2.1. Defining the model

The letter z will denote a fixed non-negative real number throughout Section 2. As usual, we denote $\exp\!(z)\;:\!=\;\textrm{e}^z$ . For a prime p and a positive integer m, we define

\begin{equation*}\delta_{p,z}(m) \;:\!=\;\begin{cases} 1, & \text{ if } p\mid m \text{ and } m \text{ is not divisible by any prime in } (p,p^{\exp\!(z) }],\\[5pt] 0, & \text{ otherwise.} \end{cases}\end{equation*}

In particular, $\omega_z(m) = \sum_{p} \delta_{p,z}(m)$ , where the sum is over all primes. Our plan, initially, is to follow the Kubilius model idea (see Billingsley [Reference Billingsley1, equations (1.8),(1.9)]) to define Bernoulli random variables $B_p$ that model the behaviour of $\delta_{p,z}$ . For this, we use the random variables $X_p$ as follows: for every prime p the random variable $X_p$ is defined so that

\begin{equation*} P[X_p=1]=\frac{1}{p} ,\quad P[X_p=0]=1-\frac{1}{p} \end{equation*}

and such that $X_p$ are independent. In particular, the mean $E[X_p]$ equals $1/p$ , thus, $X_p=1$ models the event that a random integer m is divisible by a fixed prime p. Let $$[\,\cdot\,]$$ denote the integer part. The independence of $X_p$ is related to the Chinese Remainder Theorem.

To model $\delta_{p,z}$ , we must also take into account the fact that each prime q in the range $(p, p^{\exp\!(z)}]$ must not divide m. Thus, we are naturally led to define

(2.1) \begin{equation}{B_p\;:\!=\;X_p \prod_{\substack{ q \textrm{ prime } \\ p<q\leqslant p^{\exp\!(z)} } }(1-X_q).}\end{equation}

We will later prove that $\sum_pB_p$ is a good model for $\omega_z=\sum_p\delta_{p,z}$ in the sense that their moments agree asymptotically.

2.2. Distribution and moments of the model via Stein’s method

For any positive N, we define

\begin{equation*}S_N=\sum_{p\leqslant N } B_p\end{equation*}

and denote its expectation and variance, respectively, by

\begin{equation*}c_{N}\;:\!=\;E\!\left[S_{N}\right] \textrm{ and }\; s_{N}^{2}\;:\!=\;\textrm{Var}\!\left[S_{N}\right].\end{equation*}

Our goal in this section is to prove that $(S_N-c_N)/s_N$ converges in law to the standard normal distribution as $N\to \infty $ and that its moments are asymptotically Gaussian. This will be done, respectively, in Propositions 2.5 and 2.7. We first need a few preparatory estimates.

Lemma 2.2. We have

(2.2) \begin{equation}{E[B_p]=\frac{\textrm{e}^{-z} }{p}+O\left( \frac{1}{p \log p }\right) ,}\end{equation}

(2.3) \begin{equation}{c_N=\textrm{e}^{-z}\log \log N+O(1)}\end{equation}

and

(2.4) \begin{equation}{ s_N^2 =\left(1-\frac{2z}{ \textrm{e}^z} \right) \frac{\log \log N}{\textrm{e}^z} +O(1) .}\end{equation}

Proof. Recall that Mertens’ theorem states that $\sum_{p\leqslant T} 1/p= \log \log T+c+O(1/\log T)$ for some constant c. The independence of $X_p$ yields

\begin{equation*}E[B_p]=\frac{1}{p} \prod_{p<q\leqslant p^{\exp\!(z) }} \left(1-\frac{1}{q} \right) ,\end{equation*}

which, by the approximation $1-\varepsilon=\exp\!(\!-\!\varepsilon+O(\varepsilon^2)) $ for $| \varepsilon| \leqslant 1 $ and Mertens’ theorem is

\begin{equation*} \frac{1}{p}\exp\!\left(-\sum_{p<q\leqslant p^{\exp\!(z) }}\frac{1}{p} +O\left(\sum_{p<q\leqslant p^{\exp\!(z) }} \frac{1}{p^2}\right)\right)=\frac{\exp\!(\!-\!\log \log p^{\exp\!(z)}+\log \log p +O(1/\log p ))}{p}.\end{equation*}

Since $\exp\!\left(O(1/\log p )\right)=1+O(1/\log p)$ , this is sufficient for (2.2). The estimate (2.3) is directly deduced from it and the fact that $\sum_p \!(p \log p)^{-1} $ converges. Next, denoting $h_p=E[B_p]$ we have

\begin{equation*}s_N^2= \sum_{p\leqslant N} E\!\left[(B_p-h_p)^2\right] +2\sum_{p< q \leqslant N} E\!\left[(B_p-h_p)(B_q-h_q) \right] .\end{equation*}

First note that $ E\!\left[(B_p-h_p)^2\right]= E\!\left[B_p \right]-h_p^2=h_p(1-h_p) $ . Further, if $q>p^{\exp\!(z)}$ then $B_p$ and $B_q$ are independent, hence, $E\!\left[(B_p-h_p)(B_q-h_q) \right]=0$ . If $p<q\leqslant p^{\exp\!(z)}$ then $E[B_p B_q]$ vanishes, hence

\begin{equation*}E\!\left[(B_p-h_p)(B_q-h_q) \right]=-E\!\left[B_p \right]h_q-h_pE\!\left[B_q \right]+ h_p h_q=-h_ph_q .\end{equation*}

We obtain

\begin{align*}s_N^2&= \sum_{p\leqslant N} h_p(1-h_p) -2 \sum_{\substack{ p< q \leqslant \min \{ N , p^{\exp\!(z)} \}} } h_p h_q \\ &=c_N-\sum_{p\leqslant N} h_p^2-2 \sum_{\substack{ p \leqslant N^{\exp\!(-z)} \\ p< q \leqslant p^{\exp\!(z)} } } h_p h_q-2 \sum_{\substack{ N^{\exp\!(-z)}< p \leqslant N \\ p< q \leqslant N } } h_p h_q.\end{align*}

By (2.2) we have $h_p\ll 1/p$ , hence, $\sum_{p } h_p^2 =O(1)$ and

\begin{equation*} \sum_{\substack{ N^{\exp\!(-z)}< p \leqslant N \\ p< q \leqslant N } } h_p h_q\ll\left( \sum_{ N^{\exp\!(-z)}< p \leqslant N } \frac{1}{p}\right)^2 =O(1) .\end{equation*}

Hence, (2.3) gives

(2.5) \begin{equation}{s_N^2=\textrm{e}^{-z}\log \log N-2 \sum_{ p \leqslant N^{\exp\!(-z)} }h_p \sum_{ p< q \leqslant p^{\exp\!(z)} } h_q+O(1).}\end{equation}

Using (2.2) we see that

\begin{equation*}\sum_{ p \leqslant N^{\exp\!(-z)} }h_p \sum_{ p< q \leqslant p^{\exp\!(z)} } h_q=\sum_{ p \leqslant N^{\exp\!(-z)} }h_p \sum_{ p< q \leqslant p^{\exp\!(z)} }\! \left(\frac{\textrm{e}^{-z} }{q}+O\left(\frac{1}{q\log q }\right)\right),\end{equation*}

which, by Mertens’ theorem and $\sum_{q>t}(q\log q)^{-1} \ll (\!\log t )^{-1}$ , equals

\begin{align*}\sum_{ p \leqslant N^{\exp\!(-z)} }h_p \left(\frac{z}{\textrm{e}^z }+O\left(\frac{1}{\log p} \right) \right)&=\sum_{ p \leqslant N^{\exp\!(-z)} } \left(\frac{\textrm{e}^{-z} }{p}+O\left(\frac{1}{p\log p} \right)\right) \left(\frac{z}{\textrm{e}^z }+O\left(\frac{1}{\log p} \right) \right)\\[5pt]&=\frac{z}{\textrm{e}^{2z} }\left(\sum_{ p \leqslant N^{\exp\!(-z)} } \frac{1}{p}\right)+O (1)=\frac{z}{\textrm{e}^{2z} }(\!\log \log N)+O (1).\end{align*}

Injecting this into (2.5) concludes the proof.

Lemma 2.3. For all $u\in \mathbb{N}, \textbf{r}\in \mathbb{N}^u$ and primes $p_1,\ldots,p_u$ , we have

\begin{equation*} E\!\left[\prod_{i=1}^u |B_{p_i} - E[B_{p_i}]|^{r_i} \right]=O_{\textbf{r} }\!\left(\frac{1}{ \textrm{rad} (p_1\cdots p_u )}\right),\end{equation*}

where rad denotes the radical.

