2 Preliminary lemmas and notation

We need the following two lemmas from Luca and Shparlinski [Reference Luca and Shparlinski3].

Lemma 2.1. Let $f(X) \in \mathbb {Z}[X] \setminus \{ 0 \}$ . Then, there exists a positive integer $n_0 \geqslant 2$ , which depends only on f, such that the equation

$$ \begin{align*} f(n) \prod_{i=1}^{k} (n+i) - f(n+k) = 0 \end{align*} $$

does not have any integer solution $(n,k)$ with $n \geqslant n_0$ and $k \geqslant 1$ . Moreover, for any integer $n \geqslant n_0$ , we have ${n!}+f(n)> 1$ and $f(n) \neq 0$ .

Lemma 2.2 [Reference Luca and Shparlinski3, Lemma 4]

Let $f(X) \in \mathbb {Z}[X] \setminus \{ 0 \}$ . Then, there exists a positive constant $C_0$ , which depends only on f, such that for any prime number p and any interval ${J \subset [1,p)}$ with length $|J| \geqslant 1$ ,

$$ \begin{align*} \# \{ n \in \mathbb{Z}_{>0} \cap J \mid p \text{ divides } {n!}+f(n) \} \leqslant C_0|J|^{{2}/{3}}. \end{align*} $$

From now on, we fix a polynomial $f(X) \in \mathbb {Z}[X]\setminus \{ 0 \}$ and let $n_0$ be the constant as stated in Lemma 2.1. We also fix an $\varepsilon _0 \in (0,1/100)$ .

We use the Vinogradov symbol $\ll $ as well as the Landau symbols O and o. Throughout this paper, all the constants implied by $\ll $ or O depend at most on f and $\varepsilon _0$ (in particular, these implicit constants do not depend on x, a parameter we introduce later). For convenience, we specify some constants; one of them is $C_1$ . We fix a nonnegative integer $C_1$ such that

(2.1) $$ \begin{align} |f(n)| \leqslant n^{C_1} \text{ for any integer } n \geqslant 2. \end{align} $$

Let $\lambda := 1+9\log 2 - 100\varepsilon _0$ . To prove Theorem 1.1, we will show that the set

$$ \begin{align*} B(\lambda) := \{ n \in \mathbb{Z}_{\geqslant n_0} \mid P({n!}+f(n))> \lambda n \} \end{align*} $$

has lower asymptotic density $\geqslant \varepsilon _0$ . In fact, we will show that for any sufficiently large integer x,

$$ \begin{align*} |B(\lambda) \cap \{1,2,\ldots,x\}| \geqslant \varepsilon_0 x. \end{align*} $$

We argue by contradiction. In the following, we fix a sufficiently large integer x and suppose that $|B(\lambda ) \cap \{1,2,\ldots ,x\}| < \varepsilon _0 x$ . Define

(2.2) $$ \begin{align} Z := \prod_{\substack{n=\lceil \varepsilon_0 x \rceil \\ n \notin B(\lambda)}}^{x} ({n!}+f(n)). \end{align} $$

(We assume x is large enough so that $\varepsilon _0 x> n_0$ . For $n \geqslant \varepsilon _0 x$ , we have ${n!}+f(n)>1$ and $f(n) \neq 0$ .) By Stirling’s formula, $\log ({n!}+f(n)) = n\log n +O(n)$ , so that

$$ \begin{align*} \log Z> \bigg( \kern-2pt\sum_{n = \lceil \varepsilon_0 x \rceil}^{x } ( n\log n + O(n) ) \bigg) - \varepsilon_0 x ( x\log x + O(x) ). \end{align*} $$

Therefore, under the assumption that x is sufficiently large,

(2.3) $$ \begin{align} \log Z> ( \tfrac{1}{2} - 2\varepsilon_0 ) x^2 \log x. \end{align} $$

We denote by $|I|$ the length of an interval I. The following notation will be frequently used in this paper.

Definition 2.3. Let $n_1,n_2,\ldots ,n_t$ be distinct positive integers ( $t \geqslant 2$ ). We define $\mathcal {I}(n_1,n_2,\ldots ,n_t)$ to be the following set of intervals. If $\sigma $ is the unique permutation on the index set $\{1,2,\ldots ,t\}$ such that $n_{\sigma (1)} < n_{\sigma (2)} < \cdots < n_{\sigma (t)}$ , then

$$ \begin{align*} \mathcal{I}(n_1,n_2,\ldots,n_t) := \{ I \mid I = (n_{\sigma(i)},n_{\sigma(i+1)}] \text{ for some } i \in \{ 1,2,\ldots,t-1 \} \}. \end{align*} $$

We also define

$$ \begin{align*} \mathcal{I}_{\text{good}}(n_1,n_2,\ldots,n_t) &:= \{ I \in \mathcal{I}(n_1,n_2,\ldots,n_t) \mid |I| \leqslant {x^{0.99}}/{t}\\ &\!\qquad \text{or } I \text{ contains at least } 2C_1+1 \text{ prime numbers} \}. \end{align*} $$

Note that $\mathcal {I}(n_1,n_2,\ldots ,n_t)$ is determined solely by the set $\{ n_1,n_2,\ldots ,n_t \}$ . However, $\mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ depends also on t, x and the constant $C_1$ from (2.1). Clearly, $\# \mathcal {I}(n_1,n_2,\ldots ,n_t) = t-1$ and the intervals in $\mathcal {I}(n_1,n_2,\ldots ,n_t)$ are disjoint. It is straightforward to check from the definition that

$$ \begin{align*} \# ( \mathcal{I}(n_1,n_2,\ldots,n_t) \setminus \mathcal{I}(n_1,n_2,\ldots,n_{t-1}) ) \leqslant 2 \end{align*} $$

for any integer $t \geqslant 3$ and any t distinct positive integers $n_1,\ldots ,n_t$ . This fact will be used later in Lemma 2.7.

