# 2013th powered sequence

Let $a_1$, $a_2$, ... be a sequence of integers defined recursively by $a_1=2013$ and for $n \ge 1$, $a_{n+1}$ is the sum of the 2013th power of the digits of $a_n$. Do there exist distinct positive integers $i$, $j$ such that $a_i=a_j$?

Yes because if $n$ is the number of digits, then $$9^{2013}+9^{2013}+\ldots+9^{2013}=n\cdot9^{2013}<10^n$$ for sufficiently large $n$, above that point the number of digits won't increase anymore.