Do we know a transcendental number with a proven bounded continued fraction expansion?

Question

The simple continued-fraction-expansion for the transcendental number $e$ is known to be unbounded. What about bounded continued fractions ?

Do we know any transcendental number for which it is proven that the simple continued-fraction-expansion is bounded ?

It is conjectured that the simple continued-fraction-expansion of the algebraic numbers with minimal polynomial degree greater than $2$ are unbounded.

If this would be true, every bounded non-periodic infinite simple continued-fraction-expansion would correspond with a transcendental number.

But to my knowledge, it was not proven for a single algebraic number with minimal polynomial degree greater than $2$, that its simple continued-fraction-expansion is unbounded.

Answer 1

Yes, but the transcendentals that this answer describes are quite unnatural.

Fix any noncomputable bounded sequence of positive integers $\alpha$, and let $r_\alpha$ be the real number whose continued fraction expansion is given by $\alpha$. Then - since the continued fraction expansion of a computable real is computable, and every algebraic real is computable - $r_\alpha$ is transcendental. Note that the nontrivial part here is proving transcendentality!

Answer 2

These are fairly easy to construct a non-quadratic though. Take the Thue-Morse Sequence: 0,1,1,0,1,0,0,1,.... The sequence has several definitions including being the parity of the integers in base two. It's clearly aperiodic. Next add 1 to each element 1,2,2,1,2,1,1,2,2,1,1,2,1,2,2,1.... This sequence is bounded by 2; it's aperiodic by construction; thus it can be used for the partial quotients of a continued fraction. The fraction isn't finite nor is it periodic; thus it represents a non-quadratic irrational. (Not necessarily transcendental.) Thanks to Qiaochu Yuan for the corrections. (If this answer needs deletion, OK. It does give a non-quadratic construction with bounded partial quotients.)