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 1-6 of 6 results.

A279596 Partition an n X n square into multiple integer-sided rectangles where no one is a translation of any other; a(n) is the least possible difference between the largest and smallest area.

Original entry on oeis.org

2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 4, 6, 6, 6, 6, 6, 6, 7, 7, 7, 8, 7, 6, 6, 8, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9
Offset: 3

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Author

Ed Pegg Jr, Dec 15 2016

Keywords

Comments

Similar to the Mondrian Art sequence (A276523), but allowing repetition of rectangles with different orientations.
Proved optimal to a(45) by R. Gerbicz. Best values known for a(46)-a(96): 10, 12, 11, 12, 12, 8, 12, 12, 13, 12, 12, 14, 14, 15, 12, 15, 14, 15, 14, 16, 16, 15, 16, 16, 16, 17, 16, 17, 14, 17, 18, 16, 18, 16, 18, 15, 16, 18, 18, 16, 18, 17, 19, 20, 17, 17, 21, 20, 20, 21, 22.
Seems to be bounded above by ceiling(n/log(n)). The currently verified distances from this bound are 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 3, 1, 2, 2, 2, 2, 2, 2 (A279848).

Examples

			The 9 X 9 square can be divided into non-translatable rectangles with
aaaaaaaab
ddddddeeb
fggghheeb
fggghheeb
fiiihheeb
fiiijjjjb
fiiijjjjb
fkkkkkkkb
ccccccccc
		

Crossrefs

Extensions

Moved terms to A279848, expanded best values known
a(28)-a(45) from Robert Gerbicz, Jan 01 2017

A278970 Partition an n X n square into multiple non-congruent integer-sided rectangles. a(n) is ceiling(n/log(n)) + 3 - the least possible difference between the largest and smallest area.

Original entry on oeis.org

4, 2, 3, 2, 2, 1, 2, 0, 2, 1, 1, 3, 1, 1, 2, 2, 2, 1, 1, 2, 3, 2, 1, 2, 2, 3, 3, 1, 2, 3, 1, 1, 2, 2, 3, 4, 3, 2, 2, 3, 3, 3, 2, 3, 4, 2, 4, 3, 2, 4, 3, 2, 3, 3, 3, 3, 4, 3, 3, 5, 4, 4, 4
Offset: 3

Views

Author

Ed Pegg Jr, Dec 02 2016

Keywords

Comments

If ceiling(n/log(n)) + 3 is an upper bound for the Mondrian Art Problem (A276523), a(n) is the amount by which the optimal value beats the upper bound.
Terms a(86) and a(139) are at least 5. Term a(280) is at least 7.
Term a(138) is at least 9, defect 22 (1200-1178) with 16 rectangles.
Best values known for a(66) to a(96): 3, 1, 1, 2, 1, 2, 0, 1, 1, 1, 2, 1, 0, 1, 2, 1, 2, 2, 1, 1, 5, 1, 3, 1, 0, 1, 2, 2, 0, 0, 1.

Crossrefs

Extensions

a(45)-a(57) from Robert Gerbicz added/corrected, updated best known values to a(96), Ed Pegg Jr, Dec 28 2016
a(58)-a(65) from Michel Gaillard added by Ed Pegg Jr, Sep 02 2021

A385151 a(n) is the least possible difference between the largest and smallest volumes of distinct three-cuboid combination filling an n X n X n cube.

Original entry on oeis.org

6, 24, 20, 48, 42, 80, 54, 140, 99, 192, 143, 252, 150, 352, 238, 432, 304, 520, 294, 660, 437, 768, 525, 884, 486, 1064, 696, 1200, 806, 1344, 726, 1564, 1015, 1728, 1147, 1900, 1014, 2160, 1394, 2352, 1548, 2552, 1350, 2852, 1833, 3072, 2009, 3300, 1734
Offset: 3

Views

Author

Janaka Rodrigo, Jun 19 2025

Keywords

Comments

Developed as the three dimensional extension of the Mondrian Art Problem.
Alternatively, a(n) is the optimal solution when an n X n X n cube is partitioning into 3 cuboids of different dimensions.
Let elements of the unordered integer triplet (x,y,z) be the dimensions of cuboid in a set of three cuboids.
Let V(x,y,z) = x*y*z be the volume and for a given set of triplets S, Min(S) = min{V(x,y,z):(x,y,z) in S}, Max(S) = max{V(x,y,z):(x,y,z) in S}, and defect(S) = Max(S)-Min(S).
a(n) is the least possible value of the defect as S runs over the possible partitions of the n X n X n cuboid into 3 cuboids of different dimensions.

Examples

			4 X 4 X 4 cube can be partitioned in three different ways and defects of sets are calculated as follows:
{(4,3,3), (4,3,1), (4,4,1)}: defect = max(36,12,16)-min(36,12,16) = 36-12=24,
{(4,2,1), (4,3,2), (4,4,2)}: defect = max(8,24,32)-min(8,24,32) = 32-8=24,
{(4,4,3), (4,3,1), (4,1,1)}: defect = max(48,12,4)-min(48,12,4) = 48-4=44.
Therefore, a(4) = min{24, 24, 44} = 24.
		

Crossrefs

A385153 a(n) is the least possible difference between the largest and smallest volumes of distinct four-cuboid combinations filling an n X n X n cube.

