The first negative Fourier coefficient of an Eisenstein series newform

Abstract

There have been a number of papers on statistical questions concerning the sign changes of Fourier coefficients of newforms. In one such paper, Linowitz and Thompson gave a conjecture describing when, on average, the first negative sign of the Fourier coefficients of an Eisenstein series newform occurs. In this paper, we correct their conjecture and prove the corrected version.

1 Introduction

For a Dirichlet character $\chi$ and a positive integer N, we will denote by $M_k(N,\chi )$ the vector space of modular forms on $\Gamma _0(N)$ of weight k, level N and character $\chi$. Let $E_k(N,\chi )$ be the subspace of Eisenstein series and $S_k(N,\chi )$ the subspace of cusp forms. For a prime p, we let $T_p$ be the pth Hecke operator.

Let $H_k^{*}(N)$ be the subspace of $S_k(\chi _0,N)$ of newforms with trivial character $\chi _0$. Given a newform $f\in H_k^{*}(N)$, let $\lambda _f(p)$ be the eigenvalue of f with respect to the Hecke operator $T_p$. The restriction to the trivial character ensures that the sequence $\{\lambda _f(p)\}$ is real. Many authors have studied the sequence of signs of the Hecke eigenvalues of f. For example, one could pose questions such as:

• (i) Are there infinitely many primes p such that $\lambda _f(p)>0$ (or $\lambda _f(p)<0$)?

• (ii) What is the first change of sign? More specifically, what is the smallest $n\geqslant 1$ (or prime p) such that $\lambda _f(n)<0$ (or $\lambda _f(p)<0$)? This is an analogue of the least quadratic non-residue problem.

• (iii) Given an arbitrary sequence of signs $\varepsilon _p\in \{\pm 1\}$, what is the number of newforms f (in some family) such that ${{\,\textrm{sgn}\,}}\lambda _f(p)=\varepsilon _p$ for all $p\leqslant x$?

In the cusp form setting, questions (i) and (ii) are answered in [5, 6], and [9]. In this paper, we focus on (iii). Kowalski et al. [6] obtained a lower bound for the proportion of newforms $f\in H_k^{*}(N)$ whose sequence of eigenvalues $\lambda _f(p)$ has signs coinciding with a prescribed sequence $\{\varepsilon _p\}$:

Theorem 1.1

(Kowalski, Lau, Soundararajan, Wu, 2010) Let N be a squarefree number, $k\geqslant 2$ an even integer, and $\{\varepsilon _p\}$ a sequence of signs. Then, for any $0<\varepsilon <\frac{1}{2}$, there exists some $c>0$ such that

$$\begin{aligned} \frac{1}{|H_k^{*}(N)|}\# \{f\in H_k^{*}(N) \;: \; {{\,\textrm{sgn}\,}}\lambda _f(p)=\varepsilon _p \text{ for } p\leqslant z,\; p\not \mid N\} \geqslant \Big (\frac{1}{2} - \varepsilon \Big )^{\pi (z)} \end{aligned}$$

for $z=c\sqrt{\log {kN}\log \log {kN}}$ provided kN is large enough. Here $\pi (z)$ is the number of primes less than or equal to z.

Now, let $\chi _1$, $\chi _2$ be Dirichlet characters modulo $N_1, N_2$ and for an integer $k>2$, define the following variant of the sum of divisors function:

$$\begin{aligned} \sigma _{\chi _1,\chi _2}^{k-1}(n)=\sum _{d|n} \chi _1\big (\frac{n}{d}\big )\chi _2(d)d^{k-1}. \end{aligned}$$

(1.1)

Now assume that $\chi _1$ and $\chi _2$ are not simultaneously principal $\ (\textrm{mod}\ 1)$. It is well known (see, for example, [3]) that if $\chi _1\chi _2(-1)=(-1)^k$, then the function

$$\begin{aligned} E_k(\chi _1,\chi _2,z):=\frac{\delta (\chi _1)}{2}L(1-k,\chi _2) + \sum _{n\geqslant 1}\sigma _{\chi _1,\chi _2}^{k-1}(n)q^n, \end{aligned}$$

is an Eisenstein series of weight k, level $N_1N_2$ and character $\chi _1\chi _2$. Here $q={{\,\textrm{e}\,}}^{2\pi i z}$ and

$$\begin{aligned} \delta (\chi _1)={\left\{ \begin{array}{ll} 1, &{} \text{ if } \chi _1 \text{ is } \text{ principal } \\ 0, &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

In 1977, Weisinger [11] developed a newform theory for $E_k(N,\chi )$ analogous to the one developed by Atkin and Lehner [1] for cusp forms. In this theory, we have:

• The newforms of $E_k(N,\chi )$ are functions of the form $E_k(\chi _1,\chi _2,z)$ for which $N=N_1N_2$, $\chi =\chi _1\chi _2$, and $\chi _1,\chi _2$ are primitive.

