I asked Longest known sequence of identical consecutive Collatz sequence lengths? some time ago, but I don't feel like it really got to the bottom of things.

See, in the answers lopsy find a sequence in the range of $596310 ... 596349$ and makes a heuristic argument:

There's nothing special about these numbers, as far as I can see. In fact, there are probably arbitrary long sequences of consecutive numbers with identical Collatz lengths. Here's a heuristic argument:

A number $n$ usually takes on the order of ~$\text{log}(n)$ Collatz steps to reach $1$.

Suppose all of the numbers between $1$ and $n$ have random Collatz lengths between $1$ and ~$\text{log}(n)$. Then, if we choose a starting point at random, the probability that the next $X$ consecutive numbers all have the same Collatz length is ~$\text{log}(n)^X$. There are ~$n$ possible starting points, so we want $X$ so that the probability is $\text{log}(n)^X \cong > \frac{1}{n}$. Then I'd expect the longest sequence to have around $X$ consecutive numbers.

As it turns out, $X=\frac{\text{log}(n)}{\text{log}\text{log}(n)}$ does the trick.

Except in the comments exchanged afterwards I point out that:

log(596349)/log(log(596349)) ~ 7, not 40

(that should be log(596310)/log(log(596349)) but the comment holds)

So Heuristically we're expecting 33 fewer consecutive sequences of the same length, seem like a massive outlier.

So perhaps these sequences are special; what causes them to to arise?

  • $\begingroup$ I don't think that approximating the length with a uniform random variable between $1$ and $\log n$ is accurate. For starters we have a lower bound $\log n / \log 2$. $\endgroup$ – mercio Apr 20 '15 at 19:44
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    $\begingroup$ Also, the trajectories tend to coalesce with a lot more regularity than independant random variables do. As the examples of 100 and 101 show, the sequences for $4k$ and $4k+1$ always have the same length (and you can make up a lot more relationships like this) , so you shouldn't model the length of adjacent numbers with independant random variables. $\endgroup$ – mercio Apr 20 '15 at 19:51
  • $\begingroup$ @mercio That's a good critique of Lopsy's answer...but unfortunately doesn't get close to answering my question. $\endgroup$ – Pureferret Apr 20 '15 at 20:04
  • $\begingroup$ I never claimed this was an answer or else i would have written one. I actually don't know what kind of answer you're expecting here. "special" is a rather vague term, especially when we are dealing with heuristic arguments. Do you want us to fabricate a "heuristic argument" that goes with a formula that gives the right order of magnitude in this particular case ? $\endgroup$ – mercio Apr 20 '15 at 20:18
  • $\begingroup$ @mercio Something that people have noticed and is of some significance? Also I'm trying to dissuade answers that don't get to the point. Apologies if I came across as harsh? $\endgroup$ – Pureferret Apr 20 '15 at 20:24

If we look only at the occurences of two consecutive numbers $n,n+1$ with the same Collatz-trajectory-length $cl$ (I call this $cecl$ meaning "consecutive equal collatz lengthes", here $cecl=2$) then we can observe periodicity in $n$ with cycle-lengthes of perfect powers of $2$ and odd and even starting values. For instance, all numbers of the form $n=8k+4$ (with $k$) show a $cecl=2$ property. We might express this by an analytical formula. For the even number $n$ we look at the head of the trajectory $$ n_0 \to n_1={n_0\over 4} \to n_2=3n_1+1 = 3{n_0 \over 4} +1= {3 \over 4} n_0 +1 \tag 1$$ and for the odd number $m_0=n_0+1$ we get $$ m_0=n_0+1 \to m_1 = 3m_0+1 \to m_2={m_1 \over 4}={3(n_0+1) +1\over 4}= {3 \over 4} n_0 +1 \tag 2$$ which result in equality $n_2=m_2$ and from there the same trailing trajectory follows. Because the leading trajectory $n_0 \to n_2$ is 1 "3n+1" step and 2 "/2" steps and accordingly for $m_0$ the two overall lengthes are equal. Because in (1) we see, that $n_0$ must be divisible by $4$ but not by $8$ we can conclude, that for each number $n = 8k+4$ with some $k_{min}$ and $k \ge k_{min}$ we shall have a cecl of at least 2. The first occurence is $n_0=12,m_0=13$ which give $n_2=5=m_2$ and from that the trailing trajectory is equal. The next occurence is $n_0=20,m_0=21$ and we get $n_2=16$ and $m_2=16$ and from there equality of lenghtes.
For all residual classes modulo 8 we find such cyclic cecl=2 occurences, however with perfect powers of 2 as cycle-lengthes, so for instance $n_{0,k}=37+2^5k$ give $cecl=2$, as well as $45+2^6k$ and $29+2^7k$ and so on.
After we find, that a number $n_{0,k}=4+2^3k $ has a $cecl=2$ and $m_{0,j}=29+2^7j$ then if $k$ and $j$ are selected such that $m_{0,j}=n_{0,k}+1$ then the two $cecl$'s join to build a $cecl=3$.

Accordingly this can happen if three or more conditions/cycles meet at some $n_0,n_0+1,n_0+2,...$ - then we can have $cecl$'s of arbitrary lengthes.
The occurence of such longer $cecl$'s are thus themselves periodic, and result from some multiplicative expression of modular conditions with cycle-lengthes of $2^w$ and odd and even initial values, which meet at consecutive numbers $n_0,n_0+1,...$

In my other answer in the linked question I've given a picture/table which illustrates that joins of consecutive numbers with each $cecl=2$-property.


Here at $n_0=28,m_0=29$ and on $n_0=68,m_0=69$ (and so on) join two consecutive $cecl=2$s to make one of $cecl=3$.

  • $\begingroup$ I think I follow....Maybe? $\endgroup$ – Pureferret May 29 '15 at 8:25

The number really has to do with how many $2^k-1$ odds on the path. 27 is bad because it goes through 31. You basically are growing and whittling – hence hailstone problem – I like Josephus with adding people!

You add $\frac{3n+1}2$ multiplicatively for the power of 2 closest to the odd number 27 is 3 so you increment up twice then eliminate even powers of 2. Then you hit 31, then you increment 5 times before whittling the evens. Any MAP of these shouldn't be by counting in the conventional sense, but by an alternate method of I3E3I5 until you get to 1.


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