# Odd Chebyshev, part 2

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).

• Just in case, let me mention that I just have fixed a silly typo (twice): now I have the correct Euler $e$ (i.e. $\exp(1)$) instead of accidental $\pi$. Jan 16, 2017 at 18:57

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.
• I've recognized your answer. Isn't your constant $\ 0.195...\$ the extra part over 1? I.e. shouldn't your constant, as asked in the **q.**, be $\ 1.195...$ ? Jan 16, 2017 at 19:05