Iterated Logarithmic Functions

Given any two binary operations a(x,y) and b(x,y), suppose there is a
function f such that  
                        f(a(x,y)) = b(f(x),f(y))

A simple example is a(x,y)=x+y, b(x,y)=xy, f(x)=exp(x).  Anyway, given
these two operations and the function f, define the auxiliary functions

                 u(x) = a(x,f(x))         v(x) = b(x,f(x))

It's easy to see that u and v are "conjugates", in the sense that

                         f(u(x)) = v(f(x))

Slightly less obvious is the fact that the nth compositions of u(x) 
and v(x), denoted by u_n(x) and v_n(x) respectively, are also 
conjugates, i.e.,
                       f(u_n(x)) = v_n(f(x))

This identity is sometimes useful in simplifying computations. To
illustrate, consider the case a(x,y)=xy, b(x,y)=x+y, f(x)=ln(x).  Here 
we have  u(x) = x ln(x)  and  v(x) = x + ln(x).  Taking an initial x 
value of 10, the first few iterations of u are

             10.000000   23.025851   72.223287   309.098522

Beginning with an initial value of f(10)=2.302585 the first few 
iterations of v are

              2.302585    3.136617    4.279762     5.733660

and we see that f(309.098522) = 5.733660.

Another set of functions to which this method can be applied is

                                 x + y
   a(x,y) = x + y      b(x,y) = -------      f(x) = tan(x)
                                1 - xy

In general, any function f() with an "addition rule" can be used.
For example, we can use the functions

      a(x,y) = x+y    
                  _______        _______
      b(x,y) = x /1 - y^2  +  y /1 - x^2

      f(x) = sin(x)


This can also be applied to number-theoretic functions.  For example,
consider the case a(x,y)=xy, b(x,y)=x+y, f(x)=sopf(x), where sopf is 
the "sum-of-prime-factors" function.  In this case the auxiliary 
functions are u(x) = x sopf(x)  and v(x) = x + sopf(x).  Beginning 
with an initial value of 10, the first few iterations of u are

               10    70    980    22540   1036840

Beginning with an initial value of f(10)=7, the first few iterations 
of v give
                7    14     23       46        71

and we see that sopf(1036840) = 71.