Number of steps in an upward sequence revealed by starting numbers

[Lolomet67]

One of a way to write a number in a binary compact format is

Y2^{n+1}+2^n-1 where Y belong to N_0

In this case n=0 stand for pair numbers. For other positive various values of n the equation will generate all odd numbers.
This is very useful way suggest by **Cadogan ** to extract info to know how many numbers of steps are part of a unique upward sequence perform by a repeat Crandall function with e=1. Below, we will use upper suffix to indicate how many times the C_1 function can be applied.

C_1^{n-1}(Y2^{n+1}+2^n-1)=Y2^23^{n-1}+2^13^{n-1}-1

where n is equal or greater than 2. Why? Remember the parametric equation for e odd. We rewrite it here

C_{2m-1}(2^{2m}r+2^{2m-1}+\sum_{j=0}^{m-1}2^{2j})=6r+5

if we set m=1, we get

C_1(2^2r+2+1)= 6r+5 where r belong to N_0. Since

2^2r+2+1=Y2^{n+1}+2^n-1

we get r=Y2^{n-1}+2^{n-2}-1

in order to reach r=0, we will have Y=0 and n=2(the lowest permit value of n).

Going back to our main topic, we say we reach the top of the upward sequence at

Y2^23^{n-1}+2^13^{n-1}-1\equiv1(mod4)

because

C_2(4m+1)=\frac{12m+4}{2^2}=6m+1

Not only this formula provides the number of steps of the sequence, it jump right away to the top number of the sequence and give as well as a relative (and potentially the absolute) maximum of the trajectory in one shot, such avoiding computation at each step!

In fact, if restricted to congruence in mod 12 as did Puddu, it can be shown that start numbers are some of the 3(mod12), 7(mod12) and 11(mod12) and upper sequences will always end with a 5(mod12). Intermediate values in between the start and the end are all 11(mod12).

For reference
**Cadogan, C. C.**A note on the 3x+1 problem, Caribb. J. Math. 3 (1984), p. 67-72.
Puddu, S., The Syracuse Problem, Notas Soc. Mat. Chile (1986), p.199-200.

[Lolomet67]

Correction: replace “pair numbers” by “even numbers”.

[demoncherub]

I can’t help but see some overlap between what you have posted here and the stuff I have been working on. I have been exploring the Collatz Conjecture using a binary tree structure. Every one of the vertices in the tree can be described using the path equation:

\frac{(3^n)N+f}{2^m}=M

Where n is the number of odd steps in the path, m is the path length, N is the seed value (starting value), and M is the value the path ends on. f is a path-dependent variable. f only increases during an odd step, by the amout 3(f) + 2^m. An odd step is defined as \frac{3N+1}{2 } and an even step is \frac{N}{2}.

In the case of a path with all odd steps (OOOO…O), n and m are equal. This gives us this path equation:

\frac{(3^m)N+(3^m-2^m)}{2^m}=M

Where N=2^mk+2^m-1, and M=3^mk+3^m-1. And the path’s highest value is 2M. The value of m is the path length, the total number of odd steps. It can be any natural number from 0 to infinity. I just thought I would drop this here since it appears. we arrived at the same place via slightly different routes.

[Lolomet67]

Hello Demoncherub, your are quite right. May I suggest one thing if the idea did not already occur to you, it is the choice of your starting number in your binary tree: why not picking the relative maximum numbers of your upward sequence?

[demoncherub]

I am not sure i follow you. As with any Collatz Sequence the starting number N is arbitrary. You can stat with anything you want. I am less concerned with actual values for N and M as i with the path itself. The starting values that satify any given path equation change as the path length increases.

[Lolomet67]

I am sorry, I may have skip a few chapters, so here is the background. In the Syracuse problem, there are subsets of the set of positive integers who can by coalescence generate all possible integers through the usual Crandall function or operation 3X+1 etc.
except of course for multiples of 3. You can filter them down by different ways and the one I am most familiar with is using congruence namely integers equivalent with 1(mod 12), 5(mod 12) and 9(mod 12). Pick one out of the three and you could generate the whole set except multiples of 3. When you reach the top of an upward segment the top integer is always equivalent to a 5 (mod 12). Notice the reverse is not true. So for instance starting with 15 you get the odd sequence 15, 23, 35, 53 afterward you are going to 5. Notice that 53\equiv5 (mod 12).
So selecting any 5 (mod 12) is interesting for two reasons, first you save some tedious operations on your starting integer, second if you can prove the validity of the Collatz Conjecture with this subset then you are done (Champagne, party, esta la fiesta baby!).

