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A279909 Number of steps to reach 1 or a cycle in the Collatz-like problem '3x/2' and '(x-1)/2'.

%I A279909 #23 Jun 16 2025 09:12:05
%S A279909 0,2,1,3,3,3,2,8,3,4,4,7,4,6,3,11,9,6,4,15,5,12,5,11,8,6,5,7,7,14,4,
%T A279909 18,11,10,10,11,7,9,5,11,16,6,6,15,13,12,6,17,12,9,9,11,7,11,6,19,8,8,
%U A279909 8,11,15,14,5,24,19,14,11,13,11,13,11,16,12,8,8,10,10,10,6,16,11,17,17,18,7,26,7,24,16,11,14,13,13,15,7,23,18,14,13,23
%N A279909 Number of steps to reach 1 or a cycle in the Collatz-like problem '3x/2' and '(x-1)/2'.
%C A279909 This Collatz-like problem is as follows: start with any number n. If n is even, divide it by 2 and multiply by 3, otherwise subtract 1 and divide it by 2.
%C A279909 The iteration always reach {1} or the cycles {4, 6, 9} and {16 , 24 , 36 , 54 , 81 , 40 , 60 , 90 , 135 , 67 , 33}.
%H A279909 Luca Petrone, <a href="/A279909/b279909.txt">Table of n, a(n) for n = 1..19999</a>
%H A279909 Luca Petrone, <a href="/A279909/a279909.pdf">Log Plot for the first million terms</a>
%o A279909 (Python)
%o A279909 def a(n):
%o A279909     if n==1: return 0
%o A279909     l=[n]
%o A279909     while True:
%o A279909         if n%2==0: n=(n//2)*3
%o A279909         else: n = (n - 1)//2
%o A279909         if not n in l:
%o A279909             l+=[n]
%o A279909             if n<2: break
%o A279909         else: break
%o A279909     if l[-1]==1: return len(l)-1
%o A279909     return len(l)
%o A279909 for n in range(1, 20001):
%o A279909     print(n, a(n)) # _Indranil Ghosh_, Apr 13 2017
%Y A279909 Cf. A006577.
%K A279909 nonn,easy
%O A279909 1,2
%A A279909 _Luca Petrone_, Apr 11 2017