2.1 Notation

For convenience, we first give some notation:

• • Let $\mathbb{N}$ be the set of positive integers, and let ${\mathbb{N}}_0={\mathbb{N}}\bigcup \{0\}$ . For any positive integers $i, j$ with $i\leqslant j$ , let $[i, j]=\{m\in {\mathbb{N}}\,:\, i\leqslant m\leqslant j\}$ and $[j]=\{m\in {\mathbb{N}}\,:\, 1\leqslant m\leqslant j\}$ .

• • $C_j(\pi )$ is the number of cycles of length $j$ in the permutation $\pi$ , and $C_I(\pi )$ is the number of cycles whose lengths lie in the set $I$ .

• • $\beta | \pi$ means that $\beta$ is a fixed set or divisor of the permutation $\pi$ , i.e., a product of some subset of the cycles of $\pi$ .

• • $|\beta |$ is the size of $\beta$ .

• • For any positive integer $t$ with $t\geqslant r$ , let

  \begin{equation*}{\boldsymbol{{\mathcal{C}}}}_{r, t}=\{(c_1,\ldots ,c_r,\ldots , c_t)\in {\mathbb{N}}_0^t: c_1=\cdots =c_{r-1}=0\}.\end{equation*}

• • $H_m=1+1/2+\cdots +1/m$ is the $m$ -th harmonic sum.

• • $\unicode {x1D7D9}(S)$ is the indicator function of statement $S$ ; $\unicode {x1D7D9}(S)=1$ if $S$ is true, and $\unicode {x1D7D9}(S)=0$ if $S$ is false.

• • ${\mathbb{P}}(A)$ is the probability of the event $A$ .

• • ${\mathbb{E}}(X)$ is the expectation of the random variable $X$ .

• • ${{\mathcal{C}}}_i^{(n)}$ is the set of cycles of length $i$ in ${{\mathcal{S}}}_n$ .

• • For a vector ${\mathbf{c}}=(c_1,\ldots , c_k)$ of nonnegative integers, let

  \begin{align*} {\mathscr{L}}\,({\mathbf{c}})=\{m_1+2m_2+\cdots +k m_k: 0\leqslant m_j\leqslant c_j\ \text{for}\ j=1, 2,\ldots , k\}\ \text{and}\ S({\mathbf{c}})= \max {\mathscr{L}}\,({\mathbf{c}}). \end{align*}

• • Let ${\mathbf{X}}_{r, k}$ denote the random vector $(X_r,\ldots , X_k)$ , where $X_r,\ldots , X_k$ are independent Poisson random variables with parameters $1/r,\ldots , 1/k$ , respectively. Furthermore, we write

\begin{equation*}S({\mathbf{X}}_{r, t})= \max {\mathscr{L}}\,({\mathbf{X}}_{r, t})=rX_r+\cdots +tX_t.\end{equation*}

2.2 Preliminary lemmas

We first record the inequalities

(2.1) \begin{align} \log (m+1)\leqslant H_m\leqslant 1+\log m, \ (m\geqslant 1). \end{align}

Moreover, we state Stirling’s formula as follows.

Stirling’s formula. For all sufficiently large positive integers $n$ ,

\begin{equation*} n!=\sqrt {2\pi n}\left ( \frac {n}{{{\mathrm{e}}}}\right )^n \left ( 1+O\left (\frac {1}{n}\right )\right ), \end{equation*}

or equivalently,

\begin{equation*} n!\sim \sqrt {2\pi n}\left ( \frac {n}{{{\mathrm{e}}}}\right )^n \quad (n\rightarrow \infty ). \end{equation*}

The following result is an analogue for permutations of a basic lemma from sieve theory. The readers can refer to [Reference Eberhard, Ford and Green5] or [Reference Granville16].

Lemma 2.1. Suppose that $m, n$ are integers with $1\leqslant m\leqslant n$ . Let $\pi \in {{\mathcal{S}}}_n$ be chosen uniformly at random. Then

\begin{equation*} \frac {1}{2m}\leqslant {\mathbb{P}}(\pi \ \text{has no cycle of length}\ \lt m)\leqslant \frac {1}{m}. \end{equation*}

We also need standard bounds on the Poisson distribution, see, for example, [Reference Norton19, Section 4]. To facilitate use, we employ the following simplified version proposed by Ford [Reference Ford13, Lemma2.4].

Lemma 2.2. Uniformly for $h\leqslant m\leqslant x$ ,

\begin{equation*} \sum _{h\leqslant k\leqslant m}\frac {x^k}{k!}\asymp \min \left(\sqrt {x}, \frac {x}{x-m}, m-h+1\right)\frac {x^m}{m!}. \end{equation*}

The following lemma is Cauchy’s classical formula. One may refer to [Reference Ford14, Theorem 1.2] or [Reference Stanley24, Proposition 1.3.2].

Lemma 2.3. Suppose that $\pi$ is chosen uniformly at random. If $m_1+2m_2+\cdots +nm_n=n$ , then

\begin{equation*} {\mathbb{P}} \big (C_1(\pi )=m_1,\ldots , C_n(\pi )=m_n\big )=\prod _{j=1}^n \frac {(1/j)^{m_j}}{m_j!}. \end{equation*}

We will also use the following estimate of Eberhard, Ford, and Green [Reference Eberhard, Ford and Green5], which applies to permutations whose cycle structure is constrained only at small lengths.

