Waring−Goldbach problem for one prime power and four prime cubes under Riemann Hypothesis

Abstract

Let k⩾1 be an integer. Assume that RH holds. In this paper we prove that a suitable asymptotic formula for the average number of representations of integers n=p_1^k+p_2^3+p_3^3+p_4^3+p_5^3, where p_1,p_2,p_3,p_4,p_5 are prime numbers. This expands the previous results.

Keywords

Hardy−Littlewood method / Waring−Goldbach problem / Riemann Hypothesis / short intervals

1 Introduction

Let N,k_1,k_2,\dots ,k_r be integers with 2⩽k_1⩽k_2⩽\dots ⩽k_r. The Waring-Goldbach problem for unlike powers of primes concerns the representation of N as the form

\begin{array}{l}n=p_1^{k_1}+p_2^{k_2}+\dots +p_r^{k_r}\end{array}

is classical. These topics have attracted mathematicians attentions [1-3, 5-9, 11].

Recently Feng and Ma [4] considered the exception set for the following problem,

(1)\begin{array}{l}n=p_1^k+p_2^3+p_3^3+p_4^3+p_5^3,\end{array}

with k⩾4. But under the condition that Riemann Hypothesis holds, there is no relevant result. In this paper, under the assumption that the Riemann Hypothesis holds, we reconsider the exceptional set of (1) from another perspective, that is, we give the asymptotic formula of (1) in short intervals.

Let

r_k(n)=\sum _{n=p_1^k+p_2^3+p_3^3+p_4^3+p_5^3}\log p_1\log p_2\log p_3\log p_4\log p_5.

We have the following theorem.

Theorem 1　 Let ϵ>0,N⩾2,1⩽H⩽N,k⩾4 be integers and assume the Riemann Hypothesis holds. Then there exists a suitable positive constant B such that

\begin{array}{rl}\sum _{n=N+1}^{N+H}{r}_k(n)=& \frac{\mathrm{\Gamma }\left(\frac{1}{k}\right)\mathrm{\Gamma }{\left(\frac{4}{3}\right)}^4}{k\mathrm{\Gamma }\left(\frac{4}{3}+\frac{1}{k}\right)}H{N}^{\frac{1}{3}+\frac{1}{k}}\\ & +O_k(H^2{N}^{\frac{1}{k}-\frac{2}{3}}+H^{\frac{1}{2}}{N}^{\frac{5}{6}+\frac{1}{2k}}{L}^B),\end{array}

as N\to \mathrm{\infty }, uniformly for \mathrm{\infty }(N^{1-\frac{1}{k}}{L}^{2B})⩽H⩽o(N) where f=\mathrm{\infty }(g), means g=o(f) and \mathrm{\Gamma } is Euler's function.

From Theorem 1 we can say that, in the interval [N,N+H], an integer can be represented as a sum of one prime k power and four prime cubes, where \mathrm{\infty }(N^{1-\frac{1}{k}}{L}^{2B})⩽H⩽o(N). So under the condition that Riemann Hypothesis holds, the exception set of (1) is \mathrm{\infty }(N^{1-\frac{1}{k}}{L}^{2B}). The proofs of Theorem 1 use the original Hardy-Littlewood circle method and the strategies adopted in the works of Languasco and Zaccagnini [6-8].

2 Preliminaries

Let e(\alpha )={\mathrm{e}}^{2\pi \mathrm{i}\alpha },\alpha \in [-\frac{1}{2},\frac{1}{2}],L=\log N,z=\frac{1}{N}-2\pi \mathrm{i}\alpha,

{\tilde{S}}_l(\alpha )=\sum _{n=1}^{\mathrm{\infty }}\mathrm{\Lambda }(n){\mathrm{e}}^{-\frac{n^l}{N}}e(n^l\alpha )

and

{\tilde{V}}_l(\alpha )=\sum _{p=2}^{\mathrm{\infty }}(\log p){\mathrm{e}}^{-\frac{p^l}{N}}e(p^l\alpha ).

We have

(2)\begin{array}{l}|z{|}^{-1}\ll min(N,|\alpha {|}^{-1}).\end{array}

We also set

U(\alpha ,H)=\sum _{m=1}^{H}e(m\alpha ).

We have

(3)\begin{array}{l}|U(\alpha ,H)|⩽min(H;|\alpha {|}^{-1}),\end{array}

see, e.g., Montgomery [10, p 39].

