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I've been down that road, and it's unfortunately a dead end. You can generate an infinite number of reducing forms, each of which itself covers an infinite num
by kr99x 1y ago
I've been down that road, and it's unfortunately a dead end. You can generate an infinite number of reducing forms, each of which itself covers an infinite number of integers, like 4k + 5 → 3k + 4. Each one covers a fraction of the integers 1/(2^x) where x is the number of division steps in its reducing sequence (and the right hand side is always 3^y where y is the number of multiplying steps). You can't just make 1/2 + 1/4 + 1/8 and so on though (the easy path to full coverage) because sometimes the power of 3 overwhelms the power of 2. There is no 8k → 9k form, because that's not a reduction for all k, so you instead have to go with 16k → 9k. This leaves a "gap" in the coverage, 1/2 + 1/4 + 1/16th. Fortunately, when this happens, you start to be able to make multiple classes for the same x and y pair and "catch up" some, though slower. As an amateur I wrote a whole bunch about this only to eventually discover it doesn't matter - even if you reach 1/1th of the integers by generating these classes out to infinity, it doesn't work. An infinite set of density 1 implies a complementary set of density 0, but a set of density 0 doesn't have to be empty! There can still be finitely many non-reducing numbers which are not in any class, allowing for alternate cycles - you would only eliminate infinite growth as a disproof option.
Mind you, it's almost certain Collatz is true (generating these classes out to 3^20 nets you just over 99% coverage, and by 3^255 you get 99.9999999%) but this approach doesn't work to PROVE it.