2 Setting the table

Let $\wp $ be the set of all primes. From here on, unless indicated otherwise, the letters p, q and $\pi $ will stand for primes, whereas $\pi (x)$ will stand for the number of primes not exceeding x. Also, we will be using the logarithmic integral ${\mbox {li}(x):= \int _{2}^{x} {dt}/{\log t}}$ . Finally, given a real number $x\ge e^{e^{e}}$ , we set

(2.1) $$ \begin{align} Y_{1}: = Y_{1}(x)= \exp\{ (\log x)^{\varepsilon(x)} \} \quad \mbox{and} \quad Y_{2} := Y_{2}(x)= \exp\{ (\log x)^{1-\varepsilon(x)} \}, \end{align} $$

where

$$ \begin{align*}\varepsilon(x):= \frac 12 \frac{\log \log \log x}{\log \log x}. \end{align*} $$

Let ${\cal B}$ be a subset of $\wp $ whose counting function $B(x):=\#\{p\le x: p\in {\cal B}\}$ is such that, for some real number $\beta>0$ ,

(2.2) $$ \begin{align} B(x) = \beta\,\mbox{li}(x) + O \bigg( \frac x{\log^{3} x} \bigg), \end{align} $$

so that in particular, for some constant $\beta _{2}$ ,

(2.3) $$ \begin{align} \sum_{\substack{p\le x \\ p\in {\cal B}}} \frac 1p & = \beta \log\log x + \beta_{2} + O \bigg( \frac x{\log x} \bigg),\\ \sum_{\substack{u<p<v\\ p\in {\cal B}}} \frac{\log p}p & = \beta \log \frac vu +O(1), \nonumber\\ \prod_{\substack{u<p<v\\ p\in {\cal B}}} \bigg( 1 - \frac 1p \bigg) & = \bigg( \frac{\log u}{\log v} \bigg)^{\,\beta} \bigg(1+O\bigg( \frac 1{\log u} \bigg) \bigg).\nonumber \end{align} $$

Any such set ${\cal B}$ satisfying (2.2) will be called a B-set.

Observe that given a B-set and its associated function ${\omega _{\cal B} (n):= \sum _{p\mid n,\, p\in {\cal B}} 1}$ , it is immediate that

$$ \begin{align*} \sum_{n\le x} \omega_{\cal B} (n) = \beta x \log \log x +O(x) \end{align*} $$

and one can show that it follows from the Turán–Kubilius inequality that

$$ \begin{align*}\sum_{n\le x} ( \omega_{\cal B}(n) -\beta \log \log n )^{2} \ll x\, \log \log x,\end{align*} $$

implying that the average value and the normal order of the function $\omega _{\cal B}(n)$ are each $\beta \log \log n$ . Moreover, one can establish that, for some positive constant $\beta _{3}$ ,

$$ \begin{align*}\lim_{x\to \infty} \frac 1x \#\bigg\{ n\le x: \frac{\omega_{\cal B}(n) - \beta \log \log n}{\beta_{3}\! \sqrt{\log \log n}} < y \bigg\} = \Phi(y),\end{align*} $$

where ${\Phi (y):= (1/\!{\sqrt {2\pi }}) \int _{-\infty }^{y} e^{-t^{2}/2}\,dt}$ .

Now, given positive real numbers $u<v$ , let us set

$$ \begin{align*}Q(u,v) := \prod_{\substack{u<p<v \\ p\in \wp}} p .\end{align*} $$

Also, given an integer $n>1$ and a prime divisor p of n which is smaller than $P(n)$ , the largest prime divisor of n, we set

$$ \begin{align*} \nu_{p}=\nu_{p}(n):= \min\{ q\mid n: q>p\}, \end{align*} $$

so that

$$ \begin{align*} \bigg( \frac n{p \nu_{p}} , Q(p,\nu_{p}) \bigg) =1.\end{align*} $$

Now, consider the arithmetic function

$$ \begin{align*}U_{\lambda,{\cal B}} (n) := \sum_{\substack{p\mid n \\ p\in {\cal B} \\ {\log p}/{\log \nu_{p}(n)}< \lambda}} 1 .\end{align*} $$

The technique we used in [Reference De Koninck and Kátai3] to prove (1.1) can also be used to determine the mean value of the function $U_{\lambda ,{\cal B}} (n)$ and in fact the following more accurate result.

