# Prim numbers formula

1. Jul 5, 2008

Last edited by a moderator: Apr 23, 2017
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

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?

21. Jul 8, 2008

### CRGreathouse

1. There are lots of formulas that generate primes; creating a new one won't guarantee that you can get a paper published, much less than make you rich.
2. Many encryption algorithms rely on the hardness of factoring large semiprimes; factoring primes is, of course, trivial.

22. Jul 8, 2008

### CRGreathouse

I don't think there's any push to find (general) prime numbers -- there are too many -- but to learn how to quickly factor a number into its prime factors. This is interesting because such a factorization is unique (up to order and units), though there is still interest in factoring into coprimes and partial factorization in general.

23. Jul 9, 2008

### DeaconJohn

That's a good question, Kurret (IMHO), and that's a good reply, Greathouse (IMHO).

Kurret,

A few more thoughts, adding to what Greathouse said.

One reason there is a "focus" on "prime numbers" these days is the relation with the Riemann hypothesis. For example, if we can show that the error term in the prime number theorem grows at the order of the square root of n, then we will know that the Riemann Hypothesis is true. (ref: Barry Mazur's "error term" article in this issue of the BAMS; the equivalence was shown about 100 years ago).

So, this fact reduces your question to "why is there so much interest in the Riemann hypothesis"?

Well, first of all, the RH is a very old problem.
Second, if it is true, many intereesting and beautiful facts about nature will follow. (My viewpoint is that interesting mathematics describes interesting things about nature.)
Third, everything that is known about the RH gives the impression that we a re very close to solving it.
Fourth, it's been that way for about 100 years now, so why haven't we solved it? (Riemann even wrote down an exact formula for the error term in the prime number theorem. It would seem that all we have to do is analyze that formula.)
Fifth, numerical studies of the behavour of the zeroes of the zeta function on the critical line indicate that current techniques are inadequate to solve it. The behaviour of those zeroes is just too strange and too sbutle to get a handle on. Therefore, a solution would likely result in a significant advance in our understanding of mathmatics in general.
Sixth, there is a million dollar prize for the person who resolves it.
Seventh, the name of the person who solves it is very likely to go down in history for a very long time.

On the other hand, as Greathouse pointed out, the question you ask is really of very high interest to today's mathematicians. One reason for that is IF we could find an effectgive algorithm to answer your question, THEN we would be a lot closer to finding an effective algorithm to factoring the product of two large primes in a short period of time. This would break RSA cryptography. It is highly likely that the proof would carry over to break discrete logarithm cryptography. It is quite possible that the proof would allow us to break elliptic curve cryptography.

The underlying mathematical meta-principle that you are touching on is that it is a lot easier to ask hard questions than it isw to ask "good" quesitons. A "good" question is a question that we have a hope of solving; that we feel that if we worked on it, we should be able to come up with something.

A small group of very good mathematicians are currently investigating the properties of numbers that are the product of a large number of small primes. The fact that that investigation is difficult (but doable) underlines just how hard the question that you ask probably is.

Another good example of a "hard" question is the question of whether or not there are an infinite number of Mersenne primes. For that one, there even exists a heuristic argument that there are an infinite number and it yields an asympototic estimate for how many there are!

And yet, as far as I know, nobody has any idea about how to get started on actually proving that there are an infinite number of them. Richard Guy also said this in his book "Open Problems in Number Theory" or something like that.

Oh, oh, oh, except the female star of the science fiction movie called "Proof." The "science fiction" in the movie is that she resolved the question about the finiteness of the Mersenne primes.

DJ

P.S. I welcome corrections, or even "that doesn't sound right" comments. That's how I learn.

Last edited: Jul 9, 2008
24. Jul 11, 2008