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Karatsuba's divisor problem and related questions M. R. Gabdullina, S. V. Konyagina, V. V. Iudelevichb a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia b Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 7, DOI: https://doi.org/10.4213/sm9815 
Bibliographic databases:
    § 1. IntroductionIn 2004 Karatsuba posed the following problem at his seminar “Analytic number theory and applications”: find the asymptotic behaviour of the sum as $x\to\infty$, where $\tau(n)=\sum_{d\mid n} 1$ is the divisor function and the sum is taken over the prime numbers not exceeding $x$. This is a natural ‘hybrid’ of the following two classical problems in analytic number theory.The first (so-called Titchmarsh divisor problem) asks for the asymptotic behaviour of the sum It is well known (see [1]–[4]) that
$$
\begin{equation*}
D(x) \sim \frac{\zeta(2)\zeta(3)}{\zeta(6)}x, \qquad x\to\infty,
\end{equation*}
\notag
$$
where $\zeta(s)$ denotes the Riemann zeta function. The other problem is to find the asymptotics for the sum This problem was solved by Ramanujan [5], who showed that
$$
\begin{equation}
T(x)=c_0\frac{x}{\sqrt{\log x}}\biggl( 1+O\biggl(\frac{1}{\sqrt{\log x}} \biggr)\biggr) ,
\end{equation}
\tag{1.1}
$$
where
$$
\begin{equation*}
c_0=\frac{1}{\sqrt{\pi}} \prod_{p} \sqrt{p^2 - p} \log \frac{p}{p-1}= 0.5486\dots\,.
\end{equation*}
\notag
$$
The sum $\Phi(x)$ was studied previously. In the recent work [6] it was shown that
$$
\begin{equation*}
\Phi(x)\leqslant 4K\frac{x}{(\log x)^{3/2}}+O\biggl( \frac{x\log\log x}{(\log x)^{5/2}}\biggr),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
K=\frac{1}{\sqrt{\pi}}\prod_{p}\sqrt{\frac{p}{p-1}}\biggl(p\log\frac{p}{p-1} -\frac{1}{p-1}\biggr)=0.2532\dots\,.
\end{equation*}
\notag
$$
Note that the bound $\Phi(x)\ll \frac{x}{(\log x)^{3/2}}$ follows from Corollary 1.2 in [7] and can also be derived by arguing similarly to the proof of the upper bound in Theorem 1.2 below. We make the conjecture that
$$
\begin{equation*}
\Phi(x) \sim K\frac{x}{(\log x)^{3/2}}, \qquad x\to\infty,
\end{equation*}
\notag
$$
but this is probably hard to show. In this paper we prove that the aforementioned upper bound for $\Phi(x)$ is sharp up to a constant. Thus, we find the correct order of magnitude of the sum $\Phi(x)$: which answers Karatsuba’s question in the first approximation.In addition to $\Phi(x)$, we consider the sum and establish the following.Now we discuss the main ideas of the proofs. In the sum $\Phi(x)$, for each prime number ${p\leqslant x}$ we write $p-1$ in the form $ab$, where $a$ consists of the prime factors not exceeding $z$, and $b$ has only prime factors greater than $z$; here $z=x^\varepsilon$ for some small fixed $\varepsilon>0$. Now $\Phi(x)$ can be rewritten as
$$
\begin{equation}
\Phi(x)=\sum_{\substack{a \leqslant x \\ p\mid a \ \Rightarrow\ p \leqslant z}} \frac{1}{\tau(a)} \sum_{\substack{b \leqslant (x-1)/a \\ p\mid b \ \Rightarrow\ p>z \\ ab +1\text{ is prime}}} \frac{1}{\tau(b)},
\end{equation}
\tag{1.2}
$$
and, since $\tau(b)=O_\varepsilon(1)$, we have
$$
\begin{equation*}
\Phi(x)\gg_\varepsilon \sum_{a\leqslant x^\varepsilon} \frac{1}{\tau(a)} \sum_{\substack{b \leqslant (x-1)/a \\ p\mid b \ \Rightarrow\ p>x^\varepsilon \\ ab +1\text{ is prime}}} 1.
\end{equation*}
\notag
$$
Using the Brun-Hooley sieve one can estimate the inner sum from below by a quantity of order where the contribution from $R(x;a)$ to the outer sum is negligible. Combining these estimates, we see that
$$
\begin{equation*}
\Phi(x) \gg_\varepsilon \frac{x}{(\log x)^2}\sum_{a\leqslant x^\varepsilon} \frac{1}{a\tau(a)}\gg_\varepsilon \frac{x}{(\log x)^{3/2}},
\end{equation*}
\notag
$$
which gives us the required lower bound. Note that this argument does not yield any upper bound for $\Phi(x)$, since the estimation of the contribution of the numbers $a>x^{1-\varepsilon}$ to (1.2) is actually equivalent to the initial problem. A lower bound for the sum $F(x)$ is obtained in a similar way. The upper bound for $F(x)$ follows from Theorem 1 in [8]. We also note that the same upper bound can be derived from the inequality $\tau(n)\geqslant 2^{\omega(n)}$ (here $\omega(n)$ denotes the number of distinct prime divisors of $n$) and a bound for the number of integers $n\leqslant x$ with a fixed value of $\omega(n^2+1)$. For completeness, we provide this argument. We observe that the methods used in this work can also be applied to other functions similar to $\tau(n)$. For instance, let $\tau_k(n)$ be the generalized divisor function, $\tau_k(n)=\sum_{n=d_1 d_2\dotsb d_k}1$; then one can show that
$$
\begin{equation*}
\sum_{p\leqslant x}\frac{1}{\tau_k(p-1)}\asymp_k x (\log x)^{1/k-2}.