Proof. We write the factorisation into prime powers of $\prod_{i=1}^ u p_i^{r_i} $ as $\prod_{j=1}^{v} q_j^{s_j}$ , where $q_j$ are v distinct primes. This implies that

\begin{equation*}E\!\left[\prod_{i=1}^u |B_{p_i} - E[B_{p_i}]|^{r_i} \right]=E\!\left[\prod_{j=1}^v |B_{q_j} - E[B_{q_j}]|^{s_j} \right].\end{equation*}

Using $|B_{q_j} - E[B_{q_j}]|\leqslant B_{q_j} +E[B_{q_j}]\leqslant X_{q_j} +E[X_{q_j}]=X_{q_j} +1/{q_j}$ and the binomial theorem yields

\begin{equation*}|B_{q_{j}} - E[B_{q_{j}}]|^{s_j}\leqslant\left(X_{q_{j}} +1/{q_{j}} \right)^{s_{j}}=\sum_{t_{j}\in [0,s_{j}]}\! \left(\begin{array}{c}{s_j}\\[2pt] {t_j}\end{array}\right) \frac{X_{q_j}^{t_j} }{q_j^{s_j-t_j} } \ll_{\textbf{s} }\max_{t_j\in [0,s_j] } \frac{X_{q_j}^{t_j} }{q_j^{s_j-t_j} },\end{equation*}

hence,

\begin{equation*}E\!\left[\prod_{j=1}^v |B_{q_j} - E[B_{q_j}]|^{s_j} \right] \ll_{\textbf{s} }\max_{\textbf{t} \in [0,s_1] \times \cdots \times [0,s_v] } E\!\left[\prod_{j=1}^v \frac{X_{q_j}^{t_j} }{q_j^{s_j-t_j} } \right].\end{equation*}

By the independence of the $X_q$ , we infer that

\begin{equation*} E\!\left[\prod_{j=1}^v \frac{X_{q_j}^{t_j} }{q_j^{s_j-t_j} } \right]= \prod_{j=1}^v \frac{ E [ X_{q_j}^{t_j} ] }{q_j^{s_j-t_j} }=\prod_{\substack{ j=1 \\ t_j =0 }}^v \frac{ 1 }{q_j^{s_j } }\prod_{\substack{ j=1 \\ t_j \geqslant 1 }}^v \frac{ E [ X_{q_j} ] }{q_j^{s_j-t_j} }\leqslant\prod_{\substack{ j=1 \\ t_j =0 }}^v \frac{ 1 }{q_j }\prod_{\substack{ j=1 \\ t_j \geqslant 1 }}^v E [ X_{q_j} ]=\prod_{ j=1 }^v \frac{ 1 }{q_j }. \end{equation*}

The proof now concludes by noting that $\prod_{ j=1 }^v q_j$ is the radical of $\prod_{i=1}^ u p_i^{r_i} $ .

The following lemma is the main tool in the proof of Theorem 1.1. It is due to Stein [Reference Stein18, Corollary 2, p. 110].

Lemma 2.4 (Stein). Let T be a finite set, and for each $t \in T$ , let $Z_t$ be a real random variable and $T_t$ a subset of T such that $E[Z_t]=0$ , $E[Z_t^4]<\infty$ and $E[\sum_{t\in T }Z_t \sum_{s\in T_t} Z_s]=1$ . Then for all real b,

(2.6) \begin{equation}{\left|P\left[\sum_{t\in T} Z_t\leqslant b\right] - \frac{1}{\sqrt{2\pi}}\int_{-\infty}^b\textrm{e}^{-t^2/2}\textrm{d} t\right|\leqslant4 (\Psi_1+\Psi_2+\Psi_3),}\end{equation}

where the terms $\Psi_i$ are defined through

\begin{equation*}\Psi_1 = E\!\left[\sum_{t \in T}\left|E\!\left[Z_t | Z_s, s\notin T_t\right]\right|\right],\Psi_2^2 = E\!\left[\sum_{t\in T}|Z_t|\left(\sum_{s\in T_t} Z_s \right)^2\right]\end{equation*}

and

\begin{equation*}\Psi_3^2=E\!\left[\left\{\sum_{t\in T}\sum_{s\in T_t}(Z_t Z_s-E[Z_s Z_t])\right\}^2\right].\end{equation*}

Proposition 2.5. Fix $z\geqslant 0 $ and $b \in \mathbb{R}$ . For any $N\in \mathbb{N}$ , we have

\begin{equation*}\left|P\left[ S_N\leqslant c_N +bs_N\right]- \frac{1}{\sqrt{2\pi}}\int_{-\infty}^b\textrm{e}^{-t^2/2}\textrm{d} t\right|\ll_z (\!\log \log N)^{-1/4} ,\end{equation*}

where the implied constant depends at most on z. In particular, $ (S_N- c_N)/s_N $ converges in law to the standard normal distribution as $N\to \infty$ .

Proof. We will apply Lemma 2.4 with

• T being the set of primes in [2, N],

• $T_p$ being the set of primes in $[p^{\exp\!(-z) },p^{\exp\!(z) }] \cap [2,\,N]$ ,

• $Z_p=(B_p - E[B_p] )/s_N$ for $p\in T$ .

Let $Y_p\;:\!=\;B_p -E[B_p]$ . Note that if $q \notin T_p$ then $Z_q$ and $Z_p$ are independent, hence, $E[Y_pY_q]=0$ . Therefore,

\begin{equation*}s_N^2 =\sum_{p,q\leqslant N }E[Y_p Y_q ] =\sum_{\substack{p\leqslant N \\ q\in T_p }}E[Y_p Y_q ] ,\end{equation*}

which verifies $E\!\left[\!\sum_{p\in T }Z_p \sum_{q\in T_p} Z_q\right]=1$ . We next observe that since for every $q \notin T_p$ , the random variables $Z_q$ and $Z_p$ are independent; one obtains $E\!\left[Z_p | Z_q, q\notin T_t\right]=E\!\left[Z_p \right]=0,$ therefore

(2.7) \begin{equation}{\Psi_1=0.}\end{equation}

Next, we use Lemma 2.3 to obtain

\begin{align*}\Psi_2^2 s_N^{3} =&\sum_{\substack{p\leqslant N ,q\in T_p }} E\!\left[| Y_p |Y_{q} ^2\right]+ 2 \sum_{\substack{ p\leqslant N, q_1<q_2 \in T_p}} E\!\left[| Y_p | Y_{q_1} Y_{q_2}\right]\\[5pt]\ll&\sum_{\substack{p\leqslant N, q\in T_p}} \frac{1}{p q}+\sum_{\substack{p\leqslant N, q_1,q_2 \in T_p}}\frac{1}{p q_1 q_2 }.\end{align*}

The sum $\sum_{q\in T_p }1/q $ is bounded only in terms of z by Mertens’ theorem. It shows that

(2.8) \begin{equation}{ \Psi_2^2 \ll s_N^{-3} \sum_{p\leqslant N} \frac{1}{p} \ll (\!\log \log N)^{-1/2},}\end{equation}

owing to (2.4).

To bound $\Psi_3$ , we write $\mathcal{C}_{p}\;:\!=\;\sum_{q\in T_{p}}\!\left(Y_{p}Y_{q}-E[ Y_pY_q ]\right)$ to obtain

(2.9) \begin{equation}{\Psi_3^2s_N^{4}= \sum_{p \leqslant N }E\!\left[\mathcal{C}_p^2 \right]+2\sum_{p_1 < p_2 \leqslant N }E\!\left[\mathcal{C}_{p_1}\mathcal{C}_{p_2}\right].}\end{equation}

Furthermore, $E\!\left[\mathcal{C}_p^2 \right]$ can be written as

\begin{equation*}\sum_{q\in T_p}E\!\left[\left(Y_pY_q-E[ Y_pY_q ]\right)^2\right]+2\sum_{q_1 <q_2 \in T_p }E\!\left[\left(Y_pY_{q_1}-E[ Y_pY_{q_1} ]\right)\left(Y_pY_{q_2}-E[ Y_pY_{q_2} ]\right)\right],\end{equation*}

which can be seen to be

\begin{equation*}\ll\sum_{q\in T_p}\frac{1}{pq}+\sum_{q_1 <q_2 \in T_p }\frac{1}{pq_1 q_2}\end{equation*}

by Lemma 2.3. Alluding to $\sum_{q\in T_p }1/q \ll 1 $ shows that

(2.10) \begin{equation}{ \sum_{p \leqslant N }E\!\left[\mathcal{C}_p^2 \right]\ll \sum_{p\leqslant N} \frac{1}{p}\ll \log \log N.}\end{equation}

Let us now observe that if $p_2>p_1^{\exp\!(2z)}$ then $T_{p_1} \cap T_{p_2} =\emptyset $ , therefore $\mathcal{C}_{p_1}$ and $\mathcal{C}_{p_2}$ are independent. Since for every p we have $E[\mathcal{C}_p]=0$ by definition, we get $E\!\left[\mathcal{C}_{p_1}\mathcal{C}_{p_2}\right]=\prod_{i=1}^2E\!\left[\mathcal{C}_{p_i}\right] =0$ . Thus,

