# Small prime factors of odd non-deficient numbers

Joshua Zelinsky

###### Abstract

Let $\sigma(n)$ to be the sum of the positive divisors of $n$. A number is
non-deficient if $\sigma(n)\geq 2$. We establish new lower bounds for the
number of distinct prime factors of an odd non-deficient number in terms of
its second smallest, third smallest and fourth smallest prime factors. We also
obtain tighter bounds for odd perfect numbers. We also discuss the behavior of
$\sigma(n!+1)$, $\sigma(2^{n}+1)$, and related sequences.

Hopkins School

<EMAIL_ADDRESS>
We will write $\sigma(n)$ to be the sum of the positive divisors of $n$.
Recall, that $\sigma(n)$ is a multiplicative function. Recall further that
numbers where $\sigma(n)=2n$ are said to be perfect, numbers where
$\sigma(n)<2n$ are said to be deficient, and numbers where $\sigma(n)>2n$ are
said to be abundant. A large amount of prior work has been done on estimating
the density of abundant numbers. See for example [6] and [7].

We will throughout this paper write $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{k}^{a_{k}}$ where $p_{1}\cdots p_{k}$ are primes satisfying
$p_{1}<p_{2}\cdots<p_{k}$. The main results of this paper are lower bounds on
$k$ in terms of $p_{2}$ and $p_{3}$ when $n$ is odd and not deficient. We
prove similar bounds with $p_{4}$ when $n$ is odd and perfect, and a slightly
weaker result when $n$ is a primitive non-deficient number. Using the same
techniques we will also produce estimates for $\sigma(m)$ when $m$ is near but
not equal to $n!$.

Kishore [5] proved that if $n$ is an odd perfect number with $k$ distinct
prime factors, and $2\leq i\leq 6$, then one must have

$p_{i}<2^{2^{i-1}}\left(k-i+1\right).$

Note that these bounds are all linear in $p_{i}$ and $k$. In this paper, for
$i$ in the range $2\leq i\leq 4$, we prove improve on Kishore’s results,
giving better than linear bounds. Moreover, for $i=2$ and $i=3$, our results
apply not just to odd perfect numbers but also to odd abundant numbers.

For $p_{1}$, better than linear bounds are known, due to [9] and the
author[13]. Most of the results in this paper rely on the same fundamental
techniques as in [9] and [13]. The first section of this paper extends those
techniques to these small prime divisors. The second section uses the same
techniques to look at a related problem involving the sum of the divisors of
numbers near $n!$.

We will write $h(n)=\sigma(n)/n$. Note that $h(n)$ is frequently called the
_abundancy_ of $n$, but it has also been referred in the literature as the
_abundancy index_ of $n$ as well as just the _index_ of $n$. Note that a
number is perfect if and only if $h(n)=2$, is abundant if and only $h(n)>2$,
and is deficient if and only $h(n)<2$.

One has that

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

from which it follows that $h(mn)>h(n)$ whenever $m>1$. Motivated by this
inequality, there is a notion of a _primitive abundant number_ which is a
number $n$ which is abundant but where proper all divisors of $n$ are not
abundant. There is some inconsistency in the literature here; some authors
define primitive abundant numbers as including perfect numbers where all
proper divisors are deficient. In some respects, this is a more natural set to
investigate. Others term this set to be the set of _primitive non-deficient
numbers_ , which is the term we will use. The first substantial work on this
topic was Dickson who proved that given a fixed $k$, there are only a finite
number of primitive non-deficient numbers with $k$ distinct prime factors.[2]
Prior work on these sets have looked at their density. See especially [1] as
well as [8] and [12].

We will write $H(n)=\Pi_{i=1}^{k}\frac{p_{i}}{p_{i}-1}$. It is well-known and
not hard to show that $h(n)\leq H(n)$ (and this inequality is strict as long
as $n>1$). In fact,

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

Thus, $H(n)$ is the best possible upper bound on $h(n)$ when we know only the
distinct prime divisors of $n$, but have no other information about $n$.

We will write $P_{j}$ as the $j$th prime number (note the capital letter
distinguishing this from the $p_{i}$).

Given a prime $P_{j}$ and a real number $\alpha>1$, we will write
$b_{\alpha}(P_{j})$ to be the smallest positive integer $b$ such that

$\alpha<\left(\frac{P_{j}}{P_{j}-1}\right)\left(\frac{P_{j+1}}{P_{j+1}-1}\right)\cdots\left(\frac{P_{j+b-1}}{P_{j+b-1}-1}\right)$

We will similarly define $a_{\alpha}(j)$ as $b_{\alpha}(P_{j})$.

These two functions, $a_{\alpha}(j)$, and $b_{\alpha}(P_{j})$ are
generalizations of the functions labeled as $a(j)$, and $b(P_{j})$ in [9] and
[13]. In particular, $a(j)=a_{2}(j)$ and $b(j)=b_{2}(j)$. The use of
$\alpha=2$ arises naturally in the context when one is interested in bounding
$k$ in terms of the smallest prime divisor of $n$, but examining other prime
divisors requires looking at other values of $\alpha$. Note that if $P_{j}$ is
the smallest prime divisor of an odd perfect or or odd abundant number $n$
then $k\geq a(j)$.

Norton proved

######  Theorem 1.

Let $t$ be a positive integer then:

$a(t)>t^{2}-2t-\frac{t+1}{\log t}-\frac{5}{4}-\frac{1}{2t}-\frac{1}{4t\log
t}.$ (1)

In fact, $a(t)$ is asymptotic to $\frac{1}{2}t^{2}\log t$. When Norton was
writing, bounds on the prime number theorem and related functions were not
strong enough to easily give an explicit lower bound on $a(t)$ which had the
correct lead term. Subsequently, the author proved [13]

######  Theorem 2.

For all $t>1$ we have

$a(t)\geq\frac{t^{2}}{2}\left(\log t-\frac{3}{2}\log\log
t+\frac{1}{20}+\frac{\log\log t}{\log t}\right)-t+1.$ (2)

Notice that this theorem has the correct first lead term.