\end{equation*}
\notag
$$
§ 2. NotationBy $\varphi(n) = \#\{k \leqslant n\colon (k,n) = 1\}$ we denote the Euler totient function, and we use $P^{+}(n)$ and $P^{-}(n)$ for the least and greatest prime divisors of the integer $n>1$, respectively; by convention, $P^+(1)=0$ and $P^-(1)=\infty$. We denote by $\pi(x)$ the number of primes not exceeding $x$, and $\pi(x;q,a)$ is the number of primes not exceeding $x$ and belonging to the arithmetic progression $a\ (\operatorname{mod}q)$; we also set $R(x;q,a)=\pi(x;q,a)-{\pi(x)}/{\varphi(q)}$. The notation $f(x)\ll g(x)$, as well as $f(x) = O(g(x))$, means that $|f(x)|\leqslant C g(x)$ for some absolute constant $C>0$ and all possible values of $x$. We write $f(x)\asymp g(x)$ if $f(x)\ll g(x)\ll f(x)$, and we write $f(x)\ll_k g(x)$ if we want to stress that the implied constant depends on $k$. Now we recall some notation from the sieve methods. Let $\mathcal{A}$ be a finite set of positive integers and $\mathcal{P}$ be a finite set of prime numbers. We define $P = \prod_{p \in \mathcal{P}} p$ and $S(\mathcal{A},\mathcal{P}) = \#\{a \in \mathcal{A}\colon (a,P)=1 \}$. Also let $\mathcal{A}_d=\#\{a \in \mathcal{A}\colon a \equiv 0 \ (\operatorname{mod}d)\}$. We assume that, for all $d \mid P$, where $g(d)$ is a multiplicative function such that $0<g(p)<1$ for $p \in \mathcal{P}$ and $g(p)=0$ for $p \notin \mathcal{P}$.Further, let the set $\mathcal{P}$ be divided into disjoint subsets $\mathcal{P}_1, \mathcal{P}_2, \dots, \mathcal{P}_t$, and let $P_j= \prod_{p \in \mathcal{P}_j} p$. Finally, let $\{k_j\}_{r=1}^t$ be a sequence of even integers. Then we set
$$
\begin{equation*}
\begin{gathered} \, V_j=\prod _{p \in \mathcal{P}_r} (1-g(p)), \qquad L_j=\log V_j^{-1}, \qquad E=\sum _{j=1}^{t} \frac{e^{L_j} (L_j)^{k_j+1}}{(k_j+1)!}, \\ R=\sum _{\substack{d_j\mid P_j\\ \omega(d_j) \leqslant k_j}} |r_{d_1 \dotsb d_t}| \quad\text{and}\quad R'=\sum _{l=1}^t \sum _{\substack{d_j\mid P_j \\ \omega(d_j) \leqslant k_j,\, j \ne l \\ \omega(d_l)=k_l+1}} |r_{d_1 \dotsb d_t}|. \end{gathered}
\end{equation*}
\notag
$$
§ 3. Auxiliary resultsWe need the following version of the Brun-Hooley sieve, which is due to Ford and Halberstam [9]. Theorem 3.1. Let assumption (2.1) be met. Then and For a proof, see [9]. Lemma 3.2. Let $a \leqslant x^{1/40}$ be an even positive integer, and let Proof of Lemma 3.2. We apply Theorem 3.1 to the sets
$$
\begin{equation*}
\mathcal{A}=\biggl\{n \leqslant \frac{x-1}a\colon an + 1\text{ is prime}\biggr\}
\end{equation*}
\notag
$$
and $\mathcal{P} = \{p \leqslant z\}$, where $z=x^{1/40}$ (the other parameters will be chosen later). Then $F_a(x)=S(\mathcal{A}, \mathcal{P})$, and for each $d\mid P=\prod_{p\leqslant z}p$ we have
$$
\begin{equation*}
\mathcal{A}_d=\#\biggl\{k\leqslant \frac{x-1}{ad}\colon adk+1\text{ is prime}\biggr\} =\pi(x; ad,1)=\frac{\pi(x)}{\varphi(ad)}+R(x; ad,1).
