<s>[INST] Why $\forall n\gt2\in\Bbb N:\lfloor\sqrt{n\cdot\sqrt{n\cdot\lfloor\sqrt{...\lfloor\sqrt{n\cdot\lfloor\sqrt{n}\rfloor}}\rfloor\rfloor}}\rfloor=n-2$? I have found this just by chance, but can not understand why the limit turns out to be $n-2, \forall n\gt 2 $.

Def:
$a_0 = \lfloor \sqrt{n} \rfloor$
$a_1 = \lfloor \sqrt{n \cdot a_0} \rfloor$
...
$a_k = \lfloor \sqrt{n \cdot a_{k-1}} \rfloor$

And apparently:

$\forall n\gt 2:\ lim_{k\to \infty} a_k = \lfloor \sqrt{n \cdot \sqrt{ n \cdot \lfloor \sqrt{...\lfloor \sqrt{n\cdot \lfloor \sqrt{n} \rfloor }} \rfloor \rfloor}} \rfloor = n-2$

I understand that $(n-2) \cdot n = n^2-2n \lt (n-1)^2 = n^2+1-2n$ and for that reason if one of the elements of the sequence is $n-2$ then can not grow up more because $\lfloor \sqrt {n \cdot (n-2)} \rfloor = n-2$. But what I do not understand is why the sequence exactly arrives to $n-2$. Why not $n-1$ or $n-3$ for instance?
I would like to ask the following questions:


*

*Is the observation correct? Are there counterexamples?


*Why does exactly arrive to $n-2$ instead of any other value like, for instance, $n-3$? Probably the reason is quite trivial but I can not see the property behind the behavior. Thank you!

 [/INST] All $n\gt 2$ converge to $n-2$ as you surmise.  
The easiest way to see it is to look what happens if $a_n=n-3$.  Then $a_{n+1}=\lfloor \sqrt {n(n-3)} \rfloor = \lfloor \sqrt {n^2-3n} \rfloor $.  If $n \ge 4, n^2-3n \ge n^2-4n+4 = (n-2)^2$ so you will climb from $n-3$ to $n-2$ As you have pointed out, once you get to $n-2$ you have stability because $n^2-2n \lt n^2-2n+1=(n-1)^2$  This argument does not work for $n=3,$ but $\lfloor \sqrt 3 \rfloor =1$ and we start out at $n-2$
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