5

In the Collatz sequence, starting from an odd $n_0$, you can find any odd successor $n$ by applying successively the "condensed" Collatz function:

$n_1 = \frac{3}{2^{m_1}}\cdot n_0 + \frac{1}{2^{m_1}}$

$n_2 = \frac{3}{2^{m_2}}\cdot n_1 + \frac{1}{2^{m_2}} = \frac{3^2}{2^{m_1+m_2}}\cdot n_0+\frac{3^1}{2^{m_1+m_2}}+\frac{3^0}{2^{m_1}}$

$n_i = \frac{3^i}{2^{m_1+m_2+...+m_i}}\cdot n_0+\frac{3^{i-1}}{2^{m_1+m_2+...+m_i}}+\frac{3^{i-2}}{2^{m_1+m_2+...+m_{i-1}}}+...+\frac{3^0}{2^{m_1}}$

Where $i$ is the number of time the $(3n+1)$ function was applied, and $j=\sum m_i$ is the number of time the $(\frac{n}{2})$ function was applied.

From here, we'll look at the "Glide" sub-sequence (we stop at the first $n_i<n_0$).

From the equation above $n_i=\frac{3^i}{2^j}\cdot n_0+\delta$ you can find out that $(n_i+k\cdot3^i)=\frac{3^i}{2^j}\cdot (n_0+k\cdot2^j)+\delta$, or in short, you'll reach $(n_i+k\cdot3^i)$ from $(n_0+k\cdot2^j)$ with exactly the same steps as when you reach $n_i$ from $n_0$, which is well known.

In many papers i see that it is taken for granted that since both sub-sequences follow the same steps, both sequences have the same Glide. But as far as i know, there is no proof of it yet. Or is there (that is my question)?

Edit: Some e.g. of what i am looking for. Imagine this sequence:

$n_6=\frac{3^6}{2^{10}}\cdot n_0+\frac{3^5}{2^{10}}+\frac{3^4}{2^9}+\frac{3^3}{2^8}+\frac{3^2}{2^7}+\frac{3^1}{2^5}+\frac{3^0}{2^2}$ (with $n_6$ first $n_i<n_0$)

It could be possible that all intermediate value $n_5, n_4, n_3, n_2, n_1$ are all greater than $n_0$ ($n_6$ is the first value lower than $n_0$), but that the intermediate value are not all greater than $n_0+2^{10}$ in the other sequence starting with $(n_0+k\cdot2^j)$ (e.g. $n_5+3^5\cdot2^2<n_0+2^{10}$, which would mean that the Glide is 5 and not 6 in this "other" sequence)

There is no known $n_0$ which leads to such a case, but my understanding is that there is no proof there is none. Since there is no known integer for this, i'll take $n_0=4,33333...$ as an exemple, just to make it clearer (...eventually):

The Glide sub-sequence for $n_0=4,3333..$ would be $4,3333\to7\to11\to17\to13\to5\to4$, and the equivalent sub-sequence for $n_0=4,3333..+2^{10}$ would be $1028,3333\to1543\to2315\to3473\to2605\to977\to733$ where $977$ is already smaller than $1028,3333...$

Note - if you stick to the definition of the Glide, the first $n_i<n_0$ is not necessarly odd (when you look at both odd/even $n$), but this has no importance here. You could stop at $n_6=1$ instead of $n_6=4$ and change every $2^x$ by $2^{x+2}$ in the above formula (top of my edit).