\end{equation*}
\notag
$$
Now we write a number $d\mid P$ in the form $d=d_1d_2$, where $(d_1,a)=1$ and $d_2$ consists only of primes dividing $a$. Then, since $\varphi(ad_2)=ad_2\prod_{p\mid a}(1-1/p)=d_2\varphi(a)$, we find that
$$
\begin{equation*}
\frac{1}{\varphi(ad)}=\frac{1}{\varphi(d_1)\varphi(ad_2)} =\frac{1}{\varphi(d_1)d_2\varphi(a)},
\end{equation*}
\notag
$$
hence (2.1) holds for $X= {\pi(x)}/{\varphi(a)}$, the multiplicative function $g$ defined by and In addition, we have with an absolute implied constant.
Now we choose a partition of $\mathcal{P}$ and define the numbers $\{k_j\}_{j=1}^t$. Set $z_j= z^{2^{1-j}}$ and then $t$ is the unique positive integer such that $z_{t+1} < 2 \leqslant z_t$. Let $k_j = b + 2(j-1)$, where $b \geqslant 2$ is even and will be chosen later. We also set
$$
\begin{equation*}
C=\prod_{p>2}\biggl(1+\frac{1}{p^2-2p}\biggr)\leqslant 1.52.
\end{equation*}
\notag
$$
By [10], Theorem 7 (see (3.26) and (3.27) in that paper), for any $x>1$,
$$
\begin{equation}
\frac{e^{-\gamma}}{\log x}\biggl(1-\frac{1}{\log^2x}\biggr) < \prod_{p\leqslant x}\biggl(1-\frac1p\biggr) < \frac{e^{-\gamma}}{\log x}\biggl(1+\frac{1}{2\log^2x}\biggr),
\end{equation}
\tag{3.1}
$$
where $\gamma$ is the Euler constant. Hence, for any $z\geqslant \sqrt2$ we have
$$
\begin{equation}
\prod_{z<p\leqslant z^2}\biggl(1-\frac1p\biggr)^{-1}\leqslant 3;
\end{equation}
\tag{3.2}
$$
indeed, this follows from (3.1) for $z\geqslant 4$ and can be checked manually for ${\sqrt{2}\,{\leqslant}\,z\,{<}\,4}$. Thus,
$$
\begin{equation*}
V_j^{-1}=\prod_{z_{j+1} < p \leqslant z_j} (1 - g(p))^{-1} \leqslant C\prod_{z_{j+1}<p\leqslant z_j}\biggl(1-\frac1p\biggr)^{-1} \leqslant 3C\leqslant 5.
\end{equation*}
\notag
$$
Hence $L_j = \log V_j^{-1}\leqslant L=\log 5$ and
$$
\begin{equation*}
E=\sum_{j=1}^t \frac{e^{L_j} L_j^{k_j+1}}{(k_j+1)!} \leqslant e^L\sum_{j=1}^t\frac{L^{b+2j-1}}{(b+2j-1)!}.
\end{equation*}
\notag
$$
Now we estimate $R+R'$. If an integer $d$ corresponds to a summand in $R$, then $d=d_1 \dotsb d_t$, where $d_j \mid {P}_j$ and $\omega(d_j) \leqslant k_j=b+2(j-1)$. Therefore,
$$
\begin{equation*}
d \leqslant z_1^{k_1} \dotsb z_t^{k_t} \leqslant z^{b+(b+2)/2+(b+4)/4+\dotsb}=z^{2b+4}.
\end{equation*}
\notag
$$
If a number $d$ corresponds to a summand in $R'$, then we find similarly that ${d\leqslant z^{2b+4} z = z^{2b+5}}$. Since the numbers $d$ corresponding to the sums $R$ and $R'$ are distinct and do not exceed $z^{2b+5}$, we see that Now we take $b=4$. Then $2b+5=13$, and The claim follows. We set
$$
\begin{equation*}
\mathcal{M}=\bigl\{ a \geqslant 1\colon a \mid n^2 + 1 \text{ for some } n \geqslant 1 \bigr\}.
\end{equation*}
\notag
$$
It is well known that $a \in \mathcal{M}$ if and only if $4 \nmid a$ and $p \nmid a$ for any $p \equiv 3\ (\operatorname{mod}4)$. Lemma 3.4. Let $2\leqslant a, z\leqslant x^{1/30}$, $a \in \mathcal{M}$ and $P^{+}(a)\leqslant z$. Let Then Proof. Setting $\mathcal{A}=\{k\colon ak=n^2+1 \text{ for some } n\leqslant x\}$ and
$$
\begin{equation*}
\mathcal{P}= \begin{cases} \{p \leqslant z\colon p\equiv 1\ (\operatorname{mod}4)\} &\text{if $a$ is even}, \\ \{2\}\cup \{p \leqslant z\colon p\equiv 1\ (\operatorname{mod}4)\} &\text{if $a$ is odd}. \end{cases}
\end{equation*}
\notag
$$
We have $W_a(x,z) = S(\mathcal{A}, \mathcal{P})$. As before, we write each $d$ dividing $ P=\prod_{p\in\mathcal{P}}p$ as $d=d_1d_2$, where $(d_1,a)=1$ and $d_2$ consists only of primes dividing $a$. Suppose that one of the integers $d$ and $a$ is even. Then by the Chinese remainder theorem and the fact that the congruence $x^2+1\equiv0\ (\operatorname{mod}p)$ has two solutions for $p\equiv 1\ (\operatorname{mod}4)$ and one solution for $p=2$, we obtain
$$
\begin{equation*}
\begin{aligned} \, \mathcal{A}_d &=\#\{n\leqslant x\colon n^2+1 \equiv 0 \ (\operatorname{mod}ad) \} \\ &= \frac{x2^{\omega(ad)-1}}{ad}+O(2^{\omega(ad)}) =\frac{x2^{\omega(a)+\omega(d_1)-1}}{ad_1d_2}+O(2^{\omega(ad)}). \end{aligned}
\end{equation*}
\notag
$$
If both $a$ and $d$ are odd, then a similar argument shows that
$$
\begin{equation*}
\mathcal{A}_d= \frac{x2^{\omega(a)+\omega(d_1)}}{ad_1d_2}+O(2^{\omega(ad)}).