Theorem 2.1. Given $\lambda \in (0,1]$ and a set of primes ${\cal B}$ satisfying (2.2),

(2.4) $$ \begin{align} \sum_{n\le x} U_{\lambda,{\cal B}} (n) = (1+o(1)) \lambda \,\beta x\log \log x \quad (x\to \infty) \end{align} $$

and, for an arbitrarily small $\varepsilon>0$ ,

$$ \begin{align*} \lim_{x\to \infty} \frac 1x \#\bigg\{ n\le x: \bigg| \frac{U_{\lambda,{\cal B}} (n) }{\omega(n)} - \lambda \,\beta \bigg| \ge \varepsilon \bigg\} =0. \end{align*} $$

Estimates (1.1) and (1.2) can be modified to hold for shifted primes. Indeed, we also have the following result.

Theorem 2.2. Given $\lambda \in (0,1]$ , a set of primes ${\cal B}$ satisfying (2.2) and a fixed integer $a\neq 0$ ,

$$ \begin{align*} \frac 1{\pi(x)} \sum_{p\le x} U_{\lambda,{\cal B}} (p+a)= (1+o(1)) \lambda \,\beta \log \log x \quad (x\to \infty) \end{align*} $$

and, moreover, for any arbitrarily small $\varepsilon>0$ ,

$$ \begin{align*} \lim_{x\to \infty} \frac 1{\pi(x)} \#\bigg\{ p\le x: \bigg| \frac{U_{\lambda,{\cal B}} (p+a) }{\omega(p+a)} - \lambda \,\beta \bigg|\ge \varepsilon \bigg\} =0. \end{align*} $$

3 An extension of Theorem 2.1

Given $\lambda \in (0,1]$ and a set of primes ${\cal B}$ satisfying (2.2), we introduce the arithmetic function $\widetilde {U}_{\lambda ,{\cal B}} (n)$ which counts the number of prime divisors p of n which belong to ${\cal B}$ and for which the next prime divisor q of n also belonging to ${\cal B}$ satisfies ${{\log p/\log q<\lambda }}$ . In short, setting ${ Q_{\cal B}(u,v) := \prod _{u<p<v,\, p\in {\cal B}} p}$ , we can write

$$ \begin{align*} \widetilde{U}_{\lambda,{\cal B}} (n):= \sum_{\substack{p\mid n \\ \log p /\!\log q<\lambda \\ p,q\in {\cal B} \\ ( n/{pq},Q_{\cal B}(p,q))=1 }} 1. \end{align*} $$

In the case of the function $\widetilde {U}_{\lambda ,{\cal B}} (n)$ , we can obtain an asymptotic formula for its average value with greater accuracy than the one obtained for the function $U_{\lambda ,{\cal B}} (n)$ (whose average value was revealed through the estimate (2.4)). Indeed, we can prove the following result.

Theorem 3.1. Let $\lambda \in (0,1]$ and ${\cal B}$ be a set of primes satisfying (2.2). Then

(3.1) $$ \begin{align} \sum_{n\le x} \widetilde{U}_{\lambda,{\cal B}} (n) = \beta\lambda^{\,\beta} x \log \log x +O(x\log \log \log x). \end{align} $$

Moreover, for any arbitrarily small $\varepsilon>0$ ,

(3.2) $$ \begin{align} \lim_{x\to \infty} \frac 1x \#\bigg\{ n\le x: \bigg| \frac{\widetilde{U}_{\lambda,{\cal B}}(n)}{\omega(n)} - \beta\lambda^{\,\beta} \bigg| \ge \varepsilon \bigg\} =0. \end{align} $$

Before we begin the proof of Theorem 3.1, let us recall a lemma from our recent paper [Reference De Koninck and Kátai4].

Lemma 3.2 [Reference De Koninck and Kátai4, Lemma A].

