A270205 Number of 2 X 2 planar subsets in an n X n X n cube.

0, 0, 6, 36, 108, 240, 450, 756, 1176, 1728, 2430, 3300, 4356, 5616, 7098, 8820, 10800, 13056, 15606, 18468, 21660, 25200, 29106, 33396, 38088, 43200, 48750, 54756, 61236, 68208, 75690, 83700, 92256

Offset: 0

Author

Craig Knecht, Mar 13 2016

Keywords

• nonn
• easy

Comments

William H. Press looked at the hybrid structure of a most-perfect magic square and the Hilbert space filling curve and thought it might be the "most uniform" way of putting the consecutive integers in a 2-d square. He thought a definition of "most uniform" would be useful.

Al Zimmermann suggested this: Start by defining the "non-uniformity of a distribution of integers among the cells of a square [or cube or hypercube]" to be the standard deviation of the sums of the 2 X 2 planar subsets. Then define a "most uniform distribution of integers" to be a distribution with the smallest non-uniformity. For both the most-perfect square and most-perfect cube the non-uniformity is 0 and so each is a most uniform distribution. (Of course, you'd want a better word for "non-uniformity". Skewness?) Perhaps use "2 X 2 planar subset" instead of "2 X 2 partition"?

Comment from Dwane Campbell: For cubes, the definition of compact is that all 2 X 2 X 2 subcubes add to the same sum. That definition also includes wrap around. Your most perfect space cube is compact. It has the additional constraint that each orthogonal plane is also compact. There are 64 2 X 2 X 2 subcubes that add to 260 and 192 2 X 2 subsquares that add to 130 in your cube. I did not think either result was possible. Congratulations!

The most-perfect order 4 cube and the reversible order 4 cube are the new findings to look at in the link section.

Most-perfect magic squares require every 2 X 2 cell block to have the same sum. This sequence looks at that same subset in the cube.

Most-perfect space is defined as a structure where all these 2 X 2 subsets have the same sum.

What structure provides the most uniform distribution of integers in a cube?

a(n+1) is the number of unit faces required to make an n X n X n cubic lattice. Number of unit edges required for the same is A059986(n). - Mohammed Yaseen, Aug 22 2021

a(n-3) is the maximum sigma irregularity over all maximal 3-degenerate graphs with n vertices. The extremal graphs are 3-stars (K_3 joined to n-3 independent vertices). (The sigma irregularity of a graph is the sum of the squares of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 14 2023

Examples

			The 2 X 2 X 2 cube labeled with the integers 1 to 8 has the following six 2 X 2 planar subsets each containing 4 cells: 1,2,3,4; 5,6,7,8; 1,2,5,6; 3,4,7,8; 1,4,5,8; 2,3,6,7.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
• Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
• Craig Knecht, Cube assembly from different 2x2 planar criteria.
• Craig Knecht, F1 code most-perfect magic cube 960 examples.
• Craig Knecht, F1 code reversible cube 960 examples.
• Craig Knecht, magic space.
• Craig Knecht, Most-perfect space.
• Walter Trump, Most-Perfect magic cube.
• Walter Trump, 6 unique neighbors for the most-perfect magic cube.
• Wikipedia, Most-perfect magic square translated to a cube via the Hilbert space filling curve.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A002411, A059986, A118659.

Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

Programs

Magma

[3*n^3 - 6*n^2 + 3*n: n in [0..50]]; // Wesley Ivan Hurt, Mar 13 2016

Maple

A270205:=n->3*n^3-6*n^2+3*n: seq(A270205(n), n=0..50); # Wesley Ivan Hurt, Mar 13 2016

Mathematica

Table[3*n^3 - 6*n^2 + 3*n, {n, 0, 50}] (* Wesley Ivan Hurt, Mar 13 2016 *)
CoefficientList[Series[(6 (x^2 + 2 x^3))/(-1 + x)^4, {x, 0, 32}], x] (* Michael De Vlieger, Mar 15 2016 *)

PARI

concat([0, 0], Vec(6*x^2*(1+2*x)/(x-1)^4 + O(x^100))) \\ Altug Alkan, Mar 14 2016

PARI

a(n) = 3*n^3 - 6*n^2 + 3*n \\ Charles R Greathouse IV, Mar 15 2016

Formula

a(n) = 3*n^3 - 6*n^2 + 3*n.

From Wesley Ivan Hurt, Mar 13 2016: (Start)

G.f.: 6*x^2*(1+2*x)/(x-1)^4.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)

E.g.f.: 3*x^2*(1+x)*exp(x). - G. C. Greubel, May 10 2016

a(n) = 6 * A002411(n-1) for n>=1. - Joerg Arndt, May 11 2016

a(n) = A118659((n-1)^3), n>1. - Mohammed Yaseen, Aug 22 2021

From Amiram Eldar, Jul 02 2023: (Start)

Sum_{n>=2} 1/a(n) = Pi^2/18 - 1/3.

Sum_{n>=2} (-1)^n/a(n) = Pi^2/36 - 2*log(2)/3 + 1/3. (End)