cp's OEIS Frontend

Partitioning n X n X n cube is done by decomposing the triplet (n,n,n) into five unordered integer triplets of distinct element composition in three different stages and applying the inclusion-exclusion principle to obtain all geometrically feasible triplets without repetitions.

First stage:

Generating the sequence of sets A(n) of five triplets of distinct element composition by decomposing the triplet (n,n,n).

The number of sets in each term describes the number of ways a cube can be partitioned into five cuboids by cutting one of the cuboids in the previous stage into two cuboids. The algorithm explained in A381847 (partitioning into three cuboids) and A384311 (partitioning into four cuboids). As an example: A(3) = {{(3,3,1), (3,2,2), (3,1,1), (2,1,1), (1,1,1)}, {(3,2,2), (3,2,1), (2,2,1), (3,1,1), (2,1,1)}}.

Therefore the number of sets in A(3) is |A(3)| = 2.

Second stage:

Generating the sequence of sets B(n) of five triplets of distinct element composition by decomposing the triplet (n,n,n).

The number of sets in each term represents number of distinct five-cuboid combinations filling the n X n X n cube with three full-length axial spanning sharing only two cube corners each.

Sets of five distinct triplets of the n-th term are defined by {(n,n-x,y), (n,n-y,z), (n,n-z,x), (x,y,z), (n-x,n-y,n-z)}, where 0 < x,y,z < n.

The triplet (3,3,3) can be decomposed by the rule only in one way giving, B(3) = {{(1,3,1), (3,1,2), (2,2,3), (2,1,1), (1,2,2)}}. Therefore |B(3)| = 1.

Third stage:

Generating the sequence of sets C(n) of five triplets of distinct element composition by decomposing the triplet (n,n,n).

The number of sets in each term represents number of different ways to partition n X n X n cube into five distinct cuboids such that all five cuboids going across the cube parallel to each other avoiding any cut-plane that cuts through the whole pile.

Sets of five distinct triplets of the n-th term are defined by {(n,x,y), (n,p,y+q), (n,n-p,q), (n,n-p-x,n-q), (n,p+x,n-y-q)}, where p+x < n,q+y < n and 0 < x,y,p,q < n.

Triplets (1,1,1), (2,2,2) and (3,3,3) cannot be decomposed by this rule and the triplet (4,4,4) has only one way of decomposing, C(4) = {{(1,1,4), (1,2,4), (1,3,4), (2,2,4), (2,3,4)}}. Therefore |C(4)| = 1.

Since there are no intersections between B(n) and C(n), the number of ways to partition n X n X n cube is given by: a(n) = |A(n) union B(n) union C(n)| = |A(n)| + |B(n)| + |C(n)| - |A(n) intersection B(n)| - |A(n) intersection C(n)|.