Two sequences A and B of non-negative integers are called additive complements, if their sum contains all suffciently large integers. Let A(x) and B(x) be the counting functions of A and B, respectively. In 1994, Sárközy and Szemerédi proved that, for additive complements A and B, if limsup A(x)B(x)=x ≤ 1, then A(x)B(x)-x→+∞ as x→+∞. In this paper, motivated by a recent result of Ruzsa, we prove the following result: for additive complements A, B with Narkiewicz's condition: A(2x)=A(x)→ as x→+∞, we have A(x)B(x)-x>(1+o(1))a(x)=A(x) as x→+∞, where a(x) is the largest element in A⋂[1,x]. Furthermore, this is the best possible. As a corollary, for additive complements A, B with Narkiewicz's condition: A(2x)=A(x)→1 as x→+∞ and any M >1, we have A(x)B(x)-x>A(x)M for all suffciently large x.

中文翻译：

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Narkiewicz 条件的添加剂补充

非负整数的两个序列 A 和 B 称为加法补码，如果它们的和包含所有足够大的整数。设 A(x) 和 B(x) 分别是 A 和 B 的计数函数。1994 年，Sárközy 和 Szemerédi 证明，对于加性补码 A 和 B，如果 limsup A(x)B(x)=x ≤ 1，则 A(x)B(x)-x→+∞ as x→+∞ . 在本文中，受 Ruzsa 近期结果的启发，我们证明了以下结果：对于具有 Narkiewicz 条件的加性补 A、B：A(2x)=A(x)→ as x→+∞，我们有 A(x) B(x)-x>(1+o(1))a(x)=A(x) as x→+∞，其中a(x)是A⋂[1,x]中最大的元素。此外，这是最好的。作为推论，对于具有 Narkiewicz 条件的加性补码 A、B：A(2x)=A(x)→1 as x→+∞ 并且任何 M >1，我们有 A(x)B(x)-x>A (x)M 对于所有足够大的 x。