\end{equation*}
\notag
$$
In both cases we have
$$
\begin{equation*}
\mathcal{A}_d=\frac{x2^{\omega(a)+\omega(d_1)-\mathbb I(2\mid ad)}}{ad_1d_2}+O(2^{\omega(ad)}) =\frac{x2^{\omega(a)-\mathbb I(2\mid a)}}{a}\frac{2^{\omega(d_1)-\mathbb I(2\mid d_1)}}{d_1d_2}+O(\tau(ad)),
\end{equation*}
\notag
$$
where $\mathbb I(2\mid l)$ is equal to one if $l$ is even and to zero otherwise. Thus we see that condition (2.1) holds for $X= (x2^{\omega(a)-\mathbb I(2\mid a)})/a$, the multiplicative function $g$ defined on the prime numbers in $\mathcal{P}$ by
$$
\begin{equation*}
g(p)=\begin{cases} \dfrac{2}{p} &\text{if } p\nmid a, \\ \dfrac{1}{p} &\text{if } p\mid a \end{cases}
\end{equation*}
\notag
$$
(and also $g(2)=1/2$ in the case of odd $a$), and It is well known that
$$
\begin{equation}
\prod _{\substack{p\leqslant x \\ p\equiv 1\, (\operatorname{mod}4)}}\biggl(1-\frac{1}{p} \biggr)\asymp \frac{1}{\sqrt{\log x}}
\end{equation}
\tag{3.3}
$$
(see [11]). Thus, in both cases in the definition of $\mathcal{P}$ we have for an absolute implied constant.
Now we choose a partition of $\mathcal{P}$ and define the numbers $\{k_j\}_{j=1}^t$. Again, we set $z_j= z^{2^{1-j }}$ and where $t$ is the unique positive integer such that $z_{t+1} < 2 \leqslant z_t$ and $k_j = b + 2(j-1)$ for some even $b \geqslant 2$. Now (3.2) implies that
$$
\begin{equation*}
\begin{aligned} \, V_j^{-1} &=\prod_{z_{j+1} < p \leqslant z_j} (1 - g(p))^{-1} \leqslant 2\prod_{z_{j+1}<p\leqslant z_j, \, p\neq2}\biggl(1-\frac2p\biggr)^{-1} \\ &=2\prod_{z_{j+1}<p\leqslant z_j, \, p\neq2}\biggl(1-\frac1p\biggr)^{-2}\frac{(1-1/p)^2}{1-2/p} \leqslant 18C\leqslant 28. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $L_j = \log V_j^{-1}\leqslant L=\log 28$ and
$$
\begin{equation*}
E=\sum_{j=1}^t \frac{e^{L_j} L_j^{k_j+1}}{(k_j+1)!} \leqslant e^L\sum_{j=1}^t\frac{L^{b+2j-1}}{(b+2j-1)!}.
\end{equation*}
\notag
$$
Now we take $b=10$; then $E\leqslant 0.43$ and $2b+5=25$. As in the proof of Lemma 3.2, we obtain
$$
\begin{equation*}
R+R' \leqslant \sum_{d \leqslant z^{2b+5}} |r_d|\ll \tau(a)\sum_{d\leqslant z^{25}}\tau(d)\ll x^{1/30}z^{25} \ll x^{26/30},
\end{equation*}
\notag
$$
and the main term is at least
$$
\begin{equation*}
X\prod_{p\in \mathcal{P}}(1-g(p)) \asymp\frac{x2^{\omega(a)}}{\varphi(a)\log z} \gg\frac{x^{1-1/30}}{\log x}.
\end{equation*}
\notag
$$
An application of Theorem 3.1 completes the proof. The next lemma is a special case of a result due to Tenenbaum [12]. For completeness, we provide the proof. Lemma 3.5. Let $R>0$ be fixed. Then there exist positive constants $A=A(R)$ and $B=B(R)$ such that, for all $1\leqslant k \leqslant R\log\log x$, Proof. For $m\geqslant 2$ and $y\geqslant2$ we define the ‘$y$-smooth part’ of $m$ by
$$
\begin{equation*}
d(m,y)=\max\bigl\{d\mid m\colon P^+(d)\leqslant y\bigr \}.
