A set $A$ of natural numbers possesses property ${\mathcal{P}}_h$, if there are no distinct elements $a_0,a_1,\dots ,a_h\in A$ with $a_0$ dividing the product $a_1{a}_2\dots a_h$. Erdős determined the maximum size of a subset of $\{1,\dots ,n\}$ possessing property ${\mathcal{P}}_2$. More recently, Chan et al. (2010) solved the case $h=3$, finally the general case also got resolved by Chan (2011), the maximum size is $\pi \left(n\right)+{\Theta }_h\left(\frac{n^{2∕(h+1)}}{{(logn)}^2}\right)$.

In this note we consider the counting version of this problem and show that the number of subsets of $\{1,\dots ,n\}$ possessing property ${\mathcal{P}}_2$ is $T\left(n\right)\cdot e^{\Theta (n^{2∕3}∕logn)}$ for a certain function $T\left(n\right)\approx {(3.517\dots )}^{\pi \left(n\right)}$. For $h>2$ we prove that the number of subsets possessing property ${\mathcal{P}}_h$ is $T\left(n\right)\cdot e^{\sqrt{n}(1+o\left(1\right))}$.

This is a rare example in which the order of magnitude of the lower order term in the exponent is also determined.

一套 $\mathrm{一种}$ 自然数拥有财产 ${\mathcal{P}}_H$，如果没有不同的元素 ${\mathrm{一种}}_0，{\mathrm{一种}}_{\mathrm{1个}}，\dots ，{\mathrm{一种}}_H\in \mathrm{一种}$ 与 ${\mathrm{一种}}_0$ 划分产品 ${\mathrm{一种}}_{\mathrm{1个}}{\mathrm{一种}}_2\dots {\mathrm{一种}}_H$。Erdős确定了一个子集的最大大小$\{\mathrm{1个}，\dots ，ñ\}$ 拥有财产 ${\mathcal{P}}_2$。最近，Chan等。（2010）解决了此案$H=3$，最后一般案例也由Chan（2011）解决，最大尺寸为 $\pi （ñ）+{\Theta }_H（\frac{{ñ}^{2∕（H+\mathrm{1个}）}}{{（日志ñ）}^2}）$。

在本说明中，我们考虑此问题的计数形式，并表明 $\{\mathrm{1个}，\dots ，ñ\}$ 拥有财产 ${\mathcal{P}}_2$ 是 $Ť（ñ）\cdot {Ë}^{\Theta （{ñ}^{2∕3}∕日志ñ）}$ 对于某些功能 $Ť（ñ）\approx {（3。517\dots ）}^{\pi （ñ）}$。对于$H>2$ 我们证明拥有属性的子集的数量 ${\mathcal{P}}_H$ 是 $Ť（ñ）\cdot {Ë}^{\sqrt{ñ}（\mathrm{1个}+Ø（\mathrm{1个}））}$。

这是一个罕见的示例，其中还确定了指数中低阶项的数量级。