Let k1, k2 be nonzero integers with（k1, k2） = 1 and k1k2≠-1. Let Rk1,k2（A, n）be the number of solutions of n = k1a1 ＋ k2a2, where a1, a2 ∈ A. Recently, Xiong proved that there is a set A Z such that Rk1,k2（A...Let k1, k2 be nonzero integers with（k1, k2） = 1 and k1k2≠-1. Let Rk1,k2（A, n）be the number of solutions of n = k1a1 ＋ k2a2, where a1, a2 ∈ A. Recently, Xiong proved that there is a set A Z such that Rk1,k2（A, n） = 1 for all n ∈ Z. Let f ： Z-→ N0∪ {∞} be a function such that f-1（0） is finite. In this paper, we generalize Xiong＇s result and prove that there exist uncountably many sets A Z such that Rk1,k2（A, n） = f（n） for all n ∈ Z.展开更多
基金Supported by the National Natural Science Foundation of China(Grant No.11471017)
文摘Let k1, k2 be nonzero integers with（k1, k2） = 1 and k1k2≠-1. Let Rk1,k2（A, n）be the number of solutions of n = k1a1 ＋ k2a2, where a1, a2 ∈ A. Recently, Xiong proved that there is a set A Z such that Rk1,k2（A, n） = 1 for all n ∈ Z. Let f ： Z-→ N0∪ {∞} be a function such that f-1（0） is finite. In this paper, we generalize Xiong＇s result and prove that there exist uncountably many sets A Z such that Rk1,k2（A, n） = f（n） for all n ∈ Z.
