Japan Mathematical Olympiad

Let $m$ be a positive integer and denote by $a_m$ the number of possible ways of writing down a sequence of positive integers satisfying the following, which we call the conditions $C_m$: $C_m$: Start with writing down the number $m$, and end up with writing down the number $1$, and after writing down a number $n$, follow with writing a positive integer less than or equal to $\sqrt{n}$. Then, we have $$a_1=1,a_m=a_1+a_2+\cdots +a_k,$$ where $k$ is a positive integer satisfying $k^2+1\le m\le (k+1{)}^2$.

$$a_1=1,$$ $$a_2=a_3=a_4=a_1=1,$$ $$a_5=a_6=\cdots =a_9=a_1+a_2=2,$$ $$a_{10}=a_{11}=\cdots =a_{16}=a_1+a_2+a_3=3,$$ $$a_{17}=a_{18}=\cdots =a_{25}=a_1+a_2+a_3+a_4=4,$$ $$a_{26}=a_{27}=\cdots =a_{36}=a_1+a_2+a_3+a_4+a_5=6$$ $$a_{37}=a_{38}=\cdots =a_{44}=a_1+a_2+a_3+a_4+a_5+a_6=8.$$ From these we finally obtain $$\begin{array}{rl}{a}_{2012}& =a_1+a_2+\cdots +a_{44}\\ & =(a_1+\cdots +a_4)+(a_5+\cdots +a_9)+(a_{10}+\cdots +a_{16})\\ & +(a_{17}+\cdots +a_{25})+(a_{26}+\cdots +a_{36})+(a_{37}+\cdots +a_{44})\\ & =4\times 1+5\times 2+7\times 3+9\times 4+11\times 6+8\times 8\\ & =4+10+21+36+66+64=201.\end{array}$$

So, $201$ is the desired answer.