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Let an odd integer be represented as ₍ ₌ 2ⁿ + 2ᵐ - 1 for m 1. To transform 2ᵐ - 1 according to the Collatz rules, the 3x + 1 step is applied, resulting in the even integer 2^m + 1 + 2ᵐ - 2. This even integer, being exactly once divisible by 2, becomes the next odd integer 2ᵐ + 2^m - 1 - 1. This represents the growth phase, as exactly one even step follows an odd step. However, while the overall value of the integer increases, terms with decreasing indices are generated after each odd-even cycle. After m such cycles, a term 2^m - m (which equals 1) is generated, canceling the negative 1. Additional even steps are then performed until the integer becomes odd again, which occurs when the next smallest index becomes zero. The positive 1 thus obtained is written as 2¹ - 1. If uninterrupted, the term 2¹ - 1 has the trivial cycle 2¹ - 1 2² 2¹ 2¹ - 1. This represents the shrinkage phase, as two even steps follow an odd step. To interrupt the trivial cycle, a sequence such as 2¹ + 2² + must be available to combine with 2¹ - 1 and produce 2^M - 1. This starts another growth phase that lasts for M odd-even cycles. Finally, there is an attempt to craft an integer that can escape the shrinkage phase in the Collatz dynamics.
Gaurav Goyal (Tue,) studied this question.
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