$k$-cycles in Collatz functions

Question

I have a couple of questions, but I need to give some quote and some reasoning before I ask.

Quote from Wikipedia:

A $k$-cycle is a cycle that can be partitioned into $2k$ contiguous subsequences: $k$ increasing sequences of odd numbers alternating with $k$ decreasing sequences of even numbers. For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a $1$-cycle.

This is not the exact definition of a cycle loop. We just have to include that it is a loop. If you are not very familiar with the Collatz you might be confused by cycles and must know in which context it is used.

A cycle is defined as being just increasing odds vs decreasing evens. This is one of those things that got me confused while reading about cycles in the Collatz. But I think I finally understand it, please correct me if I am wrong. I thought a cycle was the actual repeating of the numbers in the cycle. But it is not, it needs that extra loop. In some other context some cycles can be loops, but in the Collatz context, there is only one known special case where a cycle is looping, is when it hits $1$ and repeats the process. But when I think of it in another way, this I might be wrong. Maybe odds and even $k$-cycles can happen in a large sequence, where there are several $k$'s for $k$-cycles in a single trajectory. Not taking "loop" into the context. I find the definitions that I read a bit fuzzy and I feel it might get mixed up. Or is there one formal definition, just poorly explained?

So from what I understand there are only one type of cycle, the $k$-cycle, but in addition it can be extended to a loop. And the only known loop is the trivial $4,2,1$ or $2,1$ closed loop cycle. I.e. it jumps from $1$ to $2$ or $4$. Other jumps to previous numbers are not known to exist. The above definition is not telling anything about the nature of the numbers in the subsequences.

But there's a version that is called the Syracuse (also known as the Reduced Collatz) where there are no trivial loops, the sequence jumps to $1$ immediately without the $2$- or $2,4$-subsequence parts. Then it repeats the $1$'s only.

Let me quote from Wikipedia again:

For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a $1$-cycle

No $1$-cycle exists in this reduced form (the Syracuse). So, what is it? Is it a $0$-cycle?

I make an example with the number $27$ for the Reduced function:

$27\rightarrow41\rightarrow31\rightarrow47\rightarrow71\rightarrow107\rightarrow161\rightarrow121\rightarrow91\rightarrow137\rightarrow103\rightarrow155\rightarrow233\rightarrow175\rightarrow263\rightarrow395\rightarrow593\rightarrow445\rightarrow167\rightarrow251\rightarrow377\rightarrow283\rightarrow425\rightarrow319\rightarrow479\rightarrow719\rightarrow1079\rightarrow1619\rightarrow2429\rightarrow911\rightarrow1367\rightarrow2051\rightarrow3077\rightarrow577\rightarrow433\rightarrow325\rightarrow61\rightarrow23\rightarrow35\rightarrow53\rightarrow5\rightarrow1$

I make a list when the next number in the sequence is higher than the previous I write an 'u' (for up), when the next number is lower I write a 'd' (for down):

uduuuudduduuduuudduududuuuuuduuuddddddudd

I don't know if this is the way it is interpreted in formal literature.

From the list I just made I can't seem to make any sense of this and the definition of a $k$-cycle. So my last question is, is $k$-cycles relevant in the Reduced Collatz function? But would it make sense to make such a list in the Formal- and Shortcut Collatz sequence?

Clarification: Seems some of my explanation wasn't clear enough, here's a minor modification: ud is $27\rightarrow41\rightarrow31$, and so on. Starting on $27$. So the letters represent increase or decrease. I should have used up down arrows, like this:

$27\uparrow41\downarrow31\uparrow47\uparrow71\uparrow107\uparrow161\downarrow121\downarrow91\uparrow137\downarrow103\uparrow155\uparrow233\downarrow175\uparrow263\uparrow395\uparrow593\downarrow445\downarrow167\uparrow251\uparrow377\downarrow283\uparrow425\downarrow319\uparrow479\uparrow719\uparrow1079\uparrow1619\uparrow2429\downarrow911\uparrow1367\uparrow2051\uparrow3077\downarrow577\downarrow433\downarrow325\downarrow61\downarrow23\uparrow35\uparrow53\downarrow5\downarrow1$

Additional note of why cycles definition are confusing:

Here 's a subsequence from the reduced collatz function: there are many $1$-cycles; $251,377,283,425,319...$ which is $\uparrow\downarrow\uparrow\downarrow$.. (index 19 to 23) or is there? Just as a thought experiment, we could trade the numbers into $2$'s and $1$'s. $1,2,1,2,1,..$ and if this repeated it would be a $1$-cycle according to the definition. But the numbers in the subsequence is not the short-cut map, and so there are no even's and a cycle does not exist in the definition prescribed.