Lemma 2.4. Let $1\leqslant m\lt n$ and $c_1,\ldots , c_m$ be nonnegative integers satisfying

\begin{equation*}c_1+2c_2+\cdots +mc_m\leqslant n-m-1.\end{equation*}

Suppose that $\pi \in {{\mathcal{S}}}_n$ is chosen uniformly at random. Then

\begin{equation*} \frac {1}{2m+2}\prod _{i=1}^m \frac {(1/i)^{c_i}}{c_i!}\leqslant {\mathbb{P}} \big (C_1(\pi )=c_1,\ldots , C_m(\pi )=c_m\big )\leqslant \frac {1}{m+1}\prod _{i=1}^m \frac {(1/i)^{c_i}}{c_i!}. \end{equation*}

Applying Lemmas 2.3 and 2.4, we obtain a crude lower bound for ${{\mathcal{R}}}_r(n, k)$ .

Lemma 2.5. For any positive integers $r, k, n$ with $1\leqslant r\leqslant k\leqslant n/2$ , we have

\begin{equation*} {{\mathcal{R}}}_r(n, k)\geqslant \frac {1}{2k(k+1)}. \end{equation*}

Proof. If $n=2k$ , then we obtain by Lemma 2.3 that

\begin{equation*} {{\mathcal{R}}}_r(n, k)\geqslant {\mathbb{P}}_{\pi \in {{\mathcal{S}}}_n}\big ( C_1(\pi )=\cdots =C_{k-1}(\pi )=0, C_k(\pi )=2\big )=\frac {1}{2k^2}\geqslant \frac {1}{2k(k+1)}. \end{equation*}

If $n\gt 2k$ , we then have by Lemma 2.4 that

\begin{equation*} {{\mathcal{R}}}_r(n, k)\geqslant {\mathbb{P}}_{\pi \in {{\mathcal{S}}}_n}\big (C_1(\pi )=\cdots =C_{k-1}(\pi )=0, C_k(\pi )=1\big )\geqslant \frac {1}{2k(k+1)} \end{equation*}

as desired.

By using Lemma 2.1, we derive a crude upper bound for ${{\mathcal{R}}}_r(n, k)$ as follows.

Lemma 2.6. For any positive integers $r, k, n$ with $2\leqslant r\leqslant k\leqslant n/2$ , we have

\begin{equation*} {{\mathcal{R}}}_r(n, k)\leqslant \frac {1}{r^2}. \end{equation*}

Proof. For convenience, let

(2.2) \begin{align} {{\mathcal{S}}}_n^{(r)}(k)\,:\!=\,\{\pi \in {{\mathcal{S}}}_n^{(r)}\,:\, \exists \ \sigma |\pi \ \text{with}\ |\sigma |=k\}, \end{align}

where ${{\mathcal{S}}}_n^{(r)}$ is defined as in (1.3). We then obtain by Lemma 2.1 that

\begin{align*} \begin{split} \big |{{\mathcal{S}}}_n^{(r)}(k)\big |&\leqslant \sum _{I\subseteq [n]\atop |I|=k}\big |\{ \pi _1\in {{\mathcal{S}}}_{I}\,:\, C_{[r-1]}(\pi _1)=0\}\big |\big |\{ \pi _2\in {{\mathcal{S}}}_{[n]\setminus I}: C_{[r-1]}(\pi _2)=0\}\big |\\ &\leqslant \sum _{I\subseteq [n]\atop |I|=k} \frac {k!}{r}\frac {(n-k)!}{r}=\frac {n!}{r^2}, \end{split} \end{align*}

where ${{\mathcal{S}}}_I$ denotes the symmetric group consisting of all permutations of $I$ , and for $j=1, 2$ , the symbol $C_{[r-1]}(\pi _j)$ counts the number of cycles of $\pi _j$ whose lengths belong to the set $[r-1]=\{1,\ldots , r-1\}$ . Consequently, we derive that

\begin{equation*}{{\mathcal{R}}}_r(n, k)\leqslant \frac {1}{r^2}\end{equation*}

as required.

Moreover, we also need the cycle lemma from combinatorics. See [Reference Ford, De Koninck, Granville and Luca11, Lemma 2.4], for example.

Lemma 2.7. For positive real numbers $x_1,\ldots , x_t$ with product $A$ , let $x_{t+i}=x_i$ for $i\geqslant 1$ . Then

\begin{equation*} \frac {1}{\max (1, A)} \leqslant \sum _{j=0}^{t-1}\left( \sum _{h=1}^t x_{1+j}\cdots x_{h+j}\right)^{-1}\leqslant \frac {1}{\min (1, A)}. \end{equation*}

Finally, we refer to a result derived by Ford [Reference Ford10, Lemma 4.9] for the volume of a specific complex region in ${\mathbb{R}}^t$ with $t\in {\mathbb{N}}$ .

Lemma 2.8. Let $M$ be a sufficiently large integer. Suppose $v\geqslant 1$ , $10M\leqslant t\leqslant 100(v-1)$ , $s\geqslant M/2+1$ and $0\leqslant t-v\leqslant s-M/3-1$ . Let $Y_t(s, v)$ be the set of $\boldsymbol{\xi }=(\xi _1,\ldots , \xi _t)\in {\mathbb{R}}^t$ satisfying the following:

1. (i) $0\leqslant \xi _1\leqslant \cdots \leqslant \xi _t\leqslant 1$ ;

2. (ii) For $1\leqslant i\leqslant \sqrt {t-M}$ , $\xi _{M+i^2}\gt i/v$ and $\xi _{t+1-(M+i^2)}\lt 1-i/v$ ;

3. (iii) $\sum _{j=1}^t 2^{j-v\xi _j}\leqslant 2^s$ .

