Inequality from Erathostenes’ sieve

52 Views Asked by At

Let $a_1<a_2<a_3$ be three positive integers. Let $n$ be another positive integer, and $I$ an interval of $n$ successive integers. Denote by $\mu$ the number of integers in $I$ not divisible by any of $a_1,a_2,a_3$.

From a probabilistic viewpoint, if the $a_i$ are mutually coprime, divisibility by one $a_i$ should be “independent” from divisibility by another $a_j$. So one expects $\mu$ to be near the estimate

$$ \mu_0=n\bigg(1-\frac{1}{a_1}\bigg)\bigg(1-\frac{1}{a_2}\bigg)\bigg(1-\frac{1}{a_3}\bigg) $$

My question is : is it always true (whether the $a_i$ are coprime or not) that

$$ \mu \geq \mu_0- 3\bigg(1-\frac{1}{a_1}\bigg)\bigg(1-\frac{1}{a_2}\bigg) \ ? $$

1

There are 1 best solutions below

1
On BEST ANSWER

$a_1=3$, $a_2=4$, $a_3=5$, $n=4$, $I=\{{3,4,5,6\}}$; $\mu=0$, $\mu_0=(4)(2/3)(3/4)(4/5)=8/5$, $0\lt8/5-3(2/3)(3/4)=8/5-3/2=1/10$.