Does this recursive sequence have an explicit formula?

Question

My recursive sequence is defined as $f\left(2n\right)=\frac{f\left(2n-1\right)}{2}$, $f\left(2n+1\right)=f\left(2n\right)+\frac{1}{2}$, and $f\left(1\right)=0$. I see that the $\liminf=\frac{1}{2}$ and the $\limsup=1$, but I do not know if that helps me find an explicit formula for the sequence.

Answer 1

$$f(n) = \begin{cases} n \text{ even } ~&:~ \frac 12 f(n-1) \\ n \text{ odd } ~&:~ f(n - 1) + \frac 12 \end{cases}$$

So for odd $n$,

$$f(n) = \frac 12 f(n - 2) + \frac 12$$

Which is just an affine equation, so yes it has a simple solution. If you want to convert it to a linear equation, then

$$f(n + 2) = \frac 12 f(n) + \frac 12$$

And subtract the last 2 equations to get:

$$f(n + 2) - \frac 32 f(n) + \frac 12 f(n - 2) = 0$$