Answer 1

Your ud definition does not match the idea behind the ud in the Steiner/Simons-analyses - perhaps from a correction of this you'll get some clarification.

Assume the u one step (on odd numbers) $a_{k+1}=(3a_k+1)/2$ then $a_{k+1}$ is obviously larger than $a_k$ and thus we may call it u-step because it is upwards. (Not your compressed transformation!)

Assume the d one step (on even numbers) $a_{k+1}=a_k/2$ then $a_{k+1}$ is obviously smaller than $a_k$ and thus we may call it d-step because it is downwards.

(Note there is one 1-u-step-cycle, but only in the negative integers $-1=(3 \cdot -1 +1)/2$)

The trivial cycle on $1$ is by this steps-definition ud : $1\to2$ is up and $2\to 1$ is down.

Now we might assume some general cycle with the following up-and-downs: uuududuuudud and it is easy to find numerically that there is no solution in integer elements. With assumed cycles with number $N$ odd-steps (u) with $N$ large this becomes uneasy or impossible for availabe computing power.

So there is some idea lurking around to think about looking at "easier" subsets of cycles, or types of cycles - about which we could possibly say more, and more effective.

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For instance we look first at cycles of the form uuu...u ddd...d which means one sequence of upward steps and then one sequence of downward steps and whether we can find a solution in integer numbers for the elements of this to make it a proper cycle. The formulae which occur with the analysis of this simpler problem give strong evidence that no such type of cycle exist - and indeed R.Steiner succeeded to prove that indeed fitting the numerical increase of one sequence of stepping upwards with the numerical decrease of one sequence of stepping downwards would never come back to the initial integer value: so no cycle of this form (of any length!) can exist.

Let's denote for shortness sequences of length $K$ of consecutive u by U(K) and sequences of length $L$ of consecutive d by D(L). Then the so-called "1-cycle" has the form U(K) D(L) with any $K$ and $L$ . Of course the value $L$ must approximately match that of $K$ such that we arrive at least in the near of the initial value.
If we would actually find a combination of $K$ and $L$ such that the resulting transformation U(K) D(L) would transform from some $a_1$ back to $a_1$ again then this would be called a "1-cycle". (The trivial 1-cycle is U(1) D(1) satisfied by the initial element $a_1=1$ and a "1-cycle" in the negative integers exist as U(2) D(1) being a cycle with $a_1=-5$ ).

From this the "k-cycle" concept is possibly a bit more transparent: A transformation U(K1)D(L1), U(K2)D(L2) would be a "2-cycle" if for any $K1,K2,L1,L2$ this would become a cycle when applied at some initial element $a_1$
(Again there is a solution in the negative integers: beginning at $a_1=-17$ we arrive at $a_1=-17$ again by structure U(4)D(1), U(3)D(3))

