A385151 a(n) is the least possible difference between the largest and smallest volumes of distinct three-cuboid combination filling an n X n X n cube.

6, 24, 20, 48, 42, 80, 54, 140, 99, 192, 143, 252, 150, 352, 238, 432, 304, 520, 294, 660, 437, 768, 525, 884, 486, 1064, 696, 1200, 806, 1344, 726, 1564, 1015, 1728, 1147, 1900, 1014, 2160, 1394, 2352, 1548, 2552, 1350, 2852, 1833, 3072, 2009, 3300, 1734

Offset: 3

Author

Janaka Rodrigo, Jun 19 2025

Keywords

• nonn

Comments

Developed as the three dimensional extension of the Mondrian Art Problem.

Alternatively, a(n) is the optimal solution when an n X n X n cube is partitioning into 3 cuboids of different dimensions.

Let elements of the unordered integer triplet (x,y,z) be the dimensions of cuboid in a set of three cuboids.

Let V(x,y,z) = x*y*z be the volume and for a given set of triplets S, Min(S) = min{V(x,y,z):(x,y,z) in S}, Max(S) = max{V(x,y,z):(x,y,z) in S}, and defect(S) = Max(S)-Min(S).

a(n) is the least possible value of the defect as S runs over the possible partitions of the n X n X n cuboid into 3 cuboids of different dimensions.

Examples

			4 X 4 X 4 cube can be partitioned in three different ways and defects of sets are calculated as follows:
{(4,3,3), (4,3,1), (4,4,1)}: defect = max(36,12,16)-min(36,12,16) = 36-12=24,
{(4,2,1), (4,3,2), (4,4,2)}: defect = max(8,24,32)-min(8,24,32) = 32-8=24,
{(4,4,3), (4,3,1), (4,1,1)}: defect = max(48,12,4)-min(48,12,4) = 48-4=44.
Therefore, a(4) = min{24, 24, 44} = 24.
		

Links

• Janaka Rodrigo, Python program

Crossrefs

Cf. A276523, A381847.