What's The Minimum Number Of Prime Factors Needed To Replace "3x+1" With Any Linear ("mx+b") Function And Still Work Like The Collatz Conjecture?

Question

Apologies; I know there are a few assumptions used to pose this question, namely:

1): That yes, any mx+b function can work like the infamous "3x+1," problem...

• ...Provided, that you give it enough prime factors to whittle it down.
• (And that neither "m" or "b" aren't both even numbers, since the output of such a process would be "trivial?")

And:

2): That there are multiple assortments of prime factors that work with each mx+b function...

I made those assumptions based on what I've discovered, while looking into the possibility that there might be other (2n+1)x+1 functions that behave like 3x+1. I'm not surprised that there have been others¹ asking about them, but it is hard for me to accept that I'm the only one who's wondered about adding MORE "rules" into the operation...

It's rather straightforward to get 5x+1 and 7x+1 to behave like 3x+1—just throw in all the prime factors fewer than the coefficient—That I find it incredible if it's been genuinely overlooked all this time:

$$ \_{2}(x) = \begin{cases} \ 5x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$ $$ \_{3}(x) \begin{cases} 7x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} \ $$

...But that doesn't happen again until "19x+1," which is the largest odd number "m" that I tested:

$$ \_{9}(x) \begin{cases} 19x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{17} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{13} & \text{ if } x ≡ 0\ (mod\ 13), \\\ \frac{x}{11} & \text{ if } x ≡ 0\ (mod\ 11), \\\ \frac{x}{7} & \text{ if } x ≡ 0\ (mod\ 7), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

All the other odd numbers in-between (9, 11, 13, 15, 17—which almost does the same thing but overshoots) required finding seemingly random prime numbers that miraculously "come in clutch" to converge any starting (positive, whole-number) X to 4,2,1, or at least to 1, every time. The complete list of the first ten -(functions/systems?/sets?/series?/sequence?/sequences?)², therefore, appears to be:

$$ \_{0}(x) = \begin{cases} x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{1}(x) = \begin{cases} 3x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{2}(x) = \begin{cases} 5x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{3}(x) = \begin{cases} 7x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{4}(x) = \begin{cases} 9x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{79} & \text{ if } x ≡ 0\ (mod\ 73), \\\ \frac{x}{19} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{5}(x) = \begin{cases} 11x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{17} & \text{ if } x ≡ 0\ (mod\ 17), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{6}(x) = \begin{cases} 13x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{173} & \text{ if } x ≡ 0\ (mod\ 73), \\\ \frac{x}{19} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{11} & \text{ if } x ≡ 0\ (mod\ 11), \\\ \frac{x}{7} & \text{ if } x ≡ 0\ (mod\ 7), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{7}(x) = \begin{cases} 15x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{101} & \text{ if } x ≡ 0\ (mod\ 101), \\\ \frac{x}{47} & \text{ if } x ≡ 0\ (mod\ 47), \\\ \frac{x}{29} & \text{ if } x ≡ 0\ (mod\ 29), \\\ \frac{x}{19} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{11} & \text{ if } x ≡ 0\ (mod\ 11), \\\ \frac{x}{7} & \text{ if } x ≡ 0\ (mod\ 7), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{8}(x) = \begin{cases} 17x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{19} & \text{ if } x ≡ 0\ (mod\ 73), \\\ \frac{x}{17} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{13} & \text{ if } x ≡ 0\ (mod\ 13), \\\ \frac{x}{11} & \text{ if } x ≡ 0\ (mod\ 11), \\\ \frac{x}{7} & \text{ if } x ≡ 0\ (mod\ 7), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

$$ \_{9}(x) = \begin{cases} 19x+1 & \text{ if } x ≡ 1\ (mod\ 2), \\\ \frac{x}{17} & \text{ if } x ≡ 0\ (mod\ 19), \\\ \frac{x}{13} & \text{ if } x ≡ 0\ (mod\ 13), \\\ \frac{x}{11} & \text{ if } x ≡ 0\ (mod\ 11), \\\ \frac{x}{7} & \text{ if } x ≡ 0\ (mod\ 7), \\\ \frac{x}{5} & \text{ if } x ≡ 0\ (mod\ 5), \\\ \frac{x}{3} & \text{ if } x ≡ 0\ (mod\ 3), \\\ \frac{x}{2} & \text{ if } x ≡ 0\ (mod\ 2) \end{cases} $$

Now, I only tested these on the first thousand numbers for each of these (Here's my gist of each sequence for every number from 1-1,000), and while there are some absurdly long sequences, each function held out thus far. I'm not confident I have the computing power to test for numbers above one-thousand, but I have even lower confidence that I could find a proof for any of these statements. I'm not even sure how to write this as a article/journal for publication—which is part of why I'm putting these open questions here, instead.