Original entry on oeis.org

5, 16, 16, 36, 30, 60, 48, 100, 83, 96, 123, 182, 130, 264, 182, 324, 224, 280, 259, 484, 369, 576, 449, 676, 423, 560, 528, 900, 598, 1008, 638, 1054, 859, 864, 979, 1330, 884, 1620, 1054, 1764, 1152, 1364, 1185, 2116, 1553, 2304, 1713, 2500, 1513, 1924, 1760
Offset: 3

Views

Author

Janaka Rodrigo, Jun 19 2025

Keywords

Comments

Developed as the three-dimensional extension of the Mondrian Art Problem.
Alternatively, a(n) is the minimum defect when an n X n X n cube is partitioning into four cuboids of different dimensions.
Let elements of the unordered integer triplet (x,y,z) be the dimensions of a cuboid in a set S of four cuboids.
Let V(x,y,z) = x*y*z be the volume and for a given element of S. Define min(S) = min{V(x,y,z): (x,y,z) in S}, max(S) = max{V(x,y,z): (x,y,z) in S}, and defect(S) = max(S)-min(S).
a(n) is the smallest value of the defect(S) across all possible partitions of the n X n X n cuboid into four cuboids of different dimensions.

Examples

			The 4 X 4 X 4 cube has 12 different ways of partitioning into four distinct cuboids each giving a defect as the difference between the largest volume and the smallest volume relevant to the four cuboids of the set.
The optimal solution is given by the set {(4,2,1), (4,2,2), (4,3,2), (4,4,1)} because it has the minimum defect.
The least possible defect = max(8,16,24,16)-min(8,16,24,16) = 24-8 = 16.
Therefore, a(4) = 16.
		

Crossrefs

Extensions

More terms from Sean A. Irvine, Jul 16 2025

A385154 a(n) is the least possible difference between the largest and smallest volumes of distinct five-cuboid combinations filling an n X n X n cube.

Original entry on oeis.org

10, 12, 21, 28, 28, 62, 42, 80, 57, 112, 114, 143, 90, 156, 191, 288, 184, 224, 252, 396, 299, 288, 315, 504, 414, 546, 462, 720, 529, 816, 616, 837, 609, 648, 777, 1140, 858, 1260, 874, 1596, 1237, 1155, 810, 1554, 1468, 2064, 1118, 1950, 1343, 2080, 1590, 2268
Offset: 3

Views

Author

Janaka Rodrigo, Jun 19 2025

Keywords

Comments

Developed as the three-dimensional extension of the Mondrian Art Problem.
Alternatively, a(n) is the optimal solution when an n X n X n cube is partitioning into five cuboids of different dimensions.
Let elements of the unordered integer triplet (x,y,z) be the dimensions of a cuboid in a set of five cuboids and volume V(x,y,z) = x*y*z; cuboids have five values for each set of five triplets S, produced by the union of A(n), B(n), C(n), where A(n), B(n), and C(n) are sequences of sets as introduced in A384479.
Define min(S) = min{V(x,y,z):(x,y,z) in S} and max(S) = max{V(x,y,z):(x,y,z) in S}, then defect(S) = max(S) - min(S).
a(n) is the smallest possible value of defect(S) where S runs over all possible ways of partitioning the n X n X n cube into five cuboids of different dimensions.

Examples

			4 X 4 X 4 cube has 31 different ways of partitioning into five distinct cuboids and only two sets producing the minimum defects as calculated below:
{(3,2,1), (3,3,2), (4,1,2), (4,2,2),(4,4,1)} has minimum defect = max(6,18,8,16,16) - min(6,18,8,16,16) = 18 - 6 = 12.
{(4,4,1), (4,3,1), (2,3,3), (2,3,2), (2,3,1)} has minimum defect = max(16,12,18,12,6) - min(16,12,18,12,6) = 18 - 6 = 12.
Therefore a(4) = 12.
		

Crossrefs

Extensions

a(11)-a(20) from Sean A. Irvine, Jul 26 2025
a(21)-a(54) from Jinyuan Wang, Aug 04 2025

A279848 Partition an n X n square into multiple integer-sided rectangles where no one is a translation of any other. a(n) is ceiling(n/log(n)) - the least possible difference between the largest and smallest area.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 4, 4, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 3
Offset: 3

Views

Author

Ed Pegg Jr, Dec 21 2016

Keywords

Comments

If ceiling(n/log(n)) is an upper bound for the Mondrian Art Problem variant (A279596), a(n) is the amount by which the optimal value beats the upper bound.
Terms a(3) to a(45) verified optimal by R. Gerbicz.
Term a(103) is at least 9, defect 14 (630-616) with 17 rectangles.
Best values known for a(46) to a(96): 3, 1, 2, 1, 1, 5, 2, 2, 1, 2, 2, 1, 1, 0, 3, 0, 2, 1, 2, 0, 0, 1, 1, 1, 1, 0, 1, 1, 4, 1, 0, 2, 0, 3, 1, 4, 3, 1, 1, 4, 2, 3, 1, 0, 4, 4, 0, 1, 1, 0, 0.

Crossrefs

Extensions

a(28)-a(45) from Robert Gerbicz, Jan 01 2017
Showing 1-6 of 6 results.