# Prove that $\frac{2(\pi(p)-2)}{p-1} \leq 0.6$ for all prime numbers $p \geq 31$

Let $$f(p) = \frac{2(\pi(p)-2)}{p-1}$$ where $$\pi$$ is the prime counting function, prove that $$f(p) \leq 0.6$$ for all prime numbers $$p \geq 31$$.

Below are some examples of $$f(p)$$. Well, removing multiples of $$2,3,5$$ between $$n+1$$ and $$n+30$$ allows you to bound $$\pi(n+30)\leq \pi(n)+8$$ for $$n>5$$, and $$\frac{16}{30}<0.6$$. So it remains to check $$2\pi(n)-2\leq0.6\times(n-1)$$ for $$31\leq n\leq 60$$ explicitly.
• We are going to show $2(\pi(p)-2)/(p-1)<0.6$ holds for all integers $p\geq 31$, not just primes. Being nonmultiples of 2, 3 or 5 is a necessary but not sufficient condition for being a prime > 5. So $\pi(n+30)\leq\pi(n)+8$ for $n>5$ (in fact also for $n=5$ too). Check that it does give the induction step (that is what the $16/30<0.6$ is doing if you check what the induction step from $n$ to $n+30$ needs), so all you have to do is to fill in the base case(s). – user10354138 Jun 9 at 2:28
Note that $$\phi(30)=8$$ so among $$30$$ consecutive integers only $$8$$ are coprime to $$30$$ and thus cannot be prime (unless $$p=2$$, $$3$$ or $$5$$). Therefore we have $$\pi(n)-3 <\frac{8}{30}n$$, can you finish from here?