Show that $3P_{\lceil n \rceil}-2=\sum_{k=1}^{A}\left(4-\left\lceil \frac{\pi(k)}{n}\right\rceil^2\right) $ We proposed a formula for calculating nth prime number using the prime counting function.
Where $\lfloor x\rfloor$ is the floor function and $\lceil x\rceil$ is a ceiling function.
$\pi(k)$ is prime counting function and $ P_n $ the nth prime number.
Let $A=\lfloor 2n\ln(n+1)\rfloor$
Where $n \ge 0.9 $ (n MUST be a decimal number). [why n must be a decimal number I don't know, we will leave that to the authors to explain to us the reason for it]
Formula for calculating the nth prime,
$$ 3P_{\lceil n \rceil}-2=\sum_{k=1}^{A}\left(4-\left\lceil \frac{\pi(k)}{n}\right\rceil^2\right)  $$
 A: Dusart showed that $\pi(n) \ge n (\log n + \log \log n - 1)$ for $n\ge 2$.  From it's not too hard to calculate that for any $n\ge 2$,
$$\pi(2n) \ge 2n (\log 2n + \log \log 2n - 1) \ge 2n( \log n + \log \log 2n - 0.307) > A.$$
In particular (ignoring very small values of $n$), for all values of $k$ in the sum, $\pi(k) < 2n$, which means the summand is never negative.  It is also precisely $0$ whenever $n < \pi(k) < 2n$.  So the only contribution is from $\pi(k) \le n$.
Since $n$ is fractional, $\pi(k)$ is never exactly $n$.  The first value of $k$ for which $\pi(k) > n$ is $P_{\lceil n \rceil}$, so the last contributing term is $k=P_{\lceil n \rceil}-1$.  Meanwhile $\pi(k)$ is always positive except for the first term $\pi(1)=0$.  So the RHS simplifies to
$$4 + \underbrace{3 + 3 + \cdots + 3}_{P_{\lceil n \rceil}-2} + 0,$$
which is a very boring sum that simplifies to $3P_{\lceil n \rceil} - 2$ for elementary reasons that have nothing to do with number theory.  Consequently, this "formula" for the $n$th prime provides no insight whatsoever into the distribution of primes because it is essentially the trivial identity that $3p = \sum_1^p 3$.
