fun question

**snowtea** · January 7th 2011, 08:41 PM

All hexagons can be made by cutting away 3 equilateral triangles from a larger equilateral triangle.

All possible n is of the form:
$n = k^2 - k_1^2 - k_2^2 - k_3^2$
where $k_i + k_j + 1 \leq k$ for $i\neq j$

e.g. $6 = 3^2 - 1^2 - 1^2 - 1^2$
$10 = 4^2 - 2^2 - 1^2 - 1^2$
$13 = 4^2 - 1^2 - 1^2 - 1^2$

The next few would be
$5^2 - 2^2 - 2^2 - 2^2 = 13$
$5^2 - 3^2 - 1^2 - 1^2 = 14$
$5^2 - 2^2 - 2^2 - 1^2 = 16$
$5^2 - 2^2 - 1^2 - 1^2 = 19$
$5^2 - 1^2 - 1^2 - 1^2 = 22$
...

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