Sums of powers of primes II

Lawrence C. Washington
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Abstract

For a real number k, define \(\pi _k(x) = \sum _{p\le x} p^k\). When \(k>0\), we prove that

$$\begin{aligned} \pi _k(x) - \pi (x^{k+1}) = \Omega _{\pm }\left( \frac{x^{\frac{1}{2}+k}}{\log x} \log \log \log x\right) \end{aligned}$$

as \(x\rightarrow \infty \), and we prove a similar result when \(-1<k<0\). This strengthens a result in a paper by Gerard and the author and it corrects a flaw in a proof in that paper. We also quantify the observation from that paper that \(\pi _k(x) - \pi (x^{k+1})\) is usually negative when \(k>0\) and usually positive when \(-1<k<0\).

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素数幂之和 II
对于实数k,定义(pi _k(x) = \sum _{p\le x} p^k\)。当(k>0)时,我们证明 $$\begin{aligned}\pi _k(x) - \pi (x^{k+1}) = \Omega _{\pm }\left( \frac{x^{frac\{1}{2}+k}}{\log x} \log \log x\right) \end{aligned}$$as \(x\rightarrow \infty \),当(-1<k<0)时,我们证明了类似的结果。这加强了杰拉德和作者论文中的一个结果,并纠正了该论文证明中的一个缺陷。我们还量化了那篇论文中的观察结果,即当\(k>0\)时,\(\pi _k(x) - \pi(x^{k+1})\)通常是负值,而当\(-1<k<0\)时,通常是正值。
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