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Remarks on number theory. III. (Hungarian. English, Russian summaries) Zbl 0123.25503
Summary: Let $$a_1 < a_2 < \cdots$$, $$A(x) = \sum_{a_i \leq x} 1$$ be an infinite sequence for which (1) $$a_k=a_{i_1}+\cdots +a_{i_r}$$, $$i < \cdots < i_r < k$$, is not solvable. I prove that $$A(x)/x \to 0$$ and that $$\sum {1 \over a_i} < 103.$$ Further I show that $$A(x) = o(x)$$ is best possible, but there always exists a sequence $$x_i \to \infty$$ for which (2) $$A(x) < Cx_i^{(\sqrt5-1)/2}$$. On the other hand, there exists a sequence $$A$$ for which (1) has no solutions, but $$A(x)>cx^{2/7}$$ for every $$x$$. Perhaps (2) can be improved, but the exponent can certainly not be made smaller than $$2/7$$. Consider now the sequences $$A$$ for which the equation (1’) $$a_{r_1}+\cdots+a_{r_{s_1}}=a_{l_1}+\cdots+a_{l_{s2}},$$ $$r_1 <\cdots< r_{s_1}$$; $$l_1 <\cdots<l_{s_2}$$; $$s_1 \neq s_2$$, is not solvable for every choice of $$s_1 \neq s_2$$. There exists such a sequence with $$A(x) > c_x^{\alpha}$$ for every $$x$$ if $$\alpha$$ is sufficiently small. On the other hand, I show by using Rényi’s strengthening of the large sieve of Linnik that if $$A$$ is such that (1’) has no solutions, then $$A(x) < cx^{5/6}$$ for every $$x$$ if $$c$$ is a sufficiently large absolute constant. Perhaps the exponent $$5/6$$ can be improved, but I have not succeeded in doing this.

##### MSC:
 11B75 Other combinatorial number theory
##### Keywords:
upper estimate; sum of reciprocals; sum-free sequence