The reader may be interested to note that there is a closely related
recurrence that we can solve exactly and not just for powers of
two.
Suppose we put
$$f(n) = 3 f(\lfloor n/2 \rfloor) - 2 f(\lfloor n/4 \rfloor).$$
This requires a value for $f(0)$ so we set
$$f(n) = 2n+1 \quad\text{when} \quad n<3.$$
Now observe that the generating function $$g(z) =
\frac{1}{1-(3z-2z^2)}$$ encodes the tree of values visited during the
computation of $f(n)$ in a natural way.
Let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$ be the
binary representation of $n$. We now compute a closed form expression
for $f(n)$ when $n\ge 2.$
Case A. The leading digits are two one digits, i.e.
$(d_{\lfloor \log_2 n \rfloor} d_{\lfloor \log_2 n \rfloor-1})_2 =
(11)_2.$ Then the two terminal values for the recursion are $n=3$ and
$n=1$.
The case $n=3$ has $f(n)=7$ and thus gives the contribution
$$7\times
[z^{\lfloor \log_2 n \rfloor - 1}] \frac{1}{1-(3z-2z^2)}.$$
The case $n=1$ has $f(n)=3$ but the last step must have been on the
$\lfloor n/4 \rfloor$ branch so as not to be routed through $n=3$,
giving
$$3\times (-2)\times
[z^{\lfloor \log_2 n \rfloor - 2}] \frac{1}{1-(3z-2z^2)}.$$
Case B. The leading digits are a one digit followed by a zero
digit i.e. $(d_{\lfloor \log_2 n \rfloor} d_{\lfloor \log_2 n
\rfloor-1})_2 = (10)_2.$ Then the terminal values for the recursion
are $n=2$ and $n=1$.
The case $n=2$ has $f(n)=5$ and thus gives the contribution
$$5\times
[z^{\lfloor \log_2 n \rfloor - 1}] \frac{1}{1-(3z-2z^2)}.$$
The case $n=1$ has $f(n)=3$ but the last step must have been on the
$\lfloor n/4 \rfloor$ branch so as not to be routed through $n=2$,
giving
$$3\times (-2)\times
[z^{\lfloor \log_2 n \rfloor - 2}] \frac{1}{1-(3z-2z^2)}.$$
Evaluation. Note that by partial fractions we have that
$$[z^q] \frac{1}{1-(3z-2z^2)} = 2^{q+1}-1$$
so that we get for case A
$$ 7 \times 2^{\lfloor \log_2 n \rfloor} - 7
- 6 \times 2^{\lfloor \log_2 n \rfloor-1} + 6
= (14-6) \times 2^{\lfloor \log_2 n \rfloor-1} - 1
= 2^{\lfloor \log_2 n \rfloor+2} - 1$$
and for case B
$$ 5 \times 2^{\lfloor \log_2 n \rfloor} - 5
- 6 \times 2^{\lfloor \log_2 n \rfloor-1} + 6
= (10-6) \times 2^{\lfloor \log_2 n \rfloor-1} + 1
= 2^{\lfloor \log_2 n \rfloor+1} + 1 $$
This produces the sequence for $n\ge 2$
$$5, 7, 9, 9, 15, 15, 17, 17, 17, 17, 31, 31, 31, 31, 33,\ldots$$
which perfectly matches $f(n).$
Observe that we may say that $f(n)\in\Theta(n)$ in certain sense
($2^{\lfloor \log_2 n \rfloor}\in\Theta(n).$)
This MSE link shows a more sophisticated application of the trick with the generating function.
