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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))
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    $\begingroup$ Everyone knows about the existence of the cycle. But no one knows now that it's go to $+\infty$ or not. $\endgroup$ Commented Jun 28, 2021 at 15:47
  • $\begingroup$ @lonestudent what do you mean? Have you read the question? $\endgroup$ Commented Jun 28, 2021 at 15:48
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    $\begingroup$ Repeating cycles in the $3n-1$ problem $\endgroup$
    – mathlove
    Commented Jun 28, 2021 at 15:57
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    $\begingroup$ "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? $\endgroup$
    – rukhin
    Commented Jul 4, 2021 at 18:45
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    $\begingroup$ Imo, samerivertwice's comment is the most understandable. Do others agree with his second sentence though? $\endgroup$ Commented Mar 25, 2023 at 23:34

2 Answers 2

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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.

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    $\begingroup$ $3n+1$ is definitely not equal to the absolute value of $3n-1$ "for negative numbers". I don't know who upvoted this nonsense. $\endgroup$
    – Alex M.
    Commented Aug 22, 2022 at 15:10
  • $\begingroup$ Yeah, $3(-1) + 1\neq 3(-1) -1.$ $\endgroup$ Commented Aug 22, 2022 at 15:13
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    $\begingroup$ I believe the intended but inelegantly expressed point being derogated might be $|-3\cdot |n| -1| =3\cdot |n|+1$ $\endgroup$ Commented Aug 22, 2022 at 15:32
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    $\begingroup$ $$ -b = 3(-a)+1 <--> b=-(3(-a)+1) <--> b=3a-1 $$ would have been a better formulation of @moi's answer. No need to become heated... $\endgroup$ Commented Dec 6, 2022 at 15:21
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No. Short answer. But there are infinite loops. I am currently working on a proof of the Collatz. If you want to properly apply the collatz to the negative integers you must adjust the seed function of 2x+1 that feeds into n to 2x-1. so you would end up with 3(2x-1)-1. then you will replicate the Collatz sequences exactly in the negative integers. Why this works is another story, but the runaway infinite sequence doesn't exist. An infinite loop does with your version, but not when the process correctly negated. So there are infinite terms, but nothing that runs away to an unlimited or infinite value, just infinite steps as a recursive loop.

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