TITLE: Is any particular algebraic number known to have unbounded continued fraction coefficients?
QUESTION [24 upvotes]: The continued fraction 
$$[1;1,2,3,4,5,\dots]=1+\cfrac{1}{1+\cfrac{1}{2+\cdots}}, $$ for instance, is known explicitly as a ratio of Bessel function values and is (I believe - SS) known to be transcendental.  Similarly, $[1,2,2^2,2^3,2^4,2^5,\dots] $ is surely transcendental (and likely related to Liouville numbers).  Are there any explicitly known algebraic numbers with unbounded continued fraction coefficients?  (Of course, such a number must have degree greater than $2$, by all the usual theorems on continued fraction representations of quadratics.)  Any references would be greatly appreciated.

REPLY [4 votes]: I have one thought on this.  But it would take some work to carry out.
(*) Perron (Die Lehre von den Ketterbrüchen) in Kap. 11 evaluates certain continued fractions with polynomial entries in terms of hypergeometric functions.
(**) There is an algorithm to determine when a hypergeometric function is an algebraic function.
Maybe these can be combined to get an example of an algebraic number with polynomial (and thus unbounded) denominators.
added:  Examples of these two parts...  
(*) The Perron method evaluates the (non-simple) continued fraction
$$
b_0+\frac{a_1}{b_1+\frac{a_2}{b_2}+\ddots}\qquad\text{with}\qquad
a_n=18n^2-33n, b_n=3n-2
$$
as
$$
\frac{-\frac{9}{2}}{{}_2F_1\left(-\frac{5}{6},1;\frac{1}{2};\frac{1}{3}\right)}
$$
Here, $a_n$ and $b_n$ are polynomials in $n$ which is what I meant by "polynomial entries" above.
(**) Unexpectedly,
$$
{}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right)
$$
is an algebraic function of $z$, see LINK
So, the (somewhat remote) hope would be that a case of continued fraction with $a_n=1$ evaluates to something in terms of hypergeometric functions on the list of known algebraic cases.