I have found this just by chance, but can not understand why the limit turns out to be $n-2, \forall n\gt 2 $.


$a_0 = \lfloor \sqrt{n} \rfloor$

$a_1 = \lfloor \sqrt{n \cdot a_0} \rfloor$


$a_k = \lfloor \sqrt{n \cdot a_{k-1}} \rfloor$

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:

  1. Is the observation correct? Are there counterexamples?

  2. 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!


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 $n-2$

  • $\begingroup$ thank you for the explanation, I like the word you have wisely used: "stability". Just one point: I am specifying in the first paragraph $\forall n \gt 2$ so when I ask for a counterexample I meant for $n \gt 2$. $\endgroup$ – iadvd Sep 20 '16 at 3:21
  • 1
    $\begingroup$ I missed that about $n \gt 2$. I will fix it. $\endgroup$ – Ross Millikan Sep 20 '16 at 3:22

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