Is there a general way to compute a continued fraction, knowing the pattern?

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Finding the value of some continued fractions is easy. For instance, $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\ddots}}}}}$.

To evaluate that, we can let the expression be $x$, so what we have actually is $1+\frac{1}{x}$. Solving $x=1+\frac{1}{x}$, we get $x=\varphi$ (golden ratio).

Another example, $1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{\ddots}}}}}$ can be easily (with the same way above) found, that is $\sqrt{2}$.

I am not asking to evaluated the following expressions. But I do not know an easy way to compute them, and I provided them as examples only. (So this post does not have many problems to solve).

$1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{\ddots}}}}}$ (natural numbers)

$1+\frac{1}{2+\frac{1}{4+\frac{1}{8+\frac{1}{16+\frac{1}{\ddots}}}}}$ (powers of $2$)

$\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{7+\frac{1}{11+\frac{1}{13+\frac{1}{\ddots}}}}}}}$ (prime numbers)

$\frac{1}{1!+\frac{1}{2!+\frac{1}{3!+\frac{1}{4!+\frac{1}{\ddots}}}}}$ (factorials)

$\frac{1}{2+\frac{3}{4+\frac{5}{6+\frac{7}{8+\frac{9}{\ddots}}}}}$ (odd with even naturals)

$\frac{1}{1+\frac{1}{10+\frac{1}{11+\frac{1}{100+\frac{1}{101+\frac{1}{110+\frac{1}{\ddots}}}}}}} \approx 0.9098\dots$ (binary numbers but in base $10$)

It would be appreciated if you help me with a good way (if there is). THANKS.