A375416 Number of order n magic triangles composed of the numbers from 1 to n^2 in which the sum of each 2 X 2 subtriangle is the same, counted up to rotations and reflections.

1, 4, 144, 38336, 539904, 87249024

Offset: 1

Author

Donghwi Park, Aug 15 2024

Keywords

• nonn
• more
• hard

Comments

An order n triangle contains binomial(n,2) upright 2 X 2 subtriangles and binomial(n-2,2) inverted 2 X 2 subtriangles. In total, there are n^2-3*n+3 subtriangles.

It seems that the sequence is likely finite. Considering each of the n^2! possibilities of arranging 1..n^2, for each of the (n^2-3n+3) subtriangles only one choice for the central value can give the magic sum. We should, therefore, divide (n^2)! by (n^2)^(n^2-3*n+3) to calculate an estimation of a(n). For n >= 16, (n^2)!/(n^2)^(n^2-3*n+3) < 1.

For n >= 3, a(n) is a multiple of 8, because swapping between a corner triangle and an edge-adjacent triangle generate different examples,

Disregarding corner swap, a(3) to a(6) would be "18, 4792, 67488, 10906128"

Examples

			a(1)=1 because there is only the trivial case without any subtriangle.
a(2)=4 because we can choose only the number in the central triangle.
a(3)=18, which is same for A342467(4)*8. Trotter's order 4 magic triangle can be transformed to this order-3 magic triangle disregarding corner swap.
For n = 3, numbers 1..9 are placed inside the triangles shown:
        o
       / \
      o-- o
     / \ / \
    o---o---o
   / \ / \ / \
  o---o---o---o
An example with magic sum=17:
        9
        5
      1   2
      6   4
   7    3    8
This corresponds to the magic perimeter triangle (A342467):
     1 9 5 2
      7   4
       6 8
        3
		

Links

• Donghwi Park, Source code for a(4).
• Donghwi Park, Source code for a(5).

Crossrefs

Cf. A006052, A342467, A355256.