[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:

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

> 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:

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