Norton also proved that

$a(n)=\frac{1}{2}n^{2}\log n+\frac{1}{2}n^{2}\log\log
n-\frac{3}{4}n^{2}+\frac{n^{2}\log\log n}{2\log n}+O\left(\frac{n^{2}}{\log
n}\right).$ (3)

Thus, Inequality 2 is not as strong as it could be although it has the correct
lead term, the later terms do not match the asymptotic. Improving these later
terms to get an explicit function which agrees out to the $O(\frac{n^{2}}{\log
n})$ term appears to be non-trivial at this time.

In the final section of this paper we then use the same techniques to give a
new proof of a theorem about the abundancy of numbers of the form $m!+z$ where
$z$ is a fixed non-zero integer. We will prove

######  Theorem 3.

Let $z$ be a non-zero integer. Then

$\lim_{m\rightarrow\infty}h(m!+z)=h(|z|).$

Theorem 3 appears to be a known folklore result; we are uncertain if it has
been previously published.

##  1 Small prime factors of non-deficient numbers

We require the following explicit version of Mertens’ Theorem:

######  Lemma 4.

We have

$e^{\gamma}(\log x)\left(1-\frac{1}{2\log^{2}x}\right)<\prod_{p\leq
x}\frac{p}{p-1}<e^{\gamma}(\log x)\left(1+\frac{1}{2\log^{2}x}\right).$ (4)

Here $\gamma$ is Euler’s constant, with the lower bound valid for $x>3$ and
the upper bound valid for $x\geq 286.$

###### Proof.

This is proven in [10]. ∎

Note that it is a consequence of this explicit version of Merten’s Theorem and
the Prime Number Theorem that

######  Theorem 5.

Let $\alpha>1$ be fixed. Then

$b_{\alpha}(p)\sim\frac{p^{\alpha}}{\alpha\log p},$

and

$a_{\alpha}(n)\sim\frac{n^{\alpha}\log n}{\alpha}.$

Much of our work will be trying to make Theorem 5 explicit for certain
specific values of $\alpha$. We will for our purposes only need to construct
explicit lower bounds, but it may be interesting to construct similarly
explicit upper bounds.

We will write $S(x)=\prod_{p\leq x}\frac{p}{p-1}$, and will write

$Q(x,y)=\prod_{y<p\leq x}\frac{p}{p-1}=\frac{S(x)}{S(y)}$

and will when we write this assume that $x>y>3$.

######  Lemma 6.

Let $A$, $B$ and $\alpha$ be real numbers. Assume that $A>B>6$, and that
$3\geq\alpha>1$. Assume further that

$A+\frac{1}{2A}\geq\alpha\left(B-\frac{1}{2B}\right).$

Then $A\geq\alpha B-\frac{2}{B}$.

###### Proof.

We note that $A+\frac{1}{A}$ is an increasing function in $A$ for all $A>1$.
So we just need to verify that $\left(\alpha
B-\frac{2}{B}\right)+\frac{1}{2(\alpha
B-\frac{2}{B})}\leq\alpha(B-\frac{1}{2B}).$ This is the same as verifying that

$\alpha B-\frac{\alpha}{2B}-\left(\left(\alpha
B-\frac{2}{B}\right)+\frac{1}{2(\alpha B-\frac{2}{B})}\right)\geq 0.$

We can rewrite the expression on the right-hand side above as

$\frac{\left(-\alpha+4\right)(\alpha B+\frac{2}{B})-B}{2B\left(\alpha
B-\frac{2}{B}\right)}$

which has a positive denominator (since $B>\sqrt{2}$), so we just need to look
at

$(-\alpha+4)(\alpha B+\frac{2}{B}-B).$

Since $\alpha\leq 3$, we have

$(-\alpha+4)(\alpha B+\frac{2}{B}-B)\geq(\alpha
B+\frac{2}{B})-B>\frac{2}{B}>0.$

∎

We will need a few explicit estimates of certain functions of primes. We have
from [3] that

$\frac{x}{\log x}\left(1+\frac{0.992}{\log
x}\right)\leq\pi(x)\leq\frac{x}{\log x}\left(1+\frac{1.2762}{\log x}\right),$
(5)

with the lower bound valid if $x\geq 599$ and the upper bound valid for all
$x>1$.

We also need from [13] that for $n\geq 2$,

$P_{n}\geq n\left(\log n+\log\log n-1+\frac{32}{31(\log n)^{2}}\right)$ (6)

######  Theorem 7.

Let $\alpha$ be a real number such $1<\alpha\leq 3$ and let $p$ be a prime
with $p\geq 853$. Then

$b_{\alpha}(p)\geq\frac{p^{\alpha}}{\alpha\log
p}-\frac{1.669p^{\alpha}}{\alpha\log^{2}p}-\frac{0.662p^{\alpha}}{\alpha\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+1.$

###### Proof.

Let $p\geq 601$ be a prime. Let $q$ be the smallest prime such that
$Q(p,q)\geq\alpha$.

We have from Lemma 4 that

$\alpha\leq Q(q,p)\leq\frac{e^{\gamma}(\log
q)\left(1+\frac{1}{2\log^{2}q}\right)}{e^{\gamma}(\log
p)\left(1-\frac{1}{2\log^{2}p}\right)}=\frac{(\log
q)\left(1+\frac{1}{2\log^{2}q}\right)}{(\log
p)\left(1-\frac{1}{2\log^{2}p}\right)}.$ (7)

If we set $A=\log q$, and $p=\log B$, we have

$A+\frac{1}{2A}\geq\alpha B(1-\frac{2}{B^{2}}),$

so we may invoke 6 to get that $\log q\geq\alpha\log p-\frac{2}{\log p}$. We
have then $q\geq pe^{\frac{2}{\log p}}$.

