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I made a program that calculates the ratio between the primes and the geometric means of earlier primes.

5 / square root of (2X3)

7 / cubic root of (2X3X5)

Is there a pattern?

I figured that the primes were multiples of geometric means of earlier primes ?

This is silly ?

A hug.

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closed as off topic by Andreas Thom, Yemon Choi, Felipe Voloch, Gerry Myerson, Chris Godsil Apr 15 '12 at 23:05

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I'm not sure this is appropriate. I do like the hug though. – Woett Apr 15 '12 at 21:57
You might try the product of the first n primes compared with the nth power of n as n grows large. Gerhard "Ask Me About System Design" Paseman, 2012.04.15 – Gerhard Paseman Apr 15 '12 at 22:05

Yes, there is a pattern. The fraction converges to $e$. This is essentially a consequence of the prime number theorem (in a sufficiently precise form). [Sorry for the initial totally messed up version, where I also overlooked something; and while the answer below is not hard either it is slightly more subtle than I thought at first, so first no details, now too on average alright.]

Let $p_n$ denote the $n$th prime.

The question asks for the quantity

$$ \frac{p_{n+1}}{ (\prod_{i=1}^n p_i)^{1/n} } $$

Now, $\log(\prod_{i=1}^n p_i)$ is Chebychev's Theta of $p_n$ and as such well-studied. In partiular one knows (see for instance Dusart's thesis page 15) that it is $n (\log n + \log \log n - 1 + o(1))$. So the denominator in the expression above is

$$ \exp (\log n + \log \log n - 1 + o(1)) = n (\log n) e^{-1} exp(o(1)) $$

while the numerator is $p_n \sim n (\log n)$.

Therefore the fraction converges to $e$.

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You mean the fraction converges to $e$? It seems to be the case. What is the reference? – Mark Sapir Apr 15 '12 at 22:30
@Mark, the product of the primes up to $x$ is asymptotic to $e^x$. This is at the level of the Prime Number Theorem, and is in many Analytic Number Theory texts. I think this is the result quid has in mind. – Gerry Myerson Apr 15 '12 at 23:01
@Gerry: Thank you! – Mark Sapir Apr 15 '12 at 23:15
@Mark Sapir: I should have meant that, yes. Sorry for the confusing (actually wrong) initial version. @Gerry Myerson, yes I meant this (and thought it instantly answers the question). – user9072 Apr 16 '12 at 0:33
@quid: thank you for your answer! – Mark Sapir Apr 18 '12 at 9:48

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