# On $k$-layered numbers

Farid Jokar<EMAIL_ADDRESS>
###### Abstract.

A positive integer $n$ is said to be $k$-layered if its divisors can be
partitioned into $k$ sets with equal sum. In this paper, we start the
systematic study of these class of numbers. In particular, we state some
algorithms to find some even $k$-layered numbers $n$ such that $2^{\alpha}n$
is a $k$-layered number for every positive integer $\alpha$. We also find the
smallest $k$-layered number for $1\leq k\leq 8$. Furthermore, we study when
$n!$ is a $3$-layered and when is a $4$-layered number. Moreover, we classify
all $4$-layered numbers of the form $n=p^{\alpha}q^{\beta}rt$, where $\alpha$,
$1\leq\beta\leq 3$, $p$, $q$, $r$, and $t$ are two positive integers and four
primes, respectively. In addition, in this paper, some other results
concerning these numbers and their relationship with $k$-multiperfect numbers,
near-perfect numbers, and superabundant numbers are discussed. Also, we find
an upper bound for the differences of two consecutive $k$-layered numbers for
every positive integer $1\leq k\leq 5$. Finally, by assuming the smallest
$k$-layered number, we find an upper bound for the difference of two
consecutive $k$-layered numbers.

##  1\. Introduction

A perfect number is a positive integer that is equal to the sum of its proper
positive divisors. In 2003, the idea of Zumkeller numbers, which is a
generalization of perfect numbers, was first introduced by Zumkeller in the
Online Encyclopedia of Integer Sequences (OEIS) [14] (A083207).

######  Definition 1.1.

A positive integer $n$ is said to be Zumkeller if the set of all positive
divisors of $n$ can be partitioned into two subsets with the same sum. A
Zumkeller partition for a Zumkeller number $n$ is a partition
$\\{A_{1},A_{2}\\}$ of the set of all positive divisors of $n$ such that
$A_{1}$ and $A_{2}$ sums to the same value.

Shortly thereafter, Clark et al. [12] announced several results and
conjectures related to Zumkeller numbers. Peng and Bhaskara Rao [18] found
some other results about Zumkeller numbers; they also studied the relationship
between practical numbers (to be defined later in the paper) and Zumkeller
numbers. They also settled a conjecture from [12]. Moreover, they made
substantial contributions towards the second conjecture of Clark et al. [12].
Recently, the author studied a generalization of Zumkeller numbers called
$k$-layered numbers [5] which is the principal object of study in this paper.

######  Definition 1.2.

A positive integer $n$ is said to be $k$-layered if the set of all positive
divisors of $n$ can be partitioned into $k$ subsets with the same sum. A
$k$-layered partition for a $k$-layered number $n$ is a partition
$\\{A_{1},A_{2},\dots,A_{k}\\}$ of the set of all positive divisors of $n$
such that for every $1\leq i\leq k$, each of $A_{i}$ sums to the same value.

In [9], the authors found some new results related to Zumkeller numbers and
$k$-layered numbers. It is clear from the definition of $k$-layered numbers
that all Zumkeller numbers are $2$-layered numbers. Another generalization of
perfect numbers which we will consider are multiperfect numbers or
$k$-multiperfect number and near-perfect numbers. A positive integer $n$ is
called a multiperfect number or a $k$-multiperfect number if $n$ satisfies the
equation $\sigma(n)=kn$, for some positive integer $k$. It is easy to check
that every perfect number is Zumkeller; this immediately gives us a further
motivation to define $k$-layered numbers as all multiperfect numbers will be
$k$-layered numbers.

This paper is arranged as follows: in Sections 2 and 3, we recall and
generalize some results related to $k$-layered numbers, taking as basis the
results already proved for Zumkeller numbers. In Section 4, we find the
smallest $k$-layered number for $1\leq k\leq 8$. In Section 5, we state some
algorithms to find some even $k$-layered numbers $n$ such that $2^{\alpha}n$
is $k$-layered for every positive integer $\alpha$. In addition, we prove that
$n!$ is a $3$-layered number if and only if $n\geq 5$ and $n\neq 10$.
Moreover, we prove that $n!$ is a $4$-layered number if and only if $n\geq 9$.
In Section 6, we prove that all known $3$-multiperfect numbers are
$3$-layered. Furthermore, by assuming Conjecture 6.2, we prove that all known
$4$-multiperfect numbers and all known $5$-multiperfect numbers are
$4$-layered and $5$-layered, respectively. In addition, we state some
algorithms to find some more $k$-multiperfect numbers that are $k$-layered. In
Section 7, we find some relationship between Near-perfect numbers and
Zumkeller numbers. In Section 8, we find an upper bound for the differences of
two consecutive $k$-layered number for $1\leq k\leq 5$. Lastly, by assuming
the smallest $k$-layered number, we find an upper bound for the differences of
two consecutive $k$-layered numbers. There is a generous sprinkling of open
questions and conjectures in this paper, which the author hopes subsequent
mathematicians will take up.

##  2\. Some preliminary results on $k$-layered numbers

Let $n$ be a positive integer and $\sigma(n)$ denote the sum of positive
divisors of $n$. We recall that the abundancy of $n$ is defined to be
$I(n)=\dfrac{\sigma(n)}{n}$. Also, A number $n$ is said to be abundant,
perfect, and deficient if $I(n)>2$, $I(n)=2$, and $I(n)<2$, respectively.
Table 1 lists the smallest $k$-layered number for every positive integer
$2\leq k\leq 8$ (for more details, see Section 4 of this paper).

$k$ | Smallest $k$-layered number |  Number of positive divisors
---|---|---
$1$ | $1$ |  $1$
$2$ | $6$ |  $4$
$3$ | $120$ |  $16$
$4$ | $27720$ |  $96$
$5$ | $147026880$ |  $896$
$6$ | $130429015516800$ |  $18432$
$7$ | $1970992304700453905270400$ |  $1474560$
$8$ | $1897544233056092162003806758651798777216000$ |  $1245708288$
Table 1. Smallest $k$-layered number for $1\leq k\leq 8$.

######  Remark 2.1.

According to the number of divisors of the smallest 8-layered number $n$, it
seems that it is not easy to find an 8-layered partition for $n$. The author
applied an advanced algorithm to find an 8-layered partition for $n$.

In the remainder of this paper, if $A$ is a set of all positive integers, then
we define $S(A)$ to be the sum of the integers in $A$. We do not prove the
results in this section because the proofs are exactly similar to the proofs
of the corresponding results for Zumkeller numbers in [18]. The following
proposition gives a necessary and sufficient condition for a positive integer
$n$ to be $k$-layered.

######  Proposition 2.2.

A number $n$ is $k$-layered if and only if we can find $k-1$ disjoint subsets
$A_{1},A_{2},\dots,A_{k-1}$ of positive divisors of $n$ so that for every
$1\leq i\leq k-1$, $A_{i}$ sums to the $\dfrac{\sigma(n)}{k}$.

Proposition $2$ from [18] gives some necessary conditions for a Zumkeller
number. We can generalize this proposition for $k$-layered number.

######  Proposition 2.3.

If $n$ is a $k$-layered number, then the following is true:

$\mathrm{(i)}\ $ $k\mid\sigma(n)$;

$\mathrm{(ii)}\ $ If $k$ is even, then the prime factorization of $n$ must
include at least one odd prime to an odd power;

$\mathrm{(iii)}\ $ $\sigma(n)\geq kn$; this concludes that $I(n)\geq k$.

###### Proof.

The proof is identical to the proof of Proposition $2$ of [18]. ∎

######  Remark 2.4.

Let $m$ be the smallest $9$-layered number. We can see that the number

$n=4368924363354820808981210203132513655327781713900627249499856876120704000$

is the smallest number such that $I(n)\geq 9$ (A023199). Therefore, by
Proposition 2.3, $m\geq n$.

Although it is an open question that for every positive integer $k$, there
exists a $k$-layered numbers, but Table 1 and Remark 2.4 lead us to the
following conjecture.

######  Conjecture 2.5.

Size of the smallest $k$-layered number grows exponentially with respect to
$k$.

We close this section with two propositions which are a generalization of
Corollary 5 and Proposition 6 of [18], respectively.

######  Proposition 2.6.

If the integer $n$ is $k$-layered and $w$ is relatively prime to $n$, then
$nw$ is a $k$-layered number.

###### Proof.

The proof is identical to the proof of Corollary $5$ of [18]. ∎

######  Proposition 2.7.

Let $n$ be a $k$-layered number and $p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots
p_{m}^{a_{m}}$ be the prime factorization of $n$. Then for any positive
integers $l_{1},l_{2},\ldots,l_{m}$, the number

$p_{1}^{a_{1}+l_{1}(a_{1}+1)}p_{2}^{a_{2}+l_{2}(a_{2}+1)}\cdots
p_{1}^{a_{m}+l_{m}(a_{m}+1)}$

is $k$-layered.

###### Proof.

The proof of the above result is similar to the proof of Proposition 6 in
[18]. ∎

##  3\. Almost practical numbers and $k$-layered numbers

In this section we discuss the connection between $k$-layered numbers and
almost practical numbers, which are a generalization of practical numbers. The
results discussed here will be a generalization of the results connecting
Zumkeller numbers and practical numbers that were found by Peng and Bhaskara
Rao [18].

First, we recall the definition of practical numbers.

######  Definition 3.1.

A positive integer $n$ is said to be a practical number if every positive
integer less than $n$ can be represented as a sum of distinct positive
divisors of $n$.

Now we recall two results about practical numbers.

######  Proposition 3.2 (Corollary 1, [3]).

A positive integer $n$ with the prime factorization
$p_{1}^{t_{1}}p_{2}^{t_{2}}\cdots p_{m}^{t_{m}}$ and $p_{1}<p_{2}<\dots<p_{m}$
is a practical number if and only if $p_{1}=2$ and
$p_{i+1}\leq\sigma(p_{1}^{t_{1}}\cdots p_{i}^{t_{i}})+1$ for $1\leq i\leq
m-1$.

######  Proposition 3.3 (Proposition 8, [18]).

A positive integer $n$ is a practical number if and only if every integer less
than or equal to $\sigma(n)$ can be written as a sum of distinct divisors of
$n$.

We now define almost practical numbers, which are a generalization of
practical numbers. These numbers were first studied by Stewart [3].

######  Definition 3.4.

A positive integer $n$ is called an almost practical number if all of the
numbers $j$ such that $2<j<\sigma(n)-2$ can be written as a sum of distinct
divisors of $n$.

######  Remark 3.5.

It is clear that every practical number is an almost practical number.

Now we recall some results related to almost practical numbers from [3].

######  Proposition 3.6 (Theorem 3, [3]).

Let $n\neq 3$ be an odd positive integer and $1=d_{1}<d_{2}<\dots<d_{k}=n$ are
the divisors of $n$. We also define $\sigma_{i}=d_{1}+d_{2}+\cdots+d_{i}$.
Then, $n$ is an almost practical number if and only if $d_{2}=3,d_{3}=5$ and
for $i\geq 3$, at least one of the following is true:

  1. (1)

$d_{i+1}\leq\sigma_{i}-2$ and $d_{i+1}\neq\sigma_{i}-4$,

  2. (2)

$d_{i+1}=\sigma_{i}-4$ and $d_{i+2}=\sigma_{i}-2$.

######  Remark 3.7.

If $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{m}^{\alpha_{m}}$ is an odd
almost practical number where $p_{1}<p_{2}<\dots<p_{m}$ are all prime factors
of $n$, then by Proposition 3.6, it is clear that $p_{3}=7$.

######  Theorem 3.8 (Theorem 4, [3]).

Let $n\neq 3$ be an odd almost practical number and $p$ be an odd prime, then
$pn$ is an almost practical number if and only if $2p\leq\sigma(n)-2$ and
$2p\neq\sigma(n)-4$.

######  Remark 3.9.

By Propositions 3.6 and 3.2, if $n$ is an odd almost practical number, then
$2n$ is a practical number.

######  Proposition 3.10 (Corollary 4, [3]).

Let $n\neq 3$ be an odd almost practical number and $p$ be a prime dividing
$n$, then $pn$ is an almost practical number.

######  Example 3.11.

By Proposition 3.6, one can check directly that $3^{3}\times 5\times 7$ is an
almost practical number. Therefore, by Proposition 3.10, if
$n=3^{\alpha_{1}}\times 5^{\alpha_{2}}\times 7^{\alpha_{3}}$ such that
$\alpha_{1}\geq 3$, $\alpha_{2}$, and $\alpha_{3}$ are positive integers, then
the number $n$ is an almost practical number.

######  Example 3.12.

Let $k\geq 4$ be a positive integer. Also, let $p_{i}$ denotes $i$th prime
number. By Bertrand postulate, for every integer $i\geq 4$, we have

$2p_{i}<4p_{i-1}\leq\sigma(p_{2}p_{3}\cdots p_{i-1})-4.$

Thus, by Example 3.11 and Theorem 3.8, if
$m=p_{2}^{\alpha_{1}}p_{3}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k-1}}$ such that
$\alpha_{1}\geq 3$, $\alpha_{2},\ldots,\alpha_{k-1}$ are positive integers,
then $m$ is an almost practical number.

