A385154 - OEIS

A385154

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

10, 12, 21, 28, 28, 62, 42, 80, 57, 112, 114, 143, 90, 156, 191, 288, 184, 224, 252, 396, 299, 288, 315, 504, 414, 546, 462, 720, 529, 816, 616, 837, 609, 648, 777, 1140, 858, 1260, 874, 1596, 1237, 1155, 810, 1554, 1468, 2064, 1118, 1950, 1343, 2080, 1590, 2268

(list; graph; refs; listen; history; text; internal format)

OFFSET

3,1

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 five cuboids of different dimensions.

Let elements of the unordered integer triplet (x,y,z) be the dimensions of a cuboid in a set of five cuboids and volume V(x,y,z) = x*y*z; cuboids have five values for each set of five triplets S, produced by the union of A(n), B(n), C(n), where A(n), B(n), and C(n) are sequences of sets as introduced in A384479.

Define min(S) = min{V(x,y,z):(x,y,z) in S} and max(S) = max{V(x,y,z):(x,y,z) in S}, then defect(S) = max(S) - min(S).

a(n) is the smallest possible value of defect(S) where S runs over all possible ways of partitioning the n X n X n cube into five cuboids of different dimensions.

LINKS

Table of n, a(n) for n=3..54.

Janaka Rodrigo, Python Code for Minimum defects of A(n)

Janaka Rodrigo, Python Code for Minimum defects of B(n)

Janaka Rodrigo, Python Code for Minimum defects of C(n)

EXAMPLE

4 X 4 X 4 cube has 31 different ways of partitioning into five distinct cuboids and only two sets producing the minimum defects as calculated below:

{(3,2,1), (3,3,2), (4,1,2), (4,2,2),(4,4,1)} has minimum defect = max(6,18,8,16,16) - min(6,18,8,16,16) = 18 - 6 = 12.

{(4,4,1), (4,3,1), (2,3,3), (2,3,2), (2,3,1)} has minimum defect = max(16,12,18,12,6) - min(16,12,18,12,6) = 18 - 6 = 12.

Therefore a(4) = 12.

CROSSREFS

Column k=5 of A386297.