User blog:Edwin Shade/An Interesting Property

For Hypercubes and their Duals
The following is a table which describes how many m-dimensional facets are contained within an n-dimensional cube. The row number corresponds to the dimension of the cube, and the column number corresponds to the dimension of the facet.

0     1       2       3       4      5      6     7     8    9   10  0      1  1      2      1  2      4      4       1  3      8     12       6       1  4     16     32      24       8       1  5     32     80      80      40      10      1   6     64    192     240     160      60     12      1  7    128    448     672     560     280     84     14     1  8    256   1024    1792    1792    1120    448    112    16     1  9    512   2304    4608    5376    4032   2016    672   144    18    1 10   1024   5120   11520   15360   13440   8064   3360   960   180   20    1

Although these numbers may appear somewhat spontaneous, there is a definite pattern. To calculate the value of any subsequent cell, let's call it x, add together the cell to the top left of x and twice the value of the cell directly above cell x. Like so.

a  b       x     x=a+2b

This is very similar to Pascal's triangle, but with a different rule.

Summing up the rows from 0 to 10 we get the numbers 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, and 59049; which happen to be the powers of 3 from 3^0 to 3^10 ! The powers of three are evidently "contained" in this table. Also, summing up the diagonals going from down to top we have the sequence 1, 2, 5, 12, 29, 70, 169, 408,.., or the Pell numbers, which describe the denominators of the closest rational approximations to $$\sqrt{2}$$.

Now let's reverse the rows in the table while preserving the all 1's diagonal.

0    1      2      3      4       5      6       7      8      9   10  0      1  1      2     1  2      4     4      1  3      6    12      8      1  4      8    24     32     16      1  5     10    40     80     80     32       1   6     12    60    160    240    192      64      1  7     14    84    672    280    560     448    128       1  8     16   112    448   1120   1792    1792   1024     256      1  9     18   144    672   2016   4032    5376   4608    2304    512      1 10     20   180    960   3360   8064   13440  15360   11520   5120   1024    1

This now gives us a table describing the number of m-dimensional facets of a n-dimensional octahedron, which is the dual of the cube ! Summing the bottom to top diagonals of this table from left to right as we did with the prior table gives us the sequence 1, 2, 5, 10, 21, 42, 85, 170,..., or the nth number whose binary representation contains no consecutive equal binary digits.

My conjecture is, that for any polyhedron whose general form may be continued into an arbitrarily high number of dimensions, the corresponding table of facets has a rule for continuing the table given a finite number of entries.