2 Terminology

To begin with, we establish some notation and terminology. For any positive integer n, put $\underline {n}=\{1,2,\ldots ,n\}$ . For positive integers $a_{1}<a_{2}<\cdots <a_{k}$ , where $k\ge 1$ , we consider the symmetric difference of the sets $a_{i}\underline {n}=\{a_{i},2a_{i},\ldots ,na_{i}\}$ , $i=1,2,\ldots ,k$ , that is, $\Delta _{i=1}^{k}a_{i}\underline {n}=a_{1}\underline {n}{\mathbin {\Delta }} a_{2}\underline {n}{\mathbin {\Delta }}\cdots {\mathbin {\Delta }} a_{k}\underline {n}$ . Throughout the article, the k integers $a_{1},a_{2},\ldots ,a_{k}$ will be referred to as the multipliers. By using standard counting techniques based on the inclusion–exclusion principle, it follows that the cardinality of $\Delta _{i=1}^{k}a_{i}\underline {n}$ is given by

(2.1) $$ \begin{align} \xi_{n,k}(a_{1},a_{2},\ldots,a_{k})=\sum_{r=1}^{k}\sum_{1\le i_{1}<i_{2}<\cdots<i_{r}\le k}(-1)^{r-1}2^{r-1}\bigg\lfloor\frac{a_{i_{1}}\!\cdot{n}}{[a_{i_{1}},a_{i_{2}},\ldots,a_{i_{r}}]}\bigg\rfloor. \end{align} $$

Here, and throughout the rest of our discussion, we use $[x,y,z,\ldots ]$ for the least common multiple of the integers $x,y,z,\ldots .$ Likewise, we use $(x,y,z,\ldots )$ for the greatest common divisor of $x,y,z,\ldots .$

The conjecture asserts that $\xi _{n,k}(a_{1},a_{2},\ldots ,a_{k})\ge n$ for any $n\ge 1$ and any sequence $a_{1}<a_{2}<\cdots <a_{k}$ of k multipliers, for any $k\ge 1$ . Our aim is to prove this conjecture for the cases $k=1,2,3$ .

3 The cases $k=1$ and $k=2$

The conjecture is trivially true when $k=1$ , since $\xi _{n,1}(a_{1})=n$ , which is the cardinality of $a_{1}\underline {n}=\{a_{1},2a_{1},\ldots ,na_{1}\}$ .

For $k=2$ , consider the multipliers $a_{1}<a_{2}$ . Since $a_{1}<a_{2}\le [a_{1},a_{2}]$ , and since ${a_{1}\mid [a_{1},a_{2}]}$ , we have $[a_{1},a_{2}]\ge 2a_{1}$ . So, ${a_{1} n}/{[a_{1},a_{2}]}\le {n}/{2}$ , from which it follows that

$$ \begin{align*} \xi_{n,2}(a_{1},a_{2})=2n-2\bigg\lfloor\frac{a_{1}\cdot n}{[a_{1},a_{2}]}\bigg\rfloor\ge2n-2\bigg\lfloor\frac{n}{2}\bigg\rfloor\ge n, \end{align*} $$

since $\lfloor {n}/{2}\rfloor \le {n}/{2}$ .

4 The case $k=3$

The conjecture is known to be true for $1\le n\le 6$ [Reference Pilz4, Corollary 2]. Hence, for the remainder of this paper, we will assume that $n\ge 7$ . To avoid unnecessary subscripts, we will simply denote the multipliers $a_{1}<a_{2}<a_{3}$ by $a<b<c$ in this section. Furthermore, since $a=1$ does not have any prime divisors, it turns out that we should treat this case separately.

Hence, we will assume first that the multipliers are $1<b<c$ . Here we want to show that

(4.1) $$ \begin{align} \frac{\xi_{n,3}(1,b,c)-n}{2}=n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor+\bigg\lfloor\frac{bn}{[b,c]}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor\ge0. \end{align} $$

This will be investigated by considering two sub-cases.

