3n+1 problem: $f^x(n) < n$ finitely many steps

Question

Consider the shortcut function: $$f(n)=\begin{cases}(3n+1)/2\} & \text{if } (n\bmod 2)\equiv1 \\ \{n/2\} & \text{if } (n\bmod 2)\equiv0 \end{cases}$$

Define $f^x$ as the $x$'th iterate of $f$.

There is an initial number $n_0$ (i.e. a constant in the inequality equation) such that for all $n \in\mathbb{N}$ there is some $x$ such that $f^x(n) < n_0$. If someone confirmed/proved that this were true, would that mean that the sequence would never reach $\infty$, but might be open wether it would enter a non-trivial repeating cycle?

I.e. for all $n \in\mathbb{N}$ will there always exist some $x$ where $f^x$ is strictly less than $f^0$? I am just wondering if this have been proven. If so, is the proof simple? If my math-writings are wrong, please correct me. If someone proved it, would we rule out that it would enter infinity, or am my assumptions wrong?

Illustration

Set $n_0 = 27$. Start the iterate with $f(n_0)$.

We get: $27, 41, 62, 31, 47, 71, 107, 161, 242, 121, 182, 91, 137, 206, 103, 155, 233, 350, 175, 263, 395, 593, 890, 445, 668, 334, 167, 251, 377, 566, 283, 425, 638, 319, 479, 719, 1079, 1619, 2429, 3644, 1822, 911, 1367, 2051, 3077, 4616, 2308, 1154, 577, 866, 433, 650, 325, 488, 244, 122, 61, 92, 46, 23,...$

After $59$ steps we get number $23$ which is less than the initial number $27$.

So for $n_0=27$ there is some $x$ where $f^x < f^0$. In this case $f^{59} < 27$.

Answer 1

It is not known if any Collatz sequences go to infinity. It is not known if there are any cycles apart from the trivial $1\to4\to2\to1$. It is possible that no Collatz sequence goes to infinity but that some non-trivial cycle exists anyway.

To prove Collatz, you need only prove your statement. That is, if all starting seeds return a Collatz sequence that is eventually smaller than the seed, you're done. No cycle could possibly exist. (Consider the Collatz sequence beginning at the minimum of the cycle.)

On the other hand, if you show that some fixed integer exists that is greater than some element of every Collatz sequence, then no Collatz sequence diverges but a cycle might exist.

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To elaborate:

As you state, for all $x \in \mathbb{N}$, $f^x(1) \not < 1$. So the inequality does not hold for $n_0 = 1$. If it holds for $n_0 = 2$, we can be sure that the orbit starting at $2$ eventually goes to a number smaller than $2$: it must go to $1$. (Note it cannot go to zero or a negative number; these numbers are not the odd parts of any positive number.) If it holds for $n_0 = 3$, we can be sure that orbits starting at $3$ eventually go to $1$ or $2$ and all sequences that go to $1$ or $2$ go to $1$. Proceeding in this way, if the inequality holds for all $n_0 \in \mathbb{N}$ then all sequences go to $1$.

Well if $1$ is part of a cycle, why can't some other cycle exist?

By way of contradiction, lets assume your inequality is true and that some cycle exists: $$a_0 \to a_1 \to \dots \to a_{100} \to a_0$$ This Collatz orbit must have a smallest element. Call it $a_0$. Then for all elements in this cycle $a_k, k\not= 0$, the orbit starting at $a_k$ goes to $a_0 < a_k$. But element $a_0$ is the minimum element of a cycle, and thus never reaches a smaller element. Therefore either your inequality is not true for all starting seeds or there cannot be a cycle.