bounds. For example, in Acquaah
and Konyagin’s proof of an upper bound on the largest prime factor, the most
difficult situation is when $p_{j}$ is the largest prime factor. Similar
remarks apply to the bounds on the second and third largest prime factor.
There are other bounds in the literature which bound $p_{1}p_{2}\cdots p_{k}$
above such as those of Klurman [11] and similar bounds by Luca and Pomerance
[12]. In both of those papers, the situation where $p_{j}$ is the largest
prime factor gives the weakest bounds.

One final remark: For any positive integer $n$,

$D(n)\leq\frac{n\log_{p_{1}}n}{p_{1}}.$ (31)

See Dahl, Olsson, and Loiko [7] for this and related bounds. One obvious
question is how much Inequality (31) can be tightened under the assumptions
that $n$ is an odd perfect number or the weaker assumption that $n$ is a
primitive non-deficient number. The author was unable to obtain non-trivial
improvements of this inequality under these assumptions, but it seems
plausible that such improvements are possible.

Acknowledgements Pace Nielsen and Eve Zelinsky both made helpful corrections
to the manuscript.

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2020 AMS Subject Classification: Primary 11A25; Secondary 11N64 Keywords:
Perfect number, non-deficient number, abundant number, sum-of-divisors