NJAS>See one of the comments in http://oeis.org/A002487 for one possible answer. The reference is to D. E. Knuth, Problem 10906, American Mathematical Monthly; solution by Moshe *Newman*, American Mathematical Monthly, 110 (No. 7, 2003), 642-643. On Sat, Oct 12, 2013 at 12:15 PM, Warren D Smith <warren.wds@gmail.com>wrote: What is "Newman's rational counter"? Apparently, Moshe Newman discovered that iterating 1/(2 Floor[#] + 1 - #)& starting with 0 enumerates the nonnegative rationals. I was using the reverse sequence In[1106]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, 3/8, 22] Out[1106]= {3/8, 7/3, 4/7, 5/4, 1/5, 4, 3/4, 5/3, 2/5, 5/2, 3/5, 4/3, 1/4, 3, 2/3, 3/2, 1/3, 2, 1/2, 1, 0, Indeterminate, Indeterminate} So 3/8 is the "20th rational". as the actual counter, which I should have called n(r). E.g., n((k-1)/(k+1))/2 - 1 = 2^(k/2-2) (3 + √2 + (-)^k (3 - √2)) goes 1, 3, 2, 6, 4, 12, 8, 24, 16, ... (alternately multiplying by 3 and 2/3, LinearRecurrence[{0, 2}, {1, 3},9], A164073), in case you ever need slower-than-fibonacci exponential positive integers. For monotone, 2^((k-5)/2) (4 + 3 Sqrt[2] - (-1)^k (4 - 3 Sqrt[2])), 2, 3, 4, 6, 8, 12, 16, 24, 32,... which is n(1/(8/(3 + (-1)^k) + (k-5)/2)). (You can go way slower with those Pisot-Vijayaraghavan numbers.) --rwg