Now we are going to investigate the relation between almost practical numbers
and Zumkeller numbers. The following proposition is a generalization of
Proposition 10 of [18].

######  Proposition 3.13.

Let $n\neq 3$ be an almost practical number. Then, $n$ is Zumkeller if and
only if $\sigma(n)$ is even.

###### Proof.

The proof is similar to the proof of Proposition 10 of [18]. ∎

######  Example 3.14.

Let $k\geq 4$ be a positive integer, and also let $p_{i}$ denotes $i$th prime
number. Now let $m=p_{2}^{\alpha_{1}}p_{3}^{\alpha_{2}}\cdots
p_{k}^{\alpha_{k-1}}$ such that $\alpha_{1}\geq 3$, $\alpha_{2},\ldots$ and
$\alpha_{k-1}$ are positive integers. Thus, by Example 3.12 and Proposition
3.13, $m$ is a Zumkeller number if and only if we can find an integer $1\leq
i\leq k-1$ such that $\alpha_{i}$ is an odd number. In other words, the number
$m$ is Zumkeller if and only if $m$ is not square.

Now we state a proposition which is a generalization of Proposition $13$ of
[18].

######  Proposition 3.15.

Let $\alpha$ be a positive integer. Also, let $n$ be non-$k$-layered number
and $p$ be a prime number with $\gcd(n,p)=1$. If $np^{\alpha}$ is a
$k$-layered number, then $p\leq\sigma(n)$. Moreover, if $q$ is prime factor of
$k$ such that $q\nmid\sigma(n)$, then $q\mid\sigma(p^{\alpha})$.

###### Proof.

The proof is identical to the proof of Proposition $13$ of [18]. ∎

In Proposition 20 of [18], the authors by a simple method proved that every
odd Zumkeller number possesses at least three distinct prime factors. On the
other hand, we know that for every positive integer $t$, the number $2^{t}$ is
a deficient number. Therefore, by Propositions 2.3 and 3.15, if $p$ is a prime
number such that $\sigma(2^{t})<p$, then the number $2^{t}p^{\alpha}$ fails to
be a Zumkeller number for every positive integer $\alpha$. Also, by
Proposition 3.13, if $p\leq\sigma(2^{t})$ and $\alpha$ is an odd number, then
the number $2^{t}p^{\alpha}$ is Zumkeller. This was also proved in [9] (see
Theorem 2.6) by a different method. Thus, we have the following corollary.

######  Corollary 3.16.

Let $p_{1}$ and $p_{2}$ be prime numbers. Also, let $\alpha_{1}$ and
$\alpha_{2}$ be positive integers. The number
$p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}$ is Zumkeller if and only if $p_{1}=2$,
$p_{2}\leq\sigma(2^{\alpha_{1}})$ and $\alpha_{2}$ is odd.

######  Remark 3.17.

Let $p$ be an odd prime number. Also. let $\alpha$ and $\beta$ be positive
integers. One can check that the number $2^{\alpha}p^{\beta}$ is abundant if
and only if $p\leq\sigma(2^{\alpha})$ (for more details, see the proof of
Theorem 2.6 of [9]). Thus, if $n$ is an abundant number that possesses two
factors, then $n$ is Zumkeller if and only if $n$ satisfies the conditions
$\mathrm{(i)}\ $ and $\mathrm{(iii)}\ $ of Proposition 2.3.

##  4\. The smallest $k$-layered number for $1\leq k\leq 8$

We start this section with two definitions.

######  Definition 4.1.

A positive integer $n$ is said to be $k$-abundant if $I(n)\geq k$.

######  Definition 4.2.

A positive integer $n$ is said to be good $k$-abundant if $I(n)\geq k$ and
$k\mid\sigma(n)$.

######  Remark 4.3.

It is clear that a positive integer $n$ is 2-abundant if and only if $n$ is
either perfect or abundant.

Let $i$ be a positive integer. In remainder of this section, we define
$a_{i}$, $g_{i}$, and $k_{i}$ as the smallest $i$-abundant number, the
smallest good $i$-abundant number, and the smallest $i$-layered number,
respectively. In [14], Pegg listed $a_{i}$ for $1\leq i\leq 13$ (A023199). Let
$i$ be a positive integer. The following simple proposition states a
relationship between $a_{i}$, $g_{i}$, and $k_{i}$.

######  Proposition 4.4.

For positive integers $i$, $a_{i}\leq g_{i}\leq k_{i}$.

###### Proof.

By Proposition 2.3, it is obvious. ∎

We recall that a positive integer $n$ is said to be superabundant if
$I(n)>I(m)$ for all positive integers $m<n$. Now we recall a well-know theorem
about superabundant numbers of [2].

######  Theorem 4.5.

If $n$ is a superabundant number, then there exists a positive integer $k$
such that $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{k}^{\alpha_{k}}$,
where $p_{i}$ is $i$th prime number and
$\alpha_{1}\geq\alpha_{2}\geq\dots\geq\alpha_{k}\geq 1$. Also, $a_{k}=1$
except when $n$ is $4$ or $36$.

Thus, by previous theorem, every superabundant is almost practical (see
Example 3.12). Therefore, as a direct consequence of the previous theorem and
Proposition 3.13, we have the following corollary.

######  Corollary 4.6.

A positive integer $n$ is superabundant if and only if one of the following
holds:

$\mathrm{(i)}\ $ $n$ is Zumkeller;

$\mathrm{(ii)}\ $ $n=4$ or $n=36$.

A list of all superabundant numbers less than $10^{1200}$ can be found in
[13]. The following proposition states one of the special properties of
superabundant numbers.

######  Proposition 4.7.

Let $m_{1}<m_{2}$ be two consecutive superabundant numbers. For every positive
integer $t<m_{2}$, $I(t)\leq I(m_{1})$, and the equality holds when $t=m_{1}$.

###### Proof.

If $t\leq m_{1},$ then $I(t)\leq I(m_{1})$. So, we assume $m_{1}<t_{1}<m_{2},$
such that $I(t_{1})>I(m_{1}).$ Since $m_{1}$ and $m_{2}$ are consecutive
superabundant numbers, therefore $t_{1}$ is not a superabundant number. So,
there exists $t_{2}<t_{1}$ such that $I(t_{2})>I(t_{1})$. Hence
$I(t_{2})>I(t_{1})>I(m_{1}),$ which implies $m_{1}<t_{2}<t_{1}$. We once again
conclude that there exists an integer $t_{3}$ where $m_{1}<t_{3}<t_{2}<t_{1}$
such that $I(m_{1})<I(t_{1})<I(t_{2})<I(t_{3})$. Thus, for every positive
integer $r\geq 3$, by this algorithm, inductively, we can find distinct
positive integers $t_{1},t_{2},\dots,$ and $t_{r}$ such that
$m_{1}<t_{r}<t_{r-1}<\dots<t_{1}<m_{2}$ and
$I(m_{2})>I(t_{r})>I(t_{r-1})>\dots>I(t_{1})>I(m_{1})$; this contradicts the
finiteness of the set $\\{a:a\in\mathbb{N},m_{1}<a<m_{2}\\}$. ∎

In remainder of this section, we define $s_{i}$ as the $i$th superabundant
number. As a direct consequence of Propositions 4.4 and 4.7, we have the
following corollary.

######  Corollary 4.8.

For positive integer $i$, there exists a unique positive integer $t_{i}$ such
that satisfies the following two conditions:

$\mathrm{(i)}\ $ $I(s_{\ell})\geq i$ for integers $\ell\geq t_{i}$,

$\mathrm{(ii)}\ $ $I(s_{j})<i$ for integers $j<t_{i}$.

Also, if a positive integer $t$ satisfies conditions $\mathrm{(i)}\ $ and
$\mathrm{(ii)}\ $ of the proposition, then $s_{t}=a_{i}$. Furthermore,
$a_{i}=g_{i}$ or $s_{t+1}\leq g_{i}$.

Let $n$ be a positive integer. In the remainder of this paper, we define
$D_{n}$ as the set all positive divisors of $n$. Proposition 4.4 and corollary
4.8 motivate an algorithm to find $a_{i}$, $g_{i}$, and $k_{i}$ for positive
integers $i$.

######  Example 4.9.

By the list of the smallest $k$-abundant numbers in [14] (A023199) and the
list of superabundant numbers [13], $a_{5}=s_{43}=122522400=2^{5}\times
3^{2}\times 5^{2}\times 7\times 11\times 13\times 17$ and
$s_{44}=147026880=2^{6}\times 3^{3}\times 5\times 11\times 13\times 17$. One
can check directly that $5\nmid\sigma(s_{43})$. Therefore, $a_{5}\neq g_{5}$.
Also, $5\mid\sigma(s_{44})$. Thus, by corollary 4.8, $g_{5}=s_{44}$. Now we
prove that $g_{5}$ is $5$-layered. We define

$A_{1}=\\{$ $3$, $2^{6}$, $2^{6}\times{}3$, $3^{2}\times{}5\times{}17$,
$2^{6}\times{}11\times{}13$, $2^{6}\times{}3\times{}5\times{}7\times{}17$,
$2^{6}\times{}3^{3}\times{}11\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$$\\},\hfil\penalty
0\hfilneg$

$A_{2}=\\{$ $2^{2}\times{}17$, $2^{6}\times{}7\times{}13$,
$2^{2}\times{}3^{3}\times{}5\times{}7\times{}11\times{}17$,
$2^{6}\times{}3\times{}5\times{}7\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}5\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{2}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{4}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$$\\},\hfil\penalty
0\hfilneg$

$A_{3}=\\{$ $5$, $2^{4}$, $3^{2}\times{}5\times{}7$, $2^{6}\times{}5$,
$2^{6}\times{}3^{2}$, $2^{3}\times{}3\times{}5\times{}7$,
$2^{6}\times{}3^{3}$, $2^{4}\times{}3\times{}7\times{}11\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}13$,
$2^{6}\times{}7\times{}11\times{}13\times{}17$,
$2^{4}\times{}3^{2}\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{2}\times{}5\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}7\times{}11\times{}17$,
$2^{4}\times{}3^{2}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}7\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{3}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}7\times{}11\times{}13\times{}17$$\\},\hfil\penalty
0\hfilneg$

$A_{4}=\\{$ $2^{2}$, $2^{4}\times{}3\times{}11$,
$2\times{}3^{3}\times{}5\times{}7\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}13\times{}17$,
$2^{3}\times{}3^{3}\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{2}\times{}5\times{}7\times{}11\times{}17$,
$2^{4}\times{}3\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13$,
$2^{6}\times{}3^{2}\times{}5\times{}7\times{}13\times{}17$,
$2\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{2}\times{}7\times{}11\times{}13\times{}17$,
$2^{4}\times{}3^{3}\times{}5\times{}11\times{}13\times{}17$,
$2^{6}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}5\times{}7\times{}11\times{}17$,
$2^{3}\times{}3^{2}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{5}\times{}3^{3}\times{}5\times{}7\times{}13\times{}17$,
$2^{4}\times{}3^{3}\times{}7\times{}11\times{}13\times{}17$,
$2^{5}\times{}3\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13$,
$2^{2}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$,
$2^{6}\times{}3^{2}\times{}5\times{}7\times{}11\times{}13\times{}17$$\\},\hfil\penalty
0\hfilneg$

and $A_{5}=D_{g_{5}}\setminus(A_{1}\cup A_{2}\cup A_{3}\cup A_{4})$, then one
can check directly that $\\{A_{1},A_{2},A_{3},A_{4},A_{5}\\}$ is a $5$-layered
partition for $g_{5}$. Thus, by Proposition 4.4, $k_{5}=g_{5}$.

######  Remark 4.10.

One can check directly that $3\mid\mid a_{2}$, $5\mid\mid a_{3}$, $7\mid\mid
a_{4}$, $29\mid\mid a_{6}$, $13\mid\mid a_{7}$, $23\mid\mid a_{8}$,
$53\mid\mid a_{9}$, $269\mid\mid a_{10}$, $461\mid\mid a_{11}$, $47\mid\mid
a_{12}$, and $1481\mid\mid a_{13}$. This concludes that if $i$ is a positive
integer such that $i\neq 1,5$ and $2\leq i\leq 13$, then there exists a prime
number $q_{i}$ such that $q_{i}\mid\mid a_{i}$ and $q_{i}\equiv i-1$ (mod
$i$). Thus, if $i$ is a positive integer such that $i\neq 5$ and $1\leq i\leq
13$, then $i\mid\sigma(a_{i})$. In other words, for integers $1\leq i\leq 13$
except $i=5$, $g_{i}=a_{i}$.

Remark 4.10 and Example 4.9 lead us to the following open question.

######  Open question 4.11.

For positive integers $k\neq 5$, dose $a_{i}$ equal $g_{i}$?

In the following, we prove that $g_{i}=k_{i}$ for $1\leq i\leq 8$.

######  Proposition 4.12.

$g_{i}=k_{i}$ for integers $1\leq i\leq 8$.

###### Proof.

Case 1: $k=1$. It is clear.

Case 2: $k=2$. We know that $a_{2}=g_{2}=6$ is the smallest Zumkeller number.

