cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

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Author

Janaka Rodrigo, May 25 2025

Keywords

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.
		

Crossrefs

A384479 a(n) is the number of ways to partition n X n X n cube into 5 noncongruent cuboids.

Original entry on oeis.org

0, 0, 2, 31, 209, 560, 1561, 2852, 5894, 9093, 15778, 22304, 34805, 46194, 67399, 85610, 118461, 146218, 194363, 233941, 302115, 356523, 448831, 522153, 643781, 738976, 896563, 1017619, 1216376, 1368946, 1615516, 1803028, 2106173, 2333683, 2699501, 2974256, 3410946
Offset: 1

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Author

Janaka Rodrigo, May 30 2025

Keywords

Comments

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)|.

Examples

			a(1) = a(2) = 0, since the cubes 1 X 1 X 1 and 2 X 2 X 2 can never be partitioned into five cuboids.
a(3) = |A(3)| + |B(3)| + |C(3)| - |A(3) intersection B(3)| - |A(3) intersection C(3)|= 2 + 1 + 0 - 1 - 0 = 2.
a(4) = |A(4)| + |B(4)| + |C(4)| - |A(4) intersection B(4)| - |A(4) intersection C(4)|= 30 + 3 + 1 - 2 - 1 = 31.
		

Crossrefs

Extensions

a(11)-a(16) from Sean A. Irvine, Jul 26 2025
More terms from Jinyuan Wang, Aug 03 2025

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

Views

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
Showing 1-3 of 3 results.