# Weighted Versions of the Arithmetic-Mean-Geometric Mean Inequality and
Zaremba’s Function

Tim<EMAIL_ADDRESS>Joshua Zelinsky222Hopkins School, New
Haven, CT, USA<EMAIL_ADDRESS>
###### Abstract

We use the weighted version of the arithmetic-mean-geometric-mean inequality
to motivate new results about Zaremba’s function, $z(n)=\sum_{d|n}\frac{\log
d}{d}$. We investigate record-setting values for $z(n)$ and related functions.
We prove there are only finitely many record-setting values for
$v(n)=\frac{z(n)}{\log\tau(n)}$ where $\tau(n)$ is the number of divisors of
$n$, and we list all record setters for $v(n)$. Closely connected inequalities
motivate the study of numbers which are pseudoperfect in a strong sense.

##  1 Introduction

Given a positive integer $n$, we will write $\sigma_{k}(n)$ as the sum of the
$k$th powers of the divisors of $n$. We will write $\sigma(n)=\sigma_{1}(n)$
and write $\tau(n)$ for the number of divisors of $n$, noting that
$\tau(n)=\sigma_{0}(n)$. Unless otherwise noted, when referring to ‘divisors’
of a number, we will mean just the positive divisors.

Considerations of the means of the divisors of a number is an old problem. It
is well known that if $n$ is a natural number, then the geometric mean of the
divisors of $n$ is $\sqrt{n}$. The arithmetic mean of the set of divisors is
just $\frac{\sigma(n)}{\tau(n)}$, and it is straightforward to see that the
harmonic mean of the divisors of $n$ is $\frac{n\tau(n)}{\sigma(n)}$. These
seem to have been first explicitly noted by Ore [14]. Ore also noted the
following nice property: Let $D$ be the set of divisors of $n$, and let $h$ be
the harmonic mean of $D$, and $a$ be the arithmetic mean of $D$. Then the
geometric mean of $h$ and $a$ is itself the geometric mean of $D$. This
property does not hold in general if we replace $D$ with a general set of
positive integers not arising from a set of divisors, Consideration of these
means and when they take on integer values have been extensively studied. See
for example [9].

One might hope for something interesting from applying the arithmetic-mean-
geometric-mean-harmonic-mean inequality (AM-GM-HM) to these divisor sets.
Recall, the AM-GM-HM says that given a finite list of positive numbers, their
arithmetic mean is at least their geometric mean which is at least their
harmonic mean. Equality occurs exactly when all numbers in the list are equal.
Applying this to the divisors of a number the bounds leads to the inequality:

$\frac{\sigma(n)}{\tau(n)}\geq\sqrt{n}\geq\frac{n\tau(n)}{\sigma(n)}.$ (1)

But Equation 1 is essentially trivial and gives extremely weak bounds for all
but very small $n$ since $\tau(n)$ grows slower than $n^{\epsilon}$ for any
$\epsilon>0$ . At some level, the triviality should not be surprising; these
inequalities hold for any positive set of numbers, and the inequalities do not
“know” that we are dealing with sets of natural numbers, much less sets of
divisors. So the room to get number theoretic content out of these is small.

However, there is a weighted version of the AM-GM-HM. A question then arises
whether by using carefully chosen weights we can get statements with some
number theoretic content. The main goal of this paper is to show that the
answer is “yes, but only weakly.”

Recall, the weighted arithmetic-geometric mean inequality says that

######  Lemma 1.

if $\alpha_{1},\alpha_{2}\cdots\alpha_{k}$ are real numbers, with
$\alpha_{1}+\alpha_{2}\cdots\alpha_{k}=1$, and $x_{1},x_{2},\ldots x_{k}$ are
positive real numbers, then

$\prod x_{i}^{\alpha_{i}}\leq\sum\alpha_{i}x_{i}.$

In what follows, we will refer to Lemma 1 as simply the weighted AM-GM
inequality.

The goal of this paper is to use the weighted AM-GM inequality, and various
tightened versions of it, to obtain inequalities which while, not strong, are
to our knowledge not in the literature. Some of our bounds with the weighted
AM-GM inequality will be surpassed by other bounds in the paper which use
other methods, and thus one way of thinking about the use of the weighted
inequality here is as a motivator for examining the relationship between
different functions. In the course of doing so, we will also introduce a new
class of pseudoperfect numbers which will match almost precisely the broadest
context where some of our proofs apply.

Many of our results will concern Zaremba’s function, which is defined as
$z(n)=\sum_{d|n}\frac{\log d}{d}$. Notice that
$z(n)=\log\prod_{d|n}d^{\frac{1}{d}}$. This function was first introduced by
Stanisław Zaremba who used it to study certain lattices relevant for numerical
integration [25]. Subsequently, the function was investigated by Erdős and
Zaremba [7], who proved that

$\limsup\frac{z(n)}{(\log\log n)^{2}}=e^{\gamma},$

where $\gamma$ is Euler’s constant.

The average value of $z(n)$ does not appear to be explicitly noted in the
literature, but it is straightfoward to find.

######  Proposition 2.

We have

$\sum_{n\leq x}z(n)=x(-\zeta^{\prime}(2))+O(\log^{2}x).$

###### Proof.

We have

$\sum_{n\leq x}z(n)=\sum_{d\leq x}\lfloor\frac{x}{d}\rfloor\frac{\log
d}{d}=\sum_{d\leq x}\left(\frac{x}{d}+O(1)\right)\frac{\log d}{d}=\sum_{d\leq
x}\frac{\log d}{d^{2}}+O\left(\sum_{d\leq x}\frac{\log d}{d}\right).$ (2)

We now note that

$\sum_{d\leq x}\frac{\log d}{d}=O((\log x)^{2}),$

and that

$x\sum_{d\leq x}\frac{\log d}{d^{2}}=x\left(\sum_{d=1}^{\infty}\frac{\log
d}{d^{2}}+O\left(\frac{\log x}{x}\right)\right)=x(-\zeta^{\prime}(2))+O(\log
x).\qed$

Weber [23] is also notable for examining a variety of functions whose behavior
is similar but not exactly the same as $z(n)$, such as replacing the $\log d$
in the definition with $(\log d)(\log\log d)$. There is also an upper bound on
$z(n)$ as a function of which primes divide $n$.

######  Lemma 3.

(Lemma 4.3 from [23]) For all $n$,

$z(n)\leq\left(\prod_{p|n}\frac{p}{p-1}\right)\left(\sum_{p|n}\frac{\log
p}{p-1}\right).$

Here $p$ ranges only over primes.

In what follows, we will also write
$h(n)=\sigma_{-1}(n)=\sum_{d|n}\frac{1}{d}$, and note that
$h(n)=\frac{\sigma(n)}{n}$. Recall that a number is said to be perfect if
$\sigma(n)=2n$, or equivalently that $h(n)=2$. A number is said to be abundant
if $\sigma(n)>2n$, and deficient if $\sigma(n)<2n$.

We will write $H(n)=\prod_{p|n}\frac{p}{p-1}$. We will set $M(x)=\prod_{p\leq
x}\frac{p}{p-1}$. Also, given a positive integer $n$, with $t$ positive
divisors, we will write the divisors of $n$ in increasing order as $d_{0}$,
$d_{1},\cdots d_{t-1}$. Note that we have started our index at $0$ and end at
$t-1$. This notation will be convenient since often we will want to take a sum
which runs of over all divisors of an integer which are greater than 1, rather
than a sum over all positive divisors.

We recall here a few well-known facts which will be useful to us. In
particular

######  Lemma 4.

$h(n)\leq H(n)$ with equality if and only if $n=1$. And for any $n$,

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

######  Lemma 5.

(Mertens’ theorem) We have $M(x)\sim e^{\gamma}\log x$ where $\gamma$ is
Euler’s constant.

######  Lemma 6.

(Euclid-Euler Theorem). If $n$ is even, then $n$ is perfect if and only if
$n=(2^{p}-1)(2^{p-1})$ where $2^{p}-1$ is prime.

Note that if $2^{p}-1$ is prime, then $p$ is prime, but the converse is not
true. This fact motivates our choice of the letter $p$ in the above lemma.

##  2 Upper bounds for $z(n)$ in terms of $\tau(n)$

One gets a particularly nice result from applying the weighted version of the
AM-GM inequality to perfect numbers. Let be $n$ is a perfect number. We will
write $d_{1},d_{2}\cdots d_{k}$ as the positive divisors of $n$ greater than 1
in increasing order. We may apply the weighted AM-GM inequality with
$a_{i}=\frac{1}{d_{i}}$, and $x_{i}=d_{i}$. Note that $\sum_{i=1}^{k}a_{i}=1$
and so we may apply the weighted AM-GM inequality using the $a_{i}$ as our
weights. (This is why we need to often throw 1 out as one of our divisors,
because otherwise our weights will sum to 2.)

We then apply the weighted AM-GM inequality to obtain that

######  Theorem 7.

If $n$ is a perfect number, then

$\prod_{d|n}d^{\frac{1}{d}}<\tau(n)-1.$

Equivalently, $z(n)<\log(\tau(n)-1).$

Note that this inequality is false for some non-perfect $n$. For example, it
fails whenever $n$ is prime, and it also fails for $n=24$. Thus, $n$ being
perfect is really necessary in Theorem 7. Note also that when $n$ is perfect,
$z(n)$ is the Shannon entropy of the probability distribution where each
divisor $d>1$ has probability $\frac{1}{d}$.

There’s still not that much number theoretic content in this form, since we’re
just using that the sum of $\frac{1}{d_{i}}$ has the correct sum. In fact, we
can make a version of this for deficient numbers which is almost as tight. Let
$h(n)=\frac{\sigma(n)}{n}.$ Then as before we set for each $i\geq 1$,
$\alpha_{i}=\frac{1}{d_{i}}$. We set $x_{i}=d_{i}$, for all $0\leq i\leq t$
(including 1). However, we set $\alpha_{0}=2-h(n)$. Notice that we do in fact
have $\sum\alpha_{i}=1$, with $\alpha_{0}$ essentially making up for the
deficiency. We then apply the weighted AM-GM inequality, note that $1$ to any
power is 1 and obtain the following.

######  Theorem 8.

If $n$ is a deficient number then

$\prod_{d|n}d^{\frac{1}{d}}\leq\tau(n)+2-h(n).$

Equivalently,

$z(n)\leq\log\left(\tau(n)+2-h(n)\right).$

But even this might overstate the number-theoretic content. We could have
chosen the $x_{i}$ to be any set of real numbers all greater than 1 whose sum
of reciprocals is at most 1 and gotten an essentially similar result.

In fact, the situation is slightly worse than that. Recall, a number $n$ is
said to be pseudoperfect if the sum of some subset of its proper divisors adds
up to $n$. Trivially, any perfect number is also pseudoperfect. Theorem 7
applies just as well to pseudoperfect numbers where we replace $\tau(n)-1$
with the number of elements in the relevant divisor sum. In what follows, most
but not all of our results about perfect numbers will apply to pseudoperfect
numbers.