Case 3: $k=3$. By the list of the smallest $k$-abundant numbers in [14]
(A023199), the list of superabundant numbers in [13], and Corollary 4.8,
$s_{10}=120=a_{3}=g_{3}$. Now if we define $A_{1}=\\{$$20$, $40$, $60$$\\}$,
$A_{2}=\\{$$1$, $2$, $3$, $4$, $5$, $6$, $8$, $10$, $12$, $15$, $24$,
$30$$\\}$, and $A_{3}=\\{$$120$$\\}$, then $\\{A_{1},A_{2},A_{3}\\}$ is a
$3$-layered partition for $g_{3}$. Thus, by Proposition 4.4, $k_{3}=g_{3}$.

Case 4: $k=4$. By the same method that we used in Case 3, We can see
$s_{23}=a_{4}=g_{4}=27720=2^{3}\times 3^{2}\times 5\times 7\times 11$. Now we
define

$A_{1}=\\{$ $2^{3}\times{}3^{2}\times{}5$,
$2^{3}\times{}3^{2}\times{}5\times{}7\times{}11$$\\},\hfil\penalty 0\hfilneg$

$A_{2}=\\{$ $2$, $2^{3}\times{}3$, $2^{2}\times{}3\times{}5$,
$2\times{}5\times{}7$, $2^{2}\times{}3\times{}5\times{}11$,
$2\times{}5\times{}7\times{}11$, $2^{2}\times{}3^{2}\times{}5\times{}7$,
$2\times{}3^{2}\times{}7\times{}11$, $2^{2}\times{}5\times{}7\times{}11$,
$2^{3}\times{}3\times{}7\times{}11$, $2^{2}\times{}3^{2}\times{}5\times{}11$,
$2^{2}\times{}3\times{}5\times{}71$,
$2^{2}\times{}3^{2}\times{}5\times{}7\times{}11$$\\},\hfil\penalty 0\hfilneg$

$A_{3}=\\{$ $1$, $2\times{}3^{2}$, $2\times{}3\times{}5\times{}7\times{}11$,
$2^{2}\times{}3^{2}\times{}7\times{}11$, $2^{3}\times{}5\times{}7\times{}11$,
$3^{2}\times{}5\times{}7\times{}11$, $2^{3}\times{}3^{2}\times{}5\times{}11$,
$2^{3}\times{}3^{2}\times{}7\times{}11$,
$2\times{}3^{2}\times{}5\times{}7\times{}11$$\\},\hfil\penalty 0\hfilneg$

and $A_{4}=D_{g_{4}}\setminus{(A_{1}\cup A_{2}\cup A_{3})}$, then one can
check directly that $\\{A_{1},A_{2},A_{3},A_{4}\\}$ is a $4$-layered partition
for $g_{4}$.

Case 5: $k=5$. By Example 4.9, it is clear.

Case 6: $6\leq k\leq 8$. In [6], the author stated $6$-layered partition,
$7$-layered partition, and $8$-layered partition for
$g_{6}=130429015516800=2^{7}\times 3^{3}\times 5^{2}\times 7^{2}\times
11\times 13\times 17\times 19\times 23\times
29,g_{7}=1970992304700453905270400=2^{7}\times 3^{4}\times 5^{2}\times
7^{2}\times 11\times 13\times 17\times 19\times 23\times 29\times 31\times
37\times 41\times 43\times 47\times 53$, and
$g_{8}=1897544233056092162003806758651798777216000=2^{10}\times 3^{5}\times
5^{3}\times 7^{2}\times 11^{2}\times 13\times 17\times 19\times 23\times
29\times 31\times 37\times 41\times 43\times 47\times 53\times 59\times
61\times 67\times 71\times 73\times 79\times 83\times 89$, respectively. ∎

The previous proposition leads us to the following conjecture.

######  Conjecture 4.13.

$g_{i}=k_{i}$ for every positive integer $i$.

##  5\. Some more results about $k$-layered numbers

In [18], the authors if a number $n$ is Zumkeller, then the number
$2^{\alpha}n$ is half-Zumkeller 111A positive integer $n$ is said to be a
half-Zumkeller number if the proper positive divisors of n can be partitioned
into two disjoint subsets of equal sum. A half-Zumkeller partition for a half-
Zumkeller number $n$ is a partition $\\{A_{1},A_{2}\\}$ of the set of proper
positive divisors of $n$ such that $S(A_{1})=S(A_{2})$. for every positive
integer $\alpha$. By the same method, in [9], the authors proved that if a
number $n$ is Zumkeller, then the number $2^{\alpha}n$ is Zumkeller for every
positive integer $\alpha$. This does not generalize to all $k$-layered
numbers. For instance, the number $m=2^{5}\times 3\times 7$ is a $3$-layered
number because if we define $A_{1}=\\{2^{5}\times 3\times 7\\}$,
$A_{2}=\\{2^{4}\times 3\times 7,2^{3}\times 3\times 7,2^{2}\times 3\times
7,2\times 3\times 7,3\times 7,2\times 7,2\times 3,1\\}$, then one can check
directly that $A_{1}=A_{2}=\dfrac{\sigma(m)}{3}$. This concludes that the
number $m$ is $3$-layered. However, the number $2m$ fails to be $3$-layered
since $\sigma(2m)$ is not divisible by $3$. By Proposition 2.6, we know that
if $n$ is an odd $k$-layered number, then for positive integer $\alpha$ the
number $2^{\alpha}n$ is a $k$-layered number. This leads us to the following
definition.

######  Definition 5.1.

An even $k$-layered number $n$ is said to be a good $k$-layered number if the
number $2^{\alpha}n$ is $k$-layered for every positive integer $\alpha$.

The following proposition states a necessary condition for the $k$-layered
number $n$ to be a good $k$-layered.

######  Proposition 5.2.

Let $\alpha$ be a positive integer, and let $m$ be an odd positive integer. If
$n=2^{\alpha}m$ is a good $k$-layered number, then $k\mid\sigma(m)$.

###### Proof.

Let $n$ be a good $k$-layered number such that $k\nmid\sigma(m).$ Then there
exists at least one factor $p^{\gamma}$ of $k$ such that
$p^{\gamma}\nmid\sigma(m)$ and $p^{\gamma}\mid\sigma(2^{\alpha}),$ where $p$
is a prime number and $\gamma$ is a positive integer. If $p=2$, then this is a
contradiction. If $p$ is odd, then this implies that $p^{\gamma}\nmid
2^{\alpha+1}.$ This concludes that $p^{\gamma}\nmid\sigma(2^{\alpha+1})$.
Thus, $p^{\gamma}\nmid\sigma(2n),$ which is a contradiction. ∎

In the following, we state two crucial theorems that help us to find a subset
of good $k$-layered numbers. For better understanding, first, we state a
proposition, which is a special case of one of these two theorems.

######  Proposition 5.3.

Let $n$ be an odd positive integer and $\alpha$ be a positive integer. Let
$m=2^{\alpha}n$ be a $3$-layered number with $3$-layered partition
$\\{A_{1},A_{2},A_{3}\\}$. Now let $A^{\prime}_{1}$ be a subset of the set of
all positive divisors of $n$ such that $A^{\prime}_{1}$ sums to
$\dfrac{2\sigma(n)}{3}$. If $A^{\prime}=\\{2^{\alpha}d:d\in A^{\prime}_{1}\\}$
and $A^{\prime}\subset A_{1}\cup A_{2}$, then $m$ is a good $3$-layered
number.

###### Proof.

We define $M_{1}=A^{\prime}\cap A_{1}$ and $M_{2}=A^{\prime}\cap A_{2}$. Now
we define

$M^{\prime}_{1}=\\{2d:d\in M_{1}\\},M^{\prime}_{2}=\\{2d:d\in
M_{2}\\},M_{3}=\\{2^{\alpha+1}d:d\in(D_{n}\setminus A^{\prime}_{1})\\}.$

We can see that

$S((A_{1}\setminus M_{1})\cup M^{\prime}_{1})=S(A_{1}\cup M_{1}).$

Thus, we see

$S(A_{1})+\dfrac{2^{\alpha+1}\sigma(n)}{3}=S((A_{1}\setminus M_{1})\cup
M^{\prime}_{1}\cup M_{2})=S(A_{1}\cup M_{1}\cup M_{2})=S(A_{2}\cup M_{2}\cup
M_{1})\\\ =S((A_{2}\setminus M_{2})\cup M^{\prime}_{2}\cup
M_{1})=S(A_{2})+\dfrac{2^{\alpha+1}\sigma(n)}{3}.$

Therefore, by definition of $A^{\prime}$, one can check directly that

$\\{(A_{1}\setminus M_{1})\cup M^{\prime}_{1}\cup M_{2},(A_{2}\setminus
M_{2})\cup M^{\prime}_{2}\cup M_{1},A_{3}\cup M_{3}\\}$

is a $3$-layered partition for $2m$. Thus, by the same method, inductively, we
can prove that $\ell=2^{t}n$ is a $3$-layered number for every integer
$t\geq\alpha$. ∎

By Proposition 5.3, we have the following corollary.

######  Corollary 5.4.

If $n\neq 3$ is an odd almost practical number such that $6\mid\sigma(n)$,
then the number $2n$ is a good $3$-layered number.

###### Proof.

First, we prove that $2n$ is a $3$-layered number. By Proposition 3.13, $n$ is
a Zumkeller number; this concludes that $D_{n}$ can be partitioned into two
subsets $A_{1}$ and $A_{2}$ such that each of them sums to
$\dfrac{\sigma(n)}{2}$. Now we define $B_{1}=\\{2d:d\in A_{1}\\}$ and
$B_{2}=\\{2d:d\in A_{2}\\}$. We know that $n$ is an odd number. Therefore, for
every integer $a\in B_{1}\cup B_{2}$, $a\notin D_{n}$. Thus,
$\\{D_{n},B_{1},B_{2}\\}$ is a $3$-layered partition for $2n$. Also, we know
that $n$ is an almost practical number such that $6\mid\sigma(n)$ and
$2\neq\dfrac{2\sigma(n)}{3}\neq\sigma(n)-2$; this concludes that there exists
$A^{\prime}\subset D_{n}$ so that $A^{\prime}$ sums to
$\dfrac{2\sigma(n)}{3}$. Now let $A^{\prime\prime}=\\{2d:d\in A^{\prime}\\}$.
We know that $A^{\prime\prime}\subset B_{1}\cup B_{2}$. Then, by Proposition
5.3, $2n$ is a good $3$-layered number. ∎

Now we present a proposition generalizing Proposition 5.3 for $4$-layered
numbers.

######  Proposition 5.5.

Let $n$ be an odd Zumkeller number with Zumkeller partition
$\\{A^{\prime}_{1},A^{\prime}_{2}\\}$. For positive integer $\alpha$, we
define $A^{\prime\prime}_{1}=\\{2^{\alpha}d:d\in A^{\prime}_{1}\\}$ and
$A^{\prime\prime}_{2}=\\{2^{\alpha}d:d\in A^{\prime}_{2}\\}$. Now let
$m=2^{\alpha}n$ be a $4$-layered number with $4$-layered partition
$\\{A_{1},A_{2},A_{3},A_{4}\\}$. If $A^{\prime\prime}_{1}\subset(A_{1}\cup
A_{2})$ and $A^{\prime\prime}_{2}\subset(A_{3}\cup A_{4})$, then the number
$m$ is good $4$-layered.

###### Proof.

We define

$M_{1}=A^{\prime\prime}_{1}\cap A_{1},M_{2}=A^{\prime\prime}_{1}\cap A_{2}$

$M_{3}=A^{\prime\prime}_{2}\cap A_{3},M_{4}=A^{\prime\prime}_{2}\cap A_{4}$

Now for every integer $1\leq i\leq 4$, we define $M^{\prime}_{i}=\\{2d:d\in
M_{i}\\}.$ One can check directly that

$\\{(A_{1}\setminus{M_{1}})\cup M^{\prime}_{1}\cup M_{2},(A_{2}\setminus
M_{2})\cup M^{\prime}_{2}\cup M_{1},(A_{3}\setminus M_{3})\cup
M^{\prime}_{3}\cup M_{4},(A_{4}\setminus{M_{4}})\cup M^{\prime}_{4}\cup
M_{3})\\}$

is a $4$-layered partition for $m$. Thus, by applying the same method like
before, inductively, we can prove that $\ell=2^{t}n$ is a $4$-layered number
for every integer $t\geq\alpha$. ∎

The following two results are a generalization of Propositions 5.3 and 5.5,
respectively.

######  Theorem 5.6.

Let $k\geq 3$ and $n$ be two odd positive integers such that $k\mid\sigma(n)$.
Let $A^{\prime}_{1},A^{\prime}_{2},\dots,A^{\prime}_{\frac{k-1}{2}}$ be
disjoint subsets the set of all positive divisors of $n$ so that
$A^{\prime}_{i}$ sums to $\dfrac{2\sigma(n)}{k}$ for every integer $1\leq
i\leq\dfrac{k-1}{2}$. Let $\alpha$ be a positive integer and
$A^{\prime\prime}_{i}=\\{2^{\alpha}d:d\in A^{\prime}_{i}\\}$ for every integer
$1\leq i\leq\dfrac{k-1}{2}$. Now let $m=2^{\alpha}n$ be a $k$-layered number
with $k$-layered partition $\\{A_{1},A_{2},\dots,A_{k}\\}$ such that
$A^{\prime\prime}_{i}\subset A_{2i-1}\cup A_{2i}$ for every integer $1\leq
i\leq\dfrac{k-1}{2}$. Then, $m$ is a good $k$-layered number.

