On the Ces\`aro average of the numbers that can be written as sum of a prime and two squares of primes

TL;DR

This paper derives an asymptotic formula for the sum of the von Mangoldt-weighted count of numbers that are sums of a prime and two prime squares, extending understanding of their distribution.

Contribution

It provides a new asymptotic expansion for the weighted count of numbers expressible as a prime plus two prime squares, involving specific parameters and error bounds.

Findings

Asymptotic formula for the sum involving r_{SP}(n) and (N-n)^k

Identification of main terms M_i(N,k) in the expansion

Error term bounded by O(N^{k+1})

Abstract

Let $\Lambda\left(n\right)$ be the von Mangoldt function and $r_{SP}\left(n\right)=\sum_{m_{1}+m_{2}^{2}+m_{3}^{2}=n}\Lambda\left(m_{1}\right)\Lambda\left(m_{2}\right)\Lambda\left(m_{3}\right)$ be the counting function for the numbers that can be written as sum of a prime and two squares of primes. Let $N$ a sufficiently large integer, let $k>3/2$ and let $M_{i}\left(N,k\right),\, i=1,\dots,4$ suitable parameters depending on $\Gamma(s)$. We prove that $$\sum_{n\leq N}r_{SP}\left(n\right)\frac{\left(N-n\right)^{k}}{\Gamma\left(k+1\right)}=M_{1}\left(N,k\right)+M_{2}\left(N,k\right)+M_{3}\left(N,k\right)+M_{4}\left(N,k\right)+O\left(N^{k+1}\right).$$

Equations

$$\sum _{n\le N}{r}_{SP}\left(n\right)\frac{{\left(N-n\right)}^k}{\Gamma \left(k+1\right)}=\frac{N^{k+2}\pi }{4\Gamma \left(k+3\right)}+E\left(N,k\right)$$

$$r_{SP}\left(n\right)=\sum _{m_1+m_2^2+m_3^2=n}\Lambda \left(m_1\right)\Lambda \left(m_2\right)\Lambda \left(m_3\right)$$

$$M_1\left(N,k\right)=$$

$$M_2\left(N,k\right)=$$

$$M_3\left(N,k\right)=$$

$$+$$

$$M_4\left(N,k\right)=$$

$$\sum _{n\le N}{r}_{SP}\left(n\right)\frac{{\left(N-n\right)}^k}{\Gamma \left(k+1\right)}=M_1\left(N,k\right)+M_2\left(N,k\right)+M_3\left(N,k\right)+M_4\left(N,k\right)+O\left(N^{k+1}\right)$$

$$\overline{\beta }:=\sup \left\{\beta :\text{Re}\left(\rho \right)=\beta \right\}.$$

$$M_2\left(N,k\right){\ll }_k{N}^{k+3/2+\overline{\beta }/2}$$

$$M_3\left(N,k\right){\ll }_k{N}^{k+1+\overline{\beta }}$$

$$M_4\left(N,k\right){\ll }_k{N}^{k+2\overline{\beta }}.$$

$$A=\left\{n\in \mathbb{N}:n\equiv 1\mathrm{mod}2;n\not\equiv 2\mathrm{mod}3\right\};$$

$$E\left(N\right)\ll N^{1/3+ϵ}$$

$$r\left(n\right):=\sum _{p_1+p_2^2+p_3^2=n}\log \left(p_1\right)\log \left(p_2\right)\log \left(p_3\right)$$

$$\frac{1}{2\pi i}{\int }_{\left(a\right)}{v}^{-s}{e}^{v}dv=\frac{1}{\Gamma \left(s\right)}$$

$${\tilde{S}}_1\left(z\right)=$$

$${\tilde{S}}_2\left(z\right)=$$

$${\tilde{S}}_1\left(z\right)=\frac{1}{z}-\sum _{\rho }{z}^{-\rho }\Gamma \left(\rho \right)+E\left(a,y\right)$$

$$E\left(a,y\right)\ll {\left|z\right|}^{1/2}\{\begin{array}{ll}1,& \left|y\right|\le a\\ 1+{\log }^2\left(\left|y\right|/a\right),& \left|y\right|>a.\end{array}$$

$$\left|\sum _{\rho }{z}^{-\rho }\Gamma \left(\rho \right)\right|=\left|\frac{1}{z}-\tilde{S}\left(z\right)+E\left(\frac{1}{N},y\right)\right|\ll N+\frac{1}{\left|z\right|}+\left|E\left(\frac{1}{N},y\right)\right|$$