(1) Assume $b\mid c$ , say $c=tb,\ t\ge 2$ . Then,

$$ \begin{align*} n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor+\bigg\lfloor\frac{bn}{[b,c]}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor & =n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor+\bigg\lfloor\frac{bn}{c}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{n}{c}\bigg\rfloor\\ & =n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{tb}\bigg\rfloor+\bigg\lfloor\frac{n}{t}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{n}{tb}\bigg\rfloor\\ & =n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{t}\bigg\rfloor\bigg)+\bigg\lfloor\frac{n}{tb}\bigg\rfloor\\& \ge n-2\bigg\lfloor\frac{n}{2}\bigg\rfloor+\bigg\lfloor\frac{n}{tb}\bigg\rfloor,\text{ since }b\ge2\text{ and } t \ge 2\\ & \ge \bigg\lfloor\frac{n}{tb}\bigg\rfloor\ge0. \end{align*} $$

So (4.1) holds in this case.

(2) Assume $b\nmid c$ . Before we proceed with this case, we first mention three results.

1. (a) For real numbers x and y, it is well known that $\lfloor x\rfloor +\lfloor y\rfloor \le \lfloor x+y\rfloor $ .

2. (b) For a real number x, we have $\lfloor -2x\rfloor =-2\lfloor x\rfloor +\delta $ , where $\delta \in \{-2,-1,0\}$ .

   Proof. Consider three cases.

   1. (i) $x=m\in \mathbb {Z}$ . Then, $\lfloor -2x\rfloor =-2m=-2\lfloor m\rfloor =-2\lfloor x\rfloor $ , which gives $\delta =0$ .

   2. (ii) $x=m+\epsilon $ , where $m\in \mathbb {Z}$ and $\epsilon \in \mathbb {R}$ with $0<\epsilon <\tfrac 12$ . Then, $-2x=-2m-2\epsilon =-2m-1+(1-2\epsilon )$ with $0<1-2\epsilon <1$ . It follows that $\lfloor -2x\rfloor =-2m-1=-2\lfloor x\rfloor -1\text {, giving }\delta =-1.$

   3. (iii) $x=m+\epsilon $ , where $m\in \mathbb {Z}$ and $\epsilon \in \mathbb {R}$ with $\tfrac 12\le \epsilon <1$ . As in Case (ii), we see that $\lfloor -2x\rfloor =-2\lfloor x\rfloor -2$ , so that $\delta =-2$ .

3. (c) Consider the function $f(x,y)={1}/{x}+{1}/{y}-{2}/{xy}$ , where x and y are real variables with $x\ge 2$ and $y\ge 3$ . Then the maximum value of $f(x,y)$ is given by $f(2,y)={1}/{2}$ for any $y\ge 3$ .

   Proof. $f(x,y)=({1}/{x})(1-{2}/{y})+{1}/{y}$ , and since $1-{2}/{y}>0$ , $f(x,y)$ achieves its maximum value when x is as small as possible, that is, $x=2$ . However, then $f(2,y)={1}/{2}$ for any $y\ge 3$ .

We are now ready to proceed with Case (2), where $b\nmid c$ . We have $b<c<[b,c]$ , and from $b\mid [b,c]$ and $c\mid [b,c]$ , we get $3b\le [b,c]$ . Hence, ${n}/{3}\ge {bn}/{[b,c]}$ , from which it follows that $-\lfloor {bn}/{[b,c]}\rfloor \ge -\lfloor {n}/{3}\rfloor $ . So we see that

$$ \begin{align*} n-\bigg(\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor +\bigg\lfloor\frac{bn}{[b,c]}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor \ge n-\bigg\lfloor\frac{n}{b}\bigg\rfloor-\bigg\lfloor\frac{n}{c}\bigg\rfloor-\bigg\lfloor\frac{n}{3}\bigg\rfloor+2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor. \end{align*} $$

It therefore suffices to show that

(4.2) $$ \begin{align} \bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor-2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor\le n-\bigg\lfloor\frac{n}{3}\bigg\rfloor. \end{align} $$

