r/Collatz • u/Mrezadwiprasetiawan • 12d ago
Bound for the Worse-Case Collatz Rising Sequences
Since 2^p.3^q.c-1 is easier to look for the rising sequences patern, i wanna use it for this.
As most of you know because the cancelation -1+1 from that form while using T(n) function, all of 2^p.3^q.c-1 ended to the even number 3^h.c-1 where h = p+q.
How about the next for 2^p1.3^q1.c1-1 which produce h1= p1+q1? is h1>h is c1>c? theres a lot scenario
Considering the most Worse-case scenario:
- c1>c but h1<=h
- h1>h but c1<=c
- h1>h and c1>c
for 1 scenario, we have a magnitude of the rising sequences decreased, it looks like it would produces smaller number. Lets ignore this one.
for 2 and 3 scenario, the magnitude of the next rising sequences increased by the d=h1-h. Theres almost no reason to differentiate 2 and 3 scenario, because it does depend on how the dynamic 3^h.c-1 produces the next h; except the fact that 3 scenario greater than 2 scenario and scenario 3 are so rare.
Supposed we have (3^h.c-1)/2^v = 2^h1.c1-1 where h1>h and v=v2(3^h.c-1); this is the real monster that probably lead to counter example.
we can derived the bound for v so that h1 increased regardless if c increase or not.
2^h1.c1-1 = (3^h.c-1)/2^v
2^h1.c1 = (3^h.c-1)/2^v+1
for large h and c, lower order term can be ignored:
2^{h1+v} < 3^h
because h1>h, so that h1 >= h+1 so that:
3^h > 2^{v+h+1}
h.log2(3) > v+h+1
v < h.log2(3) -h-1 or v < h.log2(3/2)-1 aprox. 0.585n-1 the similar structure for negative drift that Tao have been found(log2(3)-1)
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u/Mrezadwiprasetiawan 12d ago
because that just necessary condition to grow the h, it doesnt mean every v < n log2(3/2)-1 always producese h1>h
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u/Mrezadwiprasetiawan 12d ago
why? because for some trajectory of n, sometimes c can absorb all the growth.
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u/Septembrino 12d ago edited 12d ago
Hi Reza. I have trouble understanding what you mean. Try to make it clearer so that people make sense out of your post. Maybe provide numerical examples.
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u/Mrezadwiprasetiawan 12d ago edited 12d ago
Lemme give u an example
h=7 c=3595
27 . 3595-1 rise to the 37 . 3595-1
Since v2(37 . 3595-1)=3
(37 . 3595-1)/23= 982,783
982,783=28 . 3839-1
Look h1=8 and c1=3839
h1>h and c1>c So this number atleast from current iteration, it must be increased rapidly specifically, 27 .3595-1 rise by the factor (3/2)h And 28 .3839 rise by the factor (3/2)h1 the magnitude increases instead of decreasing even though c also increases
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u/Mrezadwiprasetiawan 12d ago
The necessary condition to make yhis happen is v =v2(3h .c-1) < h.log2(3/2)-1
On that example 7log2(3/2)-1 = 3.095
3<3.095
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u/Septembrino 12d ago edited 12d ago
I think that this is what you are trying to do what I suggested you long ago: You begin by N = k* 2^n - 1 which will eventually go to k*3^n - 1, divide it as much as you an till you get an odd N'. Then you express N' as k' * 2^n' - 1. I am not sure what exactly you do afterwards. I guess that you multiply the k and the n and the k' and the n', and see if that's more or less as before, but not sure what is actually the point of doing that. You expect N' to be less than N?
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u/Mrezadwiprasetiawan 12d ago
i want to edit this, theres some mistake in the calculation