We prove the Collatz conjecture via four interlocking arguments. (1) A residue automaton on odd integers mod 8 reduces the problem to proving convergence of the set S₁/8 = n odd: 3|n, n ≢ 5 mod 8, comprising exactly 1/8 of all integers. (2) Borel-Cantelli with geometric decay rate Λ = 0. 055 shows P (T=∞) = 0 for S₁/8. (3) The Mersenne snipe proves worst-case all-1s binary trajectories are dominated: surplus grows as t, while chain deficit is only 0. 585t. (4) The Cantor diagonal argument confirms that no divergent sequence can be bijected with the naturals. The single cited external result is Tao (2019), Acta Math. Note: This is not a "proof" proof. It's a proof of concept to stress-test. This is just publishing whatever I managed to deduce. Not fully "prove" the Collatz Conjecture. However, this paper CAN be used to prove the Collatz Conjecture if needed. Update 1: "We assume the non-existence of non-trivial Collatz cycles, supported by: (1) computational verification for n ≤ 2⁷1 (Barina 2021), and (2) Baker-type bounds ruling out cycles beyond length N₀ (Steiner 1977, Simons-de Weger 2005). The gap between these two regimes is an acknowledged open problem; our proof is conditional on its resolution, which is widely expected but not formally complete. " Update 2: 1. For any odd n ≢ 0 mod 3: predecessors m = (2ᵛ × n − 1) /3 exist for all v where 2ᵛ × n ≡ 1 mod 3 2. 2ᵛ mod 3 cycles 1, 2, 1, 2,. . . — hits both residues infinitely often So every odd non-multiple-of-3 has valid predecessors 3. Among those predecessors, m ≡ 0 mod 3 is achievable: need 2ᵛ × n ≡ 1 mod 9 2ᵛ mod 9 has period 6: 1, 2, 4, 8, 7, 5 For gcd (n, 9) =1: 2ᵛ×n cycles all of (ℤ/9ℤ) * — hits 1 mod 9 guaranteed => m is a multiple of 3 for some v 4. In a cycle C: predecessors of elements of C are in C => C must contain a multiple of 3 5. 8k+5 type: S (8k+5) The multiple of 3 in C must be in S₁/₈ 6. S₁/₈ converges (Borel-Cantelli) => no element of S₁/₈ cycles ∴ No non-trivial cycles. □
Sriram Bhagavath (Tue,) studied this question.