# Collatz-like : $(5 x \pm 1)/3$ [closed]

For a motivating start, recall the definition of the usual Collatz iteration on the integers (binding exactly one halving with each tripling), rewritten in a (slightly) unusual way.

The Collatz iteration rule : For all integer $x$, let

$$T(x) = \left\{ \begin{array}{cc} \frac{x}{2} & \text{ if } x \text{ is even}\\ \lceil \frac{3x}{2} \rceil & \text{otherwise}\end{array}\right.$$

Here $\lceil \text{ } \rceil$ is the function which Rounds to the next integer UP (towards +infinity).

Comment 1 : since we (Collatz et al.) would rather T be a definite function, we had to choose how to 'round' half-integers towards integers. The choice made was to round quantities up, as expressed by the plus sign in the usual formulation $C(x) = 3x+1$ .

There had to be made an (arbitrary, somewhat) choice of a rounding rule because the divisor in Collatz formulae happened to be $2$.

Comment 2 : the ($x/2$ or $3x/2$) rule of Collatz is appealing, in part, because heuristically the "multiplicative coefficient" when going from a random largeish x to its successor is approx. $\sqrt{\frac{3}{4}}$ which is less than one, but not so much so, whence - intuitively at least - individual sequences starting at any positive x 'should be' eventually attracted to $1$, or to some other finite cycle (which as we know is only conjectured not to exist), rather than "escape" to $\infty$.

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After these possibly boring preliminaries (for which I apologise) and notations, introducing a new Collatz-like iteration ('new'as in: I don't remember seeing it explicitly in Lagarias's summaries, etc. Not claiming priority). I'll use the same style as above.

Our iteration rule : For all integer $x$,

$$f(x) = \left\{ \begin{array}{cc} \frac{x}{3} & \text{if } 3 \text{ divides }x\text{ evenly} \\ \mbox{Round} \frac{5x}{3} & \text{Otherwise} \end{array}\right.$$

Please observe the definition of $f$ is closely modelled upon that of T, but with Collatz multiplicator $5$ (instead of $3$), divisor $3$ (rather than $2$) and rounding rule to nearest (instead of up), which is well defined because the quantity to be cast to an integer now can't fall half-way between $2$ integers.

Referring to Comment 2, a similar argument to the $\frac{3x+1}{2}$ case implies for the new function a multiplicative coefficient of: $\sqrt[3]{\frac{25}{27}}$, which is "good" - for fun, at least - since it is both $<1$, but significantly nearer "escape velocity" one than the original Collatz one.

Obviously an equivalent presentation of this iteration rule is

$$f: 3x \mapsto x; 3x + 1 \mapsto 5x + 2; 3x - 1 \mapsto 5x - 2$$

Note $f$ is an odd (antisymetric) function, so unlike Collatz, it suffices to explore $f$ on the positive domain.

In one's head or with paper n' pencil one immediately finds low hanging cycles: $(1,2,3)$ and $(4,7,12)$. Are these the only cycles ?

Starting from $x = 5$, though, falls back to $1$ after a surprisingly long excursion. Not so surprisingly maybe, considering the value of the multiplicative coefficient for this iteration rule which is numerically close to $0.975$ (vs $0.866$ for Collatz $\frac{3x + 1}{2}$) . We may expect numbers to take a larger number of iterates before falling into one of the cycles (known or unknown), and we may fear, or hope, to see the iterates for initial value escape to infinity more readily than in the Collatz case, unless some fundamental number-theoretic reason be found which explains why escape can't happen !

Questions : Most questions applying to the classical Collatz problem make sense with this iterative process as well, mutatis mutandis. Primarily, Are cycles in finite number ? Are the $2$ trivial cycles the only ones ? Can iterates, from some initial values, grow unbounded ?

Have some of you considered, read or heard of others considering this particular Collatz-like puzzle? Do you think this to be worth further investigation, by machine and man ? Can its study help to isolate what's similar, and what's different here than in the standard problem ? Does the use of Round-to-nearest make the analysis more or less difficult (at 1st sight, getting closed formulas for iterates does seem more difficult here) ? And does this very technicality change the dynamics of the iteration ?

Additionally one can be tempted to change the parameters (the multiplier $5$ and divisor $3$) further in the $f$ formula, while keeping the overall coefficient near to but below $1$. Or even, maybe, just a little above $1$ ! Would that be over-generalisation in your opinion ?

I think I'm being overly talkative and so I'd better stop questioning for now .

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## closed as not a real question by Pedro Tamaroff♦, Amzoti, Chris Godsil, Lord_Farin, vadim123Jun 5 '13 at 14:01

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

Always start a long question with the question, and then detail the motivation and what you've tried, etc. Don't make people wade through to find the question, only to find they can't help you. that wastes people's time. –  Thomas Andrews Jun 5 '13 at 13:38
I don't know whether this exact question has been studied, but this kind of question certainly has; you have two or more linear functions, and the one you apply depends on the residue of the argument to some particular modulus. All such problems seem as hard as the original (except when they are trivial), and so far they have not shed much light on the original. –  Gerry Myerson Jun 5 '13 at 13:39
It is known that there are questions of this type that are undecidable --- see Conway's paper on unpredictable iterations. –  Gerry Myerson Jun 5 '13 at 13:40
@Eric, is Collatz the simplest iteration of its kind (for analysis of the dynamics)? Say that the iteration included a case $x/5$ if $x$ is divisible by $5$, would this make it easier to show that it is driven toward $1$, or complicate things more? The OP is suggesting that the "multiplicative constant" is a measure of difficulty. –  zyx Jun 9 '13 at 21:37
The book is called The Ultimate Challenge: The $3x+1$ Problem, published by the American Math Society. Some of the papers may be tough going, but others should be quite accessible. –  Gerry Myerson Jul 9 '13 at 0:21