(2.11) \begin{equation}{\sum_{p_1 < p_2 \leqslant N } E\!\left[ \mathcal{C}_{p_1} \mathcal{C}_{p_2} \right]= \sum_{\substack{ p_1 < p_2 \leqslant N \\ p_2\leqslant p_1^{\exp\!(2z)} } }\sum_{\substack{ q_1\in T_{p_1} \\ q_2\in T_{p_2}}} E\!\left[ \left( Y_{p_1} Y_{q_1} - E[ Y_{p_1} Y_{q_1} ] \right) \left( Y_{p_2} Y_{q_2} - E[ Y_{p_2} Y_{q_2} ] \right) \right].}\end{equation}

By Lemma 2.3, this is

\begin{equation*}\ll \sum_{ p_1 \leqslant N }\sum_{p_1 < p_2 \leqslant p_1^{ \exp\!(2z)}}\sum_{ q_1\in T_{p_1} }\sum_{ q_2\in T_{p_2} } \frac{1}{ \textrm{rad}(p_1p_2 q_1 q_2) }. \end{equation*}

For any positive integer c and prime q, we have $\textrm{rad}(c q)=\textrm{rad}(c)\frac{q}{\gcd\!(q,c)}$ . Hence, the sum over $q_2 $ is

\begin{equation*} \frac{1}{ \textrm{rad}( p_1 p_2 q_1 ) }\sum_{\substack{ q_2\in T_{p_2} \\ q_2\in \{p_1,p_2,q_1 \} } } 1+ \frac{1}{ \textrm{rad}( p_1 p_2 q_1) }\sum_{\substack{ q_2 \in T_{p_2} \\ q_2\notin \{p_1,p_2,q_1 \} } } \frac{1}{q_2 }\leqslant \frac{3+\sum_{q\in T_{p_2} } 1/q}{ \textrm{rad}( p_1 p_2 q_1) }\ll_z \frac{1}{ \textrm{rad}( p_1 p_2 q_1) }\end{equation*}

by Mertens’ theorem. Hence, (2.11) is

\begin{equation*}\ll\sum_{ \substack{ p_1 \leqslant N \\ p_1 < p_2 \leqslant p_1^{ \exp\!(2z)} } } \sum_{ q_1\in T_{p_1} } \frac{1}{ \textrm{rad}( p_1 p_2 q_1) }=\sum_{ \substack{ p_1 \leqslant N \\ p_1 < p_2 \leqslant p_1^{ \exp\!(2z)} } }\frac{1}{ \textrm{rad}( p_1 p_2 ) } \left\{\sum_{ \substack{ q_1\in T_{p_1} \\ q_1 \in \{p_1,p_2\} } } 1+\sum_{ \substack{ q_1\in T_{p_1} \\ q_1 \notin \{p_1,p_2\} } } \frac{1}{q_1}\right\}.\end{equation*}

The two sums over $q_1 $ in the right-hand side are both bounded only in terms of z. This can be proved similarly as before with the sum over $q_2$ . We obtain the bound

\begin{equation*}\ll\sum_{ \substack{ p_1 \leqslant N \\ p_1 < p_2 \leqslant p_1^{ \exp\!(2z)} } }\frac{1}{ \textrm{rad}( p_1 p_2 ) }=\sum_{ p_1 \leqslant N }\frac{1}{p_1 }\sum_{ p_1 < p_2 \leqslant p_1^{ \exp\!(2z)} } \frac{1}{p_2 }\ll \sum_{ p_1 \leqslant N }\frac{1}{p_1 } \ll \log \log N.\end{equation*}

This shows that the quantity in (2.11) is $\ll \log \log N$ , which, when combined with (2.10), can be fed into (2.9) to yield $\Psi_3^2 s_N^{4} \ll \log \log N$ . Invoking (2.4) provides us with $\Psi_3 \ll1/\sqrt{\log \log N}$ . Together with (2.7)–(2.8), it implies that

\begin{equation*}\left|P\left[ S_N\leqslant c_N +bs_N\right]- \frac{1}{\sqrt{2\pi}}\int_{-\infty}^b\textrm{e}^{-t^2/2}\textrm{d} t\right| \ll_z (\!\log \log N)^{-1/4} \end{equation*}

owing to Stein’s bound (2.6). Finally, letting $N\to \infty$ shows that $(S_N-c_N)/s_N$ converges in law to the standard normal distribution.

Proposition 2.7. Fix $z\geqslant 0 $ and a positive integer r. Then we have

\begin{equation*}\lim_{N\to \infty}E\!\left[\left(\frac{S_N-c_N}{s_N}\right)^r \right]=\mu_r,\end{equation*}

where $\mu_r$ is the r-th moment of the standard normal distribution.

Proof. Take 2k to be the least strictly positive integer with $r<2k $ , so that Proposition 2.5 [Reference van der Vaart19, Example 2.21] implies that it suffices to prove that

\begin{equation*}\sup_{N\geqslant 1 }\left|E\!\left[\left(\frac{S_N-c_N}{s_N}\right)^{2k } \right]\right|\end{equation*}

is bounded only in terms of k and z. Equivalently, by (2.4) it suffices to show

\begin{equation*} E\!\left[\left( S_N-c_N \right)^{2k } \right] = E\!\left[\left(\sum_{p\leqslant N} (B_p-E[B_p] )\right)^{2k} \right] \ll_{k,z} (\!\log \log N)^{k} .\end{equation*}

The left side equals

\begin{equation*}\sum_{u=1}^{2k } \sum_{\substack{ \textbf{r} \in \mathbb{N}^u \\ 2k=\sum_{i=1}^u r_i }} \frac{(2k ) !}{r_1! \cdots r_u! }\sum_{p_1<\ldots < p_u\leqslant N} E\!\left[\prod_{i=1}^u (B_p-E[B_p])^{r_i} \right].\end{equation*}

Using Lemma 2.3, we see that the contribution of the terms with $ u \leqslant k $ is

\begin{equation*}\ll_k\max_{1\leqslant u \leqslant k }\left(\sum_{p\leqslant N}\frac{1}{p}\right)^u\ll_k (\!\log \log N)^{k }.\end{equation*}

Therefore,

(2.12) \begin{equation}{E\!\left[\left( S_N-c_N \right)^{2k } \right] \ll \max_{\substack{u\in [k+1,2k] \\ \textbf{r} \in \mathbb{N}^u\,:=\,\sum_i r_i =2k } } \sum_{p_1<\ldots < p_u\leqslant N} E\!\left[\prod_{i=1}^u (B_p-E[B_p])^{r_i} \right]+ (\!\log \log N)^{k} ,}\end{equation}

with an implied constant that is independent of N.

For given $u \in \mathbb{N} $ , $z\geqslant 0 $ and primes $p_1<\ldots < p_u$ , we say that two consecutive integers $ i, i+1$ in [1, u] are linked if and only if $p_{i+1}\leqslant p_{i}^{\exp\!(z)}$ . In particular, $p_{i+1}$ lies in a relatively small interval; hence, its contribution will be small. Denote the number of linked pairs $(i,i+1)$ by $\ell(\textbf{p} ) $ . By Lemma 2.3, we obtain

\begin{equation*}\sum_{p_1<\ldots < p_u\leqslant N} E\!\left[\prod_{i=1}^u (B_p-E[B_p])^{r_i} \right]\ll_z\left( \sum_{p\leqslant N} \frac{1}{p}\right)^{u-\ell(\textbf{p} ) }\ll (\!\log \log N)^{u-\ell(\textbf{p} ) },\end{equation*}

where we used the estimate $\sum_{p_i<p_{i+1} <p_i^{\exp\!(z)}}1/p_i \ll_z 1 $ whenever i and $i+1$ are linked. Hence, the contribution of all prime vectors $(p_1,\ldots, p_u)$ with at least $ \ell(\textbf{p} )\geqslant u-k $ linked pairs is at most

\begin{equation*} \ll (\!\log \log N)^{u-\ell(\textbf{p} ) } \ll (\!\log \log N)^{ k },\end{equation*}

which is acceptable. By (2.12), we obtain

(2.13) \begin{equation}{E\!\left[\left( S_N-c_N \right)^{2k } \right] \ll \max_{\substack{u\in [k+1,2k] \\ \textbf{r} \in \mathbb{N}^u\,:=\,\sum_i r_i =2k } }\ \sum_{\substack{ p_1<\ldots < p_u\leqslant N\\ \ell(\textbf{p} ) < u-k } } E\!\left[\prod_{i=1}^u (B_p-E[B_p])^{r_i} \right]+ (\!\log \log N)^{k} ,}\end{equation}

We will now show that every sum over $p_i$ in (2.13) vanishes. Denoting the cardinality of $1\leqslant i\leqslant u $ with $r_i=1$ by a, we see that the number of i with $r_i \geqslant 2$ is $u-a$ . Since $2k =\sum_{i=1}^u r_i$ , we get $2k \geqslant a+ 2(u-a)$ . Equivalently, $ 2 (u-k)\leqslant a $ , hence, by $ \ell(\textbf{p} ) < u-k$ one gets

(2.14) \begin{equation}{2\ell(\textbf{p} ) < \sharp\{ i \in [1,u]\;:\; r_i=1 \} .}\end{equation}

We now partition the integers in [1, u] into disjoint subsets ${\mathcal{A}}_1,\ldots,{\mathcal{A}}_r$ using the following rules:

• if i and $i+1$ are in $S_j$ then they are linked,

• if $i \in S_a$ and $i+1 \in S_b$ for some $a\neq b$ then i and $i+1$ are not linked.