Let $x\ge e^{e^{100}}$ and let $Y_{1}=Y_{1}(x)$ and $Y_{2}=Y_{2}(x)$ be the functions defined in (2.1). Let $ \pi _{1} < \pi _{2}< \cdots < \pi _{s}$ be s primes located in the interval $(Y_{1},Y_{2})$ . Write their product as $R=\pi _{1}\pi _{2}\cdots \pi _{s}$ . Further, set

$$ \begin{align*}\eta:= \sum_{i=1}^{s} \frac 1{\pi_{i}} \end{align*} $$

and

$$ \begin{align*}S_{R}(x) : = \sum_{\substack{n\le x \\ (n,R)=1}} 1.\end{align*} $$

Assume that $\eta \le K$ , where K is an arbitrary number, and let h be a positive integer satisfying $h\ge 3e^{2} K$ . Then, letting $\phi $ stand for the Euler totient function,

$$ \begin{align*} \left| S_{R}(x) - \frac{\phi(R)}R x \right| \le x\,(3e)^{-h} +2Y_{2}^{h}, \end{align*} $$

so that in particular, choosing $h=\lfloor \log \log \log x \rfloor $ , there exists a positive constant c such that

$$ \begin{align*} \left| S_{R}(x) - \frac{\phi(R)}R x \right| \le \frac{c\,x}{(\log \log x)^{2}}. \end{align*} $$

We are now ready for the proof of Theorem 3.1.

Proof of Theorem 3.1.

Let $(p,q)\in {\cal B} \times {\cal B}$ , with $p<q$ , be a pair of prime divisors of an integer n which satisfy

$$ \begin{align*}\bigg( \frac n{pq} , Q_{\cal B} (p,q) \bigg) = 1.\end{align*} $$

Let us first assume that $Y_{1}<p<q<Y_{2}$ , where $Y_{1}=Y_{1}(x)$ and $Y_{2}=Y_{2}(x)$ are the functions defined in (2.1). Using Lemma 3.2 and estimate (2.3),

(3.3) $$ \begin{align} \nonumber \sum_{n\le x} \widetilde{U}_{\lambda,{\cal B}}(n) & = \sum_{\substack{Y_{1}<p<q<Y_{2} \\ p,q\in {\cal B} \\ {\log p}/{\log q}<\lambda} } \frac x{pq} \prod_{\substack{p<\pi<q \\ \pi\in {\cal B}} }\bigg( 1 - \frac 1{\pi} \bigg) \bigg( 1 +O \bigg( \frac 1{\log p} \bigg) \bigg) \\ \nonumber & = \sum_{\substack{Y_{1}<p<q<Y_{2} \\ p,q\in {\cal B} \\ {\log p}/{\log q}<\lambda} } \frac x{pq} \bigg( \frac{\log p}{\log q} \bigg)^{\,\beta} \bigg( 1 +O \bigg( \frac 1{\log p} \bigg) \bigg) \\ & = x \sum_{\substack{Y_{1}<p<q<Y_{2} \\ p,q\in {\cal B} \\ {\log p}/{\log q}<\lambda} } \frac{(\log p)^{\,\beta}}p \cdot \frac 1{q(\log q)^{\,\beta}} \bigg( 1 +O \bigg( \frac 1{\log p} \bigg) \bigg). \end{align} $$

First assume that $\beta <1$ . In this case, fixing q and summing over p,

(3.4) $$ \begin{align} \nonumber \sum_{\substack{Y_{1} < p <q^{\lambda} \\ p\in {\cal B}}} \frac{(\log p)^{\,\beta}}p & = \int_{Y_{1}}^{q^{\lambda}} \frac{(\log u)^{\,\beta}}u\,dB(u) \\[-6pt] \nonumber & = \beta \int_{Y_{1}}^{q^{\lambda}} \frac{(\log u)^{\,\beta}}{u\log u}\,du +\int_{Y_{1}}^{q^{\lambda}} \frac{(\log u)^{\,\beta}}{u}\,d(B(u)-\beta\mbox{li}(u)) \\ & = I_{1}(x) + I_{2}(x), \end{align} $$

say. On the one hand, recalling that $\beta>0$ (see (2.2)),

(3.5) $$ \begin{align} I_{1}(x) = \beta\int_{\log Y_{1}}^{\lambda \log q} v^{\,\beta-1}\,dv =\lambda^{\,\beta} (\log q)^{\,\beta} + O( (\log Y_{1})^{\,\beta} ). \end{align} $$

