# Prime Number Formula XXII

$p_n=1+\displaystyle\sum_{k=1}^{2 \cdot (\lfloor n\ln(n) \rfloor+1)}\left(1-\left\lfloor \frac{1}{n} \cdot \displaystyle\sum_{j=2}^k \left\lceil \frac{3-\displaystyle\sum_{i=1}^j \left\lfloor \frac{\left\lfloor \frac{j}{i} \right\rfloor}{\left\lceil \frac{j}{i} \right\rceil} \right\rfloor}{j} \right\rceil \right\rfloor\right)$

Try it !

## 4 thoughts on “Prime Number Formula XXII”

1. A. K. Heidel says:

Show the proof.

• Proof :

According to the Richard Nordquist (http://grammar.about.com/od/il/g/impersent09.htm)
“An imperative sentence typically begins with the base form of a verb, as in “Go now!” . The implied subject “you” is said to be elliptical. ”
Hence , “Show the proof.” is an imperative .
Q.E.D.

2. M says:

Hi,

Checked your formula in Matlab code for n up to 115 and works pretty good!

False answers are given for the following n:

n=49 -> p(49)=227, formula(49)=229
n=98 -> p(98)=521, formula(98)=523
n=103 -> p(103)=563, formula(103)=569
n=107 -> p(103)=587, formula(103)=593

Note: In those “false” cases, your formula gives the (n+1)th prime, instead of the (n)th prime.

Overall: Impressive formula!

Is it possible to get in contact with you?

• Try to run this maxima code :

n:49; 1+sum((1-floor((1/n)*sum(ceiling((3-sum(floor((floor(j/i))/(ceiling(j/i))),i,1,j))/j),j,2,k))),k,1,2*(floor(n*log(n))+1));

It should work fine . You probably have to set precision in Matlab to arbitrary .