We have then

$b_{\alpha(p)}\geq\pi(p^{\alpha}e^{\frac{-2}{\log p}})-\pi(p)+1.$

We note that $e^{\frac{-2}{\log p}}\geq 1-\frac{2}{\log p}$ and so

$b_{\alpha(p)}\geq\pi(p^{\alpha}(1-\frac{2}{\log p})-\pi(p)+1.$

Since $p\geq 853$, we have $p^{\alpha}(1-\frac{2}{\log p})\geq 599$, and may
apply Lemma 5 to get that

$b_{\alpha(p)}\geq\frac{p^{\alpha}(1-\frac{2}{\log p})}{\log
p^{\alpha}(1-\frac{2}{\log p})}\left(1+\frac{0.992}{\log
p^{\alpha}(1-\frac{2}{\log p})}\right)-\frac{p}{\log
p}\left(1+\frac{1.2762}{\log p}\right)+1.$

and so

$b_{\alpha(p)}\geq\frac{p^{\alpha}(1-\frac{2}{\log p})}{\alpha\log
p}\left(1+\frac{0.992}{\alpha\log p}\right)-\frac{p}{\log
p}\left(1+\frac{1.2762}{\log p}\right)+1.$

Since $\alpha\leq 3$, we have

$b_{\alpha(p)}\geq\frac{p^{\alpha}(1-\frac{2}{\log p})}{\alpha\log
p}\left(1+\frac{0.331}{\log p}\right)-\frac{p}{\log
p}\left(1+\frac{1.2762}{\log p}\right)+1$

Multiplying out the terms gives us the desired inequality. ∎

We will need some specific versions of the above theorem for certain values of
$\alpha$.

######  Theorem 8.

For all primes $p\geq 3$, one has

$b_{\frac{4}{3}}(p)\geq\frac{p^{\frac{4}{3}}}{\frac{4}{3}\log
p}-\frac{1.25175p^{\frac{4}{3}}}{\log^{2}p}-\frac{0.4965p^{\frac{4}{3}}}{\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+1,$ (8)

$b_{\frac{16}{15}}(p)\geq\frac{p^{\frac{16}{15}}}{\frac{16}{15}\log
p}-\frac{1.5646875p^{\frac{16}{15}}}{\log^{2}p}-\frac{0.620625p^{\frac{16}{15}}}{\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+1,$ (9)

and

$b_{\frac{256}{255}}(p)\geq\frac{p^{\frac{256}{255}}}{\frac{256}{255}\log
p}-\frac{1.662481p^{\frac{256}{255}}}{\log^{2}p}-\frac{0.65942p^{\frac{256}{255}}}{\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+1.$ (10)

$b_{\frac{26325}{26257}}(p)\geq\frac{p^{\frac{26325}{26257}}}{\frac{26325}{26257}\log
p}-\frac{1.6647p^{\frac{26325}{26257}}}{\log^{2}p}-\frac{0.6603p^{\frac{26325}{26257}}}{\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+1.$ (11)

###### Proof.

We will only discuss how to prove Equation 8. The other equations have very
similar proofs. Theorem 7 implies that 8 is true for all primes $p\geq 853$.
We then use a computer program to check that it is true for all primes $p$
with $3\leq p\leq 853$. The others are done similarly. ∎

######  Proposition 9.

Suppose that $N$ is an odd number which is either perfect or abundant. Let
$p_{2}$ be second smallest prime divisor of $N$. Let $k$ be the number of
distinct prime divisors of $N$. Then $k\geq b_{\frac{4}{3}}(p_{2})+1$.

###### Proof.

We have $h(N)\geq 2$. We assume that $N$ has the factorization
$p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ with
$p_{1}<p_{2}<p_{3}\cdots<p_{k}$. We have then

$2\leq
h(N)<H(n)=\frac{p_{1}}{p_{1}-1}\frac{p_{2}}{p_{2}-1}\cdots\frac{p_{k}}{p_{k}-1}.$

We have that in the most extreme case, $p_{1}=3$, in which case
$\frac{p_{1}}{p_{1}-1}=\frac{3}{2}$. We thus have

$\frac{3}{2}\frac{p_{2}}{p_{2}-1}\cdots\frac{p_{k}}{p_{k}-1}>2,$

and hence

$\frac{p_{2}}{p_{2}-1}\cdots\frac{p_{k}}{p_{k}-1}>\frac{4}{3}.$

We thus have $k-1\geq b_{\frac{4}{3}}(p_{2}).$ ∎

Applying Proposition 9 with Inequality 8 we obtain:

######  Corollary 10.

Suppose that $N$ is an odd number which is either perfect or abundant. Let
$p_{2}$ is second smallest prime divisor of $N$. Let $k$ be the number of
distinct prime divisors of $N$. Then

$k\geq\frac{p_{2}^{\frac{4}{3}}}{\frac{4}{3}\log
p_{2}}-\frac{1.25175p_{2}^{\frac{4}{3}}}{\log^{2}p_{2}}-\frac{0.4965p^{\frac{4}{3}}}{\log^{3}p_{2}}-\frac{p_{2}}{\log
p_{2}}-\frac{1.2762p_{2}}{\log^{2}p_{2}}+2.$

The bound from Corollary 10 exceeds Kishore’s bound when $p_{2}\geq 97$ and
also applies to all odd non-deficient numbers, whereas Kishore’s bound applies
only to odd perfect numbers.

We similarly have:

######  Proposition 11.

Suppose that $N$ is an odd number which is either perfect or abundant. Let
$p_{3}$ be the third smallest prime divisor of $N$. Let $k$ be the number of
distinct prime divisors of $N$. Then $k\geq b_{\frac{16}{15}}(p_{3})+3$.