On the other hand,

(3.6) $$ \begin{align} \nonumber I_{2}(x) & \ll ( B(u) -\beta\mbox{li}(u) ) \frac{(\log u)^{\,\beta}}u \bigg|_{Y_{1}}^{q^{\lambda}} - \int_{Y_{1}}^{q^{\lambda}} \frac d{du} \bigg( \frac{(\log u)^{\,\beta}}u \bigg)\,du \\ \nonumber & \ll (\log q)^{{\,\beta}-2} + \int_{Y_{1}}^{q^{\lambda}} \frac u{\log^{2} u} \cdot \frac{\,\beta(\log u)^{\,\beta-1}-(\log u)^{\,\beta}}{u^{2}}\,du \\ \nonumber & \ll (\log q)^{\,\beta-2} + \int_{\log Y_{1}}^{\lambda \log q} v^{\,\beta-2}\,dv \\ & \ll (\log q)^{\,\beta-2} + (\log q)^{\,\beta-1}\ll (\log q)^{\,\beta-1}. \end{align} $$

Gathering estimates (3.5) and (3.6) in (3.4), we obtain, in the case $\beta <1$ ,

(3.7) $$ \begin{align} \sum_{\substack{Y_{1}<p<q^{\lambda} \\ p\in {\cal B}}} \frac{(\log p)^{\,\beta}}p = \lambda^{\,\beta} (\log q)^{\,\beta} +O( (\log q)^{\,\beta-1} )+ O( (\log Y_{1})^{\,\beta} ). \end{align} $$

Finally, it is immediate that (3.7) also holds in the case $\beta =1$ .

In light of estimate (3.7), the sum on the right-hand side of (3.3) can therefore be replaced by

(3.8) $$ \begin{align} \lambda^{\,\beta}\sum_{\substack{Y_{1}<q<Y_{2} \\ q\in {\cal B}} }\frac{(\log q)^{\,\beta}}{q(\log q)^{\,\beta}} = \beta\,\lambda^{\,\beta} \log \log x +O(\log \log \log x), \end{align} $$

where we could ignore the contribution of the last error term on the right-hand side of (3.7) because

$$ \begin{align*} \sum_{Y_{1}<q<Y_{2}} \frac 1{q(\log q)^{\,\beta}} \ll \int_{Y_{1}}^{\infty} \frac 1{t(\log t)^{\,\beta+1}}\,dt \ll \frac 1{(\log Y_{1})^{\,\beta}}.\end{align*} $$

Then, in light of (3.8) and with the help of (2.3), it is clear that the estimate (3.3) can be replaced by

$$ \begin{align*} \sum_{n\le x} \widetilde{U}_{\lambda,{\cal B}}(n) = \beta \lambda^{\,\beta} x \log \log x +O(x\,\log \log \log x), \end{align*} $$

thus completing the proof of (3.1).

The proof of (3.2) rests on the evaluation of $\displaystyle { \widetilde {U}_{\lambda ,{\cal B}}(n)^{2} }$ , which can be handled using an approach similar to that used to obtain (3.1). We will therefore omit this proof.

4 Two B-sets involving congruence classes

Given an integer $k\ge 3$ , let $\ell _{1},\ldots ,\ell _{r}$ be the reduced residue system modulo k, with $r=\phi (k)$ . Then it is clear that each of the residue classes

$$ \begin{align*}{\cal B}_{\ell_{j}}:= \{p\in \wp: p\equiv \ell_{j} \ (\mathrm{mod}\ k)\} \quad (\,j=1,\ldots,r)\end{align*} $$

satisfies condition (2.2) and is therefore a B-set.

A second interesting B-set involves the sum-of-digits function $s_{q}(n)$ which stands for the sum of the base q digits of an integer n (here, $q\ge 2$ ). First note that it is known (see Mauduit and Rivat [Reference Mauduit and Rivat11] as well as Drmota et al. [Reference Drmota, Mauduit and Rivat5]) that if $(k,q-1)=1$ , then there exists a real number $\sigma =\sigma _{q,k}>0$ such that

(4.1) $$ \begin{align} \frac 1{\pi(x)} \#\{p\le x: s_{q}(p) \equiv \ell \ (\mathrm{mod}\ k)\} = \frac 1k + O_{q,k} ( x^{-\sigma} \log x ). \end{align} $$

Clearly, (4.1) guarantees that if, for a given integer $k\ge 3$ , we set

$$ \begin{align*}{\cal B}_{\ell}:=\{p\in \wp: s_{q}(p)\equiv \ell \ (\mathrm{mod}\ k)\} \quad (\ell=0,1,\ldots,k-1),\end{align*} $$

then ${\cal B}_{\ell }$ is indeed a B-set for each $\ell =0,1,\ldots ,k-1$ .

