Allan Wechsler wrote:
Aaaaaand I screwed up.
123446556666677777787; I left out a(6)=6.
My Mathematica code gives {0,1,2,3,3,5,4,4,5,5,5,5,5,6,6,6,6,6,6,7,6,7,7,7,7,7,7,8,7,7,7,8,8,7,8,8,8,9,8,8,8,9,8,8,8,8,8,9,8,8,9,9,9,10,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10,9,9,9,9,10,10,9,9,9,10,10,10,10,10,10,10,9,11,10,10,10,10,10,11,10,10,10,10} as the earliest occurrences of the respective denominators (though mathematically speaking that initial "1" is a glitch and should really be a "0"). I can easily generate more terms if anyone wants me to, or you can do it yourself with my code: FLengthen[L_] := Module[{i, M}, M = Table[0, {2 Length[L]}]; M[[1]] = Numerator[L[[1]]]/(1 + Denominator[L[[1]]]); For[i = 1, i < Length[L], i++, M[[2 i]] = L[[i]]; M[[2 i + 1]] = (Numerator[L[[i]]] + Numerator[L[[i + 1]]])/(Denominator[L[[i]]] + Denominator[L[[i + 1]]])]; M[[2 Length[L]]] = L[[Length[L]]]; Return[M]] F[n_] := F[n] = If[n == 0, {1}, FLengthen[F[n - 1]]] FEarliest[k_] := Module[{i}, For[i = 0, Length[Denom[F[i], k]] == 0, i++]; Return[i]] To reconcile this with Allan's numbers, note that I call {0/0,1/1} the 0th row, {0/1,1/2,1/1} the 1st row, etc. My rationale for this choice is that the nth row has (2^n)+1 elements under my convention. Jim