• The eigenvalue of $E_k(\chi _1,\chi _2,z)$ with respect to the Hecke operator $T_p$ is $\sigma _{\chi _1,\chi _2}^{k-1}(p)$. In other words, the eigenvalues of this type of Eisenstein series coincide with its Fourier coefficients.

By exploiting the analytical properties of $\sigma _{\chi _1,\chi _2}^{k-1}(n)$, Linowitz and Thompson [8] answered the three questions mentioned at the beginning of this article for Eisenstein series newforms.

Note that by (1.1), $\sigma _{\chi _1,\chi _2}^{k-1}(n)\in \mathbb {R}$ when $\chi _1,\chi _2$ are real characters. Since we want $E_k(\chi _1,\chi _2,z)$ to be an Eisenstein series, we exclude the case when $\chi _1$ and $\chi _2$ are principal. We call these types of characters quadratic because for every fundamental discriminant D, i.e., for each discriminant arising from a quadratic number field, we can associate a real character defined by $\chi _D(m)=\big (\frac{D}{m}\big )$. Therefore, counting Eisenstein series newforms of level $N\leqslant x$ is equivalent to counting fundamental discriminants $D_1, D_2$ with $|D_1D_2|\leqslant x$. Let

$$\begin{aligned} \mathscr {D}:=\{(D_1, D_2) \;: \; |D_1D_2|\leqslant x\}. \end{aligned}$$

Taking all of these facts into consideration, Linowitz and Thompson [8] showed:

Theorem 1.2

(Linowitz, Thompson, 2015) Let $\{p_1,\ldots , p_k\}$ be a sequence of primes and $\{\varepsilon _{p_1},\ldots , \varepsilon _{p_k}\}\in \{-1,0,1\}$ a sequence of signs. Then,

$$\begin{aligned} \frac{1}{|\mathscr {D}|}\#\{(D_1,D_2)\in \mathscr {D} \; :\;&{}{{\,\textrm{sgn}\,}}\sigma _{\chi _1,\chi _2}^{k-1}(p_i)=\varepsilon _{p_i},\; 1\leqslant i \leqslant k\} \\ {}&{}\xrightarrow [x\rightarrow \infty ]{}\prod _{\begin{array}{c} \varepsilon _{p_i}=0 \\ 1\leqslant i \leqslant k \end{array}}\frac{1}{(p_i+1)^2}\prod _{\begin{array}{c} \varepsilon _{p_i}\ne 0 \\ 1\leqslant i \leqslant k \end{array}}\frac{p_i(p_i+2)}{2(p_i+1)^2}\cdot \end{aligned}$$

Now, let $\eta (D_1,D_2)$ represent the smallest prime p such that ${{\,\textrm{sgn}\,}}(\sigma _{\chi _1,\chi _2}^{k-1}(p))=-1$. Linowitz and Thompson [8] then conjectured:

Conjecture 1.3

We have

$$\begin{aligned} \frac{\sum _{|D_1D_2|\leqslant x}\eta (D_1,D_2)}{\sum _{|D_1D_2|\leqslant x} 1} \xrightarrow [x\rightarrow \infty ]{} \theta , \end{aligned}$$

where

$$\begin{aligned} \theta :=\sum _{k=1}^{\infty }\frac{p_k^2(p_k+2)}{2(p_k+1)^2}\prod _{j=1}^{k-1}\frac{2+p_j(p_j+2)}{2(p_j+1)^2}\approx 3.9750223902\ldots \end{aligned}$$

(1.2)