The following lemma is a variation of an observation in [Reference Stewart9].

Lemma 2.4. Suppose that x is larger than some constant depending at most on f and $\varepsilon _0$ . Let t and $t'$ be integers such that $2 \leqslant t' \leqslant t$ . Let p be a prime number and $n_1,n_2,\ldots ,n_t$ be distinct positive integers in the interval $[\varepsilon _0 x,\min \{p-1,x\}]$ . Then, for any two distinct intervals $I_1, I_2 \in \mathcal {I}(n_1,n_2,\ldots ,n_{t'}) \cap \mathcal {I}_{\mathrm {good}}(n_1,n_2,\ldots ,n_t)$ with ${|I_2| \geqslant |I_1|}$ ,

(2.4) $$ \begin{align} (\log p) \cdot \min_{1 \leqslant k \leqslant t'}\{\operatorname{ord}_{p}(n_k! + f(n_k))\} \leqslant |I_1|\hspace{-1.2pt}\log\bigg( \frac{ex}{|I_1|} \bigg) + \bigg(|I_2|-|I_1| + \frac{x^{0.99}}{t} + O(1)\bigg)\log x. \end{align} $$

Proof. There exist indices $i_1,j_1,i_2,j_2 \leqslant t'$ such that $I_1 = (n_{i_1},n_{j_1}]$ and ${I}_2= (n_{i_2},n_{j_2}]$ . We denote

$$ \begin{align*} D = p^{\min_{1 \leqslant k \leqslant t'}\{\operatorname{ord}_{p}(n_k! + f(n_k))\}}. \end{align*} $$

From $D \mid (n_{i_1}!+f(n_{i_1}))$ and $D \mid (n_{j_1}!+f(n_{j_1}))$ , we obtain

(2.5) $$ \begin{align} D~ \bigg|~ \bigg( f(n_{i_1})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{j_1}) \bigg). \end{align} $$

Since $n_{j_1}>n_{i_1} \geqslant \varepsilon _0x > n_0$ , by Lemma 2.1,

$$ \begin{align*} f(n_{i_1})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{j_1}) \neq 0. \end{align*} $$

Hence, we deduce from (2.5) and (2.1) that

$$ \begin{align*} D \leqslant \bigg| f(n_{i_1})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{j_1}) \bigg| \leqslant 2x^{C_1+|I_1|}. \end{align*} $$

If $|I_1| \leqslant x^{0.99}/t$ , then $\log D \leqslant (x^{0.99}/t + O(1))\log x$ and (2.4) is true. Similarly,

(2.6) $$ \begin{align} D ~ \bigg|~ \bigg( f(n_{i_2})\prod_{k=1}^{n_{j_2}-n_{i_2}}(n_{i_2}+k) - f(n_{j_2}) \bigg) \neq 0. \end{align} $$

If $|I_2| \leqslant x^{0.99}/t$ , then (2.4) is true.

It remains to consider the case that both $|I_1|$ and $|I_2|$ are greater than $x^{0.99}/t$ . Then, by the definition of $\mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ , there exist distinct primes $q_1,q_2,\ldots ,q_{2C_1+1}$ in the interval $I_1$ , and there exist distinct primes $q_1^{\prime },q_2^{\prime },\ldots ,q_{2C_1+1}^{\prime }$ in the interval $I_2$ . By (2.5) and (2.6),

(2.7) $$ \begin{align} D~ \bigg|~ \bigg( f(n_{i_1})f(n_{j_2})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{i_2})f(n_{j_1})\prod_{k'=1}^{n_{j_2}-n_{i_2}}(n_{i_2}+k') \bigg). \end{align} $$

We claim that

(2.8) $$ \begin{align} f(n_{i_1})f(n_{j_2})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{i_2})f(n_{j_1})\prod_{k'=1}^{n_{j_2}-n_{i_2}}(n_{i_2}+k') \neq 0. \end{align} $$