Then we have

\begin{equation*} \textrm {Vol}(Y_t(s, v))\gg \frac {t-v+1}{(t+1)!}. \end{equation*}

4.1 The lower bound in Theorem 1.1

For a vector ${\mathbf{b}}=(b_1,\ldots , b_{J_2})$ of nonnegative integers, let $\mathscr{A}\,({\mathbf{b}})$ be the set of permutations $\pi \in S_n$ composed of exactly $b_j$ cycle factors with length lying in $[2^{j-1}, 2^j-1]$ for $1\leqslant j\leqslant J_2$ .

Suppose that $b_1,\ldots , b_{J_2}$ are arbitrary nonnegative integers with sum $t$ and there are exactly $b_j$ of the $a_i$ in each interval $[2^{j-1}, 2^j-1]$ , where $ b_j=0\ \text{for}\ j\lt J_1$ . Writing

\begin{equation*}{\mathscr{D}}({\mathbf{b}})=\prod _{j=J_1}^{J_2}\{2^{j-1},\ldots , 2^j-1\}^{b_j},\end{equation*}

We then have by (4.2) and (4.3) that

(4.5) \begin{align} \frac {1}{t!}\sum _{a_1,\ldots , a_t=2^{J_1-1}}^{2^{J_2}-1}\frac {|{\mathscr{L}}^{\,\,*}({{\mathbf{a}}})|}{a_1\cdots a_t} =\sum _{b_{J_1}+\cdots +b_{J_2}=t}\frac {1}{b_{J_1}!\cdots b_{J_2}!} \sum _{{\mathbf{d}}\in {\mathscr{D}} ({\mathbf{b}})}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|}{d_1\cdots d_t}. \end{align}

Combining (4.1), (4.2), and (4.5), we derive that

(4.6) \begin{align} {{\mathcal{R}}}_r(n, k)\gg \frac {1}{k^2}\sum _t \sum _{b_{J_1}+\cdots +b_{J_2}=t}\frac {1}{b_{J_1}!\cdots b_{J_2}!}\sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|}{d_1\cdots d_t}. \end{align}

We now give the estimate on $\sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|}{d_1\cdots d_t}$ as follows.

Lemma 4.1. Let $t$ be a positive integer. Then for any vector ${\mathbf{b}}=(b_1,\ldots , b_{J_2})$ of nonnegative integers such that $b_j=0$ for $j\lt J_1$ and $b_{J_1}+\cdots +b_{J_2}=t$ , we have

\begin{equation*} \sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|}{d_1\cdots d_t}\gg \frac {(2\log 2)^t}{\max \Big (\sum _{i=J_1}^{J_2} 2^{b_{J_1}+\cdots +b_i-i}, 1\Big )}. \end{equation*}

Proof. As in the proof of Lemma 4.1 in [Reference Eberhard, Ford and Green5], for given $\ell \in {\mathbb{N}}$ , let $R({\mathbf{d}}, \ell )$ be the number of $I\subseteq [t]$ with $\ell =\sum _{i\in I}d_i$ . Also, define $\lambda _i=\sum _{j=2^{i-1}}^{2^i-1}\frac {1}{j}$ for $J_1\leqslant i\leqslant J_2$ . One then has $\log 2\leqslant \lambda _i\leqslant 1$ for $J_1\leqslant i\leqslant J_2$ . Since $\sum _{\ell }R({\mathbf{d}}, \ell )=2^t$ , we obtain by Cauchy-Schwarz that

(4.7) \begin{align} \begin{split} 2^{2t}\prod _{j=J_1}^{J_2}\lambda _j^{2b_j}&=\bigg(\sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {1}{d_1\cdots d_t}\sum _{\ell } R({\mathbf{d}}, \ell )\bigg)^2\\ &\leqslant \bigg( \sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {\big (\sum _{\ell \in {\mathscr{L}}^{\,\,*}({\mathbf{d}})}R({\mathbf{d}}, \ell )^2\big )^{\frac {1}{2}}} {\sqrt {d_1\cdots d_t}}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|^{\frac {1}{2}}}{\sqrt {d_1\cdots d_t}}\bigg)^2\\ &\leqslant \bigg( \sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}}), \ell }\frac {R({\mathbf{d}}, \ell )^2}{d_1\cdots d_t}\bigg)\bigg(\sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}})}\frac {|{\mathscr{L}}^{\,\,*}({\mathbf{d}})|}{d_1\cdots d_t}\bigg). \end{split} \end{align}

Applying the argument on pages 6725–6726 in [Reference Eberhard, Ford and Green5], one can derive that

\begin{align*} \prod _{j=J_1}^{J_2}\lambda _j^{-b_j}\sum _{{\mathbf{d}}\in {\mathscr{D}}({\mathbf{b}}), \ell }\frac {R({\mathbf{d}}, \ell )^2}{d_1\cdots d_t} &\ll 2^t+2^t\sum _{i=J_1}^{J_2} 2^{b_{J_1}+\cdots +b_i-i}\ll 2^t\max \bigg(\sum _{i=J_1}^{J_2} 2^{b_{J_1}+\cdots +b_i-i}, 1\bigg). \end{align*}

Comparing with (4.7), and using again the fact that $\lambda _i\geqslant \log 2$ , we obtain the desired result.

