Finding recursive function Range A function $f: \Bbb{N^+} \rightarrow \Bbb{N^+}$ , defined on the set of positive integers $\Bbb{N^+}$,satisfies the following properties:
$$f(n)=\begin{cases} f(n/2) & \text{if } n \text{ is even}\\
f(n+5) & \text{if } n \text{ is odd} \end{cases}$$
Let $R=\{ i \mid \exists{j} : f(j)=i \}$ be the set of distinct values that $f$ takes. The maximum possible size of $R$ is ___________.
Answer of this question is $2$, and solution goes like this:-
every multiple of $5$ has same value, and every other number has same value.
I want to proof it, by NOT using examples, but some real mathematical proof, that can show us that indeed this is true.
Thanks.

Suppose that $f(1) = a$ and $f(5) = b$. It is clear that $$f(5n) = b$$ for all $n$. We'll prove by induction that for all $n \ne 5k$, $f(n) = a$. First note that
$$f(2) = f(\frac{2}{2}) = f(1) = a,$$
$$f(3) = f(3+5) = f(8) = f(4) = f(2) = a,$$
$$f(4) = f(2) = a.$$
Now suppose $n = 5k + r$, where $0 < r < 5$, and for all $m<n$ which are not divisible by $5$, $f(m) = a$.
If $n$ is odd, $f(n) = f(n-5)$, and by induction hypothesis, $f(n-5) = a$, so we get $$f(n) = a.$$ 
If $n$ is even,  $f(n) = f(n/2)$, and by induction hypothesis, $f(n/2) = a$, so we get $$f(n) = a.$$
Note that here $\frac{n}{2}$ isn't divisible by $5$.
