Is the time complexity of the Fibonacci sequence O(fib(n))?
$\DeclareMathOperator{\fib}{fib}$If you choose to calculate $\fib(n)$ by using the naive recursion
formula, then you need $\Theta (\fib(n))$ additions, and since the
numbers involved grow to $c\cdot n$ bits for a not very large constant
$c$, it will take $\Theta(n \cdot \fib(n))$ operations.  (I may be wrong
here because most of the numbers involved may be much smaller than $c
\cdot n$ bits).
If you are reasonably clever and calculate
$$\fib(0), \fib(1), \fib(2), \dots, \fib(n),$$
then you have $n$ additions of $c \cdot n$ bit numbers, which takes
$\Theta(n^2))$ operations.
You can use matrix exponentiation to calculate $\fib(n)$ faster for
large $n$, I'd estimate something like $M(n)$ where $M(n)$ is the time
for an $n$-bit multiplication.
Most important is that for maybe $n = 100$ or $n = 200$, your computer
has not a chance to calculate $\fib(n)$ using the naive recursive
algorithm within your lifetime, while the simple method of calculating
consecutive values will find the same result within microseconds on a
newish smartphone.