The initial condition it is not ${{a}_{1}}=\sqrt{2020}$?

no, according to my source

$a_{n}>\sqrt{2020}>44$ for every $n$ So $a_{n+1} > \sqrt{2020+44}>45$ $a_2=\sqrt{4040}<64$ $a_3<\sqrt{2020+64}<\sqrt{2116}=46$ If $a_n<46$ then $a_{n+1}< \sqrt{2020+46}<46$ So for $n\geq 3$ we have $45<a_n<46$ and so $\left\lfloor a_{2020}\right\rfloor= 45$