###### Proof.

For integers $1\leq i\leq k-1$, we define

$M_{i}=A^{\prime\prime}_{\left\lfloor\frac{i+1}{2}\right\rfloor}\cap
A_{i},M^{\prime}_{i}=\\{2d:d\in M_{i}\\},$

and

$M=\\{2^{\alpha+1}d:d\in(D_{n}\setminus A^{\prime}_{1}\cup
A^{\prime}_{2}\cup\dots,A^{\prime}_{\frac{K-1}{2}})\\}.$

Now we define

$B_{i}=\begin{cases}(A_{i}\setminus{M_{i})\cup M^{\prime}_{i}\cup
M_{i+1}},&\text{if $i$ is odd;}\\\ (A_{i}\setminus{M_{i})\cup
M^{\prime}_{i}\cup M_{i-1}},&\text{if $i$ is even.}\end{cases}$

One can check directly that $\\{B_{1},B_{2},\dots,B_{k-1},A_{k}\cup M\\}$ is a
$k$-layered partition for $2^{\alpha+1}n$. Also, by applying the previous
method, inductively, we can prove that $\ell=2^{t}n$ is $k$-layered for every
integer $t\geq\alpha$. ∎

######  Theorem 5.7.

Let $k\geq 4$ be an even positive integer. Let the odd positive integer $n$ be
a $\dfrac{k}{2}$-layered number with $\dfrac{k}{2}$-layered partition
$\\{A^{\prime}_{1},A^{\prime}_{2},\dots,A^{\prime}_{\frac{k}{2}}\\}$. Let
$\alpha$ be a positive integer and $A^{\prime\prime}_{i}=\\{2^{\alpha}d:d\in
A^{\prime}_{i}\\}$ for every integer $1\leq i\leq\dfrac{k}{2}$. Also, let
$m=2^{\alpha}n$ be a $k$-layered number with $k$-layered partition
$\\{A_{1},A_{2},\dots,A_{k}\\}$ so that $A^{\prime\prime}_{i}\subset
A_{2i-1}\cup A_{2i}$ for every integer $1\leq i\leq\dfrac{k}{2}$. Then, the
number $m$ is a good $k$-layered number.

###### Proof.

For integers $1\leq i\leq k$, we define

$M_{i}=A^{\prime\prime}_{\left\lfloor\frac{i+1}{2}\right\rfloor}\cap
A_{i},M^{\prime}_{i}=\\{2d:d\in M_{i}\\}.$

Now for every $1\leq i\leq k$, we define

$B_{i}=\begin{cases}(A_{i}\setminus{M_{i})\cup M^{\prime}_{i}\cup
M_{i+1}},&\text{if $i$ is odd;}\\\ (A_{i}\setminus{M_{i})\cup
M^{\prime}_{i}\cup M_{i-1}},&\text{if $i$ is even.}\end{cases}$

It is easy to check that the set $\\{B_{1},B_{2},\dots,B_{k}\\}$ is
$k$-layered partition for $2^{\alpha+1}n$. Also, by applying the same method,
inductively, we can prove the number $2^{t}n$ is $k$-layered for every
$t\geq\alpha$. ∎

As a consequence of Theorems 5.6 and 5.7, we have the following corollary.

######  Corollary 5.8.

For integers $2\leq k\leq 5$, the smallest $k$-layered number is a good
$k$-layered number.

###### Proof.

We investigate the five cases for $k$.

Case 1: $k=2$. This was already proved in Proposition $4.12$ of [9].

Case 2: $k=3$. Let $\\{A_{1},A_{2},A_{3}\\}$ be the $3$-layered partition for
the smallest $3$-layered number, $120$, which was defined in Section 4. Now
let $n=120=2^{3}m$. We know that $\dfrac{2\sigma(m)}{3}=16$. If we define
$A^{\prime}_{1}=\\{1,15\\}$, then it is obvious that
$A^{\prime}=\\{2^{3},2^{3}\times 15\\}\subset A_{2}\cup A_{3}$. Thus, by
Proposition 5.3, $n$ is a good $3$-layered number.

Case 3: $k=4$. Let $\\{A_{1},A_{2},A_{3},A_{4}\\}$ be the $4$-layered
partition for the smallest $4$-layered $n=27720$, which was defined in Section
4. Now let $D^{\prime}$ be the set of all positive divisors of
$m=3465=3^{2}\times 5\times 7\times 11$. We define

$A^{\prime}_{1}=\\{$ $3$, $3^{2}\times{}5$, $3\times{}7\times{}11$,
$3^{2}\times{}5\times{}7\times{}11$$\\},\hfil\penalty 0\hfilneg$

and $A^{\prime}_{2}=D^{\prime}\setminus{A^{\prime}_{1}}$.

One can check directly that for every integer $1\leq i\leq 2$,
$S(A^{\prime}_{i})=\dfrac{\sigma(m)}{2}$. Therefore,
$\\{A^{\prime}_{1},A^{\prime}_{2}\\}$ is a Zumkeller partition for $m$. We
know that $n=2^{3}m$. Also, if we define $A^{\prime\prime}_{1}=\\{2^{3}d:d\in
A^{\prime}_{1}\\}$ and $A^{\prime\prime}_{2}=\\{2^{3}d:d\in
A^{\prime}_{2}\\}$, then one can check directly that
$A^{\prime\prime}_{1}\subset A_{1}\cup A_{2}$ and $A^{\prime\prime}_{2}\subset
A_{3}\cup A_{4}$. Thus, by Proposition 5.5, $n$ is a good $4$-layered number.

Case 4: $k=5$. Let $\\{A_{1},A_{2},A_{3},A_{4},A_{5}\\}$ be the $5$-layered
partition for the smallest $5$-layered number $n=147026880$, which was defined
in Section 4. Let $n=2^{6}m$. We define

$A^{\prime}_{1}=\\{$ $1$, $3$, $11\times{}13$,
$3\times{}5\times{}7\times{}17$, $3\times{}5\times{}7\times{}13\times{}17$,
$2^{6}\times{}3^{3}\times{}5\times{}7\times{}11\times{}13\times{}17$$\\},\hfil\penalty
0\hfilneg$

and

$A^{\prime}_{2}=\\{$ $5$, $3^{2}$, $3^{3}$, $3^{3}\times{}5\times{}13$,
$7\times{}11\times{}13\times{}17$,
$3^{2}\times{}5\times{}11\times{}13\times{}17$,
$3^{3}\times{}5\times{}7\times{}11\times{}17$,
$3\times{}5\times{}7\times{}11\times{}13\times{}17$,
$3^{3}\times{}5\times{}11\times{}13\times{}17$,
$3^{3}\times{}7\times{}11\times{}13\times{}17$$\\}.$

One can check directly that
$S(A^{\prime}_{1})=S(A^{\prime}_{2})=\dfrac{2\sigma(m)}{5}$. We define
$A^{\prime\prime}_{i}=\\{2^{6}d:d\in A^{\prime}_{i}\\}$ for $i=1,2$. We know
$A^{\prime}_{1}\subset(A_{1}\cup A_{2})$ and $A^{\prime}_{2}\subset(A_{3}\cup
A^{\prime}_{4})$. Thus, by Theorem 5.6, $n$ is a good $5$-layered. ∎

Corollary 5.8, leads us naturally to the following open question.

######  Open question 5.9.

Is the smallest $k$-layered number a good $k$-layered number for every
positive integer $k\neq 1$?

It appears that the methods used in this paper are not feasible to attack this
problem at the moment.

Now we state a proposition that can be used to construct some $k$-layered
numbers.

######  Proposition 5.10.

Let $k$ and $k^{\prime}$ be positive integers. Now let $m$ be a $k$-layered
number and $n$ be a $k^{\prime}$-layered number. If $\gcd(m,n)=1$, then $mn$
is a $kk^{\prime}$-layered number.

###### Proof.

Let $\\{A_{1},A_{2},\dots,A_{k}\\}$ be a $k$-layered partition for $m$, and
$\\{B_{1},B_{2},\dots,B_{k^{\prime}}\\}$ be a $k^{\prime}$-layered partition
for n. One can check directly that $\\{A_{i}B_{j}:1\leq i\leq k,1\leq j\leq
k^{\prime}\\}$ is a $kk^{\prime}$-layered partition for $mn$. ∎

It is clear that Proposition 5.10 can be generalized.

######  Corollary 5.11.

Let $\ell_{1},\ell_{2},\dots,\ell_{r}$ be positive integers such that for
every integer $1\leq i\leq r$, $m_{i}$ is a $\ell_{i}$-layered number. If for
every distinct integers $1\leq i\neq j\leq r$, $\gcd(m_{i},m_{j})=1$, then
$m_{1}m_{2}\dots m_{r}$ is a $\ell_{1}\ell_{2}\dots\ell_{r}$-layered number.

The following example shows the power of Proposition 5.10 for finding a subset
of the set of $4$-layered numbers.

######  Example 5.12.

By Example 3.14, $n_{1}=3^{\alpha_{1}}\times 5^{\alpha_{2}}\times
7^{\alpha_{3}}$, in which $\alpha_{1}\geq 3$, $\alpha_{2}$, $\alpha_{3}$ are
positive integers, and at least one of the exponents of its factors is odd, is
a Zumkeller number. Let $t$ be a positive integer and $p$ be a prime number
such that $p\leq 2^{t+1}-1$ and $\gcd(p,n_{1})=1$. By Corollary 3.16, for
every odd number $\alpha_{4}$, the number $n_{2}=2^{t}\times p^{\alpha_{4}}$
is Zumkeller. Therefore, by Proposition 5.10, $n=n_{1}n_{2}$ is a $4$-layered
number.

In [18], the authors proved that the number $n!$ is Zumkeller for integers
$n\geq 3$ (we can also prove it in a slightly different way, see Remark 5.21).
In the following, we prove that the number $n!$ is $3$-layered for every
integer $n\geq 5$ and $n\neq 10$, and also the number $n!$ is $4$-layered for
every integer $n\geq 9$. Before that, we recall a theorem which was proved by
Breusch [11]; this theorem is a generalization of Bertrand’s postulate .

######  Theorem 5.13 (Breusch [11]).

For every integer $n\geq 7$, there are primes of the form $3k+1$ and $3k+2$
between $n$ and $2n$.

######  Theorem 5.14.

Let $n\geq 11$ be an integer. Then, the number $n!$ possesses prime
factorization $p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\dots p^{\alpha_{k}}_{k}$
such that $2=p_{1}<p_{2}<\dots<p_{k}$ and $\alpha_{k-1}=\alpha_{k}=1$. In
addition, $p_{k}\leq 2^{\alpha_{1}}$ and there exists a prime number $q_{n}$
such that $q_{n}\mid\mid n!$ and $q_{n}\equiv 2$ (mod 3).

###### Proof.

If $11\leq n\leq 13$, then one can check directly that $n!$ satisfies the
theorem. Now Let $n\geq 14$. Also, let $p^{\prime}$ be a prime number. We know
that $2p^{\prime}\leq n$ if and only if $p^{\prime 2}\mid n!$. Thus, if $p$ is
the largest prime number such that $p^{2}\mid n!$, then by definition of $n$,
it is clear that $p\geq 7$. Therefore, by Theorem 5.13, there exist at least
two distinct prime numbers $p^{\prime}_{1}$ and $p^{\prime}_{2}$ such that
$p<p^{\prime}_{1},p^{\prime}_{2}<2p$ and for at least one $i\in\\{1,2\\},$
$p^{\prime}_{i}\equiv 2$ (mod $3$). Also, by definition of $p$,
$p^{\prime}_{1}\mid\mid n!$ and $p^{\prime}_{2}\mid\mid n!$. Furthermore, if
$\nu_{2}(n!)$ denotes the exponent of the largest power of $2$ that divides
$n!$, then by Legendre’s formula, we have

$p_{k}\leq n<2^{\left\lfloor\frac{n}{2}\right\rfloor}<2^{\nu_{2}(n!)}.$

∎

Now we state and prove one of the key theorems of this section.

######  Theorem 5.15.

The number $n!$ is a $3$-layered number if and only if $n\geq 5$ and $n\neq
10$.

###### Proof.

On can check directly that for every integer $1\leq n\leq 4$, $I(n!)<3$, and
$3\nmid\sigma(10!)$. So, by Proposition 2.3, $n!$ fails to be $3$-layered for
every integer $1\leq n\leq 4$ and $n=10$.

Now we investigate the five cases for $n$:

Case 1: $n=5$. By Corollary 5.8, $n!=120$ is a $3$-layered number.

Case 2: $n=6$. Let $A_{1}=\\{2\times 3,2^{4}\times 5,2^{4}\times 3^{2}\times
5\\}$ and $A_{2}=\\{2^{3},2^{3}\times 3,2^{4}\times 3^{2},2^{3}\times 3\times
5,2^{4}\times 3\times 5,2\times 3^{2}\times 5,2^{2}\times 3^{2}\times 5\\}$.
One can check directly that $S(A_{1})=S(A_{2})=\dfrac{\sigma(n!)}{3}.$ Thus,
by Proposition 2.2, the number $n!=2^{4}\times 3^{2}\times 5$ is a $3$-layered
number.

