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 A270205 Number of 2 X 2 planar subsets in an n X n X n cube. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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? LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Craig Knecht, Cube assembly from different 2x2 planar criteria Craig Knecht, Most-perfect space Craig Knecht, F1 code most-perfect magic cube 960 examples Craig Knecht, magic space Craig Knecht, F1 code reversible cube 960 examples Walter Trump, Most-Perfect magic cube Walter Trump, 6 unique neighbors for the most-perfect magic cube Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1). 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 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 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 *) PROG (MAGMA) [3*n^3 - 6*n^2 + 3*n: n in [0..50]]; // Wesley Ivan Hurt, Mar 13 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 CROSSREFS Cf. A002411. Sequence in context: A207249 A060521 A036141 * A207443 A207437 A199243 Adjacent sequences:  A270202 A270203 A270204 * A270206 A270207 A270208 KEYWORD nonn,easy AUTHOR Craig Knecht, Mar 13 2016 STATUS approved

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