From item (c),

$$ \begin{align*}\frac{n}{2}\ge\frac{n}{b}+\frac{n}{c}-\frac{2n}{bc} \quad\text{so that}\quad \bigg\lfloor\frac{n}{2}\bigg\rfloor\ge\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor+\bigg\lfloor-\frac{2n}{bc}\bigg\rfloor,\end{align*} $$

using item (a). It follows that

$$ \begin{align*}\bigg\lfloor\frac{n}{2}\bigg\rfloor\ge\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor-2\bigg\lfloor\frac{n}{bc}\bigg\rfloor+\delta \quad\text{where } \delta\in\{-2,-1,0\},\end{align*} $$

by item (b). Since $[b,c]\le bc$ , this gives

$$ \begin{align*} \bigg\lfloor\frac{n}{2}\bigg\rfloor+2\ge\bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor-2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor. \end{align*} $$

For $n\ge 12$ , it easily follows that $n-2\ge {n}/{2}+{n}/{3}\ge \lfloor {n}/{2}\rfloor +\lfloor {n}/{3}\rfloor $ , that is, $\lfloor {n}/{2}\rfloor +2\le n-\lfloor {n}/{3}\rfloor $ . Direct checking shows that this relation is also valid for $7\le n\le 11$ . Indeed,

$$ \begin{align*} n=7 & :\quad \lfloor\tfrac{7}{2}\rfloor+2=5 \le7-2=7-\lfloor\tfrac{7}{3}\rfloor,\\ n=8 & :\quad \lfloor\tfrac{8}{2}\rfloor+2=6 \le8-2=8-\lfloor\tfrac{8}{3}\rfloor,\\ n=9 & : \quad \lfloor\tfrac{9}{2}\rfloor+2=6 \le9-3=9-\lfloor\tfrac{9}{3}\rfloor,\\ n=10 & :\quad \lfloor\tfrac{10}{2}\rfloor+2=7 \le10-3=10-\lfloor\tfrac{10}{3}\rfloor,\\ n=11 & : \quad\lfloor\tfrac{11}{2}\rfloor+2=7 \le11-3=11-\lfloor\tfrac{11}{3}\rfloor. \end{align*} $$

Hence, for all $n\ge 7$ ,

$$ \begin{align*} \bigg\lfloor\frac{n}{b}\bigg\rfloor+\bigg\lfloor\frac{n}{c}\bigg\rfloor-2\bigg\lfloor\frac{n}{[b,c]}\bigg\rfloor\le\bigg\lfloor\frac{n}{2}\bigg\rfloor+2\le n-\bigg\lfloor\frac{n}{3}\bigg\rfloor, \end{align*} $$

that is, (4.2) holds. This completes the discussion for $a=1$ .

From here on, we assume that $2\le a<b<c$ . We may also assume that $(a,b,c)=1$ (since each fraction ${a_{i_{1}}}/{[a_{i_{1}},a_{i_{2}},\ldots ,a_{i_{r}}]}$ in (2.1) remains unchanged if we cancel out any common factor between $a_{i_{1}}$ and $[a_{i_{1}},a_{i_{2}},\ldots ,a_{i_{r}}]$ .) We begin by investigating when

(4.3) $$ \begin{align} g(a,b,c):=\frac{a}{[a,b]}+\frac{a}{[a,c]}+\frac{b}{[b,c]}\le1. \end{align} $$

Note that whenever (4.3) holds,

$$ \begin{align*} \frac{\xi_{n,3}(a,b,c)-n}{2} & =n-\bigg(\bigg\lfloor\frac{a\cdot n}{[a,b]}\bigg\rfloor+\bigg\lfloor\frac{a\cdot n}{[a,c]}\bigg\rfloor+\bigg\lfloor\frac{b\cdot n}{[b,c]}\bigg\rfloor\bigg)+2\bigg\lfloor\frac{a\cdot n}{[a,b,c]}\bigg\rfloor\\ & \ge n-\bigg(\frac{a\cdot n}{[a,b]}+\frac{a\cdot n}{[a,c]}+\frac{b\cdot n}{[b,c]}\bigg)\quad\text{using Case~(2)(a)}\\ & =\big(1-g(a,b,c)\big)\cdot n \ge0, \end{align*} $$

from which it follows that $\xi _{n,3}(a,b,c)\ge n$ .