In fact, since $I_1 = (n_{i_1},n_{j_1}]$ and $I_2= (n_{i_2},n_{j_2}]$ are disjoint, there are only two cases. The first case is that $n_{i_1}<n_{j_1} \leqslant n_{i_2}<n_{j_2}$ . If (2.8) does not hold, then we deduce that $q_1^{\prime }q_2^{\prime } \cdots q_{2C_1+1}^{\prime } \mid f(n_{i_1})f(n_{j_2})$ . Since $n_{i_1},n_{j_1},n_{i_2},n_{j_2} \geqslant \varepsilon _0 x> n_0$ , we have ${f(n_{i_1})f(n_{j_2}) \neq 0}$ . Thus, $q_1^{\prime }q_2^{\prime } \cdots q_{2C_1+1}^{\prime } \leqslant |f(n_{i_1})f(n_{j_2})| \leqslant x^{2C_1}$ . However, $q_1^{\prime }q_2^{\prime } \cdots q_{2C_1+1}^{\prime } \geqslant (\varepsilon _0 x)^{2C_1+1}$ , which leads to a contradiction because x is large. The second case is that $n_{i_2}<n_{j_2} \leqslant n_{i_1}<n_{j_1}$ . If (2.8) does not hold in this case, we will arrive at a similar contradiction from $q_1q_2 \cdots q_{2C_1+1} \mid f(n_{i_2})f(n_{j_1})$ . Hence, the claim (2.8) holds.

Since $|I_2| = n_{j_2} - n_{i_2} \geqslant |I_1| = n_{j_1}-n_{i_1}$ , the factorial $(n_{j_1}-n_{i_1})!$ is a divisor of the right-hand side of (2.7). Since D is a power of the prime p and $n_{i_1},n_{j_1},n_{i_2},n_{j_2} < p$ , the factorial $(n_{j_1}-n_{i_1})!$ is coprime to D. Thus,

(2.9) $$ \begin{align} D~ \bigg|~ \frac{1}{(n_{j_1}-n_{i_1})!} \bigg( f(n_{i_1})f(n_{j_2})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{i_2})f(n_{j_1})\prod_{k'=1}^{n_{j_2}-n_{i_2}}(n_{i_2}+k') \bigg). \end{align} $$

Noting that $m! \geqslant (m/e)^m$ for any positive integer m and

$$ \begin{align*} \bigg| f(n_{i_1})f(n_{j_2})\prod_{k=1}^{n_{j_1}-n_{i_1}}(n_{i_1}+k) - f(n_{i_2})f(n_{j_1})\prod_{k'=1}^{n_{j_2}-n_{i_2}}(n_{i_2}+k') \bigg| \leqslant 2x^{2C_1 + |I_2|}, \end{align*} $$

we deduce from (2.9) and (2.8) that

$$ \begin{align*} D \leqslant 2x^{|I_2|-|I_1|+2C_1} \cdot \bigg( \frac{ex}{|I_1|} \bigg)^{|I_1|}, \end{align*} $$

which completes the proof of (2.4).

We require the intervals in $\mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ to contain at least $2C_1+1$ prime numbers solely to ensure that the left-hand side of (2.8) is nonzero. The following result of Heath-Brown will help us demonstrate that most intervals in $\mathcal {I}(n_1,n_2,\ldots ,n_t)$ are ‘good’ (when $t \leqslant C_0x^{2/3}$ ). This is the only nonelementary result that we need in this paper.

Lemma 2.5 (Heath-Brown, [Reference Heath-Brown2, Theorem 1])

Let $2=p_1<p_2<\cdots $ denote the sequence of prime numbers. For any $\varepsilon> 0$ , there exists a constant $c_{\varepsilon }$ depending only on $\varepsilon $ , such that for any $y \geqslant 2$ ,

$$ \begin{align*} \sum_{p_k \leqslant y} (p_{k+1}-p_k)^2 \leqslant c_{\varepsilon} y^{{23}/{18}+\varepsilon}. \end{align*} $$

The exponent $23/18$ in Lemma 2.5 is less than $4/3$ , which is sufficient for us to demonstrate that most intervals in $\mathcal {I}(n_1,n_2,\ldots ,n_t)$ are ‘good’ when $t \leqslant C_0x^{2/3}$ , as stated below in Corollary 2.6. The exponent $23/18$ in Lemma 2.5 has been improved to $5/4$ by Peck [Reference Peck7] (see also Maynard [Reference Maynard5]) and to $123/100$ by Stadlmann [Reference Stadlmann8]. However, these improvements to Lemma 2.5 do not yield any enhancement to Theorem 1.1.

Corollary 2.6. Suppose that x is larger than some constant depending at most on f and $\varepsilon _0$ . Let t be an integer such that $2 \leqslant t \leqslant C_0 x^{2/3}$ , where $C_0$ is the constant in Lemma 2.2. Let $n_1,n_2,\ldots ,n_t$ be any distinct integers in the interval $[\varepsilon _0 x, x]$ . Then,

$$ \begin{align*} \# \mathcal{I}_{\mathrm{good}}(n_1,n_2,\ldots,n_t) \geqslant t-2t^{0.99}. \end{align*} $$

Proof. Since x is large, we have $t \leqslant C_0x^{{2}/{3}} < x^{{2}/{3} + 0.01} < x$ . By Lemma 2.5, there exists an absolute constant C such that

(2.10) $$ \begin{align} \sum_{p_k \leqslant y} (p_{k+1}-p_k)^2 \leqslant Cy^{{23}/{18}+0.001} \quad\text{for any } y \geqslant 2. \end{align} $$

(Recall that we denote by $2=p_1<p_2<\cdots $ the sequence of prime numbers.)