Going with the assumptions I'm making being true, however, I'm confident I have something more accessible for everyone (or for everyone here, at least) to help hunt down: I made the second assumption seeing as, in finding the -Miracle primes for ₅ (and maybe one or two others which I seem to have neglected to save), the first assumption held true even with the superfluous filters used, in addition to the smaller set of the ones needed. That seems to imply you can find other varying-sizes-of sets of prime numbers that all could hold true to the first assumption. Which, if so, lead me to the question I stated in the title:

What's the smallest set of "-Miracle" primes possible, for each function?

The question is for not only these (Odd)X+1 functions, but also functions like 3x+2, 5x+3, 7x+5, 9x+11, 8x+17, and any or all other mx+b functions (not accounting for m and/or b being negative integers). Granted, I'm not sure how interested anyone else is in chartering that entire forest of linear functions, but hopefully this smaller slice of mx+1 sounds worthwhile? I'm not sure that what I've listed above are the actual, smallest possible sets for a few of them. I'd love to see if someone can find smaller working sets, especially for 15 (₇), 17 (₈), and/or 19 (₉). There's all sorts of other potential open-problem goodies in here as well, like:

• Is there a way to know the size of the smallest set, or by at least how many prime factors are needed for each function, if not the prime factors in that set themselves?

• Is there in fact a pattern which predicts these "-Miracle" primes?

• Could there be a hidden ratio between the mx+b and x/n functions? Some "threshold" that defines whether or not either function will always "out-pace" the other, resulting in either all-convergence, a mix of convergence, periodic, and/or divergence, or all-periodic (not ending with 1)/all-divergent?

Such questions are all well beyond me, but like I said, I wasn't aiming to prove any of these. I'm just happy if I gain any recognition for putting this out there. For everything else, I say:

Happy hunting!

(References?/Footnotes?/Anecdotes?):

1. Some of the other questions I glimpsed:

   • Will the Collatz conjecture work for $m \cdot (n)+1$ for an odd number, where m is any odd number?
   • Collatz Conjecture x+1 and 3x+3
   • What is known about the $7x+1$ problem?

2. I chose the symbol, "_n(x)," as it's required by my browsers-font-engine-dependency-library in Linux called: "libBecauseICanDo" (</facetious>)

   • (I've been fed up with waiting on a natural opportunity in my life for an excuse to use my "libBecauseISaid.so" joke for a while, now...)
   • But seriously, I have no idea how else to categorize these...Collatz-like functions(?)...
   • My knowledge on the terminology and language of the mathematical fields related to this problem, is rather sparse, at best (my vocabulary is already rather fragmented, as-is).
   • Had I that knowledge to wield beforehand, I not only could have better understood what to call things,
   • I could have just written and submitted that maths article/journal instead of doing all this...
   • And I wasn't sure if it was any safer to reuse any other mathematical symbols, especially the Greek-alphabet-ones, given that those tend to be (prominently?/preeminently?) recognized for their already-established usage,
   • But besides all that,
   • I feel that the C-Clef is a really cool-looking symbol, and I wish there was a broader appreciation for it as a symbol, beyond the small realms of niches which utilize sheet music notation...
   • #https://en.wikipedia.org/wiki/Attention_deficit_hyperactivity_disorder, amirite?

3. ₀(x) feels more like a demonstration of some of the basics of arithmetic with whole numbers, than something that remains to be proven...

   • I mean, it's just how even and odd numbers work, right?
   • There's always an even number one away from every odd number, and
   • There's no way x+1 could ever outpace x/2 in its growth, right?
   • I guess the spanner in the works is showing how that's true for all prime numbers, isn't it?
   • Prime numbers spoil everything...