Given a positive integer $n$, when $n$ is pseudoperfect, we will write $S_{0}$
to be a subset of its positive proper divisors which sum to $n$. We will also
write $S_{1}(n)$ to be a subset of all the divisors of $n$ (possibly including
$n$), and with the sum of $S_{1}$ adding to $2n$. Notice that any $S_{0}$
gives rise to a valid $S_{1}$ by $S_{1}\cup\\{n\\}$.) We will refer to $S_{0}$
as a divisor set of $n$. Notice that we refer to _a_ divisor set of $n$,
rather than _the_ divisor set of $n$, since there may be more than one. In
particular, if $n_{1}$ and $n_{2}$ are distinct pseudoperfect numbers, then
$n_{1}n_{2}$ is pseudoperfect with at least two different choices of $S_{0}$
one inherited from $n_{1}$ and one inherited from $n_{2}$. For example, 6 is
pseudoperfect, since $1+2+3=6$, and $20$ is pseudoperfect with $1+4+5+10=20$,
We have

$1(20)+2(20)+3(20)=6(20)=6(1)+6(4)+6(5)+6(10).$

Thus, we have as choices for $S_{0}$ both $\\{20,40,60\\}$ as well as
$\\{6,24,30,60\\}$. (Some pseudoperfect numbers have more than one choice of
$S_{0}$ even though they are not inheriting them from another set.)

With our new notation to discuss not just perfect but also pseudoperfect
numbers, Theorem 7 becomes the more general claim:

######  Theorem 9.

Let $n$ be a pseudoperfect number with divisor set $S_{0}$. Then

$\sum_{d\in S_{0}}\frac{\log d}{d}\leq|S_{0}|.$

This result is also not only true for all pseudoperfect numbers, but it also
is very weak even for even perfect numbers which we know actually exist. In
particular, a consequence of the Euclid-Euler theorem on even perfect numbers
is that if $n$ is an even perfect number, then

$z(n)<\left(\sum_{i=1}^{\infty}\frac{\log(2^{i})}{2^{i}}\right)\left(\sum_{i=1}^{\infty}\frac{\log(3^{i})}{3^{i}}\right)=\frac{3}{2}(\log
2)(\log 3).$

In fact, if there are infinitely many even perfect numbers and $P_{i}$ is the
$i$th even perfect number then

$\lim_{i\rightarrow\infty}z(P_{i})=2\log 2.$

So, at least in their case, this is a very weak bound. It seems like getting
stronger lower and upper bounds on $z(n)$ when $n$ is an odd perfect number
would be worthwhile.

One may wonder also how tightly one can bound $z(n)$ from above in terms of
$\tau(n)$ when there are no special restrictions on $n$. We first have the
following obvious bound.

######  Proposition 10.

We have $z(n)=O\left(\log(\tau(n))^{2}\right)$.

###### Proof.

Note that $\frac{\log x}{x}$ is a decreasing function for $x\geq e$, and thus
we have $z(n)\leq\sum_{i=1}^{\tau(n)+1}\frac{\log i}{i}=O((\log\tau(n))^{2}).$
∎

However, we can with a little work get a better result, and again just using
the weighted AM-GM inequality.

######  Lemma 11.

Let $n$ be a positive integer greater than 1. Then

$z(n)<\left(h(n)-1\right)\log\frac{\tau(n)-1}{h(n)-1}.$

###### Proof.

This is essentially just applying the weighted AM-GM inequality with $x_{i}$
running over $d|n$ for $d>1$ and taking as weights
$\alpha_{i}=\frac{x_{i}}{\sum_{i=1}^{\tau(n)-1}x_{i}}.$ ∎

Note also that the inequality in Lemma 11 is strongest when $h(n)-1$ is very
small or very large. This becomes essentially our earlier inequality for
perfect numbers when $h(n)=2$. We may use Lemma 11 to get a tighter bound for
$z(n)$ purely in terms of $\tau(n)$.

######  Proposition 12.

$z(n)=O\left((\log\tau(n)(\log\log\tau(n)\right)$.

###### Proof.

By Lemma 11, we need only show that $h(n)=O(\log\log\tau(n))$. To see this,
note that $n$ has at most $\log_{2}\tau(n)$ distinct prime divisors. Thus, if
$P_{j}$ is the $j$th prime number then we have

$h(n)<H(n)\leq\prod_{p\leq P_{\lfloor\log_{2}\tau(n)\rfloor}}\frac{p}{p-1}.$

We have

$\prod_{p\leq
P_{\lfloor\log_{2}\tau(n)\rfloor}}\frac{p}{p-1}=M(P_{\lfloor\log_{2}\tau()\rfloor})=O\left(\log(\log_{2}\tau(n)\log\log\tau(n))\right)=O(\log\log\tau(n)).$

where the penultimate step is an application of the Prime Number Theorem
together with Lemma 5. ∎

Comparing the size of $z(n)$ to $\tau(n)$ motivates the definitions of the
following functions. Define $v(n)$ by

$v(n)=\frac{z(n)}{\log\tau(n)}.$

Two natural sequences of integers are the record setters for $v(n)$ and
$z(n)$. By the record setters of a function $f(n)$ we mean the set $R_{f}$
defined by

$R_{f}:=\\{n\in\mathbb{N}|\forall m\in\mathbb{N},m<n\implies f(m)<f(n)\\}.$

That is, a number is a record setter for a function $f(x)$ if $f(n)$ is larger
than $f(i)$ for any $0<i<n$. Record setters for arithmetic functions have been
studied since Hardy and Ramanujan. Examples include $R_{\tau}$, called the
highly composite numbers, $R_{h}$, called the superabundant numbers, and
$R_{\sigma}$, called the highly abundant numbers.[1][17]

Neither $R_{z}$ or $R_{v}$ is currently in the OEIS. For $v(n)$, the first few
record setters are 2, 3, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720,
840, 1260, 1680, 2520, 5040, 10080, 15120.

One may wonder if $v(n)$ and $z(n)$ ever have different record setters aside
from the initial 1. And in fact, they do start off the exact same way. The
first few record setters for $z(n)$ are also 1, 2, 3, 4, 6, 12, 24, 36, 48,
60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 10080, 15120,

However, they disagree at the next term, 25200, which is a record setter for
$z(n)$ but not for $v(n).$ More terms for both are included in the Appendix.
If one looks at that data, one will notice a few patterns. In that list,
whenever $v(n)$ has a record setter, so does $z(n)$. We will show this is the
case by showing that in fact the set of record setters for $v(n)$ is a finite
list where we explicitly give all of them. It turns out that the very last
record setter for $v(n)$ is $321253732800$.

Many consecutive elements in $R_{z}$ have the same number of divisors. Can
that happen infinitely often? One might think that whenever this happens, we
might think that $v(n)$ is now also a record setter. But this can fail if
$\tau(n)$ is already too large in that range to contribute to a new record
setter. For example, 4497552259200 and 4658179125600 are both record setters
for $z(n)$ and both have 9216 divisors. However, the second is not a record
setter for $v(n)$, since although $v(4658179125600)>v(4497552259200)$, both
have smaller $v(n)$ values than the previous record setter.

The highly composite numbers, superabundant numbers, and highly abundant
numbers share the property that all of them are “waterfall numbers.”
(Sometimes called “Hardy-Ramanujan” numbers.) These numbers are $n$ of the
form $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ where $p_{i}$ is the
$i$th prime, and where $a_{b}\geq a_{c}$ whenever $b<c$. These numbers form
OEIS sequence A025487.[24]

We have a similar property for $R_{z}$ and $R_{v}$.

######  Proposition 13.

With the exception of 3, $R_{z}$ and $R_{v}$ consist only of waterfall
numbers.

###### Proof.

We sketch the proof for $R_{z}$. The proof for $v(n)$ is nearly identical. The
same basic proof as for the other record setting sets works; if a number $n$
is a record-setter and one has $a_{b}<a_{c}$ for some $b<c$, then one could
swap $a_{b}$ and $a_{c}$ to get a smaller number $n^{\prime}$ where
$z(n^{\prime})>z(n)$, and hence $n$ would not be a record setter. The only
difficulty for $n=3$ arises because $\frac{\log x}{x}$ is not a decreasing
function until $x\geq e$, and in fact $\frac{\log 2}{2}<\frac{\log 3}{3}$. To
handle this case, note that one needs only consider possible record setters
where $3||n$. If $n=3^{k}$ then it can only be a record setter when $k=0$ or
$k=1$. Thus, we may assume that $n$ is either divisible by 2 or divisible by
$p$ for some $p>3$. If $n$ is not divisible by $2$, then we may replace all
the $p$s in the prime factorization of $n$ with $2$s, and get a smaller number
with a large $z$ value. Thus, we may assume that $n$ is divisible by some
power of 2 anda single 3, and then some larger primes, possibly raised to
larger powers. But then the argument that $\frac{\log x}{x}$ is a decreasing
function may then again be applied. ∎

Since every element of $R_{v}$ and $R_{z}$ is either 3 or a waterfall number,
finding all record setters less than a given $x$ is straightforward. To do
this efficiently, one use the fact that $n$ being a waterfall numbers has the
equivalent formulation that $n$ can be represented as the products of (not
necessarily distinct) primorials.

It is not hard to show that $h(n)$ grows arbitrarily large as $n$ ranges over
elements of $R_{z}$. If $h(n)$ is bounded, then the behavior is different. In
particular,

######  Conjecture 14.

Fix a constant $C>1$, and let $n_{1},n_{2}\cdots$ be an increasing sequence of
positive integers satisfying the following conditions.

  1. 1.

For all $i$, $h(n_{i})\leq C$.

  2. 2.

$\lim_{i\rightarrow\infty}\tau(n_{i})=\infty$.

Then

$z(n_{i})=o(\log\tau(n)).$

We will prove Conjecture 14 in some limited cases.

######  Proposition 15.

Assume that there are infinitely many even perfect numbers. Then the sequence
of even perfect numbers satisfies Conjecture 14.

###### Proof.

Let $n_{i}$ be the $i$th even perfect number, and write
$n_{i}=(2^{p_{i}-1})(2^{p_{i}}-1)$ which we can do by Lemma 6. We have that
$\lim_{i\rightarrow\infty}\tau(n_{i})=\infty$, so we need only prove that
$z(n_{i})$ is bounded above by some constant. By Lemma 3,

$z(n_{i})\leq\frac{2}{2-1}\frac{(2^{p_{i}}-1)}{(2^{p_{i}}-2)}\left(\frac{\log
2}{2-1}+\frac{\log(2^{p_{i}}-1))}{2^{p_{i}}-2}\right).$

Since $\frac{logx}{x-1}$, and $\frac{x}{x-1}$ are both decreasing functions,
we have that

$\frac{2}{2-1}\frac{(2^{p_{i}}-1)}{(2^{p_{i}}-2)}\left(\frac{\log
2}{2-1}+\frac{\log(2^{p_{i}}-1))}{2^{p_{i}}-2}\right)\leq
2(\frac{3}{2})\left(\log 2+\frac{\log 3}{3}\right),$

and so we are done. ∎

Proposition 15 required the assumption that there are infinitely many even
perfect numbers. We can make a version of the statement which applies to a
broader class of numbers which does not rely on any unproven assumption. In
particular, a number is said to be primitive non-deficient if it satisfies
$h(n)\geq 2n$, and $h(m)<h(2m)$ for any $m$ where $m|n$, and $m<n$. For any
odd prime $p$, any number $n$ of the form

$n=p2^{\lfloor\log_{2}p\rfloor}$

is primitive non-deficient. Essentially, we take $p$ and we multiply it by the
smallest power of 2 which is at least $\frac{p}{2}$. Even perfect numbers are
of this form, and are essentially the case where the power of 2 is just barely
above $\frac{p}{2}$. Numbers of this form also all satisfy $h(n)\leq 3$.

We may then modify the argument from Proposition 15 to handle this specific
class of primitive non-deficient numbers.

######  Proposition 16.

Let $p_{i}$ be the $i$th odd prime. Then the sequence
$s_{i}=p_{i}2^{\lfloor\log_{2}p_{i}\rfloor}$ satisfies Conjecture 14.

In fact, for the sequence $s_{i}$, we have the stronger result that
$\lim_{i\rightarrow\infty}s_{i}=4\log 2$.

We have one other sequence of primitive non-deficient numbers which satisfy
Conjecture 14.

######  Proposition 17.

Let $p_{i}$ be the $i$th prime number and let $c_{j}=p_{j}p_{j+1}\cdots
p_{k_{j}}$ where $k_{j}$ is the smallest $k$ such that the product is not
deficient. Then $c_{j}$ satisfies Conjecture 14. In particular, we have that
$\log\tau(c_{j})$ is asymptotic to $(\log 2)\frac{p_{j}^{2}}{2\log p_{j}}$. We
also have that $j$, $z(c_{j})=O(\log p_{j})$.

###### Proof.

It follows from the techniques used in Section 5 of [26], that for large $j$,
$k_{j}$ is asymptotic to $\frac{p_{j}^{2}}{2\log p_{j}}$, so $\log\tau(c_{j})$
is asymptotic to $(\log 2)\frac{p_{j}^{2}}{2\log p_{j}}$.

That we may bound $z(c_{j})$ above then follows from applying Lemma 3, and
using that $\sum_{p\leq x}\frac{\log p}{p}=O((\log x))$. ∎

We now turn out attention to verifying that $R_{v}$ is finite, and that
$R_{v}$’s largest element is $321253732800$.

We first need a Lemma from Weber:

######  Lemma 18.

If $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$, then

$z(n)=\sum_{i=1}^{k}\sum_{j=1}^{a_{i}}\frac{j\log
p_{i}}{p_{i}^{j}}h\left(\frac{n}{p_{i}^{a_{i}}}\right).$ (3)

(Here $h(n)=\frac{\sigma(n)}{n}=\sigma_{-1}(n)$.)

Note that a similar version of this in Erdős and Zaremba is incorrect. This is
the corrected version in Weber. They have a $h(\frac{n}{p^{j}})$ term when it
should be $h(\frac{n}{p^{a_{i}}})$

######  Lemma 19.

If $n$ is a record setter for $z(n)$ or $v(n)$, and $13|n$, then $4|n$.

###### Proof.

Set $m=3(5)\cdots p_{k-1}$ where $p_{k-1}\geq 7$. Set $n_{1}=4m$, and
$n_{2}=2mp_{k}$. We have that

$\tau(n_{1})=3(2^{k-2})<2^{k}=\tau(n_{2}),$

so the Lemma will be proved if we can show that $z(n_{1})>z(n_{2}).$ This is
sufficient because record setters must be waterfall numbers, so if $a_{1}=1$,
all the other prime factors must be raised to the first also.

Our plan is to use Equation 3 to directly estimate $z(n_{1})-z(n_{2})$ by
splitting it up into two terms. One term has the contribution from both $i=1$
$i=k$, and the other term has the contribution for $i$ ranging from $2$ to
$k-1$. We will show that both are positive. We have

$z(n_{1})-z(n_{2})=A+B$

where

$A=\frac{\log 2}{2}h\left(\frac{n_{1}}{4}\right)+\frac{2\log
2}{4}h\left(\frac{n_{1}}{4}\right)-\frac{\log
2}{2}h\left(\frac{n_{2}}{2}\right)-\frac{\log
p_{k}}{p_{k}}h\left(\frac{n_{2}}{p_{k}}\right),$

and

$B=\sum_{i=2}^{k-1}\frac{\log
p_{i}}{p_{i}}\left(h\left(\frac{n_{1}}{p_{i}}\right)-h\left(\frac{n_{2}}{p_{i}}\right)\right)$

Since $h(n)$ is a multiplicative function we may rewrite $A$ as

$A=h(m)\left(\log 2-\frac{\log
2}{2}\frac{p_{k}+1}{p_{k}}-\frac{3}{2}\frac{\log p_{k}}{p_{k}}\right),$

and the quantity in the parentheses is positive as long as $p_{k}\geq 13$.

For $B$, we will show that for any $i$, where $2\leq i\leq k-1$, that

$h\left(\frac{n_{1}}{p_{i}}\right)-h\left(\frac{n_{2}}{p_{i}}\right)\geq 0.$

Since $n_{1}=4m$, and $n_{2}=2mp_{k}$, this is the same as verifying that

$\frac{7}{4}h\left(\frac{m}{p_{i}}\right)-\frac{3}{2}\left(\frac{p_{k}+1}{p_{k}}\right)h\left(\frac{m}{p_{i}}\right)\geq
0.$

This last inequality is easily seen to be true when $p_{k}\geq 7$.

∎

We can generalize this Lemma to get the following:

######  Lemma 20.

Let $p_{i}$ be the $i$th prime number, and fix some $\ell$. Assume $n$ is
record setter for $v(n)$ or $z(n)$ and $p_{k}$ satisfies

  1. 1.

$p_{k}>p_{\ell}(p_{\ell}+1)$

  2. 2.

$\log p_{\ell}+\frac{(2\log p_{\ell})}{p_{\ell}}\geq(\log
p_{\ell})\frac{p_{k}+1}{p_{k}}+\frac{\log p_{k}}{p_{k}}(p_{\ell}+1)$

  3. 3.

$p_{k}|n$.

then $p_{\ell}^{2}|n$.

###### Proof.

Assume that $p_{k}$ satisfies the above three conditions. Note that the first
condition implies $p_{k}>p_{\ell}$. Set

$m=2^{a_{1}}\cdots p_{\ell-1}i^{a_{\ell-1}}p_{i+1}\cdots p_{\ell-1}.$ (4)

Notice that $m$ does not include a $p_{i}$ term We will set
$n_{1}=p_{\ell}^{2}m$, and set $n_{2}=p_{\ell}mp_{k}$. As with the previous
Lemma, the result will be proven if we can show that $z(n_{1})\geq z(n_{2})$.
(As before we get that $\tau(n_{1})<\tau(n_{2})$ by the same logic.)

We apply Lemma 3 to get that

$z(n_{1})-z(n_{2})=A+B,$

where $A$ has the terms arising from $i=\ell$ and $i=k$, and $B$ has all the
other terms.

Then

$A=\frac{\log
p_{\ell}}{p_{\ell}}h\left(\frac{n_{1}}{p_{\ell}^{2}}\right)+\frac{2\log
p_{\ell}}{p_{\ell}^{2}}h\left(\frac{n_{1}}{p_{\ell}^{2}}\right)-\frac{\log
p_{\ell}}{p_{\ell}}h\left(\frac{n_{2}}{p_{\ell}}\right)-\frac{\log
p_{k}}{p_{k}}h\left(\frac{n_{2}}{p_{k}}\right).$

Applying that $h(n)$ is multiplicative, we may rewrite this as

$A=\frac{h(m)}{p_{\ell}}\left(\log p_{\ell}+\frac{2\log
p_{\ell}}{p_{\ell}}-(\log p_{\ell})\frac{p_{k}+1}{p_{k}}-\frac{\log
p_{k}}{p_{k}}(p_{\ell}+1)\right).$

The quantity inside the parentheses is positive by condition 2 above.

We have

$B=\sum_{i=1,i\neq\ell}^{k-1}\frac{\log
p_{i}}{p_{i}}\left(h\left(\frac{n_{1}}{p_{i}}\right)-h\left(\frac{n_{2}}{p_{i}}\right)\right).$

So we just need to verify that for $i$ with $2\leq i\leq k-1$, we have

$h\left(\frac{n_{1}}{p_{i}}\right)-h\left(\frac{n_{2}}{p_{i}}\right)\geq 0.$

We have

$h\left(\frac{n_{1}}{p_{i}}\right)-h\left(\frac{n_{2}}{p_{i}}\right)=h\left(\frac{m}{p_{i}}\right)\left(\frac{p_{\ell}^{2}+p_{\ell}+1}{p_{\ell}^{2}}-\frac{p_{\ell}+1}{p_{\ell}}\frac{p_{k}+1}{p_{k}}\right).$

Condition 3 guarantees that the above expression is positive.

∎

We have as an immediate consequence:

######  Lemma 21.

Assume $n$ is a record setter for $v$. If $17|n$ then $3^{2}|n$. If $37|n$,
then $5^{2}|n$. If $67|n$, then $7^{2}|n$.

######  Lemma 22.

If $p_{i}$ is the $i$th prime number, and $k\geq 29$ then

$\frac{2}{1}\frac{3}{2}\frac{5}{4}\cdots\frac{p_{k}}{p_{k}-1}\leq
e^{\gamma}(\log p_{k})\left(1+\frac{1}{1.2\log^{2}p_{k}}\right),$

where $\gamma$ is Euler’s constant.

###### Proof.

It follows from Theorem 8 [18] that if $p_{k}\geq 293$, that

$\frac{2}{1}\frac{3}{2}\frac{5}{4}\cdots\frac{p_{k}}{p_{k}-1}\leq
e^{\gamma}\left((\log p_{k})(1+\frac{1}{2\log^{2}p_{k}}\right).$

The last step is to then verify that the inequality for holds the primes
between 29 and 293. ∎

######  Lemma 23.

If $n$ has at least twenty distinct prime divisors, and we set $k=\omega(n)$,
and $t=\log k$ then we have

$\prod_{p|n}\frac{p}{p-1}\leq e^{\gamma}(\log\left(k\left(t+\log
t-0.5\right)\right)\left(1+\frac{1}{1.2\left(\log\left(k\left(t+\log
t-0.5\right)\right)\right)^{2}}\right)$ (5)

###### Proof.

Assume $n$ has $k=\omega(n)$ distinct prime factors. Since $\frac{x}{x-1}$ is
a decreasing function for $x>1$, we may have
$\prod_{p|n}\frac{p}{p-1}\leq\frac{2}{1}\frac{3}{2}\frac{5}{4}\cdots\frac{p_{k}}{p_{k}-1}$
where $p_{k}$ is the $k$th prime.

We have from Lemma 22 that

$\prod_{p|n}\frac{p}{p-1}\leq e^{\gamma}(\log
p_{k})\left(1+\frac{1}{1.2\log^{2}p_{k}}\right).$ (6)

Rosser and Schoenfeld showed that for $k\geq 6$ the $k$th prime satisfies

$p_{k}<k(\log k +\log\log k-0.5).$ (7)

Combining Equation 6 and 7 then gives the desired result.

∎

######  Lemma 24.

Let $p_{i}$ be the $i$th prime number. Then $\sum_{i=1}^{k}\frac{\log
p_{i}}{p_{i}-1}\leq 1+\log p_{i}$.