###### Proof.

This proof is essentially the same as the proof of Proposition 9, but we now
note that the worst case scenario is that the smallest prime is 3 and the
second smallest prime is 5. ∎

We have then as a corollary:

######  Corollary 12.

Suppose that $N$ is an odd number which is either perfect or abundant. Let
$p_{2}$ be the smallest prime divisor of $N$. Let $k$ be the number of
distinct prime divisors of $N$. Then

$b_{\frac{16}{15}}(p_{3})\geq\frac{p_{3}^{\frac{16}{15}}}{\frac{16}{15}\log
p_{3}}-\frac{1.5646875p_{3}^{\frac{16}{15}}}{\log^{2}p_{3}}-\frac{0.620625p_{3}^{\frac{16}{15}}}{\log^{3}p_{3}}-\frac{p_{3}}{\log
p_{3}}-\frac{1.2762p_{3}}{\log^{2}p_{3}}+3$

Here the inequality is better than Kishore’s inequality when $p\geq 282767$.

We now turn to smallest bounds on the fourth prime factor. Here, we must
restrict ourselves finally to odd perfect numbers. This is because there are
actually odd abundant numbers with three distinct prime factors; $N=945$ is an
example.

We will recall of the earlier results about odd perfect number is due to
Euler.

######  Lemma 13.

(Euler) If $N$ is an odd perfect number then have $N=p^{e}m^{2}$ for some
prime $p$ where $(p,m)=1$ and $p\equiv e\equiv 1$ (mod 4).

In fact, Euler’s result applies more generally to any odd number $n$
satisfying $\sigma(n)\equiv 2$ (mod 4).

We also note the following well known Lemma where for completeness we include
a proof:

######  Lemma 14.

Let $N$ be an odd perfect number. Then $105\not|N$.

###### Proof.

Let $N$ be an odd perfect number, and assume that $105|N$. Since $3\equiv 3$
(mod 4), and $7\equiv 3$ (mod 4), by Euler’s result, we must have $3^{2}|N$
and $7^{2}|N$. But $(3^{2})(5)(7^{2})$ is an abundant number. ∎

Using the same approach we can further restrict the prime divisors of an odd
perfect number.

######  Lemma 15.

Let $N$ be an odd perfect number. Assume that $3|N$, $5|N$ and $11|N$. Then
one must have $5||N$.

###### Proof.

Suppose that $N$ is an odd perfect number where $(3)(5)(11)|N$. Note that
since $3\equiv 11\equiv 3$ (mod 4), we must from Euler’s result, have
$3^{2}|N$ and $11^{2}|N$. First, we will show that $3^{4}|N$. Assume that
$3^{4}\not|N$. Then since $3\equiv 3$ (mod 4), we must have $3^{2}|N$. Thus,
$\sigma(3^{2})=13|N$. But $(3^{2})5(11^{2})(13)$ is abundant. We thus may
assume that $3^{4}|N$.

Note that $\sigma(11^{2})=(7)(19)$, so if $11^{2}||N$, we would $105|N$ which
by the prior Lemma is impossible. Thus, we conclude that $11^{4}|N$. However,
$(3^{4})(5^{2})(11^{4})$ is an abundant number. So we must have that $5|N$. ∎

######  Lemma 16.

Let $N$ be an odd perfect number. Suppose that $3|N$, $5|N$, and $13|N$. Then
we must have either $5||N$ or $3^{2}||N$.

###### Proof.

Assume that $N$ is an odd perfect number and that $3|N$, $5|n$ and $13|N$. We
note that if $13$ were the special prime, then since $7|(13+1)$, we would have
$7|N$, which is impossible since no odd perfect number is divisible by $105$.

Assume that $5^{2}|N$ and that $3^{3}|N$; the second implies that we must have
$3^{4}|N$. But $(5^{2})(3^{4})(13^{2})|N$ and then $(5^{2})(3^{4})(13^{2})$ is
abundant, which is a contradiction.

∎

######  Lemma 17.

Let $N$ be an odd perfect number, with
$N=p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}\cdots p_{k}^{a_{k}}$ with
$p_{1}<p_{2}<p_{3}\cdots<p_{k}$. Then
$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})<\frac{255}{128}$.

###### Proof.

Note that if $p_{1}\geq 5$, then
$H(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq\frac{5}{4}\frac{7}{6}\frac{11}{10}<\frac{255}{128}$.
Since $h(m)<H(m)$ for all $m>1$, this means that we would have
$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})<\frac{255}{128}$. So we may assume
that $p_{1}=3$.

Similarly, since $\frac{3}{2}\frac{7}{6}\frac{11}{10}<\frac{255}{128}$ we may
assume that $p_{2}=5$. Since
$\frac{3}{2}\frac{5}{4}\frac{17}{16}=\frac{255}{128}$, we may assume that
$p_{3}=11$ or that $p_{3}=13$.

Let us first consider the situation where $p_{3}=11$. Then from Lemma 15, we
must have $a_{2}=1$. Then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}=h(3^{a_{1}})h(5)h(11^{a_{3}})<H(3)h(5)H(11)=\frac{3}{2}\frac{6}{5}\frac{11}{10}<\frac{255}{128}.$

The situation where $p_{3}=13$ is nearly identical, but we use Lemma 16
instead of Lemma 15. ∎

######  Proposition 18.

Suppose that $N$ is an odd perfect number, and let $p_{4}$ be the fourth
smallest prime factor of $N$. Assume that $N$ has $k$ distinct prime factors.
Then

$k\geq b_{\frac{256}{255}}(p_{4})+3.$

###### Proof.

Let $N$ be an odd perfect number. Assume that $N$ has the factorization
$p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ with
$p_{1}<p_{2}<p_{3}\cdots<p_{k}$. We must have that

$h(N)=2=h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})h(p_{4}^{a_{4}}\cdots
p_{k}^{a_{k}})<\frac{255}{128}H(p_{4}^{a_{4}}\cdots p_{k}^{a_{k}}).$

We have then that $H(p_{4}^{a_{4}}\cdots p_{k}^{a_{k}})\geq\frac{256}{128}$
The result then follows from the same sort of argument as before. ∎

We then obtain:

######  Corollary 19.

Suppose that $N$ is an odd perfect number, and let $p_{4}$ be the fourth
smallest prime factor of $N$. Assume that $N$ has $k$ distinct prime factors.
Then

$k\geq\geq\frac{p^{\frac{256}{255}}}{\frac{256}{255}\log
p}-\frac{1.662481p^{\frac{256}{255}}}{\log^{2}p}-\frac{0.65942p^{\frac{256}{255}}}{\log^{3}p}-\frac{p}{\log
p}-\frac{1.2762p}{\log^{2}p}+4.$

We can make a slightly weaker version of Proposition 18 which does apply to
some non-deficient numbers. Recall a number $M$ is a primitive non-deficient
number if $M$ is not deficient and all divisors of $M$ are deficient.