Case 3: $n=7$. We proved that the number $6!=2^{4}\times 3^{2}\times 5$ is a
$3$-layered number. Thus, by Proposition 2.6, the number $7!=2^{4}\times
3^{2}\times 5\times 7$ is a $3$-layered number.

Case 4: $n=8$. We define

$A_{1}=\\{$ $2^{3}\times 3\times 5$, $2^{3}\times 3^{2}\times 5\times 7$,
$2^{5}\times 3^{2}\times 5\times 7$, $2^{7}\times 3^{2}\times 5\times 7$$\\}$

$A_{2}=\\{$ $2^{3}$, $2^{3}\times 3\times 7$, $2^{7}\times 5\times 7$,
$2^{6}\times 3\times 5\times 7$, $2^{7}\times 3^{2}\times 7$, $2^{7}\times
3\times 5\times 7$, $2^{6}\times 3^{2}\times 5\times 7$$\\}$
One can check directly that $S(A_{1})=S(A_{2})=\dfrac{\sigma(n!)}{3}$. Thus,
by Proposition 2.2, the number $n!=2^{7}\times 3^{2}\times 5\times 7$ is a
$3$-layered number.

Case 5: $n\geq 9$. Let $n=9$. Then, $n!=2^{7}m$, where $m=3^{4}\times 5\times
7$. By Example 3.11, $m$ is an odd almost practical number such that
$6\mid\sigma(m)$. Therefore, by Corollary 5.4, $n!$ is a $3$-layered number.
Now let $n>9$ and $n!=2^{\alpha_{1}}m^{\prime}$ with
$\gcd(2^{\alpha_{1}},m^{\prime})=1$ such that $m^{\prime}$ is an odd almost
practical number. Also, let $p^{\alpha_{2}}_{2}p^{\alpha_{3}}_{3}\cdots
p^{\alpha_{k}}_{k}$ be the prime factorization of $m^{\prime}$. We want to
prove that $(n+1)!=2^{\beta_{1}}m^{\prime\prime}$ with
$\gcd(2^{\beta_{1}},m^{\prime\prime})=1$ such that $m^{\prime\prime}$ is an
odd almost practical number. First, let $n+1$ be a composite number. Then, if
$p$ is a prime number such that $p\mid n+1$, then there exists a positive
integer $d\neq 1$ such that $n+1=pd$. This concludes that $p<n$. Therefore,
$p\mid n!$. Thus, $(n+1)!=2^{\beta_{1}}p^{\beta_{2}}_{2}\cdots
p^{\beta_{k}}_{k}$, where $\beta_{i}\geq\alpha_{i}$ for every integer $1\leq
i\leq k$. Therefore, by Proposition 3.10,
$m^{\prime\prime}=p^{\beta_{2}}_{2}\cdots p^{\beta_{k}}_{k}$ is an odd almost
practical number. Moreover, if $n+1=q$ is a prime number, then by Bertrand
postulate, we have

$2q<4p_{k}<\sigma(p_{2}p_{3}\cdots p_{k})-4.$

By Theorem 3.8, we once again conclude that $m^{\prime\prime}$ is an odd
almost practical number. Thus, for every integer $n\geq 9$,
$n!=2^{\alpha_{1}}t$, where $t$ is an odd almost practical number. Also, by
Theorem 5.14, there exists a prime number $q_{n}$ such that $q_{n}\mid\mid n!$
and $q_{n}\equiv 2$ (mod $3$) for all $n\geq 11$. Therefore, there exists a
positive integer $t^{\prime}$ such that $t=q_{n}t^{\prime}$ and
$\gcd(q_{n},t^{\prime})=1$; this concludes that
$6\mid\sigma(q_{n})\sigma(t^{\prime})=\sigma(t)$. Thus, by Corollary 5.4, for
every integer $n\geq 9$ except $n=10$, n! is a $3$-layered number. ∎

######  Remark 5.16.

In fact, in case 5, by Corollary 5.4, we proved that the number $n!$ is a good
$3$-layered number for $n\geq 11$. One can check that $n!$ is a good
$3$-layered number for $5\leq n\leq 9$, we omit the details here. For
instance, by Corollary 5.8, the number $5!$ is a good $3$-layered number.
Thus, we can say that the number $n!$ is $3$-layered if and only if $n!$ is a
good $3$-layered number.

######  Remark 5.17.

Let $n\geq 9$ and $n\neq 10$. Using the same arguments of Case 5 above, we get
the number $\dfrac{n!}{2^{t}}$ is a $3$-layered number for every integer
$0\leq t\leq\nu_{2}(n!)-1$. Thus, by the previous remark, if
$r=\nu_{2}(n!)-1$, then the number $\dfrac{n!}{2^{r}}$ is a good $3$-layered
number for every integer $n\geq 9$ and $n\neq 10$.

######  Theorem 5.18.

The number $n!$ is $4$-layered number if and only if $n\geq 9$.

###### Proof.

On can check directly that for every integer $1\leq n\leq 8$, $I(n!)<4$. Then,
by Proposition 2.3, $n!$ fails to be $4$-layered for every integer $1\leq
n\leq 8$. Now we want to prove that for every integer $n\geq 9$ the number
$n!$ is $4$-layered. We investigate the three cases for $n$:
Case 1: $n=9$. We define

$A_{1}=\\{$ $2^{2}\times 3$, $2^{3}\times 3^{4}$, $2^{6}\times 3\times 5\times
7$, $2^{7}\times 3^{4}\times 5\times 7$$\\}$

$A_{2}=\\{$ $2^{2}$, $2^{7}\times 7$, $2^{4}\times 3^{4}\times 5$,
$2^{6}\times 3^{3}\times 5\times 7$, $2^{7}\times 3^{3}\times 5\times 7$,
$2^{6}\times 3^{4}\times 5\times 7$$\\}$

$A_{3}=\\{$ $2^{2}\times 3^{2}$, $2^{6}\times 3^{2}\times 5$, $2^{5}\times
3^{3}\times 5\times 7$, $2^{6}\times 3^{4}\times 7$, $2^{7}\times 3^{2}\times
5\times 7$, $2^{4}\times 3^{4}\times 5\times 7$, $2^{7}\times 3^{4}\times 5$,
$2^{7}\times 3^{4}\times 7$, $2^{5}\times 3^{4}\times 5\times 7$$\\}$

One can check directly that for every integer $1\leq i\leq 3$,
$S(A_{i})=\dfrac{\sigma(n!)}{4}$. Thus, by Proposition 2.2, the number
$n!=362880=2^{7}\times 3^{4}\times 5\times 7$ is a $4$-layered number.

Case 2: $n=10$. We define

$A_{1}=\\{$$2$, $2^{5}\times 3$, $2^{4}\times 3^{3}\times 7$, $2^{7}\times
3^{2}\times 5^{2}\times 7$, $2^{8}\times 3^{4}\times 5^{2}\times 7$$\\}$

$A_{2}=\\{$$1$, $2^{4}$, $3^{2}\times 5^{2}$, $2^{6}\times 3^{2}\times 5$,
$2^{8}\times 3^{2}\times 5\times 7$, $2^{8}\times 3^{4}\times 5\times 7$,
$2^{8}\times 3^{3}\times 5^{2}\times 7$, $2^{7}\times 3^{4}\times 5^{2}\times
7$$\\}$

$A_{3}=\\{$$2\times 3^{2}$, $2^{5}\times 7$, $2^{5}\times 3^{3}\times 5^{2}$,
$2^{7}\times 3^{4}\times 5^{2}$, $2^{6}\times 3^{3}\times 5^{2}\times 7$,
$2^{7}\times 3^{4}\times 5\times 7$, $2^{8}\times 3^{2}\times 5^{2}\times 7$,
$2^{5}\times 3^{4}\times 5^{2}\times 7$, $2^{8}\times 3^{4}\times 5^{2}$,
$2^{7}\times 3^{3}\times 5^{2}\times 7$, $2^{6}\times 3^{4}\times 5^{2}\times
7$$\\}$.

One can check directly that for every integer $1\leq i\leq 3$,
$S(A_{i})=\dfrac{\sigma(n!)}{4}$. Thus, by Proposition 2.2, the number
$n!=3628800=2^{8}\times 3^{4}\times 5^{2}\times 7$ is a $4$-layered number.

Case 3: $n\geq 11$. Let $n\geq 11$ and
$2^{\alpha_{1}}p^{\alpha_{2}}_{2}p^{\alpha_{3}}_{3}\cdots p^{\alpha_{k}}_{k}$
be a prime factorization of $n!$. Now let $t=(p_{k}-1)!$. By the definition of
$n$, $(p_{k}-1)\geq 10$. Thus, there exists an odd almost practical number
$m^{\prime}$ such that $t=2^{\alpha}m^{\prime}$ (see the proof of Corollary
5.15 for Case 5). Also, we know that
$m^{\prime}=p^{\beta_{2}}_{2}p^{\beta_{3}}_{3}\cdots p^{\beta_{k-1}}_{k-1}$,
where $1\leq\beta_{i}\leq\alpha_{i}$ for every integer $1\leq i\leq k-1$.
Therefore, by Proposition 3.10, the number $m=p^{\alpha_{2}}_{2}\cdots
p^{\alpha_{k-1}}_{k-1}$ is an odd almost practical number. Also, by Theorem
5.14, $\alpha_{k-1}=1$; this concludes that $2\mid\sigma(m^{\prime})$. Then,
by Proposition 3.13, $m$ is an odd Zumkeller number. Moreover, by Theorem
5.14, $p_{k}<\sigma(2^{\alpha_{1}})$. Therefore, by Corollary 3.16, the number
$2^{\alpha_{1}}p_{k}$ is a Zumkeller number. Thus, by Proposition 5.10,
$n!=2^{\alpha}p_{k}m$ is a $4$-layered number. ∎

######  Remark 5.19.

In Case 3 above, we proved that for every integer $n\geq 11$ there exist
Zumkeller numbers $m_{1}$ and $m_{2}$ such that $\gcd(m_{1},m_{2})=1$ and
$n!=m_{1}m_{2}$. Without loss of generality, we let $m_{2}$ be odd. By Theorem
4.12 of [9], the number $2^{\ell}m_{1}$ is a Zumkeller number for every
positive integer $\ell$. Thus, by Proposition 2.6, the number $n!$ is a good
$4$-layered number for $n\geq 11$. Also, by Proposition 5.28, the number $9!$
is a good $4$-layered number. In addition, one can check directly that the
numbers $10!$ is good $4$-layered; we omit the details here. Therefore, we can
say that the number $n!$ is $4$-layered if and only if $n!$ is a good
$4$-layered number.

######  Remark 5.20.

By Theorem 5.14 and Corollary 3.16,
$2^{\left\lfloor\frac{n}{2}\right\rfloor}p_{k}$ is also a Zumkeller number.
Thus, using the same arguments of Case 3 of the above proof, we get
$\dfrac{n!}{2^{t}}$ is a 4-layered number for $n\geq 11$, where $1\leq
t\leq\sum_{i=2}^{\infty}\left\lfloor\dfrac{n}{2^{i}}\right\rfloor$. Therefore,
by the previous remark, if
$r=\sum_{i=2}^{\infty}\left\lfloor\dfrac{n}{2^{i}}\right\rfloor$, then the
number $\dfrac{n!}{2^{r}}$ is a good $4$-layered number for $n\geq 11$.

######  Remark 5.21.

By Bertrand postulate and Proposition 3.2, every primorial number 222For the
$n$th prime number $p_{n}$, the primorial $p_{n}\\#$ is defined as the product
of the first $n$ primes. $n\neq 2$ is a practical number. Also, by Proposition
3.2, if $n$ is a practical number and $d\mid n$, then $nd$ is also a practical
number. Then, By Proposition 3.13, if
$n=p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{\ell}}_{\ell}$, where
$p_{i}$ be the $i$th prime, and $\alpha_{1},\alpha_{2},\dots,\alpha_{\ell}$
are positive integers for every positive $1\leq i\leq\ell$, then the number
$n$ is a Zumkeller number if and only if $2\mid\sigma(n)$. In addition, by
Bertrand’s postulate and the same method that we used to prove Theorem 5.14,
we can prove that for integers $n\geq 3$, there exists a prime number $q_{n}$
such that $q_{n}\mid\mid n!$. Thus, $2\mid\sigma(n!)$ for integers $n\geq 3$.
Therefore, by Proposition 3.13, $n!$ is a Zumkeller number for integers $n\geq
3$.

######  Remark 5.22.

Let $n\leq 10^{10}$ be a practical number. One can check directly that if is
good $k$-abundant, then $n$ is $k$-layered.

Proposition 3.2, Theorems 5.15 and 5.18, Remarks 5.16, 5.17, 5.19, 5.20, and
5.22 leads us to the following conjecture.

######  Conjecture 5.23.

Let $n$ be a practical number. If $n$ is good $k$-abundant, then $n$ is
$k$-layered.