###### Proof.

This follows from estimates in [18]. ∎

######  Lemma 25.

If $n$ has at least $k$ distinct prime divisors for $k\geq 20$, then

$\sum_{p|n}\frac{\log p}{p-1}\leq 1+\log\left(k(\log k +\log\log
k-0.5)\right).$ (8)

######  Proposition 26.

Any record setter for $v(n)$ has at most 54 distinct prime divisors.

###### Proof.

Assume we have a record setter $n$ with at least $k\geq 19$ distinct prime
divisors. Then by Lemma 21, $4|n$, $9|n$, $25|n$, and $49|n$.

Thus $\tau(n)\geq 2^{k-2}(3^{4})$ which is the same as
$\log\tau(n)\geq(k-2)\log 2+4\log 3$.

Thus we apply Weber’s 4.3 Lemma, Lemma 5, and Lemma 8 and set $\log k=t$, and
$a=k(t +\log t-0.5)$to obtain that

$v(n)<\frac{\left(1+\log a\right)e^{\gamma}(\log
a\left(1+\frac{1}{1.2\left(\log a\right)^{2}}\right)}{(k-2)\log 2+4\log 3}.$
(9)

Call the right-hand side of Inequality 9 $f(k)$. Since we have that
$v(321253732800)=1.70595781\cdots$, if we have a record setter, we must have
$f(k)>1.7059578$. But this can only occur when $k<54.4$, and since $k$ is an
integer, we obtain that $k\leq 54$.

∎

######  Lemma 27.

Assume that $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{s}^{a_{s}}q_{1}q_{2}\cdots
q_{t}$ where the $p_{i}$ and $q_{i}$ are all distinct primes. Then

$z(n)\leq\left(\sum_{i=1}^{s}\frac{\log
p_{i}}{p_{i}-1}+\sum_{i=1}^{t}\frac{\log
q_{i}}{q_{i}}\right)\left(\prod_{i=1}^{s}\frac{p_{i}}{p_{i}-1}\right)\left(\prod_{i=1}^{s}\frac{q_{i}+1}{q_{i}}\right).$

###### Proof.

Assume we have such an $n$. We set $P=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{s}^{a_{s}}$, and $Q=q_{1}q_{2}\cdots q_{t}$. Note that $(P,Q)=1$, and that
$n=PQ$. Then, from Lemma 3, we have

$\begin{split}z(n)&=\sum_{i=1}^{s}\sum_{j=1}^{a_{i}}\frac{j\log
p_{i}}{p_{i}^{j}}h\left(\frac{PQ}{p_{i}^{a_{i}}}\right)+\sum_{i=1}^{t}\frac{\log
q_{i}}{q_{i}}h\left(\frac{PQ}{q_{i}}\right)\\\ &\leq
h(PQ)\left(\sum_{i=1}^{s}\sum_{j=1}^{a_{i}}\frac{j\log
p_{i}}{p_{i}^{j}}+\sum_{i=1}^{t}\frac{\log q_{i}}{q_{i}}\right).\end{split}$
(10)

We have that

$h(PQ)\leq\prod_{i=1}^{s}\frac{p_{i}}{p_{i}-1}\prod_{i=1}^{t}\frac{q_{i}}{q_{i}+1},$

which combines with $\sum_{j=1}^{a_{i}}\frac{j\log
p_{i}}{p_{i}^{j}}\leq\frac{\log p_{i}}{p_{i}-1}$, to give our desired
inequality. ∎

Note that when $Q=1$ (equivalently $t=0$), Lemma 27 becomes just Weber’s 4.3
Lemma.

######  Theorem 28.

The largest record-setter for $v(n)$ is 321253732800.

###### Proof.

Our method of proof can be thought of as “reverse bootstrapping.” By Lemma 26,
any record setter must have at most 54 distinct prime divisors. The 54th prime
is 251. We then apply Weber’s 4.3 Lemma; which gives that for any $n$ which is
a record setter for $v(n)$, we have $z(n)\leq 50.142$. So any new record
setter must satisfy that

$\frac{50.142}{\log\tau(n)}\geq 1.705.$ (11)

The 1.705 in Equation 11 is arises from $v(321253732800)>1.705$. Then Equation
11 is equivalent to the statement that $\tau(n)\leq e^{\frac{50.142}{1.705}}$.
Since $2^{\omega(n)}\leq\tau(n)$ we have
$\omega(n)\leq\log_{2}e^{\frac{50.142}{1.705}}<42.5$, and so $n$ must have at
most $42$ distinct prime divisors.

We repeat this process again. The 42nd prime is 181. So by Weber’s Lemma 4.3,
any record setter $v(n)$ must have $z(n)\leq 44.638$ and so we have
$\omega(n)\leq\log_{2}e^{\frac{44.638}{1.705}}<38$. Thus, $n$ has at most 37
distinct prime divisors.

The 37th prime is 157, and we get that we must have $z(n)\leq 41.982$, and so
$\omega(n)\leq 35$. The 35th prime is 149, and we have $z(n)\leq 40.843$. This
yields that $\omega(n)\leq 34$. The 34th prime is 139, and we have
$z(n)<40.265$, which unfortunately only gives us $\omega(n)\leq 34$. To
continue down, we observe that if $n$ has at least $19$ distinct prime
divisors, we must have $4|n$, $9|n$, $25|n$. and $25|n$, and so we have the
tighter inequality $2^{\omega(n)-4}3^{4}\leq\tau(n)$. Thus, we have we have
$\omega(n)\leq 4(1-\log_{2}3)+\log_{2}e^{\frac{40.265}{1.705}}<32$ This gets
us that $\omega(n)$ is at most 31.

We now repeat this process again, using this tighter inequality. The 31st
prime is 127, and so we get that $z(n)\leq 39.047$, which yields that
$\omega(n)\leq 30$. The 30th prime is 113, and this gives us that $z(n)\leq
37.782$, and so $\omega(n)\leq 29$. The 29th prime is 109, which gives us
$z(n)\leq 37.083$ This only yields $\omega(n)\leq 29$, so we need to do more
work. This should not be surprising; 109 is the last prime in the quadruplet
of very close primes $101,103,107,109$ and so the bound given by Weber’s 4.3
Lemma is larger than for what one would expect from a typical prime given just
its size.

Assume for now that $n$ has exactly 29 distinct prime factors. We must have
that $4|n$, $9|n$, $25|n$, and $49|n$. If we had $11^{2}|n$, then we would
have $\tau(n)\geq 3^{5}2^{2}4$, and so $\log\tau(n)>22.12859$, but then we
would have $v(n)\leq\frac{37.083}{22.128}<1.7$. Thus, 11 and other primes
cannot be raised to more than the first power. If 3 is raised to a power
higher than the second, then we must have $\tau(n)\geq(2^{3})(3^{3})2^{2}5$.
and so $\log\tau(n)\geq 22.7039$ , and so similarly we have then
$v(n)\leq\frac{37.083}{22.7039}$.

Thus, our only option here is that
$n=2^{2}3^{2}5^{2}7^{2}(11)(13)(17)\cdots(109)$, which we can check. We can
continue this way algorithmically at each stage identifying a finite list to
check and verifying that they do not match. The code for this, and its results
can be found at [22].

∎

##  3 Lower bounds for $z(n)$

So far, most of our results have focused on upper bounds for $z(n)$. Let us
now turn to lower bounds for $z(n)$.

Assume that $n>1$ and let $p$ be the smallest prime divisor of $n$. Then we
trivially have that

$z(n)=\sum_{d|n}\frac{\log d}{d}=\sum_{d|n,d>1}\frac{\log d}{d}\geq\frac{\log
p}{d}=(\log p)\sum_{d|n,d>1}\frac{1}{d}=(\log p)h(n).$

Here, again, the weighted AM-GM inequality can give us tighter bounds, albeit
only marginally better. As before, we will first see what we get when $n$ a
perfect number, where the form is a little nicer. We choose as our choice of
weights $\alpha_{i}=\frac{1}{d_{i}}$ with $1\leq i\leq t$. But this time we
also set $x_{i}=\frac{1}{d_{i}}$. Since $n$ is perfect, the sum of the weights
is again 1.

######  Theorem 29.

Let $n$ be a perfect number. Then,

$\log\sum_{d|n,d>1}\frac{1}{d^{2}}\geq-\sum_{d|n,d>1}\frac{\log d}{d}=-z(n)$

We have as an immediate consequence the following corrolary.

######  Corollary 30.

Let $n$ be an odd perfect number, and let $p$ be the smallest prime factor of
$n$. Then

$\log(4p)\leq\sum_{d|n}\frac{\log d}{d}.$

We get a similar statement for pseudoperfect numbers.

Note that there is a massive gap here between our lower and upper bounds for
$z(n)$ even in this case where $n$ is a perfect number. In particular, it
follows from results of Norton [13], that if $n$ is a perfect or abundant
number with smallest prime factor $p$, then $n$ has at least roughly
$\frac{p^{2}}{\log p}$ distinct prime factors, and thus we have roughly that
$2^{\frac{p^{2}}{\log p}}\leq\tau(n)$.

As before, we get a similar inequality for all $n$. Setting $\alpha_{i}$ as
running over $\frac{1}{d_{i}}$ where $d_{i}$ ranges over the divisors of $n$
which are greater than 1, and setting
$x_{i}=\frac{1}{d_{i}\left(h(n)-1\right)}$ we obtain:

######  Proposition 31.

For all positive integers, $n>1$, with smallest prime factor $p$, we have

$z(n)\geq\left(h(n)-1\right)\log\sum_{d|n,d>1}\frac{1}{d^{2}},$

with equality if and only if $n$ is prime. We also have

$z(n)\geq\left(h(n)-1\right)\log(2p)$

It is not hard to improve Proposition 31 slightly, using the Prime Number
Theorem to get:

######  Proposition 32.

For any $\epsilon>0$, if $p$ is a sufficiently large prime, then for any $n$
with smallest prime factor $p$,

$z(n)\geq\left(h(n)-1\right)\log(2p(1-\epsilon)\log p).$

Proposition 32 can be made explicit with only a little work. Unfortunately,
this is only a very tiny improvement since $(2p\log p)$ is itself inside a
logarithm. In fact, we cannot hope to improve these bounds more than a small
amount without some additional information about $n$, since in Proposition 17,
we exhibited an infinite sequence $c_{j}$ with smallest prime factor $p_{j}$
and where $z(c_{j})=O(\log p_{j}).$

Note that one cannot get a general lower bound for $z(n)$ in terms of
$\tau(n)$ even if one fixes the smallest prime factor of $n$ to be $p$. In
particular, fix $t$ to be a positive integer. Then for any $\epsilon>0$, there
is an $n$ is with $\tau(n)=t+1$ and where $z(n)\leq(1+\epsilon)z(p^{q-1})$
where $q$ is the smallest prime factor of $t$. To see this, consider numbers
$n$ of the form $p^{q-1}r^{\frac{t}{q}}$ where $r$ is a large prime.

Note also that it follows from Mertens’ theorem that for any $\epsilon>0$ and
positive constant $K$, there is an integer $n$ where $\frac{z(n)}{h(n)}>K$ and
$h(n)<1+\epsilon$.