######  Lemma 20.

Let $M$ be a primitive non-deficient number with at least four distinct prime
factors. Suppose that $3|M$, $5|M$, and $7|M$. Then one has either $3||M$ or
$3^{2}||M$. Moreover, if $3^{2}||M$ then $5||M$ and $7||M$.

###### Proof.

First, note that $(3^{3})(5)(7)$ is abundant. So one must have $3||M$ or
$3^{2}||M$.

Now assume that $3^{2}||M$. Then since $(3^{2})(5^{2})7$ and $(3^{2})5(7^{2})$
are both abundant, we must have $5||M$ and $7||M$.

∎

######  Lemma 21.

Let $M$ be a primitive non-deficient number with at least four distinct prime
factors. Suppose that $3|M$, $5^{2}|M$, and $11|M$. Then either $3||N$ or
$3^{2}||N$.

###### Proof.

This follows from noting that $(3^{3})(5^{2})(11)$ is abundant. ∎

######  Lemma 22.

Let $M$ be a primitive non-deficient number with at least four distinct prime
factors. Suppose that $3|M$, $5^{2}|M$, and $13|M$. Then $3^{5}\not|M$.

Furthermore, under the same assumptions if $3^{4}|M$, then $5^{3}\not|M$ and
$13^{2}\not|M.$

Under the same assumptions, if $3^{3}|M$, then either $5^{3}\not|M$ or
$13^{2}\not|M$.

###### Proof.

This follows from noting that $(3^{5})(5^{2})(13)$, $(3^{4})(5^{3})(13)$,
$(3^{4})(5^{2})(13^{2})$, $(3^{3})(5^{3})(13^{2})$ are all abundant numbers. ∎

######  Theorem 23.

Let $M$ be a primitive non-deficient number with at least four distinct prime
factors. Assume further that
$M=p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}p_{4}^{a_{4}}\cdots p_{k}^{a_{k}}$
where $p_{1}$, $p_{2}$, $p_{3}\cdots p_{k}$ are primes and
$p_{1}<p_{2}<p_{3}<p_{4}<\cdots p_{k}$. Then we have
$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq\frac{52514}{26325}=1.9948\cdots$.

###### Proof.

The method is essentially the same as our earlier approach. Set
$M=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{a}}$. If $p_{1}>3$, then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})<H((5)(7)(11))=\frac{5}{4}\frac{7}{6}\frac{11}{10}=\frac{77}{48}<\frac{52514}{26325}.$

We thus may assume that $p_{1}=3$. By similar logic, we may assume that
$p_{2}=5$, and that $p_{3}$ is one $7$, $11$ or $13.$

Consider the case where $p_{3}=7$. Then by Lemma 20 we have

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq\max(h((3^{2})(5)(7),h(3)H(5,7))=\max(\frac{208}{105},\frac{35}{18})<\frac{52514}{26325}.$

Consider now the case where $p_{3}=11$. Consider the possible values of
$a_{2}$. If $a_{2}=1$ (that is $5||N$), then we have

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
H(3)h(5)H(11)=\frac{99}{50}<\frac{52514}{26325}.$

Now assume $a_{2}\geq 2$. This is equivalent to $5^{2}|N$. Then by Lemma 21,
we have $a_{1}\leq 2.$ Thus

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(13)H(5)H(11)=\frac{143}{72}<\frac{52514}{26325}.$

Thus, in all situations where $p_{3}=11$ we have the desired inequality.

Finally, consider the case where $p_{3}=13$. If $a_{2}=1$, then we have

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
H(3)h(5)H(13)=\frac{3}{2}\frac{6}{5}\frac{14}{13}=\frac{39}{20}<\frac{52514}{26325}.$

Assume $a_{2}\geq 2$. Then by Lemma 22, $a_{1}\leq 4$. We will consider four
cases depending on the value of $a_{1}$.

Case I: Assume that $a_{1}=1$. Then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(3)H(5)H(13)=\frac{4}{3}\frac{5}{4}\frac{13}{12}=\frac{65}{36}<\frac{52514}{26325}.$

Case II: Assume that $a_{1}=2$. Then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(9)H(5)H(13)=\frac{13}{9}\frac{5}{4}\frac{13}{12}=\frac{845}{432}<\frac{52514}{26325}.$

Case III: Assume that $a_{1}=3$. If $a_{3}=1$, then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(27)H(5)h(13)=\frac{40}{27}\frac{5}{4}\frac{14}{13}=\frac{700}{351}<\frac{52514}{26325}.$

Therefore, we may assume that $a_{3}\geq 2$. Thus, by Lemma 22, we have that
$a_{2}\leq 2$. Thus,

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(27)h(25)H(13)=\frac{40}{27}\frac{31}{25}\frac{13}{12}=\frac{806}{405}<\frac{52514}{26325}.$

Case IV: $a_{1}=4$. Thus, by Lemma 22, we must have $a_{2}\leq 2$, and
$a_{3}=1$. Then

$h(p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}})\leq
h(81)h(25)H(13)=\frac{121}{81}\frac{31}{25}\frac{14}{13}=\frac{52514}{26325}.$

Note that it is only in Case IV above that we hit the worst case scenario. ∎

Using Theorem 23 along with Equation 11 we may use the exact same method of
proof to prove that:

