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\begin{document}$$k\ge 3$$\end{document}, a sequence A of nonnegative integers is called an APk\documentclass[12pt]{minimal}
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\begin{document}$$AP_k$$\end{document}-covering sequence if there exists an integer n0\documentclass[12pt]{minimal}
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\begin{document}$$n_0$$\end{document} such that, if n>n0\documentclass[12pt]{minimal}
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\begin{document}$$n>n_0$$\end{document}, then there exist a1∈A\documentclass[12pt]{minimal}
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\begin{document}$$a_1\in A$$\end{document}, …\documentclass[12pt]{minimal}
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\begin{document}$$\ldots $$\end{document}, ak-1∈A\documentclass[12pt]{minimal}
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\begin{document}$$a_{k-1}\in A$$\end{document}, a1<a2<⋯<ak-1<n\documentclass[12pt]{minimal}
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\begin{document}$$a_1<a_2<\cdots<a_{k-1}<n$$\end{document} such that a1,a2,…,ak-1,n\documentclass[12pt]{minimal}
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\begin{document}$$a_1,a_2,\ldots ,a_{k-1},n$$\end{document} form a k-term arithmetic progression. For k=3\documentclass[12pt]{minimal}
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\begin{document}$$k=3$$\end{document}, Kiss, Sándor and Yang observed that lim supn→∞A(n)/n≥1.77\documentclass[12pt]{minimal}
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\begin{document}$$\limsup _{n\rightarrow \infty } A(n)/\sqrt{n}\ge 1.77$$\end{document} hold for any AP3\documentclass[12pt]{minimal}
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\begin{document}$$AP_3$$\end{document}-covering sequence A. They also proved that there exists an AP3\documentclass[12pt]{minimal}
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\begin{document}$$AP_3$$\end{document}-covering sequence A such that lim supn→∞A(n)/n≤36\documentclass[12pt]{minimal}
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\begin{document}$$\limsup _{n\rightarrow \infty } A(n)/\sqrt{n}\le 36$$\end{document}. Recently, Chen proved that there exists an AP3\documentclass[12pt]{minimal}
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\begin{document}$$AP_3$$\end{document}-covering sequence A such that lim supn→∞A(n)/n=15=3.87…\documentclass[12pt]{minimal}
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\begin{document}$$\limsup _{n\rightarrow \infty } A(n)/\sqrt{n}=\sqrt{15}=3.87\ldots $$\end{document}. In this note, we prove that there exists an AP3\documentclass[12pt]{minimal}
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\begin{document}$$AP_3$$\end{document}-covering sequence A such that lim supn→∞A(n)/n=8/5=3.57…\documentclass[12pt]{minimal}
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\begin{document}$$\limsup _{n\rightarrow \infty } A(n)/\sqrt{n}=8/\sqrt{5}=3.57\ldots $$\end{document}.
