# Are there any false variants of the Collatz conjecture for which the probability heuristic works?

One of the supporting arguments for the Collatz conjecture is the probability heuristic, which states roughly that because the collatz operations tends to decrease numbers over time, it probably doesn't diverge.

Are there examples of where this isn't true, i.e. is there a variant of the collatz conjecture for which the probability heuristic holds, but not all numbers converge to a cycle? (Preferably, the set of numbers that diverge should be a non-null set.)

• How about $$f(n)= \begin{cases} n+3&\text{if}\;3{\,\mid\,}n\\ 1&\text{otherwise}\\ \end{cases}$$ – quasi Feb 22 '18 at 1:49
• The function which multiplies powers of two by 2 and all other numbers to 1 works, no? That heuristic is way too naive, methinks. – Mariano Suárez-Álvarez Feb 22 '18 at 3:05
• @MarianoSuárez-Álvarez then why do people believe in the collatz conjecture? – PyRulez Feb 22 '18 at 3:08
• Surely not because of that heuristic! I honestly do not know why people believe in the conjecture, or even if they do, but there may well be many serious reasons to believe in it that lose no value because of this naive heuristic. – Mariano Suárez-Álvarez Feb 22 '18 at 3:11
• There is a set of numbers which under Collatz operations give a power of 2. $s=[1, 5, 21, 85, . . . 4n+1]$ the sequence is infinite and all numbers lead to one of these numbers; some numbers may lead to huge numbers of this set under Collatz operations. So tending to small numbers is certain . – sirous Feb 22 '18 at 3:41

Let $A$ be any infinite subset of $\mathbb{N}{\,\setminus}\{1\}$, with positive density less than $1/2$.
For $a \in A$, let $s(a)$ be the least element of $A$ which is greater than $a$.
Define $f:\mathbb{N}\to \mathbb{N}$ by $$f(n)= \begin{cases} s(n)&\text{if}\;n\in A\\[4pt] 1&\text{otherwise}\\ \end{cases}$$ Then probabilistically, every iteration should cycle, but clearly, the iterations which start with an element of $A$, approach infinity.
• @Robert Frost: Would the following variation satisfy your concept of "decreases on average"?  Let $A$ be any infinite subset of $\mathbb{N}\setminus\{1\}$, with positive density less than $1/2$. For $a \in A$, let $s(a)$ be the least element of $A$ which is greater than $a$. Define $f:\mathbb{N}\to \mathbb{N}$ by $$f(n)= \begin{cases} s(n)&\text{if}\;n\in A\\ 1&\text{otherwise}\\ \end{cases}$$ – quasi Mar 8 '18 at 3:08