Two-layered numbers

TL;DR

This paper introduces the concept of two-layered and half-layered numbers, exploring their properties and relationships with practical and Zumkeller numbers, expanding the understanding of divisor partitioning in number theory.

Contribution

The paper defines new classes of numbers—two-layered and half-layered—and investigates their properties and connections to existing number classes.

Findings

Two-layered numbers have specific divisor partition properties.

Half-layered numbers relate to proper divisors and partitioning.

Connections between these new classes and practical and Zumkeller numbers are established.

Abstract

In this paper, first, I introduce two-layered numbers. Two-layered numbers are positive integers that their positive divisors except 1 can be partitioned into two disjoint subsets. Similarly, I defined a half-layered number as a positive integer n that its proper positive divisors excluding $1$ can be partitioned into two disjoint subsets. I also investigate the properties of two-layered and half-layered numbers and their relation with practical numbers and Zumkeller numbers.

Equations

$$p_1^{k_1+{\alpha }_1(k_1+1)}{p}_2^{k_2+{\alpha }_2(k_2+1)}\dots p_m^{k_m+{\alpha }_m(k_m+1)}$$

$$p_1^{k_1+{\ell }_1(k_1+1)}{p}_2^{k_2+{\ell }_2(k_2+1)}\dots p_m^{k_m+{\ell }_m(k_m+1)}$$

$$p_1^{k_1+{\alpha }_1(k_1+1)}{p}_2^{k_2+{\alpha }_2(k_2+1)}\dots p_m^{k_m+{\alpha }_m(k_m+1)}$$

$$p(\sum _{d\in D_1}d-\sum _{d\in D_2}d)=(\sum _{d\in D_3}d-\sum _{d\in D_4}d).$$

$$\frac{p+1}{2}(\sum _{d\in D_1}d-\sum _{d\in D_2}d)=(\sum _{d\in A_1}d-\sum _{d\in A_2}d).$$

$$p\sum _{d\in D_1}d+\sum _{d\in D_4}d=p\sum _{d\in D_2}d+\sum _{d\in D_3}d.$$

$$\frac{p+1}{2}(\sum _{d\in D_1}d-\sum _{d\in D_2}d)$$

$$=\frac{1}{2}[\sum _{d\in D_3}d-\sum _{d\in D_4}d+\sum _{d\in D_1}d-\sum _{d\in D_2}d]$$

$$=\frac{1}{2}[2(\sum _{d\in D_1\cap D_3}d)-2(\sum _{d\in D_2\cap D_4}d)]$$

$$=\sum _{d\in A_1}d-\sum _{d\in A_2}d.$$

$$\frac{p}{2}\sum _{d\in D_1}d+\frac{1}{2}\sum _{d\in A_2}+\frac{1}{2}\sum _{D_1\setminus A_1}d=\frac{p}{2}\sum _{d\in D_2}d+\frac{1}{2}\sum _{d\in A_1}d+\frac{1}{2}\sum _{d\in D_2\setminus A_2}d.$$

$$\sigma (n)=\prod _{i=1}^m\frac{p_i^{k_i+1}-1}{p_i-1}$$

$$\frac{\sigma (n)}{n}=\prod _{i=1}^m\frac{p_i^{k_i+1}-1}{p_i^{k_i}(p_i-1)}<\prod _{i=1}^m\frac{p_i}{p_i-1}$$

$$\prod _{i=1}^m\frac{p_i}{p_i-1}>2,$$

$$2p_1^{k_1}{p}_2^{k_2}\dots p_m^{k_m}=2n<\sigma (n)=\prod _{i=1}^m(\sum _{j=0}^{k_i}{p}_i^j).$$

$$2<\prod _{i=1}^m(\sum _{j=0}^{k_i}{p}_i^{j-k_i})<\prod _{i=1}^m\frac{p_i}{p_i-1}.$$

$$\prod _{i=1}^m\frac{p_i}{p_i-1}\le \frac{3}{2}\times \frac{5}{4}<2$$

$$\prod _{i=1}^m\frac{p_i}{p_i-1}\le \frac{5}{4}\times \frac{7}{6}\times \frac{11}{10}\times \frac{13}{12}\times \frac{17}{16}\times \frac{19}{18}<2.$$

$$\prod _{i=1}^m\frac{p_i}{p_i-1}\le \frac{3}{2}\times \frac{13}{12}\times \frac{17}{16}\times \frac{19}{18}\times \frac{23}{22}\times \frac{29}{28}<2.$$

$$\prod _{i=1}^m\frac{p_i}{p_i-1}\le \frac{3}{2}\times \frac{11}{10}\times \frac{13}{12}\times \frac{17}{16}<2.$$

$$\prod _{i=1}^3\frac{p_i}{p_i-1}\le \frac{3}{2}\times \frac{7}{6}\times \frac{11}{10}<2.$$

$$\prod _{i=1}^3\frac{p_i}{p_i-1}\le \frac{3}{2}\times \frac{5}{4}\times \frac{17}{16}<2.$$

