Proof for a prime number formula involving the prime counting function How would I go about proving this? I came across this when I was searching for interesting prime generating function. Or alternatively, could you kindly direct me to a source containing a complete proof for the following formula?
$$p_n=1+\sum^{2^n}_{m=1}\left\lfloor\left\lfloor\frac{n}{1+\pi(m)}\right\rfloor^{\frac{1}{n}}\right\rfloor.$$
Here $\pi(m)$ is the prime counting function.
 A: Lemma: For any $x \geqslant 0$ and $n \in \mathbb{N}_+$, $\bigl[[x]^{\frac{1}{n}}\bigr] = [x^{\frac{1}{n}}]$.
Proof: It is easy to see that $\bigl[[x]^{\frac{1}{n}}\bigr] \leqslant [x^{\frac{1}{n}}]$. Now, suppose $a = [x^{\frac{1}{n}}] \in \mathbb{N}$, then$$
x = (x^{\frac{1}{n}})^n \geqslant a^n \Longrightarrow [x] \geqslant a^n \Longrightarrow [x]^{\frac{1}{n}} \geqslant a \Longrightarrow \bigl[[x]^{\frac{1}{n}}\bigr] \geqslant a = [x^{\frac{1}{n}}].
$$
Thus, $\bigl[[x]^{\frac{1}{n}}\bigr] = [x^{\frac{1}{n}}]$.
Now back to the question. By the lemma,$$
\sum_{m = 1}^{2^n} \left[ \left[ \frac{n}{1 + π(m)} \right]^{\frac{1}{n}} \right] = \sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right].
$$
Note that $n < 2^n$ for any $n \geqslant 1$, thus for any $1 \leqslant m \leqslant 2^n$,$$
0 < \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \leqslant n^{\frac{1}{n}} < 2 \Longrightarrow \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] = 0 \text{ or } 1.
$$
By Bertrand's postulate, $p_n \leqslant 2^n$ for all $n \geqslant 1$, thus $π(m) \leqslant n - 1$ for $1 \leqslant m \leqslant p_n - 1$ and $π(m) \geqslant n$ for $p_n \leqslant m \leqslant 2^n$, which implies\begin{align*}
\sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] &= \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \leqslant n - 1}} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] + \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \geqslant n}} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right]\\
&= \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \leqslant n - 1}} 1 + \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \geqslant n}} 0 = \sum_{m = 1}^{p_n - 1} 1 + \sum_{m = p_n}^{2^n} 0 = p_n - 1.
\end{align*}
Therefore,$$
1 + \sum_{m = 1}^{2^n} \left[ \left[ \frac{n}{1 + π(m)} \right]^{\frac{1}{n}} \right] = 1 + \sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] = p_n.
$$
