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Please consider the (presumably infinite) Euler product over the twin primes:

$$ f(z) = \prod_{p\in\mathbb{P}}^{\infty} \Big( 1 - \frac{1}{(p(p+2))^ z} \Big) $$ (in which $p(p+2)$ is a divisor of $4((p-1)!+1) + p$ ).

The Euler Product is a product of a corresponding Dirichlet series. Which one is that?

Thanks in advance,

Max

Edit Update: error fixed.

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1  
Dirichlet series represent a function of a complex variable. –  Micah Milinovich Sep 20 '10 at 16:21
    
Perhaps a dumb comment, but if we had a Dirichlet series for the twin primes, couldn't we show there are infinitely many twin primes by plugging in an appropriate value to make the series infinite? I'm thinking about the particular way of showing there are infinitely many primes. –  Alex R. Sep 20 '10 at 17:09
    
Max, I am having trouble figuring out what you think an answer would look like. What kind of answer are you expecting, exactly? –  Qiaochu Yuan Sep 20 '10 at 17:15
    
@ Qiaochu Yuan: John's answer suffices. What part of the question made you confused? –  Max Muller Sep 20 '10 at 17:23
    
@ Qiaochu Yuan: Perhaps this helps: en.wikipedia.org/wiki/Twin_prime_conjecture ? All primes are twins iff they satisfy the modular relationship mentioned at the end of the introduction of that wikipedia page. In my question, that relationship is posed just a bit differently, but I think is identical. I did it because I didn't know the Latex symbol of the 'modular sign' by hard. –  Max Muller Sep 20 '10 at 17:40

1 Answer 1

up vote 5 down vote accepted

In this paper: http://arxiv.org/abs/0902.4352, the authors discuss the analytic properties of the related Dirichlet series

$$ D_{2r}(s) = \sum_{n=1}^\infty \frac{\Lambda(n)\Lambda(n+2r)}{n^s}$$

where $\Lambda(\cdot)$ is von Mangoldt's function. See section $2$.

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Thanks a lot, John! This was what I was looking for. Although I think the authors mention that it's $n^{2s}$ instead of $n^s$. –  Max Muller Sep 20 '10 at 17:57

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