######  Proposition 24.

Let $M$ be a primitive non-deficient number. (That is, $M$ is not deficient
and it has no non-deficient divisors.) And assume that $M$ has at least four
distinct prime divisors. Let $k$ be the number of distinct prime divisors of
$M$, and let $p_{4}$ be the fourth smallest prime factor of $M$. Then we have

$k\geq b_{\frac{26325}{26257}}(p_{4})+4.$

As before we have a corollary:

######  Corollary 25.

Let $M$ be a primitive non-deficient number. (That is, $M$ is not deficient
and it has no non-deficient divisors.) And assume that $M$ has at least four
distinct prime divisors. Let $k$ be the number of distinct prime divisors of
$M$, and let $p_{4}$ be the fourth smallest prime factor of $M$. Then we have

$b_{\frac{26325}{26257}}(p_{4})\geq\frac{p_{4}^{\frac{26325}{26257}}}{\frac{26325}{26257}\log
p_{4}}-\frac{1.6647p_{4}^{\frac{26325}{26257}}}{\log^{2}p_{4}}-\frac{0.6603p_{4}^{\frac{26325}{26257}}}{\log^{3}p_{4}}-\frac{p_{4}}{\log
p_{4}}-\frac{1.2762p_{4}}{\log^{2}p_{4}}+5.$

Note that the bounds in this section are weak in the following sense: Let
$\omega(n)$ be the number of distinct prime divisors of $n$. Given a function
$f(n)$ with $\lim_{n\rightarrow\infty}f(n)$, then for any positive integer
$i$, the statement “The $i$th smallest prime divisor of $n$ is at most $f(n)$”
will apply to almost all positive integers $n$ (in the sense that the
exceptional set has density zero). Similarly, any such restriction for odd
perfect numbers is a weak restriction in the sense discussed in [13].

However, it is not too hard to see that all the results in this section which
apply to non-deficient numbers in general or apply to primitive non-deficient
numbers are asymptotically best possible. Of course, for the results which
only concern odd perfect numbers, it is likely that the results are far from
best possible, since very likely no odd perfect numbers exist. This is
connected to a question raised in [13]. In particular, the author defined
$b_{o}(p)$ to be the minimum of the number of distinct prime divisors of any
odd perfect number with smallest prime divisor $p$ and set $b_{o}(p)=\infty$
when there are no odd perfect numbers with smallest prime divisor $p$. The
author asked if it was possible to prove that $b_{o}(p)>b(p)$, or in the
notation of our paper, if $b_{o}(p)>b_{2}(p)$. The notation here is less than
ideal; note that the small circle in $b_{o}(p)$ is a lower case o, not a zero.
Since we now have a numeric subscript of $b(p)$ in this paper, a slightly
better notation for our purposes is as follows: We will write $b_{(O,i)}(p)$
to be the to be the minimum of the number of distinct prime divisors of any
odd perfect number with $i$th smallest prime divisor $p$ and set
$b_{(O,i)}(p)=\infty$ when no such odd perfect number exists. The above
mentioned question then becomes whether it is possible to prove that
$b_{(O,1)}(p)>b_{2}(p)$. Following this, we can ask similar questions for
other values of $i$. In particular, can we prove any of the following:

  1. 1.

$b_{(O,2)}(p)>b_{\frac{4}{3}}(p)$.

  2. 2.

$b_{(O,3)}(p)>b_{\frac{16}{15}}(p)$.

  3. 3.

$b_{(O,4)}(p)>b_{\frac{256}{255}}(p)$.

Two other directions which may make sense for investigation are to extend
these results for larger prime factors and to extend them to multiperfect
numbers. Recall, a number $n$ is said to be $c$-multiperfect if $\sigma(n)=cN$
for some integer $c$. One expects analogous results for multiperfect numbers
in terms of $c$.

We can also extend various other results about $a(n)=a_{2}(n)$ to
$a_{\alpha}(n)$ for $\alpha>1$. In [13], it is proven that $a_{2}(n)$ is a
strictly increasing function. In general, this is false for $a_{\alpha(n)}$.
However, we strongly suspect that for any $\alpha>1$, $a_{\alpha}(n)$ is
strictly increasing for sufficiently large $n$. In fact we suspect that

######  Conjecture 26.

For any $\alpha>1$, and $C>0$, for sufficiently large $n$ one has
$a_{\alpha}(n+1)>a_{\alpha}(n)+C.$

In [13], the second difference of $a(n)$ was studied. We defined
$f(n)=a(n+1)+a(n-1)-2a(n)$. Extending this notation, let us write
$f_{\alpha}(n)=a_{\alpha}(n+1)+a_{\alpha}(n-1)-2a_{\alpha}(n)$. As before, we
have then $f_{2}(n)=f(n)$.

In that paper, it was noted that $f_{2}(n)$ is generally positive and that
negative values occurred generally at a prime gap. For example, $f(31)=-5$.
This corresponds to the fact that $30$th prime is 113, right before a record
setting gap; the next prime is 127. Motivated by this: we ask the following
questions for all $\alpha>1$:

  1. 1.

Are there infinitely many values of $n$ where $f_{\alpha}(n)$ is negative?

  2. 2.

Is the set of $n$ where $f_{\alpha}(n)<0$ a subset of those $n$ where $P_{n}$
is right after a record setting gap?

  3. 3.

Are there infinitely many $n$ where $P_{n}$ occurs at a record setting gap and
$f_{\alpha}(n)$ is positive?

  4. 4.

Does $f_{\alpha}(n)$ take on any integer value? In particular, is $f(n)$ ever
zero?

  5. 5.

Does $a_{\alpha}(n+1)-a_{\alpha}(n)$ take on every positive value?

We suspect that there are values of $\alpha$ where questions 4 and 5 are are
false. We are substantially more uncertain about the other questions.

##  2 Sums of divisors of numbers near $n!$

It is well known that for any $n>1$, any number of form $n!\pm 1$ is
deficient. We reproduce a proof here.