$$\frac{n}{2}+a=\frac{\sigma (n)-n-1}{2}.$$

$$\frac{\sigma (3^4\times 2^4)-2(3^4\times 2^4)-1}{2}=579=432+108+36+3$$

$$\frac{n}{2}+\sum _{a\in A}a=\sum _{b\in D,b\notin \{n,\frac{n}{2}\}\cup A}b.$$

$$n+\sum _{a\in A}a=\frac{n}{2}+\sum _{b\in D,b\notin \{n,\frac{n}{2}\}\cup A}b.$$

$$2(n+\frac{n}{2}+a)=\sigma (n)-1$$

$$\frac{\sigma (n)-2n-1}{2}=\frac{\sigma (n)-3n-1}{2}+\frac{n}{2}$$

$$\frac{\sigma (n)-2n-1}{2}=\frac{\sigma (n)-3n-1}{2}+\frac{n}{3}+\frac{n}{6}.$$

$$p_1^{k_1+{\ell }_1(k_1+1)}{p}_2^{k_2+{\ell }_2(k_2+1)}\dots p_m^{k_m+{\ell }_m(k_m+1)}$$

$$p(\sum _{d\in D_1}d-\sum _{d\in D_2}d)=\sum _{d\in D_3}d-\sum _{d\in D_4}d.$$

$$\frac{p+1}{2}(\sum _{d\in D_1}d-\sum _{d\in D_2}d)=\sum _{d\in A_1}d-\sum _{d\in A_2}d.$$

Full text

0. Introduction

A perfect number is a positive integer $n$ that equals the sum of its proper positive divisors. Generalizing the concept of perfect numbers, Zumkeller in [1] published a sequence of integers that their divisors can be partitioned into two disjoint subsets with equal sum. Clark et al. in [2] called such integers Zumkeller numbers and investigated some of their properties, and also suggested some conjectures about them. Peng and Bhaskara Rao in [3] introduced half-Zumkeller numbers and provided interesting results about Zumkeller numbers.

In the present paper, I define two-layered numbers based on the concept of perfect numbers and Zumkeller numbers. A two-layered number is a positive integer $n$ that its positive divisors excluding $1$ can be partitioned into two disjoint subsets of an equal sum. A partition $\{A,B\}$ of the set of positive divisors of $n$ except $1$ is a two-layered partition if each of $A$ and $B$ has the same sum.

In the first section, I investigate the properties of two-layered numbers. For a two-layered number $n$, that sum of its divisors is $\sigma(n)$, the following statements hold (See Proposition 1.4):

Let $\sigma(n)$ be the sum of all positive divisors of $n$. If $n$ is a two-layered number, then

1. (1)

$\sigma(n)$ is odd. 2. (2)

Powers of all odd prime factors of $n$ should be even. 3. (3)

$\sigma(n)\geq 2n+1$, so $n$ is abundant.

After that, In theorem 1.5, I prove that The integer $n$ is a two-layered number if and only if $\frac{\sigma(n)-1}{2}-n$ is a sum of distinct proper positive divisors of n excluding 1. I also introduce two methods of generating new two-layered numbers from known two-layered numbers. Suppose that $n$ is a two-layered number and $p$ is a prime number with $(n,p)=1$, then $np^{\alpha}$ is a two-layered number for any even positive integer $\alpha$ (See Theorem 1.7). We can also generate two-layered numbers in another way. Let $n$ be a two-layered number and $p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ be the prime factorization of $n$. Then for any nonnegative integers $\alpha_{1},\dots\alpha_{m}$, the integer

[TABLE]

is a two-layered number (See Theorem 1.8).

In the second section of the present paper, I generalize the concept of practical numbers and define semi-practical numbers. A practical number is a positive integer $n$ that every positive integer less than $n$ can be represented as a sum of distinct positive divisors of $n$ [5]. A positive integer $n$ is a semi-practical number if every positive integer $x$ where $1<x<n$ can be represented as a sum of distinct positive divisors of $n$ excluding $1$ (See Definition 2.2).

I investigate some properties of semi-practical numbers and their relations with two-layered numbers. For example, every semi-practical number is divisible by $12$ (See Proposition 2.3). I also proved that a positive integer $n$ is is a semi-practical number if and only if every positive integer $x$ where $1<x<\sigma(n)$, is a sum of distinct positive divisors of $n$ excluding $1$ (See Theorem 2.4). The most important relation between semi-practical numbers and two-layered numbers is that a semi-practical number $n$ is two-layered if and only if $\sigma(n)$ is odd (See Proposition 2.5).

In section 3, I define a half-layered number. A positive integer $n$ is said to be a half-layered number if the proper positive divisors of $n$ excluding $1$ can be partitioned into two disjoint non-empty subsets of an equal sum (See Definition 3.5). A half-layered partition for a half-layered number $n$ is a partition $\{A,B\}$ of the set of proper positive divisors of $n$ excluding $1$ so that each of $A$ and $B$ sums to the same value (See Definition 3.2).

