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

edited Nov 2 2010 at 15:46

Charles
4,159●8●26

asked Oct 30 2010 at 2:24

unknown (google)
1,009●4●11

arithboy · Accepted Answer · 2010-11-02 02:06:37Z UTC

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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

edited Nov 2 2010 at 2:06

answered Nov 2 2010 at 0:57

arithboy
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Michael Lugo · Answer 2 · 2010-10-30 02:31:47Z UTC

9

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)) ) $.

answered Oct 30 2010 at 2:31

Michael Lugo
8,385●2●21●48

	Are sharper bounds known? This is actually a pretty bad estimate for n small (less than, say, a billion). – Charles Nov 1 2010 at 16:21
	@Charles: If you assume the Riemann Hypothesis, you get roughly $\exp(n+O(\sqrt{n}))$ – S. Carnahan♦ Nov 2 2010 at 3:20
	@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. – Aaron Meyerowitz Nov 2 2010 at 4:35
	@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. – Charles Nov 2 2010 at 5:07
	@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. – Aaron Meyerowitz Nov 2 2010 at 7:24

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Jose Brox · Answer 3 · 2010-11-02 11:24:53Z UTC

2

You can also prove that $\displaystyle \ \lim_n \ \ \sqrt[p_n]{\prod_1^n p_i} = e$

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

answered Nov 2 2010 at 11:24

Jose Brox
975●1●12●22

Asymptotics of Product of consecutive primes

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