Let $d_{1}=(a,b)$ , $d_{2}=(b,c)$ and $d_{3}=(a,c)$ . Then, ${(d_{1},d_{2})=(d_{1},d_{3})=(d_{2},d_{3})=1}$ , because $(a,b,c)=1$ . So $a=d_{1}d_{3}q_{1}$ , $b=d_{1}d_{2}q_{2}$ , $c=d_{2}d_{3}q_{3}$ for some mutually relatively prime positive integers $q_{1},q_{2}$ and $q_{3}$ . It follows that

(4.4) $$ \begin{align} d_{1}d_{3}q_{1}<d_{1}d_{2}q_{2}<d_{2}d_{3}q_{3}, \end{align} $$

and

(4.5) $$ \begin{align} g(a,b,c)=\frac{1}{d_{2}q_{2}}+\frac{1}{d_{2}q_{3}}+\frac{1}{d_{3}q_{3}}. \end{align} $$

Furthermore, there exist positive integers $s_{1}$ , $s_{2}$ and $s_{3}$ such that $b=a+s_{1}d_{1}$ , $c=b+s_{2}d_{2}$ and $c=a+s_{3}d_{3}$ .

We now proceed by partitioning the possible values of the triples $(d_{1},d_{2},d_{3})$ into four different classes.

Class 1: triples $(d_{1},d_{2},d_{3})$ for which $d_{1}\ge 2$ . Here, $b=a+s_{1}d_{1}$ implies that $d_{2}q_{2}=d_{3}q_{1}+s_{1}\ge 2$ . Similarly, $c=b+s_{2}d_{2}$ implies that $d_{3}q_{3}=d_{1}q_{2}+s_{2}\ge 3$ , and also, $c=a+s_{3}d_{3}$ implies that $d_{2}q_{3}=d_{1}q_{1}+s_{3}\ge 3$ . If each of these inequalities happens to be an equality, we obtain $g(a,b,c)=\tfrac 12+\tfrac 13+\tfrac 13=\tfrac 76$ , which is greater than  $1$ . However, if any one of these inequalities becomes strict, we see that $g(a,b,c)\le 1$ . In the special event of three equalities, we must have $d_{2}q_{2}=2$ , $d_{3}q_{3}=3$ and $d_{2}q_{3}=3$ . However, this can only happen if $d_{2}=d_{3}=1$ , $q_{2}=2$ and $q_{3}=3$ , giving $c=d_{2}d_{3}q_{3}=3$ , which is not possible, since $a\ge 2$ . Therefore, $g(a,b,c)\le 1$ for all triples $(a,b,c)$ for which $d_{1}=(a,b)\ge 2$ .

Class 2: triples $(d_{1},d_{2},d_{3})$ for which $d_{1}=1$ and $d_{2}\ge 2$ . Here, $b=d_{2}q_{2}=d_{3}q_{1}+s_{1}\ge 2$ , $d_{3}q_{3}=q_{2}+s_{2}\ge 2$ and $d_{2}q_{3}=q_{1}+s_{3}\ge 2$ .

If $d_{2}=2$ , then, since $a=d_{3}q_{1}\ge 2$ and $(a,b,c)=1$ , we must have either $d_{3}\ge 3$ or $q_{1}\ge 3$ . If $d_{3}\ge 3$ , then, from the inequalities above, $b=d_{2}q_{2}=d_{3}q_{1}+s_{1}\ge 4$ , $d_{3}q_{3}=q_{2}+s_{2}\ge 3$ and $d_{2}q_{3}=q_{1}+s_{3}\ge 2$ , so that $g(a,b,c)\le \tfrac 14+\tfrac 13+\tfrac 12={13}/{12}$ . By checking the small cases, there is only one instance where $1<g(a,b,c)\le {13}/{12}$ , namely $g(3,4,6)={13}/{12}$ . In this case, referring to (4.1),

$$ \begin{align*}\frac{\xi_{n,3}(3,4,6)-n}{2}=n+\bigg\lfloor\frac{n}{4}\bigg\rfloor-\bigg\lfloor\frac{n}{2}\bigg\rfloor-\bigg\lfloor\frac{n}{3}\bigg\rfloor\ge0 \quad\text{for all } n\ge1.\end{align*} $$

