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
$$a_1 = 1, \quad a_m = a_1 + a_2 + \cdots + a_k,$$
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
$$\begin{align*} a_{2012} &= a_1 + a_2 + \cdots + a_{44} \\ &= (a_1 + \cdots + a_4) + (a_5 + \cdots + a_9) + (a_{10} + \cdots + a_{16}) \\ &\quad + (a_{17} + \cdots + a_{25}) + (a_{26} + \cdots + a_{36}) + (a_{37} + \cdots + a_{44}) \\ &= 4 \times 1 + 5 \times 2 + 7 \times 3 + 9 \times 4 + 11 \times 6 + 8 \times 8 \\ &= 4 + 10 + 21 + 36 + 66 + 64 = 201. \end{align*}$$

So, $201$ is the desired answer.