A curious reformulation of the Collatz conjecture (not a proof)
Here is quite interesting statement that follows from the Collatz conjecture (actually equivalent, I believe, but I did not bother to prove). I like it because it is simple, and at the first glance - not obviously related to CC at all.
Consider two families of infinite-dimensional vectors (or just functions from ℕ to your favorite field):
Then, for every Collatz cycle, there are subsets of vectors a and b, whose sum is equal:
aᵢ₁+aᵢ₂+...+aᵢₘ = bⱼ₁+bⱼ₂+...+bⱼₙ
One such subset is easy to notice with a naked eye: a₁ = b₁. As you could guess, it corresponds to the trivial cycle 1-4-2. Then, if we could prove that there are no other subsets with equal sum property, the no-cycles part of the Collatz conjecture would immediately follow.
It is quite obvious that vectors in the b list are somewhat more "sparse" than vectors in the a list, so it is tempting to try to find some reason why on sum of *a'*s could be equal to some sum of b's, however, they are not that easy.
For example, look at those two sums: a₄₁+a₆₈+a₃₄+a₁₇ and b₈+b₁₂+b₁₈+b₂₇+b₂₀+b₃₀+b₄₅ (edit: added forgotten term)
Here, I plotted them:
Almost the same, aren't they? They would be the same, if we shift sum of b's by 2 to the right. This near-miss actually corresponds to the cycle of the 3x-1 system that starts with 17. In general, cycles of the 3x+P systems produce sums that are equal up to shift by P-1.
How was this constructed? Actually, quite simple: by mapping integers to vectors: 1→[0000...], 2→[1000...], 3→[1100...], 4→[1110...] and so on, and then mapping edges of the shortcut Collatz graph to their differences. I can write in more details if anyone is interested, though it is quite trivial.
We can now make a purely geometrical statement from which the no-cycles part of the CC follows. Consider two cones, formed by positive linear combinations of vectors (a₂, a₃, ...) and (b₂, b₃, ...). If these two cones do not intersect (except for the apex), then there are no high cycles in the Collatz system.
I think it's neat.