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).*



Odd Chebyshev, part 2
We have
$$
I(N)=\frac{(6N-2)!(N-1)!}{2(4N-2)!(3N-1)!}.
$$
We count the exponent of a prime $p$ in $k!$ using a standard formula $[k/p]+[k/p^2]+\dots$. This gives the sum $w(p):=\sum\_{s} h(n/p^s)$ for the exponent of $p$ in the number $I(n)\frac{6n-1}{4(4n-1)}$, where we denote $h(x)=[x]+[6x]-[3x]-[4x]$. $h$ takes the values $\pm 1$ and 0. And $h(x)=-1$ only when $x$ lies between $k+3/4$ and $k+5/6$ for some integer $k$. Your number $M(n)$ equals $\prod\_p p^{\min(-w(p),0)}$. The powers $p^s$ for $s\geqslant 2$ do not affect the asymptotics (as usual).


Thus asymptotically $M(n)$ behaves as a product of primes $p$ for which $n/p$ lies in $(k+3/4, k+5/6)$ for integer $k$. By the PNT the logarithm of this product behaves as $$n\cdot \sum\_{k=0}^\infty \left(\frac1{k+3/4}-\frac1{k+5/6}\right)=n\left(\frac{\sqrt{3}-1}2\pi-\log\frac{27}4\right)=0.195\ldots \cdot n,$$
this answers all three questions.