After these definitions, I investigate the properties of half-layered numbers. For example, A positive integer $n$ is half-layered if and only if $\frac{\sigma(n)-n-1}{2}$ is the sum of some distinct positive proper positive divisors of $n$ (See Proposition 3.3). A positive even integer $n$ is half-layered if and only if $\frac{\sigma(n)-2n-1}{2}$ is the sum (possibly empty sum) of some distinct positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$ (See Theorem 3.5). If $n$ is an odd half-layered number, then at least one of the powers of prime factors of $n$ should be even (See Proposition 3.7).

Using the definition of half-Zumkeller numbers, we can derive some of the interesting properties of half-layered numbers. A positive integer $n$ is said to be a half-Zumkeller number if the proper positive divisors of $n$ can be partitioned into two disjoint non-empty subsets of an equal sum. A half-Zumkeller partition for a half-Zumkeller number n is a partition $\{A,B\}$ of the set of proper positive divisors of $n$ so that each of $A$ and $B$ sums to the same value (Definition 3 in [3]). Based on these definition, I prove that if $m$ and $n$ are half-layered numbers with $(m,n)=1$, then $mn$ is half-layered (See Proposition 3.9).

After that, I investigate some relations between half-layered and two-layered numbers. For example, let $n$ be even. Then $n$ is half-layered if and only if $n$ admits a two-layered partition such that $n$ and $\frac{n}{2}$ are in distinct subsets. Therefore, if $n$ is an even half-layered number then $n$ is two-layered (See Proposition 3.10). It is also proved that if $n$ is an even two-layered number and If $\sigma(n)<3n$, then $n$ is half-layered (See Theorem 3.11). Let $n$ be even. Then, $n$ is two-layered if and only if either $n$ is half-layered or $\frac{\sigma(n)-3n-1}{2}$ is a sum (possibly an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$ (See Proposition 3.12).

If $6$ divides $n$, $n$ is two-layered, and $\sigma(n)<\frac{10n}{3}$ , then $n$ is half-layered (See Proposition 3.13). If $n$ is two-layered, then $2n$ is half-layered (See Proposition 3.14). Let $n$ be an even half-layered number and $p$ be a prime with (n, p) = 1. Then $np^{\ell}$ is half- layered for any positive integer $\ell$ (See Proposition 3.16). Let $n$ be an even half-layered number and the prime factorization of $n$ be $p_{1}^{k_{1}}p_{2}^{k_{2}}/dotsp_{m}^{k_{m}}$ Then for nonnegative integers $\ell_{1},\dots,\ell_{m}$, the integer

[TABLE]

is half-layered (See Theorem 3.18).

1. two-layered numbers

Definition 1.1**.**

A positive integer $n$ is a two-layered number if the positive divisors of $n$ excluding $1$ can be partitioned into two disjoint subsets of an equal sum.

Definition 1.2**.**

A two-layered partition for a two-layered number $n$ is a partition $\{A,B\}$ of the set of positive divisors of $n$ excluding $1$ so that each of $A$ and $B$ sums to the same value.

Example 1.3**.**

The number 36 is a two-layered number and its two-layered partition is $\{A,B\}$, where $A=\{2,3,4,36\}$ and $B=\{6,9,12,18\}$. You can check that each of $A$ and $B$ has the sum of $45$. The numbers $72,144,$ and $200$ are also two-layered. You can find the sequence of two-layered numbers in [4].

Proposition 1.4**.**

Let $\sigma(n)$ be the sum of all positive divisors of $n$. If $n$ is a two-layered number, then

1. (1)

$\sigma(n)$* is odd.* 2. (2)

Powers of all odd prime factors of $n$ should be even. 3. (3)

$\sigma(n)\geq 2n+1$, so $n$ is abundant.

Proof.

$(1):$ If $\sigma(n)$ is even, then $\sigma(n)-1$ is odd, so it is impossible to partition the positive divisors of $n$ into two subset of equal sum.

$(2):$ using $(1)$, the number of odd positive divisors of $n$ is odd. Suppose that the prime factorization of $n$ is $2^{k_{0}}p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$. The number of odd positive divisors of $n$ is $(k_{1}+1)(k_{2}+1)\dots(k_{m}+1)$. All of $k_{i}$ should be even in order to make the product $(k_{1}+1)(k_{2}+1)\dots(k_{m}+1)$ odd.

$(3):$ Let $n$ be a two-layered number with two-layered partition $\{A,B\}$. Without loss of generality we may assume that $n\in A$, so the sum in $A$ is at least $n$ and we can conclude $\sigma(n)-1\geq 2n$. ∎

Theorem 1.5**.**

The integer $n$ is a two-layered number if and only if $\frac{\sigma(n)-1}{2}-n$ is a sum of distinct proper positive divisors of n excluding 1.

Proof.

Let $n$ be a two-layered number and its two-layered partition is $\{A,B\}$. Without loss of generality we assume that $n\in A$, so the sum of the remaining elements of $A$ is $\frac{\sigma(n)-1}{2}-n$.

Conversely, if we have a set of proper divisors of $n$ excluding $1$ that its sum is $\frac{\sigma(n)-1}{2}-n$, we can augment this set with $n$ to construct a set of positive divisors of $n$ summing to $\frac{\sigma(n)-1}{2}$. The complementary set of positive divisors of $n$ sums to the same value, and so these two sets form a two-layered partition for $n$. ∎

With the help of the next two theorems, we can generate some new two-layered numbers by knowing a two-layered number.