\end{equation*}
\notag
$$
For each $n\leqslant x$ the number $n^2+1$ can be written as the product $ab$, where
$$
\begin{equation*}
a=a(n)=\max\bigl\{d(n^2+1,y)\colon d(n^2+1,y)\leqslant x^{1/30} \bigr\}
\end{equation*}
\notag
$$
and $b=b(n)=(n^2+1)/a$; note that $(a,b)=1$. Now we set
$$
\begin{equation*}
\begin{gathered} \, A_1 =\bigl\{ n\leqslant x\colon a\leqslant x^{1/60},\, P^-(b)>x^{1/60},\, b>1\bigr\}, \\ A_2 =\bigl\{ n\leqslant x\colon a\leqslant x^{1/60},\, P^-(b)\leqslant x^{1/60},\, b>1\bigr\} \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
A_3 =\bigl\{ n\leqslant x\colon x^{1/60}< a \leqslant x^{1/30},\, b>1 \bigr\}.
\end{equation*}
\notag
$$
Since the number of integers $n\leqslant x$ for which $b(n)=1$ does not exceed $x^{1/60}$, we conclude that where
First we estimate $N_1$. In this case $1\leqslant \omega(b)\leqslant 120$; by Lemma 3.4 we have
$$
\begin{equation}
\begin{aligned} \, \notag N_1 &\leqslant \sum_{l=k-120}^{k-1}\sum_{\substack{a\leqslant x^{1/60}, \, a\in \mathcal{M} \\ \omega(a)=l }} W_a(x, x^{1/60}) \ll \sum_{l=k-120}^{k-1}\sum_{\substack{a\leqslant x^{1/60}, \, a\in \mathcal{M} \\ \omega(a)=l }}\frac{x2^{\omega(a)}}{\varphi(a)\log x} \\ &\leqslant \frac{x}{\log x}\sum_{l=k-120}^{k-1}2^l\sum_{\substack{a\leqslant x^{1/60}, \, a\in \mathcal{M} \\ \omega(a)=l }}\frac{1}{\varphi(a)}. \end{aligned}
\end{equation}
\tag{3.5}
$$
Taking the logarithm of both sides of (3.3) we obtain
$$
\begin{equation*}
\sum _{\substack{p\leqslant x \\ p\equiv 1\, (\operatorname{mod}4)}}\frac{1}{p}= \frac{1}{2}\log\log x+O(1).
\end{equation*}
\notag
$$
Hence for each $l\geqslant 0$ the inner sum in (3.5) is at most
$$
\begin{equation*}
\frac{1}{l!}\biggl(\,\sum_{p\leqslant x^{1/60},\, p\neq 3\, (\operatorname{mod}4)}\frac{1}{\varphi(p)}+\frac{1}{\varphi(p^2)}+\dotsb \biggr)^l \leqslant \frac{1}{l!}\biggl(\frac{1}{2}\log\log x+B_1\biggr)^l
\end{equation*}
\notag
$$
for some absolute positive constant $B_1$. Thus,
$$
\begin{equation}
N_1\ll \frac{x(\log\log x+B_1)^{k-1}}{(k-1)!\,\log x}\bigl(1+R+\dots+R^{120} \bigr).
\end{equation}
\tag{3.6}
$$
Now we estimate $N_2$. Let $p=P^-(b)$, and let $r$ be the largest integer such that $p^r\mid b$. Then by the definition of $a$, we have $ap^r>x^{1/30}$. Thus, for any $n\in A_2$ we have $p^r> x^{1/60}$, and since $p\leqslant x^{1/60}$, we see that $r\geqslant 2$. Let $\nu = \min\{{u\geqslant 1}$: ${p^u>x^{1/60}}\}$. Then $2\leqslant \nu\leqslant r$ and $p^{\nu-1}\leqslant x^{1/60}$. Hence ${p^\nu\leqslant x^{1/60}p\leqslant x^{1/30}}$. We conclude that for each $n\in A_2$ the integer $n^2+1$ is divisible by $p^\nu$, where $p$ is a prime number such that $x^{1/60}<p^\nu\leqslant x^{1/30}$ and $\nu\geqslant 2$. Therefore,
$$
\begin{equation*}
N_2\leqslant\sum_{x^{1/60}<p^\nu\leqslant x^{1/30},\,\nu\geqslant 2}\biggl( \frac{2x}{p^\nu}+O(1)\biggr)\ll \sum_{x^{1/60}<p^\nu \leqslant x^{1/30},\,\nu\geqslant 2}\frac{x}{p^\nu}.
\end{equation*}
\notag
$$
Note that $p^{\nu} \geqslant \max\{p^2, x^{1/60}\}$, and therefore $p^{\nu}\geqslant px^{1/120}$. It follows that
$$
\begin{equation}
N_2\ll \sum_{p\leqslant x^{1/30}}\frac{x}{px^{1/120}} \ll x^{1-1/120}\log\log x.
\end{equation}
\tag{3.7}
$$
Finally, consider $N_3$. Setting $q=P^+(a)$, we have $P^-(b)\geqslant q+1$ and
$$
\begin{equation}
\omega(b)\leqslant \frac{\log (x^2+1)}{\log (q+1)}\leqslant \frac{2\log x}{\log q}=:\eta.