To prove Theorem 1, we need some lemmas as follows.

Lemma 1 [7, Lemma 1]　 Let l⩾1 be an integer. Then we have

|{\tilde{S}}_l(\alpha )-{\tilde{V}}_l(\alpha )|{\ll }_l{N}^{\frac{1}{2l}}.

Lemma 2 [6, Lemma 2]　 Let l⩾1 be an integer, N⩾2 and \alpha \in [-\frac{1}{2},\frac{1}{2}]. Then we have

{\tilde{S}}_l(\alpha )=\frac{\mathrm{\Gamma }\left(\frac{1}{l}\right)}{l{z}^{\frac{1}{l}}}-\frac{1}{l}\sum _{\rho }{z}^{-\frac{\rho }{l}}\mathrm{\Gamma }\left(\frac{\rho }{l}\right)+O_l(1),

where \rho =\beta +\mathrm{i}\gamma runs over the non-trivial zeros of the Riemann-zeta function \zeta (s).

Lemma 3 [6, Lemma 4]　 Let N be an integer and \mu >0. Then we have

{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{z}^{-\mu }e(-n\alpha )\mathrm{d}\alpha ={\mathrm{e}}^{-\frac{n}{N}}\frac{n^{\mu -1}}{\mathrm{\Gamma }(\mu )}+O_{\mu }\left(\frac{1}{n}\right),

uniformly for n⩾1.

Lemma 4 [6, Lemma 4] [7, Lemma 1]　 Let l⩾1 be an integer, ϵ be an arbitrarily small positive constant, N be a sufficiently large integer and L=\log N. Then there exists a positive constant c_1=c_1(ϵ), which does not depend on l, such that

{\int }_{-\xi }^{\xi }|{\tilde{S}}_l(\alpha )-\frac{\mathrm{\Gamma }\left(\frac{1}{l}\right)}{l{z}^{\frac{1}{l}}}{|}^2\mathrm{d}\alpha {\ll }_l{N}^{\frac{2}{l-1}}\exp (-c_1(\frac{L}{\log L}{)}^{\frac{1}{3}})

uniformly for 0⩽\xi ⩽N^{-1+\frac{5}{(6l)}-ϵ}. Assuming RH holds we get

{\int }_{-\xi }^{\xi }|{\tilde{S}}_l(\alpha )-\frac{\mathrm{\Gamma }(\frac{1}{l})}{l{z}^{\frac{1}{l}}}{|}^2\mathrm{d}\alpha {\ll }_l{N}^{\frac{1}{l}}\xi L^2

uniformly for 0⩽\xi ⩽\frac{1}{2}.

Lemma 5 [8, Lemma 5]　 Let N be a sufficiently large integer, l,k be integers with l⩾1,1⩽k⩽l. There exists a suitable positive constant A=A(k,l), such that

{\int }_{-\frac{1}{2}}^{\frac{1}{2}}|{\tilde{S}}_l(\alpha ){|}^{2^k}\mathrm{d}\alpha {\ll }_{k,l}{N}^{\frac{2^k-k}{l}}{L}^A

and

{\int }_{-\frac{1}{2}}^{\frac{1}{2}}|{\tilde{V}}_l(\alpha ){|}^{2^k}\mathrm{d}\alpha {\ll }_{k,l}{N}^{\frac{2^k-k}{l}}{L}^A.

Lemma 6 [8, Lemma 6]　 Let N be a sufficiently large integer, ϵ>0, l>1 and \tau >0. Then we have

{\int }_{-\tau }^{\tau }|{\tilde{S}}_l(\alpha ){|}^4\mathrm{d}\alpha {\ll }_l(\tau N^{\frac{2}{l}}+N^{\frac{4}{l}-1})N^{ϵ}

and

{\int }_{-\tau }^{\tau }|{\tilde{V}}_l(\alpha ){|}^4\mathrm{d}\alpha {\ll }_l(\tau N^{\frac{2}{l}}+N^{\frac{4}{l}-1})N^{ϵ}.

