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...