Now we recall a well-known proposition which is a consequence of prime number
for arithmetic progressions.

######  Proposition 5.24.

Let $m$ be a positive integer. There exists a positive integer $s_{m}$ such
that for integers $n\geq s_{m}$, there exists a prime $p_{m,n}$ between $n$
and $2n$, where $p_{m,n}\equiv m-1\pmod{m}$.

By using the previous proposition and the same method that we used in the
proof of Theorem 5.14, we can conclude the following theorem.

######  Theorem 5.25.

Let $m$ be a positive integer. For every positive integer $n$ that is large
enough, then there exists a prime $p_{m,n}$ such that $p_{m,n}\equiv
m-1\pmod{m}$ and $p_{m,n}\mid\mid n!$.

Now we state a well-known proposition which recalls some elementary properties
of the abundancy function $I$.

######  Proposition 5.26.

The abundancy function $I$ possesses following properties.

$\mathrm{(i)}\ $ Let $p$ and $q$ be primes such that $p<q$. Let $\alpha$ and
$\beta$ be positive integers such that $\beta\leq\alpha$. Then,
$I(q^{\beta})<I(p^{\alpha}).$

$\mathrm{(ii)}\ $ Let $p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots
p^{\alpha_{k}}_{k}$ be the prime factorization of a positive integer $n$. Then

$I(n)=\prod_{i=1}^{k}I(p^{\alpha_{i}}_{i})<\prod_{i=1}^{k}\dfrac{p_{i}}{p_{i}-1}.$

Thus, by the Theorem 5.25 and some elementary properties of abundancy
function, for every positive integer $k$, there exists a positive integer
$s_{k}$ such that $k\mid\sigma(n!)$ and $I(n!)\geq k$ for integers $n\geq
s_{k}$. Then we have the following corollary.

######  Corollary 5.27.

Let $k$ be a positive integer. If Conjecture 5.23 holds, then there exists a
positive integer $s_{k}$ such that for integers $n\geq s$, the number $n!$ is
a good $k$-layered number.

Let $\alpha$ and $1\leq\beta\leq 3$ be positive integers. We close this
section by the following theorem which find all $4$-layered numbers of the
form $n=p^{\alpha}_{1}p^{\beta}_{2}p_{3}p_{4}$, where $p_{i}$ is prime for
integers $1\leq i\leq 4$.

######  Theorem 5.28.

Let $\alpha$ and $1\leq\beta\leq 3$ be positive integers. Also, let
$n=p^{\alpha}_{1}p^{\beta}_{2}p_{3}p_{4}$, where $p_{i}$ is prime for integers
$1\leq i\leq 4$. The number $n$ is 4-layered if and only if
$p_{1}=2,p_{2}=3,p_{3}=5,p_{4}=7,\alpha\geq 5,$ and $\beta=3$

###### Proof.

By some properties of the abundancy function, we know that
$I(2^{\alpha}3^{\beta}5^{2})<\dfrac{2}{2-1}\times\dfrac{3}{3-1}\times\dfrac{1+5+5^{2}}{5^{2}}<4\leq
I(p^{\alpha}_{1}p^{\beta}_{2}p_{3}p_{4})$. Thus, by Proposition 5.26, for
integers $1\leq i\neq j\leq 4$, $p_{i}\neq p_{j}$. Now we prove that
$2^{5}\mid n$ by contradiction. We let $2^{5}\nmid n$. Therefore, $I(n)\leq
I(2^{5}\times 3^{\alpha}\times 5\times 7)<4$, a contradiction. This concludes
that $p_{1}=2$ and $\alpha\geq 4$. Also, $3^{3}\mid n$, for otherwise
$I(n)\leq I(2^{\alpha}\times 3^{2}\times 5\times 7)<4$, a contradiction. By
the same method, one can prove that $p_{3}=5$, $p_{4}=7$. Thus, for completing
the proof, it is sufficient to proof that the number $n=2^{5}\times
3^{3}\times 5\times 7$ is a good $4$-layered number. Let

$A_{1}=\\{$ $2^{5}\times{}3^{3}\times{}5\times{}7$$\\},\hfil\penalty
0\hfilneg$

$A_{2}=\\{$ $2\times{}3$, $2^{4}\times{}3^{2}$, $2^{5}\times{}3\times{}5$,
$2\times{}3^{3}\times{}5\times{}7$, $2^{4}\times{}3^{2}\times{}5\times{}7$,
$2^{3}\times{}3^{3}\times{}5\times{}7$,
$2^{4}\times{}3^{3}\times{}5\times{}7$$\\},\hfil\penalty 0\hfilneg$

$A_{3}=\\{$ $3$, $2^{4}\times{}3$, $3^{3}\times{}5\times{}7$,
$2^{4}\times{}3^{3}\times{}5$, $2^{3}\times{}3^{2}\times{}5\times{}7$,
$2^{4}\times{}3^{3}\times{}7$, $2^{5}\times{}3\times{}5\times{}7$,
$2^{2}\times{}3^{3}\times{}5\times{}7$, $2^{5}\times{}3^{3}\times{}5$,
$2^{5}\times{}3^{2}\times{}5\times{}7$$\\}$,

and $A_{4}=D_{n}\setminus(A_{1}\cup A_{2}\cup A_{3})$. One can check directly
that $\\{A_{1},A_{2},A_{3},A_{4}\\}$ is a $4$-layered partition for $n$. Let
$n=2^{5}m$. Let $A^{\prime}_{1}=\\{$ $3^{3}\times 5\times 7$, $3\times 5$$\\}$
and $A^{\prime}_{2}=D_{m}\setminus A^{\prime}_{1}$. One can check directly
that $\\{A^{\prime}_{1},A^{\prime}_{2}\\}$ is a Zumkeller partition for $m$.
Also, if we define $A^{\prime\prime}_{1}=\\{2^{5}d:d\in A^{\prime}_{1}\\}$ and
$A^{\prime\prime}_{2}=\\{2^{5}d:d\in A^{\prime}_{2}\\}$, then one can check
directly that $A^{\prime\prime}_{1}\subset(A_{1}\cup A_{2})$ and
$A^{\prime\prime}_{2}\subset(A_{3}\cup A_{4})$. Thus, by Proposition 5.5, $n$
is a good $4$-layered number. ∎

##  6\. $k$-multiperfect numbers and $k$-layered numbers

We recall that a number $n$ is said to be $k$-multiperfect if $\sigma(n)=kn$
for some integer $k\geq 2$. It is clear that if $n$ is $2$-multiperfect, then
$n$ is said to be perfect. Since every perfect number is Zumkeller, this
suggests us the following open question.

######  Open question 6.1.

Is every $k$-multiperfect number $k$-layered?

In the following, we will address this open question in some special cases.
Before that, we recall that a number $n$ is said to be semiperfect if $n$ is
equal to the sum of all or some of its proper divisors. Also, the abundant
number $n$ which is not semiperfect is called weird. The existence of odd
weird numbers is still an open question. The number $70$ is the smallest weird
number. If $n$ is weird and $p$ is a prime number such that $\sigma(n)<p$,
then $np$ is a weird number (for more details, see [4]). This shows that there
exist infinity many weird numbers. Nevertheless, it is not yet known whether
there exists infinitely many primitive weird numbers 333A weird number $n$
that is not multiple of other weird number is said to be primitive weird
number. or not. Recently, a wide range of number theorists have tried to
introduce some optimal algorithms to find new primitive weird numbers. (For
instance, see [7, 8].) Fang and Beckert proved, using parallel tree search,
that there are no odd weird numbers up to $10^{21}$([15]). This leads us to
the following conjecture.

######  Conjecture 6.2.

Every odd abundant number is semiperfect.

Now we recall two well-known propositions about semiperfect numbers.

######  Proposition 6.3.

Let $m$ be a positive integer, and let $p$ be a odd prime number such that
$p\leq 2^{m+1}-1$. Then, the number $n=2^{m}p$ is a semiperfect number.

######  Proposition 6.4.

If a positive integer $m$ is divisible by a semiperfect number $n$, then $m$
is semiperfect.

######  Remark 6.5.

Let $\alpha$ and $\beta$ be positive integers. Also, let $p_{1}$ and $p_{2}$
be primes. By Remark 3.17, the number $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}$
is an abundant number if and only if $p_{1}=2$ and $3\leq p_{2}\leq
2^{\alpha+1}-1$. Also, by Propositions 6.3 and 6.4, all numbers of this form
are semiperfect. This concludes that all abundant number which possess exactly
two distinct prime factors are semiperfect.

It is believed that all $k$-multiperfect number of abundancy $3,4,5,6$ and $7$
are known. Reference [1] contains the list of all known $k$-multiperfect
numbers. According to this list, one can directly obtain the following
proposition.

######  Proposition 6.6.

$\mathrm{(i)}\ $ All the known $k$-multiperfect numbers are practical.

$\mathrm{(ii)}\ $ All the $6$ known $3$-multiperfect numbers are divisible by
$3$ or $5$. Also, all $3$-multiperfect numbers are divisible by $2^{3}$.

$\mathrm{(iii)}\ $ All the $36$ known $4$-multiperfect numbers are divisible
by $2^{2}$. In addition, if $n=2^{\alpha}t$ is one the known $4$-multiperfect,
then $\dfrac{n}{2}$ satisfies the conditions of Proposition 3.2; this
concludes that $\dfrac{n}{2}$ is a practical number.

$\mathrm{(iv)}\ $ Let $n$ be one of the known $5$-multiperfect numbers. Then,
$2^{7}\mid n$ and $5\mid n$ or $7\mid n$.

As a consequence of part $\mathrm{(i)}\ $ of the above proposition, we have
the following corollary.

######  Corollary 6.7.

If Conjecture 5.23 holds, then every known $k$-multiperfect number is
$k$-layered.

Thus, we have the following theorem.

######  Theorem 6.8.

Every known $3$-multiperfect number is $3$-layered.

###### Proof.

Let $n$ be one of the known $3$-multiperfect numbers. By Propositions 6.3 and
6.4, and part $\mathrm{(ii)}\ $ of Proposition 6.6, the number $n$ is
semiperfect. This concludes that there exists $A_{1}\subset
D_{n}\setminus\\{n\\}$ such that $S(A_{1})=n=\dfrac{\sigma(n)}{3}$. Now we
define $A_{2}=\\{n\\}$ and $A_{3}=D_{n}\setminus(A_{1}\cup A_{2})$. Thus, by
the definition of $n$, $\\{A_{1},A_{2},A_{3}\\}$ is a $3$-layered partition
for $n$. ∎

Now we state an essential and simple lemma.

######  Lemma 6.9.

Let $n=2^{\alpha}t$ is an even positive integer with $t$ odd and $I(n)\geq 4$.
Then, $I(t)>2$.

###### Proof.

We know that $I(2^{\alpha})=\dfrac{2^{\alpha+1}-1}{2^{\alpha}}<2$. This
concludes that $I(t)>2$. ∎

Now by assuming Conjecture 6.2, we state a theorem similar to Theorem 6.8
about all the known $4$-multiperfect numbers.

######  Theorem 6.10.

If Conjecture 6.2 holds, then every known $4$-multiperfect numbers are
$4$-layered.

###### Proof.

Let $n=2^{\alpha}m$ be a known $4$-multiperfect number with
$\gcd(2^{\alpha},m)=1$. By part $\mathrm{(iii)}\ $ of Proposition 6.6, the
number $t=\dfrac{n}{2}$ is a practical number. Also, by Lemma 6.9,
$\sigma(t)>n$. Thus, by definition of practical numbers, there exists
$A_{1}\subset D_{t}\setminus\\{t\\}$ such that
$S(A_{1})=n$=$\dfrac{\sigma(n)}{4}$. Moreover, by assuming Conjecture 6.2,
there exists $A\subset D_{m}\setminus\\{m\\}$ such that $S(A)=m$. Now we
define $A_{2}=\\{2^{\alpha}d:d\in A\\}$. It is clear that
$S(A_{2})=n$=$\dfrac{\sigma(n)}{4}$. Furthermore, we define $A_{3}=\\{n\\}$.
By definition of $A_{1},A_{2},$ and $A_{3}$ for every integers $1\leq i\neq
j\leq 3$, $A_{i}\cap A_{j}=\emptyset$. Then, by Proposition 2.2, $n$ is
$4$-layered. ∎

Now we proof a theorem similar to Theorem 6.10 about $5$-multiperfect numbers.

######  Theorem 6.11.

By assuming Conjecture 6.2, every known $5$-multiperfect numbers are
$5$-layered.

###### Proof.

Let $n$ be one of the known $5$-multiperfect numbers. By part $\mathrm{(iv)}\
$ of Proposition 6.6, we know there exists a prime number $p=5$ or $p=7$ where
$p\mid n$. Now let $\alpha$, $\beta$, $t$, and $m$ be positive integers such
that $\nu_{2}(n)=\alpha$, $\nu_{p}(n)=\beta$, and $n=2^{\alpha}p^{\beta}t=mt$.
we have $I(m)<\dfrac{2}{2-1}\times\dfrac{5}{5-1}=2.5$.