If $k\gt r_0$ , $r\leqslant \varepsilon k$ and $r\leqslant \sqrt {r_0}$ , then $J_2-J_1+1\asymp J_2$ . Consider those vectors ${\mathbf{b}}=(b_1,\ldots , b_{J_2})$ of nonnegative integers such that $b_j=0$ for $j\lt J_1$ and $b_{J_1}+\cdots +b_{J_2}=t_0$ . Setting $x_i=2^{b_{J_1+i-1}-1}$ , we then have $x_1\cdots x_{t_0}=1$ . And hence by Lemma 2.7, we obtain that

\begin{align*} \frac {1}{\sum _{i=J_1}^{J_2}2^{b_{J_1}+\cdots +b_i-i}}=\frac {2^{J_1-1}}{\sum _{i=1}^{t_0}x_1\cdots x_i}=\frac {2^{J_1-1}}{t_0}. \end{align*}

Using multinomial theorem and Stirling’s formula, we deduce from (4.4) that

\begin{align*} \sum _{b_{J_1}+\cdots +b_{J_2}=t_0}\frac {1}{b_{J_1}!\cdots b_{J_2}!\sum _{i=J_1}^{J_2}2^{b_{J_1}+\cdots +b_i-i}} =\frac {t_0^{t_0}}{t_0!}\frac {2^{J_1-1}}{t_0}\asymp \frac {t_0^{t_0}{{\mathrm{e}}}^{t_0}}{\sqrt {2\pi t_0}t_0^{t_0}} \frac {2^{J_1-1}}{t_0}\asymp \frac {k^{\frac {1}{\log 2}}}{(\log k)^{3/2}}. \end{align*}

It then follows from Lemma 4.1 and (4.6) that

\begin{align*} {{\mathcal{R}}}_r(n, k)\gg \frac {1}{k^2}(2\log 2)^{t_0} \frac {k^{\frac {1}{\log 2}}}{(\log k)^{3/2}}\gg \frac {1}{k^{{\mathcal{E}}} (1+\log k)^{3/2}} \asymp i(n, k) \end{align*}

as desired.

In the following, we assume that $k\gt r_0$ and $\sqrt {r_0}\lt r\leqslant \varepsilon k$ .

Take

(4.8) \begin{align} M=\frac {\log r_0}{200} \end{align}

such that $M$ is a sufficiently large constant because $r_0$ is. Choose $\varepsilon$ such that

(4.9) \begin{align} J_1\geqslant 200M,\quad J_2-J_1\geqslant 200M. \end{align}

Let $\mathscr{B}_t$ be the set of vectors ${\mathbf{b}}=(b_1,\ldots , b_{J_1},\ldots , b_{J_2})$ which satisfy the following:

1. (a) $b_1=\cdots =b_{J_1-1}=0$ ;

2. (b) $b_{J_1}+\cdots +b_{J_2}=t$ ;

3. (c) $\sum _{j=J_1}^{J_2} 2^{b_{J_1}+\cdots +b_{j}-j}\leqslant 2^{-M}$ ;

4. (d) $b_{J_1+i-1}\leqslant M+i^2\quad (i\geqslant 1)$ ;

5. (e) $b_{J_2-i+1}\leqslant M+i^2\quad (i\geqslant 1)$ .

We then obtain by Lemma 4.1 that

(4.10) \begin{align} {{\mathcal{R}}}_r(n, k)\gg \frac {1}{k^2}\sum _{t_1\leqslant t\leqslant t_2}\sum _{{\mathbf{b}}\in \mathscr{B}_t}\frac {(\log 4)^t}{b_{J_1}!\cdots b_{J_2}!} \end{align}

for any two positive integers $t_1, t_2$ with $t_1\leqslant t_2$ . Take

(4.11) \begin{align} t_2=\Big \lfloor \min \bigg (\frac {\log k}{\log 2}, 2(\log k-\log r)\bigg )-2M\Big \rfloor . \end{align}

By (4.9), we have $t_2\geqslant 100M$ . Choose $t_1$ such that

(4.12) \begin{align} 10M\leqslant t_1\leqslant t_2. \end{align}

Now let

(4.13) \begin{align} s=J_1-2-M. \end{align}

Setting

\begin{equation*}g_i=b_{J_1+i-1}\end{equation*}

for $i\geqslant 1$ , we obtain by (c) that

\begin{equation*} \sum _{i=1}^{t_0} 2^{-i+g_1+\cdots +g_i}\leqslant 2^{J_1-1}\cdot 2^{-M}=2^{s+1}, \end{equation*}

where $t_0$ is defined as in (4.4).

By (d) and (e) in the definition of $\mathscr{B}_t$ , $g_i\leqslant M+i^2$ and $g_{t_0+1-i}\leqslant M+i^2$ for every $i\geqslant 1$ . Applying the argument on top of Page 419 of [Reference Ford10], we deduce that

(4.14) \begin{align} \sum _{{\mathbf{b}}\in \mathscr{B}_t}\frac {1}{b_{J_1}!\cdots b_{J_2}!}\gg {t_0}^t \textrm {Vol}(Y_t(s, t_0)) \end{align}

for $t_1\leqslant t\leqslant t_2$ , where $Y_t(s, t_0)$ is defined as in Lemma 2.8.

