%I #41 Sep 06 2024 17:47:10
%S 1,4,144,38336,539904,87249024
%N 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.
%C 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.
%C 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.
%C For n >= 3, a(n) is a multiple of 8, because swapping between a corner triangle and an edge-adjacent triangle generate different examples,
%C Disregarding corner swap, a(3) to a(6) would be "18, 4792, 67488, 10906128"
%H Donghwi Park, <a href="https://github.com/gwahak/mathematics/blob/master/A375416/A375416%20order4%20sum%20all.ipynb">Source code for a(4)</a>.
%H Donghwi Park, <a href="https://github.com/gwahak/mathematics/blob/master/A375416/A375416%20order5%20sum%20all.ipynb">Source code for a(5)</a>.
%e a(1)=1 because there is only the trivial case without any subtriangle.
%e a(2)=4 because we can choose only the number in the central triangle.
%e 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.
%e For n = 3, numbers 1..9 are placed inside the triangles shown:
%e o
%e / \
%e o-- o
%e / \ / \
%e o---o---o
%e / \ / \ / \
%e o---o---o---o
%e An example with magic sum=17:
%e 9
%e 5
%e 1 2
%e 6 4
%e 7 3 8
%e This corresponds to the magic perimeter triangle (A342467):
%e 1 9 5 2
%e 7 4
%e 6 8
%e 3
%Y Cf. A006052, A342467, A355256.
%K nonn,more,hard
%O 1,2
%A _Donghwi Park_, Aug 15 2024