# Grander than the Riemann Hypothesis

1. Feb 15, 2009

### tgt

...But it may not exist yet.

Has any mathematician thought about producing a formula or function which spits out all the prime numbers? i.e 1->2, 2->3, 3->3, 4->5, 5->7, 6->11 etc.

What the closest that people have thought?

2. Feb 15, 2009

### csprof2000

You mean a closed-form solution, I assume? As in, piecewise with a finite number of piecewise parts?

I think it's safe to assume that nothing of the sort yet exists... if it did, I believe it would answer the Riemann hypothesis. No?

I also believe I read something somewhere about a closed-form polynomial never being able to generate "only" prime numbers.

See:

http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

3. Feb 15, 2009

### CRGreathouse

Yes. There are many of these. Ribenboim (2004) lists many examples on pp. 131 to 155, as does Guy in UPNT A17 (pp. 58-65 in the third edition). See also Prime Formulas on MathWorld.

Hardy & Wright mention this in Chapter 1 (p. 6 in the latest printing) and give several examples.

Willans, Wormell, Mináč, and Gandhi all give examples of closed-form solutions.

It would not solve the Riemann hypothesis.

A one or two-variable polynomial can't produce all the primes (and only primes). But a 26-variable one can; Jones, Sato, Wada and Wiens gives an explicit example after Matijasevič showed it was possible.

Last edited: Feb 15, 2009
4. Feb 15, 2009

### Kurret

I found this one really cool:
http://mathworld.wolfram.com/PrimeDiophantineEquations.html
How on earth do you find such a formula :p
I guess it is not a coincident that the number of variables in this diophantine equation is the same as the number of variables in the polynomial that CRGreatHouse refered to.

5. Feb 15, 2009

### CRGreathouse

Yes, that was the result I was referring to. Matijasevič showed that is was possible by showing that recursively enumerable sets are precisely Diophantine sets, and Jones, Sato, Wada and Wiens made the polynomial you mention.

Last edited: Feb 15, 2009
6. Feb 15, 2009

### Santa1

It could solve it couldn't it? If we have such a bijection $$f : \mathbb{N} \to \mathbb{P}$$ wouldn't $$f^{-1}(p)=\pi(p)$$?

7. Feb 15, 2009

### CRGreathouse

How would that help?

8. Feb 15, 2009

### Santa1

If you had a closed form for $$f^{-1}(p)$$ it might be easier extract the necessary information to show that $$f^{-1}(x) = Li(x) + \mathcal{O}(\sqrt{x}\log(x))$$, of course it is only speculation as it would depend entirely of the form of $$f(n)$$.

9. Feb 15, 2009

### CRGreathouse

OK, so let
$$F(j)=\left\lfloor{\cos^2\pi\frac{(j-1)!+1}{j}\right\rfloor.$$

Then
$$\pi(x)=\sum_{j=2}^xF(j).$$

Alternately, let
$$F(j)=\frac{\sin^2\pi\frac{(j-1)!^2}{j}}{\sin^2\frac\pi j}$$

Now you have two closed-form formulas for the prime-counting function. Does this help you prove the RH?

10. Feb 17, 2009

### Santa1

Well, no ( :þ ), because I can't show that $$\sum_{j=2}^x F(j) - Li(x)$$ is $$\mathcal{O}(\sqrt{x} \log x)$$.

What I was trying to say that if we have a "nice" form of that function it might be easier to show that it is. I'm not saying it would solve RH automatically, I'm only saying it might do it, depending of the form.