Lemma 7 [8, Lemma 7]　 Let N be a sufficiently large integer, ϵ>0, l>2,c⩾1 and N^{-c}⩽\omega ⩽N^{\frac{2}{l}-1}. We also let I(\omega ):=[-\frac{1}{2},-\omega ]\cup [\omega ,\frac{1}{2}]. Then we have

{\int }_{I(w)}|{\tilde{S}}_l(\alpha ){|}^4\frac{\mathrm{d}\alpha }{|\alpha |}{\ll }_l\frac{N^{\frac{4}{l}-1+ϵ}}{\omega }

and

{\int }_{I(w)}|{\tilde{V}}_l(\alpha ){|}^4\frac{\mathrm{d}\alpha }{|\alpha |}{\ll }_l\frac{N^{\frac{4}{l}-1+ϵ}}{\omega }.

3 Proof of Theorem 1

Let k⩾4, H⩾2, H=o(N) be an integer. We recall that we set L=\log N for brevity. From now on we assume that RH holds, we may write

\sum _{n=N+1}^{N+H}{\mathrm{e}}^{-\frac{n}{N}}{r}_k(n)={\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\tilde{V}}_k(\alpha ){\tilde{V}}_3(\alpha {)}^{4}U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha .

Recalling Lemma 2, we find it convenient to set

(4){\tilde{E}}_l(\alpha )={\tilde{S}}_l(\alpha )-\frac{\mathrm{\Gamma }(\frac{1}{l})}{l{z}^{\frac{1}{l}}}.

Then we have

(5)\begin{array}{rl}& \sum _{n=N+1}^{N+H}{\mathrm{e}}^{-\frac{n}{N}}{r}_k(n)={\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\mathrm{\Gamma }(\frac{1}{k})\mathrm{\Gamma }(\frac{4}{3}{)}^4}{k{z}^{\frac{4}{3}+\frac{1}{k}}}U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha \\ & +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\mathrm{\Gamma }(\frac{1}{k})}{k{z}^{\frac{1}{k}}}\left(S_3(\alpha {)}^4-\frac{\mathrm{\Gamma }(\frac{4}{3}{)}^4}{z^{\frac{4}{3}}}\right)U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha \\ & +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\tilde{E}}_k(\alpha ){\tilde{S}}_3(\alpha {)}^{4}U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha \\ & +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\tilde{V}}_k(\alpha )({\tilde{V}}_3(\alpha {)}^4-{\tilde{S}}_3(\alpha {)}^4)U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha \\ & +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{\tilde{S}}_3(\alpha {)}^4({\tilde{V}}_k(\alpha )-{\tilde{S}}_k(\alpha ))U(-\alpha ,H)e(-N\alpha )\mathrm{d}\alpha \\ & =:{\mathcal{L}}_1+{\mathcal{L}}_2+{\mathcal{L}}_3+{\mathcal{L}}_4+{\mathcal{L}}_5.\end{array}

Now we need to estimate these terms.

3.1 Estimate of {\mathcal{L}}_1

Using Lemma 3, a direct calculation gives

(6)\begin{array}{rl}{\mathcal{L}}_1& =\frac{\mathrm{\Gamma }(\frac{1}{k})\mathrm{\Gamma }(\frac{4}{3}{)}^4}{k\mathrm{\Gamma }(\frac{4}{3}+\frac{1}{k})}\sum _{n=N+1}^{N+H}{n}^{\frac{1}{3}+\frac{1}{k}}{\mathrm{e}}^{-\frac{n}{N}}+O_k\left(\frac{H}{N}\right)\\ & =\frac{\mathrm{\Gamma }(\frac{1}{k})\mathrm{\Gamma }(\frac{4}{3}{)}^4}{k\mathrm{\Gamma }(\frac{4}{3}+\frac{1}{k})\mathrm{e}}H{N}^{\frac{1}{3}+\frac{1}{k}}+O_k\left(H^2{N}^{\frac{1}{k}-\frac{2}{3}}+N^{\frac{1}{3}+\frac{1}{k}}\right).\end{array}

