# Is the number 3 in the Collatz conjecture arbitrary?

One of the most famous conjectures in mathematics is the Collatz conjecture also known as $$3n+1$$ but my question is why we multiply the odd number by 3? I get that the conjecture probably wants to multiply the odd number with other odd number and then add 1 to make it even and I can see why the conjecture avoided choosing the number 1 but is the number 3 arbitrary? What if we change that number to 5 or 7 or any odd number will the conjecture still hold or there will be some loops that don't got to 1 and divergent sequence? So is that $$3$$ arbitrary? Or there is a good reason for choosing it besides it is the smallest odd number bigger than 1?

Also Is that $$+1$$ arbitrary? What happen if we change it to be any other odd number? Will that affect the conjecture?

It seems like modifying the 3 to 5 will lead to a very slow sequence or divergence (the numbers that I tried didn't go to $$1$$ at all and I just assumed that the sequence is very slow ).

• People study variants of Collatz all the time. As you'll see if you play with them, most of the time it's easy to show that the analog conjecture is false. Of course, $3$ is at least a little special since, given that there is an equal chance that a randomly selected natural number (below some big $N$ say) goes to $\frac n2$ or to $\frac {3n+1}2$, the expected new value is just $n$ (in the big $N$ limit). If you expected it to be below $n$, you'd certainly expect the conjecture to hold. If it were significantly greater than $n$, you'd expect it to fail. Right at $n$, who knows?
– lulu
Commented May 27 at 14:06
• To be sure: hard problems take on a life of their own. Collatz is so simple, anyone can play with it. Sure looks like it should be easy to resolve. The fact that it really isn't is fascinating. That makes the problem interesting, even if it started out arbitrarily.
– lulu
Commented May 27 at 14:08
• Just a remark about the heuristic : That the sequence has a tendency to eventually decrease is an indicator that there are probably no divergent trajectories. This heuristic does not speak against nontrivial cycles. But concerning this , incredible progress has been made , and such a cycle is extremely unlikely as well. Commented May 27 at 14:08
• It might be worth to be mentioned that there is actually a generalization of the Collatz conjecture , known to be undecidable ! Commented May 27 at 14:12
• Should the Collatz conjecture be false , but nontrivial cycles are impossible (so we only have divergent sequences as counterexamples) , then we are probably screwed because how shall we be able to prove such a sequence divergent ? Commented May 27 at 14:15

The number $$3$$ may have been chosen somewhat arbitrarily, but it ends up being an excellent choice for fueling interesting dynamics. The addend $$1$$ is, in some ways, more arbitrary, and varying it leads one into richly structured territory.

First of all, what do I mean that $$3$$ is a good choice? Well, consider an arbitrary even number. It can be written as $$2^kq$$, where $$q$$ is odd, and applying the Collatz function to it will result in $$q$$, after $$k$$ steps.

Now, half of the time, $$k=1$$, so the even number is only divisible by $$2$$ once. The other half of the time, $$k>1$$, and the even number is divisible by 4, at least. (This remains true if we only consider even numbers of the form $$6n+4$$, which are the only ones that can result from $$3(\text{odd})+1$$.)

This means that, if we start with an arbitrary odd number $$q_0$$, and follow its Collatz sequence to the next odd number $$q_1$$, then we will have $$q_0 and $$q_0>q_1$$ occurring with roughly equal frequency. If we calculate the "average" ratio $$\frac{q_1}{q_0}$$, it comes out to $$\frac34$$, so there is a general tendency for numbers to decrease, under repeated iterations.

These dynamics – going up or down with equal probabilities, and tending downward in the long run – provide us with the structure that people find so fascinating. If we replace $$3$$ with $$5$$, for example, then our probabilities will be $$\frac34$$ for $$q_0, and only $$\frac14$$ for $$q_0>q_1$$, instead of half and half. Moreover, the average ratio $$\frac{q_1}{q_0}$$ will be $$\frac54$$, which is greater than $$1$$, and we'll see most trajectories growing very large. That just seems less fun.

Moving on, you also asked about the "$$+1$$". That can be changed, and doing so is actually the same as extending the domain of Collatz to negative and rational numbers (well, certain rational numbers).

Consider, the trajectory of $$19$$ under the "$$3n+5$$ rule":

$$19 \rightarrow 62 \rightarrow 31 \rightarrow 98 \rightarrow 49 \rightarrow 152 \rightarrow 76 \rightarrow 38 \rightarrow 19$$

Now, define a fraction with odd denominator as "even" or "odd", according to its numerator, and follow the number $$\frac{19}{5}$$ under the usual "$$3n+1$$ rule":

$$\frac{19}{5} \rightarrow \frac{62}{5} \rightarrow \frac{31}{5} \rightarrow \frac{98}{5} \rightarrow \frac{49}{5} \rightarrow \frac{152}{5} \rightarrow \frac{76}{5} \rightarrow \frac{38}{5} \rightarrow \frac{19}{5}$$

As you can see, these are the same thing. It's clear that this works because $$1=\frac55$$.