######  Lemma 27.

For any prime $p>2$, we have $b_{2}(p)\geq p$.

###### Proof.

Assume that $b_{2}(p)=m$. Then since $\frac{x}{x-1}$ is a decreasing function
for positive $x$ and the $i$th prime after $p$ is at least $p+i$, we must have

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

Thus

$2p-2<p+b(p)-1,$

and so $b_{2}(p)>p-1$, and so $b_{2}(p)\geq p$. ∎

Variants of the above lemma are very old and to date back to Servais [11] who
used it to prove that an odd perfect number with smallest prime divisor $p$
must have at least $p$ distinct prime factors. It immediately follows from
this Lemma that any non-deficient number with smallest prime factor $p$ must
have at least $p$ distinct prime divisors.

To prove from that that any number of the form $n!\pm 1$ must be deficient,
note that any such number must have smallest prime factor at least $n+1$. If
$n!\pm 1$ were not-deficient it would need to have at least $n+1$ distinct
prime factors, but then we would have $(n+1)^{n+1}\leq n!\pm 1$, which is
absurd.

In this section, we will use the same methods as earlier in this paper to
prove the related result that in fact

$\lim_{n\rightarrow\infty}h(n!\pm 1)=1.$

We will in fact prove a stronger claim:

######  Theorem 28.

Let $a$, $b$ and $c$ be positive integers. Then

$\lim_{n\rightarrow\infty}h(\frac{a}{b}n!\pm c)=h(c).$

Note that simply using the method of Servais’s Lemma is insufficient to prove
Theorem 28. Using only that method, one would just get that a number with
$h(n)\geq\alpha$ with smallest prime factor $p$ has at least about
$(\alpha-1)p$ distinct prime factors. So we really do need to use Mertens
theorem.

To prove Theorem 28, it suffices to prove the following:

######  Theorem 29.

$\lim_{n\rightarrow\infty}H\left(\frac{a}{b}n!\pm 1\right)=1.$

###### Proof.

We’ll prove the result with $\frac{a}{b}n!-1$. The proof is nearly identical
for $\frac{a}{b}n!+1$.

Fix positive integers $a$ and $b$. Fix an $\alpha>1$, and assume that there is
an increasing sequence $n_{1},n_{2}\cdots$ such that for all $k$,
$\frac{a}{b}n!$ is an integer and $H(\frac{a}{b}n_{k}!\pm 1)\geq\alpha$. For
sufficiently $n_{k}$, the smallest prime divisor of $\frac{a}{b}n_{k}!-1$ is
at least $n+1$. According to Theorem 5 for any $\epsilon>0$, for sufficiently
large $n_{k}$ we have that $\frac{a}{b}n_{k}!\pm 1$ has at least
$(1-\epsilon)\frac{(n_{k}+1)^{\alpha}}{\alpha\log(n_{k}+1)}$ distinct prime
factors. Choose $\epsilon=1/2$. Then for sufficiently large $n_{k}$, we must
have

$(n_{k}+1)^{\frac{1}{2}\frac{(n_{k}+1)^{\alpha}}{\alpha\log(n_{k}+1)}}<\frac{a}{b}n_{k}!-1<a{n_{k}}^{n_{k}}.$

This implies that

$(n_{k})^{\frac{1}{2}\frac{(n_{k})^{\alpha}}{\alpha\log(n_{k}+1)}}<\frac{a}{b}n_{k}!-1<a{n_{k}}^{n_{k}},$

and hence

$\frac{1}{2}\frac{(n_{k})^{\alpha}}{\alpha\log(n_{k}+1)}<n_{k}.$

However, this inequality cannot be true for any sufficiently large $n_{k}$
since $\alpha>1$. Thus, there are only finitely many possible values for
$n_{k}$ and the result is proved. ∎

With a little work one can use essentially the exact same method as above to
prove the following slightly more general result:

######  Theorem 30.

Let $g(n)$ be an increasing function satisfying the following conditions:

  1. 1.

If $n$ is sufficiently large then for any prime $p$ with $p\leq n$,
$(p,g(n))=1$.

  2. 2.

There is a constant $C$ such that $g(n)<n^{Cn}$.

Then

$\lim_{n\rightarrow\infty}H(g(n))=1.$

Four functions which seem natural to investigate in the context of Theorem 29,
are $A(n)=h(n!+1)$, $B(n)=h(n!-1)$, $C(n)=H(n!+1)$, and $D(n)=H(n!+1)$. We
will focus on $A(n)$, although the same remarks apply essentially to the other
three.

One might hope that $A(n)$ would be a decreasing function given our earlier
limit and the fact that $A(1)=\frac{3}{2}$, and that $A(n)>1$ for all $n>1$.
However, this is not the case. $A(3)=h(7)=\frac{8}{7}=1.1428\cdots$, and
$A(4)=h(25)=\frac{31}{25}=1.24$. It is easy to find similar values for $B(n)$,
$C(n)$, and $D(n)$.

We have

######  Conjecture 31.

There are arbitrarily large $n$ such that $A(n)<A(n+1)$.

We similarly have for the other three function:

######  Conjecture 32.

There are arbitrarily large $n$ such that $B(n)<B(n+1)$.

######  Conjecture 33.

There are arbitrarily large $n$ such that $C(n)<C(n+1)$.

######  Conjecture 34.

There are arbitrarily large $n$ such that $D(n)<D(n+1)$.

Note that standard conjectures about primes imply these conjectures. For
example, whenever $n!+1$ is prime and $(n+1)!+1$ is composite, one will have
that $A(n)<A(n+1)$ and $C(n)<C(n+1)$. Since it is believed that there are
infinitely many prime values of $n!+1$, and since there are infinitely many
values of $n!+1$ which are composite (by Wilson’s Theorem), we should strongly
believe Conjecture 31.

