Collatz conjecture but with $\ 3n-1\ $ instead of $\ 3n+1.\ $ Do any sequences go off to $\ +\infty\ $?

Question

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))

Answer 1

The sequence you mentioned is actually also a solution for the collatz problem itself, namely for the negative integer -5.

Because 3n+1 is the same as the absolute value of 3n-1 for negative numbers.

So the question remains unanswered. If you found "lots" of answers that would be interesting, since I am only aware of 5 total sequences being found in the collatz conjecture, namely 1, 0, -1, -5, -17.