Are all the numbers $\pi(n^2)/n^2\ (n=1,2,3,\ldots)$ pairwise distinct? For $x>0$ let $\pi(x)$ denote the number of primes not exceeding $x$. A well-known conjecture of Legendre states that $\pi(n^2)<\pi((n+1)^2)$ for any positive integer $n$. Here I ask the following natural question.
QUESTION: Are all the numbers $\pi(n^2)/n^2\ (n=1,2,3,\ldots)$ pairwise distinct?
Based on my computation, in 2015 I conjectured that the answer is yes. Moreover, my computation suggests that the sequence
$$\frac{\pi(n^2)}{n^2}\ \ (n=15647,15648,\ldots)$$
is strictly decreasing. For each $k=3,4,5,\ldots$, I also conjectured that the sequence
$$\frac{\pi(n^k)}{n^k}\ \ (n=2,3,\ldots)$$
is strictly decreasing. 
Any comments on the question are welcome!
 A: Edit: This is not quite an answer, leaving it up as a long comment for now after fixing some arithmetic errors.
As suggested in the comments, using Dusart's results from his paper (see here) where he proves unconditionally
$$
\pi(x)\geq \frac{x}{\log x}\left(1+ \frac{1}{\log x}+\frac{2}{\log^2 x}\right)\stackrel{\triangle}{=}xL(x), \quad x\geq 88~783,
$$
and
$$
\pi(x)\leq \frac{x}{\log x}\left( 1+\frac{1}{\log x}+\frac{2.334}{\log^2 x}\right)\stackrel{\triangle}{=}xU(x), \quad x\geq 2~953~632~287,
$$
one can obtain, for $n$ larger than the second inequality cutoffs (if I haven't made a mistake) and for $k=2$ (higher $k$ will be similar):
$$
 \frac{\pi(n^2)}{n^2} -\frac{\pi((n+1)^2)}{(n+1)^2} \geq 
L(n^2) - U((n+1)^2),\quad (0)
$$
where the right hand side equals
$$
\left(\frac{1}{2 \log n}-\frac{1}{2 \log(n+1)}\right)+
\left(\frac{1}{4 \log^2 n}-\frac{1}{4\log^2(n+1)}\right)+
\left(\frac{2.334}{8 \log^3 n}-\frac{2}{8 \log^3(n+1)}\right).\quad\quad (1)
$$
The first term in (1) can be rewritten as
$$
\frac{\log^2(n+1)-\log^2(n)}{ 2\log^2(n+1)\log^2(n)}=
\frac{(\log(n)+\log(1+\frac{1}{n}))^2-\log^2(n)}{2 \log^2(n+1)\log^2(n)}=
$$
$$
=\frac{2\log(n)\log(1+\frac{1}{n})+\log^2(1+\frac{1}{n})}{2\log^2(n+1)\log^2(n)}
\sim 
\frac{1}{ n\log^2(n+1)\log(n)}
$$
since the first term in the last fraction dominates, and it is $O(1/(n\log^3 n))$.
A similar argument for the second term in (2) shall give a positive quantity with a constant on top and an $n$ and a higher power of $\log n$ in the denominator, which will be smaller than the first term.
The third term in (2) is unfortunately negative due to the mismatched coefficients and
dominates the right hand side of (0), so this effort fails.
If that worked, it would have proved that that the numbers $\pi(n)^2/n^2$ are pairwise distinct beyond
$n\geq 2~953~632~287.$ The differences up to then can be checked computationally. There are some extensive tables of the prime counting function, e.g., see this link here. 
Actually, the Magma online calculator here can return lists of primes from $2$ to $10^9$, if they are split into sublists to avoid memory overflow as I just confirmed.
Remark: Using the simpler Dusart formulas in the paper above these don't seem to work.
