A087256 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^n.
1, 1, 1, 6, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 13, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1, 3, 1, 11, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 21, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 78, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1
Offset: 1
Keywords
Examples
n = 10: 2^10 = 1024 = peak for trajectories started with initial value taken from the list: {151, 201, 227, 302, 341, 402, 454, 604, 682, 804, 908, 1024}; a trajectory with peak=1024: {201, 604, 302, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1}
Links
- Hartmut F. W. Hoft, proof of quasi period 6
Programs
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Mathematica
c[x_]:=c[x]=(1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1);c[1]=1; fpl[x_]:=FixedPointList[c, x]; {$RecursionLimit=1000;m=0}; Table[Print[{xm-1, m}];m=0; Do[If[Equal[Max[fpl[n]], 2^xm], m=m+1], {n, 1, 2^xm}], {xm, 1, 30}]
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PARI
f(n, m) = 1 + if(2*n <= m, f(2*n, m), 0) + if (n%6 == 4, f(n\3, m), 0); a(n) = f(2^n, 2^n); \\ David Wasserman, Apr 18 2005
Formula
a(6n+4) = A105730(n). - David Wasserman, Apr 18 2005
Extensions
Terms a(19)-a(21) from John W. Layman, Jun 09 2004
More terms from David Wasserman, Apr 18 2005
Comments