For any interval $I \in \mathcal {I}(n_1,n_2,\ldots ,n_t) \setminus \mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ , we have $|I|> {x^{0.99}}/{t}$ and I contains at most $2C_1$ prime numbers. There are indices $i,j$ such that $I = (n_i,n_j]$ . Let $p_{k+1}$ be the least prime number such that $p_{k+1}> n_j$ . Then, we have $p_k \leqslant n_j \leqslant x$ and $p_{k+1}> n_j \geqslant \varepsilon _0 x$ . Note that $k> 2C_1$ since x is large ( $x>p_{2C_1+1}/\varepsilon _0$ ). Given that I contains at most $2C_1$ prime numbers, we must have $p_{k-2C_1} \leqslant n_i$ , so $I \subset (p_{k-2C_1},p_{k+1})$ .

Let

$$ \begin{align*}m = \# ( \mathcal{I}(n_1,n_2,\ldots,n_t) \setminus \mathcal{I}_{\text{good}}(n_1,n_2,\ldots,n_t) ).\end{align*} $$

For any index k such that $k>2C_1$ and $p_k \leqslant x$ , we denote by $m_k$ the number of intervals in $\mathcal {I}(n_1,n_2,\ldots ,n_t) \setminus \mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ that are contained in $(p_{k-2C_1},p_{k+1})$ . Then,

$$ \begin{align*} \sum_{\substack{p_k \leqslant x \\ k> 2C_1}} m_k \geqslant m \end{align*} $$

and

(2.11) $$ \begin{align} \sum_{\substack{p_k \leqslant x \\ k> 2C_1}} (p_{k+1} - p_{k-2C_1})^2 \geqslant \sum_{\substack{p_k \leqslant x \\ k > 2C_1}} m_k^2 \bigg(\frac{x^{0.99}}{t}\bigg)^2 \geqslant \frac{x^{1.98}}{t^2} \sum_{\substack{p_k \leqslant x \\ k > 2C_1}} m_k \geqslant m \cdot \frac{x^{1.98}}{t^2}. \end{align} $$

However, by (2.10) and the Cauchy–Schwarz inequality,

(2.12) $$ \begin{align} \sum_{\substack{p_k \leqslant x \\ k> 2C_1}} & (p_{k+1} - p_{k-2C_1})^2 \notag \\ & \leqslant (2C_1+1) \sum_{\substack{p_k \leqslant x \\ k > 2C_1}} ( (p_{k+1}-p_{k})^2 + (p_{k}-p_{k-1})^2 + \cdots + (p_{k-2C_1+1}-p_{k-2C_1})^2 ) \notag \\ & \leqslant (2C_1+1)^2 \sum_{p_{k} \leqslant x} (p_{k+1}-p_{k})^2 \notag \\ & \leqslant (2C_1+1)^2 \cdot C \cdot x^{{23}/{18}+0.001} < x^{{23}/{18}+0.01} \quad \text{(because } x \text{ is large)}. \end{align} $$

Combining (2.11) and (2.12), we obtain

$$ \begin{align*} m \leqslant \frac{t^2}{x^{{13}/{18}-0.03}} < \frac{t}{x^{{1}/{18}-0.04}} < t^{0.99}. \end{align*} $$

(We have used $t < x^{{2}/{3}+0.01}$ and $t<x$ .) Therefore,

$$ \begin{align*} \# \mathcal{I}_{\text{good}}(n_1,n_2,\ldots,n_t) = t-1-m \geqslant t - 2t^{0.99}. \end{align*} $$

The proof of Corollary 2.6 is complete.

The following lemma is the key step of this paper.

Lemma 2.7. Suppose that x is larger than some constant depending at most on f and $\varepsilon _0$ . Let p be a prime number. Let t be an integer such that

$$ \begin{align*} \bigg( \frac{10}{\varepsilon_0} \bigg)^{100} \leqslant t \leqslant C_0 x^{{2}/{3}}, \end{align*} $$

where $C_0$ is the constant in Lemma 2.2. Let J be an interval such that

$$ \begin{align*} J \subset [\varepsilon_0x, \min\{p-1,x\}]. \end{align*} $$

If $n_1,n_2,\ldots ,n_t$ are distinct integers in the interval J, then

(2.13) $$ \begin{align} \log p \sum_{t' = \lceil ({1+\varepsilon_0})t/{2}\rceil}^{t} \min_{1 \leqslant j \leqslant t'} \{ \operatorname{ord}_{p} ( n_{j}! + f(n_j) ) \} \leqslant \frac{|J|}{2} \log t + \frac{x\log x}{t^{0.98}} + O(x). \end{align} $$

Proof. Let $t_1 = \# \mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ . Obviously, $t_1 \leqslant t-1$ . By Corollary 2.6,

(2.14) $$ \begin{align} t_1 \geqslant t-2t^{0.99}. \end{align} $$

We list the lengths of all intervals in $\mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ in ascending order:

(2.15) $$ \begin{align} \gamma_1 \leqslant \gamma_2 \leqslant \cdots \leqslant \gamma_{t_1}. \end{align} $$