However, if $q_{1}\ge 3$ , then, as above, $b=d_{2}q_{2}=d_{3}q_{1}+s_{1}\ge 4$ , $d_{3}q_{3}=q_{2}+s_{2}\ge 2$ and $d_{2}q_{3}=q_{1}+s_{3}\ge 4$ , implying that $g(a,b,c)\le \tfrac 14+\tfrac 12+\tfrac 14=1$ .

Next, consider $d_{2}\ge 3$ . Using the same inequalities as above, we see that now $g(a,b,c)\le \tfrac 13+\tfrac 12+\tfrac 13=\tfrac 76$ . Again, checking small cases, there is only one instance here where $1<g(a,b,c)\le \tfrac 76$ , namely $g(2,3,6)=\tfrac 76$ . In this case,

$$ \begin{align*}\frac{\xi_{n,3}(2,3,6)-n}{2}=n-\bigg\lfloor\frac{n}{2}\bigg\rfloor\ge0 \quad\text{for all } n\ge1.\end{align*} $$

We see that, apart from these two exceptional cases (which satisfy $\xi _{n,3}(a,b,c) \ge n$ by direct checking), all the other triples $(a,b,c)$ in this class have $g(a,b,c)\le 1$ , so that $\tfrac 12{(\xi _{n,3}(a,b,c)-n)}\ge (1-g(a,b,c))n\ge 0$ .

Class 3: triples $(d_{1},d_{2},d_{3})$ for which $d_{1}=d_{2}=1$ and $d_{3}\ge 2$ . Now we have $b=q_{2}=d_{3}q_{1}+s_{1}\ge 3$ , $c=d_{3}q_{3}=q_{2}+s_{2}\ge 2$ and $c=d_{3}q_{3}=d_{3}q_{1}+s_{3}d_{3}$ , giving $q_{3}=q_{1}+s_{3}\ge 2$ . From $q_{3}\ge 2$ and $d_{3}\ge 2$ , it follows that $c=d_{3}q_{3}$ is actually greater than or equal to $4$ . For $c=4$ , there is only one possible triple $(a,b,c)$ , namely $a=2$ , $b=3$ and $c=4$ , and we already know that $\xi _{n,3}(2,3,4)\ge n$ . So we may assume that $q_{3}\ge 3$ or $d_{3}\ge 3$ . In the former case, $g(a,b,c)=g(d_{3}q_{1},q_{2},d_{3}q_{3})={1}/{q_{2}}+{1}/{q_{3}}+{1}/{d_{3}q_{3}}\le {1}/{3}+{1}/{3}+{1}/{6}={5}/{6}\le 1$ , and in the latter case, $g(a,b,c)={1}/{q_{2}}+{1}/{q_{3}}+{1}/{d_{3}q_{3}}\le {1}/{3}+{1}/{2}+{1}/{6}=1$ . As in the previous paragraphs, we conclude that $\xi _{n,3}(a,b,c)\ge n$ for all triples $(a,b,c)$ that belong to this class.

Class 4: $d_{1}=d_{2}=d_{3}=1$ . From (4.4), $2\le q_{1}<q_{2}<q_{3}$ . Then, $g(a,b,c)= g(q_{1},q_{2},q_{3}) \le \tfrac 13+\tfrac 15+\tfrac 15={11}/{15}<1$ , and we again have ${\xi _{n,3}(a,b,c)}\ge n$ .

5 Infinitely many cases where $\xi _{n,k}(a_{1},a_{2},\ldots ,a_{k})=n$ .

One should not be misled by thinking that $\xi _{n,k}(a_{1},a_{2},\ldots ,a_{k})$ would grow without bound as k gets bigger. In this section, we conclude our discussion by proving that $\xi _{n,k}(a_{1},a_{2},\ldots ,a_{k})=n$ is possible for arbitrarily large k.

