That is absolutely brilliant! It seems to resemble the process of generating a continued fraction approximation of a complex number: 1. Subtract [z] from z, where [] indicates the closest Gaussian (or Eisenstein, or whatever) integer. 2. Raise z to the power of -1 (i.e. invert in the unit circle and conjugate). 3. Repeat the first two steps ad infinitum. This differs from the ordinary CF algorithm in that the closest integer, rather than the floor, is chosen at each stage (as floor is meaningless in an unordered set such as the complex numbers). Let's see how this version produces a different CF for pi: Ordinary: [3;7,15,1,292,1,1,1,2,1,...] Modified: [3;7,16,-294,3,-4,5,-15,-3,2,...] I decided to search the OEIS for this sequence: http://oeis.org/A133593 -------- What about exp(1)? According to Mathematica, it is: [3; -4, 2, 5, -2, 7, -2, 9, -2, -11, ...] This is only slightly more complicated than the regular continued fraction for e. And exp(i) using Gaussian integers? [1+i; -2+i, 1+3i, -2, 1-5i, -2, 1+7i, -2, 1-9i, -2, 1+11i, ...] Very similar. What about exp(2)? [7; 3, -2, -3, -18, -6, 2, 6, 30, 9, -2, -9, -42, -12, 2, 12, 54, 15, -2, -15, -66, -18, 2, 18, 78, 21, -2, -21, -90, -24, ...] Excluding the obvious sign alteration, the columns are all basic arithmetic progressions. exp(3)? [20; 12, -3, -4, -4, 7, -3, -17, 2, 16, 2, 13, 14, 4, 6, 3, -2, -2, -2, -2, -3, -6, 5, -2, -68, -7, -6, 5, 3, -3, ...] Hmm, that appears completely irregular! Maybe it is still a simple interleaving of arithmetic progressions, but with a very large period. Let's try exp(1/2): [2; -3, 7, -2, -10, 2, 14, -2, -18, 2, 22, -2, -26, 2, 30, ...] Again, very regular. -------- Note that these continued fractions (of irrational numbers) do not permit the appearance of ones, in the same way that an ordinary CF cannot contain zeroes. Phi has the simplest CF; I wonder what has the 'simplest' modified CF of [2; 2, 2, 2, 2, 2, 2, ...]. Well, there's the obvious recurrence x = 2 + 1/x, which leads to the quadratic equation x^2 - 2x - 1 = 0 with positive real root 2 + sqrt(2). -------- Sincerely, Adam P. Goucher