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If $e_n$ represents the set of even numbers in the hailstone sequence starting with $n$ and $o_n$ represents the set of odd numbers in the hailstone sequence starting with $n$, plotting $\displaystyle \sum_{o \in o_n}\log_2\left({3o+1} \over {o} \right)$ in relation to $\mathbf{card}(e_n)$ gives an interesting plot, which seems to fall along a line: lack of labeled axes brought to you by out-of-date Ubuntu package for dvipng

Why does this relation appear?

Morgan H
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2 Answers2

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There is a paper on arxiv:"on the glide of 3x+1 problem". It shows that when stopping at the first $n_1$ below $n_0$, $2^{card(e_n)-1}<3^{card(o_n)}<2^{card(e_n)}$. This explains the values you found . And thank you for the warm welcome

Comment: Err... it was a sarcastic comment for thoose who deleted my previous answer. But thank you

user489810
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  • ?welcome?. The article is flagged as faulty, but there are interesting things. I wonder why the lemma is not proven yet, it does not seems that hard (adding $a\cdot2^n$ to $n_0$ gives $n_1+a\cdot3^n$, which is already known). – Collag3n Oct 10 '17 at 17:05
  • Oww...ok, I thought you removed it. Your original post seemed ok to me (?!?). Some are very picky here, don't pay attention. – Collag3n Oct 10 '17 at 18:40
  • sorry...the exponents should be read as $a\cdot2^{card(e_n)}$ and $a\cdot3^{card(o_n)}$ not $^n$ – Collag3n Oct 10 '17 at 19:09
  • The flagged "faulty" article should still be linked to. Here it is: https://arxiv.org/abs/1710.01525v1 to be able to find out about the problem oneself – Gottfried Helms Jan 14 '18 at 09:17
  • This article was published in 2019: https://www.jstage.jst.go.jp/article/transfun/E102.A/3/E102.A_613/_article – Olivier Pirson Apr 21 '20 at 21:10
  • @Olivier Pirson, thanks. Is this any different from the Arxiv version? I don't have access – user489810 May 22 '20 at 19:23
  • @user489810, the paper has been available on Sci-Hub recently. I haven't read it. – Olivier Pirson May 23 '20 at 09:51
  • Thanks :) found it – user489810 May 25 '20 at 15:19
  • This paper is cool, and I feel like it goes some of the way towards explaining what I've observed, but I'm also not sure if it can entirely explain it since it's only about how long the sequence takes to get to a number less than the starting point, while this pattern is about how long it takes to get all the way to 1 – Morgan H Aug 02 '22 at 21:07
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First thoughts, but this needs to be investigated: $\sum \log_2 (\frac{3o+1}{o}) = \log_2 \prod (3+\frac{1}{o}) \simeq log_2 (3^{card(o_n)}+\delta_1) \simeq card(o_n) \log_23+\delta_2$

$\simeq 1,5849 card(o_n)+\delta_2$.

If you apply the collatz function to a number $n_0$ until it reaches some other $n_1$, you have something like this: $n_1=\frac{3^{card(o_n)}}{2^{card(e_n)}} n_0 + \frac{3^{card(o_n)-1}}{2^{card(e_n)}} + \frac{3^{card(o_n)-2}}{2^{card(e_n)-1-\ldots}}+\ldots+\frac{3^0}{2^{\ldots}} $

The thing is that if $n_1$ is a number near $n_0$ (e.g. you start with $n_0=63$ and you stop at $n_1=61$, i think you have a $card(o_n)=34$ and $card(e_n)=54$ in that case), you'll see that

$card(e_n)\simeq card(o_n) log_23\simeq 1,5849 card(o_n)$. Indeed, the first fraction $\frac{3^{card(o_n)}}{2^{card(e_n)}} \simeq 1$

So dividing your sum with $card(e_n)$ might bring something linear, i suspect that the $\delta$ which involve fractions of successive $n_x$ is in correlation with $\frac{n_1}{n_0}$, especially if you continue to apply the collatz function until $n_1=1$. I will have a look at it later....

edit:

I didn't had much time but from what i saw, my first impression was right: $\sum \log_2 (\frac{3o+1}{o}) \simeq 1,5849 card(o_n)+\delta_2$ (and $\delta_2$ does not seems to make a difference)

and when $n_1 = 1$ , $\frac{card(e_n)}{card(o_n)}$ is floating arround 1,8 which is not far from 1,5849 when $n_1$ is near $n_0$.

Why is it "not far" and why do $card(e_n)\simeq card(o_n) log_23$ when $n_1\simeq n_0$ are interesting questions, especially for the quest of cycles.

Collag3n
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  • I tried plotting $card(e_n)$ against the sum and against $log_2(3)card(o_n)$, and the slopes for each do seem to be somewhat different; this explanation is interesting though and I'm not sure if that plotting was a valid way to do the comparison – Morgan H Aug 02 '22 at 21:10
  • oops I did my formula wrong (did a natural log instead of base 2), you were actually right – Morgan H Aug 02 '22 at 21:16