The inequality $s\leqslant 2(\!-\!1+s)$ for $s\geqslant 2$ gives

\begin{equation*}\sharp\{ i \in [1,u]\;:\; i \textrm{ linked to some index}\}=\sum_{\substack{1\leqslant j \leqslant r \\ 2\leqslant \sharp{\mathcal{A}}_j}} \sharp{\mathcal{A}}_j\leqslant\sum_{\substack{1\leqslant j \leqslant r \\ 2\leqslant \sharp{\mathcal{A}}_j}}2(\!-\!1+\sharp{\mathcal{A}}_j ) .\end{equation*}

This equals $2\ell (\textbf{p} ) $ since each ${\mathcal{A}}_j$ has $-1+\sharp{\mathcal{A}}_j$ linked pairs and the total number of links is $\ell(\textbf{p} ) $ . By (2.14), we infer that there exists an index j for which $r_j=1$ and that is not linked to any other index. This implies that the following random variables are independent:

\begin{equation*} \prod_{\substack{ 1\leqslant i \leqslant u \\ i \neq j } }(B_{p_{i}} - E[B_{p_{i}}] )^{r_i}\ \ \ \textrm{ and } \ \ (B_{p_{j}} - E[B_{p_{j}}] )^{r_{j}} = B_{p_{j}} - E[B_{p_{j}}] .\end{equation*}

Since $ E\!\left[ B_{p_j} - E[B_{p_j}] \right]=0$ , we infer that every expectation in the right-hand side of (2.13) vanishes. This concludes the proof.

2.3. Justifying the model

Let n be a positive integer and denote by $\Omega_n$ the uniform probability space $\mathbb{N}\cap [1,n]$ . Our goal now becomes to show that, as $n\to\infty $ , the moments of $\omega_z(m)$ for m in $\Omega_n$ are asymptotically the same as the moments of $S_N$ for some parameter $N=N(n)\to\infty$ . Recall (2.1). For technical reasons, we will first work with a truncated version of $\omega_z$ , namely,

(2.15) \begin{equation}{\omega_{z,N}(m)=\sum_{p\leqslant N}\delta_{p,z}(m),}\end{equation}

where $N=N(n)$ . The function $\delta_{p,z}$ imposes simultaneous coprimality conditions of m with several primes in large intervals, and to deal with this, we shall need the Fundamental Lemma of Sieve Theory [Reference Friedlander and Iwaniec8, Corollary 6.10].

Lemma 2.8 (Fundamental Lemma of Sieve Theory). Let ${\mathcal{P}}$ be a set of primes. Given any sequence $a_m\geqslant 0 $ for $m \in \mathbb{N}$ and any square-free $d\leqslant x $ that is only divisible by primes in ${\mathcal{P}}$ , we assume that

\begin{equation*} \sum_{\substack{m\leqslant x \\ m\equiv 0 \left(\textrm{mod}\ d\right)}} a_m=X g(d)+r_d \end{equation*}

for some real numbers $X, r_d$ and a multiplicative function g. Assume that $0\leqslant g (p)<1 $ and that there exist constants $K>1, \kappa>0$ such that

\begin{equation*} \prod_{\substack{w\leqslant p < y\\ p\in{\mathcal{P}} }} (1-g(p))^{-1}\leqslant K \left(\frac{\log y}{\log w } \right)^\kappa\end{equation*}

holds for all $2\leqslant w < y $ . Then for all $D\geqslant y \geqslant 2 $ , we have

(2.16) \begin{equation}{\sum_{\substack{m\leqslant x \\ p\in{\mathcal{P}}, p<y \Rightarrow p\nmid m }}a_m=X \left(\prod_{\substack{p<y \\ p\in{\mathcal{P}}}}(1-g(p))\right) \{1+O(\textrm{e}^{-s})\}+O\left(\sum_{\substack{ d < D \\ p \mid d \Rightarrow p\in {\mathcal{P}}}}\mu^2(d) |r_d |\right) ,}\end{equation}

where $s=\log D/\log y$ and the implied constants depend at most on $\kappa$ and K.

Lemma 2.9. Assume that there exists a function $N\;:\;[1,\infty)\to [1,\infty)$ satisfying

(2.17) \begin{align} \lim_{n\to\infty}N(n)=+\infty, \end{align}

(2.18) \begin{align} \limsup_{n\to\infty}\frac{(\!\log N(n))(\!\log \log \log N(n))}{\log n }\neq +\infty. \end{align}

Fix $z\geqslant 0 $ and $k\in \mathbb{N}$ . Then we have

\begin{equation*}\lim_{n\to\infty} E_{m\in \Omega_n}\left[\left(\frac{\omega_{z,N}-c_N}{s_N}\right)^k\right] =\mu_k,\end{equation*}

where $\mu_k $ is the k-th moment of the standard normal distribution.

Proof. By Proposition 2.7 and (2.17), it is sufficient to prove

(2.19) \begin{equation}{ \lim_{n\to\infty} \left( E_{m\in \Omega_n}\left[\left(\frac{\omega_{z,N}(m)-c_N}{s_N}\right)^k\right] - E \left[\left(\frac{S_N-c_N}{s_N}\right)^k\right] \right) =0.}\end{equation}

Let $r\in \mathbb{N}$ . By (2.15), the fact that $\delta_{p,z}\in \{0,1\}$ and the binomial theorem, we obtain

(2.20) \begin{equation}{E_{ m\in \Omega_n}\left[\omega_{z,N}(m)^r \right] =\sum_{u=1}^r \sum_{\substack{ r_1,\ldots, r_u \in \mathbb{N} \\ r_1+\ldots +r_u=r }} \frac{r!}{r_1!\cdots r_u!} \sum_{p_1 < \cdots <p_u \leqslant N } E_{m\in \Omega_n} \left[ \delta_{p_1,z}(m) \cdots \delta_{p_u,z}(m) \right] .}\end{equation}

Let ${\mathcal{P}}$ be the set of all primes in $\bigcup_{i=1}^u \!(p_i, p_i^{\exp\!(z)}]$ and let $a_m $ be the indicator function of integers divisible by $p_1\cdots p_u$ . In particular,

\begin{equation*}E_{m\in \Omega_n} \!\left[ \delta_{p_1,z}(m) \cdots \delta_{p_u,z}(m) \right] =\frac{1}{n}\sum_{\substack{ 1\leqslant m \leqslant n\\ p\in {\mathcal{P}} \Rightarrow p\nmid m } }a_m .\end{equation*}

We assume that $p_{i+1}>p_i^{\exp\!(z)}$ for all $i=1,2,\ldots, u-1$ since otherwise the sum clearly vanishes. We will now use Lemma 2.8 with $ X=n/(p_1\cdots p_u), g(d)=1/d, D=\sqrt n, y= N^{2\exp\!(z)}.$ If d is divisible only by primes in ${\mathcal{P}}$ , then it is coprime to $p_1\cdots p_u$ , hence,

\begin{equation*} \sum_{\substack{m\leqslant n \\ m\,\equiv\,0 \left(\textrm{mod}\ d\right) }} a_m = \left[\frac{n}{p_1\cdots p_u d }\right],\end{equation*}

thus, $|r_d|\leqslant 1 $ because $ r_d $ is the fractional part of $X/d$ . Furthermore, we can take K to be any large fixed positive constant and $\kappa =1$ , owing to

\begin{equation*}\prod_{\substack{w\leqslant p < y \\ p\in {\mathcal{P}} }} (1-g(p))^{-1} =\prod_{\substack{w\leqslant p < y \\ p\in {\mathcal{P}} }} (1-1/p)^{-1}\leqslant \prod_{w\leqslant p < y } (1-1/p)^{-1} \ll \frac{\log y}{\log w}.\end{equation*}

The bound $|r_d|\leqslant 1 $ , means that $\sum_{d\leqslant D} \mu^2(d) |r_d| \leqslant D=\sqrt n$ . Since $p_u\leqslant N$ , every prime in ${\mathcal{P}}$ is strictly smaller than y, hence, (2.16) gives

(2.21) \begin{equation}{E_{m\in \Omega_n} \left[ \delta_{p_1} \cdots \delta_{p_u} \right] = \left\{ \prod_{i=1}^u \frac{1}{p_i} \prod_{p_i<q\leqslant p_i^{\exp\!(z)}}\left( 1-1/p \right)\right\} \left\{1+O\left(\textrm{e}^{-\frac{\log n}{4\exp\!(z)\log N}}\right)\right\} +O(n^{-1/2}),}\end{equation}

where the implied constant depends at most on r and z.

