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%I #7 Feb 11 2014 19:05:33
%S 2,5,22,495,559,2972,3092,3124,3147,3153,3184,3367,3711,3748,3857,
%T 3921,3982,4450,4767,17019,17708,17769,17771,17782,17796,17825,17835,
%U 17857,17863,17892,18079,18082,18139,18298,18422,18580,18644,18688,18784
%N 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.)
%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 = 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.
%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, 18800], 2h[ #-1]==h[ # ]&]
%Y Cf. A078418, A078420.
%K nonn
%O 1,1
%A _Joseph L. Pe_, Dec 29 2002
%E Extended by _Robert G. Wilson v_, Dec 30 2002
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