Is there any way to find the product of prime numbers?

l-1j-cho
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According to the prime number theorem, the number of prime numbers that are less than N is approximately N\ln(N) for a sufficiently large N. But can we find the product of prime numbers that are less than N?
(For example, N=20 then 2x3x5x7x11x13x17x19 although I think 20 isn't large enough haha)
 
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Of course there is- multiply them!
 
If you are looking for a list of semiprimes, it exists. http://oeis.org/A001358/b001358.txt

If you are looking for a formula pik(n)=(ln(n)/n)*((n^k)/k!). Where k is the number of prime factors not necessarily unique.
 
Definition: the producht of the first n primes is called primorial and is writtn as n#

Theorem: n# < 4^{n}

Definition: \vartheta(x) := \Sigma_{p_{i}&lt;=x} Log(p_{i}) (Chebychev)

Theorem: \vartheta(x) ~ x for x -> \infty
 
For the second Theorem above from my 'numerical department':

i |p_{i} |\vartheta (p_{i}) |\vartheta (p_{i})/p_{i}

5 | 11 | 7.745 | 0.704091
26 | 101 | 88.344 | 0.874688
169 | 1009 | 963.162 | 0.954571
1230 | 10007 | 9905.202 | 0.989827
9593 | 100003 | 99696.902 | 0.996939
 
RamaWolf said:
Definition: the producht of the first n primes is called primorial and is writtn as n#

Theorem: n# < 4^{n}

Definition: \vartheta(x) := \Sigma_{p_{i}&lt;=x} Log(p_{i}) (Chebychev)

Theorem: \vartheta(x) ~ x for x -> \infty

could you let me know the name of the first theorem so i can google it?
 
l-1j-cho said:
could you let me know the name of the first theorem so i can google it?

I think, there is no sprecific name for it; but go to:

//en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate

and there look for Lemma 4
 
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