$$\ll \{\begin{array}{ll}N,& \left|y\right|\le 1/N\\ N+{\left|z\right|}^{1/2}{\log }^2\left(2N\left|y\right|\right),& \left|y\right|>1/N.\end{array}$$

$${\tilde{S}}_{\ell }\left(z\right)=\frac{\Gamma \left(1/\ell \right)}{\ell z^{1/\ell }}-\frac{1}{\ell }\sum _{\rho }{z}^{-\rho /\ell }\Gamma \left(\frac{\rho }{\ell }\right)+E_{\ell }\left(a,y\right)$$

$$E_{\ell }\left(a,y\right){\ll }_{\ell }E\left(a,y\right).$$

$${\tilde{S}}_{\ell }\left(z\right)=$$

$$=$$

$$\left|\frac{{\zeta }^{\prime }}{\zeta }\left(\ell w\right)\right|{\ll }_{\ell }\log \left(\left|t\right|+2\right)$$

$${\tilde{S}}_1\left(a\right)\sim a^{-1},\text{ when }a\to 0^{+}$$

$${\tilde{S}}_2\left(a\right)\sim \frac{\sqrt{\pi }}{2a^{1/2}},\text{ when }a\to 0^{+}.$$

$$\left|\Gamma \left(x+iy\right)\right|\sim \sqrt{2\pi }{e}^{-\pi |y|/2}{\left|y\right|}^{x-1/2}$$

$$\left|z^{-w}\right|={\left|z\right|}^{-\text{Re}\left(w\right)}\exp \left(\text{Im}\left(w\right)\arctan \left(y/a\right)\right).$$

$$\sum _{\rho ,\gamma >0}{\gamma }^{\beta -1/2}{\int }_1^{\infty }\exp \left(-\gamma \arctan \left(1/u\right)\right)\frac{du}{u^{\alpha +\beta }}$$

$$\sum _{\rho }{\left|\gamma \right|}^{\beta -1/2}{\int }_{{\mathbb{Y}}_1\cup {\mathbb{Y}}_2}\exp \left(\gamma \arctan \left(\frac{y}{a}\right)-\frac{\pi }{2}\left|\gamma \right|\right)\frac{dy}{{\left|z\right|}^{\alpha +\beta }}{\ll }_{\alpha }{a}^{-\alpha }$$

Full text

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for $k>3/2$, where $E\left(N,k\right)$ consists of lower order terms that are given in terms of $k$ and sum over the non-trivial zeros of the Riemann zeta function.

1. Introduction

We continue the recent work of Languasco, Zaccagnini and the author on additive problems with prime summands. In [12] and [13] Languasco and Zaccagnini study the Cesàro weighted explicit formula for the Goldbach numbers (the integers that can be written as sum of two primes) and for the Hardy-Littlewood numbers (the integers that can be written as sum of a prime and a square). Recently [2] the author wrote a paper regarding the Cesàro average of the integers that can be written as sum of a prime and two squares. In a similar manner, we will study a Cesàro weighted explicit formula for the integers that can be written as sum of a prime and two squares of primes. We will obtain an asymptotic formula with a main term and more terms depending explicitly on the zeros of the Riemann zeta function. This technique allow us to obtain a large number of terms in our asymptotic but unfortunately the bound $k>3/2$ seems to be very difficult to improve. We recall that, for $k=0$, the Cesàro weights vanish so a result for $k\geq 0$ would allow us to get an asymptotic for the mean of $r_{SP}\left(n\right).$

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010 Mathematics Subject Classification: Primary 11P32; Secondary 44A10, 33C10

We let

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where $\Lambda\left(n\right)$ is the Von Mangoldt function and

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The main result of this paper is the following

Theorem 1**.**

Let $N$ be a sufficient large integer. We have

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for $k>3/2$, where $\rho=\beta+i\gamma$, with or without subscripts, runs over the non-trivial zeros of the Riemann zeta function $\zeta\left(s\right)$.