Definition 1.6** (Definition 1 in [3]).**

A positive integer $n$ is said to be a Zumkeller number if the positive divisors of $n$ can be partitioned into two disjoint subsets of equal sum. A Zumkeller partition for a Zumkeller number $n$ is a partition $\{A,B\}$ of the set of positive divisors of $n$ so that each of $A$ and $B$ sums to the same value.

Theorem 1.7**.**

Let $n$ be a two-layered number and $p$ be a prime number with $(n,p)=1$, then $np^{\alpha}$ is a two-layered number for any even positive integer $\alpha$.

Proof.

Suppose that $\{A,B\}$ is a Zumkeller partition of $n$. Then $\{(A\setminus\{1\})\cup(pA)\cup(p^{2}A)\cup\dots\cup(p^{\alpha}A),(B\setminus\{1\})\cup(pB)\cup(p^{2}B)\cup\dots\cup(p^{\alpha}B)\}$ is a two-layered partition of $np^{\alpha}$. ∎

Theorem 1.8**.**

Suppose that $n$ is a two-layered number and $p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is the prime factorization of $n$. Then for any nonnegative even integers $\alpha_{1},\dots\alpha_{m}$, the integer

[TABLE]

is a two-layered number.

Proof.

If we show that $p_{1}^{k_{1}+\alpha_{1}(k-1+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ the proof will be completed. Suppose that $\{A,B\}$ is a Zumkeller partition of $n$. If $D$ is the set of positive divisors of $n$, then $(D\setminus\{1\})\cup(p_{1}^{k_{1}+1}D)\cup(p_{1}^{2(k_{1}+1)}D)\cup\dots\cup(p_{1}^{\alpha_{1}(k_{1}+1)}D))$ is the set of positive divisors of $p_{1}^{k_{1}+\alpha_{1}(k-1+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ excluding 1. Therefore a two-layered partition for $p_{1}^{k_{1}+\alpha_{1}(k-1+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is $\{A\setminus\{1\}\cup(p_{1}^{k_{1}+1}A)\cup(p_{1}^{2(k_{1}+1)}A)\cup\dots\cup(p_{1}^{\alpha_{1}(k_{1}+1)}A),B\setminus\{1\}\cup(p_{1}^{k_{1}+1}B)\cup(p_{1}^{2(k_{1}+1)}B)\cup\dots\cup(p_{1}^{\alpha_{1}(k_{1}+1)}B)\}$ and the proof is complete. ∎

2. semi-practical numbers and two-layered numbers

Practical numbers have been introduced by Srinivasan in 1948 as what follows:

Definition 2.1**.**

A positive integer $n$ is a practical number if every positive integer less than $n$ can be represented as a sum of distinct positive divisors of $n$.[5]

Because of the structure of two-layered number, if we change the definition of practical numbers and call them semi-practical numbers, we can drive some useful relation between them and two-layered numbers, so I define semi-practical numbers as what follows:

Definition 2.2**.**

A positive integer $n$ is practical if every positive integer $x$ where $1<x<n$ can be represented as a sum of distinct positive divisors of $n$ excluding $1$.

Proposition 2.3**.**

Every semi-practical number is divisible by $12$.

Proof.

Since we can not write $2,3,$ and $4$ as sums of more than one positive integer greater than $1$, they should be divisors of our semi-practical number. ∎

Theorem 2.4**.**

A positive integer $n$ is is a semi-practical number if and only if every positive integer $x$ where $1<x<\sigma(n)$, is a sum of distinct positive divisors of $n$ excluding $1$.

Proof.

Suppose that $n$ is a semi-practical number. I introduce an algorithm for writing all positive integer $x$ between $n$ and $\sigma(n)$ as sum of distinct positive divisors of $n$ excluding $1$.

First, let $x$ be $n+1$. Since $n$ is semi-practical, by Propositin 2.3, it is divisible by $n/2$ and $n/3$. Hence, $n+1=n/2+n/3+r$, where $r$ is a positive integer. By Proposition 2.3, $n>6$, so $n+1-n/2-n/3<n/3$. On the other hand, since $n$ is a semi-practical number and $r<n/3<n$, $r$ is equal to some of distinct divisors of $n$ which are less than $n/3$ and greater than $1$.

For $n+1<x<\sigma(n)$, let the positive divisors of $n$ which are greater than $1$ be written in increasing order as $m_{1}<m_{2}<\dots<m_{k}$. Now we can write $x=\sum_{i=\ell}^{k}m_{i}+r$ where $1\leq\ell\leq k$ and $0\leq r<m_{\ell-1}$. If $r=0$ then $x$ is a sum of distinct divisors of $n$. If $1<r<m_{\ell-1}$, since $n$ is semi-practical and $r<n$, then we can write $r$ as a sum of distinct divisors of $n$ which are less than $m_{\ell-1}$, so $x$ is a sum of distinct divisors of $n$. If $r=1$, then we can write $x=\sum_{i=\ell+1}^{k}+r_{1}$ where $1<r_{1}<m_{\ell}$. since $n$ is semi-practical and $r<n$, then $r_{1}$ is sum of distinct divisors of $n$ which are less than $m_{\ell}$, so $x$ is a sum of distinct divisors of $n$.