Similarly, applying a "$$3n-1$$ rule" is the same as applying the usual "$$3n+1$$ rule" to negative integers.

In this way, we can extend the Collatz function to all rational numbers with odd denominators (also known as "the ring of integers localized at $$2$$"). Indeed, we can extend it even further, to $$2$$-adic integers, but we don't find any additional loops there!

Among the rationals, denominators that are multiples of $$3$$ are not stable, but among the other odd denominators, each one gives us a world of trajectories and loops, with each world being different. If we choose $$5$$ as our denominator (or play the $$3n+5$$ game), we encounter five different loops, all in the positive numbers. If we choose $$7$$, we only encounter one loop, whether we start with a positive or negative input. The denominator $$11$$, like the denominator $$1$$, has loops in both the positive domain (two loops) and the negative domain (one loop).

All of these statements about how many loops occur under different conditions are conjectural, of course. No one knows how to prove that the Collatz function, acting only on integers, produces only one positive loop and three negative ones; corresponding statements for other denominators don't seem to be any more tractable.

Despite the fact that we've suddenly added infinitely many new questions and no new answers, this is an interesting line of inquiry. Seeing so many loops (at least 1596 with denominators under 1000) lets us think about their properties, and what properties we can expect for any loop that is a counterexample to the original conjecture. When the sizes of the numbers in loops appear to be bounded, we can ask why, and we have a lot more examples to work with. Observing denominators with only one loop, and denominators with many, we can ask what might cause that difference.

Of course, there's no telling whether this generalization will ultimately be helpful, or lead to anything productive, but it does provide us with a richly structured landscape to explore. In summary, changing the $$3$$ in "$$3n+1$$" is possible, but it changes the dynamics considerably. However, changing the $$1$$ doesn't alter the dynamics, so much as extend them to a larger domain, allowing us to study not one Collatz tree, but an entire Collatz forest.

• BTW does $7n+1$ diverge sometimes? if $n=7$ I wrote a python code and it is still growing after $10^8$ steps. If not how long is its loop ? In general does any $an+1$ where $n > 3$ diverge ?
– pie
Commented Jun 11 at 14:10
• It seems likely that some trajectories will diverge under 5n+1, 7n+1, etc., but I’m not aware of any proof. Cycles certainly can exist in such systems: 7(1)+1=8, for example. It’s hard to prove that a specific trajectory diverges, even when the multiplier is large. Commented Jun 11 at 15:01

Historically, I believe the $$3$$ case was first looked by Collatz in 1937. I would believe he only mentioned the $$3n+1$$ case but I can't definitively say he didn't also look at the $$5$$, $$7$$, ... cases (I tried finding the original paper but searching Collatz is... difficult to say the least). However, I can cite some recent research which certainly suggests there is a meaningful difference between the $$n=3$$ case and $$n>3$$ cases. M.C. Seigel is a researcher investigating $$(p,q)$$-adic analysis and found some interesting connections to the Collatz conjecture (and indeed all $$n>3$$ Collatz sequences). I want to stress that

$$1)$$ He is not claiming to be anywhere near a proof of Collatz. His work just shows an interesting link between the conjecture and $$(p,q)$$-adic analysis.

$$2)$$ I am not an expert in his research. As such, I'll simply present some equations and link to his most recent paper. I encourage anyone curious to at least give it a read.

The best example of the connection is a functional equation defined by

$$\chi_q(2\epsilon)=\frac{\chi_q(\epsilon)}{2}$$

$$\chi_q(2\epsilon+1)=\frac{q\chi_q(\epsilon)+1}{2}$$

for $$\epsilon$$ in the $$2$$-adics. $$\chi_q(\epsilon)$$ is originally defined as a certain map between the $$2$$-adics and the $$q$$-adics (hence the $$(2,q)$$-adic analysis). These equations are $$1.72$$ and $$1.73$$ in the paper. In further work (which he explains in a video) he shows that integrating this function (in a precise sense of what 'integrate' means in this context) gives

$$\int_{\mathbb{Z}_2} \chi_q(\epsilon)d\epsilon = \begin{cases} 0 & q = 3 \\ \frac{1}{3-q} & q\geq 5 \end{cases}$$

Hence the difference between $$n=3$$ and $$n\geq 3$$.

• The reason you can't find the original paper is that Collatz never published on this topic. Not having any real results, he didn't submit any paper on his conjecture. Commented Jun 11 at 13:43
• That would make sense. I had assumed his paper had gotten lost in the shuffle of Collatz research but I guess a toy problem without any results isn't worth much to a researcher. Commented Jun 11 at 13:50