A384311 a(n) is the number of ways to partition an n X n X n cube into 4 noncongruent cuboids.

0, 0, 4, 12, 47, 85, 183, 266, 466, 613, 941, 1179, 1668, 2007, 2701, 3159, 4079, 4690, 5868, 6635, 8122, 9064, 10874, 12030, 14196, 15564, 18142, 19740, 22739, 24613, 28065, 30206, 34174, 36601, 41087, 43851, 48888, 51975, 57631, 61059, 67331, 71158, 78078

Offset: 1

Author

Janaka Rodrigo, May 25 2025

Keywords

• nonn
• easy

Comments

Alternatively, a(n) is the number of ways to decompose the triplet (n,n,n) into 4 distinct unordered triplets.

Start with initial triplet (n,n,n). At each step choose a triplet from the current set and apply the rule (x,y,z) -> (x,y,z-r) and (x,y,r) for 0 < r <= z/2 (or similarly on x or y), checking to ensure the new triplets are distinct within the set.

If two duplicates (triplets with same element composition) appear, mark one for further decomposition in the next step.

Continue until reaching a set of exactly four triplets, all with distinct element composition, and with total volume (sum of the products of elements in each triplet) = n^3.

Examples

			The triplets (1,1,1) and (2,2,2) cannot be decomposed into 4 distinct triplets giving first two terms a(1) = a(2) = 0.
According to the rule there is only one way to decompose the triplet (3,3,3) into two distinct triplets, those are (3,3,1) and (3,3,2) and by applying the rule to each of the triplets at a time gives two sets of triplets {(3,3,2), (3,2,1), (3,1,1)} and {(3,3,1), (3,2,2), (3,1,1)}. Finally by repeating the process for each of the triplets of the stage three at a time gives the following sets of four distinct triplets:
   {(3,3,2), (3,2,1), (2,1,1), (1,1,1)};
   {(3,3,2), (3,1,1), (2,2,1), (2,1,1)};
   {(3,3,1), (3,2,2), (2,2,1), (2,1,1)};
   {(3,3,1), (3,2,1), (2,2,2), (2,2,1)}.
Therefore, a(3)=4.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..200
• Index entries for linear recurrences with constant coefficients, signature (-1,1,3,3,-1,-4,-4,-1,3,3,1,-1,-1).

Crossrefs

Cf. A381847, A384479, A385247.