Since $r_0$ is sufficiently large (implying that $M$ is sufficiently large) and $\varepsilon$ is sufficiently small, we obtain by (4.4), (4.9), (4.11), (4.12) and (4.13) that

\begin{align*} t_0&=J_2-J_1+1\geqslant 200M\geqslant 1,\\ 10M&\leqslant t_1\leqslant t_2\leqslant (\log 4)(J_2-J_1)=(\log 4)(t_0-1)\leqslant 100(t_0-1),\\ s&\geqslant \log r-M\geqslant M/2+1,\\ 0&\leqslant t_2-t_0\leqslant (J_2-2M)-(J_2-J_1+1)=s-M+3\leqslant s-M/3-1. \end{align*}

Therefore, it follows from (4.10), (4.14), and Lemma 2.8 that

(4.15) \begin{align} {{\mathcal{R}}}_r(n, k)\gg \frac {1}{k^2}\sum _{t=t_1}^{t_2} \frac {(t_0\log 4)^t}{t!} \bigg(\frac {t-t_0+1}{t+1}\bigg). \end{align}

We now consider the following two cases.

Case 1. $\delta \geqslant 1-1/\log 4$ . In this case, we have $\log r\geqslant (1-1/\log 4)\log k$ . By (4.11), we have

\begin{equation*} t_2=\Big \lfloor 2(\log k-\log r)-2M\Big \rfloor . \end{equation*}

Take

\begin{equation*}t_1=\frac {3t_2}{4}.\end{equation*}

It is easy to check that $t_1\geqslant 10M$ . With these choices and (4.4), we obtain for $t_1\leqslant t\leqslant t_2$ that

\begin{equation*} \frac {t-t_0+1}{t+1}\asymp 1. \end{equation*}

Applying Lemma 2.2 to the sum in (4.15), we obtain

\begin{equation*} {{\mathcal{R}}}_r(n, k)\gg \frac {(t_0\log 4)^{t_2}}{t_2!} \sqrt {t_0\log 4} \gg \frac {1}{k^2}{{\mathrm{e}}}^{t_0\log 4}\gg \frac {1}{k^2} {{\mathrm{e}}}^{2(\log k-\log r)}= \frac {1}{r^2} \end{equation*}

as desired.

Case 2. $0\lt \delta \lt 1-1/\log 4$ . By (4.11), we have

\begin{equation*} t_2=\Big \lfloor \frac {\log k}{\log 2}-2M\Big \rfloor . \end{equation*}

We now divide the proof of Case 2 into the following two subcases.

Subcase 2.1. $\frac {1}{10}\lt \delta \lt 1-1/\log 4$ . In this subcase, we take

\begin{equation*} t_1=\frac {19 t_2}{20}, \end{equation*}

so that $t-t_0+1\asymp t\asymp \log k$ for $t_1\leqslant t\leqslant t_2$ . Thus for $t_1\leqslant t\leqslant t_2$ , we have

\begin{equation*} \frac {t-t_0+1}{t+1}\asymp 1\asymp \delta . \end{equation*}

Hence, recalling the definition of ${\mathcal{E}}$ and $B(r, k)$ (i.e., (1.1) and (1.4)), we obtain by applying Lemma 2.2 and Stirling’s formula to the sum in (4.15) that

\begin{align*} {{\mathcal{R}}}_r(n, k)&\gg k^{-2}\frac {(t_0\log 4)^{t_2}}{t_2!}\min \left( (t_0\log 4)^{1/2}, \frac {t_0\log 4}{t_0\log 4-t_2}, t_2-t_1+1\right)\\ &\gg k^{-{{\mathcal{E}}}+\log (1-\delta )/\log 2}(\log k)^{-1/2}\min \Big ( (\log k)^{1/2}, \frac {1-\delta }{1-\delta -1/\log 4}+O(1/\log k)\Big )\\ &\gg k^{-{{\mathcal{E}}}+\log (1-\delta )/\log 2}(\log k)^{-1/2}(\log k)^{1/2}\min \Big ( 1, \frac {1}{(1-\delta )\log 4-1}(\log k)^{-1/2}\Big )\\ &\gg \delta B(r, k)k^{-{{\mathcal{E}}}+\log (1-\delta )/\log 2} \end{align*}

as required.

Subcase 2.2. $0\lt \delta \lt \frac {1}{10}$ . In this case, we take

\begin{equation*} t_1=\Big \lfloor \frac {\log k}{\log 2}-2M \Big \rfloor =t_2. \end{equation*}

We now have

\begin{equation*} t_2-t_0+1\asymp J_1,\quad \frac {t_2-t_0+1}{t_2+1}\asymp \frac {\log r}{\log k}=\delta , \quad B(r, k)\asymp (\log k)^{-1/2}. \end{equation*}

Applying Stirling’s formula to the sum in (4.15), we obtain that

\begin{equation*} {{\mathcal{R}}}_r(n, k)\gg \frac {\delta }{k^2} \frac {(t_0\log 4)^{t_2}}{t_2!}\gg \delta B(r, k)k^{-{{\mathcal{E}}}+\log (1-\delta )/\log 2} \end{equation*}

as desired. This completes the proof of the lower bound in Theorem1.1.

