Its hard to necessarily discover what all the connections might be with just a single read through of your post. I’ll take a deeper look.
As a note I think I must have failed to make what I was attempting to do with the forward tree structure. The idea was to visualize how the Collatz sequence for any value of N progresses independent from any specific value of N, where we only know two things about N for sure: N is a natural number and N is greater than zero. Depending on whether N is even or odd it would take either the even or odd path. So the tree is probalistic in nature, showing all possible paths. Collapsing to a specific path if a specific value of N is chosen.
My thinking was that this could be a way to possibly work out the behavior of paths further down the tree. Much farther down than we can realistically trace out individual paths. The reason I think this is that for any level of the tree there is a binomial distribution of possible paths. With two possible paths at each node of the tree.
Hello again Demoncherub,
it is interesting since yours perspective seems to be at the exact opposite of what I was trying to do as I was looking for a sort of DNA hidden in each number to explain its trajectory. Well may be the solution is somewhere in between. Anyway, what I was just trying to point out were nice shortcuts in computing your numbers.
Good luck in your research.
Thank you. I certainly appreciate any offer of help. I am sure there are all sorts shortcuts I have overlooked. I am curious as to what this “DNA” hidden inside numbers would look like.
I’m not sure if anyone noticed, but 2^71 numbers have been checked. So we can improve our loop bound from 2^68, or 114,208,327,604 with new 301994𝑎 +17087915𝑏 +85137581c to a bigger bound. I did it a bit ago so I know it’s possible but don’t remember it off the top of my head and want to make sure that everyone knows it will be correct so I think someone else should also try
once we reach 1.8459\cdot 2^{71} we can use the new bound 217,976,794,617 (Eliahou). If Barina’s project is still running, it would be reached by the end of this year.
I didn’t forget about your questions. But, when I thought it over I realized what a chore it would be to analyze a bunch of paths by hand. But, here is what I came up with:
\frac{3^nN+f}{2^m}=M
Beginning with the path equation and a parity vector, like OEOEOE we can determine the values of n. m, and f. n is just the number of odd steps and m is the path length. f is determined using the recurrence:
\sum_{i=1}^{n}3^{n-i}2^{p_i-1}
Where n is the number of odd steps, and p_i is the position of the ith odd step. After we have all the pieces. We can rearrange the path equation into the standard Diophantine form:
2^mM-3^nN=f
Which is fairly easy to solve because 2^m and 3^n are coprime and their gcd is 1. The goal here is to get general equations for M and N.
N=2^mt + b and M=3^nt+c; here t in an integer value. Setting the two equations equal to one another and solving for t. If you find an integer value of t then you have a loop.
Anyway, if automated the process with a Python script.
Parity Solver Python Script
Starting with the loop equation and setting N=M, we get:
\frac{3^nN+1}{2^m}=N
Rearranging, we get:
\frac{f}{N}=2^m-3^n
\frac{f}{N}-(2^m-3^n)=0
This is the loop condition. The quantity \delta=\frac{f}{N}-(2^m-3^n) can be used as a measure of how much M and N diverge over a given path. For a fixed path of length m, δ(N) = f/N − gap. As N increases through the residue class (N = b, b + 2^m, b + 2·2^m, …), δ decreases monotonically toward −gap. So the maximum δ in any residue class is always at the minimum seed of N = b.
This means the question “which paths come closest to δ = 0?” reduces to asking which paths have the smallest value of δ(b) = f/b − gap from above.
The paths closest to zero from above are those where:
f/b ≈ gap i.e. f ≈ gap · b
which is exactly the loop condition approached but not quite reached. These are paths where the residue f is nearly a multiple of the gap. The gaps are smallest at the convergent (m,n) pairs— the points where 3^n is closest to 2^m — so those cells naturally produce the smallest |δ| values across their entire Pascal cell, not just for individual paths.
There’s a sharper way to say this. The normalised quantity δ/2^m = (M−N)/N tells you the fractional overshoot or undershoot. For a path to have δ/2^m close to zero, it needs:
- \frac{n}{m} close to \frac{log(2)}{log(3)} — so gap is small relative to 2^m.
- \frac{f}{b} close to gap — the residue nearly divides evenly into the gap
**The Pascal group (with binomial coefficient C_m^n) where m=24,727and n=15,601, \frac{n}{m} approaches pretty close to \frac{log(2)}{log(3)}. which is precisely why it seems like an interesting candidate cell for a hypothetical new loop. I would look into those paths, except that none of my devices can handle math with numbers that big.