By the binomial theorem, we get

\begin{equation*} E\!\left[S_N^r \right]=E\!\left[\left(\sum_{p\leqslant N}B_p \right)^r \right]=\sum_{u=1}^r \sum_{\substack{ r_1,\ldots, r_u \in \mathbb{N} \\ r_1+\ldots +r_u=r }} \frac{r!}{r_1!\cdots r_u!} \sum_{p_1 < \cdots <p_u \leqslant N }E \left[ B_{p_1} \cdots B_{p_u} \right]\end{equation*}

and we note that we can restrict the sum over $p_i$ to the terms with $p_{i+1}>p_i^{\exp\!(z)}$ for all i, since otherwise $E \left[ B_{p_1} \cdots B_{p_u} \right]=0$ . Under this restriction, the random variables $B_{p_i}$ are independent, hence,

\begin{equation*} \prod_{i=1}^u \frac{1}{p_i} \prod_{p_i<q\leqslant p_i^{\exp\!(z)}}\left( 1-1/p \right) = E \left[ B_{p_1} \cdots B_{p_u} \right].\end{equation*}

We infer from (2.20) and (2.21) that

\begin{equation*}\left| E_{ m\in \Omega_n}\left[\omega_{z,N}(m)^r \right]-E\!\left[S_N^r \right] \right|\ll_r E\!\left[S_N^r \right] \textrm{e}^{-\frac{\log n}{4\exp\!(z)\log N}}+n^{-1/2} \sum_{u=1}^r\sum_{p_1<\ldots < p_u\leqslant N }1 .\end{equation*}

By (2.3), this is $\ll(\!\log \log N)^r \exp\!(\!-\!\frac{\log n}{4\exp\!(z)\log N})+n^{-1/2}N^r$ . Thus, the difference in (2.19) is

\begin{align*}\ll&s_N^{-k}\sum_{r=0}^k \left(\begin{array}{c}{k}\\[2pt] {r}\end{array}\right) (\!-\!c_N)^{k-r}\left( E_{ \Omega_n}\left[\omega_{z,N} ^r \right]-E\!\left[S_N^r \right] \right)\\[5pt] \ll&s_N^{-k}\left\{\textrm{e}^{-\frac{\log n}{4\exp\!(z)\log N}} (c_N+\log \log N)^k +n^{-1/2} (N+c_N)^k\right\}.\end{align*}

We need to show that this vanishes asymptotically, and by (2.4) and (2.17), it suffices to show

\begin{equation*} (2\log \log N)^k\leqslant \exp\!\left( \frac{\log n}{4\exp\!(z)\log N}\right)\ \ \textrm{ and } \ (2N)^{k} \leqslant n^{1/2}.\end{equation*}

Both of these inequalities can be directly inferred from (2.17) to (2.18).

Lemma 2.10. Assume that there exists a function $N\;:\;[1,\infty)\to [1,\infty)$ satisfying

(2.22) \begin{align} \lim_{n\to\infty} \frac{\log \log N(n)}{\log \log n }=1, \end{align}

(2.23) \begin{align} \limsup_{n\to\infty}\frac{(\!\log N(n))\sqrt{ \log \log n}}{\log n }= +\infty. \end{align}

Fix $z\geqslant 0 $ . Then ${s_N} ( (1-2z\textrm{e}^{-z}) {\textrm{e}}^{-z}\log \log n )^{-1/2} \to 1$ as $n\to \infty $ and

\begin{equation*}\lim_{n\to \infty} \frac{ \max \!\left\{ \left| ( \omega_{z}-\textrm{e}^{-z}\log \log n ) - ( \omega_{z,N}-c_N ) \right| :\;m \in \mathbb{N} \cap[1,n] \right\} } {\sqrt{\log \log n } }=0 .\end{equation*}

Proof. Combining (2.4) and (2.22) one immediately gets

\begin{equation*}\lim_{n\to\infty } \frac{s_N}{ ( (1-2z\textrm{e}^{-z}) {\textrm{e}}^{-z}\log \log n )^{1/2} } = 1. \end{equation*}

For any $m \in [1,n]$ , we have

\begin{equation*} \left| ( \omega_{z}-\textrm{e}^{-z}\log \log n ) - ( \omega_{z,N}-c_N ) \right|\leqslant\sum_{p>N}\delta_{p,z}(m)+|\textrm{e}^{-z}(\!\log \log n)-c_N |. \end{equation*}

Since $\delta_{p,z}$ takes only values in $\{0,1\}$ and $\delta_{p,z}(m)=1$ implies that p divides m, we see that

\begin{equation*}\sum_{p>N}\delta_{p,z}(m) \leqslant \sharp\{p\mid m \;:\; p>N\} \leqslant \frac{\log m}{\log N} \leqslant \frac{\log n}{\log N} .\end{equation*}

Furthermore, (2.3) gives

\begin{equation*} \textrm{e}^{-z}(\!\log \log n)-c_N \ll 1+ \log\frac{\log n }{\log N}\ll \frac{\log n }{\log N}.\end{equation*}

The proof now concludes by using (2.23).

2.4. Proof of Theorem 1.1

The function

\begin{equation*}N(n)\;:\!=\; n^{1/\log \log n}\end{equation*}

fulfills (2.17)–(2.18)–(2.22)–(2.23). Hence, we can apply Lemmas 2.9–2.10.

For any $r\in \mathbb{N}$ , $c\in \mathbb C$ any probability space $\Omega_n$ and any two sequences of random variables $X_n,Y_n$ satisfying $\lim_{n\to\infty } \sup_{m \in \Omega_n}|X_n(m)-Y_n(m) |=0$ and $\lim_{n\to \infty} E_{m\in \Omega_n} [X_n(m)^r]=c$ it is easy to see by the binomial theorem that $\lim_{n\to \infty} E_{m\in \Omega_n}[Y_n(m)^r]=c$ . Using this with $ \Omega_n=\mathbb{N}\cap[1,n] $ ,

\begin{equation*}X_n(m)= \frac{\omega_{z,N}(m)-c_N}{s_N}\ \ \textrm{ and } \ \ Y_n(m)=\frac{\omega_{z}(m)-\textrm{e}^{-z}\log \log n}{s_N},\end{equation*}

in combination with Lemmas 2.9–2.10, shows that for every $k \in \mathbb{N} $ one has

\begin{equation*}\lim_{n\to\infty} E_{m\in \Omega_n}\left[\left(\frac{\omega_{z}(m)-\textrm{e}^{-z}\log \log n}{s_N}\right)^k\right] =\mu_k.\end{equation*}

Given any sequence $a_n \in \mathbb{R}$ having limit 1 and any sequence of random variables $X_n$ with $E[X_n] $ having limit c, it is clear that $a_n E[X_n]$ has limit c. Using this with

\begin{equation*}a_n= \frac{s_N}{ ( (1-2z\textrm{e}^{-z}) {\textrm{e}}^{-z}\log \log n )^{1/2} }\ \ \textrm{ and } \ \ X_n(m)=\frac{\omega_{z}(m)-\textrm{e}^{-z}\log \log n}{s_N}\end{equation*}

and invoking Lemma 2.10 shows that for every $k\in \mathbb{N}$ one has

(2.24) \begin{equation}{\lim_{n\to\infty} E_{m\in \Omega_n}\left[\left(\frac{\omega_{z}(m)-\textrm{e}^{-z}\log \log n}{ ( (1-2z\textrm{e}^{-z}) {\textrm{e}}^{-z}\log \log n )^{1/2} }\right)^k\right] =\mu_k .}\end{equation}

This proves Theorem 1.1 whenever r is a positive integer and this is sufficient. To see that, take any $r\in [0,\infty)$ and note that (2.24) implies that

\begin{equation*} T_n=\frac{\omega_{z}(m)-\textrm{e}^{-z}\log \log n}{ ( (1-2z\textrm{e}^{-z}) {\textrm{e}}^{-z}\log \log n )^{1/2} }\end{equation*}

converges in law to the standard normal distribution. Taking p to be the least even integer strictly exceeding r in [Reference van der Vaart19, Example 2.21] shows that the r-th moment of $T_n$ converges to the r-th moment of the standard normal distribution.