##  4 Other weights

Another weighing option for the weighted AM-GM inequality is to restrict our
attention to perfect numbers and use all of the divisors, including $n$
itself, and then divide by $2n$. We can use as weights
$\alpha_{i}=\frac{d_{i}}{2n}$, and set $x_{i}=\frac{n}{d_{i}}$, where $i$
ranges from $0$ to $t$.

Note that to have a chance at get an interesting result here we do need to
that $n$ is a perfect number rather than just a pseudoperfect number, since
otherwise $\frac{n}{d}$ will not be guaranteed to be a divisor in our divisor
set.

From this choice of $x_{i}$ and $\alpha_{i}$ we obtain that if $n$ is a
perfect number then,

$\log\frac{\tau(n)}{2}\geq\sum_{d|n}\frac{d\log\left(\frac{n}{d}\right)}{2n}.$
(12)

If $n$ is perfect, a little work can show that Inequality 12 is equivalent to
Inequality 7 by substituting $n/d$ for $d$. However, Inequality 12 turns out
to be less interesting, because it is actually true for all $n$ for
essentially trivial reasons since the function $f_{n}(x)=x\log\frac{n}{x}$ has
its global maximum at $m=\frac{n}{e}$, where $f(m)=\frac{n}{e}$, so all the
terms on the right hand side of Inequality of 12 are less than $\frac{1}{2}$.

Some other choices of weights are more interesting. We will write $\phi(n)$ to
denote Euler’s totient function.

Given the close connection between the Euler function $\phi(n)$ and
$\sigma(n)$, applications involving $\phi(n)$ are a natural thing to
investigate. We may use the weighted AM-GM inequality with
$\alpha_{i}=\frac{d}{2n}$ and $x_{i}=\phi(d)$. This gives us:

######  Proposition 33.

If $n$ is a perfect number then

$\sum_{d|n}\frac{d}{2n}\log\phi(d)\leq\log\sum_{d|n}\frac{\phi(d)d}{2n}.$

We also note that by replacing $d$ with $\frac{n}{d}$, we obtain that
$\log\sum_{d|n}\frac{\phi(d)d}{2n}=\log\sum_{d|n}\frac{\phi(\frac{n}{d})}{2d}$
Thus, one also has:

######  Proposition 34.

If $n$ is a perfect number then

$\sum_{d|n}\frac{d}{2n}\log\phi(d)\leq\log\sum_{d|n}\frac{\phi(\frac{n}{d})}{2d}.$

We wish to extend these notions to pseudoperfect numbers. Given a positive
integer $n$ and $S$, a subset of the divisors of $n$, we will define three
functions as follows. We define

$A(S,n)=\sum_{d\in S}\frac{d}{2n}\log\phi(d),$ (13)

$B(S,n)=\log\sum_{d\in S}\frac{\phi(d)d}{2n},$ (14)

and

$C(S,n)=\log\sum_{d\in S}\frac{\phi(\frac{n}{d})}{2d}.$ (15)

We immediately have from using the same weighing as in Proposition 33 that for
any pseudoperfect $n$ and corresponding set $S_{1}$ we have $A(S_{1},n)\leq
B(S,n)$. However, it is not true that $C(S_{1},n)=B(S_{1},n)$ for any
pseudoperfect $n$ and corresponding set $S_{1}$. The proof for perfect numbers
used the replacement of $d$ with $\frac{n}{d}$ and for a general pseudoperfect
number this cannot happen. It is possible, that $d\in S$ but
$\frac{n}{d}\not\in S$. In these circumstances, $C(S_{1},n)$ can not only be
less than $A(S_{1},n)$ it can even be negative. For example,
$C(\\{13,26,,39,78,39],78)=-2.0053\cdots$.

One can reasonably object that the example of $n=78$ is extreme, and that this
number is obviously not really a perfect number since the set does not include
1. And in fact, it is not hard to show that if $1\in S_{1}$, then
$C(S_{1},n)>0$. However, there are examples of pseudoperfect $n$ which have an
$S_{1}$, where $1\in S_{1}$ and yet $A(S_{1},n)>C(S_{1},n)$. An example is
$n=90$ with the set $S_{1}=\\{1,3,5,6,30,45,90\\}.$

We can show even more pathological behavior. Consider $n=272$ with
$S_{1}=\\{1,16,1734,68,136,272\\}$. This example is interesting because $272$
is a primitive non-deficient number, which is a much stronger property than
being a primitive pseudoperfect. Of course, even primitive non-deficients are
much more common than odd primitive non-deficients. However, this example has
$A(S_{1},n)>C(S_{1},n)$.

######  Question 35.

Is there an odd primitive non-deficient $n$ which is pseudoperfect with set
$S_{1}$, with $1\in S_{1}$, and where $A(S_{1},n)>C(S_{1},n)?$

Note that $n=1575$ almost provides an answer to the above question. In
particular, $n=1575$ is an odd primitive non-deficient, pseudoperfect number
with pseudoperfect set
$S_{1}=\\{225,35,5,7,9,15,21,63,75,105,175,315,525,1575\\}$ where
$A(S_{1},n)>C(S_{1},n)$. However, $1$ is not a element of $S_{1}$

However, note if a pseudoperfect number $n$ does satisfy that $d$ is in its
sum set if and only if $\frac{n}{d}$ is in its sum set, then for that choice
of $S_{1}$ we will have $B(S_{1},n)=C(S_{1},n)$ and hence they will satisfy
that $A(S_{1},n)<C(S_{1},n)$. Motivated by this observation, we will say that
a number $n$ is _strongly pseudoperfect_ if it is a pseudoperfect number with
a divisor sum set $S_{1}$ where $d\in S_{1}$ if and only if $n/d\in S_{1}$.
The strongly pseudoperfect numbers are given by A334405 in the OEIS[6], and
the first few are

$6,28,36,60,84,90,120,156,210,216,240,252\cdots.$

Strongly pseudoperfect numbers can be thought of as a generalization of
Descartes’ “spoof perfect number.” Descartes noticed the number

$D=(3^{2})(7^{2})(11^{2})(13^{2})(22021).$

He saw that $D$ looks almost like it is an odd perfect number, in that if one
naively applies the sum of divisors formula to $D$, one has

$\sigma(D)=(3^{2}+3+1)(7^{2}+7+1)(13^{2}+13+1)(22021+1)=2(3^{2})(7^{2})(11^{2})(13^{2})(22021).$
(16)

A reader who has not previously seen Equation 16 should be immediately
suspicious because all the factors on the left-hand side are much smaller than
the supposed prime 22021. (The exception is of course $2202+1$, but $22021$
and $22021+1$ must be relatively prime.) And in fact, 22021 is not prime. But
if we pretend that it is prime, then we would have an odd perfect number.
Spoof perfect numbers are an obstruction to proving the non-existence of odd
perfect numbers because many non-trivial statements one can prove about odd
perfect numbers will also have corresponding, nearly identical statements
about spoofs. Since spoof perfect numbers exist, any collection of such
statements will then be insufficient to prove the non-existence of odd perfect
numbers.

Any spoof perfect number in the sense of Descartes will give rise to a
strongly pseudoperfect number if one expands out the factorization, and using
the factorization implied by pretending that the corresponding factors are
prime. Some authors have referred to the set of all spoof perfect numbers and
actual perfect numbers as “freestyle perfect numbers.” What precisely is meant
here has some degree of ambiguity. For example, when some $p$ is wrongly
assumed to be prime, do we wish to only do so if none of factors of $p$ appear
as primes themselves? See OEIS A174292 for some discussion.

In recent years, spoof perfect numbers have been generalized and made more
rigorous. John Voight introduced the example
$V=(3^{4})(7^{2})(11^{2})(19^{2})(-127)=2V$, where if we ignore the somewhat
glaring issue that $-127$ is a negative number when we apply the usual formula
for $\sigma$ we obtain that

$(3^{4}+3^{3}+3^{2}+3+1)(7^{2}+7+1)(11^{2}+11+1)(19^{2}+19+1)(-127+1).$

Another way we can allow for spoofs, introduced in [10], is where we are
allowed to fail to notice that we have repeated a prime factor. For example,
consider the number

$P=(3^{2})(7^{2})(7^{2})(13^{1})(-19)^{2}$

If one ignores that $7^{2}$ and $7^{2}$ are not powers of distinct primes, and
ignores that $(-19)$ is negative (as with Voight’s example), then one has that

$(3^{2}+1+1)(7^{2}+7+1)(7^{2}+7+1)(13+1)((-19)^{2}+(-19)+1)=2P.$

Continuing with this idea, [10] gave a more rigorous expansion of these spoofs
and showed that there are infinitely many and of a wide variety of forms.

In any event, 156 is strongly pseudoperfect since the sum of all its divisors,
excepting 2 and 78, adds up to 2(156), and 156 is not a freestyle perfect or
spoof for any of these definitions. Thus, being a strongly pseudoperfect is a
broader class of numbers than spoofs.

The more general notions of spoof perfect numbers due to Voight and Nielsen
can give rise to a corresponding and more general notion of strongly
pseudoperfect numbers, where one is allowed to have negative divisors in
$S_{1}$, and where $S_{1}$ rather than be a set, becomes a multiset. However,
there is some room for hope that statements proven using the sort of
techniques in this paper may still be of some interested for the perfect
number problem. In particular, the weighted AM-GM inequality only applies to
positive numbers, and Zaremba’s function does not make sense for negative
numbers. Although there are inequalities similar to the weighted AM-GM
inequality which allow zero, they generally lose something. See for example,
the Steffensen inequality [21]. We will not investigate this broader notion
here.

If one follows one Descartes’s lead and restricts one’s “freestyle perfect
numbers” to only being allowed to pretend that a number is prime if that
number is genuinely relatively prime to all the other primes (real or pretend)
in the factorization, and do not allow negative primes, then every spoof
perfect number gives a strongly pseudoperfect number.

It is plausible that tightened versions of the weighted AM-GM inequality or
related bounds which use that we are taking sums over genuinely distinct
integers can at least avoid the obstruction posed by the most general form of
spoofs. For example, our argument giving a lower bound for $z(n)$ used that
the sum of $\frac{1}{d^{2}}$ was ranging over distinct positive integers $d$,
and our proof would not apply if repetition of divisors is allowed.

Unfortunately, all of these results apply to all Cartesian spoofs. Worse, when
one has applied Mertens’ theorem or similar density statements, it seems
unlikely that they will be highly informative, since prime density results
cannot see if one has slipped in a single non-prime or even a few non-primes
if they are sparse enough.

With that pessimistic digression out of the way, let us at least see what we
can prove about strongly pseudoperfect numbers.

######  Proposition 36.

If $m\geq 2,$ then $n=2^{m-1}(2^{m}-1)$ then $n$ is strongly pseudoperfect.

###### Proof.

The proof is essentially the same as the proof that if $2^{p}-1$ is prime,
then $2^{p-1}(2^{p}-1)$ is perfect, but we get to ignore any unwanted
contribution from divisors of $2^{n}-1$. ∎

If $n$ is pseudoperfect, then so is $mn$ for any $m$, since one take each
element in $S_{0}1$ and multiply it by $m$. However, this is not the case for
strongly pseudoperfect numbers. The next two results show different ways this
can fail.

######  Proposition 37.

If $n$ is strongly pseudoperfect, then $n$ is not $3$ (mod 4) and $n$ is not
$2$ (mod 3).

