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Does anyone happen to have at hand a short, elegant proof that demonstrates that there do (or do not) exist one or more algebraically representable prime number generating functions?

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What do you mean by "algebraically representable"? –  Qiaochu Yuan Jun 1 '12 at 3:30
    
A reasonable version might be: there is a nonconstant bivariate polynomial $P(z,w)$ such that for some sequence $p_n$ of distinct primes, $P(n,p_n) = 0$. –  Robert Israel Jun 1 '12 at 19:04
    
I guess that wasn't precise. By 'algebraically representable' I meant: en.wikipedia.org/wiki/Algebraic_function –  enthdegree Jun 3 '12 at 0:24

2 Answers 2

up vote 2 down vote accepted

Try http://en.wikipedia.org/wiki/Prime_generating_function

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$$\pi(x)=\sum_{n\le x}\left|\operatorname{sgn}\prod_{k=2}^n\prod_{l=2}^n(n-kl)\right|$$ is a representation that can be linked fairly directly to the definition of primes. –  anon Jun 1 '12 at 1:54
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@anon I'm not sure that can be called "algebraic." –  Thomas Andrews Jun 1 '12 at 2:12

You clarified that you meant, "Is there an algebraic function which generates the primes", that is, "Is there a fixed expression using only addition, subtraction, multiplication, division, and (fixed) root extraction which generates the primes?" The answer is no, because algebraic expressions cannot have the required growth rate, lacking anything that grows logarithmically, while $p_n\sim n\log n.$

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