3

The "Path Records" page by Eric Roosendaal related to the Collatz conjecture (see here), reports a "striking" feature of the path records.

If we define $\text{MaxSeq}(M)$ (in Roosendaal it is $\text{Mx}(M)$) the maximum value reached starting the Collatz sequence from a positive integer $M$, a positive integer $N$ is called a path record if for all $M$ < $N$ we have $\text{MaxSeq}(M) \lt \text{MaxSeq}(N)$.

The "striking" feature about path records is this:

$$\log_2{\text{MaxSeq}(N)}\approx 2\log_2{N} \tag{1}$$

meaning that the maximum number of bits to store the Collatz sequence values starting from a number with $k$ bits is about $2k$ and not much more than that.

So, now this is my comment, when we start from a path record $N$ we could think it has an "energy" of about $2$, because the number of its bits will be eventually doubled, but other bigger values in the sequence toward $\text{MaxSeq}(N)$ have a lower "energy", and lower and lower when they are more near to $\text{MaxSeq}(N)$. This $2$ factor is not so big after all.

My question is: someone can help giving some probabilistic/heuristic explanation for equation $(1)$?

I tried using the fact that each odd number in the sequence is on average $3/4$ of the previous one (see here), but with no success.

Gottfried Helms
  • 34,920
Fabius Wiesner
  • 2,573
  • what is $Mx(N)$? I don't get your notation – mathworker21 Sep 19 '19 at 22:02
  • It is the maximum value reached starting from $N$, e.g. $Mx(3)=16$ ($3 \rightarrow 10 \rightarrow 5 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1$), $Mx(7)=52$, $Mx(15)=160$, $Mx(27)=9232$. Note that Eric Roosendal don't divide by $2$ in case of odd number (i.e. odd $n$ gives $3n+1$, even $n$ gives $n/2$). – Fabius Wiesner Sep 19 '19 at 22:07
  • Another way of looking at it is that for path records $Mx(N) \approx N^2$ This says the same thing, but avoids the logs. – Ross Millikan Sep 19 '19 at 22:12
  • 1
    @mbjoe ohhhhh, I thought the first $M$ was a dummy variable – mathworker21 Sep 19 '19 at 22:24
  • @mathworker21 edited the question, hope it is more clear now that $M$ is a positive integer. – Fabius Wiesner Sep 20 '19 at 06:02
  • 1
    This "feature" was predicted in Lagarias, J. C.; Weiss, A. The 3x+1 Problem: Two Stochastic Models. Ann. Appl. Probab. 2 (1992), no. 1, 229–261 using the theory for random walks. – DaBler May 24 '20 at 09:33

1 Answers1

-1

Try this and find out. $$a,n,k,\ell \in \mathbb{N}$$ $$\forall\;n_0:=2^ka-1\;\exists\;\mathcal{C}^k(n_0):\;n_k=\frac{3^ka-1}{2^\ell}$$

SKIRBY
  • 1
  • 1