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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?

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    $\begingroup$ 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 $x=3y$. $\endgroup$ – Mark Bennet Jun 19 '18 at 15:49
  • $\begingroup$ @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 $1$. $\endgroup$ – Arnaud Mortier Jun 19 '18 at 16:01
  • $\begingroup$ You might be interested in the short discussion with a slightly more generalization go.helms-net.de/math/collatz/Collatz_3x_r.pdf $\endgroup$ – Gottfried Helms Jun 19 '18 at 16:04
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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 $\lceil x/2\rceil<x$ 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 $3$.

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