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

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))

• Everyone knows about the existence of the cycle. But no one knows now that it's go to $+\infty$ or not. Commented Jun 28, 2021 at 15:47
• @lonestudent what do you mean? Have you read the question? Commented Jun 28, 2021 at 15:48
• Repeating cycles in the $3n-1$ problem Commented Jun 28, 2021 at 15:57
• "In fact, there are lots of cycles that don't contain 1 that I found with the Python code below. " Define "lots": there is another cycle (en.wikipedia.org/wiki/…) that we are aware of; have you found another one? Commented Jul 4, 2021 at 18:45
• Imo, samerivertwice's comment is the most understandable. Do others agree with his second sentence though? Commented Mar 25, 2023 at 23:34

• $3n+1$ is definitely not equal to the absolute value of $3n-1$ "for negative numbers". I don't know who upvoted this nonsense. Commented Aug 22, 2022 at 15:10
• Yeah, $3(-1) + 1\neq 3(-1) -1.$ Commented Aug 22, 2022 at 15:13
• I believe the intended but inelegantly expressed point being derogated might be $|-3\cdot |n| -1| =3\cdot |n|+1$ Commented Aug 22, 2022 at 15:32
• $$-b = 3(-a)+1 <--> b=-(3(-a)+1) <--> b=3a-1$$ would have been a better formulation of @moi's answer. No need to become heated... Commented Dec 6, 2022 at 15:21