# 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,

$$3P_{\lceil n \rceil}-2=\sum_{k=1}^{A}\left(4-\left\lceil \frac{\pi(k)}{n}\right\rceil^2\right)$$

• For most of the range of $k$, the value of $\pi(k)$ will be at most $2n$, so the summand is $\ge 4$, making the total $> 8n \log n$. This seems like a severe overestimate for $3P_n$. Do you have any examples to justify your claim? – Erick Wong May 10 '16 at 17:49
• May be there is an Incorrection in the formula I need to check – user335850 May 10 '16 at 18:31
• @ErickWong (+1) spotting the error, I have corrected it. Honest with you this formula is come about messing around with Willan and Martin Ruiz prime formulas – user335850 May 10 '16 at 18:52

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 $3p = \sum_1^p 3$.

• all I want to know is that is this formula correct yes or no. If I punch in a number onto the LHS, I will expect to get it on the RHS. – user335850 May 11 '16 at 5:20
• unfortunately this formul is a variation of Sebastian Martin Ruiz formula, if this is nonsense so are his also, so why there is paper accept by the maths arxiv, I thought they don't accept nonsense from author. – user335850 May 11 '16 at 5:34
• @pisquare arXiv has a registration process that requires some connection to academia, but its express purpose is to be a preprint repository. By definition, articles can appear there long before they are subject to peer review. There is a screening process that filters out obvious nonsense, but they can't have subject experts in every field doing thorough reviews at the volume arXiv supports. – Erick Wong May 11 '16 at 5:42
• @pisquare The formula is correct as my answer proves (for all but the tiniest values of $n$ which can be checked separately). It is also useless for any practical purpose. You will note I also explain why it only works for non-integer $n$. – Erick Wong May 11 '16 at 5:43