2. The sequence $(\left.a_{n}\right)$ is defined by:

(i) $a_{1}=1$

(ii) $\quad(\forall n \in N) a_{n+1}=a_{n}+\frac{1}{\left[a_{n}\right]}$

For which values of $n$ does $a_{n}>20$ hold? ( $[x]$ is the greatest integer less than or equal to $x$.)
Solution. The first few terms of the sequence are:

$$
a_{1}=1, a_{2}=2, a_{3}=1+\frac{1}{2}, a_{4}=3, a_{5}=3+\frac{1}{3}, a_{6}=3+\frac{2}{3}, a_{7}=4, \ldots
$$

We observe the following rule: the sequence monotonically increases and its terms have the form $m+\frac{k}{m}, 0 \leq k \leq m-1$. Let's prove that this is indeed the case. For $n=1$, the statement is true. Suppose $a_{n}=m+\frac{k}{m}, 0 \leq k \leq m-1$.

Then

$$
a_{n+1}=a_{n}+\frac{1}{\left[a_{n}\right]}=m+\frac{k}{m}+\frac{1}{m}=m+\frac{k+1}{m}, \quad a_{n+1}=m+1 \text { if } k=m-1
$$

Thus, one term of the sequence has an integer part of 1, two terms have an integer part of 2, three terms have an integer part of 3, and so on. The integer part not greater than 19 has $1+2+3+\cdots+19=190$ terms of the sequence. Then $a_{191}=20$. Therefore, $a_{n}>20$ if $n>191$.