Since $n_1,n_2,\ldots ,n_t \in J$ and the intervals in $\mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ are disjoint,

$$ \begin{align*} \sum_{k=1}^{t_1} \gamma_k \leqslant |J|. \end{align*} $$

Let

(2.16) $$ \begin{align} t_2 = \lfloor t-3t^{0.99} \rfloor. \end{align} $$

From $(t_1 - t_2+1)\gamma _{t_2} \leqslant \sum _{t'=t_2}^{t_1} \gamma _{t'} \leqslant |J|$ , and using (2.14) and (2.16), we deduce that

(2.17) $$ \begin{align} \gamma_{t_2} \leqslant \frac{|J|}{t^{0.99}}. \end{align} $$

For any $t' \leqslant t$ , from the definition of $\mathcal {I}(n_1,n_2,\ldots ,n_{t'})$ (see Definition 2.3), we immediately see that

$$ \begin{align*} \# ( \mathcal{I}(n_1,n_2,\ldots,n_{t'}) \setminus \mathcal{I}(n_1,n_2,\ldots,n_{t'-1}) ) \leqslant 2. \end{align*} $$

Let k be an integer in the range

(2.18) $$ \begin{align} 0 \leqslant k \leqslant t_3 \quad\text{where } t_3 = \lfloor \tfrac{1}{2}( t - 5t^{0.99} )\rfloor - 1. \end{align} $$

Then,

(2.19) $$ \begin{align} \# ( \mathcal{I}(n_1,n_2,\ldots,n_{t-k}) \cap \mathcal{I}_{\text{good}}(n_1,n_2,\ldots,n_t) ) \geqslant t_1 - 2k. \end{align} $$

In particular, for any k in the range (2.18), we have (recall (2.14))

$$ \begin{align*} \# ( \mathcal{I}(n_1,n_2,\ldots,n_{t-k}) \cap \mathcal{I}_{\text{good}}(n_1,n_2,\ldots,n_t) ) \geqslant 3t^{0.99}. \end{align*} $$

Let $I_1^{(k)},I_2^{(k)},\ldots ,I_{\lceil t^{0.99}\rceil + 1}^{(k)}$ be the shortest $\lceil t^{0.99}\rceil + 1$ intervals in $\mathcal {I}(n_1,n_2,\ldots ,n_{t-k}) \cap \mathcal {I}_{\text {good}}(n_1,n_2,\ldots ,n_t)$ such that

$$ \begin{align*} |I_1^{(k)}| \leqslant |I_2^{(k)}| \leqslant \cdots \leqslant |I_{\lceil t^{0.99}\rceil + 1}^{(k)}|. \end{align*} $$

Then, by (2.19) and (2.15),

(2.20) $$ \begin{align} |I_{i}^{(k)}| \leqslant \gamma_{2k+i} \quad\text{for } i=1,2,\ldots,\lceil t^{0.99}\rceil + 1. \end{align} $$

In particular, taking (2.18) and (2.17) into consideration,

(2.21) $$ \begin{align} |I_{\lceil t^{0.99}\rceil + 1}^{(k)}| \leqslant \gamma_{t_2} \leqslant \frac{|J|}{t^{0.99}}. \end{align} $$

Let k be in the range (2.18). By Lemma 2.4, for any $i \in \{1,2,\ldots ,\lceil t^{0.99}\rceil \}$ ,

(2.22) $$ \begin{align} (\log p) & \cdot\min_{1 \leqslant j \leqslant t-k}\{\operatorname{ord}_{p}(n_{j}! + f(n_{j}))\} \notag\\ & \leqslant |I_i^{(k)}|\hspace{-1.2pt}\log \bigg( \frac{ex}{|I_{i}^{(k)}|} \bigg) + \bigg(|I_{i+1}^{(k)}|-|I_i^{(k)}| + \frac{x^{0.99}}{t} + O(1)\bigg)\log x. \end{align} $$

Taking the average of (2.22) over $i \in \{1,2,\ldots ,\lceil t^{0.99}\rceil \}$ and noticing that the function $y \mapsto y\log (ex/y)$ is increasing on $y \in (0,x)$ , we deduce that

$$ \begin{align*} (\log p) & \cdot\min_{1 \leqslant j \leqslant t-k}\{\operatorname{ord}_{p}(n_{j}! + f(n_{j}))\} \\ & \leqslant |I_{\lceil t^{0.99}\rceil}^{(k)}|\hspace{-1.2pt}\log \bigg( \frac{ex}{|I_{\lceil t^{0.99}\rceil}^{(k)}|} \bigg) + \bigg( \frac{|I_{\lceil t^{0.99}\rceil + 1}^{(k)}|}{t^{0.99}} + \frac{x^{0.99}}{t} + O(1) \bigg) \log x. \end{align*} $$

Then, bounding $|I_{\lceil t^{0.99}\rceil }^{(k)}|$ by (2.20) and bounding $|I_{\lceil t^{0.99}\rceil + 1}^{(k)}|$ by (2.21), we obtain

