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A336914 Number of steps to reach 1 in '3^x+1' problem (a variation of the Collatz problem), or -1 if 1 is never reached.

%I A336914 #17 Jun 27 2021 03:40:41
%S A336914 0,1,4,2,11,2,9,5,7,5,7,5,5,5,16,3,5,3,5,3,16,3,14,3,9,3,14,3,9,3,9,
%T A336914 12,14,12,22,12,14,12,7,12,5,12,5,12,7,12,5,12,7,12,5,12,5,12,20,12,5,
%U A336914 12,16,12,5,12,14,3,12,3,5,3,14,3,5,3,14,3,5,3,5
%N A336914 Number of steps to reach 1 in '3^x+1' problem (a variation of the Collatz problem), or -1 if 1 is never reached.
%C A336914 The 3^x+1 map, which is a variation of the 3x+1 (Collatz) map, is defined for x >= 1 as follows: if x is odd, then map x to 3^x+1; otherwise, map x to floor(log_2(x)).
%C A336914 It seems that all 3^x+1 trajectories reach 1; this has been verified up to 10^9.
%H A336914 Wikipedia, <a href="http://en.wikipedia.org/wiki/Collatz_conjecture">Collatz conjecture</a>
%e A336914 For n = 5, a(5) = 11, because there are 11 steps from 5 to 1 in the following trajectory for 5: 5, 244, 7, 2188, 11, 177148, 17, 129140164, 26, 4, 2, 1.
%e A336914 For n = 6, a(6) = 2, because there are 2 steps from 6 to 1 in the following trajectory for 6: 6, 2, 1.
%o A336914 (Python)
%o A336914 from math import floor, log
%o A336914 def a(n):
%o A336914     if n == 1: return 0
%o A336914     count = 0
%o A336914     while True:
%o A336914         if n % 2: n = 3**n + 1
%o A336914         else: n = int(floor(log(n, 2)))
%o A336914         count += 1
%o A336914         if n == 1: break
%o A336914     return count
%o A336914 print([a(n) for n in range(1, 101)])
%Y A336914 Cf. A006370 (image of n under the 3x+1 map).
%Y A336914 Cf. A336913 (image of n under the 3^x+1 map).
%K A336914 nonn
%O A336914 1,3
%A A336914 _Robert C. Lyons_, Aug 08 2020