Is it possible to write every integer with only one $3$?

There's a cool math challenge about writing every number from 0 to 100 with exactly three $3$'s.

Most of the solutions use a clever way to write $50$ with only one $3$ :

$$\left \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3!!!}}}}}}}}}}\right \rfloor = 50$$

I wonder if it's possible to write every natural number with a similar way ?

Edit : To keep the spirit of the puzzle, I think only factorials, square roots and floor functions are allowed.

• I'm not sure what is says about humanity that we have time for pursuits like this. I guess a testament to how easy life has become. – Gregory Grant Nov 28 '17 at 20:26
• You'll need to define what "base operations" are ok. Otherwise, you could just let $S:\mathbb{N}\to\mathbb{N}$ be the successor function, and write $n$ as $S^{n-3}(3)$ where $S^n(k)$ is the $n$th iterate of $S$. If you don't like the $n-3$ in the "exponent", you could similarly write $S(S^{n-4}(3))$, and you've done it via a single 3. – Mark Nov 28 '17 at 20:28
• @Mark You're right, I should clarify. To keep the spirit of the puzzle, I think only factorials, square roots and floor functions should be allowed. – Pyrofoux Nov 28 '17 at 20:42
• Probably. See this related question (with 4 instead of 3): math.stackexchange.com/questions/48633/… – Michael Lugo Nov 28 '17 at 21:32