I see a pattern which I'm sharing just to make sure I'm understanding correctly. Starting with \\(2x\\) and repeatedly getting the right adjoint, I see:

\\(2x\\), \\(\lfloor\frac{n}{2}\rfloor\\), \\(2x+1\\), \\(\lfloor\frac{n-1}{2}\rfloor\\), \\(2x+2\\), \\(\lfloor\frac{n-2}{2}\rfloor\\), ...

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

\\(2x\\), \\(\lfloor\frac{n}{2}\rfloor\\), \\(2x+1\\), \\(\lfloor\frac{n-1}{2}\rfloor\\), \\(2x+2\\), \\(\lfloor\frac{n-2}{2}\rfloor\\), ...

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