Note that an upper bound for $M_{i}\left(N,k\right),\,i=2,\dots,4$ depends closely on $\beta$. Let us define

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We have that

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Note also that, if the Riemann hypothesis is true, then $M_{4}\left(N,k\right)$ can be incorporated in the error term. The problem of representing an integer as sum of a prime and two prime squares is classical. Let

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it is conjectured that every sufficiently large natural number $n\in A$ is a sum of a prime and two prime squares. Many authors studied the cardinality $E\left(N\right)$ of the set of integers $n\leq N$, $n\in A$ that are not representable as a sum of prime and two squares of primes. We recall Hua [10], Schwarz [19], Leung-Liu [16], Wang [21], Wang-Meng [22], Li [17], Harman-Kumchev [9]. Zhao [24] proved that

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and so every integer $n\in\left[1,N\right]\cap A$, with at most $O\left(N^{1/3+\epsilon}\right)$ exceptions, is a sum of a prime and two squares of primes. Letting

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Zhao also proved that an asymptotic formula for $r\left(n\right)$ holds for $n\in\left[1,N\right]\cap A$, with at most $O\left(N^{1/3+\epsilon}\right)$ exceptions. Similar averages of arithmetical functions are common in literature, see, e.g., Chandrasekharan - Narasimhan [3] and Berndt [1] who built on earlier classical work. The method we will use in this additive problem is based on a formula due to Laplace [15], namely

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with $\textrm{Re}\left(s\right)>0$ and $a>0$ (see, e.g., formula 5.4 (1) on page 238 of [5]), where the notation $\int_{\left(a\right)}$ means $\int_{a-i\infty}^{a+i\infty}$. As in [13], we combine this approach with line integrals with the classical methods dealing with infinite sum over primes and integers.

I thank A. Zaccagnini and A. Languasco for their contributions and the conversations on this topic. I also thank the referee, who pointed out further inaccuracies and suggested improvements in the presentation. This work is part of the Author’s Ph.D. thesis.

2. Preliminary definitions and Lemmas

Let $z=a+iy$, $a>0$, let

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and let us introduce the following

Lemma 2**.**

Let $z=a+iy,\,a>0$ and $y\in\mathbb{R}$. Then

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where $\rho=\beta+i\gamma$ runs over the non-trivial zeros of $\zeta\left(s\right)$ and

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(For a proof see Lemma 1 of [12]. The bound for $E\left(a,y\right)$ has been corrected in [11]). So in particular, taking $z=\frac{1}{N}+iy$ we have

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We now introduce the following

Lemma 3**.**

Let $z=a+iy,\,a>0$, $y\in\mathbb{R}$ and $\ell$ a fixed positive integer. Then

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where $\rho=\beta+i\gamma$ runs over the non-trivial zeros of $\zeta\left(s\right)$ and

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Proof.

It is well know (see for example formula 5 of [14]) that, for $\ell\in\mathbb{N}_{0}$,

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so, taking $w=-\frac{1}{2}+it$, following the proof of the Lemma 1 in [12] and observing that

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we can conclude that we may estimate the integral in (13) exactly as in [12], so the claim follows. ∎

Now we have to recall that the Prime Number Theorem (PNT) is equivalent, via Lemma 2, to the statement

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(see Lemma 9 of [8]) and from Lemma 3 we have

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For our purposes it is important to introduce the Stirling approximation (see for example §4.42 of [20])

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uniformly for $x\in\left[x_{1},x_{2}\right]$, $x_{1}$ and $x_{2}$ fixed, as well as the identity

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We now quote Lemmas 2 and 3 from [12]:

Lemma 4**.**

Let $\beta+i\gamma$ run over the non-trivial zeros of the Riemann zeta function and let $\alpha>1$ be a parameter. The series

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converges provided that $\alpha>3/2$. For $\alpha\leq 3/2$ the series does not converge. The result remains true if we insert in the integral a factor $\log^{c}\left(u\right)$, for any fixed $c\geq 0$.

Lemma 5**.**

Let $\beta+i\gamma$ run over the non-trivial zeros of the Riemann zeta function, let $z=a+iy,$ $a\in\left(0,1\right)$, $y\in\mathbb{R}$ and $\alpha>1$. We have

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where $\mathbb{Y}_{1}=\left\{y\in\mathbb{R}:\,\gamma y\leq 0\right\}$ and $\mathbb{Y}_{2}=\left\{y\in\left[-a,a\right]:\,y\gamma>0\right\}$. The result remains true if we insert in the integral a factor $\log^{c}\left(\left|y\right|/a\right)$, for any fixed $c\geq 0.$

Let us introduce another lemma

Lemma 6**.**

Let $\rho=\beta+i\gamma$ run over the non-trivial zeros of the Riemann zeta function, let $z=\frac{1}{N}+iy,$ where $N>1$ is a natural number, $y\in\mathbb{R}$, $\ell\geq 1$ an integer and $\alpha>3/2$. We have

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Proof.