We know $I(n)=I(m)I(t)=5$; this concludes that $I(t)>2$. By Proposition 6.3,
there exists $A\subset D_{m}\setminus\\{m\\}$ where $S(A)=m$. Also, by
assuming Conjecture 6.2, there exists $A^{\prime}\subset
D_{t}\setminus\\{t\\}$ where $S(A^{\prime})=t$. Now we define
$A_{1}=\\{n\\},A_{2}=\\{dt:d\in A\\},A_{3}=\\{dm:d\in A^{\prime}\\},$ and
$A_{4}=\\{d_{1}d_{2}:d_{1}\in A\land d_{2}\in A^{\prime}\\}$. One can check
directly that for every integer $1\leq i\leq 4$,
$S(A_{i})=\dfrac{\sigma(n)}{5}=n$. Thus, by Proposition 2.2, $n$ is
$5$-layered. ∎

######  Remark 6.12.

For every known 4-multiperfect number $n$, one can check that the number $m$,
which is defined in the proof of Theorem 6.9, is semiperfect. In addition, for
every known 5-multiperfect number $n$, one can check that the number $t$,
which is defined in the proof of Theorem 6.11, is semiperfect. This concludes
that the all known 4-multiperfect numbers and the all known 5-multiperfect
numbers are 4-layered and 5-layered, respectively. We do not pursue it here.

Now we state an essential proposition.

######  Proposition 6.13.

Let $n$ and $m$ be positive integers with $gcd(n,m)$=1. If $n$ is $k$-layered
and $\dfrac{\sigma(m)}{k+1}$ is a sum of a subset of the set of all positive
divisors of $m$, then $nm$ is $(k+1)$-layered.

###### Proof.

Let $\\{A_{1},A_{2},\dots,A_{k}\\}$ be a $k$-layered partition for $n$. Also,
let $A$ be a subset of the set of all positive divisors of $m$ such that
$S(A)=\dfrac{\sigma(m)}{k+1}$. We define $A^{\prime}=D_{m}\setminus A$ One can
check directly that for every integer $1\leq i\leq k$,
$S(A_{i}A^{\prime})=\dfrac{\sigma(nm)}{k+1}$. Thus, by Proposition 2.2, $nm$
is $(k+1)$-layered. ∎

As a consequence of Proposition 6.13, we have the following corollary.

######  Corollary 6.14.

Let $p$ and $n$ be a prime number and a $p$-layered number, respectively. If
$\gcd(n,p)=1$, then $pn$ is a $(p+1)$-layered.

######  Remark 6.15.

Let $p$ and $n$ be a prime number and a $p$-multiperfect number, respectively.
One can check directly that if $\gcd(n,p)=1$, then $pn$ is a
$(p+1)$-multiperfect number.

By Corollary 6.14 and Remark 6.15, we have the following theorem.

######  Theorem 6.16.

Let $p$ be a prime number and $n$ be a positive integer such that
$\gcd(n,p)=1$. If $n$ is $p$-multiperfect and $p$-layered, then $pn$ is
$(p+1)$-multiperfect and $(p+1)$-layered.

The two following examples state a application of previous theorem to find a
new family of $k$-multiperfect numbers which are $k$-layered.

######  Example 6.17.

One can check that the number $n=2^{14}\times 5\times 7\times 19\times
31\times 151$ is a $3$-multiperfect number. By Theorem 6.8, $n$ is
$3$-layered. Thus, by Theorem 6.16, $3n$ is $4$-multiperfect and $4$-layered

######  Example 6.18.

One can check that the number $n=2^{29}\times 3^{10}\times 7^{5}\times
11^{2}\times 13\times 19^{3}\times 23\times 31\times 43\times 83\times
107\times 151\times 181\times 331\times 3851$ is a $5$-multiperfect number. By
Remark 6.15, $5n$ is $6$-multiperfect. If conjecture 6.2 holds, then by
Theorem 6.10, $n$ is $5$-layered. Also, if $n$ is $5$-layered, then by
Corollary 6.14, $5n$ is $6$-layered.

Now we state a proposition by which we can find more family $k$-multiperfect
numbers which are $k$-layered.

######  Proposition 6.19.

Let $n$ and $m$ be a $k$-multiperfect number and a $(k+1)$-multiperfect
number, respectively, such that $n\mid m$. If $n$ is a $k$-layered number,
then $m$ is $(k+1)$-layered number.

###### Proof.

Let $\\{A_{1},A_{2},\dots,A_{k}\\}$ be a $k$-layered partition for the
$k$-multiperfect number $n$. Now let $dn=m$ . For every integer $1\leq i\leq
k$, we define $B_{i}=\\{ad:a\in A_{i}\\}$. Then, for every integer $1\leq
i\leq k$, $\sigma(B_{i})=m=\dfrac{\sigma(m)}{k+1}$. Thus, by Proposition 2.2,
$m$ is $(k+1)$-layered. ∎

######  Remark 6.20.

One can check directly that exactly half of known $4$-multiperfect numbers are
divisible by at least a $3$-multiperfect number. Then, by Propositions 6.8 and
6.19, at least half of known $4$-multiperfect are $4$-layered.

We close this section by an example that implies an application of Proposition
6.19 to find another new family of $k$-multiperfect number which are
$k$-layered.

######  Example 6.21.

Let $t_{1}=2\times 3,t_{2}=2^{3}\times 3\times 5,t_{3}=2^{5}\times 3^{3}\times
5\times 7,t_{4}=2^{11}\times 3^{3}\times 5^{2}\times 7^{2}\times 13\times
19\times 31,t_{5}=2^{19}\times 3^{5}\times 5^{2}\times 7^{2}\times 11\times
13^{2}\times 19^{2}\times 31^{2}\times 37\times 41\times 61\times
127,t_{6}=2^{39}\times 3^{11}\times 5^{7}\times 7^{3}\times 11\times
13^{2}\times 17\times 19^{2}\times 29\times 31^{2}\times 37\times 41\times
61\times 73\times 79\times 83\times 127\times 157\times 313\times 331\times
2203\times 30841\times 61681$. One can check that for integers $1\leq i\leq
6$, the number $t_{i}$ is a $(i+1)$-perfect number. Also, it is easy to see
that for every integer $1\leq i\leq 5$, $t_{i}\mid t_{i+1}$. We know that $6$
is Zumkeller. Thus, by Proposition 6.19, for every integer $1\leq i\leq 6$,
$t_{i}$ is a $(i+1)$-layered number.

##  7\. Near-perfect numbers and Zumkeller numbers

In this short section, we do not apply any advanced method. In fact, the
importance of Remark 7.7 motives the author to state this section; by this
remark, we can find a new family of Zumkeller numbers.

We recall that a number $n$ is said to be a near-perfect number if $n$ is the
sum of all of its proper divisors, except for one of them, which we term the
redundant divisor [10]. By definition of near-perfect numbers, we have the
following well-known proposition.

######  Proposition 7.1.

A number $n$ is a near-perfect number with redundant divisor $d$ if and only
if $d$ is a proper divisor of $n$, and $\sigma(n)=2n+d$

######  Remark 7.2.

By Proposition 7.1, if a number $n$ is an odd near-perfect number, then
$\sigma(n)$ is odd. Thus, by Proposition 2.3, $n$ fails to be Zumkeller.

Now we state a proposition similar to Proposition 3.13, about near-perfect
numbers.

######  Proposition 7.3.

Let $n$ be a near-perfect number. Then, $n$ is Zumkeller if and only if
$\sigma(n)$ is even.

###### Proof.

By Proposition 2.3, if $n$ is a Zumkeller number, then $\sigma(n)$ is even.
Now let $\sigma(n)$ is even. By the definition of $n$ and Proposition 7.1, $n$
possesses a divisor $d$ such that $2d$ is a proper divisor of $n$ and
$\sigma(n)=2n+2d$. We define $A=\\{n,d\\}$. One can check directly that
$\\{A,D_{n}\setminus A\\}$ is a Zumkeller partition for $n$. ∎

The following proposition was stated by Ren and Chen [16]; this proposition
classifies all near-perfect numbers with two distinct prime factors.

######  Proposition 7.4.

Let $n$ be a positive integer with two prime factors. The number $n$ is near-
perfect if and only if one of the following holds:

$\mathrm{(i)}\ n=2^{t-1}(2^{t}-2^{k}-1)$, where $2^{t}-2^{k}-1$ is prime;

$\mathrm{(ii)}\ n=2^{2p-1}(2^{p}-1)$, where $2^{p}-1$ is a Mersenne prime;

$\mathrm{(iii)}\ n=2^{p-1}(2^{p}-1)^{2}$, where $2^{p}-1$ is a Mersenne prime;

$\mathrm{(iv)}\ n=40$.

As a direct consequence of Propositions 7.3 and 7.4, we have the following
corollary.

######  Corollary 7.5.

Let $n$ be a near-perfect number with two distinct prime factors. The number
$n$ is non-Zumkeller if and only if there exists a Mersenne prime $p$ such
that $n=2^{p-1}(2^{p}-1)^{2}$.

Now we recall a proposition of [17]. This proposition classifies all near-
perfect numbers of the form $2^{\alpha}p_{1}p_{2}$, where $p_{1}$ and $p_{2}$
are odd primes with $p_{1}<p_{2}$.

######  Proposition 7.6.

Let $\alpha$ be a positive integer. Let $p_{1}$ and $p_{2}$ be odd primes with
$p_{1}<p_{2}$. Then, the number $n=2^{\alpha}p_{1}p_{2}$ is near-perfect if
and only if $n$ satisfies one the following conditions.

$\mathrm{(i)}\ p_{1}=\dfrac{2^{\alpha+1}-1+k}{2^{\beta}+1-k}$, where
$k=\dfrac{2^{\alpha+1}-1}{p_{2}}$ and $1\leq\beta\leq\alpha-1$.

$\mathrm{(ii)}\ $ $p_{1}=2^{\alpha+1}-1+\dfrac{2^{\alpha}-2^{\beta-1}}{k}$,
where $k$ is determined by equation

$p_{2}=(2^{\alpha+1}-1)(2k+1)-2^{\beta},1\leq\beta\leq\alpha.$

$\mathrm{(iii)}\
p_{2}=2^{\alpha+1}-1+\dfrac{2^{2\alpha+1}-2^{\alpha}-2^{\beta-1}}{k}$, where
$k=\dfrac{p_{1}-(2^{\alpha+1}-1)}{2}$ and $1\leq\beta\leq\alpha$.

######  Remark 7.7.

Let $n=2^{\alpha}p_{1}p_{2}$ be a number satisfying case $\mathrm{(i)}\ $ of
Proposition 7.6. One can check directly that $p_{1}<2^{\alpha+1}-1$. This
concludes that $2^{\alpha}p_{1}$ is a practical number. Thus, without knowing
Propositions 7.3 and 7.6, by Propositions 3.2 and 3.13, we know that all the
numbers satisfying case $\mathrm{(i)}\ $ of Proposition 7.6 are Zumekeller,
but by case $\mathrm{(ii)}\ $and case $\mathrm{(iii)}\ $ of Proposition 7.6,
we can construct a new subset of Zumkeller numbers.

In [17], the authors checked that there exists exactly $8$ near-perfect
numbers satisfying case $\mathrm{(i)}\ $of Proposition 7.6 for $\alpha<1000$,
$\beta=\dfrac{\alpha+5}{2}$, and $k=2^{\beta-2}+1$; the number $2^{9}\times
11\times 31$ is the smallest of such numbers. Also, for case $\mathrm{(ii)}\ $
of Proposition 7.6, They checked that there exist exactly $289$ near-perfect
number; the number $2^{3}\times 17\times 101$ is the smallest such number.
Furthermore, for case $\mathrm{(iii)}\ $ of Proposition 7.6, they checked that
there exist exactly $248$ near-perfect number for $\alpha<100$ and $k\leq
100$; the number $2^{3}\times 17\times 131$ is the smallest such number.

In [12], the authors raised the following conjecture.

######  Conjecture 7.8.

If $n$ is an even and Zumkeller number, then $n$ is half-Zumkeller.

In [18], the authors verified that the conjecture is true in some cases. They
proved that if $n$ is an even Zumkeller number such that $\sigma(n)<3n$, then
$n$ is a half-Zumkeller. Thus, by Remark 7.2, we have the following
proposition.

######  Proposition 7.9.

Let $n$ be a near-perfect number. If $n$ is a Zumkeller number, then $n$ is
half-Zumkeller.

##  8\. The difference of two consecutive $k$-layered numbers

In this section, we let $k,m,a,b,s,r$, and $z$ be positive integers. We also
let $p_{1},p_{2},\cdots,p_{m}$ be distinct primes such that
$p_{1}<p_{2}<\dots<p_{m}$. We start this section with two new definitions.

######  Definition 8.1.

The ascending chain $\ell=(a+t)^{z}_{t=0}$ is said to be a
$(p_{1},p_{2},\dots,p_{m})$-$\gcd$ chain if for every integer $0\leq j\leq z$,
$\gcd(a+j,p_{1}p_{2}\cdots p_{m})\neq 1$.