They gave a heuristic argument as evidence towards their conjecture, showing:

$$\begin{aligned} \frac{\sum _{|D_1D_2|\leqslant x}\eta (D_1,D_2)}{\sum _{|D_1D_2|\leqslant x} 1} \xrightarrow [x\rightarrow \infty ]{}{}&\sum _{k=1}^{\infty }p_k\, \text{ Prob }(\eta (D_1,D_2)=p_k) \\ ={}&\sum _{k=1}^{\infty }p_k\, \text{ Prob }(\varepsilon _{p_k}=-1)\prod _{i=1}^{k-1}\text{ Prob }(\varepsilon _{p_i}=0 \text{ or } 1) \\ ={}&\sum _{k=1}^{\infty }\frac{p_k^2(p_k+2)}{2(p_k+1)^2}\prod _{i=1}^{k-1}\bigg (\frac{1}{(p_i+1)^2}+\frac{p_i(p_i+2)}{2(p_i+1)^2}\bigg ), \end{aligned}$$

where the last equality follows from Theorem 1.2. The problem with this argument is that Theorem 1.2 fixes a set of primes and then lets $x\rightarrow \infty$. In this argument we need to allow the primes to tend to infinity with x. The authors stated: “[W]e have a good understanding of the effect of the small primes, but one would need to argue that the primes after some cutoff point do not make much of an impact on the average. Presumably, this would require using the large sieve”.

The goal of the present article is to correct their conjecture by proving the following result:

Theorem 1.4

We have

$$\begin{aligned} \frac{\sum _{|D_1D_2|\leqslant x}\eta (D_1,D_2)}{\sum _{|D_1D_2|\leqslant x} 1} \xrightarrow [x\rightarrow \infty ]{} {}\Theta \cdot (1-\beta )+\alpha , \end{aligned}$$

where

$$\begin{aligned} \Theta= & {} \sum _{k=1}^{\infty }\frac{p_k^2}{2(p_k+1)}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)},\\ \alpha= & {} \sum _{k=1}^{\infty }\frac{p_k^2}{2(p_k+1)^2}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}, \end{aligned}$$

and

$$\begin{aligned} \beta = \sum _{k=1}^{\infty }\frac{p_k}{2(p_k+1)^2}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}\cdot \end{aligned}$$

Numerically,

$$\begin{aligned} \Theta \cdot (1-\beta )+\alpha \approx 4.63255603509332\ldots \end{aligned}$$

The numerical computation was done using Sage. We used RIF for interval arithmetic and we truncated at $k=1000$.

2 Main Tools

First we will need asymptotic estimates for some sets of fundamental discriminants. It is well known (see, for example, [2]) that

$$\begin{aligned} \sum _{|D|\leqslant x} 1 \sim \frac{x}{\zeta (2)}, \end{aligned}$$

(2.1)

where D runs over all fundamental discriminants with $|D|\leqslant x$. Here $\zeta$ is the Riemann zeta function. Now, let $n_1(m)$ be the smallest integer $n\geqslant 1$ relatively prime to m such that the congruence $x^2\equiv n\ (\textrm{mod}\ m)$ has no solutions. Even though Vinogradov’s conjecture remains open, it is possible to show that large values of $n_1(p)$ are rare. More specifically, using the large sieve, Linnik [7] showed that for all $\varepsilon >0$, we have

$$\begin{aligned} \#\{p\leqslant x \;: \; n_1(p)>x^{\varepsilon }\}\ll _{\varepsilon } 1. \end{aligned}$$

Using similar ideas to the ones from Linnik’s paper, Erdős [4] obtained a result concerning the average of $n_1(p)$ as p varies over prime numbers less than or equal to x:

$$\begin{aligned} \frac{1}{\pi (x)}\sum _{p\leqslant x}n_1(p)\xrightarrow [x\rightarrow \infty ]{}\sum _{k=1}^{\infty }\frac{p_k}{2^k}, \end{aligned}$$

(2.2)

where $p_k$ is the kth prime and $\pi (x)$ is the prime counting function. In a similar fashion, Pollack [10] considered a variation of (2.2). We summarize his result in the following theorem:

Theorem 2.1

(Pollack, 2012) For each fundamental discriminant D, let $\chi _D$ be the associated Dirichlet character, i.e., $\chi _D(m):=\big (\frac{D}{m}\big )$. For each character $\chi$, let $n_{\chi }$ denote the least n for which $\chi (n)\notin \{0,1\}$. Finally, let $n(D):=n_{\chi _D}$. Then

• (i) Uniformly in k such that the kth prime satisfies $p_k\leqslant (\log {x})^{\frac{1}{3}}$, we have

  $$\begin{aligned} \# \{|D|\leqslant x \; : \; n(D)=p_k\}=\frac{p_k}{2(p_k+1)}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}\frac{x}{\zeta (2)}+ O(x^{\frac{2}{3}}). \end{aligned}$$
• (ii)
  $$\begin{aligned} \sum _{\begin{array}{c} |D|\leqslant x \\ n(D)>(\log {x})^{\frac{1}{3}} \end{array}}n(D)=o(x). \end{aligned}$$