Put $a=\frac{1}{N}.$ Using the identity (17), (16) and

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we get that the left hand side in the statement above is

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The case $\ell=1$ has already been discussed in Lemma 6 of [2]. For $\ell>1$, observing Lemmas 2 and 3 of [12] and Lemma 6 [2], we can conclude that the presence of $\ell$ does not alter the proofs. Hence using the same argument of Lemma 6 of [2] we have the convergence for $\alpha>3/2$. ∎

3. Setting

From (6) and (7) it is not hard to see that

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so let $z=a+iy$ and $a>0$ and let us consider

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Now we prove that we can exchange the integral with the series. From (14) and (15) we have

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hence

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assuming $k>0$, so we have that

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Now from (8), (11), (14) and (15) and observing that, for $\ell\geq 1,$

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we have

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Now let us consider $l,m,r,s\geq 1$ integers. From (9) and (12) we have that

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[TABLE]

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assuming $k>\frac{r+s}{2}$. We now have to deal with the terms in (21) and (22): taking $a=1/N$ we can observe that

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and

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hence the Cesàro average of $r_{SP}\left(n\right)$ can be broken down as

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say. In the next sections we will prove that $I_{1}=M_{1}\left(N,k\right)$, $I_{2}+I_{3}=M_{2}\left(N,k\right)$, $I_{4}+I_{5}=M_{3}\left(N,k\right)$ and $I_{6}=M_{4}\left(N,k\right).$

4. Evaluation of $I_{1}$

From $I_{1}$ we will find the main term. If we put $Nz=s$ we get

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using (5). Then $I_{1}=M_{1}\left(N,k\right)$.

5. Evaluation of $I_{2}$ and $I_{3}$

We have

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and

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We want to exchange the integral with the series, then we will prove the absolute convergence for a suitable choice of $k.$ Hence we have to study

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and

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and from Lemma 6 we have the convergence for $k>-1/2$ and $k>-1$ respectively. So we can switch the integral and the series and get

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and

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then $I_{2}+I_{3}=M_{2}\left(N,k\right)$.

6. Evaluation of $I_{4}$

We have to evaluate

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We want to switch the integral with two series so we will prove the absolute convergence of

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and

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Now we have to introduce some notations, which is necessary since the evaluation of the integrals depends strictly on the sign of $y$ and the sign of the imaginary part of $\rho.$ Assume that $A_{m,n}:=\int_{\left(1/N\right)}\dots\left|dz\right|.$ Hereafter we will use the symbol

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and

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From (11) we can see that

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Let us consider $y\leq 0$ and, recalling the notation $\rho_{j}=\beta_{j}+i\gamma_{j}$, the notation (25) and assuming $\gamma_{1}>0$ for symmetry, we have to study

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from Lemma 5, assuming that $k>0.$ Note that we have to split the integral since, from (18) and (26), we have different evaluations if $\left|y\right|\leq 1/N$ or $\left|y\right|>1/N.$ Now let us consider $y>0$. Recalling (24), we have to study

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say, and we have that

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from Lemma 5 and

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from Lemma 4, assuming $k>1/2$. Now let us consider

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By symmetry, it suffices to consider only the cases $\gamma_{1},\gamma_{2}>0$ and $\gamma_{1}>0,$ $\gamma_{2}<0$. As in (24) and (25) we have to introduce some new notations since the evaluation depends on the sign of the product $\gamma_{1}\gamma_{2}$ and the sign of $y.$ Hereafter we will use the symbol $B_{m,n}$ when we consider $A_{m,n}$ with the assumption $\gamma_{1},\gamma_{2}>0$ and the symbol $C_{m,n}$ when we consider $A_{m,n}$ with the assumption $\gamma_{1}>0,$ $\gamma_{2}<0$. Since

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and recalling (25), we have

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Now let us consider $y>0$.We have

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say. If $y\in\left(0,1/N\right]$ we obviously have $\arctan\left(Ny\right)-\frac{\pi}{2}\leq-\frac{\pi}{4}$ and so

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For $\mathcal{A}_{4}$ we can see, following the proof of the Lemma 4, that we have

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and observing that

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we get

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and so we proved the convergence if $k>1/2$ using the Riemann - Von Mangoldt formula. Let us consider the case $\gamma_{1}>0$, $\gamma_{2}<0$ (and so we will use the symbol $C_{m,n}$) and let $y\leq 0$. Using again (27) we have to study