5 Two B-sets involving fractional parts

We first identify a B-set involving fractional parts of prime powers. Let $\alpha ,\sigma \in (0,1)$ . In what follows, $\{y\}$ stands for the fractional part of y. Using the method of exponential sums initiated in 1940 by Vinogradov [Reference Vinogradov13], one can prove that

(5.1) $$ \begin{align} \sum_{\substack{p\le x \\ \{p^{\alpha}\}<\sigma}} 1 = \sigma\,\pi(x) +R(x) \end{align} $$

(see the recent paper of Shubin [Reference Shubin12] for a thorough discussion), where $R(x)\ll x^{\theta (\alpha )+\varepsilon }$ with

$$ \begin{align*}\theta(\alpha) = \begin{cases} 1- 2\alpha/{15} & \mbox{ if } 0<\alpha \le 3/5,\\ {(4+\alpha)}/5 & \mbox{ if } 3/5<\alpha <1. \end{cases}\end{align*} $$

In light of (5.1), it is clear that indeed, given $\alpha ,\sigma \in (0,1)$ , the set

$$ \begin{align*}{\cal B} = \{ p\in \wp: \{p^{\alpha}\}<\sigma\}\end{align*} $$

is a B-set.

Another B-set involving fractional parts is as follows. Let $\alpha $ be an irrational number. As was shown in 1940 by Vinogradov [Reference Vinogradov13],

$$ \begin{align*}\lim_{x\to \infty} \frac 1{\pi(x)} \sum_{p\le x} e^{2\pi i \alpha p} =0.\end{align*} $$

Consequently, the sequence $(\alpha p)_{p\in \wp }$ is uniformly distributed modulo 1. Thus,

$$ \begin{align*}E_{\alpha}(x):= \sup_{0\le u<v<1} \bigg|\frac 1{\pi(x)} \sum_{\substack{p\le x \\ \{\alpha p \} \in [u,v)}} 1 -(v-u) \bigg| \to 0 \quad \mbox{ as }x\to \infty.\end{align*} $$

Our next theorem will depend on the following result of Harman.

Theorem 5.1 [Reference Harman10, Theorem 2.2].

Let $\alpha \in {\mathbb R}$ and suppose that, for integers $a,q$ with $(a,q)=1$ , the inequality

$$ \begin{align*}|q\alpha -a|<\frac 1q\end{align*} $$

holds. Let $\delta \in (0,1]$ be given. Write

$$ \begin{align*}\chi(\theta)= \begin{cases} 1 & \mbox{ if } \|\theta\| < \delta,\\ 0 & \mbox{ otherwise.} \end{cases}\end{align*} $$

Then, for any real $\beta $ and positive integers $N, H$ such that $N^{\varepsilon } \ll q \ll N^{1-\varepsilon }$ for some $\varepsilon>0$ , we have

$$ \begin{align*} \sum_{n\le N} \Lambda(n) \chi(n\alpha+\beta) & = 2 \delta \sum_{n\le N}\Lambda(n) + O \bigg( \frac{N\delta}H \bigg) \\ &\quad + O \bigg( N(\log H) (\log N)^{7} \bigg( \frac{q\delta}N + \frac 1q + \frac 1{N^{1/2}} + \frac{\delta}{N^{1/3}} \bigg)^{1/2} \bigg). \end{align*} $$

(Here $\Lambda (n)$ stands for the von Mangoldt function. Also $\|\theta \|:=\min _{m\in {\mathbb Z}} |\theta -m|$ .)

From Theorem 5.1, one can easily deduce the following result.

Theorem 5.2. Let $\varepsilon>0$ be an arbitrarily small number. Let $\alpha \in (0,1)$ be an irrational number such that, for every real $x>x_{0}(\alpha )$ , there exists a positive integer $q\in [x^{\varepsilon },x^{1-\varepsilon }]$ for which $\|q\alpha \|<1/q$ . Then

$$ \begin{align*}E_{\alpha}(x) \ll \exp\bigg\{ -\frac{\varepsilon}2 \log x \bigg\}.\end{align*} $$

Given $0\le u < v <1$ , let $\alpha $ be an irrational number satisfying the conditions of Theorem 5.2 and set ${\cal B}:=\{p\in \wp : \{\alpha p\}\in [u,v)\}$ . Then it follows from Theorem 5.2 that ${\cal B}$ is a B-set.