Therefore, using (2.1), we have

$$\begin{aligned} \frac{\sum _{|D|\leqslant x}n(D)}{\sum _{|D|\leqslant x} 1} \xrightarrow [x\rightarrow \infty ]{} \Theta , \end{aligned}$$

(2.3)

where

$$\begin{aligned} \Theta :=\sum _{k=1}^{\infty }\frac{p_k^2}{2(p_k+1)}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}\approx 4.9809473396\ldots \end{aligned}$$

We will also need the following lemma from Linowitz and Thompson [8]:

Lemma 2.2

Let $\mathbb {P}(\varepsilon , p)$ denote the proportion of fundamental discriminants D with $\big ( \tfrac{D}{p}\big )=\varepsilon$. Then, we have

$$\begin{aligned} \mathbb {P}(\varepsilon , p)={\left\{ \begin{array}{ll} \frac{p}{2p+2}, &{} \text{ if } \varepsilon \in \{\pm 1\} \\ \frac{1}{p+1}, &{} \text{ if } \varepsilon =0. \end{array}\right. } \end{aligned}$$

3 Proof of Theorem 1.4

Let $\chi _1, \chi _2$ be Dirichlet characters associated with the fundamental discriminants $D_1$ and $D_2$. For a prime p,

$$\begin{aligned} \sigma _{\chi _1,\chi _2}^{k-1}(p)=\sum _{d|p} \chi _1\big (\frac{p}{d}\big )\chi _2(d)d^{k-1}=\chi _1(p)+\chi _2(p)p^{k-1}, \end{aligned}$$

so that

$$\begin{aligned} {{\,\textrm{sgn}\,}}{\sigma _{\chi _1,\chi _2}^{k-1}(p)}={\left\{ \begin{array}{ll} \chi _1(p), &{} \text{ if } p|D_2 \\ \chi _2(p), &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

(3.1)

Proof of Theorem 1.4

By (2.1),

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x} 1=\sum _{|D_1|\leqslant x}\sum _{D_2\leqslant \frac{x}{|D_1|}}1\sim \frac{x}{\zeta (2)}\sum _{|D_1|\leqslant x}\frac{1}{ |D_1|}\cdot \end{aligned}$$

Let $A(x):=\sum _{|D|\leqslant x} 1$ and $f(x):=\frac{1}{x}$. Since $A(x)\sim \frac{x}{\zeta (2)}$, then by partial summation

$$\begin{aligned} \sum _{|D_1|\leqslant x}\frac{1}{|D_1|}={}&A(x)f(x)-A(1)f(1)-\int _{1}^{x}A(t)f'(t)\textrm{d}t\nonumber \\ \sim {}&\frac{1}{\zeta (2)}-1+\int _{1}^{x}\frac{\textrm{d}t}{\zeta (2)t} \nonumber \\ \sim {}&\frac{\log {x}}{\zeta (2)}\cdot \end{aligned}$$

(3.2)

Hence

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x} 1\sim \frac{x\log {x}}{\zeta (2)^2}\cdot \end{aligned}$$

(3.3)

Now let us estimate the numerator. For the sake of simplicity, let $\eta :=\eta (D_1,D_2)$. Then,

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x}\eta =\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}\eta +\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta \not \mid D_2 \end{array}}\eta . \end{aligned}$$

If $\eta | D_2$, then by (3.1), $\eta$ is the smallest prime p such that $\chi _1(p)\notin \{0,1\}$, and with the notation of Theorem 2.1, this means that $\eta =n(D_1)$. Similarly, if $\eta \not \mid D_2$, then $\eta =n(D_2)$. Therefore,

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x}\eta =\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_1)+\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta \not \mid D_2 \end{array}}n(D_2). \end{aligned}$$

Now,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta \not \mid D_2 \end{array}}n(D_2)=\sum _{|D_1D_2|\leqslant x }n(D_2)-\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta | D_2 \end{array}}n(D_2), \end{aligned}$$

so that

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x}\eta =\sum _{|D_1D_2|\leqslant x }n(D_2)+\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_1)-\sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_2). \end{aligned}$$