4.2 The upper bound in Theorem 1.1

By the discussion in the first paragraph of this section, we only need to deal with the case that $k\gt r_0$ and $r\leqslant \varepsilon k$ . If $\delta =\frac {\log r}{\log k}\geqslant (1-1/\log 4)$ , then by Lemma 2.6, one can get

\begin{equation*}{{\mathcal{R}}}_r(n, k)\ll \frac {1}{r^2},\end{equation*}

which is just the desired upper bound in Theorem1.1.

From now on, we assume that

\begin{equation*}\delta =\frac {\log r}{\log k}\lt 1-1/\log 4.\end{equation*}

We divide our proof into the following two cases.

Case 1. $1/10\lt \delta \lt 1-1/\log 4$ . By the definition of ${\mathscr{L}}^{\,\,*}({{\mathbf{a}}})$ (see (4.3)), we have

(4.16) \begin{align} \Big | {\mathscr{L}}^{\,\,*}({{\mathbf{a}}})\Big |\leqslant \min (2^t, a_1+a_2+\cdots a_t+1). \end{align}

It then follows from (4.1), (4.2), and (4.16) that

(4.17) \begin{align} \begin{split} {{\mathcal{R}}}_r(n, k)&\ll \frac {1}{k^2} \sum _{t\geqslant 1}\frac {1}{t!}\sum _{a_1,\ldots , a_t=r}^k \frac {|{\mathscr{L}}^{\,\,*}({{\mathbf{a}}})|}{a_1\cdots a_t}\\ &\leqslant \frac {1}{k^2}\sum _{t\leqslant J_2}\frac {1}{t!}\sum _{a_1,\ldots , a_t=r}^k \frac {2^t}{a_1\cdots a_t}+\frac {1}{k^2}\sum _{t\gt J_2}\frac {1}{t!}\sum _{a_1,\ldots , a_t=r}^k \frac {a_1+\cdots +a_t+1}{a_1\cdots a_t}\\ &\ll \frac {1}{k^2}\sum _{t\leqslant J_2} \frac {(2\log \frac {k}{r})^t}{t!}+ \frac {1}{k^2}\sum _{t\gt J_2}\frac {t(k-r+1)(\log \frac {k}{r})^{t-1}}{t!}. \end{split} \end{align}

Since $J_2=\lfloor \frac {\log k}{\log 2}\rfloor$ and $1/10\leqslant \delta =\frac {\log r}{\log k}$ , we have $ \sqrt {2}\log \frac {k}{r}\leqslant J_2\leqslant 2\log \frac {k}{r}$ . So we derive from Lemma 2.2 and Stirling’s formula that

(4.18) \begin{align} \begin{split} \frac {1}{k^2}\sum _{t\leqslant J_2} \frac {(2\log \frac {k}{r})^t}{t!} &\ll \frac {1}{k^2} \frac {(2\log \frac {k}{r})^{J_2}}{J_2!}\min \Bigg( \sqrt {2\log \frac {k}{r}}, \frac {2\log \frac {k}{r}}{2\log \frac {k}{r}-J_2}, J_2\Bigg)\\ &\ll k^{-{{\mathcal{E}}}+\frac {\log (1-\delta )}{\log 2}} \min \Big (1, (\log k)^{-\frac {1}{2}}\big ((1-\delta )\log 4-1\big )^{-1}\Big ). \end{split} \end{align}

Since $J_2\gt \sqrt {2}\log \frac {k}{r}$ , we obtain by Stirling’s formula that

(4.19) \begin{align} \frac {1}{k^2}\sum _{t\gt J_2}\frac {(k-r+1)t(\log \frac {k}{r})^{t-1}}{t!} \ll \frac {k}{k^2}\sum _{t\geqslant J_2}^\infty \frac {(\log \frac {k}{r})^{t}}{t!} \ll \frac {1}{k}\frac {(\log \frac {k}{r})^{J_2}}{J_2!}\ll k^{-{{\mathcal{E}}}+\frac {\log (1-\delta )}{\log 2}}(\log k)^{-\frac {1}{2}}. \end{align}

So we derive from (4.17), (4.18), and (4.19) that

\begin{align*} {{\mathcal{R}}}_r(n, k)\ll \delta B(r, k) k^{-{{\mathcal{E}}}+((\log (1-\delta ))/\log 2)} \end{align*}

as required in this case.

Case 2. $0\lt \delta \leqslant 1/10$ . For any positive integer $t$ , let

(4.20) \begin{align} G_t= \sum _{a_1,\ldots , a_t=r}^k \frac {|{\mathscr{L}}^{\,\,*}({{\mathbf{a}}})|}{a_1\cdots a_t}, \end{align}

where ${\mathscr{L}}^{\,\,*}({{\mathbf{a}}})$ is defined as in (4.3). Then by (4.1) and (4.2), we have

(4.21) \begin{align} {{\mathcal{R}}}_r(n, k)\ll \frac {1}{k^2}\sum _{t=1}^\infty \frac {G_t}{t!}. \end{align}

