Collatz conjecture but with $\ 3n-1\ $ instead of $\ 3n+1.\ $ Do any sequences go off to $\ +\infty\ $?
$$$$Background (not necessary to answer my question):
Considering the following operation on an arbitrary positive integer:
- If the number is even, divide it by two.
- If the number is odd, triple it and add one.
The Collatz conjecture is: This process will eventually reach the number $1$, regardless of which positive integer is chosen initially.
If the Collatz conjecture is false, then either there will be cycles that don't contain the number $\ 1,\ $ or there will be a (at least one) sequence that goes off to $\ +\infty.$
My question:
Considering the following operation on an arbitrary positive integer:
- If the number is even, divide it by two.
- If the number is odd, triple it and take away one.
An analogue to the Collatz conjecture with these rules fails, because $\ 5\to 14\to 7\to 20\to 10\to\ 5\ $ is a cycle that does not contain $\ 1.\ $ In fact, there are lots of cycles that don't contain $\ 1\ $ that I found with the Python code below.
My question is do any sequences with this $\ 3n-1\ $ rule go off to $\ +\infty,\ $ or not?
It seems "less likely" than the likelihood Collatz sequences will go off to $\ +\infty,\ $ but proving such a thing seems hard.
Edit: I have checked all numbers up to $\ 5000\ $ using the code below and every sequence either goes to $\ 1\ $ or is in a loop. Also, there are no really long sequences (relative to number size) as opposed to some small starting numbers in the Collatz conjecture, like $\ n=27,\ $ which has $\ 111\ $ steps. This seems to suggest that no sequence goes off to infinity, and there should be some (relatively simple?) number theory proof for this.
$$$$
def collatz2(n): if n % 2 == 0: return int(n/2) else: return 3*n-1
def collatz_sequence2(n): sequence = [n] while n != 1: n = collatz2(n) sequence += [n] if n in sequence[:-1]: print(sequence[0], "is in a loop not containing 1:",) break return sequence
for i in range(1,100): print(i, ':', collatz_sequence2(i))
