Your method should work after using a little bit of the theory of Diophantine approximation. I think that for example Baker's theory of linear forms of logarithms should be way more than enough to handle this (especially seeing as his theory is meant to apply to more general questions involving logarithms of large collections of algebraic numbers.)tomtom2357 wrote:For anyone who has not heard of the collatz conjecture before, the collatz conjecture states that if you take any number, and if it is even, divide by 2, and if it is odd, multiply by three, add 1, then divide by two, and if you repeat this procedure long enough, eventually any positive integer will end up at 1. There are two ways this could fail, by the sequence of a number going to infinity, or by the number entering a cycle of finite length. I am investigating the latter option. In particular, I am trying to prove the non-existence of a 1-cycle, where there are m iterations of x->(3x+1)/2, and n applications of x->x/2. I have proved that the existence of this cycle for a given starting number x implies that 2m+n/3n<=1+(1/x)<=1+(1/(2m-1)) (because x>=2m-1). Taking the logarithm of both sides to base 2 gives: m+n-log2(3)n<=m-log2(2m-1)<1/(2m-1). Is this possible? If I can prove that this is impossible, then I will have disproved the existence of a 1-cycle.
tomtom2357 wrote:m+n-log2(3)n<=m-log2(2m-1)<1/(2m-1). Is this possible?
tomtom2357 wrote:Choosing a multiple of three would be a good idea.
CCC wrote:tomtom2357 wrote:Choosing a multiple of three would be a good idea.
Hmmm. A multiple of three cannot be part of a loop; 3x+1 can never lead to a multiple of 3, and x/2 can only lead to a multiple of 3 if starting from a multiple of three. Thus, no multiple of 3 can be reached from any smaller number at any point, nor can it be reached from itself; it can only be reached from higher, even multiples of three. I'm not entirely sure how that helps, though.
I've been putting a but more thought into the special case of a number that's a continual string of ones in binary; that is, a number of the form 2n-1 for arbitrary n. It is odd, so multiplying by 3 gives 3*2n-3; add one, 3*2n-2; halve it, 3*2n-1-1. Continuing this process n times leads me to a result of 3n-1. This is itself even, so the next step is to halve it, but after that it gets harder to predict (for n=2, you halve it until it hits one; for n=3, you jump around a bit before hitting 1; for n=4, it collapses quickly; for n=5 you've got somewhere around 96 steps to go before getting there).
In base 3, this number takes the form of n 2s in a row. Halve that; and the number consists of n ones in a row.. If n is odd, this is an odd number; multiply by 3 and add 1 (in base 3, this leads to a string of (n+1) ones). You now have a number consisting of an even number of ones all in a row in base 3; halve it, to get a (n-1)-digit (or n-digit if n was odd) number of the form 202020....020202. This can be halved again: 101010...010101. This may be odd or even, depending on the number of ones; it's an odd base, and therefore the number is even if and only if the sum of the digits is an even number...
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