Primorial: asymptotics and bounds

[CRGreathouse]

I was wondering if anyone knew of good approximations of the primorial function for large numbers, or of reasonable bounds for it. By primorial I mean:

n\#=\prod_{p\le n}p for p prime.

All I know is n\#\sim e^n and the trivial \pi(n)!\le n\#\le n!.

For small numbers (n < 1,000,000,000), e^n is larger than n#. Does this hold generally?

Obviously, the crude upper limit can be hacked slightly: take out the evens from the factorial and multiply by 2, take out the multiples of 3 and divide by 3, etc. In the extreme case this is just calculating the primorial itself, so not very useful. The lower bound can be improved to

p_n\#\le n!\prod^n_{k=5}\ln k+\ln\ln k-1

using lower bounds on the primes, but this is unwieldy, and I'm pretty sure \lim_{n\rightarrow\infty}\frac{p_n\#}{n!\Pi}=0 where the product is as above. Using the best bounds out there (by Dusart) improves the result slightly at the cost of complicating it further.

 

[CRGreathouse]

Well, I did some 'research' online:

Chris Caldwell's page says that n\#\sim e^n is equivilent to the PNT. I knew the formula already, but at least this gives me a different way of thinking about it.

I found the asymptotic expression \exp((1 + o(1))\cdot n\cdot\log(n))=n^n(1+o(1))^{n\log(n)} on Sloane's OEIS. This is larger than e^n, which already appears to be an upper bound... any ideas about this?

 

[CRGreathouse]

I forgot that Pierre Dusart worked on this already. He shows that \vartheta(x)&lt;x for x < 10^{11} (100 billion, better than I was able to calculate by a factor of 50) and gives some explicit bounds.

 

[shmoe]

Chris Caldwell's page says that n\#\sim e^n is equivilent to the PNT. I knew the formula already, but at least this gives me a different way of thinking about it.

This isn't true though. if n#~e^n then n#/e^n->1, which means log n#-n->0, i.e. \vartheta(n)-n\rightarrow 0. This is false, the difference gets arbitrarily large.

What is true is log(n#)~n is equivalent to the prime number theorem.

That asymptotic from Sloanes appears to be for a(n), the nth primorial, a(1)=1,a(2)=2,a(3)=6,a(4)=30 etc. I don't believe this asymptotic for the same reasons, unless I misunderstand what they mean.

 

[CRGreathouse]

This isn't true though. if n#~e^n then n#/e^n->1, which means log n#-n->0, i.e. \vartheta(n)-n\rightarrow 0. This is false, the difference gets arbitrarily large.

Sorry, I've been converting between the two much too readily. This is the second mistake I've made between theta and # in as many days (the last was writing one for the other!).

That asymptotic from Sloanes appears to be for a(n), the nth primorial, a(1)=1,a(2)=2,a(3)=6,a(4)=30 etc. I don't believe this asymptotic for the same reasons, unless I misunderstand what they mean.

I don't trust it. I can get the upper bound

x\#&lt;x^{\frac{x}{\ln x-1.1}} for x >= 4

from Dusart's bounds on primes and the trivial observation that \vartheta(x)\le\pi(x)\ln x.

 

[shmoe]

Sorry, I've been converting between the two much too readily. This is the second mistake I've made between theta and # in as many days (the last was writing one for the other!).

I'd be lying if I said I've never made this mistake myself.

I don't trust it. I can get the upper bound

x\#&lt;x^{\frac{x}{\ln x-1.1}} for x >= 4

from Dusart's bounds on primes and the trivial observation that \vartheta(x)\le\pi(x)\ln x.

It looks fine if you take logs, log a(n)~n*log(n). Remember a(n)=p(n)#, where p(n) is the nth prime and p(n)~nlog(n).