[tex]\it u(2^{4k})=u\Big((2^4)^k\Big)=u(16^k)=u(6^k)=6\\ \\\\ u(2^n)=\begin{cases}\it 6,\ dac\breve a\ n=4k\\ \\ \it 2,\ dac\breve a\ n=4k+1\\ \\ \it 4,\ dac\breve a\ n=4k+2\\ \\ \it 8,\ dac\breve a\ n=4k+3\end{cases}[/tex]
[tex]\it u(2^{2013})=u(2^{2012+1})=u(2^{4k+1})=2[/tex]
