How do I find an upper bound on this recurrence

Question

$f(n)=f(n-\sqrt{n})$

I believe $f(n)\in O(\sqrt{n})$

However I cannot seem to prove it, my intuition comes from the fact that we can remove $\sqrt{n}$ exactly $\sqrt{n}$ times, but if $n$ shrinks then does anything change?

Answer 1

For definiteness, let's assume that the recurrence is $$ f(n) = \begin{cases} f(\lfloor n-\sqrt{n} \rfloor) + 1 & n > 0 \\ 0 & n = 0. \end{cases} $$ (We assume that the domain is the non-negative integers.)

Since $\lfloor n-\sqrt{n} \rfloor > n-\sqrt{n}-1$, we see that $$ f(n) \geq \frac{n}{\sqrt{n}+1} = \Omega(\sqrt{n}). $$ In the other direction, let us note that $f(n)$ is monotone and that $$ f(n) \leq f(\lfloor n/2 \rfloor) + \frac{\lceil n/2 \rceil}{\sqrt{\lceil n/2 \rceil}}. $$ This is because until the parameter gets down to $\lfloor n/2 \rfloor$ (or smaller), the decrease at every step is at least $\sqrt{\lceil n/2 \rceil}$. This shows that $$ f(n) \leq f(\lfloor n/2 \rfloor) + \sqrt{\lceil n/2 \rceil} \leq f(\lfloor n/2 \rfloor) + \sqrt{n/2} + 1. $$ Iterating this $\log_2 n+1$ times, we get $$ \begin{align*} f(n) &\leq f(\lfloor n/2 \rfloor) + \sqrt{n/2} + 1 \\ &\leq f(\lfloor n/4 \rfloor) + \sqrt{n/2} + \sqrt{n/4} + 2 \\ &\leq \cdots \\ &\leq f(0) + \sqrt{n/2} + \sqrt{n/4} + \cdots + \sqrt{n/2^{\log_2 n + 1}} + \log_2n + 1 \\ &\leq \sqrt{n} \sum_{t=1}^\infty \frac{1}{\sqrt{2}^t} + O(\log n) \\ &= O(\sqrt{n}). \end{align*} $$ We conclude that $f(n) = \Theta(\sqrt{n})$.