63rd Czech and Slovak Mathematical Olympiad

For each natural $k$, the equalities $l(2k)=l(k)$ and $l(2k-1)=2k-1$ are clearly valid. Thus we can add the values $l(n)$ over groups of numbers $n$ lying always between two consecutive powers of $2$. In this way we will prove by induction the formula $$s(n)=l(2^{n-1}+1)+l(2^{n-1}+2)+l(2^{n-1}+3)+\cdots +l(2^n),(1)$$ for $n=1,2,3,\dots$. The case $n=1$ is trivial. If $s(n)=4^{n-1}$ for some $n$, then $$\begin{array}{rl}s(n+1)& =l(2^n+1)+l(2^n+2)+l(2^n+3)+\cdots +l(2^{n+1})\\ & =[(2^n+1)+(2^n+3)+\cdots +(2^{n+1}-1)]+s(n)\\ & =\frac{2^n}{2}(2^n+1+2^{n+1}-1)+4^{n-1}=2^{n-2}\cdot 3\cdot 2^n+4^{n-1}=4^n.\end{array}$$

(We have used the fact the number of all odd numbers from $2^n+1$ to $2^{n+1}-1$ [including both limits] equals $2^{n-1}$.) The proof of (1) by induction is complete. Using the formula (1) we compute the requested sum as follows: $$\begin{array}{rl}l(1)+l(2)+l(3)+\cdots +l(2^{2013})& =l(1)+s(2)+s(3)+\cdots +s(2013)\\ & =1+1+4+4^2+4^3+\cdots +4^{2012}=1+\frac{4^{2013}-1}{3}=\frac{4^{2013}+2}{3}.\end{array}$$