One obvious question is if there are infinitely many $n$ where $n!+1$ is
composite but $A(n)<A(n+1)$. This question seems difficult but we would
tentatively conjecture that it is also yes, since if $n!+1$ is a semiprime
with both prime factors near the square root, and $n!+1$ has a small prime
divisor substantially smaller than its square root, we will have
$A(n)<A(n+1)$.

Let $S_{A}$ be the set of positive integers such that $A(n)<A(n+1)$. Define
$S_{B}$, $S_{C}$ and $S_{D}$ similarly. What is the density of these sets? The
obvious conjectural natural densities for $S_{A}$ is 1/2 given the apparent
slow rate which $A(n)$ tends towards 1. We do not see any way of even proving
that this density exists. Curiously, we do not at this time see a way of even
ruling out $S_{A}$ has density 1. If that were the case, it would mean that
$A(n)<A(n+1)$ on almost all $n$ but that $A(n)$ takes large drops at isolated
$n$ which drive it downwards overall. This seems like it should be obviously
false but we do not see any easy proof.

It seems likely that

######  Conjecture 35.

$\lim_{n\rightarrow\infty}H\left(\left(n!\right)^{n}+1\right)=1.$

Unfortunately, Merten’s Theorem with by itself does not seem to be strong
enough to prove Conjecture 35. It would likely require some result of the sort
that $n!+1$ must have many fewer distinct prime factors than $\log_{n}(n!)$.

We use this as an opportunity to ask similar questions about $h(f(n))$ for
other somewhat well-behaved fast growing functions $f(n)$. Let us consider the
functions $J(n)=h(2^{n}-1)$ and $K(n)=h(2^{n}+1)$, $L(n)=H(2^{n}-1)$, and
$M(n)=H(2^{n}+1)$. Unlike with our earlier quadruplet, the behavior in the
$-1$ case here is substantially different than the $+1$ case. It is a
straightforward consequence from Fermat’s Little Theorem, and the fact that
the sum of the reciprocals of the primes diverges, that $J(n)$ can be
arbitrarily large.

In contrast, it is less obvious that $K(n)$ can be arbitrarily large.

######  Theorem 36.

$K(n)$ can be arbitrarily large.

###### Proof.

Consider a prime $p\equiv 3$ (mod 8). Note that $2$ is a quadratic non-reside
mod $p$. Thus, by Gauss’s Lemma, $2^{\frac{p-1}{2}}\equiv-1$ (mod $p$). So,
whenever $n$ is odd, and $\frac{p-1}{2}|n$, we have $p|2^{n}+1$. Now note that
by the strong form of Dirichlet’s theorem, the sum of the reciprocals of the
primes which are 3 mod 8 diverges. Thus, to make $h(2^{n}+1)$ arbitrarily
large, we just need to set $n$ to be the odd part of $k!$ for large $k$. ∎

One natural question in this context is to look at record setting values of
$K(n)$. This becomes an analog of the superabundant numbers restricted to just
looking at one more than a power of 2. More generally, given an increasing
function $f(n)$, we’ll define a number to be a _$f$ -superabundant_ if
$h(f(n))>h(f(i))$ for any $i<n$. Note that the usual superabundant numbers are
just the $f$-suberabundant numbers where $f(x)$ is the identity function.

In our context then, we are interested in the $K$-superabundant numbers. We’ll
write the $n$th $K$-superabundant number as $a_{n}$. The first few are $1$,
$3$, $5$, $9$, $15,$ $45$, $135\cdots$. Note that the first few $a_{i}$ are
odd. It seems likely that all the $a_{i}$ are odd because if $n$ is even, then
$2^{n}+1$ is not divisible by 3, and that makes making $2^{n}+1$ large
substantially more difficult. Proving that they are all odd seems difficult. A
related open question is whether for any odd prime $p$, if $p|a_{n}$ for all
arbitrarily large $n$?

Note that the $J$-superabundant numbers behave very differently. They to be
always even except the first term, and seem to grow much slower. The first few
are 1, 2, 4, 6, 8, 12, 24, 36, 60, 120, and 180. (Here we are using the
convention that $\sigma(0)=0$.)

## References

  * [1] M. Avidon On the distribution of primitive abundant numbers. Acta Arith. 77 (1996), 2, 195–205.
  * [2] L. E. Dickson, Finiteness of the Odd Perfect and Primitive Abundant Numbers with n Distinct Prime Factors. Amer. J. Math. 35 (1913), 4, 413–422.
  * [3] P. Dusart, The $k$th prime is greater than $k(\ln k+\ln\ln k-1)$ for $k\geq 2$. Math. Comp. 68 (1999), 411–-415.
  * [4] P. Ellia, A remark on the radical of odd perfect numbers, Fibonacci Quarterly 50 3 (2012) 231–234
  * [5] M. Kishore, On Odd Perfect, quasiperfect and almost perfect numbers. Math. Comp. 36 (1981), 154, 583-586
  * [6] Klyve, Dominic; Pidde, Melissa; Temple, Kathryn Estimating the density of the abundant numbers. Integers 19 (2019), Paper No. A16.
  * [7] Kobayashi, Mitsuo A new series for the density of abundant numbers. Int. J. Number Theory 10 (2014), 1, 73–84.
  * [8] J. Lichtman, Jared Duker The reciprocal sum of primitive nondeficient numbers. J. Number Theory 191 (2018), 104–118
  * [9] K. Norton, Remarks on the number of factors of an odd perfect number. Acta Arith. 6 (1960/1961), 365–374.
  * [10] J. Rosser, L. Schoenfeld, Sharper bounds for the Chebyshev functions $\theta$(x) and $\Psi$(x). Math. Comp. 29 (1975), 243–269.
  * [11] C. Servais, Sur les nombres parfaits, Mathesis 8 (1888), 92–93.
  * [12] Schinzel, Andrzej Erdős’s work on the sum of divisors function and on Euler’s function. Erdös centennial, 585–610, Bolyai Soc. Math. Stud., 25, 2013.
  * [13] J. Zelinsky, On the total number of prime factors of an odd perfect number. Integers, to appear.