Proof. For the sake of this theorem, we denote

$$ \begin{align*}D_{n,k}^{r}(a_{1},a_{2},\ldots,a_{k}) = \Delta_{i=1}^{k}a_{i}\underline{n}=\{a_{1},\text{ }2^{2^{r}}\!\cdot a_{1},\text{ }3^{2^{r}}\!\cdot a_{1},\text{ }\dots,\text{ }n^{2^{r}}\!\cdot a_{1}\}.\end{align*} $$

The proof is by induction on r. For $r=0$ , take $k=1\text { and }a_{1}=1$ so that

$$ \begin{align*} D_{n,1}^{0}(1)=\{1,2,3,\dots,n\}=\{1,\text{ }2^{2^{0}}\!\cdot1,\text{ }3^{2^{0}}\!\cdot1,\dots,\text{ }n^{2^{0}}\!\cdot1\}. \end{align*} $$

Assume that the statement is true for some $r\geq 0$ , and $a_{1}<a_{2}<\dots <a_{k}$ , where $k>r$ , are the multipliers used to produce the symmetric difference

$$ \begin{align*} D_{n,k}^{r}(a_{1},a_{2},\ldots,a_{k})=\{a_{1},\text{ }2^{2^{r}}\!\cdot a_{1},\text{ }3^{2^{r}}\!\cdot a_{1},\text{ }\dots,\text{ }n^{2^{r}}\!\cdot a_{1}\}. \end{align*} $$

Then, for each m, $1\le m\le n$ ,

$$ \begin{align*} {\mathbin{\Delta}}_{j=1}^{k}(m^{2^{r}}a_{j}\cdot{\underline{n}}) & =m^{2^{r}}{\mathbin{\Delta}}_{j=1}^{k}(a_{j}\underline{n})\\ & =\{m^{2^{r}}\!\cdot a_{1},\text{ }m^{2^{r}}(2^{2^{r}}\!\cdot a_{1}),\text{ }m^{2^{r}}(3^{2^{r}}\!\cdot a_{1}),\text{ }\dots,\text{ }m^{2^{r}}(n^{2^{r}}\!\cdot a_{1})\}\\ & =m^{2^{r}}D_{n,k}^{r}(a_{1},a_{2},\ldots,a_{k}). \end{align*} $$

Now take the symmetric difference between the n sets:

Due to the symmetry of this ‘matrix’, the symmetric difference is the ‘diagonal’,

$$ \begin{align*} D=\{a_{1},\ 2^{2^{r+1}}\!\cdot a_{1},\ 3^{2^{r+1}}\!\cdot a_{1},\dots,\ n^{2^{r+1}}\!\cdot a_{1}\}. \end{align*} $$

This symmetric difference uses the list of multipliers

$$ \begin{align*} a_{1},a_{2},\dots,a_{k};\text{ }2^{2^{r}}\!\cdot a_{1},2^{2^{r}}\!\cdot a_{2},\dots,2^{2^{r}}\!\cdot a_{k};\text{ }\dots;\text{ }n^{2^{r}}\!\cdot a_{1},n^{2^{r}}\!\cdot a_{2},\dots,n^{2^{r}}\!\cdot a_{k}. \end{align*} $$

There may be duplicates in this list. Since two identical entries will not have any effect on D, our final list of multipliers $a^{\prime }_{1}<a^{\prime }_{2}<\cdots <a^{\prime }_{k'}$ is given by

$$ \begin{align*} A'={\mathbin{\Delta}}_{m=1}^{n}(m^{2^{r}}A)=\{a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{k'}\} \end{align*} $$

and

$$ \begin{align*} D_{n,k'}^{r+1}(a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{k'})=D. \end{align*} $$

Since $n>1$ , there is a prime p with ${n}/{2}<p\leq n$ . Then, $\{a_{1},p^{2^{r}}a_{1},p^{2^{r}}a_{2},\dots ,p^{2^{r}}a_{k}\}\subseteq A'$ , and so $r+1<k+1\leq k'$ . The induction is complete.