[Alex Varga wrote:](https://forum.azimuthproject.org/discussion/comment/16316/#Comment_16316)

> So Puzzle 12 would be \$$\lfloor\frac{n}{2}\rfloor\$$ and Puzzle 13 would be \$$\lfloor\frac{n+1}{2}\rfloor\$$. Am I crazy?

For Puzzle 13 you seem to be saying that for natural numbers \$$m\$$ and \$$n\$$,

$$\lfloor \frac{m+1}{2}\rfloor \le n \textrm{ if and only if } m \le 2n .$$

Is that what you're saying? It seems that [Patrick O'Neill](https://forum.azimuthproject.org/profile/1926/Patrick%20O%27Neill) is claiming this solution to Puzzle 13:

$$\lfloor \frac{m}{2}\rfloor \le n \textrm{ if and only if } m \le 2n .$$