Conversely, if every positive integer less than $\sigma(n)$ excluding $1$, is a some of distinct positive divisors of $n$ excluding $1$, it is clear that $n$ is semi-practical.

∎

Proposition 2.5**.**

A semi-practical number $n$ is two-layered if and only if $\sigma(n)$ is odd.

Proof.

If $n$ is two-layered number, then $\sigma(n)$ is odd by Proposition 1.4. Conversely, if $\sigma(n)$ is odd, then $\frac{\sigma(n)-1}{2}$ is a positive integer smaller than $\sigma(n)$. Since $n$ is a semi-practical number, using Proposition 2.4. ∎

Theorem 2.6**.**

Let $n$ be a positive integer and $p$ be a prime with $(n,p)=1$. Let $D$ be the set of all positive divisors of $n$ including $1$. The following conditions are equivalent:

1. (1)

$np$* is two-layered.* 2. (2)

There exist two partitions $\{D_{1},D_{2}\}$ and $\{D_{3},D_{4}\}$ of $D\setminus\{1\}$ such that

[TABLE] 3. (3)

There exists a partition $\{D_{1},D_{2}\}$ of $D\setminus\{1\}$ and subsets $A_{1}\subseteq D_{1}$ and $A_{2}\subseteq D_{2}$ such that

[TABLE]

Proof.

It is clear that $(pD)\cup(D\setminus\{1\})$ is the set of all positive divisors of $np$ excluding $1$.

$(1)\Rightarrow$ (2). Suppose that $np$ is two-layered. Hence, there is a two-layered partition $\{A,B\}$ of $(pD)\cup(D\setminus\{1\})$. Let $D_{1}=\frac{1}{p}(A\cap(pD))$, $D_{2}=\frac{1}{p}(B\cap(pD))$, $D_{3}=B\cap(D\setminus\{1\})$, $A\cap(D\setminus\{1\})$, then

[TABLE]

and the proof is complete.

$(2)\Rightarrow(3)$. Let $A_{1}=D_{1}\cap D_{3}$ and $A_{2}=D_{2}\cap D_{4}$. We have

[TABLE]

$(3)\Rightarrow(1)$. We can rewrite the equation in $(3)$ as follows:

[TABLE]

By multiplying this by $2$, we obtain the two-layered partition $\{(pD_{1})\cup A_{2}\cup(D_{1}-A_{1}),(pD_{2})\cup A_{1}\cup(D_{2}-A_{2})\}$ for $np$, so $np$ is a two-layered number. ∎

Proposition 2.7**.**

Let the positive divisors of $n$ excluding $1$ be written in increasing order as follows: $a_{1}<a_{2}<\dots<a_{k}=n.$ If $a_{i+1}<2a_{i}$ for all $1\leq i<k$ and $\sigma(n)$ is odd, then $n$ is two-layered.

Proof.

Let $b_{i}=a_{i}$ or $−a_{i}$ for each $i$. I will explain how to chose the sign of $b_{i}$ precisely. Then I show that $\sum_{i=1}^{k}b_{k}=0$. Hence, it will imply that $\sigma(n)-1$ can be partitioned into two equal-summed subsets.

Let $b_{k}=a_{k}=n$ and let $b_{k−1}=−a_{k−1}$. Note that $0<b_{k}+b_{k−1}<a_{k−1}$ since $a_{k}<2a_{k−1}$. Since the current sum $b_{k}+b_{k−1}$ is positive, we assign the negative sign to $b_{k−2}$. Then $b_{k−2}<b_{k}+b_{k−1}+b_{k−2}<a_{k−1}−a_{k−2}<a_{k−2}$ since $a_{k−1}<2a_{k−2}$. If $b_{k}+b_{k−1}+b_{k−2}\geq 0$, we assign the negative sign to $b_{k−3}$; Otherwise, we assign the positive sign to $b_{k−3}$. Let $s_{i}$ be $\sum_{j=1}^{k}b_{j}$. In general, the sign assigned to $b_{i−1}$ is the opposite of the sign of $s_{i}$ . Let us show inductively that $|s_{i}|<a_{i}$ for $1\leq i\leq k$. It is true for $i=k$. Assume that $|s_{i+1}|<a_{i+1}$. Since the sign of $b_{i}$ is opposite of the sign of $s_{i+1}$, $|s_{i}|=||s_{i+1}|−a_{i}|$. Note that $−a_{i}<|s_{i+1}|−a_{i}<a_{i+1}−a_{i}<a_{i}$ since $a_{i+1}<2a_{i}$ . Therefore $|s_{i}|<a_{i}$. So $|s_{1}|<a_{1}=1$. Since $\sigma(n)-1$ is even, $s_{1}$, which is obtained by assigning a positive or negative sign to each of the terms in $\sigma(n)-1$ is even as well. So $s_{1}=0$. This implies that $\sigma(n)-1$ can be partitioned into two equal-summed subsets. Hence it is two-layered. ∎

