Now to the structural character, literally, of the cube group.
This is a multiplex:
The formula for a computable number and hence any code that can be written by a machine (like Enigma) is
$$ (2i -1)n + \sum_{r=1}^{\infty} (2c_r -1)(\frac {2}{3})^r $$
Suppose a recursive number $$ N $$, and an algebraic character $$ r\; =\ 2 \sqrt {\frac {3} {2}} $$, not the same as r in the above formula (hence a correspondence is needed).
$$ N $$ has a commutative relation to k, the
length of any algorithm which can be written with a general structure $$ [(AB)^n + (A'B')^n] $$ and the formula
$$ N_k = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4 $$, where $$ N $$ is any character (in a standard basis of the sliced cube) generated of length k.
N the slice number is the 'computational surface' over which $$ N_k $$ is iterated.
Suppose also the sequence $$ s_0, s_1, ..., s_n $$ is the formula: $$s_n\; = 12.s_{n-1} + 18.s_{n-2} $$, in respect of n. There are two numbers $$ \alpha, \beta $$ that start equal at n = 1. The recurrence formula in n for these is:
$$ \alpha_1= \beta_1 = 9,\; \alpha_{n+1} = 6.\alpha_n + 9.\beta_n,\; \beta_{n+1} = 6.\alpha_n + 6.\beta_n $$
Note the 3/2 nature of the character's symmetry. (hope i got this right)