Suppose $a_1=1$ and $a_n=\sqrt{n-a_{n-1}}$ $\forall n \in\mathbb{Z_{\ge 2}}$. What is the least integer $n$ for which $a_n>4$? What is $\left \lfloor a_{100} \right \rfloor$, where $\left \lfloor u \right \rfloor$ is the greatest integer that is less than or equal to $u$?
As an example, the $n^\text{th}$ Fibonacci number can be found without finding the preceding $(n-1)$ numbers.
I am not sure if we can solve my problem efficiently, easily, and without a calculator.
Using a computer calculator, I found that
$\sqrt{19-\sqrt{18-\sqrt{17-\sqrt{\cdots-1}}}}<4$ and $\sqrt{20-\sqrt{19-\sqrt{18-\sqrt{17-\sqrt{\cdots-1}}}}}>4$
If we can not solve this easily without a calculator, then can we just get an upper limit and a lower limit "good guesses"?
Here, "good guesses" like when we try to find $\sqrt[3]{50}$, we know that $3^3=27$ and $4^3=64$, therefore, $\sqrt[3]{50}$ is a value between $3$ and $4$. So, to get a good approximation, we do not need to look at any value $\notin (3,4)$.
I am interested in finding an explicit method rather than guessing.
Your help would be appreciated. THANKS!
Your second question
Your first question
$\forall n \geqslant 2 \ , \ a_n<\sqrt{n}$
$\forall n \geqslant 3 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1}}$
$\forall n \geqslant 4 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1-\sqrt{n-2}}}$
$\forall n \geqslant 5 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1-\sqrt{n-2-\sqrt{n-3}}}}$
We can show that $(a_n)_{n\geqslant 1}$ is increasing.
$a_{19} < \sqrt{19-\sqrt{18-\sqrt{17}}} < 4$
$a_{20} > \sqrt{20-\sqrt{19-\sqrt{18-\sqrt{17}}}} >4$