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arXiv 1809.05463math.NT
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关于可被其非零基数$b$乘积整除的数

On numbers divisible by the product of their nonzero base $b$ digits

Carlo Sanna

摘要:

对于每个整数$b \geq 3$和每个$x \geq 1$,令$\mathcal{N}_{b,0}(x)$是正整数$n \leq x$的集合,这些正整数可以被它们的非零基$b$数字的乘积整除。我们证明的形式$x^{ρ_{b,0} + o(1)} < \#\mathcal{N}_{b,0}(x) < x^{η_{b,0} + o(1)}$,作为$x \to +\infty$,其中$ρ_{b,0}$和$η_{b,0}$是常数${]0,1[}$只依赖于$b$的界限。特别是,我们表明,$x^{0.526} < \#\mathcal{N}_{10,0}(x) < x^{0.787}$,所有足够大的$x$。这改进了由De Koninck和Luca证明的界$x^{0.495} < \#\mathcal{N}_{10,0}(x) < x^{0.901}$。

英文摘要:

For each integer $b \geq 3$ and every $x \geq 1$, let $\mathcal{N}_{b,0}(x)$ be the set of positive integers $n \leq x$ which are divisible by the product of their nonzero base $b$ digits. We prove bounds of the form $x^{ρ_{b,0} + o(1)} < \#\mathcal{N}_{b,0}(x) < x^{η_{b,0} + o(1)}$, as $x \to +\infty$, where $ρ_{b,0}$ and $η_{b,0}$ are constants in ${]0,1[}$ depending only on $b$. In particular, we show that $x^{0.526} < \#\mathcal{N}_{10,0}(x) < x^{0.787}$, for all sufficiently large $x$. This improves the bounds $x^{0.495} < \#\mathcal{N}_{10,0}(x) < x^{0.901}$, which were proved by De Koninck and Luca.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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