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and using Lemma 4, Lemma 5 and the identity $\arctan\left(x\right)+\arctan\left(1/x\right)=-\pi/2,\,x<0$ we have

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for $k>-1/2$. If $y>0$ we have essentially the same situation exchanging the role of $\gamma_{1}$ and $\gamma_{2}$. So we can switch the integral with the series and get

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7. Evaluation of $I_{5}$

We have to evaluate

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and we can see that the argument used in $I_{4}$ works also in this case since the presence of $\beta_{1}/2$ instead of $\beta_{1}$ does not alter the validity of the proof. So repeating the reasoning we can obtain the convergence for $k>1/2$ and so

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[TABLE]

then $I_{4}+I_{5}=M_{3}\left(N,k\right)$.

8. Evaluation of $I_{6}$

We have to evaluate

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We want to switch the integral with three series, so we will prove the absolute convergence of

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and

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Let us consider

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and we assume, by symmetry, that $\gamma_{1}>0$. Let $y\leq 0.$ From (27) and recalling the notation (25) we have that

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which is bounded by

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for $k>1$. Let $y>0$. We have

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From Lemma 5 we have

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for $k>0$ so

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and using the well known identity

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and placing $Ny=u$ we get

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from Lemma 4, assuming $k>3/2$. Now we have to study

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and, by symmetry, we can consider the cases $\gamma_{1},\gamma_{2}>0$ or $\gamma_{1}>0$, $\gamma_{2}<0$. Let $\gamma_{1},\gamma_{2}>0$ and $y\leq 0$. From (27) we have

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for $k>1/2$. Let $y>0,$ and so the symbol $B_{m,n}$. We recall again that we have to split the integral for $y\in\left(0,1/N\right)$ and $y\in\left(1/N,\,\infty\right)$ since, by (18) and (26), we have different estimation in these two set. We have that

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which is bounded by

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and again from (28) and placing $Ny=u$ we get

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and from the proof of Lemma 4 we have

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and observing that

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we get

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and so the convergence if $k>3/2$. Let us assume that $\gamma_{1}>0$, $\gamma_{2}<0$ and $y\leq 0$. From (27) we have

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hence

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for $k>1.$ If $y>0$ we have essentially the same calculations exchanging the role of $\gamma_{1}$ and $\gamma_{2}$. So we have to consider

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It is sufficient to consider the cases $\gamma_{i}>0,\,i=1,2,3$, $\gamma_{1},\gamma_{2}>0$ and $\gamma_{3}<0$ and lastly $\gamma_{1}>0,$ $\gamma_{2},\gamma_{3}<0$. We will use the symbol $D_{6,3}$ when we consider $A_{6,3}$ with the assumption $\gamma_{i}>0,\,i=1,2,3$, the symbol $E_{6,3}$ when we consider $A_{6,3}$ with the assumption $\gamma_{1},\gamma_{2}>0$ and $\gamma_{3}<0$ and $F_{6,3}$ when we consider $A_{6,3}$ with the assumption $\gamma_{1}>0,$ $\gamma_{2},\gamma_{3}<0$. From (27) we have

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for $k>1$. Let $y>0$. We have

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and from the proof of the Lemma 4 we get

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and observing that

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we get

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and from AM-GM inequality we get

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for $k>3/2$. Let $\gamma_{1},\gamma_{2}>0,\,\gamma_{3}<0$ (and so the symbol $E_{m,n}$) and $y\leq 0$. From (27) we have

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from the proof of Lemma 4, for $k>1/2$. If $y>0$ we have essentially the same calculations exchanging the role of $\gamma_{2}$ and $\gamma_{3}$. Let $\gamma_{2},\gamma_{2}<0$ , $\gamma_{1}>0$ and $y<0$. Recalling (25) we have

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using Lemma 4, for $k>1/2$. Let $y>0$. Observing that $-\left(\frac{\left|\gamma_{2}\right|+\left|\gamma_{3}\right|}{2}\right)\left(\arctan\left(Ny\right)+\frac{\pi}{2}\right)\leq-\left(\frac{\left|\gamma_{2}\right|+\left|\gamma_{3}\right|}{2}\right)\frac{\pi}{2}$ we have

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from Lemma 4 for $k>1/2.$ Now we can exchange the integral with the series and get

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then $I_{6}=M_{4}\left(N,k\right)$.