Proposition 2.8** (Proposition 1 in [3]).**

Let the prime factorization of $n$ be $\prod_{i=1}^{m}p_{i}^{k_{i}}$. Then

[TABLE]

and

[TABLE]

Proposition 2.9**.**

Let the prime factorization of an odd number $n$ be $p_{1}^{k}p_{2}^{k}\dots p_{m}^{k_{m}}$, where $3\leq p_{1}<p_{2}<\dots<p_{m}$. If $n$ is two-layered, then

[TABLE]

and $m$ is at least $3$. In particular:

1. (1)

If $m\leq 6$, then $p_{1}=3$, $p_{2}=5$, $7$ or $11.$ 2. (2)

If $m\leq 4$, then $p_{1}=3$, $p_{2}=5$ or 7. 3. (3)

If $m=3$, then $p_{1}=3$, $p_{2}=5$, and $p_{3}=7$ or $11$ or $13$.

Proof.

If $n$ is two-layered, then by Propositions 1.4 and 2.8,

[TABLE]

Dividing both sides by $p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$, we get

[TABLE]

If $m\leq 2$, then

[TABLE]

Therefore $m\geq 3$. The parts of $1-3$ follows by verifying the condition $\prod_{i=1}^{m}\frac{p_{i}}{p_{i}-1}>2$ directly as given below.

1. Let $m\leq 6$. If $p_{1}\neq 3$, then $p_{1}\geq 5$ and

[TABLE]

Therefore, $p_{1}=3$. If $p_{2}>11$, then $p2\geq 13$ and

[TABLE]

Hence, $p_{2}\leq 11$. This implies that $p_{2}=5$, $7$ or $11$.

2. Let $m\leq 4$. By $1$, $p_{1}=3$. If $p_{2}>7$, then $p_{2}\geq 11$, so

[TABLE]

Therefore, $p_{2}\leq 7$. This implies that $p_{2}=5$ or $7$.

3. Let $m=3$. By $1$, $p_{1}=3$. If $p_{2}\neq 5$, then $p_{2}\geq 7$ and $p3\geq 11$. So

[TABLE]

Hence $p_{2}=5.$

If $p_{3}\geq 17$, then

[TABLE]

Hence, $p_{3}<17$ and consequently $p_{3}=7$, $11$ or $13$. ∎

3. half-layered numbers

Definition 3.1**.**

A positive integer $n$ is said to be a half-layered number if the proper positive divisors of $n$ excluding $1$ can be partitioned into two disjoint non-empty subsets of equal sum.

Definition 3.2**.**

A half-layered partition for a half-layered number $n$ is a partition $\{A,B\}$ of the set of proper positive divisors of $n$ excluding $1$ so that each of $A$ and $B$ sums to the same value.

Proposition 3.3**.**

A positive integer $n$ is half-layered if and only if $\frac{\sigma(n)-n-1}{2}$ is the sum of some distinct positive proper positive divisors of $n$.

Example 3.4**.**

In Example 1.3, we saw that $36$ was a two-layered number. It is also a half-layered number and its half-layered partition is $\{A,B\}$, where $A=\{2,3,4,18\}$ and $B=\{6,9,12\}$. You can check that each of $A$ and $B$ has the sum of $27$. The numbers $72,105,$ and $144$ are also half-layered. You can find the sequence of half-layered numbers in [6].

Theorem 3.5**.**

A positive even integer $n$ is half-layered if and only if $\frac{\sigma(n)-2n-1}{2}$ is the sum (possibly empty sum) of some distinct positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$.

Proof.

An even number $n$ is half-layered if and only if there exists a which is the sum (possibly empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$ such that

[TABLE]

Therefore, $a=\frac{\sigma(n)-2n-1}{2}.$ ∎

Example 3.6**.**

The number $3^{4}\times 2^{4}$ is a half-layered number, since

[TABLE]

is a sum of positive divisors of $3^{4}\times 2^{4}$ excluding $3^{4}\times 2^{4}$ , $3^{4}\times 2^{3}$, and $1$. Hence, by Theorem 3.5, it is a half-layered number.

Proposition 3.7**.**

If $n$ is an odd half-layered number, then at least one of the powers of prime factors of $n$ should be even.

Proof.

If n is odd and half-layered, then $\sigma(n)−n-1$ must be even and $\sigma(n)$ must be even. Let the prime factorization of $n$ be $\prod_{i=1}^{m}p_{i}^{k_{i}}$. Then $\sigma(n)=\prod_{i=1}^{m}(\sum_{j=0}^{k_{i}}p_{i}^{j})$. If $\sigma(n)$ is odd, then there exists one $k-i$ which is odd. ∎

Definition 3.8** (Definition 3 in [3]).**

A positive integer $n$ is said to be a half-Zumkeller number if the proper positive divisors of $n$ can be partitioned into two disjoint non-empty subsets of an equal sum. A half-Zumkeller partition for a half-Zumkeller number n is a partition $\{A,B\}$ of the set of proper positive divisors of $n$ so that each of $A$ and $B$ sums to the same value.

