# Collatz Conjecture x+1 and 3x+3

First Question

$\bullet$ For the Collatz Conjecture as $ax+b$, when $a=1$ and $b=1$, is it a safe assumption to say that this variant of the conjecture will always reach $1$, since dividing by $2$ will always lower the number further than it will rise?

Second Question

$\bullet$ Has anyone done any heavy work on the $3x+3$ variant, where all numbers will eventually reach the loop at 3 and 1?

• Since you never divide by $3$ this is a factor of all late elements (once you have multiplied by $3$ once, the factor never disappears). The behaviour of $x$ can then be discerned from the behaviour of $y$ with $x=3y$. – Mark Bennet Jun 19 '18 at 15:49
• @MarkBennet It is true that you will keep the factor $3$ but it doesn't mean that $y$ and $3y$ necessarily lie in distinct orbits: $9\to 30\to 15\to 48\to 24\to 12\to 6\to 3$. This is because when you do $+3$, you keep a multiple of $3$ but you take the $3$-valuation down to $1$. – Arnaud Mortier Jun 19 '18 at 16:01
• You might be interested in the short discussion with a slightly more generalization go.helms-net.de/math/collatz/Collatz_3x_r.pdf – Gottfried Helms Jun 19 '18 at 16:04

You are right about the $$x+1$$ variant. Since the add-$$1$$ process will always be followed by the divide-by-$$2$$ process—$$x+1$$ is even when $$x$$ is odd—we may formulate the rule as $$x\mapsto\begin{cases}(x+1)/2 & \text{if x is odd}\\ x/2 & \text{if x is even}\end{cases}$$ which is equivalent to the rule $$x\mapsto\lceil x/2\rceil$$. Since $$\lceil x/2\rceil for all $$x>1$$, the sequence of iterates goes to $$1$$ in a straightforward way, essentially by repeated halving.
The $$3x+3$$ variant is connected with the usual $$3x+1$$ problem to such an extent that, in contrast to, say, the $$3x+5$$ variant, you won't learn much new by studying it. To see this, compare the action of the $$3x+1$$ process on the starting value $$n$$ with the action of the $$3x+3$$ process on the starting value $$3n$$. If $$n$$ is even, then so is $$3n$$, so in both processes halving will occur. If $$n$$ is odd, then so is $$3n$$. Applying the $$3x+3$$ process to $$3n$$ yields $$9n+3=3(3n+1)$$, which is precisely $$3$$ times the result of applying the $$3x+1$$ process to $$n$$. The result is that each of the iterates of the $$3x+3$$ process applied to $$3n$$ will be $$3$$ times the corresponding iterate of the $$3x+1$$ process applied to $$n$$. An example: \begin{aligned} (3x+1): & 26\mapsto13\mapsto\ \ 40\mapsto20\mapsto10\mapsto\ \ 5\mapsto16\mapsto\ \ 8\mapsto\ \ 4\mapsto2\mapsto1\mapsto\ \ 4\mapsto2\mapsto\ldots\\ (3x+3): & 78\mapsto39\mapsto120\mapsto60\mapsto30\mapsto15\mapsto48\mapsto24\mapsto12\mapsto6\mapsto3\mapsto12\mapsto6\mapsto\ldots \end{aligned} Now combine this with the observation that, regardless of starting value, the $$3x+3$$ process will eventually produce an iterate that is a multiple of $$3$$. This will, in fact, happen the first time the $$3x+3$$ operation is applied, that is, immediately following the first odd iterate. In summary, the $$3x+3$$ process quickly and inevitably yields a multiple of $$3$$, after which the iterates of the process are the same of those of the $$3x+1$$ process scaled by a factor of $$3$$.