3.1. Proof of Theorem 3.2

For a prime p, we define

\begin{equation*}\sigma_p\;:\!=\;\frac{\sharp\big\{x \in \mathbb{P}^n(\mathbb{F}_{p})\;:\; f^{-1}(x) \mbox{ is non-split}\big\}}{\sharp\mathbb{P}^n(\mathbb{F}_p)},\end{equation*}

where we use the term “non-split” in the sense of Skorobogatov [Reference Skorobogatov17, Def. 0.1]. We then introduce the random variable $\widetilde{X}_p$ so that

\begin{equation*} P[\widetilde X_p=1]=\sigma_p , P[\widetilde X_p=0]=1-\sigma_p \end{equation*}

and such that $\widetilde X_p$ are independent. We then define

\begin{equation*}\widetilde B_p\;:\!=\;\widetilde X_p \prod_{\substack{ q \textrm{ prime } \\ p<q\leqslant p^{\exp\!(z)} } }(1-\widetilde X_q).\end{equation*}

Furthermore, for any positive N, we define

\begin{equation*}\widetilde S_N=\sum_{p\leqslant N } \widetilde B_p, \ \ \ \ \widetilde c_N\;:\!=\;E\!\left[\widetilde S_N\right]\ \ \textrm{ and } \ \ \widetilde s_N^2\;:\!=\;\textrm{Var}\!\left[\widetilde S_N\right].\end{equation*}

Using [Reference Loughran and Sofos14, Proposition 3.6] instead of Mertens’ theorem and the estimate $\sigma_p \ll 1/p$ from [Reference Loughran and Sofos14, Lemma 3.3], the arguments in Lemma 2.2 can be modified to yield

(3.1) \begin{equation}{E[\widetilde B_p]=\exp\!( \!-\!z\Delta(f) ) \sigma_p +O\left( \frac{1}{p \log p }\right) ,}\end{equation}

(3.2) \begin{equation}{\widetilde c_N=\Delta(f)\exp\!( \!-\!z\Delta(f) )\log \log N+O(1)}\end{equation}

and

(3.3) \begin{equation}{ \widetilde s_N^2 =\left(1-\frac{2\Delta(f)z}{ \exp\!(\Delta(f)z)} \right) \frac{\Delta(f)\log \log N}{ \exp\!(\Delta(f)z)} +O(1) .}\end{equation}

Next, the proof of Lemma 2.3 goes through easily upon replacing $B_p $ by $\widetilde B_p$ owing to the inequality $E[\widetilde B_p]\leqslant E[\widetilde X_p] =\sigma_p \ll 1/p$ . Replacing $S_N$ by $\widetilde S_N$ in the statement of Proposition 2.5, we see that the proof goes through by replacing $Z_p$ by $\widetilde Z_p\;:\!=\;(\widetilde B_p-E[\widetilde B_p])/\widetilde s_N$ . Finally, using all the analogues of results in Section 2 that we mentioned so far allows one to modify the arguments of the proof of Proposition 2.7 to obtain the following result:

Proposition 3.5. Fix $z\geqslant 0 $ and a positive integer r. Then we have

\begin{equation*} \lim_{N\to \infty}E\!\left[\left(\frac{\widetilde S_N-\widetilde c_N}{\widetilde s_N}\right)^r \right] =\mu_r,\end{equation*}

where $\mu_r$ is the r-th moment of the standard normal distribution.

This concludes the probabilistic part of the proof of Theorem 3.2. The number–theoretic part requires the Fundamental lemma of sieve theory and the following:

Proposition 3.6. Keep the setting of Theorem 3.2. Then there exist constants $\delta >1, A>0$ that depends on V and f with the following property. Let $Q \in \mathbb{N}$ with $p \nmid Q$ for all $p \leqslant A$ . Then for all $\varepsilon>0$ and $Q\leqslant B^{1/6}$ , we have

\begin{equation*}\sharp\left\{x\in \mathbb{P}^n(\mathbb{Q})\;:\; \begin{array}{l} H(x)\leqslant B, f^{-1}(x) \, smooth\\[5pt] f^{-1}(x)(\mathbb{Q}_p)=\emptyset \ \forall \ p\mid Q \end{array} \right\} = c_n B^{n+1}\prod_{p \mid Q} \sigma_{p} + O \left( \frac{\delta^{\omega(Q)} B^{n+1}}{Q\min \{p\mid Q\} } \right), \end{equation*}

where the implied constant is independent of B and Q.

Proof. By [Reference Loughran and Sofos14, Proposition 3.4], there exist $\alpha >0 , d\in \mathbb{N} $ such that the left-hand side is at most

\begin{equation*} c_nB^{n+1} \Big(\prod_{p \mid Q} (\sigma_{p} + \alpha/p^2)\Big) + O\left( (4d)^{\omega(Q)} ( Q^{2n+1} B + Q B^n(\!\log B)^{[1/n]} ) \right). \end{equation*}

while, it exceeds a similar quantity with $\alpha $ replaced by $-\alpha$ . As shown in [Reference Loughran and Sofos14, Lemma 3.7], we have

\begin{equation*} \prod_{p \mid Q} (\sigma_{p} + \alpha/p^2)= \prod_{p \mid Q} \sigma_{p} + O\left( \frac{(2\alpha d)^{\omega(Q)}}{Q \min\{p\;:\; p\mid Q\}} \right) .\end{equation*}

This is satisfactory by defining $\delta=2+\max\{4d, 2\alpha d\}$ . Finally,

\begin{equation*} Q^{2n+1} B + Q B^{n} \log B \ll \frac{B^{n+1}}{Q^2} \leqslant \frac{B^{n+1}}{Q \min\{p\mid Q\}}\end{equation*}

owing to $Q\leqslant B^{1/6}$ .

Our next task is to show that the moments of a truncated version of $\omega_{f,z}$ are asymptotically Gaussian. For this we shall follow the arguments in Section 2.3, where $\Omega_n =\mathbb{N}\cap [1,n]$ is replaced by the uniform discrete probability space

\begin{equation*}\widetilde \Omega_B=\{x\in \mathbb{P}^n(\mathbb{Q})\;:\; H(x) \leqslant B, f^{-1}(x) \textrm{ smooth}\}\end{equation*}

for $B>0$ . The condition that $f^{-1}(x) \textrm{ smooth}$ is included in the definition of $\widetilde \Omega_B$ to make $\omega_{f,z}(x)$ well-defined for each $x\in \widetilde \Omega_B$ . Choosing a polynomial which vanishes on the singular locus of f, we see that

\begin{equation*}\sharp\{x\in \mathbb{P}^n(\mathbb{Q})\;:\; H(x) \leqslant B, f^{-1}(x) \textrm{ not smooth}\}=O(B^n).\end{equation*}

Then the standard result

\begin{equation*} \sharp\{x\in \mathbb{P}^n(\mathbb{Q})\;:\; H(x) \leqslant B\}=c_nB^{n+1}+O(B^n(\!\log B)^{[1/n]}) ,\end{equation*}

where $c_n=2^n/\zeta(n+1)$ , shows that

\begin{equation*} \sharp\widetilde \Omega_B=c_n B^{n+1}+O\!\left(B^n (\!\log B)^{[1/n] }\right) .\end{equation*}

We furthermore let for $x\in \mathbb{P}^n(\mathbb{Q})$ ,

\begin{equation*}\widetilde \delta_{p,z}(x) \;:\!=\;\begin{cases} 1, & \text{ if } f^{-1}(x)(\mathbb{Q}_p)=\emptyset \text{ and } f^{-1}(x)(\mathbb{Q}_q)\neq \emptyset \text{ for every prime } q\in (p,p^{\exp\!(z) }],\\[5pt] 0, & \text{ otherwise.} \end{cases}\end{equation*}

We shall choose any two functions $t_0,t_1\;:\;(0,\infty )\to (0,\infty )$ satisfying

(3.4) \begin{equation}{1<t_0(B)<t_1(B)<B,\lim_{B\to\infty} t_0(B)=\lim_{B\to\infty} t_1(B)=\infty.}\end{equation}

They will be chosen optimally later. The analogue of (2.15) in our setting is defined as

(3.5) \begin{equation}{\widetilde \omega_{z,B}(x)=\sum_{t_0(B)<p\leqslant t_1(B)}\widetilde \delta_{p,z}(x).}\end{equation}

We obtain for $r\in \mathbb{N}$ ,

(3.6) \begin{equation}{E_{ x\in \widetilde \Omega_B}\left[\widetilde \omega_{z,B}(x)^r \right] =\sum_{u=1}^r \sum_{\substack{ r_1,\ldots, r_u \in \mathbb{N} \\ r_1+\ldots +r_u=r }} \frac{r!}{r_1!\cdots r_u!}\sum_{\substack{ t_0(B)< p_1 < \cdots <p_u \leqslant t_1(B)\\ p_{i+1} >p_i^{\exp\!(z) } \ \forall i}} E_{ x\in \widetilde \Omega_B}\!\left[\widetilde \delta_{p_1,z}(x) \cdots \widetilde\delta_{p_u,z}(x) \right] ,}\end{equation}

where we added the assumption $p_{i+1} > p_i^{\exp\!(z) } \ \forall i$ since otherwise the expectation in the right-hand side vanishes.