(2.23) $$ \begin{align} (\log p) & \cdot\min_{1 \leqslant j \leqslant t-k}\{\operatorname{ord}_{p}(n_{j}! + f(n_{j}))\} \notag\\ & \leqslant \gamma_{2k+\lceil t^{0.99}\rceil} \log \bigg( \frac{ex}{\gamma_{2k+\lceil t^{0.99}\rceil}} \bigg) + \bigg( \frac{|J|}{t^{1.98}} + \frac{x^{0.99}}{t} + O(1) \bigg) \log x. \end{align} $$

Summing (2.23) over k in the range (2.18),

(2.24) $$ \begin{align} \log p & \sum_{k=0}^{t_3} \min_{1 \leqslant j \leqslant t-k}\{\operatorname{ord}_{p}(n_{j}! + f(n_{j}))\} \notag \\ &\leqslant \bigg( \sum_{k=0}^{t_3} \gamma_{2k+\lceil t^{0.99}\rceil} \log \bigg( \frac{ex}{\gamma_{2k+\lceil t^{0.99}\rceil}} \bigg) \bigg) + \bigg( \frac{|J|}{t^{0.98}} + x^{0.99} + O(t) \bigg) \log x \notag \\ &\leqslant \bigg( \sum_{k=0}^{t_3} \gamma_{2k+\lceil t^{0.99}\rceil} \log \bigg( \frac{ex}{\gamma_{2k+\lceil t^{0.99}\rceil}} \bigg) \bigg) + \frac{x\log x}{t^{0.98}} + O(x), \end{align} $$

where in the last inequality, we have used $|J| < x$ and $t \leqslant C_0 x^{{2}/{3}}$ .

Finally, since the function $y \mapsto y\log (ex/y)$ is concave and increasing on $y \in (0,x)$ , and since

$$ \begin{align*} \sum_{k=0}^{t_3} \gamma_{2k+\lceil t^{0.99}\rceil} \leqslant \frac{1}{2} \sum_{k=0}^{t_3} ( \gamma_{2k+\lceil t^{0.99}\rceil} + \gamma_{2k+\lceil t^{0.99}\rceil+1} ) < \frac{1}{2} \sum_{k=1}^{t_1} \gamma_k \leqslant \frac{|J|}{2}, \end{align*} $$

we have (by Jensen’s inequality and monotonicity)

(2.25) $$ \begin{align} \sum_{k=0}^{t_3} \gamma_{2k+\lceil t^{0.99}\rceil} \log \bigg( \frac{ex}{\gamma_{2k+\lceil t^{0.99}\rceil}} \bigg) &< (t_3+1) \cdot \frac{|J|/2}{t_3+1} \log \bigg( \frac{ex}{(|J|/2)/(t_3+1)} \bigg) \notag \\ &< \frac{|J|}{2} \log \bigg( \frac{ext}{|J|/2} \bigg) = \frac{|J|}{2} \log t + \frac{|J|}{2} \log \bigg( \frac{ex}{|J|/2} \bigg) \notag \\ &< \frac{|J|}{2} \log t + x, \end{align} $$

where the last inequality holds because $|J|/2<x$ and the function $y \mapsto y\log (ex/y)$ is increasing on $y \in (0,x)$ .

By (2.24), (2.25) and the fact that $t-t_3 < \lceil ({(1+\varepsilon _0)t}/{2}) \rceil $ (because $t> (10/\varepsilon _0)^{100}$ and $t_3 = \lfloor ( t - 5t^{0.99} )/2 \rfloor - 1$ ), we complete the proof of (2.13).

3 Proof of Theorem 1.1

By the definitions of $B(\lambda )$ and Z (see (2.2)), we have $P(Z) \leqslant \lambda x$ . For any prime number p, we define the set

$$ \begin{align*} N_p := \{ n \in \mathbb{Z} \cap [\varepsilon_0x,x] \mid n \notin B(\lambda) \text{ and } p \mid ({n!}+f(n)) \}. \end{align*} $$

Then,

(3.1) $$ \begin{align} (\log p)\operatorname{ord}_p(Z) = \log p \sum_{n \in N_p} \operatorname{ord}_p({n!}+f(n)). \end{align} $$

If $n \geqslant p$ and $p \mid ({n!}+f(n))$ , then $f(n) \equiv 0 \pmod {p}$ . Let $a \neq 0$ be the leading coefficient of $f(X)$ . If $p>|a|$ , then $f(X) \pmod {p}$ has degree $\deg f$ . Hence, for any two real numbers $y_2>y_1 \geqslant p$ , we have $\# (N_p \cap [y_1,y_2]) \leqslant ((y_2-y_1)/p + 1)\deg f$ . That is,

(3.2) $$ \begin{align} \# (N_p \cap [y_1,y_2]) \ll \frac{y_2-y_1}{p} + 1 \quad\text{for any } y_2>y_1 \geqslant p. \end{align} $$

If $p \leqslant |a|$ , then $\# (N_p \cap [y_1,y_2]) \leqslant y_2-y_1+1 \leqslant |a|(y_2-y_1)/p + 1$ , so (3.2) still holds.