6 Disjoint classification of primes

Given an integer $k\ge 2$ , we will now be interested in sets of prime numbers $\wp _{1},\wp _{2},\ldots ,\wp _{k}$ such that

$$ \begin{align*} \wp= \wp_{1} \cup \wp_{2} \cup \cdots \cup \wp_{k} \cup {\cal D}\quad \mbox{and}\quad \wp_{i}\cap \wp_{j} = \emptyset\quad \mbox{if}\ i\neq j, \end{align*} $$

where ${\cal D}$ is a finite (perhaps empty) set of primes. We assume that, for each ${j\in \{1,2,\ldots ,k\}}$ ,

$$ \begin{align*} \pi_{j}(x):=\sum_{\substack{p\le x \\ p\in \wp_{j}}} 1 = \beta_{j} \, \mbox{li}(x) + O\bigg( \frac{x}{\log^{3} x} \bigg), \end{align*} $$

where each $\beta _{j}\in (0,1]$ and $\sum _{j=1}^{k}\beta _{j}=1$ . Such a collection of subsets of primes is called a disjoint classification of primes, a notion we introduced in [Reference De Koninck and Kátai2] a decade ago in order to create large families of normal numbers. Here, we use this concept for a different purpose.

For each $j=1,2,\ldots ,k$ , let $E_{j}$ be a subinterval of $[0,1]$ . Given a large integer n, we will now count those k-tuples of primes $(q_{1},q_{2},\ldots ,q_{k})$ which satisfy the conditions

(6.1) $$ \begin{align} \nonumber \ q_{1}q_{2}\cdots q_{k}\mid n,\quad \bigg(\dfrac n{q_{1}q_{2}\cdots q_{k}},Q(q_{1},q_{k})\bigg)=1,\qquad\qquad \\ q_{j}\in \wp_{\ell_{j}},\ q_{j+1}\in \wp_{\ell_{j+1}},\ \log q_{j}/\log q_{j+1}\in E_{j}\quad (\,j=1,2,\ldots,k-1). \end{align} $$

In fact, let $E(n)$ be the number of these primes $q_{1}$ , that is, primes $q_{1}$ for which there exist primes $q_{2},\ldots ,q_{k}$ satisfying the conditions listed in (6.1).

By using our usual technique and writing $\lambda (I)$ for the length of the interval I, we can prove that, if we set

$$ \begin{align*}T(\ell_{1},\ell_{2},\ldots,\ell_{k}):=\beta_{\ell_{1}}\beta_{\ell_{2}}\cdots \beta_{\ell_{k}}\lambda(E_{1})\lambda(E_{2})\cdots \lambda(E_{k-1}),\end{align*} $$

then,

$$ \begin{align*} \frac 1x \sum_{n\le x} E(n) = T(\ell_{1},\ell_{2},\ldots,\ell_{k}) \log \log x +O(\log \log \log x), \end{align*} $$

and moreover that, given any small number $\varepsilon>0$ ,

$$ \begin{align*} \lim_{x\to \infty} \frac 1x \# \bigg\{ n\le x: \bigg| \frac{E(n)}{\omega(n)} - T(\ell_{1},\ell_{2},\ldots,\ell_{k}) \bigg| \ge \varepsilon \bigg\} =0. \end{align*} $$

An analogous result for shifted primes can also be proved, namely that, for any fixed integer $a\neq 0$ ,

$$ \begin{align*} \lim_{x\to \infty} \frac 1{\pi(x)} \# \bigg\{ p\le x: \bigg| \frac{E(p+a)}{\omega(p+a)} - T(\ell_{1},\ell_{2},\ldots,\ell_{k}) \bigg| \ge \varepsilon \bigg\} =0. \end{align*} $$

7 The sequence $n^{2}+1$

It is known that if $p\mid n^{2}+1$ , then either $p=2$ or $p\equiv 1 \ (\mathrm {mod}\ 4)$ (since $({-1}/p)=1$ if and only if $p\equiv 1 \ (\mathrm {mod}\ 4)$ ). On the one hand,

(7.1) $$ \begin{align} \sum_{n\le x} \omega(n^{2}+1) = x \log \log x + O(x). \end{align} $$