In this case, we bound $G_t$ in a manner similar to that in [Reference Eberhard, Ford and Green5]. Let $\widetilde {a_1}, \widetilde {a_2}, \ldots , \widetilde {a_t}$ be the increasing rearrangement of the sequence $\{a_i\}_{i=1}^t$ , so that $\widetilde {a_1}\leqslant \widetilde {a_2}\leqslant \cdots \leqslant \widetilde {a_t}$ . For any $j$ satisfying $0\leqslant j\leqslant t$ , we have

\begin{align*} {\mathscr{L}}^{\,\,*}({{\mathbf{a}}})\subseteq \left\{ m+\sum _{i\in I}\widetilde {a_i}: r\leqslant m\leqslant \sum _{i=1}^j \widetilde {a_i}, I\subseteq \{j+1,\ldots , t\} \right\}, \end{align*}

which implies that

(4.22) \begin{align} |{\mathscr{L}}^{\,\,*}({{\mathbf{a}}})|\leqslant F({{\mathbf{a}}}), \end{align}

where

(4.23) \begin{align} F({{\mathbf{a}}})=\min _{0\leqslant j\leqslant t} 2^{t-j}(\widetilde {a_1}+\cdots +\widetilde {a_j}+1). \end{align}

It is reasonable to expect that

(4.24) \begin{align} \sum _{a_1,\ldots , a_t=r}^k \frac {F({{\mathbf{a}}})}{a_1\cdots a_t} \thicksim \int _r^k \cdots \int _r^k \frac {F({{\mathbf{x}}})}{x_1\cdots x_t} d{{\mathbf{x}}} =\big (\log \frac {k}{r}\big )^t \int _0^1\cdots \int _0^1 F(r{{\mathrm{e}}}^{\xi _1\log \frac {k}{r}},\ldots , r{{\mathrm{e}}}^{\xi _t\log \frac {k}{r}})d{\boldsymbol \xi }. \end{align}

We may also prove an approximate version of (4.24) as in [Reference Eberhard, Ford and Green5].

Lemma 4.2. Suppose that $k, r$ are sufficiently large integers, with $\log r\leqslant (1-1/\log 4)\log k$ . Let

(4.25) \begin{align} v_0=\Big \lfloor \frac {\log k-\log r}{\log 2}\Big \rfloor , u_0=\Big \lfloor \frac {\log r}{\log 2}\Big \rfloor . \end{align}

Then for any $t\geqslant 1$ , we have

\begin{equation*} G_t\ll \Big (2\log \frac {k}{r}\Big )^t t! U_t(v_0, u_0), \end{equation*}

where

(4.26) \begin{align} U_t(v, u)\,:\!=\, \int _{\Omega _t} \min _{0\leqslant j\leqslant t} 2^{-j}\Big (2^{v\xi _1+u}+\cdots +2^{v\xi _j+u}+1\Big )d{\boldsymbol \xi } \end{align}

with

\begin{equation*}\Omega _t=\{{\boldsymbol \xi }=(\xi _1,\ldots , \xi _t)\in {\mathbb{R}}^t: 0\leqslant \xi _1\leqslant \xi _2\leqslant \cdots \leqslant \xi _t\leqslant 1\}. \end{equation*}

Proof. As in [Reference Eberhard, Ford and Green5], define the product sets

\begin{equation*} R({{\mathbf{a}}})=\prod _{i=1}^t[\exp (H_{a_i-1}), \exp (H_{a_i})]. \end{equation*}

Then we obtain by (4.22) that

\begin{align*} G_t\leqslant \sum _{a_1,\ldots , a_t=r}^k \frac {F({{\mathbf{a}}})}{a_1\cdots a_t}=\sum _{a_1,\ldots , a_t=r}^k F({{\mathbf{a}}})\int _{R({{\mathbf{a}}})}\frac {d{{\mathbf{x}}}}{x_1\cdots x_t}. \end{align*}

For ${{\mathbf{x}}}\in R({{\mathbf{a}}})$ , we write $\widetilde {x_1}\leqslant \widetilde {x_2}\leqslant \cdots \leqslant \widetilde {x_t}$ for the non-decreasing rearrangement of the components of the vector ${\mathbf{x}}$ . We then derive from (2.1) that

\begin{equation*} \widetilde {a_i}\leqslant \exp (H_{\widetilde {a_i}-1})\leqslant \widetilde {x_i}\leqslant \exp (H_{\widetilde {a_i}})\ \text{for}\ 1\leqslant i\leqslant t. \end{equation*}

So we have

\begin{align*} F({{\mathbf{a}}})\leqslant \min _{0\leqslant j\leqslant t} 2^{t-j}(\widetilde {x_1}+\cdots +\widetilde {x_j}+1)=F({{\mathbf{x}}}) \end{align*}

for all ${{\mathbf{x}}}\in R({{\mathbf{a}}})$ . It follows that

\begin{equation*} \sum _{a_1,\ldots , a_t=r}^k F({{\mathbf{a}}}) \int _{R({{\mathbf{a}}})}\frac {d{{\mathbf{x}}}}{x_1\cdots x_t} \leqslant \sum _{a_1,\ldots , a_t=r}^k \int _{R({{\mathbf{a}}})}\frac {F({{\mathbf{x}}})}{x_1\cdots x_t}d{{\mathbf{x}}} =\int _{\exp (H_{r-1})}^{\exp (H_k)}\cdots \int _{\exp (H_{r-1})}^{\exp (H_k)}\frac {F({{\mathbf{x}}})}{x_1\cdots x_t}d{{\mathbf{x}}}. \end{equation*}

