smc

Define $f(2)=\log (2)$, and $f(n)=\log (n+f(n-1)),\text{ for }n>2.$ We are looking for $f(2013)$. First we show Lemma. For any integer $n>2$, if $n<10^k-k$ then $f(n)<k$. Proof. First note that $f(2)<1$. Let $n<10^k-k$. Then $n+k<10^k$, so $\log (n+k)<k$. Suppose the claim is true for $n-1$. Then $f(n)=\log (n+f(n-1))<\log (n+k)<k$. The Lemma is thus proved by induction. Finally, note that $2012<10^4-4$ so that the Lemma implies that $f(2012)<4$. This means that $f(2013)=\log (2013+f(2012))<\log (2017)$, which leaves us with only one option $\boxed{(\log 2016,\log 2017)}$.