\end{equation}
\tag{3.8}
$$
Lemma 3.4 implies that
$$
\begin{equation}
\begin{aligned} \, N_3 &\leqslant \sum_{\substack{q\leqslant x^{1/30} \\ q\not{\equiv} 3\, (\operatorname{mod}4)}}\sum_{k-\eta\leqslant l\leqslant k-1}\sum_{\substack{x^{1/60}<a\leqslant x^{1/30} \\ P^+(a)=q, \, \omega(a)=l}}W_a(x,q) \nonumber \\ &\ll x\sum_{\substack{q\leqslant x^{1/30} \\ q\not{\equiv} 3\, (\operatorname{mod}4)}}\frac{1}{\log q}\sum_{k-\eta\leqslant l\leqslant k-1}2^l\sum_{\substack{x^{1/60}<a\leqslant x^{1/30} \\ P^+(a)=q, \, \omega(a)=l}}\frac{1}{\varphi(a)}. \end{aligned}
\end{equation}
\tag{3.9}
$$
Let
$$
\begin{equation}
\delta=\frac{C}{\log q}, \quad\text{where } C=120\log(R+2)+60.
\end{equation}
\tag{3.10}
$$
Let us denote by $N_3^{(1)}$ and $N_3^{(2)}$ the contributions from $q\leqslant e^{2C}$ and $q>e^{2C}$, respectively. Since
$$
\begin{equation*}
\varphi(a)\gg\frac{a}{\log\log a}\gg\frac{x^{1/60}}{\log\log x},
\end{equation*}
\notag
$$
we have
$$
\begin{equation}
N_3^{(1)}\ll x^{59/60}\log\log x\sum_{q\leqslant e^{2C}}\sum_{1\leqslant l\leqslant \pi(q)}2^l\sum_{\substack{a\leqslant x^{1/30} \\ P^+(a)=q}}1 \ll x^{59/60}(\log x)^{c_R},
\end{equation}
\tag{3.11}
$$
where $c_R>0$ depends only on $R$. Now we turn to $N_3^{(2)}$. For fixed $q$ and $l$ we see that
$$
\begin{equation*}
\begin{aligned} \, S_{q,l} &:=\sum_{\substack{x^{1/60}<a\leqslant x^{1/30} \\ P^+(a)=q, \, \omega(a)=l}}\frac{1}{\varphi(a)} \leqslant \sum_{\substack{a\leqslant x^{1/30} \\ P^+(a)=q, \, \omega(a)=l}}\biggl(\frac{a}{x^{1/60}}\biggr)^{\delta}\frac{1}{\varphi(a)} \\ &\leqslant x^{-\delta/60}\frac{1}{(l-1)!}\biggl(\sum_{\substack{p<q \\ p\not{\equiv} 3\, (\operatorname{mod}4)}}\frac{p^{\delta}}{\varphi(p)} +\frac{p^{2\delta}}{\varphi(p^2)}+\dotsb \biggr)^{l-1}\biggl(\frac{q^{\delta}}{\varphi(q)} +\frac{q^{2\delta}}{\varphi(q^2)}+\dotsb\biggr). \end{aligned}
\end{equation*}
\notag
$$
Note that
$$
\begin{equation*}
\frac{q^{\delta}}{\varphi(q)}+\frac{q^{2\delta}}{\varphi(q^2)}+\dotsb \ll_R \frac1q \quad\text{for } q>e^{2C},
\end{equation*}
\notag
$$
and the sum over the primes $p$ is (since $e^u\leqslant 1+O_K(u)$ for $0\leqslant u\leqslant K$) equal to
$$
\begin{equation*}
\begin{aligned} \, &\sum_{\substack{p<q \\ p\not{\equiv} 3\, (\operatorname{mod}4)}}\frac{p^\delta}{p}+O_R(1) \\ &\qquad=\sum_{\substack{p<q \\ p\not{\equiv} 3\, (\operatorname{mod}4)}}\frac{1}{p}+ O_R\biggl(\delta\sum_{\substack{p<q \\ p\not{\equiv} 3\, (\operatorname{mod}4)}}\frac{\log p}{p}+1\biggr)\leqslant \frac{1}{2}\log\log x+B, \end{aligned}
\end{equation*}
\notag
$$
where $B=B(R)>0$. Thus,
$$
\begin{equation*}
S_{q,l}\ll_R \frac{x^{-\delta/60}}{q(l-1)!}\biggl(\frac{1}{2}\log\log x+B\biggr)^{l-1}
\end{equation*}
\notag
$$
and
$$
\begin{equation}
\begin{aligned} \, N_3^{(2)} &\ll_R x\sum_{\substack{e^{2C}<q\leqslant x^{1/30} \\ q\not\equiv 3\, (\operatorname{mod}4)}}\frac{x^{-\delta/60}}{q\log q}\sum_{k-\eta\leqslant l\leqslant k-1}\frac{2^l(1/2\log\log x+B)^{l-1}}{(l-1)!}\notag \\ &\ll \frac{x(\log\log x+2B)^{k-1}}{(k-1)!}\sum_{q\leqslant x^{1/30}}\frac{x^{-\delta/60}}{q\log q}(1+R+\dots +R^{[\eta]}). \end{aligned}
\end{equation}
\tag{3.12}
$$
Using the inequality and the fact that (which follows from the definitions (3.8) and (3.10) of $\eta$ and $\delta$, respectively), we see that the sum over primes $q$ in (3.12) is at most
$$
\begin{equation*}
\begin{aligned} \, &(R+2) \sum_{q\leqslant x^{1/30}}\frac{(R+2)^{\eta}x^{-\delta/60}}{q\log q} \leqslant (R+2)\sum_{q\leqslant x^{1/2}}\frac{x^{-1/\log q}}{q\log q} \\ &\qquad\leqslant \frac{R+2}{\log x}\sum_{j\geqslant 1}\sum_{q\in (x^{2^{-(j+1)}}, \,x^{-2^j}]}\frac1q \,2^{j+1}\exp(-2^j) \ll_R \frac{1}{\log x}\,. \end{aligned}
\end{equation*}
\notag
$$
Substituting this into (3.12) and taking (3.11) into account we find that
$$
\begin{equation}
N_3 \ll_R \frac{x(\log\log x+2B)^{k-1}}{(k-1)!\,\log x}.