For your example with $a_1=27$ you have given the trajectory-structure as $\qquad$ uduuuudduduuduuudduududuuuuuduuuddddddudd
but this is not meant with the u's and d's according to the "k-cycle"-idea. The structure should have been described with the indicated correct interpretation of u and d as at the top of my answer which would then give
$\qquad$ uud uuuuud ud uud uuud uuuud udd uuud uud uuuuuudd uuuuddd ud ud uddd udd uuudddd uddd
and which would become
U(2)D(1), U(5)D(1), U(1)D(1), U(2)D(1), U(3)D(1), U(4)D(1), U(1)D(2), U(3)D(1), U(2)D(1), U(6)D(2), U(4)D(3), U(1)D(1), U(1)D(1), U(1)D(3), U(1)D(2), U(3)D(4), U(1)D(3).
_{Numerically this would be
$\qquad \small{ 27}$ $ \small{ \nearrow 41 \nearrow 62 \searrow 31 \\ \nearrow 47 \nearrow 71 \nearrow 107 \nearrow 161 \nearrow 242 \searrow 121\\ \nearrow 182 \searrow 91 \\ \nearrow 137 \nearrow 206 \searrow 103 \\ \nearrow 155 \nearrow 233 \nearrow 350 \searrow 175 \\ \nearrow 263 \nearrow 395 \nearrow 593 \nearrow 890 \searrow 445 \\ \nearrow 668 \searrow 334 \searrow 167 \\ \nearrow 251 \nearrow 377 \nearrow 566 \searrow 283 \\ \nearrow 425 \nearrow 638 \searrow 319 \\ \nearrow 479 \nearrow 719 \nearrow 1079 \nearrow 1619 \nearrow 2429 \nearrow 3644 \searrow 1822 \searrow 911 \\ \nearrow 1367 \nearrow 2051 \nearrow 3077 \nearrow 4616 \searrow 2308 \searrow 1154 \searrow 577 \\ \nearrow 866 \searrow 433 \\ \nearrow 650 \searrow 325 \\ \nearrow 488 \searrow 244 \searrow 122 \searrow 61 \\ \nearrow 92 \searrow 46 \searrow 23 \\ \nearrow 35 \nearrow 53 \nearrow 80 \searrow 40 \searrow 20 \searrow 10 \searrow 5\\ \nearrow 8 \searrow 4 \searrow 2 \searrow 1} $.
Note the occurence of the even numbers in that so-defined trajectory.}

Having 17 pairs of U()D() this would be a "17-cycle" if it were a cycle at all : if it were transforming from $27 \to 27$. But actually it transforms $27 \to 1$.
So to be able to mention something about this "hilliness" of a trajectory (which happen to not to close up to become a cycle) one could introduce the term "k-trajectory" or even better because more explicite "k-peak-trajectory".
So your examples in the comment were "2-peak-trajectories" (because they do not cycle)

The impressive work that Simons/deWeger have done was to prove, that this structure cannot form a cycle in integers with any combination of $K$'s and $L$'s that you like to try. (Of course the sum of all $K$ is the number of odd-steps, which I usually denote by $N$)
Due to their work we know that a possible nontrivial cycle must have a structure of U()D(),...,U()D() with more than $72$ pairs of $U()D()$ .

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Perhaps a more enlightening treatize is this very basic (and old, and amateurish, my sorry...:) ) on my webspace about the concept of k-cycles.

Answer 2

Here is an excerpt of the original work of Simons/de Weger in 2005 (pg 55). $K$ means here the number of odd steps $(3x+1)/2$ , $L$ the number of even steps $x/2$ . Perhaps this is more enlighting what "m-cycles" or "k-cycles" and "68-cycle" and "72-cycle" could mean:

Because we have higher lower bounds for $x_\min$ meanwhile, more "m-cycles" might be disprovable/disproved now (likely something for "m" in between $73$ and $90$.

Answer 3

A cycle and a loop are the same thing, but I understand what you are talking about.

I too have stumbled upon the "hilliness" of several Collatz trajectories, and what I do is I break them down into U(n)D(n) maps and compare different trajectories. However, I group the U(n)D(n) notation differently by including information on the number of times I divide by two.

Instead of keeping track of U's and D's, I broke it down my notation to how many times you divide by two after multiplying by three and adding one. For example, I would write UUDD, the trajectory of $7$, as $ß(2) \rightarrow 3ß(1) \rightarrow 4ß(1)$. #$ß$ means how may divisions of two were done, and the number in the parenthesis means how many times the operation was done consecutively.

7 ß(1) [?]
22
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11 ß(1) [u]
34
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17 2ß(1)[u]
52
26
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13 3ß(1) [d]
40
20
10
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5  4ß(1) [d]
16
8
4
2
1

With this notation, $nß(1)$ is equal to one odd number, so your "k-cycles" are really the number of odd numbers in any given trajectory. The U(n)D(n) notation will always be one letter short than the total number of odd numbers however. I bring up my notation because it gives more information on what "D" is and it's much easier for me to find patterns between various Collatz trajectories.