What if you change the algorithm in the Collatz conjecture by saying that if the number is divisible by four, just add two. Otherwise use the normal formula (divide by two if even).Does this version have a loop like the original algorithm? And do all starting numbers eventually return to it?
>>17017425every sequence would run off to infinity, since every two steps would get you from x to either 3x+3 or (3x+3)/2, both bigger than x
>>17017431correction, (3x+1)/2 or (3x+3/2
>>17017425I'm pretty sure the solution to this problem involves like hundreds of pages of complicated schizo math lol.
>>17017425The whole "point" of the Collatz conjecture is that even numbers are "on average" divisible by two factors of 2, so you're "on average" going to divide by 2 twice every time you multiply by 3, so heuristically you should consistently decrease. What you're proposing will force a multiplication by 3 for every division by 2, so of course it'll exponentially increase, like the other anon said.The question of "to what extent can we treat divisibilities as independent random events" is one of the huge research mines of modern analytic number theory, and these sorts of questions are often notoriously extremely difficult.
When you divide a number it goes down. Truly revolutionary.
>>1701746650% of even numbers are divisible by 4. 50% of even numbers are only divisible by 2.Therefore, on average, most even numbers are divisible by [math] \frac{2+4}{2}=3 [/math]Prove me wrong.
