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On the wikipedia page for the Collatz conjecture, there is a statement of the Collatz conjecture "in reverse" by growing a graph where R(n) = {2n, (n-1)/3} for
by guskel 5y ago
On the wikipedia page for the Collatz conjecture, there is a statement of the Collatz conjecture "in reverse" by growing a graph where R(n) = {2n, (n-1)/3} for n ≡ 4 mod 6, and R(n) = 2n for n ≡ 0,1,2,3,5 mod 6.
I wonder how many attempted proofs attempt to solve through this bottom up approach rather than top-down.
- mynegation 5y agoAbout a week ago, I spent several hours after midnight in this ultimate nerd trap. I started with powers of 2 that obviously reduce to the cycle and tried to apply these transitions inferring larger and larger sets of numbers whose binary representation satisfies specific regular expressions on 0 and 1s, but got lost pretty quickly. My intuition is that, given that several generalizations of Collatz Conjecture are undecidable (equivalent to a halting problem), this process is in the territory of being not yet Turing-complete but already undecidable. But I am pretty sure many people way smarter than me tried this approach as well.