For the sake of completeness, this can be proved by first observing that

(7.2) $$ \begin{align} \sum_{n\le x} \omega(n^{2}+1) = \bigg\lfloor \frac{x+1}2 \bigg\rfloor +\sum_{\substack{p\le x \\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \sum_{\substack{n\le x \\ n^{2}+1\equiv 0\ (\mathrm{mod}\ p)}} 1 +O(x), \end{align} $$

where the last term accounts for the contribution of those primes $p\in (x,x^{2}+1]$ . Since for each prime $p\equiv 1 \ (\mathrm {mod}\ 4)$ , the congruence $n^{2}+1\equiv 0\ (\mathrm {mod}\ p) $ has exactly two solutions, the number of positive integers $n\le x$ for which $n^{2}+1\equiv 0 \ (\mathrm {mod}\ p)$ is equal to $2x/p+O(1)$ . Hence,

$$ \begin{align*} \sum_{\substack{p\le x \\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \sum_{\substack{n\le x \\ n^{2}+1\equiv 0\ (\mathrm{mod}\ p)}} 1 & = \sum_{\substack{p\le x \\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \bigg( \frac{2x}p+O(1) \bigg) = 2x \sum_{\substack{p\le x \\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \frac 1p +O\bigg( \frac x{\log x} \bigg) \\ & = 2x \bigg( \frac 12 \log \log x +O(1) \bigg) +O\bigg( \frac x{\log x} \bigg) = x \log \log x +O(x), \end{align*} $$

which, combined with (7.2), proves (7.1).

On the other hand, using estimate (7.1), it follows from the Turán–Kubilius inequality that

(7.3) $$ \begin{align} \sum_{n\le x} \left( \omega(n^{2}+1) -\log \log n\right)^{2} \ll x\,\log \log x. \end{align} $$

In light of (7.1) and (7.3), we may conclude that both the average value and the normal order of the function $\omega (n^{2}+1)$ are $\log \log n$ .

For convenience, we will now be using the notation

$$ \begin{align*}\Lambda(p,q):= \frac{\log p}{\log q} \quad \mbox{and} \quad Q^{(1)}(p,q)= \prod_{\substack{p<\pi<q \\ \pi\equiv 1 \ (\mathrm{mod}\ 4)}} \pi.\end{align*} $$

Given a small number $\varepsilon>0$ and $\lambda \in (\varepsilon ,1)$ , let us count those pairs of primes $p<q$ such that

(7.4) $$ \begin{align} pq\mid n^{2}+1,\quad \varepsilon\le \Lambda(p,q)<\lambda,\quad p\equiv q\equiv 1 \ (\mathrm{mod}\ 4),\quad \bigg( \frac{n^{2}+1}{pq},Q^{(1)}(p,q)\bigg)=1. \end{align} $$

For each such pair of primes $p,q$ , let $E_{p,q}$ be the number of those positive integers $n\le x$ for which conditions (7.4) hold. Using Lemma 3.2, we may write

(7.5) $$ \begin{align} \nonumber \sideset{}{^{\prime}}\sum_{Y_{1}<p<q<Y_{2}} E_{p,q} & = \sideset{}{^{\prime}}\sum_{Y_{1}<p<q<Y_{2}} \frac x{pq} \prod_{\substack{p<\pi<q \\ \pi\equiv 1 \ (\mathrm{mod}\ 4)}} \bigg( 1-\frac 1{\pi} \bigg) +O(x)\\ & = \sum_{\substack{Y_{1}<q<Y_{2}\\ q\equiv 1 \ (\mathrm{mod}\ 4)}} \frac 1{q\log q} \sum_{\substack{q^{\varepsilon}<p<q^{\lambda}\\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \frac{\log p}p +O(x) , \end{align} $$

where the prime ( $^{\prime }$ ) on the above sums indicates that the summation runs over those pairs of primes $p,q$ which satisfy the conditions listed in (7.4).

