In a previous topic, I describe a way to know how to extract information to zip to the top of an upward segment of a trajectory. Following the same line, I will dig here a little deeper. Going from a start number to the top, by subtracting the two following equations
2^23^{n-1}Y+2^13^{n-1}-1
minus
2^{n+1}Y+2^n-1
we get the gap
\Delta=(4Y+2)(3^{n-1}-2^{n-1})
Would there be other upward segments giving the same gap? It would have to solve
(4Y'+2)(3^{m-1}-2^{m-1})=(4Y+2)(3^{n-1}-2^{n-1}).
Without lost of generality, stating Y'\gt Y and m\lt n then we can write
4Y'+2=(4Y+2)\frac{3^{n-1}-2^{n-1}}{3^{m-1}-2^{m-1}}
where n and m \ge2
In order for solution to exist, the last fraction need to give an integer. Luckily there is an known identity to our help:
(3^r-2^r,3^s-2^s)= 3^{(r,s)}-2^{(r,s)}
where (a , b) stands for the greatest common divisor of a and b.
Therefore to solve the equation, m-1 must divide n-1. Conclusion:
1.If n-1 is a prime number, there is no solution and the gap is unique.
2. If n-1 is a composite number then there as many equal gaps as the number of divisors of n-1 (less n-1 itself and 1 of course).
Voilà!
I think it’s true that no two consecutive integers can merge before reaching the same top, while also satisfying your equations for equal gaps.
Hello HungryMonkey7,
I am not quite sure what you mean by two consecutive integers. Could you please provide an exemple or a counter exemple?
Like 28 and 29, two numbers are consecutive if one of them is one larger than the other. Two consecutive numbers merge when their paths yield the same number after the same number of steps. Like in the example for 28 and 29, after 3 total steps, they each reach 22 after one odd step and two even steps. It turns out that for consecutive merges, it must have equal even counts and equal odd counts.
So if they have the same gap, by satisfying your equation, they can’t merge before or at the same top node because the same gap would imply that they end up one apart. But I think it also reaches a contradiction (not sure) if the gaps are offset by one in the case where they should merge.
Sorry HungryMonkey7, perhaps I have not been clear enough.
Firstly this post is dealing with odd numbers only by using the Crandall function . Secondly for an odd integer to be a member of an upward segment it must satisfy the following property:
X (odd) is such that \frac{3x+1}{2}= odd integer
so 28 or 29 do not belong to an upward segment.
The purpose of the Crandall function is getting rid of the pair numbers and also pruning multiple division by 2 in one operation like
C_3(29)=\frac{3(29)+1}{2^3}=11.
If you need more explanations, it will be my pleasure to answer to your request.
Ok, thanks for explaining, I didn’t understand how your function worked. I guess that’s why it didn’t work for me. I just was trying to connect different formulas to learn something new, because I don’t have ideas for better progress fast enough for the rate of progress that inspired me when I started this problem.
There is a big mistake in my previous conclusion because if n-1 is a primary number the gap is not unique!
In fact in
4Y'+2=(4x+1)\frac{3^{n-1}-2^{n-1}}{3^{m-1}-2^{m-1}}
where m and n \ge2. In order for the solution to exist,
the above fraction does not need to be an integer.
However the identity (3^r-2^r,3^s-2^s)= 3^{r,s}-2^{r,s} is still useful because the second conclusion in the first post of this series is valid. This very one property may lead to some interesting results.
Just consider m-1=2, just by playing with Y’, we can generate all the gap with n-1 been and even number since the fraction simply give an integer.
Along the same line, any composite number which is a multiple of primary numbers can have all his gaps generated with only one of his primary integers. For exemple in the case n-1=15, you could pick m-1=3 or 5.
This how primary numbers can communicate together through common composite. For instance let n-1=6, we can use m-1=2 and get Y_2 or use m-1=3 and get Y_3 so
4Y_2+2=(4X+1)\frac{3^6-2^6}{3^2-2^2}
and
4Y_3+2=(4X+1)\frac{3^6-2^6}{3^3-2^3}
or
(4Y_2+2)(3^2-2^2)=(4X+1)(3^6-2^6)=(4Y_3+2)(3^3-2^3)
The End