Post History

## Rust, 35 bytes

```rust
|n:f64|(12.*n-3.).sqrt().fract()>0.
```

Counterintuitively, this outputs `false` for a hex number, and `true` otherwise, but this is consistent with the output requirement:

> - One of 2 distinct values, indicating whether the input is a hex number.

This allows using `>0` rather than `==0` to save 1 byte.

[All test cases on Rust Playground](https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=95e6fbc442632795c89a73ed31812ef0)

## Breakdown of the code
```rust
|n:f64|(12.*n-3.).sqrt().fract()>0.
|n    |                             // Anonymous function with input n
  :f64                              // Input is 64 bit floating point
       (12.*n                       // Multiply input by 12
             -3.)                   // Subtract 3 ("." denotes float)
                 .sqrt()            // Calculate square root
                        .fract()>0. // Is fractional part non-zero?
                                    // Implicitly return true or false
```

## Explanation
From [Wikipedia](https://en.wikipedia.org/wiki/Centered_hexagonal_number#Formula), the following formula gives the index $n$ of a hex number $H$ (it gives $1$ for the first hex number, $2$ for the second, and so on):

$$n=\frac{3+\sqrt{12H-3}}{6}$$

This $n$ is an integer if and only if $H$ is a hex number. Rather than calculate the whole formula, the code only calculates the square root:

$$R=\sqrt{12H-3}$$

because $n$ is an integer if and only if $R$ is an integer.

## Reasoning
Reasons for asserting that $n$ is an integer if and only if $R$ is an integer. Since all the test cases pass, this section is just out of interest, to show that this also holds outside the test cases.

### When $R$ is not an integer, $n$ is not an integer
If the square root $R$ is not an integer, then $n=\frac{3+R}{6}$ cannot be an integer.

### When $R$ is an integer, $n$ is an integer
$n=\frac{3+R}{6}$ is an integer if and only if $R$ is an odd multiple of $3$.

Need to show that $R\in \Bbb Z\implies R\equiv 3\pmod 6$ (that is, whenever $R$ is an integer, it is also an odd multiple of $3$).

$$R\in\Bbb Z\implies R=6x+c, x\in\Bbb Z,c\in\{0,1,2,3,4,5\}$$

Need to show that $R\in\Bbb Z\implies c=3$

$$\begin{align}R&=\sqrt{12H-3}, H\in\Bbb Z\\
R^2&=12H-3\\
H&=\frac{R^2+3}{12}\\
H&=\frac{(6x+c)^2+3}{12}\\
H&=\frac{36x^2+12xc+c^2+3}{12}\\
H&=3x^2+xc+\frac{c^2+3}{12}\\
H-3x^2-xc&=\frac{c^2+3}{12}\\
\end{align}$$
$$H-3x^2-xc\in\Bbb Z\therefore \frac{c^2+3}{12}\in\Bbb Z$$
$$c\in\{0,1,2,3,4,5\}, \frac{c^2+3}{12}\in\Bbb Z\therefore c=3$$
$$\therefore R\in\Bbb Z\implies R\equiv 3\pmod 6\implies n\in\Bbb Z$$