

A270205


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


3



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
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OFFSET

0,3


COMMENTS

William H. Press looked at the hybrid structure of a mostperfect magic square and the Hilbert space filling curve and thought it might be the "most uniform" way of putting the consecutive integers in a 2d square. He thought a definition of "most uniform" would be useful.
Al Zimmermann suggested this: Start by defining the "nonuniformity 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 nonuniformity. For both the mostperfect square and mostperfect cube the nonuniformity is 0 and so each is a most uniform distribution. (Of course, you'd want a better word for "nonuniformity". 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 mostperfect order 4 cube and the reversible order 4 cube are the new findings to look at in the link section.
Mostperfect magic squares require every 2 X 2 cell block to have the same sum. This sequence looks at that same subset in the cube.
Mostperfect 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


LINKS



FORMULA

a(n) = 3*n^3  6*n^2 + 3*n.
G.f.: 6*x^2*(1+2*x)/(x1)^4.
a(n) = 4*a(n1)  6*a(n2) + 4*a(n3)  a(n4) for n>3. (End)


EXAMPLE

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.


MAPLE



MATHEMATICA

CoefficientList[Series[(6 (x^2 + 2 x^3))/(1 + x)^4, {x, 0, 32}], x] (* Michael De Vlieger, Mar 15 2016 *)


PROG

(PARI) concat([0, 0], Vec(6*x^2*(1+2*x)/(x1)^4 + O(x^100))) \\ Altug Alkan, Mar 14 2016


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



