Elegant solution to the Magic Hexagon problem

Question

The Magic Hexagon Problem

A magic hexagon of order $n$ is an arrangement of close-packed hexagons containing the numbers $1, 2, ..., H_{n-1}$, where $H_n$ is the $n^{th}$ hex number such that the numbers along each straight line add up to the same sum. (Here, the hex numbers are i.e., $1, 7, 19, 37, 61, 91, 127$, ...; Sloane's A003215. In the above magic hexagon of order $n=3$, each line (those of lengths $3$, $4$, and $5$) adds up to $38$.

From Wolfram's Magic Hexagon.

Here is the programming puzzles entry to the Magic Hexagon problem. As you can see, the solutions there use appropriate systems of linear equations, and then try all the possibilities until it finds one. Boring, but effective; and can't be done by a human in a reasonable amount of time.

This is de la Loubere's method for magic squares. Much cleaner and more beautiful, and a human can do this process easily.

I want to know, is there such a solution / methodology to the Magic Hexagon problem that has the elegance of de la Loubere's method? How would you begin the process of coming up with one?

Answer 1

According to the article, the Magic Hexagon is the only one that exists (aside from n=1). There wouldn't be a simpler method for generating one, because no other hexagons of order 2 or >3 would have a solution. Quoting from the last paragraph:

Trigg showed that the magic constant for an order n hexagon would be (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)), the first few of which are 1, 28/3, 38, 703/7, 1891/9, 4186/11, ... (Sloane's A097361 and A097362), which requires 5/(2n-1) to be an integer for a solution to exist. But this is an integer for only n=1 (the trivial case of a single hexagon) and Adams's n=3 (Gardner 1984, p. 24).

If I had to guess, I might claim the n=3 magic hexagon working at all is just a coincidence.