We then obtain by making the change of variables $x_i=r{{\mathrm{e}}}^{\xi _i(H_k-H_{r-1})}$ that

\begin{align*} G_t &\leqslant (2(H_k-H_{r-1}))^t t! \int _{\Omega _t} \min _{0\leqslant j\leqslant t}2^{-j} \Big ( r{{\mathrm{e}}}^{\xi _1(H_k-H_{r-1})}+\cdots + r{{\mathrm{e}}}^{\xi _j(H_k-H_{r-1})}+1\Big )d{\boldsymbol \xi }\\ &\ll \big (2\log \frac {k}{r}\big )^t t! \int _{\Omega _t} \min _{0\leqslant j\leqslant t}2^{-j} \Big (r\big (\frac {k}{r}\big )^{\xi _1}+\cdots +r\big (\frac {k}{r}\big )^{\xi _j}+1\Big )\textrm {d}{\boldsymbol \xi }\\ &\ll \big (2\log \frac {k}{r}\big )^t t! U_t(v_0, u_0) \end{align*}

as desired.

To bound $U_t(v_0, u_0)$ , we apply the following estimate established by Ford [Reference Ford13].

Lemma 4.3. Suppose that $t, u, v$ are integers satisfying $1\leqslant t\leqslant 10v$ and $u\geqslant 1$ . Then

\begin{equation*} U_t(v, u)\ll \frac {u(1+(t-u-v)^2)}{(t+1)!(2^{t-v-u}+1)}. \end{equation*}

Since $v_0=\lfloor \frac {\log k-\log r}{\log 2}\rfloor$ , we have

\begin{equation*} \big (\frac {k}{r}\big )^{\xi _i}\asymp 2^{v_0\xi _i}. \end{equation*}

If $0\lt \delta \leqslant 1/10$ , then by (4.4) and (4.25), we have $J_2=v_0+u_0+O(1)$ and $8J_2\leqslant 10v_0$ . Hence $2\log \frac {k}{r}=2(\log k-\log r)\geqslant 2\times 0.9\log k\gt {1.2J_2}$ . Note that $\frac {\log k-\log r}{J_2}\lt \frac {5}{7}$ . Hence, using Lemmas 4.2 and 4.3, we obtain by (4.25) that

(4.27) \begin{equation} \begin{split} \sum _{1\leqslant t\leqslant 8J_2}\frac {G_t}{t!} &\ll \sum _{1\leqslant t\lt J_2}\left(2\log \frac {k}{r}\right)^t U_t(v_0, u_0)+\sum _{J_2\leqslant t\leqslant 8J_2}\left(2\log \frac {k}{r}\right)^t U_t(v_0, u_0)\\ &\ll \sum _{1\leqslant t\lt J_2}\frac {u_0(1+(t-J_2)^2)}{(t+1)!(2^{t-J_2}+1)}\left(2\log \frac {k}{r}\right)^t\\ &\quad +\sum _{J_2\leqslant t\leqslant 8J_2}\frac {u_0(1+(t-J_2)^2)}{(t+1)!(2^{t-J_2}+1)}\left(2\log \frac {k}{r}\right)^t\\ &\ll u_0 \frac {\big(2\log \frac {k}{r}\big)^{J_2-1}}{J_2!}+u_0\sum _{l=0}^{7J_2}\frac {1+l^2}{(J_2+1+l)! 2^l}\left(2\log \frac {k}{r}\right)^{J_2+l}\\ &\ll u_0 \frac {(2\log \frac {k}{r})^{J_2-1}}{J_2!}+ u_0\frac {(2\log \frac {k}{r})^{J_2}}{(J_2+1)!}\\ &\ll (\log r)\frac {(2\log \frac {k}{r})^{J_2}}{(J_2+1)!}. \end{split} \end{equation}

If $t\gt 8J_2$ , one can infer from (4.26) that $U_t(v_0, u_0)\leqslant \int _{\Omega _t} 1 d{\boldsymbol \xi }=\frac {1}{t!}$ . Moreover, one can derive from $2\log \frac {k}{r}\leqslant 2\log k$ and $J_2=\lfloor \frac {\log k}{\log 2}\rfloor$ that $\frac {2\log \frac {k}{r}}{8J_2}\leqslant 1/4$ . Hence we have by Lemma 4.3 that

(4.28) \begin{align} \sum _{t\gt 8J_2}\frac {G_t}{t!}\ll \sum _{t\gt 8J_2}\frac {\big (2\log \frac {k}{r}\big )^t }{t!} \ll \frac {(2\log \frac {k}{r})^{8J_2}}{(8J_2)!}\ll \frac {(2\log \frac {k}{r})^{J_2}}{(J_2+1)!}. \end{align}

If $0\lt \delta \leqslant \frac {1}{10}$ , then by (1.4) we have $B(r, k)\asymp (\log k)^{-1/2}$ . So we derive from (4.21), (4.27), (4.28), and Stirling’s formula that

\begin{align*} {{\mathcal{R}}}_r(n, k) \ll \frac {\log r}{k^2}\frac {(2\log \frac {k}{r})^{J_2}}{(J_2+1)!} \ll \frac {\log r}{(\log k)^{3/2}}k^{-{{\mathcal{E}}}+((\log (1-\delta ))/\log 2)} \ll \delta B(r, k)k^{-{{\mathcal{E}}}+((\log (1-\delta ))/\log 2)} \end{align*}

as desired if $0\lt \delta \leqslant \frac {1}{10}$ .