Recall that $|f(n)| \leqslant n^{C_1}$ for all $n \geqslant 2$ . Since $f(n) \neq 0$ for $n \geqslant n_0$ , we have $\operatorname {ord}_p(f(n)) \leqslant C_1 \log n/\log p$ . We can take a large constant $C_2$ depending only on $C_1$ and $n_0$ (so $C_2$ depends only on f) such that

$$ \begin{align*} \begin{cases} C_2 - {C_1} \log C_2/{\log 2}> C_1+1, \\ C_2 > {C_1}/{\log 2}, \\ C_2 > n_0. \end{cases} \end{align*} $$

(For example, $C_2 = 100(C_1+1)^2 + n_0$ is sufficient.) Then, for any prime p and any integer $n \geqslant C_2p$ , we have $n/p-1> C_1 \log n/\log p$ . Since $\operatorname {ord}_p(n!)> n/p - 1$ , we deduce that $\operatorname {ord}_p({n!}+f(n)) = \operatorname {ord}_p(f(n)) = O(\log n/\log p)$ for any prime p and any $n \geqslant C_2p$ . Since $\{ n \in N_p \mid n \geqslant C_2p \} \subset [C_2p,x]$ , it follows that $\#\{ n \in N_p \mid n \geqslant C_2p \} \ll (x/p+1)$ by (3.2). Therefore, for any prime p,

(3.3) $$ \begin{align} \log p \sum_{\substack{n \in N_p \\ n \geqslant C_2p}}\operatorname{ord}_p({n!}+f(n)) \ll \bigg( \frac{x}{p} + 1 \bigg) \cdot \log x \ll x\log x. \end{align} $$

By (3.2), $|N_p \cap [p,C_2p)| = O(1)$ . Since $\log ({n!}+f(n)) = O(x\log x)$ for $n_0 < n \leqslant x$ ,

(3.4) $$ \begin{align} \log p \sum_{\substack{n \in N_p \\ p \leqslant n < C_2p}}\operatorname{ord}_p({n!}+f(n))\ll 1 \cdot x\log x \ll x\log x. \end{align} $$

For any $n \in N_p \cap [1,p)$ , by the definition of $B(\lambda )$ , we have $p \leqslant \lambda n$ . Consider the interval

$$ \begin{align*} J_p := [\max\{p/\lambda, \varepsilon_0 x\}, \min\{ p-1,x \}]. \end{align*} $$

Then,

$$ \begin{align*} N_p \cap [1,p) \subset J_p. \end{align*} $$

We claim that

(3.5) $$ \begin{align} \log p \sum_{\substack{n \in N_p \\ n<p}}\operatorname{ord}_p({n!}+f(n)) \leqslant \frac{1}{9\log (2/(1+\varepsilon_0))}|J_p| \mathrm{log}^2\ x + O(x\log x). \end{align} $$

If $N_p \cap [1,p) = \emptyset $ , then there is nothing to prove. In the following, we assume that $T:= \#(N_p \cap [1,p)) \geqslant 1$ and $N_p \cap [1,p) = \{ n_1,n_2,\ldots ,n_{T} \}$ . Relabelling if necessary, we may assume that

(3.6) $$ \begin{align} \operatorname{ord}_{p}(n_1! + f(n_1)) \geqslant \operatorname{ord}_{p}(n_2! + f(n_2)) \geqslant \cdots \geqslant \operatorname{ord}_{p}(n_T! + f(n_T)). \end{align} $$

If $|J_p| \geqslant 1$ , then Lemma 2.2 implies that $T \leqslant C_0 |J_p|^{{2}/{3}} < C_0 x^{{2}/{3}}$ . If $|J_p| < 1$ , then ${T \leqslant 1 < C_0 x^{{2}/{3}}}$ . Therefore, we always have

(3.7) $$ \begin{align} T < C_0 x^{{2}/{3}}. \end{align} $$

Let $k^{*}$ be the least nonnegative integer such that

$$ \begin{align*} \bigg\lfloor \bigg( \frac{1+\varepsilon_0}{2} \bigg)^{k^*} T \bigg\rfloor < \bigg( \frac{10}{\varepsilon_0} \bigg)^{100}. \end{align*} $$

We denote

$$ \begin{align*} T_k := \bigg( \frac{1+\varepsilon_0}{2} \bigg)^{k} T \quad\text{for } k=0,1,\ldots,k^{*}. \end{align*} $$

Note that

(3.8) $$ \begin{align} k^{*} = \frac{\log T}{\log (2/(1+\varepsilon_0))} + O(1). \end{align} $$

For $0 \leqslant k < k^{*}$ , by Lemma 2.7 (with $t = \lfloor T_k \rfloor $ and $J = J_p$ ) and (3.6),

(3.9) $$ \begin{align} \log p \sum_{i = \lfloor T_{k+1} \rfloor + 1}^{\lfloor T_k \rfloor} \operatorname{ord}_{p}(n_i! + f(n_i)) \leqslant \frac{|J_p|}{2}\log T_k + \frac{x\log x}{\lfloor T_k \rfloor^{0.98}} + O(x). \end{align} $$

Since $\log ({n!}+f(n)) = O(x\log x)$ for $n_0 < n \leqslant x$ and $T_{k^{*}} = O(1)$ ,

