For which values of $n$ does the following inequality hold for? 



> 
> $$2(n-2) < Ap\_n\prod\_{i=3}^n \left(\frac{p\_i-1}{p\_i}\right)$$
> 
> 
> 


$p$ are prime numbers and the notation $p\_i$ indicates the $i$-th prime number as per convention. 


$A$ is defined by the following relationship $p\_n = 6A + r$ and $0\leq r<6$




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**Context**


The inequality comes from this question (<https://math.stackexchange.com/questions/1718774/a-necklace-problem-related-to-the-modulo-function>)



The values of $n$ which satisfy an inequality about prime numbers
Here is a proof that the inequality holds for $n\geq 9$. I let you verify the remaining cases $n\leq 8$.


Let $x:=p\_n$ so that $x\geq 23$ by $n\geq 9$. Then the inequality can be rewritten as
$$ 2\bigl(\pi(x)-2\bigr)<\left\lfloor\frac{x}{6}\right\rfloor x \prod\_{3\leq p\leq x}\frac{p-1}{p}.$$
By increasing the LHS and decreasing the RHS, we get the following stronger inequality:
$$ 2\pi(x)<\frac{(x-5)x}{6}\prod\_{3\leq p\leq x}\frac{p-1}{p}.$$
We shall further increase the LHS and decrease the RHS by using the bounds in the famous paper of Rosser-Schoenfeld (1961): Approximate formulas for some functions of prime numbers. Using (3.6) and (3.27) in this paper, we see that it suffices to show
$$ \frac{8 x}{3\log x}<\frac{(x-5)x}{6\log x}\left(1-\frac{1}{\log^2 x}\right),$$
that is,
$$ 16 < (x-5) \left(1-\frac{1}{\log^2 x}\right).$$
However, this last inequality is clear by $x\geq 23$. Done.


**Added.** The OP mentioned in a comment the following [related problem](https://math.stackexchange.com/questions/1720346/which-values-of-n-is-this-inequality-related-to-prime-numbers-true-for). This can be solved in much the same way, using that
$$ \left(\frac{p-1}{p}\right)^3<\frac{p-2}{p}. $$
So, unless $x$ is very small, the sharper inequality in the above related problem is also true.