Let us now define the function $m_B\;:\;\mathbb{P}^n(\mathbb{Q})\to \mathbb{N}$ given by

\begin{equation*}m_B(x)\;:\!=\;\prod_{\substack{t_0(B)<p \leqslant t_1(B) \\ f^{-1}(x)(\mathbb{Q}_p)=\emptyset }} p.\end{equation*}

Letting $\widetilde x$ be the product of all primes $p\leqslant t_1(B)$ we note that $m_B(x) \leqslant \widetilde x$ . Now let ${\mathcal{P}}$ be the set of all primes in $\bigcup_{i=1}^u \!(p_i, p_i^{\exp\!(z)}]$ and for $m\in \mathbb{N}$ let

\begin{equation*}\widetilde a_m\;:\!=\; \begin{cases} \sharp\{x\in \widetilde\Omega_B\;:\; m_B(x)=m\}/\sharp\widetilde \Omega_B, & \text{ if } m\equiv 0 \left(\textrm{mod}\ {p_1\cdots p_u}\right),\\[5pt] 0, & \text{ otherwise.} \end{cases}\end{equation*}

This gives

\begin{equation*} E_{ x\in \widetilde \Omega_B}\!\left[\widetilde \delta_{p_1,z}(x) \cdots \widetilde\delta_{p_u,z}(x) \right]= \sum_{ \substack{ m\leqslant \widetilde x \\ p\in {\mathcal{P}} \Rightarrow p\nmid m }}\widetilde a_m .\end{equation*}

We shall use Lemma 2.8 with $a_m \;:\!=\;\widetilde a_m , \kappa=\Delta(f) $ ,

\begin{equation*} X= \prod_{i=1}^u \sigma_{p_i} ,\ \ g(d)=\prod_{p\mid d }\sigma_p, \ \ D=B^{1/10}, \ \ y= t_1(B)^{2\exp\!(z)}. \end{equation*}

The assumption $0\leqslant g(p)<1$ is satisfied here due to $\sigma_p\ll 1/p$ and $p>t_0(B) \to \infty $ . Note that for square-free d that is only divisible by primes in ${\mathcal{P}}$ , we have

\begin{equation*}r_d=-g(d)X + \sum_{\substack{m\leqslant \widetilde x \\ m\,\equiv\,0 \left(\textrm{mod}\ d\right) }} \widetilde a_m .\end{equation*}

Assuming that

(3.7) \begin{equation}{\log t_1(B)=o(\!\log B)}\end{equation}

we see that when $d\leqslant D$ , one has

(3.8) \begin{equation}{dp_1\cdots p_u \leqslant d t_1(B)^u \leqslant Dt_{1}(B)^u=B^{\frac{1}{10}+\frac{\log t_1(B)}{\log B}}\leqslant B^{1/6}}\end{equation}

for all large B. This allows us to employ Proposition 3.6 with $Q=dp_1\cdots p_u$ to obtain

\begin{equation*}\sum_{\substack{m\leqslant \widetilde x \\ m\equiv 0 \left(\textrm{mod}\ d\right) }} \widetilde a_m =\frac{c_nB^{n+1} X g(d) }{\sharp\widetilde\Omega_B}+O\left(\frac{\delta^{u+\omega(d) }B^{n+1} }{\sharp\widetilde\Omega_B d p_1^2 p_2 \cdots p_u }\right)=Xg(d) +O_u\left(\frac{(\!\log B)^{[1/n]}}{B}+\frac{\delta^{ \omega(d) } }{ d p_1^2 \prod_{i=2}^u p_i }\right),\end{equation*}

where we used $\sharp\widetilde\Omega_B=c_n B^{n+1}+O(B^n(\!\log B)^{[1/n]})$ and $X g(d) \ll 1 $ . The inequality (3.8) shows that

\begin{equation*}\frac{ d p_1^2 \prod_{i=2}^u p_i }{\delta^{ \omega(d) } }\leqslant \left(d \prod_{i=1}^u p_i \right)^2 \leqslant B^{1/3} \leqslant \frac{B}{\log B},\end{equation*}

hence,

\begin{equation*}r_d=-Xg(d)+\sum_{m\leqslant \widetilde x, m\equiv 0 \left(\textrm{mod}\ d\right) }\widetilde a_m \ll \frac{\delta^{ \omega(d) } }{ d p_1^2 \prod_{i=2}^u p_i }.\end{equation*}

This shows that the error term occurring in (2.16) is

\begin{equation*}\ll\frac{X}{ \textrm{e}^{s}}\prod_{\substack{p<y \\ p\in{\mathcal{P}}}}(1-\sigma_p)+ \sum_{\substack{d\leqslant B \\ p\mid d \Rightarrow p\in {\mathcal{P}} }} \frac{|\mu(d)| \delta^{ \omega(d) } }{ d p_1^2 \prod_{i=2}^u p_i }\leqslant\frac{X}{ \textrm{e}^{s}}+\frac{1}{p_1^2 \prod_{i=2}^u p_i } \prod_{p\in {\mathcal{P}} } \left(1+\frac{1}{p}\right)^\delta .\end{equation*}

The product over $p\in {\mathcal{P}}$ equals

\begin{equation*}\prod_{i=1}^{u} \prod_{p_i < p \leqslant p_i ^{\exp\!(z) } } \left(1+\frac{1}{p}\right)^\delta\ll \prod_{i=1}^{u}\left(\frac{\log \!(p_i ^{\exp\!(z) })}{\log p_i }\right)^\delta \ll 1 .\end{equation*}

Furthermore, the estimates $p_1 >t_0(B)$ and $X\ll 1/(p_1\cdots p_u) $ show that

\begin{equation*}\frac{X}{ \textrm{e}^{s}}+\frac{1}{p_1^2 \prod_{i=2}^u p_i } \prod_{p\in {\mathcal{P}} } \left(1+\frac{1}{p}\right)^\delta \ll \frac{1}{p_1\cdots p_u } \frac{1}{\min \{ \textrm{e}^{s} , t_0(B) \} } .\end{equation*}

The main term occurring in Lemma 2.8 is

\begin{equation*}X\prod_{p\in {\mathcal{P}} }(1-\sigma_p)=\prod_{i=1}^u\left(\sigma_{p_i} \prod_{p_i< p \leqslant p_i^{\exp\!(z) }} (1-\sigma _p ) \right),\end{equation*}

hence, the expectation $E_{ x\in \widetilde \Omega_B}\left[\widetilde \delta_{p_1,z}(x) \cdots \widetilde\delta_{p_u,z}(x) \right]$ in the right-hand side of (3.6) equals

\begin{equation*}\prod_{i=1}^u\left(\sigma_{p_i} \prod_{p_i< p \leqslant p_i^{\exp\!(z) }} (1-\sigma _p ) \right)+O\left(\frac{1}{p_1\cdots p_u } \frac{1}{\min \{ \textrm{e}^{s} , t_0(B) \} }\right) .\end{equation*}

Injecting this into (3.6) produces the error term

\begin{equation*}\ll_r\frac{1}{\min \{ \textrm{e}^{s} , t_0(B) \} }\sum_{u=1}^r \sum_{ p_1 < \cdots <p_u \leqslant t_1(B) } \frac{1}{p_1\cdots p_u } \ll_r \frac{(\!\log \log t_1(B) )^r}{\min \{ \textrm{e}^{s} , t_0(B) \} } .\end{equation*}

Following arguments similar to the ones in the proof of Lemma 2.8, the main term is

\begin{equation*} \sum_{u=1}^r \sum_{\substack{ r_1,\ldots, r_u \in \mathbb{N} \\ r_1+\ldots +r_u=r }} \frac{r!}{r_1!\cdots r_u!}\sum_{\substack{ t_0(B)< p_1 < \cdots <p_u \leqslant t_1(B)\\ p_{i+1} >p_i^{\exp\!(z) } \ \forall i}}\prod_{i=1}^u \sigma_{p_i} \prod_{p_i< p \leqslant p_i^{\exp\!(z) }} (1-\sigma _p ) =E\!\left[ T_B^r\right],\end{equation*}

where

\begin{equation*} T_B\;:\!=\;\sum_{t_0(B)< p \leqslant t_1(B) } \widetilde B_p.\end{equation*}

