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

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

Here's one for you:

$\begin{align} K &= \sum^\infty_{n=0}10^{-2^{n}} \\ &= 10^{-1}+10^{-2}+10^{-4}+10^{-8}+10^{-16}+10^{-32}+10^{-64}+\ldots \\ &= 0.\mathbf{1}\mathbf{1}0\mathbf{1}000\mathbf{1}0000000\mathbf{1}000000000000000\mathbf{1}0000000000000000000000000000000\mathbf{1}\ldots \end{align}$

a constant with 1's in positions corresponding to an integer power of two and zeros everywhere else.

K has a canonical continued fraction expansion of:

$\left[0; 9, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 8, 10, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, \ldots\right]$

After calculating the first 1000000 terms on Wolfram Cloud, I'm fairly certain that (except for the first term which is 0 and the second term which is 9) all of the terms are 8, 10, or 12. (Maybe someone can prove this)

Looking at the terms themselves, the position numbers of the 12's seem to all be congruent to 2 or 7 mod 8, and even after 10000 terms there seems to be nothing special as to their ordering. And the positions of the eights (5, 9, 12, 17, 21, 24, ...) are all congruent to 1 or 0 mod 4. But it seems that there is a particular order as to which of the positions are which. I was also able to use Wolfram Alpha to find a function that was able to correctly evaluate the positions of all the 8's for the first 10000 terms. And after unsuccessfully trying to find a formula for the 10's, here is what the structure of the continued fraction appears to look like:

$K=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{a_4+\frac{!}{a_5+\ldots}}}}}$

where

$\forall~n\in\mathbb{Z}_{\geqslant 0},~a_n=\begin{cases} 0 & n=0 \\ 8 & n\in\left\{\frac{8m+\left(\frac{-1}{m-1}\right)+1}{2}~:~m\in\mathbb{Z}^{+}\right\} \\ 9 & n=1 \\ 10 & \text{otherwise} \\ 12 & n\equiv 2\left(\operatorname{mod}8\right)\text{or}~7\left(\operatorname{mod}8\right) \end{cases}$

where $\left(\frac{n}{m}\right)$ is the Jacobi symbol.

So there we have it. A transcendental number whose continued fraction has bounded terms.

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!