Let $$ \forall_{n=1\ 2\ \ldots}\quad I(n)\ :=\ \frac {(6\cdot n-3)!!}{(2\cdot n-1)!!\cdot(4\cdot n-3)!!} $$

Let $\ M(n)\ $ be the smallest natural number such that

$$ M(n)\cdot I(n)\ \in\ \mathbb Z $$

is an integer. What is the behavior of sequence $\ M(n)$? As a minimum, the Prime Number Distribution Theorem tells us that $\ M(n)\rightarrow\infty.\ $ But I'd like to ask about more:

**Q1** Is $\ M(n) < \left(\frac 43\right)^{3\cdot n}\ $ for all large $\ n?$

**Q2** What is the least non-negative real $\ a\ $ such that
for every $\ \epsilon > 0\ $ the inequality $\ M(n) < (a+\epsilon)^n\ $
holds for all large n. (*Actually,* $\ a \ge \frac {4\cdot e^2}{27})$.

**Q3** Would above constant $\ a\ $ be $\ a=\frac {4\cdot e^2}{27}$? *(Highly unlikely).*