Membership in 1, 5, 2, 13, 10, ... (recursively defined sequence)
Your sequence contains all positive integers not divisible by 3. Let us prove this by induction.
Suppose first that $x$ is divisible by 3, and $f(n) = x$. Considering the definition, we see that $x = 2f(n/2) + 3$, and so $f(n/2) = \tfrac{x-3}{2}$, which is also divisible by 3. By induction, this is impossible.
Suppose next that $x&gt;0$ is not divisible by 3. If $x=1$ then $x$ appears in the sequence since $f(1) = 1$, so we can assume that $x \geq 2$. We now consider two cases: $x$ even and $x$ odd.
If $x$ is even then $\tfrac{x}{2}&lt;x$ is positive not divisible by 3, and so by induction, $\tfrac{x}{2} = f(m)$ for some $m$. Then $f(2m) = x$.
Similarly, if $x$ is odd then $\tfrac{x-3}{2} &lt; x$ is not divisible by 3. Furthermore, since $x \geq 5$, it is furthermore positive. By induction, $\tfrac{x-3}{2} = f(m)$ for some $m$. Then $f(2m+1) = x$.