(3.10) $$ \begin{align} \log p \sum_{i=1}^{\lfloor T_{k^{*}} \rfloor} \operatorname{ord}_{p}(n_i! + f(n_i)) = O(x\log x). \end{align} $$

Summing (3.9) over $k=0,1,\ldots ,k^{*}-1$ and adding (3.10),

(3.11) $$ \begin{align} \log p & \sum_{i=1}^{T} \operatorname{ord}_{p}(n_i! + f(n_i)) \notag\\ & \leqslant \frac{|J_p|}{2} \sum_{k=0}^{k^{*}-1} \log T_k + x\log x \sum_{k=0}^{k^{*}-1} \frac{1}{\lfloor T_k \rfloor^{0.98}} + O(x\log x) + O(k^{*}x). \end{align} $$

(If $k^{*} = 0$ , the empty sum $\sum _{k=0}^{k^{*}-1}$ is zero.) We estimate each term on the right-hand side of (3.11) as follows. Since

$$ \begin{align*} \sum_{k=0}^{k^{*}-1} \log T_k &= \sum_{k=0}^{k^{*}-1} \bigg( \log T - k\log\bigg( \frac{2}{1+\varepsilon_0} \bigg) \bigg) = k^{*}\log T - \frac{(k^{*}-1)k^{*}}{2} \log \bigg( \frac{2}{1+\varepsilon_0} \bigg) \\ &= \frac{1}{2\log(2/(1+\varepsilon_0))}\log^2 T + O(\log T) \quad (\text{by (3.8)}) \\ & \leqslant \frac{2}{9\log(2/(1+\varepsilon_0))} \log^{2} x + O(\log x) \quad (\text{by (3.7)}), \end{align*} $$

we have

$$ \begin{align*} \frac{|J_p|}{2} \sum_{k=0}^{k^{*}-1} \log T_k \leqslant \frac{1}{9\log(2/(1+\varepsilon_0))} |J_p|\hspace{-1.2pt}\log^2 x + O(x\log x). \end{align*} $$

Since $T_k$ decreases exponentially with respect to k,

$$ \begin{align*} \sum_{k=0}^{k^{*}-1} \frac{1}{\lfloor T_k \rfloor^{0.98}} \ll \frac{1}{\lfloor T_{k^{*}-1} \rfloor^{0.98}} \ll 1. \end{align*} $$

By (3.8) and (3.7),

$$ \begin{align*} k^{*}x = O(x\log x). \end{align*} $$

By substituting these estimates into (3.11), we deduce that

$$ \begin{align*} \log p \sum_{i=1}^{T} \operatorname{ord}_{p}(n_i! + f(n_i)) \leqslant \frac{1}{9\log (2/(1+\varepsilon_0))}|J_p|\hspace{-1.2pt}\log^2 x + O(x\log x), \end{align*} $$

which is exactly the statement in (3.5) that we claimed.

Combining (3.1), (3.3), (3.4) and (3.5),

(3.12) $$ \begin{align} (\log p) \operatorname{ord}_p(Z) \leqslant \frac{1}{9\log (2/(1+\varepsilon_0))}|J_p| \mathrm{log}^2\ x + O(x\log x), \end{align} $$

which holds for any prime number p. Since $P(Z) \leqslant \lambda x$ and $|J_p| \leqslant \min \{ p,x \} - p/\lambda $ , summing (3.12) over primes $p \leqslant \lambda x$ , we obtain

$$ \begin{align*} \log Z \leqslant \frac{1}{9\log (2/(1+\varepsilon_0))} \bigg( \sum_{p \leqslant x} (1-1/\lambda)p + \sum_{x<p\leqslant \lambda x} (x - p/\lambda) \bigg) \log^2 x + O(x^2), \end{align*} $$

where for the error term, we have used $\pi (\lambda x) \ll x/\mathrm{log}\ x$ . Now, we note that because $\pi (y) = (1+o(1))y/\mathrm{log}\ y$ and

$$ \begin{align*} \sum_{p \leqslant y} p = \bigg( \frac{1}{2}+o(1) \bigg) \frac{y^2}{\log y} \quad \text{as } y \rightarrow +\infty, \end{align*} $$

we can deduce that

(3.13) $$ \begin{align} \log Z \leqslant \bigg( \frac{1}{9\log (2/(1+\varepsilon_0))}\frac{\lambda - 1}{2} + o(1) \bigg)\, x^2 \log x \quad \text{as } x \rightarrow +\infty. \end{align} $$

By comparing (2.3) with (3.13),

$$ \begin{align*} \frac{1}{9\log (2/(1+\varepsilon_0))}\frac{\lambda - 1}{2} \geqslant \frac{1}{2} - 2\varepsilon_0, \end{align*} $$

which implies that

$$ \begin{align*} \lambda \geqslant 1+9(1-4\varepsilon_0)\log\bigg( \frac{2}{1+\varepsilon_0} \bigg). \end{align*} $$

However, this contradicts the assumptions $\lambda = 1+9\log 2 - 100\varepsilon _0$ and $\varepsilon _0 \in (0,1/100)$ .

Therefore, the lower asymptotic density of $B(\lambda )$ must be greater than or equal to $\varepsilon _0$ . The proof of Theorem 1.1 is complete.