A177903 Consider the weighted Farey tree A177405/A177407; a(n) = row at which the denominator 2n+1 first appears (assumes first row is labeled row 0).
0, 1, 2, 2, 2, 3, 3, 4, 3, 3, 4, 4, 4, 3, 4, 4, 5, 5, 5, 5, 4, 4, 5, 4, 5, 6, 4, 4, 6, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 5, 6, 7, 6, 6, 6, 6, 6, 5, 6, 5, 6, 6, 6, 6, 5, 6, 7, 6, 6, 6, 6, 6, 6, 6, 5, 6, 5, 7, 7, 6, 6, 7, 7, 6, 7, 6, 6, 6, 5, 5, 7, 6, 6, 6, 7, 7, 7, 6, 6, 6, 7, 7, 6, 7, 7, 7, 6, 7, 7
Offset: 0
Keywords
References
- Based on postings by Richard C. Schroeppel and James Propp to the Math Fun Mailing List, Dec 15 2010.
Programs
-
Mathematica
Denom[L_, k_] := Module[{M, i}, M = {}; For[i = 1, i <= Length[L], i++, If[Denominator[L[[i]]] == k, M = Append[M, L[[i]]]]]; Return[M]] Earliest[k_] := Module[{i}, For[i = 1, Length[Denom[WF[i], k]] == 0, i++]; Return[i]] Latest[k_] := Module[{i}, For[i = 1, Length[Denom[WF[i], k]] < EulerPhi[k], i++]; Return[i]] Table[Earliest[2 n + 1], {n, 1, 100}] (* James Propp *)
Comments