# Does this iterating process always returns to 0 for positive $a_0$?

Given $$a_0$$ be an positive integer, define

$$a_{n+1} = \begin{cases} 8a_n, & \text{if a_n is odd} \\ \lfloor a_n/3\rfloor, & \text{if a_n is even} \end{cases}$$

Now form the sequence $$(a_n)_{n\in \mathbb{N}}$$ by performing this operation repeatedly.

Example: $$a_0=11,(a_n)_{n\in\mathbb{N}}=(11,88,29,232,77,616,205,1640,546,182,60,20,6,2,0,0,...)$$

1 Question: For every $$a_0,(a_n)_{n\mathbb{N}}$$ will eventually reach the number $$0$$?

The above claim is true for all $$a_0$$ up to $$10^6$$.

Keith Matthews enlarge conjecture to trajectories starting with a negative integer. Go through Keith Matthews programming to insure mapping, link

Extended Claim

Let $$a_0$$ be an positive integer and $$k$$ be any odd positive integer greater than $$1$$, define

$$a_{n+1} = \begin{cases} (k^2-1)a_n, & \text{if a_n is odd} \\ \lfloor a_n/k\rfloor, & \text{if a_n is even} \end{cases}$$

Now form the sequence $$(a_{n,k})_{n\in \mathbb{N}}$$ by performing this operation repeatedly

Example $$a_0=11,k=5,(a_{n,5})_{n\in\mathbb{N}}=(11,264,52,10,2,0,0,...)$$

2 Question: For every $$a_0$$ and odd $$k\ge 3$$ the sequence $$(a_{n,k})_{n\mathbb{N}}$$ will eventually reach the number $$0$$?

Programming for extended claim by Keith Matthews, including negative values, link

Extended Collatz conjecture

Let $$a_0$$ be an positive integer and $$k$$ be any even positive integer, define

$$a_{n+1} = \begin{cases} (k+1)a_n+1, & \text{if a_n is odd} \\ \lceil a_n/k\rceil, & \text{if a_n is even} \end{cases}$$

Now form the sequence $$(a_{n,k})_{n\in \mathbb{N}}$$ by performing this operation repeatedly

Example $$a_0=5,k=4,(a_{n,k})_{n\in\mathbb{N}}=(5,26,7,36,9,46,12,3,16,4,1,...)$$

again we can ask about, every sequence will eventually reach at the number $$1$$?

Programming for extended Collatz conjecture by Keith Matthews, including negative values, link

More on extended Collatz conjecture

First: Let $$a_0,t$$ be an positive integer and $$k$$ be any even positive integer, define

$$a_{n+1} = \begin{cases} (k^t+k^{t-1})a_n+k^{t-1}, & \text{if a_n is odd} \\ \lceil a_n/k\rceil, & \text{if a_n is even} \end{cases}$$

Now form the sequence $$(a_{n,k,t})_{n\in \mathbb{N}}$$ by performing this operation repeatedly then sequence will eventually reach at the number $$1$$

Second: Let $$a_0$$ be an positive integer and $$k$$ be any even positive integer greater than $$2$$ and $$t\in\mathbb{Z}_{\ge 2}$$, define

$$a_{n+1} = \begin{cases} (k^t+k^{t-2})a_n+k^{t-2}, & \text{if a_n is odd} \\ \lceil a_n/k\rceil, & \text{if a_n is even} \end{cases}$$

Now form the sequence $$(a_{n,k,t})_{n\in \mathbb{N}}$$ by performing this operation repeatedly then sequence will eventually reach at the number $$1$$

I again extend claim on odd $$k$$

First: Let $$a_0$$ be an positive integer and $$k$$ be any odd positive integer greater than $$1$$ and $$t\in\mathbb{Z}_{\ge 2}$$, define

$$a_{n+1} = \begin{cases} (k^t-k^{t-2})a_n, & \text{if a_n is odd} \\ \lfloor a_n/k\rfloor, & \text{if a_n is even} \end{cases}$$

Then the sequence $$(a_{n,k,t})_{n\in \mathbb{N}}$$ will eventually reach at the number $$0$$

Second: Let $$a_0$$ be an positive integer and $$k$$ be any odd positive integer greater than $$1$$ and $$t\in\mathbb{Z}_{\ge 1}$$, define

$$a_{n+1} = \begin{cases} (k^t-k^{t-1})a_n, & \text{if a_n is odd} \\ \lfloor a_n/k\rfloor, & \text{if a_n is even} \end{cases}$$

Then the sequence $$(a_{n,k,t})_{n\in \mathbb{N}}$$ will eventually reach at the number $$0$$

• This does not seem any easier than the Collatz conjecture. Dec 6 '20 at 20:39
• Without getting into exact definitions, the problem "Given a 'Collatz-like' function $g$, does $g^k(n)$ eventually equal $1$ for all $n \in \mathbb{N}$?" is undecidable, due to Kurtz & Simon. Your question is in some sense akin to asking whether a specific group presentation has decidable word problem, or whether a fixed Turing machine $T$ halts. That could be interesting (certainly it is interesting e.g. when that Turing machine encodes the Collatz problem itself) but as it currently stands, it's just a question in an infinite family of questions, with nothing to distinguish it from the rest. Dec 7 '20 at 9:01
• @Carl-Fredrik, I thought that result was due to Conway ("Unpredictable iterations"). Dec 7 '20 at 12:18
• @GerryMyerson Conway showed it is undecidable uniformly in $g$ and $n$. That is, Conway's input is a fixed Collatz-like function $g$ and a natural number $n$. Kurtz & Simon just take a function $g$ as input. Dec 7 '20 at 13:09
• @Carl-FredrikNybergBrodda To clarify, I believe you mean this question is akin to whether a fixed Turing machine halts on every input, is that right? Dec 7 '20 at 17:43