Is there a maximum number of consecutive decreasing steps a Collatz cycle can have?

Question

If we take a look into the (known) cycles of the Collatz Conjecture when all integers are included, we get 4 cycles:

1 → 4 → 2 → 1 …

−1 → −2 → −1 …

−5 → −14 → −7 → −20 → −10 → −5 …

−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …

The 1 and -5 cycles have at most 2 consecutive decreasing steps (n/2), the -1 cycle has just 1, and the -17 cycle has 4, which brings me to the question:

If there was any other cycle, would there be a maximum of consecutive decreasing steps it can have? And if so, would this maximum be different between a negative and a positive cycle?

Also, do we know for example why the -17 cycle reaches a point with 4 consecutive decreasing steps? As in, do we know why any given number reaches for example 4 consecutive decreasing steps?

And lastly, would having a maximum number of consecutive decreasing steps higher than 2 for a positive cycle prove the existence of another cycle different than the trivial 4:2:1 ?

Sorry for the amount of followup questions.

Answer 1

The number of decreasing steps is just the power of $2$ in the first even integer in the run. Note that $272=2^4\cdot 7$, so we can divide by $2$ four times. I don't know why there is a multiple of $16$ in that cycle and not the others.

Yes, if we found a positive cycle with more than two decreasing steps it would have to be a different cycle than $4,2,1$. You can have many more decreasing steps before you reach a cycle. To have $k$ steps at the start take a number that is a multiple of $2^k$.

Answer 2

The answer is, that the maximum consecutive decreasing steps is depending on the length of a cycle - if one exists at all.

A bit of formalism before: let's write one transformation from odd to odd as $$ b = {3a+1\over 2^A } \tag 1$$ then, if continued, $c = {3b+1\over 2^B }$. One more, $d = {3c+1\over 2^C }$. One could write the whole concatenated trajectory as $$ d=\mathcal C(a;[A,B,C]) \tag 2$$ where we use small numbers $a,b,c,..$ for odd numbers to be transformed and capital letters $A,B,C,...$ for the exponents. Here, moreover $N$ should be reserved for indication of length of the trajectory(number-of-odd-steps) and $S$ for the sum-of-exponents.

Then we can introduce even longer trajectories, using indexes, $$ a_N=\mathcal C(a_0;[A_0,A_1,A_2,...,A_{N-1}])\qquad \text{with} \quad S=\sum_{k=0}^{N-1}A_k \tag 3$$ If this should be a cycle, we had $a_0=a_N$ and to see the full formula for this we could expand that short formula for $\mathcal C()$.

Now, only if some $A_k=1$ we could have an increasing step there, and only if $A_j>1$ we have a decreasing step (let the exception $A_k=2$ and $a_k=1$ aside as the "trivial cycle"). And because we want to have a cycle, the increasing steps and the decreasing steps must somehow be balanced, so if we have, say, $4$ increasing steps $e=\mathcal (a,[1,1,1,1]) \approx 1.5^4 \cdot a \implies e \approx 5.06 \cdot a$ and we can have at best $a = \mathcal C(e,[4])$ which means $a ={ 3 e+1\over 16} \approx {15 a+1 \over 16} \approx a $.

So if you allow a $N=5$-step transformation, then the maximum number of consecutive division-by-$2$-substeps is $4$ and I think it is obvious how this generalizes to longer transformations.

You ask here for the maximum possible exponent $A_N$ assumed some $N$-step cycle (don't forget: "step" means here from odd-to-odd number) and this is always of the form $$ a_N= a_0 = \mathcal C ( a_0,[1,1,1,1,...,1,A_{N-1}) \tag 4$$ and the required $A_{N-1}$ can thus be estimated by the overall length of the assumed cycle.

But note, that for this special type of cycle it has been proven that no such can exist at all (except $a_0=1$ or $a_0=-1$ (1-step-cycles) or $a_0=-5$ (2-step-cycle)) (by R. Steiner, 1986) and even that cycles with more descendents interspersed between the ascendents (which smoothes the up-and-downs of the numbers too) up to $75$ subtrajectories cannot exist at all (irrespective of length) (besides the single cycle beginning at $a_0=-17$). (This is by John Simons & Benne de Weger in 2000 and later)

However, it is unknown whether for a 100-billion-step with more than, say, 100 up-and-downs a cycle-solution exists. With such a structure (and even more steps) you can configure any size for one $A_{N-1}$ (just increase the number of increasing steps having exponents $A_k=1$) until formal approximate fit. Then see, whether some integer $a_0$ fits that whole cycle-transformation-formula exactly...

(For references to Steiner/Simons see external links in the wikipedia-article)

Answer 3

So first off. Whether there is a maximum number of decreasing steps is an unknown problem. The issue there is that we don't even know whether the number of cycles is finite or infinite. However, even if there are an infinite number of cycles there will still even then potentially be an upper bound. However, such an upper bound has not been proven.

Of course given any particular cycle there would be an upper bound to consecutive decreasing steps within that cycle. For instance the $1 \to 2 \to 4 \to 1$ has a local maximum decrease of $2$. However, this is a trivial observation due to a cycle requiring a finite number of elements.

Now you also ask if proving an upper bound other than $2$ for positive sequences forces there to be a sequence with more than $2$ consecutive decreases. No, of course not. If you prove the least upper bound is that, then that would force there to be another sequence because it being the least upper bound of an integer set means that some cycle has to have that number of decreasing elements. However, if you have any set of numbers with a maximum $N$ then $N+1$ is an upper bound along with all numbers $N+M$ where $M$ is a non-negative real number. So no, I can say with certainty that if there are no other positive cycles then $100$ is an upper bound on the size of the cycles and the greatest number of decreasing sequences.

Now you also ask why the $-17$ contains $4$ decreasing elements. The short answer is that there is a number divisible by $16$. The longer more complicated answer would be to analyze the numbers in base $2$ and observe visually the patters that occur at each element in the cycle to see why it ends up returning to $-17*16$. However, that I have no clear reasoning on why other than that it simply does. Most likely it has more decreasing elements because it is a larger initial value. That's my hypothesis.