I am looking for the asymptotic growth of product of consecutive primes. Is there anything that is known about this growth?
[Math] Asymptotics of Product of consecutive primes
analytic-number-theorynt.number-theoryprime numbers
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Forgetting the squarefree condition for a moment, the number of integers up to $N$ that are not divisible by any primes less than $N^{1/k}$ is asymptotic to $$ \omega(k) \frac N{\log N} \sim e^\gamma \omega(k) N \prod_{p\le N^{1/k}} \bigg( 1-\frac1p \bigg), $$ where $\omega$ is the Buchstab function. (In particular, the first half of the density you derive in your answer is not correct: there is a correction factor of the form $e^\gamma \omega(k)$.)
Now the number of integers up to $N$ that are divisible by the square of a particular prime $p$ is at most $N/p^2$. So the number of integers up to $N$ that are divisible by the square of a prime greater than $N^{1/k}$ is at most $$ \sum_{p>N^{1/k}} \frac N{p^2} < N \sum_{n>N^{1/k}} \frac1{n^2} < N \cdot \frac1{N^{1/k}}. $$ Therefore the above asymptotic also holds for squarefree numbers not divisible by small primes.
The question asks how many primes $p_n \le x$ are there such that $p_n + p_{n+1}+p_{n+2}$ is also prime. This is beyond our reach to answer, but one can use Hardy-Littlewood type heuristics to attack this. Since $p_n + p_{n+1}+ p_{n+2}$ is roughly of size $x$, it has about $1/\log x$ chance of being prime, and so one should expect the answer to be on the scale $\pi(x)/\log x \approx x/(\log x)^2$. But the asymptotic will be a little different, since $p_n + p_{n+1} +p_{n+2}$ is not quite random -- for example it will always be odd (omitting $2+3+5$).
Let's flesh out the usual heuristic. To be prime, a number must be coprime to each prime number $\ell$. A random integer has a chance $(1-1/\ell)$ of being coprime to $\ell$. What is the chance that $p_n+ p_{n+1} + p_{n+2}$ is coprime to $\ell$? Each of the primes $p_n$, $p_{n+1}$, $p_{n+2}$ itself lies in some reduced residue class $\mod \ell$. Assuming these possibilities are equally distributed, one should get the chance $$ \frac{1}{(\ell-1)^3} \sum_{\substack{a, b, c=1 \\ (a+b+c,\ell)=1}}^{\ell-1} 1. $$ If $a$ and $b$ are given with $a\not\equiv -b \mod \ell$, then $c$ has $\ell-2$ possible residue classes, and if $a\equiv -b \mod \ell$ then $c$ has $\ell-1$ possible residue classes. So the answer is $$ \frac{1}{(\ell-1)^3} \Big( (\ell-1)(\ell-2)(\ell-2) + (\ell-1)(\ell-1)\Big)= \frac{\ell^2-3\ell+3}{(\ell-1)^2}. $$ So we adjust the random probability at $\ell$ by the factor $$ \Big(1-\frac {1}{\ell}\Big)^{-1} \frac{(\ell^2-3\ell+3)}{(\ell-1)^2}= 1 +\frac{1}{(\ell-1)^3}. $$ For example, when $\ell=2$, this adjustment factor is $2$ reflecting the fact that $p_n+p_{n+1}+p_{n+2}$ is guaranteed to be odd.
Then the analog of the Hardy--Littlewood conjecture will be $$ \sim \prod_{\ell} \Big(1+\frac{1}{(\ell-1)^3} \Big) \frac{x}{(\log x)^2}. $$ The constant in the product is approximately $2.3$; definitely not $e$.
There are interesting biases concerning why this conjecture would be an underestimate. Work of Lemke Oliver and Soundararajan makes precise conjectures on the distribution of consecutive primes in arithmetic progressions. For example, if we look at $\ell=3$, the sum $p_n + p_{n+1} +p_{n+2}$ can be a multiple of $3$ only if all three primes are congruent to each other $\mod 3$. This configuration is not preferred, and there are significant biases against it for small numbers! Similarly for other $\ell$ also, there are biases that indicate that the actual probability would be a tiny bit larger (with the effect wearing off slowly for large $x$). In other words, one might be able to identify lower order terms on the scale of $x(\log \log x)/(\log x)^3$, which might explain why the numerics suggest a larger value for the constant.
Best Answer
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