TITLE: What is the geometric average of 2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10...?
QUESTION [5 upvotes]: It is known that the Khinchin constant is not the geometric mean of the first $n$ coeffecients, as $n$ approaches infinity, of the continued fraction of e, which is $$[2; 1, 2, 1, 1, 4, 1,1,  6, 1, 1, 8, 1, 1, 10, 1, 1, 12,\ldots]$$
Seeing the graph of $f(n)$, the geometric mean of the first $n$ numbers, it seems like there would be a fairly elementary function $g(n)$ which follows and approximates it.
Since $2\cdot4\cdot6\cdot8\cdot10\cdots = 2^n\cdot n!$, we can say
$$f(n)=\sqrt[n]{\left\lfloor \frac{n}{3} \right\rfloor !\cdot2^{\lfloor n/3 \rfloor + 1}}$$
And since the floor function doesn't matter much on a larger scale, and to be more consistent with the $\frac{n}3+1$, we can essentially rewrite the function as
$$f(n)=\sqrt[n]{\Gamma\left(\frac{n}{3}+1\right)\cdot2^{n/3+ 1}}$$After graphing this function and reading a list of identities of the gamma function I could not simplify this, even though the graph looks fairly straightfoward.
My question is, is there a simpler, or perhaps constant $g(n)$ where $$\lim_{x\to \infty} \frac{f(n)}{g(n)} = 1 \text{ ?}$$

REPLY [5 votes]: By Stirling's approximation, we know that $(n!)^{1/n} \sim n/e$ as $n\to\infty$, so
$$
\left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{1/n} \;=\; \left( \left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{3/n}\right)^{1/3} \;\sim\; \left(\frac{n}{3e}\right)^{1/3},
$$
and hence
$$
f(n) \;\sim\; \left(\frac{2n}{3e}\right)^{1/3}
$$
as $n\to\infty$.