SAT1000 - P893

We are given that $a_n = n^2$ .

Let $c_n = \{(a_n)^{*}\}$ be a sequence. Then,

$c_n =$ No. of $m$ 's such that $a_m < n$ .

So solving for given $n$ ,

$\displaystyle \quad a_m < n\newline \Rightarrow m^2 < n\newline \Rightarrow m < \sqrt{n}\newline \Rightarrow 1 \leq m \leq \lceil \sqrt{n}\rceil - 1$ where $\lceil . \rceil$ is ceiling function.

So, no. of $m$ 's for given $n$ is $\displaystyle \lceil \sqrt{n}\rceil - 1$

Hence we get,

$\displaystyle c_n = \lceil \sqrt{n}\rceil - 1$

Now we are given that $\displaystyle b_n = \{((a_n)^{*})^{*}\}$ . So $b_n = \{(c_n)^{*}\}$ .

Solving similarly for no. of $m$ 's such that $c_m < n$ for given n,

$\displaystyle \quad c_m < n\newline \Rightarrow \lceil \sqrt{m}\rceil - 1 < n\newline \Rightarrow \lceil \sqrt{m}\rceil < n + 1\newline \Rightarrow \lceil \sqrt{m}\rceil \leq n \newline \Rightarrow \sqrt{m} \leq n \newline \Rightarrow 1 \leq m \leq n^2$ .

So, no. of $m$ 's for given $n$ is $\displaystyle n^2$

Hence we get,

$\displaystyle b_n = n^2$

So, $\displaystyle b_{2020} = 2020^2 = \boxed{4080400}$

It's pretty obvious that $\{(a_n)^*\}$ is a monotonous sequence because if $a_m < n$ then $a_m < n+1$ . Moreover, $\{(a_n)^*\}$ rises either by 1 or 0 since $\{a_n = n^2\}$ is strictly monotonous. That means that $b_n$ is one less than the index of the first $(a_n)^* \geq 2020$ . So we need a number that is bigger than the first 2020 $a_n$ which is obviously is $2020^2+1$ . Then $b_n = 2020^2 = \boxed{4080400}$