(3.4)

By (2.3), we have

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x }n(D_2)=&{}\sum _{|D_1|\leqslant x}\sum _{|D_2|\leqslant \frac{x}{|D_1|}}n(D_2) \nonumber \\ \sim&{}\Theta \frac{ x}{\zeta (2)}\sum _{|D_1|\leqslant x}\frac{1}{|D_1|} \nonumber \\ \sim&{} \Theta \frac{x\log {x}}{\zeta (2)^2}, \end{aligned}$$

(3.5)

where the final estimate follows from (3.2). Now, by Lemma 2.2, the proportion of fundamental discriminants such that p|D is $\frac{1}{p+1}\cdot$ Hence,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_1)={}&\sum _{|D_1|\leqslant x}n(D_1)\sum _{\begin{array}{c} |D_2|\leqslant \frac{x}{|D_1|} \ \\ n(D_1)|D_2 \end{array}}1 \\ ={}&\sum _{|D_1|\leqslant x}\frac{n(D_1)}{n(D_1)+1}\sum _{|D_2|\leqslant \frac{x}{|D_1|}}1 \\ \sim {}&\frac{x}{\zeta (2)}\sum _{|D_1|\leqslant x}\frac{n(D_1)}{|D_1|(n(D_1)+1)}\cdot \end{aligned}$$

To find an asymptotic for the last sum we again use partial summation. Let

$$\begin{aligned} B(x):=\sum _{|D_1|\leqslant x}\frac{n(D_1)}{n(D_1)+1}\cdot \end{aligned}$$

Then, by (i) of Theorem 2.1,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1|\leqslant x\\ n(D_1)\leqslant (\log {x})^{\frac{1}{3}} \end{array}}\frac{n(D_1)}{n(D_1)+1}={}&\sum _{\begin{array}{c} k=1 \\ p_k\leqslant (\log {x})^{\frac{1}{3}} \end{array}}^{\infty }\frac{p_k}{p_k+1}\# \{|D_1|\leqslant x \; : \; n(D_1)=p_k\} \\ \sim {}&\alpha \frac{x}{\zeta (2)} \end{aligned}$$

where

$$\begin{aligned} \alpha = \sum _{k=1}^{\infty }\frac{p_k^2}{2(p_k+1)^2}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}\cdot \end{aligned}$$

Now, by (ii) of Theorem 2.1,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1|\leqslant x\\ n(D_1)>(\log {x})^{\frac{1}{3}} \end{array}}\frac{n(D_1)}{n(D_1)+1}\leqslant \sum _{\begin{array}{c} |D_1|\leqslant x\\ n(D_1)>(\log {x})^{\frac{1}{3}} \end{array}}n(D_1)=o(x). \end{aligned}$$

Hence,

$$\begin{aligned} B(x)= \sum _{\begin{array}{c} |D_1|\leqslant x\\ n(D_1)\leqslant (\log {x})^{\frac{1}{3}} \end{array}}\frac{n(D_1)}{n(D_1)+1} + \sum _{\begin{array}{c} |D_1|\leqslant x\\ n(D_1)>(\log {x})^{\frac{1}{3}} \end{array}}\frac{n(D_1)}{n(D_1)+1} \sim \alpha \frac{x}{\zeta (2)}\cdot \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{|D_1|\leqslant x}\frac{n(D_1)}{|D_1|(n(D_1)+1)}=B(x)f(x)-B(1)f(1)-\int _{1}^{x}B(t)f'(t)\textrm{d}t\sim \alpha \frac{\log {x}}{\zeta (2)}, \end{aligned}$$

so that

$$\begin{aligned} \sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_1)\sim \alpha \frac{x\log {x}}{\zeta (2)^2}\cdot \end{aligned}$$

(3.6)

Finally,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_2)=\sum _{|D_2|\leqslant x}n(D_2) \sum _{|D_1|\leqslant \frac{x}{|D_2|}}\frac{1}{n(D_1)+1}\cdot \end{aligned}$$

To get an estimate for the inner sum, let

$$\begin{aligned} C(x):=\sum _{|D_1|\leqslant \frac{x}{|D_2|}}\frac{1}{n(D_1)+1}\cdot \end{aligned}$$