We have shown that for all $r\in \mathbb{N}$ one has

\begin{equation*}\left|E_{ x\in \widetilde \Omega_B}\!\left[\widetilde \omega_{z,B}(x)^r\right]-E\!\left[ T_B^r\right] \right|\ll_r \frac{(\!\log \log t_1(B) )^r}{\min \{ \textrm{e}^{s} , t_0(B) \} } .\end{equation*}

Noting that $T_B= \widetilde S_{t_1(B)}-\widetilde S_{t_0(B)}$ gives

\begin{equation*}E\!\left[ T_B^r\right]=E\!\left[ \widetilde S_{t_1(B)}^r\right]+O_r\!\left(\max_{0\leqslant k \leqslant r-1 } E\!\left[ \widetilde S_{t_0(B)}^{r-k} \widetilde S_{t_1(B)}^k\right]\right).\end{equation*}

and the Cauchy–Schwarz inequality shows that

\begin{equation*}E\!\left[ T_B^r\right]=E\!\left[ \widetilde S_{t_1(B)}^r\right]+O_r\left(\max_{0\leqslant k \leqslant r-1 } E\!\left[ \widetilde S_{t_0(B)}^{2( r-k ) } \right]^{1/2} E\!\left[ \widetilde S_{t_1(B)}^{2 k } \right]^{1/2}\right).\end{equation*}

Since $0\leqslant \widetilde B_p\leqslant \widetilde X_p $ , we infer that $0\leqslant \widetilde S_N \leqslant \sum_{p\leqslant N } \widetilde X_p$ , hence,

\begin{equation*} E\!\left[ \widetilde S_{N}^r\right] \leqslant E\!\left[ \left( \sum_{p\leqslant N } \widetilde X_p \right)^r\right]=\sum_{u=1}^r \sum_{\substack{ r_1,\ldots, r_u \in \mathbb{N} \\ r_1+\ldots +r_u=r }} \frac{r!}{r_1!\cdots r_u!}\sum_{\substack{ p_1 < \cdots <p_u \leqslant N }} \prod_{i=1}^u E[\widetilde X_p^{r_i} ]. \end{equation*}

But $E[\widetilde X_p^{r_i} ] = E[\widetilde X_p ] =\sigma_p\ll 1/p$ , hence, $E\!\left[ \widetilde S_{N}^r\right] \ll (\!\log \log N)^r $ . Hence,

\begin{equation*}\max_{0\leqslant k \leqslant r-1 }E\!\left[ \widetilde S_{t_0(B)}^{2( r-k ) } \right]^{1/2}E\!\left[ \widetilde S_{t_1(B)}^{2 k } \right]^{1/2}\ll\max_{0\leqslant k \leqslant r-1 } (\!\log \log t_0(B))^{r-k} (\!\log \log t_1(B) )^k, \end{equation*}

which is $\ll (\!\log \log t_0(B)) (\!\log \log t_1(B) )^{r-1}.$ Hence,

\begin{equation*}\left|E_{ x\in \widetilde \Omega_B}\left[\widetilde \omega_{z,B}(x)^r\right]-E\!\left[ \widetilde S_{t_1(B)}^r\right] \right|\ll_r \frac{(\!\log \log t_1(B) )^r}{\min \!\left\{ \frac{\log \log t_1(B) }{\log \log t_0(B)},\textrm{e}^{s} , t_0(B) \right\} } .\end{equation*}

Therefore,

\begin{equation*} \left| E_{ \widetilde \Omega_B}\left[\left(\frac{\widetilde \omega_{z,B}(x)-\widetilde c_{t_1(B)} }{\widetilde s_{t_1(B)} }\right)^k\right] - E \left[\left(\frac{S_N-\widetilde c_{t_1(B)} }{\widetilde s_{t_1(B)} }\right)^k\right] \right|\end{equation*}

is

\begin{equation*} \ll (\widetilde s_{t_1(B)})^{-k}\sum_{r=0}^k \left|\widetilde c_{t_1(B)} \right| ^{k-r}\left( E_{ \widetilde \Omega_B}\left[\widetilde \omega_{z,B} ^r \right]-E\!\left[S_{\widetilde t_1(B)}^r \right] \right),\end{equation*}

which, by (3.2)–(3.3) is

\begin{equation*} \ll (\!\log \log t_1(B) )^{-r} \frac{(\!\log \log t_1(B) )^r}{\min \!\left\{ \frac{\log \log t_1(B) }{\log \log t_0(B)},\textrm{e}^{s} , t_0(B) \right\} }\ll\frac{1}{\min \!\left\{ \frac{\log \log t_1(B) }{\log \log t_0(B)},\textrm{e}^{s} , t_0(B) \right\} }. \end{equation*}

This vanishes asymptotically as long as we assume that

(3.9) \begin{equation}{\log \log t_0(B) =o(\!\log \log t_1(B)).}\end{equation}

This is due to (3.7) which implies that

\begin{equation*}s=\frac{\log D}{\log y }=\frac{1}{20 \exp\!(z)} \frac{\log B}{\log t_1(B)}\to+\infty.\end{equation*}

We have therefore shown that, subject to (3.4)–(3.7)–(3.9), one has

\begin{equation*}E_{ \widetilde \Omega_B}\left[\left(\frac{\widetilde \omega_{z,B}(x)-\widetilde c_{t_1(B)} }{\widetilde s_{t_1(B)} }\right)^k\right] \to \mu_k. \end{equation*}

The concluding arguments follow those in Lemma 2.10, the only difference being dealing with primes $p\leqslant t_0(B)$ . Recall from [Reference Loughran and Sofos14, Lemma 3.2, part (2)] that there exists a constant $A>0$ and a homogeneous $F\in \mathbb{Z}[x_0,\ldots, x_n]$ (both of which depend only on f) with the property that for all primes p and $x\in \mathbb{P}^n(\mathbb{Q})$ with $f^{-1}(x) $ smooth and $f^{-1}(x) (\mathbb{Q}_p)=\emptyset$ , one has $p\mid F(x)$ . Then

\begin{equation*}0\leqslant \omega_{f, z}(x)- \widetilde \omega_{ z,B}(x) \leqslant \sum_{p\leqslant t_0(B) } 1+\sum_{\substack{p> t_1(B) \\ f^{-1}(x) (\mathbb{Q}_p)=\emptyset}} 1 \leqslant t_0(B) + \sharp\{p\mid F(x)\;:\; p>t_1(B) \} .\end{equation*}

For $z>1$ and $m \in \mathbb N$ , we have $\sharp\{p\mid m \;:\; p>z\} \leqslant (\!\log m )/(\!\log z)$ . For $x\in \widetilde \Omega_B$ , we have $H(x)\leqslant B$ , thus, $\log |F(x)|\ll \log B$ . In particular,

\begin{equation*} \omega_{f, z}(x)= \widetilde \omega_{ z,B}(x) +O\left(t_0(B)+\frac{\log B}{\log t_1(B)}\right),\end{equation*}

where the implied constant is independent of B, z and x. Combined with arguments similar to the ones in Lemma 2.10, we obtain

\begin{equation*}\lim_{B\to \infty} \frac{ \max \!\left\{ \left| ( \omega_{f, z}(x)-\Delta(f)\textrm{e}^{-z\Delta(f) }\log \log B ) - ( \widetilde \omega_{ z,B}(x)-\widetilde c_{t_1(B)} ) \right| \;:\;x\in \widetilde \Omega_B\right\} } {\sqrt{\log \log B } }=0 ,\end{equation*}

as long as

(3.10) \begin{equation}{t_0(B)=o\!\left(\sqrt{\log \log B } \right)\ \ \textrm{ and } \ \ \frac{\log B}{\log t_1(B)}=o\!\left(\sqrt{\log \log B } \right).}\end{equation}

The proof of Theorem 3.2 concludes by adapting the arguments in Section 2.4 to the current setting. This can be achieved as long as we assume that

(3.11) \begin{equation}{\frac{\log \log t_1(B)}{\log \log B}\to 1\ \ \textrm{ and } \ \ \log \log B-\log \log t_1(B)=o( \sqrt{\log \log B} ) }\end{equation}

and it now remains to find functions $t_0(B)$ and $ t_1(B)$ that satisfy all assumptions (3.4)–(3.7)–(3.9)–(3.10)–(3.11). This can be done by choosing $t_0(B)$ and $t_1(B)$ so that

\begin{equation*}t_0(B)=\log \log \log B\ \ \textrm{ and } \ \ \frac{ \log t_1(B) }{\log B}=\frac{\log \log \log B}{\sqrt{\log \log B}}.\end{equation*}