######  Definition 8.2.

The chain $\ell=(a+t)^{z}_{t=0}$ is said to be a
$(p_{1},p_{2},\dots,p_{m})$-$\gcd$-max chain if $\ell$ satisfies the following
conditions:

$\mathrm{(i)}\ \ell$ is a $(p_{1},p_{2},\dots,p_{m})$-$\gcd$ chain;

$\mathrm{(ii)}\ $ If $\ell^{\prime}=(a+t)^{s}_{t=0}$ is a
$(p_{1},p_{2},\dots,p_{m})$-$\gcd$ chain, then $z\geq s$.

We also define $L_{max}(p_{1},p_{2},\dots,p_{m})$ as the cardinality of
$\ell$.

We recall a standard well-known proposition, which is a generalization of the
Chinese reminder theorem.

######  Proposition 8.3.

Let $n_{1},n_{2},\dots,n_{k},t_{1},t_{2},\dots,$ and $t_{k}$ be positive
integers. Then, the simultaneous set of congruences $x\equiv
t_{1}\pmod{n_{1}}$, $x\equiv t_{2}\pmod{n_{2}}$, $\dots$, and $x\equiv
t_{k}\pmod{n_{k}}$ has a solution if and only if for every integers $1\leq
i\neq j\leq k$, $\gcd(n_{i},n_{j})\mid t_{i}-t_{j}$.

The following proposition finds a lower bound for
$L_{max}(p_{1},p_{2},\dots,p_{m})$.

######  Proposition 8.4.

$L_{max}(p_{1},p_{2},\dots,p_{m})\geq m$.

###### Proof.

By Proposition 8.3, we can find a positive integer $a$ such that $a\equiv 0$
(mod $p_{1}$), $a\equiv-1$ (mod $p_{2}$), $\dots$, and $a\equiv-(m-1)$ (mod
$p_{m}$). Then, the chain $\ell=(a+t)^{m-1}_{t=0}$ is a
$(p_{1},p_{2},\dots,p_{m})$-$\gcd$ chain. ∎

We know that $L_{max}$ can be known as a function defined on the set of all
positive square-free integers. The following proposition state another
property of this function.

######  Proposition 8.5.

$L_{max}(p_{1},p_{2},\dots,p_{m})<L_{max}(p_{1},p_{2},\dots,p_{m},p_{m+1})$.

###### Proof.

Let $\ell=(a+t)^{s}_{t=0}$ be a $(p_{1},p_{2},\dots,p_{m})$-$\gcd$ chain. This
concludes that for every integer $0\leq t\leq s$, there exists positive
integer $1\leq i_{t}\leq m$ such that $p_{i_{t}}\mid a+t$. Thus, by
Proposition 8.3, if $0\leq t\neq t^{\prime}\leq s$, then
$\gcd(p_{i_{t}},p_{i_{t^{\prime}}})\mid t-t^{\prime}$. Therefore, by the
definition of $p_{m+1}$ and Proposition 8.3, there exists a positive integer
$b$ such that $b\equiv 0$ (mod $p_{i_{0}}$), $b\equiv-1$ (mod $p_{i_{1}}$),
$\dots$, $b\equiv-s$ (mod $p_{i_{s}}$), and $b\equiv-(s+1)$ (mod $p_{m+1}$).
Thus, $\ell^{\prime}=(b+t)^{s+1}_{t=0}$ is a
$(p_{1},p_{2},\dots,p_{m},p_{m+1})$-$\gcd$ chain. This completes the proof. ∎

In remainder of this note, we define $\phi$ to be the Euler’s totient
function. Let $n$ be a positive integer with prime factorization
$p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{m}}_{m}$. We recall
that $\phi(n)=n\prod_{i=1}^{m}(1-\dfrac{1}{p_{i}})$. In Proposition 8.4, we
found a lower bound for $L_{max}(p_{1},p_{2},\dots,p_{m})$. Now we find an
upper bound for $L_{max}(p_{1},p_{2},\dots,p_{m})$.

######  Proposition 8.6.

$L_{max}(p_{1},p_{2},\dots,p_{m})\leq p_{1}p_{2}\cdots
p_{m}-(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1).$

###### Proof.

Let $t=p_{1}p_{2}\cdots p_{m}$. Let $r$ be a non-negative integer such that
$0\leq r<t$. We know that for every positive integer $j$, $\gcd(jt+r,t)=1$ if
and only if $\gcd(r,t)=1$. Therefore, between $0$ and $t$ there exist exactly
$\phi(t)$ distinct integers $r_{1},r_{2},\dots,$ and $r_{\phi(t)}$ such that
$1=r_{1}<r_{2}<\dots<r_{\phi(t)}<t$ and $\gcd(jt+r_{i},t)=1$ for every
integers $1\leq i\leq\phi(t)$ and $j\geq 0$. This concludes that between $t$
and $2t$ there exist exactly $\phi(t)$ positive integers
$n_{1}<n_{2}<\dots<n_{\phi(t)}$ such that $\gcd(n_{i},t)=1$ for every integer
$1\leq i\leq\phi(t)$. Thus, $L_{max}(p_{1},p_{2},\dots,p_{m})\leq t-\phi(t).$
This completes the proof. ∎

The following proposition calculates the function $L_{max}$ at some value.

######  Proposition 8.7.

$\mathrm{(i)}\ L_{max}(3,5,7,11)=6$.

$\mathrm{(ii)}\ L_{max}(3,5,7,11,13,17)=12$.

###### Proof.

$\mathrm{(i)}\ $ By the Chinese remainder theorem, there exists a positive
integer $a$ such that $a\equiv 0$ (mod $5$), $a\equiv-1$ (mod $3$),
$a\equiv-2$ (mod $7$), and $a\equiv-3$ (mod $11$). So, $a\equiv-4$ (mod $3$)
and $a\equiv-5$ (mod $5$). Therefore, $\ell=(a+t)^{5}_{t=0}$ is a $(3,5,7,11)$
\- $\gcd$ chain. Now it is sufficient to prove that for every positive integer
$a^{\prime}$, there exists an integer $0\leq i\leq 6$ such that
$\gcd(a^{\prime}+i,3\times 5\times 7\times 11)=1$. Let
$\ell^{\prime}=(a^{\prime}+t)^{6}_{t=0}$ be a chain of positive integers, and
let $D=\\{3,5,7,11\\}$. For every integer $d\in D$, We define $A_{d}$ as the
set of all numbers $b$ of chain $\ell^{\prime}$ such that $d\mid b$. Then,
$|A_{d}|\leq\left\lceil\dfrac{7}{d}\right\rceil$. We investigate the two cases
for $a^{\prime}$:

Case 1: $3\mid a^{\prime}$. We know that $|A_{5}|\leq 2$. Also, one can check
directly that in this case, if $|A_{5}|=2$, then $|A_{5}\cap A_{3}|=1$. This
concludes that $|A_{3}\cup A_{5}|\leq 4$. Thus, $|A_{3}\cup A_{5}\cup
A_{7}\cup A_{11}|\leq 6$.

Case 2: $3\nmid a^{\prime}$. In this case, $|A_{3}|\leq 2$. Thus, one can
check directly that $|A_{3}\cup A_{5}\cup A_{7}\cup A_{11}|\leq 6$.

This completes the proof.

$\mathrm{(ii)}\ $ By Chinese remainder theorem, there exists a positive
integer $a$ such that $a\equiv 0$ (mod $11$), $a\equiv-1$ (mod $3$),
$a\equiv-2$ (mod $7$), $a\equiv-3$ (mod $5$), $a\equiv-5$ (mod $13$), and
$a\equiv-6$ (mod $17$). One can check directly that $\ell=(a+t)^{11}_{t=0}$ is
a $(3,5,7,11,13,17)$ \- $\gcd$ chain. Now it is sufficient to prove that for
every positive integer $a^{\prime}$, there exists an integer $0\leq i\leq 12$
such that $\gcd(a^{\prime}+i,3\times 5\times 7\times 11\times 13\times 17)=1$.
Let $\ell^{\prime}=(a^{\prime}+t)^{12}_{t=0}$ be a chain of positive integers,
and let $D=\\{3,5,7,11,13,17\\}$. For every integer $d\in D$, We define
$A_{d}$ as the set of all numbers $b$ of chain $\ell^{\prime}$ such that
$d\mid b$. Then, $|A_{d}|\leq\left\lceil\dfrac{13}{d}\right\rceil$. We
investigate the two cases for $a^{\prime}$:

Case 1: $3\mid a^{\prime}$. We know that $|A_{11}|\leq 2$. Also, one can check
directly that in this case, if $|A_{11}|=2$, then $|A_{11}\cap A_{3}|=1$.
Then, $|A_{3}\cup A_{11}|\leq 6$. Also, we know that $|A_{5}|\leq 3$, and if
$|A_{5}|=3$, then $|A_{3}\cap A_{5}|=1$. This concludes that $|A_{3}\cup
A_{5}\cup A_{11}|\leq 8$. Thus, $|A_{3}\cup A_{5}\cup A_{7}\cup A_{11}\cup
A_{13}\cup A_{17}|\leq 12$.

Case 2: $3\nmid a^{\prime}$. By the definition of $a^{\prime}$, $|A_{3}|\leq
4$. Once again, we know that if $|A_{5}|=3$, then $|A_{3}\cap A_{5}|=1$.
Therefore, $|A_{3}\cup A_{5}|\leq 6$. Thus, $|A_{3}\cup A_{5}\cup A_{7}\cup
A_{11}\cup A_{13}\cup A_{17}|\leq 12$.

This completes the proof. ∎

We can find some more results about the function $L_{max}$. But we do not
pursue this here. Now we state a proposition that finds a lower density for
the set of $k$-layered numbers by having the smallest $k$-layered number.

######  Proposition 8.8.

Let $k$ be a positive integer, and $n$ be the smallest $k$-layered number with
prime factorization $p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots
p^{\alpha_{m}}_{m}$, where $p_{1}<p_{2}<\cdots<p_{m}$. Then
$(L_{max}(p_{1},p_{2},\dots,p_{m})+1)n$ is an upper bound for the difference
of two consecutive $k$-layered numbers.

###### Proof.

Let $a$ and $b$ be two consecutive $k$-layered numbers. By the division
algorithm, there exist non-negative integers $s\neq 0$ and $r$ such that
$a=sn+r$ and $0\leq r<s$. By the definition of $L_{max}$ and Proposition 2.6,
we know that there exists an integer $\ell$ with $s+1\leq\ell\leq
s+1+L_{max}(p_{1},p_{2},\dots,p_{m})$ such that $\ell n$ is $k$-layered. Then
$b-a\leq\ell n-sn\leq(L_{max}(p_{1},p_{2},\dots,p_{m})+1)n$. ∎

######  Corollary 8.9.

Let $n$ be the smallest $k$-layered number. If $n$ with prime factorization
$p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots p^{\alpha_{m}}_{m}$ is a good
$k$-layered number, then $b-a\leq(L_{max}(p_{2},p_{3},\dots,p_{m})+1)n.$

By Propositions 8.6 and 8.8, we have the following corollary.

######  Corollary 8.10.

Let $k$ be a positive integer, and $n$ be the smallest $k$-layered number. Let
$a$ and $b$ be two consecutive $k$-layered numbers. Then,

$|b-a|\leq(p_{1}p_{2}\cdots p_{m}-(p_{1}-1)(p_{2}-1)\cdots(p_{m}-1)+1)n.$

######  Remark 8.11.

Let $k1$ be a positive integer. Corollary 8.10 gives a lower density for the
set of $k$-layered numbers by having the smallest $k$-layered number.

Let $k$ be a positive integer such that $2\leq k\leq 5$, and $n$ be the
smallest $k$-layered number. In Section 4, we showed that $n$ possesses a
prime factorization $p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdots
p^{\alpha_{m}}_{m}$, where $p_{i}$ is the $i$th prime number for integers
$1\leq i\leq m$. Then, by Corollaries 5.8 and 8.9, we conclude that
$(L_{max}(p_{2},\dots,p_{m})+1)n$ is an upper bound for the difference of two
consecutive $k$-layered number. Thus, by Proposition 8.7, we have the
following table.

$k$ |  The smallest $k$-layered number |  An upper bound for the difference of two consecutive $k$-layered numbers
---|---|---
$1$ |  $1$ |  $0$
$2$ |  $6$ |  $12$
$3$ |  $120$ |  $360$
$4$ |  $27720$ |  $194040$
$5$ |  $147026880$ |  $1911349440$
Table 2. An upper bound for the difference of two consecutive $k$-layered for
$1\leq k\leq 5$.

######  Remark 8.12.

The numbers $282$ and $294$ are two consecutive Zumkeller numbers. Thus, by
Table 2, the number $12$ is a sharp upper bound for difference of two
consecutive Zumkeller numbers.

## Acknowledgements

I thank Pankaj Jyoti Mahanta and Prof. Manjil P. Saikia for reading the whole
of this paper and pointing out some mirror mistakes, and also for very
important suggestions for improving the structure of the paper. I also thank
Prof. Michael Filaseta, Prof. Steven J. Miller, and Roohallah Mahkam for their
comments that helped me improve the paper.

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2020 Mathematics Subject Classification. 11A25; 11D99; 11Y99.

_Keywords:_ $k$-layered numbers, $k$-multiperfect numbers, Near-perfect
numbers, Perfect numbers, Practical numbers, Superabundant numbers, Zumkeller
numbers.