3.2 Estimate of {\mathcal{L}}_2

According to the definition of {\mathcal{L}}_2 and (4), we have

(7)\begin{array}{rl}& {\mathcal{L}}_2\ll {\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{|{\tilde{S}}_3(\alpha ){|}^2}{|z{|}^{\frac{1}{3}+\frac{1}{k}}}|{\tilde{E}}_3(\alpha )||U(-\alpha ,H)|\mathrm{d}\alpha +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{{\tilde{E}}_3(\alpha )}{|z{|}^{1+\frac{1}{k}}}|U(\alpha ,H)|\mathrm{d}\alpha \\ & +{\int }_{-\frac{1}{2}}^{\frac{1}{2}}{N}^{\frac{2}{3}}\frac{|{\tilde{E}}_3(\alpha ){|}^2}{|z{|}^{\frac{1}{k}}}|U(\alpha ,H)|\mathrm{d}\alpha \\ & =:{\mathcal{L}}_1+{\mathcal{L}}_2+{\mathcal{L}}_3.\end{array}

Let

\psi ={\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{|{\tilde{E}}_3(\alpha ){|}^2}{|z{|}^{\frac{1}{k}}}|U(-\alpha ,H)|\mathrm{d}\alpha .

Using Lemma 4, (3) and integration by parts, we have

(8)\begin{array}{rl}\psi & \ll H{N}^{\frac{1}{k}}{\int }_{-\frac{1}{N}}^{\frac{1}{N}}|{\tilde{E}}_3(\alpha ){|}^2\mathrm{d}\alpha +2H{\int }_{\frac{1}{N}}^{\frac{1}{H}}\frac{|{\tilde{E}}_3(\alpha ){|}^2}{{\alpha }^{\frac{1}{k}}}\mathrm{d}\alpha +2{\int }_{\frac{1}{H}}^{\frac{1}{2}}\frac{|{\tilde{E}}_3(\alpha ){|}^2}{{\alpha }^{1+\frac{1}{k}}}\mathrm{d}\alpha \\ & {\ll }_{k}H{N}^{\frac{1}{k}-\frac{2}{3}}{L}^2+H^{\frac{1}{k}}{N}^{\frac{1}{3}}{L}^2{\ll }_k{H}^{\frac{1}{k}}{N}^{\frac{1}{3}}{L}^2.\end{array}

Then we have

(9){\mathcal{L}}_3{\ll }_k{H}^{\frac{1}{k}}N{L}^2.

By the Cauchy−Schwarz inequality, (2), (3) and (9), we have

(10)\begin{array}{rl}{\mathcal{L}}_2& \ll {\psi }^{\frac{1}{2}}{\left({\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{|U(-\alpha ,H)|}{|z{|}^{2+\frac{1}{k}}}\mathrm{d}\alpha \right)}^{\frac{1}{2}}\\ & \ll {\psi }^{\frac{1}{2}}{\left(H{N}^{2+\frac{1}{k}}{\int }_{-\frac{1}{N}}^{\frac{1}{N}}\mathrm{d}\alpha +2H{\int }_{\frac{1}{N}}^{\frac{1}{H}}\frac{\mathrm{d}\alpha }{{\alpha }^{2+\frac{1}{k}}}+2{\int }_{\frac{1}{H}}^{\frac{1}{2}}\frac{\mathrm{d}\alpha }{{\alpha }^{3+\frac{1}{k}}}\right)}^{\frac{1}{2}}\\ & \ll H^{\frac{1}{2}+\frac{1}{(2k)}}{N}^{\frac{2}{3}+\frac{1}{2k}}L.\end{array}

By the Cauchy−Schwarz inequality, Lemma 5, (2), (3) and (9) we have

(11)\begin{array}{rl}{\mathcal{L}}_1& \ll {\psi }^{-\frac{1}{2}}{\left({\int }_{-\frac{1}{2}}^{\frac{1}{2}}|{\tilde{S}}_3(\alpha ){|}^8\mathrm{d}\alpha \right)}^{\frac{1}{4}}{\left({\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{|U(-\alpha ,H){|}^2}{|z{|}^{\frac{4}{3}+\frac{2}{k}}}\mathrm{d}\alpha \right)}^{\frac{1}{4}}\\ & \ll H^{\frac{1}{2k}}{N}^{\frac{1}{6}}L{N}^{\frac{5}{12}}{L}^{\frac{A}{4}}(H^2{N}^{\frac{4}{3}+\frac{2}{k}}{\int }_{-\frac{1}{N}}^{\frac{1}{N}}\mathrm{d}\alpha \\ & +2H^2{\int }_{\frac{1}{N}}^{\frac{1}{H}}\frac{\mathrm{d}\alpha }{{\alpha }^{\frac{4}{3}+\frac{2}{k}}}+2{\int }_{\frac{1}{H}}^{\frac{1}{2}}\frac{\mathrm{d}\alpha }{{\alpha }^{\frac{10}{3}+\frac{2}{k}}}{)}^{\frac{1}{4}}\\ & \ll H^{\frac{1}{2}+\frac{1}{2k}}{N}^{\frac{2}{3}+\frac{1}{2k}}{L}^{\frac{1+A}{4}}.\end{array}