###### Proof.

We will prove that $n$ is not 2 (mod 3). The proof for $3$ (mod 4) is similar.
Suppose that $n$ is strongly pseudoperfect with divisor set $S$, and that $n$
is 2 mod 3. Since $n$ cannot be a perfect square, we may pair up the divisors
$d$ in $S$, as $(d,\frac{n}{d})$. Note, if $d\equiv 1$ (mod 3), then
$\frac{n}{d}\equiv 2$ (mod 3), and that if $d\equiv 2$ (mod 3), then
$\frac{n}{d}\equiv 1$ (mod 3). No divisor can be $0$ (mod 3), because $n$ is
not divisible by $3$. Thus, for each such pair we have $d+\frac{n}{d}\equiv 0$
(mod 3), and so the total sum of the divisors is $0$ (mod 3). This is a
contradiction since the sum of all the divisors is $2n$ which is $1$ (mod 3).
∎

Proposition 37 shows, incidentally, that if $n$ is either an odd strongly
pseudoperfect number, or is a strongly pseudoperfect number which is not
divisible by 3, then a positive proportion of its multiples will fail to be
strongly pseudoperfect. We suspect that the modulus restrictions in
Proposition 37 are the only modulus restrictions on strongly pseudoperfect
numbers.

If the Artin primitive root conjecture is true, (in particular if $2$ is a
primitive root for a positive proportion of primes), then Proposition 36 would
imply that a positive proportion of primes have a strongly pseudoperfect
number in at least half their congruence classes.
One may be tempted to wonder if $n$ is a strongly pseudoperfect number which
is divisible by 6, will all multiples of $n$ be strongly pseudoperfect? The
next result shows that this is not the case.

######  Proposition 38.

Let $n$ be a positive integer and let $P$ be a prime with $P>\sigma(n)$. Then
$nP$ is not strongly pseudoperfect.

###### Proof.

Note that we have

$\sum_{d|nP,(d,P)=1}d=\sigma(n).$

Thus, if $nP$ is strongly pseudoperfect with set $S_{1}$, then at least one
element in $S_{1}$ must be divisible by $p$, and $S_{1}$ must contain at least
one divisor of $n$. We can write

$\sum_{d\in S_{1}}d=\sum_{d\in S_{1},(d,P)=1}d+\sum_{d\in S_{1},P|d}d$

where both sums on the right-hand side are positive. But the second sum is
divisible by $P$ whereas the first is not, so the sum of all elements in
$S_{1}$ cannot be divisible by $P$, hence $\sum_{d\in S_{1}}d\neq nP.$ ∎

An open question of Erdős [8] is whether there is a constant $C$ such that if
$h(n)>C$, then $n$ must be pseudoperfect. Proposition 38 shows that this is
not the case for strongly pseudoperfect numbers. The pseudoperfect numbers
have positive natural density, but it seems likely that the strongly
pseudoperfect numbers have natural density zero.

Aside from the Euclid-like strongly pseudoperfect numbers, explicit families
of strongly pseuoperfect numbers seem hard to come by. However, we note that
60, 120, 240, 480, 960, 1920, 3840, 7680, 15360 are all strongly pseuoperfect.

######  Question 39.

For any $n\geq 0$, is $60(2^{n})$ strongly pseudoperfect?

There may be other similar families. However, any $m$ where $m(2^{n})$ is
strongly pseudoperfect for all $m$ must have $3|m$. To see this, note that if
$m\not\equiv 0$ (mod 3) then either $m\equiv 2$ (mod 3), or $2m\equiv 2$ (mod
3). Thus at least one of this pair of numbers must fail to be strongly
pseuoperfect. So hope for any similar family where we replace $60$ with
another $m$ must at least have $m\equiv 0$ (mod 3). This condition is
necessary but it is not sufficient: 90 is strongly pseudoperfect but 180 is
not.

Note also that our definition of strongly pseudoperfect does not insist that
$n\in S_{1}$. Any genuinely perfect number will have $1$ and $n$ in its
corresponding set of divisors which sums to $2n$. Define a number to be
extremely strongly pseudoperfect if it is strongly pseudoperfect for some set
$S_{1}$ with $n\in S_{1}$.

######  Question 40.

Is every strongly pseudoperfect number extremely strongly pseudoperfect?

Pomerance and Shevelev [15] defined a class of pseudoperfect numbers they
called “near-perfect numbers.” A number $n$ is said to be near-perfect number
when the sum of all $n$’s proper divisors except a single divisor $k$ adds up
to the number. Further investigation of near perfects includes [16]. An
example is 12, where the proper divisors of $12$ are $1,2,3,4,$ and $6$, and
one may omit $k=4$ to have $1+2+3+6=12$. One might expect that no strongly
pseudoperfect number can be a near perfect number; here one removes only a
single divisor and a strongly pseudoperfect number must lose at least two.

However, there are two problems with this line of reasoning. First, it may be
that one has a near perfect number $n=k^{2}$ where the single divisor omitted
is $k$, in which case the same set would give a strongly pseudoperfect sum for
$n$. Second, it may be that the omitted divisor is itself the sum of a pair of
divisors, and so we can take out that pair of divisors instead of $k$. Notice
that for the near perfect number 12, one could instead of omitting 4, omit the
pair of divisors $1$ and $3$ to get a pseudoperfect sum for 12. However, since
$(1)(3)\neq 12$, this does not give rise to a strongly pseudoperfect
representation.

Pomerance and Shevelev constructed what is likely an infinite family of
perfect square near-perfect numbers, since they showed that
$2^{p-1}(2^{p}-1)^{2}$ is near perfect, whenever $(2^{p}-1)$ is prime.
However, the omitted divisor $k$ is $2^{p}-1$, not the square root. This
prompts three questions.

######  Question 41.

Is there any near-perfect number $n$ which is a perfect square, the omitted
divisor $k$ is in fact the square root of $n$? Equivalently, does there exist
an $m$ such that $\sigma(m^{2})=2m^{2}+m$?

######  Question 42.

Are there infinitely many strongly pseudoperfect perfect squares?

######  Question 43.

Is there any number which is both near-perfect and strongly pseudoperfect?

Note that an affirmative answer to Question 41 would imply an affirmative
answer to Question and 43.

It is known that pseudoperfect numbers have positive natural density and that
perfect numbers have zero density. Let $SP$ be the set of strongly
pseudoperfect numbers.[1]

######  Question 44.

Does $SP$ have positive natural density?

Question 44 is connected to another obstruction to proving that no odd perfect
numbers exist. In [26], the second author introduced the following idea. Given
a set of natural numbers $S$, we will write $S(x)$ as the number of elements
in $S$ which are at most $x$. Euler proved that if $n$ is an odd perfect
number, then $n=q^{s}m^{2}$ for some prime $q$, where $q\equiv s\equiv 1$ (mod
4), and where $(q,m)=1$. Let be $E$ the set of all numbers of that form. Given
a statement $S$ we will write $E_{S}$ to be the set of elements of E which
satisfy $S$. The second author defined a property $s$ to be a weak property if

$\lim_{x\rightarrow\infty}\frac{E_{S}(x)}{E(x)}=1.$

Observe that if one has a finite list of properties which an odd perfect
number must satisfy, and all of them are weak properties, then they cannot
suffice to prove there are no odd perfect numbers. That is almost all elements
in $E$ will not have been ruled out, as the intersection of all the relevant
sets will still have positive density.

In that context, it seems natural to ask which properties proven here are weak
or strong properties. For example, simply being non-deficient is not a strong
property, but it is not weak either. In particular, let $D_{0}$ be the
statement “N is not deficient.” Then it is not too hard to show that.

$0<\lim_{x\rightarrow\infty}\frac{E_{D_{0}}(x)}{E(x)}<1.$

We then ask:

######  Question 45.

Is being a strongly pseudoperfect number a strong restriction, a weak
restriction, or neither?

Similarly, let $D_{1}$ be the statement “n is pseudoperfect with some $S_{1}$
where $A(S_{1},n)\leq C(S_{1},n)$.”

######  Question 46.

Is $D_{1}$ a strong property, a weak property, or neither?

Another variant of perfect numbers is that of a _primary pseudoperfect
number_.[19] A number $n$ is said to be a primary pseudoperfect number if $n$
satisfies the equation

$\frac{1}{n}+\sum_{p|n}\frac{1}{p}=1,$

where the sum is taken over all prime divisors of $n$. For example, $n=42$ is
primary pseudoperfect since one has

$\frac{1}{42}+\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=1.$

It is not hard to see that if $n$ is primary pseudoperfect and $n$ is not $1,$
$2,$ or $6$, then $n$ must be abundant. Primary pseudoperfect numbers are very
rare. We suspect the following.

######  Conjecture 47.

6 is the only primary pseudoperfect numbers which is also strongly
pseudoperfect.

A number is called a _weak primary pseudoperfect numbers_ if it satisfies the
weaker condition that

$\frac{1}{n}+\sum_{p|n}\frac{1}{p}\in\mathbb{N}.$

Note that now the sum is allowed to be any integer, not just 1. Other than
$n=1$, there is no known example of a weak primary pseudoperfect number that
is not in fact a primary pseudoperfect number. It may seem safe then to also
make what seems on its face to be only a slightly stronger conjecture.

######  Conjecture 48.

6 is the only weak primary pseudoperfect numbers which is also strongly
pseudoperfect.

At first glance, Conjecture 48 to be only a hairsbreadth stronger than
Conjecture 47, but this may not be the case. It is plausible that there are
many weak primary pseudoperfect numbers that are not primary pseudoperfect as
one gets to large numbers. In any event, we suspect that that Conjecture 48,
if true, will be substantially more difficult to prove than Conjecture 47.

Closely related to the weak primary pseudoperfect numbers are the _Giuga
numbers_ , which satisfy the similar equation

$\frac{-1}{n}+\sum_{p|n}\frac{1}{p}\in\mathbb{N}.$

Giuga numbers were extensively studied in [4]. The original motivation for
studying Giuga numbers was the Agoh–Giuga conjecture which states that if
$n>1$, then $n$ is prime if and only if $nB_{n-1}\equiv-1$ (mod $n$). The
Agoh–Giuga conjecture is equivalent to the statement that if $n>1$, then $n$
is prime if and only if

$1^{n-1}+2^{n-1}+\cdots+(n-1)^{n-1}\equiv-1\pmod{n}.$

Any counterexample to the Agoh-Giuga conjecture must be a Giuga number; this
fact was the original motivation for studying Giuga numbers. We suspect that
there are no strongly pseudoerfect Giuga numbers.

Recently, Grau, Oller-Marcén, and Sardonil[5], generalized the notion of weak
primary pseudoperfect numbers and of Giuga numbers to what they called a
_$\mu$ -Sondow number_, which is a number $n$ along with an integer $\mu$,
such that

$\frac{\mu}{n}+\sum_{p|n}\frac{1}{p}\in\mathbb{N}.$ (17)

Equation 17 is equivalent to the statement that

$\sum_{p|n}\frac{n}{p}\equiv-\mu\pmod{n},$ (18)

Therefore every $n$ is a $\mu$-Sondow number for some choice of $\mu$.

