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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?

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AncientSwordRage
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  • 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 $\log n / \log 2$. – mercio Apr 20 '15 at 19:44
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    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. – mercio Apr 20 '15 at 19:51
  • @mercio That's a good critique of Lopsy's answer...but unfortunately doesn't get close to answering my question. – AncientSwordRage Apr 20 '15 at 20:04
  • 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 ? – mercio Apr 20 '15 at 20:18
  • @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? – AncientSwordRage Apr 20 '15 at 20:24

2 Answers2

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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.

image

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 $cecl=3$.

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Gottfried Helms
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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.

George Law
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Mark
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