The don't-remove-common-factors version of my question is the more natural one, as Bill Thurston's posting makes clear. But the less-natural remove-common-factors version still looks like it might generate all rationals with odd denominators in [0,1]. Here's some Mathematica code I just wrote: Lengthen[L_] := Module[{i, M}, M = Table[0, {3 Length[L]}]; M[[1]] = Numerator[L[[1]]]/(2 + Denominator[L[[1]]]); M[[2]] = 2*Numerator[L[[1]]]/(1 + 2 Denominator[L[[1]]]); For[i = 1, i < Length[L], i++, M[[3 i]] = L[[i]]; M[[3 i + 1]] = (2 Numerator[L[[i]]] + Numerator[L[[i + 1]]])/(2 Denominator[L[[i]]] + Denominator[L[[i + 1]]]); M[[3 i + 2]] = (Numerator[L[[i]]] + 2 Numerator[L[[i + 1]]])/(Denominator[L[[i]]] + 2 Denominator[L[[i + 1]]])]; M[[3 Length[L]]] = L[[Length[L]]]; Return[M]] WF[n_] := WF[n] = If[n == 0, {1}, Lengthen[WF[n - 1]]] WF[n] consists of all the non-zero rational numbers in [0,1] obtained after n iterations of the insertion-of-weighted-medians procedure. I find that for all n up through 15 (the largest value of n that I tried), WF[n] contains all rationals in [0,1] with denominator 1/(2n+1). E.g., Table[MemberQ[WF[15], k/31], {k, 1, 30}] returns a list whose 30 elements all equal True, because WF[15] contains all proper fractions with denominator 31. Can anyone prove or disprove that this holds for all n? Jim Propp