Grau, Oller-Marcén, and Sardonil conjectured that for any $\mu$, there is a
$\mu$-Sondow number greater than $|\mu|$ but noted that for any fixed $\mu$,
$\mu$-Sondow numbers are rare. In this context, we have the following
conjecture:

######  Conjecture 49.

For any $\mu$, there are only finitely many $\mu$-Sondow numbers which are
strongly pseudoperfect.

Note also that 11025 is a curious number for being odd, strongly pseudoperfect
and for being primitive non-deficient (that is, it is not deficient and has no
non-deficient proper divisors). Are there infinitely many such numbers?

For much of this paper, we have assumed that $n$ is perfect or pseudoperfect
number and used that fact to construct a choice of weights. However, one of
the basic facts about $\phi(n)$ is that for any $n$, it satisfies
$\sum_{d|n}\phi(d)$. In that context, another choice of weights is
$\alpha_{i}=\frac{\phi(d)}{n}$ where $d$ ranges over the divisors of $n$. If
one uses this choice with $x_{i}=d$, then the weighted AM-GM inequality yields
the following:

######  Proposition 50.

For any positive integer $n$,

$\sum_{d|n}\frac{\phi(d)\log d}{n}\leq\log\sum_{d|n}\frac{\phi(d)d}{n}.$

Note that this means that when $n$ is perfect, we have two natural sets of
numbers which both sum 1, namely $\frac{\phi(d)}{n}$ and $\frac{1}{2d}$. One
can think of these as two different probability distributions on set of
divisors, and can use ideas from probability to derive other inequalities. The
Kullback–Leibler divergence is one of the more obvious choices here, and a
sequel paper to this one will discuss the consequences of applying it in this
and related contexts.

##  5 Tightened versions of the AM-GM Inequality

Most of the bounds here rely on appropriately chosen weights for the weighted
AM-GM inequality. This raises a question: There are many tightened versions of
the inequality which specifically apply when we know more about the set of
numbers and choice of weights. Can we take advantage of them here? The answer
appears to be “yes but not by much.” However, it is plausible that further
tightened versions that use more information about the sets in question, would
work. But the ones in the literature currently are not well-optimized for our
purposes. We will discuss some of these, and present a modified versions
specifically using information from the situations of interest. However, none
of these will add more than a small improvement to our inequalities from
earlier without some further insight.

We list one potential exception here. Aldaz [2] has the following version of
the weighted AM-GM inequality.

$\prod_{i=1}^{n}x_{i}^{\alpha_{i}}\leq\sum_{i=1}^{n}{\alpha_{i}x_{i}}-\sum_{i=1}^{n}\alpha_{i}\left(x_{i}^{\frac{1}{2}}-\sum_{j=1}^{n}\alpha_{j}x_{j}^{\frac{1}{2}}\right)^{2}$
(19)

When apply Equation 19 with our earlier weighing to with
$\alpha_{i}=\frac{1}{d_{i}}$ where we range over the non-1 divisors of a
perfect number $n$, we obtain

$\prod_{d|n}d^{\frac{1}{d}}<\tau(n)-1-\sum_{d|n,d>1}\frac{1}{d}\left(d^{\frac{1}{2}}-\sigma_{\frac{-1}{2}}(n)+1\right)^{2}.$
(20)

It seems plausible that a non-trivial estimate of the last term could lead to
an interesting improvement in the general inequality for the perfect case. A
somewhat sharper inequality also due to Aldaz is in [3].

One approach of interest is to say that the weighted AM-GM inequality is just
essentially Jensen’s inequality with $f(x)=\log x$. We can tweak our choice of
$f(x)$, using that for our purposes, $f(x)$ need only obey a concavity or
convexity requirement in a specified interval. Since $n$ is generally much
larger than its smallest prime factor $p$, if one is going to use that all
non-1 divisors are in the range $[p,n]$, using the lower bound on the domain
is more helpful than using the upper bound.

We will therefore choose $f(x)=\log x+\frac{A}{2}\frac{1}{x}$ which is chosen
so that it is concave in the interval $[A,\infty]$.

######  Lemma 51.

Let $A$ be a positive real number, and let $f(x)=\log
x+\frac{A}{2}\frac{1}{x}$. Then $f^{\prime\prime}(x)\leq 0$ for all $x$ in the
interval $[A,\infty).$

###### Proof.

This follows immediately from the fact that
$f^{\prime\prime}(x)=\frac{-1+\frac{A}{x}}{x^{2}}$. ∎

We may then apply Jensen’s inequality to $f(x)$ obtain the following.

######  Lemma 52.

Assume that $\alpha_{1},\alpha_{2}\cdots\alpha_{k}$ are positive real numbers
where $\alpha_{1}+\alpha_{2}\cdots+\alpha_{k}=1$. Assume that
$x_{1},x_{2}\cdots x_{n}$ are real numbers in the interval $[A,\infty]$. Then

$\log(\sum_{i=1}^{k}\alpha_{i}x_{i})+\frac{A}{2}\left(\sum_{i=1}^{k}\alpha_{i}x_{i}\right)^{-1}\geq\sum_{i=1}^{k}\alpha_{i}\log
x_{i}+\frac{A}{2}\sum_{i=1}^{k}\frac{\alpha_{i}}{x_{i}}.$ (21)

From which we can obtain the following using our earlier weighing choice of
$\alpha_{i}=\frac{1}{d_{i}}$, and $x_{i}=d_{i}$ to obtain:

######  Proposition 53.

If $n$ is an odd perfect number with smallest prime factor $p$, then
$\log(\tau(n)-1)\geq z(n)+\frac{1}{2p}$.

There is also some hope of using existing versions of the weighted AM-GM
inequality which are already in the literature.

For example, Theorem 1 of [2] is, after taking logs:

######  Lemma 54.

Suppose that $n\geq 2$, and $x_{1},x_{2}\cdots x_{k}$ are non-negative real
numbers, with at least one $x_{i}>0$, and suppose that
$\alpha_{1},\alpha_{2}\cdots\alpha_{k}$ are positive real numbers where
$\alpha_{1}+\alpha_{2}\cdots+\alpha_{k}=1$. Then

$\log\sum_{i=1}^{k}\alpha_{i}x_{i}\geq\sum_{i=1}^{n}\alpha_{i}\log
x_{i}+2\left(1-\frac{\sum_{i=1}^{k}\alpha_{i}x_{i}^{\frac{1}{2}}}{\left(\sum_{i=1}^{k}\alpha_{i}x_{i}\right)^{\frac{1}{2}}}\right).$

So as before we obtain:

######  Lemma 55.

Suppose that $x_{1},x_{2}\cdots x_{n}$ are positive real numbers where
$\sum_{i=1}^{n}\frac{1}{x_{i}}=1$. Then

$\log n\geq\sum_{i=1}^{n}\frac{\log
x_{i}}{x_{i}}+2\left(1-\frac{\sum_{i=1}^{n}x_{i}^{-\frac{1}{2}}}{\sqrt{n}}\right).$

Another variant of the weighted AM-GM inequality in the literature is in [20]:

######  Lemma 56.

Assume that $\alpha_{1},\alpha_{2},\cdots\alpha_{n}$ are positive numbers with
$\alpha_{1}+\alpha_{2}\cdots+\alpha_{n}=1$. Set $\alpha_{\mathrm{min}}$ to be
the minimum of the $\alpha_{i}$, and $\alpha_{\mathrm{max}}$ to be the maximum
of the $\alpha_{i}$. Let $x_{1},x_{2}\cdots x_{n}$ be positive numbers, and
let $A$ be the unweighted arithmetic mean of the $x_{i}$, and $G$ be the
unweighted geometric mean of the $x_{i}$ Then we have

$\left(\frac{A}{G}\right)^{n\alpha_{\mathrm{min}}}\prod_{i=1}^{n}x_{i}^{\alpha_{i}}\leq\sum_{i=1}^{n}\alpha_{i}x_{i}\leq\left(\frac{A}{G}\right)^{n\alpha_{\mathrm{min}}}\prod_{i=1}^{n}x_{i}^{\alpha_{i}}.$
(22)

Using the same method as before, We obtain from Lemma 56 the following.

######  Proposition 57.

If $n$ is a strongly pseudoperfect number with divisor set $S_{1}$ then then

$\frac{|S_{1}|\log n}{3n}+\sum_{d\in S_{1}}\frac{\log
d}{d}\leq\log(|S_{1}|-1)$

###### Proof.

We apply Lemma 56 with $\alpha_{i}$ running over $d_{i}$ and
$x_{i}=\frac{1}{d}$ as before, and the proof is essentially identical to our
earlier result, except we also use that the geometric mean of the elements in
$S_{1}$ must be $\sqrt{N}$. Note that the proof that the geometric mean of the
divisors of a number is equal to the square root was essentially using fact
that divisors for non-squares can be paired up via $(d,\frac{n}{d})$. Thus,
the same proof still goes through for the geometric mean of the $S_{1}$ set
for a strongly pseudoperfect number, which completes the proof. ∎

In the specific case of perfect numbers Proposition 57 becomes:

######  Corollary 58.

If $n$ is a perfect number then

$\frac{\tau(n)\log n}{3n}+z(n)\leq\log(\tau(n)-1).$

But this is a very weak improvement. In particular, $\frac{\tau(n)\log n}{3n}$
goes to zero as $n$ goes to infinity.

Another inequality which may be of interest is the weighted Ky Fan’s
inequality. This stronger version of the weighted AM-GM inequality states that
if $x_{1},x_{2}\cdots x_{n}$ are all non-negative numbers at most
$\frac{1}{2}$, and $\alpha_{1},\alpha_{2}\cdots\alpha_{n}$ are non-negative
real numbers satisfying our usual condition that
$\alpha_{1}+\alpha_{2}\cdots\alpha_{n}=1$, then

$\frac{\prod_{i=1}^{n}x_{i}^{\alpha_{i}}}{\prod_{i=1}^{n}(1-x_{i})^{\alpha_{i}}}\leq\frac{\sum_{i=1}^{n}\alpha_{i}x_{i}}{\sum_{i=1}^{n}\alpha_{i}(1-x_{i})}.$
(23)

However, we have been unable to obtain a non-trivial result using the Ky Fan
inequality which is substantially tighter than any of our results here. It is
possible that someone more skilled withthe manipulation of inequalities may
find something more interesting. Similarly, a far reaching generalization of
the Ky Fan inequality is Levinson’s inequality.

######  Theorem 59.

(Levinson)[11] Let $a$ be a positive constant, and let $f(x)$ be a function
whose third derivative is positive in $(0,2a)$. Assume further that
$\alpha_{1},\alpha_{2}\cdots\alpha_{n}$ are positive real numbers with
$\alpha_{1}+\alpha_{2}\cdots+\alpha_{n}=1$, and that $x_{1},x_{2}\cdots x_{n}$
are in the interval $(0,a)$. Then

$\sum_{i=1}^{n}\alpha_{i}f(x_{i})-f\left(\sum_{i=1}^{n}\alpha_{i}x_{i}\right)\leq\sum_{i=1}^{n}\alpha_{i}f(2a-x_{i})-f\left(\sum_{i=1}^{n}\alpha_{i}(2a-x_{i})\right).$
(24)

The conditions for Levinson’s inequality have been further generalized in
[12].

