A078420 Numbers n such that h(n) = 3 h(n-1) where h(n) is the length of the sequence {n, f(n), f(f(n)), ...., 1} in the Collatz (or 3x + 1) problem. (The earliest "1" is meant.)
105, 548, 683, 1508, 3652, 4278, 4295, 8145, 8150, 9417, 9419, 18247, 18287, 18370, 18433, 18586, 18695, 18706, 18742, 18945, 22024, 22140, 22311, 22324, 22708, 22714, 25336, 25681, 25771, 25777, 25785, 25814, 44545, 44593, 46505, 46847
Offset: 1
Keywords
Examples
n, f(n), f(f(n)), ...., 1 for n = 105, 104, respectively, are: 105, 316, 158, 79, 238, 119, 358, 179, 538, 269, 808, 404, 202, 101, 304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1; 104, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, of lengths 39 = 3 x 13 and 13, respectively. Hence 105 belongs to the sequence.
Programs
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Mathematica
f[n_] := If[EvenQ[n], n/2, 3n+1]; h[n_] := Module[{a, i}, i=n; a=1; While[i>1, a++; i=f[i]]; a]; Select[Range[2, 47000], 3h[ #-1]==h[ # ]&] Flatten[Position[Partition[Table[Length[NestWhileList[If[EvenQ[#],#/2, 3#+1]&, n, #>1&]],{n,50000}],2,1],?(3#[[1]]==#[[2]]&),1,Heads-> False]]+1 (* _Harvey P. Dale, Apr 07 2018 *)
Extensions
Extended by Robert G. Wilson v, Dec 30 2002
Comments