%I
%S 105,548,683,1508,3652,4278,4295,8145,8150,9417,9419,18247,18287,
%T 18370,18433,18586,18695,18706,18742,18945,22024,22140,22311,22324,
%U 22708,22714,25336,25681,25771,25777,25785,25814,44545,44593,46505,46847
%N Numbers n such that h(n) = 3 h(n1) 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.)
%C Recall that f(n) = n/2 if n is even; = 3n + 1 if n is odd.
%e 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.
%t 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[ # ]&]
%t 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 *)
%Y Cf. A078418, A078419.
%K nonn
%O 1,1
%A _Joseph L. Pe_, Dec 29 2002
%E Extended by _Robert G. Wilson v_, Dec 30 2002