It is easily established that

(7.6) $$ \begin{align} \sum_{\substack{q^{\varepsilon}<p<q^{\lambda} \\ p\equiv 1 \ (\mathrm{mod}\ 4)}} \frac{\log p}p = \frac{\lambda-\varepsilon}2 \log q + O \bigg( \frac 1{\log q} \bigg). \end{align} $$

Using (7.6) in (7.5), we obtain

(7.7) $$ \begin{align} \frac 1x \sideset{}{^{\prime}}\sum_{Y_{1}<p<q<Y_{2}} E_{p,q} = \frac{\lambda-\varepsilon}2 \sum_{\substack{Y_{1}<q<Y_{2}\\ q\equiv 1 \ (\mathrm{mod}\ 4)}} \frac 1q = \frac{\lambda-\varepsilon}4 \log \log x + O(\log \log \log x). \end{align} $$

Denote by $K_{\varepsilon ,\lambda }(n)$ the number of those primes p for which there exists a prime $q>p$ satisfying the list of conditions (7.4). Since the contribution of those primes $p<Y_{1}$ and $q>Y_{2}$ to the above sums is relatively small, it follows from (7.7) that

(7.8) $$ \begin{align} \frac 1{x\log \log x} \sum_{n\le x} K_{\varepsilon,\lambda}(n) = \frac{\lambda-\varepsilon}4 + O \bigg( \frac{\log \log \log x}{\log \log x} \bigg). \end{align} $$

Since $\varepsilon $ can be chosen arbitrarily small, it follows from (7.8) that

$$ \begin{align*} \lim_{x\to \infty} \frac 1{x\log \log x} \sum_{n\le x} K_{0,\lambda}(n) = \frac{\lambda}4. \end{align*} $$

Moreover, one can show that, given any arbitrarily small number $\delta>0$ ,

$$ \begin{align*} \lim_{x\to \infty} \frac 1x \#\bigg\{ n\le x: \bigg| \frac{K_{0,\lambda}(n)}{\log \log n} - \frac{\lambda}4 \bigg| \ge \delta \bigg\} =0. \end{align*} $$

We can also prove an analogue of [Reference De Koninck and Kátai3, Theorem 5]. Indeed, let $\delta _{0},\delta _{1},\ldots ,\delta _{k-1} \in (0,1)$ and set $H:=\delta _{0}\delta _{1}\cdots \delta _{k-1}$ . Further, let ${\cal F}$ be the set of all $(k+1)$ -tuples of primes $(p_{0},p_{1},\ldots ,p_{k})$ where $p_{0}<p_{1}<\cdots <p_{k}$ satisfy

$$ \begin{align*}1-\delta_{j} < \Lambda(p_{j},p_{j+1})<1 \quad (\,j=0,1,\ldots,k-1).\end{align*} $$

Also let $V_{\delta _{0},\delta _{1},\ldots ,\delta _{k-1}}(n)$ stand for the number of those prime divisors $p_{0}$ of $n^{2}+1$ for which

$$ \begin{align*}p_{1}p_{2}\cdots p_{k}\mid n^{2}+1\quad \mbox{and}\quad \bigg( \frac{n^{2}+1}{p_{0}p_{1}\cdots p_{k}}, Q(p_{0},p_{k}) \bigg)=1 \quad\mbox{for } (p_{0},p_{1},\ldots,p_{k})\in {\cal F}.\end{align*} $$

Then we have the following result.

Theorem 7.1. Let $\delta _{0},\delta _{1},\ldots ,\delta _{k-1}$ , H and ${\cal F}$ be as above. Then, as $x\to \infty $ ,

$$ \begin{align*} \sum_{n\le x} V_{\delta_{0},\delta_{1},\ldots,\delta_{k-1}}(n) & = (H+o(1)) x\,\log \log x,\\ \sum_{n\le x} V_{\delta_{0},\delta_{1},\ldots,\delta_{k-1}}^{2}(n) & = (H^{2}+o(1)) x\,(\log \log x)^{2}. \end{align*} $$

Remark 7.2. It follows from Theorem 7.1 that

$$ \begin{align*}V_{\delta_{0},\delta_{1},\ldots,\delta_{k-1}}(n) = (1+o(1)) H \log \log n \quad \mbox{for almost all }\textit{n}.\end{align*} $$

On the other hand, most likely, for some positive constant c, the function

$$ \begin{align*}\frac{ V_{\delta_{0},\delta_{1},\ldots,\delta_{k-1}}(n) -H \log \log n}{c\sqrt{\log \log n}}\end{align*} $$

is distributed according to the normal law. However, in order to prove this, one would need to compute the asymptotic value of the sum $\sum _{n\le x} V_{\delta _{0},\delta _{1},\ldots ,\delta _{k-1}}(n)^{k}$ with a good remainder term, for every k. We were not successful in doing so.