[demoncherub]

I am a bit confused. The reason i started with N=2^mk+2^m-1 (which can be 3,7,or 11 (mod 12)) is because they are the only values that lead to back to back odd functions (3N+1)/2. Notice 53 under the odd operation yields an even number. So i am confused as to what your objective is here. I thought it was to look at the trajectory of numbers under successive odd functions. But, i must be mistaken. It is quite interesting that a 5 (mod12) shows up at the end of a chain of odd functions. But hardly surprising, since probability is working against long chains of odd functions as m increases.

[Lolomet67]

Thank you demoncherub,
you are helping me to get more accurate in my definition
of an integer being a member of and upward segment so now on

DEFINITION
An integer is a member of an upward segment when it is filling at least one of the following conditions:

1. the integer is greater than is immediate predecessor,
2. the integer is the immediate predecessor of a greater integer.

In other word the integer X_i is a member of an upward segment if it is included in the Crandall function with e=1 (C_1) as in

C_{1}(X_n)=\frac{3X_n+1}{2^1}=X_{n+1}

With this definition 53 is the member at the top of the segment.
For the reminder of my suggestions, just ignore it at this time and let me go into your original pod for a week or two so we can speak the same language.

[demoncherub]

It stands to reason that the member at the end of an upward segment would be the largest. But, i am assuming you are discounting even numbers, since 80 meets that definition. Its larger than its predecessor (53). 53 is the point where that uoward segment shifts direction however.

[Lolomet67]

Exactly, with Crandall function you are operating only on odd integers as it compress all n tied divisions by 2 by 2^n.

[Lolomet67]

Correction

replace C_2(4m+1)=\frac{12m+4}{2^2}= 6m+1

by obviously C_2(4m+1)=\frac{12m+4}{2^2}= 3m+1

[jeff317]

Jeff here. I read the discussion and would like to offer a graphical layout that may clarify an analysis of both upward runs of odd values and downward runs of even values for all natural numbers. All even numbers run down a branch of the Collatz tree by a length controlled by a binary sieve. Numbers of 4m+2 format have a single step. Numbers of 8m+4 have two steps etc. Odd values link up by a 3N+1 rule to nodes of 6k+4 format.

Switching the upward runs to steps of (3N+1)/2, run lengths are also governed by a binary sieve.

The length of an odd run matches the length of the following even run. This is driven by a shadow flight of steps starting from the same even number as the following even run. Long runs are separated by shorter runs by the binary sieve.

This is just a small part of a longer analysis of families of flights. These images show duplicates of the top part of long flights repeated as shorter runs starting at a larger value on the number line. Downward runs are shown as shorter runs duplicated at earlier positions. Displaying only the flights and removing the duplicates, a fractal arrangement of flights is generated. The first family of flights have bases at 2^k-1, are the longest flights in a local area and contain a growing self-similar fractal arrangement of flights That grow in the intervals defined by the first family, Infill flights are generated by flight seed that occur in pairs at 12k-3 and 12k+1. This is just a small sample of a longer proof of the Collatz conjecture. All flights in a family have a separation of 2k+1 from the prior flight or the seed of the family.

The first seed pair (9,13) generate flights 19>29 and 27>41 in the following interval. The next flights for these families are in the following interval as 39>59>89 and 55>83>125.
The Collatz tree contains a huge amount of structural regularity that guides the orbits that are alternating runs down a branch and up a flight. At a higher hierarchy the orbits are runs of 4k+1 downlinked branch and 4k+3 uplinked branches as alternations of a run down a subtree and up a flight of steps. Every step on a flight is a root of a subtree. The flights come in families as pairs with seeds at 12k-3 and 12k+1.
This is a fully deterministic, discrete, digital, chaotic, fractal, tree of natural numbers that has lost monotonicity, and uses modular arithmetic of bases 2, 3, 4, 6, 12 and 18 to connect branches of three types as a series of downlinked subtrees of 4k+1 branch types, built in layers, with roots that are steps of 4k+3 uplinked flights that come in infinitely many families of flights, each with infinitely many flights of all possible finite lengths. The tree contains many substructures including a binary tree and self-similar growing fractal shapes with bilateral symmetry. There are processes of a binary sieve, a binomial like transition of powers of 2 to powers of 3 that explains the log(3)/log(2) relationship in shadow flights that are parallel to real flights. Many patterns have exceptions that lead to new layers of structure. This system has many hidden structural relationships.

Would you be interested in a full proof of the conjecture?