Examine if $\exists M$ such that $\forall n>M$, $\pi(2n)$ $-$ $\pi(c_n)$ $>$ $0$

Question

The problem is-

Examine if $\exists M$ such that $\forall n>M$, $\pi(2n)$ $-$ $\pi(c_n)$ $>$ $0$. Also find a value of such $M$ for which the theorem is true.

Though I haven't still given it a serious thought, but I think that for the second part under the constraint the value of $M$ should be the least possible value satisfying the first part of the theorem, the problem becomes difficult. It is under this special restriction that I am trying to solve the problem.

The first part of the problem is easy and easily solvable by Prime Number Theorem. But solving the second part problem using Prime Number Theorem is so far an unsuccessful attempt.

Any help will be appreciated.

Answer 1

To fix ideas let us recall from earlier questions of yours that $c_n$ is the $n$-th composite number, and $\pi$ is the prime counting function.

Recall a result of Nagura (1952)

For $x\ge 25$ there is a prime between $x$ and $6x/5$.

This can be proved by elementary means.

Since every element congruent $2,3,4,6$ modulo $6$ is composite except for $2,3$, we see that below $3n/2+6$ there are certainly at least $n-2$ composite numbers. And, having $25$ and $35$ in addition as compositite numbers, we conclude that $c_n \le 3n/2+6$ for $n \ge 36$.

Now, applying Nagura's result with $x_0=3n/2+6$ we get a prime between this $x_0$ and $6x_0/5=6(3n/2+6)/5\le 18n/10 +7 < 2n$ for $n \ge 36$.

So, for $n \ge 36$, we always have a prime between $c_n$ and $2n$ and thus $\pi(2n) - \pi(c_n)> 0$.

This is likely not the samllest possible choice, not only but also, as we were a bit sloppy on the way, but one could check the few remaining small cases by hand.