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A232462 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006666(n), the initial number n is not counted.

%I A232462 #29 Oct 03 2024 15:13:53
%S A232462 0,1,4,2,3,0,6,5,8,4,5,7,7,8,8,3,9,9,9,1,1,6,6,6,4,6,7,8,8,8,11,4,10,
%T A232462 5,5,9,9,9,7,7,7,7,2,8,8,8,11,9,10,10,10,9,9,12,12,9,7,9,7,9,9,7,7,1,
%U A232462 10,10,10,6,6,6,11,4,0,4,6,4,4,7,7,6,4,7,7,6,6,2,2,8,2,8,8,8,8,11,11,5,7,10,10,10,10,10,10,5,7,5,2,5,5,5,9,9,5,7,7,9,9,7,7,9
%N A232462 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006666(n), the initial number n is not counted.
%C A232462 The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. As multiplying by 3 and adding 1 always gives an even number, the next step is always a division by 2. So there already exists a shortened version of Collatz problem: If n is even, multiply it by three, add one and divide the result by two, else divide it by two. A006666(n), which is the Number of halving steps to reach 1 in '3x+1' problem, also is the number of all steps for the shortened form to reach 1. If you map this function onto itself, you'll get a sequence that ends with 1 or 6 (for 6, 20, 21, 43, 64, 86, ...) or 73 (for 73, 216, 218, ...).
%H A232462 Christoph Neubauer, <a href="/A232462/b232462.txt">Table of n, a(n) for n = 1..10000</a>
%e A232462 For n = 3, the mapping of the shortened Collatz sequence is [3, 5, 4, 2, 1], its number of steps is 4. For n=6 the sequence is [6], the number of steps is 0.
%o A232462 (Python)
%o A232462 #!/usr/bin/env python
%o A232462 # output to
%o A232462 outputStr = "b232462.txt"
%o A232462 # max index
%o A232462 maxIndex = 10000
%o A232462 # goodies
%o A232462 sep=" "
%o A232462 eol="\n"
%o A232462 # subroutine for collatz sequence; classical and shortened
%o A232462 def collatz (n, k):
%o A232462     # put starting number to list
%o A232462     list = [n]
%o A232462     # collatz assumption used as end criterion
%o A232462     while (n > 1):
%o A232462         # collatz formula
%o A232462         if (n%2 == 0):
%o A232462             n = n // 2
%o A232462         else:
%o A232462             n = (3*n + 1) // k  # if k==2 --> shortened
%o A232462         # put new number to list
%o A232462         list.append(n);
%o A232462     # return complete list
%o A232462     return list
%o A232462 # subroutine for collatz composition; classical and shortened
%o A232462 # composition:  collatz (collatz (collatz (...)))
%o A232462 def composition (n, k):
%o A232462     # put starting number to list
%o A232462     list = [n]
%o A232462     # calculate collatz(starting number)
%o A232462     l = len (collatz(n, k))-1
%o A232462     # while we do not have a cycle
%o A232462     while ((l >= 1) and (not l in list)):
%o A232462         # put new number to list
%o A232462         list.append(l)
%o A232462         # get next collatz number
%o A232462         l = len (collatz(l, k))-1
%o A232462     # return complete list
%o A232462     return list
%o A232462 # open output
%o A232462 output = open (outputStr, 'w')
%o A232462 # for index 1 to maxIndex
%o A232462 for i in range (1, maxIndex+1):
%o A232462     # compute complete composition sequence
%o A232462     list = composition(i, 2)
%o A232462     # get number of steps
%o A232462     l = len(list)-1
%o A232462     # write to file
%o A232462     output.write (str(i)+sep+str(l)+eol)
%o A232462 # close output
%o A232462 output.close ()
%K A232462 nonn
%O A232462 1,3
%A A232462 _Christoph Neubauer_, Nov 24 2013