What would be the standard way to show for $n \ge 148, \pi(n) < \dfrac{n}{4}$

Question

As I understand it, using the Prime Number Theorem, it is well known that for any positive integer $x$, there exists an integer $N$ such that for all $n \ge N, \pi(n) < \dfrac{n}{x}$

For many proofs, it is helpful to provide a lemma as a way to shorten the proof (for example, see the proof of the Sylvester-Schur Theorem by Paul Erdos).

Is there a standard way to show the proof for $x=4$ or $x=5$?

It seems quite straight forward for $x=2$ or $x=3$. I could not find a straight forward argument for $x=4$ or $x=5$. Am I missing the obvious or does the argument get significantly more complicated with $x \ge 4$?

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Edit: Added proofs for $x=2, x=3$

Here are the straight forward proofs for $x=2, x=3$.

Lemma 1: For $n \ge 8, \pi(n) \le \dfrac{n}{2}$

(1) It is true for $\pi(8)$ and $\pi(9)$ so we can assume it is true up to some $n-1 \ge 9$ such that $\pi(n-1) = \dfrac{n-1}{2}$

(2) if $n$ is not prime, $\pi(n) = \pi(n-1) \le \dfrac{n-1}{2} < \dfrac{n}{2}$ If $n$ is prime, it is odd. So, $\pi(n) \le \dfrac{n-2}{2} + 1 = \dfrac{n}{2}$

Lemma 2: For $n \ge 37, \pi(n) < \dfrac{n}{3}$

(1) It is true for $\pi(37)$ so we can assume that it is true up to some $n-1 \ge 37$ such that $\pi(n-1) < \dfrac{n-1}{3}$

(2) We can assume $n$ is prime since if $n$ is not prime, $\pi(n) = \pi(n-1) < \dfrac{n-1}{3} < \dfrac{n}{3}$

(3) We can assume $n-2$ is prime since if $n-2$ is not prime, $\pi(n) \le \pi(n-4) + 1 < \dfrac{n-4}{3} + 1 = \dfrac{n-1}{3} < \dfrac{n}{3}$

(4) Since $n$ and $n-2$ are prime, $n-4$ is divisible by $3$ and $\pi(n) \le \pi(n-6) + 2 < \dfrac{n-6}{3} + 2 = \dfrac{n}{3}$

Answer 1

Given any number $a$, the number of primes less than $ab$ cannot be greater $b\phi(a)$ (where $\phi$ is the totient function). This is because there are $a-\phi(a)$ numbers that are less than $a$ and share a prime factor with $a$, and if they share a prime factor with $a$, then they will share a prime factor with $a+ka$ for any $k$. The set of numbers less than $ab$ can be written as $$(0+1a,0+2a ... 0+(b-1)a ),$$ $$(1+0a,1+1a,1+2a ... 1+(b-1)a),$$ $$(2+0a,2+1a,2+2a ... 2+(b-1)a), $$ $$....$$ $$((a-1)+0a,(a-1)+1a,(a-1)+2a ... a+(b-1)a) )$$ We have $a$ parentheses of $b$ numbers. $a-\phi(a)$ of these parentheses consist entirely of composite numbers. Thus, we have at least $b(a-\phi(a))$ composite numbers (there are some off-by-one issues that I'm sweeping under the rug, but those aren't too difficult to deal with), so at most $b\phi(a)$ are prime. So given any $x$, we just have to find an $a$ for which $\phi(a)/a$ is sufficiently low. For instance, $\phi(420)=96$, so fewer than 96 numbers less than 420 are prime (and 96 is less than 420/4=105). There will also be fewer than 96 prime numbers between 420 and 2*420=840, if $p$ is such a prime number, then $p-420$ would have to be co-prime with 420, and there are only 96 such numbers. In general, the number of prime numbers less than $420b$ will be less than $96b$.

There are further issues such as that the prime factors of $a$ will be in this set of number not coprime to $a$, yet will prime numbers for $k=0$. Another issue is that my argument deals with $n$ in the form $n=ab$, which leaves out $n$ between $ab$ and $a(b+1)$. For large $n$ these issues disappear. For instance, if $b=100$, then the initial prime numbers will make up only 4 out of the 100*420 numbers. For a number between 100*420 and 101*420, we have that at most 4+96*101 numbers less than 101*420 are prime, and 4+96*101 is less than $4n$ for any $n$ between 100*420 and 101*420. Thus, we can prove the inequality by inspection for small $n$, and then use the asymptotic argument for large $n$. (Note that 420 and 100 aren't the smallest numbers possible; I just picked them for easy calculation. We can also decrease the size of $b$ required by presenting upper bounds for the number of primes between $ab$ and $ab+c$ better than just $\phi(a)$).