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Both the $5x+1$ and $7x+1$ variant of the Collatz sequence are conjectured to have large number of divergent trajectory. Here, i combined the two. As always, when you encounter even $x$, you apply $x\rightarrow x/2$, but if you encounter odd $x$, you have the option of applying either $x \rightarrow 5x+1$ or $x \rightarrow 7x+1$. My conjecture is that you can always reach $1$ from any positive integer starting point. It's very counterintuitive, but i tested the conjecture on integers where $5x+1$ and $7x+1$ are conjectured to diverge on, and yet the conjecture still holds for those integers. Is there heuristic argument that can explain why this happen?

Bryle Morga
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    Is it so surprising? Lots of values converge to $1$ for either of those processes. If you have a divergent path for one of them, it doesn't seem so surprising that it contains a convergent point for the other, at least in examples. – lulu Mar 02 '24 at 15:05
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    If you get to choose $5x+1$ or $7x+1$ then the resulting next odd integer after dividing by all powers of 2 will be smaller on average than having a fixed choice. It's sufficient to make the average ratio of consecutive odd integers less than 1. If $5x+1 mod 4 = 0$ (only a single divide by 2) then $7x+1 mod 4 > 0$. For $5x+1$ the average ratio of consecutive odd integers is roughly $5/4$ (with some assumptions) so it only takes a little bit to get the ratio < 1. – Eric Shumard Mar 02 '24 at 18:21

1 Answers1

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This is a long comment, not an answer

Nice question. It seems it is a bit sharper than the comparable $5x \pm 1$ and $7x \pm 1$ problems (where the $ 5x \pm1$ is easily solvable) , and the statistical formula for the average increase/decrease is a bit more difficult - I would like to see it explicitely.

Here is some short heuristic, using the basic formula $ {m\cdot a_k +1\over 2^{A_k} } \to a_{k+1}$, where $m \in \{5,7\}$ finding this regular pattern when we observe the $ a_1 \pmod 8$

a1    m:A1->a2   m:A2->a3   m:A3->a4   m:A4->a5  m:A5->a6
----------------------------------------------------------------
 3   (5:4)  1   (7:3)  1   (7:3)  1   (7:3) 1   (7:3) 1    ---   
11   (5:3)  7   (5:2)  9   (7:6)  1   (7:3) 1   (7:3) 1    ---    
19   (5:5)  3   (5:4)  1   (7:3)  1   (7:3) 1   (7:3) 1    ---   
27   (5:3) 17   (7:3) 15   (5:2) 19   (5:5) 3   (5:4) 1    ---
35   (5:4) 11   (5:3)  7   (5:2)  9   (7:6) 1   (7:3) 1    ---



1   (7:3)  1   (7:3)  1   (7:3) 1   (7:3) 1   (7:3) 1
 9   (7:6)  1   (7:3)  1   (7:3) 1   (7:3) 1   (7:3) 1--- 
17   (7:3) 15   (5:2) 19   (5:5) 3   (5:4) 1   (7:3) 1---
25   (7:4) 11   (5:3)  7   (5:2) 9   (7:6) 1   (7:3) 1---
33   (7:3) 29   (7:2) 51   (5:8) 1   (7:3) 1   (7:3) 1---


=========================================================================


7   (5:2)  9   (7:6)  1   (7:3)  1   (7:3)  1   (7:3)  1---

15   (5:2) 19   (5:5)  3   (5:4)  1   (7:3)  1   (7:3)  1---

23   (5:2) 29   (7:2) 51   (5:8)  1   (7:3)  1   (7:3)  1--- 
31   (5:2) 39   (5:2) 49   (7:3) 43   (5:3) 27   (5:3) 17---
39   (5:2) 49   (7:3) 43   (5:3) 27   (5:3) 17   (7:3) 15---




5   (7:2)  9   (7:6)  1   (7:3)  1   (7:3)  1   (7:3)  1---
13   (7:2) 23   (5:2) 29   (7:2) 51   (5:8)  1   (7:3)  1---
21   (7:2) 37   (7:2) 65   (7:3) 57   (7:4) 25   (7:4) 11---
29   (7:2) 51   (5:8)  1   (7:3)  1   (7:3)  1   (7:3)  1---
37   (7:2) 65   (7:3) 57   (7:4) 25   (7:4) 11   (5:3)  7---

If my quick-check are not completely messed, I get for the statistical average growthrate $q$ in each step $$(5/4)^{1/8}\cdot (5/8)^{1/16}\cdot \ldots \cdot (7/4)^{1/8}\cdot \ldots... \\ \vdots \\ q = (35/64)^{1/4} \approx 0.85994 \\ $$ which means decrease in the long run (except for possible cycles).

Gottfried Helms
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