You forgot a close-parenthesis in the $f(f(n-1)$ term. (I figure it's there since the other way of balancing them gives a false statement.) Anyways, this problem is silly. We show by strong induction that $f(n) - f(n-1) \in \{0,1\}$. The base cases 1 through $k$ are given to us. Suppose it is true for all values less than $n+1$. If $f(n) - f(n-1) = 0$, Then $$\begin{align*} f(n+1) - f(n) &= f(f(n)) + f(n+1-f(n)) - f(f(n-1)) - f(n-f(n-1)) \\ &= f(n-f(n)+1) - f(n-f(n)) \in \{0,1\}. \\ \end{align*}$$ Otherwise, if $f(n) - f(n-1) = 1$, $$\begin{align*} f(n+1) - f(n) &= f(f(n)) + f(n+1-f(n)) - f(f(n-1)) - f(n-f(n-1)) \\ &= f(f(n-1)+1) - f(f(n-1)) \in \{0,1\}. \\ \end{align*}$$ This completes the induction. All that is left is to show $f$ is unbounded. Suppose there exist $a,b$ with $f(a), f(a+1), f(a+2), \ldots$ all equal to $b$. Then putting $n = a+b$ into the given equation, we have $$f(a+b) = f(f(a+b-1)) + f(a+b-f(a+b-1)) = f(b) + f(a+b-b) = f(b) + b > b,$$ a contradiction.

MellowMelon wrote: You forgot a close-parenthesis in the $f(f(n-1)$ term. (I figure it's there since the other way of balancing them gives a false statement.) Thanks, fixed..