5
$\begingroup$

I am looking for the asymptotic growth of product of consecutive primes. Is there anything that is known about this growth?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
7
$\begingroup$

Denote by $$\Pi(x)=\prod_{p\leqslant x}p,$$ thus $$\log\Pi(x)=\sum_{p\leqslant x}\log p:=\theta(x)\sim x,$$ which is known as the Prime Number Theorem. You may find further information in http://en.wikipedia.org/wiki/Prime_number_theorem

Improve this answer
$\endgroup$
12
$\begingroup$

I think you're asking about the primorial function $n\sharp$, the product of all the primes less than or equal to $n$. This satisfies $ n \sharp = \exp( n(1+o(1)) ) $.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Are sharper bounds known? This is actually a pretty bad estimate for n small (less than, say, a billion). $\endgroup$
    – Charles
    Commented Nov 1, 2010 at 16:21
  • $\begingroup$ @Charles: If you assume the Riemann Hypothesis, you get roughly $\exp(n+O(\sqrt{n}))$ $\endgroup$
    – S. Carnahan ♦
    Commented Nov 2, 2010 at 3:20
  • $\begingroup$ @Charles: Why do you say it is bad? Up to the (very) small bound of 350000 I find results such as [ 82619, .99728] [119549, .99537] [155893, .99855] [302831, .99671] [338477, .99898] meaning that at the prime n=82619 it is about $\exp( 0.99728n )$. etc. After a relatively large prime gap the exponent will be lower and after a relatively prime dense interval it will be higher. $\endgroup$
    – Aaron Meyerowitz
    Commented Nov 2, 2010 at 4:35
  • $\begingroup$ @Aaron Meyerowitz: exp(1e6)/1e6# is about 2e658. It's not so much that I mind being off by a factor of googol^6, but I wanted to know if more asymptotic terms were known. $\endgroup$
    – Charles
    Commented Nov 2, 2010 at 5:07
  • $\begingroup$ @Charles OK so you are telling me that $10^6 \sharp$ is about $exp(10^6\cdot 0.99934)$. Working from that, I then get [1090697, .9991] ,[1195247, 1.00026] and [1243337, .99948]. I can't vouch for all those decimal places but the relative fluctuations should be pretty accurate. I can't imagine that asymptotic terms are going to account for fluctuations like that. The best you should expect is to get $\ln(n \sharp)$ about right and I think the answers above do that. $\endgroup$
    – Aaron Meyerowitz
    Commented Nov 2, 2010 at 7:24
5
$\begingroup$

You can also prove that $$\displaystyle\lim_n\left(\prod_1^n p_i\right)^{1/p_n} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .