A317888 - OEIS

%I #28 Jan 19 2022 21:26:53

%S 6,120,362880,355687428096000,8683317618811886495518194401280000000,

%T 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000

%N The minimum value in the central cell of an n-dimensional cube, three cells wide in each dimension, such that each cell is the product of its adjacent cells, each cell is a whole number, and no two cells have the same value.

%C We can prove that the corner cells are filled with values from 2 to 2^n+1, because in order to prove that none of the corner cells have to skip a value so as not to repeat a value within any of the other cells, we must prove that the non-corner cells all have values greater than 2^n+1 in at least one arrangement of corner cell values. 2(2^n-n+2) can describe the lowest value of cells in between two corner cells in every dimension from one to infinity, excluding the third dimension, because in the ideal arrangement of the three-dimensional grid, 12 must appear between 3 and 4, while the formula returns the value 14. Regardless, 12 is still greater than 2^n+1, which is 9 in three dimensions. Every value for 2(2^n-n+2) is greater than 2^n+1 as can clearly be observed from a graph. The formula 2(2^n-n+2) was derived from the fact that the smallest non-corner values appear next to 2 in every dimension except the third, and next to 3 in the first three dimensions. So with the exclusion of the third dimension due to its limited number of corner cells, these values must be equal to the product of 2 and the lowest number that is paired with 2. The lowest number that is paired with 2 in an ideal arrangement of values is always equal to one less than the number of dimensions, n-1, subtracted from the highest corner value, 2^n+1, thus resulting in 2^n+1-(n-1), or in its simplified form, 2^n-n+2. When multiplying this by two, we receive the lowest non-corner value possible in every dimension except the third.

%H Aidan Clarke, <a href="/A317888/b317888.txt">Table of n, a(n) for n = 1..8</a> (shortened by _N. J. A. Sloane_, Jan 17 2019)

%F a(n) = (2^n+1)!.

%F a(n) = A000142(A000051(n)). - _Michel Marcus_, Aug 11 2018

%e One arrangement for n=2 is:

%e    2   10   5

%e    8  120  15

%e    4   12   3

%e a(2) = 120 because this is the minimum possible value for the central cell.

%Y Cf. A000051, A000142.

%K nonn,easy

%O 1,1

%A _Aidan Clarke_, Aug 10 2018