Find all $n$ so that $c_n$ $>$ $\pi(n^2)$

Question

Find all $n$ $\in$ $\mathbb{N}$ so that $p_{c_n}$ $>$ $n^2$ where $p_n$ denotes the $n$-th prime and $c_n$, the $n$-th composite.

I have tried doing the problem using The stronger version of Roger's Theorem, i.e., by the inequality, $$\ln n + \ln \ln n > \dfrac {p_n}{n} > \ln n + \ln \ln n - 1$$ but in vain.

I think that the problem is true for all sufficiently large values of $n$. Even a proof of only this claim will also be appreciated.

Answer 1

From equation $(6)$ of this MathSciNet page we have

$$ \pi(n) > \frac{n}{\log n} $$

for $n \geq 17$, so for $n \geq 5$ we have

$$ \pi(n^2) > \frac{n^2}{2\log n}. $$

Also,

$$ c_n \leq 2n+2 $$

for all $n \geq 1$.

Since $n^2/(2\log n)$ is increasing for $n > \sqrt{e}$ and strictly convex and

$$ 2n+2 |_{n=11} < \left.\frac{n^2}{2\log n}\right|_{n=11}, $$

we know that

$$ 2n+2 < \frac{n^2}{2\log n} $$

and hence that

$$ c_n < \pi(n^2) $$

for $n \geq 11$. Thus the largest value of $n$ satisfying $c_n > \pi(n^2)$ must be less than or equal to $10$, and one can check by hand that it is $n = 6$.