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A381847 a(n) is the number of ways to partition an n X n X n cube into 3 noncongruent cuboids.

Original entry on oeis.org

0, 0, 2, 3, 8, 11, 19, 23, 35, 40, 55, 62, 80, 88, 110, 119, 144, 155, 183, 195, 227, 240, 275, 290, 328, 344, 386, 403, 448, 467, 515, 535, 587, 608, 663, 686, 744, 768, 830, 855, 920, 947, 1015, 1043, 1115, 1144, 1219, 1250, 1328, 1360, 1442, 1475, 1560
Offset: 1

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Author

Janaka Rodrigo, May 06 2025

Keywords

Comments

Alternatively, a(n) is the number of ways to decompose the triplet (n,n,n) into 3 distinct unordered triplets.
Initial stage: One triplet of the form (n,n,n).
Second stage: Decompose the original triplet into two distinct triplets by splitting one of the elements of (x,y,z) into two parts at a time according to the following rule; (x,y,z) is replaced by (x,y,z-r), (x,y,r), where 0 < r <= z/2. Each resulting triplet must be distinct in element composition when comparing with the rest of the triplets in the set. Sets including the same element composition including the case r = z/2 are named as duplicates and set aside to reconsider when forming the next term.
Third stage: Apply the same rule to one of the two triplets of the second term at a time to create another two distinct triplets. At this stage consider the duplicates of the second term and apply the same rule to decompose one of the two identical triplets into two triplets of distinct element composition.

Examples

			(1,1,1) and (2,2,2) cannot be decomposed in three distinct triplets giving first two terms; a(1) = a(2) = 0.
(3,3,3) can be decomposed into two triplets (3,3,2), (3,3,1) by applying the rule and repeat the process for one the triplets at a time for obtaining the following two sets:
   {(3,3,2), (3,2,1), (3,1,1)},
   {(3,3,1), (3,2,2), (3,2,1)}.
Therefore, a(3) = 2.
		

Crossrefs

Formula

G.f.: x^3*(2+3*x+4*x^2+3*x^3+2*x^4)/((1+x+x^2)*(1+x)^2*(1-x)^3). - Jinyuan Wang, Aug 03 2025

Extensions

More terms from Sean A. Irvine, May 11 2025