Any idea about Images of a Function?

The first step is to show that $f(2n+1) = 3f(n)+1$ and $f(2n) = 3f(n)$ . Certainly any function of this form obeys the conditions.

We get $f(1)^2 = f(0)^2+6f(0)+1 = (f(0)+3)^2-8$ , and the only positive integer solution to $x^2-y^2 = 8$ is $x=3, y=1$ (factor), so $f(0) = 0$ and $f(1) = 1$ .

Now note that $(f(2n+1)-f(2n))(f(2n+1)+f(2n)) = 6f(n)+1$ and the two factors differ by at least $2f(2n) \ge 2f(n)$ . So the second factor is at least $2f(n)+1$ , and so the first factor is at most $\frac{6f(n)+1}{2f(n)+1} < 3$ .

But the first factor can't be $2$ , so it must be $1$ , and the second factor is $6f(n)+1$ . So we get $f(2n+1) = 3f(n)+1$ and $f(2n) = 3f(n)$ . Note that together with the values of $f(0)$ and $f(1)$ , this uniquely determines $f$ .

One way to describe $f(n)$ is that it equals the binary expansion of $n$ translated into ternary. So e.g. $f(5) = 10$ because $5$ is $101$ in binary, and $101$ in ternary is $10$ . It's pretty clear that this function satisfies the conditions in the first step, so it equals $f$ .

So the answer is the number of numbers less than 2015 whose ternary expansion is all $1$ s and $0$ s. This is just every number with such a ternary expansion that has up to $7$ digits. So there are $2^7 = \fbox{128}$ such numbers.