# Are there any patterns in simple continued fraction expansions of algebraic real numbers?

As we know there are patterns in simple continued fraction expansion of quadratic algebraic numbers,are there any patterns in simple continued fraction expansions of other algebraic real numbers?Or any law in them?or is there any universal algorithm to compute the integer sequence in simple continued fraction expansions,with the equation whose one solution is the real algebraic number

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By the way,can we extend continued fraction expansions of real number to complex numbers?especially to $i$?By $$\sqrt{x}=1+\frac{x-1}{2+\frac{x-1}{2+\frac{x-1}{2+\ddots}}}$$ one can know that continued fraction of $i$ is not convergent –  XL _at_China Jul 30 '14 at 11:40
To the best of my knowledge, no pattern and no law has ever been found in the continued fraction of any algebraic number of degree exceeding 2. –  Gerry Myerson Jul 30 '14 at 12:56
@GerryMyerson,philosophically,there does exist some law,since we know that all algebraic numbers are computable,but no law no pattern maybe imply those law and pattern are difficulty to be found.And we know there are patten even in simple continued fraction of $e$ –  XL _at_China Jul 30 '14 at 14:50

For a "formula" for the continued fraction of algebraic numbers, in particular $2^{1/3}$, see Bombieri and van der Poorten. It's just not a simple pattern.

EDIT: Actually there's an error in the formula in the middle of page 152 there: it should be

$$\pmatrix{p_{h+1} & q_{h+1}\cr p_h & q_h\cr} = \pmatrix{c_{h+1} & 1\cr 1 & 0\cr} \pmatrix{p_h & q_h\cr p_{h-1} & q_{h-1}\cr}$$

That is, the recurrence for the continued fraction $1 + \frac{1}{c_1 + \frac{1}{c_2 + \ldots}}$ of $2^{1/3}$ is

\eqalign{ c_{h+1} &= \left\lfloor {\frac { 3 \;\left( -1 \right) ^{h+1}{p_{{h}}}^{2}}{q_{{h} } \left( {p_{{h}}}^{3}-2\,{q_{{h}}}^{3} \right) }}-{\frac {q_{{h-1}}}{ q_{{h}}}}\right\rfloor\cr p_{h+1} &= c_{h+1} p_h + p_{h-1}\cr q_{h+1} &= c_{h+1} q_h + q_{h-1}\cr} with initial values $p_0 = 1, q_0 = 1, p_{-1} = 1, q_{-1} = 0$.

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thanks,I have just browsed it,and have not gotten it or what formula for continued fraction of algebraic numbers. –  XL _at_China Jul 30 '14 at 15:16
I think the formula for $2^{\frac{1}{3}}$ is interesting –  XL _at_China Aug 24 '14 at 14:06

Continued fractions for complex numbers can be studied. However, there is no "canonical" choice of which possible continued fraction to use. And, for that matter, why use the Gaussian integers $\mathbb Z[\sqrt{-1}\;]$ and not some other chioce?

See HERE for an article:
Convergence of Complex Continued Fractions
by John Marafino (James Madison University) and Timothy J. McDevitt (James Madison University)