, 2013

Starting with $2013$ contestants $1 + \frac{2012}{2} = 1007$ of them go to the second round and $1 + \frac{1006}{2} = 504$ make it to the third round. For the fourth and the fifth round we get $\frac{504}{2} = 252$ and $\frac{252}{2} = 126$ contestants, respectively. For the sixth and the seventh round we have $\frac{126}{2} = 63$ and $1 + \frac{62}{2} = 32$ contestants. For the eighth and the ninth round $\frac{32}{2} = 16$ and $\frac{16}{2} = 8$ contestants remain and for the tenth round there are $\frac{8}{2} = 4$. Finally, in the eleventh round only $2$ contestants remain and compete for the first place. So, $f(2013) = 11$.

If the number of contestants is less than or equal to $1024 = 2^{10}$ then the tournament will consist of at most $10$ rounds. Indeed, after the $j$th round there will be at most $\frac{2^{10}}{2^j} = 2^{10-j}$ contestants left. So, after $10$ rounds we can have at most $2^{10-10} = 1$ and the tournament ends. Let us show that $f(1025) = 11$. Since $1025 = 1 + 2^{10}$ there are $1+2^9$ contestants in the second round, $1+2^8$ in the third, $1+2^7$ in the fourth and so on, all the way to the tenth round, where there are $1+2^1 = 3$ contestants. The eleventh round will see two of these three contestants competing for first place. So, the smallest natural number such that $f(n) = f(2013)$ is $1025$.