\end{equation}
\tag{3.13}
$$
Combining (3.4), (3.6), (3.7) and (3.13) we complete the proof. § 4. Proof of Theorem 1.1Writing $p-1=ab$, where $P^+(a)\leqslant x^{1/40}$ and $P^-(b)>x^{1/40}$, we can rewrite the sum $\Phi(x)$ as
$$
\begin{equation*}
\Phi(x)=\sum_{\substack{a \leqslant x \\ P^+(a) \leqslant x^{1/40}}} \frac{1}{\tau(a)} \sum_{\substack{b \leqslant (x-1)/a \\ P^-(b) > x^{{1}/{40}} \\ ab +1\text{ is prime}}} \frac{1}{\tau(b)}.
\end{equation*}
\notag
$$
Let $b = p_1^{\alpha_1} p_2^{\alpha_2} \dotsb p_s^{\alpha_s}$ be the prime factorization of $b$. Then
$$
\begin{equation*}
x^{ (\alpha_1 + \alpha_2 + \dots + \alpha_s)/40} < b \leqslant x,
\end{equation*}
\notag
$$
so that $\alpha_1+\dots+\alpha_s < 40$ and
$$
\begin{equation*}
\tau(b)=(\alpha_1+1) \dotsb (\alpha_s+1) \leqslant 2^{\alpha_1+\dots+\alpha_s} < 2^{40}.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\Phi(x) \geqslant 2^{-40} \sum_{\substack{a \leqslant x \\ P^+(a) \leqslant x^{1/40}}} \frac{1}{\tau(a)} \sum_{\substack{b \leqslant (x-1)/a \\ P^-(b) > x^{1/40} \\ ab +1\text{ is prime}}} 1 \geqslant 2^{-40} \sum_{\substack{a \leqslant x^{1/40} \\ a \text{ is even}}} \frac{1}{\tau(a)}F_a(x) .
\end{equation*}
\notag
$$
Lemma 3.2 implies that
$$
\begin{equation*}
\Phi(x) \geqslant \frac{c_2 \pi(x)}{\log x}\sum_{\substack{a \leqslant x^{1/40} \\ a \text{ is even}}} \frac{1}{\tau(a)\varphi(a)}-R_2,
\end{equation*}
\notag
$$
where $c_2>0$ and
$$
\begin{equation*}
0\leqslant R_2 \leqslant \sum_{a\leqslant x^{1/40}}\sum_{d\leqslant x^{13/40}}|R(x;ad,1)| \leqslant \sum_{q\leqslant x^{7/20}}\tau(q)|R(x;q,1)|.
\end{equation*}
\notag
$$
Using the trivial bound $|R(x;q,1)|\ll x/q$ and the Cauchy-Schwarz inequality we obtain
$$
\begin{equation*}
\begin{aligned} \, R_2 &\ll x^{1/2} \sum_{q \leqslant x^{0.35}} \frac{\tau(q)}{q^{1/2}} (|R(x; q, 1)|)^{1/2} \\ &\leqslant x^{1/2} \biggl(\sum_{q \leqslant x^{0.35}} \frac{\tau^2(q)}{q}\biggr)^{1/2} \biggl(\sum_{q \leqslant x^{0.35}} |R(x; q, 1)|\biggr)^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
Applying the bound
$$
\begin{equation*}
\sum_{n\leqslant x}\frac{\tau^2(n)}{n} \asymp (\log x)^4,
\end{equation*}
\notag
$$
which follows from $\sum_{n\leqslant x}\tau^2(n)\asymp x(\log x)^3$ (see [13], Ch. III, Exercise 7) by partial summation, and the Bombieri-Vinogradov theorem we find that, for any fixed ${A>0}$,
$$
\begin{equation*}
R_2 \ll_A x^{1/2}(\log x)^2 \frac{x^{1/2}}{(\log x)^{A/2}}= \frac{x}{(\log x)^{A/2-2}}.