In the cases of both Ky Fan’s inequality and Levinson’s inequality, there is a
large amount of flexibility in the choices involved, and we suspect that it
should be possible to use these inequalities to incorporate more number
theoretic information into inequalities to get tighter results. One potential
goal for future is under the assumption that $n$ is a perfect to get a lower
bound for $z(n)$ in terms of the number of distinct prime divisors of $n$.

##  Appendix A Algorithms and list of record setters

In order to find all record-setters for $z(n)$ and $v(n)$ up to a maximum n,
we generated all waterfall numbers up to n and then iterated through them in
order, computing z and v. For this task, we found it most effective to treat
waterfall numbers as a product of primorial numbers. Any product of primorials
is a waterfall number and every waterfall number can be represented as a
product of primorials, so algorithmically a waterfall number can be
represented by a list of exponents of the successive primorials. For example,
the list $[3,0,1]$ would represent $(2^{3})(6^{0})(30^{1})=240$ The empty list
represents the empty product, or 1.

The code starts with an empty list and uses a recursive scheme to construct
the waterfall numbers from 0 up to and possibly including a max-$n$. Each call
iteratively increments the last exponent in the list and outputs the resulting
products until the product would exceed max-n. For each pass through the loop,
the code also makes a recursive call that adds a 0 to the end of the list
(until it reaches a primorial larger than max-$n$). In this way, the code
generates every possible list of primorial exponents that represents a
waterfall number that is within range. For details, see [22].

$n$ | $z(n)$ | $\tau(n)$ | $v(n)=\frac{z(n)}{\log\tau(n)}$ | Record type
---|---|---|---|---
1 | 0 | 1 | - | z
2 | 0.34657359028 | 2 | 0.5 | Both
3 | 0.36620409622 | 2 | 0.52832083357 | Both
4 | 0.6931471805599453 | 3 | 0.6309297535714574 | Both
6 | 1.0114042647073518 | 4 | 0.7295739585136225 | Both
12 | 1.5650534091363246 | 6 | 0.8734729387592397 | Both
24 | 1.9574025114441351 | 8 | 0.9413116320946855 | Both
36 | 2.0693078747916944 | 9 | 0.9417825998016082 | Both
48 | 2.2113393276447026 | 10 | 0.9603724676117413 | Both
60 | 2.6291351167661694 | 12 | 1.0580418049066245 | Both
120 | 3.1536019699500124 | 16 | 1.1374214807461371 | Both
180 | 3.296829727702829 | 18 | 1.140624806721235 | Both
240 | 3.485150114597928 | 20 | 1.1633716889070718 | Both
360 | 3.897045010404812 | 24 | 1.2262363126738316 | Both
720 | 4.272243596316464 | 30 | 1.2560998721657932 | Both
840 | 4.438078029538006 | 32 | 1.2805586328585488 | Both
1260 | 4.611033134346248 | 36 | 1.286733295829273 | Both
1680 | 4.844788911264986 | 40 | 1.3133497506571945 | Both
2520 | 5.3572240096668935 | 48 | 1.3838661424637553 | Both
5040 | 5.816137669682257 | 60 | 1.4205296064593296 | Both
10080 | 6.08850361086746 | 72 | 1.4236565215589612 | Both
15120 | 6.248037988991858 | 80 | 1.425832955818364 | Both
25200 | 6.387503013476975 | 90 | 1.4195051527608638 | z
27720 | 6.653923297464633 | 96 | 1.4578036158378216 | Both
50400 | 6.675036433693168 | 108 | 1.4256406131942176 | z
55440 | 7.181545678560509 | 120 | 1.5000643477637126 | Both
110880 | 7.492166808574835 | 144 | 1.5075348623687792 | Both
166320 | 7.674162694562543 | 160 | 1.5120984963170123 | Both
277200 | 7.832742267027409 | 180 | 1.5083395629764706 | z
332640 | 7.99772386924453 | 192 | 1.5212041672534768 | Both
554400 | 8.160359559426864 | 216 | 1.518127795531729 | z
665280 | 8.183509553747768 | 224 | 1.5122033989903259 | z
720720 | 8.560085828505317 | 240 | 1.5618773555849463 | Both
1441440 | 8.907925303844221 | 288 | 1.5730156223216971 | Both
2162160 | 9.111778860078964 | 320 | 1.5796240997550626 | Both
3603600 | 9.288911580881782 | 360 | 1.5781086320000497 | z
4324320 | 9.473895419732957 | 384 | 1.5920793991588504 | Both
7207200 | 9.655498762787248 | 432 | 1.5911044178397928 | z
8648640 | 9.680805342657756 | 448 | 1.5857712086389766 | z
10810800 | 9.870365907305127 | 480 | 1.598754109907839 | Both
21621600 | 10.251889488064096 | 576 | 1.6129194210253008 | Both
36756720 | 10.418517980548113 | 640 | 1.6124072263755087 | z
43243200 | 10.469364642977977 | 672 | 1.608133517192604 | z
61261200 | 10.611850914128109 | 720 | 1.6129268471701907 | Both
73513440 | 10.814367000246168 | 768 | 1.6277407074294288 | Both
122522400 | 11.012526770001736 | 864 | 1.628693078747944 | Both
147026880 | 11.039663838081106 | 896 | 1.6239718455034093 | z
183783600 | 11.247419532881505 | 960 | 1.6379101218780654 | Both
367567200 | 11.664231546026192 | 1152 | 1.6546758215862982 | Both
698377680 | 11.725712741610478 | 1280 | 1.6389019055063054 | z

$n$ | $z(n)$ | $\tau(n)$ | $v(n)=\frac{z(n)}{\log\tau(n)}$ | Record type
---|---|---|---|---
735134400 | 11.90092229151731 | 1344 | 1.6521244370715231 | z
1102701600 | 11.928538234423074 | 1440 | 1.6402481809316225 | z
1163962800 | 11.93491248056824 | 1440 | 1.6411246794127394 | z
1396755360 | 12.154635463940904 | 1536 | 1.6566362239305026 | Both
2205403200 | 12.168527631038101 | 1680 | 1.6385170975662962 | z
2327925600 | 12.369007879661227 | 1728 | 1.659218315869329 | Both
2793510720 | 12.397909801933391 | 1792 | 1.6550213366391964 | z
3491888400 | 12.623535748416932 | 1920 | 1.6697620887531766 | Both
6983776800 | 13.074932764428576 | 2304 | 1.6887437164315817 | Both
13967553600 | 13.330404681822575 | 2688 | 1.688129641564845 | z
20951330400 | 13.359790283619853 | 2880 | 1.6771971437444477 | z
27935107200 | 13.473027345213657 | 3072 | 1.6778189601814566 | z
41902660800 | 13.618787163390186 | 3360 | 1.6772532788317212 | z
48886437600 | 13.661035931587017 | 3456 | 1.6766395074626765 | z
80313433200 | 13.898496037013476 | 3840 | 1.6840073524996981 | z
97772875200 | 13.9238684365927 | 4032 | 1.6771667750746457 | z
146659312800 | 13.95391871073356 | 4320 | 1.6669335689665656 | z
160626866400 | 14.381230476504792 | 4608 | 1.7048362983235954 | Both
321253732800 | 14.653665601698984 | 5376 | 1.70595781024433 | Both
481880599200 | 14.684621625013545 | 5760 | 1.6959397963111271 | z
642507465600 | 14.805417117020358 | 6144 | 1.697240007810422 | z
963761198400 | 14.96078376959392 | 6720 | 1.697611432956445 | z
1124388064800 | 15.005991789605918 | 6912 | 1.6973156317382547 | z
1927522396800 | 15.114528244214418 | 7680 | 1.6894584161675266 | z
2248776129600 | 15.286212094623194 | 8064 | 1.6993809575575591 | z
3373164194400 | 15.317866917412017 | 8640 | 1.6899382326482713 | z
4497552259200 | 15.442133591869041 | 9216 | 1.6916033729345097 | z
4658179125600 | 15.532888047165569 | 9216 | 1.701545039464922 | z
6746328388800 | 15.601905131290353 | 10080 | 1.6924910964782531 | z
9316358251200 | 15.819921876477741 | 10752 | 1.7042101167364407 | z
13974537376800 | 15.85220556834494 | 11520 | 1.695089488250149 | z
18632716502400 | 15.979508397400322 | 12288 | 1.6969908838209302 | z
27949074753600 | 16.143138304865357 | 13440 | 1.698206842426234 | z
32607253879200 | 16.19090273009808 | 13824 | 1.698198918206643 | z
55898149507200 | 16.304808192777614 | 15360 | 1.6914540141556873 | z
65214507758400 | 16.486083129236103 | 16128 | 1.701646565312295 | z
97821761637600 | 16.519094363887366 | 17280 | 1.6929976353281253 | z
130429015516800 | 16.65002989232633 | 18432 | 1.695204142143814 | z
144403552893600 | 16.681117338585267 | 18432 | 1.698369275660548 | z
195643523275200 | 16.818268468291016 | 20160 | 1.6968514958550285 | z
288807105787200 | 16.982546580718836 | 21504 | 1.7023412570717529 | z
391287046550400 | 16.984348388710064 | 23040 | 1.690828290229381 | z
433210658680800 | 17.01612849485301 | 23040 | 1.6939920679206526 | z
577614211574400 | 17.14984918244781 | 24576 | 1.6964049353083852 | z
866421317361600 | 17.321625217537015 | 26880 | 1.69834211149153 | z
1010824870255200 | 17.37191487994591 | 27648 | 1.69858126294547 | z
1732842634723200 | 17.49109979271339 | 30720 | 1.6927958764301567 | z
2021649740510400 | 17.681845203355937 | 32256 | 1.7032138358195825 | z
3032474610765600 | 17.716182715340032 | 34560 | 1.6952551274755545 | z
4043299481020800 | 17.85369455966352 | 36864 | 1.6979277485105486 | z

$n$ | $z(n)$ | $\tau(n)$ | $v(n)=\frac{z(n)}{\log\tau(n)}$ | Record type
---|---|---|---|---
6064949221531200 | 18.030279139683728 | 40320 | 1.700231428280654 | z
10685862914126400 | 18.034756898282858 | 43008 | 1.69036627988754 | z
12129898443062400 | 18.204352248079783 | 46080 | 1.695299364613963 | z
21371725828252800 | 18.208916713887955 | 49152 | 1.6855936510511806 | z
24259796886124800 | 18.299901250390008 | 51840 | 1.6857074987195217 | z
30324746107656000 | 18.32631342665929 | 53760 | 1.6825040355068184 | z
32057588742379200 | 18.38794334401685 | 53760 | 1.6881621611891613 | z
37400520199442400 | 18.44048746990043 | 55296 | 1.6886188378525169 | z
60649492215312000 | 18.502217106680238 | 61440 | 1.6780813764210956 | z
64115177484758400 | 18.5643532977372 | 61440 | 1.6837168948248216 | z
74801040398884800 | 18.763548743910583 | 64512 | 1.694285814101721 | z
112201560598327200 | 18.799052019527668 | 69120 | 1.686982036850719 | z
149602080797769600 | 18.942419905102103 | 73728 | 1.6900594936024502 | z
224403121196654400 | 19.12643161225541 | 80640 | 1.6929416451262043 | z
448806242393308800 | 19.30760643717392 | 92160 | 1.6890150485603024 | z
897612484786617600 | 19.406936363910518 | 103680 | 1.6803903332985408 | z

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