More results: The function f that maps the stern-brocot tree onto the tree of surreals is closely related to the Minkowski questionmark function. For x in [1,2], f(x) = ?(x-1). Define the pseudo-lg as plg(x) = 2^k x - k - 1, 2^(-k) <= x < 2^(-k+1), k>=0 a linear interpolation between powers of 1/2. Let f' = inverse(f). Then f'(plg(f(x))) = x-1 for x in [1,2]. For x>=2, f'(f(x)-1) = x-1. And of course, for x>0, -f(1/x) = x. Using f, we can compute homomorphisms of continued fractions. For example, what is the point corresponding to the golden ratio? Since plg(x) = 2x-2 for all the convergents, we solve x=-plg(x)=-(2x-2) and find that f(phi) = 2/3. What about if we use the binary logarithm lg instead of plg? f'(lg(f(x))) >= x-1 on [1,2]. It's a bumpy fractal with equality at powers of 1/2 and some extra points I haven't characterized yet. The golden ratio in this case is x=-lg(x)=-log(x)/log(2) x=lambertw(log(2))/log(2) What kind of tree has the homomorphism g(x), where g satisfies g'(lg(g(x))) = x-1 on [1,2]? lg(g(x)) = g(x-1) g(x) = exp(g(x-1)) Looks like a job for continuous tetration (flow stuff) to me. -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039