Then, by (i) of Theorem 2.1,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1|\leqslant \frac{x}{|D_2|}\\ n(D_1)\leqslant (\log {x})^{\frac{1}{3}} \end{array}}\frac{1}{n(D_1)+1}={}&\sum _{\begin{array}{c} k=1 \\ p_k\leqslant (\log {x})^{\frac{1}{3}} \end{array}}^{\infty }\frac{1}{p_k+1}\# \{|D_1|\leqslant \frac{x}{|D_2|} \; : \; n(D_1)=p_k\} \\ \sim {}&\beta \frac{x}{\zeta (2)|D_2|} \end{aligned}$$

where

$$\begin{aligned} \beta = \sum _{k=1}^{\infty }\frac{p_k}{2(p_k+1)^2}\prod _{j=1}^{k-1}\frac{p_j+2}{2(p_j+1)}\cdot \end{aligned}$$

On the other hand,

$$\begin{aligned} \sum _{\begin{array}{c} |D_1|\leqslant \frac{x}{|D_2|}\\ n(D_1)>(\log {x})^{\frac{1}{3}} \end{array}}\frac{1}{n(D_1)+1}\leqslant {}&\sum _{|D_1|\leqslant \frac{x}{|D_2|}}\frac{1}{(\log {x})^{\frac{1}{3}}+1} \\ \sim {}&\frac{x}{|D_2|\zeta (2)((\log {x})^{\frac{1}{3}}+1)} \\ \leqslant {}&\frac{x}{(\log {x})^{\frac{1}{3}}+1} \\ ={}&o(x). \end{aligned}$$

Hence,

$$\begin{aligned} C(x)= \sum _{\begin{array}{c} |D_1|\leqslant \frac{x}{|D_2|}\\ n(D_1)\leqslant (\log {x})^{\frac{1}{3}} \end{array}}\frac{1}{n(D_1)+1} + \sum _{\begin{array}{c} |D_1|\leqslant \frac{x}{|D_2|}\\ n(D_1)>(\log {x})^{\frac{1}{3}} \end{array}}\frac{1}{n(D_1)+1} \sim \beta \frac{x}{\zeta (2)|D_2|}\cdot \end{aligned}$$

From this we see that

$$\begin{aligned} \sum _{\begin{array}{c} |D_1D_2|\leqslant x \\ \eta |D_2 \end{array}}n(D_2)\sim \sum _{|D_2|\leqslant x}\beta \frac{n(D_2)x}{\zeta (2)|D_2|}\sim \Theta \beta \frac{x\log {x}}{\zeta (2)^2}, \end{aligned}$$

(3.7)

where the last estimate follows from partial summation and applying Theorem 2.1. Therefore, plugging (3.5), (3.6) and (3.7) into (3.4) shows that

$$\begin{aligned} \sum _{|D_1D_2|\leqslant x}\eta \sim (\Theta +\alpha -\Theta \beta )\frac{x\log {x}}{\zeta (2)^2}\cdot \end{aligned}$$

This together with (3.3) completes the proof. $\square$

Remark 3.1

We can give the following explanation of why Linowitz and Thompson’s Conjecture 1.3 was slightly off from the correct number: the result from Theorem 1.2 is not uniform in k for the choice of the $p_k$ (we fix a set of primes beforehand), while the result from Theorem 2.1 is uniform in k satisfying $p_k\leqslant (\log {x})^{\frac{1}{3}}$. In order to make Linowitz and Thompson’s heuristic argument rigorous we would first need to show that Theorem 1.2 holds uniformly in k such that $p_k\leqslant f(x)$ for some function f with $f(x) \xrightarrow [x\rightarrow \infty ]{} \infty$. Then,

$$\begin{aligned} \frac{\sum _{|D_1D_2|\leqslant x}\eta (D_1,D_2)}{\sum _{|D_1D_2|\leqslant x} 1} =\sum _{p_k\leqslant f(x)}{}&p_k\, \text{ Prob }(\eta (D_1,D_2)=p_k) \\ {}&+ \sum _{p_k>f(x)}p_k\, \text{ Prob }(\eta (D_1,D_2)=p_k) \\ \xrightarrow [x\rightarrow \infty ]{}{}&\theta + \mu , \end{aligned}$$

where $\theta$ is the conjectured constant (1.2) and

$$\begin{aligned} \mu =\lim _{x\rightarrow \infty } \sum _{p_k>f(x)}p_k\, \text{ Prob }(\eta (D_1,D_2)=p_k). \end{aligned}$$

Linowitz and Thompson conjectured that $\mu =0$, but according to Theorem 1.4, $\mu$ does make a small contribution.