Proposition 3.9**.**

If $m$ and $n$ are half-layered numbers with $(m,n)=1$, then $mn$ is half-layered.

Proof.

Let $M$ be the set of proper positive divisors of $m$ and let $\{M_{1},M_{2}\}$ be a half-Zumkeller partition for $m$. Let $N$be the set of proper positive divisors of $n$ and let $\{N1,N2\}$ be a half-Zumkeller partition for $n$. Since $(m,n)=1$, then the set of proper positive divisors of $mn$ is $(MN)\cup(nM)\cup(mN)$. Observe that $\{(M_{1}N\setminus\{1\})\cup(mN_{1})\cup(nM_{1}),(M_{2}N\setminus\{1\})\cup(mN_{2})\cup(nM_{2})\}$ is a half-layered partition for $mn$. Therefore $mn$ is half-layered. ∎

Proposition 3.10**.**

Let $n$ be even. Then $n$ is half-layered if and only if $n$ admits a two-layered partition such that $n$ and $\frac{n}{2}$ are in distinct subsets. Therefore, if $n$ is an even half-layered number then $n$ is two-layered.

Proof.

Let $n$ be even. Let $D$ be the set of all positive divisors of $n$ excluding $1$. The number $n$ is half-layered if and only if there exists $A\subset D\setminus\{n,\frac{n}{2}\}$ such that

[TABLE]

That is,

[TABLE]

This is equivalent to saying that $n$ admits a two-layered partition such that $n$ and $\frac{n}{2}$ are in distinct subsets.

∎

Theorem 3.11**.**

Let $n$ be an even two-layered number. If $\sigma(n)<3n$, then $n$ is half-layered.

Proof.

Since $n$ and $\frac{n}{2}$ together sum to more than $\frac{\sigma(n)}{2}$ , they must be in different subsets in any two-layered partition for $n$. Therefore, by Proposition 3.10, $n$ is half-layered. ∎

Proposition 3.12**.**

Let $n$ be even. Then, $n$ is two-layered if and only if either $n$ is half-layered or $\frac{\sigma(n)-3n-1}{2}$ is a sum (possibly an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$.

Proof.

Let $n$ be even. If $n$ is two-layered but not half-layered, then by Proposition 3.10, any two-layered partition of the positive divisors of $n$ must have $n$ and $\frac{n}{2}$ in the same subsets. In other words, there exists a which is a sum (possibly an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$ such that

[TABLE]

So, $a=\frac{\sigma(n)-3n-1}{2}$. Therefore, the number $\frac{\sigma(n)-3n-1}{2}$ is a sum (possibly an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$.

If $n$ is half-layered, then $n$ is two-layered by Proposition 3.10. If $\frac{\sigma(n)-3n-1}{2}$ is a sum (possibly an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$, then

[TABLE]

is a sum of some positive divisors of $n$ excluding $n$, and $1$. By Theorem 1.5, the number $n$ is two-layered. ∎

Proposition 3.13**.**

If $6$ divides $n$, $n$ is two-layered, and $\sigma(n)<\frac{10n}{3}$ , then $n$ is half-layered.

Proof.

If $n$ is not half-layered, by Proposition 3.12, $\frac{\sigma(n)-3n-1}{2}$ is a sum (might be an empty sum) of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$. Then,

[TABLE]

Since $\sigma(n)/n<\frac{10}{3}$ we have that $\frac{\sigma(n)-3n-1}{2}<\frac{n}{6}$. Hence $\frac{\sigma(n)-2n-1}{2}$ is a sum of some positive divisors of $n$ excluding $n$, $\frac{n}{2}$, and $1$. By Proposition 3.3, $n$ is half layered. This is a contradiction. ∎

Proposition 3.14**.**

If $n$ is two-layered, then $2n$ is half-layered.

Proof.

Let $n=2^{k}L$ with $k$ a nonnegative integer and $L$ an odd number, be a two-layered number. Then all positive divisors of $n$ excluding $1$ can be partitioned into two disjoint equal-summed subsets $D_{1}$ and $D_{2}$. Observe that every positive divisor of $2n$ which is not a positive divisor of $n$ can be written as $2^{k+1}\ell$ where $\ell$ is a positive divisor of $L$. Observe that $2^{k}\ell$ is either in $D_{1}$ or $D_{2}$. Without loss of generality, assume that $2^{k}\ell$ is in $D_{1}$. In this case, we move $2^{k}\ell$ to $D_{2}$ and add $2^{k+1}\ell$ to $D_{1}$. Perform this procedure to all positive divisors of $2n$ which are not positive divisors of $n$ except $2n$ itself. This procedure will yield an equal-summed partition of all positive divisors of $2n$ except $2n$ itself. This shows that $2n$ is half-Zumkeller. ∎

Corollary 3.15**.**

Let $n$ be even and the prime factorization of $n$ be $2^{k}p_{1}^{k_{1}}\dots p_{m}^{k_{m}}$. If $n$ is two-layered but not half- layered, then $2^{i}p_{1}^{k_{1}}\dots p_{m}^{k_{m}}$ is not two-layered for any $i\leq k-1$, and $2^{i}p_{1}^{k_{1}}\dots p_{m}^{k_{m}}$ is half-layered for any $i\geq k+1$.

