A105730 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^(6n+4).
6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 21, 8, 6, 78, 8, 6, 9, 13, 6, 15, 8, 6, 16, 8, 6, 9, 20, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 32, 8, 6, 9, 32, 6, 23, 8, 6, 24, 8, 6, 9, 14, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 19, 8, 6, 9, 13, 6, 80, 8, 6, 29, 8, 6, 9, 18, 6, 12, 8, 6, 13, 8, 6, 9, 11
Offset: 0
Examples
a(1) = 12, i.e. the number of initial values for 2^10, since 804 -> 402 -> 201 -> 604 -> 302 -> 151 -> 454 -> 227 -> 682 -> 341 -> 1024 and 908 -> (454 -> ... -> 1024) are the two maximal trajectories containing all 12 initial values. a(8) = 11 since 2^(6*8+4) has 11 different initial values for Collatz trajectories leading to it. - _Hartmut F. W. Hoft_, Jun 24 2016
Links
- Hartmut F. W. Hoft, identities to prove for quasi period 9
Programs
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Mathematica
trajectory[start_] := NestWhileList[If[OddQ[#], 3#+1, #/2]&, start, #!=1&] fanSize[max_] := Module[{active={max}, fan={}, current}, While[active!={}, current=First[active];active=Rest[active]; AppendTo[fan, current]; If[2*current<=max, AppendTo[active, 2*current]]; If[Mod[current, 3]==1 && OddQ[(current-1)/3] && current>4, AppendTo[active, (current-1)/3]]]; Length[fan]]/;max==Max[trajectory[max]] a105730[low_, high_] := Map[fanSize[2^(6#+4)]&, Range[low, high]] a105730[0,89] (* Hartmut F. W. Hoft, Jun 24 2016 *)
Formula
a(n) = A087256(6n+4).
Comments