TITLE: Is there any pattern to the continued fraction of $\sqrt[3]{2}$?
QUESTION [26 upvotes]: Is there any pattern to the continued fraction of $\sqrt[3]{2}$ ?  Wolfram Alpha returns for cube root of 2:

$\sqrt[3]{2}=$ [1; 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14,
  3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1, 1, ...]

So the answer is likely no.  There certainly won't be any repeating pattern such as:
$$ 1+\sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots }}}$$
But I was hoping maybe for a pattern like we might find for the number $e = 2.718\dots$:
$$ e = [2; 1, \color{blue}{2}, 1, 1, \color{blue}{4}, 1, 1, \color{blue}{6}, 1, 1, \color{blue}{8}, 1, 1, \color{blue}{10}, 1, 1, \color{blue}{12}, 1, 1, \color{blue}{14}, 1, 1, \color{blue}{16}, 1, 1, 1, ...]$$
So in fact $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ has a nicer continued fraction pattern than $\sqrt[3]{2}$.  
You can actually derive the continued fraction of $e$ (Henry Cohn) - but I haven't worked through it yet.

Proposal: Tree-Like Continued Fraction
If we try to derive a continued fraction for cube roots let's see how we get in trouble:
$$ \sqrt[3]{2} \approx 1 $$
That's really lousy guess but let's see how far we are off:
$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \tag{$\square$}$$
so let's backtrack we actually needed two pieces of information:
$$ 
\sqrt[3]{2} \approx 1 
\hspace{0.25in}
\textbf{and}
\hspace{0.25in} 
\sqrt[3]{4} \approx 1 $$
The cube root of 4 is still not quite $2 = \sqrt[3]{8}$.  Now let's try:
$$ \sqrt[3]{4} - 1 = \frac{3}{1 + \sqrt[3]{4} + \sqrt[3]{16}}
=  \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}}
\tag{$\Delta$}
$$
And then run all of this back into the equation we started from:
$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}
= \cfrac{1}{1 + \left(1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \right) + \left( 1 + \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}}\right)}$$
Let's make it look a little bit cleaner but the same:
$$ \sqrt[3]{2} = 1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}
= 1 + \cfrac{1}{3 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}  +  \frac{3}{1  + 2\sqrt[3]{2}+ \sqrt[3]{4}}}$$
Using the $\square$ and $\Delta$ pattern we can get a regular infinite pattern this way.

Clarification
I am being repeatedly asked to clarify the question what do I mean by "pattern" the truth is I don't know.  To this end I make two points:

the are transcedental numbers with clear patterns if their continued fraction such as $e$.  We find that every third digit is an even number starting with $2$ and increasing by $+2$.  Every other digit is $1$.
I have already proposed a "tree-like" generalized continued fraction that may have recursive properties similar to what repeated patterns havefor numberes like $\sqrt{n}$
This paper by Yann Bugeaud states continued fraction digits definitely cannot have any pattern:


Let $\alpha = [0; a_1, a_2,...]$ be an algebraic
  number of degree at least three. One of our criteria implies that the
  sequence of partial quotients $(a_n)_{n≥1}$ of $\alpha$ cannot be generated by a finite automaton, and that the complexity function of $(a_n)_{n≥1}$ cannot increase too slowly.

Personally it's hard for me to fathom a number that doesn't satisfy any pattern.  There must be some pattern.  My question is then what could we try instead?
Lastly, if the cubic case is this intractable, I might shift my attention to the quadratic case where more results are known...

REPLY [5 votes]: This is the Hermite Problem: https://en.wikipedia.org/wiki/Hermite%27s_problem
The Jacobi-Perron algorithm attempts to solve this.
Mittal and Gupta have defined Bifurcating Continued Fractions- http://arxiv.org/abs/math/0002227 http://arxiv.org/abs/math/0008060
Specifically, they say the pair $2^{1/3}$ and $2^{2/3}$ has the expansion:
$1,\overline{1,2} $ and $ \overline{1,0}$
Lehmer talks of Ternary Continued Fractions - https://oeis.org/A000962/a000962.pdf
How it all fits together I have no idea.