A078419
Numbers n such that h(n) = 2 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.)
Original entry on oeis.org
2, 5, 22, 495, 559, 2972, 3092, 3124, 3147, 3153, 3184, 3367, 3711, 3748, 3857, 3921, 3982, 4450, 4767, 17019, 17708, 17769, 17771, 17782, 17796, 17825, 17835, 17857, 17863, 17892, 18079, 18082, 18139, 18298, 18422, 18580, 18644, 18688, 18784
Offset: 1
n, f(n), f(f(n)), ...., 1 for n = 22, 21, respectively, are: 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1; 21, 64, 32, 16, 8, 4, 2, 1. Hence h(22) = 16 = 2 * 8 = h(21) and 22 belongs to the sequence.
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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, 18800], 2h[ #-1]==h[ # ]&]
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.)
Original entry on oeis.org
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
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.
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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 *)
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