My thinking is that every even number is a product of a unique odd number multipled by a particular power of 2.
And that the n/2 part of the Collatz causes us to get swept down to the odd number at the "base" of each of these "grooves" that can be represented by (particular odd number)*2^(m [where m represents all possible non-negative integer powers]).
So, I think there's a way through to the proof if I can:
1. Prove that (so long as the odd number is not 1 [perhaps this will have to do with primes, and the fundamental theorem of arithmetic]) applying 3*n+1 to an odd number with always cause a switch of "groove" to a groove not previously visited in the sequence.
2. Eventually the even number that the groove switch (3*n+1) causes us to land on will be in the "1 groove" (1*2^m [where m is all possible natural numbers]).
So I'll do some more calculation, but could y'all folkx, you esteemed and lofty scholars. You KINGS. Assist me in some way?
168.19
And that the n/2 part of the Collatz causes us to get swept down to the odd number at the "base" of each of these "grooves" that can be represented by (particular odd number)*2^(m [where m represents all possible non-negative integer powers]).
So, I think there's a way through to the proof if I can:
1. Prove that (so long as the odd number is not 1 [perhaps this will have to do with primes, and the fundamental theorem of arithmetic]) applying 3*n+1 to an odd number with always cause a switch of "groove" to a groove not previously visited in the sequence.
2. Eventually the even number that the groove switch (3*n+1) causes us to land on will be in the "1 groove" (1*2^m [where m is all possible natural numbers]).
So I'll do some more calculation, but could y'all folkx, you esteemed and lofty scholars. You KINGS. Assist me in some way?
168.19
