Classifications of squares and cubes



searching the internet for magic squares one gets really confused on the terminology used. some hold a magic square with off both diagonals, others a magic square with the subdiagonal off,
the regular magical cubes seems to be called SemiPerfect, those with no off sums therefore called perfect. The term perfect is also in use for cubes which are mere pantriagonal
Because of the above I here summarize the terminology I use:
squares: generalized: square with doubly appearing / missing numbers singular/regular: all horizontal vertical and both diagonal sums correct semi-magic: only both diagonals are off semi magic: vertical sums are off by a constant, horizontal sums are off by twice that constant pandiagonal: also all broken (sub) diagonals sum correct n-pandiagonal: the given square can be rotated n times into another square cubes: generalized: cube with doubly appearing / missing numbers singular/regular: all horizontal, vertical, transversal and all 4 triagonal sums correct semi-magic: only the four triagonal sums are off semi magic: all four triagonal sums correct the other off sums situated as C(2;1) ie: 1 2 | 5 6 3 4 | 7 8 pandiagonal: also all broken diagonals sum correct pantriagonal: also all broken triagonals sum correct n-pandiagonal: the given cubes planes can be rotated n times into another cube n-pantriagonal: the given cube can be rotated n times into another cube
since by default al squares and cubes follow the definition of the 'regular' magic square/cube, I use the term 'singular' to express that the cube stands alone. The classification pandiagonal is a stronger classification than the classifications truly, either truly or pandiagonal can be combined with pantriagonal to denote cubes with satisfy both conditions. the conditions n-pandiagonal and n-pantriagonal are not tested yet in my classification procedures, they result however quite naturally from my multiplication theorema,
The square T(n,(T(m)i,j-1)n2) is m-pandiagonal
the cube C(n,(C(m)i,j-1)n3) is m-pantriagonal,
whether a cube can be n-pandiagonal (n != 1) is yet unclear
Pandiagonal cubes is a quite common phenomenon
Hendricks established a treshold order 2d for a d dimensional hypercube to be pan-r-agonal for all r == 1 .. d



further classifications around:

The term 'most perfect' was defined for squares of order n by: Kathleen Ollerenshaw and David Brée
as each two by two square sums to 2 * (n2+1) and each pair (n/2 apart) on all (broken) diagonals sum up to (n2+1)
I found that augmenting the power of n from 2 to 3 and diangular in triangular 'most perfect' was simular defined for cubes, 'most perfect' is tested for by my qualification procedure and used troughout this site in conjunction with the other classifications.
the term pandiagonal cube is recently added to cubes classification procedure, therefore: cubes classified as truly might in fact be pandiagonal (this will be corrected in the near future)

M Suzuki uploaded 'Complete' magic squares of order 5 wherein the corners of all possible squares with there center sum up to the magic number see bottom of Directory listing of pages on magic squares and cubes.
I don't know wheter this can be usefully defined in higher dimension, it might be tested for in future work