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.)

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

Author

Joseph L. Pe, Dec 29 2002

Keywords

• nonn

Comments

Recall that f(n) = n/2 if n is even; = 3n + 1 if n is odd.

Examples

			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.
		

Crossrefs

Cf. A078418, A078420.

Programs

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, 18800], 2h[ #-1]==h[ # ]&]

Extensions

Extended by Robert G. Wilson v, Dec 30 2002