\end{equation*}
\notag
$$
Taking $A = 24$, say, we obtain Now we turn to the sum
$$
\begin{equation*}
T_1=\sum_{\substack{a \leqslant x^{1/40} \\ a \text{ is even}}} \frac{1}{\tau(a) \varphi(a)}=\sum_{l \leqslant 0.5x^{1/40}} \frac{1}{\tau(2l)\varphi(2l)}.
\end{equation*}
\notag
$$
Since $\tau(mn) \leqslant \tau(m)\tau(n)$ and $\varphi(n) \leqslant n$, we see that
$$
\begin{equation*}
T_1\gg \sum_{l \leqslant 0.5x^{1/40}} \frac{1}{\tau(l)l}.
\end{equation*}
\notag
$$
By (1.1), using partial summation we have Hence
$$
\begin{equation*}
\Phi(x)\gg \frac{\pi(x)}{(\log x)^{1/2}}-O\biggl( \frac{x}{(\log x)^{10}}\biggr)\gg \frac{x}{(\log x)^{3/2}}.
\end{equation*}
\notag
$$
Theorem 1.1 is proved. § 5. Proof of Theorem 1.2First we prove the lower bound. For any $z\geqslant2$,
$$
\begin{equation*}
F(x)=\sum_{n \leqslant x} \frac{1}{\tau(n^2+1)}=\sum_{\substack{ab=n^2 + 1,\,n \leqslant x\\ P^+(a) \leqslant z,\, P^-(b) > z}} \frac{1}{\tau(ab)}.
\end{equation*}
\notag
$$
Setting $z = x^{1/30}$ we have $\tau(b) \leqslant 2^{60}$, and now by Lemma 3.4,
$$
\begin{equation*}
\begin{aligned} \, F(x) &\geqslant 2^{-60} \sum_{\substack{a \leqslant x^{1/30} \\ a \in \mathcal{M}}} \frac{1}{\tau(a)} W_a(x, x^{1/30})\gg \frac{x}{\log x} \sum_{\substack{a \leqslant x^{1/30} \\ a \in \mathcal{M}}} \frac{2^{\omega(a)}}{\tau(a)\varphi(a)} \\ &\geqslant \frac{x}{\log x}\sum_{\substack{a \leqslant x^{1/30} \\ a \in \mathcal{M} \\ a\text{ is square-free}}}\frac{1}{a}= \frac{x}{\log x}\sum_{\substack{a \leqslant x^{1/30} \\ a \in \mathcal{M}}} \frac{1}{a}\,\sum_{\delta^2\mid a}\mu(\delta). \end{aligned}
\end{equation*}
\notag
$$
Changing the order of summation and using the fact that we have
$$
\begin{equation}
F(x)\gg \frac{x}{\log x}\sum_{\substack{\delta \leqslant x^{1/60} \\ \delta \in \mathcal{M}}} \frac{\mu(\delta)}{\delta^2} \sum_{\substack{a \leqslant x^{1/30}\delta^{-2} \\ a \in \mathcal{M} }} \frac{1}{a} \geqslant \biggl( 2-\frac{\pi^2}{6}\biggr)\frac{x}{\log x}\sum_{\substack{a \leqslant x^{1/30} \\ a \in \mathcal{M}}} \frac{1}{a}.
\end{equation}
\tag{5.1}
$$
It is well known (see [14], § 183) that for $y\geqslant 2$ one has
$$
\begin{equation*}
\# \{a\leqslant y\colon p\mid a \ \Rightarrow\ p\equiv 1\ (\operatorname{mod} 4)\} \asymp \frac{y}{(\log y)^{1/2}}.
\end{equation*}
\notag
$$
Using partial summation we obtain
$$
\begin{equation*}
\sum_{\substack{a\leqslant y \\ p\mid a \ \Rightarrow\ p\equiv 1\, (\operatorname{mod} 4) }}\frac{1}{a}\asymp (\log y)^{1/2}.
\end{equation*}
\notag
$$
Applying this to (5.1) we find that as required. Now we prove the upper bound. By Lemma 3.5,
$$
\begin{equation*}
\begin{aligned} \, F(x)&\leqslant \sum_{n\leqslant x}\frac{1}{2^{\omega(n^2+1)}} \\ &=\sum_{k\leqslant 2\log\log x}\frac{\#\{n\leqslant x\colon \omega(n^2+1)=k\}}{2^k} + O\biggl(\sum_{k>2\log\log x}\frac{x}{2^k}\biggr) \\ &\ll \frac{x}{\log x}\sum_{k=1}^{+\infty}\frac{((1/2)\log\log x+(1/2)B)^{k-1}}{(k-1)!} + O\biggl(\frac{x}{\log x}\biggr)\ll \frac{x}{(\log x)^{1/2}}. \end{aligned}
\end{equation*}
\notag
$$
Theorem 1.2 is proved. 
Citation in format AMSBIB
\Bibitem{GabKonIud23} |
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