Summing up by (7)−(11), we have

(12){\mathcal{L}}_2{\ll }_k{H}^{\frac{1}{2}+\frac{1}{2k}}{N}^{\frac{2}{3}+\frac{1}{2k}}{L}^{1+\frac{A}{4}},

for every k⩾4.

3.3 Estimate of {\mathcal{L}}_3

Using the Cauchy−Schwarz inequality, (2), Lemma 4 and Lemma 5, we can get

(13)\begin{array}{rl}{\mathcal{L}}_3& {\ll }_k{\left({\int }_{-\frac{1}{2}}^{\frac{1}{2}}|{\tilde{S}}_3(\alpha ){|}^8\mathrm{d}\alpha \right)}^{\frac{1}{2}}{\left({\int }_{-\frac{1}{2}}^{\frac{1}{2}}|{\tilde{E}}_k(\alpha ){|}^2|U(\alpha ,H){|}^2\mathrm{d}\alpha \right)}^{\frac{1}{2}}\\ & {\ll }_k{N}^{\frac{5}{6}}{L}^{\frac{A}{2}}{\left(H^2{\int }_{-\frac{1}{H}}^{\frac{1}{H}}|{\tilde{E}}_k(\alpha ){|}^2\mathrm{d}\alpha +2{\int }_{\frac{1}{H}}^{\frac{1}{2}}|{\tilde{E}}_k(\alpha ){|}^2\frac{\mathrm{d}\alpha }{{\alpha }^2}\right)}^{\frac{1}{2}}\\ & {\ll }_k{H}^{\frac{1}{2}}{N}^{\frac{5}{6}+\frac{1}{2k}}{L}^{1+\frac{A}{2}}.\end{array}

3.4 Estimate of {\mathcal{L}}_4

From Lemma 1, f^2-g^2=2f(f-g)-(f-g{)}^2 and {\tilde{V}}_k(\alpha ){\ll }_k{N}^{\frac{1}{k}}, we have

\begin{array}{rl}{\tilde{V}}_k(\alpha )({\tilde{V}}_3(\alpha {)}^4-{\tilde{S}}_3(\alpha {)}^4)& {\ll }_k|{\tilde{V}}_k(\alpha )||{\tilde{V}}_3(\alpha )-{\tilde{S}}_3(\alpha )|(|{\tilde{V}}_3(\alpha )|+|{\tilde{S}}_3(\alpha )|{)}^3\\ & {\ll }_k{N}^{\frac{1}{6}+\frac{1}{k}}max(|{\tilde{V}}_3(\alpha ){|}^3,|{\tilde{S}}_3(\alpha ){|}^3).\end{array}

Then we have

(14)\begin{array}{rl}{\mathcal{L}}_4& {\ll }_k{N}^{\frac{1}{6}+\frac{1}{k}}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}(|{\tilde{V}}_3(\alpha ){|}^3+|{\tilde{S}}_3(\alpha ){|}^3)|U(-\alpha ,H)|\mathrm{d}\alpha \\ & ={:}_k{N}^{\frac{1}{6}+\frac{1}{k}}(J_1+J_2).\end{array}

By (2), Lemma 6 and Lemma 7, we have

(15)\begin{array}{rl}{J}_2& \ll H{\int }_{-\frac{1}{H}}^{\frac{1}{H}}|{\tilde{S}}_3(\alpha ){|}^3\mathrm{d}\alpha +{\int }_{I(1,H)}|{\tilde{S}}_3(\alpha ){|}^3\frac{\mathrm{d}\alpha }{|\alpha |}\\ & \ll H^{\frac{3}{4}}{\left({\int }_{-\frac{1}{H}}^{\frac{1}{H}}|{\tilde{S}}_3(\alpha ){|}^4\mathrm{d}\alpha \right)}^{\frac{3}{4}}+{\left({\int }_{I(1,H)}|{\tilde{S}}_3(\alpha ){|}^4\frac{\mathrm{d}\alpha }{|\alpha |}\right)}^{\frac{3}{4}}{\left({\int }_{I(1,H)}\frac{\mathrm{d}\alpha }{|\alpha |}\right)}^{\frac{1}{4}}\\ & \ll H^{\frac{3}{4}}{N}^{\frac{1}{4}+ϵ}+(H{N}^{\frac{1}{3}+ϵ}{)}^{\frac{3}{4}}{L}^{\frac{1}{4}}\\ & \ll H^{\frac{3}{4}}{N}^{\frac{1}{4}+ϵ},\end{array}

