# Prim numbers formula

1. Jul 5, 2008

Is any formula that produces nth prime ?
if not visit this web site

2. Jul 5, 2008

### Santa1

Doesnt m=10 give 21=3*7 ??

3. Jul 5, 2008

### morphism

According to Maple, H(10)=2.

If you look at the formula closely, you'll notice that H(m) is going to be an odd prime whenever 2m+1 is a prime, and otherwise it's going to be 2. This follows from Wilson's theorem (2m+1 is a prime iff (2m)! + 1 = 0 (mod 2m+1)) and the behaviour of $\left\lfloor\lfloor x \rfloor / x \right\rfloor$. It's really nothing special.

4. Jul 5, 2008

Please work on this formula and find all gaps in order to fin a formula for primes

5. Jul 5, 2008

### Hurkyl

Staff Emeritus
Ah, so it's just a clever way to obscure a definition by cases, and is really just saying

$$H(m) = \begin{cases} 2m + 1 & 2m + 1 \textbox{\ is\ prime} \\ 2 & \textbox{otherwise} \end{cases}$$

6. Jul 5, 2008

### Santa1

I guess you're right, I must've had an error somewhere.

7. Jul 5, 2008

### HallsofIvy

Staff Emeritus
Are you saying then that one can test whether an odd number is prime by calculating this? Wouldn't that be faster than, say, trial division by primes less than its square root?

8. Jul 5, 2008

### huba

As long as (2m)! can be calculated (although it should be difficult for larger values of m), there is no need for the proposed formula, because Wilson's theorem already tests primality: if 2m+1 divides (2m)!+1 then 2m+1 is prime.
When (2m)! is difficult to calculate then the formula becomes difficult as well.

Last edited: Jul 5, 2008
9. Jul 5, 2008

### huba

Using Hurkyl's definition for H(m),

H(m)=2 if 2m+1 is composite.
H(4) = 2, and for all k, H(4+3k)=2.
H(12)=2, and for all k, H(12+5k)=2,
H(24)=2, and for all k, H(24+7k)=2, etc.

This is Eratosthenes' sieve, and I know about one exact formula for primes which utilizes it: Riemann's solution for the prime counting function. What other formula can be found?

10. Jul 5, 2008

### Hurkyl

Staff Emeritus
Why would you think any of those claims are true? Did you actually try to evaluate them for any value of k?

Sure doesn't look like it.

11. Jul 5, 2008

The range of H is
{3,5,7,2,11,13,2,17,19,...}={2,3,5,...}
so there is no problem.am I right?

12. Jul 5, 2008

### huba

H(m) = 2 if 2m+1 is composite.

When m = 4 + 3k, then 2m + 1 = 3(2k + 3) which is composite for all k. In fact, 3(2k + 3) gives all odd numbers > 9 that are divisible by 3.

Similarly for m = (p^2-1)/2 + pk where p is prime, because then 2m + 1 = p(2k + p), and so 2m+1 is composite, and so H(m) = 2.

13. Jul 5, 2008

### Hurkyl

Staff Emeritus
Good call; I had the wrong definition of H in my head when I responded.

14. Jul 5, 2008

### huba

The formula is interesting (it has certainly attracted many responses in a short time), but it does not produce the nth prime as effectively suggested in the first entry in this thread.
For a given n, we don't know for what value of m we get H(m) = the nth prime, until finding it by trial and error by calculating H(m) for a certain range of m. For instance, if the first prime is 2 (n=1), we find it (first) when m=4.

15. Jul 5, 2008

### CRGreathouse

Trial division on n takes $$O(\sqrt n)$$ arithmetic operations, or $$O(\sqrt n/\log n)$$ with precomputation of primes. With Newton's method and Karatsuba, this takes $$O(n^{0.8})$$ bit operations. Without pecomputation, this method takes only logarithmic space -- log n space for n, log n space for the calculation, and 1/2 log n space for a counter.

A quick way to calculate $$\lfloor(2m)!/(2m+1)\rfloor$$ is noting that if 2n+1 is composite, $$\lfloor(2m)!/(2m+1)\rfloor=(2m)!/(2m+1)$$. But this is what we're trying to determine, so slower methods must prevail. Now there are methods for computing n! in $$O\left(n(\log n)^{2+\varepsilon}\right)$$ bit operations, but these require factorization of n -- again, what we want to avoid. So binary splitting gives the solution in time $$O(n^{1.585})$$, though with lots of logarithmic factors hidden in the big O.* After that the exponentiation must be performed, at cost $$O(\log n!)=O(n\log n)$$. The space requirement for a literal computation is huge: [log]O(\log (2n)!)=O(n\log n)[/tex], an order of magnitude larger than trial division.

In conclusion, trial division is far superior, taking roughly the square root of the time and the logarithm of the space of this method.

* At several points I'm able to hide logarithmic factors between lg 3 = 1.58496... and the 1.585 I display; sneaky, eh?

16. Jul 7, 2008

### majesticman

guys if you all were able to generate a formula that gives the nth prime...you will be rich!!!...correct me if i am wrong but aren't security for top secret files as well as e-commerce based on the fact that really big prime numbers (they are set as encryption keys) cannot be factorized. So good luck...i thnk it is RSA algorithm that relies on prime numbers

17. Jul 7, 2008

### patrcik

While I doubt you would get rich(If you were thinking about the RSA factorization prizes, I'm sorry to tell you, but they stopped giving them out in 2007), but you would secure your place in history.
Yes, the RSA uses prime numbers to encrypt. SSL uses prime numbers as well. If some smart guy(or gal) figures out the distribution of prime numbers those would be dropped pretty quickly.

18. Jul 7, 2008

### Hurkyl

Staff Emeritus
There already exist explicit formulas for the n-th prime, and a variety of algorithms for computing the n-th prime.

I should hope not -- I can factor such numbers in my head, no matter how big they are.

19. Jul 7, 2008

### Hurkyl

Staff Emeritus
It's already known. The primes occur precisely when we have an integer that is not a multiple of smaller integers. :tongue: Aside from that obvious fact, there are bewildering array of bulk statistics known about the primes, and a great many more that are consequences of the Riemann hypothesis.

20. Jul 8, 2008

### Kurret

Why is the mathematical world so focused on prime numbers, why not the general case, to find all numbers with n prime factors?