If I have the following iterated function: $$ L_{i+1} = \left\lceil \frac{N}{\left\lceil \frac{N}{L_{i}} \right\rceil-1} \right\rceil, L_{0} = 1 $$
I am trying to find $k$ such that $L_{k} = N$. Is there a way to derive a closed form expression for $k$ as a function of $N$ or an upper limit on $k$ which is tighter than $N$?
I tried to get an estimate of how $k$ increases as $N$ grows exponentially from $10^1 .. 10^q$ and it seems to have an order of $C(q)^{\log{q}}$ with $C(q)$ decreasing in $q$, but I could not find a way to prove it by using the ceiling inequality $\lceil x \rceil = y \rightarrow y-1 < x \leq y$. I am not very knowledgable with discrete maths, so I am asking if you have some ideas that can help with this nested ceiling expression.
Edit: We know now that $k = \lceil 2\sqrt{N} - 1 \rceil$, so the question is how to get there from the iterated function?
$k$ will be close to $2\sqrt N$. When $L_i$ is less than $\sqrt N$, we have $L_{i+1}=L_i+1$. When $L_i$ is well greater than $\sqrt N, \frac N{L_i}$ decreases by $1$ each time. Each part takes about $\sqrt N$ steps. The ceilings make it hard to do a solid analysis.