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

%I A384311 #24 Aug 18 2025 03:10:25
%S A384311 0,0,4,12,47,85,183,266,466,613,941,1179,1668,2007,2701,3159,4079,
%T A384311 4690,5868,6635,8122,9064,10874,12030,14196,15564,18142,19740,22739,
%U A384311 24613,28065,30206,34174,36601,41087,43851,48888,51975,57631,61059,67331,71158,78078
%N A384311 a(n) is the number of ways to partition an n X n X n cube into 4 noncongruent cuboids.
%C A384311 Alternatively, a(n) is the number of ways to decompose the triplet (n,n,n) into 4 distinct unordered triplets.
%C A384311 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.
%C A384311 If two duplicates (triplets with same element composition) appear, mark one for further decomposition in the next step.
%C A384311 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.
%H A384311 Sean A. Irvine, <a href="/A384311/b384311.txt">Table of n, a(n) for n = 1..200</a>
%H A384311 <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (-1,1,3,3,-1,-4,-4,-1,3,3,1,-1,-1).
%e A384311 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.
%e A384311 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:
%e A384311    {(3,3,2), (3,2,1), (2,1,1), (1,1,1)};
%e A384311    {(3,3,2), (3,1,1), (2,2,1), (2,1,1)};
%e A384311    {(3,3,1), (3,2,2), (2,2,1), (2,1,1)};
%e A384311    {(3,3,1), (3,2,1), (2,2,2), (2,2,1)}.
%e A384311 Therefore, a(3)=4.
%Y A384311 Cf. A381847, A384479, A385247.
%K A384311 nonn,easy
%O A384311 1,3
%A A384311 _Janaka Rodrigo_, May 25 2025