Collag3n
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    Hmm, I think you know my http://go.helms-net.de/math/collatz/Collatz061102.pdf pg.5 (and there refering to pg 3&4) Doesn't fit this your question? Or did I overlook perhaps some subtility? – Gottfried Helms Nov 27 '17 at 15:12
  • Not sure, or perhaps i missed something. I gave an exemple in my "Edit" where i applied this transformation $T(a;1,1,1,2,3,2)$ and i looked at value modulo $2^S=2^{10}$. I know my exemple does not fit, it is just to get the idea, and since i also looked at even numbers in the exemple (at least for the target $b_k=n_i$), i took $k=1$ – Collag3n Nov 27 '17 at 18:29
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    Ah,, I see. If we have with $b_1 = T(a_1;A,B,C,D)$ and $a_1>b_1$ (which is what we want for a "glide" ) then increasing $a_2=a_1+22^S$ as well as $b_2 = b_1+23^N$ *keeps* the greater/smaller relation ($a_2>b_2$ if $2 ^S> 3^N$) and (possibly) *inverts* it when $2^S<3^N$ . However, to make $b_k < a_k$ requires that $2^S>3^N$ - so the relation should stay stable in general - except for possibly few configuartions. I'll have to think more about this... – Gottfried Helms Nov 27 '17 at 18:43
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    Yes that's the greater/smaller relation that should be kept (and probably is kept) but is not proven. At least that's what i understood from the comment on this paper https://arxiv.org/abs/1710.01525 – Collag3n Nov 27 '17 at 18:46
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    If the iteration were exactly $x \to 3/2^A x$ then the $2^S/3^N$ - relations should always be consistent I think; but because we have one unit more in the numerator $(3x+1)/2^A$ that might introduce a small -but possibly effective- distortion of the expectation. I think I'll need now some time to look at it. So this is now becoming a really interesting question... – Gottfried Helms Nov 27 '17 at 19:13
  • My issue is, it is easy to show that $b_k<a_k$ implies $2^S>3^N$ (from $b_k=\frac{3^N}{2^S}\cdot a_k+\delta$) and $2^S<3^N$ implies $b_k>a_k$, but in between, it is hard to prove that $2^S>3^N$ implies $b_k<a_k$. In my example, we see that with the same $2^S>3^N$ we could have $b_k<a_k$ and $b_k>a_k$ ($b_k$ or $b_{k-1}$ in this case) – Collag3n Nov 27 '17 at 19:31
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    To have this distortion, we must be in a case where $\frac{3^N}{2^S}\cdot a_0<a_0$, but adding $\delta$ to it make it greater than $a_0$, but with a greater value $a_k$, adding the same small $\delta$ is not enought to make it greater than $a_k$. So it probably comes to prove that this situation is not possible. – Collag3n Nov 27 '17 at 19:47
  • Sitiations like this might occur when $N,S$ is a pair of the continued fraction of $\log_2(3)$, for instance $(N,S)=(5,8)$, or $(N,S)=(41,65)$ or one of that pairs which I've listed in my online "$2^S/3^N$"-article. (It's near midnight here, I'll return to this tomorrow evening) – Gottfried Helms Nov 27 '17 at 21:34
  • Indeed, $\frac{3^N}{2^S}$ must be smaller but very close to $1$, and $(N,S)=(5,8)$ is a good exemple: $\frac{3^5}{2^8}\cdot (4,333...)+\frac{3^4}{2^8}+\frac{3^3}{2^7}+\frac{3^2}{2^6}+\frac{3^1}{2^5}+\frac{3^0}{2^3}=5$ which is greater, but $\frac{3^5}{2^8}\cdot (1028,333...)+\frac{3^4}{2^8}+\frac{3^3}{2^7}+\frac{3^2}{2^6}+\frac{3^1}{2^5}+\frac{3^0}{2^3}=977$ which is smaller. I didn't found any working integer exemple of this, i guess we need to search exponent paterns e.g. $(1,1,1,2,3)$ for $(N,S)=(5,8)$ in numbers bellow the treshold ($a_0$ for which $\delta$ is enought to invert $b_0<a_0$) – Collag3n Nov 28 '17 at 07:32
  • The treshold is $\frac{\delta\cdot2^S}{(2^S-3^N)}$. I still have to look but i don't think i'll find a value. – Collag3n Nov 28 '17 at 17:54
  • Interesting. While searching for $(N,S)$ combinations where $0,9989<\frac{3^N}{2^S}<1$ i found these: (306,485) (971,1539) (1636,2593) (2301,3647) (2966,4701) (3631,5755) (4296,6809) (4961,7863) (5626,8917) (6291,9971) (6956,11025) (7621,12079) (8286,13133) (8951,14187) (9616,15241) (10281,16295) (10946,17349) (11611,18403) (12276,19457)....or more generaly $N=665i+306$ and $S=1054i+485$. Pretty regular – Collag3n Nov 28 '17 at 19:35
  • The regularity is "temporary" - they follow the "generalized convergents" ot he continued fraction between $(N,S)=(306,485),(971,1539)...(12276,19457))$ (taken from your comment) and the next entry. Perhaps the effect might be even stronger at $N=15601,47468,79335,...$ or $N=190537,...$ with the appropriate differences. I'm curious of what this gives... – Gottfried Helms Nov 28 '17 at 19:51
  • Indeed, $N=190537$ is almost 1. But yes, at $16266$, the ratio drop to $0.5000$. I just saw that if you take any number (not just 306), it gives a series with almost the same ratio. Probably because $\frac{1054}{665}\approx\frac{ln(3)}{ln(2)}$, and even more with $\frac{50508}{31867}$ – Collag3n Nov 28 '17 at 20:00
  • and $15601$ is the last working value in my serie, and the first of your serie (with greater precision)..I guess this patern is also regular. I think i need a supercomputer – Collag3n Nov 28 '17 at 20:11
  • No, no supercomputer needed. Just, for instance, Pari/GP and a absolutely common laptop newer than, say, 2000. But Pari/GP with such integers would have run even on my Win95 Laptop in the 90ies... Everyone can get a free installation of Pari/GP; btw you can take the next handful of convergents of continued fractions from my page http://go.helms-net.de/math/collatz/2hochS_3hochN_V2.htm (slightly updated) – Gottfried Helms Nov 29 '17 at 02:45
  • Yes, you are right. What i need more is time, not a computer. I hope to find some this week-end, when I canl try Pari/GP. I took a quick look at you lower enveloppe convergence page yesterday and that is exactly what is needed here, thanks. In the 2.1 table, there are sub-series with identical intervals (2, 12, 53, 665,...) and i noticed that substracting an interval with the previous one always yield a multiple of the last element of the sub-serie (e.g. $665-53=2\cdot306$). Knowing that in a "glide" situation, $\delta<\frac{N}{3}$, and predicting the lower bounds might give something... – Collag3n Nov 30 '17 at 05:17

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