I'm trying to find out the growth rate of an algorithm that grows with $p_{n}\#$ where $p_{n}\#$ is the smallest primorial that is larger than the input: $x\in\mathbb{Z^{+}}$.
Other research has established an asymptotic bound of $p_{n}\# = e^{(1 + o(1))n\log{n}}$ (as per https://en.wikipedia.org/wiki/Primorial)
Any clue on how to put the above bound in terms of x?
Let $\theta(y)=\sum_{p\le y} \log p$, so that $e^{\theta(y)}$ is a primorial $p_n\#$ for $n=\pi(y)$. Let $z$ be the smallest integer (necessarily prime) such that $\theta(z) \ge \log x$. Since $\theta(y)\sim y$ by the prime number theorem, we see that $z=(1+o(1))\log x$. But also $\theta(z) = \theta(z-1) + \log z \le \log x + \log z$ since $z$ is the smallest such integer, and so $\theta(z) \le \log x + \log\log x + o(1)$. It follows that the corresponding primorial $e^{\theta(z)}$ is $$ \le e^{\log x + \log\log x + o(1)} = (1+o(1)) x\log x. $$ (And of course $\ge x$ is trivial.)