provided that H\gg N^{\frac{1}{3}}. Similarly, we have

(16)J_1\ll H^{\frac{3}{4}}{N}^{\frac{1}{4}+ϵ}.

Finally, by (14)−(16), we have

(17){\mathcal{F}}_4{\ll }_k{H}^{\frac{3}{4}}{N}^{\frac{5}{12}+\frac{1}{k}+ϵ},

provided that H\gg N^{\frac{1}{3}}.

3.5 Estimate of {\mathcal{L}}_5

By Lemma 1, Lemma 6, Lemma 7, (2) and a partial integration, we have

(18)\begin{array}{rl}{\mathcal{L}}_5& {\ll }_{k}H{N}^{\frac{1}{2k}}{\int }_{-\frac{1}{H}}^{\frac{1}{H}}|{\tilde{S}}_3(\alpha ){|}^4\mathrm{d}\alpha +N^{\frac{1}{2k}}{\int }_{\frac{1}{H}}^{\frac{1}{2}}\frac{|{\tilde{S}}_3(\alpha ){|}^4}{|\alpha |}\mathrm{d}\alpha \\ & {\ll }_{k}H{N}^{\frac{1}{2k}}\left(\frac{N^{\frac{2}{3}+ϵ}}{H}+N^{\frac{1}{3}+ϵ}\right)+H{N}^{\frac{1}{2k}}{N}^{\frac{1}{3}+ϵ}\\ & {\ll }_k{N}^{\frac{1}{2k}}\left(N^{\frac{2}{3}+ϵ}+H{N}^{\frac{1}{3}+ϵ}\right),\end{array}

provided that H\gg N^{\frac{1}{3}}.

3.6 Completion of the proof

By (2), (3) and (6)−(18), there exists B=B(A) such that for H⩾N^{\frac{1}{3}},

(19)\begin{array}{rl}\sum _{n=N+1}^{N+H}{\mathrm{e}}^{-\frac{n}{N}}{r}_k(n)=& \frac{\mathrm{\Gamma }(\frac{1}{k})\mathrm{\Gamma }(\frac{4}{3}{)}^4}{k\mathrm{\Gamma }(\frac{4}{3}+\frac{1}{k})\mathrm{e}}H{N}^{\frac{1}{3}+\frac{1}{k}}\\ & +O_k(H^2{N}^{\frac{1}{k}-\frac{2}{3}}+H^{\frac{1}{2}}{N}^{\frac{5}{6}+\frac{1}{2k}}{L}^B),\end{array}

where \mathrm{\infty }(N^{1-\frac{1}{k}}{L}^{2B})⩽H⩽o(N). From {\mathrm{e}}^{-\frac{n}{N}}={\mathrm{e}}^{-1}+O(\frac{H}{N}), for n\in [N+1,N+H], 1⩽H⩽N, we can get

\begin{array}{rl}\sum _{n=N+1}^{N+H}{r}_k(n)=& \frac{\mathrm{\Gamma }(\frac{1}{k})\mathrm{\Gamma }(\frac{4}{3}{)}^4}{k\mathrm{\Gamma }(\frac{4}{3}+\frac{1}{k})}H{N}^{\frac{1}{3}+\frac{1}{k}}\\ & +O_k(H^2{N}^{\frac{1}{k}-\frac{2}{3}}+H^{\frac{1}{2}}{N}^{\frac{5}{6}+\frac{1}{2k}}{L}^B)\\ & +O_k\left(\frac{H}{N}\sum _{n=N+1}^{N+H}{r}_k(n)\right).\end{array}

Using {\mathrm{e}}^{\frac{n}{N}}⩽{\mathrm{e}}^2 and (19), the last error term is dominated by all of the previous ones. Thus Theorem 1 follows.

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