Proposition 3.16**.**

Let $n$ be an even half-layered number and $p$ be a prime with (n, p) = 1. Then $np^{\ell}$ is half- layered for any positive integer $\ell$.

Proof.

Since $n$ is an even half-layered number, the set of all positive divisors of $n$, excluding $1$, denoted by $D_{0}$ can be partitioned into two disjoint subsets $A_{0}$ and $B_{0}$ so that the sums of the two subsets are equal and $n$ and $\frac{n}{2}$ are in distinct subsets (by Proposition 3.10).

Group the positive divisors of $np^{\ell}$ except $1$ into $\ell+1$ groups $D_{0},D_{1},\dots D_{\ell}$ according to how many positive divisors of $p$ they admit, i.e., $D_{i}$ consists of all positive divisors of $np^{\ell}$ admitting $i$ positive divisors of $p$. Then each $D_{i}$ can be partitioned into two disjoint subsets so that the sums of the two subsets are equal and $np^{i}$ and $\frac{np^{i}}{2}$ are in distinct subsets according to the two-layered partition of the set $D_{0}$. Therefore all positive divisors of $np^{\ell}$ excluding $1$ can be partitioned into two disjoint subsets so that the sum of these two subsets equal and $np^{\ell}$ and $\frac{np^{\ell}}{2}$ are in distinct subsets. By Proposition 3.10, $np^{\ell}$ is half- layered. ∎

Corollary 3.17**.**

If $n$ is an even half-layered number and $m$ is a positive integer with $(n,m)=1$, then $nm$ is half-layered.

Theorem 3.18**.**

Let $n$ be an even half-layered number and the prime factorization of $n$ be $p_{1}^{k_{1}}p_{2}^{k_{2}}/dotsp_{m}^{k_{m}}$ Then for nonnegative integers $\ell_{1},\dots,\ell_{m}$, the integer

[TABLE]

is half-layered.

Proof.

It is sufficient to show that $p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is half-layered if $p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is an even half-layered number. Assume that $p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is even and half-layered, then the set of all positive divisors of $n$ excluding $1$, denoted by $D_{0}$ can be partitioned into two disjoint subsets $A_{0}$ and $B_{0}$ so that the sums of the two subsets are equal and $n$ and $\frac{n}{2}$ are in distinct subsets (by Proposition 3.10). Note that the positive divisors of $p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ excluding $1$ can be partitioned into $\ell_{1}+1$ disjoint groups $D_{i},0\leq i\leq\ell_{1}$, where elements in $D_{i}$ are obtained by multiplying $p_{1}^{i(k_{1}+1)}$ with elements in $D_{0}$. Using the partition $A_{0},B_{0}$ of $D0$ we can partition every $D_{i}$ into two disjoint subsets $A_{i}$ and $B_{i}$ so that the sums of the corresponding subsets are equal and $np_{1}^{i(k_{1})+1}$ and $\frac{np_{1}^{i(k_{1})+1}}{2}$ are in distinct subsets. Therefore, the set of all positive divisors of $p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ excluding $1$ can be partitioned into two disjoint equal-summed subsets and $p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ and $\frac{p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}}{2}$ are in distinct subsets. By Proposition 3.10, $p_{1}^{k_{1}+\ell_{1}(k_{1}+1)}p_{2}^{k_{2}}\dots p_{m}^{k_{m}}$ is half-layered. ∎

Theorem 3.19**.**

Let $n$ be an even integer and $p$ be a prime with $(n,p)=1$. Let $D$ be the set of all positive divisors of $n$ excluding $1$. Then the following conditions are equivalent:

1. (1)

$np$* is half-layered.* 2. (2)

There exist two partitions $\{D_{1},D_{2}\}$ and $\{D_{3},D_{4}\}$ of $D$ such that $n$ is in $D_{1}$, $\frac{n}{2}$ is in $D_{2}$ and

[TABLE] 3. (3)

*There exists a partition $\{D_{1},D_{2}\}$ of $D$ and subsets *$A_{1}\subseteq D_{1}$and $A_{2}\subseteq D_{2}$ such that $n$ is in $D_{1}$, $\frac{n}{2}$ is in $D_{2}$ and

[TABLE]

Proof.

By Proposition 3.10, $np$ is half-layered if and only if there is a two-layered partition $\{A,B\}$ of $(pD)\cup D$such that $n\in A$ and $\frac{n}{2}\in B$. The rest of the proof follows along the lines of the proof of Theorem 2.6. ∎

Proposition 3.20**.**

If $a_{1}<a_{2}<\dots<a_{k}=n$ are all positive divisors of an even number $n$ excluding $1$ with $a_{i+1}<2a_{i}$ for all $i$ and $\sigma(n)$ is odd, then $n$ is half-layered.

Proof.

Note that in the proof of Proposition 2.7, $b_{k}=n$ and $b_{k−1}=-\frac{n}{2}$ have different signs. So we get a two-layered partition of $n$ such that $n$ and $\frac{n}{2}$ are in distinct subsets. By Proposition 3.10, $n$ is half-layered. ∎